STRUCTURAL ENGINEERING 
 
 By 
 J. E, KIRKHAM 
 
 Professor of Structural Engineering, Iowa State College. Consulting 
 
 Bridge Engineer, Iowa Highway Commission. Formerly 
 
 Designing Engineer with American Bridge Co. 
 
 FIRST EDITION 
 THIRD IMPRESSION 
 
 McGRAW-HILL BOOK COMPANY, INC. 
 NEW YORK: 370 SEVENTH AVENUE 
 
 LONDON : 6 & 8 BOUVERIE ST., E. C, 4 
 
 1914 
 
COPYRIGHT 1914 
 
 BY 
 
 THE MYRON C. CLARK PUBLISHING CO. 
 CHICAGO 
 
TABLE OF CONTENTS 
 
 CHAPTER I. 
 Preliminary. 
 
 PAGES 
 
 Definitions, structural material, riveting, method of procedure in designing, and 
 
 fabrication 18 
 
 CHAPTER II. 
 Structural Draughting 1 . 
 
 Equipment, free-hand lettering, drawings, and hints regarding shop drawings 
 
 and bills 9-13 
 
 CHAPTER III. 
 Fundamental Elements of Structural Mechanics. 
 
 Definitions, forces, elasticity, distortion, equilibrium polygon, center of gravity, 
 
 moment of inertia, radius of gyration 14-50 
 
 CHAPTER IV. 
 Theoretical Treatment of Beams. 
 
 Shearing stresses, bending stresses, reactions and deflections of cantilever, simple, 
 over-hanging, fixed at one end and supported at other, fixed and continuous 
 beams 51-105 
 
 CHAPTER V. 
 Theoretical Treatment of Columns. 
 
 Rankine's formula, straight line formula, eccentrically loaded columns, and ex- 
 amples 106-114 
 
 CHAPTER VI. 
 Rivets, Fins, Rollers and Shafting*. 
 
 Bearing, shearing and bending stresses on rivets and pins, allowable pressure 
 
 on rollers and stresses on shafting 115-123 
 
 CHAPTER VII. 
 
 Maximum Reactions, Shears, and Bending Moments on Beams and Trusses and 
 
 Stresses in Trusses. 
 
 Maximum reactions, shears and bending moments on simple beams and trusses 
 due to both uniform and concentrated loads, and stresses in simple trusses 
 due .to panel loads 124-143 
 
 iii 
 
j v CONTENTS 
 
 CHAPTER VIII. 
 Graphic Statics. 
 
 PAGES 
 
 Definition and limitation, graphical extermination of reactions and bending 
 moments on simple beams, graphical determination of stresses in trusses 
 and drawing of an equilibrium polygon through two ana three points. .. .144-161 
 
 CHAPTER IX. 
 Influence Lines, 
 
 Definition, influence lines for determining reactions shears arid moments on 
 
 simple beams and trusses and stresses in trusses 162-171 
 
 CHAPTER X. 
 Desig-n of I-Beams and Plate Girders. 
 
 Data and method of procedure, section modulus, economic depth of girders, 
 designing of flanges and webs, length of cover plates, flange increments, 
 rivet spacing in flanges, web splice, stiffening angles, and examples 172-104 
 
 CHAPTER XL 
 Design of Simple Railroad Bridges. 
 
 Types, loadings, design of beam, plate girder and truss bridges and viaducts and 
 
 deflection and camber of trusses . . 195-532 
 
 CHAPTER XII. 
 Design of Simple Highway Bridges. 
 
 Types, loadings, specifications and design of beam, pony truss and high truss 
 
 bridges 583-569 
 
 CHAPTER XIII. 
 
 Skew Bridges, Bridges on Curve, Economic Height and Length of Trusses and 
 
 Stresses in Portals. 
 
 Types, determination of stresses in skew bridges, determination of stresses due 
 to centrifugal force, economic depth and length of ordinary trusses and 
 determination of stresses in various types of portals 570-582 
 
 CHAPTER XIV. 
 Design of Buildings. 
 
 Description, loadings, and designing of various types of mill buildings and the 
 
 analysis and design of high buildings 583-652 
 
PREFACE 
 
 This book is intended as a textbook for college students and as a 
 self-explanatory manual of structural engineering for practical men. 
 
 During the author's nineteen years of engineering experience (the 
 greater part of which was spent in actual practice) it fell to his lot to 
 "break in" students from most all of our engineering schools, and he has 
 no apology to offer for the elementary mechanics given in this volume, as 
 he is convinced that no matter how thorough the course in mathematics 
 and theoretical mechanics may be it is quite desirable that students have 
 a short review of the materialistic phase of mechanics at the beginning 
 of the subject of structures to whet their appetites for the work. 
 
 As regards the designing given, the author has endeavored to present 
 the usual methods, and this in such a fashion that the average engineering 
 student as well as the practical man can read, understandingly, without 
 having a dictionary, glossary, or a compendium on theoretical mechanics 
 at his elbow. 
 
 The author's aim has been to arrange for college use the drawing 
 room exercises throughout the text, so that the work in the classroom 
 and drawing room will go hand in hand. The appreciation of this 
 feature will depend a great deal upon the allotment of hours for the two 
 classes of work. The ratio of two hours recitation to six in drawing is 
 recommended. "Drawing Room Exercise No. 1," at the end of Chapter 
 II, can be started at the very first drawing room period without wasting 
 any time. Chapters I, II, VII, X, and Chapter XI as far as "Deck 
 Plate Girder Bridges," should be read by the time "Drawing Room 
 Exercise No. 1" is finished, so that "Drawing Room Exercise No. 2" can 
 be taken up. During the time spent on "Drawing Room Exercises No. 2, 
 No. 3, and No. 4" ample time will be found for the necessary advance 
 reading up to "Through Plate Girder Bridges" and the reading of 
 Chapters III to VI and also VIII and IX. Beyond this the author 
 refrains from making suggestions as to assignments as he has confidence 
 in the ability of instructors to assign correctly the work to suit conditions. 
 
 The practical men of limited theoretical training and others desiring 
 a review of the subject treated, will find the book well suited to their case 
 if the chapters be taken up in consecutive order. 
 
 The designs given in this book are entirely the work of the author 
 and are designed especially for this work, yet he has endeavored to make 
 them as general as possible. 
 
 This book, which treats only of simple structures, is the author's 
 first volume on Structural Engineering. A second volume, which will 
 be known as Higher Structures, is in preparation; this will treat of 
 Movable Bridges, Cantilever, Arch, and Suspension Bridges, Secondary 
 Stresses, etc. 
 
 The author desires to acknowledge his appreciation of the work of 
 his assistant, Mr. B. S. Myers, in reading and checking proof and pre- 
 paring and checking drawings, and to thank Prof. F. O. Dufour for 
 valuable suggestions and criticisms. J. E. KIRKHAM. 
 
 Ames, Iowa, Sept. 5, 1914. 
 
CHAPTER I 
 
 PRELIMINARY 
 
 1. Structural Engineering. The part of Civil Engineering per- 
 taining to the designing of steel structures, such as bridges, buildings, 
 towers, etc., is known as Structural Engineering. The work involved 
 consists, principally, in determining the stresses, selecting the material 
 to be used, known as sections, and the contriving and drawing of the 
 details. But, in addition to this work, the structural engineer has, as a 
 rule, the designing of a great deal of incidental construction, such as 
 foundations, concrete floors, roofs, etc. 
 
 In order to design structures properly, one must be perfectly 
 familiar with the material used in their construction and have a clear 
 understanding of the manner in which their manufacture and erection 
 are accomplished as well as have a thorough knowledge of the mechanical 
 principles involved throughout. A properly designed structure is one 
 wherein no mechanical principles are seriously violated and which at the 
 same time is economic in material and easily manufactured and erected. 
 
 2. Structural Material. Steel, concrete, stone, and wood are the 
 principal materials used by the structural engineer in the construction of 
 modern structures. However, cast iron, wrought iron, and a few other 
 materials are used in a few cases, as will be designated when these cases 
 present themselves in the text. Concrete and stone will not be treated in 
 this book further than to designate certain allowable pressures. 
 
 3. Manufacture Of Steel. In describing in a general way the 
 usual process of making steel, we can say that steel is made from iron ore 
 which is mined and carried to a blast furnace through which it is run 
 together with coke and limestone and thus converted into cast iron. The 
 cast iron is then taken to either an open-hearth furnace or to a Bessemer 
 furnace, usually spoken of as a Bessemer converter, and there converted 
 into steel. When the conversion is complete, the molten steel is run into 
 ingot molds, which are rectangular cast-iron molds, and molded into 
 ingots. These ingots are allowed to cool to some extent. Then the molds 
 are pulled off and the ingots are taken to a rolling mill, which is known 
 as a slab or blooming mill, where they are heated in what is known as a 
 soaking pit until they have the proper temperature for rolling. Then 
 they are rolled to a convenient rectangular cross-section and cut up into 
 convenient-sized pieces known as slabs or billets. These billets or slabs, 
 as the case may be, are then taken to another rolling mill and reheated 
 very much the same as the ingots just described, and then rolled into 
 structural shapes, plates, rails, etc., ready for the market. 
 
 Sometimes the cast iron from the blast furnace is cast into rough 
 bars, known as pig iron, which may be stored and converted into steel 
 or molded into castings at any desired time, or it may be shipped to some 
 distant point to be utilized in the same manner. At the most modern 
 plants the molten metal is taken directly from the blast furnace and 
 converted into steel without letting it cool to any great extent. 
 
 1 
 
2 STRUCTURAL ENGINEERING 
 
 The steel produced in an open-hearth furnace is known as "open- 
 hearth steel/' while the steel produced in a Bessemer converter is known 
 as "Bessemer steel." Open-hearth steel is considered more reliable in 
 every respect than Bessemer,, and, consequently, the Bessemer steel is 
 being fast replaced by the open-hearth product. In fact, most engineers 
 at present exclude the use of Bessemer steel in structural work altogether. 
 
 4. Grades of Common Structural Steel. There are three recog- 
 nized grades of common structural steel: Medium Steel; Soft Steel; 
 Rivet Steel. 
 
 Medium Steel is a medium-hard steel which has practically replaced 
 all other grades of steel in structural work. It has an ultimate strength 
 of 60,000 to 70,000 pounds per square inch, and an elastic limit of 30,000 
 to 35,000 pounds per square inch. 
 
 Soft Steel is softer than medium steel. It was the grade of steel 
 first used in structural work, but has been practically replaced by medium 
 steel. It has an ultimate strength of 52,000 to 62,000 pounds per square 
 inch and an elastic limit of 26,000 to 31,000 pounds per square inch. 
 
 Rivet Steel is a very soft steel used almost exclusively for making 
 rivets and bolts. It has about the same chemical composition as wrought 
 iron, but has a higher ultimate strength and elastic limit. It has an 
 ultimate strength of 48,000 to 58,000 pounds per square inch and an 
 elastic limit of 24,000 to 29,000 pounds per square inch. 
 
 5. Nickel Steel is a steel containing from 2 per cent to 3J per cent 
 of nickel. It has an ultimate strength of 80,000 to 112,000 pounds per 
 square inch and has a very high elastic limit something like 60,000 
 pounds per square inch. This metal has been used recently in some of 
 the large bridges in this country. 
 
 6. High Carbon Steel is a hard steel which contains more com- 
 bined carbon than the ordinary medium steel. It has practically as high 
 ultimate strength and elastic limit as nickel steel. It is used almost exclu- 
 sively in the form of rods to reinforce concrete. 
 
 7. Wood, or timber, as it is usually spoken of, is used in structural 
 work principally for floor and roof covering. The timber mostly used is 
 white oak and yellow pine, both of which have an average ultimate crush- 
 ing strength of about 7,000 pounds per square inch, and an ultimate 
 tensile strength of about twice that amount. 
 
 8. A Casting is made by running molten metal into an impression 
 made in sand by means of a wooden pattern which is shaped to the size of 
 the metal piece desired. The castings used in structural work are made 
 of either cast iron or steel. The castings made of steel are much stronger 
 than those made of cast iron but cost more. The steel used for making 
 castings has a somewhat different chemical composition than that of rolled 
 steel, referred to above. It is produced in the same way, however, the 
 chemical composition being controlled mostly in the mixing of the charge. 
 
 9. Structural Steel Shapes. The I-beam, channel, angle, Z-bar, 
 and T-shape, shown in Fig. 1, are the principal steel shapes used in 
 structural work. The T-shape is not used very extensively except for 
 very special work. Plates, while they are not classified as structural, 
 shapes, are really used in a more general way than the shapes. 
 
PRELIMINARY 
 
 Web 
 
 I- Beam 
 
 Channel 
 
 The tables in the back of this book give the properties, gauges, etc., 
 of the shapes and plates in general use. The use of these tables will be 
 explained as the occasion for their 
 use occurs. Either a Carnegie or 
 Cambria handbook is a convenient 
 book to have in fact, practically 
 indispensable in structural work 
 but the student must bear in 
 mind that he is not at liberty to 
 use just any section therein that 
 he happens to pick out, as a great 
 many specials which are not in 
 general use are listed, which will 
 be furnished only when the order 
 for such becomes large enough to 
 warrant the rolling, and, conse- 
 quently, in the case of ordinary 
 orders containing these special 
 sections, an unusual delay in 
 obtaining the material will likely 
 be experienced. 
 
 10. Rivets and Riveting. 
 
 Rivets are used in structural work 
 
 to connect the structural shapes and plates together. They are simple bits 
 
 of metal in themselves, but, indirectly, they are the source of a great deal 
 
 . Length 
 
 Shank 
 
 Shank 
 
 Button Headftive/- 
 
 Countersunk /-/eaa '/?/ vef' 
 
 Fig. 2 
 
 of trouble. Practically, there are but two kinds: the buttonhead rivet 
 which has a hemispherical head, and the countersunk rivet which lias a 
 countersunk head. Each kind is shown in Fig. 2, including the names of 
 their parts. The buttonhead rivet is the kind most used; in fact, the 
 countersunk rivet is never used except where it is absolutely necessary 
 to do so. 
 
 As a rule, each structural company manufactures its own rivets. 
 They are made by feeding red-hot rods into a machine which cuts the rods 
 into pieces of proper length and at the same time forms a head on each 
 piece, thus completing the rivets as they are shown in Fig. 2. The way 
 in which this is done can be seen by referring to Fig. 3, where the parts 
 of the machine that directly form the rivets are shown. Let us first 
 consider the plan diagram at (a), where A represents the "header," 
 B a fixed die, C a moving die, D a forked bar, and S a bar known as the 
 gauge. While the parts are in the position shown at (a) the heated rod 
 is pushed into the machine, passing through the slot in the bar D until it 
 comes in contact with the gauge S, as indicated by the dotted outline 
 e-n-o-k. Then the die C moves toward B and cuts the rod off at o by 
 
STRUCTURAL ENGINEERING 
 
 shearing with the bar D, and at the same time the piece of the rod thus 
 cut off is caught in the cylindrical grooves in the dies (see section m-m) 
 and held firmly while the gauge S moves up out of the way of the header 
 A which then moves until it ^ 
 
 comes in contact with the dies 
 B and C, upsetting the part e-n 
 of the piece of the rod into the 
 
 D) 
 
 k 
 
 1 
 
 
 I s 
 
 n 
 
 
 c 
 
 m 
 
 
 
 A 
 
 (a) 
 
 (b) 
 
 , ,. 
 
 Sechon-mrn. 
 
 Fig. 3 
 
 hemispherical cup in the header 
 A, thus forming the head of the m I 
 rivet. The parts of the ma- t 
 chine are then in the position c 
 
 shown at (6). The header A 
 and then the die C and the 
 gauge S move back to the posi- 
 tion shown at (a), while the 
 rivet just formed drops into a 
 cooling basin. Then to obtain 
 the next rivet, the rod is fed 
 into the machine as before. 
 Thus rivets are made at the 
 rate of about one per second. 
 
 The rivets formed by the machine just described are the buttonhead 
 rivets. However, countersunk rivets can be made on the same machine by 
 replacing the header shown by a header having a plane end (without any 
 cup), and the dies shown by dies which have the cylindrical groove 
 chamfered out on the end next to the header so as to form the counter- 
 sunk head. In that case the heads are formed in the dies instead of in 
 the header as in the case of the buttonhead rivet. 
 
 After the rivet holes have been either punched or drilled into the 
 plates and shapes to be riveted together, and the same have been 
 assembled, each in its proper place, rivets are heated, placed in the holes, 
 and "driven." The driving of a rivet consists principally of forming a 
 head on the plane end, usually the same as the one on the other end 
 formed by the rivet machine just described. 
 
 Rivets are driven by machines known as "riveters" except in a few 
 cases where it is necessary to drive them "by hand." The riveter, or 
 rivet-driving machine, has two headers known as tools, one being fixed 
 and the other movable, each of which is very similar to the header used 
 in the rivet-manufacturing machine described above. 
 
 In order to show the working of 
 a riveter, let m (Fig. 4) represent an 
 angle which is to be riveted to a plate n. 
 The rivet holes are either punched or 
 drilled in the two so as to match. Then 
 
 Sf- . n \\ft / rjj the plate and the angle are placed to- 
 
 ( C// \H> ' gether as shown, and the driving of the 
 
 rivets proceeds as follows: A rivet is 
 heated and placed in one of the holes, as 
 shown at (a). Then the fixed tool B of the riveter is placed against the 
 head of the rivet and the power is applied (which may be air, steam or 
 water) which moves the tool A toward B whereby the projecting part of 
 
 Fig. 4 
 
PRELIMINARY 
 
 m 
 
 T 
 
 (d) 
 
 m 
 
 r 
 
 m 
 
 n 
 
 Fig. 5 
 
 the rivet is upset into the cup in the end of A, thus forming the head on 
 that side. When the tool A moves until it practically touches the plate n, 
 the rivet is fully driven, as shown at (6). The tool A is then brought 
 back to its former position,, and the riveter is placed on the next rivet by 
 either moving the pieces being riveted together, or by moving the riveter, 
 and the operation of driving is repeated, and so on. 
 
 In case a countersunk rivet is to be driven, the rivet hole is either 
 punched or drilled the same as though a buttonhead rivet were to be used. 
 Then the hole is countersunk, which means that it is reamed out cone- 
 shaped on the side where the countersunk head is to come so that the 
 head will just fit in. The countersinking of the hole is done, usually, 
 with a flat, diamond point drill, as shown at (a), Fig. 5. The driving of 
 a countersunk rivet with a riveter is very much the same as driving a 
 buttonhead rivet. For example, 
 suppose a rivet having a counter- 
 sunk head is to be used to connect 
 the angle m to the plate n as shown 
 at (6), Fig. 5, where the counter- 
 sunk head is on the same side as 
 the angle. The rivet hole will be 
 countersunk in the angle as shown. 
 The countersunk rivet is heated 
 and placed in the hole from the 
 same side as the angle and 
 driven just the same as was explained above in the case of the button- 
 head rivet, but the tool held against the countersunk head would be a 
 plane tool without a cup. If the other tool has a cup in it, the head on 
 that side would be an ordinary buttonhead, as shown at (c), Fig 5. If it 
 were necessary to have a countersunk head on each side, the hole would 
 be countersunk in the plate as well as in the angle, as shown at (d), 
 Fig. 5. The rivet before driving would have a countersunk head, the 
 same as before, and the driving would take place in the same manner, but 
 each of the tools would have a plane end. 
 
 In cases where it is either impossible or impracticable to use the 
 riveter, the rivets are driven by hand. This, in the case of a buttonhead 
 rivet, consists of heating the rivet and placing it in the hole, and while 
 the end of a round bar which contains a cup (known as a dolly bar) is 
 held firmly (by hand) against the head, the projecting end of the rivet is 
 battered down with a hammer and then formed into a finished head by a 
 heading hammer (known as a snap) which is held against the battered 
 end of the rivet and struck by a sledge. A heading hammer, or snap, is 
 really a two-faced hammer with a cup in one end used to form the head 
 of the rivet. In the case of a countersunk rivet, the driving by hand is 
 practically the same as in the case of the buttonhead rivet, but the tools 
 used on the countersunk heads will be plane, that is, without cups. 
 
 There are but very few cases where it is necessary to drive rivets by 
 hand in the shop, but practically all of the rivets connecting the individual 
 members in a structure to one another are driven by hand as the structure 
 is being erected. Such rivets are known in structural engineering as 
 "field rivets." Holes left open in the shop for such rivets are known as 
 open holes or field holes, 
 
6 STRUCTURAL ENGINEERING 
 
 The thickness of metal connected by a rivet is known as the grip of 
 the rivet. The diameter of the shank of a rivet is known as the diameter 
 of the rivet. It is also known as the size of the rivet. 
 
 The part of the rivet projecting beyond the material to be riveted 
 together is just a little longer than that necessary to form a head,, for, in 
 driving, some of this is taken up in upsetting the shank of the rivet into 
 the hole which is always about -fa in. larger in diameter than the shank. 
 
 The size of the rivets used depends upon the material to be connected. 
 A common rule is that no rivet shall be less in diameter than the thickness 
 of any single piece connected. Rivets vary in size from J in. in diameter 
 to 1 ins. The rivets most often used are the J- and J-in. diameter. 
 
 11. Method of Procedure in the Designing of Steel Structures. 
 As a rule, parties desiring structures built specify the location and pur- 
 pose, and limit to some extent the amount they shall cost. The engineering 
 work really begins with the survey of the site, which, however, does not 
 necessarily require the services of a structural engineer, yet a structural 
 engineer should know how to do such work as it is sometimes required of 
 him ; and also such knowledge is often essential in working out the design 
 of a structure. The structural engineering work really begins after the 
 drawings showing the site are made from which the structural engineer 
 works out the preliminary plans of the structure. These preliminary 
 plans show, in a general way, what is desired. They usually consist of 
 what are known as "Stress Sheets" and "General Drawings." A Stress 
 Sheet is usually a single-line drawing of the structure wherein each 
 member is represented by a single line upon which the corresponding 
 stresses and sections are written, and the general dimensions of the struc- 
 ture are given in reference to centers of bearing and to centers of gravity. 
 A General Drawing is usually intended to show the structure as a whole. 
 In many cases it is a general picture of the structure wherein the details 
 are shown in a general way but not fully dimensioned. In the case of 
 very important structures, the general drawings usually include a general 
 picture of the proposed structure showing no specific details at all. Such 
 drawings may be prepared by an engineer or by an architect. The 
 purpose is to present the general appearance of the structure. That such 
 drawings are included does not lessen in the least the necessity of the 
 other general drawings. 
 
 After the general drawings have been completed to the satisfaction 
 of all parties concerned, they are used as a guide in making the "Shop 
 Drawings" (sometimes called "Working Drawings") which show the 
 spacing of rivets and all holes, cuts, etc., for each individual member of 
 the structure. After these shop drawings are made and thoroughly 
 checked and are approved by all parties concerned, they are sent to the 
 structural shops where each member of the structure is fabricated accord- 
 ingly. 
 
 All of the drawings referred to above are, as a rule, made on tracing 
 cloth and blueprints are made from these which are furnished to all 
 parties concerned instead of actual drawings. These prints, as a rule, 
 are what are referred to as drawings. 
 
 12. A Bridge Company is an organization which designs, manu- 
 factures and erects structural work. However, the name is appropriated 
 
PRELIMINARY 7 
 
 by various concerns which in many cases are capahle of doing only part 
 of this work. A full organization really consists of an operating depart- 
 ment and an engineering department. The operating department has the 
 general management of the company,, especially the commercial end,, 
 while the engineering department attends to everything pertaining to the 
 engineering. It suffices here for us to consider only the engineering 
 department. This department consists of a designing and estimating 
 department, a draughting department, a shop organization, and an erect- 
 ing department. 
 
 The work of the designing and estimating department consists mainly 
 in determining the stresses in structures, the selecting of the required 
 sections, drawing up the stress sheets showing the same, making prelimi- 
 nary estimates of the weight of the material to be used, and computing 
 the cost of the work proposed to be manufactured by the company. 
 
 The work of the draughting department consists of the working out 
 of the complete details of structures, using the stresses and sections as 
 specified on the stress sheets (which are furnished by either the designing 
 and estimating department of the company or by outside parties) and 
 making the necessary general and shop drawings. In addition, the 
 draughting department makes out bills of the material required from 
 which the material is ordered from the rolling mills. These bills when 
 completed are known as "Shop Bills" and are sent to the shops along 
 with the shop drawings. 
 
 The work of the shop organization consists in fabricating the 
 individual members of the structures in the shops according to the shop 
 drawings furnished by the draughting department. 
 
 The work of the erecting department consists in erecting the 
 structures in their final position after the individual members are fully 
 fabricated in the shops. 
 
 The shops are divided into departments which are referred to as 
 separate shops. These different shops, or departments, are as follows : 
 Templet Shop, Pattern Shop, Laying-off Shop, Punch Shop, Rivet Shop, 
 Finishing Shop, Forge Shop, Machine Shop, Paint Shop, Foundry, etc., 
 the name of each clearly indicating the nature of the work done therein. 
 Some departments occupy separate buildings, while in some cases several 
 departments are under one roof. 
 
 13. Fabrication of Steel Structures. The draughting depart- 
 ment of a bridge company usually orders the material for any structure 
 that the company is to fabricate just as soon as the preliminary work on 
 the shop drawings is far enough along. The steel mills roll this material 
 and ship it to the shops where it is unloaded and placed in the "receiving 
 yards," where it remains until the shops need it. After the shop drawings 
 are completed, blueprints are made of them which are sent to the shops, 
 each department receiving the prints showing the part of the work 
 required of it. The fabrication of riveted work, as a rule, begins in the 
 templet shop where full-sized wooden templets are made for most of the 
 shapes and plates in the structure. These templets are made so that each 
 hole and cut shown on the shop drawings can be located on the actual 
 pieces of metal used. These templets, as a rule, are made of one-inch 
 white pine boards, well seasoned. The templets are either made of one 
 
g STRUCTURAL ENGINEERING 
 
 board or of two or more boards connected together, quite often forming 
 a frame. Pasteboard is being used of late, to some extent, in making 
 templets. 
 
 The templets are taken to the laying-off shop where they are clamped 
 to the material for which they were made to lay off. Then the workmen 
 mark where each hole is to be in the metal by placing a center punch in 
 each hole in the templet and striking the punch with a hammer, and 
 indicate the cuts required by marking the outline of the templet on the 
 metal. Then the templets are undamped and thrown to one side and the 
 material thus laid off is taken to the shearing and punch shop where all 
 cuts are made in accordance with the marks, and the rivet holes are 
 punched or drilled as indicated. After this work is completed, the 
 material is taken to the assembling shop where the pieces are assembled 
 so as to form individual members of the structure and bolted together 
 temporarily, just a few bolts being used in each member. Then these 
 members are taken to the rivet shop where the rivet holes are reamed, if 
 such is called for, and all the rivets indicated on the shop drawings are 
 driven. Then the members are taken to the finishing shop where pin holes 
 are bored and all surfaces and joints are finished as called for on the shop 
 drawings. When this work is completed the members are taken to the 
 paint shop where they are cleaned and painted. Then they are taken to 
 the loading yards where they are loaded on cars for shipment to the site. 
 Thus the fabrication is completed. 
 
 Usually when plates and shapes are duplicated a great many times, 
 templets are not used. The duplicated pieces, in that case, are run 
 through a multiple punch which is so constructed that the operator can 
 punch the rivet holes by referring directly to the shop drawings. 
 
 In addition to the riveted work referred to above, there are usually 
 castings, forged work and machine-shop work included in each structure 
 which are gotten out by the foundry, forge shop and machine shop 
 independently of the other shops, except that the patterns used in 
 molding the castings are made in the pattern shop, which is usually very 
 closely associated with the templet shop. This work, when completed, is 
 loaded on cars and shipped to the site, the same as the riveted work, often 
 going on the same cars. 
 
CHAPTER II 
 
 STRUCTURAL DRAUGHTING 
 
 14. Preliminary. It is very essential that all young engineers 
 should be good draughtsmen,, as they are usually called upon to do 
 considerable draughting, and good draughting is highly appreciated 
 throughout the engineering profession. Good draughting is an accom- 
 plishment; however, one sometimes hears of a "natural born" draughts- 
 man. A newspaper reporter while talking to one of our greatest inventors 
 referred to him as a genius. The inventor retorted that genius and 
 perspiration were synonymous terms. The inventor may have been 
 slightly mistaken in that particular case, but in this case there is no 
 question: anyone can acquire the art of making good drawings, that is, 
 good, neat, plain, practical drawings, nothing of an artistic nature being 
 considered, by simply going about it with a will, being careful, each time 
 trying to do a little better than before. 
 
 Structural drawings consist of lines, letters, and figures. The lines 
 should be true and distinct. The letters and figures should be plain, neat, 
 free-hand letters and figures. Each can be practiced independently of 
 the others, but after a fair amount of such practice, good draughtsmanship 
 can be most readily acquired by making exact copies of some good draw- 
 ings, providing the letters and figures be of the same type as those being 
 practiced by the student, as it is not advisable to shift from one type to 
 another until one type is fairly mastered. 
 
 15. The Necessary Equipment for Structural Draughting 
 
 consists of a drawing board, T-square, two triangles, one decimal scale., 
 one duodecimal scale, one first class, medium size, right line drawing pen, 
 one first class, small ink compass, one large combined ink and pencil com- 
 pass, one pair of medium size dividers, a slide rule, either a Carnegie or 
 Cambria handbook, one small bottle of black waterproof drawing ink, a 
 writing pen and holder, a scratch pad, drawing pencils, thumb tacks, 
 erasers, drawing paper, and tracing cloth. 
 
 There are several very useful things which could be added to the 
 above list, such as logarithmic tables, book of squares, beam compass, etc. 
 There are no objections to a full set of drawing instruments instead of the 
 few instruments mentioned above, but in either case, the instruments 
 should be first class. It is better to have a few good instruments than a 
 full set of poor ones. 
 
 16. Free-Hand Letters and Figures. Some distinct character- 
 istics will always be seen in the free-hand lettering of each individual the 
 same as in the case of ordinary writing, but the type of letters and figures 
 should always conform to some recognized standard. Poor lettering 
 should not be recognized as a characteristic, but as an indication of lack 
 of practice. 
 
 9 
 
10 STRUCTURAL ENGINEERING 
 
 Most of the free-hand lettering on drawings is done on tracing cloth, 
 so it is best to practice upon the same. The small remnants of cloth 
 which would otherwise be wasted can often be utilized for that purpose. 
 The student can have the necessary material, which includes a small 
 
 y$fem of Lettering 
 
 SmaJI LeM&ns 
 a(c>) 
 
 c 
 
 s(s) 
 
 Figures 
 
 /2 t 2 ' 23 '* 
 
 / un/ess 
 
 nof-ecf. /?// rivets soft sfee / ar>c/ -j ' c//a. 
 All rivet ho/es punch&c/ ' 
 
 for 
 2-2SO-0 "5wg/e Track 7ft ra fto Con. Spans 
 
 Jfo. 
 
 Figr- 6 
 
 bottle of drawing ink, writing pen, and tracing cloth, at his private room 
 and practice during spare moments as a diversion. Opinions will diffei 
 somewhat as to the kind of pen to use in making free-hand letters and 
 figures. The author prefers Gillott's No. 303. 
 
 The letters and figures shown in Fig. 6 are of about the type used on 
 structural drawings. Just how each letter and figure is made is fullv 
 indicated, the arrows indicating the direction of 
 the strokes, and the small figures above the 
 letters and figures indicating the order in which 
 
 the strokes are made . 
 
 Fig. 7 The student can best practice making the 
 
 letters and figures shown in Fig. 6 by making as 
 
 nearly as possible exact copies of the work shown there. In addition, at 
 intervals he should practice drawing parallel lines (free hand) having the 
 same slope as the letters, and curves which form the letter O, all of which 
 is outlined in Fig. 7. If there is any letter or figure which gives the 
 
STRUCTURAL DRAUGHTING 
 
 11 
 
 student particular trouble, he should keep practicing upon it at intervals 
 until he has mastered it. 
 
 17. Size of Drawings. The usual practice in structural work is 
 to make the drawings 23 x 35 inches inside the border line with a one-half 
 inch margin on all sides. This, however, is not an absolutely fixed size. 
 In some cases it is necessary to make larger drawings, and in other cases 
 it is convenient to make smaller ones. Drawings 18 x 24 inches is a very 
 convenient size in some cases. The border line should be an ordinary 
 plain, single line. 
 
 18. Tracings. It is the usual practice to pencil out all drawings 
 upon ordinary drawing paper, and then make tracings of these pencil 
 drawings upon tracing cloth. The drawing on the tracing cloth should be 
 upon the dull side. Before starting the tracing, the cloth should be 
 rubbed over with talcum powder or with chalk, which should then be 
 brushed off with a cloth or brush. A knife should never be used to make 
 erasures on the cloth. Repeated erasures can be made by using a 
 comparatively soft rubber, as the Ruby Eberhard Faber No. 112. No 
 other than this quality of eraser should be used. 
 
 19. General Hints Regarding Shop Drawings. The shop draw- 
 ings for a structure are usually the last drawings made., but in order to 
 work up the preliminary drawings satisfactorily, .knowledge of shop draw- 
 ings is indispensable, which makes it necessary that the student take up the 
 making of some simple shop drawings as preliminary work. As stated 
 above, the shop drawings show the complete details of the individual 
 members of structures, that is, all rivets, holes, and cuts are located, and 
 sizes of all shapes and plates are given, and in addition, the kinds and 
 
 SHOP R/YETS 
 
 HELD R/VET3 
 
 " 
 
 P/o/n Countersunk F/a//ened fog 
 
 <fc 
 
 Countersunk Plain 
 
 
 
 
 ll. , '^ :-: 
 
 Fig. 8 
 
 sizes of rivets and holes and also the nature of the machine work desired 
 are indicated. 
 
 In order to indicate the kind of rivets desired, certain conventional 
 signs are used, most of which are shown in Fig. 8. Shop rivets are those 
 driven by riveters while the members are in the shops, and the field rivets 
 are those driven at the site as the structures are being erected. Counter- 
 sunk rivets are used on the account of clearance, but sometimes sufficient 
 
12 
 
 STRUCTURAL ENGINEERING 
 
 clearance will be obtained by just mashing down the heads of the rivets. 
 In such cases the heads are said to be flattened. 
 
 Rivet holes are usually -f% inch larger than the rivets to be driven 
 into them. The sizes of the heads of rivets are given in Fig. 9. The 
 sizes of the heads shown on drawings should correspond with the sizes 
 given here. 
 
 In locating rivets, the following rules should be practically followed : 
 Never space rivets closer together, that is, center to center, than three 
 
 !L" 
 
 I6_ 
 
 icJB?* 
 
 Ib 
 
 
 m 
 
 i-iier/ \ 'IMF/ \ 
 
 \ ,2.\ I ,9,,\ 
 
 \./8\ 
 
 Fig. 9 
 
 times the diameter of the rivets, nor farther apart than six inches. Never 
 space rivets closer to the edge or end of a shape or plate than two times 
 their diameter. These rules are not iron-clad, but the deviation from 
 them should be slight. For example : in practice it is customary to space 
 J-inch rivets 1^ inches from the edge or end of a shape or plate, and f-inch 
 rivetis 1^ inches. It is also customary to limit the minimum distance 
 between f-inch rivets to 3 inches, and f-inch rivets to 2J inches. 
 
 The spacing of rivets or holes in reference to one another along one 
 direction is known as the pitch. Rivets or holes passing through the 
 flange of any shape are located in reference to the shape by what is known 
 as the gauge. For example, the distance between any two consecutive 
 rivets or holes along the angles shown in Fig. 10 is the pitch at that 
 point, while the distance g from the back of the angles to the line o-o, 
 passing through the rivets, is the gauge. The distances marked e are 
 
 Fig. 10 
 
 known as the end distances. In case the flange of a shape has double 
 gauge lines, the pitch is the distance between the rivets measured along 
 the piece and not the distance between the rivets on one gauge line. 
 
 The standard gauges for shapes are given in the tables in the back 
 of this book. The same are to be found in the various "standards" and 
 handbooks gotten out by structural companies. 
 
 No rivet or hole should be located in reference to the edge of a 
 flange, as would be done by giving the distance x in Fig. 10. 
 
 All lines upon which rivets are located are known, in general, as 
 rivet lines, and the lines used in giving the spacing of the rivets, as well 
 as all other distances, are known as dimension lines. All rivet lines 
 (including gauge lines and dimension lines) should be light in comparison 
 
a 
 
 p yneqo/wu/Ag 
 
 
 .tf.fr.ff.J 
 
 fr fr .%* . 
 
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 * 
 
 Sv 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 *o'-f~&.-fy'enct& 
 *- ^ 
 
 
 
 ! 
 
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 *x. 
 
 3 
 
 ;" 
 
 V, 
 S 
 
 
 
 X 
 
 s 
 
 < 
 
 Off. 
 
 /* 
 . 
 
 \ 
 
 1 
 
 UJS 
 
 \ 
 
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 r 
 i 
 
 ! 
 
 V 
 
 
 $ 
 
 ( 
 
 j 
 
 
 
 
 
 V 
 
 
 
 
 
 
 
 > 
 
 
 
 Is 
 
 
 
 ^ 
 
 
 r 
 
 
 
 
 
 
 ^? 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 ^ 
 
 
 
 
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 i 
 
 
 
 
 
 
 
 
 
 
 
 
 
 x 
 
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 f*r- 
 
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 4- 
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 CL 
 
STRUCTURAL DRAUGHTING 13 
 
 with the lines used to outline the members. All should be in black ink. 
 
 The duo-decimal scale is used in laying out structural drawings. 
 The usual scale for shop drawings is three-quarters inch equals one foot, 
 but in some cases a scale of one-half inch equals one foot is used. In 
 light work it is sometimes advisable to use a larger scale, in which case 
 a scale of one inch equals one foot or one and one-half inches equals one 
 foot is used. It is advisable for the student to use a large scale at the 
 beginning. 
 
 In making structural drawings, it is customary to represent feet by 
 putting one dot above the figures specifying the number of feet, and two 
 dots in the case of inches. Thus 4' is read as four feet, and 4" is read as 
 four inches. 2' -2" is read as two feet and two inches; 6'-7-|" is read 
 as six feet and seven and one-half inches, and so on. Square feet and 
 square inches are represented by placing a square in front of the dots. 
 Thus 4 a ' is read as four square feet, and 4^" is read as four square 
 inches. 
 
 In specifying the material on structural drawings it is customary to 
 represent the shapes by symbols instead of writing the name. Thus, the 
 I-beam, channel, Z-bar, and angle are represented by I, [, Z, and L, 
 respectively, or in case of plural, as Is, [s, Zs, and Ls, respectively. 
 
 All dimensions and sizes given on structural drawings are in feet 
 and inches and fractional parts of inches. The fractional parts are 
 expressed in halves, fourths, eighths, sixteenths, thirty-seconds, and 
 sometimes in sixty-fourths of an inch. 
 
 Depths, widths, and thicknesses are always expressed in inches, 
 while length is always expressed in feet and inches. 
 
 In the case of I-beams and channels, the number required, depth, 
 weight and length are given thus: 
 
 1 I 12"x30#x20'-4"; 1 [ 15"x33#x6'-9 T y ; 
 
 2 Is 15"x42x21'-8i"; 6 [s 12" x 35* x 12'-6J"; etc.; etc. 
 
 In the case of angles, the number required, width of flanges (known 
 as legs), thickness of metal, and length are given, thus: 
 
 1__ L 5" x 31" x i" x 6'- W ; 1 L 6" x 6" x f " x 9'-7l" ; 
 
 4 Ls V x 4" x J" x 12'-8i" ; 7 Ls 3" x V x f " x 4'-0" ; etc. 
 
 In case of plates, the number required, width, thickness, and length 
 are given, thus: 
 
 1 Plate 36" Xl L"xl4'-2i"; 6 Plates 14" x T y x 4'-l". 
 
 In detailing a member, it is the usual practice to show an elevation, 
 top view, and a section through the member, looking downward. How- 
 ever, there should always be as many sections and views taken as is 
 necessary to clearly show the member in detail. 
 
 DRAWING ROOM EXERCISE NO. 1 
 
 Reproduce Plates 1, 2, and 3 each to a scale of 1J" = 1'-0". Each 
 drawing should first be reproduced in pencil upon an 18 x 24-inch sheet 
 of ordinary drawing paper, then traced upon a sheet of tracing cloth of 
 the same size. In making these drawings, the student should study each 
 detail as he goes along. All necessary information concerning the 
 material shown on these drawings will be found in the tables in the back 
 of this book or in a Carnegie or Cambria handbook. 
 
CHAPTER III 
 
 FUNDAMENTAL ELEMENTS OF STRUCTURAL MECHANICS 
 
 20. Structural Mechanics. Mechanics, in general, is the science 
 which treats of the effects produced upon bodies by force, while Structural 
 Mechanics, as the name would imply, is simply that part of mechanics 
 that applies directly to structures. 
 
 21. Body. In common usage, the word "body" refers to what is 
 known as an ordinary solid body. But in reality we are unable to think 
 of material (matter) at all other than as bodies. This is most readily 
 realized by attempting to conceive of material in a rarefied state, as in the 
 composition of a gas. We can conceive of it only as being made up of 
 very small bodies, known as particles, molecules, etc. Neither can we 
 conceive of the composition of solid bodies or liquids otherwise if thought 
 of in detail. Such being the case, it is evident that any part of a recog- 
 nized body, to us, is really a body in itself; and as we are not limited as to 
 subdivision, it is evident that any part of any body may, at will, be 
 treated as an independent body. 
 
 22. Force. In mechanics, the word "force" refers to that which 
 we recognize as the push or pull that bodies exert upon each other, which 
 invariably produces motion or a tendency of motion of the bodies con- 
 cerned. 
 
 Forces exerted by bodies at rest produce only a tendency to motion, 
 and are known as static forces, and the part of mechanics treating of such 
 forces is known as Statics. Forces exerted by moving bodies are known 
 as dynamic forces, and the part of mechanics treating of such forces is 
 known as Dynamics. Most of the mechanics known as Structural 
 Mechanics can be classed under the head of Statics. 
 
 Force is known by various names, as pressure, load, stress, reaction, 
 etc. These names are used to indicate certain existing conditions; how- 
 ever, force is the same wherever we find it. 
 
 23. Measure of Forces. Measuring forces is the same as measur- 
 ing any other thing; it is simply a matter of comparison, that is, we com- 
 pare forces with some one force which is taken as the basis. We are all 
 familiar with the practice of measuring forces in terms of the common 
 gravity unit, in which case the pull of gravity upon a certain body selected 
 as a standard is taken as the basis of comparison and designated as a 
 pound and known as the unit of weight. It is readily seen that this unit 
 is an arbitrary one, as the pull of gravity upon most any body could have 
 been selected as the unit, although in most cases convenience would have 
 been sacrificed. 
 
 In speaking of a force, we say a force of so many pounds, or the 
 intensity of a force is so many pounds. But what we really mean is that 
 the push or pull (as the case may be) exerted upon a certain body by 
 
 14 
 
FUNDAMENTAL ELEMENTS 15 
 
 some other body is so many times greater or less than the pull of gravity 
 upon the standard body, which pull, as stated above, is known as a pound. 
 It is evident that any static force is directly measurable as compara- 
 tive weight, but dynamic forces are more readily determined by consider- 
 ing the change of motion produced by them. It is known by experiment 
 that the force of gravity will cause the velocity of a body falling freely 
 to increase about 32.2 feet per second for each second of time. This 
 increment of velocity, 32.2 feet, is known as the acceleration of gravity 
 and is usually represented by the letter "g." Then using the force of 
 gravity expressed in the pound unit as our basis of comparison (as in 
 static forces) we can say that the intensity of any dynamic force is to the 
 force of gravity as the acceleration produced by the dynamic force is to 
 the acceleration produced by gravity, regardless of whether the body 
 considered moves horizontally, vertically, or in any other direction. For 
 example, suppose a dynamic force of an unknown intensity F by acting 
 continuously upon a body weighing W pounds increases its velocity a feet 
 per second for each second of time. Then by direct proportion we have 
 
 F : W :: a : g. 
 
 Expressing this in words we say: The unknown force (F) in pounds is 
 to the force of gravity (W) in pounds as the acceleration (a) in feet per 
 second caused by the unknown force (F) is to the acceleration (g) caused 
 by gravity. From which we have 
 
 F=j (A), 
 
 which is one of the fundamental equations of mechanics. The equivalent 
 formula is often written : F = ma, where m = W /g, and is known as the 
 mass of the body considered. But Formula (A) is the one mostly used 
 in practical work. The mass of a body is the quantity of material it 
 contains ; but in a practical sense, the mass of a body is equivalent to what 
 its weight would be if the acceleration of gravity were one foot (consider- 
 ing a foot as unity) per second instead of 32.2 feet, which it really is. 
 Then by direct proportion,, using the force of gravity as our basis as 
 before, we have 
 
 W 
 
 m : W : : 1 : g, or m - 
 
 t/ 
 
 which is the expression for mass given above. 
 
 The above discussion refers to constant forces, as the forces encoun- 
 tered in structural engineering are mostly constant forces. In case of 
 variable forces it is necessary to know the rate and the limit of variation 
 first, then the intensity can be determined. 
 
 Problem 1. A body weighing 200 Ibs. is acted upon by a constant 
 force which increases its velocity from to 64.4 ft. in 1 second of time. 
 What is the intensity of the force? 
 
 Solution: Substituting in Formula (A), we have 
 
 F = ^x 64.4 -400 Ibs. 
 32.2 
 
 Problem 2. What would be the intensity of the force in Problem 1 
 if the velocity of the body were increased in 5 seconds from 50 ft. per 
 second to 150 ft. per second? 
 
16 STRUCTURAL ENGINEERING 
 
 Solution: Here the increment of velocity or acceleration per second 
 is (150 50) -h 5 = 20 ft. Then substituting in Formula (A), we have 
 
 20 = 124 Ibs. (about). 
 
 Problem 3. If the body in Problem 1 had a velocity of 150 ft. per 
 second,, what would be the intensity of a constant force required to stop 
 the body in 3 seconds? 
 
 24. Direction of Action of a Force. Whenever a force is applied 
 to a body it either moves the body or has a tendency to move it. The 
 direction in which the body moves or has a tendency to move is said to be 
 the direction of action of the force causing the motion or tendency. 
 
 25. Line of Action of a Force. Whenever a body receives a force 
 by coming into contact with another body, which is the most common case, 
 there is always a surface upon which the force is exerted. W T e cannot 
 conceive of this force being transmitted to the surface other than that 
 each infinitesimal area of the surface receives a small amount of the total 
 force transmitted. So it is evident that what we usually consider a force 
 is really made up of an infinite number of forces. But it would be 
 impossible to deal with forces in mechanics when considered in this way, 
 since there would be an infinite number of forces to consider in every case. 
 We avoid this difficulty by assuming the total force to be applied along a 
 line which passes through the surface at the center of mean intensity of 
 the force and in the direction of its action. This line is known as the line 
 of action of the force. The line of action of a force is unlimited in length, 
 and the force may be considered to be applied at any point along the line 
 of action as far as motion or tendency of motion of the body upon which 
 it acts is concerned. 
 
 When a force is distributed over quite a large surface it is usually 
 necessary to consider the surface divided up into small portions, say, 
 portions one foot square, or one inch square, as the case may require, and 
 to treat the part of the total force exerted upon each portion as a separate 
 force. This means that the force exerted upon any portion will have a 
 line of action of its own passing through that portion at the center of the 
 mean intensity of the force exerted upon it. In any case we can obtain 
 only an approximation, but the approximation should come within the 
 limits of practicability. 
 
 In case of such forces as gravity and magnetism, where each particle 
 of material is assumed to be equally affected, the line of action passes 
 through what is known as the center of gravity of the body. However, if 
 the bodies be large, as is the usual case of structures, it often becomes 
 necessary to consider the body to be divided up into smaller ones and to 
 consider that the force acting upon each of the smaller bodies has a line 
 of action of its own, similar to the case of surfaces. 
 
 The plane in which the line of action of a force lies is known as the 
 plane of the force, and the direction of the line of action is known as the 
 direction of the force, and the direction of the action along the line of 
 action is known as the direction of the action of the force. Any number 
 of forces could act upon a body and each force could be in a different 
 plane, but in most of the problems in structural engineering the conditions 
 are such that we can consider them as acting in the same plane. 
 
FUNDAMENTAL ELEMENTS 17 
 
 26. Graphical Indication of Forces. When considering the effects 
 produced upon a body by forces, for convenience we graphically represent 
 the forces applied without reference to the bodies applying them. This is 
 done by drawing ail arrow in the line of action of each force. The arrow 
 point in each case indicates the direction 
 
 of action., and the point upon the body \<p\ &/ ~j 
 
 where the arrow touches indicates the \ V ?/ 
 
 point of application. Thus in Fig. 11, the ^1 5 ~~ b~ c~j 
 
 arrows PI, P2, P3 and P4 indicate forces ~i ' 
 
 applied upon the body AB at the points ^/ 
 
 a, b, c, and d, respectively. The arrows Fjg n 
 
 would be spoken of as forces PI, P2, etc. 
 
 And we would say that the forces PI, P2, and P4 act toward the body, 
 
 while the force P3 acts away from it, this being indicated in each case 
 
 by the arrow points. 
 
 27. Applied Forces and Reactions. An applied force is any force 
 which we conceive of as being applied to a body from without by some 
 other body, and at the same time as being the initiative of the outward 
 activity. A reaction is the same as an applied force, except we conceive 
 of it as resisting the activity instead of being the cause of it. Both are 
 spoken of as external forces. For example, a body placed at C upon the 
 bar AB (Fig. 12) by virtue of its own weight will exert a force P (pres- 
 
 sure) upon the bar at that point. This 
 r force P will cause the bar to have a ten- 
 
 A Li * ijl B dency to move downward, which is the ac- 
 
 JRI JR2 tivity produced. Actual motion is prevented 
 
 Fig. 12 by the supports at A and B exerting the 
 
 forces Rl and R2 acting upward on the 
 
 bar. We would call P an applied force, as we conceive of it as being 
 
 the initiative of the activity of the bar, while we would call Rl and R2 
 
 reactions, as we conceive of them as resisting the activity, which, in this 
 
 case, as stated above, is the tendency of motion caused by the force P. 
 
 But regardless of name, all three of these forces are exactly the same 
 
 in character. This can be seen very readily 
 
 by imagining the bar to be turned upside 
 
 down and supported only at C, while all 
 
 three of the forces continue to act the same 
 
 in reference to the bar. Then the case 
 
 would be as shown in Fig. 13. This could Fig. 13 
 
 be actually accomplished by placing a body 
 
 weighing the same in pounds as the reaction Rl where Rl is indicated 
 
 to act and another body weighing the same in pounds as the reaction R2 
 
 where R2 is indicated to act. Then Rl and R2 would become applied 
 
 forces, while P would become a reaction. 
 
 28. Stress. Stress is the push or pull which the constituent parts 
 of a body exert upon each other due to the application of external forces. 
 Stresses are known as internal forces, while applied forces and reactions 
 are known as external forces. 
 
 As a simple case, suppose a steel rod having a uniform cross-section 
 of A square inches is suspended from one end and a weight of P pounds 
 is hung on the other. It is known from experience that the pull P exerted 
 
18 STRUCTURAL ENGINEERING 
 
 at the lower end of the rod by the weight is in turn exerted at the uppe"r 
 end. So this pull is, as we say, transmitted through the rod, and we 
 cannot conceive of this transmission taking place other than that the 
 material particles exert a pull upon each other in the direction of the 
 length of the rod. A practical idea of the stress in the rod can be obtained 
 by first imagining the weight P to be supported by a very small wire, so 
 small in cross-section that the wire would be really a row of individual 
 molecules ; then we can conceive of the pull P being transmitted along the 
 wire from molecule to molecule, thus producing a stress of P pounds upon 
 each molecule. Next imagine another wire of the same size suspended 
 along the side of the wire just considered, and suppose this second wire 
 now supports half of the weight P: then the stress on the molecules in 
 either wire would be P/2 ; and by adding another wire of the same size 
 the stress on the molecules in each wire would be P/3; and by adding 
 another wire it would be P/4; and so on. So if there be n number of 
 wires, the stress on each would be P/n. Now imagine the rod made up 
 of n number of such wires and let da be the area of the cross-section of 
 each wire in square inches, and let s be the stress per square inch on each 
 wire; also let p be the stress per square inch on the rod at any cross- 
 section. It is obvious that the stress per square inch on the rod at any 
 cross-section will be p = P/A. Then, likewise, the stress on each wire 
 being P/n, the stress per square inch in each wire will be 
 
 s= , = 
 
 da nda 
 
 But nda = A; 
 
 therefore, we have 
 
 P 
 
 s = -=p. 
 
 That is, the stress on the rod per square inch is the same whether we 
 consider the entire cross-section or a portion or a mere molecule in the 
 cross-section. When we say that the stress on a body is so much per 
 square inch, we do not necessarily mean that there is a square inch of 
 material in the body that actually has that stress. We may simply mean 
 that there is some material in the body that has that stress ; it may be a 
 square foot or a millionth part of a square inch. The stress per square 
 inch on a body is known in structural mechanics as Unit Stress. 
 
 If the above rod were stood up vertically say, upon a floor and 
 the weight placed upon the top end, the direct stress produced by the 
 weight would be the same in intensity as if suspended as above, but the 
 stress would be produced by a push instead of a pull. When the stress in 
 a body is produced by a direct pull, it is known as Tensile Stress; but 
 when it is produced by a direct push, it is known as a Compressive Stress. 
 
 Whenever a body is subjected to one kind of stress uniformly dis- 
 tributed over its cross-section throughout its length, as was considered in 
 the case of the above rod, the stress is known as Simple Stress. The unit 
 stress in that case at any cross-section is always equal to the total stress 
 on the cross-section divided by the area of the cross-section, or, in general, 
 we have p = P/A, as given above. 
 
FUNDAMENTAL ELEMENTS 19 
 
 From the above discussion of stress one may be led to think of all 
 bodies as being fibrous in structure. It is true thst wood, wrought iron, 
 and some other materials manifest this structure to some extent, but not 
 wholly, while some other materials, as steel and concrete for example, 
 seem to be devoid of it. But in any case we .cannot conceive of the 
 arrangement of the ultimate parts of a body other than in some consecu- 
 tive order, and consequently our demonstration is not faulty, as we only 
 conform to that conception. 
 
 Stress, known as shear, cross-bending, and torsion, will be treated 
 later under these specific heads as the occasion for their treatment occurs, 
 
 29. Elasticity. It is an observed fact that whenever a force is 
 applied to a body, the dimensions of the body are changed, and that when 
 the force causing such a change is released, the body has a tendency to 
 regain its original dimensions, and will regain them unless the force be so 
 great as to break or injure the body. This property of regaining their 
 original dimensions which bodies manifest is known as their elasticity. 
 We can conceive of this regaining of dimensions as being due to a stress- 
 resisting force inherent in the material and which we can designate as the 
 force of elasticity, yet the nature of the force is unknown. As an example, 
 suppose a steel rod to be in tension. We can conceive of the material 
 particles exerting a pull upon each other. This we call stress. And we 
 can conceive of this pull being resisted by a force inherent in each particle. 
 This inherent force is the force of elasticity, the intensity of which, in 
 accordance with Newton's Law, is necessarily equal to the stress which it 
 resists. 
 
 30. Distortion. In this book the total change of form of a body 
 due to external forces will be known as distortion, while the amount of 
 change per unit of dimension of a body will be known as unit distortion. 
 What really takes place in every case is cubic distortion, but, in deter- 
 mining the distortion of structures, it is usually necessary to consider only 
 the distortion of the length of their parts, and hereafter the word dis- 
 tortion in this book will refer only to the distortion of length unless other- 
 wise stated. Suppose a steel rod, ten feet long, is suspended from one 
 end and a weight is hung on the other end. If the weight causes the rod 
 to stretch one-tenth of an inch in length, the distortion of its length or 
 simple distortion, in that case, is one-tenth of an inch, while the unit 
 distortion, considering an inch as the unit of length, is 1/120 of the 
 distortion, the rod being ten feet long or 120 inches. 
 
 31. Elastic Limit. Whenever a body is distorted the maximum 
 amount that it will sustain and yet return to its original form when 
 permitted to do so, it is said to be distorted to the elastic limit. If a body 
 is distorted beyond the elastic limit, it will remain shorter or longer than 
 its original length. Then we say that the body has taken set. If a body 
 be in tension, its length would remain longer, and if in compression it 
 would remain shorter. 
 
 32. Relation of Stress and Distortion. Hooke (in 1678) was the 
 first to state that the distortion of a body is directly proportional to 
 the stress. This is known as Hooke's Law, and has been proven by 
 experience to be practically true, provided the distortion does not exceed 
 the elastic limit. Whenever a body is distorted beyond the elastic limit, 
 the distortion increases more rapidly than the stress. 
 
20 STRUCTURAL ENGINEERING 
 
 Suppose a steel rod having a length L and a uniform cross-section 
 A is distorted an amount D when subjected to a simple stress P; then, 
 according to Hooke's Law, if the stress were 2P, the corresponding dis- 
 tortion of its length would be 2D; if 3P, it would be 3D; and so on. 
 While this direct proportion holds good in all cases as long as the stresses 
 are within the elastic limit, yet the actual value of D in any case will 
 depend upon the value of L and A, being directly proportional to L,, and 
 inversely proportional to A. This is readily seen, for it is obvious that 
 if the rod were two feet long it would be distorted twice as much when 
 subjected to the same stress as a rod of the same section only one foot 
 long, since each foot of length would be distorted the same in amount in 
 either case. 
 
 In regard to the cross-section, it is obvious that if the area is two 
 square inches, the distortion of the rod will be only one-half as much as 
 when its area is one square inch, as the actual stress on the material in the 
 rod in the first case is only one-half as much as it is in the second case. 
 33. Modulus of Elasticity and Determination of Simple Dis- 
 tortion. According to Hooke's Law, if we know the distortion of any one 
 piece of material subjected to a known stress, we can determine the 
 distortion of any other piece of the same kind of material subjected to any 
 known stress simply by direct proportion. 
 
 Suppose it is found by experiment that a rod of some unknown 
 material, having a length of L inches and a uniform cross-section of A 
 square inches, is distorted D inches in length when subjected to a simple 
 stress of P pounds. What would be the distortion of a rod of the same 
 kind of material having a length of L' inches and a uniform cross-section 
 of A' square inches, if subjected to a simple stress of P' pounds? Let 
 D' be the distortion required. Reducing everything concerned to its 
 lowest terms, which is done for convenience, we have 
 
 The unit stress in the first rod = p = P/A, 
 while the unit distortion in inches = d = D/L. 
 The unit stress in the second rod = p f P f /A' 9 
 while the unit distortion in inches = d' = D' /L'. 
 By direct proportion we have 
 
 Expressing this in words, we say: The stress (p) per square inch 
 in the first rod is to the stress (p') per square inch in the second rod as 
 the distortion (d) of the first rod per inch of length is to the distortion 
 (d') of the second rod per inch of length; from which we have d' = dp'/p t 
 which is the distortion of each inch of the second rod ; then, of course, the 
 total distortion of the rod would be L' times this, so we have 
 
 D' = d'L'=- p'L' . .............................. (3). 
 
 P 
 
 Now d/p is the ratio of the unit distortion to the unit stress given 
 for the first rod, but according to Hooke's Law, this ratio is the same for 
 any piece of this same kind of material. So if this ratio is determined 
 for any one piece of this material, the elastic property of the material is 
 known, and the distortion of any piece of the material can then be deter- 
 mined if its length, area of cross-section, and stress are known. 
 
FUNDAMENTAL ELEMENTS 21 
 
 The ratio of the unit distortion to the unit stress, as expressed above, 
 would be a very small fraction in any case, so, for convenience, we can 
 invert the expression so as to have a whole number. Then we have 
 p/d p f /d' E, which is a constant for any piece of the same kind of 
 material composing the two rods. Now substituting 1/-E for d/p in 
 equation (2), we have D' = p'L'/E. E would be known as the Modulus 
 of Elasticity of the material composing the two rods. For the modulus of 
 elasticity of any material in general we have 
 _, p unit stress 
 
 Jl, ~ - - - y 
 
 a unit distortion 
 
 which is usually designated as Young's Modulus. It varies with the 
 different kinds of material, but is practically constant for all pieces of 
 the same kind. For the simple distortion of any piece of any kind of 
 material having a uniform cross-section and being subjected to a simple 
 stress, we have the general equation 
 
 where D = distortion of the piece in inches (or feet if L be taken in feet) ; 
 P = total stress on the cross-section of the piece in pounds ; 
 A = area of cross-section in square inches ; 
 L = total length of piece in inches or feet; 
 p = unit stress on the cross-section of the piece; 
 E = modulus of elasticity of the material composing the piece. 
 The modulus of elasticity of a material is determined by experiment. 
 The average values of the moduli of elasticity of the principal materials 
 used in structural engineering have been found to be as follows : 
 29,000,000 for steel; 
 26,000,000 for wrought iron; 
 30,000,000 for cast steel; 
 18,000,000 for cast iron; 
 
 720,000 for long-leaf yellow pine; 
 555,000 for white oak. 
 
 In determining the modulus of elasticity, it is laboratory practice to 
 measure the distortion in inches, in which case the length of the test piece 
 must always be reduced to inches; but in using Formula (B), L may be 
 taken either in feet or inches. If taken in feet, the distortion will be 
 given in feet, and if taken in inches, it will be given in inches. 
 
 Problem 4. Suppose a steel rod having a length of 10 feet and a 
 uniform cross-section of 2 square inches is suspended from one end and 
 supports a weight of 30,000 pounds hung on the lower end: 
 
 (a) What will be the distortion of the rod due to the 30,000 pounds ? 
 Referring to Formula (B), given above, we have in this case 
 
 P = 30,000; p = 30,000 *2 = 15,000; E- 29,000,000. Substituting these 
 values in the formula, we have for the distortion 
 
 15,000x10 f -t . nch (about) 
 
 29,000,000 
 
 (b) What will be the distortion of the rod due to its own weight? 
 In that case the stress varies from zero at the bottom of the rod to a 
 
 maximum at the top. For convenience, instead of using the numerals 
 
22 STRUCTURAL ENGINEERING 
 
 given above, let L be the length of the rod and A the cross-section; and 
 let w be the weight of the rod per foot of length. Then the stress on the 
 cross-section at a point b (Fig. 14), x feet from the 
 lower end of the rod, will be P = wx, which is really the 
 weight of the rod below b. 
 
 For the distortion of the length of the material in 
 the rod between the cross-section at b and the cross- 
 section at c, which is an infinitesimal distance dx above 
 b, we have 
 
 wxdx 
 
 which corresponds to Formula (B), given above. 
 
 Now as x is a variable, varying from to L, the last expression will 
 give the distortion of any infinitesimal part of the length of the rod any- 
 where between the ends. Then the total distortion of the rod will be equal 
 to the summation of the distortion of these parts ; so, for the total distor- 
 tion of the rod, we have 
 
 wxdx 1 wL 2 
 
 /I 
 
 AE ~ 2 AE 
 
 Now let wL, which is the total weight of the rod, be represented by P. 
 Then substituting in (c) we have 
 
 This shows that the distortion of the rod due to its own weight is one-half 
 of what it would be for an equal weight suspended from the lower end. 
 The weight of the rod = 6.8 X 10 = 68 Ibs. Substituting in (d) we have 
 for the distortion 
 
 Problem 5. What will be the distortion of length of a solid, circular, 
 cast-iron column having a diameter of 6 ins. and a length of 18 ft., when 
 supporting a load of 282,000 Ibs.? 
 
 Problem 6. What will be the distortion of an eye-bar 8" x 1 J" x 39'-0" 
 (c.c. end holes) when subjected to a stress of 16,000# n "? 
 
 34. Motion. Strictly speaking, a body can have but two distinct 
 kinds of motion, one known as translation and the other as rotation. 
 When a body moves as a whole along an imaginary line, known as its 
 path, its motion is translation, particularly in reference to any relatively 
 fixed position in the immediate vicinity of the body; but if it either moves 
 around or turns about an imaginary line, known as its axis, its motion is 
 rotation in reference to that line. 
 
 There is one fact regarding motion which should never be overlooked, 
 that is, motion is absolutely relative, and that the kind of motion is always 
 dependent upon the position of reference. For instance, the motion of a 
 particle in a body rotating about an axis is rotation in reference to that 
 axis; yet at the same time its motion (the particles in the axis excepted) 
 is translation when referred to any relatively fixed position in the imme- 
 diate vicinity of the particle. 
 
FUNDAMENTAL ELEMENTS 
 
 23 
 
 Fig. 15 
 
 In reference to time, motion is either uniform or variable; in refer- 
 ence to the character of path described it is either rectilinear or curvi- 
 linear, being rectilinear when the path is a straight line and curvilinear 
 when a curve. 
 
 35. Moment of a Force. The moment of a force about any point 
 is a measure of its tendency to produce rotation of the body upon which 
 it acts about the point chosen and is equal to the perpendicular distance 
 from the point to the line of action of the force multiplied by the intensity 
 of the force. The point is known as the center of moment, while the 
 perpendicular distance is known as the lever arm of the force, or simply 
 "arm." 
 
 In the case shown in Fig. 15, the moments of the forces PI, P2, and 
 P3 about the point O would be expressed as aPl, bP2, and cP3, respect- 
 ively. As is readily seen, the forces PI 
 and P2 would have a tendency to rotate 
 the body A clock-wise about 0, while P3 
 would have a tendency to rotate it counter 
 clock-wise. Now, taking one direction as 
 minus and the other direction as plus, say, 
 counter clock-wise minus, and letting M 
 be the algebraic sum of the moments of 
 the three forces about 0, we have M = 
 P1 + bP2 - cP3, which is the equation 
 of moments of the three forces 
 about 0. 
 
 In case of forces not in the same plane, the moments are usually 
 taken about a line, in which case the center of moment of each force is the 
 point on the line nearest the force. The tendency of rotation about the 
 line is what is really measured in that case. 
 
 It is obvious that the moment of a force about any point in its line 
 of action is zero, for in that case its arm is zero. If the arm is taken in 
 feet and the force in pounds, the moment will be expressed in what is 
 known as foot pounds, but if the arm is taken in inches and the force in 
 pounds, the moment will be expressed in what is known as inch pounds. 
 
 36. A Couple and Moment of Same. Two equal and opposite 
 parallel forces acting upon a body (other than along the same line of 
 action) form what is known as a couple. Thus in Fig. 16, the force P 
 applied at c upon the body AB and the equal and opposite force P applied 
 at b form a couple. 
 
 The moment of a couple is equal to the distance between the forces 
 
 multiplied by the intensity of one of the $ 10 
 
 forces. For example, the moment of the 
 
 couple shown in Fig. 16 is M = Pd. j 
 
 The moment of a couple is not changed 
 by changing the center of moments, that 
 is, the moment about one point is the same ft I \t> 
 as about any other point in the plane of t p 
 
 the forces. This is readily seen, for tak- 
 ing moments about e or g, we have M = ' )U 
 (es + sg)P Pd; taking moments about s, 
 
 iLJB 
 
 Fig. 18 
 
24 STRUCTURAL ENGINEERING 
 
 we have M = P(se) + P(sg) = P(se + sg) = Pd; and, again, taking 
 moments about o, we have M = P(ho) - P(ko) = P(ho - ko) = P(hk) = 
 Pd. Now, evidently, if the moment of a couple is the same for all points 
 in the plane of the forces, a couple can be considered to be applied 
 anywhere in the plane of its forces, which means that we can consider 
 any couple as being moved bodily to any desired position in the plane of 
 the forces. Of course, the forces must act upon the body concerned. 
 
 Any two couples are equivalent when their moments are equal. But 
 the forces and arms in the two cases are not necessarily equal. Rotation 
 or tendency of rotation is always due to the action of a couple. No single 
 force can produce rotation. 
 
 37. Concurrent and Non- Concurrent Forces. When the lines of 
 
 action of two or more forces acting upon a body meet at a common point, 
 the forces are said to be concurrent, while if their lines of action do not 
 so meet, the forces are said to be non-concurrent. 
 
 38. Conspiring Forces. When the lines of action of two or more 
 forces acting upon a body coincide, the forces are known as conspiring 
 forces. 
 
 39. Resultant and Component Forces and Graphical Repre- 
 sentation of Same. A body acted upon by two or more forces may, 
 thereby, have a tendency to move in more than one direction at the same 
 instant ; but as a body cannot occupy two positions at the same time, it is 
 evident that motion can take place in but one direction, which is known as 
 the direction of the resultant of the two or more forces. Now, it is evident 
 that a single force acting in this direction with a certain required intensity 
 could produce. the same effect as that produced by the two or more forces 
 combined. Such a force would be known as the resultant of the two or 
 more forces, while each of the two or more forces would be known as a 
 component of the resultant force. 
 
 For example, if two concurrent forces PI and P2 (Fig. 17) be 
 
 applied simultaneously to a small 
 body resting upon a smooth, horizon- 
 tal plane at A, we can readily see that 
 the force PI would have a tendency 
 to move the body along its line of 
 action Ac, while at the same time the 
 force P2 would have a tendency to 
 move the body along its own line of 
 Fig. 17 action Ab. Now we know, from 
 
 reasons given above, that the body 
 
 cannot move along both of the lines at the same time, and we can readily 
 see that it would not move along either of the lines, as the two tendencies 
 of motion are not in the same direction; so, undoubtedly, if the two forces 
 moved the body, the motion would take place along some line different 
 from either of the two. 
 
 Suppose the force PI, acting alone and continuously for one second, 
 moved the body from A to B, so that at the end of the second the body 
 would be at B. Then suppose the force PI ceased to act at that instant, 
 and suppose the body stopped instantly at that point and the force P2 
 were then applied and in the same manner moved the body from B to C in 
 
FUNDAMENTAL ELEMENTS 25 
 
 one second,, so that at the end of the next second the body would be at C. 
 Then, the two forces PI and P2, each acting separately for one second, 
 would have moved the body from A to C in two seconds, but if they had 
 acted simultaneously upon the body they would undoubtedly have accom- 
 plished the same thing in one second, for the simple reason that each 
 force would have been acting for one second, as in the case when acting 
 separately, and with the same intensity, for otherwise they would lose 
 their identity. By this it is not meant that the two forces acting simul- 
 taneously would have moved the body along the same paths as when 
 acting separately, but that they would have moved it from A to C along 
 some line in one second. We cannot conceive of these forces doing any- 
 thing more than they would have to do in moving the body from A to C; 
 then, undoubtedly, they would move it along the straight line AC joining 
 the two positions, as that would be the least required of them in the 
 transaction. Then the line AC would be the direction of the resultant of 
 the forces PI and P2. 
 
 Now it is evident that the two forces PI and P2 acting simulta- 
 neously upon the body at A would have the same effect as a single force 
 R (Fig. 17) would have if acting along the line AC with such an intensity 
 as to move the body from A to C in one second. Therefore, the two forces 
 PI and P2 could be replaced by this single force R, which is their 
 resultant. Each of the forces PI and P2 are, in that case, components 
 of the resultant force R. 
 
 Assuming the resistance to be the same everywhere on the plane 
 which was really the assumption at the start the distance AB will be 
 directly proportional to the intensity of force PI, which moved the body 
 over that distance, and, likewise, the distance BC will be directly propor- 
 tional to the force P2. Then as the sides AB and BC of the triangle 
 ABC (Fig. 17) are directly proportional to the forces PI and P2, 
 respectively, it is evident that we can construct a similar triangle such 
 that the sides will represent to convenient scale the intensities of these 
 forces. Such a triangle is shown in Fig. 18, where the side PI, parallel 
 to the side AB in Fig. 17, is drawn to represent the intensity of the force 
 PI to scale (so many pounds per inch) and the side P2 parallel to the 
 side BC in Fig. 17 is drawn to represent the 
 intensity of the force P2 to the same scale as PI. 
 Then the side R, which will be parallel to the 
 side AC in Fig. 17, will represent the intensity 
 of the resultant force R to the same scale as PI 
 and P2. Now it is evident that any two concur- Fig. 18 
 
 rent forces with their resultant can be repre- 
 sented in magnitude and direction by the sides of a triangle which is 
 known as a Force Triangle. It is important to note that the forces act 
 in one direction around the triangle, while their resultant acts in the 
 opposite direction. 
 
 Suppose, instead of the two forces PI and P2 acting upon the small 
 body shown in Fig. 17, that there be four concurrent forces PI, P2, P3, 
 and P4, as shown in Fig. 19. The resultant R of the two forces PI and 
 P2 is determined by constructing the force triangle ABC in the same 
 manner as was shown above. Then as R is the resultant of the two forces 
 
26 
 
 STRUCTURAL ENGINEERING 
 
 Fig. 19 
 
 PI and P2, it will replace them, and the remaining forces to be considered 
 are P3, P4, and R. Now constructing the force triangle A CD (Fig. 19) 
 
 for the force R and P3, we have their re- 
 sultant 121. Next, constructing the force 
 triangle ADE for the two remaining forces, 
 121 and P4, we have their resultant, R2, 
 which is the final resultant, or, in other 
 words, is the resultant force of all the 
 forces PI, P2, P3, and P4, and as such can 
 replace them. We now have a polygon 
 ABCDEA wherein the forces are seen to 
 act in the same direction around the poly- 
 gon, while their final resultant 122 acts in 
 the opposite direction. Such a polygon is 
 known as a Force Polygon. The force 
 polygon ABCDEA in this case was con- 
 structed by combining force triangles; but 
 it is evident that the same could have been 
 constructed by laying off the forces PI, P2, P3, and P4 in consecutive 
 order, thus obtaining the part ABCDE of the polygon, and then joining 
 EA for completion as shown in Fig. 20. 
 
 While it is advisable to begin with some force as PI and lay the 
 forces off in consecutive order around the body as we have done here, yet 
 
 it is not absolutely neces- 
 
 sary to do so, for it will 
 
 ,E be seen upon inspection 
 
 that the forces may be 
 taken in any order (ex- 
 cept that they must be 
 laid off so that they act 
 in the same direction 
 around the polygon) as 
 shown in Fig. 21, where 
 the line AE, which is the 
 final resultant, is the 
 same in length and direc- 
 tion as the final resultant 
 AE shown in Fig. 19 and 
 Fig. 20. 
 
 Now it is evident from the above that any number of concurrent 
 forces with their resultant can be represented in magnitude and direction 
 by the sides of a closed polygon known as a force polygon, wherein the 
 forces act in one direction around the polygon while their resultant acts 
 in the opposite direction, the direction of action in each case being indi- 
 cated by the arrow points. The force triangle is really a force polygon 
 wherein the component forces are only two in number. 
 
 As any number of concurrent forces with their resultant can be 
 represented in magnitude and direction by a closed polygon, it follows 
 that any force may be resolved into any number of components, the only 
 requirements being that the force and its components form a closed 
 polygon. Thus we can construct any number of components as A a, ab, 
 
 Fig. 20 
 
 Fig. 21 
 
FUNDAMENTAL ELEMENTS 
 
 27 
 
 be, cd, and dB for the force AB, as shown in Fig. 22, by simply drawing 
 the components practically at will, the only restrictions being that they 
 form a closed polygon with the force AB. Here the force AB is consid- 
 ered as a resultant, which is obviously permissible in any case. We would 
 say that the force AB was resolved into its component forces Aa, ab, be, 
 cd, and dB. Resolving forces into components is known in mechanics as 
 the Resolution of Forces. The most convenient components into which a 
 force can be resolved are two components whose directions are at right 
 b 
 
 Fig. 22 
 
 Fig. 23 
 
 
 B 
 
 angles to each other, and are known as rectangular components. Thus 
 AD and DB or AE and EB (Fig. 23) are rectangular components of the 
 force AB. It is obvious that a force can be resolved into any number of 
 pairs of rectangular components or any number of any other kind. 
 
 It will now be shown that the above is as true for non-concurrent 
 forces as it is for concurrent forces. Let PI, P2, and P3 (Fig. 24) 
 represent three non-con- 
 current forces acting s\ V^L, 
 upon the body AB. As p X \ \ 
 far as motion or ten- ^^i 
 dency of motion are con- 
 cerned, any of these 
 forces may be considered 
 as applied at any point 
 along their line of action ; 
 however, in making such 
 assumptions in any case, 
 we are compelled to con- 
 ceive that there always 
 exists the necessary ma- 
 terialistic connection be- 
 tween such points of ap- 
 plication and the body 
 concerned. Then, if we 
 prolong the lines of action of the two forces PI and P2 until they inter- 
 sect at 0, we can consider them as being two concurrent forces applied 
 at 0, and by constructing the force triangle OCD (where OC = P2 and 
 CD - PI to scale) we have their resultant given in amount and direction 
 by the line OD. Now this resultant, which we will designate as 121, can be 
 considered as being applied at any point along yy, its line of action (the 
 same as any other force). Then prolonging the line of action yy of this 
 resultant 121 until it intersects the line of action of the third force P3 at 
 0', and constructing the force triangle O'EF, we have the resultant 122, 
 which is the final resultant, or, in other words, the resultant of all three 
 
 Fig. 24 
 
28 STRUCTURAL ENGINEERING 
 
 forces, PI, P2, and P3. This final resultant R2 can be considered as 
 being applied at any point upon its line of action xx. 
 
 From the above it is evident that the resultant of any number of 
 intersecting non-concurrent forces can be represented graphically by first 
 selecting any two forces and prolonging their lines of action until they 
 intersect; then constructing a force triangle and obtaining the resultant 
 of these two forces ; then prolonging the line of action of this resultant 
 until it intersects with the line of action of a third force; and then again 
 constructing a force triangle and obtaining the resultant of the first 
 resultant and the third force; and again prolonging the line of action of 
 this second resultant until it intersects the line of action of a fourth force ; 
 and so on until all of the forces are combined. The last resultant will be 
 the resultant of all of the forces, or the final resultant. 
 
 As an additional example, let PI, P2, P3, P4, and P5 represent five 
 non-concurrent forces acting upon the body AB, as shown in Fig. 25 (a). 
 Let us first take the two forces PI and P2 and prolong their lines of action 
 until they intersect at 0. Then by constructing the force triangle ODC 
 we obtain their resultant JR1. Then by prolonging the line of action of 
 this resultant until it intersects the line of action of a third force P3 at 0', 
 and constructing the force triangle O'EF, we obtain R2, the resultant of 
 Rl and P3. Then prolonging the line of action of R2 until it intersects 
 the line of action of a fourth force P4 at 0", and constructing the force 
 
 (a) 
 
 Fig. 25 
 
 triangle 0"GH, we obtain 123, the resultant of R2 and P4. Then pro- 
 longing the line of action of R3 until it intersects the line of action of a 
 fifth force P5 at 0" ', and constructing the force triangle 0" ' KL, we 
 obtain the resultant .R4, which is the final resultant, or, in other words, 
 the resultant of all the forces PI ... P5. 
 
 So far our treatment of non-concurrent forces does not seem to have 
 much in common with our treatment of concurrent forces, but let us go a 
 little further and see what we can observe. The force triangles ODC, 
 O'EF, 0"GH, and O" ' KL are the same in kind as we found in our 
 treatment of concurrent forces. Now suppose the force triangle ODC in 
 Fig. 25 (a) be moved bodily and parallel to itself to the position ODC in 
 Fig. 25 (6). This would not change the triangle in the least. Next 
 suppose the force triangle O'EF be moved in the same manner to Fig. 25 
 (6), taking the position OCF f the vertex 0' coinciding with in Fig. 
 25 (6) and E with C. Next, suppose the force triangles 0"GH and 
 0" ' KL (in Fig. 25 (a)) to be moved in the same manner to the position 
 in Fig. 25 (6) so that 0" and 0" ' would fall at O. Then these two 
 triangles would take the positions OFK and OKL, respectively. By thus 
 grouping the force triangles we have constructed the force polygon 
 
FUNDAMENTAL ELEMENTS 29 
 
 ODCFKLO, which is the same kind of a force polygon as was constructed 
 for concurrent forces in Fig. 19, and it can be drawn in the same manner. 
 This shows that any number of non-concurrent forces with their resultant, 
 the same as concurrent forces, forms a closed polygon, known as a force 
 polygon, and hence the resultant of any number of any kind of external 
 forces acting upon any body whatever can be determined simply by con- 
 structing a force polygon representing each in intensity and direction, at 
 the same time placing the forces so that they act in the same direction 
 around the polygon. 
 
 As a general case, let Fig. 26 represent a body acted upon by four 
 non-concurrent forces PI . . . P4. We can determine the resultant of 
 these four forces by simply taking any one of them, say, P4, and drawing 
 a line as AB equal and parallel to it. Then from B draw BC equal and 
 parallel to P3. Then from C draw CD equal and parallel to P2. Then 
 from D draw DE equal and parallel to PI. 
 Then joining E and A we have EA, which /r,/ 
 represents the resultant of the four forces in \J IRsV^ 
 
 intensity and in direction. This same method JL * j j 
 
 of procedure holds for both concurrent and 
 
 non-concurrent and also for conspiring forces. 
 The conspiring forces would simply be laid 
 off in one straight line. Their resultant would 
 be represented by the distance from the last 
 force laid off to the starting point. 
 
 40. Analytical Representation of Resultant and Component 
 Forces. From the force triangle shown in Fig. 18 (Art. 39) -we have 
 
 R 2 = Pl 2 + P2 2 + 2 cos <f> PlxP2 (1), 
 
 where </> is the angle which the lines of action of the forces PI and P2 
 make with each other. Any one of the forces PI, P2, R, and the angle <f> 
 can be determined from the above equation, provided the other three are 
 given. When < = 90, the above equation becomes 
 
 which is the "rectangular equation" that results whenever a force is 
 resolved into rectangular components. As a rule, it is much more con- 
 venient to express these rectangular components in terms of the force 
 and angles which they make with the force. Thus, from Fig. 27, we have 
 
 12 cos = PI; 12 = =Plxsec0; 
 
 Rcos6 = P2; cos</> 
 
 Rsind> = P2', P2 
 
 The most useful thing to be observed from the above is that the 
 rectangular component of a force along any line is equal to the force 
 multiplied by the cosine of the angle which the force makes with the line. 
 
 In the 'case of conspiring forces, it is evident that the resultant of 
 two or more conspiring forces is equal to their algebraic sum, for we can 
 consider the forces acting in one direction as plus and those acting in the 
 opposite direction as minus. Then by adding all of the plus forces 
 together and all of the minus forces together and subtracting the lesser 
 
30 
 
 STRUCTURAL ENGINEERING 
 
 from the greater sum, we shall thus obtain the intensity and direction of 
 
 action of their resultant. 
 
 The resultant of three or more forces 
 can be determined analytically most readily 
 by resolving the forces into their horizontal 
 and vertical components, thus obtaining a 
 rectangular equation where the square of 
 the algebraic sum of the horizontal com- 
 ponents plus the square of the algebraic 
 sum of the vertical components is equal to 
 
 the square of the resultant. Thus from Fig. 19 (Art. 39) we have 
 
 Fig. 27 
 
 But aE = ab + be 4 cE = P2cos6 l + P3cos0, + P4cos0 3 , 
 
 which is the algebraic sum of the vertical components of the forces 
 PI, P2, P3, andP4; 
 
 and 
 
 = AB + Bd-Cf-Dc = PI + P2sm0 1 - P3sin0 2 - P4sin<9 3 , 
 
 which is the algebraic sum of the horizontal components of the same 
 forces, as is readily seen from Fig. 19. 6 13 6 2 , etc., are the angles which 
 the forces make with the vertical line aE. PI has no vertical component, 
 as the cosines of the angles it makes with the vertical is O the angles 
 being 90 and, the sine of 90 being 1, it is evident that its horizontal 
 component is equal to PI. 
 
 41. Proposition. The moment of the resultant of two or more 
 forces about any point in the plane of the 
 forces is equal to the algebraic sum of the 
 moments of the 'two or more forces them- 
 selves, about the same point. 
 
 As the resultant of two or more forces 
 is a force which will produce the same effect 
 (as far as motion or tendency of motion are 
 concerned) as the two or more forces com- 
 bined, it is practically self-evident that the 
 above proposition is true. However, it is 
 seen to be true from the following demon- 
 stration : 
 
 Let R (Fig. 28) represent the resultant of two concurrent forces PI 
 and P2, acting upon a body at A, and let the intensity of these forces and 
 their resultant be represented by the sides of the force triangle ABC. 
 Select any point as the center of moments. Then we are to prove that 
 the moment of R about O is equal to the algebraic sum of the moments of 
 the component forces PI and P2 about the same point. Now, it makes no 
 difference where is taken, we can always draw a line through parallel 
 to the line of action of the resultant R. Then in this case draw the line 
 OO' parallel to R and from A draw h' perpendicular to the line 00'; 
 then the moment of the resultant R about O' is the same as it is about O, 
 since h' = h. 
 
 For the equation of moments about 0', if the above proposition be 
 true, we have Rh' = a'Pl + d'P2; but a' = h'cos<j> and d' = h'cosO, and 
 substituting we have Rh' = (Plcos<j>-\-P2cosO)h' t or R = Plcos<j> + P2eos6. 
 
 Fig . 2 s 
 
FUNDAMENTAL ELEMENTS 31 
 
 But from the force triangle ABC we have R = Plcos <J> + P2cos 6; then 
 substituting this value for R in the last equation, we have (Plcos<-f 
 P2cos#) (Plcos< + P2cos#) an identity,, which proves the proposition 
 when 0' is the center of moments. Now,, as the moment of R about is 
 the same as it is about 0' , it remains for us to prove that the change in 
 the algebraic sum of the moments of PI and P2, due to the changing of 
 the center of moments from 0' to is zero. That is, we are to prove that 
 (a - a')Pl - (d' -d)P2 = 0, or (a - a')Pl = (d f - d)P2. Now from Fig. 28 
 we have a a f c sin < and d' -d = c sin 6. Then substituting these 
 values, and cancelling c, we have 
 
 which is readily seen to be true from the force triangle ABC, where we 
 have 
 
 So we have thus proven our proposition for the case of two forces. Now 
 as it was shown in Art. 39 that the force polygon representing any 
 number of forces was really made up by combining force triangles, it is 
 obvious that we could show that the moment of the final resultant of any 
 number of forces about any point is equal to the algebraic sum of the 
 moments of the forces about the same point by simply considering the 
 resultants and components in each triangle in consecutive order. 
 
 42. Condition of Equilibrium. We say a body is in equilibrium 
 whenever it is either at rest or in uniform motion. The only state of 
 equilibrium here considered shall be that of rest, as uniform motion is of 
 little interest to us in structural mechanics. As there are but two kinds 
 of motion, translation and rotation, it is evident that a body will be at rest 
 if neither of these motions takes place, and that the problem regarding 
 equilibrium thus reduces to the investigation of the conditions conducive 
 to these two motions. 
 
 Our only conception of equilibrium of forces is that of two equal and 
 opposite conspiring forces balancing each other through the body upon 
 which they act. Then, in order that a body be in equilibrium, it is obvious 
 that the forces acting upon it must reduce to that condition. The forces 
 themselves, or resultants, or a combination of the forces and resultants, 
 may form a pair or several pairs of equal and opposite conspiring forces, 
 so that all of the forces acting upon a body will thus be balanced, or, as 
 we say, be in equilibrium, and such being the case, the body upon which 
 they act will be in equilibrium. It makes no difference what combination 
 we choose to consider the forces forming, for so long as we can show that 
 the whole system of forces acting upon a body reduces to equal and 
 opposite conspiring forces, we are assured that the body upon which the 
 system acts is in equilibrium, and of course the converse will be true. 
 
 Referring to Fig. 17 (Art. 39), it is obvious that if a force equal in 
 intensity to the resultant of the forces PI and P2 be applied to the body 
 at A so that it would act along the line AC in the direction from C to A, 
 it would balance the forces PI and P2 and the body would thus be in 
 equilibrium under the action of these three forces. Here it is seen that 
 the resultant of the two forces PI and P2 and the balancing force would 
 be two equal and opposite conspiring forces to which the system reduces 
 when the body is in equilibrium. 
 
32 STRUCTURAL ENGINEERING 
 
 And, referring to Fig. 19 (Art. 39), it is obvious that if a force equal 
 in intensity to the final resultant R2 be applied to the body at A so that 
 it would act along the line EA in the direction from E to A, it would 
 balance all of the forces PI, P2, P3, and P4, and thus the body would be 
 in equilibrium under the action of the five forces PI, P2, P3, P4, and the 
 balancing force. Here it is seen that the resultant of the forces PI, P2, 
 P3, and P4 and the balancing force are two equal and opposite conspiring 
 forces to which the system reduces when the body is in equilibrium. 
 
 Further, referring to Fig. 25 (a) (Art. 39), it is obvious that if a 
 force equal and opposite to R4: be applied along the line S'O" ' it would 
 balance all of the forces PI . . . P5, acting upon the body AB, and thus 
 the body would be in equilibrium. Here again it is seen that the system 
 reduces to equal and opposite conspiring forces when equilibrium exists. 
 So it is in all cases. 
 
 It will be observed from the above that any force acting upon a body 
 in equilibrium will always form an equal and opposite conspiring force 
 with the resultant of all the other forces acting upon the body. 
 
 Beyond a question, a body will move in translation if the forces 
 acting upon it have a resultant. Then, evidently, a body will be in 
 equilibrium, as far as translation is concerned, whenever the forces acting 
 upon it have no resultant. Then as the forces acting upon a body in 
 equilibrium must have no resultant in order that no translation may 
 take place it is obvious that the force polygon representing them, instead 
 of having a final resultant, as R 2 in Fig. 19, will simply extend on around 
 and close on the starting point without a resultant. This would give us a 
 force polygon wherein all the forces are indicated to act in the same 
 direction around the polygon. From this the following practical state- 
 ment can be made: A body will be in equilibrium as far as translation is 
 concerned whenever the forces acting upon it (when represented graph- 
 ically to scale) will form a closed polygon wherein all of the forces act 
 in the same direction around the polygon, or, in other words, the force 
 polygon will close without a resultant, and the converse is just as true, 
 that is, if a body be in equilibrium the polygon will close. This is true 
 regardless of whether the forces be non-concurrent, conspiring, or con- 
 current forces, for it is evident that a body will have motion of translation 
 if a resultant exists, as the resultant is equivalent to a single force, in 
 which case motion is inevitable; and if no resultant exists, motion of 
 translation will not take place, as there would be no force to produce it. 
 All this is manifestly true regardless of condition. 
 
 As the resultant of concurrent or conspiring forces always passes 
 through a common point of action, and the resultant of all the forces 
 except one always forms an opposite and equal conspiring force with the 
 remaining one, in case of equilibrium, and the moments about any point 
 of these two equal and opposite conspiring forces are equal and have 
 opposite signs, it is evident that rotation will not take place if the 
 equilibrium polygon closes. So in the case of concurrent or conspiring 
 forces, if the force polygon closes, we need go no further in our investi- 
 gations regarding equilibrium, for if no resultant exists, no motion of 
 rotation, as well as no motion of translation, will take place. 
 
 But in regard to non-concurrent forces the case is different, for here 
 the forces may reduce to two equal and opposite forces, which indicates 
 
FUNDAMENTAL ELEMENTS 33 
 
 that no resultant exists, and thus the force polygon would close, indicating 
 no motion of translation ; but at the same time the two forces might not be 
 conspiring forces, in which case, while motion of translation would not 
 take place, motion of rotation would, as the two forces would form a 
 couple which would produce motion of rotation. For example, the forces 
 acting upon the body AB shown at (a) (Fig. 29) could form a closed 
 polygon as shown at (fr), which indicates 
 that no resultant exists, and such being ~\ /P2 
 
 the case, no translation would take place ; pi * -. 
 
 but by mere inspection of the figure at $J i ' 
 
 (a) we can see that the forces would () ' \P4 
 
 form a couple which would rotate the 
 
 body AB counter clock-wise and hence Fig. 29 
 
 equilibrium would not exist. So to state 
 
 the condition of equilibrium fully, we say : A body is in equilibrium when 
 
 both the resultant and the algebraic sum of the moments about every point 
 
 of the external forces acting upon it, equal zero. 
 
 In addition to the above statement, we may add that a body is in 
 equilibrium, as far as motion of translation is concerned, when the 
 algebraic sum of both the horizontal and vertical components of the forces 
 acting upon it is equal to zero, for it was shown in Art. 40 that the sum 
 of the squares of the horizontal and vertical components of any number 
 of forces is equal to the square of their resultant, and, of course, if the 
 components are equal to zero, the resultant will be equal to zero. Further, 
 we can state that the sum of the components of the forces acting upon a 
 body in equilibrium is zero along any line whatsoever, for it is evident 
 that if a component exists along any line, a resultant would exist in some 
 direction, and hence motion of translation would surely take place in that 
 direction. 
 
 43. Point of Application of a Force in Relation to Stress and 
 
 Equilibrium. In order to determine the nature of the stress produced in 
 any case it is absolutely necessary to know the actual point of application 
 of the force producing the stress. Each force in that case is indicated to 
 act at some particular point as shown in Art. 26, while in reference to 
 equilibrium (as stated in Art. 39), a force may be considered as being 
 applied at any point along its line of action, in which case, of course, we 
 are compelled to conceive of there always being the necessary materialistic 
 connection between the point of application and the body concerned. 
 
 In the case of simple stress, whenever a force acts toward a body, 
 the stress produced in the body by the force will always be compression, 
 but if a force acts away from a body, the stress will be tension. 
 
 As a simple case, let AB (Fig. 30) represent a steel rod acted upon 
 by two equal and opposite conspiring forces, PI and P2. If PI be 
 applied at a and P2 at b, as indicated, it is obvious that the rod would be 
 in tension. Now let us suppose the forces interchanged, PI being applied 
 at b and P2 at a, then it is obvious that the rod would be in compression. 
 Thus, we see that by changing the point of application of the forces, the 
 stress in the rod would be completely reversed, while the state of 
 equilibrium would not be disturbed. 
 
 Then again, suppose PI be applied at a and P2 at c, then it is 
 obvious that the stress in the rod between a. and c would be the same as 
 
34 STRUCTURAL ENGINEERING 
 
 when P2 is applied at b and PI at a, while the stress in the rod between 
 c and b would be zero. So it is readily seen that any change in the point 
 of application of either of the forces will produce a corresponding change 
 
 Fig. 30 
 
 upon the rod in reference to the stress, but it is just as readily seen that 
 any such change will riot disturb the state of equilibrium, for it is evident 
 that the forces PI and P2 will balance each other so long as their points 
 of application remain in the body anywhere on their common line of action 
 between the points a and b. As for that, either of the forces can even be 
 considered as being applied at any point, as o, off of the body altogether, 
 in which case, of course, we would have to consider the rod as extending 
 to o or being rigidly connected to it in some way. 
 
 As another case, let three non-concurrent forces, PI, P2, and P3 
 (Fig. 31) be applied to the body ZZ at the points a, b, and c, respectively. 
 Now, as far as equilibrium is concerned, we can 
 consider any of these forces as being applied at 
 any point along their line of action, so prolong 
 the lines of action of the two forces PI and P2 
 (see Art. 39) until they meet at A. Then we can 
 consider these two forces as two concurrent forces 
 applied at A, and constructing the force triangle 
 ABC, we have the intensity and direction of their 
 resultant Rl, which can be considered as a force 
 being applied at any point along its line of action 
 Cy. Then prolong the line of action of the force 
 P3 until it meets the line Cy at D. Then we can 
 rig. 31 consider that Rl and P3 are two concurrent 
 
 forces applied at D. Then constructing the force 
 
 triangle DEF, we have their resultant R2, which is the final resultant of 
 the three forces PI, P2, and P3. 
 
 Now it is evident that if a force equal to Rl be applied to the body 
 along the line Cy it would produce the same motion as PI and P2 if it 
 acted in the same direction as indicated by Rl, but if it acted in the 
 opposite direction it would balance them. Then again, it is evident that if 
 a force equal to the resultant R2 be applied to the body along the line Dx 
 it would produce the same motion of translation as the forces PI, P2, and 
 P3 combined, providing it acted along the line Dx, in the same direction 
 as indicated by R2; but if it acted in the opposite direction, it would 
 balance the three forces PI, P2, and P3. 
 
 Now in regard to stresses, it is obvious that the stress produced upon 
 the body ZZ by the action of the three forces PI, P2 and P3 will be of a 
 very complicated nature. We can readily see that a force equal to the 
 resultant 121, if applied at o or at any other point in or on the body along 
 the line Cy, would not produce the same stress as the two forces them- 
 selves produce. And it is more readily seen that a force equal to the 
 resultant R2 applied along the line Dx would not produce the same stress 
 in the body as the forces PI, P2, and P3 actually produce. So it is 
 
FUNDAMENTAL ELEMENTS 35 
 
 evident that a force representing the resultant of two or more forces, 
 while it will produce the same effect as the forces themselves as far as 
 motion or tendency of motion is concerned, yet will not necessarily produce 
 the same effect in reference to stress. However, in the case of concurrent 
 forces, we assume the stress produced by a force representing a resultant 
 to be the same as that produced by the components, especially when the 
 forces meet at the surface of the body considered. Some cases are very 
 complicated and require very careful consideration, but we can always 
 obtain a reasonable approximation which will come within the limits of 
 practicability, and with this we must be contented. 
 
 44. Equilibrium of Couples. In regard to equilibrium of couples, 
 there is one important fact that should always be borne in mind, and that 
 is, it always requires another couple to balance a couple, and the couples 
 must be equivalent to each other and act in opposite direction to each 
 other. As to the proof that it requires another couple to balance a couple, 
 we have the following: 
 
 We know from the definition that a couple can have no single 
 resultant, as the resultant always reduces to zero, and therefore would 
 not balance a single force, and hence any single force or any system of 
 forces reducing to a single resultant could not balance a couple. Then, 
 evidently, the forces balancing a couple must have a resultant equal to 
 zero and an equal and opposite moment, which would undoubtedly con- 
 stitute another couple. 
 
 45. Graphical Determination of the Resultant of Parallel Forces 
 or Those Nearly Parallel. First Method: In the case of intersecting 
 forces, the "final resultant can be fully determined in its true position by 
 prolonging in consecutive order the lines of action of both the forces and 
 their successive resultants to intersection, and constructing a force triangle 
 at each intersection, as shown in Art. 39, but it is obvious that in the case 
 of parallel forces or forces nearly parallel the same method will not apply 
 as such forces either do not intersect at all or their intersection is beyond 
 practical limits. However, by resorting to the resolution of forces, the 
 resultant of parallel forces and those nearly parallel can be determined 
 as readily in its true position as the resultant of intersecting forces. As 
 an example, let AB at (a) in Fig. 32 represent a body acted upon by 
 three non-concurrent forces PI, P2, and P3, which are nearly parallel. 
 Select some convenient point on the line of action of one of the forces, 
 say, point on the line of action of PI. Then at this point resolve PI 
 into any two component forces as c and cl by constructing (at will) a 
 force triangle as abO. This is accomplished by assuming the intensity of 
 one of the components, say c, and laying off aO from in any convenient 
 direction to represent this intensity, and then from a laying off ab equal 
 and parallel to PI, and drawing bO to complete the triangle, we have cl, 
 the other component, fully given by this last line bO. Then from pro- 
 long the line of action of the component cl until it intersects the line of 
 action of the force P2 at 0' '. Then at this point resolve P2 into two 
 components such that one of them will be equal and opposite to the com- 
 ponent cl at and act along the same line 00'. Then there will be a 
 component force cl at and an equal and opposite component force cl 
 at 0', which will balance each other. The component cl at 0' being 
 designated, we obtain the other component of P2 which we will call c2, 
 
36 
 
 STRUCTURAL ENGINEERING 
 
 by constructing the force triangle deO' ', which is constructed by laying off 
 dO' equal and parallel to cl and de equal and parallel to P2, and drawing 
 eO' for completion. Then c2 is given by this last line eO' '. Now from 
 0' prolong the line of action of the component force c2 until it intersects 
 the line of action of the force P3 at 0". Then at this point (O") resolve 
 P3 into two components such that one of them will be equal and opposite 
 
 Fig. 32 
 
 to the component c2 at 0' and act along the same line O'O". Then there 
 will be a component force c2 at 0' and an equal and opposite component 
 force c2 at 0" which will balance each other. The one component force 
 c2 at 0" being designated, the other component of P3, which we will call 
 c3, we obtain by constructing another force triangle ghO" ', where c3 is 
 given by the line hO". 
 
 Now it will be observed that the three forces PI, P2, and P3 at (a) 
 
FUNDAMENTAL ELEMENTS 37 
 
 have each been resolved into two component forces which replace the 
 original force in each case,, and that these component forces are all mutu- 
 ally balanced,, except c (at 0) and c3 (at 0"), so that the three original 
 forces,, PI, P2, and P3, are really replaced by these two component forces, 
 c and c3. So, evidently, if we determine the resultant of these two com- 
 ponent forces in its true position, we will have the resultant of the three 
 original forces, PI, P2, and P3, fully determined in its true position, as 
 evidently the resultant in the two cases must be identical. 
 
 Now as the two component forces c and c3 are intersecting forces, 
 their resultant is readily determined in its true position by prolonging 
 their lines of action until they intersect at / and constructing the force 
 triangle Ihm, where their resultant is fully given in its true position by 
 the line Im. This resultant is the same in every respect as the resultant 
 of the original forces PI, P2, and P3, and it may be assumed to be applied 
 at any point along its line of action yy. 
 
 Another way of determining the resultant is as follows: 
 
 Instead of resolving the forces PI, P2, and P3 into components, as 
 was done above, we can select any point on the line of action of one of the 
 forces, as point S (really the same as we did before) on the line of action 
 of PI (at (a)) and apply two forces / and /I such that the force PI will 
 be balanced by them. This is accomplished by assuming the intensity 
 and direction of one of the forces and then constructing a force triangle 
 to determine the intensity and direction of the other. Thus, beginning at 
 S, lay off Sn (at will) to represent / in intensity and direction. Then 
 from n lay off no equal and parallel to PI and draw oS for completion of 
 the triangle and we have /I represented in intensity and direction by this 
 last line oS. As the three forces at S are in equilibrium, they will act in 
 the same direction around the triangle Sno as indicated, which is in 
 accordance with Art. 42. Next, from S prolong the line of action of the 
 force /I until it intersects the line of action of the force P2 at S'. Then 
 at this point apply two forces such that they will balance P2 and one of 
 them be equal and opposite to the force /I at S and act along the same 
 line SS'. Then there will be a force /I at S and an equal and opposite 
 force fl at S' which will balance each other. The force fl at S' being 
 known, the other balancing force at S', which we will designate as /2, we 
 determine by constructing the force triangle S'pr in the same manner as 
 was explained in the case of triangle snO at S. Then from S' prolong 
 the line of action of the force f2 until it intersects the line of action of 
 P3 at S". Then at this point (") apply two forces such that they will 
 balance P3 and one of them be equal and opposite to f2 at S' and act 
 along the same line S'S". Then there will be a force f2 at S' and an 
 equal and opposite force f2 at $" which will balance each other. The 
 force f2 at S" being known, the other balancing force, which we will 
 designate as /3, at S", we determine, by constructing another force 
 triangle S"st, where /3 is given by the line S"t. 
 
 Now it will be observed that each of the three forces PI, P2, and P3 
 is balanced, that is, held in equilibrium by two forces applied in each 
 case for that purpose, and such being the case, the system will be in 
 equilibrium. But it will be observed that all of the balancing forces are 
 mutually balanced except the force / at S and /3 at S". Then, evidently, 
 these two forces, / and /3, hold the three forces PI, P2 } and P3 in 
 
38 STRUCTURAL ENGINEERING 
 
 equilibrium, in which case the resultant of the forces / and /3 will be 
 an equal and opposite conspiring force to the resultant of the forces 
 PI, P2, and P3 (see Art. 42). So if we determine the resultant of the 
 two forces / and /3 in its true position, we will have a force the same in 
 position and in intensity as the resultant of PI, P2, and P3, but which 
 acts in the opposite direction, so that the resultant of the forces PI, P2, 
 and P3 will thus be relatively determined in its true position. To deter- 
 mine the resultant of the forces / and /3, prolong their lines of action 
 until they intersect at /' and construct the force triangle I'k'm' and we 
 have their resultant fully represented by the line I'm'. 
 
 The only difference in the two methods shown above is due to the com- 
 ponents being reversed in direction in the second method so that they are 
 really balancing forces instead of actual components. Otherwise the two 
 methods are identical. Either of the broken lines xOO'0"z or vSS'S"u 
 would be known as an Equilibrium Polygon, and the lines xO, 00', 
 O'O", and 0"z, or vS, SS', S'S", and S"u would be known as their 
 respective segments. 
 
 It is evident that the resultant of any number of such forces as PI. 
 P2 and P3 can be determined in its true position in the same manner as 
 shown above for these three forces, but in practice the same thing is 
 accomplished with less work by a somewhat different method of pro- 
 cedure, as will now be shown. 
 
 Imagine the triangle abO, at (a) Fig. 32, moved bodily and parallel 
 to itself to the position abP at (b}. Next imagine the triangle deO' at 
 () moved in the same manner to the position beP, d coinciding with b 
 and 0' with P, and likewise imagine the triangle ghO" moved in the same 
 manner to the position ehP, g coinciding with e and 0" with P. Now we 
 have the three force triangles abO, deO' , and ghO" collected into a single 
 diagram PabehP (at (&)) which we will call a Ray Diagram, where the 
 line abeh is the load line, and the lines aP, bP, eP, and hP are the rays, 
 and the point P is the pole. It will be observed that these rays, aP, bP, 
 eP, and hP are parallel, respectively, to the segments xO, 00', O'O", 
 and 0"z of the equilibrium polygon xOO'0"z, at (a). So, evidently, if 
 this ray diagram, PabehP, had been drawn beforehand, the equilibrium 
 polygon xOO'0"z could have been constructed by beginning at and 
 drawing xO parallel to the ray aP, then from drawing 00' parallel to 
 the ray bP, and likewise from 0' drawing O'O" parallel to the ray eP, 
 and then from 0" drawing 0"z parallel to the ray hP. Then after 
 having drawn the equilibrium polygon xOO'0"z in this manner, the 
 segments xO and zO" could be prolonged until they intersect at /, and 
 thus the point 7 would be located, which is one point upon the line of 
 action of the desired resultant. But the things lacking would be the 
 intensity, direction, and direction of action of the resultant. But it will 
 be observed at (b) that by drawing the line ah we will have a force 
 polygon abeha which represents the forces PI, P2, and P3, and their 
 resultant, each in intensity, direction, and direction of action. Then the 
 intensity, direction, and direction of action of the desired resultant is 
 given by the line ah. So by drawing through / a line parallel to this line 
 ah we would have the line of action of the desired resultant, and hence 
 the resultant would be fully determined, as its intensity and direction of 
 
FUNDAMENTAL ELEMENTS 39 
 
 action, as stated above, as well as its direction, are given by the line ah 
 in the force polygon abeha at (6). 
 
 Now, according to Art. 39, in any case the force polygon represent- 
 ing any two or more forces and their resultant can always be drawn. So 
 suppose in the above case the force polygon abeha at (6) be the very 
 first thing drawn. Then we would have the intensity, direction, and 
 direction of action of the resultant of the three forces PI, P2, and P3 
 given by the line ah, and the only thing lacking would then be the loca- 
 tion of the line of action of the resultant, which can be obtained as above 
 by constructing an equilibrium polygon. But we would have so far only 
 the force polygon abeha constructed. Now if the point P at (&) and the 
 point at (a) were given, we could draw the rays aP, bP, etc., and 
 beginning at we could draw the equilibrium polygon xOO'0"z, as 
 explained above, and thus obtain the desired line of action yy. However, 
 it is not necessary to have the location of either of these particular points 
 given, as the location of was arbitrarily assumed in the first place, and 
 the position taken by the equilibrium polygon xOO'0"z was governed by 
 the arbitrarily constructed triangle abO. So, evidently, we might just as 
 well assume the pole P and draw the rays of a ray diagram and construct 
 an equilibrium polygon accordingly by starting at any convenient point 
 on the line of action of one of the forces. 
 
 As an example, let AB, Fig. 33, represent a body acted upon by four 
 forces, PI, P2, P3, and P4. First construct the force polygon abcdea, at 
 (fr),by drawingafc equal and parallel 
 to PI, and be equal and parallel to 
 P2, and so on as explained in Art. 39, 
 and we have the resultant of the four 
 forces PI ... P4 given in intensity, 
 direction, and direction of action by 
 the line ae. Then to obtain its line 
 of action, take any convenient point 
 P as a pole (at (&)) and draw the 
 rays aP, bP, etc., thereby obtaining 
 the ray diagram PabcdeP. Then 
 construct a corresponding equilibrium 
 polygon at (a) as xOO'0"0' "z, 
 which is accomplished in the follow- 
 ing manner: First select any con- 
 venient point on the line of action of 
 one of the forces, say, point on the Fig. 33 
 
 line of action of PI. Then, from this 
 
 point draw the segments xO and 00' parallel to the rays aP and bP, 
 respectively. Then from 0' draw the segment O'O" parallel to the ray 
 cP. Then from 0" draw the segment 0"0' " parallel to the ray dP. 
 Then from 0' " draw the segment zO f " parallel to the ray eP. Then 
 prolong the segments xO and zO' " until they intersect at I, and through 
 this point I draw the line yy parallel to the line ae in the force polygon 
 abcdea, and it will be the desired line of action of the resultant of the 
 four forces, and hence the resultant is fully determined. 
 
 If the forces be absolutely parallel, the load line will be a straight 
 
40 
 
 STRUCTURAL ENGINEERING 
 
 line,, and the line representing their resultant will coincide with this line, 
 but, however, the work of determining the resultant is the same as shown 
 above. For example, let AB, at (a) Fig. 34, represent a body acted upon 
 by four parallel forces PI ... P4. We construct the load line abode at 
 (b) by laying off the forces in consecutive order as shown. Then assume 
 the pole P and construct the ray diagram PabcdeP and draw the 
 equilibrium polygon xOO'O"0' " 2 at (a) as before and locate the point 
 I, through which draw the line of action yy of the resultant parallel to the 
 load line ae. 
 
 
 
 1 \" i 
 
 K 
 
 Al 
 
 
 ^ 
 
 ID 
 
 
 
 f\ 1 s 
 
 \ 1 x 
 
 / 
 
 
 /' ^o* * 
 
 >v x N s 
 
 
 'y (a) 
 
 Fig. 34 
 
 The method of procedure just shown is quite satisfactory in practice, 
 but the student should not acquire the habit of constructing the force 
 polygon and the ray diagram and then the corresponding equilibrium 
 polygon without fully recognizing the exact significance of each step. 
 
 In Fig. 32 the force triangles were constructed on the segments of 
 the equilibrium polygons, while in Figs. 33 and 34 the force triangles 
 were constructed on the load lines of the force polygons, but the principle 
 involved is just the same in either case. In the case shown in Figs. 33 
 and 34 we really resolved each of the forces PI, P2, P3, and P4 into two 
 components by constructing, respectively, the triangles abP, bcP, cdP, 
 and deP. However, it is easy to overlook this fact when the triangles 
 are constructed by merely drawing the rays aP, bP, etc. If we consider 
 the forces resolved into components, as is the usual custom, the com- 
 ponent in each case will act around the force triangle in the opposite 
 direction to that of the force, while if we reverse the direction of the 
 components, they become balancing forces, in which case they will act in 
 the same direction around the triangle as the force (see Arts. 39 and 42). 
 Take, for example, the case shown in Fig. 33 ; the two sides aP and bP of 
 the force triangle abP, at (6), give directly either the intensity and 
 direction of two components of PI or of two balancing forces. If we 
 assume components, the one represented by the side bP will act from P 
 to b, but if we assume balancing forces, the one represented by the side 
 aP will act from P to a, while the one represented by the side bP will act 
 from b to P. The same is true of all the other force triangles bcP, cdP, 
 and deP shown at (b). In constructing an equilibrium polygon it really 
 makes no difference whether we consider the forces resolved into com- 
 ponents or balanced by forces applied for that purpose, as the final result 
 will be just the same. 
 
FUNDAMENTAL ELEMENTS 41 
 
 In constructing an equilibrium polygon after the ray diagram is 
 completed, we virtually transfer the two components or the two balancing 
 forces, as the case may be, of each force to a point upon its actual line of 
 action. In order to show what is really involved in the construction, take 
 for an example the case shown in Fig. 33. As the forces here were 
 assumed to be resolved into components, the components will in each case 
 act around the force triangle in the opposite direction to that of the force. 
 Now beginning with PI, we can assume its own components, represented 
 by the sides aP and bP of the force triangle abP at (6), as applied at any 
 point upon the line of action of PI. Then as the component represented 
 by the side aP acts from a to P, a line drawn from in this direction and 
 parallel to aP will represent that component as being applied at 0, and 
 we thus obtain the segment xO of the equilibrium polygon. The other 
 component of PI, which is represented by the side bP, acts from P to b. 
 Then a line drawn from in this direction and parallel to bP will repre- 
 sent that component as being applied at 0, and we thus obtain the seg- 
 ment 00' , which is really the line of action of this other component of PI 
 drawn only to 0', where it intersects the force P2. We can assume the 
 two components of P2, represented by the sides bP and cP of the force 
 triangle bcP at (6), applied at any point upon the line of action of P2 the 
 same as in the case of the two components of PI. Then, evidently, we 
 can assume them applied at 0' as well as at any other point. Now the 
 component of P2 represented by the side bP of the force triangle bcP 
 at (6) acts from b to P. Then a line drawn from 0' in this direction and 
 parallel to bP will represent that component of P2 as being applied at 
 O f ', and we thus obtain the segment 00' again. The other component of 
 P2 represented by the side cP acts from P to c. Then a line drawn in 
 this direction from 0' and parallel to cP will represent that component 
 applied at 0', and we thus obtain the segment O'O" , which is really the 
 line of action of this other component of P2 drawn only to 0" where it 
 intersects the force P3. The two components of P3, represented by the 
 sides cP and dP of the force triangle cdP at (6), can be assumed to be 
 applied at 0" ', and then the two components of P4, represented by the 
 sides dP and eP of the force triangle deP, can be assumed to be applied 
 at 0' ", and the discussion of the previous cases will apply to each of 
 these cases. 
 
 The direction of action of the components of each force can be 
 indicated on the rays of the ray polygon, and on the segments of the 
 equilibrium polygon as in Fig. 33, but usually we omit such indications, 
 because such are usually not necessary to any great extent however, that 
 is a minor item. 
 
 After an equilibrium polygon is drawn for a system of forces as 
 shown in Fig. 33, the resultant of any number of these forces, if taken in 
 consecutive order, can be determined as well as the resultant of all of 
 them. For example, take the three consecutive forces PI, P2, and P3 
 (Fig. 33). By drawing the line ad (at (b)), we have the force polygon 
 dbcda, wherein the resultant of the three forces is given in intensity, 
 direction, and direction of action by this line ad. By prolonging the 
 segment 0"0' " until it intersects the segment xO at /', we have one 
 point on the line of action of the resultant of these three forces, PI, P2, 
 and P3. Then by drawing a line through /' parallel to the line ad, we 
 
42 
 
 STRUCTURAL ENGINEERING 
 
 obtain the desired line of action of their resultant, and thus we have the 
 resultant of the three forces determined. As another case, the resultant 
 of the two forces P2 and P3 can be determined by drawing a line bd 
 (at (fc)) p^d prolonging the segments 0"0' " and 00' until they intersect 
 at I". 
 
 Second Method: The following graphical method, which may be 
 designated as the Proportional Triangle Method, can be used to advantage, 
 quite often, in determining the resultant of parallel forces : 
 
 Let PI and P2 (Fig. 35) represent any two parallel forces, and let 
 
 Fig. 35 
 
 R represent their resultant. Let a line ab be drawn perpendicular to the 
 two forces and their resultant, intersecting the resultant at o. As there is 
 no question as to the intensity of the resultant being equal to the sum of 
 the forces, that is, ^tJ = Pl + P2, and the direction being the same as the 
 forces, it remains only to determine the point o through which the line of 
 action of the resultant passes. This can be obtained by the following 
 construction: From b draw a line bx, making any convenient angle with 
 the line ab. Then from b lay off, to any convenient scale, bc = Pl; and 
 from c lay off, to the same scale, cd=P2. Then draw ad and through c 
 draw a line parallel to ad, and where it intersects the line ab will be the 
 required point o through which the resultant passes. The same can be 
 accomplished by drawing the line ay from a and laying off the loads as 
 indicated and then drawing the lines fb and eo. 
 
 The above is readily seen to be true, for taking moments about b, 
 according to Art. 41, we have 
 
 obxR = Plxab or ob = 
 
 which is the distance of the point o from b; but from similar triangles, 
 bco and bda t we have 
 
 ob cb PI 
 
 or 
 
 Plxab 
 
 and thus we have an identity. 
 
FUNDAMENTAL ELEMENTS 43 
 
 In case there are more than two forces, the resultant of any two is 
 determined as above and this resultant is then considered with one of the 
 remaining forces, and so on. For example, let PI, P2, and P3 (Fig. 36) 
 
 represent three parallel forces. First draw a line as abc perpendicular to 
 the forces. Then from one of the points a, b, or c, say a, draw a line ax t 
 making any convenient angle with the line abc. Then lay off ae = P2 and 
 eg = Pl. Then draw gb and through e draw a line parallel to gb, and the 
 point k where this line intersects the line abc is a point on the line of 
 action of the resultant of the two forces PI and P2. Then from this 
 point draw a line ky, making any convenient angle with the line abc. 
 Then lay off kh = P3 and hm = Pl+P2. Then draw cm and through h 
 draw a line parallel to cm, and the point o where this line intersects the 
 line abc is a point on the line of action of the resultant of the three 
 parallel forces. 
 
 It is important to note that the force from which the diagonal line 
 is drawn is laid off last on that line. Thus in Fig. 36, eg = PI and 
 hm = PI + P2. 
 
 In cases where the forces are in groups, as locomotive wheel loads, 
 the method is a very convenient one, as the resultant of each group is 
 known from observation and thus the forces really considered in the con- 
 struction are quickly reduced to only a few. 
 
 46. Analytical Determination of the Resultant of Parallel 
 Forces. It is evident that the resultant of two or more parallel forces 
 must act in the same direction as the forces, and with an intensity equal 
 to their algebraic sum; otherwise it would not produce the same effect as 
 the forces themselves. So the intensity, direction, and direction of action 
 of the resultant of two or more parallel forces are really known directly 
 from the forces themselves, and such being the case, the problem involved 
 in determining their resultant really reduces to the locating of the line of 
 action of the resultant. 
 
 Let PI . . . P5 represent five parallel forces acting upon the body 
 AB (Fig. 37), and let R represent their resultant. Then we have 
 # = P1+P2 + P3+P4 + P5. According to Art. 41, the moment of this 
 resultant about any point is equal to the algebraic sum of the moments of 
 the five forces about the same points. Then taking moments about some 
 point 0, we have 
 
 aPl + bP2 + cP3 + dP4: + eP5 = R*, 
 from which we get 
 
 aPl + bP2 + cP3 + dP4 4- eP5 
 
 R P1+P2 + P3 + P4 + P5 
 
44 
 
 STRUCTURAL ENGINEERING 
 
 So then we have the line of action of R located in reference to O. 
 
 In practice it is usual to take moments about one of the forces, 
 
 At 
 
 Fig. 37 
 
 thereby shortening the work by eliminating the moment of that force. 
 Thus taking moments about PI (Fig. 38), we have 
 
 _ 
 " 
 
 In case the moments about one point are known, the moments about any 
 other point can be determined by either increasing or diminishing the 
 known moments by the sum of the forces multiplied by the common dif- 
 ference of lever arms. For example, the moments about the point z 
 (Fig. 38) are equal to the moments about PI plus the sum of the forces 
 
 
 d 
 
 
 
 
 
 
 
 
 
 
 
 oc 
 
 p b . 
 
 ._*. 
 
 
 * 
 
 p 
 
 I 
 
 * 
 
 \ 
 
 \ 
 
 * 
 
 ^ 1 n 
 
 
 
 Fig. 38 
 
 Fig. 39 
 
 PI . . . P5 multiplied by k. That is, the moments about z are equal to 
 6P2-l-cP3 + dP4 + eP5+(Pl+P2 + P3 + P4 + P5)Ar. This is readily seen 
 to be true, for let M be the moment of a force P about A (Fig. 39) and let 
 M' be its moment about B; then we have M aP and M' = P(a + b) = 
 aP + bP = M + bP. It is readily seen that this is as true for a number 
 of forces as it is for one. The product of the sum of the forces and 
 common difference of lever arms would be subtracted from the moments 
 in case the lever arms became shortened by the transferring of the center 
 of moments. 
 
 47. Center of Gravity. The most common case of parallel forces 
 is that of gravity, which acts with equal intensity upon each ultimate 
 part of material contained in a body, regardless of the kind of material. 
 In case of an individual body, the line of action of the resultant of the 
 gravity forces passes through what is known as the center of gravity of 
 
FUNDAMENTAL ELEMENTS 45 
 
 the body, but when more than one body is considered, the center of gravity 
 of the bodies considered as a group is the same as the resultant of so many 
 parallel forces, the weight of each body being considered as a force acting 
 through the center of gravity of the respective body. 
 
 We can consider symmetrical bodies as being made up of parallel 
 layers of homogeneous material wherein the centers of gravity of the 
 layers coincide each with each, and such being the case, we need treat 
 only one layer, which we may consider as a plane without thickness, from 
 which results the common practice of treating area instead of volume, 
 weight, or mass. 
 
 In case the material composing a body be symmetrical about a plane, 
 there is no question but that the center of gravity of the body will be in 
 that plane; and hence, if the material be symmetrical about three or more 
 planes intersecting in a point, their point of intersection will undoubtedly 
 be the center of gravity of the body. So it is evident that the centers of 
 gravity of many bodies and plane figures are known from mere inspection 
 of their state of symmetry. Thus, the center of gravity of a sphere is at 
 its center, a line at its middle point, and we know the center of gravity 
 of a circle, cylinder, cube, an ellipse, etc., from the same deduction. In 
 fact, whatever the method employed in determining the centers of gravity 
 of bodies and plane figures, we assume the center of gravity of certain 
 elements as being predetermined by mere symmetry. 
 
 The determination of the center of gravity of loads in groups and the 
 center of gravity of the cross-section of individual members of structures 
 are the principal center of gravity problems occurring in structural 
 engineering. 
 
 We can determine the center of gravity of any group of loads in the 
 same manner as the resultants of parallel forces were determined in Arts. 
 
 o 
 
 Fig. 40 Fig. 41 
 
 45 and 46. The analytical method outlined in Art. 46 is most used as it is 
 usually more convenient than any other method. As an example, let 
 PI . . . P4 (Fig. 40) represent four loads. Taking moments about one 
 of the end loads, either PI or P4, say P4, we have 
 
 cPl+bP2+aP3 
 "P1 + P2 + P3 + P4 ' 
 
 where x is the distance the center of gravity of the four loads is from P4. 
 In case of the cross-section of a member, we deal with the area 
 instead of the weight. For example, let the sketch in Fig. 41 represent 
 the cross-section of a plate and an angle which is riveted to the plate. 
 Let A be the area of the plate, and let A' be the area of the angle, and let 
 a and b be the distance from the bottom edge of the plate to the center 
 of gravity of the plate and angle, respectively. Then taking moments 
 about the bottom edge of the plate, we have 
 
46 
 
 STRUCTURAL ENGINEERING 
 
 or 
 
 A'b ) A a 
 - ~r= x > 
 A -{-A 
 
 where x is the distance from the bottom edge of the plate to the center of 
 gravity of the total cross-section of the plate and angle combined. The 
 work could be shortened by taking moments about the center of gravity 
 of the plate or angle, thus following out the same scheme as was used in 
 the case of loads (Fig. 40). 
 
 The center of gravity of plain areas can be determined quite readily 
 by the aid of calculus, wherein the general formula is 
 
 For example, let Fig. 42 represent a rectangle having a height h and 
 width 6. Taking moments about the lower edge of the rectangle, we have 
 
 * h bydy = bh 2 _ bh 2 h 
 ~~A 2A ~2bh~2' 
 
 That is, the center of gravity of the rectangle from its lower edge is equal 
 to one-half of its height, which we really knew from mere inspection. 
 
 Fig. 42 
 
 Fig. 43 
 
 As another example, let Fig. 43 represent a triangle. Taking 
 moments about a line AB parallel to its base, we have 
 
 ~ = [yda _ 
 
 2bh s 2 
 
 h A 
 
 which is the distance of the center of gravity of the triangle from the 
 axis AB. 
 
 Problem 7. Determine the center of gravity in reference to the 
 horizontal axis of the section shown in Fig. 44, which is composed of 
 314" x \" plates and 26" x 6" x \" angles. 
 
 Solution: Taking moments of the areas about the center of gravity 
 of the top plate, we have 
 
 32.5 * = 0.5 x 7 + 1 x 7 + 2.93 x 11.5, 
 - 44.19 
 
 or 
 
 x = 
 
 32.5 
 
 = 1.36". 
 
 Hence the center of gravity of the section is 1.36" below the center of the 
 top plate, or 0.11" below the back of the angles. The numerical quanti- 
 
FUNDAMENTAL ELEMENTS 
 
 47 
 
 ties in this and the preceding problem can be verified by referring to the 
 tables in the back of this book. 
 
 Problem 8. Determine the center of gravity in reference to the 
 horizontal axis of the section shown in Fig. 45, which is composed of 
 2 15" x 33* channels and one cover plate 22" x J". 
 
 3- 
 
 I* ^ 
 
 1 K* 
 
 33*-" 
 
 Fig. 44 
 
 V 
 
 Fig. 45 
 
 Solution: Taking moments of the areas about the center of gravity 
 of the cover plate, we have 
 
 30.8^=7.75x19.8, 
 - 7.75x19.8 
 
 or 
 
 x 
 
 30.8 
 
 = 4.98". 
 
 That is, the center of gravity of the entire section is 4.98" below the 
 center of the cover plate, or 4.73" from the back of the top flanges of the 
 channels. The center of gravity here lies on the horizontal line marked 
 gg, which is known as the gravity line or gravity axis. 
 
 It is readily seen that the center of gravity of the above section in 
 reference to the vertical axis would lie on the vertical line vv passing 
 through the center of the cover plate; however, the center of gravity of 
 any section in reference to a vertical axis can be obtained by taking 
 moments about some vertical line in the same manner as shown above for 
 the horizontal axis. 
 
 48. Inertia. It is an observed fact that it always requires a force 
 to move a body when at rest, to stop it when in motion, or to change its 
 motion in any respect. This would be true even if friction and all other 
 external resistances were removed. Hence, it is evident that a body in 
 itself offers resistance to every change of motion; otherwise, it would 
 require no force to change the motion of the body if all the external 
 resistances were absent. This property which bodies have in themselves 
 of offering resistance to change of motion is known as Inertia. As an 
 example, a body lying upon an absolutely smooth horizontal plane would 
 offer resistance to any change of motion along the plane simply by virtue 
 of its inertia. The actual force exerted would be directly proportional 
 to the mass of the body concerned, and to the change of motion, or in 
 other words to the acceleration produced. Let W be the weight of the 
 body concerned, and let a be the acceleration produced in feet per second. 
 
48 
 
 STRUCTURAL ENGINEERING 
 
 Then for the intensity of the force (in pounds) exerted, we have the 
 proportion 
 
 F: W:: a: g, 
 from which we obtain 
 
 8 
 
 which is Formula (A) given in Art. 23. 
 
 49. Moment of Inertia and Radius of Gyration. Let AB (Fig. 
 46) represent a very thin rectangular plate of homogeneous material. 
 
 Suppose the plate starts to rotating about 
 a vertical axis YY. The resistance offered 
 by all such strips as cd (which will be 
 due to their inertia) will be directly pro- 
 portional to their distance out from the 
 axis YY, as their relative increment of 
 velocity or acceleration will be directly 
 proportional to that distance. Then the 
 intensity of the resistance of any strip 
 will be directly proportional to the dis- 
 tance of the strip from the axis and to the 
 mass of the strip. Suppose the plate made 
 up of infinitesimal strips as cd, and let m be the mass of each strip. Then 
 the force resisting the motion of the strip out unit distance from the axis 
 will be (m) 1, while the force out x distance will be mx, and the moment of 
 it about the axis would be (mx) (x)=mx 2 . Then evidently the moments 
 about the axis YY of all the resisting forces of these strips will be ^mx 2 
 where m is the mass of each strip, which was assumed to be the same for 
 each and every one, and x the distance out to each strip, no two x's being 
 the same. The expression 2moj 2 is known as the Moment of Inertia, which 
 is usually indicated by I, and we have in general 
 
 Fig. 46 
 
 In the case of the above plate it is evident that all of its mass could 
 be concentrated at some point out from the axis YY so that its moment 
 of inertia about the axis would be the same as that of the plate. This 
 distance out would be known as the Radius of Gyration of the plate in 
 reference to the axis YY. Then we have the general equation 
 
 7 = Mr 2 , 
 
 where M is the total mass of the plate and r the radius of gyration. 
 
 The mass of any one of the infinitesimal strips of the above plate is 
 directly proportional to da, the area of the strip; then if we imagine the 
 plate to diminish in thickness until it becomes a plane without thickness, 
 we have for its moment of inertia 
 
 from which we obtain 
 
FUNDAMENTAL ELEMENTS 
 
 Let a be the length of the plate and b its width ; then considering the 
 plate as a plane, its moment of inertia about the axis YY is 
 
 1= 
 
 Knowing its moment of inertia, we can then determine its radius of 
 gyration from the above formula, I = Ar 2 . Thus substituting, we have 
 
 - = bar 2 or r 
 o 
 
 The determination of the moment of inertia and radius of gyration 
 of the cross-section of individual members of structures are the most 
 common problems under this head met with in structural engineering. It 
 is really the case of determining the moment of inertia and the radius of 
 gyration of a plane, or as we may put it, the material cut by a plane. 
 There are a few cases, however, where it is necessary to consider the 
 mass, but in such cases the student will have no trouble if the general 
 formula, I Mr 2 , given above, is applied. 
 
 $ 
 
 Fig. 47 
 
 Fig. 48 
 
 Fig. 49 
 
 PROBLEMS 
 
 (1) Moment of inertia of a rectangle in reference to an axis GG 
 (Fig. 47) through its center of gravity. 
 Solution: Here we have 
 
 -II 
 
 '24 ^24 
 
 bh? 
 12 
 
 (2) Moment of inertia of a triangle about its base. (Fig. 48.) 
 Here we have 
 
 b f h _bh* 
 
 hjo 12 
 
 In case the axis is perpendicular to the plane, we have what is known 
 as the Polar Moment of Inertia. For example, for the polar moment of 
 inertia of a circle about its center we have from Fig. 49 
 
 rR CR 
 
 2 = I dax 2 =27r I x 
 
 Jo *Jo 
 
 A more general case of the polar moment of inertia would be that of 
 a rectangle. Here let (Fig. 50) be the axis and draw XX and YY 
 through at right angles to each other. Then we have 
 
50 
 
 STRUCTURAL ENGINEERING 
 
 which is the moment of inertia of the rectangle about the axis XX plus 
 the moment of inertia about the axis YY. This gives the general formula 
 which will apply to any plane figure, and further discussion of the polar 
 moment of inertia is unnecessary. Its principal use is in the designing of 
 shafting. The polar moment of inertia is usually designated by the letter 
 J instead of /, which is known as the Rectangular Moment of Inertia. 
 50. Proposition. The rectangular moment of inertia of any plane 
 figure about any axis in the same plane as the figure is equal to its moment 
 of inertia about a parallel axis through its center of gravity , plus its area 
 multiplied by the square of the distance between the two axes. As proof 
 of this, let ABCD (Fig. 51) represent a rectangle whose area is A=bh 
 
 -9 
 
 e 
 
 r 
 
 
 
 
 
 & 
 
 
 k 
 
 X 
 
 O 
 
 /x. 1 
 
 
 x 
 
 
 
 
 
 -n 
 
 
 Y 
 
 Fig. 50 
 
 A 
 
 3 
 
 d 
 
 a 
 
 
 
 & 
 
 - 
 
 -dx 
 
 oc 
 
 L h 
 
 e 
 
 
 
 
 Q 
 
 
 
 fe 
 
 | 
 
 9 
 
 F 
 
 ig. 51 
 
 and whose moment of inertia about its gravity axis gg is I' ' . Then, 
 according to the above, the moment of inertia of the rectangle about any 
 axis as YY is / = I' + d 2 A. This is shown to be true in the following 
 manner: The moment of inertia of the rectangle about the axis YY is 
 
 fo --,'). .. (1) . 
 
 /a 
 1 
 
 o 
 
 Now from Fig. 51 we have a = d+ (h/2) and e = d- (h/2). Substituting 
 these values of a and e in (1) we have 
 
 But bh*/12 equals the moment of inertia of the rectangle about its gravity 
 axis gg. Then substituting /' for bh 3 /12 and A for bh, we have 
 
 which proves the above proposition in the case of a rectangle. 
 
 The above proposition is true of any plane figure. This can be shown 
 by following out the same scheme as above. In case of bodies, the only 
 thing different is that we would consider mass instead of area and write 
 the formula 
 
 instead of 
 
 = I' + Ad 
 
CHAPTER IV 
 
 THEORETICAL TREATMENT OF BEAMS 
 
 51. Classification. Beams are, as a rule, classified according co 
 the number and kind of supports they have. Whenever a beam simply 
 rests upon a support, the support is known as a simple support, but when 
 the beam is held rigidly by a support, the support is known as a fixed 
 support. 
 
 Figures 52 to 55 show the most common types of beams. The canti- 
 lever beam shown in Fig. 52 has one fixed support, while the fixed beam 
 
 Fig. 52 
 
 Fig. 53 
 
 Fig. 54 
 
 Fig. 55 
 
 shown in Fig. 53 has two fixed supports. The simple beam shown in 
 Fig. 54, which is the most common type, has two simple supports. The 
 continuous beam shown in Fig. 55 has four simple supports; however,, any 
 beam with more than two supports of any kind is a continuous beam. 
 
 There are various other types of beams, such as overhanging, inclined, 
 etc., but they are all really modifications of the above types and will be 
 readily recognized as such without any preliminary description. 
 
 Beams are usually supported in a horizontal position and support 
 vertical loads which produce vertical reactions. This will be understood 
 to be the case unless otherwise stated. 
 
 52. Shearing Stress on Beams. Let AB (Fig. 56) represent an 
 ordinary rectangular wooden cantilever beam supported in a horizontal 
 position by having the end B built into a wall. Let P represent a load at 
 A which the beam supports in addition to its own weight. Let ab and cd 
 represent the traces of two imaginary planes passing perpendicularly to 
 the longitudinal axis of the beam, mentally separating the very short 
 rectangular block abed from the other parts of the beam, so that we can 
 think of this block as being a separate body. 
 
 By imagining the beam to be cut off instantly along the section ab we 
 readily realize that the part abA of the beam has a tendency to move 
 down vertically and the load P with it. As the material in the block abed 
 prevents the motion, evidently the part abA of the beam exerts a down- 
 ward force along the section a b upon the block equal to the combined 
 weight of the part abA of the beam and the load P. Let S be this force. 
 
 51 
 
52 STRUCTURAL ENGINEERING 
 
 Now it is evident that the part cdB of the beam to the right of the block 
 must exert an equal force S (neglecting the weight of the block abed) 
 upward along the section cd upon the block in order to prevent the block 
 from being pulled downward by the force S acting downward upon it 
 along the section ab. It is readily seen that the downward force S acting 
 along the section ab upon the block and the equal and opposite force S 
 acting upward along the section cd have a tendency to break the material 
 in the block off vertically. This action is most readily 
 s ^ comprehended by imagining the sections ab and cd to 
 
 -^ WMk. approach each other until the block abed is really a 
 
 single layer of molecules. Then there would be a 
 force S acting downward along ab upon the left side 
 
 56 of the molecules and an equal and opposite force S 
 
 acting upward along cd upon the right side of the 
 molecules, thus tending to break the molecules off crosswise instead of 
 tending to crush or pull them apart as in the case of simple stress, and 
 the stress thus produced would be known as the shearing stress on the 
 beam at that point. Whenever forces act in this crosswise manner, in any 
 case, the stress directly produced is known as shearing stress. 
 
 In speaking of the shear on a beam, or on any other body, we refer 
 to it as being x pounds along a certain section of x pounds cut by an 
 imaginary plane; but what we really mean is that a stress of x pounds is 
 produced upon a strip of material infinitesimal in thickness adjoining 
 that section. 
 
 It is evident that in the above case the shearing stress on the section 
 ab (really the shear on the block abed) of the beam is equal to the load P 
 and to the weight of the part abA of the beam. The part of the shear due 
 to the load P is the same for all vertical sections of the beam between A 
 and.B, while the part due to the weight of the beam will evidently be 
 directly proportional to the length of the part abA. Then, if w be the 
 weight of the beam per foot of length, the shear at any vertical section x 
 feet from the end A will be S = P + wx. Here it is seen that the shear on 
 any vertical section of the above beam is equal to the algebraic sum of the 
 forces between the section and the end. If additional forces were applied, 
 they would likewise be included in the summation, provided they were to 
 the left of the section considered. As the forces upon the part of the 
 beam to the right of the block abed must exert an upward force upon the 
 block along the section cd equal to the downward force along ab exerted 
 by the forces to the left, it is evident that the shear on the block is equal to 
 the algebraic summation of the forces on either side of it. So is the case 
 of all beams, where the forces are applied perpendicularly to the longi- 
 tudinal axis, and hence we have in general: The shear on any section of 
 a beam perpendicular to its longitudinal axis is equal to the algebraic 
 summation of the forces on either side of the section, provided all of the 
 forces act perpendicularly to such axis. 
 
 This is readily seen to be true. For, let AB (Fig. 57) represent a 
 beam in equilibrium acted upon by six forces PI ... P6, no reference 
 being made as to which are reactions; let abed be an imaginary block 
 cut through the beam, the same as the block abed shown in Fig. 56 : It is 
 obvious that the forces PI and P2 would cause the part abA of the beam 
 
THEOEETICAL TREATMENT OF BEAMS 53 
 
 to exert a force S upon the block along ab as shown, and it is readily seen 
 
 that this force S, which is the shear on the block (neglecting the weight of 
 
 the beam) would be equal to PI + P2, that is, their sum. As the block is 
 
 in equilibrium, undoubtedly there must be an equal and opposite force S 
 
 exerted upon the block along cd, as shown, by the part cdB of the beam 
 
 due to the forces acting upon that part. Then we have P1+P=+P6 
 
 P5+P4-P3 = S; that is, the shear on the block 
 
 5. i i ps abed is equal to the algebraic summation of the 
 
 old J 1 forces on either side of it. Likewise, the shear on 
 
 8 any section between P6 and P5 is equal to either 
 P6 P6 or P1+P2-P3 + P4-P5; between P5 and 
 FIg . 57 P4 it is equal to either P6-P5 or P1+P2-P3 + 
 
 P4; and between P4 and P3 it is equal to either 
 
 P6-P5+P4 or P1+P2-P3. This means that the shear at any cross- 
 section of any beam (as stated above) is equal to the algebraic summa- 
 tion of the forces on either side of the section, provided the forces are 
 perpendicular to the beam. 
 
 It is common practice to assume this shearing stress to be uniformly 
 distributed over the cross-section of the beam. Then according to this 
 assumption, the shearing stress per square unit on any cross-section of 
 the beam is equal to the shear at the section divided by the area of the 
 cross-section. So if p the shearing stress in pounds per square inch of 
 cross-section, S = the total shear on the cross-section in pounds, and A = 
 the area of the cross-section in square inches, we have for the shearing 
 stress per square inch the general formula 
 
 The assumption that the shearing stress is uniformly distributed over the 
 cross-section is not absolutely true, as will be shown later; however, it 
 meets the requirements of practical designing in most cases. 
 
 53. Bending Stress on Beams. Imagine for the time being that 
 the block abed (Fig. 56) be removed and suppose instead the parts abA 
 and dcB of the beam to be connected by a hinge placed in the center of 
 the cross-section of the beam as shown in Fig. 58. Now it is evident that 
 this hinge could be so constructed that it would prevent the part abA of 
 the beam from moving down vertically. This means 
 that the hinge would resist the shearing force or 
 "shear" which was resisted by the block abed, but 
 it is evident that the part abA would now fall by 
 rotating downward to the left about the hinge. So it 
 is seen that the part abA of the beam has a tendency 
 to rotate about the block abed, and hence will be subjected to stresses 
 therefrom. All stresses produced in this manner in beams, or any other 
 bodies, are known as bending stresses, or as stresses due to cross bending. 
 
 To determine the bending stress in the block abed of the above beam 
 (Fig. 56), let us assume all of the block removed except a thin horizontal 
 strip ad at the top of the beam and an equal strip be at the bottom, as 
 shown in Fig. 59. For the sake of simplicity let us first consider the 
 stress on these strips due to the load P only. The load P would cause the 
 
54 
 
 STRUCTURAL ENGINEERING 
 
 P 
 
 d 
 
 X 
 
 \ h Q 
 
 i 
 
 s 
 
 Fig. 59 
 
 part abA of the beam to exert a downward vertical force s upon each strip 
 (see Fig. 59) which must be resisted by an upward force s exerted upon 
 the part abA in turn by each strip. These two upward forces s upon the 
 part abA and the load P form a couple whose moment is 2sx or Px, as 
 2s P. This couple tends to rotate the part abA of the beam counter 
 clock-wise, but actual rotation is prevented by the strips ad and be, and 
 consequently each will be subjected to a stress. Now as it requires a 
 couple to balance a couple (see Art. 44) the stress produced upon the two 
 strips must form a couple and the stress in strip 
 ad must act to the right, while the stress in the 
 strip be must act to the left upon the part abA of 
 the beam. Let F be the stress produced on each 
 strip and let h be the vertical distance between 
 their centers of cross-section. Then we have Fh = 
 Px, from which we obtain F= (Px*)/h as the stress 
 on each strip. So far our problem is very simple. 
 But suppose we add two other strips, mm and nn, the same in section as 
 the two just considered and which are also symmetrically arranged in 
 reference to the longitudinal axis of the beam as shown in Fig. 60. Now 
 it is evident that the additional strips will help to resist the moment of 
 the couple Px, so that the stress F on the two strips just considered will 
 be reduced. Let / and /' now be the stresses on the strips, and h and h' 
 their lever arms, as indicated in Fig. 60. Then we have Px = fh + f'h'. 
 Here it is seen that we have one equation and two 
 unknown quantities, / and /', and it is readily seen 
 that for each additional pair of such strips com- 
 posing the block added, another unknown force 
 will occur, so our problem is not as simple as it 
 first appeared. However, we get out of our diffi- 
 culty by resorting to Hooke's Law (see Art. 32). 
 It is seen that the strips at the top of the beam are 
 
 in tension while those at the bottom are in compression, consequently the 
 strips at the top will be lengthened while those at the bottom will be 
 shortened, and the block abed, which is rectangular when not subjected to 
 bending stress, will really be wedge-shaped as shown in Fig. 61. As the 
 top strips are in tension and the bottom ones in compression, it is evident 
 that there must be a horizontal strip of fibers somewhere between the top 
 and bottom of the beam which has neither tension nor compression; that 
 
 is, no stress at all. Let og be the 
 location of this strip which is known 
 as the neutral plane or neutral axis. 
 As all the strips above og are in ten- 
 sion and all below are in compression, 
 it is obvious that g is the center of 
 tendency of rotation of the part abA 
 of the beam. Then, evidently, the 
 distortion of the strips both above and below og will vary directly as their 
 distance out from og. Now as the stress on any strip, according to 
 Hooke's Law, will be directly proportional to its distortion, the stress per 
 square inch on any strip will be directly proportional to its distance out 
 from og t and hence the stress on the strips may be represented by the 
 
 Fig. GO 
 
 
 Fig. 61 
 
THEOEETICAL TREATMENT OF BEAMS 55 
 
 arrows enclosed in the triangles nkr and mpr as shown in Fig. 61, where 
 the lines kr and mr have the same slope with the vertical. 
 
 Let / be the stress per square inch on any horizontal strip of fibers 
 y distance out from og, and let /' be the stress per square inch on a strip 
 out unit distance from og. Then we have the proportion 
 
 1 = 9 f f- 
 from which we get 
 
 /'=-? 
 
 y 
 
 Now as f/y is the stress on a strip out unit distance from og, the stress on 
 a strip out any distance z from og would be z times f/y or (f/y)z, and 
 the moment of it about g would be s times this or (f/y)z 2 . Now / was 
 taken as so many pounds per square inch, but as the stress varies con- 
 tinuously from og outwardly, there will not be a square inch of material 
 anywhere having the same stress, so that each strip must really be con- 
 sidered as being a horizontal element; that is, a mere plane of fibers 
 having an infinitesimal area of cross-section da. For the sake of concep- 
 tion we may say that da is 1/1,000,000 of a square inch. Then the actual 
 stress or force out z distance from og would be 1/1,000,000 of (f/y)z or 
 (ffy)zda (da = dzb) (see section Fig. 61), and the moment of it about g 
 would be z times that, or (f/y)z 2 da. Now, undoubtedly the summation of 
 these moments (f/y)z 2 da about g for all of the strips in the block abed, 
 which we can express as (f/y)*%z 2 da, must be equal to the moment of the 
 vertical couple Px. So we have 
 
 But 2z 2 da = I 
 
 (see Art. 49), the moment of inertia of the cross-section of the beam. 
 Then we have the equation 
 
 p*=li. 
 
 y 
 
 It will be observed very readily that Px is simply the moment of the 
 load P about the section ab (ab being considered vertical, and practically 
 it is) ; that is, Px is the measure of the tendency of rotation of the part 
 abA of the beam about the section ab due to the load P. Then evidently 
 the moment of any other load to the left of the section ab about the section 
 would be the measure of the tendency of rotation of the part abA of the 
 beam about the section due to any such loads. Then if ^Px be the alge- 
 braic summation of the moments of any number of loads on the left of the 
 section about the section, we have the formula 
 
 y 
 
 from which the stress / on any horizontal element of the block abed due to 
 the loads can be computed by substituting for y the distance of the element 
 out from og, and / would be known as the bending stress on the element, 
 the block abed being considered infinitesimal in length. 
 
 Now it is obvious that the above formula will apply to any section 
 of the above beam if SPj? be taken as the algebraic summation of the 
 
56 STRUCTURAL ENGINEERING 
 
 moments of the loads to the left of that section about the section. But it 
 is readily seen that the ^Px on one side of any section must be equal and 
 opposite to the ^Px on the other side in order that the beam may not 
 rotate about the section. This is undoubtedly true in the case of any 
 beam whatever. That means that the above formula applies to any 
 section of any beam if ^Px be taken as the algebraic summation of the 
 moments of the forces (both applied forces and reactions) on either side 
 of the section about the section. This summation is known as the bend- 
 ing moment at the section and is usually represented by the letter M. 
 Then we can write the general equation as 
 
 and by transposing we have 
 
 (D), 
 
 which is the stress on any horizontal element y distance from the neutral 
 axis either above or below it. It will be observed that the neutral axis 
 extends all the way along the beam. 
 
 It is seen that y varies from zero to either +e or -t (Fig. 61) and 
 that / will be a maximum when y is a maximum, as M and / are constants 
 for each specific section. So if y be taken as the distance to the farthest 
 element out from g the maximum value of / will be obtained. This is what 
 is usually desired, and for that reason y is usually spoken of as the dis- 
 tance to the extreme fiber and / as the stress on the extreme fiber. 
 
 The summation of the bending stresses above the neutral axis is 
 always equal to the summation of the bending stresses below it, but the 
 sums of their moments about the neutral axis are not equal except in the 
 case of symmetrical section. But each force on one side is always paired 
 with an opposite and equal force on the other side, thus forming couples 
 throughout the cross-section. 
 
 54. Reaction on Simple Beams. In order to determine the shear- 
 ing and bending stress in a beam, it is first necessary to know all of the 
 external forces acting upon it. The applied forces, or loads as they are 
 known, are usually given, but the reactions are 
 
 <J. j usually unknown and hence have to be deter- 
 
 H - c - -J mined before the stresses in the beam can be 
 
 determined. We usually resort to the equations 
 of moments to determine the reactions on beams. 
 However, they can be graphically determined, 
 as will be shown later. 
 
 Fig. 02 Let AB (Fig. 62) represent a beam which 
 
 is supported at A and B and which in turn sup- 
 
 ports the loads PI, P2, P3, and P4. The reaction R at A due to the 
 above loads can be obtained by taking moments about B, and the reaction 
 wRl at B can be obtained by taking moments about A. Thus, taking 
 moments about B f we have 
 
THEOKETICAL TKEATMENT OF BEAMS 57 
 
 from which we obtain 
 
 aPl + bP2 + cP3 + dP4: 
 
 -TT 
 
 and by taking moments about A we can determine R~L in the same manner. 
 It will be observed that by taking moments about the line of action of one 
 of the reactions we eliminate that reaction from the equation of moments, 
 thus reducing the number of unknowns in the equation to one. The sum 
 of the moments being equal to zero is, in accordance with the laws of 
 equilibrium, that the moments of the external forces acting upon a body in 
 equilibrium about every point must be equal to zero (see Art. 42). The 
 above method is quite general, provided the loads are fully known and the 
 reactions are limited to two in number, which is always the case for 
 simple beams. 
 
 55. Graphical Representation of Shear on Simple Beams. 
 The shear on a beam at any section is equal to the algebraic sum of the 
 loads aad reaction (as stated in Art. 52), summed up to the section, begin- 
 ning at either end. For example, the shear at every section between the 
 loads P3 and P2 (Fig. 62) is equal to either R - P4 - P3 or R1-P1-P2; 
 between A and P4 it is equal to either R or Rl -PI -P2-P3-P4; 
 between B and PI it is equal to either Rl or R -P4-P3-P2-P1. 
 
 This shear on the above beam can be graphically represented in the 
 following manner: 
 
 Draw a horizontal line ab (Fig. 63) equal to L, the length of the 
 beam (between centers of end bearings), to some convenient scale. Then 
 draw the lines of action of the forces in their relative positions as shown 
 at 1, 2, 3, and 4. Then beginning at one end, 
 say, at a, draw ac = R. Then through c and 
 parallel to ab draw cd and from d lay off de = 
 P4. Then from e draw ef and from / lay off 
 fg = P3, and so on, thus obtaining the zigzag line 
 cde . . . n. Any ordinate as rr from the line 
 ab to the zigzag line represents the shear on 
 Fig. 63 the beam at that point. 
 
 The value of the shear on any vertical sec- 
 tion o-f a beam will depend upon the weight of the load or loads it sup- 
 ports and upon the position they occupy in reference to the section. 
 Loads are known as dead load and live load. 
 
 The dead load upon a beam is the weight it supports that is fixed in 
 position, while the live load is the weight it supports that is not fixed in 
 position but may move to any position upon it. The dead load of struc- 
 tures usually consists of the weight of the structure itself, while the live 
 load consists of loads which it supports but which at the same time move 
 over it, as a locomotive, train of cars, wagons, etc. The dead load is 
 usually considered as a uniformly distributed load. The live load is 
 sometimes considered as a uniformly distributed load, but more often as 
 concentrated loads. 
 
 If the load be a uniformly distributed dead load, each reaction will 
 be R wL/2, where L is the length of the span and to the weight of the 
 dead load per foot of span, and the shear out any distance z from either 
 end will be S = R-zw=wL/2-zw. It is seen that when z = L/2 the 
 
 P4- 
 
 P2 
 
58 
 
 STRUCTURAL ENGINEERING 
 
 shear is equal to zero, and when 2 = the shear is equal to wL/2. This 
 shows that the shear due to such a load is a maximum at the ends and 
 zero at the center. 
 
 The above equation for shear, S = wL/2-zw, is an equation to a 
 straight line, so evidently the shear can be represented graphically by the 
 ordinates to a straight line, as acb (Fig. 64), where AB represents the 
 
 length of span and aA and bB are equal to 
 tvL/2. The shear, beginning at A, decreases 
 until c, the center of the span, is reached, 
 where it is zero ; then beyond that point, still 
 summing up to the right, the shear will be 
 minus, increasing toward B, being equal to 
 wL/2 at B. So it is seen that the ordinates 
 Fig. 64 to the line acb truly represent shear for dead 
 
 load uniformly distributed. For example, 
 
 the ordinate fe represents the shear at /, which is positive, while kh rep- 
 resents the negative shear at k. The sign of the shear has no practical 
 significance other than algebraic. 
 
 In case of uniform live load moving over a beam, the shear at any 
 section will be a maximum when the load just extends from one of the 
 ends up to the section. For example, suppose a live load of p pounds per 
 foot extends from B up to some section 0, a point x feet from B. Then 
 the shear at is equal to the reaction at A, which expressed in terms of 
 the load is (px)x/2 + L = px 2 /2L. It is evident that the shear produced 
 at section 0, as the load moves from B to 0, keeps increasing as the load 
 approaches 0, as it is equal to the reaction at A, which keeps increasing 
 as the load moves from B to 0. Now if the load extends past to any 
 section m, the reaction at A will be increased a certain amount by the 
 additional load py, but in obtaining the shear at we subtract all of the 
 load py from the reaction at A, and as all of the additional load py is not, 
 as we say, transmitted to the end A, it is evident that the load extending 
 beyond will diminish the shear at that section. Therefore the above 
 statement is shown to be true, and the equation S = px 2 /2L is the general 
 equation for the maximum shear produced on a beam by a uniform live 
 load moving over it. The equation is that of a parabola. Then, evidently, 
 if we construct a parabola as auB so that Aa is equal to pL/2, the shear 
 at any section will be represented by the ordinate to the parabola at that 
 section. For example, the shear at t is represented by the ordinate ut. 
 The shear curve auB (Fig. 64), as stated above, is an absolute 
 parabola for a uniform live load, but for such loads as the wheels of a 
 locomotive it would be a curve made up of a series of straight lines, 
 however, if the loads are nearly equal in weight and quite uniform in 
 spacing, the curve will approach a parabola, and in most cases it is near 
 enough to be considered one for practical purposes. 
 
 56. Graphical Representation of Bending Moments on Simple 
 Beams. 
 
 Case I. When the load is uniformly distributed over the entire 
 length. 
 
 Let AB (Fig. 65) represent a beam supported at A and B which 
 in turn supports a uniform load of w pounds per lineal foot of length. 
 The bending moment at any section ss, x feet from A due to this load is 
 
THEORETICAL TREATMENT OF BEAMS 
 
 But 
 
 so we have 
 
 R = 
 
 wLx 
 
 was* 
 
 for the bending moment at any section of the beam. When x = L/2, we 
 have 
 
 wL 2 wL 2 
 
 from which we obtain 
 
 (E), 
 
 which is a formula of great practical value as it expresses the moment at 
 the center of any simple beam uniformly loaded, which is the maximum 
 moment that can occur under such loading, and usually this is what we 
 desire to know. 
 
 The above formula, M = wLx/% - wx z /2 ) is an equation to a 
 parabola. Then, evidently, the bending moments at the different sections 
 along the beam vary as the ordinates to a par- 
 abola. It is seen from the equation that M 
 when x = Q, and also when x L,; and when x 
 = L/2, M = wL 2 /S. Then if we draw CO equal to 
 wL 2 /S to scale, and perpendicular to the beam 
 at its center and pass a parabola through A, O, 
 and B as shown in Fig. 65, any ordinate as rr 
 will represent the bending moment on the beam 
 
 Fig. 65 
 
 When the loads are concentrated at different points along 
 
 at that point. 
 
 Case II. 
 the beam. 
 
 Let AB (Fig. 66) represent a beam supported at A and B which 
 in turn supports the loads PI . . . P5, as shown. The bending moment 
 anywhere between A and P5, due to these loads, is M = Rx, which is an 
 equation to a straight line. When ,r = 0, M = 0, and when x a, M = Ra. 
 Then, if we draw cd equal to Ra, to scale, and 
 join A and d, any ordinate as ss to the line Ad 
 will represent the bending moment on the beam 
 at that point. The bending moment anywhere 
 between Po and P4 is M = .R(a + ,r') P5#', 
 which is also an equation to a straight line. When 
 #' = 0, M = Ra, so, evidently, one point in the 
 line represented by the equation M R ( + #') 
 
 - PS*' is d, as cd = Ra. When x' = b, M = R (a + 6) - P56. So if we draw 
 ef equal to R(a + b)-P5b to scale, using the same scale as in the case 
 of cd, we obtain the line df, the ordinates to which evidently represent 
 the bending moments between P5 and P4. Now it is evident that the 
 lines fg, gh, hk, and TcB can be similarly constructed, the ordinates to 
 
60 
 
 STRUCTURAL ENGINEERING 
 
 which represent the bending moments between P4 and P3, P3 and P2, 
 P2 and PI, PI and B, respectively. So it is seen that the bending 
 moments on a simple beam due to any number of concentrated loads 
 can be graphically represented by the ordinates to a series of straight 
 lines forming a closed polygon with the ends of the beam. The nearer 
 the loads approach a uniform load, the nearer this polygon approaches 
 a parabola. 
 
 It is important to note that the maximum moment always occurs 
 under a load. This is readily seen, for the maximum ordinate to any 
 segment of the above polygon is under a load and hence the maximum 
 of them all will be under a load. 
 
 57. Graphical Representation of the Bending Moments and 
 Shears on Cantilever Beams. 
 
 Case I. When the load is uniformly distributed 
 over the entire length. 
 
 Let AB (Fig. 67) represent a cantilever beam 
 supporting a uniform load of w pounds per foot of 
 length. The bending moment at any section ss, x feet 
 from A, is M= (wx)x/% = wx 2 /2, which is an equation 
 to a parabola. Making x = L, we have M = wL 2 /%, 
 which is the bending moment at B. Then if we draw 
 the horizontal line ab = L, and the vertical line cb 
 wL 2 /2, to scale, and construct the parabola ac, the 
 ordinates between the line ab and the curve ac will 
 graphically represent the bending moments on the beam due to the above 
 load. For example, the moment at the section ss is represented by the 
 ordinate rr. 
 
 The shear at any section ss is S = wx, which is an equation to a 
 straight line. When x = 0, S = 0, and when x L, t S = wL, which is the 
 shear at B. Then, if we draw bo equal to 
 wL, to scale, and join a and o, the shear at 
 any section in the beam is represented by 
 the ordinate between the line ab and ao 
 immediately under the section. For exam- 
 ple, the shear at the section ss is repre- 
 sented by the ordinate er. 
 
 Case II. When the loads are concen- 
 trated at different points along the beam. 
 Let AB (Fig. 68) represent a canti- 
 lever beam supporting the concentrated 
 loads PI . . . P4 as shown. As a general 
 example, the bending moment due to the 
 above loads at a section as ss, expressed 
 analytically, is M = nPl + bP2 + tP3. If we 
 proceed in the same manner as in the case 
 of a simple beam, making the horizontal 
 line ab = L, and the verticals wx, w'x', 
 etc., under each load, equal to the bending Fig. cs 
 
 moment about each, respectively, making 
 
 bv equal to the moment at B, and drawing the broken line aww'w"v, 
 we have the bending moments on the beam graphically represented by 
 
'THEORETICAL TREATMENT OF BEAMS 
 
 61 
 
 the ordinates between the line ab and this broken line. For example, 
 the moment at the section ss is represented by the ordinate ro. The 
 shear at the section ss is S = P1+P2 + P3, expressed analytically. Con- 
 structing the zigzag line acd . . . m, the same as in the case of simple 
 beams (Art. 55)., we have the shear at any section represented by the 
 ordinate between the line ab and this zigzag line. 
 
 58. Graphical Construction of a Parabola. The following method 
 of constructing a parabola will be found to be quite convenient. It really 
 consists in passing a parabola through two points, when one of the points 
 is taken at the vertex. Let A and B (Fig. 69) be any two given points 
 and let A be at the vertex of the parabola desired. Draw the line XX 
 through A as the X-axis of the parabola. Then draw AC perpendicular 
 
 to XX and BC perpendicular to AC. Divide AC and BC into an equal 
 number of equal parts, say, six, as shown. Then from A draw the radial 
 lines \-A, 2-A, etc., and where the radial line \-A intersects the vertical 
 aa is one point on the curve of the required parabola, and where the next 
 radial line 2-A intersects the next vertical bb is another point on the 
 curve, and so on. The other side AB' of the parabola can be constructed 
 by laying off AC' -AC and dividing it into the same number of equal 
 parts as AC and drawing verticals as shown and then projecting the 
 corresponding points over from the curve AB just constructed, or by 
 dividing B'C' into the same number of equal parts as AC' and proceeding 
 as explained for the other side of the curve. 
 
 59. Relation of Shear to Bending Moment. Let AB (Fig. 70) 
 represent a simple rectangular wooden beam supported at A and B and 
 let P be a load which the beam supports at mid-span, and let R and R\ 
 be the reactions (which will be equal) at A and B, respectively, due 
 to this load P. Suppose the reactions applied exactly at the ends of 
 the beams as indicated, which is not at all an imaginary case, and imagine 
 the beam divided up into short rectangular blocks as shown, by imaginary 
 planes passing perpendicularly to the longitudinal axis of the beam. Let 
 A,r be the length of each block. The reaction R at A applied along the 
 end of the first block, beginning at A, tends to move that block upward. 
 If the block does not move undoubtedly the second block prevents it by 
 exerting an equal and downward force R where the two blocks join as 
 indicated. As the first block is held entirely by the second block it is 
 obvious that the first block will exert an upward force upon the second 
 block where the two blocks join, and if the second block does not move 
 upward undoubtedly the third block prevents it by exerting a downward 
 
62 
 
 STRUCTURAL ENGINEERING 
 
 Fig. 70 
 
 force /2 where the second and third blocks join, and hence it is seen that 
 the blocks to the left of the load resist each other in transmitting the 
 shear, each thereby receiving an upward force R upon one end and an 
 
 equal downward force upon the other end 
 as is indicated. The same is true of the 
 blocks to the right of the load P except 
 the forces are reversed in order. These 
 vertical forces not only tend to shear the 
 blocks off vertically but tend to rotate them 
 as well, as the two forces on each block 
 form a vertical couple. 
 
 Now beginning at A and considering the blocks to the left of P only, 
 the bending moment at the right end of the first block is RAx ; at the right 
 end of the second block it is R(2kx) ; and at the right end of the third 
 block it is .R(3A,r) ; and so on, being R(nAx) at the right end of the nth 
 block from A, which is seen at a glance to be nothing more than the 
 summation of the moments of the vertical couples on the blocks beginning 
 at A. This means that the bending moment beginning at the end A 
 increases toward the load P by one Rkx at each block. For example, the 
 bending moment at kk is one Rkx greater than it is at hh. So, evidently, 
 R&x is the increment of the bending moment to the left of the load P in 
 the case of the above beam. By the same reasoning we have Rl&x as the 
 increment of the bending moment to the right of the load P. 
 
 It will be observed that all of the blocks to the left of the load P 
 tend to rotate clock-wise while all of the blocks to the right of P tend to 
 rotate in the opposite direction or counter clock-wise, so that the sign of 
 the increments of the bending moment on one side will be different from 
 those on the other side of the load. The bending moment at any point 
 to the left of the load P is equal to the sum of the increments R&x 
 between A and the point and the bending moment at any point to the 
 right of the load is equal to the sum of the increments R\kx between B 
 and the point, although the bending moment at any point is equal to the 
 sum of the increments summed up algebraically from either end of the 
 beam. For example, the bending moment at gm is equal to the sum of 
 the increments to the right of gm or to the sum of the increments to the 
 left of P minus the sum of the increments between P and gm, as the signs 
 are different. 
 
 In the case shown in Fig. 70 the increments of the bending moment 
 are all equal, but suppose we consider the case shown in Fig. 71 where 
 R and Rl are the reactions at A and B, respectively, due to the three loads 
 PI, P2, and P3. Here it is readily seen 
 that the increment of the bending moment 
 between A and PI due to PI, P2, and P3 
 is flAx; between PI and P2 it is (R-Pl) 
 Ax; between P2 and P3 it is (R-P1-P2) 
 Ax; and between P3 and B it is (R -PI 
 -P2-P3)Ax. That is, the increment of 
 the bending moment at any point is equal to the shear at that point multi- 
 plied by Ax. It will be observed that this was true in the case shown in 
 Fig. 70, for R is the shear to the left of P and Rl the shear to the right 
 of P, which are equal in that case. Now, the vertical couples on any such 
 
 Fig. 71 
 
THEOKETICAL TEEATMENT OF BEAMS 63 
 
 blocks or strips, as here considered, in any beam, could not be anything 
 other than the vertical shear couples on them, so we have, in general, 
 
 for the increment of the bending moment, where S is the shear on the 
 block in question and A,r its length. Now, A.r could have any practical 
 value but the ultimate increment will be when A,r is an infinitesimal. 
 Then assigning the smallest possible value to A,r making it an infinitesimal 
 dx, we obtain 
 
 dM = Sdx 
 
 as the ultimate increment of the bending moment, as it is the smallest 
 increment possible. That is, the differential of the bending moment at 
 any section of any beam is equal to the vertical shear couple on an 
 imaginary vertical block of infinitesimal length. The preceding formula 
 is usually written 
 
 
 
 Expressing this in words, we say that the first derivative of the bending 
 moment in reference to x is equal to the shear, where x is the distance 
 from the end of the beam to the section considered. 
 
 The above equation (F) is sometimes quite useful, as the determining 
 of the shear from the bending moment is sometimes desirable. The 
 equation, or formula, is very readily applied, as is seen from the follow- 
 ing, although its true worth is more readily recognized in more compli- 
 cated cases. 
 
 Considering both the load P and the weight of the beam acting upon 
 the beam shown in Fig. 70, the bending moment at any section x distance 
 from the end A is 
 
 wx 2 
 M = Rx- 
 
 10 
 
 where w represents the weight of the beam per foot. Differentiating 
 both sides of the equation and dividing through by dx we have 
 
 dM 
 
 -j-^R-wx, 
 dx 
 
 which, as is readily seen, is the shear at any cross-section x distance from 
 the end A and to the left of the load. 
 
 The bending moment at any section x distance from the end of a 
 simple beam having a length L and supporting a uniform load of w pounds 
 per lineal foot is 
 
 wLx wx 2 
 
 M= -~r 
 
 (see Art. 56, Case I). Differentiating both sides of this equation, 
 dividing through by dx, we have 
 
 dM wL 
 
 = - wx, 
 dx 2 
 
64 STRUCTURAL ENGINEERING 
 
 which, as is readily seen, is the general expression for the shear at any 
 section of the beam. If ,r = 0, the shear is equal to wL/2, and if x I^ y it 
 becomes -(wL/2), each of which is recognized as an end shear or reac- 
 tion. If x = L/2 the shear is equal to 0, all of which goes to show how 
 readily Formula (F) can be applied, even for ordinary cases. 
 
 60. Increment of the Bending Stress. Let ab (Fig. 70) represent 
 a very thin horizontal strip through the beam y distance above the neutral 
 axis oo. The end block at A, owing to its tendency to rotate clock- wise, 
 due to the vertical shear couple Rkx, will exert a force to the right upon 
 all horizontal elements in the second block above the neutral axis and a 
 force to the left upon all such elements in the second block below the 
 neutral axis. Then undoubtedly the first block will exert a force to the 
 right upon the strip ab. Let A/ be this force. The end block at B has 
 the same tendency to rotate as the end block at A but in the opposite 
 direction, and hence the end block at B will exert a force A/ to the left 
 upon the strip ab. Then the strip ab will have a compressive stress of 
 A/ in it from c to d due to the end blocks. The second block from the 
 end A having the same tendency to rotate as the end block will likewise 
 exert a force A/ to the right upon the strip ab and the second block from 
 the end B will exert a force A/ to the left upon the strip ab, and hence the 
 strip ab will have a compressive stress of 2A/ from e to / due to the first 
 and second blocks from the ends. The third block from the end A will 
 exert a force A/ to the right upon the strip ab and the third block from B 
 will exert a force A/ in the opposite direction upon the strip ab and hence 
 the strip ab will have a compressive stress of 3A/ from I to g due to the 
 first, second, and third blocks from the ends. Thus it is seen that the 
 stress in the strip ab is increased at each block by one A/ as we pass from 
 either end toward the load P where the stress is a maximum. So, 
 evidently, A/ is the increment of the bending stress in the strip ab. As 
 the strip ab could be any horizontal element out any distance y from the 
 neutral axis, either above or below, it is evident that the bending stress on 
 any horizontal element in the beam will vary by such increments as A/ 
 the same as in the case of the strip ab; of course, their real values will be 
 different owing to their distance out from the neutral axis being different. 
 For the value of A/ at any block to the left of the load P (Fig. 70) we 
 have 
 
 and for its value to the right we have 
 
 where I is the moment of inertia of the cross-section of the beam in 
 reference to the horizontal gravity axis of the beam; y the distance out 
 from the neutral axis to the strip or element considered; and Rkx and 
 /21A,r the vertical couple in each respective case. But as the vertical 
 couple on any imaginary vertical block or strip in any beam whatever is 
 simply the shear on the block multiplied by its length, we can write the 
 general expression for the increment of the bending stress at any point in 
 
THEOEETICAL TEEATMENT OF BEAMS G5 
 
 any beam as 
 
 A /= j-^- (a). 
 
 Then beginning at either end of a beam we can obtain the stress on any 
 horizontal element x distance from the end by summing up the stress 
 increments as expressed by (a). This would give us 
 
 But SA/ = / and ^S^x-M, the bending moment (see Art. 59) at a point x 
 distance from the end, when A# becomes an infinitesimal. Then substi- 
 tuting / for 2A/ and M for 2$A,r in the last equation we have 
 
 My 
 
 r ~ I > 
 which is Formula (D), Art. 53. 
 
 Owing to the shear being constant the increments of the bending 
 stress in the beam shown in Fig. 70 are constant throughout for each 
 horizontal element and are equal to 
 
 But in such cases, as shown in Fig. 71, they will be different owing to the 
 variation of the shear. The increment between A and the load PI, in 
 that case, due to the loads PI, P2, and P3 is expressed as 
 
 (flAg)y. 
 
 / 
 between PI and P2 as 
 
 (fl-Pl)Asy. 
 
 I 
 between P2 and P3 as 
 
 (fl-Pl-P2)A*y , 
 
 A/- - 
 
 and between P3 and B as 
 
 (fl-P'l-P2-P3)Aary (JRlAjQy 
 
 It is seen from the above that the increment of the bending stress 
 varies directly as the shear and hence in most cases the greatest increment 
 will be at the ends of a beam, especially when the weight of the beam is 
 considered. 
 
 61. Horizontal Shearing Stress in Beams. Let AB (Fig. 72) 
 represent a portion of the same beam shown in Fig. 56 (Art. 52), which 
 is here drawn so as to show the imaginary block abed to a large scale. 
 Conceive of the original imaginary block abed sub-divided into smaller 
 imaginary blocks as egkh, gmnk, etc. We can now conceive of the shear- 
 ing force exerted upon the block abed along ab and cd as being distributed 
 to each of the smaller blocks, whereby each is seen to have a vertical 
 couple which we can conceive of as being an increment couple of the 
 
66 
 
 STRUCTURAL ENGINEERING 
 
 vertical couple acting upon the block abed as a whole. Now, evidently, 
 each of the smaller blocks has a tendency to rotate owing to this vertical 
 couple, and if rotation does not take place, evidently it is prevented by 
 the material joining the blocks horizontally. Then the forces resisting the 
 vertical couple on each of the smaller blocks must act through this mate- 
 rial, forming a horizontal couple on each block as shown, which must 
 necessarily be equivalent and opposite to the vertical couple that it resists. 
 Now, it is readily seen that these horizontal couples tend to shear the 
 smaller blocks off horizontally the same as the vertical couples tend to 
 shear the same blocks off vertically. If a block be square^ it is readily 
 seen that the vertical and horizontal couples on it will not only be 
 equivalent, but will be equal and opposite, which means that each of the 
 forces in the horizontal couple is equal to each of the forces in the vertical 
 couple; that is, the four forces acting upon the block are equal. This 
 means that the horizontal shear on the smaller blocks is equal to the 
 vertical shear on the same. But if the smaller blocks are considered 
 
 O d 
 
 e 
 
 ~=^~ 
 
 * 1 
 
 9 
 
 -*s- 
 
 * 
 
 m 
 
 1 
 
 ' 
 
 b c 
 
 Fig. 72 
 
 m o 
 
 Fig. 73 
 
 shorter or longer in one direction than in the other, the unit shear is not 
 changed. The larger sides having the greater area will simply have a 
 greater force acting, but the force per square unit will be just the same, 
 as our mentally changing the block will not alter the unit shear. As far 
 as the two couples on any block remaining equivalent is concerned, it is 
 evident they will, for as the force on one side is increased, the lever arm 
 of the other is increased, so that the moments of the couples are con- 
 tinually equivalent. So it really does not matter what shape our fancy 
 molds the imaginary parts, for it remains evident that the horizontal and 
 vertical shear on any particle in any beam are equal. Then, if we know 
 the horizontal shear on any particle, we know the vertical, as the two are 
 equal, and vice versa. The determining of the total vertical shear at any 
 vertical section of a beam is an easy matter, but just what the intensity 
 of it is on any particular particle is another thing. What we really do is 
 to find the horizontal shear and consider the vertical shear equal to the 
 same. In order to do this, let abed, Fig. 73, represent the imaginary block 
 abed of Fig. 56, as an independent body where the other parts of the beam 
 are replaced by the forces which those parts exert upon the block. Let 
 go be the neutral plane. The part of the beam to the left of the block 
 will exert upon it the bending stresses represented by the arrows in 
 triangles agm and bgm', and also the vertical shearing force S along ab, 
 while the part of the beam to the right will exert upon it the forces 
 
THEORETICAL TREATMENT OF BEAMS <J7 
 
 represented by the arrows in the triangles dok and cok' ', which resist the 
 bending stresses represented by the arrows in the triangles agm and bgm' ', 
 and hence are equal and opposite to them, and also the shearing force S 
 along dc, which is equal and opposite to the force S acting along ab, and 
 also the bending stress- increments represented by .the arrows in the 
 triangles don and con'. 
 
 The bending stresses represented by the arrows in triangles agm and 
 bgm' are transmitted directly through the block and are balanced directly 
 by the forces represented by the arrows in the triangles dok and cok' , so, 
 evidently, none of these forces causes horizontal shear on the block and 
 can be disregarded. The force S acting along ab and. the equal and 
 opposite force S acting along dc and the bending stress increments repre- 
 sented by the arrows in the triangles don and con' are the only forces 
 remaining. But as each of the forces S acts vertically, it is evident that 
 the horizontal shear in the block is due entirely to the forces represented 
 by the arrows in the triangles don and con' ; that is, to the increments of 
 the bending stresses on the block, and hence it remains for us to consider 
 these forces only. 
 
 Now, evidently, the horizontal shear on any element as ee of the 
 block abed is equal to the algebraic summation of these increment forces 
 on either side of ee, the same as in the case of any body. Therefore, the 
 horizontal shear at ee is equal to the sum of these forces between d and e, 
 or between e and c. 
 
 The moment of these increment forces is equal to Skx, which they 
 resist. Then it is readily seen that the stress at d, represented by nd, is 
 
 and at e it is 
 
 where I is the moment of inertia of the cross-section of the beam at the 
 block abed. Then the average increment stress between d and e is 
 
 . A/ + A/' (S*x\(h + y\ 
 
 f ~ir -\T P /vT/ 
 
 Multiplying this by bxde, where 6 = width of beam, we obtain the total 
 horizontal shear along ee, which is 
 
 Now this shear is distributed over the horizontal strip through the block 
 at ee. If b be the width of the beam, the area of this horizontal strip will 
 be bbx. Then the shear per square inch upon it will be 
 
 from which we obtain the formula 
 
STRUCTURAL ENGINEERING 
 
 From Formula (G) the horizontal shear (and incidentally the vertical) 
 can be obtained at any point in any rectangular beam either above or 
 below the neutral axis. 
 
 It is readily seen that Formula (G) is an equation to a parabola, 
 which means that the horizontal shear on any rectangular beam, and 
 likewise the vertical shear, varies on any vertical section as the ordinates 
 to a parabola, as represented by the horizontal ordinates to the curve 
 axb shown in Fig. 73. If y = h, the shear S^ is equal to 0. If # = 0, 
 S 1 f S/A f which is the maximum, where A = area of cross section 
 of beam. So it is seen that both the vertical and horizontal shear is at 
 the top and vlikewise at the bottom of any beam, and a maximum at the 
 neutral axis. The same is true of all beams. 
 
 In case the width of a beam varies the general equation 
 
 m 
 
 8 
 
 can be used for determining the horizontal shear, where m = the moment 
 about the neutral axis of the part of the cross section above ee or below 
 in case ee is below the neutral axis and 6 = width of beam at ee. 
 
 62. Maximum Stress. It is seen from the preceding article that 
 every particle in a loaded beam, except the particles at the top and bottom 
 edges, and those in the neutral axis, is subj ected to horizontal and vertical 
 shear and to a direct horizontal stress which may be 
 either tension or compression, depending upon the 
 position of the particle in the beam. In the case of 
 a cantilever beam the particles at the top edge are 
 subjected to direct tension only, while all particles 
 between the top edge and neutral axis are subjected 
 to horizontal and vertical shear and also to direct 
 Fig. 74 tension, and the particles in the neutral axis are sub- 
 
 jected only to horizontal and vertical shear. At the 
 bottom edge of a cantilever beam the particles are 
 subjected to a direct compression only, while the 
 particles between the bottom edge and neutral axis 
 are subjected to horizontal and vertical shear and 
 also to direct compression. In the case of a simple 
 beam we have exactly the reverse of a cantilever 
 beam. As an illustration, let AB (Fig. 74) represent 
 a short portion of a simple beam where oo represents 
 the neutral axis. The particles at the top edge are subjected to direct 
 compression only, as indicated at a, while the particles between the top 
 edge and neutral axis are subjected to horizontal and vertical shear and 
 also to direct compression as indicated at 6, the particles in the neutral 
 axis are subjected only to horizontal vertical shear as indicated at c. At 
 the bottom edge the particles are subjected to direct tension only, as in- 
 dicated at e, while the particles between the bottom edge and the neutral 
 axis are subjected to horizontal and vertical shear and also to direct 
 tension, as indicated at d. 
 
 It is seen from Fig. 74 that the intensity of the maximum stress on 
 any particle at the top or bottom edge of the beam is simply the direct 
 Stress on it, but the intensity of the maximum stress on any other particle 
 
THEORETICAL TREATMENT OF BEAMS 69 
 
 is not so evident,, for here the maximum stress is due to the combined 
 action of the shearing forces and direct stress. Let Fig. 75 represent a 
 small imaginary block taken from a simple beam, corresponding to the 
 block at d, Fig. 74. For convenience consider the block to be a cube 
 having sides of unit length, and let h represent the horizontal shear and 
 v the vertical shear. It is readily seen that the h along fe and the v along 
 ek acting against the v along fg and the h along kg produce tension upon 
 all such strips through the block as fk, while the h along fe and the v 
 along fg acting against the v along ek and the h along kg will produce 
 compression upon all such strips through the block as eg. Now it is 
 evident that the maximum tension and compression as the case may be, 
 due to the shearing forces, will be upon strips perpendicular to the 
 resultant of the shearing forces. As the horizontal and vertical shear are 
 equal it is obvious that their resultant will always be at 45 with the 
 horizontal and vertical and hence the maximum tension or compression 
 stresses due to these forces will be upon strips perpendicular to this 
 direction, that is, 45 from the direction of the shearing forces them- 
 selves. It is readily seen that the direct tensile stress indicated as p on 
 the block will increase the tension due to the shearing forces on the strip 
 fk and decrease the compression on the strip eg; while if p were a com- 
 pressive stress just the reverse would be true. While the tensile stress p 
 would increase the tension on the strip fk, yet this would not be the 
 maximum tension on the block, for evidently the maximum would be on a 
 strip through the block perpendicular to the resultant of the stress p and 
 the shearing forces v and h. This resultant would have some position as 
 Rt, if p were tension, and Re, if'p were compression, and the maximum 
 tensile stress then would be upon strips through the block perpendicular 
 to Rt; and if p were compressive the maximum compressive stress would 
 be upon strips through the block perpendicular to Rc> 
 
 If the resultant of the shearing forces on the particles of any beam 
 were graphically combined throughout the beam with the direct stresses 
 on the particles, the balanced resultants obtained would form curves 
 known as the lines of maximum 
 stress. As an example, let AB (Fig. L, pg P3 
 
 76) represent a simple beam. By * * _ . 
 
 combining the forces at e, d, c, etc., 
 
 as indicated at e' ', d' ', and c', the/\ 
 
 curve edc would be obtained and by 
 
 combining the forces on the particles 
 
 throughout the entire beam we would 
 
 obtain some such lines as t, tl, t2, 
 
 etc., representing the direction of the 
 
 action of the maximum tensile 
 
 stresses in the beam. Now, evi- Fig. 76 
 
 dently, if any particle in the beam 
 
 failed in tension the failure would take place perpendicular to these lines 
 
 of maximum tension. 
 
 If the maximum compressive stresses were platted in the same 
 manner they would take some such position as indicated by the dotted 
 lines on the beam, and if any particle failed in compression the failure 
 
70 STRUCTURAL ENGINEERING 
 
 would evidently take place perpendicular to these lines of maximum 
 compression. 
 
 It is readily seen from Fig. 75 that the shear due to the shearing- 
 forces v and h along a strip as eg, 45 with the horizontal or vertical, 
 will be zero, for the two shearing forces h and v on each side will just 
 balance each other when resolved along that direction. Then the only 
 shear along that direction will be due to the direct stress p. If we 
 imagine the strip rotated about o it is readily seen that the shearing 
 forces will produce shear along the strip when it slopes other than 45 
 with the horizontal and the maximum shear on it due to the shearing 
 forces and direct stress combined will occur when the angle of slope has 
 some particular value. But this angle of slope depends upon the relative 
 intensity of the shearing forces and the direct stress. The direction of 
 maximum tension and compression stresses also depends upon the relative 
 intensity of the shearing forces and direct stress. So we will now deter- 
 mine these relations. 
 
 Let Fig. 77 represent a material particle assumed rectangular, 
 subjected to horizontal and vertical shear of s pounds per square inch 
 and to a tensile stress of / pounds per square inch. Suppose the particle 
 to be one unit in thickness, and let a be its width 
 and b its length. Then sa will be the total vertical 
 shear and sb the total horizontal shear, and fa the 
 total tensile stress, as indicated in the figure. Let 
 dd be an imaginary diagonal strip through the par- 
 ticle. If the shearing and tensile forces be resolved 
 Fig. 77 perpendicularly and parallel to this imaginary strip 
 
 dd, the components perpendicular to the strip will 
 
 produce tension on it, while the components along the strip will produce 
 shear. Let t be the tensile stress per square inch perpendicular to the 
 strip dd and let v be the shearing stress per square inch along the strip, 
 and let 9 be the angle that the strip dd makes with the horizontal axis 
 of the particle; then, considering the forces on one side of dd only, we 
 have 
 
 (1) tdd = fa sinO + sb sinO + sa cos0 for the tension, 
 and 
 
 (2) vdd = fa cosO + sb cos6-sa sin# for the shear. 
 
 Dividing (1) and (2) by dd and substituting sin0 for a/dd and cos0 for 
 b/dd t we have 
 
 (3) t = f sin 2 + 2s cosOsinO = ~ (1 - cos20) + * sin20, 
 and 
 
 (4) v = f sinOcosO + s (cos 2 - sin 2 ^) = sin2^ + s cos20. 
 
 It is evident that t will be a maximum when 6 has a certain value and v a 
 maximum when 6 has a certain other value. Differentiating (3) we have 
 
 dt = f sinZOdO + 2s cos26d6. 
 
THEOEETICAL TREATMENT OF BEAMS 71 
 
 Now, t will be a maximum when 
 
 jg = f sin20 + 2s cos20 = 0, 
 from which we obtain 
 
 (5) tan 26 = - j- 
 
 as the value of when t is a maximum. 
 
 Treating (4) in the same manner we obtain 
 
 (6) tan 20 = 
 
 as the value of when v is a maximum. Expressing the value of the 
 tangent in (5) in terms of the sine, we have 
 
 sin 20 _2s 
 
 Vl-sin 2 20~ T' 
 Squaring and reducing, we have 
 
 (a) sin 20- 9 * 
 
 Then expressing the value of the tangent in (5) in terms of the cosine, 
 we have 
 
 Vl-cos 2 20 2s 
 
 
 cos 26 f 
 
 from which we obtain 
 
 (6) cos20= 
 
 V/ 2 + 4s 2 
 
 Referring to equation (5), as tan 20 = -sin 20/cos 26, a minus quantity, it 
 is evident that sin 26 and cos 26 will have opposite signs, that is, when 
 one is plus the other will be minus. Then substituting the value of sin 26 
 and cos 26 given in (a) and (6), respectively, in (3), taking one plus and 
 the other minus, we obtain 
 
 By treating (6) in the same manner as (5) was treated above, we obtain 
 
 n2 _/ 
 
 "V/ 2 
 
 and 
 
 ~ V/f + 4* 2 
 Substituting these values, which will have like sign, in (4), we obtain 
 
 The maximum or minimum tensile or compressive stresses on any 
 material particle subjected to shear and direct stress can be computed 
 from (H).. If / be tension, t will be tension when the radical is taken as 
 
STRUCTURAL ENGINEERING 
 
 Fig. 78 
 
 plus and compression when taken as minus. Thus both tension and 
 compression stresses are obtained. If / be compression, by using the plus 
 sign before the radical we obtain the compressive stress and by using the 
 minus sign we obtain the tension. As the plus -sign in (H) gives the 
 maximum stress in all cases it is rarely necessary to use the minus sign. 
 The maximum shear on any material particle subjected to shear and 
 direct stress can be obtained from (I). It is immaterial whether the 
 plus or minus sign before the radical be used as the two values will be 
 the same. 
 
 The direction of the maximum shearing stress can be obtained from 
 (6) and a direction perpendicular to the maximum and minimum tensile 
 and compressive stresses can be obtained from (5). 
 
 There are only a few cases where the above formulas need to be 
 applied. They need not be applied in the case of ordinary beams, for 
 here both the maximum tension and compression stresses occur on the 
 outer elements where the shearing stress is zero, and the maximum shear 
 occurs at the neutral axis where the direct stress is zero. 
 
 63. Deflection of Beams. Beams 
 supporting loads deflect or bend ow- 
 ing to the distortion of their parts which 
 results from the loading. Let Fig. 78 rep- 
 resent an unloaded rectangular wooden 
 cantilever beam which we will assume to 
 be straight and absolutely horizontal. 
 Imagine this beam divided into very short 
 rectangular blocks by imaginary planes 
 passing perpendicularly to its longitudinal 
 axis. Now if loads were applied to this 
 beam, these rectangular blocks would be- 
 come wedge-shaped as shown in Fig. 79, 
 and the imaginary planes dividing the beam 
 into blocks instead of being parallel would 
 become tangent to some curve as ss's", and 
 the neutral axis oo instead of being straight 
 would become a curve which would be 
 known as the "elastic curve" of the beam. 
 The distance from the intersection of any 
 two adjacent imaginary planes to the neu- 
 tral axis oo would be known as the radius 
 of curvature of the elastic curve at the in- 
 tercepted block. The vertical distance that 
 any point on the neutral axis would move 
 down from its original position would be 
 known as the deflection of the point and of 
 the beam at that point, and the deflection 
 of any one point on the neutral axis in 
 reference to another would be the differ- 
 ence of their deflections. For example, the 
 distance y (Fig. 79) from the horizontal 
 line Ho to the point g would be the deflec- 
 Fig. 79 tion of point g, and likewise the deflection 
 
THEORETICAL TREATMENT OF BEAMS 
 
 73 
 
 of the beam at that point,, while the distance z would be the deflection of 
 point g in reference to point k. 
 
 It is readily seen that the curving of the neutral axis results from 
 the distortion of the blocks. Then it is evident that if the slope of the 
 blocks in reference to one another be determined, an equation of the 
 elastic curve expressing this relation can be derived, from which the 
 deflection of the beam at any point in reference to any other point can be 
 computed. So we shall now proceed to derive such an equation of the 
 elastic curve. 
 
 Let Fig. 80 represent any four consecutive distorted blocks of the 
 beam shown in Fig. 79. By drawing the center line of each block we 
 have the broken line curve BabcA which has for its limit the elastic 
 curve SS as the lengths of the blocks decrease; that is, become infini- 
 tesimal. It is readily seen that the radius of curvature at any block is 
 perpendicular to its center line 
 at midpoint. So the slope of the 
 center line of any block with the 
 horizontal or X-axis is the same 
 as that of the tangent of the 
 elastic curve at that point. In 
 fact, the two coincide. 
 
 Let HB be a horizontal 
 line. Then </> is the angle that 
 the tangent to the elastic curve 
 at the center of the block vurw 
 makes with the horizontal or X- 
 axis. Now, as the length of the 
 block vurw decreases, the center 
 line aB comes nearer and nearer 
 to coinciding with the arc of the 
 elastic curve passing through the 
 block. So if the length of the 
 block be infinitesimal the points 
 a and B will be on the elastic 
 curve and the vertical distance 
 ad will be the deflection of point 
 a in reference to B, and ad 
 would be a first differential of the vertical or y ordinate to the elastic 
 curve at that point. Likewise, the distance eb, gc, and JcA will each be 
 a first differential of a y ordinate to the elastic curve when the length of 
 each of the other blocks becomes infinitesimal. Letting A represent the 
 deflection of point A in reference to B, we have 
 
 Fig. 80 
 
 which is simply the summation of the first differentials summed up from 
 B to A. Let dB = ea = gb = kc = dx, an infinitesimal, which would really 
 be the case when the lengths of the blocks are infinitesimal. Then by 
 prolonging aB, the center line of block vurw, to / we have ef = ad. Now 
 as ad and eb are each a first differential of a y ordinate to the elastic 
 curve, fb is a second differential of the same, being the difference between 
 two consecutive first differentials; and similarly he and mA are also 
 
74 
 
 STRUCTURAL ENGINEERING 
 
 second differentials of y ordinates to the elastic curve. It is readily seen 
 that gc = ad + fb + hc and that kA=ad + fb + hc + mA, etc.; that is, each 
 first differential is equal to the summation of the second differentials plus 
 the first differential ad summed up from B to the point considered. Now 
 if the summation be from A to B instead of B to A, as above, we have the 
 case shown in Fig. 81, where ce, bf, ak, and Bn are first differentials, and 
 gb, ha, and mB are second differentials. Taking A as the origin, we have 
 ak = ce-gb-ha, and Bn = ce-gb-ha-mB; that is, any first differential 
 of a y ordinate to the elastic curve is equal to the summation of the second 
 differentials summed up from the origin to the point considered, plus the 
 first differential ce. (Fig. 81.) The second differentials are minus in the 
 last case, as they are measured down to the curve; while the first dif- 
 ferentials are measured up to the curve. For the deflection of point A in 
 
 reference to B we have A = ce + bf 
 + ak + Bn, which is simply the sum- 
 mation of the first differentials 
 summed up from A to B. Now, as 
 stated above, any first differential 
 at any point is equal to the summa- 
 tion of the second differentials 
 summed up from A to the point, 
 plus the first differential ce, while 
 in the case shown in Fig. 80 the 
 first differential at any point is 
 equal to the summation of the sec- 
 ond differentials summed up from B 
 
 Fig. si to the point, plus the first differen- 
 
 tial ad. It is readily seen that as 
 
 far as the deflection of point A in reference to point B is concerned, 
 both of the first differentials ad (Fig. 80) and ce (Fig. 81) are 
 constants, and being such they will appear as constants of integration in 
 the summation of the second differentials. It is first necessary to derive 
 an expression for the second differential of any y ordinate to the elastic 
 curve in terms of known quantities. Referring to Fig. 80, if the length 
 of the two blocks stuv and vurw be infinitesimal, the angle subtended by 
 their radii of curvature will be an infinitesimal angle d6, and as one radius 
 is perpendicular to aB, the center line of block vurw, and the other one 
 to ab, the center line of block stuv, the angle fab will be equal to that 
 subtended by the radii, or dO. Then, as the angle dO (Fig. 80) is 
 infinitesimal, we have 
 
 = abd6 or 
 
 (1). 
 
 The permissible deflection of a beam in any case is very small compared 
 to its length (1/1,000), so that no appreciable error will be made if we 
 assume the slope distance ab = ea = dx. Then by substituting in (1), we 
 have 
 
 d 2 y = dxdO. 
 
 (2). 
 
 Now dO is the slope that the block stuv makes with the block vurw, 
 which results entirely from the distortion of the block stuv. Then, 
 evidently, the value of dO will be directly proportional to the bending 
 
THEOEETICAL TREATMENT OF BEAMS 75 
 
 moment at the block stuv and inversely proportional to the modulus of 
 elasticity of the material composing the block and also inversely propor- 
 tional to the moment of inertia of its cross-section. Then dO can be 
 expressed in terms of these quantities. 
 
 Through a (Fig. 80) draw the line s't' parallel to st. Then the 
 angle s'a^-dQ. Let a = the longitudinal distortion of an element of the 
 block (stuv) distance 2 above ab. Then we have 
 
 But, according to Art. 33, 
 
 Mz ab Mzdx 
 
 (Mz/I stress on element) (see Formulas B and D, Arts. 33 and 53), 
 where M = the bending moment of the block stuv and / = moment of 
 inertia of the cross-section of the block and E the modulus of elasticity 
 of the material composing it. Now, combining these two equations, we 
 have 
 
 Mdx 
 
 Substituting this value of dO in (2) we have 
 
 AMH 
 
 -IE" 
 
 which is the expression for the second differential of the y ordinate to the 
 elastic curve at block stuv in terms of known quantities, but which at the 
 same time is really the expression for the vertical drop of the elastic 
 curve from a to b due wholly to the distortion of block stuv. Now, this 
 same expression will hold in the case of any of the blocks, as no special 
 case was taken. Then, evidently, by substituting the proper value of M 
 in the expression and integrating twice, the deflection of point A in 
 reference to B will be obtained. It is evident that the above discussion 
 will apply to any number of blocks as well as to four, so our expression 
 for the second differential of the y ordinates to the elastic curve will apply 
 to the entire beam or any part of it, as A and B can be any two points, 
 and as the bending in the case of any beam can be nothing different from 
 the bending of a cantilever beam here considered, the expression is 
 evidently applicable in the case of any beam whatever, loaded in any 
 manner, and is hence a general differential equation of the elastic curve 
 of any beam or any body whatever, subjected to bending stresses. For 
 convenience, the equation is usually written 
 
 Referring to Fig. 79, it will be seen that the curve ss's" is an 
 evolute, while the elastic curve oo is an involute. In case of a simple 
 beam there would be two branches to the evolute. 
 
 64. Deflection of Cantilever Beams. 
 
 Case I. When the beam supports a single load at its free end. 
 
76 
 
 STRUCTURAL ENGINEERING 
 
 Let AB (Fig. 82) represent the beam of length L, P the load, and 
 let x and y be the co-ordinates to the elastic curve at any point when the 
 origin is taken at B, and x, and y, the co-ordi- 
 nates when the origin is taken at A. 
 
 First, taking B as the origin, we have MP 
 (L-x) for the bending moment at any point x 
 distance from B. Now substituting this value of 
 M in Formula (K) (Art. 63), we have 
 
 Fig. 82 
 
 Integrating once, we have 
 
 dx 
 
 2 
 
 Now, dy,/dx, is the expression for the tangent of the angle that the elastic 
 curve makes with the horizontal or x axis at any point x distance from B. 
 It is readily seen that dy/dx = at B, the origin, as the elastic curve is 
 horizontal at that point. But x also at that point. Then substituting 
 for dy/dx and for x in the above equation, we have C^O. Now as 
 Ci = 0, the preceding equation becomes 
 
 which is the general equation for the slope of the elastic curve at any 
 point between B and A with B as the origin. From this equation, the 
 slope of the beam ( = slope of the elastic curve) at any point can be 
 computed by substituting for x its numerical value. For example, the 
 slope of the beam at a point b distance from B is 
 
 dy 
 dx 
 
 El 
 
 which is the tangent of the angle that the tangent to the elastic curve at 
 that point makes with the .r-axis. 
 Next integrating (2) we have 
 
 Now at B both y and x = 0. Then substituting for x and also for y, we 
 have C 2 = 0. Hence, we have 
 
 which is the algebraic equation of the elastic curve when the origin is at 
 B. From this equation the deflection of the beam at any point can be 
 computed by substituting for x its numerical value. It is readily seen 
 that the deflection will be a maximum when x = L. Let A = the maximum 
 deflection; then we have 
 
THEORETICAL TREATMENT OF BEAMS 77 
 
 PL 3 
 3El' 
 
 Now taking A as the origin, we have for the bending moment at any 
 point x, distance from A, M = Px,. Then substituting this value of M in 
 Formula (K) (63), we have 
 
 Integrating once, we have 
 
 Now, dy,/dx, when x = L. Then substituting these values for dy,/dx f 
 and ,r, respectively, in the preceding equation, we have C' = -PL 2 /2. 
 Then substituting this value of C" in the equation, we have 
 
 Px. 
 
 dx. 
 
 for the general equation for the slope of the elastic curve, from which the 
 slope of the curve at any point x, distance from A can be computed by 
 substituting for x, its numerical value. 
 Next, integrating (4), we have 
 
 P 3 pr 2 
 
 rx, fL, x, ~ ff 
 
 6 2 
 
 Now, when x, - 0, y, - 0. Substituting for x, and also for y, we have 
 c" = 0. Then we have 
 
 _ P /x? L 2 x\ 
 ~/\6 >. / 
 
 (5) 
 
 for the equation of the elastic curve when the origin is at A, from which 
 the vertical ordinate y, at any point x, distance from A can be computed 
 by substituting for x, its numerical value. Th'en, by subtracting this y, 
 ordinate from the maximum deflection A the deflection of the point in 
 question is obtained. It is readily seen from Fig. 82 that the y, ordinate 
 is equal to A when x, - L. So substituting in (5) we have 
 
 P /L 3 L 3 \ PL 3 
 
 to the 
 
 3EI 
 
 which is the same as found above except for the sign, which is due 
 ordinates being measured in the opposite direction. 
 
 Case II. When the beam supports two loads. 
 
 Let AB (Fig. 83) represent the beam, P 
 and P' the loads, and let a be the distance that 
 load P is from B, b the distance between the 
 loads and d the distance that the load P' is from 
 A, the end of the beam. Take B as the origin, 
 and let x and y represent the co-ordinates to 
 the elastic curve. When x is less than a, we 
 have for the bending moment at a point x dis- Fig. 83 
 
78 STRUCTURAL ENGINEERING 
 
 tance from B 
 
 M=(a-x}P+(b + a-x}P', 
 
 therefore, EI^^ (a- x}P + (b + a- x}P'. 
 
 Integrating, we have 
 
 . 
 
 2 2 
 
 But dy/dx = when x - ; therefore, C" = 0, and we have 
 
 2 
 Now integrating (6), we have 
 
 r Pa,- 3 P'bx 2 P'ax 2 P' 
 
 (6). 
 
 But y = when x = ; therefore, C" = 0, and we have 
 
 y r 7 
 
 *'ax 2 P'x'\ 
 2 6 / 
 
 6 2 
 
 Now when x is greater than a and less than a + b, we have 
 
 for the bending moment at any point between P and P', .r distance from 
 B. Therefore, 
 
 EI d ^L=P 
 
 dx 2 
 
 Integrating, we have 
 
 dii P'x 2 
 
 EI-=P'ax + P'bx - -+ C, ........................ (8). 
 
 6/iT 2 
 
 Now it is readily seen that equations (6) and (8) are equal when 
 x a in each. So, substituting a for x in each, equating and cancelling, 
 we have 
 
 Substituting this value of C, in (8) we have 
 
 . 
 
 Integrating, we have 
 
 P'ax* P'bx 2 P'x* PaT-x 
 
 Now it is readily seen that equations (7) and (10) are likewise 
 equal when x = a in each. So substituting a for x in each, equating and 
 cancelling, we have 
 
 Pa* 
 C "~ ~6~* 
 
THEORETICAL TREATMENT OF BEAMS 
 
 79 
 
 Substituting this value of C,, in (10), we have 
 
 1 IP 
 
 ? 2 P'bx 2 
 
 - + 
 
 P'* 3 Pa*x 
 
 
 
 Now the slope of the elastic curve at any point between B and the 
 load P can be computed from equation (6) and at any point between the 
 loads from (9), while the deflection at any point between B and the load 
 P can be computed from equation (7) and at any point between the loads 
 from (11) by substituting for x its numerical value in each case. 
 
 The maximum deflection will be at A. This can be determined in 
 the following way: 
 
 First compute the deflection of the beam at the load P' from equa- 
 tion (11) (substituting (a + fe) for x). Then compute the slope, that is, 
 the tangent of the slope angle at that point from (9) and multiply this 
 slope by the distance d, which will give the deflection of the point A in 
 reference to the point at P'. Then add this deflection to the deflection of 
 the beam at P' ', and we will have the total deflection at A. 
 
 Case III. When the beam supports a uniform 
 load. 
 
 Let AB (Fig. 84) represent the beam support- 
 ing a uniform load of w pounds per foot of length. 
 Take B as the origin, and let x and y represent the 
 co-ordinates to the elastic curve. Then the bending 
 moment at any point x distance from B is 
 
 Pigr. 84 
 
 therefore, 
 
 Integrating once, we have 
 
 When x = 0, dy/dx = ; therefore, C' = 0, and we have 
 
 Integrating again, we have 
 
 Now, when a? = 0, = 0; therefore, C" = 0, and we have 
 
 . 
 
 y ~ 
 
 2EI 2 
 
 (13). 
 
 From equation (12) the slope at any point of the beam can be 
 computed, and from (13) the deflection at any point can be computed by 
 substituting for x its numerical value in each case. It is evident that the 
 maximum deflection will be at A, the free end. For that point x = L. 
 
30 STRUCTURAL ENGINEERING 
 
 Then substituting L for x in equation (13), we have 
 
 wL* 
 
 y ~ ~ 
 
 for the maximum deflection. 
 
 65. Deflection of Simple Beams. 
 
 Case 1. Vhen the beam supports a single load at mid-span. 
 
 L Let AB (Fig. 85) represent a simple beam of 
 
 ^ ~~*| length L supporting a single load P at mid-span. 
 
 ^J^^_.n p ^-~---\ 3 Let R and R' represent the reactions at A and B, 
 ^ ^ ^^ \ K respectively, which are equal in this case. Take A 
 Fig. 85 as the origin, and let x and y represent the co-ordi- 
 
 nates to the elasti/ 1 curve. For the bending moment 
 at any point to the left of the load, x distance from A, we have 
 
 therefore, we have 
 Integrating once, we have 
 
 dx 4 
 
 Now dy/dx = when x L/2. Substituting this value for x, we have 
 
 16 
 
 and substituting this value of C" in the above equation, we have 
 EI dy_Px* PL- 
 
 >L T x -^r TG~- 
 
 Integrating this equation, we have 
 
 Px' P^x 
 
 ~- " 
 
 Now, y = when x ; therefore, C" - 0, and we have 
 
 which is the equation of the elastic curve to the left of the load. As the 
 elastic curve is symmetrical about the load, there is no need for deriving 
 the equation for the curve to the right of the load. However, this can be 
 readily accomplished by substituting in Formula (K) (Art. 63), the 
 expression for the bending moment to the right of the load, which is 
 
 
 It is readily seen that the maximum deflection will occur under the 
 load. So letting A represent the maximum deflection, and substituting 
 L/2 for x in equation (2), we have 
 
THEORETICAL TREATMENT OF BEAMS 81 
 
 PL 3 
 
 The slope of the elastic curve at any point between A and the load can be 
 computed from equation (1). 
 
 Case II. When the beam supports a single load at any point. 
 
 Let AB (Fig. 86) represent a simple beam 7 
 
 supporting a load P at any point z distance 
 from A. Let R and R' represent the reactions 
 at A and B, respectively, due to this load P. 
 Take A as the origin, and let x and y represent p . 
 
 the co-ordinates to the elastic curve. For the 
 bending moment at any point to the left of the load, we have 
 
 j 
 
 ~~j 
 therefore, we have 
 
 Integrating, we have 
 
 Integrating again, we have 
 
 Px* Pzx 3 
 
 
 Now, y = when x - ; therefore, C" - 0, and we have 
 
 Now for the bending moment at any point to the right of the load, that is, 
 when #>, we have 
 
 +Pz, 
 i 
 
 therefore, we have 
 
 Integrating, we have 
 
 Integrating again, we have 
 
 ,, + C 2 ........................ ..(6). 
 
 It is readily seen that equations (3) and (5) are equal when x z 
 in each. Then substituting z for x in each equation, equating and 
 cancelling, we have 
 
STRUCTURAL ENGINEERING 
 
 Now substituting this value of C x in (6), we have 
 
 Pvr- Prr s p^- r 
 
 ' ^-'-^ .................... w- 
 
 It is readily seen that equations (4) and (8) are equal when x z 
 in each case. Then substituting z for x in each, equating and cancelling, 
 we have 
 
 Now it is readily seen that y in equation (8) equals zero when 
 x L. Then substituting L for x and Pz 3 /6 for C 2 in that equation, 
 equating and reducing, we have 
 
 Pz- PzL > 
 
 ; 
 
 
 Substituting this value of C" in equations (3) and (4), we have, 
 respectively, 
 
 dy __Px* Pzx* Pz* PzL 
 
 ^'- ~~ ~ ~~ ~~ 
 
 Pz-x PzxL Pz z x 
 
 Now, from equation (11) the slope of the elastic curve at any 
 point to the left of the load can be computed, while the deflection at any 
 point to the left of the load can be computed from (12). 
 
 From equations (7) and (10) we have 
 
 PzL Pz* 
 
 Cl ~ 3 " 6L' 
 and from (9) we have 
 
 c - p * 3 - 
 
 C2 ~~6~ 
 Substituting these values in equations (/5) and (G), we have, respectively, 
 
 EI d -S.-P-, ^ (13) 
 
 dx~ ~ 2L 3 ~ 6L ' 
 
 Pzx* Pzx 3 PzxL Ps*x Pz* 
 
 Now, from equation (13) the slope of the elastic curve at any point to 
 the right of the load can be computed, while the deflection at any point to 
 the right of the load can be computed from (14). 
 
 Equations (11) and (12) apply only to the part of the elastic curve 
 to the left of the load, while equations (13) and (14) apply only to the 
 
THEORETICAL TREATMENT OF BEAMS 83 
 
 part to the right of the load, an<jl hence we can consider the elastic curve 
 to be composed of two separate curves which are tangent at the load. 
 Now, if the beam supported two loads some distance apart, the elastic 
 curve would be composed of three separate curves; and if it supported 
 three loads, the elastic curve would be composed of four separate curves; 
 and so on. That is to say, if there be n loads, there will be (w + 1) 
 separate curves composing the elastic curve. The equations for each of 
 these curves could be derived as readily as the above equations, and then 
 from these equations the slopes and deflections of the elastic curve could 
 be computed. The main labor is the determining of the constants of 
 integration. It will be observed that, at the point of maximum deflection, 
 dy/dx = 0, as the tangent to the curve will be horizontal at that point. 
 
 Case III. When the beam supports a uni- 
 form load. 
 
 Let AB (Fig. 87) represent a simple beam, 
 length L, supporting a uniform load of w pounds 
 per foot of length. Let R and Rl represent the Fi& 87 
 
 reactions at A and B, respectively, and let x and 
 y represent the co-ordinates of the elastic curve, A being taken as the 
 origin. Then the bending moment at any point x distance from A is 
 
 therefore, 
 
 d' 2 y __ 
 
 ( *to? s 
 
 Integrating once, we have 
 
 _ . 
 
 dx 4 . 6 
 
 Now, dy/dx - when x - L/2, as the tangent to the elastic curve is 
 horizontal at that point. Now substituting this value of x, we have 
 
 and substituting this value of C' in the above equation, we have 
 
 dy _wLx 2 wx 5 wL 3 
 ' 1 ~~~ '~~~~' 
 
 Integrating this equation, we have 
 
 wLx 3 wx* wL 3 x 
 
 
 But y - when x - ; therefore, C" = 0, and we have 
 
STRUCTURAL ENGINEERING 
 
 y EI\ 13 24 " 24 /; 
 
 ,r. ->,,., 
 (2), 
 
 from which the deflection at any point can be computed. 
 
 It is evident that the deflection will be a maximum at midspan ; that 
 is, when x L/2. So letting A represent the maximum deflection, and 
 substituting L/2 for a; in (2), we have 
 
 66. Proposition. The bending moment at any point in a beam is 
 equal to the bending moment at any other point plus the shear at the 
 other point multiplied by the distance between the points, plus the 
 algebraic sum of the moments of the forces between the points about the 
 point in question. 
 
 Let AB (Fig. 88) represent a simple beam supporting the loads 
 P, PI, and P2 as shown. Now, according to the above proposition, the 
 
 t 
 
 Fig. 89 
 
 bending moment at D is equal to the bending moment at E, plus the shear 
 at E multiplied by c, plus the sum of the moments of the forces between 
 the points about D. This is readily seen to be true, for, taking moments 
 about E, and letting R represent the reaction at A due to the three loads, 
 we have 
 
 for the bending moment at that point, and taking moments about D, we 
 have 
 
 dPl- hP2 
 
 -dPl-hP2, 
 
 which is seen to agree with the above proposition, as the part R(a + b) - 
 Pb is the bending moment at E, RP is the shear at E, which is multi- 
 plied by c, the distance between E and D, and -dPl - hP2 is the sum of 
 the moments of the forces between E and D, about D. The minus signs 
 in the last are simply the signs of the moments. 
 
 The above proposition is readily proven in the case of any beam 
 whatever, but regardless of its simplicity, it is quite useful. 
 
 67. Reactions, Shears, and Bending Moments on Overhanging 
 Beams. Let ABC (Fig. 89) represent an overhanging beam supported 
 at B and C and which in turn supports the loads P and P', as shown. 
 The part AB is known as the cantilever arm, while the part BC is known 
 as the anchor arm. Let R represent the reaction at B due to the two 
 loads, P and P', and let R' represent the reaction at C due to the same. 
 Now, taking moments about C, we have 
 
THEORETICAL TREATMENT OF BEAMS 
 
 from which we obtain 
 
 aP + bP' 
 ~~ 
 
 Then taking moments about B, we have 
 
 R' 
 from which we obtain 
 
 If eP be greater than dP' it is evident that R' will act downward 
 upon the beam, and hence the beam would pull upward upon the support 
 at C, which would require that the beam be anchored in some manner to 
 the support. In all such cases the reaction is spoken of as being negative. 
 If dP' be greater than eP, of course R' will be positive, the same as for 
 simple beams. 
 
 In case of a greater number of loads, the reactions are determined 
 in the same manner as shown above for the two loads. We simply include 
 the moment of each load in the equation of moments. In the case of a 
 uniform load the reactions are determined practically in the same manner, 
 except that we use average lever arms for the uniform load. For example, 
 suppose the beam ABC (Fig. 89) supports a uniform load of w pounds 
 per foot of length, extending from A to C. Let R, represent the reaction 
 at B due to this uniform load, and let R,, represent the reaction at C due 
 to the same. Then taking moments about C, we have 
 
 
 and taking moments about B, we have 
 
 ' >v ...,.' JB " = i(V-V)'\, .;;...;'";/"' . 
 
 In case the beam overhangs two supports the reactions are deter- 
 mined as above by taking moments about the supports and simply 
 including the moment of each force acting upon 
 the beam in the equation of moments. As an a ^ b d 
 
 example, let ABCD (Fig. 90) represent a beam 9^f>. 
 
 overhanging two supports. Let P, P f , and P" ^ c 
 
 represent three loads supported by the beam as 
 
 shown, and let R represent the reaction at B Fig 90 
 
 due to these three loads, and let R' represent 
 
 the reaction at C due to the same. Now, taking moments about B, we have 
 
 aP - bP' R'L - (e + L)P" = 0, 
 from which we obtain 
 
 R' = 
 
86 STRUCTURAL ENGINEERING 
 
 Then taking moments about C, we have 
 
 -(L + a)PRL-dP' + eP" = Q 
 from which we obtain 
 
 (L + a)P + dP'-eP" 
 
 - 
 
 Now suppose the beam represented in Fig. 90 supports a uniform 
 load of w pounds per foot of length extending from A to -D, and let R, 
 represent the reaction at B due to this uniform load and let R,, represent 
 the reaction at C due to the same. Then taking moments about B, we 
 have 
 
 _ 
 
 2L 
 
 and taking moments about C, we have 
 
 _ 
 
 2L 2L 
 
 The determining of the shear at any section of an overhanging 
 beam is just the same as for any other beam; that is, the shear at any 
 section is equal to the algebraic summation of the forces on either side of 
 the section summed up from the end of the beam in each case. For 
 example, the shear at section ss of the beam ABCD (Fig. 90) due to the 
 three loads P, P' , and P", is 
 
 S = PRorP"R' + P'. 
 
 The determining of the bending moment at any section of an over- 
 hanging beam is just the same as for any other beam; that is, the bending 
 moment at any section is equal to the algebraic summation of the moments 
 of the forces on either side of the section about the section. For example, 
 the bending moment about the section ss of the beam ABCD (Fig. 90) 
 due to the three loads P, P', and P" is 
 
 All of the above holds as well for cantilever arms as for anchor arms, 
 but it is more convenient to treat the cantilever arms as independent 
 cantilever beams. 
 
 68. Reactions, Shears, and Bending Moments on Beams Fixed 
 at One End and Supported at the Other. Let AB (Fig. 91) repre- 
 sent such a beam which has the simple support at A and the fixed support 
 at B. As the beam is fixed at B it is evident that any 
 load on the beam will produce tension in the elements 
 above the neutral axis and compression below the 
 neutral axis at that point. But it is equally evident 
 that the reverse is true just to the right of the simple 
 Fig. 91 support at A; that is, the top elements of the beam just 
 
 to the right of A will be in compression and the bot- 
 tom ones in tension. Then, evidently, there will be some section be- 
 tween A and B where the bending reverses; that is, a section where 
 there is no bending, and hence the bending moment at that point of the 
 
THEOEETICAL TEEATMENT OF BEAMS 87 
 
 beam will be equal to zero. Let S be such a point. Then the part SB of 
 the beam,, to the right of S, would be simply a cantilever beam, while the 
 part SA of the beam., to the left of S, would be a simple beam, whence the 
 bending or curving of the beam due to loads would be as indicated in 
 Fig. 91. The point S, where the bending moment equals zero, would be 
 known as the point of "contra-flexure," also as the "point of inflection." 
 As a case of concentrated loads, let R (Fig. 91) represent the 
 reaction of A due to a single load P at any distance kL from A, k being 
 any fraction less than unity. Owing to the unknown forces acting upon 
 the beam to the right of B, this reaction R, at A, cannot be determined in 
 the usual way by taking moments about B. Take A as the origin, and let 
 x and y represent the co-ordinates to the elastic curve. Then, for the 
 bending moment at any point to the left of the load, that is, when x is 
 less than kL, we have 
 
 M = Rx, 
 
 therefore, 
 
 dx* 
 Integrating once, we have 
 
 Integrating again, we have 
 
 But, y = when x = ; therefore, C" = 0, and we have 
 
 C'x .................................... (2). 
 
 Now for the bending moment at any point to the right of the load, 
 that is, where x is greater than kL, we have 
 
 M^Rx 
 
 therefore, 
 
 El ^=Rx- 
 dx- 
 
 Integrating once, we have 
 
 du Rx 2 Px 2 
 
 But, dy/dx = when x - L. Then substituting L for x, and equating, 
 we have 
 
 RL> PL* 
 ._ +_-/, -C r 
 
 Substituting this value of C l in the last equation, we have 
 
88 STRUCTURAL ENGINEERING 
 
 Then integrating (3), we have 
 
 Rx 3 Px 3 PkLx 2 RL 2 x PL 2 x 
 
 2 
 
 But here y = when x - L. Then substituting L for x, equating and 
 cancelling, we have 
 
 PL 3 
 
 3 3 2 
 
 Substituting this value of C 2 in the last equation, we have 
 Rx 3 ^ Px^ PkLx 2 RL 2 x PL 2 x 
 
 RL 3 PL 3 PkL 3 
 
 It is readily seen that equations (1) and (3) are equal when x = kL 
 in each. Then substituting kL for x in each equation, equating and 
 reducing, we have 
 
 Then substituting this value of C" in (2) we have 
 Rx 3 Pk 2 L 2 x RL 2 x PL 2 x 
 
 It is readily seen, also, that equations (4) and (5) are equal when 
 x kL in each. Then substituting kL for x in each of these equations, 
 equating and reducing, we have 
 
 B=f(*-3* + 2) ................................... (6), 
 
 r i * * ^ r> 
 
 from which the reaction at A due to a load at any point on the beam can 
 be computed. For example, suppose the load P is at mid-span. Then 
 k = ^. Substituting this ^ in equation (6) we have 
 
 Again, suppose the load is at the quarter point nearest A. Then 
 k = . Substituting this J for A: in (6) we have 
 
 If there be more than one load on the beam, the reaction at A due to 
 each load would be computed separately and all added together and we 
 would thus obtain the reaction at A due to all the loads. Of course the k 
 for each would be different from the others; that is, no two A;'s would be 
 the same. 
 
 In case the beam supports a uniform load of w pounds per foot of 
 length extending over the full length of the beam, that is, from A to B, 
 
THEOEETICAL TREATMENT OF BEAMS 
 we have 
 
 _ . 
 
 for the moment at any point x distance from A, where R represents the 
 reaction at A due to this uniform load. Therefore, we have 
 
 /*f = R*_*!. 
 
 dx 2 2 
 
 Integrating once, we have 
 
 . 
 
 dx 2 6 
 
 But dy/dx when x = L. So substituting L for x in the preceding 
 equation, we have 
 
 RL* wL 
 
 1~ + ~T' 
 
 Then substituting this value of C" in the last equation, we have 
 
 dy_R^ RL* 
 
 >L dx~ 2 ' 6 2 6 
 
 Integrating this equation, we have 
 
 But here y when x ; therefore, C" 0, and we have 
 
 Now y in (8) when x - L. Then substituting L for x (8), we have 
 
 flL 3 wL 4 flL 3 wL 4 
 
 ~6~ ~~W ' "2" "6~ J 
 from which we obtain 
 
 That is, when a beam supported at one end and fixed at the other 
 supports a uniform load, the reaction at the supported end is three- 
 eighths of the total load. 
 
 After the reaction at the supported end due to the loads supported 
 is determined, the shear and bending moment at any point in the beam 
 are obtained just the same as in the case of a simple beam, except we deal 
 only with the supported end. The shear at any point is equal to the 
 reaction at the supported end minus all intervening loads, while the 
 bending moment at any point is equal to the reaction at the supported 
 end multiplied by its lever arm, minus the moment of all intervening 
 loads about the point. For example, the shear at any point x distance 
 from the supported end due to a uniform load of w pounds per foot of 
 span is 
 
 g 
 
 S- wL-wx, 
 8 
 
90 STRUCTURAL ENGINEERING 
 
 while the bending moment is 
 
 3 tvx 
 M wLx 
 
 o & 
 
 If x - Li, we have 
 
 which is the bending moment at the support B or the fixed end. 
 
 Deflections can be computed from equations (4), (5), and (8). 
 69. Shears and Bending Moments on Fixed Beams. Let AB 
 (Fig. 92) represent a fixed beam. The supports being fixed, any load on 
 the beam, as is seen, will produce tension in the 
 top elements of the beam and compression in the 
 bottom elements of the beam at the supports, 
 while the reverse will be the case out some dis- 
 tance from the supports. Then, evidently, there 
 Fig 92 will be two points of contra-flexure and the 
 
 bending of the beam due to loads supported will 
 be as indicated in Fig. 92, where S' and S are the points of contra-flexure. 
 It is evident that the parts AS' and SB are cantilevers, while the part 
 S'S is a simple beam. In the case of fixed beams, the point of applica- 
 tion of the reactions cannot be definitely fixed, so we deal with the end 
 shears instead of the reactions. 
 
 As a case of concentrated loads let P represent a single load upon 
 the beam at any distance kL from A. Now, when we proceed to deter- 
 mine the shears and bending moments on the beam due to this load P, we 
 quickly realize that we have but little to start with, and it is only through 
 the application of the general differential equation of the elastic curve 
 (K) that we are able to obtain either. Let V' and V represent the end 
 shears at A and B, respectively, due to the load P, and let M' and M" 
 represent the bending moments at A and B, respectively. Take A as the 
 origin and let x and y represent the co-ordinates to the elastic curve. 
 Then, according to Art. 66, the bending moment at any point to the left 
 of the load any distance x from A is 
 
 M = A 
 Therefore, we have 
 
 ~dx*~ 
 and integrating once, we have 
 
 J-i M. , *'.". W | | V 
 
 dx 2 
 
 But dy/dx = when x = ; therefore, C" = 0, and we have 
 
 Integrating this equation, we have 
 
 M'x- V'x* 
 
THEOEETICAL TREATMENT OF BEAMS 91 
 
 But here y when x ; therefore, C" = 0, and we have 
 M'x 2 V'x* 
 
 
 Now the bending moment at any point to the right of the load 
 distance from A is 
 
 Therefore, 
 
 and integrating this equation, we have 
 
 E I = M' + V'x -Px + PkL, 
 ax 
 
 Now, dy/dx - when x = L. So substituting L for x and equating, we 
 have 
 
 Then substituting this value of C x in the last equation, we have 
 
 Integrating (3), we have 
 
 V'x* Px* PkLx 2 V'L 2 x PL 2 * 
 
 _ ._ __ _ -M'Lx 
 
 
 But y - when x = L. So, substituting L for x in the last equation, 
 equating and reducing, we have 
 
 M'L 2 V'L* PL* PkL* 
 
 ~~ ~~ ~~ ~~ 
 
 Then substituting this value of C 2 in the last equation, we have 
 
 M'x 2 V'x* Px* PkLx 2 __, r V'L*x 
 
 ___ + _ __ 4 ____M'L,--^ + 
 
 M'L 2 V'L* PL* PkL* 
 
 .............. (4). 
 
 Now it is evident that equations (1) and (3) are equal when x = kL 
 in each. Then by substituting kL for x in each, equating and reducing, 
 we have 
 
 Then by substituting this value of M' in each of the equations (2) 
 and (4) and then substituting kL for x in each, we have the two equal. 
 Then by equating these equations each to each and reducing, we have 
 
 V = P(l + 2& 8 - 3fc 2 ) ................................ (6). 
 
92 STRUCTURAL ENGINEERING 
 
 By substituting this value of V in (5), and reducing, we have 
 
 M' = PL(2k 2 - P - /c) ................................ (7). 
 
 Then, having V and M' determined from (6) and (7) for the point 
 A, the bending moment at any other point in the beam, due to the load P, 
 can be determined according to Art. 66, and the shear at any point can 
 be determined, as it is equal to the summation of the forces summed up 
 from A beginning with V '. 
 
 If there be more than one load on the beam, the shear and bending 
 moment at A due to each can be determined separately from (6) and (7). 
 Then adding these shears together and these bending moments together, 
 we obtain the shear and bending moment at A due to all of the loads. 
 Then the bending moment and shear at any point of the beam, due to all 
 of the loads, can be determined. That is, if the shear and bending 
 moment at one support of a fixed beam are known, the shear and bending 
 moment at any other point can be determined, and hence the equations 
 (6) and (7) are all that are absolutely necessary. But from these 
 equations we can readily derive equations for the shear and bending 
 moment at the other support, if such be desired. For example, the shear 
 at B, Fig. 92, is equal to the load P, producing the shear, minus the 
 shear at A. Then subtracting the value of V , given in (6), from P, we 
 have 
 
 and reducing, we have 
 
 V = P(2P - 3k 2 ) ......................... . . . . ....... (8). 
 
 The bending moment at B (Fig. 92) is 
 
 M" = AT + V'L - P(L - jfcL). 
 
 Then by substituting the value of V and M' given in (6) and (7), 
 respectively, we have 
 
 M" = PL(2k 2 -k*-k)+PL(l+2k z -3k 2 )-P(L-kL), 
 and reducing, we have 
 
 M" = PL(k 3 - F) . . ..................... ............ (9). 
 
 As an example of application, suppose the load P (Fig. 92) to be at 
 mid-span. Then k = 1/2. Now substituting this 1/2 for k in (6) and 
 (7), we have 
 
 - 
 
 respectively, which is the shear and bending moment at A. For the 
 bending moment under the load we have 
 
THEOEETICAL TREATMENT OF BEAMS 93 
 
 In case a uniform load extends over the entire length of a fixed beam, 
 the end shears are each equal to one-half of the total load on the beam. 
 Suppose the beam' shown in Fig. 92 supports a uniform load of w pounds 
 per foot of length extending over the entire span. Let V and V ff repre- 
 sent the end shears at A and B, respectively, due to this uniform load, 
 and let M' and M" represent the bending moments at A and B f respect- 
 ively, due to the same. 
 
 Taking A as the origin as before, we have for the bending moment 
 at any point x distance from A 
 
 therefore, 
 
 Integrating this equation, we have 
 
 i . 
 
 dx 46 
 
 But dy/dx = when x = ; therefore, C l - 0, and we have 
 
 Integrating again, we have 
 
 M'x 2 wLx 3 wx* 
 
 But y = when x - ; therefore, C 2 = 0, and we have 
 
 a 
 
 But y also when x = L. Then substituting L for x in (11), and 
 reducing, we have 
 
 which is the bending moment at A. Then knowing the bending moment 
 and shear at A, the bending moment and shear at any other point in the 
 beam can be determined. For example, the bending moment at the 
 center is 
 
 To determine the points of contra-flexure, we simply write the 
 equation for the bending moment at any point and equate it to zero. For 
 example, the bending moment at any point x distance from A, when the 
 beam supports a uniform load of w pounds per foot (Fig. 92), is 
 
 wx 2 wL 2 wLx wx* 
 
 M= ^ +rx ___ = __ + __ __. 
 
94 
 
 STRUCTURAL ENGINEERING 
 
 Then by equating this to zero, we have 
 
 w x- 
 
 ~ 
 
 and solving for x, we have 
 * = - 
 
 /v 
 
 = 0.21 Lor 0.79 L (about). 
 
 That is, one point of contra-flexure is 0.21 of L from A considered 
 in practice as being J of L and the other 0.79 of L from A considered 
 in practice as being J of L. The points of contra-flexure can be deter- 
 mined in a similar manner in case of any other kind of loading. 
 
 Deflections, if desired, can be computed from equations (2), (4), 
 and (11). 
 
 70. Bending Moments, Shears, and Reactions on Continuous 
 
 Beams. Let ABC (Fig. 93) represent two consecutive spans of a con- 
 tinuous beam of n spans. The bending or curving of the beam in the two 
 
 J=L. 
 
 Fig. 93 
 
 spans is shown to be similar to that of two fixed beams, each span being 
 considered a beam. If the spans were equal in length and symmetrically 
 loaded, this would practically be the case, but otherwise it would not be, 
 for it is readily seen that the loads in one span would tend to produce 
 reverse bending in the adjacent span, which would curve it up instead 
 of down. For example, the loads in span AB could be such that the span 
 BC would be curved upward instead of downward, as it is shown. So it is 
 evident that the tangent to the elastic curve at any support is horizontal 
 only when the spans are equal in length and rigidity, and symmetrically 
 loaded. Hence the slope of the elastic curve at the supports cannot, in 
 general, be utilized in preliminary investigations, as in the case of fixed 
 beams. 
 
 In the following treatment it is assumed that the beams have a 
 uniform cross-section and are homogeneous throughout, and are supported 
 upon simple supports of equal elevation. This is an assumption usually 
 made in practice. The equation employed in analyzing continuous beams 
 is known as the "Three-Moment Equation/' which we shall now proceed 
 to derive. 
 
 Case I. When the beam supports concentrated loads. 
 
 Referring to Fig. 93 let L be the length of the span AB and L, the 
 length of the span BC. Let P represent a load at any point kL distance 
 from A in span AB and let P' represent a load at any point k,L, distance 
 from B in span BC. Let M' ', M" , and AT" represent the bending 
 
THEOEETICAL TEEATMENT OF BEAMS 95 
 
 moments at the supports A, B, and C, respectively, due to the loads P 
 and P', and let V and V represent the shears just to the right of the 
 supports A and B, respectively, due to these same loads. 
 
 Considering the span AB, and taking A as the origin, the bending 
 moment (according to Art. 66) at any point to the left of the load P, x 
 distance from A, is 
 
 therefore, 
 
 Integrating once, we have 
 
 EI = M ' x + ir + c ' (1) - 
 
 Integrating again, we have 
 
 But y when x = ; therefore, C" 0, and we have 
 
 M'r 2 T7r 3 
 
 *- + C'i .............................. (2). 
 
 Now, the bending moment at any point to the right of the load P, 
 x distance from A, is 
 
 therefore, 
 
 Integrating once, we have 
 
 El = AP + Vx - Px + PkL. 
 
 dx 2 
 
 (3). 
 
 Integrating again, we have 
 
 M'x 2 Vx z Px z PkLx 2 
 EIy = ~^-- + --g- + g + C /a r + C ............... (4). 
 
 It is readily seen that equations (1) and (3) are equal when x-'kL 
 in each. Then substituting kL for x in each and equating, we have 
 
 6 6 
 
 from which we obtain 
 
 Now substituting this value of C' in (2), we have 
 M'x* Vx* P 
 
96 STRUCTURAL ENGINEERING 
 
 Now in equation (4) y = when x = L. Then substituting L for x 
 in that equation, we have 
 
 M'L 2 FL 3 __PL 3 _ PkL 3 
 
 from which we obtain 
 
 M'i* rL* PL* PkL* 
 
 ~~ ~~ 
 
 ' 
 
 Substituting this value of C,, in equation (4), we have 
 M'x 2 Fx 3 Px 3 PkLx 2 
 
 M'L 2 FL 3 PL 3 PkL 3 
 
 IT 
 
 Equation (6) applies to the part of the elastic curve to the left of the 
 load P, while (8) applies to the part of the curve to the right. Then these 
 two equations are equal when x = kL in each. So substituting kL for x in 
 each, equating and reducing, we have 
 
 r _ Pk 3 L 2 M'L FL 2 PL 2 PkL 2 
 
 6 2 6 ~6~ ~T~ 
 
 Then substituting this value of C, in (5), we have 
 
 , Pk*L* Pk 3 L 2 M'L FL 2 PL 2 PkL 2 
 
 " = 2~ -IT TT Tr*nr "T" 
 
 and substituting the value of C, given by (9) in (7) and reducing, we have 
 
 So we have thus determined all of the constants of integration so far 
 involved. 
 
 Now substituting in (1) the value of C" given in (10), we have 
 
 Fj^lL^M' PPL2 PM2 M/L 
 
 dx~ '2 2 6 2 
 
 FL 2 PL* PkL 2 
 
 + 
 
 662 
 and substituting the same value of C' in (2), we have 
 
 M'x 2 vx 3 ^ Pk 2 L 2 x _ Pk 3 L 2 x _ M f Lx 
 2 + 6 2 ~T~ 2 
 
 PkL 2 x 
 
 6 +6 2~ W' 
 
 Next substituting in (3) and (4) the value of C, and C,, , respectively, 
 given in (9) and (11), we have 
 
THEOKETICAL TKEATMENT OF BEAMS 97 
 
 , T dy Vx 2 Px 2 Pk s L 2 
 EI-^=M'x + - + PkLx 
 
 dx 22 6 
 
 --?**?-*? (> 
 
 and 
 
 _M'x 2 Vx 3 Px 3 PkLx 2 Pk*L 2 x 
 EIy ~~2~ '~Q~ ~T ~T~ ~6~ 
 
 M'Lx VL 2 x PL 2 x PkL 2 x 
 
 + 
 
 26626 
 
 Equations (12) and (13) apply to the part of the elastic curve to the 
 left of the load P, while equations (14) and (15) apply to the part of the 
 elastic curve to the right of the load P. Now it is evident that by pro- 
 ceeding in the same manner with span BC, four equations corresponding 
 to (12), (13), (14), and (15) could be derived for the elastic curve in 
 that span. But it is readily seen that the equations would differ from the 
 above equations only in the marking of the M's, V's, P's, k's, and Us. So 
 for span BC, we can write 
 
 dx 226 
 
 _ M"L, _ V'L, 2 P f L 2 P'k,L, 2 
 
 2662 
 
 (16). 
 
 6 O 4> O 
 
 M"L,x V'L, 2 x P'L, 2 x P'lc,L*x 
 
 2662 
 for the part of the elastic curve to the left of the load P' , and 
 
 F7 d y-M X J' X * P ' X * P'kLx P '*' L > 
 ( J&4* ~ ~2~ ~^~ ' ' ~Mp 
 
 M"L, V'L;< P'L 2 P'k,L, 2 
 
 (17) 
 
 M"x 2 V'x* P'x* P'k,L,x 2 P'Jc, 3 L,"x M"L,x 
 
 ~Y~ ~6~ ~6~ ~T~ ~Q~ ~~2~ 
 
 V'L,*x P'L, 2 x P'k,L, 2 x P'k*L, s 
 + --- H - + - - 
 
 
 
 6 o 2 
 
 for the part of the elastic curve to the right of the load P'. 
 
 Now taking moments about B, according to Art. 66, we have 
 
 from which we obtain 
 
98 STRUCTURAL ENGINEERING 
 
 Then substituting this value of V in (14) and reducing, we have 
 ^dy M"x 2 M'x 2 Pkx 2 Pk*L* 
 
 EI fx= Mx+ ^r-*r- 
 
 M'L M"L M'L PkL 2 
 
 2 6 ~~6~ ~T~ 
 
 Likewise, taking moments about C, we have 
 
 M'" = M"+F'L,-P' (!,, 
 from which we obtain 
 
 M ' " - M" 
 
 (22). 
 
 Then substituting this value of V in (16), we have 
 
 (23). 
 2662 
 
 Equation (21) applies to the part of the elastic curve to the right of 
 the load P in span AB, and equation (23) applies to the part of the 
 elastic curve to the left of the load P' in span BC. Now it is readily seen 
 that these two equations are equal when x L in (21) and in (23), as 
 the two curves have a common tangent at B. Then substituting L for x 
 in (21), and for x in (23), equating and reducing, we have 
 
 M"L M'L, PkL 2 Pk*L 2 M"L, M 
 
 3 6 6 6 3 6 
 
 P'k, 2 L, 2 P'k, 3 L, 2 P'k,L, 2 
 
 "T~ "6" I" 
 from which we obtain 
 
 M'L + 2M"(L + L,)+M"'L,= -PL 2 (k-k z )-P'L, 2 (2k,-3k, 2 + k*) . (L), 
 
 which is the three-moment equation when the loads are concentrated loads. 
 The more general equation, as usually written, is 
 
 Case II. When the beam supports uniform loads. 
 
 Let w be the uniform load per foot on span AB (Fig. 93) and let w f 
 be the uniform load per foot in span BC. Let M' , M" , and M' " now 
 represent the bending moments at the supports A, B, and C, respectively, 
 due to the above uniform loads, and let V and V now represent th*- 
 shears just to the right of supports A and B, respectively, due to these 
 same loads. Considering span AB, and taking A as the origin, the bend- 
 ing moment at any point x distance from A is 
 
THEORETICAL TREATMENT OF BEAMS 99 
 
 therefore, 
 
 Integrating once, we have 
 
 Ei i= M '^ F - 
 
 Integrating again, we have 
 
 M'x 2 Fx 5 wx 
 
 6 M 
 But y when x ; therefore, C" = 0, and we have 
 
 M'x 2 Fx* wx* 
 
 But also y = when x L. Then substituting L for x in the last 
 equation, we have 
 
 ML* FL* 
 
 -r- + ^r- 
 
 from which we obtain 
 
 M'L F 
 
 
 Substituting this value of C in equation (24), we have 
 
 Similarly, if the origin be taken at the support B, for span BC, we have 
 for the bending moment at any point x distance from B 
 
 therefore, 
 
 EI dx~< 
 
 Integrating once, we have 
 
 7 ^ = M", + ^-^ + C ........................ (36). 
 
 dx 26 
 
 Integrating again, we have 
 
 M"x 2 V'x* w'x* 
 
 -^r + -6---w + 
 
 But x - when y - ; therefore, C,, = 0, and we have 
 Af"a? 2 F'a s w'a; 4 
 
 "^ " 
 
 But y = when a? = L, . Then substituting L, for j7 in the last 
 equation, we have 
 
100 STRUCTURAL ENGINEERING 
 
 wi* VL* L* 
 
 2 6 2 
 
 from which we obtain 
 
 M"L, V'L? w'L? 
 2 ~6~ ~W 
 Now substituting this value of C, in equation (26), we have 
 
 .. , 
 
 dx 2 6 2 fi 24 
 
 It is readily seen that equations (25) and (27) would be equal 
 when x = L in (25) and x - in (27), the two elastic curves having a 
 common tangent at B. Then substituting L for x in (25) and for x in 
 (27), equating and reducing, we have 
 
 12M'L + 8FL 2 - 3wL 5 = -12M"L, - 4F'L, 2 + w'L, s ..... (28). 
 
 Now (referring to span AB) taking moments about the support B, 
 we have, according to Art. 66, 
 
 Transposing and dividing by L, we have 
 M-M> wL 
 
 L 2 C '* 
 
 Next (referring to span BC) taking moments about C, we have 
 
 M' " = M" + V'L, - - '- 
 Transposing and dividing by L, , we have 
 
 Now substituting the values of V and V as given by (29) and (30), 
 respectively, in equation (28), and reducing, we have 
 
 M'L + 2M"(L + L,) + M'"L, = _!!_ ^! 
 
 which is the three-moment equation when the loads are uniformly 
 distributed. 
 
 The three-moment equation, as is readily seen, is an expression for 
 the bending moments at any three consecutive supports in terms of the 
 
 M' 
 
 Fig. 94 Fig. 95 
 
 lengths of the two intervening spans and the loads supported in these 
 spans. The lengths and loads, of course, are known quantities. Now 
 for any continuous beam of n spans, there are always (n + 1) supports, 
 
THEOEETICAL TEEATMENT OF BEAMS 101 
 
 and for each three consecutive supports a three-moment equation can be 
 written. Then, (n 1) three-moment equations can be written for any 
 continuous beam. For example, let abcdef (Fig. 94) represent a con- 
 tinuous beam of five spans. Here we can write four (which is (n 1)) 
 three-moment equations: one for supports a, b, and c; one for supports 
 b, c, and d; one for supports c, d, and e; and one for supports d, e, and /. 
 Now the bending moment at each of the supports would be included in 
 these four three-moment equations, being (n -f 1) moments in all. But, 
 as the bending moments at both a and / are equal to zero, the four 
 unknown moments at b, c, d, and e can be determined, as we have four 
 equations and four unknown moments. Then, after these unknown 
 moments at the supports are determined, the reactions at each support 
 can be determined and then the shear and the bending moment at any 
 point of the beam can be determined. 
 
 As an example, let ABC (Fig. 95) represent a continuous beam 
 having three supports which would be known as a beam continuous over 
 three supports. Let M' ', M" , and M' " represent the bending moments 
 at the supports A, B, and C, respectively, due to any loads that we choose 
 to consider, and let R\ 9 R2, and R3 represent the reactions at the sup- 
 ports A, B, and C, respectively, due also to any loads that we choose to 
 consider. The intensity of these moments and reactions would, of course, 
 vary with the loads. 
 
 Let us first consider two concentrated loads P and P' on the beam as 
 shown in Fig. 95. Now, as there are but three supports, only one three- 
 moment equation can be written. So writing this one, which is really 
 Formula (L), given on page 98, we have 
 
 M'L + %M"(L + L,) + M' "L, = -PL 2 (k - P) - P'L, 2 (2k, - 3k, 2 + k, 3 ). 
 But in this case M' = and M' " 0. So the equation becomes 
 2M"(L + L,) = -PL 2 (k - P) - P'L 2 (2k, - 3k 2 + A;, 3 ), 
 
 from which we obtain 
 
 PL*(k-k*)-P'L?(2k,-3k;- + k,) 
 
 Now considering points A and B, and taking moments about B, 
 according to Art. 66, we have 
 
 M"=RlL-P(L-kL). 
 Transposing and reducing, we have 
 
 Now, by substituting the value of M" ', as given in equation (31), in 
 this last equation, the value of the reaction R1 at A is obtained. Then, 
 after Rl is known, the bending moment at any point in the span AB is 
 obtained as readily as in the case of a simple beam, as it is equal to the 
 algebraic sum of the moments about the point considered of the forces to 
 the left of the point, which are now all known. 
 
102 STRUCTURAL ENGINEERING 
 
 Now, similarly, considering points B and C, and taking moments 
 about B, we have 
 
 Transposing and reducing, we have 
 
 Then by substituting the value of M" , as given in equation (31), in this 
 last equation, the value of the reaction R3 at C is obtained. Then after 
 R3 is known, the bending moment at any point in the span BC is readily 
 obtained, as it is equal to the algebraic sum of the moments about the 
 point considered of the forces to the right of it, which are now all known. 
 The shear at any point in either span is obtained by simply adding up 
 (algebraically in each case) the forces between the end and the point 
 considered. R2, as is readily seen, is equal to the sum of the loads minus 
 the two reactions R\ and R3. It is also equal to the shear just to the 
 right of B plus the shear just to the left of B. By taking moments about 
 both A and B, equations can be derived from which these shears can be 
 obtained, and consequently R2. 
 
 Now it is evident that the bending moments, shears, and reactions on 
 the above beam due to any number of such loads could be determined by 
 considering the loads in pairs, as in the above case. But it is more con- 
 venient to derive equations expressing the values of these for one single 
 load at any point. So let P (Fig. 95) be the only load on the beam. 
 Then P f and k, will not occur in the three-moment equation, that is, they 
 will be equal to zero, and hence the three-moment equation becomes 
 
 from which we obtain 
 * 
 ' 
 
 Now taking moments about B, considering points A and B, we have 
 M" = LRl-P(L-kL). 
 
 Then substituting this value of M" , as given in (33), in this last equation, 
 and reducing, we have 
 
 .......................... (34) - 
 
 Then taking moments about B, considering points B and C, we have 
 
 Substituting this value of M", as given in (33), in this last equation, and 
 dividing by L, , we have 
 
 ' PL'(k-V) 
 
 -ZL,(L + L.) ................................. (3 ' 
 
 Now, Rl and R3 can be determined for any load in the span AB 
 from equations (34) and (35), respectively, and likewise for any load in 
 
THEOEETICAL TEEATMENT OF BEAMS 103 
 
 span EC by interchanging the R's, that is, #1 would in that case be at C 
 and R3 at A or just turn the beam end for end. If there are several 
 loads in the two spans, the reactions R\ and R3 due to each can be deter- 
 mined separately and added algebraically (as R3 is always minus), and 
 thus the total reaction at both A and C would be determined. Then the 
 bending moment and shear at any point in the beam is readily determined, 
 as explained above. 
 
 Now if the two spans are equal, we have L = L, in both equations 
 (34) and (35). Then (34) reduces to 
 
 Rl = : ? (4 - 5fc + P) ................................ (36), 
 
 and (35) reduces to 
 
 R3 = - ? (k - P) .................................. (37). 
 
 Then adding (36) and (37) and subtracting the result from the load P 
 and reducing, we have the formula 
 
 R2 =? (3k - P) .................................... (38). 
 
 </ 
 
 Now the reactions due to a load at any point on any beam continuous 
 over three supports and having equal spans, can be determined from 
 equations (36), (37), and (38). Then the bending moments and shears 
 are readily determined as explained above. 
 
 As a practical example, let L and L, in Fig. 95, each be equal to 12 
 feet, and let P be 4 feet from A and let P' be 6 feet from C. Then 
 k = 4/12 = 1/3, and k, = 6/12 = 1/2. 
 
 First considering P alone and substituting 1/3 for k in (36) and 
 reducing, we have 
 
 for the value of the reaction at A due to the load P. Next substituting 
 1/3 for k in (37) and reducing, we have 
 
 for the reaction at C, which is minus ; that is, it pulls down upon the beam 
 instead of pushing up. Now after these reactions are determined, the 
 bending moment and shear at any point in the beam due to the load P 
 alone can be determined as explained above. The bending moments in 
 span EC would be negative, as #3 is negative; that is, the top elements in 
 the beam would be in tension while the bottom ones would be in com- 
 pression. 
 
 Now considering P' alone, and substituting 1/2 for k ( = &,) in (36) 
 and reducing, we have 
 
104 STRUCTURAL ENGINEERING 
 
 which is the reaction at C due to load P' in span BC. Next, substituting 
 1/2 for k in (37) and reducing, we have 
 
 which is the reaction at A due to the load P' in span BC. Then the 
 reaction at A due to the two loads is 
 
 _ 
 
 27 32 ' 
 
 and the reaction at C due to the same is 
 
 13 13 
 32 54 
 
 After these reactions are thus determined, the bending moment and shear 
 at any point in the beam are readily determined, as explained above. 
 
 Now instead of the concentrated loads just considered, suppose the 
 beam shown in Fig. 95 supported a uniform load of w pounds per foot in 
 span AB and w' pounds per foot in span BC, extending over the entire 
 span in each case. In this case, as the loads are uniformly distributed, 
 we would use the three-moment equation (M), given above, which, as M f 
 and M" are equal to zero, reduces to 
 
 8M"<L + L,)=-- ......................... (39). 
 
 Then transposing and dividing through by 2(L + L,), we have 
 
 Now taking moments about B, and considering points A and B, according 
 to Art. 66, we have 
 
 Then substituting the value of M", as given in (40), in this last equation, 
 transposing, and reducing, we have 
 
 wL wL 3 + w'L* n 
 
 ~-~- ' ......................... 
 
 Now taking moments about B, and considering points B and C, we have 
 
 and substituting the value of M", as given in (40), in this last equation, 
 transposing and reducing, we have 
 
 
 After these reactions are obtained, the bending moment and shear at any 
 
 point in the beam can be determined very readily as explained above. 
 
 Suppose the spans to be equal in length and suppose the loads are 
 
THEORETICAL TREATMENT OF BEAMS 105 
 
 equal. Then L = L, and w = w' , and substituting L for L, and w for w' 
 in both (41) and (42), and reducing, we have 
 
 3 = ?-wL, 
 
 o 
 
 that is, the reactions at A and C are equal to f of the load on one span. 
 Then, evidently, the shear just to the right and also just to the left of B 
 is equal to f (wL} or f of the load on one span, and hence for the reaction 
 at B we have 
 
 Substituting L for L, and w for w' in (39) and reducing, we have 
 
 ' 
 
 which is the bending moment at B, the central support. This being minus 
 shows that the top elements of the beam are in tension while the bottom 
 ones are in compression. This same moment can be obtained by taking 
 moments about B and considering either span, say, span AB. Then we 
 have 
 
 But Rl = f (te>L), as shown above. So substituting this value for Rl in 
 this last equation, and reducing, we have 
 
 Now taking moments about the center of span AB, we have 
 
 Substituting f (wL) for Rl, we have for the bending moment at mid-span 
 
 which is one-half of what it is at B, the central support. 
 
 Beams having four or more supports can be analyzed in a similar 
 manner. Of course, there would be more three-moment equations involved 
 being (w 1) in each case. However, the same general method as 
 outlined above would hold for all cases. 
 
CHAPTER V 
 
 THEORETICAL TREATMENT OF COLUMNS 
 
 L_ 
 
 71. Preliminary. Let CB (Fig. 96) represent a round steel rod 
 of uniform cross-section A and length L standing vertically upon a 
 smooth, rigid base at B and supporting a load P at C which is sym- 
 metrically applied in reference to the longitudinal axis of 
 the rod ; that is, the load is applied in the center of gravity 
 of the cross-section of the rod. It is evident that the load 
 P will produce a direct simple compressive stress P at 
 every cross-section of the rod throughout its length, and 
 hence the direct compressive unit stress at every section 
 will be equal to P/A. All bodies having a length much 
 greater than their width, as the above rod, and loaded in 
 the same manner, thereby being in compression, are known 
 in general as columns. 
 
 Now suppose the above rod to be 2 inches in diameter 
 and, say, 4 inches long. Then the unit compressive stress 
 would be equal to 
 
 P P 
 
 R 
 Fig. 96 
 
 7T 
 
 which corresponds to P/A given above. It is evident that this short rod 
 would support a very heavy load without failure, in fact, the compressive 
 stress P/TT could be as great as if it were a tensile stress. 
 But suppose the rod to be 20 feet long instead of 4 inches. 
 We know that the rod, if 20 feet long, would not support as 
 heavy a load as it would if only 4 inches long, as the longer 
 rod would be too limber, that is, it would fail by bending 
 transversely; of course, the direct compressive stress would 
 be acting just the same in either case. 
 
 If a column were absolutely straight and the load 
 applied absolutely in the longitudinal axis, the bending re- 
 ferred to above would not occur. But such conditions do not 
 exist, as we know from experience. So in the designing of 
 columns the stress due to this bending must be taken into 
 account in addition to the direct compressive stress. 
 
 72. Rankine's Formula. Let ED (Fig. 97) repre- 
 sent a column which bends as indicated under a load P. It 
 is evident that the elements on the concave side of the column 
 are in compression, owing to the bending, while the elements 
 on the other side are in tension, exactly as in the case of a 
 loaded beam. Now it is readily seen that the maximum stress 
 
 106 
 
 Fig. 07 
 
THEOEETICAL TREATMENT OF COLUMNS 107 
 
 on the column will occur on the compression or concave side, for here the 
 direct and bending stresses combine, while the tensile stress on the other 
 side reduces the direct stress. This maximum compressive stress will, as 
 is readily observed, occur at the middle of the column, where the deflection 
 is greatest, which is represented as A. The bending moment on the 
 column at any point x distance from E due to the load is 
 
 M = Pz, 
 
 where z is the deflection. This moment will be a maximum when z = A. 
 Then we have 
 
 M = PA 
 
 for the maximum bending moment on the column. But from (D), Art. 
 53, we have / = My /I, and substituting PA for M, we have 
 
 (where y = the distance from neutral axis of column to the extreme 
 elements) for the maximum compressive stress on the column due to 
 bending. Now if we add this to the direct stress, P/A, we have 
 
 P 
 
 which is the maximum compression unit stress on the column. 
 
 Now as is proven by experiment A will vary directly as L 2 and 
 inversely as y; that is, 
 
 y 
 
 Then if L 2 /y be multiplied by some constant c, we have 
 
 L 2 
 A= c -- 
 
 y 
 
 Substituting this value of A in (1), we have 
 
 P PcL 2 
 
 But 7 = Ar 2 (Art. 49). Then substituting this value of I in the last 
 equation and reducing, we have 
 
 This can be written in the form 
 P x 
 
 ....................................... (O), 
 
 which is known as Rankine's Formula, where 
 
 p maximum compressive unit stress on the column; 
 
 P = load on the column; 
 
 L = length of column in inches 3 ' 
 
108 STRUCTURAL ENGINEERING 
 
 A = area of cross-section of the column in square inches ; 
 r least radius of gyration of the cross-section in inches in 
 
 reference to the neutral axis of the column; 
 c = a constant which is determined by experiment, and depends 
 upon the material composing the column and upon end 
 conditions. 
 
 The maximum unit stress on any known column due to a given load 
 can be determined from Formula (N) and the allowable direct or average 
 compressive unit stress from (O), providing the proper value of c is 
 known, which is determined from experiment. 
 
 The following practical values of c are recommended: 
 
 For timber columns c 4/3,000; 
 For cast-iron columns c 4/5,000 ; 
 For (medium) steel columns c - 1/11,000. 
 (American Bridge Co.) 
 
 If p be taken as the elastic limit of the material in the column, P/A 
 will be the direct compressive unit stress that the column would be sub- 
 jected to when the elastic limit of the column was reached, but if p be 
 taken as the ultimate strength of the material, P/A would be the direct 
 compressive unit stress that the column would be subjected to at failure. 
 If p be given either of these values, a factor of safety must be used in 
 determining the "working stress" for any column. If 32,000 Ibs. be 
 taken as the value of p at the elastic limit of steel, a factor of safety of 
 2 should be used, and if 64,000 Ibs. be taken as the value of p when the 
 column fails, a factor of 4 should be used. For practical designing, it is 
 more convenient to take p as the allowable or working stress of the 
 material in tension. This for medium steel may be taken as 16,000 Ibs., 
 in accordance with the specifications of A. R. E. Ass'n. Now, substituting 
 this value of p and the above value of c in Rankine's Formula (O), we 
 have 
 
 P 16,000 
 
 11,000 
 
 This Formula (P) may be used for designing any steel column, but 
 a Straight Line Formula is preferable, as it is more readily applied and 
 the results obtained are practically as accurate. 
 
 73. Straight Line Formula. The ordinates to the curve abed 
 (Fig. 98) measured from the horizontal line OH give the value of P/A 
 corresponding to the different values of L/r, as shown. This curve is 
 obtained by substituting the different values of (L/r) 2 in Formula (P). 
 For example, the ordinate eb is equal to 
 
 P 16,000 
 
 A ~ 1 + OT^ 30)2 " ' 
 
 It is readily seen from Formula (P) that the ordinate Oa is equal to 
 16,000 Ibs., as L/r = 0. Now, if some line as ak be drawn such that the 
 ordinates to it will be practically the same as to the curve, it is obvious 
 
THEOEETICAL TREATMENT OF COLUMNS 
 
 109 
 
 _ -^_ ._ 80 9O 1OO //O t2O 
 
 Vafvesof A (zo">3 C afe) 
 
 Fig. 98 
 
 that the equation to this line can be used as a column formula instead of 
 Formula (P), and the equation to the line would be known as a "Straight 
 Line Formula." 
 
 The ordinate Hk to this line is taken here as 7,600 Ibs., and the 
 ordinate Oa is 16,000 Ibs. Then, as OH is known, the slope of the line ak 
 can be determined, and consequently its equation in reference to can be 
 derived. 
 
 Let 6 be the angle that the line makes with the horizontal. Then we 
 have any ordinate y, distance x from 0, given by the formula 
 
 y = 16,000 - .rtantf ................................... (1). 
 
 Nowtan6=(Oa-Hk) + OH = 0.14:, as a and Hk are equal to 16,000 and 
 7,600 Ibs., respectively, and OH, to the same scale, is equal to 60,000 Ibs. 
 Now for x, in terms of L/r, we have 
 
 (Q), 
 
 Then substituting these values of tan0 and x in (1), we have 
 
 = = 16,000 - 
 
 which is the straight line formula as given in the A. R. E. Ass'n Speci- 
 fications. 
 
 As is obvious, the author just deliberately took the ordinate Hk so 
 that the above formula would result. If the line ak be drawn to conform 
 to mere observation, a slightly different formula would, very likely, be 
 obtained. 
 
 It is readily seen that a straight line formula for any other material 
 can be determined in the same manner by simply using the proper c and 
 working stress in each case. 
 
 74. Examples in the Application of Column Formulas. 
 
 Example 1. What will be the maximum compressive unit stress 
 produced in a wooden column 10 x 10 ins. in cross-section and 12 ft. 
 long by a load of 20,000 Ibs.? 
 
 Here A = 100, L = 144, r = 2.88, and c = 4/3,000. The value of p is 
 desired. Then substituting the above values in Formula (N-), we have 
 
 20,000 r 4 / 144V] 
 P = -m-[ 1+ 3Mo\2A8) \ = 
 
 In this example, the direct unit stress is 20,000 -=- 100 = 200 Ibs. So 
 the stress due to cross bending is 868 - 200 = 668 Ibs. The working unit 
 
11C STRUCTURAL ENGINEERING 
 
 stress on timber should be taken at about 1,200 Ibs. So the above column 
 will safely support more than 20,000 Ibs. 
 
 Example 2. What direct unit stress would the above column safely 
 sustain if the working unit stress be taken as 1,200 Ibs.? Here P/A is 
 desired. Then using Formula (O), we have 
 
 P 1,200 
 
 144 i = per sq ' m ' 
 
 4 / 144 yi 
 
 + 3,000 \2.88/ J 
 
 Then the safe load, or P, would be 277 x 100 = 27,700 Ibs. 
 
 What is usually given is the load and the length of the column. In 
 that case the problem is to design a column outright that will safely carry 
 this known load. In all such cases it is practically a matter of first 
 guessing a column which we think will satisfy the conditions and then 
 modifying our guess until the conditions are satisfied, as will be shown in 
 the following problems. 
 
 Example 3. Design a timber column 10 ft. long to carry a load of 
 20,000 Ibs. Taking 1,200 Ibs. as the working unit stress, then p will be 
 1,200. Using Formula (N), we have 
 
 Here it is seen that A and r are unknown. First assume that a column 
 9x9 ins. will do. Then A = 81 and r = 2.6. Substituting these values in 
 the last equation, we have 
 
 which is too small, or, in other words, the column is too large. So next 
 try an 8 x 8-in. section. Then we have 
 
 which is too large, that is, this column is too small. So the correct size is 
 between the two. However, a 9 x 9-in. section would be used, as more 
 than likely nothing between the two could be obtained. 
 
 In case of steel columns, the straight line Formula (Q) will be used, 
 although the Formula (P) could be used. 
 
 Example 4- Design a steel column 25 ft. long to support a load of 
 160,000 Ibs. Using Formula (Q), we have 
 
 p = 16,000 -70. 
 
 The direct unit stress in accordance with economy should not be less than 
 9,000 Ibs., and it rarely ever exceeds 14,000 Ibs. So, guessing in this 
 case, as the column is short, that it will be 13,000 Ibs., and dividing the 
 load by that, we have 
 
 160,000 
 
 From Table 3 it is seen that 2 [s 12" x 20.5 # have about this 
 
THEOEETICAL TEEATMENT OF COLUMNS HI 
 
 section. The r for these two channels in reference to the gravity axis 
 perpendicular to their webs will be the same for the two as it is for one 
 channel, which is given in the same table as 4.61. The radius in ref- 
 erence to the gravity axis parallel to the webs can always be made equal 
 to the one given in the table by moving the channels far enough apart. 
 Now substituting the above value of r in the above formula, we have 
 
 p = 16,000 - 70 j = 11,450 Ibs. 
 
 Then dividing this into the load, we have 
 
 160,000 
 
 '11450 =14 ' sq * m ' ( about )> 
 
 which is more than the area of the channels assumed. So try 
 2[ s 12" x 25# = 14.7 n ". Then r, from Table 3, is 4.43, and our 
 formula becomes 
 
 p = 16,000 -70 - -11,260 Ibs. 
 
 Then dividing this into the load, we have 
 160,000 
 
 which agrees very closely with the area of the 2 [s 12" x 25 #, which 
 would therefore be used. The channels composing a column as just 
 designed would be latticed together. 
 
 The method of applying the column formula in the last example is 
 general, but the form of column is only one of many. It is beyond the 
 limits of practicability to have the radii of gyration tabulated for each 
 form. However, the approximate value of the radius for any form in 
 general use can be obtained from Table 10, where it is given in terms of 
 the height and width of the section, as shown. After the approximate 
 area of the cross-section is obtained by dividing the load by an assumed 
 intensity, as above, the section can be selected and then the corresponding 
 approximate radius of gyration can be obtained from Table 10. In this 
 way the assumed section can be tested, and, if found to be about correct, 
 the exact radius can be computed and substituted in the column formula, 
 and the section modified slightly, if necessary, to satisfy this last require- 
 ment. The form of steel columns depends upon the kind of structure and 
 the position they occupy in the structure as well as to the loads they 
 support. 
 
 The value of p for the different values of L/r could be taken from 
 such a diagram as shown in Fig. 98 if it be drawn to a sufficiently large 
 scale, or a table giving the same can be computed. The latter is often 
 used in designing offices. 
 
 In case a column is fixed at the ends, one-half of the length of the 
 column is taken as L, and if fixed at one end only, two-thirds of the 
 length is taken as L in the column formulas. The reason for so doing is 
 readily seen, as the actual length of the column as far as bending is 
 concerned in the first case is the distance between the points of contra- 
 flexure, which is about one-half of the length of the column; and in the 
 
112 
 
 STEUCTUEAL ENGINEEEING 
 
 st-cond case it is the distance from the free end to the point of contra- 
 flexure, which is about three-fourths of the length of the column, the same 
 as in the case of beams. (See Arts. 68 and 69.) 
 
 75. Columns Eccentrically Loaded. Whenever 
 
 possible, the load on any column should be applied in the 
 center of gravity of its cross-section. However, in some 
 cases this is not possible. In such cases the bending 
 stresses due to the eccentric loading must be considered. 
 Let AB (Fig. 99) represent a column, which bends 
 as indicated, due to the eccentrically applied load P. Let 
 e represent the eccentric distance, A the maximum deflec- 
 tion, and x and y the co-ordinates to the elastic curve re- 
 ferred to A as origin. Now the bending moment at any 
 point x distance from A is equal to 
 
 *&=-*<+> 
 
 For convenience let k = yP/EI. Then we have 
 d~v 
 
 Fig. 
 
 Now multiplying both sides of the equation by 2dy, we have the form 
 
 dx* 
 Integrating (dx constant), we have 
 
 (1) 
 
 Now, as is readily seen., dy/dx = when y = A. Therefore, C, = k*(e + A) 2 . 
 Then substituting this value of C, in (1), we have 
 
 Now extracting the square root and transposing, we have 
 
 Integrating this equation, we have 
 
 Now y = A when x = L/2; therefore, C,, = (L/2 - ir/2k). Then substi- 
 tuting this value in (2), we have 
 
 Now y = when x = 0. So, substituting in (2) for both y and x, we 
 
THEORETICAL TREATMENT OF COLUMNS 113 
 
 e,,=-r 
 
 Then by equating the two values of C,, , we have 
 
 _l/ 5in -i * \ /_,r\ 
 k\ e + &J \2 2k J' 
 
 from which we obtain 
 
 cos 
 
 Then substituting this value of A in (3) , we have 
 
 Now transposing and reducing, we have 
 
 kL 
 
 But,, according to trigonometry, 
 
 kL 
 
 e + y kL 
 
 = COS T- 
 
 But, cos ir/2 0, and sin ?r/2 = 1. Then the last equation reduces tc 
 
 / kL\ e + y kL 
 cos (kx--)= cos-. 
 
 But again, according to trigonometry, 
 
 /, *L\ kL , kL e + y kL 
 
 cos ( kx - 1 = cosAr^cos -^- + smfcxsiu - = ^-cos -. 
 
 \ a I <> a e fy 
 
 Dividing through by coskL/2, we have 
 
 kL e + y 
 coskx + sinAr^rtan - = - - . 
 
 But, tan kL/2 = 1 - cosfcL/sin&L. Then substituting this value of 
 tan kL/2 in the last equation, and reducing, we have 
 
 el sink (L - x ) + si 
 - 
 
 . ir 
 sinArL 
 
 (5). 
 
 Now from this last equation, y can be computed for any point along 
 the column, and then the lever arm (e + y) is known and hence the stress 
 due to bending at any point along the column can be determined, as we 
 have 
 
J14 STRUCTURAL ENGINEERING 
 
 according to Art. 53. (d is here the distance from the neutral axis to the 
 outermost element in compression.) 
 
 Then by adding the unit compressive stress (/) thus obtained to the 
 direct unit stress (P/A), we have the maximum unit stress at the point 
 considered. This, of course, will be a maximum when y = A. Now 
 y = A when x - L/2. Then, substituting this value of x in (5) and 
 reducing, we have 
 
 IcT 
 e(sec - 1) = A ................................... (6). 
 
 This last equation (6) can be used for determining A, the maximum 
 deflection. After this is determined, the maximum stress can readily be 
 determined as stated above. For the maximum compressive stress we 
 then have the formula 
 
 P= 
 
 where p is the maximum compressive unit stress in pounds, 7 the moment 
 of inertia of the cross-section of the column, A the area of the cross- 
 section in square inches (which is assumed to be constant), and P, e> and 
 A are as stated above, 
 
CHAPTER VI 
 
 j: 
 
 RIVETS, PINS, ROLLERS, AND SHAFTING 
 
 RIVETS 
 
 76. Kinds of Stress. Rivets may fail by shearing off transversely, 
 by bearing, that is, by zrushing against the metal through which they 
 pass, or by bending, as a beam. So, therefore, we have to consider shear- 
 ing, bearing, and bending stresses on them. 
 
 77. Shearing Stress. The shear on a rivet at any cross-section is 
 equal to the algebraic sum of the forces on either side of the section 
 just the same case as a beam. Let A and B (Fig. 100) represent two 
 bars held together by one rivet, as shown. Let P be the tensile stress on 
 each bar. Then the bar B would exert forces along 
 
 ab upon the rivet, the sum of which (resolved along the 
 
 bar) would be equal to P, while the bar A would exert 
 
 forces upon the rivet along cd, the sum of which (re- 
 
 solved along the bar) would be equal to P. The forces 
 
 along ab, acting against the forces along cd, tend to 
 
 shear the rivet off transversely. This shear, as is 
 
 readily seen, varies from at each end of the shank to 
 
 a maximum at the cross-section be. If it were possible 
 
 to determine the intensity of these forces at all points, 
 
 the shear at any cross-section along the rivet could be 
 
 obtained by beginning at either end of the shank and 
 
 simply summing up the forces to the section consid- 
 
 ered. However, it is not practical to do so, neither is 
 
 it necessary, for it is readily seen that the maximum 
 
 shear occurs just between the two bars at cross-section 
 
 be. This shear, as is evident, is equal to P, and as it 
 
 is the maximum, it is the only shear we need consider. 
 
 The shear between the pieces connected is what is generally referred to as 
 
 the shear on rivets, and it is what we shall consider as the shear. 
 
 The unit shearing stress, that is, the stress per square inch on a rivet, 
 is obtained by dividing the shear (in pounds) by the area (in square 
 inches) of the cross-section of the rivet. Thus if S be the shear in pounds 
 at a given section of a rivet having a cross-section of A square inches, and 
 V the unit shearing stress, we have 
 
 8 
 
 (p) 
 
 t P 
 
 Fig. 100 
 
 The allowable shear on a given rivet is what we usually desire. This is 
 obtained by multiplying the allowable intensity per square inch by the 
 area of the cross-section of the rivet. We shall take this allowable 
 intensity as 12,000 pounds per square inch for shop rivets and 10,000 
 pounds for field rivets, as specified by the A. R. E. Ass'n in their speci- 
 
 115 
 
116 
 
 STRUCTURAL ENGINEERING 
 
 fications for railroad bridges. Then if s be the allowable shearing stress, 
 we have s = 12,000 x A for shop rivets, and * = 10,000 x A for field rivets. 
 Substituting in these formulas we have for the allowable shear on a J" 
 shop rivet, 5-12,000x0.6013 = 7,216 Ibs., and for a J" field rivet, we 
 have s,= 10,000x0.6013 = 6,013 Ibs. These values for the different size 
 rivets are given in column 3 of Table 11. 
 
 A rivet is in single shear when there is (so to speak) one tendency 
 to shear it off, double shear when there are two tendencies, triple shear 
 when there are three, and so on. If two pieces be riveted together, as 
 indicated in the sketch at (a) (Fig. 101), the rivets will be in single 
 shear; and if three pieces be riveted together as indicated in the sketch 
 at (6), the rivets will be in double shear; and if there were four pieces, 
 the rivets would be in triple shear. If the pieces riveted together are 
 properly designed, rivets in double shear will have twice the allowable 
 
 P b ^ 
 
 (O) (b) 
 
 . Fig. 101 
 
 Fig. 102 
 
 shearing strength of rivets in single shear, and three times in the case of 
 triple shear, and so on. For example, the shear on the rivet shown at 
 (a) (Fig. 101) would be equal to P, while in the case shown at (6) it 
 would be equal to 'P/2, providing e and d have equal cross-section. 
 
 78. Bearing Stress on Rivets. Let Fig. 102 represent a bar 
 pulling -wdth a force of P pounds against a rivet which passes through it. 
 The force transmitted to the rivet thereby will be a normal force 
 uniformly distributed from b around to a, as indicated. Let t be the 
 thickness of the bar, D the diameter of the rivet, and let p be the uniform 
 bearing force per square inch. Now, the bearing force on any infini- 
 tesimal strip through the plate, as ids, where t is the thickness of the 
 plate, would be ptds. Now this force, resolved along the bar, is equal to 
 ptds cos#, where 6 is the angle the normal force makes with that direc- 
 tion, as indicated. Then if the normal forces on all such strips, from 
 i around to a, be resolved along the bar and summed up, we would have 
 pfZdscosQ = P. But 2dscos6 = ab = D, the diameter of the rivet. So we 
 have 
 
 ptD = P. 
 
 (1). 
 
 To obtain the value of P, such that the rivet would not be too highly 
 stressed, we would substitute the working value of p in the above formula, 
 in which case p would be known as the allowable unit-bearing stress on 
 the rivet and P as the allowable bearing of the rivet on the plate of 
 thickness t. For the unit value we will take 24,000 Ibs. per square inch 
 for shop rivets and 20,000 Ibs. for field rivets, as specified by the A. R. E. 
 Ass'n in their specifications for railroad bridges. 
 
 Now suppose the rivet shown in Fig. 102 to be a J" shop rivet, and 
 
RIVETS, PINS, ROLLERS, AND SHAFTING 117 
 
 suppose the plate to be J" thick. Then substituting in (1), we have 
 
 24,000 x *- x 1 = 10,500 Ibs. = P, 
 A o 
 
 that is, the allowable bearing of a j" rivet on a \" plate is 10,500 Ibs. 
 If the thickness of the plate be -J" instead of J", we would have 
 
 i-x^ 
 4 8 
 
 In this way the allowable bearing for rivets of different diameters on 
 different thicknesses of plates can be computed. These values are given 
 in Table 11. 
 
 79. Bending Stress On Rivets. If rivets are properly driven, 
 bending stresses can be ignored, except in the case of loose fillers as is 
 illustrated in Fig. 103. Here the bars c, d, and e exert double shear on 
 the rivet just the same as the case shown at e f 3 / 
 
 (b), Fig. 101, and the bearing stresses are "" r ~* > 
 
 just the same, but the idle fillers, / and g, 
 
 hold the bars apart so that the rivet is really Fig. 103 
 
 a beam loaded in the center and supported at 
 
 the ends. Let k be the distance from the center of the bar c to the center 
 
 of the bar d, and h the distance from the center of the bar c to the center 
 
 of the bar e. Then the maximum bending moment on the rivet is (P/2)k 
 
 or (P/2)h. Now, according to (D), Art. 53, the maximum bending 
 
 stress on the rivet per square inch is 
 
 My 
 
 ' " I 
 
 64 
 
 where D is the diameter of the rivet. 
 
 The bending stress on a rivet should always be combined with the 
 shearing stress according to Art. 62 so as to obtain the maximum stress. 
 
 Details should be so contrived as to avoid bending on rivets. In 
 fact, there is practically no excuse for placing rivets so that they are 
 subjected to bending of any consequence. 
 
 80. Examples in Determining Number of Rivets. Suppose 
 that each of the bars a and b, shown at (o), Fig. 101, be J" thick, and 
 suppose the stress P to be 50,000 Ibs., how many }" shop rivets would be 
 required to transmit this stress? 
 
 The allowable single shearing stress on one f " shop rivet, from Art. 
 77, is 12,000 x 7r(|) 2 -r 4 = 12,000 x 0.4418 = 5,301 Ibs. Then the number 
 of rivets required for shear is 50,000 ^ 5,301 = 9.4 rivets (use 10). 
 
 The allowable bearing stress on one f" shop rivet, from (1), Art. 78, 
 is 24,000 x J x f = 4,500 Ibs. Then the number required for bearing is 
 50,000 4- 4,500 = 11 rivets (about). So 11 rivets is the number required. 
 
 Now suppose the bars to be f " thick instead of J". The allowable 
 shearing stress on each rivet would not change. So 10 rivets would be 
 required for shear the same as before. But the allowable bearing stress 
 is now 24,000 x J x } = 6,750 Ibs. Then the number of rivets required 
 
STRUCTURAL ENGINEERING 
 
 for bearing is 50,000 -f 6,750 = 7.4 (use 8). So in that case 10 rivets, as 
 required for shear, would be used, as it is the greater number. 
 
 As another example, suppose the bar c, shown at (6), Fig. 101, to be 
 y thick, and each of the bars d and e to be " thick, and suppose the 
 stress P to be 60,000 Ibs. ; how many f " shop rivets would be required to 
 transmit this stress? The allowable double-shearing stress on one f" 
 shop rivet is 2 x 12,000 x 0.3068 = 7,364 Ibs. Then the number of rivets 
 required for shear is 60,000 -^ 7,364 = 8.15 rivets (use 9). The bearing 
 here, in one direction, is on a \" bar, while in the other direction it is on 
 two f " bars, which is equivalent to one f " bar. So the bearing stress on 
 the \" bar (c) will be the greater. The allowable bearing stress on one 
 rivet is 24,000 x -J x f = 7,500 Ibs. So the number of rivets required for 
 bearing is 60,000 -j- 7,500 8 rivets. Here the number of rivets used 
 would be 9, the number required for shear. 
 
 The allowable bearing and shearing intensities for rivets are given 
 in Table 11. However, the student should know how to compute these 
 intensities. 
 
 PINS 
 
 81. Kinds of Stress. The same kind of stresses occur on pins as 
 on rivets. However, in the case of pins, the bending stresses are, as a 
 rule, the most serious. 
 
 82. Shearing and Bearing' Stresses. The shearing and bearing 
 stresses on pins are determined exactly as on rivets. The allowable unit 
 intensities for shear and bearing are 12,000 and 24,000 Ibs., respectively 
 the same as for shop rivets. The shearing stress very rarely affects 
 the design of a pin, while the same is practically true of the bearing stress. 
 However, the bearing affects the details of the other parts of a structure 
 as each member connected must have the proper thickness of bearing on 
 the pin. As an example, what would be the required thickness of bearing 
 to transmit a stress of 250,000 Ibs. to a 6-in. pin? The allowable bearing 
 of a plate 1 in. thick on this pin, according to Art. 78, is 24,000 x 1 x 6 = 
 144,000 Ibs. Then the thickness of the bearing required is 250,000 
 -f 144,000 = 1.76 ins. 
 
 83. Bending Stresses on Pins. The forces applied to pins are 
 assumed to be applied at the center of bearings of the pieces connected. 
 
 The bending moments on pins can be computed most 
 readily by utilizing the proposition in Art. 66. For ex- 
 ample, let AB (Fig. 104) represent a pin acted upon by 
 forces F a , F^, etc., applied at sections a, b, c, etc., as 
 shown, and let M a , M b , etc., and S a , Si>, etc., represent 
 the bending moments and shears, respectively, at the sec- 
 tions a, b, c, etc. It is obvious that the bending moments 
 at sections a and g are equal to zero. Then, by starting 
 at either of these sections we can determine the bending 
 moment at each of the other sections, according to 
 Art. 66. 
 So, starting from a, the bending moment at 6 is 
 
 = (a6)F ; 
 
RIVETS, PINS, EOLLERS, AND SHAFTING 119 
 
 at c it is 
 
 M c 
 at d it is 
 
 M d 
 at e it is 
 
 M e 
 at / it is 
 
 M f = M c + (ef)S e = M e + ef(F a -F b + F c - F d 
 
 In determining the bending moments in this manner, the maximum 
 is readily observed. Care should be taken in regard to algebraic signs. 
 The shear in some cases will be minus, as is readily seen, and consequently 
 the moments in some cases may be minus. 
 
 As a numerical example, let the lever arms be as follows : 
 
 ab = 2 in., de = l in., 
 
 6c = lJ in., ef=ll> in., 
 
 cd = 1 in., /<7 ^ in- ; 
 
 and let 
 
 F a = 150,000 Ibs., F e = 125,000 Ibs., 
 
 F 6 --200,000 Ibs., F f = -200,000 Ibs., 
 
 F c -125,000 Ibs., F, = 150,000 Ibs.; 
 F d = -150,000 Ibs., 
 then 
 
 S a = 150,000 Ibs., S d = -75,000 Ibs., 
 
 b = -50,000 Ibs., S e = 50,000 Ibs., 
 
 S e = 75,000 Ibs., 5; = -150,000 Ibs. 
 
 Now, by substituting these values in the above equations, we have 
 
 M 6 = 2 x 150,000 =300,000 in. Ibs., 
 
 M c = 300,000- 75,000 = 225,000 in. Ibs., 
 M d = 225,000+ 75,000 = 300,000 in. Ibs., 
 M e = 300,000- 75,000 = 225,000 in. Ibs., 
 M f = 225,000+ 75,000 = 300,000 in. Ibs., 
 M g = 300,000 - 300,000 = 0. 
 
 Here it is seen that the maximum moment on the pin would be 
 300,000 inch pounds, which happens to occur at three points b, d, and /. 
 
 After, the maximum bending moment is thus obtained, the bending 
 stress is computed according to (D), Art. 53, just the same as if the pin 
 were a beam. Thus, in the last example, suppose the pin to be 6" in 
 diameter. Then we would have 
 
 / = 300,000x3 + 7r = 14,150 Ibs. (about). 
 
 The allowable bending stress intensity for pins, as specified by the 
 A. R. E. Ass'n, is 25,000 Ibs. per square inch. 
 
 Let D be the diameter of any pin, M, the bending moment, and we 
 have 
 
120 STRUCTURAL ENGINEERING 
 
 2 T 64 
 Transposing, we have 
 
 (i) 
 
 Now by substituting 25,000 Ibs. for / in this last equation, the diameter 
 of pin required for any given bending moment can be determined. The 
 diameters required for given moments are given in Table 12. Here the 
 moments are given in inch pounds, and / taken as 25,000 pounds. 
 
 Often the members bearing on a pin act upon it from different 
 directions, as shown in Fig. 105. In such cases the stresses in the 
 members are resolved horizontally and vertically and the bending moments 
 determined in each plane, and the maximum or resultant moment is then 
 obtained by extracting the square root of the sum 
 of the squares of these maximum horizontal and 
 vertical moments. Thus, in the case shown in Fig. 
 F2 105, F3 would be resolved horizontally and ver- 
 Fig. 105 tically. Then the horizontal component would be 
 
 included with the forces Fl and F2 in the computa- 
 
 tions for the horizontal moments, and the vertical component would be 
 included with the force F4 in the computations for the vertical moments. 
 
 Let Mh represent the maximum horizontal moment, and let Mv 
 represent the maximum vertical moment; then the maximum or resultant 
 moment would be 
 
 for which the pin would be designed. 
 
 ROLLERS 
 
 84. Allowable Pressure. In modern practice, the allowable pres- 
 sure on rollers is obtained by the use of an empirical formula, which has 
 the form 
 
 p = cD ............... .............................. (1), 
 
 where p = the allowable pressure per lineal inch of roller and D = the 
 diameter of the roller in inches, and c = a constant. 
 
 Prof. A. Marston* found that for the elastic limit of soft steel rollers 
 c was about 880. (See paper Trans. A. S. C. E., Vol. 32.) For the 
 allowable pressure, c should be taken as about one-half of this. Then for 
 the allowable pressure per lineal inch on soft steel rollers, we have 
 
 p = 440 D .......................................... (2). 
 
 The formula, for medium steel rollers as specified by the A. R. E. 
 Ass'n, is 
 
 p = 600 D .......................................... (3). 
 
 This is the formula now in general use, as rollers are, as a rule, made 
 of medium steel. 
 
 *Dean, Engineering Division, Iowa State College. 
 
RIVETS, PINS, ROLLERS, AND SHAFTING 121 
 
 The formula is very readily applied. As a practical example, sup- 
 pose a load of 600,000 pounds be supported upon six 6" rollers; what will 
 be the required length of each? 
 
 Substituting in Formula (3), we have p = 600 x 6 = 3,600 Ibs. Then 
 the total linear inches required is 600,000 -=- 3,600 = 166.6 ins.; and as 
 there are six rollers, the length of each will be 166.6 * 6 = 27.7 ins. 
 
 Rollers should always be as large in diameter as is consistent with 
 accompanying details. 
 
 SHAFTING 
 
 85. Stress Due to Torsion. Shafting used to transmit power for 
 operating movable bridges, as well as in all other machinery, is subjected 
 to a twisting about the longitudinal axis known as torsion. The stresses 
 resulting therefrom are known as torsion stresses. This stress is the same 
 as the shearing stress in a beam, except the action on each particle is 
 perpendicular to a radius through the center of rotation, and the stress 
 varies directly as the distance out from this center of rotation. The 
 difference results owing to the stress being caused by a tendency of 
 rotation instead of a tendency of translation, as in the case of a beam. 
 
 Let AB (Fig. 106) represent a round steel shaft subjected to torsion. 
 Suppose this shaft be cut off at ab and spliced by means of inserted steel 
 pins which fit tightly into drilled holes. 
 Now it is readily perceived that the tor- 
 sion (twisting of the shaft) tends merely 
 to shear each of these pins off transversely 
 and perpendicularly to a radius through 
 the center of rotation, and as far as the Fig. IOG 
 
 torsion is concerned, the pins have only a 
 
 direct shearing stress on them. Now, evidently, the material particles in 
 every cross-section of the shaft are subjected to exactly the same kind of 
 stress from torsion as the pins, as the action causing the stress is the same. 
 
 This shearing stress varies directly as the distance out from the 
 center of the shaft. This is readily seen by considering again the pins at 
 section ab. Each pin will be distorted a certain amount, and as the 
 tendency of rotation is about the center of the shaft, it is evident that the 
 distortion of the pins will vary directly as their distance out from the 
 center of the shaft, and hence the stress on them will so vary. Then, 
 evidently, if the stress on these pins varies directly as their distance out 
 from the center of the shaft, the stress on the material particles at every 
 cross-section of the shaft will vary directly as their distance out from the 
 center of the shaft. 
 
 So we have thus far shown that the stress on the shaft due to torsion is 
 a transverse shearing stress acting perpendicularly to the radii of rotation, 
 and that it varies directly as the distance out from the center of the shaft. 
 Now it remains yet to determine the intensity of the stress. The stress, 
 of course, will depend upon the torsion, which is directly proportional to 
 the moment about the center of the shaft of the force producing the 
 torsion. Let P represent the force producing the torsion of the shaft AB 
 (Fig. 106) and let d be its lever arm. Then the moment of this force, 
 which produces the torsion, is Pd. This would be known as the "torque." 
 Now it is evident that this moment Pd, or torque, must be balanced at 
 
122 STRUCTURAL ENGINEERING 
 
 each cross-section of the shaft by the moments of the torsion stresses, at 
 each section, about the center of the shaft. 
 
 Let Fig. 107 represent an enlarged cross-section of the shaft AB 
 (Fig. 106). As the torsion stresses vary directly as the distance out from 
 the center of the shaft, the maximum stress will evidently occur at the 
 outermost elements. Let S be this stress per square inch, 
 and let r be the radius of the shaft. Then the stress per 
 square inch out unit distance will be S/r, and the stress per 
 square inch out any distance z will be (S/r)z. But as the 
 stress varies continuously outward from the center of the 
 shaft it is evident that there will not be a square inch of 
 Fig. 107 material out z distance having a stress of (S/r)z, but only 
 an infinitesimal area da of material having that stress. So 
 the actual stress or force out any distance z is (S/r)zda, and, of course, 
 the moment of this about the center of the shaft is (S/r)z 2 da. Now it is 
 evident that the stress is the same on all material equal distances out from 
 the center of the shaft, so da can be taken as a circular strip of width dz 
 and length 2irz. Then we have da = 2-rrzdz. Then the moment of the tor- 
 sion stress out z distance from the center of the shaft is 
 
 z 2 da = 27T z 3 dz, 
 
 and summing up this for the entire cross-section, we have 
 
 [l 
 
 Jo 
 
 (1) 
 
 Integrating, we have Pd = S7rr z /2, and transposing, 
 2Pd 
 
 from which the maximum torsion stress per square inch on any round 
 solid shaft, as the above, can be computed. S = stress per square inch; 
 Pd = torque ; and r = radius of the shaft. 
 
 In the case of a hollow shaft, the integration of (1) would be between 
 the limits of the internal and external radii. Thus, let r be the external 
 radius and r, the internal radius of a hollow shaft. Then we would have 
 
 T" 7 \ I Q 7 
 
 pd-2TT I z*dz. 
 
 Integrating, we have 
 
 Pd = ~ 
 Transposing, we have 
 
 (3), 
 
 Kt) 
 
RIVETS, PINS, ROLLERS, AND SHAFTING 123 
 
 from which the maximum torsion stress per square inch on any hollow 
 shaft can be computed. $ = stress per square inch ; r = external radius ; 
 r, - internal radius ; and Pd the torque. 
 
 In case of a shaft having a square cross-section, da 
 would be taken as any infinitesimal rectangular area. 
 For instance, let Fig. 108 represent the cross-section of a 
 square shaft, the center of which is at 0. Then, for the 
 moment of the torsion stress on an infinitesimal area da 
 out any distance z from 0, we have (S/e)z 2 da. Here *V 
 is the stress on the outermost element, which is distance Fig 108 
 
 e from 0. Let Pd be the torque, and we have 
 
 e 
 Expressing z in terms of the rectangular co-ordinates x and y, we have 
 
 Pd = - (2x 2 da + 2y 2 da). 
 
 But (^x~da + Hy 2 da) is the polar moment of inertia of the cross-section, 
 which is designated as J in Art. 49. Then we have 
 
 Transposing, we have 
 
 S=l> .......................................... (4), 
 
 J 
 
 from which the maximum torsion stress, not only on square shafting but on 
 any shaft whatever, can be computed. S - stress per square inch; Pd = 
 torque ; J = polar moment of inertia ; and e = the distance to the outermost 
 element. Formula (4) is a general formula, as it applies to any form of 
 cross-section. 
 
CHAPTER VII 
 
 MAXIMUM REACTIONS, SHEARS, AND BENDING MOMENTS 
 
 ON SIMPLE BEAMS AND TRUSSES, AND 
 
 STRESSES IN TRUSSES 
 
 86. Maximum Reactions on Simple Beams. The reactions on 
 simple beams in all cases can be determined by taking moments about the 
 supports as explained in Art. 54. In the case of uniform dead load, the 
 two reactions are equal and each is known directly as being equal to half 
 of the dead load supported. In case the dead load is not uniformly 
 distributed, the reactions are obtained by taking moments about the 
 supports, as stated above. In all cases of dead load, the reactions are 
 fixed in intensity, as the loads are fixed in position, but for live load the 
 case is different as the reactions vary with the position of the loads, and 
 the maximum reaction occurs when the loads are in one certain position. 
 If the live load be a uniform load, the reactions at the two supports will 
 be a maximum at the same time and that will be when the load extends 
 over the full length of the beam, in which case each reaction is known 
 directly as being equal to half of the load supported. The live load often 
 consists of wheel loads, as in the case of locomotives, trains, wagons, 
 traction engines, etc., where the loads or wheels are a fixed distance apart. 
 The maximum reaction due to such loading will occur, as is readily seen, 
 when the beam is fully loaded and the heaviest loads are as near as 
 possible to the support considered. 
 
 87. Maximum Shear on Simple Beams. The reactions on simple 
 
 beams, in all cases, can be determined by taking moments about the 
 
 L 
 
 Fig. 109 
 
 supports, as stated in the preceding article, and after the reactions are 
 known the shear at any vertical section of a beam is determined by simplv 
 adding up the forces on either side of the section, beginning at the end of 
 the beam, as explained in Art. 52. This much applies in all cases whether 
 
 124 
 
MAXIMUM REACTIONS, ETC. 125 
 
 the shear be a maximum or not. The maximum shear on any vertical 
 section of a beam, due to dead load, is readily obtained as such loads are 
 fixed in position and hence the shear obtained by adding up the forces 
 on either side of any section is the maximum for that section, but in the 
 case of live load the maximum shear is not so readily obtained for the 
 shear on every vertical section of a beam changes continually as such 
 loads move over the beam, and hence the maximum shear will occur when 
 the load is in one certain position. As an example of live load, let AB 
 (Fig. 109) represent a simple beam supporting a single moving load 
 P, and let R and Rl represent the reactions at A and B, respectively, due 
 to this load when at any point on the beam. 
 
 The shear on any short strip through the beam, as abed, due to P, 
 can be expressed as 
 
 when the load is at any point to the right of the strip. The shearing 
 force exerted upon the strip will then act as indicated. It is readily seen 
 that this shear on the strip varies directly as x and hence will be a 
 maximum when x m. Then if we lay off the vertical line gh equal (by 
 scale) to Pm/L and draw the line hk, any ordinate to this line hk graph- 
 ically represents the shear on the strip abed when the load is directly over 
 the ordinate. For example, the ordinate y represents the shear on the 
 strip when the load is in the position shown, and any other ordinate as 2 
 represents the shear on the strip when the load is over that ordinate. 
 When the load is at any point to the left of the strip, the shear on 
 the strip can be expressed as 
 
 but the shearing forces exerted upon the strip will then act in the oppo- 
 site direction to those shown. This change in the direction of action of 
 the shearing forces takes place the instant the load passes the strip. We 
 would say that the shear changes signs as the load passes the strip and 
 that the shear on the strip is positive when the load is on one side, and 
 negative when on the other. 
 
 It is seen, from the last equation, that AS" will be a maximum when 
 R has its least value. This, as is seen, occurs when the load is just to 
 the left of the strip, that is, when x = m + dx, but, as dx can be taken as 
 an infinitesimal, we would say when x m. Then if we lay off the vertical 
 line fg equal to (Pm/L) -P=(R-P) and draw the line fe we have the 
 shear on the strip graphically represented for the load P at any point to 
 the left of the strip, as any ordinate r, to the line fe, represents the shear 
 on the strip when the load is over that ordinate. Then we have the variation 
 of the shear on the strip abed due to the load P, as it moves over the beam 
 from B to A, fully represented by the ordinates to the broken line efhk. 
 It is seen from this that the loads on the two sides of any vertical section 
 of a beam really neutralize each other as regards the shear they produce 
 on the section, and hence it appears that the maximum shear would occur 
 when the loads are on one side only and extending on that side from the 
 
126 
 
 STRUCTURAL ENGINEERING 
 
 end of the beam up to the section. This is absolutely true in the case of a 
 uniform live load, as shown in Art. 55, and is practically true for prac- 
 tically all classes of loading, yet it is not absolutely true in some cases of 
 concentrated live loads, for sometimes the maximum shear on a vertical 
 section of a beam, due to such loads, occurs when the loading extends 
 from one end of the beam to a little beyond the section considered. The 
 most common instance of this is that of the typical system of locomotive 
 wheel loads. In such cases the exact position required to produce maxi- 
 mum shear must be determined by trial. 
 
 '*' 
 
 As a general case of uniform load, 
 let it be required to determine the maxi- 
 mum positive and negative shear on a 
 beam 12 ft. long at a point 4 ft. from 
 the end due to a uniform dead load of 
 Q 100 Ibs. per foot and a uniform live load 
 of 1,500 Ibs. per foot. Let AB (Fig. 
 Rl 110) represent the beam, and let ba 
 represent the section at which the shear 
 is required. 
 
 As the dead load is symmetrically 
 located in reference to the center of the 
 beam, the two reactions due to dead 
 load are equal to each other, and each 
 is equal to 6x100 = 600 Ibs. Then for 
 the dead-load shear at the section ba, adding up from A and taking R 
 (= 600) as positive, we have 
 
 600 -(4x100)= + 200 Ibs. 
 
 When the uniform live load extends from the end B up to the section, as 
 shown, one of the maximum live-load shears will occur. Taking moments 
 about B when the load is in this position, we have 
 
 (O) 
 
 Fig. 110 
 
 1,500x8x4 
 12 
 
 = + 4,000 Ibs. 
 
 for the live-load reaction at A. Then for the shear due to this load at the 
 section ba, adding up from A, we have +4,000 Ibs. as there is no inter- 
 vening live load. This is the maximum positive shear at the section ba 
 due to the live load. It will be readily seen that the dead load and the 
 live load, where the live load extends from B to the section ba, both exert 
 shearing forces at the section which act in the direction shown at (a), 
 hence the two will add, and we have 200 + 4,000 = + 4,200 Ibs. for the 
 maximum positive shear at the section ba, due to the dead and live load 
 combined. Now if the live load extends from A to the section ba instead 
 of from B, the reaction at A, due to the live load in that position, would be 
 
 R = 
 
 1,500x4x10 
 12 
 
 = 5,000 Ibs. 
 
 Then for the shear at the section ba due to this load, adding up from A, 
 we have 5,000- 6,000 = - 1,000 Ibs. which is the maximum negative shear 
 at the section due to the live load. 
 
MAXIMUM REACTIONS, ETC. 127 
 
 It will be readily seen that the live load when extending from A to 
 the section ba will exert shearing forces at the section which will act as 
 shown at (6), and as this direction is opposite to the direction of action 
 of the shearing forces exerted by the dead load, the shear at the section, 
 due to dead and live load combined, will undoubtedly be equal to the 
 difference of the two shears. So we have 200 -1,000 =-800 Ibs. for the 
 maximum negative shear at the section ba due to dead and live load 
 combined. 
 
 Now it is seen that in the above example the maximum positive and 
 negative shear at the section ba, due to dead and live load combined, is 
 simply the algebraic summation of the two simultaneous shears in each 
 case. This will hold in all cases provided the adding up of the forces be 
 in reference to the same end of the beam in each case. As an illustration, 
 let us add up the forces in the above example from the end B instead of 
 from A, as we did above. The reaction at B, due to the dead load, is 
 600 Ibs. Then for the dead-load shear at the section ab, adding up from 
 B, we have, 600- 8 x 100 = -200 Ibs. 
 
 For the live-load reaction at B, when the live load extends from B to 
 the section ba, we have 
 
 12 
 
 Then for the shear at the section, adding up from B, we have 8,000- 
 12,000 = -4,000 Ibs. Adding this to the dead-load shear, we have 
 -200 -4,000 = -4,200 Ibs., which is the same as we obtained by adding 
 up from A, except the sign is negative instead of positive. If the live 
 load extends from A to the section ba instead of from B to the section, 
 we have 
 
 1,.500X4 X2 
 
 for the reaction at B due to the live load. Then for the shear at the 
 section ba, adding up from B, we have + 1,000 Ibs. Now adding this to 
 the dead load, as just determined by adding up the forces from B, we 
 have -200 + 1,000 = + 800 Ibs., which is the same as we obtained above 
 by adding up from A, except that the sign is positive instead of negative. 
 Thus we see that by adding up the forces from one end we obtain 
 positive shear and by adding up the forces from the other end we obtain 
 negative shear. There is no mystery about this at all, as is readily 
 perceived by referring to Fig. Ill where AB represents a simple beam 
 supporting a number of loads, and R and Rl represent the reaction at 
 A and B, respectively, due to the loads, and abed represents a very short 
 strip through the beam. If we take the reactions as positive, then any 
 force acting upward will be positive. By adding up the forces from A to 
 the strip abed we really obtain the shearing force acting upon the strip 
 along ab and by adding up the forces from B to the strip we really obtaiff 
 the shearing force acting upon the strip along dc. As these two shearing 
 forces on any vertical section of a beam are always equal and opposite, it 
 is evident that the sum obtained by adding up the forces from one end 
 will be positive, while the equal sum obtained by adding up the forces 
 from the other end will be negative. 
 
STRUCTURAL ENGINEERING 
 
 There will be no confusion as regards the sign of the shear on any 
 se/'tion of a beam in any case, providing the adding up of the forces, that 
 is. the algebraic summation, be started from the same support. 
 
 i lad I I 1 
 
 R 
 
 Rl 
 
 B 
 
 Fig. Ill 
 
 88. Maximum Bending Moments on Simple Beams. The bend- 
 ing moment at any cross-section of a beam, in all cases, is equal to the 
 algebraic sum of the moments about the section, of the forces on either 
 side of the section. In case the load be uniformly distributed over the 
 entire length of the beam, the maximum moment will occur at the center 
 of the span, and will be equal to wL 2 /8 as shown in Art. 56. This is true 
 for all uniformly distributed loads, either live or dead. In the case of 
 fixed concentrated loads, the maximum moment will occur under one of 
 the loads which can be ascertained readily by trial. However, it can be 
 determined otherwise as shown later. 
 
 If the live load be wheel loads, it is always necessary to first deter- 
 mine the position of the loads on the beam for maximum moment. As 
 wheel loads roll over a beam it is evident that the moment at every section 
 is continually changing, and it is the absolute maximum of these various 
 moments produced that we desire, although it may last only for an instant 
 during the passage of the load. 
 
 Let AB (Fig. 112) represent 
 a simple beam supporting a system 
 of wheel loads. According to Art. 
 56, the maximum moment will occur 
 under a wheel. Let this wheel be 
 the one marked C. Let W be the 
 weight of all the wheels on the 
 beam, and x the distance from the 
 right support to their center of 
 gravity. Further, let P be the 
 
 weight of the wheels to the left of wheel C, and b the distance of their 
 center of gravity from wheel C, and let 2 be the distance from the left 
 support to wheel C. Then the bending moment under wheel C can be 
 expressed as 
 
 Fig. 112 
 
 Now from Fig. 112 it is seen that x = L-a-z. Substituting this 
 v&me of ,r in (1), we have 
 
 za-^)-Pb (2). 
 
 ZJ 
 
 Now any slight movement of tlie loads either to the right or left will 
 cause a small change in 2 and M. Let dz and dM represent this incre- 
 
MAXIMUM REACTIONS, ETC. 129 
 
 ment of z and M, respectively, due to a slight movement of the loads. 
 Adding the increments in equation (2), we have 
 
 W 
 
 ~[ 
 Li 
 
 and reducing we have 
 
 W W 
 
 M + dM = ~ [^L - az - z 2 ] + -^ [ dzL - adz - 2zdz] - Pb. 
 Li Li 
 
 Now subtracting (2) from this last equation and reducing we obtain 
 
 W W 
 
 dM=Wdz--^- adz- 2zdz .......................... (3) 
 
 Li Li 
 
 which is undoubtedly the expression for the increment of the bending 
 moment under wheel C due to any slight movement of the loads. It is 
 readily seen that if wheel C is to the right of the point of maximum 
 moment any slight movement of the loads to the left will increase the 
 moment M by the amount dM in other words, dM will be positive and 
 that all such movements of the loads to the left will increase M the amount 
 dM until wheel C arrives at the point of maximum moment and from there 
 on any slight movement of the loads to the left will decrease M by the 
 amount dM; in other words, dM is negative in that case. Evidently the 
 increment dM is zero just as wheel C arrives at the point of maximum 
 moment as it is positive when the wheel is just to the right of the point 
 of maximum moment and negative when to the left and hence changes 
 signs at the point. So we have from (3), when M is a maximum, 
 
 dM = Wdz-~ adz-~2zdz = 0. 
 Li Li 
 
 Reducing, we obtain 
 
 which is the value of z when the maximum bending moment occurs under 
 wheel C. This last equation shows that the center of the beam bisects 
 the distance a. From this we see, that the point of maximum moment and 
 the center of gravity of all the wheels are equidistant from the center of 
 the beam and on opposite sides of the center. The maximum moment will 
 occur under one of the two wheels adjacent to the center of gravity, 
 practically always under the wheel nearer the center of gravity. So, as a 
 rule, all we have to do to locate a system of wheels on a beam for 
 maximum moment is to find their center of gravity and then place the 
 center of the beam half way between this center of gravity and the wheel 
 nearest to it. For example, take the case shown in Fig. 112. By taking 
 moments about either of the end wheels, either wheel D or E, we can 
 determine the center of gravity of all the wheels which comes nearest to 
 wheel C and hence the center of the beam will come half way between 
 the center of gravity and this wheel. In any case where the center of 
 gravity comes near half way between two wheels, the moment under each 
 of the wheels should be determined in order to ascertain which is the 
 maximum. 
 
130 
 
 STRUCTURAL ENGINEERING 
 
 Usually the only bending moment used in designing beams is the 
 absolute maximum referred to above, yet there are a few cases where the 
 maximum at other points, other than the point where the absolute maxi- 
 mum occurs, is needed, and which will occur when the wheels are in one 
 certain position in reference to the point considered. 
 
 Let AB (Fig. 113) represent a simple 
 beam supporting a system of wheel loads 
 as shown, and let D be a point distance e 
 from A where the maximum bending mo- 
 ment, due to these wheels as they move 
 over the beam, is desired. Let W repre- 
 ss sent the weight of all of the wheels on the 
 beam and x the distance of their center of 
 gravity from B. Further, let P represent 
 the weight of the wheels to the left of 
 point D and 2 the distance of their center 
 of gravity from D. 
 
 According to Art. 56, the maximum moment at D will occur when 
 a wheel is at that point. So let a wheel be at the point as shown. Now 
 taking moments about D and considering the forces to the left, we have 
 
 W 
 
 e 
 
 L-e 
 
 
 
 
 z 
 
 \ 
 
 oc 
 
 ( 
 
 OOO 
 
 
 
 
 
 If 
 
 R 
 
 'P 
 
 Fig 
 
 u 
 \ 
 r, i 
 
 / 
 
 V 
 
 13 
 
 JRI 
 
 = ex-Pz=(Re-Pz) 
 
 for the bending moment at that point. Now, if the wheels are in the 
 position for maximum moment at D, the increment of the moment that 
 would be caused by the wheels moving to the right or left an infinitesimal 
 distance would be equal to zero. Then assuming we have such a move- 
 ment, M, x, and s will have corresponding increments which we will 
 designate, respectively, as dM, dx, and dz. Now adding these in equation 
 (1), we have 
 
 W W 
 
 and subtracting equation (1), we have 
 
 But, as is readily seen, dz dx, so we have 
 
 for the increment of the bending moment, which will be equal to zero 
 when the loads are in the position for maximum moment at D. Then we 
 have 
 
 Reducing, we have 
 
 W P 
 ~L = 7 
 
 (2), 
 
MAXIMUM REACTIONS, ETC. 
 
 131 
 
 In the last equation we have the total load on the beam divided by 
 the length of the beam equal to the load on the left of the point D 
 divided by the distance from the point to the left support. That is, we 
 have the average unit load on the beam equal to the average unit load to 
 the left of the point. Then, to obtain the maximum moment at any point 
 in a simple beam, due to wheel loads, the loads must be so placed that 
 there will be a load at the point considered and at the same time the 
 average unit load on the entire beam will be equal to the average unit 
 load to the left of the point. 
 
 It is shown, in Art. 59, that the first derivative of the bending 
 moment, in terms of x, at any section of a beam, is equal to the shear. 
 But at the point of maximum moment the first derivative of the bending 
 moment will be equal to zero. So, it follows that the shear at the point 
 of maximum bending moment will be zero. Then, as the point of maxi- 
 mum bending moment is at the point of zero shear, or where the shear 
 changes signs, which is the same thing, the point of maximum moment 
 can be obtained by adding up the forces from one end of the beam until 
 the point where the shear is zero, or changes signs, is reached, and this 
 point will be the point of maximum bending moment. This method of 
 determining the point of maximum moment is quite convenient in the case 
 of fixed loads, especially where the loads are mixed, uniform, and 
 concentrated. 
 
 As an example, let it be required to determine the point of maximum 
 bending moment on the beam shown in Fig. 114, due to the concentrated 
 and uniform load indicated. After determining the reactions shown at 
 A and B, by taking moments about the supports, we can begin at one end 
 of the beam and ascertain the point of zero shear by adding up the forces 
 
 3' ._.*' . S' ,.. Z' . ?- 
 
 * T x 
 
 B 
 
 Fig. 114 
 
 and noting the sign of the shear as we progress. Thus, beginning at A, 
 we have 10,766- 1,800 -3,000 = +5,966 Ibs. for the shear just to the right 
 of the 3,000-pound load, and 10,766-3,000 - 7,000 = +766 Ibs. for the 
 shear just to the right of the 4,000-pound load. This last shear shows us 
 that the point of zero shear is just a little to the right of the 4,000-pound 
 load. Let x be the distance. Then we have 
 
 from which we obtain 
 
 600 x = 766, 
 766 
 
 600 
 
 = 1.27 ft. 
 
132 
 
 STRUCTURAL ENGINEERING 
 
 So the point of maximum moment is 1.27 ft. to the right of the 4,000- 
 pound load. If the 4,000-pound load were something over 766 pounds 
 heavier, it is evident that the point of maximum moment would be under 
 that load. For example, suppose that load weighed 6,000 pounds instead 
 of 4,000 ; then, all the other loads remaining the same, the reaction at A 
 would be 11,932 Ibs. instead of 10,766. Adding up from A we have 
 11,932 -3,000 -3,000 =+5,932 Ibs. for the shear just to the left of the 
 second load from A, which is now 6,000 Ibs., and 11,932-3,000-3,000- 
 6,000 = -68 Ibs. for the shear just to the right of the load. As the shear 
 is positive on one side of the load and negative on the other, the shear 
 undoubtedly passes through zero at the load, and hence the point of 
 maximum moment is at the load. 
 
 89. Maximum Reaction on Simple Trusses. The loads on trusses 
 are transmitted to the joints, either directly or indirectly; directly when 
 applied at the joints, and indirectly when applied between the joints, 
 that is, in the panels. As an example, let the diagram in Fig. 115 
 
 Fig. 115 
 
 represent a truss of length L supporting two loads, PI and P2, as shown. 
 These loads, as is readily seen, would be transmitted to the joints c, d, 
 and e. Let rl, r2, and r3 represent the amount transmitted to each as 
 indicated. These would be known as concentrations or panel loads. The 
 intensity of rl and r3 can be determined by taking moments about d and 
 treating cd and de as simple beams. Thus we have 
 
 hPl kP2 
 
 rl = 7- and r6 = - 
 cd de 
 
 By taking moments about e we can determine the concentration at d due 
 to P2, and by taking moments about c we can determine the concentration 
 at d due to PI. Then, by adding these two concentrations together, we 
 obtain the concentration r2. Thus we have 
 
 .9- 
 
 cd-h 
 
 P1 + 
 
 de-k 
 de 
 
 \P2. 
 
 In this manner the concentrations on the joints of any truss, loaded in 
 any manner, can be determined. As the concentrations are really com- 
 ponents of the loads, either may be used in determining reactions on 
 trusses. Thus, for the reaction at A (Fig. 115), due to the two loads 
 PI and P2, we have 
 
 H 
 
 ePl + yP2 
 
 or Li = 
 
 (Bd)r 
 
MAXIMUM REACTIONS, ETC. 133 
 
 As a rule, the loads are used instead of the concentrations (owing to 
 convenience), in which case any concentration at the end point due to 
 loads in the end panel must always be subtracted from the reaction found. 
 For instance, if there were loads in the panel Ab, we would take moments 
 about b and determine the concentration at A due to the loads in that 
 panel, and subtract this from the reaction found at A by taking moments 
 of all the loads about B. Other than this one step, the determination of 
 reactions on trusses is exactly the same as for beams. 
 
 Dead load (which is usually taken as a uniform load) and uniform 
 live load are always considered as concentrated at the joints. In such 
 cases we deal only with joint loads, or panel loads, which is the same 
 thing. If the panels be of equal length, the panel loads will all be equal, 
 and in such cases the reactions can be determined practically by inspec- 
 tion. For example, let the diagram in Fig. 116 represent a truss of 6 
 
 equal panels supporting a uniform load of w pounds per foot of truss. 
 Let L be the length of the truss and d the length of each panel. If the 
 load extends over the full length of the span there would be five panel 
 loads, represented as P, each equal to wd. In that case we can readily 
 see that each reaction would be equal to 2J P. This, of course, is mere 
 inspection. In case the panel loads are unequal or some of the joints 
 not loaded, we can obtain the reactions by simply taking moments about 
 the supports just the same as in the case of beams, except as stated above, 
 but if the panels are of equal length it is more convenient to deal with 
 panel lengths instead of feet. For example, to obtain the reaction at A 
 due to a load at /, we would take moments about B and the load at / 
 would be multiplied by d and divided by 6d, and hence the reaction at 
 A would be J of the load. Similarly a load at e would be multiplied by 
 2d and divided by Gd, and hence the reaction at A, due to this load, 
 would be f of the load at e, and likewise the reaction at A, due to a load 
 at d, would be of the load at d, and and f for a load at c and b, 
 respectively. From this it is seen that the reactions on trusses of equal 
 panels can be obtained directly by numbering the panels 1, 2, 3, etc., as 
 shown (just below the diagram) in Fig. 116, and multiplying the load on 
 each j oint by the number under it, and then adding all the results together 
 and dividing by the number of panels in the truss. If the reaction at the 
 other support be desired, the numbers would simply be reversed in order. 
 If the panel loads are all equal, the work will be shortened, of course, as 
 the numbers under all the joints loaded can be added together and multi- 
 plied by one panel load, divided by the number of panels. As an example, 
 suppose the joints /, e, d, and c each support a load P. The reaction at 
 A due to these four loads would be 
 
 ?10. 
 6 
 
134 STRUCTURAL ENGINEERING 
 
 If the joint b were loaded also, we would have 
 
 If joints e, c, and b alone were loaded, we would have 
 
 P P 
 
 ^ = -(2 + 4 + 5)=- 11. 
 
 If joint c alone were loaded, we would have 
 
 P P 
 
 and so on. In general, the maximum reaction on a truss, the same as a 
 beam, will occur when the span is fully loaded and the heaviest loads 
 are near the support considered. Specific cases will be taken up later. 
 90. Maximum Shear on Simple Trusses. The determination of 
 the shear on trusses is the same as that of beams, except the loads are con- 
 sidered to be applied to trusses only at the panel points, or joints. Let 
 the diagram in Fig. 117 represent a simple truss supporting a load P at 
 each bottom joint, as indicated, and let R and Rl represent the reactions 
 due to these loads. Then the shear in the first panel from the left end is 
 Rj in the second it is R-P; in the third it is R-2P; in the fourth 
 it is R - 3P; and so on. This case is that of dead load or a uniform live 
 
 load extending all the way across the span. In case of dead load, this 
 shear would be a maximum. It is customary to assume that a uniform 
 live load moving over a truss increases by panel loads, in which case the 
 maximum shear in any panel will occur when the joints from one end of 
 the truss up to the panel considered are loaded. By loading from one 
 end we obtain, as we may say, the maximum positive shear, and loading 
 from the other end we obtain the maximum negative shear. For example, 
 one maximum shear would occur in the panel be of the truss shown in 
 Fig. 118, when the joints e, d, and c alone are loaded, and the other 
 maximum would occur when the joints a and b alone are loaded. Instead 
 of considering the load to move on from the two directions, we usually 
 consider it to move on from only one direction, and by determining the 
 shear in each panel as the load reaches it we obtain all the shears we 
 need, as the positive shears are determined in the panels on one side of 
 the center of the truss while the negative shears are determined in the 
 corresponding panels on the other side of the center. 
 
 As the maximum shear in any panel, due to a uniform live load, 
 occurs when the load just reaches that panel, it is readily seen that this 
 shear, in all cases, will be equal to the reaction at the unloaded end of the 
 truss. Thus, the maximum shear in panel ed (Fig. 118) due to a uniform 
 live load moving on from the right, will be equal to the reaction at A due 
 
MAXIMUM REACTIONS, ETC. 
 
 135 
 
 to the load at e; and, similarly, the maximum shear in the panel dc will 
 be equal to the reaction at A due to the loads at e and d; and the maxi- 
 mum shear in panel cb will be equal to the reaction at A due to the loads 
 at e, d, and c; and so on for the other panels. Now if the panels be of 
 equal length, these reactions can be determined by numbering the joints 
 1, 2, 3, etc., and adding these together for the loaded joints and multiply- 
 ing the sum by one panel load divided by the total number of panels in 
 the truss, as explained in the last article. As an example, let the diagram 
 in Fig. 118 represent a truss of six equal panels, each of length d (the 
 
 Fig; 118 
 
 length of the span then will be L Qd}, and let it be required to deter- 
 mine the maximum shear in each panel due to a uniform live load of w 
 pounds per foot of truss moving over it from right to left. Let W (=wd) 
 represent the panel load. Then numbering the joints 1, 2, 3, etc., from 
 right to left, as shown, we have 
 
 for the maximum shear in the panel ed, when joint e alone is loaded, 
 
 for panel dc when joints e and d are loaded, 
 
 for panel cb when joints e, d, and c are loaded, 
 
 W 
 
 6 
 
 + 2 + 3 + 4) 
 
 for panel ba when joints e, d, c, and b are loaded, and 
 
 for panel aA when joints e, d, c, b, and a are loaded. 
 
 From this it is seen that the maximum shear on any truss of equal 
 panels, due to a uniform live load moving over the truss, can be expressed 
 by the general formula, 
 
 +O-1)], 
 
 where n is the number of panels in the truss considered. 
 
 It will be found quite convenient to write the summation of the 
 
136 
 
 STRUCTURAL ENGINEERING 
 
 numbers in the parentheses under the corresponding panel points as 
 shown in Fig. 118 where the incircled numbers are the respective summa- 
 tions. These summations, as is readily seen, can be made in any case by 
 simply referring directly to the diagram of the truss considered. Then 
 the shear in the panels can be read off of a slide rule directly as, 
 
 W 
 
 5(3), E 
 
 ^ TO 
 
 
 
 and so on. 
 
 In the case of uniform live load, just considered, we usually assume, 
 as stated above, that the load is applied to the joints of a truss in maxi- 
 mum panel loads. This is not possible, however, as such loads are con- 
 tinuous and must necessarily extend beyond the last loaded joint in order 
 to fully load it. For example, to fully load the joints e and d (Fig. 118) 
 the uniform live load would really have to extend from B to the joint c, 
 as is readily seen. However, as will be shown later, the maximum shear 
 would occur in panel dc when the load extended a short distance z beyond 
 joint d. In that case there would be less than a full panel load at d and 
 a small concentration at c due to the load in the panel dc, and the shear 
 in the panel dc would really be equal to the reaction at A minus the 
 concentration at c. The error resulting from the assumption stated above 
 is small in the case of a uniform live load, as will be seen later, and hence 
 the assumption in that case is permissible, but in the case of wheel loads 
 no such assumption can be made as any slight shifting of the loads may 
 materially change the shear in a panel and consequently it is necessary 
 for us to determine the exact position of such loads for maximum shear. 
 As a general case, let the diagram in Fig. 119 represent a truss 
 
 ) O O\ A o o r /h O O O O 
 
 Fig. 119 
 
 supporting a system of wheel loads. Now let it be required to determine 
 the position of these wheels for maximum shear in panel be. Let W be 
 the weight of all the wheels and x the distance from B to their center of 
 gravity, and let P be the weight of the loads in the panel be, and z the 
 distance from c to their center of gravity. Now if R be the reaction at A 
 due to W, and r the concentration at b due to P, the shear in the panel 
 be can be expressed as 
 
MAXIMUM REACTIONS, ETC. 137 
 
 = R-r ........................................... (1). 
 
 W r 
 
 But 
 
 L a 
 
 Then substituting these values, we have 
 
 With P and W constant, any shifting of the wheels to the right or left 
 will affect R and r, not the same, but similarly. The weight of P could 
 be such that any movement to the left would increase the value of R 
 more than r. Then, evidently, any such movements to the left would 
 continue to increase the shear S until another load rolled past c into the 
 panel be, when the rate of increase of r would be greater than just before 
 the load passed joint c; but R would continue to increase all the while, 
 and, consequently, the shear S would continue to increase, unless the load 
 P were increased so much by the additional load that r would be increased 
 at a greater rate than R for each movement to the left. So it is seen that 
 the shear S will be a maximum when the total load in the panel be is such 
 that the increment of R is equal to the increment of r. That is, any 
 shifting of the loads with W and P constant (for that instant) would not 
 affect the shear S in panel be; or, in other words, the increment of the 
 shear would be zero. Now, suppose the loads move an infinitesimal 
 distance to the left; then equation (2) would become 
 
 Wdx Pz Pdx 
 
 and subtracting (2) from this, we have 
 
 which is the increment of the shear S in panel be due to the movement. 
 But this increment will be zero when S is a maximum, and then we 
 would have 
 
 It is readily seen that this occurs when W/L = P/d, that is, the shear in 
 panel be is a maximum when the average unit load on the truss is equal to 
 the average unit load in the panel. This will hold in the case of any truss 
 of the type shown as the case being considered is general. In the case of 
 a truss of n equal panels, each of length d, we have L = nd. Substituting 
 this value of L in the last equation, and reducing, we have 
 
 That is, the shear in any panel of a truss of n equal panels will be a 
 maximum when the load in the panel is equal to the total load on the 
 truss divided by the number of panels in the truss. This will hold for 
 any class of live load either concentrated or uniform. 
 
 It is readily seen that the increment of the shear in panel be, 
 expressed by equation (3), will pass through zero only as a load passes 
 
138 
 
 STRUCTURAL ENGINEERING 
 
 joint c, so there will be a wheel at joint c when the maximum shear in 
 panel be occurs. Thus it is in all cases a wheel will be at the joint on 
 the loaded side of the panel considered when the maximum shear in the 
 panel occurs. 
 
 91. Maximum Bending Moments on Simple Trusses. In the 
 case of bending moments on trusses, the moments are usually taken about 
 the joints only, otherwise the case is the same as that of beams. As a gen- 
 eral case of dead load or uniform live load, let the diagram in Fig. 120 
 represent a truss of six panels supporting a load at each bottom joint as 
 
 q . 
 
 
 i 
 h \ 
 
 
 \ 
 
 \ 
 
 \k 
 
 PI 
 
 c \ a 
 
 p^ PS 
 
 Fig. 120 
 
 indicated. Let R and Rl represent the reaction at A and B, respectively, 
 due to these loads. Then the bending moment at joint b or g is M' = Rx; 
 at joint c or h it is M"=R(x + y)-Ply ; at joint k or d it is M" f - 
 R(x + y + z)-Pl(y + z)-P2(z); and so on. 
 
 In the case of uniform live load, the maximum moment at any joint 
 will occur when the load extends the full length of the truss, and hence 
 the moment at all joints, the same as in the case of dead load, is a 
 maximum. In the case of wheel loads it is necessary to determine the 
 exact position of the wheels for the maximum bending moment at each 
 joint. Let the diagram in Fig. 121 represent a truss of six panels 
 supporting a system of wheel loads, as indicated. Now let it be required 
 to determine the position of these wheels for maximum bending moment 
 at joint d. 
 
 Let W be the weight of all the wheels on the truss, and x the 
 distance from B to their center of gravity, and let P be the weight of the 
 
 Fig. 121 
 
 wheels to the left of joint d and z the distance from d to their center of 
 gravity, and also let k be the distance from A to d and R the reaction 
 at A due to all the wheels when in any position. Then the bending 
 moment at d can be expressed as 
 
 M = Rk-Px ..,, (1). 
 
MAXIMUM REACTIONS, ETC. 139 
 
 But | B=*i - 
 
 LJ 
 
 so, substituting this value of R, we have 
 
 W r 
 
 M = k~-Pz ..................................... (2). 
 
 Now as the loads move to the left,, R will be increased, and likewise 
 the moment M, until P becomes so great that the rate of increase of Pz 
 is greater than the rate of increase of (Wx / L)k; then, evidently, the 
 moment M at d will be a maximum when its increment is zero, which, as is 
 readily seen, will occur when the increment of (Wx / L)k is equal to the 
 increment of Pz. Now suppose the loads move an infinitesimal distance to 
 the left, then equation (2) becomes 
 
 Then subtracting (2) we have 
 
 which is the increment of the bending moment at d due to the movement. 
 But this increment will be equal to zero when M is a maximum, and hence 
 we would have 
 
 (4) 
 
 It is readily seen that this occurs when 
 
 that is, the bending moment at joint d is a maximum when the average 
 unit load on the truss is equal to the average unit load on the left of the 
 joint d. Now, as the above is a general case, we have: The maximum 
 bending at any joint of a truss of the type shown will occur when the 
 average unit load on the truss is equal to the average unit load to the left 
 of the joint considered. It will be seen that this is the same as found for 
 beams in Art. 88. 
 
 92. Stresses in Trusses. As the loads supported by trusses are 
 applied at the joints, the stress produced in each member is simple 
 stress, and consequently the unit stress in any member is equal to the 
 total stress divided by the area of its cross section. The stresses in the 
 individual members can readily be obtained by resorting to the resolution 
 of forces and to the equation of moments. 
 
 As an example, let the diagram at (a) in Fig. 122 represent a truss 
 of six equal panels, supporting a load at each lower joint as indicated. 
 Let d be the length of each panel, h the height of the truss center to 
 center of chords, L the total length of span center to center of end 
 bearings, and let R and Rl represent the reactions as indicated. To 
 
140 
 
 STRUCTURAL ENGINEERING 
 
 obtain the stress in the end post LOU1 and in the bottom chord LOL1, 
 let S and SI be the stress in each, respectively, and suppose the two 
 members cut off along the section mm, and imagine the part of the truss 
 to the left of this section moved bodily to (6) without any stresses or 
 forces being affected in the least. Then we simply have an independent 
 
 m"' 
 
 Fig. 122 
 
 structure at (&) held in equilibrium by the three external forces R, S, 
 and SI the stresses S and $1 being unknown. Now, as these three 
 forces are in equilibrium, the algebraic sum of their components along 
 any direction is equal to zero. So resolving the forces vertically, we have 
 
 ....................................... (1), 
 
 from which we obtain 
 
 (2) ; 
 and resolving the forces horizontally, we have 
 
 SI -Ss'm6 = ...................................... (3), 
 
 from which we obtain 
 
 SI = SsinO ......................................... (4). 
 
MAXIMUM REACTIONS, ETC. 141 
 
 But, according to (2), S = R/cos6. Then, substituting in (4), we have 
 
 S1=R -^ = Rtan6 ................................ (5). 
 
 cos<9 
 
 Equation (2) shows that the stress S in LOUl is equal to the shear 
 (=#) in the end panel multiplied by the secant of the angle 6 which the 
 member makes with the vertical; and equation (5) shows that the stress 
 SI in LOZ/1 is equal to the shear in the end panel multiplied by the 
 tangent of angle 6. It should be observed that R has no horizontal com- 
 ponent while $1 has no vertical the cosine of 90 Q being zero. 
 
 Again referring to the figure at (b), by taking moments about any 
 point O on the line of action of S, we have 
 
 Rx 
 from which we obtain 
 
 But it is obvious that it is more convenient to take moments about the 
 panel point 71, for in that case the lever arms are directly known, and 
 we have 
 
 Rd-hSl = 0, 
 from which we obtain 
 
 (6). 
 
 The stress S could be computed by taking moments about a point in the 
 line of action of the stress 1, but it is obvious that S can be computed 
 more readily from equation (2). 
 
 It is also obvious that the stress in the hanger U1LI is equal to the 
 load P which it supports directly at LI. 
 
 To obtain the stress in the chords 71 72, L1L2, and diagonal 71L2, 
 let S2, S3, and $4 be the stress in each, respectively, and suppose the 
 three members cut off along the section m'm' , and imagine the part of 
 the truss to the left of this section moved bodily to (c) without any 
 stresses or forces being affected in the least. Then we have an inde- 
 pendent structure at (c) held in equilibrium by the external forces R, P, 
 S2, S3, and 4 the stresses S2, S3, and S4 being unknown. 
 
 As S2 and 4 intersect at 71, S3 can be determined by taking 
 moments about that point. So taking 71 as the center of moments 
 we have 
 
 Rd-hS3 = 0, 
 from which we obtain 
 
 (7), 
 
 which is the same as we obtained for the stress in LOL1 in equation (6). 
 As 4 and S3 intersect at L2, S2 can be determined by taking moments 
 about L2. Therefore, taking L2 as the center of moments, we have 
 
 2dR-dP-hS2 = Q 
 
142 STRUCTURAL ENGINEERING 
 
 from which we obtain 
 
 S2 = 2^R~^P (8). 
 
 4, the stress in the diagonal 71L2, can be determined most readily 
 from the algebraic sum of the vertical components of the five forces, for 
 which we have 
 
 from which we obtain 
 
 (9)- 
 
 This shows that the stress $4 in the diagonal U1L2 is equal to the shear 
 in the panel L1L2 multiplied by the secant of angle 6. 
 
 To obtain the stress in the vertical post U2L2, let 6 be the stress, 
 and suppose the three members 71172, U2L2, and L2L3 be cut off along 
 the section m"m" , and imagine the part of the truss to the left of this 
 section moved bodily to (e) without any forces or stresses being affected 
 in the least. Then we have an independent structure at (e) held in 
 equilibrium by six external forces #, 7, 6, S5, P, and PI. The 
 stresses S7, S6, and S5 may all be unknown. As *S7 and S5 have no 
 vertical components (being horizontal members), S6 can be determined 
 directly from the algebraic sum of the vertical components of all the 
 forces, for which we have 
 
 R-2P-S6 = 0, 
 from which we obtain 
 
 S6 = R-2P ....................................... (10). 
 
 This shows that 6 is equal to the shear in the panel L2L3. 
 
 To obtain the stress in the chords 72 73, L2L3, and the diagonal 
 72L3, let $10, 8, and 9 be the stress in each, respectively, and sup- 
 pose the three members cut off along the section m" 'm ff ', and imagine 
 the part of the truss to the left of this section moved bodily to (/) with- 
 out any forces or stresses being affected in the least. Then we have an 
 independent structure at (/) held in equilibrium by six forces, R, *S10, 
 S9, SS, P, and PI, the stresses S10, 59, and S8 being unknown. Taking 
 moments about 72, we have 
 
 2dR-dP-hS8 = Q, 
 from which we obtain 
 
 Then taking moments about L3, we have 
 
 3dR - 2dP - dPl - hSIQ = 0, 
 from which we obtain 
 
 (11). 
 
 (12). 
 
MAXIMUM REACTIONS, ETC. 143 
 
 Resolving the forces vertically, we have 
 
 R-P-Pl-S9cos6 = 
 from which we obtain 
 
 COSC7 
 
 =sec0(J2-P-Pl) .................... (13). 
 
 The above loads do not produce any stress in the post U3L3. This 
 member could be omitted as far as loads applied at the lower panel points 
 are concerned, but it will support directly any load applied at [73. 
 
 Stress in the members to the right of C73L3 can be determined in 
 the same manner as shown above for the members to the left. 
 
 The above shows how readily the stresses in a truss can be deter- 
 mined by considering portions of the truss as independent structures, the 
 portion being selected each time to suit the case considered; the portion 
 may be a single joint or several panels of a structure. This manner of 
 treatment is known as the "Method of Section." It will apply in the case 
 of any truss and may be carried to any desired extent, but after one 
 becomes familiar with a certain type of truss, the analysis is often made 
 without any thought of the method of section, as will be seen later. 
 However, the method is really implied. 
 
 Problem 1. Determine the stresses in the members of the truss 
 shown at (a), Fig. 120, due to a load of 20,000 Ibs., on each lower panel 
 point, assuming the length of each panel to be 25'-0", and the height of 
 the truss 30'-0", center to center of chords. 
 
CHAPTER VIII 
 
 GRAPHIC STATICS 
 
 93. Definition and Limitation of Graphic Statics. Graphic 
 
 Statics is that part of Mechanics which treats of the graphical determina- 
 tion of static forces. It results from the fact that forces acting upon a 
 body in equilibrium will form a closed polygon (see Art. 42) wherein all 
 the forces act in the same direction around the polygon. This knowledge 
 applied to geometrical construction constitutes the method. 
 
 A force is fully known when its point of application, its intensity, 
 the direction and location of its line of action, and its direction of action 
 (along the line of action) are known. 
 
 The determination of the point of application does not come within 
 the scope of Graphic Statics, for it is one of the many things that can be 
 ascertained only from knowledge of actual conditions, and consequently 
 is a presupposed knowledge in graphic problems. The intensity, direc- 
 tion, and location of the line of action and the direction of action can be 
 graphically determined. However, the method is practically limited to 
 the two following cases: 
 
 Case I. The intensity and direction of action of two forces can be 
 graphically determined provided all of the other forces acting upon the 
 same body are fully known and the lines of action of the two are known. 
 This will hold in the case of one unknown force as well as for two. 
 
 Case II. The intensity, direction of action, and the direction and 
 the location of the line of action of one force can be graphically deter- 
 mined provided all of the other forces acting upon the same body are 
 fully known. 
 
 Concisely expressed, the intensity and direction of action of two 
 forces can be found when the lines of action of all the forces are known 
 and the intensity and direction of action of all except those two are 
 known ; and the intensity, direction of action, and the direction and 
 location of the line of action of one force can be found when all the other 
 forces are known. As a rule, the intensity and direction of action are all 
 that are desired. 
 
 94. Preliminary Application. Let AB, at (a), Fig. 123, represent 
 a body in equilibrium under the action of the five forces PI P5. Sup- 
 pose all of the forces are known except P4 and P5, which are unknown 
 in intensity and direction of action only, which means their lines of action 
 are known, so that the problem comes under Case I. 
 
 To determine their intensity and direction of action, select some 
 convenient scale and lay off AB at (6), equal and parallel to PI; and 
 from B lay off BC equal and parallel to P2 ; and from C lay off CD equal 
 and parallel to P3. Then, to complete the polygon, P4 and P5 must 
 close on D and A. So from D draw Dy parallel to P4 and from A draw 
 A z parallel to P5, intersecting Dy at E. Then the length of the lines 
 DE and AE represents the intensity of P4 and P5, respectively. Now, as 
 
 144 
 
GRAPHIC STATICS 
 
 all of the forces will act in the same direction around the polygon. P4 
 will act in the direction from D to E and P5 from E to A, as the arrow 
 points indicate. So we have their direction of action, which is toward 
 the body in each case. Then the forces P4 and P5 are fully known. The 
 two forces P4 and P5 could close on A and D, as AO and DO, just as 
 well as DE and AE. Either is correct. 
 
 It is usually convenient to go 
 around a body laying off the forces 
 in consecutive order as was done at 
 (6), but in some cases it is not pos- 
 sible to do so, and in fact it is not 
 at all necessary. For example, in 
 the above case we could lay off the 
 forces as they are shown at (c) by 
 laying off PI first, then P3, then P2, 
 and close on K and H with P4 and 
 P5 ; or we could lay off either P2 or y 
 P3 first, just so the forces are joined / 
 so that they act in the same direc- * 
 tion around the polygon, which is the 
 important thing to watch. However, 
 any mistake in that respect is readily 
 detected. For instance, suppose we 
 were to lay off PI as AB (at d) and 
 then lay off P3 from A; we could 
 readily see that the forces would not 
 be acting in the same direction 
 around the polygon being constructed, and consequently the polygon 
 would not be correct. 
 
 Now suppose P2 and P4 at (a) are the two forces unknown in 
 intensity and direction of action instead of P4 and P5 as considered 
 above. In that case there would be a known force in between the two 
 unknown forces, and consequently it would be impossible to lay off the 
 forces in consecutive order as we pass around the body. However, the 
 case presents no trouble at all, as we can simply lay off the known forces 
 in any order, so long as they are placed so that they all act in the same 
 direction around the polygon. For example, to determine the intensity 
 and direction of action of P2 and P4, we could lay off P5 as AB at (e), 
 then from B lay off PI as EC, and from C lay off P3 as CD, then closing 
 on A and D with P4 and P2 by drawing Dx parallel to P4 and Ay 
 parallel to P2, intersecting Dx at N. Then DN and NA represent the 
 intensity of P4 and P2, respectively, and following around the polygon 
 from A to B, C, etc., it is readily seen that P4 acts from D to N, and 
 P2 from N to A. 
 
 It is evident that if only one of the five forces at (a) were unknown 
 in intensity and direction of action, our problem would be quite simple. 
 For example, suppose P5 were the only unknown force. We would simply 
 construct a polygon as ABCDE, as shown at (6), and join E to A and 
 P5 would thus be determined in intensity and direction of action. 
 
 As an example, under Case II, suppose all of the forces shown at 
 (a), Fig. 123, are completely known except P5, which we will assume is 
 
 Fig. 123 
 
146 
 
 STRUCTURAL ENGINEERING 
 
 completely unknown. By constructing the force polygon ABCDEA at 
 {f) we have P5 given in intensity, direction, and direction of action by 
 rhe line EA, but the location of its line of action is unknown. To locate 
 its line of action select any point 0, at (/), as a pole, and draw the ray 
 diagram as shown, and construct at (a) the equilibrium polygon fabcdf 
 (as explained in Art. 45). Then prolonging the segments df and fa 
 until they intersect at 7 and through I drawing Im parallel to EA, we 
 have the line of action of the resultant of the four forces PI, P2, P3, 
 and P4. Now, as all five of the forces are in equilibrium, the resultant 
 just found must balance P5, which requires that P5 must act along the 
 same line as the resultant. (See Art. 42.) Therefore, the line Im is 
 the line of action of P5, and thus P5 is fully determined. P5 may be 
 applied upon either side of the body as far as equilibrium is concerned. 
 Referring to the force polygon at (6), Fig. 123, it is evident that 
 if P4 and P5 were parallel, a straight line joining A and D would 
 represent the direction and intensity of both, but it would be impossible 
 
 W \ 
 
 Fig. 124 
 
 to determine the intensity of either directly from the polygon, as there 
 would be no break in the line indicating where the two join. However, 
 their intensities are readily determined by the aid of the equilibrium 
 polygon. For example, let AB, at (a), Fig. 124, represent a body held 
 in equilibrium by the five forces PI. . .P5 as indicated. Let PI and P5 
 be the two unknown parallel forces. By constructing the force polygon 
 ABCDA, at (6), we have the combined intensities of the two given by the 
 line DA. Next select any point as 0, at (6), as a pole and draw the ray 
 diagram ABCDOA. Then, beginning at any point on the line of action 
 of one of the known forces, say at b on the line of action of P2, construct 
 the equilibrium, polygon fbcdf, as explained in Art. 45. Then the three 
 known forces P2, P3, and P4 are replaced by components, all of which 
 are balanced except / at b and /' at d. As the next step, prolong the 
 
GRAPHIC STATICS 
 
 147 
 
 segment bf from b until it intersects the line of action of the unknown 
 force PI at a, and likewise prolong the segment df from d until it 
 intersects the line of action of the other unknown force Po at e. In order 
 to extend the equilibrium polygon farther, it is necessary to resolve PI 
 into two components such that one of them will be equal and opposite to 
 the known component / at b, and P5 into two components such that one 
 of them will be equal and opposite to the known component /' at d. This 
 could be readily accomplished if the forces PI and P5 were known, but 
 as they are unknown we have only one component in each case to start 
 with ; that is, / at a and /' at e. 
 
 But as the five forces are in equilibrium, their components will be in 
 equilibrium. Therefore, the unknown component /" of PI must balance 
 the unknown component of P5, and consequently they will act along the 
 same line. So, by drawing the line ea, we have the line of action of the 
 two unknown components, as this is the only line that both could act 
 along. Now at e we have one known force /' and the direction of the 
 two unknown forces Po and /" ', each of which can be graphically deter- 
 mined by drawing the force triangle DEO as shown at (c). PI and /" 
 at a can be graphically determined in the same way. But it is not neces- 
 sary to construct the triangle at (c) as /' is given in the diagram at (6) 
 by the line OD, and by drawing OE parallel to the line ae we would have 
 the same forces given by the equal triangle ODE at (b), and the forces 
 at a by the triangle OEA, as indicated; and thus PI and P5 are fully 
 determined as EA and ED, respectively, in the diagram at (b). 
 
 Referring to Fig. 124, it will be observed that the equilibrium 
 polygon closes, which always occurs in the case of any system of forces 
 in equilibrium. The line ea is known as the closing line. 
 
 95. Graphical Determination 
 of Reactions and Bending Mo- 
 ments on Simple Beams. Let AB, 
 at (a), Fig. 125, represent a simple 
 beam supporting the four vertical 
 loads PI . . .P4 as indicated, and sup- 
 pose the reactions R and Rl, due to 
 these loads, are unknown in intensity. 
 First lay off the load line AB as 
 shown at (b). Then select any point 
 as a pole and draw the ray dia- 
 gram as shown and construct the 
 equilibrium poly gonA'abcdB' at (c). 
 Then from O draw the line OE par- 
 allel to the closing line A'B', and, 
 according to the last article, we have 
 R given by the line AE and Rl by 
 the line BE, and thus the reactions 
 are determined. 
 
 Conceive of the closing line A 'B' 
 at (c) as being a beam and the 
 
 other part of the equilibrium polygon as a rope suspended from the ends 
 of this beam, and imagine the loads supported upon the rope as indicated. 
 Then the equilibrium polygon really becomes a structure upon which the 
 
 , _ 
 
 } S| \ S| \ S| \ 
 
 \B 
 
 Fig. 125 
 
148 STRUCTURAL ENGINEERING 
 
 bending moment at any vertical section will be the same as for the beam, 
 and consequently it may be treated instead of the beam for the purpose 
 of finding these moments. 
 
 Then to obtain the bending moment at the section mm of the beam, 
 extend this section on down to the equilibrium polygon, cutting it off 
 along the section ee' ' . Then imagine the part of the equilibrium polygon 
 to the left of this section moved bodily to (d) without any loads or forces 
 being affected in the least. Then taking moments about e, we have 
 
 aR-bPl-cP2 = M. 
 Now, this moment is balanced by the moment of f2 about e, then we have 
 
 and taking moments about e' , we have 
 
 M = k(f5). 
 
 Both f2 and /5 (f2 being the stress in the rope from b to c, and /5 the 
 stress in the beam throughout) are given in the ray diagram, and d and k 
 can be ascertained by scale. A better way to obtain the moment is to 
 resolve f2 and /5 into horizontal and vertical components as shown, and 
 then taking moments at either e or e' ' , we have 
 
 H is the same for all segments, as is seen from the ray diagram at (6), 
 and is known as the "pole distance." Then, evidently, the bending 
 moment at any vertical section of the beam is equal to this pole distance 
 (H ) multiplied by the vertical ordinate between the closing line and the 
 broken line of the equilibrium polygon at the same vertical section. For 
 example, the bending moment at the section m'm' is equal to Hy, at 
 m"m", Hz, and so on. The ordinates are measured in feet or inches to 
 the same scale as the beam, while H is in pounds, so many thousand 
 pounds per inch. In laying out the ray diagram, H can always be taken 
 as some convenient number, as 100,000 Ibs. 
 
 96. Graphical Analysis of Trusses consists in graphically deter- 
 mining the reactions and stresses. The reactions are determined by 
 means of the equilibrium polygon, while the stresses are determined as 
 concurrent forces at the joints by means of force polygons, one polygon 
 for each joint, the joints being considered in the order that the applica- 
 tion of the method requires, being usually the consecutive order. 
 
 Example 1. Let the diagram at (a), Fig. 126, represent a truss 
 supporting the three loads PI, P2, and P3, as indicated, and let R and 
 Rl represent the reaction at A and B, respectively, due to the three loads. 
 
 Let it be required to determine the reactions on the above truss and 
 the stresses in the members due to the three loads PI, P2, and P3. 
 
 To determine the reactions, first construct the ray diagram at (6), 
 which is accomplished by laying off the line 1-2 to any convenient scale, 
 equal and parallel to PI, and line 2-3 equal and parallel to P2, and line 
 3-4 equal and parallel to P3, thus obtaining the load line 1-2-3-4, and 
 then taking any point as a pole and drawing the rays 1-0, 2-O, etc. 
 Then construct the equilibrium polygon CghkD, which is accomplished 
 
GRAPHIC STATICS 
 
 149 
 
 by taking any point on the line of action of one of the forces, say point 
 g, on the line of action of PI, and resolving PI into two components by 
 drawing Cg parallel to the ray 1-0 and gh parallel to 2-0 and virtually 
 resolving each of the other two loads into two components by drawing hk 
 parallel to 3-0 and kD parallel to 4-0, and drawing the closing line CD 
 
 Fig. 126 
 
 for completion. Then by drawing OE parallel to the closing line CD, 
 we have the reaction R given by the line 1-E and Rl by the line 4-. 
 These lines can be scaled and thus the reactions will be fully known as 
 we know their direction of action will be upward, in order to balance the 
 loads. 
 
 After the reactions are determined all of the external forces are 
 known and we can proceed to determine the stresses in the truss members. 
 
 Any joint of any truss can be considered as being a small body 
 acted upon by the truss members, meeting at the joint, and any load 
 there applied. The details should be such that these will really be con- 
 current forces in every case and are so considered in graphic analysis. 
 
 Then at j oint A we have three concurrent forces : the reaction R, the 
 force exerted by the member Ac, and the one exerted by the member Ab. 
 At joint c we have four concurrent forces, the load PI and one for each 
 of the three members meeting at that joint; at joint b we have four con- 
 current forces, one for each of the members meeting at that joint; at joint 
 d we have five concurrent forces, the load P2 and one for each of the 
 
150 STRUCTURAL ENGINEERING 
 
 members meeting at that joint; and likewise at joints e, B, and / we 
 have, respectively, four, three, and four concurrent forces, as is readily 
 seen. 
 
 The intensity of the force exerted by any member on a joint will be 
 equal to the stress in the member exerting the force, and if the force acts 
 toward the joint the member exerting it will evidently be in compression 
 and hence the stress in the member will be compression, while if the force 
 acts away from the joint the member exerting it will be in tension and 
 hence the stress in the member will be tension. So if the force exerted 
 at a joint by a member be known, the stress in the member will be known. 
 
 Now it is evident that the graphical determination of the stresses 
 in the truss members is simply a matter of determining the concurrent 
 forces at each joint, which we can do by simply drawing a force polygon 
 for each joint. However, we cannot begin with just any joint, as the 
 application of the method is limited, as stated in Art. 86. It will be seen 
 upon inspection of the diagram at (o) that the only joints that can be 
 analyzed at the beginning are joints A and B. Here we have, in each 
 case, all forces known in direction and only two unknown in intensity and 
 direction of action. So we have simply Case I, Art. 86. Then let us take 
 joint A to begin with: 
 
 Laying off 1-2 at (c), to any convenient scale, equal and parallel to 
 R, and drawing 2-3 parallel to the member Ac and 1-3 parallel to the 
 member Ab, we have the force diagram 1-2-3-1 representing the concur- 
 rent forces at joint A, where 2-3 represents the force exerted by the 
 member Ac, and 1-3 represents the force exerted by the member Ab. 
 By scaling 2-3 and 1-3 we obtain the intensity of these forces, respect- 
 ively. We know that R acts upward and hence will act from 1 to 2 in 
 the force polygon. Then as all of the forces must act in the same direc- 
 tion around the polygon, we have the force exerted by Ac acting from 
 2 to 3 (in the polygon) and the one exerted by Ab acting from 3 to 1. 
 Now transferring these directions to joint A we have the force exerted 
 by the member Ac acting away from the joint and the force exerted by 
 the member Ab acting toward it, as indicated. Then the stress in Ac will 
 be tension and the stress in Ab compression, and as the intensity of the 
 first is given by the line 2-3, and the intensity of the second by the line 
 1-3, in the force polygon, we have the stresses in the two members Ac 
 and Ab fully determined. 
 
 Knowing the force exerted by the member Ac at joint A, we know 
 the force it exerts at joint c, as the two will, undoubtedly, be equal and 
 opposite, and hence the one at c will act away from the joint, as indicated, 
 and likewise, knowing the force exerted at joint A by the member Ab, 
 we know the force it exerts at joint b, which acts toward b. 
 
 Joint b cannot be analyzed as yet for there are still three unknown 
 forces acting at that joint, but joint c can be, as there are only two 
 unknown forces there. Then laying off 2-3 at (d) equal and parallel to 
 the line 2-3 at (c) and 2-4 equal and parallel to PI and drawing 4-5 
 parallel to the member cd and 3-5 parallel to cb, we have the force 
 polygon 2-3-5-4-2 representing the forces at joint c, where the force 
 exerted by the member cd is represented by the line 4-5, and the force 
 exerted by the member cb is represented by the line 3-5, which is equiva- 
 lent to saying that the intensity of the stress in cd is given by the line 4-5, 
 
GRAPHIC STATICS 
 
 and the intensity of the stress in cb is given by the line 3-5,, and by 
 scaling these lines we have the intensity of the stress in each of the two 
 members. As the force exerted at joint c by the member Ac acts to the 
 left from the joint, it will act from 3 to 2 in the polygon at (d), and as 
 the other forces must act in the same direction around the polygon, it is 
 readily seen that the forces exerted by the members cd and cb both act 
 away from the joint, and, hence, the stress in each of the two members 
 will be tension and thus we have the stresses in the members cd and cb 
 fully determined. 
 
 Now as the force exerted at c by the member cb is fully determined, 
 the force exerted at b by the same member is known and acts away from 
 the joint, as indicated. Then we have only two unknown forces at joint b 
 and hence the joint can now be analyzed. Then laying off 1-3 at (e), 
 equal and parallel to the line 1-3 at (c), and 3-5 equal and parallel to 
 the line 3-5 at (c?), and drawing 5-6 parallel to the member bd, and 1-6 
 parallel to the member be, we have the force polygon 1-3-5-6-1 repre- 
 senting the forces at joint b, where 5-6 and 1-6 represent the intensity of 
 the force exerted by the member bd and be, respectively. Then by 
 scaling these lines we obtain the intensity of the stress in the members 
 bd and be. 
 
 Now the force exerted upon the joint b by the member Ab, as 
 previously stated, acts toward the joint and the force exerted by cb acts 
 away from the joint. Then transferring these directions to the polygon 
 at (e), it is readily seen that the forces will act around the polygon, as 
 indicated, and that the force exerted by the member bd acts in the polygon 
 from 5 to 6, while the force exerted by the member be acts from 6 to 1, 
 and transferring these directions to joint b, we have the first acting away 
 from the joint and the second toward it, hence the stress in bd is tension 
 while the stress in be is compression, and thus we have the stresses in the 
 two members bd and be fully determined. 
 
 The force exerted at b by the member bd being determined, the force 
 exerted by the same member at d is fully known and the same is true ol 
 the force exerted at d by the member cd. Then there are but two unknown 
 forces at d and hence the joint can be analyzed by drawing the force 
 polygon shown at (/) and thus the stresses in the members de and df can 
 be determined. Then next the joints / and e can be analyzed in the same 
 way and thus the stresses in all the members in the truss can be deter- 
 mined, but one continuous diagram as shown at (/i), wherein the forces at 
 each joint and likewise the stress in each member are represented, can be 
 constructed more readily than the separate diagrams, providing we pass 
 around all the joints in the same direction. Either direction, clock- wise 
 or counter clock- wise, can be taken for the first joint analyzed, but when 
 the direction is once taken we should pass around each of the other joints 
 in the same direction, for otherwise there is apt to be confusion. 
 
 For convenience, let S, 1, S2, and so on, represent the stresses in 
 the various members of the truss, as indicated at (a). 
 
 To construct the continuous diagram at (&), let us begin at joint A, 
 as before, and pass around the joint counter clock- wise. Beginning with 
 the known force R we lay off 1-2 at (h) to any convenient scale, equal 
 and parallel to R. The next force we come to is the one exerted by the 
 member Ac. So from 2 draw a line %-x parallel to the member Ac. Then 
 
152 STRUCTURAL ENGINEERING 
 
 the next force we come to is the one exerted by the member Ab. So from 
 1 draw a line parallel to this member, intersecting the line 2-x at 3, and 
 we have the force polygon 1-2-3-1 representing the forces at joint A, 
 where the line 2-3 represents the intensity of the stress S in the member 
 Ac and the line 1-3 represents the intensity of the stress SI in the 
 member Ab. As R acts upward the force exerted by Ac acts from 2 to 3 
 (in the polygon), and the force exerted by the member Ab acts from 
 3 to 1, all in the same direction around the polygon. Transferring these 
 directions to joint A, we see that S is tension and S\ compression, as 
 explained above, and thus we have S and S~L fully determined. Passing 
 on to joint c, we have the force exerted by the member Ac represented by 
 the line 2-3 and acting from 3 to 2 as S is tension. The next force we 
 come to, passing around the joint counter clock- wise, is PI, which we lay 
 off downward as 2-4. Then from 4 draw a line 4-2 parallel to the 
 member cd, and from 3 draw a line parallel to the member cb, and we 
 have the force polygon 3-2-4-5-3, representing the forces at joint c, where 
 the line 4-5 represents the intensity of the stress S2 in the member cd, 
 and the line 5-3 represents the intensity of the stress S3 in the member 
 cb. As the force exerted by the member Ac acts (in the diagram) from 
 3 to 2, and PI from 2 to 4, the force exerted by the member cd will act 
 from 4 to 5 and the force exerted by the member cb will act from 5 to 3. 
 Transferring these directions to joint c, we have the force exerted by the 
 member cd acting away from the joint and the force exerted by the 
 member cb acting away from it also. Then S2 and S3 are both tension 
 and thus we have both fully determined. At j oint b the force exerted by 
 the member Ab is known, as its intensity was determined at joint A, and 
 its direction of action at b will be in the opposite direction to that of the 
 equal force exerted by the same member at A, and hence toward the joint 
 b, as indicated, and the force exerted at b by the member cd is known as 
 its intensity was determined at joint c, and its direction of action will be 
 in the opposite direction to that of the equal force exerted by the same 
 member at joint c, and hence away from joint b. Then only the forces 
 exerted by the members bd and be remain to be determined at joint b. 
 Beginning with the force exerted by the member Ab we have its intensity 
 represented by the line 1-3 and it acts from 1 to 3 in reference to joint b. 
 Passing on around the joint counter clock- wise, we have next the force 
 exerted by the member cb, the intensity of which is represented by the 
 line 5-3, and it acts from 3 to 5. The next force we come to is the one 
 exerted by the member bd. So from 5 draw a line 5-r parallel to the 
 member bd and close the polygon by drawing a line from 1 parallel to 
 the member be, intersecting the line 5-r at 6, and we have the polygon 
 1-3-5-6-1 representing the forces at joint b. As the force exerted by the 
 member Ab acts (in the polygon) from 1 to 3 and the one exerted by cb 
 acts from 3 to 5, the force exerted by the member bd will act from 5 to 6, 
 and the one exerted by the member be from 6 to 1. Transferring these 
 directions to joint b we have the force exerted by bd acting away from 
 the joint and the force exerted by be acting toward it. Hence the stress 
 4 in bd is tension and the stress S5 in be\ is compression, and as the 
 intensity of 4 is given by the line 5-6 and that of *S5 by the line 6-1, we 
 have the stress in the members bd and be fully determined. 
 
 Passing on now to joint d we have the forces exerted by the members 
 
GRAPHIC STATICS 153 
 
 bd and cd and also the load PI known, to determine the forces exerted by 
 the members de and df. Beginning with the force exerted by the member 
 bd, and passing around the joint counter clock- wise, we have that force 
 represented by the line 6-5 and it acts from 6 to 5, and the force exerted 
 by the member cd represented by the line 5-4 and it acts from 5 to 4. 
 Now, as P2 acts downward, from 4 lay off 4-7 downward and equal and 
 parallel to P2 and from 7 draw a line 7-n parallel to the member df and 
 close the polygon by drawing from 6 a line parallel to the member de 
 intersecting the line 7-w at 8, and we have the polygon 6-5-4-7-8-6 repre- 
 senting the forces at joint d. Following around this polygon in the 
 direction of the action of the known forces, that is, from 6 to 5, from 
 5 to 4, on around to 6, we see that the forces exerted by the member df 
 and de each act away from joint d and hence the stress in each of the 
 members df and de will be tension, and, as the intensity of the stress SQ 
 in the member df is given by the line 7-8 and the intensity of the stress 
 S7 in the member de by the line 8-6, we have the stress in each of the 
 members df and de fully determined. 
 
 Passing on to joint / we have the forces exerted by the member df 
 and the load P3 known, to determine the forces exerted by the members 
 fB and fe. The force exerted by the member df is represented by the 
 line 8-7 and acts from 8 to 7. Then from 7 lay off 7-9 downward and 
 equal and parallel to P3 and drawing 9-10 from 9 parallel to the member 
 fB and 8-10 from 8 parallel to the member fe, we have the force polygon 
 8-7-9-10-8 representing the forces at joint / where the line 9-10 repre- 
 sents the force exerted by the member fB and 10-8 represents the force 
 exerted by the member fe. Following around the polygon in the direction 
 of the action of the known forces, that is, from 8 to 7, 7 to 9, on around 
 to 8, we have the force exerted by the member fB and by the member fe 
 both acting away from the joint /, and hence the stress SS in fB and the 
 stress S9 in fe are both tension and as the line 9-10 gives the intensity of 
 S8, and 10-8 the intensity of *S9, we have the stresses in the two members 
 fB and fe fully determined. 
 
 Now, passing on to joint e, we have here all of the forces known 
 except the one exerted by the member eB. Starting with the force 
 exerted by the member be, and passing around the joint counter clock- 
 wise, we have the force exerted by the member (be) given by the line 1-6 
 and it acts from 1 to 6 ; the force exerted by the member de given by the 
 line 6-8 and it acts from 6 to 8 ; and the force exerted by the member fe. 
 given by the line 8-10 and it acts from 8 to 10. Then a line drawn from 
 10 parallel to the member eB should close the polygon 1-6-8-10-1 repre- 
 senting the forces at joint e. If this polygon should not close it shows 
 that the continuous diagram has not been accurately drawn and hence 
 must be wholly redrawn to insure that the intensity of the stresses thus 
 obtained is correct. The intensity of the stress 10, in the member eB, 
 is given by the line 10-1, and as the force exerted by that member at joint 
 e acts from 10 to 1 in the polygon 1-6-8-10-1, 10 is seen to be compres- 
 sion and hence is fully determined, and thus the stresses in all of the truss 
 members are found. 
 
 For joint B we have the force polygon 10-9-1-10, where the line 9-1 
 represents the reaction Rl, and 1-10 and 10-9 the stress in eB and fB, 
 respectively. The continuous diagram at (h) would be known as a 
 
154 
 
 STRUCTURAL ENGINEERING 
 
 "stress diagram/' in fact all such diagrams representing stresses are 
 known as stress diagrams. 
 
 It will be observed that all of the external forces,, which consist of 
 the two reactions and the three loads, are laid off on the same line 2-9, 
 which is known as the load line; yet these forces, as they hold the truss 
 in equilibrium, really form a closed polygon, so to speak, where the 
 reaction R is laid off from 1 to 2 and the load PI from 2 down to 4, P2 
 from 4 to 7, P3 from 7 to 9, and R~L from 9 back to 1 the starting point 
 all acting, as we may say, in the same direction around the polygon 
 1-2-4-7-9-1, which would really be a polygon if the forces were not 
 parallel. In constructing any stress diagram the force polygon, known 
 as the load line, formed by the external forces can be laid off first and the 
 remainder of the diagram added to this polygon, as shown in the follow- 
 ing example: 
 
 Example 2. Let it be required to determine the stresses in the 
 members of the truss shown at (a), Fig. 127, due to the three loads PI, 
 P2, and P3. Let R and Rl represent the reaction at A and B, respect- 
 ively, due to the three loads. 
 
 (a) 
 
 Beginning with R and passing around the truss as a whole counter 
 clock- wise, and taking the forces in consecutive order, we obtain at (6) 
 the force polygon 1-2-3-4-5-1 by laying off 1-2 equal and parallel to R, 
 2-3 equal and parallel to PI, 3-4 equal and parallel to P2, 4-5 equal and 
 parallel to P3, and 5-1 equal and parallel to 121, thus closing the polygon. 
 As is readily seen, the forces act in the order 1-2-3-4-5-1. 
 
 To obtain the stresses, we can begin at either A or B, say, B, and 
 passing around that joint counter clock- wise we obtain the polygon 
 1-11-5-1, representing the forces at that joint, by drawing 1-11 parallel 
 to gB and 5-11 parallel to fB. The line 1-11 gives the stress in gB and 
 5-11 the stress in fB. As R\ acts upward from 5 to 1, the force exerted 
 by gB will act from 1 to 11, and hence toward the joint, and the force 
 exerted by fB will act from 11 to 5, and hence away from the joint. 
 Thus it is seen that the stress $12, in the member gB, is compression, 
 while the stress 10, in fB, is tension. 
 
 Passing on to joint /, we obtain the polygon 4-5-11-10-4 by drawing 
 11-10 parallel to gf and 10-4 parallel to fd. We really begin with the 
 load P3, which acts from 4 to 5. Then the force exerted by fB comes 
 next, which acts from 5 to 11, and next is the force exerted by the 
 
GRAPHIC STATICS 155 
 
 member gf which we can lay off from 11 only as a line parallel to gf, 
 and its length is then obtained by drawing a line from 4 parallel to fd, 
 thus closing the polygon. In a similar manner the polygon 10-11-1-9-10 
 for joint g is obtained, and likewise the polygon 9-1-8-9 for joint e t 
 3-4-10-9-8-7-3 for joint d, 1-8-7-6-1 for joint b, and 3-7-6-2-3 for joint c. 
 By scaling in each case the line representing the intensity of stress and 
 observing at the same time the direction of action of the force exerted at 
 the joints, the stress in each member is fully determined. 
 
 If the direction around the joints in examples 1 and 2 be taken 
 clock-wise instead of counter clock-wise, as above, the stress diagram 
 would simply fall on the left side of the load line instead of the right as 
 shown at (k), Fig. 126, and (6), Fig. 127. 
 
 Stress diagrams should be drawn without the aid of any letters or 
 marking of any kind other than the forces on the load line, and the 
 stresses in the members may be indicated in the way shown above; how- 
 ever, after a little practice, even this marking will be found to be 
 superfluous. The student should not acquire the habit of laying off the 
 load line and drawing the stress diagram without fully analyzing each 
 joint of the truss, and if this is done, no marking of the diagram is needed. 
 
 Example 3. So far we have considered loads only at the bottom 
 of the trusses, which is often the case for live load, but for dead load, 
 especially, we have loads applied at both the top and bottom of the 
 trusses as shown at (a), Fig. 128. Let it be required to determine the 
 reactions on the truss shown there, and also the stresses in the members, 
 due to the six loads indicated. Let R and R1 represent the reaction at 
 A and B, respectively, due to these loads. 
 
 The most practical way of determining the reactions is to add the 
 top load at each panel point to the bottom load and treat the sum as a 
 single load in each case. Then the ray diagram is constructed as shown 
 at (b) and an equilibrium polygon as shown at (c) can be drawn. How- 
 ever, the loads can be laid off on a load line in any order and a ray 
 diagram constructed and a corresponding equilibrium polygon drawn. 
 For example, suppose the loads be laid off on the load line in consecutive 
 order, passing around the truss, as a whole, counter clock-wise, as shown 
 at (d). Then taking G as a pole and drawing the ray diagram at (d) and 
 beginning at any convenient point on the line of action of one of the 
 forces, say, point 1, on the line of action of PI, resolve PI into two 
 components /I and /2, which is accomplished by drawing /I (at point 1) 
 parallel to the ray 1-G and /2 parallel to ray 2-G. As fl and /2 are 
 components of PI, the first will act in the ray diagram from 1 to G and 
 the second from G to 2, and transferring these directions to point 1 we 
 have them acting as shown. By prolonging the line of action of /2 from 
 point 1 till it intersects the line of action of P2 at point 2, and resolving 
 P2 into two components /2 and /3 by drawing /2 parallel to the ray 2-G 
 and /3 parallel to the ray 3-G, we have the component at point 2 balanc- 
 ing the component /2 at point 1, and drawing the line 2-3 parallel to the 
 ray 3-G, 3-x parallel to the ray 4-G, 3-4 parallel to the ray 5-G, and so 
 on, we obtain the polygon ?/-l-2-3-,r-3-4-5-.2r, wherein the forces, or com- 
 ponents, are all balanced except fl and /7. Then prolonging the line of 
 action of fl until it intersects the line of action of #1 at K and prolonging 
 the line of action of /7 until it intersects the line of action of R at H, and 
 
156 
 
 STRUCTURAL ENGINEERING 
 
 drawing the closing line HK, we have the equilibrium polygon K-y-1-2-3- 
 x-3-4:-5-H-K closed, and by drawing E'-G, at (d), parallel to the closing 
 line HK we have the reaction R given by the line E'-l and the reaction 
 Rl by the line E'-7. The polygon could be closed just as well by draw- 
 ing z-K' and 1-H' ', and then H'-K' , which is the closing line in that case, 
 and which is parallel to H-K. 
 
 As another case, suppose the loads are laid off on the load line, as 
 we may, just in any order as shown at (e). Then constructing the ray 
 diagram as shown we can draw the equilibrium polygon ^-6-7-8-9-10- 
 t-W-U-F as shown at (/), where UV is the closing line. Then drawing 
 
 Fig. 128 
 
 E"-P parallel to U-V we have the reaction R given by the line E"-F and 
 Rl by the line E"-T. 
 
 Complicated diagrams and polygons should be avoided as much as 
 possible. We can accomplish this result only by judicious selection of 
 order and position, for which no fixed rule can be given, and one must be 
 guided by experience and common judgment. Complicated diagrams and 
 polygons are more objectionable on account of inaccuracy, as a rule, than 
 on account of difficult construction. For example, it is obvious that there 
 is more chance of error in constructing the polygon at (/) than there is 
 in the case of the one at (c). 
 
 The most practical way of obtaining the stresses in the truss shown 
 at (a) is to lay off the load line as shown at (#), which is accomplished 
 
GRAPHIC STATICS 
 
 157 
 
 by beginning with R and passing around the truss counter clock-wise, 
 taking the forces in consecutive order. Thus we lay off at (gr) the line 
 1-2 upward,, equal and parallel to R. Then from 2 lay off 2-3 downward, 
 equal and parallel to PI, 3-4 equal and parallel to P2, 4-5 equal and 
 parallel to P3, and from 5 lay off 5-6 upward, equal and parallel to R1, 
 and from 6 lay off 6-7 downward equal and parallel to P4, 7-8 equal and 
 parallel to P5, and the line 8-1 closing the polygon 1-2-3-4-5-6-7-8-1 
 should be equal and parallel to P6. After the load line is laid off, we 
 can start either from joint A or B, and, passing around each joint counter 
 clock- wise, the stress diagram shown at (#) can be readily constructed. 
 97. Oblique Reactions. As a rule, the dead and live load upon 
 structures act vertically, producing vertical reactions, but the pressure 
 due to the wind, known as wind load, produces reactions which are not 
 vertical. However, the method of determining such reactions is prac- 
 
 (b) 
 
 Fig. 129 
 
 tically the same as for vertical reactions. As an example, let the diagram 
 at (a), Fig. 129, represent a truss supporting five wind loads, PI P5, 
 applied normally to the sloping side of the truss as indicated. In case 
 the truss be equally fixed at its supports A and B, the reactions R and Rl 
 will be parallel to the loads as indicated, or in case the loads are not all 
 parallel, the reactions will be parallel to their resultant. 
 
 To determine the reactions R and #1, first lay off the load line CD 
 at (b) and draw the ray diagram as shown and construct the equilibrium 
 polygon obcdega at (a) the same as for any other forces. Then by 
 drawing OE in the ray diagram parallel to the closing line ag of the 
 
158 
 
 STRUCTURAL ENGINEERING 
 
 equilibrium polygon, we have the reaction R given by the line EC and .Rl 
 by the line ED. 
 
 There is nothing about the equilibrium polygon at (a) out of the 
 ordinary, except at point a. At e the force P5 is resolved into two com- 
 ponents,, and the line of action of the one to the right (/o) is prolonged 
 until it intersects the line of action of J?l at g. Now, at a exactly the 
 same method of procedure is followed, but as PI and R have the same 
 line of action, the segment parallel to the ray CO does not appear as its 
 length is zero. 
 
 After the reactions are known, the stress in the individual members 
 of the truss can be graphically determined, as explained above, by begin- 
 ning at either joint A or B. 
 
 In case one end of a truss be supported upon rollers, the reaction at 
 that end will be vertical, as the rollers will not resist any horizontal force. 
 In such cases we always know the direction of the one reaction, but, as a 
 rule, the intensity of that one and the intensity and direction of the other 
 remain to be determined. 
 
 Let the diagram at (a), Fig. 130, represent the same truss and loads 
 as represented at (a), Fig. 129. First, suppose the truss supported upon 
 
 Fig. 130 
 
 rollers at B and upon a fixed bearing at A. Then, the end B being upon 
 rollers, the reaction there, indicated by R, will act vertically and, conse- 
 quently, we know its line of action, while the reaction at A, indicated as 
 Rl, will evidently act obliquely, as it resists the horizontal component of 
 the wind; but, further than this, R\ is unknown. 
 
 Lay off the load line CD at (6), Fig. 130, and draw the ray diagram 
 the same as in any other case, and construct the equilibrium polygon 
 AabcdeA at (a) by beginning at A, the point of application of the 
 oblique reaction. Then treating this equilibrium polygon as a truss, we 
 can obtain R by analyzing joint e, as the stress in the member ed is 
 given by ray 6 in the ray diagram at (b). Then by drawing from D, at 
 (k), the line DE parallel to R and closing on with line OE drawn from 
 O parallel to the member eA (at e), we have the intensity of R given 
 by the line DE. Then, as the forces PI P5 and the two reactions R 
 and Rl are a system in equilibrium, they will form a closed polygon, 
 
GRAPHIC STATICS 
 
 159 
 
 B 
 
 DECD, wherein the forces act in the same direction around the polygon. 
 So, evidently, the line EC, at (6), represents the other reaction Rl in 
 intensity and direction. However, the analysis of joint A will show this, 
 for beginning with PI, we have, at (6), the force polygon CmOEC. 
 
 In case the truss were supported upon rollers at A and upon a fixed 
 bearing at B, we would begin at B, as that would be the point of applica- 
 tion of the oblique reaction, and construct an equilibrium polvgon as 
 Ba'b'c'd'e'H' and then analyze the points H f and B to obtain the reac- 
 tions R2 and R3, whence their intensity and direction would be given, at 
 (6), by the lines CF and FD, respectively. 
 
 98. Method of Drawing an Equilibrium Polygon through Two 
 and Three Given Points. As a case of two points, let PI, P2, and P3, 
 Fig. 131, represent three given forces and let A and B be the two given 
 points. Imagine the line AB, joining the two points, as being a beam 
 supporting the three forces. Then the reactions on this beam, due to the 
 three forces, will be parallel to the resultant of the forces. By con- 
 structing the force polygon CabDC, at (6), we have the resultant of the 
 forces given by the line CD. Then the reaction at A and B, represented, 
 respectively, as R and 721, can 
 be drawn, as shown, parallel to 
 the line CD. Taking any point 
 O, at (&), as a pole and con- 
 structing the ray diagram as 
 shown, we can draw the equi- . 
 librium polygon A-l-2-3-B'-A, 
 as shown at (a). Then drawing 
 OE parallel to the closing line 
 AB' we have R given by the 
 line EC and Rl by the line ED. 
 These reactions will, of course, 
 be constant regardless of the 
 slope of the closing line of the 
 equilibrium polygon used to de- 
 termine them. So, regardless of 
 the location of the pole in the 
 ray diagram, the line drawn 
 through it and parallel to the 
 closing line of the correspond- 
 ing equilibrium polygon will 
 
 pass through E. Then evidently the pole of any ray diagram, where the 
 corresponding equilibrium polygon passes through both A and B will be 
 on a line through E parallel to AB as Ex, and as any equilibrium polygon 
 drawn from A, having a closing line parallel to AB, will pass through 
 both A and B, it follows that any point on the line Ex can be taken as a 
 pole and the corresponding equilibrium polygon will pass through the two 
 points A and B. Thus, taking 01 on the line Ex, at (&), as a pole, and 
 constructing the ray diagram, as shown, the equilibrium polygon J -4-5-6-5 
 can be drawn, and taking O2 as a pole and constructing the ray diagram 
 to the left of the load line, as shown, the equilibrium polygon A-7-8-9-B 
 can be drawn. 
 
 Fig. 
 
160 
 
 STRUCTURAL ENGINEERING 
 
 As a case of three points, let PI P5, shown at (a), Fig. 132, repre- 
 sent five given forces and let A, B, and C be three given points through 
 which an equilibrium polygon is to be drawn. 
 
 First construct the ray diagram at (b) by laying off the forces in 
 consecutive order, to any convenient scale, thus obtaining the load line 
 DabGcF, and taking any point as a pole and drawing the rays. 
 
 Imagine the line AB, joining the two points A and B, as being a 
 beam supporting the three forces PI, P2, and P3 (as loads). Then the 
 reactions on beam AB, at A and B, due to the three forces, will be 
 parallel to the resultant of the three forces, which is represented by the 
 line DG in the ray diagram. Then by drawing a line, as yy, through 
 each of the points A and B, parallel to DG, we have the lines of action 
 of the two reactions on the beam AB which are indicated as R and Rl. 
 
 We can consider the part 01) G of the ray diagram as pertaining to 
 beam AB. By drawing the equilibrium polygon A-l-2-3-m-A, as shown 
 
 Fig. 132 
 
 at (a), and then drawing OE parallel to the closing line Am, and from 
 E drawing a line parallel to AB, we have the line Ex, any point of which 
 could be taken as a pole for a ray diagram, and the corresponding 
 equilibrium polygon would pass through the points A and B as shown 
 above for the case of two points. 
 
 Next, imagine the line BC as being a beam supporting the two forces 
 P4 and P5. Then the reactions on this beam at B and C would be 
 parallel to GF (shown at (6)) and by drawing the lines it through B 
 and C parallel to GF we have the lines of action of the reactions on the 
 beam BC which are represented as R2 and R3. The part OGF of the ray 
 diagram can be considered as pertaining to the beam BC. Then drawing 
 the equilibrium polygon n-4-5-A-n and next the line OEl parallel to the 
 closing line nh and from El drawing a line parallel to BC we have the 
 
GRAPHIC STATICS 
 
 line ~E\-z, any point of which could be taken as a pole for a ray diagram, 
 and the corresponding equilibrium polygon would pass through the two 
 points B and C, provided, of course, that the polygon be started from 
 one of the points. But, as any point on the line E-x can be taken as a 
 pole for a ray diagram, and the corresponding equilibrium polygon would 
 pass through points A and B, provided it be started from one of the 
 points, undoubtedly if the point T, the intersection of E-x and El-z, be 
 taken as a pole of a ray diagram, the corresponding equilibrium polygon 
 will pass through all three points A, B, and C, if started at one of the 
 points. 
 
 Problem 1. Determine graphically the bending moment on the beam 
 AB, as shown in Fig. 125, at load P3 due to the loads PI, P2, P3, and P4, 
 assuming that each of the loads weighs 8,000 Ibs. and that all are spaced 4 
 feet apart along the beam and load PI 4 feet from R, and load P4 5 feet 
 from Rl. 
 
 Problem 2. Determine graphically the stress in the members of the 
 truss shown in Fig. 126 due to the loads indicated. Assuming: 
 
 PI = 8,000 Ibs., P2 = 10,000 Ibs., and P3 = 12,000 Ibs. 
 
 Length of span = 4 panels @ 25' = 100'. 
 
 Height of truss = 28'. 
 
 Problem 3. Determine graphically the stress in the members of the 
 truss shown in Fig. 127 due to the loads indicated. Assuming: 
 
 PI = 20,000 Ibs., P2 = 30,000 Ibs., and P3 = 20,000 Ibs. 
 
 Length of span = 4 panels @ 26' = 104'. 
 
 Height at c and / = 28'. 
 
 Height atd = 34'. 
 
 Problem 4* Determine graphically the stress in the members of the 
 truss shown in Fig. 128 due to the loads indicated. Assuming: 
 
 PI = 22,000 Ibs., P2 = 24,000 Ibs., P3 = 25,000 Ibs. 
 
 P4 = 8,000 Ibs., P5 = 7,000 Ibs., P6 = 9,000 Ibs. 
 
 Length of span = 4 panels @ 26' = 104'. 
 
 Height of truss = 31'. 
 
CHAPTER IX 
 
 INFLUENCE LINES 
 
 99. Definition. An influence line is a line showing the intensity 
 and variation of reactions, shears, moments, and stresses, produced on 
 beams or trusses by a single moving load. Owing to convenience the single 
 moving load is taken as a unit load. 
 
 100. Influence Line for Reactions and Shears on a Simple 
 Beam. Let AB, Fig. 133, represent a simple beam supporting a single 
 moving load P and let R and Rl represent the reaction at A and B, 
 respectively, due to this load P when at any point on the beam. 
 
 Suppose that the load P moves over the 
 beam from B to A; the shear on all vertical 
 sections of the beam between A and the load P, 
 due to that load, at all times will be equal to R, 
 which varies, of course, with the position of 
 the load. Then for the shear at all points be- 
 tween A and the load, including the point at 
 the load, we have R = Px/L, which is an equa- 
 tion to a straight line. 
 
 Fig. 133 From the equation we have R = P when 
 
 x = L, and R = Q when ,r = 0. Then if we draw 
 
 the horizontal line ab (=L) and erect the perpendicular ac equal to P 
 (by scale) and draw the line cb, any ordinate as y to the line cb will be 
 equal to the reaction at A and also to the shear on the vertical section of 
 the beam just over the ordinate when the load P is at that point, and from 
 this it is seen that the reaction R and the shear in the beam vary from 
 to P as the load moves from B to A. So then the line cb is the influence 
 line for the reaction at A and also for the shear in the beam when the 
 single moving load moves from B to A. In case the load moved from A 
 to B, the line ad would be the influence line for the reaction at B and for 
 the shear in the beam if bd be laid off equal to P. 
 
 As an example of application, suppose we wish to determine the 
 reaction at A and B and the shear on the beam due to a load of 20,000 
 Ibs. at C and a load of 30,000 Ibs. at M. For the sake of illustration, 
 suppose P, the single moving load referred to above (which is really a 
 mythical load and is not on the beam at all and is used only to construct 
 the influence line), is equal to 1,000 Ibs. Then the ordinate 2 represents 
 (by scale) what the reaction R would be if the 1,000-lb. load were at C. 
 Then, evidently, the reaction at A, due to the 20,000-lb. load at C, would 
 be 20 times as much or 20s, and similarly the reaction at A due to the 
 30,000-lb. load at M will be 30s, and hence the reaction at A due to both 
 the 20,000- and 30,000-lb. loads is equal to 20* 4- 30*, which is also equal 
 to the shear anywhere between A and C, while the reaction at B and also 
 the shear anywhere between B and M is equal to 20v + 3Qt and the shear 
 
 162 
 
INFLUENCE LINES 
 
 163 
 
 anywhere between M and C is equal to 20* + 30* - 20,000 or 2Qv + 30* - 
 30*000. 
 
 The value of any ordinate, as z, s, etc., will always be given to the 
 same scale as that used in laying off the single moving load P. It is 
 always convenient to take this single moving load P as unity, for in that 
 case it does not appear in the computations at all. 
 
 Example 1. Determine the maximum reaction and shear on a simple 
 beam 30 ft. long due to the wheel loads as per diagram. 
 
 The placing of the wheels in 
 order to obtain the maximum re- 
 action or shear on a simple beam is 
 really a matter of trial, yet in most 
 cases we can determine the position 
 by mere inspection. As in this case, 
 
 we can readily see that the maximum reaction will occur when wheel 2 
 is at the end of the beam and wheel 1 off of the beam altogether. Then, 
 to obtain the maximum reaction, draw the line ab (Fig. 134) to represent 
 the length of the beam to some convenient scale, say, -jV' = l'-0", and 
 space the loads to the same scale, placing wheel 2 at a. Then wheels 2, 3, 
 4, 5, 6, and 7 will be on the beam as shown in Fig. 134. Next draw the 
 vertical ordinate ac to represent 1 Ib. to a convenient scale, say, 1 in. = 
 1 Ib., and then draw the influence line cb. Next draw the vertical lines 
 through the loads as shown and we are then ready to find the reaction at a, 
 which we do in the following manner : 
 
 Take a pair of dividers, put one of the points at c and bring the other 
 point to a, then we have the length ac on our dividers ; then set one point 
 at e and rest the other point at o (eo = ca) ; hold the point firmly at o and 
 
 open the dividers, bringing 
 the point e to /; then we will 
 have the length of ca and ef 
 on our dividers. Then put 
 one point of the dividers at g 
 and rest the other point at o' 
 (o'g = ca + ef of) ; holding 
 the point firmly at o' bring 
 the point at g down to h. 
 Then put one point of the 
 dividers at k and rest the 
 other at o" (o"k = ac + ef + 
 gh) ; holding the point firmly 
 at o" bring the point at k 
 down to m, and the distance 
 o"m on the dividers equals 
 the sum of the ordinates ac, 
 ef, gh, and km, each of which 
 is an ordinate under a 20,000- 
 Ib. load. Then lay the di- 
 viders on a scale divided to 1/10 inches and suppose we find that we have 
 3.03 inches (=o"m). Then the reaction at a due to the four 20,000-lb. 
 loads is equal to 20,000x3.03 = 60,600 Ibs. Then with our dividers we 
 get, in the same manner, the distance nt and rp, the sum of which suppose 
 
 Fig. 134 
 
164 
 
 STRUCTURAL ENGINEERING 
 
 we find is 0.25 of an inch. Then the reaction at a due to the two 13,000- 
 Ib. loads is equal to 13,000 x 0.25 = 3,250 Ibs. Then the total reaction at a 
 is 60,600 + 3,250 = 63,850 Ibs. In the case of wheel loads, as a rule, time 
 can be saved by using the ordinates at the centers of gravity of the dif- 
 ferent groups of wheels instead of the ordinates directly under the wheels. 
 For example, the maximum reaction in the above case is equal to 
 
 80,000 x/ + 26,000 x#". 
 
 The maximum reaction at b due to the above wheel loads would be 
 the same as at a, and would be determined in the same manner, but of 
 course the loads and the influence line would be reversed, end for end, 
 from the position shown in Fig. 134. 
 
 To determine the maximum shear at any intermediate point in the 
 beam, the first thing to do after constructing the influence line is to place 
 some wheel which we think will be at the point in question when the 
 maximum shear occurs. Then determine the end reaction in the same 
 manner as was shown above and subtract the intervening loads, if any, 
 from the reaction, and the difference will be the shear at the point. 
 
 Thus, let ab (Fig. 135) represent a beam to a convenient scale, and 
 let it be required to find the maximum shear at point d due to a system of 
 wheel loads as shown. First construct the influence line cb, making ac 
 
 L 
 
 cL O Oi (ft ($7ir?l7rf^ 
 
 Fig. 135 
 
 Fig. 136 
 
 equal unity. Say we place wheel 2 at d, as shown, and find the reaction at 
 a in the same manner as was shown above. Then subtract wheel 1 from 
 this reaction and the difference will be the shear at d. If there be any 
 question as to this being the maximum shear at d, wheels 1 and 3, and 
 possibly 4, should each in turn be placed at d and the shear computed. 
 In this way we can readily ascertain the wheel to place at d for maximum 
 shear at that point. 
 
 101. Influence Line for Bending Moments on Simple Beams. 
 As a general case, let AB (Fig. 136) represent a simple beam of length 
 L and let cc be any section a distance from A and b distance from B. 
 Imagine a single load P moving over the beam from B to A. The reaction 
 at A due to this load P, at any time, will be R = Px/L. When P is to the 
 right of CC) the moment at that section is 
 
 which is an equation to a straight line wherein M r when x 0, and 
 M r = P(b/L)a when x = b. Then drawing A'B' (=L) as a reference 
 
INFLUENCE LINES 
 
 line, and laying off om = P(b/L)a, we can draw the line mB f , which 
 evidently is the influence line for the bending moment at the section cc 
 for loads on the right of the section, as any ordinate y represents the 
 bending moment at the section cc due to P when P is directly over that 
 ordinate, and the moment for any other load can be obtained by direct 
 proportion. When P is to the left of cc, the moment at that section is 
 
 which is also an equation to a straight line wherein M t = when x L and 
 MI = P(b/L)a when x b. Then, evidently, the line mA' is the influence 
 line for the bending moment at the section cc for loads to the left of the 
 section, as any ordinate y' represents the bending moment at the section cc 
 due to P when P is directly over that ordinate, and the moment for any 
 other load can be obtained by direct proportion. 
 
 The two lines B'm and mA' combined would be known as the 
 influence line, which would be referred to as influence line A'mB'. By 
 prolonging the line mB' until it intersects the line of action of R at n, we 
 have two similar triangles, A'B'n and oB'm. From these similar triangles 
 we have 
 
 A'n A'B' 
 om oB' 
 
 or 
 
 A'n L 
 ~Ib~ = b 
 
 from which we obtain A'n Pa. But if P unity, we would have A'n a, 
 and the influence line for unit load could then be constructed by making 
 A'n = a, drawing B'n, dropping the perpendicular om, and drawing A'm. 
 Or the same thing could be accomplished by laying off B'r = b, drawing 
 A'r, dropping the perpendicular om, and drawing mB f . 
 
 Example 1. Let it be required to determine the maximum bending 
 moment on a 40-ft. girder due to the loading given in Example 1 of the 
 preceding article. Wheels 1 to 6 are the heaviest that can be placed on 
 the girder, and consequently will very likely produce the maximum 
 moment this much being determined by mere inspection. It is first 
 necessary to determine the center of gravity of these wheels in order to 
 place them on the girder so as to satisfy the position for maximum 
 moment. (See Art. 88.) In such problems it is quite convenient to 
 determine the center of gravity by means of proportional triangles. (See 
 Art. 45.) This is accomplished in the following manner: 
 
 First lay out the diagram of the loads to a convenient scale (say, 
 y l'-0") as shown on line AB in Fig. 137. We know from observation 
 that the center of gravity of wheels 2 to 5 inclusive is at d. Then the 
 center of gravity of wheels 2 to 5 inclusive being at d, the problem reduces 
 to finding the center of gravity of wheels 1, 6, and the 80,000 pounds 
 acting through d. From wheel 1 draw the line ax, making a convenient 
 angle with line AB and lay off ab = 80,000 Ibs. and be = 10,000 Ibs. 
 (weight of wheel 1) to any convenient scale. Then join c and d and 
 
166 
 
 STRUCTURAL ENGINEERING 
 
 through b draw a line parallel to cd and the point e where it intersects 
 the line AB will be the center of gravity of all of the wheels from 1 to 5 
 inclusive. Next, from e draw the line ex f at random, making any con- 
 venient angle with AB, and lay off ef = 13,000 Ibs. (weight of wheel 6) 
 
 Fig. 137 
 
 and /<7 = 90,000 Ibs. (weight of wheels 1 to 5). Then join-gr and h, and 
 through / draw a line parallel to gh and the point k where it intersects the 
 line AB is the center of gravity of all of the wheels from 1 to 6 inclusive, 
 which was desired. 
 
 Now, as this point k is nearest wheel 4, the maximum moment will 
 (very likely) occur under that wheel, and the line SS bisecting the dis- 
 tance from k to wheel 4 will be the center of the span. Then, by laying 
 
 off 20 ft. on each side of this line 
 SS to the same scale as the load 
 diagram, we have the girder in 
 position represented as DE. By 
 dropping the perpendicular from 
 wheel 4 we have the point o on the 
 beam where the maximum moment 
 occurs. The influence line DmE 
 for the moment at that point is then 
 constructed by laying off EF = oE 
 and drawing FD, om, and mE, as 
 explained above. Then by drop- 
 ping perpendiculars down from the loads, we have the ordinates 1, 2, 3, 
 ... 6, and by multiplying each by the load above it, and adding these 
 products, we will have the bending moment at o, which is the maximum 
 bending moment on the beam. The sum of the ordinates 2, 3, 4, and 5 
 
 Fig. 138 
 
INFLUENCE LINES 167 
 
 can be obtained by the use of dividers as explained in the preceding 
 article. 
 
 Example 2. As a case of uniform load, let AB (Fig. 138) represent 
 a simple beam supporting a uniform load of p pounds per foot, and let o 
 be any section a distance from A and b distance from B. The influence 
 line AmB for the bending moment at o is constructed in the same manner 
 as explained above by laying off An = Ao = a, then drawing nB, om, and 
 Am in consecutive order. From proportional triangles we obtain 
 
 b 
 
 om a. 
 L-J 
 
 We can consider the uniform load as made up of infinitesimal loads 
 each weighing pdx pounds. Then for the bending moment at o due to any 
 such infinitesimal load at x distance to the right of o we have 
 
 dM = pdxy (1). 
 
 But from proportional triangles we have 
 
 y _b-x 
 
 (I) 
 
 from which we obtain 
 
 Now substituting this value of y in (1), we have 
 
 dM = a(b -x^dx .................................... (2). 
 
 Li 
 
 Then for the bending moment at o due to all such loads to the right, we 
 have 
 
 But ^(b/L}axb = area of the triangle omB, since the ordinate om 
 (b/L)a. So we have the bending moment at o due to the uniform load to 
 the right equal to the area of the triangle omB multiplied by the uniform 
 load per linear foot of span. In a similar manner it can be shown that 
 the bending moment at o due to the uniform load to the left of o is equal 
 to the area of the triangle omA multiplied by the uniform load per linear 
 foot. Then letting M represent the total bending moment at o due to the 
 total uniform load on the girder, we have 
 
 = area of triangle AmB x p ...... (3), 
 
 that is, the total bending moment at o is equal to the area of the triangU 
 AmB (formed by the influence line and the reference line) multiplied by 
 the uniform load per linear foot of span. 
 
168 
 
 STRUCTURAL ENGINEERING 
 
 If the section be taken at the center of the span, we have b = a = L/2. 
 Substituting in equation (3), we have 
 
 L L 
 
 which is readily recognized as the formula for the bending moment at the 
 center of a beam uniformly loaded. 
 
 102. Influence Lines for Shears and Bending- Moments on 
 Trusses. Let the diagram at (a), Fig. 139, represent a simple truss and 
 let it be. required to construct an influence line for the shear in any panel 
 as CD. It is evident that any load as W when at D or to the right of D 
 will produce tension in the diagonal UD, while any load as W\ when at 
 C or to the left of C will produce compression in the same diagonal. 
 
 Then evidently there must be some point between C and D where, if a 
 load were placed, the stress in the diagonal due to the load would be zero. 
 Then evidently the influence line for the shear in the panel will be of the 
 form EdeF shown at (6), which is readily constructed for a unit load by 
 laying off En and Fn' each equal to unity and dropping the perpen- 
 diculars ef and hd under the panel points and drawing ed. The shear 
 in the panel producing tension in the diagonal UD, due to any load as W 
 when at any point x distance from B, is equal to Wy lf y l being the 
 ordinate under the load as shown, while the shear in the panel producing 
 compression in the diagonal due to any load as W\ is equal to W\y 2 . 
 
 It is evident then that to obtain the maximum shear in panel CD 
 producing tension in the diagonal UD, which we will call positive shear, 
 there should be no loads to the left of 0, and to obtain the maximum 
 shear producing compression in the same diagonal, which we will call 
 negative shear, there should be no loads to the right of 0. That is, one 
 kind of shear will be a maximum when the loads extend from B to 0, and 
 the other kind will be a maximum when the loads extend from A to O. 
 
INFLUENCE LINES 169 
 
 In case of a uniform live load, the maximum positive shear would be equal 
 to the area of the triangle Fke multiplied by the load per foot, while the 
 maximum negative shear would be equal to the area of the triangle Edk 
 multiplied by the load per foot. In case of uniform dead load, the shear 
 is equal to the difference of the areas of the two triangles Edk and Fke 
 multiplied by the dead load per foot. In the case of concentrated live 
 loads, as wheel loads, to obtain the maximum shear in the panel the loads 
 would be so placed that a load would be at the panel point and the front 
 load as near the point as possible (not beyond). For example, to obtain 
 the maximum positive shear in panel CD a load would be placed at D, the 
 one that would bring the front load closest to the point 0, and to obtain 
 the maximum negative shear the load would be placed at C that would 
 bring the front load as near as possible. The same result is obtained 
 in this manner (graphically) as would be obtained by satisfying the 
 criterion of Art. 90. 
 
 The influence line for the shear in each of the other panels can be 
 drawn on the same preliminary construction at (&) as used for panel CD. 
 For example, the influence lines for panel MC, LN, NB are Ed'e'F, 
 Ed"e"F, and Ed" ' F, respectively. The construction of each is accom- 
 plished in the same manner as explained above in the case of panel CD. 
 
 In the case of trusses, the bending moments are desired only at the 
 panel points. The influence lines for the moments at these points are 
 constructed the same as though the points were on a simple beam, which 
 was treated in the preceding article. For example, the influence line for 
 bending moment at the panel point C of the truss represented at (a) 
 (same for point U) is constructed as shown at (c) by laying off Gs = a, 
 the distance of the panel point from A, and drawing sH, mm' , and mG 
 in consecutive order, as explained in the last article for simple beams. 
 The influence line for the bending moment at any other panel point would 
 be constructed in the same manner. The moment at any panel point 
 would be obtained in the same way as explained in the preceding article 
 for the case of any point on a simple beam. The exact placing of concen- 
 trated loads for maximum moments at the different panel points would be 
 according to Art. 91. 
 
 103. Influence Lines for Stresses in Truss Members. Instead 
 of constructing influence lines for the shears and moments as in the last 
 article and then determining the stresses in the truss members from these, 
 it is more convenient to construct influence lines for the stresses in the 
 members, thereby obtaining the maximum stresses directly from the 
 influence lines without dealing with either shears or moments. 
 
 Let the diagram at (a), Fig. 140, represent a truss and let it be 
 required to construct an influence line for the stress in the diagonal ED. 
 First draw the line A'B' at (b) (=L) and lay off A'n and B'n' each equal 
 to unity and draw nB' and A'n'. Then by drawing the ordinates de and 
 fb and the line db we have the influence line A'bdB' for the shear in the 
 panel CD the same as in the preceding article. The ordinate ed is equal 
 to the positive shear that would be produced in the panel by a unit load 
 at D, and the ordinate fb the negative shear in the panel that would be 
 produced by a unit load at C. Now, as the stress in the diagonal ED 
 varies as the shear in the panel, it is evident that if the ordinates de and 
 bf were laid off to represent the stress in the diagonal that would be 
 
170 
 
 STRUCTURAL ENGINEERING 
 
 produced by a unit load at these same points and an influence line be 
 constructed accordingly, we would have an influence line for the stress in 
 the diagonal. 
 
 For example, if e'd' ', at (c), were equal to the stress produced in the 
 diagonal ED by a unit load at D, and f'b' were equal to the stress pro- 
 duced in the same by a unit load at C, the line A"b'd'B" would be the 
 influence line for the stress in the diagonal. And likewise, if ft", at (c), 
 and r"e' were equal, respectively, to the stress produced in the post EC 
 by a unit load at C and D, the line A"t"r"B" would be the influence line 
 for the stress in the post. 
 
 Laying off B'k at (6), equal to 
 BD and drawing kA', ea, and aB' in 
 consecutive order, we have the influ- 
 ence line for the bending moment at 
 panel point D, as explained in the 
 last article. Now, as the stress in 
 the top chord EF varies directly as 
 this moment, it is evident that if ev, 
 at (6), were equal to the stress pro- 
 duced in the chord by a unit load at 
 D, the line A'vB' would be the influ- 
 ence line for the stress in the chord 
 EF. From the above it is evident 
 that an influence line for the stress in 
 any truss member can be drawn as 
 readily as the influence lines for the 
 moments and shears on the truss by 
 computing the maximum stress in the 
 member considered, due to a unit 
 load and using this value as the ordi- 
 nate in establishing the influence line. 
 
 The stresses due to the unit load can be determined most readily by 
 graphics. 
 
 As an example, let us first construct the influence line for the stress 
 in the top chord EF. The first thing to do is to determine the stress in 
 the member due to a unit load at D. Imagine OO' (drawn through the 
 top chord) to be a beam, and each of the lines OD and DO' to be a rod 
 or a rope. Then we would have a structure OO'D such that the stress in 
 the member 00', produced by a unit load at D, is the same as the stress 
 produced in the top chord EF by a unit load at the same point. Now this 
 structure OO'D is very readily analyzed graphically. The reaction at 
 due to a unit load at D is given by the ordinate de, at (6). Then con- 
 sidering the point and drawing from d a line parallel to 00' and from 
 e a line parallel to OD intersecting the first line at r, we have the stress 
 in 00' given by the line dr which is also the stress in the top chord EF. 
 Then from e lay off ev equal to dr, just found, and the influence line 
 B'vA' for the stress in the top chord EF is drawn. 
 
 Next, let us construct the influence line for the stress in the diagonal 
 ED. The stress in the diagonal corresponding to positive shear due to a 
 unit load at D is given by the line rs, at (6), which is drawn from r 
 parallel to the diagonal, and as bf is equal to the negative shear in the 
 
 Fig. 140 
 
INFLUENCE LINES 171 
 
 panel that would be produced by a unit load at C, by drawing the lines 
 OC and CO' and br' and r'f parallel, respectively, to 00' and CO', and 
 r's' parallel to the diagonal, we will have the stress in the diagonal that 
 would be produced by a unit load at C given by this last line r's'. Then 
 by laying off e'd' = rs, at (c), and fb' = r's', the influence line A"b'd'B" 
 is readily constructed. In like manner the influence line A"t"r"B" for 
 the stress in the post EC is readily constructed by laying off the ordinates 
 e'r" = rt (given at (6)) and t"f = r'f, as it will be readily seen that rt 
 is equal to the stress in the post due to a unit load at D, and r't' is equal 
 to the stress in the same due to a unit load at C. The influence line for 
 the stress in any member of any truss can be readily constructed in the 
 same manner as shown for the above cases. 
 
 There are other methods employed to determine the stresses in the 
 truss members due to the unit load, and these will be presented later as 
 the practical application of influence lines is taken up. 
 
CHAPTER X 
 
 DESIGN OF I-BEAMS AND PLATE GIRDERS 
 
 104. General Data. 
 
 Specifications,, A. R. E. Ass'n.* (For steel railroad bridges.) 
 Allowable bending stress on beam and girder flanges . 16,000 Ibs. per sq. in. 
 
 Allowable shearing stress on shop rivets 12,000 Ibs. per sq. in. 
 
 Allowable shearing stress on field rivets and webs. . . . 10,000 Ibs. per sq. in. 
 
 Allowable bearing stress on shop rivets 24,000 Ibs. per sq. in. 
 
 Allowable bearing stress on field rivets 20,000 Ibs. per sq. in. 
 
 Allowable bearing on masonry 600 Ibs. per sq. in. 
 
 DESIGN OF I-BEAMS 
 
 105. Outline of Usual Method of Procedure. In designing 
 I-beams the first thing to do is to determine the maximum bending mo- 
 ment in inch pounds. Then, using Formula C, Art. 53, we have the 
 bending moment 
 
 M=tL; 
 
 y 
 
 dividing through by /, we have 
 
 M 7 
 
 / ' y 
 
 The quantity I/y is known as the "Section Modulus/' which is given in 
 Tables 1 and 2 in the back of this book, for practically all I-beams, and 
 the same will be found in structural handbooks, such as Carnegie, 
 Cambria, etc. So the size of an I-beam required to resist a known 
 bending moment can be determined by simply dividing the bending 
 moment (in inch pounds) by the allowable stress, which is specified above 
 as 16,000 Ibs., thus obtaining the required section modulus, and from the 
 tables we find the lightest I-beam having this or approximately this 
 modulus and that will be the beam to use. 
 
 For example, suppose the bending moment for a given span is 
 433,000 inch pounds. Then the section modulus of the beam required 
 for the span is 
 
 433,000 
 16.000 
 
 Glancing over column 10 of Table 1 we find that a 10-inch by 30-pound 
 beam, which has a section modulus of 26.8, is the nearest to the required 
 beam, and hence would be used. 
 
 After an I-beam is designed to resist the maximum bending moment, 
 it can be tested for shear, by dividing the maximum shear by the area of 
 
 * Those specifications can be obtained from the Secretary. American Railway En- 
 gineering Association. 900 South Michigan Avenue Chicago, 111. Twenty-five cents per 
 copy, or less if several copies are ordered at one time. 
 
 172 
 
DESIGN OF I-BEAMS AND PLATE GIRDEHS 
 
 173 
 
 the cross-section of the beam. If this should exceed 10,000 Ibs. per 
 square inch., the permissible intensity, a heavier beam should be used or 
 the web of the beam should be reinforced. However, there are but very 
 few cases where the shear affects the design of an I-beam, owing to the 
 webs of I-beams being comparatively thick. 
 
 An I-beam can be designed by using Formula D, 
 
 My 
 
 T ~ I ' 
 
 and finding values for y and I that will give / = 16,000 (approximately). 
 But, as is readily seen, this is not a very convenient method of procedure. 
 106. Example 1. Design an I-beam of 15-ft. span to support a 
 total uniform load of 800 Ibs. per foot. 
 
 The maximum bending moment (=pL 2 /8) = |x800x!5 2 x!2 = 
 270,000 inch Ibs. 
 
 Then, 
 
 270,000 
 
 = 16.8 =section modulus. 
 
 16,000 
 
 The I-beam having a section modulus nearest this value is the 8-in. 
 by 25.5-lb., but the 9-in. by 21-lb. would be used as it is lighter. It is 
 seen that the 8-in. beams are all too small except the 25.5-lb. beam. 
 
 107. Example 2. Determine the size of I-beam required in the 
 case of the simple beam shown in Fig. 141, where AB represents the 
 
 Al 
 
 1 
 
 1 
 
 *l 
 
 J 
 
 
 W////////A 
 
 '////////////A 
 
 '(//fido^ffii 
 
 &//////////M 
 
 t 
 
 ,3'- 
 
 Rl 
 
 Fig. 141 
 
 beam supporting the loads indicated. By taking moments about B we 
 find 12 = 19,630 Ibs. Adding up the forces, beginning at A, we find that 
 the shear passes through zero at the 2,000-lb. load, hence the maximum 
 bending moment occurs at that load. (See Art. 88.) Then for the 
 maximum bending moment we have 
 
 M=[19,630x6-(l,600x6x3)-(9,000x3)]12 = 743,760 inch Ibs. 
 
 Then, 743,760 
 
 =46.4 = Section Modulus. 
 
 1 o,0 00 
 
 So a 15-in. by 42-lb. I-beam would be used. The 12-in. by 40-lb. is too 
 light and 12-in. by 45-lb. is heavier than the 15-in. by 42-lb. beam. 
 
 108. Example 3. Determine the size of I-beam required in the 
 case of the simple beam shown in Fig. 142, where AB represents the 
 beam supporting a load which varies from at A to 2,000 Ibs. at B, as 
 indicated. 
 
 For convenience, let p (=2,000 Ibs.) represent the intensity of the 
 
174 
 
 STRUCTURAL ENGINEERING 
 
 load at B. Then for the total load on the beam (neglecting the weight 
 of the beam) we have 
 
 Fig. 142 
 
 and for the intensity of the load at any point a; distance from A we have 
 
 pz 
 L ' 
 
 Now, taking moments about B we have 
 
 W- 
 
 ~ 3 W pL 
 
 ~ =r: ~~ 
 
 for the reaction at A. 
 
 Now for the bending moment at any point x distance from A, we have 
 
 When this moment is a maximum 
 dM pL px 2 
 
 = ~~ 
 
 (which is obtained by differentiating (1)). This, as is readily seen, will 
 occur when L 2 = 3x 2 , from which we obtain 
 
 ' = LAJ =0.58L (approximately). 
 
 So the point of maximum bending moment is 0.58Z, from A. Then 
 substituting 0.58L for x in equation (1) we have for the maximum 
 bending moment 
 
 M= (- J0.58L 2 - (- j0.19L 2 = ^ (0.39L 2 ), 
 
 and substituting the numerical value of L and p, and multiplying by 12, 
 to reduce the moment to inch pounds, we have 
 
 2,000 
 
 (0.39 x 144 x 12)= 224,600 inch Ibs. 
 
DESIGN OF I-BEAMS AND PLATE GIRDERS 175 
 
 for the maximum bending moment. 
 
 Then, -7^= 14 = section modulus. 
 
 lo,UUU 
 
 So an 8"xl8# I-beam would be used. 
 
 In case the weight of the beam were considered, we would simply 
 include the moment of this weight in an equation for the bending moment, 
 corresponding to (1). Thus, if the weight of the beam be w pounds per 
 foot, the equation for the bending moment at any point x distance from A 
 would then be 
 
 __ pL px* wL tvx z 
 
 M= -- + *- ......................... 3 - 
 
 Then by placing dM/dz = Q, the point of maximum moment can be 
 determined and then the maximum bending moment at that point due to 
 the combined loads can be obtained and the beam designed accordingly. 
 
 109. Example 4. Determine the size of I-beam required in the 
 case of the overhanging beam shown in Fig. 143, where ABC represents 
 the beam supporting the loads indicated. 
 
 Fig. 143 
 
 In this case there are two bending moments to consider; one at B 
 and the other at the point of zero shear in the span BC. 
 For the bending moment at B we have 
 
 M= (4,000x3 + 600x5x2.5) 12 = 234,000 inch Ibs. 
 Then taking moments about C we have 
 
 R = ~ (600x15x7.5 + 8,000x4 + 4,000x13)= 15,150 Ibs. 
 
 for the reaction at B. Then for the reaction at C we have the total load 
 on the beam minus R, that is, 
 
 Rl = 21,000 -15,150 = +5,850 Ibs. 
 
 Now beginning at C and adding up the forces toward the left we 
 find that the shear changes signs, that is, passes through zero, at the 
 8,000-lb. load. So the maximum bending moment in span BC occurs at 
 that load. Then taking moments about that load and considering the 
 forces to the right of it, we have 
 
 M'=(5,850x4- 600x4x2) 12 = 223,200 inch Ibs. 
 
 for the maximum bending moment in span BC. The bending moment at 
 B is the greater and hence the I-beam must be designed for that moment. 
 
176 
 
 STRUCTURAL ENGINEERING 
 
 So, using the moment at B, we have 
 
 234,000 
 
 = 14.0 = section modulus, 
 
 -L O .U \J \) 
 
 and hence an 8-in. by 18-lb. I-beam would be used. 
 
 110. Example 5. Determine the size of I-beam required in the 
 case of the continuous beam shown in Fig. 144, where ABC represents the 
 beam supporting 4 ft. of uniform load in span AB and a single load of 
 18,000 Ibs. in span BC. 
 
 DC <?=kl) . 
 
 vc 
 
 ikL, = c) 
 
 
 \PI 
 
 W 
 
 m 
 
 4 
 
 Fig. 144 
 
 Applying the three-moment equation, (L), 
 
 given in Art. 70, we have both M' and M" ' 0, and the quantity 
 
 and the quantity 
 
 -P'L l(2k 1 -3kl 
 
 Then substituting these values in the general three-moment equation, 
 we have 
 
 Now everything in this equation is known except M", the bending moment 
 at support B. Then by substituting the numerical values of the known 
 quantities, as given in Fig. 144, and reducing, we have 
 
 M"=-25,363 ft. Ibs. =-304,356 inch Ibs. 
 
 for the bending moment at B. 
 
 Then taking moments about B, we have 
 
 -25,363 = 10121 - 4,000 x 4, 
 from which we obtain 
 
 721 =-936 Ibs. 
 
 for the reaction at A, and taking moments about B and considering span 
 BC, we have 
 
 - 25,363 = 12 R3 - 18,000 x 6, 
 
DESIGN OF I-BEAMS AND PLATE GIRDEES 17 r , 
 
 from which we obtain 
 
 R3= +6,884 Ibs. 
 
 for the reaction at C. 
 
 Now, as the three reactions must be equal to the total load on the 
 beam, we have 
 
 22,000 =-936 + #2 + 6,886, 
 from which we obtain 
 
 2 =+16,050 Ibs. 
 
 for the reaction at B. 
 
 Now, as all of the reactions are determined, the moments and stresses 
 on the above beam can be determined as readily as for any beam. There 
 are three bending moments to consider in the above case ; the maximum in 
 each of the spans and the one at support B. 
 
 The reaction Rl being minus, it is readily seen that the maximum 
 bending moment in span AB will occur at support B, and as span BC 
 supports only the one load it is evident that the maximum bending 
 moment in that span will occur under the load. Then we really have only 
 the moment under the 18,000-lb. load yet to determine as the moment at 
 B is determined above. 
 
 Then, taking moments about the 18,000-lb. load, we have 
 
 A^ = 6x^3 = 6x6,884 = 41,316 foot Ibs. = 41,316x12 = 495,792 inch Ibs. 
 
 for the bending moment at that load, which is greater than the moment at 
 B and hence is the maximum bending moment on the beam, and conse- 
 quently the beam must be designed to resist this moment. Thus we have 
 
 495,79.2 
 
 =3 1.0 = section modulus. 
 
 16,000 
 
 So a 12-in. by 31.5-lb. I-beam would be used. 
 
 DESIGN OF PLATE GIRDERS 
 
 111. Description. A plate girder is, for the most part, a built 
 I-beam composed of a plate web and angle flanges which are riveted to 
 the edges of the plate as shown at (a), Fig. 145. Yet in addition to these 
 parts there are usually vertical angles, known as stiffeners, riveted to the 
 web, as shown at (6), to stiffen the web against buckling. The stiffeners 
 are either bent around the flange angles as shown at (c) or filler plates 
 are placed between them and the web, as shown at (d). When they are 
 bent around the flange angles we say they are crimped. It is practice to 
 limit the thickness of metal to about f of an inch and when the area 
 required in a flange is greater than that of two 6" x 6" x f " Z s, plates are 
 riveted to the backs of the outstanding legs of the flange angles as shown 
 at (e), to provide for the additional area required. These plates are 
 known as cover plates, or flange plates. The area of the cover plates in 
 any flange should never be much greater than that of the two flange 
 angles, and when they exceed the area of the angles very much, plates, 
 known as side plates, are placed between the flange angles and the web 
 
178 
 
 STRUCTURAL ENGINEERING 
 
 as shown at (/) so that the area of the cover plates can be reduced. IE 
 this case the side plates are considered as part of the flange. There are 
 other types of flanges used occasionally which will be shown later. 
 
 Web 
 
 (a) 
 
 Shffeners 
 
 o 0)0 ooo 
 
 CJ / 
 II 
 
 3H^O-e\o -<j 
 
 U. ^ 
 
 OOQO O <) 
 o 
 
 Web {] 
 
 o 
 
 h n n n on 
 
 
 O 
 
 
 
 
 (c) (d) 
 
 (b) 
 
 Cover Plafes 
 
 Fig. 145 
 
 112. Stress and Area in Flange. In case of a beam composed of 
 one piece, as an I-beam, the stress due to cross bending is obtained 
 through the application of For- 
 mula D, Art. 53, but in the case r 
 of a plate girder, while the same 
 method would apply, quite a dif- 
 ferent one is used, wherein the web 
 and flanges are treated separately. 
 The cross bending is resisted by 
 the flanges and web combined, but 
 the resistance of the web is ignored 
 by some engineers, while others 
 consider it. So we really have two 
 cases to consider. 
 
 In case the resistance of the 
 web be ignored, the stresses in the 
 two flanges form a couple which 
 will be equal and opposite to the 
 
 algebraic sum of the external cou- Fig . i 46 " 
 
 pies on either side of any cross-section which is the same thing as the 
 bending moment. 
 
DESIGN OF I-BEAMS AND PLATE GIRDERS 179 
 
 Let Fig. 146 represent a portion of a plate girder: 
 Let M - bending moment at section cc; 
 
 F = total stress in each flange at that section ; 
 d = distance between the centers of gravity of the flanges, 
 which is known as the effective depth. 
 
 Then, in accordance with the condition of equilibrium, we have 
 
 M = Fd or F = ^- .................................... (1). 
 
 If / be the allowable unit stress, the area required for each flange will be 
 
 This is assuming that the unit stress is the same over the entire cross- 
 section of each flange an assumption invariably made, although not 
 absolutely true in any case, but quite accurate enough for practical 
 designing. 
 
 In case the resistance of the web be considered, the bending moment 
 will be resisted by the flanges, the same as shown above, aided by the 
 web, which is really a rectangular beam. Then for the bending moment 
 we have 
 
 2 
 
 where /' represents the unit stress on the extreme elements of the web, 
 and h and / represent, respectively, the height and moment of inertia of 
 the web, while the other letters signify the same as they do in equation 
 (2), except F, of course, is less. 
 
 Now, let t be the thickness of the web, A' its area of cross-section 
 and A the area of each flange; then we have 
 
 A'=th,I = -th s 
 
 IxJ 
 
 and as the unit stress on the extreme elements of the web is practically 
 the same as that of the flanges, we have /'=/ and also F fA =fA. Then 
 by substituting these values in (3), we have 
 
 But h = d, practically, so we have 
 
 fA'd ,t A A'\ 
 M = fAd + ' =fd lA + \ 
 
 from which we obtain 
 
 which shows that one-sixth of the area of the web appears as flange area, 
 but owing to the moment of inertia of the web being reduced in most cases 
 
180 STRUCTURAL ENGINEERING 
 
 by rivet holes, one-eighth is assumed in practice instead of one-sixth, in 
 which case we have 
 
 A' M 
 A + ^ = fd ....................... .............. 
 
 from which we obtain 
 
 for the area required in each flange. 
 
 113. Economic Depth. It is seen from the preceding article that 
 the area, and consequently the weight, of the flanges of a plate girder 
 varies inversely as the depth of the girder and it is evident that the 
 weight of the web, stiffeners, and fillers varies directly as the depth. Then 
 evidently a plate girder will have a theoretical economic depth when the 
 weight of the two flanges equals the combined weight of the web, stif- 
 feners, and fillers. There are really two cases to consider, which are as 
 follows : 
 
 Case I. When the web is not considered to resist cross bending. 
 Let M = maximum bending moment in inch pounds ; 
 L = length of girder in feet; 
 / = allowable unit stress on flanges ; 
 t = thickness of web in inches; 
 W = total weight of girder in pounds ; 
 
 x depth of girder in inches back to back of flange angles. 
 Girders without cover plates. For the area of the cross-section of 
 the two flanges we have 
 
 2M 
 
 f* 
 
 assuming depth and effective depth as being equal, and for the weight of 
 the two flanges we have 
 
 2fx3.4L. 
 fx 
 
 As a bar of steel one square inch in cross-section and one foot long 
 weighs 3.4 pounds, the total weight of the web alone is equal to ZALtx 
 and the total weight of the stiffeners, fillers, splices, etc., usually runs 
 about 60 per cent of the weight of the web, so the total weight of the web, 
 stiffeners, fillers, etc., can be taken as 1.6(3.4Lto). 
 
 Now adding the weight of the web, stiffeners, fillers, etc., to that of 
 the flanges, we have 
 
 W = 2-x 3.4L + 1.6(3.4Lte) .......................... (1) 
 
 for the total weight of the girder. 
 This will be a minimum when 
 
 = ~ 2 x 3AL + 1 - 6 ( 3 - 4L *)= o. 
 
 which is obtained by differentiating (1). 
 
DESIGN OP I-BEAMS AND PLATE GIRDEKS 
 From this we obtain for the economic depth 
 
 ............ ......... (2). 
 
 Girders with cover plates. In this case the area of the flanges varies 
 from the center to the ends of the span. If the cover plates have theo- 
 retical lengths, the average cross-section of each flange will be about 0.75 
 of the maximum area. So for the total weight of the two flanges we have 
 
 Now substituting this instead of 2(M/fx) x (3.4L) in (1) and differ 
 entiating and reducing we obtain for the economic depth 
 
 O /QN 
 
 Case II. When the web is considered to resist bending moment. 
 
 By assuming the web to resist bending moment each flange can be 
 reduced in area to the amount of one-eighth of the area of the cross- 
 section of the web, and hence the total weight of the girder would be 
 reduced an amount equal to 2(^tx)3AL. 
 
 Then subtracting this from (1) we have 
 
 W = 2 j 3AL + 1.6(3.4Lte)- i* 3.4L .............. (4) 
 
 for the total weight of the girder. Differentiating this equation and 
 placing the first derivative = 0, and reducing, we obtain 
 
 /KN 
 
 for the economic depth of girders without cover plates, and by substituting 
 1.5(M//j?) x (3.4L) for 2(M/fx) x (3.4L) in (4) and differentiating and 
 placing the first derivative = 0, and reducing, we obtain 
 
 x = 
 
 ( 6 ) 
 
 for the economic depth of girders with cover plates. 
 
 The theoretical economic depth of plate girders can be computed 
 from the above formulas. An inch or two either way from the theoretical 
 depth will affect the design but little. 
 
 114. Length of Cover Plates. A flange of a plate girder varies in 
 area along the girder practically as the ordinates of a parabola, the same 
 as the bending moment. (See Art. 56.) Then, if AB (Fig. 147) repre- 
 sents the length of a plate girder and OC the total area of the cross- 
 section of one flange at the center of the span, any ordinate x to the 
 parabola ABC will represent (approximately) the area of the flange at 
 that point. 
 
 Suppose the flange of this girder to be composed of two angles and 
 three cover plates, 
 
182 
 
 STRUCTURAL ENGINEERING 
 
 Let al = area of top plate; 
 
 a2 = area of second plate from the top; 
 03 = area of third plate from the top; 
 a4 = area of the two angles ; and 
 A = total area of the flange. 
 
 Theoretically, net areas should be used throughout and one-eighth of 
 the area of the web should be included with the area of the angles, but to 
 provide against discrepancies that may result by assuming that the bend- 
 ing moment varies along the girder as the ordinates to a parabola (which 
 is not absolutely true in the case of concentrated loads), we will use gross 
 areas throughout and neglect the one-eighth of the web in determining 
 the length of cover plates. 
 
 2-6 
 
 oz 
 
 ^L 
 
 03 
 
 4 
 
 ^ 
 
 
 B 
 
 Fig. 147 
 
 Now, if these areas be laid off to scale, on the line OC (Fig. 147), 
 the theoretical diagram of the cover plates can be drawn as shown and 
 then the theoretical length of each plate can be determined by scale. In 
 this manner the length of the cover plates on any plate girder can be 
 graphically determined, but in practice the lengths are usually computed 
 from the formulas given below, as that is the more convenient way. 
 
 In accordance with the properties of the parabola we have, referring 
 to Fig. 147, 
 
 A 
 
 from which we obtain 
 
 for the half length of the top cover plate, and multiplying this by 2, we 
 have 
 
 (i) 
 
DESIGN OF I-BEAMS AND PLATE GIRDERS 183 
 
 for the full theoretical length. Likewise we have 
 
 #" al + a2 
 
 (I) 
 
 from which we obtain 
 
 for the half length of the second cover plate from the top, and multiplying 
 this by 2, we have 
 
 lal + a2 
 
 for the full theoretical length of that plate. 
 In the same manner we obtain 
 
 al + a2 + a3 ,o\ 
 
 -J- 
 
 for the full length of the third cover plate from the top, and further, we 
 have 
 
 al + a2 + a3 + a4 T 
 
 ~ ~ =L 
 
 for the full length of the angles. 
 
 Now from the above it is readily seen that the general formula for 
 the theoretical length of cover plates can be written as 
 
 - ' +an ............................ (a). 
 
 A 
 
 Then for the length of the first or outside plate, in either the top or bottom 
 flange, we have 
 
 for the second plate from the top or bottom we have 
 
 "_ r 
 
 and for the third we have 
 
 and so on. 
 
 The theoretical lengths of cover plates can be determined very 
 readily by the use of the ordinary slide rule. When a slide rule is used, 
 Formula (a) should be squared, thereby obtaining 
 
 . . . +an). 
 
184 
 
 STRUCTURAL ENGINEERING 
 
 Then, first the quantity L?/A can be set off on the upper scale, once for all. 
 By multiplying this by al, (al + a%), (al + a2 + a3), and so on, we obtain, 
 respectively, the square of the lengths of the first, second, third, etc., 
 cover plate, and at the same time the square root of this in each case 
 (which is the length desired) is read off on the bottom scale. Thus the 
 length of each cover plate on a girder can be determined by one setting of 
 the rule. 
 
 115. Increment of the Flange Stress. Imagine the web of the 
 girder shown at (a), Fig. 145, to be made up of vertical strips each of 
 
 r 
 
 a d 
 
 ot.-oi-o-r-o i o 
 
 J J i 
 
 o o 
 
 o o r 
 
 O i O JO 
 
 Kl^e 
 
 4 I E i 
 
 , oi o; 
 
 B 
 
 Rl 
 
 Fig. 148 
 
 which is connected to each flange by one rivet, as shown in Fig. 148. 
 
 Suppose the girder supports a number of loads and let R and 721 be 
 the reactions due to these loads and for the sake of simplicity suppose 
 both of the reactions and all of the loads to be applied directly to the web. 
 
 The loads tend to move the girder downward as a whole ; this motion 
 is prevented by the reactions, and through the balancing of this activity 
 results first a vertical shearing couple on each of the imaginary strips 
 which tends to rotate each strip, but rotation of each is prevented by the 
 rivets connecting it to the flange angles, whereby the flange stress results. 
 To show this, let us first consider the end strip at A. The shearing couple 
 on that strip tends to rotate it clock-wise which causes the strip to exert 
 (through the connecting rivet) a force to the right upon the top flange 
 angles and an equal force to the left upon the bottom flange angles, and 
 the same is true of the second strip from the end A, and of the third, and 
 so on, while, as is readily seen, the shearing couple on each of the strips 
 near end B tends to rotate each of those strips in the opposite direction, or 
 counter clock-wise, and hence the force exerted on the top flange angles 
 by each will be to the left, while the equal force in each case will be 
 exerted to the right upon the bottom flange angles. 
 
 The end rivet in the top flange angles at A and the corresponding end 
 rivet at B, acting toward each other, produce a simple compressive stress 
 in the flange angles throughout the distance between these rivets, and, 
 similarly, the second rivets from the ends, acting toward each other, will 
 produce a like stress in the angles throughout the distance between those 
 rivets, and the third rivets from the ends will produce a like compressive 
 stress in the flange angles from one rivet to the other, and so on. The 
 same is true of the bottom flange except the direct stress produced there is 
 tension. 
 
 Thus it is seen that the stress in the flanges of a simple plate girder 
 
DESIGN OP I-BEAMS AND PLATE GIRDERS 185 
 
 is increased by each flange rivet as we pass from either end up to the 
 point of maximum moment. (See Art. 60.) This increase at each rivet 
 is the increment of the flange stress, which is often referred to as the 
 "flange increment." The summation of these increments between any two 
 points would be known as the increment of the flange stress between the 
 two points. 
 
 116. Spacing of Rivets in the Vertical Legs of Flange Angles, 
 The rivets in a plate girder are practically always the same size 
 throughout the girder, in which case the rivets in the vertical legs of the 
 flange angles should be so spaced that the pressure against the flange 
 angles would be the same for each rivet throughout the girder,, and hence 
 the stress on each rivet would be the same throughout. So the problem 
 involved is to determine the pitch of the rivets so that the increment of the 
 flange stress is just equal to the allowable stress on the flange rivet at all 
 points along the flange. 
 
 Let S be the shear on any strip abed (Fig. 148) of longitudinal 
 length p, and let r represent the increment of the flange stress at that 
 strip. Now as the couple rh is equal to the equal and opposite couple 
 reacting on the strip and balancing the shearing couple Sp, we have 
 
 Sp = rh, 
 from which we obtain 
 
 rh n ^ 
 
 P=^ ............................................. (1) - 
 
 From this the length of any strip can be computed for any desired value 
 of r. But the length of any strip can, in all cases, be taken as the pitch 
 of the rivets at the same point and hence if r be taken as the allowable 
 stress on one rivet, the required pitch of the rivets in the vertical legs of 
 the flange angles at any point can be computed from this formula. 
 
 It is readily seen (from Fig. 148) that these rivets would fail either 
 in double shear or in bearing on the web, and r, in any case, should be 
 taken equal to whichever is the least. The bearing on the web is usually 
 the least. 
 
 There are really four cases to consider: 
 
 Case I. When the loads are applied directly to the web and the 
 resistance of the web to bending is neglected. In this case the pitch of 
 the rivets at any point of the flange is determined from the above equation, 
 
 where p = pitch ; 
 
 r- allowable stress on one rivet in either double shear or bearing 
 
 on the web, whichever is the least ; 
 
 h = vertical distance between the rivets in the two flanges ; 
 S - shear on the girder at the point considered. 
 
 Case II. When the loads are applied directly to the web and the 
 resistance of the web to bending is considered. In this case a certain part 
 of the shear couple Sp is resisted by the web. 
 
 Let S = shear on the girder at any point ; 
 
 A =area of the cross-section of one flange at any point; 
 
386 
 
 STRUCTURAL ENGINEERING 
 
 A f = area of the cross-section of the web; 
 f = stress per square inch transmitted to each flange by each 
 
 rivet ; 
 
 r pressure exerted upon the flange by each rivet ; 
 h = vertical distance between the rivets in the two flanges ; 
 p = pitch of flange rivets. 
 
 It is shown in Art. 112 that the resistance of the web to cross bending 
 can be accounted for by considering one-eighth of its area as being con- 
 centrated in each flange. So the portion of the shearing couple Sp resisted 
 by the web can be approximately expressed as fA'h/% and the remaining 
 portion of the couple which is resisted by the flanges can be approximately 
 expressed as fAh. Then adding these two expressions, we have 
 
 But, 
 
 then substituting this value of / in the last equation and reducing, we have 
 rhi A" 
 
 -*V+a, 
 
 P 
 
 for the pitch of the rivets. 
 
 Case III. When the loads are applied di- 
 rectly to the flange angles and the resistance of 
 the web to bending is neglected. In this case the 
 loads will exert a vertical force upon the rivets in 
 addition to the horizontal force considered above. 
 Let v (Fig. 149) represent the vertical force per 
 linear inch of flange and r the horizontal force on 
 the rivet (the same as above). Then for the 
 resultant force we have 
 
 But from (2) we have 
 
 Fig. 149 
 
 -* 
 
 and substituting this in the last equation, we have 
 
 Then transposing and extracting the square root, we have 
 
 R 
 P = 
 
 S 
 h 
 
 for the pitch of the rivets, where 
 
 p = pitch ; 
 
 S = shear on the girder at point considered; 
 
 v load on the flange angles per linear inch ; 
 
 (4) 
 
DESIGN OF I-BEAMS AND PLATE GIRDERS 
 
 187 
 
 R = stress on rivet, which should be taken as the allowable stress on 
 
 one rivet; 
 h = vertical distance between the rivets in the two flanges. 
 
 Case IV . When the loads are applied directly to the flange angles 
 and the resistance of the web to bending is considered. We have here, 
 the same as in Case III, 
 
 except 
 
 r 
 
 Sp 
 
 which is obtained from (3). 
 
 Then substituting this value of r, we have 
 
 from which we obtain 
 
 (5) 
 
 for the pitch, where the letters signify the same as specified above. 
 
 In case the flange angles have double rows of rivets, h should be 
 taken as the mean vertical distance between the rivets in the two flanges. 
 
 117. Web Splice. It is always desirable that the web of any plate 
 girder be one continuous piece, but the length of such plates is limited by 
 the steel mills and whenever the length of a web exceeds these limits it 
 becomes necessary to splice it. 
 
 8 
 
 Fig. 150 
 
 In practice there are two standard ways of splicing a web, both of 
 which are shown in Fig. 150. The splice shown at (a) is made up of six 
 
188 STRUCTURAL ENGINEERING 
 
 splice plates, three on each side of the web, while the splice shown at (6) 
 is made up of only two plates, one on each side of the web. The splice 
 plates should have sufficient material to transmit the shear at the point of 
 splice, and also the cross bending on the web. The maximum shearing 
 and maximum bending stress at any point in a simple girder do not occur 
 at the same time, as is readily seen, yet in designing web splices it is 
 practice to consider the two occurring simultaneously at the splice which 
 is an assumption entirely on the side of safety. 
 
 In the case of the splice shown at (a) the plates marked pi are 
 usually assumed to resist only the cross bending on the web, while the 
 plates marked p2 are assumed to resist only the shear on the web. That 
 is, the plates pi and their connection to the web are designed as if no 
 shear occurred at the splice and plates p2 and their connection to the web 
 are designed as if no cross bending occurred. To illustrate this method of 
 designing the splice: 
 
 Let / = allowable stress per square inch in each flange; 
 A' = area of the cross-section of the web; 
 
 d = vertical distance between the centers of gravity of the flanges ; 
 d' vertical distance center to center of plates pi ; 
 F = stress in each pair of plates pi ; 
 S = maximum shear at the splice. 
 
 Then the resistance of the web to bending is represented by the 
 couple (fA'/8)d which must be equal to Fd' and hence we have 
 
 8 
 
 for the direct stress in each pair of plates pi, due to the cross bending on 
 the web. Now, evidently, these plates should be designed to take this 
 stress F and the number of rivets on each side of the splice connecting 
 each pair of these plates to the web should be sufficient to transmit the 
 stress F in either double shear or bearing on the web, whichever requires 
 the greater number of rivets. However, as the bending stress varies 
 directly as the distance out from the center of the web, the intensities used 
 in designing the plates pi and also the stress allowed on the rivets con- 
 necting them to the web must be proportional to the intensities allowed in 
 the flange. For example, if the allowable stress in the flange is / per 
 square inch, the intensity to allow on the plates pi would be (/) d' /d, 
 and if r is the stress allowed on a rivet, say, in bearing on the web at the 
 flange, the same size rivet through the plates would have an allowable 
 bearing on the web of (r) d f '/d. 
 
 Then for the net area of cross-section of each pair of plates pi we 
 have 
 
 ..I,-' 
 
 And for the number of rivets required on each side of the splice, we have 
 
 F 
 
DESIGN OF I-BEAMS AND PLATE GIRDEKS 189 
 
 In designing the plates p2, all we have to do is to select two plates 
 that have sufficient net area along the vertical section to transmit the 
 shear and wide enough to admit the necessary rivets on each side of the 
 splice. However,, as a matter of fact, there is,, as a rule, more metal in 
 the two splice plates than is necessary to carry the shear, as each plate is 
 usually as thick as the web, in which case the two plates would have twice 
 the net cross-section of the web. The number of rivets connecting the 
 plates p2 to each side of the splice should be sufficient to transmit the 
 shear. If r be the allowable stress on each rivet and we will assume 
 that each resists the same amount of stress we have 
 
 for the number of rivets on each side of the splice in plates p2. The 
 value of r in this case is the full allowable intensity on a rivet in double 
 shear or bearing on the web, using, of course, whichever is the least. 
 
 The assumption that plates pi resist only the cross bending on the 
 web and plates p2 only the shear is quite a reasonable assumption as 
 plates p2 are too near the center of the web to resist so very much of the 
 cross bending and plates pi are really so narrow and too far from the 
 center of the web to transmit so very much of the shear. As a matter of 
 fact, however, the plates p2 do resist some of the cross bending and plates 
 pi resist some of the shear. 
 
 In the case of the splice shown at (6) the two plates should have 
 sufficient net area of cross-section along the line cc or c f c f to resist the 
 cross bending on the web and also the shear. 
 
 Let /' = stress per square inch on the outer edges of the splice plates ; 
 
 / = the allowable stress per square inch in the flanges; 
 A f area of cross-section of the web; 
 h height of each splice plate; 
 
 7 = net moment of inertia of the two splice plates along the 
 section cc or c' ' c f (deducting the moment of inertia of the 
 rivet holes). Then we have 
 
 for the stress per square inch on the outer edges of the splice plates. This 
 stress should not exceed fh/d. As a matter of fact it never does, as the 
 two splice plates, as a rule, have at least twice as much area of cross- 
 section as the web. However, the stress /' should be considered in doubt- 
 ful cases. The net area of the cross-section of the two splice plates along 
 section cc or c'c' should be sufficient to transmit the shear. That is, the 
 shear divided by the net area should not exceed the allowable unit shear- 
 ing stress on the web, say 10,000 Ibs. per square inch, as specified in the 
 A. R. E. Ass'n Specifications. 
 
 In case the web be spliced as shown at (6) we can not well consider 
 other than that the maximum stress on each rivet is due to the cross 
 bending on the web and shear combined. The cross bending produces a 
 horizontal force on each rivet while the shear produces a vertical force 
 and, as is evident, the maximum stress on each rivet will be the resultant 
 of these two forces. The vertical shearing stress will be practically the 
 
190 STRUCTURAL ENGINEERING 
 
 same for each rivet, but the stress due to cross bending, which (as stated 
 above) is transmitted to each horizontally, will vary directly as their 
 distance out from the center of the web. 
 
 Let s be the horizontal stress due to cross bending on each of the 
 rivets farthest out, as indicated at (6), and let e be the distance of these 
 rivets from the center of the web. Then if there were a rivet out unit 
 distance from the center of the web the stress on it would be equal to s/e 
 and evidently the stress on any rivet out 2 distance from the center of the 
 web would be (s/e) z and the moment of this force or stress about the 
 center of the web would be 
 
 s \ i 
 -z )= - 
 e J \t 
 
 - r 
 
 Now it is evident that if this moment be determined for each and 
 every rivet on one side of the splice, and these be added together, their 
 sum would be equal to the couple (fA'/8)d. So we have 
 
 A* = 
 
 From this equation s can be determined and if v be the vertical shear on 
 each rivet, which is readily computed, we have 
 
 for the maximum resultant stress on each outer rivet which is the absolute 
 maximum. 
 
 As an example, let a, b, c, d, and e represent, respectively, the dis- 
 tance of the first, second, third, fourth, and fifth horizontal row of rivets 
 out from the center of the web, shown at (6),, Fig. 150, then we have 
 
 n /2sa 2 2sb 2 2sc 2 2sd* \ ifA'\. 
 
 2 (- -+- - + - - + - - + 2se) = ' )d. 
 \ e e e e / \ 8 / 
 
 From this we obtain 
 
 fA'd 
 
 * = 
 32 
 
 /a 2 b 2 c 2 d 2 \ 
 
 I + 4- + + e ) 
 \e e e e I 
 
 Then combining this with the vertical shear, as shown above, the maximum 
 stress on the rivets can be obtained. However, s should not exceed the 
 allowable at the flange multiplied by 2e/d. 
 
 The maximum stress on the rivets in the type of splice shown at (a) 
 can be determined in the same manner. In fact, the rivets in all web 
 splices should be tested in this manner. 
 
 118. The Stiffening Angles on a plate girder really have two func- 
 tions: one is to stiffen the web against buckling, and the other is to 
 transfer loads from the flanges directly to the web. It is obvious that the 
 closer the stiff eners are spaced along a web the stronger the web is 
 against buckling, and consequently the higher the web can be stressed. 
 So the design of the web is influenced by the spacing of the stiffening 
 angles. 
 
 There is no theoretical way of determining the spacing of stiff eners. 
 A practical rule is to space them so that their distance apart along the 
 
DESIGN OF I-BEAMS AND PLATE GIRDEHS 191 
 
 girder is equal to the depth of the girder, but in no case farther apart 
 than five feet or a little over. However,, to satisfy modern requirements, 
 it is usually necessary to space them closer together near the ends of the 
 girders than this practical rule calls for in order to obtain an economic 
 web. The allowable spacing of the stiffeners for a given stress on the 
 web is specified as 
 
 ,,) .................................... (1) 
 
 in the Specifications prepared by the A. R. E. Ass'n, 
 where d = distance between the stiffeners (as shown at (6), Fig. 145); 
 thickness of web; 
 
 6' = shearing stress per square inch in the web, which is obtained 
 by dividing the shear at the section considered by the gross 
 area of the cross-section of the web. 
 Equation (1) is really an empirical formula. 
 
 It is not economic, as a rule, to space stiffeners closer together than 
 half the depth of the girder. When a spacing less than that at the ends 
 of a girder is given by the above formula, a thicker web should be used. 
 There should always be stiffeners at any point where excessive 
 concentrated loads are applied. These stiffeners should be designed as a 
 column to support the load, the effective length being taken as half the 
 depth of the girder. 
 
 119. Example 1. Let it be required to design a girder 30 ft. long 
 (c.c. end bearings), to support a uniform dead load of 400 Ibs. per linear 
 foot of girder, and a live load of 4,000 Ibs. per foot. 
 
 For the maximum bending moment we have from dead load 
 
 M = -i x 400 x 30 2 x 12 = 540,000" Ibs. 
 
 8 
 
 and from live load 
 
 M' = - x 4,000 x"30 2 x 12 - 5,400,000" Ibs., 
 
 8 . 
 
 making a total of 5,940,000" Ibs. bending moment. 
 
 Let us assume the web to be f " thick, and that the loads are applied 
 directly to the top flange, and that the web resists bending. Having made 
 these assumptions we can proceed with the determination of the economic 
 depth. Formula (5), Art. 113, will be used, as girders of such short 
 lengths usually have no cover plates. So, substituting in this formula we 
 have 
 
 1,940,000 OOA ing 
 
 >,000 x 3/8 
 
 for the economic depth of the girder. So we will assume a 38" x f " web. 
 The next thing, we will test this web to see if it is satisfactory. For the 
 maximum end shear we have from dead load 
 
 = 400x15 = 6,000 Ibs. 
 
192 STRUCTURAL ENGINEERING 
 
 and from live load 
 
 ' = 4,000x15 = 60,000 Ibs., 
 
 making a total of 66,000 Ibs. end shear. 
 
 Now dividing this by the area of cross-section of the web 
 (=38"xf"),wehave 
 
 for the actual average unit shearing stress on the web. 
 
 Next, substituting this in Formula (1), Art. 117, we have 
 
 d = ^r (12,000 -4,630)= 69.2 ins. 
 
 for the maximum theoretical distance between stiffeners at the ends of the 
 girder; and as this spacing is not less than half the depth of the girder 
 the web itself is satisfactory, but common practice is not to permit the 
 distance between the stiffeners to be greater than the depth of the girder. 
 So we will space them so as not to violate this practice, that is, the clear 
 horizontal distance between the stiffeners will not be made more than 38" 
 in any case. 
 
 Then substituting 38 for d in Formula (1), Art. 117, we have 
 
 38=1(12,000-.), 
 
 from which we obtain 
 
 s = 7,946 Ibs. 
 
 as the allowable unit shearing stress on the web. 
 Now dividing this into the shear we have 
 
 66,000 
 
 for the required area (of cross-section) of the web, which is 5.95 n " 
 (=14.25-8.3) less than the area of the assumed web. So theoretically 
 the web can be thinner than ", but we will use the assumed web as the 
 specifications limit the thickness to f". However, in the cases of build- 
 ings and highway bridges, the web would likely be reduced in thickness, 
 but in no case should it be thinner than \" . 
 
 As an example in such cases, let us assume a 38" x J" web which has 
 an area of cross-section of 9.5 n " (=38"x J"). Then for the actual unit 
 shearing stress on the web we have 
 
 = 6,953 lb, 
 J.o 
 
 Substituting this in Formula (1), Art. 117, we have 
 d- i (12,000 -6,952)= 31.5 ins. 
 
 for the required spacing of the stiffeners at the end of the girders, and as 
 
DESIGN OF I-BEAMS AND PLATE GIRDERS 193 
 
 this is not less than half the depth of the girder the web is satisfactory 
 and hence would be used. 
 
 We will next take up the designing of the flanges. Let us first try 
 2 Ls 6" x 6" x \" for each flange. From Table 6 we find that the dis- 
 tance from the back of these angles to their center of gravity is 1.68". 
 So if the girder be made 38]" deep back to back of flange angles, that is, 
 J" greater than the depth of the web, which is usual practice, we have 
 38.25-1.68x2 = 34.89" for the effective depth. Now, by dividing this 
 into the maximum bending moment we have 
 
 for the stress in each flange, and by dividing this by 16,000, the allowable 
 unit stress, we have 
 
 170,000 
 
 10.6sq.ms. 
 
 for the required net flange area. 
 We have 
 
 2_[_s 6" x 6" x i" = 11.50 - 1 = 10.50 n " net 
 \ of web = (14.25 x)= 1.78 n "net 
 
 12.28 n "net 
 This is too large, so let us try 
 
 2__ [_s 6" x 6" x T y = 10.12 - .87 = 9.25 n " net 
 1 ! r ofweb = 1.78 n "net 
 
 11.03 n "net 
 
 This flange section is about right, being only about 0.4 n " more than 
 required, but the specifications require that the thickness of these flange 
 angles be at least one-twelfth of the width of the outstanding legs which 
 are 6". So these T V angles are too thin and consequently if 6" x 6" 
 angles be used at all the 6" x 6" x -J" angles would have to be used, which 
 gives an excess of metal. 
 
 Let us try 2 Ls 6"x4"x^" (the 4" legs outstanding) for each 
 flange. 
 
 The distance from the back of the 4" leg to the center of gravity of 
 each of these angles is (given in Table 4) 1.99", say, 2". Then for fc 
 effective depth we have 38.25-2x2 = 34.25". 
 
 Now dividing this into the maximum bending moment we have 
 
 5,940,000 
 
 34.25 
 
 for the stress in each flange, and dividing this by 16,000 we have 
 173,000 
 
 = 173,000 Ibs. 
 [ividing this fr 
 = 10.8 sq. ins. 
 
 16,000 
 
 for the required net flange area. 
 We have 
 
 g_Ls 6"x4"xi" = 9.50 - 1 = 8.50 n "net 
 ofweb= 1.78Q"net 
 
 10.28 n "net. 
 
194 STRUCTURAL ENGINEERING 
 
 This flange section is 0.52 n " less than the required, so let us try the 
 
 following: 
 
 2 Ls 6" x 4" x T V = 10.62 - 1.12 = 9.50 n " net 
 J- of web- 1.78 n "nct 
 
 11.28 n "net. 
 
 This section is 0.4S n " more than required. So either of the last two 
 sections may be used. We will use the latter (6"x 4" x -f$' f angles) as 
 that is on the side of safety. 
 
 The size of the intermediate stiffeners is governed practically by the 
 specifications which require that they be no less than " thick and the 
 width of their outstanding legs be no less than -$ of the depth of the 
 girder plus 2 inches. So in this case we will use 3J" x 3 J" x f " angles 
 for stiffeners throughout as this is the minimum size allowed. Thickness 
 of the end stiffeners would very likely be iV' or V > depending upon 
 conditions. If the girder be supported upon masonry, these stiffeners 
 would be designed as columns to take the end shear or reaction, one-half 
 of the depth of the girder being taken as the length of the column. 
 
 If the girder be riveted at the ends to other girders or to columns 
 which wholly support the girder, the end stiffeners in that case would be 
 at least -fa ff thick to provide for the facing of the ends of the girder 
 which is usually required in modern practice. 
 
 To obtain the spacing of the flange rivets at the ends of the girder, 
 we have 
 
 R = 7,880 Ibs. allowable bearing of a J" rivet on the f" web; 
 
 t> = 333 Ibs. = 4,000 H- 12; 
 
 = 66,000 Ibs.; 
 
 h = 31.5 = (38.25 - 6.75). 
 
 Then substituting these values in Formula (4), Art. 116, (the formula 
 required by the specifications), we obtain 
 
 7,880 
 
 *""" '66,000 
 
 .or the pitch of the flange rivets at the ends of the girder. This pitch 
 should be used for a distance out from the ends of the girder equal to the 
 depth of the girder and varied from there on toward the center of the 
 span to suit the shear at the different points, no pitch, however, being 
 greater than 6" in accordance with the specifications. 
 
 From the above calculations the necessary information for making 
 the detail drawing of the girder is obtained, and when this drawing is 
 finished the design of the girder is complete. All plate girders are 
 designed in this manner. 
 
CHAPTER XI 
 
 DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 120. Types. Simple railroad bridges can be divided into four 
 general types: beam bridges, plate girders, viaducts, and truss bridges. 
 Plate girders and truss bridges can be further divided into deck and 
 through bridges; deck when the track is supported on the top of the 
 structure; and through when the track is between the main girders or 
 trusses. Beam bridges and viaducts are always deck bridges. 
 
 121. The Specifications prepared by the American Railway Engi- 
 neering Association, referred to in the preceding chapter, will be used 
 throughout this work. 
 
 122. The Live Load usually specified for railroad bridges consists 
 of two typical consolidated locomotives and tenders coupled in tandem 
 followed by a uniform train load, and a special load concentrated on two 
 axles which is used in case of very short spans. The most common load- 
 ing of this type is that known as "Cooper's Loading," devised by Theodore 
 Cooper, M. A. Soc. C. E. Cooper's loadings vary in weight and are desig- 
 nated, beginning with the lightest, as 30, 35, 40, 45, 50, 55, and 
 60. These loadings vary by a certain ratio so that if any stress, shear, 
 bending moment, etc., due to any one of the loadings be known the intensi- 
 ties of the same due to any one other of the loadings can be obtained by 
 direct proportion, and hence any tables or diagrams giving the shears, 
 moments, etc., for any one of the loadings can be used for any of the other 
 loadings. The figures following the letter are the indices of the ratio 
 referred to above. Thus, for example, if any stress, shear, etc., due to 
 40 be known, the intensity of the same for 50 will be 50/40 of that 
 due to 40; 60/40 for 60; 35/40 for 35; and so on. Most of the 
 railroads in this country have adopted some one of Cooper's loadings, 
 either exactly or slightly modified. The trend has been toward the using 
 
 Fig. 151 
 
 of the heavier loadings. At present 50, represented in Fig. 151, is used 
 quite extensively. The spacing of the wheels is the same for all of the 
 loadings. 
 
 The 40 loading is the most convenient to use owing to the loads 
 being in even thousands of pounds. 
 
 Table A* gives for this loading moments about any wheel of the 
 wheels to the right or to the left of it, as will be seen upon inspection. 
 
 * The student should verify the results given In this table. 
 
 195 
 
196 STRUCTURAL ENGINEERING 
 
 This table is quite convenient in determining the centers of gravity, 
 shears, and moments, as will be shown later. 
 
 123. "An Equivalent Uniform Live Load" is sometimes used 
 instead of the wheel loads described above. In the case of beams and deck 
 girders this will give exactly the same results as the wheel loads, while in 
 the case of trusses and through girders the results obtained are only 
 approximately the same as for the wheels. 
 
 To obtain an equivalent uniform live load for the bending moment 
 on a beam or deck girder of length L, the maximum bending moment M 
 due to the wheel loads is computed, and then we have 
 
 from which we obtain 
 
 8M 
 
 **# 
 
 for the equivalent uniform load per foot which will produce the same 
 maximum moment on the beam or girder as the wheels. 
 
 To obtain an equivalent uniform live load for the end shear or 
 reaction on a beam or deck girder of length L, the maximum reaction R 
 due to the wheel loads is computed and then we have 
 
 from which we obtain 
 
 for the equivalent uniform live load which will produce the same maxi- 
 mum end shear or reaction as the wheels. 
 
 In practice the equivalent uniform live load for trusses is usually 
 taken either as the uniform load that will produce as great a bending 
 moment at the quarter point of the span as the wheel loads or the uniform 
 load that will produce as great a shear in the end panel as the wheel 
 loads. 
 
 The first is obtained by computing the maximum bending moment 
 M' at the quarter point due to the wheel loads, the span being considered 
 as a simple deck beam, and then we have 
 
 -= 
 
 from which we obtain 
 
 32M' 
 
 for the equivalent uniform live load per foot where L is the length of 
 span in feet. 
 
 The second equivalent load is obtained by computing the maximum 
 shear S in the end panel due to the wheel loads and then we have 
 
 -' (> 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 197 
 
 from which we obtain 
 
 S 
 P=L^d 
 
 2 
 
 for the equivalent uniform live load where L is the length of span and d 
 the panel length in feet. 
 
 The first loading gives stresses too low for web members and chords 
 near the end and too high for the chords near the center of the span, 
 while the second loading gives stresses about correct in all web members, 
 but considerably too high in the chord members, except the chords in the 
 end panels. To correct for this discrepancy the author has proposed the 
 reduction of the stresses in the chords (except the chords in the end 
 panels) by the per cent given by the following empirical formula:* 
 
 where L is the length of span in feet. 
 
 For example, the chord stresses in a 300-ft. span would be reduced 
 by 5.5 per cent. This does not apply to chords in end panels. 
 
 124. Dead Load consists of the weight of the metal in the structure 
 (except the metal at the supports), and the weight of the track, known 
 as the deck. The approximate weight of metal per foot of single-track 
 bridges designed for Cooper's E50 loading can be obtained from the 
 following formulas wherein 
 
 p = weight of metal per foot of span, 
 L = length of span in feet, c. c. end bearing: 
 For beam spans without lateral bracing, and stringers, 
 
 p = l2L + 100 (1), 
 
 For deck girder spans and beam spans with lateral bracing, 
 
 p = 12L + 150 (2), 
 
 For through plate girder spans, 
 
 p = 13L + 600 (3), 
 
 For truss spans, 
 
 p = VL + 660 (4) . 
 
 The weight obtained from the above formulas is for the metal alone, 
 and to this must be added the weight of deck which in the case of ordinary 
 wooden decks can be taken at 400 Ibs. per foot of track. The weight 
 obtained from the above formulas is usually correct enough for computing 
 stresses, as 10 per cent variation is permissible, but it should not be used 
 in making estimates of cost. 
 
 The approximate dead weight of metal in bridges designed to carry 
 Cooper's E6Q loading is about one-eighth more than given by the above 
 formulas, and the weight of those designed to carry the 40 loading is 
 about one-eighth less than that given by the above formulas. 
 
 The weight of metal in double-track bridges depends upon their 
 construction. Their weight as for metal is about 70 per cent heavier than 
 that of single-track bridges if three or two main girders or trusses are 
 used, but about 100 per cent if four trusses or girders are used. 
 
 * See Engineering News, September 13, 1906. 
 
198 STRUCTURAL ENGINEERING 
 
 In case of concrete or metal floors the floor should be designed and 
 then the weight of it computed, and then the weight of the main girders 
 or trusses can be assumed and added to this weight and in this way the 
 approximate dead load can be determined, which should be within 10 per 
 cent of the actual weight; if not, the dead load stresses should be redeter- 
 mined, using a revised dead load. 
 
 125. Impact. Rapidly moving trains will produce greater stresses 
 in bridges than the same load when simply standing on a structure, as we 
 consider it when computing the live-load stress, and to provide for this 
 additional stress a certain amount of the live-load stress in each member 
 is taken as the impact stress which is added to the corresponding live-load 
 ^tress. The impact stress is obtained by simply multiplying the maximum 
 live-load stress by a coefficient obtained from an empirical formula. The 
 following formula is the one most used: 
 
 _300_ 
 "L + 300 
 
 Then for the impact stress in any member we have the formula 
 
 r ._o/ 300 \ 
 \L + 300/ 
 
 as specified by the A. R. E. Ass'n in their Specifications referred to above, 
 where 
 
 7 = impact stress, 
 
 S = maximum computed live-load stress, 
 
 L = length of track (in feet) loaded when the maximum live-load 
 stress occurs. 
 
 126. Wind Loads. The horizontal pressure exerted on bridges by 
 the wind is known as the "Wind Load." As a rule this load can be safely 
 taken at 30 Ibs. per square foot of the horizontal projection of both the 
 structure and the live load carried. But, in addition to this, some pro- 
 vision should be made for lateral vibration due to the live load. To pro- 
 vide for this and the wind load a greater pressure than 30 Ibs. probably 
 should be used in some cases. 
 
 The lateral force, which includes the wind pressure, specified in the 
 A. R. E. Ass'n Specifications, will be used in this book. 
 
 BEAM BRIDGES 
 
 127. Preliminary. Beam bridges are used only for short spans, 
 rarely ever exceeding 20 ft. in length. There are usually at least two 
 I-beams under each rail connected to each other by diaphragms, similar to 
 that shown m Figs. 158 and 159, thus forming a compound girder. These 
 compound girders should be braced to each other by diagonal and trans- 
 verse bracing, as shown in Fig. 159, whenever the span exceeds 12 ft. 
 When the span length exceeds 15 ft., four panels of bracing should be 
 used, in which case a horizontal diaphragm is needed at each lateral 
 connection. In case there be more than two beams in each compound 
 girder the beams can be placed somewhat closer together than in the case 
 of two beams. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 199 
 
 Complete Design of a 10-ft. Span 
 
 128. Data.- 
 
 Length = 10' c.c. end bearings. 
 Dead load = 220 + 400 = 620 Ibs. per foot of span. 
 
 (From (1), Art. 123.) 
 
 Live load,, Cooper's E5Q (shown in Fig. 151). 
 Specifications A. R. E. Ass'n. 
 
 129. Calculations. For the maximum bending moment due to dead 
 
 load we have 
 
 = i x -^f- x 10 2 x 12 = 46,500 inch Ibs. 
 
 o & 
 
 From mere inspection of the live-load diagram, in Fig. 151, it is seen 
 that the maximum moment will occur either when one 62,500-lb. axle load 
 (of the special load) is on the span or when two of the 50,000-lb. axle 
 loads are on. So we have two cases to try. 
 
 The maximum moment due to the 62,500-lb. axle, one-half of which 
 goes to each girder, will occur (according to Art. 88) when the load is at 
 the center of the span, as shown in Fig. 152. Each of the reactions will 
 
 
 2.5' 
 
 
 
 ?! J 
 
 31 
 
 PB 
 
 ,t 
 
 /O' 
 
 Fig. 153 
 
 be equal to one-half of the load on each girder, or 15,625 Ibs. Then we 
 have 
 
 15,625x5x12 = 937,500 inch Ibs. 
 
 for the maximum bending moment due to this load. 
 
 The maximum due to the two 50,000-lb. axles will occur when the 
 loads are in the position shown in Fig. 153 (according to Art. 88) and it 
 will occur under the wheel marked 3. 
 
 Taking moments about A (Fig. 153) we have 
 
 Rl x 10 - 25,000 x 1.25 - 25,000 x 6.25 = 0, 
 from which we obtain the reaction 
 
 Rl = 18,750 Ibs. 
 
 Then for the bending moment at wheel 3 (which is the maximum) 
 we have 
 
 M' = 18,750x3.75x12 = 843,700 inch Ibs., 
 
 which is less than that produced by the one wheel of the special load, as 
 
200 STRUCTURAL ENGINEERING 
 
 is seen above, and hence the moment due to the special load will be used. 
 Then for the maximum impnct moment we have 
 
 = 937,500 =907,000 inch Ibs. 
 
 1U 
 
 Now adding the maximum dead- and live-load moments and impact 
 together we have 
 
 46,500 + 937,500 + 907,000 = 1,891,000 inch Ibs., 
 
 which is the total moment which the girder under each rail must be 
 designed to resist. 
 
 Dividing this by 16,000 (see Art. 105) we have 
 
 1,891,000 
 
 TTTTTTTT; =llo. tor the section modulus. 
 
 From Table I we find that this calls for 2 Is 15" x 42 # under 
 each rail. 
 
 We can now make a preliminary estimate of the dead load. The 
 weight of the four Is is 168 Ibs. per foot of span, and the details can be 
 neglected as the channel diaphragm (see Fig. 158) at the center of the 
 span is all the detail there is to consider. Then adding this 168 Ibs. to 
 the weight of the deck we have 568 Ibs., which is 52 Ibs. less than the 
 assumed dead load, but as this is within 10 per cent of the assumed load 
 no recalculations are necessary. 
 
 For the end shear or reaction due to dead load we have 
 
 R = x = 1 550 Ibs 
 
 The maximum end shear or reaction due to the live load will occur 
 when the two 62,500-lb. axles are on the span and in the position indi- 
 cated in Fig. 154. 
 
 Taking moments about the support B we have 
 
 R : x 10 - 31,250 x 10 - 31,250 x 3 = 0, 
 
 from which we obtain 
 
 ^ = 40,600 Ibs., /T) 
 
 D T 
 
 which is the maximum live-load end shear or reaction. T 
 
 For the impact we have Fig. 154 
 
 Now adding the above reactions and impact together we have 
 
 1,550 + 40,600 + 39,300 = 81,450 Ibs. 
 for the total end shear or reaction. Dividing this by 600 we have 
 
 81,450 
 
 - - =136 sq. ins. 
 600 
 
 for the required area of bearing on the masonry for each of the four 
 supports. This completes the necessary calculations, and next the general 
 drawing, as shown in Fig. 158, or a shop drawing can be made for the 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 201 
 
 span. The details shown in Fig. 158 should be studied by the student 
 until thoroughly understood. 
 
 Complete Design of a 15-Ft. Span 
 Data. 
 
 130. 
 
 131. 
 
 Length = 15' c.c. end bearings. 
 Dead load = 330 + 400 = 730 Ibs. per ft. of span. 
 
 (From (2), Art. 123.) 
 Live load, Cooper's 50. 
 Specifications, A. R. E. Ass'n. 
 Calculations. For the maximum bending moment due to dead 
 
 load we have 
 
 = - x x 15 2 x 12 = 123,000 inch Ibs. 
 
 8 2 
 
 , I 
 
 Oil <Nll 
 
 pms; 
 
 R]2.5'| 
 
 s' s' 
 
 2.SJK' 
 
 r 
 
 IS' 
 
 -J 
 
 It is readily seen, from the diagram 
 in Fig. 151, that the heaviest loading pos- 
 sible for this span is three 50,000-lb. axle 
 loads, and, according to Art. 88, these will 
 produce the maximum moment when 
 placed in the position shown in Fig. 155, 
 where the loads are indicated for one 
 girder only. 
 Then taking moments about B (Fig. 155) we have 
 
 R x 15 - 25,000 x 2.5 - 25,000 x 7.5 - 25,000 x 12.50 = 0, 
 from which we obtain 
 
 # = 37,500 Ibs. 
 for the reaction at A. Then taking moments about the center load we have 
 
 M' = (37,500 x 7.5 - 25,000 x 5) 12 = 1,875,000 inch Ibs., 
 and for the impact we have 
 
 300 
 
 7=1,875,000 
 
 1Kn 
 lo + oUU 
 
 =1,785,000 inch Ibs. 
 
 Now adding the dead and live moments and impact together we have 
 3,783,000 inch Ibs. for the maximum bending moment. Then dividing 
 by 16,000 we have 
 
 3,783,000 _ 
 16,000 
 
 for the section modulus which calls for 2 Is 20" x 70 # under each rail. 
 This beam appears a little heavy, from the section modulus, but the 
 material cut out of the web along the vertical row of rivets at the center 
 of the span must be taken into account. This is done by subtracting the 
 moment of inertia of the material cut out of the web from the moment of 
 
202 STRUCTURAL ENGINEERING 
 
 inertia of the beam. Then substituting this net moment of inertia in the 
 formula 
 
 f= My 
 
 which is Formula D, Art. 53, we obtain the maximum stress on the beam, 
 which should not exceed 16,000 Ibs. 
 
 The area of cross-section cut out of the web at each rivet hole, 
 assuming that |" rivets are used, is it x A = 0.46 square inches. Then 
 taking moments about the neutral axis, which we will assume is at the 
 third rivet from the bottom of the beam, which is only approximately true 
 (see drawing, Fig. 160), we have 
 
 0.46x6.75 +0.46x2.5 =23.8 
 
 for the moment of inertia of the material cut out above the neutral axis,, 
 and 
 
 0.46x^5 +0.46 x6 2 = 19.4 
 
 for the moment of inertia of the material cut out below the neutral axis. 
 Then adding these together we have 43.2 for the total moment of inertia 
 of the material cut out. Subtracting this from the moment of inertia of 
 one beam we have 
 
 1219.9-43.2 = 1176.7 
 
 for the net moment of inertia of one beam. 
 
 Now substituting in the above formula we have 
 
 1 3,783,000x10 
 /- 8 X " 1176.7 = 16 > 1001bs "> 
 
 which is very nearly the allowable stress. So the beam has practically 
 the correct section. 
 
 The rivet holes in the beams shown in Fig. 158 are so near the 
 neutral axis that the moment of inertia is affected but little, and hence 
 the material cut out was not considered in designing those beams. 
 
 We can now make a preliminary estimate of the dead weight. 
 
 The four beams will weigh 280 Ibs. per foot of span and the laterals 
 and details, including lateral connections and diaphragms, will be about 
 30 Ibs. per foot of span, so the total effective weight of metal per foot is 
 about 280 + 30 = 310 Ibs. Adding this to the weight of the deck we have 
 710 Ibs., which is 20 Ibs. less than the assumed dead weight, but this is 
 much less than the allowable 10 per cent deviation, so recalculation is 
 unnecessary. 
 
 The weight of details can be correctly assumed only by experienced 
 designers. It is necessary that the student draw out the details to some 
 extent in order to get the weight of metal to any reasonable degree of 
 accuracy. 
 
 For the maximum end shear or reaction due to dead load we have 
 
 rvqn 
 
 R = - i -x 7.5 = 2,740 Ibs. 
 II 
 
 The maximum live-load end shear or reaction will occur at A when 
 the wheels are iiv the position shown in Fig. 156. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 203 
 
 Then taking moments about B we have 
 
 R' x 15 - (25,000 x 15 + 25,000 x 10 + 25,000 x 5) = 0, 
 from which we obtain 
 
 fl' = 50,000 Ibs. 
 for the maximum live-load end shear or reaction. For the impact we have 
 
 = 50,000 
 
 / 300 \ 
 \15 + 300/ 
 
 47,600. 
 
 Adding these reactions and impact together we have 
 
 2,740 + 50,000 + 47,600 = 100,340 Ibs. 
 
 for the maximum end shear or reaction. 
 Dividing this by 600 we have 
 
 100,340 
 
 600 
 
 = 167.2 sq. ins. 
 
 for the required bearing on the masonry, for each of the four supports 
 According to the A. R. E. Ass'n Specifications the lateral force 01 
 the laterals will be 
 
 200 + 5,000x0.10 = 700 Ibs. per foot of span. 
 
 This force is applied to the laterals only at t!*e central connection. Let 
 Fig. 157 represent the plan of the laterals. There will be a load of; 
 700 x 7.5 = 5,250 Ibs. applied at the central connection which must be 
 
 Fig. 156 
 
 Fig. 157 
 
 transmitted to the ends by the laterals, one-half going each way. So, 
 evidently, the maximum shear in each panel will be 2,625 Ibs., which is 
 one-half of 5,250 Ibs. The stress in each lateral will then be equal to 
 2,625 xsectf. Sec(9 is equal to about 2.6. Then we have 2,625x2.6 = 6,825 
 Ibs. for the stress in each lateral. If this stress be tension, as it would be 
 in the case shown in Fig. 157, the net area required in each lateral would 
 be equal to 
 
 6,825 
 
 but when the lateral force is exerted in the opposite direction the laterals 
 would be subjected to stress of the same intensity, but it would be com- 
 pression, in which case the area required would be equal to 
 
 6,825 
 ( 16,000 -70- 
 
204 STRUCTURAL ENGINEERING 
 
 So the laterals must be designed as compression members as well as 
 tension members. Now as the stress is so very small the ratio of length 
 to radius of gyration will really govern. This should not be over ISO. 
 As regards construction, a single angle for each lateral is the best section. 
 Then if we use an angle the first question is which radius of gyration 
 shall we use in designing it? The laterals are fixed at their ends in the 
 horizontal plane so that if the radius about the vertical axis be used, only 
 one-half of the length of each lateral should be taken as the length. In 
 the vertical plane the laterals are only partially fixed at their ends, as the 
 connection plates resist bending upward and downward only to a limited 
 extent so that if the radius about the horizontal axis be taken, the full 
 length of the lateral should be taken, which, to be sure, is on the side of 
 safety. As the laterals are fixed in the horizontal plane and partially 
 fixed in the vertical plane they cannot fail readily about the 45 axis, so 
 it would not be correct to use the least radius of gyration of an angle. 
 From this it is seen that the radius about the horizontal axis is the 
 one to use. The length of each of the laterals is about 8 ft. or 96 ins. 
 Then the radius of gyration about the horizontal axis must not be less than 
 
 which permits the use of a 3" x 3" angle, but the specifications limit use 
 to 3J" x 3J" x f " angles, so this size will be used. 
 
 In the designing of the transverse struts, sense of fitness must govern 
 to some extent as rigidity is the thing sought. However, the stress is 
 readily determined. The one at the center of the span has the greatest 
 load which is 5^250 Ibs., which produces a compression stress in the strut 
 of that amount, as is readily seen from Fig. 157. A 9" [ is the maximum 
 size that will give us a convenient three-rivet connection. This size also 
 looks about correct as far as rigidity is concerned. 
 
 The length of these struts is about 42 ins. and the least radius of 
 gyration of the 9" [ is 0.64. Then we have 
 
 which is quite low. So the 9" [ appears to be satisfactory, and we will 
 use 1 [ 9"x20# for each transverse strut. 
 
 This completes the necessary calculations and next a general draw- 
 ing as shown in Fig. 159 can be made. This drawing shows the design only 
 in a general way and the bridge could not be fabricated from it. Such 
 drawings are usually made in railroad offices and in the offices of con- 
 sulting engineers. A bridge company would make a shop drawing as 
 shown in Fig. 160, and, in addition, the necessary shop bills from which 
 the bridge would be fabricated. 
 
 The following shop bills are about what a bridge company would 
 require for the work. However, bridge companies use printed forms. 
 
 Page No. 1 of the shop bills, which would accompany the shop 
 drawing shown in Fig. 160, is principally for the templet and bridge shop. 
 
 Pages No. 2 and No. 5 are principally for the forge shop. 
 
 Page No. 3 is principally for the pattern shop and foundry. 
 
 Pages No. 4 and No. 5 would be used by the shipping department. 
 
205 
 
Y////////////// 
 
 206 
 
207 
 
6 Eh 
 
 HAM E OF STRUCTURE / 
 
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 208 
 
GENERAL BRIDGE COMPANY 
 MISCELLANEOUS 
 
 NAME OF STRUCTURE I- Beam Bridge- for A.MK Y.R.ft. 
 
 
 
 Sketch 
 
 Maferial C a 
 
 c y Order 
 
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 Description 
 
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GENERAL BRIDGE COMPANY 
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 HAM EOF STRUCTURE 
 
 SENERAL BRIDGE COMPANY 
 
 SHIPPING BILL 
 
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 NAMEOFSTRUGTU 
 
 GENERAL BRIDGE COMPANY 
 
 SUMMARY OF FIELD RIVETS AMD BOLTS 
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 NAMEOF STRUCTURE 
 
 JEMERAL BRIDGE COMPANY 
 rORSLISTOF FIELD RIVETS AND BOLTS 
 
 I- Beam Bridge for AM&Y ft/?. 
 
 
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 213 
 
214 
 
 STRUCTURAL ENGINEERING 
 
 The superintendent of the shops, the inspectors, the erectors and 
 others would receive the shop drawing and bills, depending upon the way 
 the work is handled. The different companies do not have exactly the 
 same kind of bills nor do they handle the work exactly in the same way. 
 The above bills are intended only as a fair example of shop bills. 
 
 DRAWING ROOM EXERCISE NO. 2 
 
 Design a 12-ft. I-beam span and make a general drawing to a 1" 
 scale and a tracing of the same. The details to be similar to those shown 
 in Fig. 159. 
 
 Data: 
 
 Length = 12' 0" c.c. end bearings. 
 
 Width- 5' 0" c.c. of girders. 
 
 Dead load to be assumed. 
 
 Live load, Cooper's E5Q. 
 
 Specifications, A. R. E. Ass'n. 
 
 Fig. 161 
 
 A layout to a 1J" or 3" scale similar to the one shown in Fig. 161 
 should be made before beginning the final drawing for the span in order 
 to determine the correct dimensions of details. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 215 
 
 DECK PLATE GIRDER BRIDGES 
 
 132. Preliminary. Deck plate girder bridges are used for spans 
 ranging from 25 ft. to 110 ft. There have been a few spans longer than 
 110 ft. built, but such lengths are not economic, as a rule. 
 
 A single-track deck plate girder bridge is composed of two main 
 girders connected to each other by cross frames and laterals. An isometric 
 view of a typical single-track span is shown in Fig. 162 where the names 
 of the different parts of the structure are given. 
 
 Double-track deck plate girder spans are, as a rule, composed of two 
 single-track spans placed side by side. 
 
 Fig. 162 
 
 Complete Design of 50-ft. Single-Track Deck Plate Girder Span 
 133. Data. 
 
 Length = 50'-0" c.c. end bearings. 
 Width = 6'-6" c.c. girders. 
 
 Dead load = (750 + 400) = 1,150 Ibs. per ft. of span (Art. 124). 
 
 Live load, Cooper's ESQ. 
 Specifications, A. R. E. Ass'n. 
 
216 STRUCTURAL ENGINEERING 
 
 134. Calculations for Main Girders. For the maximum bending 
 moment due to dead load we have 
 
 M = i x 575 x~50 2 x 12 = 2,156,000 inch Ibs. 
 
 o 
 
 From inspection of the diagram of the loading in Table A, we can see 
 that the maximum moment due to live load will, very likely, occur under 
 one of the heavy wheels, either wheel 12 or 13. Let us assume wheel 13, 
 and let us consider wheels 9 to 16 on the span. Taking moments about 
 wheel 16 (using Table A) we have 
 
 for the distance that the center of gravity of these wheels (9 to 16 
 inclusive) is to the left of wheel 16. This shows that the center of gravity 
 comes 2.24 ft. to the left of wheel 13. Then the maximum bending 
 moment under wheel 13 will occur when the loads are in the position 
 shown in Fig. 163. (See Art. 88.) 
 
 2-14 , ' 
 
 Q 12 i i?s$\*\-+i i? 4.88 
 
 8' 8' 5>!5 
 
 25' 
 
 Rl 
 
 Fig. 163 
 
 Taking moment about B (Fig. 163), and using Table A, we have 
 R x 50 - 2,740 - (155 - 26) 4.88 = 0, 
 
 from which we obtain 
 
 #=67,390 Ibs. 
 
 for the reaction at A. 
 
 Then taking moments about wheel 13, considering the forces to the 
 left, we have 
 
 M' = 67,390x26.12 -818,000 = 942,200 foot Ibs. 
 
 or 11,306,000 inch Ibs. for the bending moment at that wheel when wheels 
 9 to 16 are on the span. Next, suppose wheels 10 to 16 on the span. 
 Taking moments of these wheels about 16 (using Table A) we have 
 
 - / 2,155 ' 
 
 #- = 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 217 
 
 for the distance that the center of gravity of the wheels (10 to 16 
 inclusive) is to the left of wheel 16. Therefore, the center of gravity is 
 0.43 ft. to the right of wheel 13. But when these wheels are placed on 
 the span for maximum moment, wheel 17 comes on from the right. So 
 we will consider wheels 10 to 17 on the span. Taking moments about 
 wheel 17 we find that the center of gravity of wheels 10 to 17 is 2.9 ft. 
 to the right of wheel 13. So this group of wheels will be in the position 
 shown in Fig. 164 when the maximum under wheel 13 occurs. 
 
 Fig. 164 
 
 Then taking moments about B (Fig. 164) (using Table A) we have 
 
 fix 50 -2,851 -(142 -13) 1.45-0, 
 from which we obtain 
 
 fi = 60,760 Ibs. for the reaction at A. 
 Then taking moments about wheel 13 we have 
 
 M"= 60,760x23.55 -480,000 = 950,900 foot Ibs. 
 
 or 11,411,000 inch pounds for the maximum bending moment under wheel 
 13 for this group of wheels. This is the absolute maximum as will be 
 found by further trials. It is seen that the center of gravity is a little 
 nearer to wheel 14 than it is to 13, yet the maximum moment occurs under 
 wheel 13. This is one of the few cases where the maximum does not 
 occur under the wheel nearer the center of gravity. (See Art. 88.) 
 Generally, the position of the wheels for maximum moment can be ascer- 
 tained the first trial. The above span is about as troublesome as any 
 found. 
 
 For the maximum impact we have 
 
 /= 11,411,000 x 
 
 300 
 
 L 
 
 \50 + 300 
 
 780,000 inch Ibs. 
 
 Now multiplying the above maximum moment and impact each by 
 50/40, as the loading specified is 50, and adding these results to the 
 dead-load moment, we have 
 
 2,156,000+14,263,000 + 12,220,000 = 28,639,000 inch Ibs. 
 for the total maximum moment on the span. 
 
 The next thing is to determine the economic depth. Assuming the 
 thickness of the web to be f" and substituting in equation (6), Art. 113, 
 we have 
 
 = 1.055 
 
 /28,< 
 
 Vie: 
 
 639,000 
 
 000 x 
 
 = 72.7 ins., 
 
 so we will use a 72" x f " web. 
 
218 STRUCTURAL ENGINEERING 
 
 We can now determine the flange area required. It is readily seen 
 that the effective depth will likely be a little less than the depth of the 
 web, so let us assume 71" as the effective depth. Then we have 
 
 38,639,000 + 71 = 403,000 Ibs. 
 for the flange stress. 
 Then 403,000 + 16,000 = 25.2 sq. ins., 
 
 the net area required for each flange. 
 Using the following: 
 
 2 Ls 6"x6"x T y = 12.86 - 2.25= 10.61 n " net 
 1 cover plate 14" x J" = 7.00 - 1 = 6.00 n " net 
 1 cover plate 14"x T V'= 6.12 - 0.87 - 5.25 n " net 
 I of the web = 3.37 n " net 
 
 25.23" net 
 
 for each flange we have practically the required area. 
 
 We can now determine the actual effective depth. Let Fig. 165 
 represent the cross-section of the above flange. 
 
 _ _ r^ 
 
 ^p .11. / M . a *. 
 
 NjJ 
 
 oiar" /?* 
 
 21$ 6*6 *j 
 
 Fig. 165 
 
 Then taking moments about the center of the top cover plate (see 
 Art. 47), we have 
 
 -_ 12.86x2.42 + 7.00x0.46 _ 
 
 for the distance from the center of the top cover plate to the center of 
 gravity of the flange. 
 Then we have 
 
 1.32-0.71 = 0.61 ins. 
 
 as the distance from the back of the angles to the center of gravity of the 
 flange. 
 
 The web is assumed 72" deep and according to practice the vertical 
 distance from the back of top flange angles to the back of the bottom 
 flange angles will be 0.25" more or 72.25". Then for the actual effective 
 depth we have 
 
 d= 72.25 - (0.61 x 2) = 71.03 ins., 
 
 which is almost what we assumed, so recalculation of the flange is 
 unnecessary. As a rule, the assumed effective depth does not come so 
 close to the actual ; however, \" either way will not materially affect final 
 results. If a greater difference than -J" is obtained the stress and section 
 of the flange should be recalculated, using the computed effective depth. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 219 
 
 The thickest cover plates should always be placed next to the flange 
 angles. So in this case the \" plate will be next to the angles, as shown 
 in Fig. 165. 
 
 For the length of the -}$' cover plate, or outside plate, we have 
 
 1 50^5=24.3 ft. ;;' ' ' 
 
 (see Art. 114) and for the length of the \" cover plate we have 
 
 It is customary to make cover plates from two to three feet longer 
 than the theoretical length, so we will make the above plates 27 and 37 
 feet long, instead of 24.3 and 35.5. 
 
 The next thing to determine is the stiffening angles and the web. To 
 do this the first thing is to calculate the maximum end shear. 
 
 For maximum reaction or end shear due to dead load we have 
 
 The maximum live-load reaction or end shear will occur when the 
 wheels are in the position shown in Fig. 166. (See Art. 86.) 
 
 Z' 
 
 Fig. 166 
 
 Taking moments about B, Fig. 166, and using Table A, we have 
 
 R'x50-4,072-142x2 = 0, 
 from which we obtain 
 
 R' = 87,100 Ibs. 
 
 for the maximum reaction or end shear at A due to JE40 loading. Then 
 for 50 we have 
 
 87,100x^| =108,900 Ibs. 
 For impact we have 
 
 I = 108,900 x = 93,400 Ibs. 
 o50 
 
 Now adding the above dead- and live-load reactions and impact 
 together we have 
 
 14,400 + 108,900 + 93,400 = 216,700 Ibs. 
 
 for the maximum reaction or end shear on each girder. 
 
 According to the specifications the outstanding legs of intermediate 
 stiffeners must not be less than one-thirtieth of the depth of the girder 
 plus two inches. So we have 
 
 Z + 2 = 4.4 ins. 
 
220 STRUCTURAL ENGINEERING 
 
 for the required width of their outstanding legs. The standard angle 
 coming nearest to this requirement is a 5" x 3J" ', 
 which will be used throughout. 
 
 ^ ie enc ^ stiff 61161 * 8 m ust be designed to transmit 
 the total maximum reaction, each pair being consid- 
 ered as a column having a length equal to one-half 
 the depth of the girder. (See specifications.) Then 
 assuming Ls 5" x 3J" x \" used, we have for each 
 pair a column having a cross-section as shown in 
 Fig. 167. For the radius of gyration of this column 
 in reference to axis y-y, we have 
 
 r= /8. x2.41 2 + 19.98 
 
 Then substituting this radius and 36" for the length in Formula Q, 
 Art. 73, we have 
 
 16,000-70^ = 15,100 Ibs. 
 
 for the allowable compressive unit stress on the end stiffeners. 
 Now dividing this stress into the maximum reaction we have 
 
 216,700 
 
 = 14.00 sq. ins. 
 
 for the required area of cross-section of the end stiffeners. So we will 
 use two pairs, or 
 
 4 Ls 5"x3i"xl" = 16.0 sq. ins., 
 
 the 5 x 3 J x -f^ angles being a little too small. 
 
 There is no theoretical way of computing the area of the inter- 
 mediate stiffeners. It is practice to make them as thin as the specifica- 
 tions will permit. So we will use 5" x 3J" x f " angles throughout for 
 intermediate stiffeners. 
 
 Using the assumed web, 72" x f ", we have 
 
 216,700 
 s= - =8,020 Ibs. 
 
 for the maximum average unit shearing stress in the web. 
 Then substituting in Formula (1), Art. 118, we have 
 
 40 ' > = 37 ins ' ( about ) 
 
 for the required distance between the stiffeners near the ends of the 
 girders. Now as this spacing is not less than half the depth of the web 
 the assumed web is economic and hence will be used. 
 For the bearing on the masonry we have 
 
 216,700 
 
 sq.ms. 
 
 Each support must be so designed that there will be at least this much 
 area of bearing on the masonry. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 221 
 
 This completes the necessary calculations for the main girders, as 
 far as the general design is concerned, and the next thing is the designing 
 of the lateral bracing, that is, the laterals and frames. 
 
 135. Calculations for Lateral Bracing. The lateral bracing 
 should always be symmetrical about the center of the span. The laterals 
 should have a slope as near 45 as is practicable. The distance between 
 cross-frames should never be over 15 ft. In accordance with this there 
 must be three intermediate cross-frames in a 50-ft. span, as a less number 
 would place them more than 15 ft. apart. So the lateral bracing will be 
 as shown in Fig. 168. 
 
 so' 
 
 [ 
 
 & 
 
 r 
 
 4- 3 
 
 @ 
 
 Fig. 168 
 
 r 
 
 / 
 
 
 
 According to the specifications the laterals must resist a uniform 
 lateral force of 200 + 0.10 (5,000) = 700 Ibs. per ft. of span, considered 
 as a moving load. 
 
 Suppose this force or load acts from the direction indicated by the 
 arrows, and suppose it moves on to the span from the right, as a uniform 
 live load. The panel load at any point h, g, etc., will be 
 
 P = 700x6.2 = 4,340 Ibs. (about). 
 
 Then for the maximum shear in the different panels, according to 
 Art. 90, we have 
 
 Shear in panel hg = - x 1= 540 Ibs. ; 
 
 x 3= 1,630 Ibs.; 
 x 6= 3,260 Ibs.; 
 xlO= 5,420 Ibs.; 
 x!5= 8,140 Ibs.; 
 x 21 = 11,400 Ibs.; 
 
 < 
 
 28 = 15,200 Ibs. 
 
 Now if each of these shears be multiplied by the secant of the angle 
 marked Q we shall obtain the stress in the corresponding diagonals (see 
 Art. 92). 
 
222 STRUCTURAL ENGINEERING 
 
 The tangent of angle 0, for the purpose of determining stresses, 
 can be taken as 
 
 || = 0.962 (about), 
 
 and the corresponding secant can then be found in almost any table of 
 natural trigonometrical functions (see Carnegie or Cambria handbook) ., 
 or the secant can be determined directly by arithmetic. 
 For the above assumed figures we have 
 
 Sec 6 = 1.39. 
 
 Then for the maximum stress in the diagonals we have 
 
 5,420x1.39=- 7,500 Ibs. for diagonal nd, 
 
 8,140x1.39 = +11,300 Ibs. for diagonal dm, 
 
 11,400x1.39 = -15,800 Ibs. for diagonal mb, 
 
 15,200x1.39 =+21,000 Ibs. for diagonal bl. 
 
 These are all of the stresses that we need determine with the load 
 applied as indicated as the corresponding diagonals in the right half of 
 the span will have the same stress when the force moves on to the span 
 from left to right. 
 
 If the lateral force comes from the other direction than that indi- 
 cated by the arrows, the panel loads will be twice as great as given above, 
 as there are only half as many panels considered. 
 
 Then the stress in each of the diagonals nd and dm is 
 
 4 ' 34 X 2 x 3 x 1.39 = 9,050 Ibs. 
 
 4- 
 and. -~ x 6 x 1.39 = 18,100 Ibs. 
 
 in each of the diagonals mb and bl. 
 
 We now have the maximum stresses in the diagonals determined 
 which are indicated in the diagram in Fig. 168. 
 
 The plus and minus signs above signify compression and tension, 
 respectively. 
 
 It is seen from the above that the diagonals have to resist both 
 tension and compression. Compression will likely govern. 
 
 Let us try a single angle, say, 1 L 3J" x 3 " x f", as this is the 
 smallest that the specifications will permit. From Table 6, or from a 
 Carnegie or Cambria handbook, we have 1.07 for the radius of gyration 
 of this angle about the horizontal axis (see Art. 130) and the length of 
 each lateral will be about 8 ft. or 96 ins., which can be determined accu- 
 rately enough by scale. Then substituting in Formula Q, Art. 73, we 
 have 
 
 Qfi 
 16,000-70^=9,700 Ibs. 
 
 for the allowable unit stress. 
 
 Dividing the greatest compressive stress, which occurs in the lateral 
 bl, by this intensity we have 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 223 
 
 for the required cross-section of the lateral, and the assumed angle has 
 2.48 n ". Then, as the ratio of L/r is only about 90, the assumed angle is 
 safe in the case of the greatest compressive stress. 
 
 As is seen above the greatest tension is 18,100 Ibs., which occurs also 
 in lateral bl. Then we have 
 
 18,100 
 lMOO =1 - 138q - m8 - 
 
 for the required net cross-section, and as the angle assumed has 
 [2.48 -(f"xl")] 2.11 n " it is quite safe for the greatest tension. So 
 we will use 1 L 3-J" x 3^" x f " for each end lateral. But as this angle 
 is the smallest permitted in this work the laterals throughout will each 
 be made of 1 3J" x 3 J" x f" angle. 
 
 The intermediate cross-frames are not subject to theoretical analyses 
 and hence their design is governed principally by experience and sense of 
 fitness. 
 
 The end cross-frames can be fairly well analyzed. The stress in the 
 top angle of the frame can be taken equal to one-half of the lateral force 
 per foot of span multiplied by one-half the length of the span, and this 
 force multiplied by the secant of the slope of the diagonals in the frame 
 is equal to the stress in each of these diagonals. The bottom angle of the 
 frame has no stress (theoretically). 
 
 Then according to the above we have 
 
 -^2-x 25 = 8,750 Ibs. 
 
 for the stress in the top angle of the end frame, and as the diagonals of 
 the frame slope about 45 with the horizontal we have 
 
 8,750x1.4 = 12,250 Ibs. 
 
 for the stress in the diagonals of the end frame. From this it is seen that 
 very small angles are required theoretically for the end frames which 
 are subjected to greater stresses (apparently) than the intermediate 
 frames, but as 3J" x 3|" x f " angles are practically the minimum size 
 used in railroad bridges we will use this size in all of the frames. 
 
 This completes the necessary preliminary calculations except for the 
 preliminary estimate of the dead weight. The stress sheet for the span, 
 as shown in Fig. 169, can be drawn from the information given in the 
 above calculations. The estimate of the dead weight in this case will be 
 deferred until after the detail shop drawing of the span is considered so 
 as to familiarize the student with details before taking up preliminary 
 estimates of dead weight. 
 
 136. Making of the Shop Drawing. After the stress sheet 
 (Fig. 169) is completed the shop drawing for the span, as shown in Fig. 
 171, can be made. This work is known as detailing and includes not only 
 the drawing but the calculations for the details as well. 
 
 To evolve this drawing (Fig. 171) the details are first drawn in 
 pencil to a f " scale upon a 24" x 36" sheet of ordinary drawing paper 
 After selecting the relative positions for the different views, so that no 
 part of the sheet will be crowded, we start the drawing, as shown in Fig. 
 170, by drawing the center line CC of the span. Then scaling off 25 ft., 
 
224 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 
 
 225 
 
 the line BB through the center of bearing is drawn. Next, the line EE, 
 through the end of the span, must be located and drawn. The end 
 stiffeners should be placed 6" back to back as shown, this being about the 
 minimum distance permissible on account of driving rivets in the out- 
 standing legs of the stiffeners connecting to the end cross-frames. Then 
 allowing, say, 2" from the end pair of stiffeners to the end of the girders, 
 we have 3" + 3J" + 2" = 8 \" for the distance from the line BB to line 
 
 ?-> - ^a--te-^&aiE<^^=^^ss~j^ 
 
 
 Fig. 170 
 
 EE, and the line EE locating the end of the span can then be drawn. 
 Then the top plan of each girder is drawn as shown in Fig. 170. The 
 next thing after that is to locate the cross-frames as shown, so that the 
 lateral bracing will be symmetrical about the center line CC, and the 
 laterals will all have the same length (thus saving templets). Now there 
 will be a pair of stiffeners on each girder at each cross-frame, one of 
 which, in each case, will be connected to a frame, and when the position 
 of the cross-frames is fixed these stiffeners to which the frames connect 
 are also fixed in position. Then the elevation of one of the girders, 
 always the far girder, can be drawn and the drawing completed as far as 
 is shown in Fig. 170. 
 
 The next thing is to draw the laterals and the lateral plates connect- 
 ing the laterals to the girders (as shown in Fig. 171). As the laterals 
 are all of the same section and the strength of each is greater than is 
 required by the stress, each end connection must contain enough rivets 
 to develop the strength of the lateral in compression (see specifications). 
 Now, as shown above, each lateral will safely resist 9,700 Ibs. per square 
 inch of cross-section. Then, as 2.48 n " is the area of cross-section of 
 each lateral, we have 2.48x9,700 = 24,000 Ibs. for the total allowable 
 compressive stress in each lateral. Then using " field rivets for the 
 
226 
 
GEM 
 NAME OF STRUCTURE J?t 
 
 ERAL BRIDGE COMPANY 
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GENERAL BRIDGE COMPANY 
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 NAMEOF STRUCTURE Soft. Deck Plate. Girder Bridge forAtf.fyYKft. 
 
 
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GENERAL BRIDGE COMPANY 
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 NAME OF STRUCTURE fi. 
 
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GENERAL BRIDGE COMPANY 
 SUMMARY OF FIELD RIVETS >\ND BOLTS 
 
 NAMEOF 5TRUCTUF? E Sott.S.T Deck P/afe G/'rder Bridge for A //,&)(/?. /?. 
 
 
 
 No. 
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 232 
 
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 fl/WEOrSTRUCTURE . 
 
 GENERAL BRIDGE CoMPAhY 
 
 SHIPPING BILL 
 
 J^y?v5: 1 Deck Plate 6/rderBr/c/qe for A.rt&rftK 
 
 
 Member 
 
 Material /v 
 
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 Contract No.-3^ . 
 Made by -&kA-J]&JS!e~ Checked by M%L. Page No..Z_._. 
 
 233 
 
234 STRUCTURAL ENGINEERING 
 
 lateral connections, the value of which in single shear will govern, 6,000 
 Ibs. being the allowable value of each, we have 
 
 24,000 _ 
 6,000 = 
 
 for the number of rivets required in each end of each lateral. 
 
 Now, the number of rivets connecting the lateral plates to the girders 
 must be sufficient to take the component along the girder of the lateral or 
 laterals. Twenty-four thousand pounds is the strength of each lateral, 
 and using J" field rivets in single shear, the number of rivets for inter- 
 mediate points will be 
 
 24,000 xsinA / 24,000 x .695 
 
 6,000 / 
 
 The number used would be not less than 6, or 3 for each lateral. 
 
 The laterals and lateral plates can now be drawn in the top plan 
 (shown in Fig. 171), giving the spacing of the rivets at each connection, 
 and determining clearances and the sizes of plates by scale as the work 
 progresses, using separate sketches for each point drawn, say, to 1-J" or 
 3" scale. After this work is completed the stiffeners between the cross- 
 frames can be drawn on the elevation of the girder, care being taken to 
 space the stiffeners so as to miss the lateral plates as much as possible. 
 Then the longitudinal section showing the bottom flange can be drawn. 
 
 After this the next thing is to space the rivets in the vertical legs of 
 the flange angles. As the live load is applied directly to the top flanges 
 and the specifications ignore the one-eighth of the web in determining the 
 spacing of flange rivets, the problem of determining the spacing comes 
 under Case III, Art. 116. 
 
 We assume that each wheel of the live load is distributed over three 
 ties and assuming ties 8" wide and spaced 6" apart (face to face) we 
 have 42" for the length over which each wheel is distributed. Then for 
 the greatest vertical load per linear inch of flange due to the live load 
 (Cooper's 50), when the maximum shear occurs, we have 
 
 = 595 Ibs.; say 600 Ibs., 
 
 / 
 
 per linear inch, and adding a 100 per cent for impact will make a total 
 of 1,200 Ibs. per inch. 
 
 Now referring to Case III, Art. 116, we have 
 
 v= 1,200 Ibs. 
 R= 7,880 Ibs., which is the allowable bearing of a J" shop 
 
 rivet on the f " web. 
 S= 216,700 Ibs., which is the maximum shear at either end of 
 
 each girder. 
 h= 66 ins. 
 
 Then substituting these values in Formula (4), Art. 116, we have 
 
 7 880 
 p= . =2.27 ins. (about) 
 
 / 216,700 \ 2 
 
 ^/1,200 2 ' 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 235 
 
 for the theoretical spacing of the rivets near the ends of the girders, and 
 the theoretical spacing at other points along the girders can be deter- 
 mined by computing the maximum shear at each of the points and apply- 
 ing Formula (4), Art. 116, in each case. However, the following graph- 
 ical method of determining the spacing is preferable as it saves time and 
 is accurate enough for practical work: 
 
 The shear due to dead load is quite small compared with that due to 
 live load and impact, so that the maximum shear can be assumed to vary 
 across the span as the ordinates to a parabola (see Art. 55). 
 
 Now the above spacing (2.27") gives about 5.3 rivets per foot of 
 girder. Then if we lay off a vertical line AB (Fig. 172) equal to 5.3", 
 and taking a pair of dividers divide the line AB into any convenient 
 
 Fig. 172 
 
 number of equal parts and from A step off the same number of equal 
 parts on the line AC (the equal parts on the line AC can be taken any 
 length, just so A to C is a convenient distance), and construct the 
 parabola CDB (see Art. 58) and draw the horizontal lines to the curve 
 through the points 1, 2, 3, etc., we have a diagram from which the 
 required spacing of the rivets at any point between the end and the 
 center of the span can be obtained, as the distance AC corresponds to 
 the length of the span and the line AB gives the shear (correspondingly) 
 in number of rivets per foot of girder. From the diagram (Fig. 172) it 
 is seen that about one and one-third rivets per foot are required at the 
 center of the span, and practically three rivets per foot at the quarter 
 point. The required spacing at all other points from A to the center of 
 the span is seen at a glance. 
 
 It is not practical, if not impossible, to space the rivets throughout 
 a girder according to the theoretical requirements, as no two spaces 
 would be the same (which is impractical) and the spaces next to stif- 
 feners can not be less than about 8J", which in some cases is greater than 
 the required; so the best we can do is to fit the rivets in between the 
 stiffeners to about the theoretical spacing. 
 
236 STRUCTURAL ENGINEERING* 
 
 The theoretical spacing of the flange rivets as determined in the 
 manner shown above will satisfy the requirements of the specifications, 
 which are as follows: 
 
 "The flanges of plate girders shall be connected to the web with 
 sufficient number of rivets to transfer the total shear at any point in a 
 distance equal to the effective depth of the girder at that point combined 
 with any load that is applied directly to the flange." 
 
 This is simply a practical manner of obtaining approximately the 
 theoretical spacing. To show the justification of the above requirement 
 given in the specifications let AB, Fig. 173, represent a plate girder. 
 Let M and S be the bending moment and shear, respectively, at section 
 C. Then for the bending moment at section D we have 
 
 M' = M + Sx-2Pz. (See Art. 66.) 
 
 Now, Sx - 2 Pz is the difference between the moments at the two sections. 
 If the intervening loads be neglected, which is an error on the side of 
 
 safety, the difference between the two 
 moments is Sx, and if this be divided by 
 the effective depth the result would be 
 the difference between the flange stresses 
 at the two sections, which would be 
 equal to the shear when x is equal to the 
 Fig. 173 effective depth, as is readily seen. This 
 
 difference of flange stress is simply the 
 
 increment of the flange stress between the two sections, and of course 
 there should be a sufficient number of rivets to transfer this from the 
 web to the flange and at the same time support whatever vertical load 
 there is applied to the flange. This is entirely in accord with Arts. 115 
 and 116. 
 
 Now, having the stiffeners spaced as shown in Fig. 171 we can begin 
 spacing the rivets in the vertical legs of the flange angles. Beginning at 
 the end of the girder, the theoretical spacing as given above is 2.27", but 
 we will use 2J" in order to have practical spaces. So we obtain the 
 spacing from the end of the girder to the first intermediate stiffener as 
 shown in Fig. 171. Between the first and second stiffeners we can 
 increase the spacing a little as shown by the diagram in Fig. 172, and 
 between the second and third stiffeners we can increase the spacing a 
 little more and so on towards the center of the span until the limit of 6" 
 spacing is reached, which according to the specifications must not be 
 exceeded. It is practice to limit the fractions in the spacing to no less 
 than y. If necessary the stiffeners can be shifted slightly to suit the 
 spacing. 
 
 After the rivets are spaced in the vertical legs of the flange angles 
 the rivets in the horizontal legs, connecting the cover plates to the angles, 
 can be spaced. The theoretical requirement in this case is that the rivets 
 be spaced so that the number of rivets per foot will be sufficient to 
 transmit the part of the flange increment taken by the cover plates. The 
 part of the flange increment taken at any point by the cover plates will 
 be to the total flange increment at that point as the area of the cross- 
 section of the cover plates is to the total area of the cross-section of the 
 flange at that point. As an example, let M be the bending moment found 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 237 
 
 at section C (Fig. 173) and AT the bending moment at section D, which 
 is a short distance to the right of C (say, a foot). Then for the total 
 flange increment between the two sections we have 
 
 where h is the effective depth of the girder. Now it is obvious that there 
 must be a sufficient number of rivets in the vertical legs of the flange 
 angles between the two sections to take all of this increment as the total 
 increment is transmitted thrpugh these rivets, but as the cover plates 
 take only their proportional part of the increment, the number of rivets 
 connecting them to the flange angles need be only sufficient to take their 
 part. So if r be the allowable stress on each rivet connecting the cover 
 plates to the flange angles, and A the total area of cross-section of the 
 flange between section C and D, and A", the area of the cover plates, we 
 have 
 
 F A" 
 n = xr 
 r A 
 
 for the number of rivets required in the cover plates between the two 
 sections and the number of rivets required to connect the cover plates to 
 the flange angles at any other point along the girder can be determined 
 in the same manner, but the same result can be obtained in any case with 
 much less work by using such a diagram as shown in Fig. 172. So we 
 will use the above diagram (Fig. 172) in this case. Beginning at the 
 center of the span, the number of rivets per foot required to take the total 
 flange increment is 1J. Then as the net area of cross-section of the 
 flange at that point, not including the one-eighth of the web, is 21.86 n ", 
 and the net area of the cover plates is 11.25 n ", we have 
 
 ' " ' Hxgff = 0.68 (about) 
 
 for the number of rivets per foot in the cover plates, which gives a 
 spacing of 17.6". This would be the spacing if the allowable stress on 
 the rivets in the horizontal legs were the same as in the vertical. But as 
 the rivets in the horizontal legs must be considered in single shear and 
 the rivets in the vertical legs in bearing on the f " web, the above number 
 (0.68) must be multiplied by 7,880/7,200. So we have 
 
 11.25 7,880 _ 
 
 for the number of rivets per foot in the cover plates at the center of the 
 span, which gives (12/0.75) 16" spacing, but as- there are two rows of 
 rivets in the cover plates the spacing of the rivets in each flange angle 
 would really be twice this, or 32". 
 
 Again, from the diagram (Fig. 172) the required number of the 
 rivets in the vertical legs of the flange angles at the quarter point is 3 per 
 foot. Then for the number of rivets required at that point in the cover 
 plate (as the T V' plate only extends to about that point), we have 
 6.00 7,880 
 
238 STRUCTURAL ENGINEERING 
 
 which gives a spacing of 10.1" or 20.2" along each angle. Now, if 
 other points along the girder be considered it will likewise be found that 
 the required spacing of the rivets in the cover plate or plates will exceed 
 the 6" maximum allowed. So that if we were to space the rivets in the 
 cover plates, using the 6" maximum allowable throughout, there would yet 
 be an excess of rivets. Now the spacing of the rivets in the lateral plates 
 is determined when the laterals are detailed, as stated above, and as 
 these" rivets must be spaced to suit those details it only remains to fit the 
 rivets in between the lateral connections using as near the maximum 6" 
 spacing as will fit in nicely without interfering with the stiffeners. So 
 we obtain the spacing shown in the plan of the top flanges (Fig. 171) and 
 the same spacing can be used in the bottom flange, as shown in the longi- 
 tudinal section below the girder. As is evident, the spacing selected at 
 the different points depends entirely upon individual judgment and hence 
 no two persons are likely to make the same selection. However, this is 
 not a serious matter at all as any reasonable variation is permissible. 
 
 We shall next space the rivets in the stiffeners. Now according to 
 shop practice the rivets in the stiffeners should all line up horizontally 
 all the way across the girders. 
 
 There is no definite rule for determining the number of rivets 
 required in the intermediate stiffeners other than that there should be a 
 sufficient number to clamp the stiffeners firmly to the web, and the 
 maximum 6" spacing usually suffices for this, but the end stiffeners should 
 be connected with a sufficient number to transmit the total end shear 
 from the web to the masonry, as these stiffeners are really columns bear- 
 ing against the bottom flange of the girder and really transmit this force. 
 It is customary to space the rivets in the end stiffeners so that about 
 every other rivet of this spacing can be omitted for the spacing in the 
 intermediate stiffeners whereby an economic spacing that lines up hori- 
 zontally throughout the full length of the girder is obtained. It is 
 economic to crimp all stiffeners on all girders over three feet deep, but it 
 is usual shop practice to put fillers under end stiffeners and under the 
 stiffeners at all points where cross-frames connect. So we shall put 
 fillers under the end stiffeners and under the stiffeners at cross-frames. 
 The rivets in the stiffeners are in double shear and bearing on the f " web. 
 Then, using J" shop rivets, we have 
 
 12,000x0.6x2 = 14,400 Ibs. 
 for the allowable shear on each rivet and 
 
 24,000 xjx 1 = 7,880 Ibs. 
 
 for the allowable bearing on each rivet. So the bearing governs the 
 number of rivets. 
 
 Now, taking the case of the end stiffeners, the maximum end shear 
 being 216,700 Ibs., we have 
 
 216,700 
 
 =27.5, say, 28, 
 
 for the number of rivets required in the two pairs of end stiffeners at each 
 support. But, as the end stiffeners are on fillers there must be an excess 
 of 50 per cent, according to the specification, so 42 rivets will be used as 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 239 
 
 shown (Fig. 171), and by omitting every other rivet in the spacing shown 
 in the end stiffeners we obtain the spacing shown in the intermediate 
 stiffeners. The rivets in the stiffeners can be made to line up horizontally 
 without any trouble if 3" spacing be used in the end stiffeners. In case 
 this spacing gives more rivets in the end stiffeners than are required, 
 some of the rivets can be omitted, thus making some 6" spaces. 
 
 The distance from the toe of any flange angle to the first rivet out 
 (in the stiffener) from the flange, according to shop practice, should be 
 equal to not less than twice the thickness of the flange angle plus 1J". 
 So in this case we have 1 J" + 1J" + 1-J" = 4-J" for the minimum allowable 
 distance from the outer gauge line in the flange angles to the first rivet 
 out from the flange. These distances are given in the spacing at the end 
 of the girder (Fig. 171) as 4-f", which exceeds the minimum allowed 
 by \" . These spaces should not be less in any case nor much more than 
 J" greater than the minimum allowed in conformity with the practice 
 mentioned above. 
 
 After the rivets are spaced in the stiffeners the web splices can be 
 calculated and drawn to conform to this spacing. The number of web 
 splices is always limited in all cases to as few as is possible, which 
 depends altogether upon the maximum length of the web plates obtainable 
 from the steel mills. Each splice should be at a stiffener. The maximum 
 length of 72" x f" plates is given in Table 9 as 30 ft. So one splice is 
 necessary in the above 50-ft. span. 
 
 Let us use the type of splice shown at (a), Fig. 150, Art. 117. The 
 first thing to do is to determine the size of the longitudinal splice plates 
 next to the flanges. These plates should have at least three horizontal 
 rows or rivets in them in order to get a good hold on the web. This much 
 is simply a matter of judgment. So let us assume that there are three 
 horizontal rows of rivets in each pair, then the distance from the center 
 of gravity of the rivets in the top pair of plates to the center of gravity 
 of the rivets in the bottom pair is equal to the distance from the second 
 rivet, from the top flange, to the second rivet from the bottom flange, 
 which, as seen from the spacing on the girder (Fig. 171) is 4 ft. or 48". 
 Now, according to Art. 117, using the figures given above, we have 
 
 from which we obtain 
 
 F = 79,600 Ibs. 
 
 for the direct longitudinal stress on each pair of plates. 
 
 Then we have 79,600 -H (16,000) 48/71 = 7.35 n " for the required net 
 area of each pair of plates, or about 3.67 n " for each plate. Allowing \" 
 clearance between these plates and the flange angles and the same clear- 
 ance between them and the vertical splice plates, we obtain 10J" for 
 their width. Now, let us assume that they are \" thick, and that their 
 cross-section is reduced by three rivet holes (for J" rivets), then we have 
 
 for the net area of cross-section of each plate, which is about equal to the 
 required area. But if the plates be \" thick a -fa" filler would be 
 
240 STRUCTURAL ENGINEERING 
 
 required under each of the stiffeners at the splice and to avoid these thin 
 fillers we will make the plates as thick as the flange angles, which is iV'- 
 This, of course, gives a little excess area, but the result obtained justifies 
 its use. 
 
 The next thing is to determine the number of rivets in these longi- 
 tudinal plates. As the number of rivets depends upon their allowable 
 bearing upon the f " web, which is 
 
 (7,880) M = 5,330 Ibs., 
 
 we have 
 
 79,600 
 
 = 15 (about) 
 
 5,330 
 
 for the number of rivets required on each side of the splice in each pair of 
 plates ; and spacing them in the plates, so as to conform to the spacing in 
 the stiffeners and flange angles previously established, we obtain the 
 spacing shown in Fig. 171. 
 
 The next thing is to determine the size of the vertical splice plates 
 and the number of rivets required in them to transmit the maximum shear 
 at the splice as explained in Art. 117. 
 
 /' 
 
 ,- ci 
 
 & 
 
 so' 
 
 a 
 
 Fig. 174 
 
 Placing the wheel loads on the span so that wheel 2 will be at the 
 splice, as shown in Fig. 174, and taking moments about the support B 
 (using Table A), we have 
 
 50 
 for the reaction at A, then, we have 
 
 (34,900-10,000)^=31,200 Ibs. 
 
 for the maximum live-load shear at the splice. 
 For the impact we have 
 
 Now, as the dead-load shear at the splice is zero, we have 
 31,200 + 28,800 = 60,000 Ibs. 
 
 for the total maximum shear at the splice. 
 Then we have 
 
 60,OOP. 
 
 IoW = 6 ' 0s * ins - 
 
 for the required net area of the vertical section of the vertical splice 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 plates, but as the splice plates are made $" thick in order to avoid the 
 Q" fillers under the stiffeners, referred to above, there will be consider- 
 able more metal in them than is required for shear, so they will be amply 
 strong as far as shear is concerned. This is practically always the case, 
 for the splice plates cannot, according to the specifications, be less than 
 f" in thickness and the two of them will always be thicker than the web 
 they splice. The number of rivets required to connect these plates to the 
 web will depend upon their allowable bearing on the f" web as shown 
 above. So we have 
 
 60,000 
 
 =7.6, say, 8, 
 
 for the required number of rivets on each side of the splice, but a greater 
 number will be used as there must be two rows of rivets on each side of 
 the splice in order to securely clamp the plates to the web and the spacing 
 must conform to that already established in the stiffeners. So following 
 out these requirements without spacing the rivets farther apart than 6", 
 we obtain the spacing shown in Fig. 171. 
 
 Now to test the splice as a whole, let S = the horizontal stress on each 
 of the rivets in the horizontal rows farthest out from the center of the 
 web, that is, in the rows next to the flanges, and let V = the vertical shear 
 on each rivet, which we will assume to be the same for each rivet in the 
 splice, that is, 
 
 Now taking moments about the center of the web, as explained in 
 Art. 117, we have 
 
 16,000x3.37x71, 
 from which we obtain 
 
 5 = 5,118 Ibs. 
 
 Then for the maximum resultant stress which occurs on the outer 
 rivets, we have 
 
 ^ = V / 5,118 2 + 1,764: 2 = 5,413 (about), 
 
 which is about 567 Ibs. less than the allowable, which is 5,980 = (7,880) 
 27x2/71. So the splice is all right and the drawing of it can be com- 
 pleted as shown. 
 
 It is seen from the above equation that the rivets near the center of 
 the web are of little value, in resisting bending. 
 
 We will next take up the detailing of the cross-frames. The first 
 thing to do is to draw the vertical lines representing the centers of the 
 girders as shown in the detail to the right of the girder. Next the top 
 and bottom horizontal angles should be located and the position of the 
 diagonal determined. Then the next thing is to determine the size of 
 each of the connection plates marked ok. The size of each of these plates 
 will depend upon the number of rivets in the ends of the diagonals. As 
 the diagonals take both tension and compression we consider them as 
 
242 STRUCTURAL ENGINEERING 
 
 compression numbers, and assuming them independent of each other, each 
 has a length of 97" (about). Then we have 
 
 07 
 16,000-70-^=9,650 Ibs. 
 
 for the allowable unit compressive stress on each diagonal, 
 and 9,650 x 2.48 = 23,900 Ibs. 
 
 for the allowable stress on each. Then using J" shop rivets, we have 
 
 23,900 
 
 7,220 
 
 = 3.3, say, 4, 
 
 for the number of rivets required in each end of the diagonals. 
 
 Now a large scale drawing of one of the four plates ak, showing the 
 entire detail at that point (girder and all) can be made from which the 
 location of the rivets and all necessary clearances can be determined. 
 Then the cross-frame and the cross-section of the girder can be drawn as 
 shown and the pencil drawing is then complete and the tracing of the 
 same can be made and by adding the required list and writing on the 
 title and general notes we have the shop drawing for the span completed 
 except for the anchor bolts and cast pedestals which are detailed on 
 sketch sheets included in the above shop bills which are made after 
 the shop drawing is completed. After the shop drawing and bills are 
 checked and approved blueprints are made of them which are used in the 
 fabricating, shipping, and erecting of the span. 
 
 137. Estimate of Dead Weight and Cost of Span. From the 
 shop drawing (Fig. 171) we have the following weight of metal: 
 
 Estimate of the weight of main girders. 
 
 l_ we b 72 x f x 91.8 Ibs. x 51.4' 4,718 Ibs. 
 
 4 Ls 6x6x^x21.9 Ibs. x 51.4' 4,503 Ibs. 
 
 2 cov. pis. 14 x T Vx 20.82 Ibs. x 27.0' (top and bot.) 1,124 Ibs. 
 
 1 cov. pi. 14 x i x 23.80 Ibs. x 51.4' (top) 1,223 Ibs. 
 
 1 cov. pi. 14 x 4 x 23.80 Ibs. x 37.0' (bot.) 881 Ibs. 
 
 8 Ls 5 x 3i x J x 13.6 Ibs. x 6.0' (end stiff.) 653 Ibs. 
 
 26 Ls 5 x 3 x | x 10.4 Ibs. x 6.0' (int. stiff.) 1,622 Ibs. 
 
 12 fills. 3Jx T %x6.7 Ibs. x 5.0' 402 Ibs. 
 
 2 sp. pis. 16 x T % x 30.6 Ibs. x 3.2' 196 Ibs. 
 
 4 sp. pis. 10x T %x 19.14 Ibs. x 2.1' 161 Ibs. 
 
 2 sole pis. 14xf x35.71 Ibs. x 1.3' . 93 Ibs. 
 
 15,576 Ibs. 
 
 1,015* rivets @ .44 Ibs. per pair of heads ( J" rivets) 446 Ibs. 
 
 Total weight of one main girder ; .16,022 Ibs. 
 
 Total weight of two main girders 32,044 Ibs. 
 
 Total weight of main girders per foot of span equals 
 
 =641 Ibs. 
 
 *For weight of rivet heads see Carnegie Handbook. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 243 
 
 It Till be seen from the above that the weight of the rivet heads is 
 about 3 per cent of the weight of the other material in the girders. 
 
 Estimate of the weight of laterals and lateral plates. 
 
 8 Ls 3J x 31 x f x 8.5 Ibs. x 8.2' 558 Ibs. 
 
 7 pis. 12 x f x 15.3 Ibs. x 2.6' 278 Ibs. 
 
 2 pis. 12 x x 15.3 Ibs. x 2.0' 61 Ibs. 
 
 2 pis. 12 x f x 15.3 Ibs. x 1.3' 40 Ibs. 
 
 9 pis. 12 x f x 15.3 Ibs. x 1.0' 137 Ibs. 
 
 4 pis. 12xf x 15.3 Ibs. x 0.8' 49 Ibs. 
 
 Total weight of laterals and lateral plates 1,123 Ibs. 
 
 Estimate of the weight of cross-frames. 
 
 2 Ls 3^ x 3J x f x 8.5 Ibs. x 6.2' 105 Ibs. 
 
 2 Ls 3| x 31 x f x 8.5 Ibs. x 7.3' 124 Ibs. 
 
 4 pis. 14 x f x 17.86 Ibs. x 1.1' 78 Ibs. 
 
 1 pi. 8J x f x 10.84 Ibs. x 0.8' 9 Ibs. 
 
 316 Ibs. 
 33 rivets @ 0.44 Ibs . 14 Ibs. 
 
 Total weight of one frame 330 Ibs. 
 
 Total weight of five frames 1,650 Ibs. 
 
 Total weight of lateral system = 1,123 + 1,650 = 2,773 Ibs. 
 
 p lyiVQ 
 
 Total weight of lateral system per foot =77; = 55 Ibs. per ft. of span. 
 
 oO 
 
 Estimate of the total effective dead load per ft. of span. 
 
 2 girders 641 Ibs. 
 
 Lateral system 55 Ibs. 
 
 Deck (track) 400 Ibs. 
 
 Total 1,096 Ibs. 
 
 Difference between the assumed dead load and estimated = 1,150 - 
 1,096 = 54 Ibs., which is less than 10 per cent, so no recalculation is 
 necessary on account of error in assumed dead load. 
 
 Summary of total weight of metal in span. 
 
 2 girders , 32,044 Ibs. 
 
 Lateral system 2,773 Ibs. 
 
 4 pedestals @ 360 Ibs 1,440 Ibs. 
 
 Anchor bolts and nuts 61 Ibs. 
 
 Total 36,318 Ibs. 
 
 Approximate Cost of Span Erected: 36,318 Ibs. @ 3^ = $1,180. 
 Three and one-quarter cents is only a fair average price for this class of 
 work erected. The price will vary from 2^ to 4^. It depends upon the 
 market price of steel and the freight. 
 
244 STRUCTURAL ENGINEERING 
 
 It will be seen from the above tli.it the weight of rivet heads (= 1,000 
 Ibs., approximately) is equal to about 3 per cent of the weight of the 
 other metal (=32,700 Ibs.) in the span, not considering the pedestals. So 
 in making preliminary estimates of such structures the weight of the rivet 
 heads can be taken at about 3 per cent of the w r eight of the other metal. 
 
 138. Hints Regarding Shop Bills, The above shop bills are for 
 the most part self-explanatory. Pages Nos. 1 and 2 are for the structural 
 material proper. The finished material is given on the material side of 
 the bills exactly as it appears on the shop drawing. The length of 
 material as obtained from the rolling mills is likely to vary slightly from 
 the length ordered, so in case of long pieces where an under run would be 
 objectionable a \" or so is added in the length given on the mill order side 
 as shown. In case of short pieces having a length of 10 ft. and under, 
 and having the same section, they are ordered in multiple, that is, they 
 are combined and ordered as one or more pieces, as is seen. 
 
 All pieces planed (marked fin.) on one side are ordered -j\r'' thicker 
 than the finished thickness and J" thicker if planed on two opposite sides. 
 All pieces planed on the ends are ordered \" to \" longer than the 
 finished length. This is in addition to the amount added for under runs. 
 
 The dimensions appearing on the mill order side are only those 
 differing from the finished dimensions. In cases where the material is to 
 be ordered exactly as the finished material, it is not repeated on the mill 
 order side, as a rule. 
 
 The cast-steel pedestals are detailed on page No. 4. The required 
 area of bearing on the masonry of each pedestal is given in Art. 134 as 
 361 n ". The actual area is 
 
 20^x18 = 364.5 sq. ins., 
 
 which is about the correct area. The thickness of metal in such pedestals 
 should not be less than 1J" in order to insure the molds to properly fill. 
 The top of each pedestal (where the girder rests) should be as small as is 
 consistent with good details. 
 
 Everything shown on such bills as the above is, as a rule, written on 
 by the draughtsman making the shop drawing. The item numbers are 
 given by the order clerk as the final mill order blanks are made out in the 
 order department, which, as a rule, is a separate department from the 
 drawing room. The bills not mentioned are considered as being entirely 
 self-explanatory. 
 
 Partial Design of a 75-Ft. Single-Track Deck Plate Girder Span 
 
 139. Data. 
 
 Length = 75' 0" c.c. end bearings. 
 
 Dead Load = 12 x 75 + 550 = 1,450 Ibs. per ft. of span. 
 
 Live Load, Cooper's 50. 
 
 Specifications, A. R. E. Ass'n. 
 
 140. Calculations. In a manner similar to that shown in Art. 134 
 for the 50-ft. span, we obtain the following for the 75-ft. span: 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 245 
 
 Maximum End Shear: 
 D= 27,200 Ibs. 
 L = 147,200 Ibs. 
 7=117,700 Ibs. 
 
 292,100 Ibs. 
 
 Maximum Moment: 
 
 /) = 6,117,000 in. Ibs. 
 
 L = 28,875,000 in. Ibs. 
 
 7 = 23,100,000 in. Ibs. 
 
 58,092,000 in. Ibs. 
 
 Assuming the web to be -$ in. thick we have 
 
 ,- 1055 
 
 / 58,092, 
 
 VAX 16, 
 
 ,000 
 000 
 
 for the economic depth. So we will try a 96 x $" web. Using the 
 ordinary flange the effective depth is about 95" (merely an assumption). 
 Then we have 58,092,000 + 95 = 611,500 Ibs. for the flange stress, and 
 611,500 -r 16,000 = 38.2 n " for the net area of the flange. As stated 
 before, about half of this area (minus J of the area of the web) should 
 be in each pair of flange angles. This would require angles which would 
 be entirely too thick if 6" x 6" angles be used. So either 6" x 6" angles 
 with side plates and cover plates, or 8" x 8" angles with cover plates 
 should be used. The last-mentioned section is really the ordinary flange 
 wherein the angles are 8" x 8". 
 
 Let us first try the flange made up of 6" x 6" angles, side plates 
 and cover plates. As the effective depth in this case is less than in the 
 case of ordinary flanges, the area of the flange will be a little larger than 
 indicated above. So let us assume the following section: 
 
 2 Ls 6" x 6" x f" 
 2 Side plates 13" x 
 1 Cover plate 16" x 
 1 Cover plate 16" x 
 J of web 
 
 Gross Area Net Area 
 
 = 14.22 n "-2.5 n" = 11.72" 
 = 13.00 n "-3.0 n" = io.00 n " 
 = 8.00 n " -1.0 n "= 7.00 n " 
 = 7.00 n "-0.87 n "= 6.13 n " 
 = 5.25 n " 
 
 42.22 n " 
 
 40.10 n " 
 
 The flange is assembled as shown in Fig. 175. 
 
 By taking moments about the center of the top cover plate we obtain 
 
 -_ 8x1 + 14.22x2.48 + 13x7^ 
 42.22 
 
 for the distance from the center of the top 
 cover plate to the center of gravity of the 
 flange. Then we have 
 
 3.24 -0.75 = 2.49 ins. 
 
 for the distance from the back of the flange 
 angles to the center of gravity of the 
 flange, and 
 
 96.25-4.9S = 91.27ins. 
 for the effective depth. 
 
 = 3.24 ins. 
 
 Fig. 175 
 
246 STRUCTURAL ENGINEERING 
 
 Then using this effective depth we have 
 
 58,092,000 
 
 91.27 
 for flange stress, and 
 
 = 637,000 Ibs. (about) 
 
 637,000 __ _ . 
 
 = 39.8 sq. ins. 
 
 16,000 
 
 for the required net area of each flange. This shows that the above 
 assumed flange is about correct. 
 
 Next let us try a flange made up of 8" x 8'" angles and cover plates. 
 
 First assume the following section: 
 
 Gross Area Net Area 
 
 2_ Ls g" x 8" x ii" = 21.06 n " - 2.75 n " - 18.31" 
 
 2 cov. pis. 18" x J" = 18.00 n " - 2.00 n " = 16.00 n " 
 
 J of web _ = 5.25 n " 
 
 39.06 n " 39.56 n " 
 
 Taking moments about the center of the top cover plate we obtain 
 
 jj 
 
 for the distance from the center of the top cover plate to the center of 
 gravity of the flange, and hence we have 
 
 1.73 - 0.75 = 0.98 ins., say 1 in., 
 
 for the distance from the back of the angles to the center of gravity of 
 the flange. This gives 96.25 - 2 = 94.25" for the effective depth. Then 
 we have 
 
 58,092,000 
 94.25 x 16,000 = 38 ' 55sq ' m 
 
 for the required net area of each flange. This shows that the section 
 assumed above is a little too large. By making one of the cover plates 
 T V' thick (instead of ") the section will be about correct. 
 
 It is seen from the above that the flange made up of 8" x 8" angles 
 and cover plates is more economic than the flange made up of 6" x 6" 
 angles, side plates, and cover plates, and hence apparently should be 
 used. However, this will depend mostly upon whether the 8" x 8" angles 
 can be readily obtained from the rolling mills at the same price as the 
 other sections. In case several spans are required it would very likely 
 pay to use 8" x 8" angles, but in case of only one span it would likely 
 be best to use the 6" x 6" angles, side plates, and cover plates. 
 
 The calculations for the remainder of this span would be quite 
 similar to that shown above for the 50-ft. span, and when completed the 
 corresponding stress sheet, detail shop drawing, and shop bills could be 
 made. Fig. 176 shows a partial detail of the girder where the flanges 
 are made up of 6" x 6" angles, side plates, and cover plates. 
 
 The student will have no trouble in designing and detailing this type 
 of girder if the outline given above for the 50-ft, span be followed. 
 
Or'OO O-i; O O O-O-e-O Q O O 0:^ 
 
 
 
 I 
 
 \ H tT"n 
 
 /-Cov.P/. /x/i 
 
 
 e 
 
 -e o e e- 
 
 e e e< 
 
 Med holes &*/;> "M/s end 
 
 / Tap bo/ fa topped /'nto so 
 
 Fig. 176 
 
 247 
 
248 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 249 
 
 Partial Design of a 90-Ft. Single-Track Deck Plate Girder Span 
 
 141. Data. 
 
 Length = 90'-0" c.c. end bearings. 
 
 Dead Load = 12 x 90 + 550 = 1,630 Ibs. per ft. of span. 
 
 Live Load, Cooper's E50. 
 
 Specifications, A. R. E. Ass'n. 
 
 142. Calculations. For the main girders of this span we have: 
 
 Maximum End Shear: Maximum Moment: 
 
 D= 36,700 Ibs. D= 9,902,000 in. Ibs. 
 
 L= 171,500 Ibs. L = 40,057,000 in. Ibs. 
 
 7=132,000 Ibs. 7=30,813,000 in. Ibs. 
 
 340,200 Ibs. 80,772,000 in. Ibs. 
 
 Assuming the web to be T V thick, we have 
 
 ., AKK 80,772,000 
 
 .r = 1.055 A / ' ' = 113 ins. (about) 
 
 \ AX 16,000 
 
 for the economic depth. But, as a few inches either way from the theo- 
 retical depth in case of such deep girders will not materially affect the 
 design, it will be better to take 108" as the depth, as this will give us a 
 108" web, which is a very common plate, much more so than the 112" 
 or 113". So we will assume a 108" x T y web. 
 
 There are three types of flanges suitable for this girder: One made 
 up of 6" x 6" angles, side plates, and cover plates, as shown in Fig. 176; 
 one made up of 8" x 8" angles and cover plates, as ordinary flanges; or 
 one made up of 4 6" x 6" angles and side plates for the top flange and 
 8" x 8" angles and cover plates for the bottom flange, as shown in 
 Fig. 177. 
 
 The flanges made up of 8" x 8" angles and cover plates the author 
 thinks preferable. However, the design shown in Fig. 177 has some 
 desirable features. For instance, there are no rivet heads on the top 
 flange to interfere with the ties, as there are no cover plates and the top 
 laterals are connected to the bottom angles of the top flange. 
 
 The stress sheet and detail shop drawing and bills for this span 
 can be worked up in the same manner as shown above for the 50-ft. span. 
 
 All spans over 75 ft. long should be supported at one end upon 
 rollers. Fig. 178 shows the general details for such a bearing, which is 
 for the above 90-ft. span. Fig. 179 shows the general details for the 
 corresponding fixed support for the same span. Figures 180 and 181 
 show the shop details for the girder shoe, roller shoe, roller nest, and 
 pedestal. 
 
 In designing and drawing up such a bearing as that shown in Fig. 
 178, the first thing is to determine the space taken up by the rollers. For 
 the above 90-ft. span we have 340,200* for the maximum reaction. The 
 minimum size of rollers is limited by the specifications to 6" diameter, 
 and as plate girder bridges are comparatively small bridges we will use 
 the minimum size roller. The allowable pressure per linear inch on 
 
Fig. 178 
 
 c 
 
 o r 
 
 
 
 { 
 \ 
 
 V 
 
 
 
 
 * 
 
 i 
 
 
 ] 
 
 
 
 *o 
 
 % 
 
 
 
 I 
 
 5^ 
 
 M> 
 
 *t XL* 
 
 > *o Q D o 
 
 L 
 
 ! 
 
 4 
 
 2- O" 
 
 Fig. 179 
 
 250 
 
 
 
 s 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 251 
 
 these rollers is 600 x 6 = 3,600#. (See specifications, also Art. 84.) 
 We then have 
 
 340,200 
 3,600 
 
 = 94.5 
 
 for the linear inches of roller required. If we use five rollers we have 
 
 94.5 
 
 ' = 19 ins. (about) 
 
 for the required net length of each roller, and adding 4^" to each to 
 provide for the slots in tdie ped- 
 estal we have 23%" for the total 
 length of each roller, as shown in 
 Fig, 178. All rollers having a 
 diameter of 6" and over are 
 usually flattened on the sides like 
 those shown in Figs. 178 and 181, 
 and are known as segmental 
 rollers. These rollers should be 
 placed quite close to each other 
 so that their sides will come to- 
 gether whenever they are rolled 
 beyond the amount desired. Thus 
 the rollers are prevented from fall- 
 ing over. The distance between 
 the rollers (between the sides) is 
 usually about J", as shown in Fig. 
 181. After determining the size 
 of the roller nest, a preliminary 
 drawing of the shoes and pedestal 
 can be made similar to that shown 
 in Fig. 178, wherein the general 
 dimensions are made to conform 
 to those of the roller nest, while 
 the correct size of the pin passing 
 through the shoes and the thick- 
 ness of metal throughout are de- 
 termined as the work progresses. 
 We begin by deciding upon the 
 general form of the shoes. Then 
 we assume the size of the pin passing through them. Next we calculate 
 the required thickness of the bearings of the shoes for the pin assumed. 
 If this seems satisfactory we proceed with the work of calculating the 
 stress on the pin to determine whether the pin assumed is of correct size, 
 and if the size is found to be correct we proceed farther with the design, 
 but if found to be incorrect we start over and make some new assumption, 
 and so on, until the design is worked out satisfactorily. 
 
 ' 2-8" 
 
 r ?- o" 
 
 la' 2- 4k" % 
 
 E- Roller Shoes- RS. 
 
 ^ //* 3"SloH- e dho/ es . . 
 
 
 P 
 
 (Gost&ee/) 
 
 1 1 i 
 
 
 
 ... J 
 
 \ 
 
 j 
 
 A \ 
 
 
 
 tf 
 
 t 
 
 \ 
 
 
 
 i 
 
 
 $ 
 
 Fi*. 180 
 
252 
 
 STRUCTURAL ENGINEERING 
 
 . 8-*.ii5feeip, l > 5 .tohoK</rnmiqM; l > r oiier* Taking the case shown in Fig. 
 
 *N^ *N. "^. -1 fvr~* 1 . .1 ii// 
 
 178, let us assume the pin to be 
 in diameter. The total thickness of 
 the required bearing on each shoe is 
 then 
 
 340.200 
 
 24,000 
 
 - 
 
 .|]f Drilled holes in bars. 
 
 J 
 
 2> 2- Roller tfest-RM. 
 
 
 Ur /-/ 
 
 ^" 
 
 - 
 
 5p 
 
 
 
 //-"Corec//>o/es - 
 
 --:---:::-.-.:--.-;.-.-- 
 
 Ff= 
 
 
 
 S 
 
 ill 1 
 
 
 Z- Pedestal* -RP 
 
 (Cast- Sfee/.j 
 
 Fig. 181 
 
 The central bearing of each shoe 
 will likely be subjected to a little 
 more pressure than either of the side 
 bearings. So we will make them a 
 little thicker than the side bearings. 
 By making the central bearing of 
 each shoe 1J" thick and each side 
 bearing 1-|" we have 3J" for the 
 total thickness of bearing on each 
 shoe, which is about equal to that 
 required, and as the side bearings 
 are of minimum thickness for such 
 castings these thicknesses are quite 
 satisfactory and hence will be used. 
 The maximum shear and cross 
 bending on the pin occur at the side 
 bearings. For the maximum shear 
 we have 
 
 340,200 x if = 109,300 Ibs. 
 03 
 
 and for the maximum shearing unit stress on the 4J" pin we have 
 109,300 109,300 
 
 14.18 
 
 - 7,700 Ibs. 
 
 For the maximum cross bending on the pin we have 
 
 109,300 x If "=150,000 inch Ibs. (about) 
 
 and for the maximum fiber stress due to cross bending we have 
 .-_ 0,000 x 2J =2(M) 
 
 Now, as the allowable shear on pins is 12,000 Ibs. per square inch and 
 the allowable fiber stress is 24,000 Ibs. per square inch (see specifica- 
 tions), it is seen from the above that the 4J" pin assumed is amply strong; 
 in fact it could be reduced in size, but as the saving in cost would be 
 insignificant we will use the size assumed. It also appears to be about 
 the correct size. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 253 
 
 In addition to having sufficient bearing on the pin the shoes should 
 be deep enough and contain enough metal to resist the cross bending to 
 which they are subjected. 
 
 Considering the girder shoe (GS) we have a case of cross bending 
 as indicated in Fig. 182. For the maximum moment we have 
 
 340,200 
 
 inch lbs ( a b ou t), 
 
 which occurs at the vertical transverse section c-c through the pin hole. 
 The moment of inertia of this cross-section of the shoe (not subtracting 
 the pin hole) is about 390. Then for the maximum compressive unit 
 stress due to cross bending we have 
 
 638,000 x 6| 
 390 
 
 = 12,200 lbs. 
 
 and for the maximum tensile unit stress we have 
 
 638,000 x 3j 
 
 390 
 
 -5,500 lbs. 
 
 This shows that the girder shoe is amply strong, as far as cross bending 
 is concerned, as 16,000 lbs. unit stress is permissible. The bending on 
 the roller shoe (RS) can be determined in the same manner. Further 
 designing of the shoes is very much a matter of common judgment. 
 
 Regarding the above calculations 
 it may first appear that the material cut 
 out of the vertical section c-c by the pin 
 hole should be deducted from the sec- 
 tion in determining the moment of 
 inertia. But this is far from being cor- 
 rect, as the top half of the pin is 
 assumed to transmit 24,000 lbs. per 
 square inch against the shoe this we 
 can rely upon and owing to the lateral 
 deformation of the pin there will be 
 some horizontal pressure (perhaps equal 
 to Poisson's ratio) ; so, taking all 
 in all, the above assumption is fairly 
 
 The designing of the pedestal (RP} is very much a matter of com- 
 mon judgment. The pedestal should be at least 4 ins. or 5 ins. high in 
 order to have the rollers above snow, slush, and dirt, and should be broad 
 and long enough to support the load coming on them without producing 
 a greater pressure upon the masonry than 600 lbs. per square inch. 
 There should be open slots extending down from the top as shown in 
 Fig. 178, so that dirt will not accumulate between the rollers. The 
 longitudinal circular holes should be cored in the casting and the slots 
 cut from the solid when the pedestals are being finished in the machine 
 shop. This will guard against the pedestal warping when cooling. The 
 details of the fixed bearing shown in Fig. 179 are considered to be self- 
 explanatory. 
 
 UUUiii UlliUJl! 
 
254 
 
 STRUCTURAL ENGINEERING 
 
 143. Flange Splices. The flanges of all plate girders over 90 ft. 
 long must be spliced, as that is about the maximum length of angles and 
 cover plates obtainable. These splices should preferably be symmetrically 
 arranged in reference to the center of the span. The flange angle splices 
 should, as a rule, be nearer to the ends of the girder than to the center 
 of the span, and only one flange angle should be cut off at a splice. That 
 is, one flange angle should be spliced on one side of the girder on one 
 side of the center of the span and the other flange angle should be 
 spliced on the other side of the girder at a corresponding point on the 
 other side of the center of the span, as shown in Fig. 183. Here the 
 near flange angle is spliced at D, one part extending from D to B, and 
 the other part from D to A, while the far flange angle is spliced at E, 
 one part extending from E to A and the other part from E to B. 
 
 Sechm-SS 
 
 Fig. 183 
 
 The splice of a flange angle should be made by means of an angle 
 having the same net area of cross-section, as the flange angle. The splice 
 angle must be ground to fit the fillet of the flange angle and its legs cut 
 so as not to project beyond the legs of the flange angle, as indicated at 
 section S-S. (Fig. 183.) It is general practice to use two splice angles 
 at each splice, one on each side of the girder. The splice angle on the 
 side of the flange angle spliced should be of sufficient cross-section in 
 every case to splice that angle. The splice angle on the other side of 
 the girder is used simply to balance the flange section at that point. 
 
 The cover plates are usually spliced by extending the adjacent cover 
 plates a short distance beyond their theoretical length, as shown in Fig. 
 183. As a rule, the cover plates next to the flange angles are the only 
 ones that need to be spliced. Cover plates, of course, can be spliced by 
 simply using a short plate having the same area of cross-section as the 
 plate spliced, but usually this presents an unsightly appearance. 
 
 As an example in figuring splices, let the two flange angles shown in 
 Fig. 183 be 2 l_s 6" x 6" x f ", as indicated. For the net area of either 
 of these flange angles we have (using J" rivets) 7.11 - 1.25 = 5.86 n// . Then 
 for the allowable stress in each flange angle we have 5.86 x 16,000 = 
 93,700*. Then for the number of J" shop rivets required on each side 
 of the splice (in the splice angle) we have 93,700 -j- 7,200 = 13 rivets (in 
 single shear). As is seen, 14 are used 7 in the vertical leg and 7 in the 
 horizontal leg. The splice angle at each splice should have 5.86 n " net. 
 Using a 6"x6"x{J" angle, and deducting 1.37 n " for rivet holes and 
 0.86 n " for the cutting of the legs and for the grinding to fit fillet, we 
 have 7.78 -1.37- 0.86 = 5.55 n " net, which is about correct. 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 
 
 255 
 
 The calculation for the splicing of a cover plate is simply a matter 
 of determining the number of rivets required in single shear to develop 
 the strength of the plate spliced. The student should have no trouble 
 in doing that. 
 
 144. Graphical Determination of Live-Load Shear and Bending 
 Moment on Deck Plate Girder Bridges. The analytical method out- 
 lined in Art. 133 is generally used, as it is simple in application and the 
 results obtained are absolutely numerically accurate, yet the graphical 
 method is fully as simple in application and the results obtained are 
 quite accurate enough. The graphical method at least affords an excel- 
 lent means of checking results obtained analytically, and if for no other 
 reason than this the engineer should be familiar with the method. In 
 fact, there are two methods: that of influence lines and the one wherein 
 the equilibrium polygon is used. 
 
 so'-o' 
 
 Fig. 184 
 
 The application of influence lines is fully given in Arts. 100 and 
 101. As an application of the equilibrium polygon let us take the case 
 of the 50-ft. span treated analytically in Art. 133, where Cooper's 50 
 loading is used. Let AB (Fig. 184) represent the span drawn, say, to 
 J" scale. From inspection of the diagram in Table A (in the back of 
 this book) we can see that the maximum end shear will occur when 
 wheel 2 is at the end of the girder and wheels 2 to 10 (inclusive) are 
 on the span, as shown in Fig. 184. After the loads are thus placed, the 
 next thing to do is to lay off the load line uv (say, to a -gSr-in. scale) at 
 (6), and construct the ray diagram, as explained in Art. 95, and draw 
 the corresponding equilibrium polygon ab...na at (a). Then by draw- 
 ing from (at 6) the line OE parallel to the closing line an, we have 
 the maximum end shear given by the part of the load line extending from 
 E to u which can be scaled, using the same unit of scale as was used in 
 laying off the load line. 
 
 To determine the shear at any point K of the girder, place the point 
 K under wheel 2 by moving the girder, so to speak, to the left, so that 
 
256 
 
 STRUCTURAL ENGINEERING 
 
 the end A will be at A' and the other end will be at B f and wheels 1 to 7 
 will be on the span. Then add wheel 1 to the load line and draw the 
 ray wO and extend the equilibrium polygon on to s and we have the 
 equilibrium polygon sta . . . fs's. Then draw OE' in the ray diagram 
 parallel to the closing line **' and we have the maximum shear at point 
 K given by the part of the load line extending from E' to u. In this 
 manner the maximum shear at any point on the span can be determined. 
 To determine the maximum bending moment place the loads on the 
 span, as shown in Fig. 185, so as to satisfy the criterion for maximum 
 bending moment as per Art. 88. Then construct the ray diagram as 
 
 Fig. 185 
 
 shown at (6) and draw the corresponding equilibrium polygon ab . . . ma, 
 and we obtain the maximum moment by multiplying the ordinate e'e 
 under wheel 13 (e'e must be scaled off in feet or inches to the same 
 scale as was used in laying off the length of span at (a) and spacing 
 the loads) by H, the pole distance which is laid off in pounds to the 
 same scale as the load line. If the ordinate e'e is taken in feet the 
 bending moment will be in foot pounds, and if taken in inches the bending 
 moment will be in inch pounds. 
 
 Such a diagram as shown in Fig. 186 is very convenient if the work 
 to be done is extensive enough to warrant the drawing of it. In offices 
 where one loading is used for all bridges it is advisable to draw up such 
 a diagram of the loading on a sheet of thick cardboard, inking in all 
 lines shown in Fig. 186 except the closing lines. The closing lines can 
 be drawn lightly in pencil as needed and then erased. In this way the 
 diagram can be used in determining the shears and moments for any 
 number of bridges. The diagram shown in Fig. 186 is for Cooper's J40 
 loading. The loads are spaced off to -^-in. scale on the horizontal line 
 at the top of the sheet. Then the load line LL is laid off to a Tsfo-in. scale 
 and the ray diagram is constructed, as shown, and the corresponding 
 equilibrium polygon ABC is drawn, and then the scale lines are drawn 
 below this, as shown. 
 
 To show the manner of using the diagram in Fig. 186, let us take 
 the case of a 50-ft. span. The maximum moment due to Cooper's loading 
 in the case of most deck plate girder bridges will occur under one of the 
 four wheels 11 to 14 (this we obtain from experience). So, beginning with 
 zero at wheel 13 we lay off the scale line HII into f)-ft. units. To determine 
 the maximum bending moment in this 50-ft. span let us start by assuming 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 257 
 
 that the span extended out. 25 ft. each side of wheel 13. Then drawing 
 the closing line 1-1 we have the equilibrium polygon l-$-l, and bv 
 scaling off the maximum ordinate between the closing line 1-1 and the 
 curve 1-5-1 and multiplying it by the pole distance given in the ray 
 diagram we obtain the maximum bending moment for the span in that 
 position. By moving the span 5 ft. to the right and drawing the closing 
 line 2-2 we have the equilibrium polygon 2-B-2 and by scaling off the 
 maximum ordinate and multiplying it by the pole distance we obtain 
 the maximum bending moment for the span in that position, and moving 
 the span to the left the equilibrium polygon 3-B-3 is drawn and the 
 maximum bending moment for the span in that position can be deter- 
 mined. Now, by simply drawing a few of these sub-equilibrium polygons 
 the absolute maximum ordinate can be ascertained from the intersection 
 of the closing lines with the verticals through the wheels. The maximum 
 
 35' 30' 25' 20' i/5' Jo' \S' \O \f' 10' IS' ZO' 2S' 
 
 ! i ' ! I 
 
 ordinate will always be under a wheel, and after a sufficient number 
 of closing lines are drawn the maximum ordinate can be quickly deter- 
 mined by the aid of a pair of dividers and then scaled off and multiplied 
 by the pole distance. In this way the maximum bending moments are 
 determined without bothering with the criterion for maximum moment, 
 as is necessary when other methods are used. 
 
 To determine the maximum end shear on the 50-ft. span, place 
 wheel 2 at the end of the span. Now using the scale line SS, the span 
 will extend 50 ft. to the right of wheel 2 and we obtain an equilibrium 
 polygon which has the closing line nn'. Then drawing Pr in the ray 
 diagram parallel to this closing line we have the maximum end shear 
 or reaction given by the part of the load line extending from r to e. By 
 
258 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 259 
 
 the following construction the end shears can be obtained more quickly 
 than if scaled from the ray diagram: From n (on the equilibrium 
 polygon) lay off the horizontal line nm equal to the pole distance EP 
 and draw the vertical line kk' through m and prolong the segment gn, 
 of the equilibrium polygon, until it intersects this vertical line at k' '. 
 Then we have the triangle nmk f equal to the triangle Pre (in the ray 
 diagram), and hence xk' is equal (by scale) to the maximum end shear 
 on the 50-ft. span. Likewise, x f h' and x"k f are equal, respectively, to 
 the maximum end shear on a 60-ft. and 75-ft. span. 
 
 The equilibrium polygon ABC can be used in general to determine 
 the shear at any point in a span in the manner explained in the case 
 shown in Fig. 184. 
 
 Problem 1. Construct a diagram for Cooper's E50 loading similar 
 to the diagram shown in Fig. 186 and determine from it the maximum 
 end shears and bending moments on a 40-ft., 50-ft., 60-ft., 70-ft., 80-ft., 
 and 90-ft. span. 
 
 DRAWING ROOM EXERCISE NO. 3 
 
 Design a 60-ft. single-track deck plate girder railroad bridge and 
 make a stress sheet for the same. The finished drawing, similar to that 
 shown in Fig. 169, is to be upon an 18" x 24" sheet of tracing cloth. 
 
 Data: 
 
 Length = 60 '-0" c.c. end bearings. 
 
 Width = 6'-6"c.c. girders. 
 
 Height, to be determined by student. 
 
 Dead Load, to be determined by the student. 
 
 Live Load, Cooper's E5Q. 
 
 Specifications, A. R. E. Ass'n. 
 
 DRAWING ROOM EXERCISE NO. 4 
 
 Make a shop drawing and shop bills for the 60-ft. bridge specified 
 in Drawing Room Exercise No. 3. The finished drawing, similar to that 
 shown in Fig. 171, is to be upon a 24" x 36" sheet of tracing cloth. 
 
 THROUGH PLATE GIRDER BRIDGES 
 
 145. Preliminary. Through plate girder bridges are used, as a 
 rule, only when the under clearance will not permit of the use of a deck 
 span owing to the cost of the through bridge being considerably more 
 than the deck bridge. The ordinary through plate girder bridge is com- 
 posed of two main girders and a floor system composed of longitudinal 
 beams (or girders), which support the ties and are known as stringers, 
 and cross beams, which support the stringers and are known as floor 
 beams. 
 
 An isometric view of an ordinary single-track through plate girder 
 span is shown in Fig. 187, where the names of the different parts of 
 the structure are given. 
 
 Double-track through plate girder bridges are usually the same in 
 construction as the single-track bridges, except the main girders in the 
 double-track structures are farther apart and the floor system provides 
 for the two tracks, which requires twice as many stringers. 
 
260 STRUCTURAL ENGINEERING 
 
 Complete Design of a 60-Ft. Single-Track Through Plate Girder Span 
 
 146. Data.- 
 
 Length = 4 panels @ 15'-0" = 60'-0" c.c. end bearings. 
 Width = IS'-G" c.c. main girders. 
 Assumed Dead Load: 
 
 For main girders, (13 x 60 + 600 4 400) = 1,780 Ibs. per ft. 
 
 of span. 
 For stringers, (12 x 15 + 100 + 400) = 680 Ibs., say 700 Ibs., 
 
 per ft. of span. 
 
 Live Load, Cooper's 50 loading. 
 Specifications, A. R. E. Ass'n. 
 
 147. Design of 15-Ft. Stringers. It is usually necessary to make 
 the stringers in through plate girders just as shallow as good details will 
 permit, owing to the under clearance being limited. This can be accom- 
 plished best by the use of I-beams, two or more under each rail. So we 
 will use I-beams. 
 
 For dead-load moment, using the load assumed in Art. 146, we have 
 
 M= -xx!5x 12 = 118,000 inch Ibs. 
 o 
 
 For live-load moment we have 
 
 M' = 1,875,000 inch Ibs. (same as Art. 130). 
 For impact we have 
 
 7=1,785,000 inch Ibs. (same as Art. 130). 
 Then we have for the total bending moment 
 
 118,000 + 1,875,000 + 1,785,000 = 3,778,000 inch Ibs. 
 Then for the section modulus we have 
 3,778,000 
 
 16,000 
 
 = 236. (See Art. 105.) 
 
 This calls for 2 Is, 20" x 70*, under each rail. That is, each stringer 
 is to be composed of 220" x 70# I-beams. 
 
 For dead-load end shear on each stringer we have 
 
 R =^T x =2,625 Ibs., say 2,600 Ibs. 
 <i 
 
 For live-load end shear we have 
 
 fl' = 50,000 Ibs. (same as Art. 130) 
 and for impact we have 
 
 I = 47,600 Ibs. 
 Then, for the total end shear we have 
 
 2,600 + 50,000 + 47,600 = 100,200 Ibs. 
 
DESIGN OF SIMPLE KAILEOAD BRIDGES 
 
 261 
 
 Preliminary estimate of weight of stringers. 
 
 4 Is 20"x70#xl5'-0" = 4,200 Ibs. 
 
 1_[ 9"x20#x3'-9" = 75 Ibs. 
 
 2 end con. Ls on 9" [s = 21 Ibs. 
 
 2 [s ' 12" x 25# x l'-3" = 62 Ibs. 
 
 Total weight of one panel = 4,358 Ibs. 
 
 Weight of stringers per ft. of span = 4,358-=- 15 = 289 Ibs. metal. 
 Weight of deck =400 Ibs. metal 
 
 Total dead load for stringers per ft. of span = 689 Ibs. metal. 
 
 This shows that the 700 Ibs. dead load assumed above is about 
 correct. 
 
 148. Design of Intermediate Floor Beams. The dead load on 
 these beams consists of the dead weight applied to them by the stringers 
 and also the weight of the floor beams themselves. 
 
 The dead load from the stringers is concentrated on the floor beams 
 at the points where the stringers are connected to them. This concen- 
 tration at each point is equal to twice the dead-load end shear on one 
 stringer which is given above as 2,600 Ibs. So each concentration is 
 2,600x2 = 5,200 Ibs. The weight of the floor beam is a uniform load. 
 Such floor beams usually weigh about 3,600 Ibs. each. So we will assume 
 that weight. Then the dead load on any intermediate floor beam will be 
 as shown in Fig. 187A, where A-B represents the beam. 
 
 It is readily seen from Fig. 187 A that 
 zero shear occurs at the center of the beam 
 and hence that is the point of maximum 
 moment. As the bending moment on the 
 floor beam due to the two concentrated loads 
 is constant between the stringers it is cus- 
 tomary to multiply one concentration by its 
 distance from the end of the beam to obtain 
 the bending moment due to these loads. 
 To show the correctness of this method let AB, Fig. 188, represent 
 a floor beam supporting the two equal concentrated loads P, as shown. 
 Taking moments about the center of the beam we have, since R = P, 
 
 1 
 
 I 
 
 
 W/////////////A 
 
 ''//360'GW/A 
 
 W//////////M 
 
 R 5-3" 
 
 C 5'-0" 
 
 
 ,s'- e 
 
 n 
 Fig. 187A 
 
 That the moment between the loads is constant can also be shown very 
 
 readily by graphics : Lay off the load line VWS 
 
 (Fig. 188) and construct the ray diagram VOS 
 
 and draw the corresponding equilibrium polygon 
 
 efghe. Now, as the reactions are equal for the 
 
 two loads, the closing line eh must be parallel to 
 
 the ray OW, and as the segment fg is parallel 
 
 to this ray also, it is seen that the ordinates of 
 
 the equilibrium polygon are constant between 
 
 the two loads, and hence the moment between 
 
 them must be constant. Now going back to the rigr. 188 
 
262 STRUCTURAL ENGINEERING 
 
 problem in hand, we have the moment 
 
 M = 5,200 x 63 = 327,600 inch Ibs. 
 from the concentrated load and 
 1 3,600 
 
 M/ =zr x ^nrr x 15 - 5 x 13 = 83 > 700 inch lbs - 
 
 8 15.5 
 
 from the weight of the floor beam, making a total maximum of 411,300 
 inch-lbs. dead-load bending moment. 
 
 It is obvious that the maximum live-load bending moment on the 
 floor beam will occur when the maximum live-load concentrations from 
 the stringers on to the floor beam occur. So it is first necessary to deter- 
 mine the position of the wheels when the maximum live-load concentra- 
 tions occur. 
 
 Let Fig. 189 represent two adjacent panels of stringers with a floor 
 beam at each of the ends, A, B, 
 
 * |*_ z .. and C. Let I be the length of 
 
 jJirX s~\ \ A"^ panel AB and r the length of 
 
 panel BC, as indicated. Let W 
 
 
 o o 
 
 Y* My be the total weight of the wheels 
 
 Fi 189 in panel AB, the center of gravity 
 
 of which we will assume is x dis- 
 
 tance from A, and let W be the total weight of the wheels in panel BC, 
 the center of gravity of which we will assume is s distance from C. 
 Let r be the concentration on the floor beam at B due to the wheels in 
 the panel AB, and let r f be the concentration due to the wheels in the 
 panel BC, and let R be the concentration on the floor beam at B from 
 the wheels in both panels. Then we have 
 
 R = r+r' ........................................... (1). 
 
 Taking moments about A we have 
 
 _Wx 
 
 I 
 
 and taking moments about C we have 
 
 W <? 
 
 V 
 Then substituting these values in (1) we have the general expression 
 
 for the concentration on the floor beam at B. Now, if the wheels roll 
 to the left, r will decrease and r' increase provided no wheels pass B, 
 and just the reverse is true if the wheels roll to the right. Now if the 
 weight of the wheels in the two panels is such that the increment of r 
 is just equal to the increment of r' for a very slight movement, it is evi- 
 dent that the increment of the concentration R will be zero and hence R 
 will then be a maximum, for otherwise it would be either increasing or 
 decreasing. If increasing, it evidently has not reached the maximum; 
 and if decreasing, it has passed the maximum. So, differentiating equa- 
 tion (2) we have 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 263 
 
 / /' 
 
 But dz dx. Then we have 
 
 dR W W W W 
 
 -37- = -7- + -77- = or -y- = ~YT - 
 dx I I I I 
 
 That is, the unit load in one panel will be equal to the unit load in the 
 other panel when the maximum concentration on the floor beam occurs. 
 If the panels are of equal length, as they are here, we have 
 
 w=w. 
 
 That is, the maximum concentration on the floor beam will occur when 
 the load in one panel is equal to the load in the other. Now the increment 
 of R can change sign only when a wheel passes the floor beam at B. 
 So there will be a load at that point when the maximum concentration 
 occurs. It is evident that this criterion for maximum concentration on the 
 floor beam, so far established, can be satisfied by most any group of 
 wheels, but it is also evident that the absolute maximum concentration will 
 occur when the heaviest group of wheels is near the floor beam. So the 
 whole criterion for maximum concentration on an intermediate floor beam 
 can be stated as follows: 
 
 The maximum concentration will occur when the heaviest group of 
 wheels is near the floor beam and one wheel at it, and when the load in 
 one panel is equal to the load in the other. 
 
 The next thing in our case is to satisfy this criterion. Referring to 
 Table A we can see that wheels 1 to 5 are as heavy a group of wheels as 
 can be placed on two adjacent panels. 
 
 Let us place them as shown in Fig. 190. Then the load in panel 
 AE- 30,000 Ibs., and that in 
 BC = 40,000 Ibs. This does not 
 seem to be very close to the re- 
 quirement that the loads in the 
 two panels be equal. But by trial 
 it will be found to be as close as 
 possible and still satisfy the first 
 part of the criterion at the same Fig. 100 
 
 time. It is not often that this 
 
 criterion can be satisfied absolutely as to the loads in the panels being 
 equal. We should, however, place the loads so that the criterion is as 
 nearly satisfied as possible. In testing for the criterion the load at the 
 floor beam can be considered as being equally divided between the two 
 panels; that is, one-half of its weight being considered in each panel, 
 or it can be ignored as was done above. It will make no difference which 
 way it is considered. 
 
 So, taking the position shown in Fig. 190 as satisfying the criterion and 
 taking moments about C, we have 
 
 20,000 pTF^I =20,000 Ibs., 
 
264 
 
 STRUCTURAL ENGINEERING 
 
 and taking moments about A we have 
 
 10,000 x 2 + 20,000 x 10 
 15 
 
 Then adding these two concentrations to the 20,000-lb. load at the floor 
 beam, we have 
 
 R = 20,000 + 14,660 + 20,000 = 54,600 Ibs. 
 
 for the maximum floor beam concentration due to Cooper's 40 loading, 
 and multiplying this by 50/40 to reduce it to 50, we have 
 
 ^x 54,600 = 68,200 Ibs. (about) 
 for the maximum concentration. 
 
 | . Then the maximum live load on the floor 
 
 a, beam will be as shown in Fig. 191. 
 
 For the maximum live-load bending mo- 
 
 , we have 
 
 M = 68,200 x 63 = 4,297,000 inch Ibs., 
 and for the impact we have 
 
 x 4,297,000 = 3,906,000 inch Ibs. 
 oU + oUU 
 
 Now adding the above maximum dead- and live-load bending moments 
 and impact together we have 
 
 411,000 + 4,297,000+3,906,000 = 8,614,000 inch Ibs. 
 
 for the total maximum bending moment on the floor beam for which the 
 beam must be designed to resist. 
 
 For the maximum end shear on the floor beam due to dead load we 
 have 
 
 
 T7K 
 
 U ~"O L 
 
 |r? rt 
 
 IS'- 6" 
 
 | 
 
 
 ~\ 
 
 Fig. 191 
 
 a 
 
 7 
 
 300 
 
 5,200 
 
 = 7,000 Ibs. (See Fig. 187A.) 
 
 From live load we have 68,200 Ibs., as shown above (Fig. 191), and from 
 impact we have 
 
 68,200 x 
 
 Then adding the above dead- and live-load end 
 shears and impact together, we have 
 
 7,000 + 68,200 + 62,000 = 137,200 Ibs. 
 
 for the total maximum end shear. 
 
 The next thing is to design the floor beam. 
 The flanges of such floor beams are usually com- 
 posed of 6" x 6" angles, and the stringers should 
 fit in between the flanges as shown in Fig. 192 so 
 as to avoid fillers. So giving a J" clearance at both the top and bottom 
 of the stringers, as shown, we obtain floor beam 22%" deep. The 
 
 Fig. 102 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 265 
 
 distance from the back of the 6" x 6" flange angles to their center of 
 gravity could be obtained from a handbook or from Table 6 provided 
 the exact weight of them were known. However, we can obtain the 
 average from this source. So let us take 1.78" as the average. Then 
 for the approximate effective depth we have 
 
 32.5 -1.78x2 = 28.94 ins. 
 Now dividing this into the maximum bending moment given above we have 
 
 8 ii 4 ; 00 = 898,000 Ibs. (about) 
 
 /CO. 94: 
 
 for the flange stress. Then we have 
 
 298,000 
 
 16,000 
 
 = 18.6 sq. ins. 
 
 for the net flange area required minus one-eighth of the area of the cross- 
 section of the web. Let us assume the web to be 32" xj". Then one- 
 eighth of the area of web = 2 n ". 
 
 Then for the make-up of each flange we have 
 
 Gross Net 
 
 2 Ls 6" x 6' x ff" = 18.18 n "-1.62 = 16.56 n " 
 area of web = 2.00 n " 
 
 18.56 n " 
 
 Now for the actual effective depth (since we know the weight of the 
 flange angles) we have 
 
 32.5-1.8x2 = 28.9 ins., 
 
 which is practically the same as assumed above. 
 
 As the floor beam has no regular stiffeners (the stringers, however, 
 stiffen it) the vertical distance between the flange angles can be taken as 
 d in Formula (1), Art. 118. Then we have 
 
 20.5 =i (12,000-*), 
 
 from which we obtain 
 
 5 = 10,360 Ibs., say, 10,000 Ibs. 
 
 Now dividing this into the maximum end shear given above we have 
 137,200 
 
 10,000 
 
 = 13.72 sq. in. 
 
 for the required area of the cross-section of the web. The web assumed 
 above has 16 n ", which is a little more than required, but if a thinner web 
 be used the flange rivets would be quite close together, perhaps too close. 
 This trouble often occurs in such shallow floor beams. So we will use the 
 l" 
 
266 STRUCTURAL ENGINEERING 
 
 Preliminary estimate of the weight of an intermediate floor beam.* 
 
 1 web 32" x y x 15.5' x 54.4# ......................... = 843 Ibs. 
 
 4 Ls 6" x 6" x if" x 15.5' x 31* ....................... =1,922 Ibs. 
 
 Gusset plates (1 plate 60" x \" x 4'-0" for the two) @ 102# 
 
 per ft ............................................ = 408 Ibs. 
 
 4 Ls 3y x 3J" x |" x 4'-0" x 8.5* (on gusset plates or 
 
 brackets) ........................................ = 136 Ibs. 
 
 2 Ls 31" x W x f" x 2'-6" x 8.5# (end con. angles) ...... = 42 Ibs. 
 
 4 special plates 16" x y x l'-8" x 27.20* ............... = 181 Ibs. 
 
 3,532 Ibs. 
 2.5% for rivets .......................................... 88 Ibs. 
 
 Total ......................................... 3,620 Ibs. 
 
 This shows that the 3,600 Ibs. assumed weight of the floor beam is 
 very close to the actual weight and hence no recalculations are necessary 
 as 10 per cent variation either way is permissible. 
 
 149. Design of End Floor Beam. The designing of end floor 
 beams is very much the same as that of intermediate floor beams. The 
 concentrations, however, are different and the weight of the beams them- 
 selves is a little less. 
 
 Each concentration from dead load is equal 
 to the dead-load end shear on a stringer and 
 each live-load concentration is equal to the maxi- 
 mum live-load end shear on the same. Then 
 assuming the weight of an end floor beam to be 
 3,000 Ibs. we have the loading on the beam as 
 shown in Fig. 193, using the end shears given 
 in Art. 146. Then we have for the dead-load 
 moment. 
 
 M = 2,600 x 63 = 163,800 inch Ibs. 
 from the concentrated dead load and 
 
 M' = ^x 3,000x15.5x12 = 69,750 inch Ibs. 
 
 from the weight of the floor beam, making a total of 233,550 inch Ibs. for 
 the total maximum dead-load bending moment. 
 For the live-load bending moment we have 
 
 M" = 50,000x63 = 3,150,000 inch Ibs., 
 and for the impact we have 
 
 7=3,150,000 1Knn = 3,000,000 inch Ibs. 
 
 + 
 
 Now adding these moments and impact together, we have 
 
 233,550 + 3,150,000 + 3,000,000 = 6,383,550 inch Ibs. 
 
 for the total maximum bending moment. 
 
 Assuming an effective depth of 29" we have 
 
 6,383,550 
 
 * These preliminary estimates are made before the detail drawings are made and 
 are intended to be only approximately correct. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 267 
 
 for the flange stress. Then we have 
 220,000 
 
 16,000 
 
 = 13.75 sq. ins. 
 
 for the net area required for each flange minus one-eighth of the area of 
 the cross-section of the web. 
 
 Assuming the web to be 32" x f " = 12", one-eighth of the area of 
 the web is 1.5 n// . Then for each flange we have 
 
 2 Ls 6" x 6" x T V = 12.86 n " - 1.12 = 11.74 n "net 
 area of web = 1.50 n "net 
 
 13.24" net 
 
 As is seen, the net area of this flange is about 0.51 n " less than required, 
 but the 6" x 6" x f " angles give a flange about 0.72 n " too large, so the 
 above will be used. 
 
 For the maximum end shear, as is readily seen from Fig. 193, we have 
 
 D = 1,500 + 2,600 = 4,100 Ibs. 
 L = 50,000 Ibs. 
 
 7 - 50,000 r = 47,600 Ibs. 
 
 Total 101,700 Ibs. 
 
 For the allowable stress on the web we have 
 
 20.5=1(18,000-.), 
 
 from which we obtain 
 
 s = 9,815 Ibs., say, 10,000 Ibs. 
 
 Then dividing this into the above shear we have 
 101,700 
 
 10,000 
 
 = 10.17 sq. ins. 
 
 for the required area of the web. So the assumed web 32" x " = 12 n " 
 is satisfactory, as the specifications permit of no thinner web. 
 
 The flange angles and web as specified above for the end floor beam 
 weigh about 2,000 Ibs., which is about 765 Ibs. less than the flange angles 
 and web in the intermediate floor beam. Then as the other parts are about 
 the same as for an intermediate floor beam we have 3,621 - 765 = 2,856 
 Ibs. for the weight of the end floor beam. So the assumed 3,000 Ibs. for 
 the weight of an end floor beam is within the 10 per cent limit. 
 
 150. Design of the Main Girders. In the case of through plate 
 girder bridges the live load is applied to the main girders only at the 
 panel points, that is, where the floor beams connect to the main girders, 
 just the same as in the case of through truss bridges. The dead load is 
 applied to the main girders in the same way except for the weight of the 
 girders themselves which is uniformly distributed along the girders. But 
 
268 
 
 STRUCTURAL ENGINEERING 
 
 it is usual practice to consider the dead load as wholly applied at the 
 panel points as about the same result is obtained as if the more exact 
 conditions were considered. 
 
 Taking the assumed dead load given in Art. 146 we have 
 
 !5 = 13,350 Ibs. 
 
 /v 
 
 for the panel load of dead load per girder. Then the dead load per girder 
 will be as shown in Fig. 194, where AB represents the girder. Now it is 
 obvious that the maximum bending moment due to this dead load will 
 occur under the load at C. So taking moments at C we have 
 
 M= (20,025x30 -13,350x15) 12 = 4,806,000 inch Ibs. 
 
 for the maximum bending moment due to dead load. 
 
 The live load will be ap- * 
 
 plied at the panel points, but 
 of course the panel loads will 
 not be equal to each other as 
 in the case of dead load. The 
 maximum bending moment 
 will occur at a panel point as 
 in the case of a truss. It is 
 readily seen that the maximum moment will occur at the panel point C, 
 at the center of the span. Then the first thing to do is to place the 
 wheel loads so as to obtain the maximum moment at that point which is 
 simply the satisfying of the criterion for maximum moment as given in 
 Art. 91. That 5s, the maximum live-load bending moment on the girder 
 (which will occur at C) will occur when the average load to the left of C 
 is equal to the average load on the span. 
 
 Now referring to Table 
 
 @ 8' @> g' X^C 9' (fi'faeffisOh A, let us try wheel 13 at C. 
 
 Then the loads will be in the 
 position shown in Fig. 195, 
 and for the average unit load 
 on the left, considering half 
 
 J <$l $. 
 
 \ 
 
 /st- o" 
 
 /S-0" F /5-0" 
 
 AS--<7" 
 
 
 6O-O" 
 
 Fig. 194 
 
 1" ,, i 
 
 15' 
 
 ic 
 
 IS' 
 
 IS' 
 
 r rt 
 
 I 0'- 0" 
 
 Fig. 195 
 
 of wheel 13 as being on the left, we have 
 
 73,000 
 30 
 
 = 2,430 Ibs. (about), 
 
 and for the average unit load on the whole span, we have 
 
 142,000 
 
 60 
 
 = 2,370 Ibs. 
 
 Now it is quickly seen that this position will give the maximum moment 
 at C. For if wheel 14 be placed at C there would be 80,000 Ibs. to the 
 left which would give 2,660 Ibs. average unit load to the left and the 
 average unit load on the span would be the same as given above, and if 
 wheel 12 be placed at C the average unit load on the left will be 53,000 -f 
 30 = 1,760 Ibs. and the average unit load on the span will remain the same 
 as before. So the wheels when in the position shown in Fig. 195 will 
 produce the maximum bending moment on the girder. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 269 
 
 Then, the next thing is to determine the bending moment about wheel 
 13 as that is the moment desired. To do this we must first determine the 
 reaction R at A which we do by taking moments about B. As wheel 18 
 just happens to come exactly at B we can obtain the moments of all the 
 wheels on the span directly from Table A. Passing down the line through 
 wheel 18 (in the table) until we come to the zigzag line, then passing to 
 the left until we come to the vertical line through wheel 9, we find the 
 moment 4,224: which is the moment in thousands of foot pounds of all the 
 wheels, 9 to 17 inclusive, about wheel 18, which happens to be at the right 
 support. 
 
 Then for the reaction R we have 
 
 70.4 (thousands). 
 
 Then taking moments about wheel 13 we have (using Table A) 
 
 M= (70.4x30 -818) 1,000x12 = 15,528,000 inch Ibs. 
 Then multiplying this by 50/40 to reduce it to Cooper's 50, we have 
 
 15,528,000x^-19,410,000 inch Ibs. 
 
 for the maximum live-load bending moment on the girder. 
 Then for the impact we have 
 
 ( at ? Q ~ } 
 
 \ 60 + oOO I 
 
 19,410,000 x at ~ = 16,175,000 inch Ibs. 
 
 Now adding the above dead- and live-load moments and impact together 
 we have 
 
 4,806,000 + 19,410,000 + 16,175,000 = 40,391,000 inch Ibs. 
 
 for the total maximum bending moment on the girder. Now, assuming 
 the web will be f " thick, we have 
 
 40,391,000 B -V. 
 = 
 
 for the economic depth. So we. will use a web 84" deep (as this is a 
 common plate) and assume it to be f " thick. 
 
 The total depth of the girder back to back of angles will then be 
 84.25". Then subtracting about an inch we have 83.2" for an assumed 
 effective depth. Dividing this into the maximum moment given above we 
 have 
 
 40,391.000 s (about) 
 
 for the flange stress, and then we have 
 
 486,000 
 lg - 656 = 30.4 sq. 
 
 for the required net area of the cross-section of each flange. 
 
 486,000 
 
 -= 30.4 sq.ms. (about) 
 
270 STRUCTURAL ENGINEERING 
 
 The area of the cross-section of the 84" x $" web = 31. 5 n ", and 
 .J = 3.9 n ". Then for the make-up of each flange we have 
 
 Gross (J" rivets) Net 
 
 2 Ls 6" x 6" x ^" = 15.56 n "-2.75 n " = 12.80 n " 
 
 1 cov. pi. 14" x f" = 8.75 n "-1.25 n " = 7.50 n " 
 
 1 cov. pi. 14" x J" = 7.00 n "-1.00 n " = 6.00 n " 
 
 area of web = 3.90 n " 
 
 31.31 n " 30.20 n " 
 
 Now taking moments about the center of the top cover plate (see 
 Fig. 196) we have 
 
 - 2.62 x 15.56 + 8.75 x 0.56 
 
 x = ~~rnr~ - = 1.46 ms. 
 
 * HR, >/ *< l>v ^ or *^e distance from the center of the top cover 
 ^f 1 '" ^ ** -~ t ' c l V ^t plate to the center of gravity of the whole flange. 
 5;L^t-2Lllp^5r.41l Then we have 
 
 1.46-0.87 = 0.59 ins. 
 
 Fig. 196 
 
 for the distance from the back of the flange 
 angles to the center of gravity of the flange. Then we have 
 
 84.25 -(0.59x2) =83.07 ins. 
 
 for the actual effective depth of the girder, which is quite close to that 
 assumed above, so no recalculation on account of the assumed effective 
 depth is necessary. 
 
 For the length of the y cover plates, which will be the outside cover 
 plates on the girder, we have 
 
 = 60 = 28 '- 6 " + 1/ - 6// = 30 '- 
 
 For the f " cover plate we have 
 
 y ' = 60 rj - 42'-6" + l'-6" = 44'. 
 
 The maximum end shear due to dead load is equal to the reaction 
 shown in Fig. 194 plus the weight of the girder for half of the length of 
 the end panel. The four flange angles of the girder, as given above, 
 weight 106* per ft. of girder, the web weighs 108* per ft., and the weight 
 of stiffeners and details at this point on the girder will be about 90 per 
 cent of the weight of the web per ft., or 97*, and the two f " cover plates 
 weigh about 60*, making in all about 370*, say, 400* per ft. of girder. 
 Then we have 
 
 5 = 20,025 + 400x7.5 = 23,000 Ibs. (about) 
 
 for the maximum end shear on the girder. 
 
 The maximum end shear on the girder due to live load is equal to 
 the maximum shear in the end panel due to that load. Now according to 
 Art. 90 the maximum shear in the end panel (the same as in a truss 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 271 
 
 bridge) will occur when the load in the panel is equal to the total load on 
 the bridge divided by the number of panels. So the first thing is to 
 satisfy this criterion for maxi- 
 mum shear in the end panel. . 
 Placing wheel 3 at the first A@- 
 
 panel point out from the end of [5 ^ i ^ i- i 
 
 the girder as shown in Fig. 197 V G0 ~" 
 
 we have (using Table A) Fig. 197 
 
 + 20,000 + 10,000-40,000 Ibs. 
 
 for the load in the end panel AD, considering one-half of wheel 3 in the 
 panel. With the wheels in this position the total load on the span is 
 152,000*. Then we have 
 
 15 *' 000 = 38,000 Ibs., 
 
 4 
 
 which to satisfy the criterion should be 40,000*. But this is as close as 
 the criterion can be satisfied, as can be verified by trial. 
 
 The next thing is to determine the shear in the end panel AD with 
 the wheels in the position shown in Fig. 197. First of all the shear in the 
 panel is equal to the reaction R at A, due to all of the loads on the span, 
 minus the concentration r from the stringers, due to the loads in the panel 
 AD. In fewer words, the shear in the end panel AD is 
 
 S' = R-r. 
 
 Taking moments about the right support B (using Table A) we have 
 fl= 4.638.000 + 158,000 x 2 =8g)3001hgj> 
 
 and taking moments about the panel point D we have 
 
 230,000 
 
 r = = 15,300 Ibs. 
 
 15 
 
 Then for the maximum shear in the end panel AD we have 
 
 S' = 82,300 - 15,300 = 67,000 Ibs., 
 and multiplying this by 50/40 we have 
 
 67,000 x|g = 84,000 Ibs. (about) 
 
 for the maximum live-load shear on the girder. 
 Then for the impact we have 
 
 Now adding together the dead-load and live-load end shears, given above, 
 and the impact, we have 
 
 23,000 + 84,000 + 73,000 = 180,000 Ibs. 
 
272 STRUCTURAL ENGINEERING 
 
 for the total maximum end shear on the girder, which is practically the 
 shear throughout the end panel. 
 
 Now for the unit-shearing stress on the assumed 84" x " web we 
 
 ) . 
 
 Then substituting this for s in Formula (1), Art. 117, we have 
 <J = A (12,000 -5,710) =59 ins. 
 
 for the maximum spacing of the stiffeners at the ends of the girders. 
 Now,, as this is not less than the half depth of the girder, the assumed 
 web is satisfactory. The specification will not permit of the web being 
 made thinner. 
 
 The outstanding flanges or legs of the stiffeners, to satisfy the speci- 
 fications,, should not be less than 1/30x84 + 2 = 4.6". This requires the 
 use of 5" x 3J" angles, the 5" leg outstanding. All intermediate stiffeners 
 can be taken as 5" x 3J" x f" angles without hesitating this being the 
 minimum thickness allowed but the end stiffeners must have sufficient 
 area of cross-section to take the maximum reaction on the girder when 
 considered as columns, as per specifications. The cross-section of a pair 
 of these angles, considered as a column, is shown in Fig. 198. 
 
 R 
 
 15' \ 
 
 15' 
 
 
 /5' 
 
 \ ,s' 
 
 
 
 
 60'- 
 
 o" 
 
 
 '9 
 
 Fig. 198 Fig. 199 
 
 Let us first assume them to be 5" x 3J" x f" angles. Then for the 
 radius of gyration about the axis y-y we have 
 
 [2 x 7.78 + 2.48^ x 6.1 . 
 
 r- \| =2.96 ms. 
 
 Then taking one-half the depth of the girder as the length, as per speci- 
 fication, we have 
 
 p = 16,000 - 70 -^ = 15,000 Ibs. (about) 
 
 for the allowable unit stress on such a column. 
 
 Now from Fig. 194 it is readily seen that the total maximum dead- 
 load reaction is 
 
 20,025 + 6,675 = 26,750 Ibs., say, 27,000 Ibs. 
 
 The maximum live-load reaction, as is readily seen, will occur when 
 the wheels are in the position shown in Fig. 199, where AB represents the 
 span. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 273 
 
 Taking moments about B (using Table A) we have 
 
 f 162 x 4 N 
 
 _ /5,208 + 
 
 R= (~ ~ 
 
 1,000- 97,600 Ibs. 
 
 for the maximum live-load reaction at A due to Cooper's 40, and multi- 
 plying this by 50/40, to reduce it to Cooper's 50 loading, we have 
 
 97,600 x jg = 122,000 Ibs. 
 
 for the maximum live-load reaction desired. 
 Then for the impact we have 
 
 =101,000 Ibs. '. '. :'.-' . 
 
 Now adding together the above maximum dead- and live-load reactions 
 and the impact we have 
 
 27,000 + 122,000 + 101,000 = 250,000 Ibs. 
 
 for the total maximum reaction. Now dividing this by the allowable unit 
 stress as found above for the end stiffeners we have 
 
 250,000 
 
 : I6 ' 6 sq< ms ' 
 
 for the required area of the end stiffeners. 
 
 It is necessary to have two pairs of end stiffeners at each end of the 
 girders in order to properly distribute the pressure over the pedestal. 
 Two pairs of the stiffeners assumed above have 12. 2 n " cross-section, 
 which is about 4 n/ ' less than required. So thicker angles than those 
 assumed will have to be used. The allowable unit stress will not be 
 materially affected by using thicker angles as the radius of gyration varies 
 but little with the thickness. So we can use any two pairs of 5" x 3J" 
 angles that have the proper area. It is seen from a handbook, or from 
 Table 4, that 4 Ls 5" x 3J" x J" have 16 n " cross-section, and that 
 4 Ls 5" x 3" x T V have 17.88 n ". As is seen, the J" angles have an 
 area of cross-section that is 0.6 n " smaller than required, and the iV' 
 angles have about 1.28 n " more than required. So let us use 4 Ls 5" x 
 3J" x \" = 16.0 n " at each support for end stiffeners. 
 
 For area of bearing on the masonry we have required 
 
 250,000 
 
 The pedestal should be made to suit. 
 
 Preliminary estimate of the weight of one main girder. 
 
 4 Ls 6" x 6" x ft" x 69'-0" x 26.5* (combined 
 
 with end Ls) .............................. 7,314 Ibs. 
 
 l_ we b 84" x f" x 62'-0" x 107.12* ............. 6,641 Ibs. 
 
 1 cov. pi. 14" x J" x 60'-0" x 23.80* (2 combined) 1,428 Ibs. 
 
 1 cov. pi. 14" x f" x 118'-0" x 29.75* .......... 3,510 Ibs. 18,893 Ibs. 
 
274 
 
 STRUCTURAL ENGINEERING 
 
 8 end stiff. Ls 5" 
 26 int. stiff. Ls 5" 
 
 x 3J" x \" x 13.6* x 7'-0" ...... 762 Ibs. 
 
 x 3J" x f" x 10.4* x 7'-0" ..... 1,893 Ibs. 
 
 7 fillers 8" x ii" x 6'-0" x 18.7* (at floor beams) . 785 Ibs. 
 
 8 splice pis. 10" x 
 4 splice pis. 14 
 3 fills. 31" x 
 
 x 2'-3" x 23.38* .......... 421 Ibs. 
 
 x x 4'-4" x 32.72* ......... 568 Ibs. 
 
 x 6'-0" x 8.18* ............... 147 Ibs. 
 
 4,576 Ibs. 
 
 3% for rivet heads 
 
 23,469 Ibs. 
 700 Ibs. 
 
 Total weight of one main girder 24,169 Ibs. 
 
 Total weight of two main girders 48,338 Ibs. 
 
 Total weight of main girders per ft. of span = 48,338/60 = 805 Ibs. 
 
 To make this estimate the student would have to sketch the details of 
 the girders to some extent. 
 
 151. Design of Lateral System. The laterals with the floor 
 beams and main girders form a horizontal double truss as shown in Fig. 
 
 200, where AB represents one main girder 
 and CD the other. The laterals, according 
 to the specifications, must be designed to 
 resist a lateral force of 200 + 5,000x0.10 
 o - 700* per ft. of span considered as a live 
 load. One system can be considered as 
 taking the force when applied from one 
 direction and the other system when it is 
 applied from the other direction. So the 
 
 
 f 
 
 F 
 
 G 
 
 ^ / 
 
 ^ / 
 
 \ / 
 
 \ / 
 
 X*-, 
 
 X 
 
 X 
 
 X 
 
 y \ 
 
 y \ 
 
 / \ 
 
 / \ 
 
 m /s' _[ is' 
 
 2 
 
 (3) fS' 
 
 / - 
 /s 
 
 t 4. Panels IS-O"= 6O- O " 
 
 Fig. 200 
 
 laterals may be considered as taking tension only. 
 For a panel load we have 
 
 P = 700x15 = 10,500 Ibs. 
 
 Then assuming the lateral force to be applied from the direction indicated 
 by the arrows and suppose it moves onto the structure from right to left, 
 loading panel points G and F, we have 
 
 S = -x 10,500 = 7,800 Ibs. 
 4 
 
 for the maximum shear in panel FE (see Art. 90), and for the maximum 
 shear in panel EA, we have 
 
 S'=-x 10,500 = 15,700 Ibs. 
 4 
 
 By multiplying each of these shears by the secant of the angle 6 we will 
 obtain the stress in the corresponding laterals. As angle is about 45 
 degrees we can take the secant as 1.4. Then for the stress in the diagonal 
 marked (a) we have 
 
 T = 15,700x1.4 = 22,000 Ibs., 
 and for the stress in the diagonal marked (6) we have 
 
 r = 7,800x1.4 = 10,900 Ibs. 
 
 If the lateral force were applied in the opposite direction to that 
 indicated by the arrows, the lateral marked (c) would be subjected to a 
 maximum stress of 22,000* and the lateral marked (d) would be subjected 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 275 
 
 to a maximum stress of 10,900* and laterals (a) and (6) in that case 
 would have no stress. The same maximum stresses would be found in the 
 laterals in the right half of the span if the lateral force were considered 
 to move onto the span from left to right. 
 
 The stresses in the floor beams due to the lateral force ( 700*) can 
 be neglected, as the stress produced is simple compression in the bottom 
 flanges, which are in tension from dead and live load. It is readily seen 
 that the stresses calculated above are all that we need for designing the 
 laterals. 
 
 For the laterals marked (a) and (c) we have 
 
 22,000 
 
 for the required area of cross-section. 
 
 Use 1 L 3" x 3" x f" = 2.30 - 0.37 = 1.93" net. This is the 
 smallest angle allowed by the specifications and therefore the same size 
 will be used for each of the other laterals. 
 
 Preliminary estimate of weight of lateral system. 
 
 8 Ls 31" x 3" x f" x 7.9* x 20' ............. 1,264 Ibs. 
 
 6 lat. pis. 17" x f " x 21.6* x 3' ............. 389 Ibs. 
 
 4 lat. pis. 17" x |" x 21.6* x 2' ............. 173 Ibs. 
 
 4 sp. pis. 9" x |" x 11.5* x 3' ............... 138 Ibs. 
 
 16 Ls6"x4"x i" x 16.2* x 1' ............... 259 Ibs. 
 
 rivets ................................... 70 Ibs. 
 
 2,293 Ibs. 
 
 Total weight of laterals per ft. of span = 2,293/60 = 38* per ft. 
 152. Preliminary Estimate of Weight and Cost. 
 
 Estimates of effective dead weight per ft. of span. 
 
 Weight of stringers per ft. of span (Art. 147) . . 289 Ibs. 
 Weight of intermediate floor beam per ft. of 
 
 span (3,621/15) (Art. 148) . . ........... 242 Ibs. 
 
 Weight of main girders per ft. of span (Art. 150) 805 Ibs. 
 
 Weight of laterals per ft. of span (Art. 151) ---- 38 Ibs. 
 
 Weight of deck per ft. of span ................ 400 Ibs. 
 
 Total weight of dead load per ft. of span. . . 1,774 Ibs. 
 As 1,780* was assumed no recalculations are necessary. 
 
 Estimate of total weight of metal in span. 
 
 2 main girders at 24,169* ................... 48,338 Ibs. 
 
 3 intermediate floor beams at 3,621* ..... . ---- 10,863 Ibs. 
 
 2 end floor beams at 2,856* ................. 5,712 Ibs. 
 
 4 panels of stringers and details at 4,358* ..... 17,432 Ibs. 
 
 8 laterals and details ...... ................. 2,293 Ibs. 
 
 4 pedestals and sole plates ................... 2,200 Ibs. 
 
 86,838 Ibs. 
 
I. i I 
 
 
 - 
 
 
 V 
 
 
 
 
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 ^H) 
 
 
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 J^-^. 
 
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 Vj 
 
 
 
 
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 <i 
 
 
 
 -5 
 
 Jfc 
 
 
 
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 <V| 
 
 
 
 
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 J L 
 
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 000)0*^?; 
 
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 1*868S* 
 3*$&^ 
 
 i:ir|S 
 
 ij%*% ? 
 
 So ooio ! ^ 
 
 -JOOOQ -,.(0 
 
 SO^Qki o-a 
 
 I^^OOMk OvJ* 
 
 _ vojth CM^ 
 
 * =8 - 
 
 | 5 v)A? 
 
 i&*& 
 
 i^^^rr- 
 
 R OOOIOK, 
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 ^OOOOQjg, ^ 
 
 |S?|H?* 
 
 **s*l|lj$*i| 
 54<s IIIIK 
 
 vSn.^^.^ 
 
 ^ 
 
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 5oo< 
 ^T o o Dig jr" 
 
 ^OOOQ * 
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 ^Q-w^ 
 
 
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 io ook -i. ;< 
 
 000 C. o 
 lOJ'o'oltO Q 
 
 276 
 
277 
 
278 
 
279 
 
280 
 
 STRUCTURAL ENGINEERING 
 
 For the cost at 3^ we have 86,838 x 0.035 = $3,039.33. 3J^ per Ib. 
 is a common price for this class of work erected. However, the pound 
 price will vary from 2-|^ to 4^. It depends upon the market price of 
 metal and freight. 
 
 This completes the necessary preliminary calculations for the span. 
 Next a stress sheet as shown in Fig. 201 can be drawn. Then the shop 
 drawings (Figs. 202 and 203) for the span can be made. 
 
 Fig. 205 
 
 The work up to the completion of the stress sheet, as a rule, is done 
 in the designing office of a bridge company, or in the office of a railroad 
 company, or in the office of a consulting engineer. Consulting engineers 
 as a rule (and sometimes railroad companies) make general drawings, 
 similar to the one shown in Fig. 204, instead of stress sheets. In that 
 case the general drawing for the span is submitted to the different bridge 
 companies for use in preparing their bids for fabricating and (usually) 
 erecting the structure. After the contract is awarded, the general draw- 
 ing, instead of a stress sheet, is used as a guide in making the shop draw- 
 
DESIGN OF SIMPLE EAILKOAD BRIDGES 281 
 
 ings by the bridge company obtaining the contract. As a rule bridge 
 companies do not make general drawings of such structures only the 
 stress sheets and shop drawings. 
 
 153. Making of the Detail Drawings. In beginning the work, 
 the first thing to do is to draw a half cross-section of the span,, as shown 
 in Fig. 205, to a large scale, say 1-|" scale. In making this sketch (Fig. 
 205) first draw the cross-section of the main girder and the stiffeners. 
 Then the next thing to do is to draw the floor beam. Counting from the 
 back of the bottom flange angles of the main girder, we have \^'' ^ or the 
 thickness of the flange angle, and f " for the thickness of the lateral plate, 
 making in all yi + t = Wif" f r tne distance from the back of the flange 
 angles of the main girder to the back of the bottom flange angles of the 
 floor beam. This distance being determined, the bottom flange of the floor 
 beam is located and can be drawn. The distance from the back of the 
 bottom flange angles to the back of the top flange angles of such shallow 
 floor beams is usually made J" more than the depth of the web. So in this 
 case, as the web is 32" deep, we have 32 J" for the distance from the back 
 of the bottom flange angles to the back of the top flange angles, and thus 
 the top flange is located and can be drawn. Next the cross-sections of the 
 two I-beams composing the stringer can be drawn at their proper location. 
 Then by drawing a horizontal line 9J" above the top of these beams 
 (using 10" ties dapped \" on the stringer) we have the base of rail 
 located, and having the base of rail located the clearance line can be drawn 
 as shown. This is about as far as we can proceed with the drawing (Fig. 
 205) until the required spacing of the rivets in the floor beam is deter- 
 mined. 
 
 For the shear on the part of the floor beam between the end and 
 where the stringer connects, we have 137,200* (see stress sheet). The 
 vertical distance from the center of rivets in the top flange to center of 
 rivets in the bottom flange is about 25.5", and the allowable bearing of a 
 f" shop rivet on the " web is 10,500*. So, substituting 137,200 for S, 
 25.5 for h, and 10,500 for r in Formula (2) (Art. 116), we have 
 
 10,500 x 25.5 
 P= 137,200 :1 ' 95inS " 
 
 say, 2", for the theoretical spacing of the flange rivets in the part of the 
 floor beam between the end and the stringer. 
 
 The shear on the part of the floor beam between the stringers is very 
 slight. In fact there is no shear at all except that due to the weight of the 
 intervening part of the floor beam itself. This being the case, the flange 
 rivets in that part of the floor beam can be spaced 6" apart, which is the 
 maximum spacing allowed. 
 
 Now beginning at the end of the floor beam (Fig. 205) we space the 
 rivets 2" apart (as required) in the flange angles out to the splice (where 
 the gusset and web meet) and the only break in the 2" spacing there is 
 that necessary to fit the details of the splice. The splice is located so 
 as to bring the bracket just inside the clearance line as shown. 
 
 The rivets in the splice should line up with those in the stringer 
 connection and at the same time with those in the end details of the floor 
 beam. 
 
 There must be enough rivets in each stringer connection to transmit, 
 
282 STRUCTURAL ENGINEERING 
 
 in bearing on the web, the maximum floor beam concentration. For the 
 concentration we have 5,200* from the dead load, 68,200* from the live 
 load as given in Art. 148, and 62,000* [=300 -=- (30 + 300) x 68,200] from 
 the impact, making a total of 135,400*. The allowable bearing of a I" 
 field rivet on the \" web is J x 20,000 x \ => 8,750*. Then for the number 
 of rivets required in each stringer connection we have 
 
 135,400 
 
 = 15.4, say 16, rivets. 
 
 It is seen from this that a single line of rivets in each of the angles 
 connecting the stringers to the floor beam is sufficient. So we have this 
 much data for later use. 
 
 Now the next thing to do is to determine the vertical spacing of the 
 rivets in the floor beam. It appears that there is no question but that the 
 rivets should be spaced as closely as is permissible in the vertical direction 
 in the case of such shallow floor beams. So we will simply try putting in 
 as many rivets in the vertical direction as will fit in. Beginning at the 
 bottom of the floor beam, we will make the first gauge in the flange angle 
 2J", as the angle is quite thick (yf"), and the second gauge 2|-", which 
 leaves If " edge distance on the angle. For the next space we have If" 
 (edge distance on the angle) plus \" (clearance) plus \\" (edge distance 
 for fillers and splice plates), making in all 3-J". Now these same spaces 
 will be used at the top of the floor beam in order to have symmetrical 
 spacing. So we have 
 
 2-f3 =17 ins. 
 
 for the remaining distance in which rivets are to be spaced. It is readily 
 seen that 3" spacing will not fit in this distance and that five spaces is the 
 maximum number possible. So we will space the rivets as shown. 
 
 We will now see how this spacing fits the detail at the end of the floor 
 beam, at the splice, and at the stringer connection. There should be 
 enough rivets in the end of the floor beam proper to transmit the greater 
 part of the maximum end shear on the beam to the main girder. In the 
 detail shown we have 8 field rivets in double shear or bearing on lyV' 
 metal. The double shear being the least we have 8 x 0.6 x (10,000 x 2) = 
 96,000 Ibs. for the amount that these rivets will be considered to transmit. 
 This leaves 41,200 Ibs. (=137,200-96,000) to be taken by the rivets in 
 the bracket above the floor beam proper. These rivets are in single shear 
 and hence the allowable stress on each is 6,000 Ibs. Then we have 
 
 41,200 
 
 ^55^=6.8 rivets, say 7, 
 
 required in the bracket above the beam proper. So the detail as shown for 
 the end of the floor beam is quite satisfactory, as the extra rivets at the top 
 of the bracket are really needed to hold the bracket which stiffens the top 
 flange of the main girder. Now by prolonging the lines giving the vertical 
 spacing on to the stringer connection, it is seen that this spacing fits nicely 
 at that point, as we obtain the detail shown. The connecting angles 
 between the I-beams are shop riveted to the floor beam to avoid difficult 
 field riveting, as it would be difficult work to drive the rivets connecting 
 these angles to the floor beams in the field. 
 
DESIGN OF SIMPLE EAILROAD BEIDGES 283 
 
 Let us now turn our attention to the splice. The most satisfactory 
 way of designing such splices is to draw in what looks to be about correct, 
 and then determine the stress on the rivets assumed, and make whatever 
 modifications are necessary. So let us assume we need three rows of rivets 
 on each side of the splice as shown. First suppose that the f" plates 
 extending over the flange angles are omitted and that there is just a single 
 f " splice plate on each side of the web. 
 
 Now, for the vertical force on each rivet due to the vertical shear we 
 have 
 
 137,200 
 v = =7,620 Ibs., 
 
 and let * be the maximum horizontal force on each of the rivets farthest 
 out from the center of the web due to cross bending. Then, according to 
 Art. 117, we have the equation 
 
 2x3(1.75 +5.25 +8.5 ) xs-f 8.5 = 1/8 area of web 
 
 multiplied by 16,000 x h (=2 x 16,000 x 29) =928,000 in. Ibs., from which 
 we obtain 
 
 , = 928.000 = lbs 
 
 72.6 
 
 This horizontal force should not exceed ($ x 24,000 x ) 17/25.5 = 7,000 
 Ibs., so the splice as just considered would not be sufficient. 
 
 As one more line of rivets on each side of the splice will not be 
 sufficient, it is obvious that it will be best to use the f" plates extending 
 over the flange angles as shown. So, considering the splice as it is shown 
 in Fig. 205, we have 
 
 = 928,000 in. Ibs., 
 
 from which we obtain 
 
 928,000 _ , , . v 
 
 1390 =6 ' y lbs< ( about ) 
 
 for the horizontal force on the outer rivets (which are those in the outer 
 gauge lines in the flange angles). These rivets are so near the top and 
 bottom edge of the beam that the vertical shear on them can be neglected 
 (see Art. 61). 
 
 The above 6,700* force can be considered as applied to the rivets by 
 the web and transmitted from there to the angles and on to the f" plates 
 producing a shear of 3,330* on each rivet at each f" plate. The flange 
 increment can be considered as being transmitted by the f" plates and 
 web combined. Taking the flange increment as 10,500* (allowable bear- 
 ing of a J" rivet on the J" web) and assuming that one-third is trans- 
 mitted to the flange angles by the web and one-third by each of the f" 
 plates, we have 
 
 + 3,330 = 6,830 Ibs. 
 
284 STRUCTURAL ENGINEERING 
 
 for the shear on each rivet at each f " plate and 
 10,500 
 
 for the bearing stress on the web. Now, as 7,200* is allowed in the 
 former case and 10,500* in the latter, the top and bottom part of the 
 splice is about correct. 
 
 The rivets connecting the splice plates to the web directly, not passing 
 through the flange angles, are in double shear and bearing on the web. 
 The allowable bearing on the web, which is 10,500 Ibs., is less than double 
 shear, and hence the maximum stress on these rivets results from the 
 bearing on the web. Now, as a test for these rivets, let us consider the 
 ones farthest out from the center of the web, which are 8^" out. Let 
 s" be the horizontal force on each. Then we have 
 
 2 I 3 (L75 2 + 5725 2 + 8J5*) +2 (TTTif + 13.75 2 )J^ =928,000 in. Ibs., 
 
 from which we obtain 
 
 928,000 
 s"= = 4,140 Ibs. (about) 
 
 (this could be as great as 7,000 Ibs.). 
 
 Then for the resultant force on each of these rivets we have 
 
 = 8,670 Ibs., 
 
 which is less than the allowable bearing of each on the \" web, and hence 
 the splice as shown is amply strong. 
 
 The splice, as shown in Fig. 205, is designed to develop the strength 
 of the beam (as should be done), while as a matter of fact the flange at 
 the point of splice is practically twice as strong as need be, as the bending 
 moment at that point is only about one-half the maximum on the beam 
 and if the splice were designed to carry only the shear no actual weakness 
 would be detected in it. 
 
 This completes the necessary calculation in connection with the 
 drawing of the general cross-section shown in Fig. 205. Next a sketch 
 to 1J" scale for each of the lateral connections, and also a sketch (to a 
 large scale) of the curved end of the main girder, should be made. After 
 these preliminary sketches are made, we can proceed with the making of 
 either the general drawing (Fig. 204) or the shop drawings (Figs. 202 
 and 203). We simply transfer the dimensions on these sketches to the 
 final drawings. 
 
 Most of the calculations for the details shown on these drawings are 
 quite similar to those given in Art. 135 for the 50-ft. deck plate girder. 
 The main difference is in the determination of the flange rivets in the 
 main girders. In the case of the through girder, here considered, the 
 loads are applied directly to the web of the main girders (by the floor 
 beams) instead of to the flanges as in the case of the deck girder treated 
 in Art. 135. 
 
 So Formula (2) of Art. 116 is used to determine the pitch of the 
 flange rivets instead of Formula (4). The shear in each panel is prac- 
 tically constant throughout the panel, and consequently the rivet spacing 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 285 
 
 should be constant throughout each respective panel. Taking the case of 
 the end panel, we have 
 
 S= 180,000* (maximum end shear); 
 
 r = 7,880* (= allowable bearing of a " rivet on the f" web); 
 ft = 77". 
 
 Now substituting these values in Formula (2) (Art. 116), we have 
 
 7,880 x 77 
 P= 180,000 =3.36 ins., say, 3i ins., 
 
 for the theoretical pitch of the flange rivets in the end panels. The shear 
 on the girder is reduced some by the weight of the girder itself, as we 
 pass from the end toward the center, and, consequently, the pitch can be 
 increased slightly as the first intermediate floor beam is approached, as is 
 shown in Fig. 202. 
 
 The dead-load shear in the second panel from the end, as seen from 
 Fig. 194, Art. 150, is 
 
 20,025-13,350 = 6,670 Ibs. 
 
 The maximum live-load shear occurs in the second panel when the 
 wheel loads are in the position shown in Fig. 206, as the criterion of Art. 
 90 is satisfied. / 
 
 Taking moments about | r Os' sss(s) 9' 5' 
 
 the end B of the span (Fig. TR , \ \ c 
 
 OA/\ 1. / m IS A IS -I- is' J. IS' 
 
 206) we have (using Ta- 6g ;_ o// * 
 
 ble A) Fig . 206 
 
 2,155,000 
 
 for the reaction at A. Then taking moments about the floor beam at C 
 we have 
 
 =5,330 Ibs. .. . : 
 
 ( _ . 
 
 for the stringer concentration on the floor beam at D. Then we have 
 R - r = 37,900 - 5,330 = 32,570 Ibs. for the live-load shear in the second 
 panel from the end of the span due to Cooper's 40; and for Cooper's 
 50 we have 
 
 32,570x|5 = 40,700 Ibs., 
 
 and for impact we have 
 
 Now adding the above dead- and live-load shears and the impact 
 together, we have 
 
 6,670 + 40,700 + 37,000 = 84,370 Ibs. 
 
 for the total maximum shear in the second panel from the end of the span, 
 Then using Formula (2) (Art. 116) we have 
 
 7,880 x 74 R Q . 
 
 ^.9 ms. (about) 
 
286 
 
 STRUCTURAL ENGINEERING 
 
 for the theoretical spacing of the flange rivets in the main girders through- 
 out the second panel from the end of the span. So 6", the maximum 
 spacing allowed, will be used. 
 
 Further analyses of the details of the above span are considered 
 unnecessary, for if the details of the deck plate girder spans previously 
 treated are thoroughly understood, the student should have little trouble in 
 making the detail drawings for through plate girder spans, and especially 
 if the drawings shown in Figs. 202, 203, and 204 be observed closely as 
 the work progresses. The shop bills for this work are practically the 
 same as for deck spans. 
 
 154. General Remarks. The general discussion given above, 
 regarding the design of flanges and end bearings for deck plate girder 
 bridges, holds in the case of through plate girder bridges. 
 
 The floor beams for through plate girder bridges are sometimes made 
 as shown in Fig. 207. This type of detail eliminates the web splice in the 
 
 Fig. 207 
 
 floor beam. The brackets are usually shipped loose and riveted to the 
 floor beams and main girders in the field, especially when the main 
 girders are quite deep. While this type simplifies the details of the floor 
 beam, it complicates the details of the main girders and there are con- 
 siderably more field rivets than there are in the type used above for the 
 60-ft. span. The details of the two types vary considerably in practice, 
 and the details given here are intended only as a fair example. 
 
 The stringers for through plate girder bridges are sometimes plate 
 girders (where the under clearance will permit), in which case two 
 stringers per track are used. 
 
 155. Graphical Determination of Live-Load Shears and Bend- 
 ing Moments on Main Girders of Through Plate Girder Bridges. 
 The "influence line" method, outlined in Arts. 100 and 101, is the most 
 convenient graphical method in the case of through plate girder bridges. 
 However, the equilibrium polygon can be used if one so desires. 
 
 As an illustration, let it be required to determine the maximum 
 reaction, end shear, and bending moment on a 75-ft. single-track through 
 plate girder span of five 15-ft. panels, due to Cooper's 50 loading. (See 
 diagram, Fig. 151.) 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 
 
 287 
 
 To determine the maximum reaction, lay off the span AB (Fig. 208") 
 to, say, i" scale, and place the loads as shown. 
 
 Then draw the base line ab arid lay off ac = V 1* and draw the 
 influence line cb, and draw the ordinates 1, 2, ... 13 under the loads, as 
 shown, and all is ready for determining the reaction. Take a pair of 
 dividers and step off the ordinates 1 to 8 as explained in Example 1, Art. 
 100, and multiply their sum, in inches, by 25,000*. Next, in the same 
 way, step off the ordinates 9 to 12 and multiply their sum by 16,250*. 
 
 Fig. 208 
 
 Then multiply the length of ordinate 13 by 12,500 and add all three of 
 the results together, and the desired reaction is thus obtained. 
 
 To determine the maximum shear in the end panel AD (Fig. 209), 
 place the wheels as shown, thus satisfying the criterion for maximum 
 shear (Art. 90). Then draw the reaction influence line cb for the span 
 and cd for the panel AD. Then by multiplying the sum of the ordinates 
 (obtained as explained above) ef, 1, 2, 3, 4, and 5 by 25,000*, 6 to 9 by 
 16,250*, and h and 10 by 12,500*, and adding these products together, 
 the maximum shear in the end panel AD is obtained. 
 
 To determine the maximum shear in any of the intermediate panels 
 
 (/) 8 (^(3)5^(5} 9' (6)5 (7)6 (8)(9) S' (fo 8 (//)*(/ 
 
 Fig. 209 
 
 as DE (Fig. 210), first draw the reaction influence lines eb and af and 
 then the influence line adcb for shear in the panel, as explained for trusses 
 in Art. 102, and place the wheels as shown for maximum shear in the 
 panel, as per Art. 90. Then by multiplying the sum of the ordinates 
 2 to 5 by 25,000*, 6 to 9 by 16,250*, and ordinate 1 by 12,500*, and 
 adding the products together, the maximum shear in the panel DE is 
 obtained. 
 
 It should be noted in the last case that the position of the wheels for 
 maximum shear in the panel can be determined directly from the influence 
 line. For, according to Art. 102, no load should pass the point (Fig. 
 210), and there must be a load at E, so by placing wheel 1 as near as 
 possible, with a wheel at E, we have the position of the wheels for 
 
268 
 
 STRUCTURAL ENGINEERING 
 
 maximum shear in the panel. The position of the wheels for maximum 
 shear in any of the intermediate panels can be determined in this manner. 
 The maximum bending moment will undoubtedly occur at the floor 
 beam nearest the center of the span, and, as there are two floor beams 
 live load (in reference to C) as shown, thus satisfying the criterion for 
 
 Fig. 210 
 
 equally near the center, the location of either beam can be taken as the 
 point of maximum moment. So, if AB (Fig. 211) represents the span, 
 either C or D can be taken as the point of maximum bending moment. 
 Let us take point C. Then to determine the maximum moment, place the 
 maximum moment given for trusses in Art. 91. Next construct the 
 influence line aOb for the bending moment at C as explained in Art. 101, 
 and also in Art. 102. Then by multiplying the sum of the ordinates 1 to 4 
 by 25,000#, 5 to 9 by 16,250#, 10 by 12,500*, and one-half of 11 by 
 
 Fig. 211 
 
 (2,500*. x 10') and adding these products together, we obtain the maxi- 
 mum bending moment on the main girders. 
 
 To determine the maximum shear in any panel, as CD (Fig. 212), 
 by means of an equilibrium polygon, first place the load for maximum 
 shear in the panel, as shown, and draw the ray diagram MNO, and the 
 corresponding equilibrium polygon a-c-b . . . a. Then by drawing OE, 
 in the ray diagram, parallel to the closing line ac we have the reaction 
 (R) at A due to the loads given by the line EM. The shear in panel CD 
 is equal to R minus the stringer reaction (r) at C due to the load in panel 
 CD. By drawing the closing line ek we have the equilibrium polygon 
 eJche for the panel CD. Then by drawing E'O (in the ray diagram) 
 parallel to the closing line eJc we have the stringer reaction r, at C, due to 
 wheel 1 (as that is the only load in the panel), given by the line E'M. 
 Then, evidently, the maximum shear in panel CD is given by the line EE'. 
 
DESIGN OF SIMPLE KAILEOAD BEIDGES 
 
 289 
 
 The maximum shear in the other panels can be determined in the same 
 manner. 
 
 To determine the maximum bending moment on the main girders by 
 means of an equilibrium polygon, first place the loads on the span for 
 
 Fig. 212 
 
 maximum moment at any panel point as C (Fig. 213), as explained above, 
 and draw the ray diagram STP and the corresponding equilibrium polygon 
 acda. Then the maximum moment at C is equal to the ordinate y multi- 
 plied by the pole distance H. H is measured in pounds to the same scale 
 as used in drawing the ray diagram and y is measured in feet or inches to 
 
 Fig. 213 
 
 the same scale as used in drawing the span AB and in spacing the loads. 
 If y is taken in feet the moment will be in foot pounds, and if taken in 
 inches the moment will be in inch pounds. 
 
 In case an equilibrium polygon similar to the one shown in Fig. 186 
 (Art. 144) be drawn the shears and bending moments for any number of 
 through plate girder spans can be determined from it. 
 
290 
 
 STRUCTURAL ENGINEERING 
 
 The shear would be determined as shown in Fig. 212, the wheels 
 being placed for maximum shear as per criterion, Art. 90. The maximum 
 moment at any panel point C can, however, be determined without refer- 
 ence to the criterion for maximum bending moment, very much the same 
 as the case of deck spans treated in Art. 144, the main difference being 
 that the moment here is for a certain point on the girder. 
 
 As an illustration let UVW (Fig. 214) represent an equilibrium 
 polygon, similar to the one shown in Fig. 186. We start, say, by first 
 placing the point C, the panel point where the moment is desired, under 
 
 wheel 11 and drawing the closing line 1-1, we obtain the ordinate y. Next 
 placing the point C under wheel 12 and drawing the closing line 2-2 we 
 obtain the ordinate y' , and placing C under wheel 13 and drawing the 
 closing line 3-3 we obtain the ordinate y" , and so on. In this manner the 
 maximum ordinate for point C is obtained by trial, and by multiplying 
 this maximum ordinate by the pole distance the maximum bending moment 
 is obtained. 
 
 The maximum bending moment at any panel point can be obtained in 
 this manner. 
 
 156. Plate Girder Bridges with Solid Floors. Plate girder 
 
 bridges, especially through spans, sometimes have solid floors, particularly 
 those over streets in cities where an ordinary open floor is undesirable 
 owing to the dripping of water from the floor upon the sidewalk and 
 street below due to rain and melting snow. 
 
 These structures, in reference to floor, taking the most common con- 
 struction, can be divided into two general types : those having solid metal 
 floors covered with concrete and ballast, and those having reinforced con- 
 crete floors covered with ballast. The most common of the solid metal 
 floors is that known as the trough floor shown in Fig. 215 where the floor 
 proper is made up of vertical and horizontal plates connected by angles 
 so as to form successive troughs. In designing such bridges the first thing 
 to do is to draw a longitudinal section of the floor, enough to include three 
 ties 18" on centers, as shown to a large scale in Fig. 215. At the start the 
 whole thing is an assumption, that is, we draw in what looks to be correct. 
 
291 
 
292 
 
 STRUCTURAL ENGINEERING 
 
 From this sketch we determine the dead weight of the floor, including 
 metal, concrete, ballast and track. Then we compute the dead- and live- 
 load stresses on this assumed floor and if necessary modify the design until 
 it agrees with the final computation. 
 
 As an example let it be required to design the through plate girder 
 span indicated in Fig. 215, using Cooper's E50 loading. Estimating from 
 the longitudinal section of the floor (Fig. 215), taking concrete to weigh 
 150* and ballast 125* per cu. ft., we have for a strip through the floor one 
 foot wide and two feet long, along the track, the following weights : 
 
 H cu. ft. of concrete at 150* 225 Ibs. 
 
 TCU. ft. of ballast at 125* 125 Ibs. 
 
 Tie (6 ft. B. M.) at 6x4^* 27 Ibs. 
 
 31 ft. of 12" x f" plates at 15.3*. . 53 Ibs. 
 
 4 "ft. 3Jx3ixf Ls at 8.5* 34 Ibs. 
 
 Rails (90*) and spikes 12 Ibs. 
 
 Tar, stone dust, and burlap 20 Ibs. 
 
 Total 496 Ibs. 
 
 for two square feet of floor, or 248* for each square 
 foot (horizontal projection). The floor can be con- 
 sidered as being made up of independent Z-shape sec- 
 tions, each composed of one vertical plate, two angles 
 and two halves of horizontal plates as indicated in 
 Fig. 216 Fig. 216. This section can be treated as an inde- 
 
 pendent transverse beam. For the moment of inertia 
 of this beam with reference to axis 0-0, we have 
 
 (43 
 
 6" x f" ............. =186 (hor.pls.) 
 
 ..................... = 54 (vert.pl.) 
 
 2 (5T25" x 5 + 2.87) .............. = 281 (angles) 
 
 521 
 ~(1.5 n " x 6J25*) (rivets staggered) .. -60 
 
 Total 
 
 461 
 
 ' I2'@248*perlin,ft. 
 
 111 
 
 IS- 6' 
 
 Fig. 217 
 
 If the main girders are 15'-6" apart 
 the dead load on our Z-beam will really be 
 as indicated in Fig. 217, as the 248-lb. load 
 computed above holds only for the part of 
 the floor directly under the track. But as 
 the final result will not be affected mate- 
 rially if we assume the 248 Ibs. to extend 
 the full length of the beam, we shall so consider it. 
 
 Then for the maximum bending moment due to dead load we have 
 
 M = J x 248 x!5^5 2 x 12 = 89,370 incVi Ibs. 
 
 The next thing is to determine the maximum bending moment due to 
 live load. In accordance with the usual practice we will assume the 
 
DESIGN OF SIMPLE EAILKOAD BEIDGES 293 
 
 maximum live load in this case to be the heaviest axle, distributed over 
 three ties. So we have (special load, see diagram, Fig. 151) 
 
 ' = 2 0,833 Ibs. 
 
 for the total load on each tie. Now from the longitudinal section of the 
 floor (Fig. 215) we can see that our Z-beams carry from half of this load 
 to practically all of it. So to be sure we will assume that each carries the 
 20,833 Ibs. uniformly distributed along the beam by the tie. 
 
 Then for the maximum bending moment due to live load we have 
 
 M' = 4 x 20,833 x 15.5 x 12 = 484,300 inch Ibs. 
 
 Now adding together the above dead- and live-load moments and allowing 
 100 per cent for impact we have 
 
 M" = 89,370 + 2 (484,300)= 1,057,970 inch Ibs. 
 
 for the total maximum bending moment on our Z-beam. 
 
 Then substituting the above values in Formula (D), (Art. 53), we 
 have 
 
 /= 1,057,970 x 6.6 = 1 
 
 for the maximum bending stress on our Z-beam which is the maximum 
 bending stress on the floor. This shows that the floor is but little heavier 
 than necessary for cross bending, 16,000 Ibs. being the allowable bending 
 stress. 
 
 For the end shear on our Z-beam we have 
 
 248x^=1,920 Ibs. 
 
 from dead load and 
 
 20,833 
 
 = 10,416 Ibs. 
 
 from live load and the same from impact, making 22,700 Ibs. in all. 
 Then .to resist this shear we should have 
 
 22,700 
 
 16^00 =*** *". 
 
 of cross-section in the vertical plate of our Z-beam. This shows that the 
 shear on the floor is amply provided for, as each vertical plate contains 
 4.5 sq. ins. 
 
 The above calculations show that the floor shown in Fig. 215 is about 
 as correct as we can design it, as f " metal is the minimum thickness 
 allowed. 
 
 Having the floor designed, the dead weight coming onto the main 
 girders from the floor can be computed and by adding this per foot of 
 girder to the assumed weight of girder itself (per ft.) we obtain the 
 dead load for determining the dead-load stress in the main girders. Then, 
 using the formula pL 2 /8, the maximum dead-load bending moment on the 
 main girders can be determined. 
 
294 
 
 STRUCTURAL ENGINEERING 
 
 To obtain the maximum live-load shear and bending moment on the 
 main girders the live load is placed just the same as in the case of deck 
 spans. 
 
 The trough floor shown in Fig. 215, while it is the most common type 
 and a good design,, is only one of several types of solid floors in use. For 
 example,, there is the buckle plate floor composed of buckle plates (see 
 manufacturers' handbooks) laid upon I-beam or plate girder stringers, 
 flat plates laid upon I-beams used either as stringers or transverse beams, 
 the rolled trough section, etc. In addition to these types there are the 
 I-beam and concrete floors shown in Fig. 218, in which case the I-beams 
 are designed to carry all of the load. All of these types are used mostly 
 
 rBollasl 
 
 Ballast 
 
 Concrete, 
 
 Fig. 218 
 
 Fig. 219 
 
 for through plate girder bridges.- The trough floor is sometimes used on 
 deck spans, in which case the troughs are placed transversely resting 
 directly upon the main girders. 
 
 The out and out reinforced concrete floors are used on deck plate 
 girders. These floors are simple concrete slabs resting directly upon the 
 top of the main girders and turned up at each end so as to hold the 
 ballast, as shown in Fig. 219. 
 
 In designing all plate girder bridges having solid floors, the first 
 thing to do is to design the floor and determine its weight. Then the 
 weight of the main girders can be assumed and added to the weight of the 
 floor and we have the dead weight for determining the dead-load stresses 
 in the main girders. The other work involved is practically the same as 
 given above for ordinary bridges. 
 
 157. Double-Track Spans. Double-track deck plate girder bridges 
 are practically always composed of two simple single-track spans placed 
 side by side, as previously stated. Double-track through spans are usually 
 composed of two main girders placed far enough apart to admit the two 
 parallel tracks 13-ft. centers. In the case of ordinary open floors there 
 are usually four lines of stringers. The load in that case on each stringer 
 is just the same as in the case of single-track spans, while the floor beams 
 have four equal concentrations, each concentration being the same as for a 
 single-track span. The dead load on the main girders, as stated in Art. 
 124, is about 70 per cent more than for single-track spans, while the live 
 load is just twice as much, two trains being considered as moving abreast 
 over the structure. In the case of double-track through plate girder 
 bridges composed of three main girders, there are two independent single- 
 track floor systems. The loads in that case carried by each floor and by 
 each outside main girder are the same as for a single-track span, while the 
 central girder carries practically twice as much as each of the outside 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 295 
 
 girders. The objection to the three-girder type is that it is usually 
 necessary to "spread" the tracks at the location of these bridges in order 
 to obtain the necessary side clearance. 
 
 DRAWING ROOM EXERCISE NO. 5 
 
 Design a 64-ft. single-track through plate girder bridge and make a 
 stress sheet for same upon a 24" x IS" sheet and tracing of same. 
 
 Length of span = 4 panels at 16'-0" = 64'-0" c.c. end bearings. 
 
 Width = 15'-S" c.c. girders. 
 
 Live load, Cooper's E50 loading. 
 
 Dead load, to be assumed by student. 
 
 Specifications, A. R. E. Ass'n. 
 
 The required work consists of making the calculations and the 
 drawing of a stress sheet for the span, similar to the one shown in Fig. 
 201 for the 60-ft. span. 
 
 VIADUCTS 
 
 158. Preliminary. The ordinary railroad viaduct is a bridge com- 
 posed of a series of deck plate girder spans supported upon towers. How- 
 ever, as a rule any bridge supported upon towers is classed as a viaduct. 
 Viaducts are used where the crossings are so deep that ordinary concrete 
 or masonry piers are impracticable as to cost. The most common type of 
 viaduct is shown in Fig. 220 (a). The type shown in Fig. 220(6) is 
 built to some extent. The principal difference in the two types is in the 
 
 Cross Section 
 
 (a} 
 
 Cross Section 
 
 Fig. 220 
 
 tower bracing. Two columns connected together transversely by bracing 
 constitutes what is known as a bent (see Fig. 221). Two bents connected 
 together by longitudinal bracing constitutes what is known as a tower. It 
 is usual practice to make the spans between the towers twice as long as 
 the tower spans, except where the height of the viaduct is 40 ft. and 
 under, in which case the spans are usually made equal in length. 
 
 From actual calculations the economic lengths of spans are found for 
 ordinary cases to be as follows: 30' and 30' for viaducts 30 to 40 ft. 
 high; 30' and 60' for a height of 40 to 80 ft.; and 40' and 80' for a 
 height of 80 ft. and over. 
 
 The cost of erection governs the lengths of spans to quite an extent. 
 
 The approximate weight of metal in towers of ordinary single-track 
 viaducts is given on the diagram in Fig. 222. This diagram is self- 
 explanatory. 
 
296 
 
 STRUCTURAL ENGINEERING 
 
 Complete Design of an Ordinary Single-Track Viaduct 
 
 159. Data. Specifications, A. R. E. Ass'n. 
 
 Live Load, Cooper's 50. 
 
 160. General Layout of Structure. The first thing the designer 
 needs is a good profile of the crossing, which is usually furnished by the 
 railroad company. Let the profile shown in Fig. 223 be such a profile of 
 the crossing for which we are to design a viaduct. 
 
 Cross Frame 
 Transv. Struf 
 
 Longitudinal View of Tower 
 
 Transverse View of Benh 
 
 Figr. 221 
 
 Usually (the first thing) we would redraw this profile carefully to a 
 convenient scale (as a rule to a ^ scale) so that it can be traced on the 
 general stress sheet for the structure. Then upon this profile the positions 
 of the towers are chosen, avoiding the stream and obtaining as many 
 towers and columns of equal length as possible, using the length of spans 
 that is economic for the height of the viaduct. Working in this manner 
 we obtain the general layout of our structure as shown at the top of the 
 general stress sheet, Fig. 224. After this preliminary work we start the 
 detail design by taking up the spans first. 
 
 161. Designing of the Spans. This work is just the same prac- 
 tically as previously outlined for deck plate girder bridges, except the 
 50-ft. spans in this case are made the same depth as the 60-ft. spans in 
 order to simplify construction, the 60-ft. spans being of economic depth. 
 The complete stress diagram, or stress sheet, as we might say, for the 
 three different lengths of spans is given on the general stress sheet 
 (Fig. 224). 
 
 162. Designing of the Columns and Tower Bracing. In design- 
 ing a column the first thing to do is to determine the maximum concentra- 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 297 
 
 tions on the top of it due to dead and live load. The total dead load 
 applied to the top of each column is equal to the sum of the dead-load 
 reactions 9n the two girders supported and the greatest live load applied 
 
 Weighiof Single TrackV/oducf-lbwens. 
 
 C o ope.r-3 E-5O L ooding 
 
 130 
 
 A.R.E.Assn. Specs. 
 
 4-0' SO' 60' 70' . 80" 90' 100' 
 
 H&(ghrof7bwens(Frombaseofrai/fo masonry) 
 Spa 
 
 8O' 9O f IOO' IIO r I2O' /3O' I4O ISO 
 
 Height of Towers (From base ofrai/ to masonry) 
 Fig. 222 
 
 to the top of each is the maximum live-load concentration coming from 
 the live load in the two adjacent spans, just the same as in the case of an 
 intermediate floor beam. (Art. 148.) These loads are the same for all 
 columns in bents 1 to 9 (Fig. 224), as each supports an end of a 30- and 
 
 uj| 
 
 '/Grade- >V 
 
 Fig. 223 
 
 GO-ft. span. The load on the columns in bent 10 is that coming from a 
 30- and 50-ft. span and from the two 50-ft. spans in the case of bent 11. 
 Using the dead-load reactions given for the girders on the stress sheet, 
 Fig. 224, we have 
 
 19,000 + 6,800 = 25,800 Ibs. 
 
 for the dead-load concentration on all columns supporting 30-ft. and 
 
ilil * Sg-s 
 
 ^Sl^S-v 
 
 ^Ei^soisrxS ? 
 
 298 
 
SIMPLE KAILKOAD BEIDGES 299 
 
 60-ft. spans, which includes the columns in bents 1 to 9, and 
 
 6,800 + 14,400-21,200 Ibs. 
 for columns in bent 10 and 
 
 14,400x2=28,800 Ibs. 
 
 for those in bent 11. 
 
 To obtain the maximum live-load concentration on a column we place 
 the loading so that the heaviest loads (wheels) are near the column with 
 one load at the column and the unit load in one adjacent span equal to 
 
 I @5@5 9' 5(7)6 f 5@ 8' @ ff' 
 
 A 
 
 Fig. 225 
 
 the unit load in the other, which is in accordance with Art. 148 ; the spans 
 in this case being of unequal length. 
 
 Taking first the columns supporting 30-ft. and 60-ft. spans, let C 
 (Fig. 225) represent the column considered. By placing the loads as 
 shown in Fig. 225, we have (see Table A) (considering one-half of wheel 
 12 in each span) 
 
 i*M= 2,533 Ib, 
 oU 
 
 for the average unit load in the 60-ft. span, and 
 
 76,000 
 
 30 
 
 = 2,533 Ibs. 
 
 for the average unit load in the 30-ft. span. 
 
 This shows that the criterion is exactly satisfied, and we need go no 
 farther with the work of satisfying the criterion. Taking moments about 
 A (using Table A) we have 
 
 for the reaction on the column at C due to loads 3 to 11 (inclusive), and 
 taking moments about B we have 
 
 = 37,300 Ibs. 
 
 for the reaction on the column at C due to wheels 13 to 16 (inclusive). 
 Now adding these two reactions and the weight of wheel 12 together we 
 have 
 
 62,000 + 37,300 + 20,000 = 119,300 Ibs. 
 
 for the maximum live-load concentration on the column for Cooper's 40 
 loading and multiplying this by 50/40 we have 
 
 119,300x55 = 149,000 Ibs. 
 
300 STRUCTURAL ENGINEERING 
 
 for the maximum live-load concentration due to the 50 loading which 
 is the concentration desired. 
 
 Then for the impact we have 
 
 Now owing to the columns sloping transversely, known as the batter, 
 the actual stress in the columns due to the above concentrations is equal to 
 the concentrations multiplied by the secant of the slope angle. The slope, 
 or batter, will be taken as 2 in 12, which is the usual batter for such 
 columns. So we have 1.014 for the secant of the slope angle. Then we 
 have 
 
 25,800x1.014 = 26,000 Ibs. 
 
 in bents 1 to 9 (inclusive) for the dead-load stress in the top portion of 
 the columns, and 
 
 149,000x1.014 = 151,000 Ibs. 
 for the live-load stress in these columns throughout their whole length and 
 
 115,000x1.014 = 116,000 Ibs. 
 
 for the impact. 
 
 Now adding these together we have 
 
 26,000 + 151,000 + 116,000 = 293,000 Ibs. 
 
 for the total maximum stress in the top sections of all columns in bents 
 1 to 9 (inclusive). 
 
 We will next determine the dead- and live-load stresses in the 
 columns in bent 10. The dead-load concentration on each of these 
 
 '$7)6' (faS 9'@5@5@.?'@ 8' > 8' 5* g'(?)5@ 
 
 ~^r^ -~J 
 
 S 6' _ 
 
 30' j SO' 
 
 SenUO 
 
 Fig. 226 
 
 columns is given above as 21,200 Ibs., and we will proceed to determine 
 the live-load concentration on them. Placing the loads as shown in Fig. 
 226 we have (considering one-half of wheel 13 in each span) 
 
 50 
 
 for the average unit load in the 50-ft. span, and 
 
 69,000 
 
 30 
 
 = 2,300 Ibs. 
 
 for the average unit load in the 30-ft. span. 
 
 This shows that the load in the 30-ft. span is a little too great. So 
 let us place wheel 14 at column C. Then we have 
 
 
DESIGN OF SIMPLE RAILROAD BKIDGES 301 
 
 for the average unit load in the 50-ft. span, and 
 
 62,000 
 
 30 
 
 = 2,066 Ibs. 
 
 for the average unit load in the 30-ft. span. So it is seen that the 
 position of the wheels shown in Fig. 226 will give the maximum concen- 
 tration on the column. So taking moments about A (using Table A) 
 we have 
 
 916 x 1.000 =30>50Qlbs ..^:: ,,'., 
 
 for the reaction on the column at C due to loads 17 to 14 (inclusive), and 
 taking moments about B we have 
 
 (2,036 + 102x8)1.000 ^^ 
 
 50 
 
 for the reaction due to wheels 6 to 12 (inclusive). 
 
 Now adding these reactions and the weight of wheel 13 together we 
 have 
 
 30,500 + 57,000 + 20,000 = 107,500 Ibs. 
 
 for the maximum live-load concentration on each column in bent 10 due 
 to the 40 loading. Then for the E50 loading we have 
 
 55x107,500 = 134,000 Ibs. 
 and for the impact we have 
 
 . .: ; hi 134 ' 000x (8o 1 Soo)= 105 ' ooolbs - ; 
 
 Now multiplying the dead- and live-load concentrations and impact 
 by the secant of the slope angle of the column, as was done above, and 
 adding all three of the results together we obtain 264,000 Ibs. for the 
 
 rH . , . .-* ts 
 
 S(j) 6 y@ 8' @ 8' 5@5@5@ 9 ' @S'6 @5@ -S 
 
 so _ _ so 
 
 Bent-// 
 Fig. 227 
 
 total maximum stress in each of the columns in bent 10. Next placing 
 the wheel loads as shown in Fig. 227 and proceeding in the same manner 
 as above, in the case of the other columns, we obtain 
 
 128,600x^ = 161,000 Ibs. 
 
302 
 
 STRUCTURAL ENGINEERING 
 
 for the live-load concentration on each column in bent 11, and 
 
 300 \ _ 
 
 161,000 
 
 121,000 Ibs. 
 
 + 300 / 
 
 for the impact concentration. 
 
 Then, taking the dead-load concentration found above for these 
 columns, we have 
 
 28,800x1.014 = 29,200 Ibs. 
 for the dead-load stress, and 
 
 161,000 x 1.014= 163,000 Ibs. 
 
 62003^ 
 
 Fig. 228 
 
 for the live-load stress, and 
 
 121,000 x 1.014 = 122,000 Ibs. 
 
 for the impact, making in all 314,000 Ibs. stress in each of the columns 
 in bent 11. 
 
 Having the stresses in the columns this far determined we can now 
 write the dead, live and impact stress on them at the cross-sections, as 
 shown in Fig. 224, increasing the dead-load stress in the lower sections 
 of the columns where the weight of the part of the tower above increases 
 the stress an appreciable amount. The dead weight per ft. of height of 
 towers 1-2 and 7-8 as seen from the diagram in Fig. 222 is about 1,000 
 
DESIGN OF SIMPLE KAILEOAD BEIDGES 303 
 
 Ibs. So the weight per ft. of column would be 250 Ibs. Then the dead- 
 load stress in the lower portion (about one-half of the height) of the 
 columns in these towers will be increased (in the worst case) about 
 250 x 28 = 7,000 Ibs., about 11,000 Ibs. in the case of towers 3-4 and 5-6. 
 In tower 9-10 and bent 11 this increased dead weight is ignored as it is 
 inappreciable. 
 
 We shall next determine the stresses in the columns and transverse 
 bracing due to wind load. According to the specifications: "Viaduct 
 towers shall be designed for a force of 50 Ibs. per sq. ft. on one and 
 one-half times the vertical projection of the structure unloaded; or 30 Ibs. 
 per sq. ft. on the same surface plus 400 Ibs. per linear ft. of structure 
 applied 7 ft. above the rail for assumed wind load on the train when the 
 structure is fully loaded with empty cars assumed to weigh 1,200 Ibs. per 
 linear ft. of track." 
 
 Let us first consider the towers 3-4 and 5-6 (see Fig. 224), and let 
 the diagram at (#), Fig. 228, represent the elevation of one of these 
 towers drawn to, say, a -j 1 ^ scale. The wind load coming onto a single 
 bent as AB would be that applied between the vertical sections SS and 
 S'S'. This load will be considered (in accordance with practice) to be 
 applied at the joints of the bents and on the girders and train as shown 
 in the cross-section at (fr), Fig. 228. To obtain the intensities of these 
 forces it is first necessary to estimate the areas of the vertical projection 
 of the structure at the different points. The area of the vertical projec- 
 tion of the train need not be considered as the load on it is given in the 
 specification as 400 Ibs. per ft. of track. However, the train is usually 
 considered to be 10 ft. high. 
 
 For the area contributing to force Fl we have 1.5 x 45 = 67 n ' in 
 round numbers from ties and guard rail; 3.5 x 30 = 105 n ' from the 60-ft. 
 girder, and 2.25 x 15 = 33' from the 30-ft. tower girder, making in all 
 205 n ' (= 67 n ' + 105 n/ + 33 n ')- (See Figs. 221 and 232.) 
 
 In the case of F2 we have the same area from the girders (= 105 + 
 33 = 138 n/ ) and 9 ft. of tower. From the tower, we have 0.8 x 15 = 12 n/ 
 from the top longitudinal strut (see detail of tower, Fig. 232, also see 
 Fig. 221), 0.5x24 = 12 n ' from one-half of the longitudinal diagonal, and 
 1.5 x 9 = 14 n ' from the column, making in all 176 n ' (= 138 n/ + 12 n/ + 
 12 n/ + 14'). 
 
 For F3 we have 1.5 x 18 = 27 n/ from the column (only). 
 
 For F4 we have 1.5 x 27.7 = 42 n ' from the column; 0.8 x 15 = 12 n ' 
 from the longitudinal strut and 0.5 X 48 = 24 n ' from the two longitudinal 
 diagonals (one-half of each), making in all 78 n/ (= 42 n/ + 12 n ' + 24 n '). 
 
 Now by increasing each of the above areas by one-half (as per 
 specification) we obtain the areas (to the nearest square feet) indicated at 
 (6). The wind load on the lower part of the bent is transmitted directly 
 to the masonry and hence is not considered. 
 
 Taking first the case of the 50-lb. load (when the train is not on the 
 structure) we obtain the forces indicated at (c) by multiplying the given 
 areas by 50. Then laying off these forces on the load line BC, at (/), 
 we obtain the diagram of the stresses in the bent, as shown, by beginning 
 at a and passing around each joint counter clock- wise. The intensities 
 of the stresses thus obtained for the different members are given on the 
 bent at (c). Next for the case of the 30-lb. load on the structure and the 
 
304 STRUCTUARL ENGINEERING 
 
 400-lb. per linear ft. on train, we have the forces shown at (d). Those 
 applied to the structure here are obtained by multiplying the correspond- 
 ing areas given at (6) by 30 and the load on the train by multiplying 
 400 by 45', the length of the train contributing pressure to bent AB. 
 By drawing on and np we have the forces acting upon a continuous frame 
 which can be graphically analyzed and the addition of the members on 
 and np will not affect the stresses in the members of the bent in the least. 
 Then by laying off the forces on the load line BD and beginning at joint 
 n and passing around each joint counter clock- wise we obtain the diagram 
 of the stresses shown at (g). The intensities of the stresses thus obtained 
 for the different members are given on the bent at (d). 
 
 This work completes the determination of the wind stresses in the 
 towers 3-4 and 5-6, and we shall next take up the determination of the 
 stresses in these same towers due to the longitudinal force from the train. 
 This force is known as traction. It is the force exerted along the rails 
 when the brakes are applied to the wheels of a moving train causing the 
 wheels to slide on the rails. The intensity of this force, as is evident, is 
 
 Fig. 229 
 
 equal to the coefficient of friction (of the wheels on the rails) times the 
 vertical load on the wheels. This coefficient is designated in the specifica- 
 tions as 0.20. We will consider the traction on each of the towers, 3-4 
 and 5-6, as coming from the loads on the girders rigidly connected to the 
 tower in each case, or, in other words, the loads between the expansion 
 points of the girders. This, as is seen from Fig. 224 (the points of 
 expansion being marked Ex), will be the loads on a 30-ft. and 60-ft. span, 
 making in all 90 ft. of continuous load. The load will be a maximum 
 when the wheels are in the position shown in Fig. 229. This gives a 
 load of 248,000 Ibs. (see Table A) for Cooper's 40 and 310,000 Ibs. 
 (= 248,000 x 50/40) for 50. Then for the traction force on each tower 
 we have 
 
 T = 310,000x0.2 = 62,000 Ibs., 
 
 which is applied at the top of the rail. By drawing mk and Jen at the top 
 of the tower as shown at (e) we have the load applied to a continuous 
 frame which can be graphically analyzed. The diagram of stresses in 
 the tower shown at (/i), due to this 62,000-lb. force is obtained by laying 
 off AB (representing the force) as a load line and then beginning at A: 
 and passing around each joint counter clock- wise. The intensities of the 
 stresses thus obtained are given on the elevation of the tower at (<?). The 
 exact values of the stresses, however, are obtained by multiplying each of 
 the given stresses by the secant of the slope angle of the columns, but as 
 each stress is increased but a small amount we will ignore this correction 
 (as is usual practice) and use the stresses shown. 
 
DESIGN OF SIMPLE EAILKOAD BRIDGES 305 
 
 We now have all of the stresses determined in the towers 3-4 and 5-6 
 except the dead, live, impact, and traction stresses in the top transverse 
 struts (see Fig. 221) of the bents. These stresses are due to the columns 
 being inclined. Owing to this inclination the columns exert a horizontal 
 thrust toward each other upon this strut causing a compressive stress in 
 it equal to the vertical load at the top of the column multiplied by the 
 tangent of the slope angle of the columns. The slope of the columns is 
 2 in 12, so the tangent of the slope angle is 0.1C66. Then for the top 
 transverse strut we have the following stresses, in round numbers: 
 
 D= 25,800x0.1666- 4,000 Ibs. ") 
 L = 149,000 x 0.1666 = 25,000 Ibs. > + 48,000 Ibs. 
 7-115,000x0.1666-19,000 Ibs. J 
 T= 88,000x0.1666-15,000 Ibs. 
 W= 26,000 Ibs. 
 
 + 89,000 Ibs. 
 
 In order to design the column bases and anchorage it is necessary to 
 know the positive and negative reactions on each column. The maximum 
 positive reaction on each is equal to the maximum combined stress in the 
 bottom section of the column divided by the secant of the slope angle of 
 the column plus one-fourth of the weight of the lower half of the tower, 
 which is about 11,000 Ibs. So we have (see Fig. 224) 
 
 +R = 11,000 + -+567,000 Ibs. 
 
 for the maximum positive reaction. 
 
 The maximum negative reaction, if there be such, will occur when 
 the structure is loaded with a train of empty box cars which is assumed 
 to weigh 1,200 Ibs. per ft. of track, as per specification. 
 
 We have -98,000 Ibs. for the negative reaction due to wind load as 
 given at (g), Fig. 228. For the traction force along each rail due to the 
 1,200 Ibs. live load we have 
 
 x 0-2 x 90 = 10,800 Ibs. 
 
 Now this 10,800-lb. force can be applied to the top of the tower as 
 was done with the 62,000-lb. force (at (e), Fig. 228) and the reaction 
 due to it determined graphically. But as the reaction due to the 62,000-lb. 
 force is already determined (at (/O), the one due to the 10,800-lb. force 
 can be determined very readily by proportion as the reactions are directly 
 proportional to the forces. Hence we have 
 
 x 160,000 = -28,000 Ibs. 
 for the negative reaction due to traction. 
 
306 
 
 STRUCTURAL ENGINEERING 
 
 For the positive reaction due to the 1,200-lb. live load we have 
 
 x 45 = +27,000 Ibs. 
 
 Now adding (algebraically) this and the dead-load reaction, which is 
 
 11,000 + 
 
 37,000 
 1.014 
 
 -48,000 Ibs. 
 
 (see cross-section of bents, Fig. 224), to the above negative reactions we 
 have 
 
 27,000 + 48,000 - 98,000 - 28,000 = -51,000 Ibs. 
 
 for the maximum negative reaction that can occur on each column and for 
 which anchorage must be provided. 
 
 This method of determining these 
 negative reactions is not absolutely correct 
 as the effect of the wind and traction forces 
 being in different planes is not taken into 
 account. The tendency of rotation of each 
 tower, as is obvious, will really be about 
 axes perpendicular to the resultant of the 
 two kinds of forces. However, owing to 
 the uncertainty of the intensities of the 
 forces in the first place and to the result- 
 ing error being small, the more exact anal- 
 ysis is not really justifiable and hence will 
 not be given. 
 
 As we now have all of the stresses determined for towers 3-4 and 5-6 
 we can write the same on the stress sheet (Fig. 224) as shown, the stresses 
 in the longitudinal bracing on the elevation of the towers and those in the 
 transverse bracing and columns on the cross-section of the bents, and we 
 can then proceed with the designing of the sections of the different mem- 
 bers in these towers. 
 
 Taking the columns first (bents 3, 4, 5, and 6) let us assume the 
 following section for the top portion of each column : 
 
 1 cov. pi. 18"xf" = 11.25 n " 
 2 [s 15" x 40* = 23.52 n " 
 
 34.77 n " 
 
 Taking moments about the center of the cover plate (see Fig. 230) we have 
 - 23.52 x 7.81 
 
 Fig. 230 
 
 34.77 
 
 = 5.29 ins. 
 
 for the distance to the center of gravity of the assumed section from the 
 .center of the cover plate (axis x-x). 
 
 Now for the moment of inertia about axis x-x we have 
 
 7 = 347.5x2 + (2.52 x 23.52) + (5.29 x 11.25) = 1,159, 
 and hence for the radius of gyration about axis x-x we have 
 
 r- 
 
 = 5.77. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 3Q7 
 
 For the moment of inertia about axis y-y we have 
 
 r = 9.39 x 2 + (6.03 x 23.52) + 303.75 = 1,178, 
 and for the radius of gyration about the same axis we have 
 
 - 
 
 The distance along the column between the points of connection of the 
 Longitudinal bracing is the greatest unsupported length (as is seen from 
 Fig. 224) and hence will be taken as the length of column. This length 
 (as obtained by scale) is about 36' or 432". Then substituting this 
 length and the radius r' (as the column would fail about axis y-y) in the 
 column formula, we have 
 
 4.00 
 p = 16,000 -70-^- =10,800 Ibs. 
 
 5.8/& 
 
 for the allowable unit stress in the case of the combined dead- and live- 
 load and impact stresses. As the maximum stresses due to live load, wind 
 and traction are not likely to occur simultaneously the unit stress can be 
 increased 25 per cent for the case of combined dead- and live-load, impact, 
 wind, and traction stresses, as per specifications. 
 So we have 
 
 293,000 f 
 
 for the area required for the combined dead- and live-load and impact 
 stresses, and for the combined dead, live, impact, wind, and traction 
 stresses we have 
 
 468,000 468,000 
 
 10,800(1 + 0.25) = T^50F = 34.66 sq.ms., 
 
 which is the actual required area, being the greater. The section assumed 
 is satisfactory as it is about as near Oe required area as we can get it. 
 Taking next the bottom portion of the column, let us assume the 
 following section: 
 
 1cov. p i. 18"xii" = 12.37 n " 
 2 [s 15" x 50* = 29.42 n " 
 
 41.79 n// 
 
 Taking moments about the center of the cover plate (see Fig. 231) we have 
 - 29.42 x 7.84 
 
 41.79 
 
 = 5.52 ins. 
 
 for the distance to the center of gravity of the section from the center of 
 the cover plate. 
 
 For the moment of inertia about axis x-x we have 
 
 7 - 402.7 x 2 + (2J32 2 x 29.42) + (5^2* x 12.37) = 1,341, 
 
308 
 
 STRUCTURAL ENGINEERING 
 
 and for the radius of gyration about the same axis we have 
 
 For the moment of inertia about axis y-y we have 
 
 /' = (6.05 x 29.42) + (11.22 x 2) + 334.13 = 1,433, 
 and for the radius of gyration we have 
 
 r = 
 
 The length of the bottom portion of the columns is about 36 ft., or 
 432 ins., and as this length is the same as regards the two axes, r the 
 smaller radius will be used in the column formula. So we have 
 
 P = 16,000 -70 = 10,680 
 
 O.OD 
 
 for the allowable unit stress. Dividing this into the combined dead, live, 
 and impact stress we have 
 
 304,000 
 
 10,680 
 
 = 28.46 sq. ins. 
 
 for the required area of the column, and by increasing this unit stress 
 25 per cent and dividing it into the combined dead, live, impact, wind, and 
 traction stress we have 
 
 564,000 
 13,360 
 
 = 42.2 sq. ins. 
 
 Fig. 231 
 
 for the required area which is the greater 
 of the two. The assumed section is about 
 as near to the required section as is pos- 
 sible to obtain, so it will be used. 
 
 We will next take up the designing 
 of the transverse bracing. Beginning 
 with the top diagonals (see cross-section of 
 bents 3, 4, 5, and 6, Fig. 224) we have 
 a stress of 38,000 Ibs., which we will 
 assume to be carried in tension by one 
 system, so we have 
 
 38,000 
 16,000 
 
 = 2.37 sq. ins. 
 
 for the required net area of each of the top diagonals. It is necessary to 
 use two angles for each diagonal in order to obtain good details. It is 
 seen that comparatively small angles could be used as far as the area of 
 cross-section is concerned, but as 3 J" x 3|" x J" angles are about as small 
 as is consistent with good practice in the design of railroad bridges we 
 will use 2 Ls 3i"x3J" xf" = 4.98 -0.37 = 4.61 sq.ins.net for each diag- 
 onal. 
 
DESIGN OF SIMPLE KAILEOAD BKIDGES 309 
 
 Now as these angles are the minimum size used and the stresses being 
 less in the other transverse diagonal than in the ones just considered, 
 2 Ls 3 y x 3J" x f " will be used for each of the other transverse 
 diagonals. 
 
 In designing the transverse struts L/r should not be greater than 
 120,, and 100 would be better. The stresses, as is seen, are so small in 
 these struts that the designing of the sections really consists in selecting 
 sections that will be sufficiently rigid and provide good details. 
 
 The bottom strut is about 31', or 372", long. So we have 
 
 373 =3.72 
 
 100 
 
 for the required radius of gyration to give L/r = 100. We will use 
 2 [s 10" x 20* which have a radius of 3.66 and considerable more 
 section than required, but it is about as satisfactory a section as is obtain- 
 able and hence will be used for the bottom transverse strut. 
 
 The second transverse strut from the bottom is about 18'-6", or 222", 
 long, so we have 
 
 for the required radius. Here we will use 2 [s 8" x 13.75* which have 
 a radius of 2.98. We will use the same section for the next strut above 
 also, in order to obtain uniform details. 
 
 Let us next take the case of the top strut. This strut has a length of 
 about 78". Let us assume 2 Ls 5" x 3J" x f" = 6.10 n ". The least 
 radius of gyration of these two angles taken as a strut is 1.60. Then 
 substituting in the column formula we have 
 
 JVQ 
 
 p = 16,000 -70 -l- = 12,600 Ibs. 
 1.60 
 
 for the allowable unit stress for the combined dead, live, and impact 
 stresses. So we have 
 
 48,000 
 
 for the required area of cross-section, and increasing the above unit stress 
 25 per cent and dividing it into the combined dead, live, impact, wind, and 
 traction stresses we have 
 
 89,000 
 
 for the required area which is the greater. So our assumed section is 
 about correct and hence will be used. 
 
 We will next take up the designing of the longitudinal bracing in 
 towers 3-4 and 5-6. Here each diagonal will be designed to carry the 
 96,000-lb. stress in tension, assuming only one system to act at a time. 
 So we have 
 
 96,000 , ^ 
 
 68q.ms. (net) 
 
DESIGN OF SIMPLE EAILBOAD BKIDGES 3H 
 
 for the required net area of cross-section of each diagonal. Then for each 
 diagonal we can use (counting out of each angle one rivet hole) 2 Ls 
 6" x 4" x f" = 7.22 - 0.75 = 6.47 sq. ins. (net) . 
 
 We will next consider the longitudinal struts, which are really 
 columns 30' or 360" long. The maximum stress of 62,000* occurs on 
 the intermediate one. Let us assume a section composed of 2 [s 10"x 
 20* = 11.76". Then we have 
 
 r 3.66 
 and we also have 
 
 p = 16,000 - 70 = 9,110 Ibs. 
 o.oo 
 
 for the allowable unit stress. Dividing this into the stress we have 
 
 62,000 
 
 for the required area which is much less than the area of the assumed 
 section, yet we will use the assumed section for each of the longitudinal 
 struts as the L/r is about correct. This completes the design for the 
 towers 3-4 and 5-6, and the sections can be written on the stress sheet as 
 shown (Fig. 224). 
 
 The other towers can be designed in the same manner as shown above 
 for towers 3-4 and 5-6 and then the stress sheet can be finished as shown 
 in Fig. 224. 
 
 163. Detail Drawings. After the stress sheet (Fig. 224) is 
 completed a general drawing of one (at least) of the towers should be 
 made in order to show the kind of details desired unless standard draw- 
 ings for such details, previously made, are available. The tower selected 
 for detailing should be one of medium height so as to obtain average 
 details. In this case we will select tower 1-2 (see stress sheet, Fig. 224). 
 The general details for the tower will be sufficiently shown by drawing 
 the details of one column and the principal details of the bracing connect- 
 ing to the column. The first thing to do is to make large scale detail 
 sketches of the column cap, base, and of the principal intermediate points. 
 From these sketches we determine the exact slopes and clearances for the 
 bracing, in fact all essential details are worked out on these sketches and 
 simply transferred to the finished general drawing. Working in this 
 manner we obtain the general drawing for the tower 1-2 shown in 
 Fig. 232. 
 
 The calculations for these details are quite simple. The cap is made 
 as small as is consistent with good details. The I-beam diaphragm at 
 the top of the column is for the purpose of distributing the loads from the 
 girders equally to the two channels. There should be a sufficient number 
 of rivets in each side of this diaphragm to transmit one-half of the 
 maximum end shear on the longer girder minus the dead-load end shear 
 on the shorter girder. So in this case we have (see stress sheet, Fig. 224) 
 
 243_,000 
 2 
 
312 STRUCTURAL ENGINEERING 
 
 for the shear that these rivets are to transmit. Dividing this by 7,200 
 (the value of a J" shop rivet in single shear) we obtain practically 16 
 rivets. Sixteen is the number used. 
 
 The bolts connecting the girders to the columns should be sufficiently 
 large to transmit the shear due to traction, neglecting the friction on the 
 expansion end. Taking the case of the 60-ft. span, the heaviest load will 
 occur when wheels 2 to 11 (inclusive) are on the span. This (for Cooper's 
 E5Q, see Fig. 151) gives a load of 205,000 Ibs. per girder. 
 
 Then, we have 0.2 x 205,000 = 41,000 Ibs. for the horizontal thrust 
 on each girder which must be taken by the bolts at the fixed end of the 
 girder. We have four bolts taking this, so each must take one-fourth of 
 it or 10,250 Ibs. This calls for 1-J" bolts, as the area of cross-section of 
 each is 0.994 n ", and stressing them 10,000 Ibs. per square inch gives 
 9,940 Ibs., which is about the required value. The same size bolts are 
 used for the tower spans in order to have uniform sizes, although the 
 shear due to traction in that case does not require them to be so large. 
 
 The column base should have sufficient area so as not to stress the 
 masonry more than 600 Ibs. per square inch (see specifications). The 
 maximum reaction on bents 1 and 2 (see stress sheet, Fig. 224) is 512,000 
 Ibs. Dividing this by 600 we obtain 853 sq. ins. for the required area of 
 bearing on the masonry. The masonry plate or base plate used has 
 (30" x 29") 870 sq. ins., which is about the correct area. The masonry 
 plate should be symmetrically placed in reference to the center of gravity 
 of the column and the details should be such that the pressure from the 
 column is quite uniformly distributed over the plate. 
 
 The anchor bolts should be large enough to take the maximum nega- 
 tive reaction (uplift) on the column at 16,000 Ibs. per square inch. The 
 splice in the column is considered to be a butt joint, that is, the top section 
 of the column is considered to bear firmly against the bottom section so 
 that the stress is transmitted from one to the other without the aid of the 
 rivets in the splice. In that case there need be only enough rivets in the 
 splice to hold the column in line. Just how many to use in such cases 
 must be determined by mere judgment. 
 
 The calculation for the other details is mostly a matter of develop- 
 ing the sections in the bracing which the student should have no trouble 
 in doing. Take, for example, the longitudinal diagonals. Each of these 
 diagonals is composed of 2 Ls 6" x 4" x f " = 7.22 - 0.75 = 6.47 sq. ins. (net) . 
 Multiplying this net area (of the two angles) by 16,000 Ibs. we have 
 
 6.47x16,000 = 103,500 Ibs. 
 
 for the tensile strength of each diagonal. Dividing this by 6,000 Ibs. we 
 obtain about 17 -J" field rivets to develop the section. Eight on a side 
 are used (in the end connections), which is about correct. 
 
 Take next the longitudinal struts which are composed of 2 [s 
 10"x20# = 11.76 n ". The strength of these struts as given in the last 
 Article is 9,110 Ibs. per square inch. Then for the total strength of each 
 we have 
 
 11.76x9,110 = 107,000 Ibs. 
 
DESIGN OF SIMPLE KAILKOAD BKIDGES 313 
 
 Dividing this by 6,000 Ibs. we obtain about 18 J" field rivets. Nine on 
 a side are used (in the end connections), which is correct. The riveting 
 in the end connections and splices of the transverse bracing is obtained in 
 the same manner. 
 
 The stress sheet (Fig. 224) and the general drawing of the tower 1-2 
 (Fig. 232) are quite sufficient for general drawings in this case as the 
 work is quite similar throughout. However, there should always be a 
 sufficient number of general drawings to fully show the character of the 
 work. All special towers should really be detailed. 
 
 From these general drawings the bridge company, obtaining the 
 contract for fabricating the structure, works up the shop drawings for 
 the work. 
 
 In working up the shop drawings usually the first thing, after large 
 scale sketches of the principal details of the towers are made, is to make 
 a drawing showing the location of the anchor bolts. This is known as the 
 masonry plan. This drawing is used by the party building the sub- 
 structure and is usually the first drawing called for. The material as a 
 rule is next ordered and line drawings of the towers are made upon which 
 the lengths of all members are given, which are usually computed by the 
 aid of logarithmic tables. After this the making of the shop drawings 
 proper (and bills) proceeds. The shop drawings for the girders are prac- 
 tically the same as previously shown for deck-plate girder bridges. The 
 shop drawings for the towers are quite easy to make ; the columns should 
 be on separate sheets from the bracing. All should be drawn as much in 
 their relative position as possible to aid in checking and also in identifying 
 the members. 
 
 We will next make a preliminary estimate of the weight and cost of 
 metal in the structure. 
 
 164. Preliminary Estimate of Weight of Metal and Cost. 
 Taking up the weight of the girders first we have the following: 
 
 Weight of metal in one 60-ft. span. 
 Weight of one girder. 
 
 1 web 84" x f" x 107.1* x 60' 6,426 Ibs. 
 
 4_ Ls 6" x 6" x I" x 24.2* x 60' 5,808 Ibs. 
 
 1 cov. pi. 14" x " x 29.76* x 60' 1,786 Ibs. 
 
 l_cov. pi. 14" x f " x 29.76* x 44' 1,309 Ibs. 
 
 2 cov. pis. 14" x i" x 23.8* x 29' 1,380 Ibs. 
 
 26 stiff. Ls 5" x 3i" x f " x 10.4* x 7' 1,893 Ibs. 
 
 8 end stiff. Ls 5" x 3J" x T V x 12.0* x 7' 672 Ibs, 
 
 4 fillers 7" x f" x 14.88* x 6' 357 Ibs. 
 
 8 sp. pis. 10" x f " x 21.25* x 2.4' 408 Ibs. 
 
 4 sp. pis. 14" x f" x 29.76* x 4.3' 512 Ibs. 
 
 2_sole pis. 12" x f " x 30.6* x 1.2' 74 Ibs. 
 
 6 fillers 3J" x f" x 7.44* x 6' 268 Ibs. 
 
 20,893 Ibs. 
 rivets, 3% (=20,890 x 0.03) 627 Ibs. 
 
 Total weight of 1 girder 21,520 Ibs. 
 
 Total weight of 2 girders. 43,040 Ibs. 
 
314 STRUCTURAL ENGINEERING 
 
 Weight of one frame. 
 
 2 Ls 3i" x 3J" x f " x 8.5* x 6.1' 104 Ibs. 
 
 2 Ls 3" x 3|" x f" x 8.5* x 8.2' 140 Ibs. 
 
 4 pis. 12J" x f " x 15.94* x 1.1' 70 Ibs. 
 
 1 pi. 9" x f" x 11.48* x 0.8' 9 Ibs. 
 
 323 Ibs. 
 rivets 22 Ibs. 
 
 Total weight of 1 frame 345 Ibs. 
 
 5 
 
 Total weight of 5 frames 1,725 Ibs. 
 
 Weight of laterals and lateral plates. 
 
 8 Ls 3J" x 3" x f" x 8.5* x 9' 612 Ibs. 
 
 7 pis. 12" x f" x 15.3* x 2.8' 300 Ibs. 
 
 15 pis. 12" x |" x 15.3* x 1.0' 230 Ibs. 
 
 2 pis. 12" x f" x 15.3* x 2.0' 61 Ibs. 
 
 rivets 27 Ibs. 
 
 Total weight of laterals and lateral plates 1,230 Ibs. 
 
 Summary of weight. 
 
 2 girders 43,040 Ibs. 
 
 5 frames 1,720 Ibs. 
 
 laterals and plates 1,230 Ibs. 
 
 Total weight of metal in 1 60-ft. span 45,990 Ibs. 
 
 Weight of metal in one 30-ft. span. 
 Weight of one girder. 
 
 1 web 54" x f" x 68.85* x 25' 1,721 Ibs. 
 
 4 Ls 6" x 6" x f" x 24.2* x 30' 2,904 Ibs. 
 
 2 end pis. 30" x f" x 38.25* x 6' 459 Ibs. 
 
 4 sp. pis. 12" x f" x 25.5* x 3.5' 357 Ibs. 
 
 4 Ls 6" x 6" x " x 19.6* x 4' (bent) 314 Ibs. 
 
 4 Ls 6" x 6" x |" x 19.6* x 7' 549 Ibs. 
 
 4 fillers 6" x f " x 12.7* x 6' 305 Ibs. 
 
 14 Ls 5" x 3" x f" x 10.4* x 4.5' 655 Ibs. 
 
 2 pis. 12" x f" x 25.5* x 2' 102 Ibs. 
 
 2 sole pis. 12" x |" x 30.6* x 1.2' . 74 Ibs. 
 
 7,440 Ibs. 
 rivets, 2% 150 Ibs. 
 
 Total weight of 1 girder 7,590 Ibs. 
 
 2 
 
 Total weight of 2 girders 15,180 Ibs. 
 
 Weight of frames. 
 
 2 end frames = 345 x 2 690 Ibs. (from 60- ft. span) 
 
 1 interior frame 320 Ibs. (computed) 
 
 Total weight of 3 frames., 1,010 Ibs. 
 
DESIGN OF SIMPLE KAILKOAD BRIDGES 315 
 
 1 230 
 Weight of laterals and plates -br~x 30 = 615* (from 60-ft. span) 
 
 Summary of weight. 
 
 2 girders 15,180 Ibs. 
 
 3 frames 1,010 Ibs. 
 
 laterals and plates 615 Ibs. 
 
 Total weight of metal in 1 30' tower span 16,805 Ibs. 
 
 Weight of metal in one 50-ft. span. 
 
 1 web 84" x f" x 107.1* x 50' 5,355 Ibs. 
 
 4_ Ls 6" x 6" x J" x 33.1* x 50' 6,620 Ibs. 
 
 22 -Ls 5" x 34" x f" x 10.4* x 7' 1,602 Ibs. 
 
 4 Ls 5" x 3J" x T y x 12.0* x 7' 336 Ibs. 
 
 4 fillers 7" x " x 20.83* x 6' 500 Ibs. 
 
 4 fillers 31" x J" x 10.42* x 6' 250 Ibs. 
 
 splice plates (see 60-ft. span) 650 Ibs. 
 
 sole plates (see 60-ft. span) 74 Ibs. 
 
 15,387 Ibs. 
 rivets, 3% 450 Ibs. 
 
 Total weight of 1 girder 15,837 Ibs. 
 
 O 
 
 Total weight of 2 girders 31,664 Ibs. 
 
 Lateral system = (about same as 50-ft. span, Art. 137) . . 2,776 Ibs. 
 Total weight of metal in 1 50-ft. span 34,440 Ibs. 
 
 Weight of metal in tower 1-2. 
 
 (Detailed, Fig. 232.) 
 
 One Column. 
 
 2 [s 15" x 33* x 23.75' 1,568 Ibs. 
 
 1cov. p i. 18" x f" x 22,95* x 23.7' 544 Ibs. 
 
 2_[ s 15" x 50* x 26.6' 2,660 Ibs. 
 
 l_ cov . p i. is" x &" x 26.78* x 26.6' 712 Ibs. 
 
 Total weight of main section 5,484 Ibs. 
 
 Cap. 
 
 1 cap plate 20J" x f" x 52.28* x 3.1' 162 Ibs. 
 
 1__I 10" x 30* x 2.3' 69 Ibs. 
 
 2 fills. 5" x i" x 4.25* x 2.3' 20 Ibs. 
 
 2_i_s 6 " x 4" x f" x 20* x 1.7' 68 Ibs. 
 
 2_Ls 6" x 4" x f" x 20* x 1.0' 40 Ibs. 
 
 *1 gus. pi. 29" x f" x 36.97* x 2.4' 89 Ibs. 
 
 1 gus. pi. 36" x'f" x 45.90* x 2.6' 120 Ibs. 
 
 2_ Ls 3" x 3-|" x f " x 8.5* x 1.4' 24 Ibs. 
 
 2 pis. 16" x f" x 20.4* x 1.75' 71 Ibs. 
 
 2_[_s 5" x 3J" x |" x 10.4* x 1.0' 21 Ibs. 
 
 684 Ibs. 
 
 * Average width and length of plates are given. 
 
316 STRUCTURAL ENGINEERING 
 
 Intermediate details. 
 
 3 tie pis. 18" x f " x 22.95* x 1.5' 103 Ibs. 
 
 2 lat. pis. 14" x f" x 17.85* x 3' 107 Ibs. 
 
 4_ sp . pig. 12" x f " x 15.3* x 1.85' 113 Ibs. 
 
 2 sp. pis. 18" x f" x 22.95* x 2.0' 92 Ibs. 
 
 2 lat. pis. 16" x f" x 20.4* x 3.0' 122 Ibs. 
 
 2 lat. pis. 20" x f" x 25.5* x 3.4' 173 Ibs. 
 
 128 ft. of 2J" x f" latt. bars @ 3.19* per ft 408 Ibs. 
 
 1,118 Ibs. 
 Base. 
 
 l_base plate 30" x f" x 89.25* x 2.5' 223 Ibs. 
 
 2_Ls 6" x 6" x f" x 24.2* x 1.1' 53 Ibs. 
 
 2_ Ls 6 " x 4" x f" x 20* x 1.6' 64 Ibs. 
 
 2_Ls 6" x 6" x f" x 24.2* x 0.4' 19 Ibs. 
 
 2_Ls 6" x 6" x f" x 24.2* x 0.8' 39 Ibs. 
 
 4 Ls 5" x 31" x f" x 10.4* x 1.0' 42 Ibs. 
 
 2 Ls 5" x 31" x 10.4* x 2.0' 42 Ibs. 
 
 1_ L 6" x 6" x f " x 24.2* x 2.4' 58 Ibs. 
 
 2 fills. W x f " x 7.44* x 1.5' 22 Ibs. 
 
 l_pl. 9" x |" x H.48* x 1.5' 17 Ibs. 
 
 2 Ls 31" x 31" x f" x 8.5* x 1.5' 26 Ibs. 
 
 1 gus. pi. 30" x f" x 38.25* x 3.3' 126 Ibs. 
 
 1 gus. pi. 28" x f " x 35.7* x 1.8' 64 Ibs. 
 
 2 gus. pis. 30" x f" x 38.25* x 1.75' . 134 Ibs. 
 
 929 Ibs. 
 
 Summary of weight of details. 
 
 Cap 684 Ibs. 
 
 Int. details 1,118 Ibs. 
 
 Base 929 Ibs. 
 
 Rivets 200 Ibs. 
 
 2,931 Ibs. 
 
 2 931 
 Percentage of details = S ,* 53% of main section. 
 
 Summary of weight of one column (bents 1 and 3). 
 
 Details 2,931 Ibs. 
 
 Main section 5,484 Ibs. 
 
 8,415 Ibs. = total weight of one column. 
 4 
 
 33,660 Ibs. = total weight of four columns. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 317 
 
 Longitudinal bracing. 
 Longitudinal strut. 
 
 2_[ s 10 " x 20* x 28.4' 1,136 Ibs. 
 
 4 tie pis. 12" x f " x 15.3* x 1.25' 77 Ibs. 
 
 164 ft. latt. bars of 21" x f" @ 2.87* per ft 470 Ibs. 
 
 rivets 53 Ibs. 
 
 (Details - 44%.) 1,736 Ibs. 
 
 6_ 
 
 Weight of metal in six struts 10,416 Ibs. 
 
 Diagonal. 
 
 2 Ls 6" x 4" x f" x 12.3* x 35.5'. . 873 Ibs. 
 
 2 tie pis. 12" x f" x 15.3* x 1.25' 38 Ibs. 
 
 48 ft. latt. bars 2J" x f" @ 2.87* per ft 138 Ibs. 
 
 rivets 20_ Ibs. 
 
 (Details = 22%.) 1,069 Ibs. 
 
 4 
 
 Weight of metal in four diagonals 4,276 Ibs. 
 
 Diagonal (spliced). 
 
 4 Ls 6" x 4" x f" x 12.3* x 17.5 ; 861 Ibs. 
 
 4 tie pis. 12" x f" x 15.3* x 1.25' 77 Ibs. 
 
 2 sp. pis. 13" x |" x 16.57* x 3.5' 116 Ibs. 
 
 44 ft. latt. bars 2J" x f" @ 2.87* per ft 126 Ibs, 
 
 rivets 24 Ibs. 
 
 (Details = 40%.) 1,204 Ibs. 
 
 4 
 
 Weight of metal in four diagonals 4,816 Ibs. 
 
 Total weight of longitudinal bracing 19,508 Ibs. 
 
 Transverse bracing (bents 1 and 2). 
 Strut (beginning at top of bent). 
 
 2_Ls 5" x 3J" x f " x 10.4* x 4.0' 83 Ibs. 
 
 7 ft. 21" x f" lat. bars @ 2.87* 20 Ibs. 
 
 rivets 2 Ibs. 105 Ibs. 
 
 (Details, 27%.) 
 Diagonal. 
 
 2_Ls 3J" x 3J" x f" x 8.5* x 12.75' 217 Ibs. 
 
 2 tie pis. 9" x f" x 11.48* x 1.0' 23 Ibs. 
 
 16 ft. lat. bars 21" x f" @ 2.87* 46 Ibs. 
 
 rivets 11 Ibs. 297 Ibs. 
 
 (Details, 36%.) 
 
318 STRUCTURAL ENGINEERING 
 
 Diagonal. 
 
 2Ls 34" x 34" x f" x 8.5* x 3.75' 64 Ibs. 
 
 2 Ls 34" x 3i" x f" x 8.5* x 8.4' 143 Ibs. 
 
 4 tie pis. 9" x f" x 11.48* x 1.0' 46 Ibs. 
 
 2 sp. pis. 12" x f" x 15.3* x 3.0' 92 Ibs. 
 
 14 ft. latt. bars 2J" x f" @ 2.87* 40 Ibs. 
 
 rivets 16 Ibs. 401 Ibs. 
 
 (Details, 93%.) 
 Strut. 
 
 2 [s 8" x 13.75* x 8.75' 240 Ibs. 
 
 4 tie pis. 12" x f " x 15.3* x 0.8' 49 Ibs. 
 
 20 ft. lat. bars 2J" x f" @ 2.87* 57 Ibs. 
 
 rivets , 12 Ibs. 358 Ibs. 
 
 (Details, 49%.) 
 Diagonal. 
 
 2 Ls 3J" x 34" x f" x 8.5* x 17.2' 292 Ibs. 
 
 2 tie pis. 9" x f" x 11.48* x 1.0' 23 Ibs. 
 
 34 ft. lat. bars 2J" x f" @ 2.87* 97 Ibs. 
 
 rivets 16 Ibs. 428 Ibs. 
 
 (Details, 46%.) 
 Diagonal. 
 
 2 Ls 34" x 34" x f" x 8.5* x 6.6' 112 Ibs. 
 
 2 Ls 34" x 34" x f" x 8.5* x 10.0' 170 Ibs. 
 
 4 tie pis. 9" x f" x 11.48* x 1.0' . : 46 Ibs. 
 
 30 ft. lat. bars 2J" x f" @ 2.87* 86 Ibs. 
 
 2 sp. pis. 12" x f" x 15.3* x 3.0' 92 Ibs. 
 
 rivets . 20 Ibs. 526 Ibs. 
 
 (Details, 87%.) 
 Strut. 
 
 2 [s 8" x 13.75* x 13.3' 366 Ibs. 
 
 4 tie pis. 12" x f " x 15.3* x 0.8' 49 Ibs. 
 
 40 ft. lat. bars 2J" x f" @ 2.87* 115 Ibs. 
 
 rivets . 40 Ibs. 570 Ibs. 
 
 (Details, 56%.) 
 Diagonal. 
 
 2 Ls 34" x 34" x f" x 8.5* x 27.2' 462 Ibs. 
 
 2 tie pis. 9" x f" x 11.48* x 1.0' 23 Ibs. 
 
 48 ft. lat. bars 2J" x f" @ 2.87* 137 Ibs. 
 
 rivets 26 Ibs. 648 Ibs. 
 
 (Details, 40%.) 
 Diagonal. 
 
 2\_ s 34" x 34" x f" x 8.5* x 10.6' 180 Ibs. 
 
 2 Ls 34" x 34" x f" x 8.5* x 16.0' 272 Ibs. 
 
 4 tie pis. 9" x f" x 11.48* x 1.0' 46 Ibs. 
 
 44 ft. lat. bars 2J" x f" @ 2.87* 126 Ibs. 
 
 2 sp. pis. 12" x f" x 15.3* x 3.0' 92 Ibs. 
 
 rivets . 30 Ibs. 746 Ibs. 
 
 (Details, 65%.) 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 319 
 
 Strut. 
 
 2 [s 10" x 20# x 20.5' 820 Ibs. 
 
 4 tie pis. 12" x |" x 15.3* x 0.8' 49 Ibs. 
 
 72 ft. lat. bars 2" x f " @ 2.87* 206 Ibs. 
 
 rivets 50 Ibs. 1,125 lbs ; 
 
 (Details, 37%.) 
 
 Weight of transverse bracing in one bent = 5,204 Ibs. 
 
 Weight of transverse bracing in two bents or tower = 10,408 Ibs. 
 
 Summary of weight of metal in tower 1-2. 
 
 4 columns 33,660 Ibs. 
 
 Long, bracing 19,508 Ibs. 
 
 Trans, bracing 10,408 Ibs. 
 
 Total 63,676 Ibs. 
 
 Proceeding in the same manner the following weights are found: 
 
 Weight of metal in each of the towers 3-4 and 5-6. 
 
 4 columns (36% details) 52,200 Ibs. 
 
 Long, bracing , 21,820 Ibs. 
 
 Trans, bracing 14,440 Ibs. 
 
 Total 88,460 Ibs. 
 
 Weight of metal in tower 7-8 64,900 Ibs. 
 
 Weight of metal in tower 9-10 42,380 Ibs. 
 
 Weight of metal in bent 11 8,420 Ibs. 
 
 Summary of weight of metal in structure. 
 
 5 60-ft. spans @ 45,990 Ibs 229,950 lbs-1 
 
 5_30-ft. spans @ 16,805 Ibs 84,025 Ibs. j> 382,855 Ibs. 
 
 2 50-ft. spans @ 34,440 Ibs 68,880 Ibs.J 
 
 l_tower (1-2) 63,676 Ibs. 1 
 
 2 towers (3-4 and 5-6) @ 88,460 Ibs 176,920 Ibs. j 
 
 l_tower (7-8) 64,900 Ibs. j> 356,296 Ibs. 
 
 1 tower (9-10) 42,380 Ibs. 
 
 1 bent (10) 8,420 Ibs.J 
 
 Total weight of metal 739,151 Ibs. 
 
 Cost of structure (superstructure). 
 
 Girders, 382,855 Ibs. @ 3^ $11,486 
 
 Towers, 356,296 Ibs. @ 3^ 12,470 
 
 Total cost of steel work (except anchor bolts) $23,956 
 
 This price is only a fair average pound price. 
 
 165. Double-Track Viaducts. The ordinary double-track viaduct 
 has a double line of spans, that is, four lines of track girders, which, as 
 a rule, are supported upon cross-girders as shown in Fig. 233 instead of 
 
320 
 
 STRUCTURAL ENGINEERING 
 
 resting directly upon the top of columns. The towers, as for general 
 design, are practically the same as for single-track viaducts. The concen- 
 trations at the top of the towers due to dead and live load and impact are 
 just twice as much as for single-track viaducts and consequently the 
 stresses in the columns due to the same will be twice as much, provided 
 of course, the columns in the two cases have the same batter. We assume 
 that two trains move abreast over the double-track structure. The trac- 
 tion is just twice as much for a double-track as it is for a single-track 
 viaduct, while the wind pressure is practically the same for the two struc- 
 tures. (See specifications for wind load.) 
 
 The method of procedure in designing double-track viaducts is the 
 same as for single-track viaducts except the transverse bracing is sub- 
 
 Fig. 233 
 
 jected to live-load stress and impact when only one track is loaded. This 
 is due to the unequal thrusts exerted by the two columns upon the top 
 strut, or cross-girder. As an illustration, suppose the right-hand track of 
 a double-track viaduct (see Fig. 233) to be loaded with the maximum live 
 load and no live load on the left-hand track. The right-hand column will 
 receive most of the load, as is evident. Let V the concentration on the 
 right-hand column and V the concentration on the left-hand column and 
 let <f> represent the slope angle of the columns. The horizontal thrust on 
 the cross-girder exerted by the right-hand column is equal to Ftan<, and 
 that exerted by the left-hand column is equal to F'tan<. Now if these 
 two thrusts were equal they would just balance each other and there 
 would be no live-load stress in the transverse bracing. But as they are 
 unequal the transverse bracing must transmit the difference (which pro- 
 duces a horizontal shear) down to the masonry. This force (= Ftan<f> - 
 F'tanc^) can be assumed as applied at the bottom flange (considered as a 
 strut) of the cross-girder (just exactly as a wind load) and the stresses 
 in the transverse bracing due to it graphically determined. 
 
DESIGN OF SIMPLE RAILEOAD BEIDGES 
 
 321 
 
 The maximum stress in the columns occurs when the two tracks are 
 fully loaded, at which time no live-load stress occurs in the transverse 
 bracing. 
 
 166. Analytical Method of Determining Stresses in Tower 
 Bracing. Let Fig. 234 represent a transverse view of a bent of an 
 ordinary viaduct, acted upon by forces as indi- 
 cated. Let us first suppose that the horizontal 
 wind forces F, Fl, F2, and F3 alone are acting 
 and that the stress in the diagonals AD, CN, and 
 in the strut CD, due to these forces, is to be deter- 
 mined. 
 
 To determine the stress in the diagonal AD,< 
 first assume the part of the bent (forces and all) 
 below the section mm removed. Then the part 
 above this section mm is held in equilibrium by 
 the forces F, Fl, F2, S, S3, and 4, where S, S3, 
 and S4: represent the stress in members AD, AC, 
 and BD, respectively. The stress S is what we 
 desire. By prolonging the lines of action of 
 the forces (stresses) S3 and $4 and taking mo- 
 ments about their point of intersection, 0, we 
 eliminate these forces from the equation of moments 
 and we have 
 
 Fig. 234 
 
 from which we obtain (in known quantities) 
 
 cF + hFl + JcF2 
 
 O = - : - 
 
 Similarly, to determine the stress S2 in diagonal CN, assume the 
 part of the bent below the section nn removed and taking moments about 
 we have 
 
 cF + hFl + kF2 + rF3 + dS2 = 0, 
 
 from which we obtain the required stress 
 cF + hFl + JcF2 + rF3 
 
 To determine the stress /SI in the strut CD assume the part of the 
 bent below the section oo removed and then taking moments about we 
 have 
 
 cF + hFl + kF2 + rF3 + rSl = 0, 
 
 from which we obtain the required stress 
 
 cF + hFl + kF2 + rF3 
 
 Sl = 
 
 r 
 
 Instead of wind loads, as just considered, suppose there be an eccen- 
 trically applied load W acting upon the bent. Then taking moments 
 about 0, as before, we have 
 
322 
 
DESIGN OF SIMPLE KAILEOAD BEIDGES 
 
 from which we obtain 
 
 We 
 
 323 
 
 and similarly we obtain 
 
 We 
 
 We 
 
 ?-?- 
 
 d 
 
 Webs- 
 
 Fig 236 
 
 Eccentrically applied loads, as just considered, occur mostly on 
 
 viaducts built on curve and on double-track viaducts when only one track 
 
 is loaded with live load. 
 
 If desired, the stresses in the columns can be determined by taking 
 
 moments about panel points. For example, the stress S in BD can be 
 obtained by passing the section mm and taking 
 moments about A. By passing section oo and 
 taking moments about D the stress S3 in AC can 
 be determined, and, by taking moments at the same 
 time about C the stress S6 in DN can be determined. 
 Similarly, by passing the section nn and taking 
 moments about N the stress S5 in CM can be 
 determined. 
 
 The only objection to the above method is that 
 
 the length of some of the lever arms is troublesome to determine unless 
 
 it be determined by scale, which, as a rule, is not a very satisfactory 
 
 manner unless special care be taken. 
 
 167. General Remarks. Viaduct columns are sometimes built 
 
 without cover plates as shown in Fig. 235. There is no question but that 
 
 a column with a cover plate is better than one without. Whenever a 
 
 greater area is required than can be obtained by using two channels and a 
 
 cover plate a section like the one shown in Fig. 236 should be used. 
 
 However, in case there be just a few column sections in a structure 
 
 requiring more area than is contained in two 
 
 channels and a cover plate, the area may be 
 
 increased so that that type of section can be 
 
 used by riveting a plate to the back of each 
 
 channel. 
 
 The batter of the columns in single- 
 
 track viaducts varies from 1 j in 12 to 3 in 12, 
 
 for very low bents. Two in 12 is a very 
 
 common batter. The batter of the columns in 
 
 double-track viaducts, as a rule, is less than 
 
 in the case of single-track structures. The 
 
 batter should be sufficient to limit the negative 
 
 reactions at the bases of the columns to a 
 
 reasonable intensity, at least. In fact it 
 
 would be better to have no negative reaction 
 
 at all, but as a rule it is not practical to have that condition in the case of 
 
 single-track viaducts. 
 
 There should be sufficient material in each pedestal (which is usually 
 
 made of concrete) to properly distribute the positive reaction of the 
 
 column supported and at the same time have sufficient weight to resist 
 
 X"C* Anchor both 
 
 Fig. 237 
 
324 
 
 STRUCTURAL ENGINEERING 
 
 the negative reaction of the column. The anchor bolts should extend well 
 down into the pedestals as indicated in Fig. 237. 
 
 In the case of viaduct towers having no horizontal struts, as shown 
 in Fig. 221, the two systems of bracing are usually considered to act 
 simultaneously, which requires that each member of the bracing be 
 designed for compression as well as for tension, but at the same time, 
 however, the stress in each is only half as much as obtains in the type 
 of bracing shown in Fig. 220, assuming only one system to act at a time. 
 
 DRAWING ROOM EXERCISE NO. 6 
 
 Design and make stress sheet of a single-track viaduct for the 
 following crossing: 
 
 Elevation of 
 Elevation of Ground 
 640 
 588 
 579 
 568 
 
 Base of Rail 
 
 648 
 648 
 648 
 648 
 
 Creek 
 
 Station 
 400 
 
 400 + 81 
 401 
 
 401 + 70 
 
 402 + 13 564 
 402 + 20 554 
 402 + 40 554 
 
 402 + 49 566 
 
 403 567 
 
 403 + 44 564 
 404+ 5 608 
 
 404 + 41 613 
 404 + 80 626 
 405+ 2 646 
 
 Data: 
 
 Live load, Cooper's jE50 loading. 
 
 Specifications, A. R. E. Ass'n. 
 
 Dead load, to be determined by student. 
 
 TRUSS BRIDGES 
 
 168. Preliminary. Truss bridges are used for spans 100 ft. in 
 length and over. With regard to the location of the track there are two 
 types of truss bridges, the through and the deck bridge, the same as in 
 
 648 
 648 
 648 
 648 
 648 
 648 
 648 
 648 
 648 
 648 
 
 Wit Truss (b) 
 
 Flff. 288 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 325 
 
 the case of plate girders. The tnrough type is used wherever the under 
 clearance will not permit of the use of the deck type. 
 
 The four trusses shown in Fig. 238 are the most common types used 
 for through bridges. Pratt trusses are used for spans up to 175 ft. in 
 length and as a rule are riveted trusses. Curved chord Pratt trusses are 
 used for spans from 200 ft. to 325 ft. long and as a rule are pin-connected 
 
 Fig. 240 
 
 trusses. Pettit trusses, both (a) and (b) types, are used for long spans, 
 350 ft. in length and over. 
 
 There are other types of bridge trusses which are used to some 
 extent. These will be considered later. 
 
 In the case of deck bridges, the Pratt truss is practically always 
 
326 STRUCTURAL ENGINEERING 
 
 used. The end panels, however, are modified either as shown at (a) or 
 (6), Fig. 239. 
 
 The diagram, Fig. 240, gives the names of the different parts of an 
 ordinary Pratt truss bridge, but the same names hold for bridge trusses 
 :n general. 
 
 Complete Design of a 150-Ft. Single-Track Through Riveted 
 Pratt Truss Span 
 
 169. Data. 
 
 Length = 6 panels @ 25'-0" = 150'-0" c.c. end pins. 
 Width = 16'-0" c.c. of trusses. 
 Height = 30'-0" c.c. chords. 
 Stringers spaced, 6'-6" c.c. 
 Live Load, Cooper's .E50 loading. 
 Specifications, A. R. E. Ass'n. 
 
 170. Design of 25-Ft. Stringers. (For detail of stringers, see 
 Fig. 290.) For dead load on stringers we have, from (1) Art. 124, 
 B>=12x25 + 100 + 400 = 8001bs. per ft. of span or 400 Ibs. per ft. of 
 stringer. Then, using this load, we have 
 
 25 2 x 12 = 375,000 inch Ibs. 
 
 12.5' 
 
 for the maximum bending moment due to dead load. 
 
 It is seen from Table A that the maximum live-load moment will 
 
 likely occur when wheels 2 to 5 are on the stringer, and, according to Art. 
 
 88, they will be in the position shown in Fig. 241. 
 
 Taking moments about A (Fig. 241), to 
 find the reaction R (using Table A), we find 
 first the moment of wheels 3, 4 and 5 about 2 
 
 *L W * WfflW J W |B by passing down the vertical line through 
 
 wheel 2 to the zig-zag line, then to the right 
 to the vertical line through wheel 5 and the 
 figure 600, just to the right of this line, is the 
 
 moment of wheels 3, 4 and 5 about 2, in thousands of foot pounds. 
 
 Then multiplying the total weight, in thousands of pounds, of the 
 
 wheels, 2 to 5, by 3.75 (see Art. 46) and adding this product to the 600, 
 
 we have the moment of the wheels, 2 to 5, about A. So we can write 
 
 the equation of moment about A as 
 
 600 + (80 x 3.75) - (R x 25) =0, 
 from which we obtain 
 
 25' 
 
 /CO 
 
 for the reaction at B. 
 
 Then taking moments about 4, as the maximum moment occurs under 
 that wheel (see Art. 88), we have the moment of R about wheel 4 minus 
 the moment of wheel 5 about wheel 4 for the maximum bending moment; 
 that is, we have 
 
 M" = 36,000 x 11.25 - 5 x 20,000 = 305,000 ft. Ibs., 
 
 for the maximum live-load bending moment due to Cooper's 40 loading, 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 327 
 
 and multiplying this by 50/40 and by 12 we have 4,575,000"* for the 
 desired maximum live-load moment in inch pounds, provided the correct 
 group of wheels were selected, which can be ascertained by trial, in case 
 of doubt. 
 
 Then for the impact moment we have 
 
 7 = 4,575,000 x - = 4,223,000 inch Ibs. 
 325 
 
 Now adding the above moments together we have 
 
 M = 375,000 + 4,575,000 + 4,223,000 = 9,173,000 inch Ibs. 
 
 for the total maximum bending moment on the stringer. 
 
 Next let us assume the web as f" thick. Then substituting in (5) 
 of Art. 113 (stringers, as a rule, never have cover plates), we have 
 
 ff=l.$ 
 
 for the economic depth. So we will use a web 48' 
 
 25' V For the dead-load end shear on the stringer. 
 
 Fig 242 WG haVC 
 
 400x12.5 = 5,000 Ibs., 
 
 and placing the wheels as shown in Fig. 242 and taking moments about 
 the support B (using Table A) we have 
 
 for the maximum end shear due to Cooper's 40 and 
 
 |5 x 56,500 = 70,650 Ibs., say 71,000 Ibs., 
 due to Cooper's E5Q loading, and for impact we have 
 71,000 x |55 = 66,000 Ibs. (about) 
 
 and adding we have 
 
 5,000 + 71,000 + 66,000 = 142,000 Ibs. 
 
 for the total maximum end shear on the stringer. 
 
 Now for the unit shear on the assumed web we have 
 
 142,000 
 
 and substituting this in (1) of Art. 118, we have 
 
 d= 1_ (12,000 - 7,888) = 39 ins. (about) 
 
 for the maximum distance allowed between the stiffeners near the ends 
 of the stringer, and, as this is more than half the depth of the girder 
 and as the minimum thickness of web allowed by the specifications is 
 f in., we will use a 48" x f" web. 
 
328 STRUCTURAL ENGINEERING 
 
 To determine the area of flanges the first thing to do is to approxi- 
 mate the effective depth of the stringer. Assuming that 6" x 6" angles 
 will be required we find from Table 6, or from some handbook, that the 
 average distance from the back to the center of gravity of these angles 
 is about 1.75 ins., and taking the depth of the girder as J" more than 
 the web, we have 
 
 48.25 - 1.75 x 2 = 44.75, say 45 ins., 
 
 for the approximate effective depth. Dividing this into the maximum 
 total bending moment, we have 
 
 = 204,000 lb, (about), 
 
 for the stress in each flange (see Art. 112), and dividing this by 16,000 
 we have 
 
 204,000 
 
 = 12.75sq. ins., 
 
 16,000 
 for the net area of each flange. Let us try 
 
 2 Ls 6" x 6" x \" = (5.75 x 2) - 1 = 10.50 n " net 
 $ area of web = 2.25 n "net 
 
 12.75" net 
 
 This gives exactly the area indicated. Now checking back, we have 
 48.25 -1.68x2 = 44.89 ins. 
 
 for the actual effective depth, which differs so little from the assumed 
 that the area of the flanges would be changed but very little, and hence 
 recalculation of flanges is unnecessary. 
 
 The stiffeners on stringers are usually made of 3J" x 3 J" angles. 
 There is no rational way of determining them. Owing to the planing off 
 of the ends of the stringers to obtain exact length, the end stiffeners are, 
 as a rule, made ^$" to -J" thicker than the intermediate stiffeners, which 
 are of minimum (f") thickness. 
 
 The laterals bracing for stringers are designed the same as for 
 any deck plate girder. (See Art. 135.) 
 
 171. Design of Intermediate Floor Beams. (For details of 
 floor beams, see Fig. 291.) The dead load on an intermediate floor beam 
 consists of the dead-load concentrations from the stringers and the weight 
 of the floor beam itself. 
 
 Each concentration from the stringers is equal to twice the end shear 
 on one stringer and such floor beams weigh, depending on details, from 
 3,000* to 3,800* each. So let us, in this case, assume 3,600* as the 
 weight of one intermediate floor beam complete, or 225* per ft. of floor 
 beam. Then, for dead load we have the case shown in Fig. 243, and 
 for the maximum bending moment due to dead load, we have 
 
 10,000 x 4.75 x 12 = 570,000 in. Ibs. 
 x225xl6 2 x!2 = 86,000 in. Ibs. 
 
 656,000 in. Ibs. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 329 
 
 and for the maximum end shear, due to the same load, we have 
 
 10,000 = 11,800 Ibs. 
 
 To determine the maximum live-load bending moment and end shear 
 on the floor beam we must first satisfy the criterion of Art. 148 for 
 maximum live-load concentration on the floor beam. That is, the loads in 
 the two adjacent panels must be as nearly equal as possible and at the 
 same time the heaviest loads must be near the floor beam, with one load 
 exactly at it. 
 
 
 Fig. 243 Fig. 244 
 
 By trial (using Table A) it will be found that the maximum con- 
 centration on the floor beams will occur when the wheels are in the 
 position shown in Fig. 244, as the criterion is most nearly satisfied with 
 them in that position, that is, the heaviest wheels are near the floor beam, 
 wheel 13 at it, and the load in panel BC, consisting of wheels 14, 15, and 
 16 (wheel 17 is exactly over the floor beam at C and hence is not in the 
 panel #C), weighs 46,000 Ibs., while the load in panel AB, consisting of 
 wheels 10, 11, and 12, weighs 50,000 Ibs. This is as close as the crite- 
 rion can be satisfied, as will be found by further investigation. So, 
 taking moments (using Table A) about A to find the concentration at B 
 due to wheels 10, 11, and 12, we have 
 
 r = (420 + 50 x 7) r = 30 > 800 lbs '> 
 
 ivO 
 
 and taking moments about C to find the concentration at B due to wheels 
 14, 15, and 16, we have 
 
 r' = x 1,000 = 24,850 Ibs. 
 & o 
 
 Now adding these concentrations and the weight of wheel 13 together, 
 we have 
 
 R = r + r' + 20,000 = 30,800 + 24,850 + 20,000 = 75,650 Ibs., 
 
 for the maximum live-load concentration due to Cooper's 40 loading, 
 and multiplying this by 50/40 we have 94,562*, say 94,500#, for the 
 maximum live-load concentration due to Cooper's -E50 loading. 
 
 Then for the live load on each intermediate floor beam we have the 
 case shown in Fig. 245, and for the maximum live-load moment, taking 
 moment about either C or D, we have 
 
 4.75 x 94,500 x 12 = 5,386,000 inch Ibs., 
 and for the impact we have 
 
 5,386,000 x = 4,616,800, say, 4,620,000 inch Ibs. 
 
330 
 
 STRUCTURAL ENGINEERING 
 
 (The load extends over two panel lengths, or 50 ft.) 
 
 Now adding the above dead, live, and impact moments, we have 
 M = 656,000 + 5,386,000 + 4,620,000 = 10,662,000 inch Ibs., 
 
 for the total maximum bending moment on the floor beam, and, as is 
 seen from Fig. 245, the maximum live-load shear is 94,500* and the 
 impact = 94,500x300/350 = 81,000*. So by adding these to the above 
 dead-load shear we have 11,800 + 94,500 + 81,000 = 187,300* for the total 
 maximum end shear on the floor beam. 
 
 The depth of the floor beam is governed practically by the depth 
 of the stringers. The bottom laterals and the bottom flanges of the floor 
 beams should be in the same plane, as the bottom flanges of the beams 
 act as struts in the bottom lateral system, and there should be sufficient 
 distance from the bottom of the stringers down to the bottom of the floor 
 beam to permit the laterals to pass beneath the stringers, which usually 
 requires from 4" to 6", depending on the size of the laterals, and the dis- 
 tance from the top of the floor beams down to the top of the stringers 
 is usually about 3". In this case we will assume 3" as the distance 
 from the top of the floor beams down to the top of the stringers and 
 5" from the bottom of the stringers down to the bottom of the floor 
 beams, which gives 56J" for the depth of the floor beam as shown in 
 Fig. 246. It will be found upon investigation that floor beams, as a rule, 
 are deeper than the economic depth. 
 
 1 
 
 e'-e 
 
 Floor Beam. 
 
 Fig. 245 
 
 Fig. 246 
 
 After having decided upon the depth of the floor beam, the next 
 thing is to approximate the effective depth. Assuming that each flange 
 of the floor beam will be composed of 2 6" x 6" angles, we can take 
 about the average distance from the back of the angles to their centers 
 of gravity, which is given in Table 6 as, say, 1.75". Then, for the 
 approximate effective depth, we have 
 
 56.25 -(1.75x2) =52.75, say 53 ins. 
 Dividing this into the above maximum bending moment, we have 
 
 10,662,000 + 53 = 201,000 Ibs. (about) 
 
 for the flange stress, and dividing this by 16,000 we have 
 201,000 -5- 16,000 = 12.56 sq. in. 
 
 for the required net area of each flange. 
 
 Assuming a 56" x T V' weD which has a section of 24.5 n " (= 56 x &), 
 
 let us try 
 
 2 Ls 6" x 6" x " = (5.75 x 2) - 1 = 10.50*" net 
 -J area of cross-section of web = 3.06 n " net 
 
 13.56 n " net 
 
DESIGN OF SIMPLE EAILKOAD BEIDGES 331 
 
 This gives l n " more than is required. 2 Ls 6" x 6" x T y have a 
 section more nearly equal to the required area, but the specifications 
 require the angles to be -J" thick if the outstanding leg is 6", so, to com- 
 ply with the specifications, the above section will be used. 
 
 It will be seen that the true effective depth here is slightly less than 
 the assumed, as the distance from the back of the 6" x 6" x \" angles is 
 1.68". So, for the actual effective depth, we have 
 
 56.25 -(1.68x2) -52.89 ins. 
 
 This, however, is so near the assumed effective depth that recalculations 
 are unnecessary. 
 
 For the maximum unit shear on the floor beam we have 
 
 1 87,300 -f 24.5 = 7,644 Ibs. per sq. in. 
 Now substituting this value for s in (1), Art. 118, we have 
 
 d = 15 (12,000 - 7,644) = 47.6 ins., 
 
 for the allowable distance between stiffeners. The shear is practically 
 zero between the stringers, so that no stiffeners are required there, and 
 the clear distance from stringer to truss is 4 / -9' / minus about 7" for 
 half width of truss and 6" (at least) for connections, leaving 3' - 8", or 
 44", for the actual value of d, while 47.6" is allowed, so the assumed 
 web is quite thick enough. In fact it could be thinner, but the rivets, 
 spaced along the flanges between the stringers and trusses, would be 
 closer than desired, if a thinner web be used, and hence the assumed 
 56" x T 7 <r" web will be used, even though it is a little thicker than 
 required. 
 
 The end stiffeners on the floor beams are usually 6" x 6" x J" angles, 
 placed upon fillers as shown in Fig. 291. 
 
 172. Design of End Floor Beams. The dead load on an end 
 floor beam consists of the two dead-load concentrations from the stringers, 
 
 ,50 f*r ft. 
 
 A 1^^^ 
 I * g 1 
 
 4.'-9" I 6- 6' 
 
 4-9' 
 
 I6-O" 
 
 Fig. 247 Fig. 248 
 
 each of which is equal to the dead-load end shear on one stringer, and the 
 weight of the floor beam itself. The dead-load end shear on a stringer, as 
 given in Art. 170, is 5,000*, and such end floor beams weigh about 2,400#, 
 or, say, 150* per foot. Then the dead weight on each end floor beam is 
 as shown in Fig. 247. 
 
 For the maximum bending moment, due to this dead load, we have 
 
 5,000x4.75x12 =285,000 inch Ibs. 
 
 | x 150 x 16~ 2 x 12 = 57,600 inch Ibs. 
 
 342,600, say 343,000 inch Ibs., 
 
332 STRUCTURAL ENGINEERING 
 
 and for the maximum end shear, due to the same dead load, we have 
 5,000 + 150 x 8-6,200 Ibs. 
 
 The maximum live-load concentrations from the stringers on an end 
 floor beam are each equal to the maximum live-load end shear on one 
 stringer, which is given in Art. 170 as 71,000*. So the live load on an 
 end floor beam is as shown in Fig. 248. 
 
 For maximum bending moment, due to this live load, we have 
 
 71,000 x 4.75 x 12 = 4,047,000 inch Ibs., 
 and for the impact we have 
 
 4,047,000 X |22 = 3,735,000 inch Ibs., 
 
 and for maximum end shear, due to live load, we have 71,000, and for 
 end shear, due to impact, we have 
 
 71,000 x = 66,000 Ibs. (about). 
 o#5 
 
 Now adding the above moments, we have 
 
 343,000 + 4,047,000 + 3,735,000 = 8,125,000 inch Ibs., 
 
 for the total maximum bending moment, and adding together the above 
 end shears we have 
 
 6,200 + 71,000 + 66,000 = 143,200 Ibs., 
 
 for the total maximum end shear on an end floor beam. 
 
 Now assuming 53" for the effective depth, the same as for the inter- 
 mediate floor beam, and dividing it into the above moment, we have 
 
 = 153,000 Ibs. 
 
 for the flange stress, and dividing this by 16,000, we have 
 
 153,000 
 
 16,000 
 
 = 9.56sq. in., 
 
 for the required net area of each flange. Now it is readily seen, from 
 Table 6, that 6" x 6" angles are too large for these flanges and hence we 
 will use 6" x 4" angles with the 6" leg along the web to provide for the 
 flange rivets. 
 
 Let us try, assuming a 56" x f " web (same depth as the intermediate 
 floor beams), 
 
 2_Ls 6"x4"x T y'=:4.18x2-0.87 = 7.49" net 
 area of web = (56 x J) -r 8 = 2.62 n "net 
 
 10.11 n "net 
 
 This flange section is about correct. It is 0.55 n " greater than called for 
 above, but the assumed effective depth is greater than the actual, which is 
 really 56.25 - (2 x 1.96) = 52,33". 
 
DESIGN OF SIMPLE EAILROAD BEIDGES 333 
 
 Dividing the moment by this, we have 
 
 8,125,000 -r 52.33 = 155,000, 
 and dividing this by 16,000*, we have 
 
 155,000 
 
 16,000 
 
 = 9.68 sq. ins. 
 
 for the actual flange area required, which is 0.43 n " less than the above 
 assumed section, but the above section will be used, as 6" x 4" x f " angles 
 are too small, giving a net area of 0.59 n " less than required. 
 
 Dividing the end shear by the area of cross-section of the web, 
 we have 
 
 = 6,819 Ibs., 
 
 21 
 
 for the actual unit shear on the web. Substituting this value for s in (1), 
 Art. 118, we have 
 
 4=sjr (12,000 -6,819) =48.5 ins., 
 
 for the allowable distance between stiffeners, but as the unsupported dis- 
 tance between stringers and truss is about 3' - 8", or 44", according to 
 the preceding article, no stiffeners are needed and the assumed 56" xf" 
 web will be used. The actual allowable stress on it is determined by sub- 
 stituting 44" for d in (1), Art. 118, and solving for s. Thus we have 
 
 44 =1 (18,000 -.), 
 
 from which we obtain 
 
 * = 7,306 Ibs., 
 
 for the actual allowable unit stress. Now dividing this into the maximum 
 end shear, we have 
 
 143,200 
 
 = 19.57 sq.ms., 
 
 for the required area of cross-section of the web. The area of the 
 56" x f" web, which we propose to use, is 21 n ", which, as is seen, is 
 more than is required, but the area of a 56"x^V web, which is the next 
 in size, contains only 17.5 n ". Therefore, a thinner web could not be 
 used even if the specifications would permit it. From this it is seen that 
 the assumed 56" x f " web is as near the correct section as is possible 
 to obtain and hence will be used. 
 
 This completes the design of the floor system, and next we will take 
 up the design of the trusses. 
 
 173. Determination of Dead-Load Stresses in Trusses. The 
 dead load of an ordinary bridge is considered as uniformly distributed 
 along the span but applied to the trusses only at the panel points or 
 joints, one- third at the top joints and two-thirds at the bottom joints, in 
 the case of through bridges, and just the reverse in the case of deck spans. 
 
334 
 
 STRUCTURAL ENGINEERING 
 
 This load consists of the weight of metal in the span and the weight 
 of the deck. 
 
 From (4), Art. 124, we have 
 
 p = 7x150 + 660 = 1,710 Ibs., 
 
 for the approximate weight of the metal per foot of span, and adding 
 400* for the weight of the deck, we have 
 
 1,710 + 400 = 2,110 Ibs., 
 
 for the total assumed dead load per foot of span, or 2,110/2 = 1,055* per 
 foot of truss. 
 
 Multiplying this by 25, the panel length in feet, we have 
 
 W= 1,055 x 25 = 26,375 Ibs., 
 
 for the panel load on each truss. One-third of this is considered as 
 applied at each top joint and two-thirds at each bottom joint. However, 
 this assumption affects only the verticals, and, as far as the stresses in 
 the other members are concerned, the full panel load could be considered 
 at the bottom joints, or top either, as for that. 
 
 After having thus computed the panel load, the next thing to do is 
 to draw a sketch of the truss, as shown in Fig. 249 (as a guide), and then 
 compute the values of tan 6 and sec 6. 
 
 9 f\ 
 
 Tan 6 = ^:= 0.8333 and sec 0=1.299, say 1.3. 
 oU 
 
 In determining the dead-load stresses in the web members (diagonals 
 and verticals), we can begin either at the center of the span or at the 
 end. The dead-load stresses in these members can really be determined 
 by mere inspection. 
 
 c/=25-0, 
 
 Fig. 249 
 
 Beginning at the center of the span and considering the post dD, it 
 is obvious that this post can do nothing more than carry the load applied 
 to the top of it, which is one-third of the panel load; so, evidently, the 
 stress in it is 26,375 -=-3 = 8,791*, say 9,000*, compression. 
 
 The one-third of a panel load transmitted by the post dD down to 
 joint D combines with the two-thirds at D and undoubtedly one-half of 
 this combination is transmitted to the two ends of the truss. Therefore, 
 the shear in panel CD is one-half of a panel load and the stress in the 
 diagonal cD is 
 
 sec 6 = 
 
 x 1.3 = 17,143, say 17,000 Ibs. tension. 
 
DESIGN OF SIMPLE EAILKOAD BEIDGES 335 
 
 The stress in the post cC is equal to the half panel load coming from panel 
 point D, or the vertical pull of diagonal cD, plus the one-third panel 
 load at c. So, for the stress in post cC, we have 
 
 ^ = 26,375 = 21,980, say 22,000 Ibs. 
 
 fy O O O 
 
 Likewise, it is evident that the diagonal bC must transmit, so to 
 speak, the \ W coming from the central panel point D, the J W from c, 
 and the f W from C, making 1J W in all, which is the shear in panel BC 
 and must, undoubtedly, be equal to the vertical component of the stress in 
 the diagonal bC, and hence the stress in the diagonal is equal to IJx 
 26,375 x sec = 51,430#, say 51,000*, tension. 
 
 The member bB, known as a hanger or hip vertical, as is readily 
 seen, can carry nothing more than the load hung on (so to speak) at the 
 joint B, and hence the stress in it is f x 26,375 = 17,582#, say 18,000#, 
 tension. 
 
 It is evident that the end post bA must transmit -J W from the cen- 
 tral panel point D, the J W from c, the f W from C, the f W from B, 
 and the J W from b, making in all 2J W, which is the shear in panel AB 
 and equal to the vertical component of the stress in this member bA. So 
 the stress in it is equal to 2J x 26,375 x sec = 85,718#, say 85,000#, com- 
 pression. 
 
 Now, if we begin at the end of the span instead of at the center, as 
 we did above, we have first the reaction R = 2% W at A. (This is obtained 
 from mere observation.) As this reaction acts vertically the end post 
 bA must resist it, as the end post is the only member at point A having a 
 vertical component. So we have 2-J W sec for the stress in the end post, 
 the same as found above for this member. The stress in the hanger bB 
 is f W, as found before. The shear in panel BC is equal to the reaction 
 R (=2J W) minus the $ W at b and f W at B, so that we have 2J W- 
 rr=lj W for the shear in panel BC. This, evidently, must be equal to 
 the vertical component of the stress in diagonal bC, as that member is the 
 only member in that panel having a vertical component force, and hence 
 the stress in diagonal bC is equal to 1^ W sec 6, the same as found before. 
 Likewise, the stress in diagonal cD is equal to the shear in panel CD 
 times sec#. The shear is equal to R-2W = $W, and hence the stress is 
 equal to -J W sec 6, the same as found before. The stress in the post cC 
 is equal to the vertical component of the diagonal cD, which is J W (the 
 shear in panel CD) plus the J W at c, making f W, the same as found 
 before, or it is equal to the vertical component of diagonal bC, which is 
 11 W (the shear in panel BC), minus the f W at C. 
 
 The expression for the stresses in the web members can be written 
 on the members, as shown on the right half of the truss (Fig. 249), and 
 the numerical results of the same may be found very quickly by the use 
 of the slide rule. 
 
 For the chord stresses we can assume the full panel loads as applied 
 at the bottom panel points, in order to simplify the work. 
 
 Let us first write out the formulas for the chord stresses. For the 
 stress S in bottom chords AB and BC, we have, taking moments about 
 point b (see Fig. 249), 
 
336 STRUCTURAL ENGINEERING 
 
 Transposing, we have 
 
 S = 2 JF| = (2J) Wtan6 ............................. (1). 
 
 For the stress S 2) hi chord be, we have,, taking moments about C, 
 
 from which we obtain 
 
 S 2 = 5W^-W^=5Wtan6-Wtan6 = 4:Wtaii 6 .......... (2) 
 
 for the stress in chord be. We obtain the same thing for bottom chord 
 CD by taking moments about c. 
 
 For the stress S 3 , in top chord cd, we have, taking moments about D, 
 
 from which we obtain 
 
 We thus have an expression for the stress in each chord member. 
 Substituting the numerical values in (1), we have 
 
 S = 2$x 26,375x0.8333 = 54,945, say 55,000 Ibs. 
 
 tension for the stress in chords AB and EC, and by substituting, like- 
 wise, the numerical values in (2) and (3) we obtain 87,672*, say 
 88,000#, for the stress in the chord members be and CD, and 98,631*, 
 say 99,000*, for the stress in the chord member cd. This completes the 
 calculations for the dead-load stresses in the trusses, as the structure is 
 symmetrical about the center of the span. 
 
 Fig. 250 
 
 It will be seen from Formulas (1), (2), and (3) that WtanQ is a 
 constant, and after the coefficients are written out the stresses in the 
 chords could be quickly determined by the use of the slide rule. 
 
 These coefficients can be written down on the chord members by 
 mere inspection. For example, let Fig. 250 represent a truss of nine 
 equal panels. 
 
 Taking moments about b for the stress in the bottom chords AB and 
 BC f we have 4Wd/h, so the coefficient for chords AB and BC is 4. 
 
 Taking moments about C for stress in top chord be, we have 
 (4:W)2d/h-Wd/h, so the coefficient for chord be is 7. It is the same for 
 the bottom chord CD, as the center of moments in that case is at c, which 
 is in the same vertical line as point C. Taking moments about either d 
 or D, for the stress in chords cd and DE, we have the 4 at A multiplied 
 by 3d minus the W at B multiplied by 2d and minus the W at C multi- 
 
DESIGN OF SIMPLE EAILKOAD BKIDGES 
 
 337 
 
 plied by Id, so the coefficients for chords cd and DE are each (4x3)- 
 2-1 = 9. 
 
 Similarly, taking moments about e or E, for the stress in chords de 
 and EF, also ef, we have (4x4) - 3 - 2 - 1 = 10 for the coefficient to be 
 used in determining the stress in these chords. 
 
 After a little practice the student can write these coefficients down 
 without hesitating. 
 
 If the slide rule is used,, about all that need be written down, out- 
 side of the stresses (as they are determined),, are the numerical values 
 of the chord coefficients and that of W, tan 6, and sec 0, and if found 
 convenient or necessary, W tan and W sec 0. 
 
 174. Determination of Live-Load Stresses in Trusses. First, 
 draw a sketch of the truss, as shown in Fig. 251, for reference. 
 
 Member bA. Suppose we start by determining the maximum stress 
 in the end post bA. As is readily seen, the stress in this member will be 
 
 b + 26OOOOC +294000 d 
 
 C D E F 
 
 e Panels @, 25-Q '= /so -o" 
 
 Fig. 251 
 
 a maximum when the shear in the panel AB is a maximum, and this will 
 occur when the wheels are in the position that most nearly satisfies the 
 criterion of Art. 90, which is: The load in the panel AB must be as 
 nearly equal as possible to the total load on the bridge divided by the 
 number of panels, and, at the same time, a wheel must be at B. Now 
 placing wheel 4 at B, as shown in Fig. 252, we have (using Table A) 
 125 - 91 = 34 ft. of uniform load on the bridge, making a total of 352,000* 
 = (284 + 34x2)1,000 per truss, and dividing by the number of panels 
 we have 
 
 The load in the panel including one-half of wheel 4 is 60,000*. So this 
 position comes as near to satisfying the criterion for shear as is possible, 
 as will be found by trying the other positions. 
 
 D 
 Fig. 252 
 
 34-' 
 
 Taking moments about G (Fig. 252), we have (using Table A) 
 
 R=( 16,364 + 284x34 + 2 
 
 !> 
 
 2 / ] 
 
 OOP 
 150 
 
 = 181,172 Ibs. 
 
 for the total reaction at A. Then taking moments about B, we hare 
 
 ,000 = 19,200 Ibs. 
 
 r= 
 
338 STRUCTURAL ENGINEERING 
 
 for the floor beam concentration at A. So we then have 
 
 tf-r= 181,1 72 -19,200 = 161,972, say 162,000 Ibs. 
 
 for the maximum shear in the end panel AB, due to Cooper's 40 loading 
 and for E 50 we have 
 
 162,000x^5 = 902,500 Ibs. 
 
 Then for the maximum live-load stress in the end post bA we have 
 202,500x1.3 = 263,250, say 263,000 Ibs. (compression) 
 
 Member bB. The maximum live-load stress on the hanger bB, as is 
 obvious, is equal to the maximum floor beam concentration found in Art. 
 171 to be 94,500*, and hence we will not recalculate it. 
 
 Tte 
 
 Fig. 253 
 
 Member bC. The maximum stress in the diagonal bC will occur 
 when the shear in panel BC is a maximum, as this member carries the 
 shear in that panel. Placing wheel 3 at panel point C, as shown in Fig. 
 253, we have 100-96 = 4 ft. of uniform load on the bridge, making in all 
 292,000*= (284 + 2x4) 1,000 on one truss. 
 
 Dividing this by the number of panels, we have 292,000 -f- 6 = 48,666*, 
 and the load in panel BC is 40,000*, including one-half of the wheel 3, 
 at C (see Fig. 253). This does not seem to satisfy the criterion very 
 closely, so we will try wheel 4 at C, in which case we have 9 ft. of uni- 
 form load on the bridge, making a total load of 302,000* per truss, and 
 dividing by the number of panels we have 302,000 -f 6 = 50,333*, and the 
 load in the panel is 60,000. This position does not satisfy the criterion 
 as closely as the first, and hence the first position, with wheel 3 at C, 
 will be taken. It is evident that no other positions need be tried, as the 
 load in the panel in the first case was too light and in the second it was 
 too heavy. 
 
 Then taking moments about G (Fig. 253), we have (using Table A) 
 
 R = f 1 6,364 + 284 x 4 + 4^ ^jj? = 116,773 
 for the reaction at A, and taking moments about C we have 
 
 =9,200 Ibs. 
 
 for the concentration at point B. 
 
 Then, for the maximum shear in the panel BC, due to the 40 
 loading, we have 
 
 R - r = 116,773 - 9,200 = 107,573, 
 
 and for the 50 loading we have 134,466, say 134,500*. Then multi- 
 plying this by sec B, we have 
 
 134,500x1.3 = 174,850, say 175,000 Ibs. 
 for the maximum tensile stress in the diagonal bC. 
 
DESIGN OF SIMPLE EAILKOAD BKIDGES 339 
 
 Member cD. The maximum live-load stress in the diagonal cD will 
 occur when the live-load shear in the panel CD is a maximum. To 
 satisfy the criterion for maximum shear in that panel, let us try wheel 3 
 at point D (see Fig. 254). This position of the wheels brings wheel 15 
 exactly at G, the right support. 
 
 Fig. 254 
 
 Then the load on the bridge (not including wheel 15) is 232,000*. 
 Dividing this by the number of panels, we have 232,000 -=-6 = 38,666#, 
 and the load in the panel is 40,000* (including one-half of wheel 3). 
 This comes as near satisfying the criterion as is possible, as will be seen 
 by trying other positions. 
 
 Now, by taking moments about G (using Table A) we obtain 
 
 R = 10,816 x = 72,107 Ibs. 
 
 lot) 
 
 for the reaction at A, and taking moments about D we obtain 
 
 r = 230 xi^ = 9,200 Ibs. 
 ^5 
 
 for the concentration at C. 
 
 Then, for the shear in panel CD, due to the 40 loading, we have 
 
 R - r = 72,107 - 9,200 - 62,907 Ibs. 
 
 and for the E50 loading we have 78,633#. Multiplying this by sec 0, 
 we have 
 
 78,633 x 1.3 = 102,223, say 102,000 Ibs. 
 
 for the maximum live-load tensile stress in the diagonal cD. 
 
 Member cC. As the live load is applied wholly on the bottom chord 
 joints, the post cC will have a maximum compression stress when the 
 diagonal cD has maximum tension, as the post takes only the vertical 
 component of the stress in that diagonal. The maximum vertical com- 
 ponent of the stress in the diagonal cD, as seen above, is equal to the 
 maximum shear in the panel CD and given there as 78,633. So the 
 maximum live-load compressive stress in the post cC is, say, 78,000*, 
 using round numbers. 
 
 Member dD. There is no live-load stress in the post dD at all, as 
 there is no member (as a diagonal) connecting to the top of it that will 
 resist vertical action. For the sake of illustration, let us assume a live- 
 load stress, S, in this post. Then the components of this stress must be 
 taken by the top chords, and we would have 
 
 S cos 90 = C = 0, as cos of 90 = 0. 
 
 Members eE and eD. If the span is loaded from the right end so 
 that maximum shear occurs in panel DE, the post eE will be subjected to 
 maximum live-load tensile stress and the diagonal eD will be subjected 
 to maximum compressive stress. 
 
340 STRUCTURAL ENGINEERING 
 
 If wheel 2 be placed at E, wheel 10 will be just 2 ft. from the right 
 end of the span, as shown in Fig. 255, and the total load on the bridge 
 will be 152,000*. 
 
 Dividing this by the number of panels we have 152,0004-6 = 25,333*, 
 and the load in the panel DE is equal to 20,000*, including one-half of 
 wheel 2. Now, if wheel 3 be placed at E, the total load on the bridge 
 
 Fig. 256 
 
 will be 152,000*. Dividing this by the number of panels, we have 
 152,0004-6-25,333*, and the load hi the panel DE is then 40,000*. 
 So it is seen that the position of the wheels, shown in Fig. 255, comes 
 nearest to satisfying the criterion for shear in panel DE. 
 Taking moments about G (Fig. 255), we obtain 
 
 = ^4, 
 
 ,632 + 152 x 2 . = 32,906 Ibs. 
 
 for the reaction at A, and taking moments about E we obtain 
 
 25 
 
 for the concentration at D. 
 
 Then, for the shear in panel DE f due to the E40 loading, we have 
 
 R - r = 32,906 - 3,200 = 29,706 Ibs., 
 
 and multiplying by 50/40 we have 37,100* for the shear due to the 50 
 loading, which is the maximum live-load tensile stress in post eE, and 
 multiplying this by sec we obtain 48,230*, say 48,000*, for the maxi- 
 mum live-load compression stress in diagonal eD. 
 
 r 
 
 Fig. 256 
 
 Member fE. Now placing wheel 2 at F, wheel 6 will be just 1 ft. 
 to the left of the right end of the span as shown in Fig. 256. Then the 
 load on the span is 103,000*. Dividing this by 6 we have 17,166*, and 
 the load in panel EF is 20,000*. This position comes the nearest to 
 satisfying the criterion, as will be found by trial. 
 
 Taking moments about G (Fig. 256), we obtain 
 
 R =(l,640 + 103^^? = H,620 Ibs. 
 
 \ / 
 
 for the reaction at A, and taking moments about F we obtain 
 
 r = 80x^5^ = 3,200 Ibs. 
 for the concentration at E. 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 341 
 
 Then,, for the shear in panel EF, due to the E5Q loading, we have 
 
 R - r ~ = ll,62Q - &20<j~ = 10,525 Ibs. 
 
 Multiplying this by sec we have 13,682*, say 14,000*, for the 
 maximum live-load compressive stress in diagonal fE. 
 
 It is seen, from Fig. 251, that diagonal bC has 175,000* tension in 
 it, while the corresponding diagonal fE t on the right half of the span, 
 has 14,000* compression, that the post cC has 78,000* compression, 
 while the corresponding post eE has 37,000* tension and diagonal cD has 
 102,000* tension while the corresponding diagonal on the right half of 
 the span has 48,000* compression. Now it is evident that, if the span 
 is loaded from the left end instead of the right end, as it was above, and 
 the maximum stresses in the web members computed, just the opposite 
 results will be obtained, that is, for example, diagonal bC would be sub- 
 jected to 14,000* compression, while diagonal fE would be subjected 
 to 175,000* tension. 
 
 So it is seen that we have now determined both maximum tensile 
 and compressive live-load stresses in the web members throughout the 
 structure, and we will next determine the chord stresses. 
 
 Members AB and BC. It is obvious that, if either of the members 
 AB or BC (Fig. 251) be cut, rotation would take place instantly about 
 
 C D ' /= 
 
 Fig. 257 
 
 joint b. Therefore these members undoubtedly prevent rotation about 
 that joint, and hence the greater the tendency of rotation the greater the 
 stress in these two members ; so undoubtedly the maximum live-load stress 
 in them will occur when the live-load bending moment about b is a 
 maximum. 
 
 According to Art. 91, the moment about b will be a maximum when 
 the average unit load on the left of the panel point is equal to the average 
 unit load on the bridge. 
 
 Placing wheel 4 at B (Fig. 257), we have 34 ft. of uniform load on 
 the bridge (see Table A). Then for the average unit load on the span 
 we have 
 
 ( 284 + 34 x2\^^- = 2,346 Ibs. 
 \ j lou 
 
 and for .the average unit load on the left of B (or b) we have 5 including 
 one-half of wheel 4, 60,000 -r 25 = 2,400*. 
 
 This position of the wheels satisfies the criterion as nearly as pos- 
 sible, as will be found by trying other positions of the wheels. 
 
 Now, taking moments about G, we have 
 
 R = 181,173 Ibs. 
 
342 
 
 STRUCTURAL ENGINEERING 
 
 for the reaction at A. Then taking moments about b we have, using 
 Table A, 
 
 181,173 x 25 - 480 x 1,000 = 4,049,300 ft. Ibs. 
 
 for the maximum bending moment about b, and multiplying this by 50/40 
 and dividing by the height of span (see Art. 92), we obtain 168,000* 
 tension for the maximum live-load stress in member AB, and also in BC. 
 Members be and CD. By imagining the top chord be (Fig. 258) to 
 be cut, it is readily observed that this member resists rotation about joint 
 C. So the live-load stress in the top chord be will, evidently, be a maxi- 
 mum when the load is in the position for maximum moment about panel 
 point C, and a maximum in the bottom chord CD when the load is in the 
 position for maximum moment about panel point c. But as joints C and c 
 are in the same vertical line the position of the wheels will be the same 
 in the two cases, and it will make no difference which joint be taken as 
 the center of moments. 
 
 28' 
 
 B 
 50' 
 
 D 
 Fig. 258 
 
 By placing wheel 7 at panel point C, there will be 28 ft. of uniform 
 load on the bridge, as shown in Fig. 258, and we have 
 
 (284,000 + 56,000) -f 150 = 2,266 Ibs. 
 for the average unit load on the bridge, and 
 
 103,000 
 
 50 = 2,190 Ibs. 
 
 for the average unit load to the left of joint C. This position of the 
 wheels comes as near to satisfying the criterion for maximum moment 
 about joint C or c as possible. Then, taking moments about G, we have 
 (using Table A) 
 
 R= ( 16,364 + 284x28 + 28 1 
 
 ^f = 167,333 Ibs. 
 loU 
 
 for the reaction at A (Fig. 258), and then taking moments about C 
 we have 
 
 167,333 x 50 - (2,155 x 1,000) = 6,211,650 ft. Ibs. 
 
 for the bending moment about joint C. Now, it is a question as to Whether 
 the actual maximum moments occur at C or at the corresponding joint E 
 on the right half of the structure, as the load passes to the left over the 
 bridge. Therefore, to make sure, we will test for joint E. 
 
 By placing wheel 14 at E, we have 20 ft. of uniform load on the 
 bridge, as shown in Fig. 259, and for the average unit load on the span 
 
DESIGN OF SIMPLE RAILKOAD BKIDGES 
 
 343 
 
 we have 
 
 ^284 + 40 )i^ = o,iGO, 
 
 and for the average unit load to the left of E we have 
 
 \ 
 
 D 
 Fig. 259 
 
 This position of the wheels comes nearest to satisfying the criterion for 
 maximum moment about E. Taking moments about G, with the wheels 
 in this position, we obtain 
 
 = 16,364 + 284 x 20 + 
 
 = 149,600 
 
 for the reaction at A, and taking moments about E of the forces to the 
 left we obtain 
 
 149,600 x 100 - 8,728 x 1,000 = 6,232,000 ft. Ibs. 
 
 for the maximum moment about E, which is greater than found above for 
 point C, and consequently will be taken as the maximum bending moment 
 for determining the stress in chords be and CD or for the corresponding 
 chords ef and DE on the right half of the bridge which amounts to the 
 same. Multiplying this moment by 50/40 and dividing by 30, the depth 
 of the truss (in feet), we have 259,666, say 260,000#, for the stress in 
 top chord be and also in bottom chord CD, being compression in be and 
 tension in CD. 
 
 It is obvious from Fig. 259 that the bending moment about E would 
 be increased if a uniform load be placed to the left of wheel 1, in which 
 case a uniform load would be preceding the engines as well as following 
 them. This loading is used in some cases, but, when such is used, the 
 impact added should be materially less than when the engines are not 
 preceded by a uniform load, for the reason that the speed of the train 
 is not likely to be very high when the engines are pushing as well as 
 pulling a load. In other words, a train so loaded is not likely to reach 
 the "critical speed." 
 
 As an illustration, let us assume wheel 13 at E and 18 ft. of uniform 
 load (2,000* per ft.) ahead of the engines, as indicated in Fig. 260. 
 
 Then, for the average unit load on the bridge, we have 
 
 36 + 284 + 30 
 
 1,000 
 
 = 2,333 Ibs., 
 
344 
 
 STRUCTURAL ENGINEERING 
 
 and for the average unit load to the left of E we have 
 
 36 + 202 
 
 \ 1,000 
 / 100 
 
 = 2,380 Ibs. 
 
 This position of the loading comes nearest to satisfying the criterion 
 for the maximum moment about E. Taking moments about G, we obtain 
 
 R = ( 36 x 141 + 16,364 + 284 x 15 + 
 for the reaction at A. 
 
 = 172,833 Ibs. 
 
 Now taking moments about E, of the forces to the left, we obtain 
 172,833 x 100 - 7,668 x 1,000 - 36 x 91 x 1,000 = 6,339,300 ft. Ibs., 
 
 for the maximum bending moment due to Cooper's 40 loading. Multi- 
 plying by 50/40 and dividing by 30 we obtain 264,137#, say 264,000*, 
 for the stress in chords be and CD, which, as is seen from the above, is 
 about 4,000* greater than for the usual load, but if the impact is reduced, 
 say, 50 per cent below the usual amount, the final result would be much 
 less than that obtained from using the usual loading. 
 
 Member cd. It is evident that the maximum live-load stress will 
 occur in the top chord cd when the live-load moment about joint D is a 
 maximum. By placing wheel 12 at D we have 35 ft. of uniform load, as 
 shown in Fig. 261, and for the average unit load on the bridge we have 
 
 = 2,360 Ibs, 
 
 and for the average unit load on the left of D we have (including one- 
 half of wheel 12) 
 
 This position of the loading comes the nearest to satisfying the crite- 
 rion for maximum moment about joint D. Taking moments about G, we 
 obtain 
 
 R= 
 
 5,364 + 284 x 35 + 35^^^ = 183,526 Ibs. 
 lou 
 
DESIGN OF SIMPLE KAILKOAD BRIDGES 
 
 345 
 
 for the reaction at A, and taking moments about joint D we obtain 
 
 183,526 x 75 - 6,708 x 1,000 = 7,056,450 ft. Ibs. 
 
 for the maximum live-load moment about that point, due to the 40 load- 
 ing. Then multiplying this by 50/40 and dividing by 30 we have 294,018, 
 say 294,000* compression stress in top chord cd. 
 
 As the bridge is symmetrical about the center of span we have all 
 of the live-load stresses now determined and we will next determine the 
 impact and make a summary of all the stresses in the trusses. 
 
 175. Summarizing of Stresses and Determination of Impact in 
 the Trusses. The dead- and live-load stresses can be written on the 
 members, as shown in Fig. 262, directly from Figs. 249 and 251, and we 
 have only the impact to determine. 
 
 + 88000*0 
 + 2GOOOO*L 
 + /7300O I 
 +3Z/OOO* C 
 
 55OOOD B - 55OOO*D 
 
 -I 68OOO*L -/68OOO*L 
 
 -I I 2000*1 -/ I 2000*1 
 
 335000* 335000* / - 52/O OO* 
 6 Pane/s() ZS-O=/BO'- o" 
 
 - 88OOO*D 
 
 - 26OOOO*L 
 -I 73000*1 
 
 Fig. 262 
 
 The live load extends practically over the entire length of the bridge, 
 as seen above, when the chords and end posts receive their maximum live- 
 load stress. So, in determining the impact stress in these members, L 
 in the impact formula (Art. 125) will be taken as the total length of the 
 span, or 150 ft. 
 
 Then we have 
 
 C = 
 
 300 
 
 150 + 300 
 
 = 0.6666 
 
 for the coefficient of impact for these members, and multiplying the live- 
 load stress in each by this coefficient we obtain the impact given on them 
 in Fig. 262. As an example, the impact in the end post bA is 263,000 x 
 0.6666 = 175,315, say 175,000*; in chord be it is 260,000x0.6666 = 173,- 
 316, say 173,000*; and so on for the other chord members. 
 
 The maximum live-load stress in hanger bB results from loads in 
 the two adjacent panels, AB and BC, alone. So the L in the impact 
 formula would be taken as 50 ft., the sum of the lengths of the two panels, 
 in determining the impact stress in that member. Then for the impact 
 stress in hanger bB we have 
 
 When the maximum live-load tensile stress occurs in the diagonal 
 bC the live load extends from the right support to a little beyond panel 
 
346 STRUCTURAL ENGINEERING 
 
 point C (see Fig. 253). In the case of a uniform live load the distance 
 from the right support C would be taken as L in the impact formula, as 
 in that case the loads are considered to be only at panel points, and as the 
 result is affected but little we will use the same distance in the case of 
 the concentrated loads. So for the impact stress in diagonal bC we have 
 
 (300 \ 
 100 + 300 ) =131,250, say 131,000 Ibs. tension. 
 
 The maximum live-load compression occurs in post cC when the load 
 extends from the right support to a little beyond panel point D, and the 
 maximum tension occurs in diagonal cD (see Fig. 254) at the same time. 
 So the distance from the right support to panel point D (75 ft.) will be 
 taken as L in determining the impact in these members. Then for the 
 impact (compression) in post cC we have 
 
 78,000 x ( 75 3 + 3 OQ )= 6 ^400, say 62,000 Ibs., 
 and for the impact (tension) in diagonal cD we have 
 
 102,000 x ( 75 30 3 00 ) =81,600, say 82,000 Ibs. 
 
 The maximum live-load tension in post cC and maximum compression 
 in diagonal cD occur when the load extends from the left support to a 
 little beyond panel point C. (See case of corresponding members (eE 
 and eD) on right half of truss, Fig. 255.) So the distance from the left 
 support to panel point C will be taken as L in determining the impact 
 in this case, and we have 
 
 37,000 x (35^55) =31,685, say 32,000 Ibs. 
 for the impact (tension) in post cC and 
 
 48,000 x ( 50 3 + 3 OQ ) =41,146, say 41,000 Ibs. 
 
 for the impact (compression) in diagonal cD. 
 
 The impact stress found above for each respective member can now 
 be written on each member, and adding them in each case to the dead- 
 and live-load stresses the total combined stresses, as shown in Fig. 262, 
 are obtained. 
 
 In the case of members cC and cD we have what is known as 
 reversal of stress. The dead-load stress, which acts at all times, is com- 
 pression in post cC and tension in diagonal cD. When the live load 
 moves onto the structure from the left a tensile stress of 37,000* occurs 
 in post cC and a compression stress of 48,000* in diagonal cD, as shown 
 for the corresponding members (eE and cD) on the right half of the 
 bridge (Fig. 251). Now it is evident that, if the dead-load compression in 
 post cC were just equal to the 37,000* live-load tension, the two stresses 
 would just balance each other and consequently the actual stress in the 
 post would be zero when the maximum live-load tension occurred in the 
 post; but, if the dead-load stress were less than the live, the actual stress 
 would be the difference between the two. In any case, before the dead 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 347 
 
 load is subtracted the impact should be added to the live-load stress. To 
 be sure that reversal is well provided for, as a rule not all of the dead- 
 load stress is subtracted from the sum of the live-load and impact stresses. 
 The A. R. E. Ass'n specification, the one we are using, calls for f. 
 Therefore, for the tensile stress in post cC, we have 
 
 -3 7,000 -32,000 + |x22,000 = -54,334, say 54,000 Ibs., 
 
 o 
 and for the compressive stress in diagonal cD we have 
 
 +48,000 + 41,000-|xl7,000 = +77,667, say 78,000 Ibs. 
 
 o 
 
 In case f of the dead-load stress in any member is greater than 
 the sum of the reverse live-load stress and impact, the member is not 
 reversed and, consequently, the reverse stress is ignored, as in the case 
 of diagonal bC. In bC we have 14,000* (see Fig. 251) live-load com- 
 pression and 14,000 x 300 +(25 + 300) = 13,000* (about) impact, making 
 in all a stress of 14,000 + 13,000 = 27,000* compression, but f of the 
 dead-load tension is 51,000x2/3 = 33,666* and hence no reversal takes 
 place, that is, the diagonal is always in tension, due to dead load, and, 
 consequently, the reversal stress can be ignored. The same is true in all 
 such cases. 
 
 176. Designing of Members in Trusses. After all of the maxi- 
 mum stresses are computed in the members of the trusses and combined 
 as shown in Fig. 262, the next thing, logically, to do is to design the 
 sections of these members. The first thing to do in such work is to 
 glance, so to speak, over the structure and associate the details of the 
 different members and, as far as possible, ascertain the governing 
 members. 
 
 In this case, as in all such bridges, the intermediate posts are the 
 governing members, as their width governs the width of practically all 
 of the other members of the trusses. So we will first design the inter- 
 mediate post cC (Fig. 262). 
 
 Member cC. (See Fig. 262.) Let us assume 
 each of these posts to be composed of 2 15" [s 
 placed, in reference to each other, as shown in Fig. 
 263. The distance d between the toes of the flanges 
 is really the governing distance. This distance 
 should not be less than 5" in any case, and a little 
 more is preferable. If this distance is less than 
 5y it is impossible to get the jaw of an ordinary 
 riveter inside of the member and consequently the 
 
 rivets would have to be hand driven, which is expensive work. From 
 table 3 it is seen that the average width of flange of 15" [s is about 3J", 
 making 7" for the two flanges, and if we make d=6J", we have 
 
 for the width of the posts. The least average radius of gyration, which 
 is in reference to axis x-x, as seen from Table 3, is about 5.5". Then 
 substituting this value of r in the column formula of Art. 73, and taking 
 
348 STRUCTURAL ENGINEERING 
 
 L as 30 ft., or 360 ins., we have 
 
 16,000 -70^ = 11,420 Ibs. 
 o.o 
 
 for the allowable unit compressive stress on the post. 
 
 For the stress in the post we have 162,000* compression and 54,000* 
 tension, as given in Fig. 262. Now, according to the specifications, each 
 of these stresses must be increased by 0.5 of the lesser. So we have 
 162,000 + 0.5x54,000 = 189,000* compression and 54,000 + 0.5x54,000 = 
 81,000* tension which the post must be designed to carry. 
 
 Taking, first, the case of compression, we have 
 
 189,000 
 11420 =16 - 6s q- ms - (about) 
 
 for the area required for compression and 
 
 81,000 
 
 16,000 
 
 = 5.06 sq. ins. 
 
 for the net area required for tension. The area, 16.6 n ", required for 
 compression governs. The nearest to this (considering 15" channels) is 
 2 [s 15"x33*, which have an area of 19.8 n// . These channels have 
 3.2 n " more area than required, so let us try 2 [s 12"x30*. Then 
 we have 
 
 16,000 - 70 252 - 10,100 Ibs. 
 
 4.,o 
 
 for the allowable unit stress on the post, and dividing this into the stress 
 we have 
 
 189,000 + 10,100 = 18.7 sq. ins. 
 
 for the required area. This shows that the 12"x30* channels are too 
 light and that the 12"x35* channels would have to be used, which are 
 heavier than the 15"x33* channels; and, besides, better details are ob- 
 tained by using the 15" channels, so we will use the 2 [s 15"x33*. 
 The radius of gyration of these channels is 5.62, within 0.12 of the 
 assumed radius, and hence recalculations are unnecessary. 
 
 Member dD. The post dD carries nothing but the 9,000* of dead- 
 load compression and theoretically could be made of very light section, 
 but to keep the floor beams the same throughout the span and other details 
 constant, as well as for general appearance, we will make this post of 
 2 [s 15"x33* (the same as post cC), that is, each of these posts will 
 be made of 2 [s 15"x33* (the lightest 15" channels) regardless of 
 the excess of metal. 
 
 Member bB. The hangers bB, or hip verticals as they are often 
 called, are not posts but wholly tension members, and it is not necessary 
 that they be of the same type of section as the posts; however, they are 
 often made so, but more often they are made of four angles built into an 
 I-section. (See Fig. 298.) The total stress, as given in Fig. 262, divided 
 by the allowable unit intensity gives 193,500 + 16,000 = 12.09" (net) 
 for the required area of cross-section. We will use 4 Ls 6" x 4" x f " = 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 349 
 
 14.44 n "-1.5 n " = 12.94 n " (net), deducting a one-inch hole (for J-in. 
 rivet) out of each angle. This type of member is cheaper than the two 
 channel sections, as used in the posts, as there are fewer details and, con- 
 sequently, less shop work, but it is not an economic column section, as the 
 radius of gyration is small in comparison with the area of cross-section 
 contained, as in comparison with two channels of an equal cross-section. 
 Member bC. The diagonal is subjected to tension only, and, using 
 the maximum stress given in Fig. 262, we have 
 
 357,000 -=- 16,000 = 22.31 sq. ins. (net) 
 
 for the required area of cross-section. We could use either four angles, 
 as in the case of hanger bB, or two channels, as the member is subjected 
 to tension only, but the two-channel section is better in this case as the 
 member is subjected to some cross bending due to its own weight, and 
 as the channel section is the more capable of resisting this it will be used. 
 Let us try 2 [s 15"x40#. We can consider either two rivet holes cut 
 out of each web or one hole cut through each flange, whichever is the 
 greater. The web of the 15"x40# [ is J" thick (see Table 3) and 
 the grip thickness (g) on the flange is J", hence the holes through 
 the flanges govern. Then we have 2 [s 15" x 40* = 23.52 -4 x 21/32 = 
 20.90 n ". As is seen, these two channels are too light. So let us try 
 2 [s 15"x45#. Jn this case the thickness of the web and the grip 
 through the flanges are the same, and hence either can be taken in deter- 
 mining the net section of member. So we have 2 [s 15"x45# = 
 26.48 -4xf = 23.98" (net), which is 1.67 n " more than required, but 
 this is the best we can do if we use channels, so we will use this section 
 for diagonal bC. 
 
 Member cD. This diagonal is subjected to 201,000* tension and 
 78,000* compression, as given in Fig v 262. Now increasing each by 0.5 
 of the lesser, as we did above in the case of post cC (according to the 
 specifications), we have 
 
 201,000 + 0.5 x 78,000 = 240,000 Ibs. tension 
 
 and 
 
 78,000 + 0.5 x 78,000 = 117,000 Ibs. compression. 
 
 Then dividing the tensile stress by 16,000* (the allowable unit stress) 
 we have 
 
 240,000 -5- 16,000 = 15.0 sq. ins. 
 
 for the net area of cross-section required to carry the tension. Now 
 assuming the member to be made of two 15" channels having a radius of 
 gyration of 5.6 (a mere guess) and taking the length of the diagonal as 
 39 ft., or 468 ins., we have 
 
 16,000 -70 y = 10,150 Ibs. 
 
 for the allowable compressive unit stress. Dividing this into the com- 
 pressive stress, we have 
 
 117,000 -r 10,150 = 11.5 sq. ins. 
 
 for the required area of cross-section to carry the compression. Deduct- 
 ing for four rivet holes, out of the flanges of the two lightest 15" channels, 
 
350 STRUCTURAL ENGINEERING 
 
 we have 2 [s 15" x 33#= 19.8-4x21/32 = 17.2 n " (about) for the net 
 area to carry tension, which is 2.2 n " more than required. The total, 
 19.8 n// is available for compression, so the section is larger than is neces- 
 sary for compression as well as for tension, but this section will be used 
 as it is desirable to have all of the diagonals of the same width. We now 
 have all of the web members designed and we will next take up the 
 designing of the chord members. 
 
 Members AB and BC. The bottom chords are purely tension mem- 
 bers; so, dividing the maximum stress in chord AB or BC by 16,000, 
 we have 
 
 335,000 -r 16,000 = 20.93 sq. ins. 
 
 for the net area required. By glancing over the table of channels (see 
 Table 3) we see that two 15" channels can be used as section for these 
 members. Deducting four rivet holes from the flanges we have 2 [s 
 15"x40# = 23.52- 2.6 = 20.9 n ", which is practically the section required, 
 and hence these channels will be used. 
 
 Member CD. For the net area of cross-section required for chord 
 CD, we have 521,000 -f 16,000 = 32.56 sq. ins. Now, from the table of 
 channels (see Table 3 or any structural handbook, as Cambria or Car- 
 negie) it is seen that no two 15" channels will have the required net sec- 
 tion. So, if channels are used, they will have to be reinforced by riveting 
 plates to their webs. Let us try the following sections: 
 
 2 [s!5"x45# =26.48-4x5/8 = 23.98 n "net 
 2 pis. 12" x 7/16 = 10.50 - 4 x 7/16 = 8.75 n " net 
 
 32.73 n " net 
 
 The net area of this is practically equal to the area of the required 
 section and hence will be used. 
 
 Top chords. The top chords and end posts of such bridges are usu- 
 ally made up of either two channels and a cover plate as shown at (a), 
 
 Fig. 264, or of two built chan- 
 nels and a cover plate as shown 
 at (6). The rolled channels are 
 used whenever that section is 
 sufficient. It is not desirable to 
 use the heaviest channels, which 
 are of -Jf" metal, as the shop 
 work is rather expensive on ac- 
 count of the thick metal, and 
 whenever the end post or heav- 
 
 . 264 iest chord members exceed the 
 
 area of the cover plate and, say, 
 
 two 15"x45# channels, a section similar to the one shown at (6), Fig. 
 264, is used throughout the structure, as it would be unsightly construction 
 to build part of the chord members of a bridge of rolled channels and 
 part of plates and angles. So, in this case, we will first ascertain as to 
 whether we can or cannot use the rolled channel section as shown at (a), 
 Fig. 264, for the top chord. The member cd, as is seen from Fig. 262, will 
 be the heaviest member, that is, the one requiring the greatest area of 
 
DESIGN OF SIMPLE KAILEOAD BEIDGES 351 
 
 cross-section. In preliminary calculations the least radius of gyration of 
 this type of section can be taken as 0.40 of its depth. Then, assuming 15" 
 channels, we have 15x0.40 = 6.0 for the approximate radius of gyration 
 about axis x-x, which is usually the least radius. Now substituting this 
 value of r and the length of the member, which is 25 ft., or 300 ins., into 
 the column formula, we have 
 
 16,000 -7o = 12,500 Ibs. 
 
 0.0 
 
 for the allowable unit stress on the member. Then dividing the maximum 
 stress, as given in Fig. 262, by this, we have 
 
 589,000 
 
 -j^- =47.1 sq. ins. 
 
 for the required area of cross-section of the member. The width of the 
 intermediate posts, as found above, is 13J", and allowing for two f" 
 gusset plates and 7" ( = 3-J-x2) for the flanges of the channels, we have 
 
 13J + lJ + 7 = 21f, say 22 ins. 
 
 for the width of the cover plate. The cover plate should be as thin as is 
 consistent with good practice. The specifications limit this to -^ of 
 the distance (shown as d at (a), Fig. 264) between the lines of rivets 
 connecting the plate to the other parts of the member. That distance 
 here, using 2" gauge in channel flanges, is 13^ + lJ + 4 = 18 j", say 19". 
 One- fortieth of this is practically J", so the cover plate would be 22" x-J". 
 Now, using the heaviest channels that it is practical to use, we have the 
 following sections : 
 
 2 [s!5"x45# = 26.48 n " 
 
 1 cov. pi. 22" x " = 11.00 n " 
 37.48*" 
 
 This is about 10 n " less than the required area found above, and hence 
 the rolled channel section has not sufficient area; and, therefore, the type 
 of section shown at (6), Fig. 264, will be used. 
 
 In designing the top chords it is best to begin with the lightest sec- 
 tion, which is the member having the least stress, and make it of minimum 
 thickness of metal and then increase the area of cross-section of the 
 other top chord members by increasing the thickness of webs, leaving the 
 other parts about constant. For by so doing the centers of gravity of 
 the chords will practically be in the same plane throughout. Therefore, 
 we will take up the designing of chord be first it being the lightest. 
 
 Member be. The maximum stress in chord be, as given in Fig. 262, 
 is 521,000*. Then assuming the allowable unit stress to be 12,500" (see 
 Example 4, Art. 74), we have 
 
 521,000 
 
 ^ 41.68 sq.ms. 
 
 tor the approximate required area. Next draw a sketch of the cross- 
 section of the chord as shown in Fig. 265. The cover plate and top 
 angles should be made as light as the specifications will permit in order 
 that the gravity axis x-x be as near the center of the web as possible. As 
 seen above, the cover plate must be about \" thick and the smallest angles 
 
352 
 
 STRUCTURAL ENGINEERING 
 
 that can be used at the top, considering details and all, are 3J" x 3J" x f ". 
 The bottom angles should be as heavy as is practical to use for the same 
 reason that the top ones are made light. So for the bottom angles we 
 will use 6"x4"xf", as that is about the maximum size and thickness 
 used for such work. The cover plate, top angles, and the bottom angles 
 are usually about constant throughout the chord. 
 
 Allowing for 2 f " gusset plates and for 2 |" webs (a mere guess), 
 we have 
 
 for the width of cover plate. To insure a neat finish, the cover plate 
 
 should project J" or \" beyond the angles. So we will make the cover 
 ^ plate 23" wide. However, the details 
 
 should always be such that the width of 
 cover plate would be in even inches. 
 
 The depth of webs should be such 
 that the radius of gyration about the 
 horizontal axis (x-x, Fig. 265) is about 
 equal to the radius about the vertical 
 axis (#-#). For preliminary calcula- 
 tions the radius of gyration in reference 
 to axis x-x of such chords can be taken 
 as 0.4 of the depth of the web, and the 
 radius in reference to axis y-y can be 
 taken as the horizontal distance out 
 
 from the axis to the outer face of the web. So, in this case, we have 
 
 (assuming \" webs) 
 
 </ 
 
 
 t 
 
 
 
 i- 
 L 
 
 JL 
 
 ^ I 
 
 ** 
 
 -fr^ 
 
 v,' 
 
 t_ 
 
 "N 
 
 k 
 
 <* 
 
 \ 
 
 
 * 
 
 i Jy 
 
 i.se V e 
 
 
 
 
 78' 
 
 
 Fig. 265 
 
 for the approximate radius about axis y-y. Then for equal radii about 
 the two axes, we have 0.4A = 7.87, from which we obtain for the depth 
 of the web 
 
 7.87 
 -T - 
 0.4 
 
 1Q ,. 
 =19. 6 ins. 
 
 As the work here is only fairly approximate this figure shows only about 
 what the depth of the web should be, and hence any web about this depth 
 can be used. A 19" web is an odd width and a 20" web appears a little- 
 deep for this section, so we will use an 18" web. Now the radius of 
 gyration of the section about the horizontal gravity axis, which is usu- 
 ally the least radius, is about 0.4 of the depth of the web, as stated above, 
 so we have 
 
 0.4x18 = 7.2 ins. 
 
 for the approximate value of the least radius of gyration of the section. 
 Then substituting this value of r in the column formula, we have 
 
 16,000 - 70 = 13,084, say 13,000 Ibs. 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 353 
 
 for the approximate allowable unit stress. Then dividing this into the 
 stress we have 
 
 521,000 + 13,000 = 40.07 sq. ins. 
 
 for the approximate required area. Taking this as a guide, let us assume 
 the following section for chord be: 
 
 1 cov. pi. 23" xj" = 11.50 n " 
 2 web pis. 18" x f" = 13.50 n " 
 2 Ls 3"x3"x" = 4.96" 
 2 Ls 6"x4"x T V' = 10.62n" 
 40.58" 
 
 The next thing in order is to determine the center of gravity, moment 
 of inertia and radius of gyration of this section and check back to see if 
 the section is actually the correct one to use. 
 
 Taking moments about the center of the cover plate (Fig. 265), 
 we have 
 
 - 13.50 x 9.37 -f- 4.96 x 1.26 + 10.62 x 16.49 
 
 -4^58~ -=7.59 ins. 
 
 for the distance from the center of the cover plate to the horizontal grav- 
 ity axis x-x, and for the moment of inertia about this axis we have the 
 following : 
 
 cover plate 11.50 x 7^59 2 + = 662.49 
 
 webs 13.50x178% 182.25x2 = 407.27 
 
 top angles 4.96 x 673? + 2.87 x 2 = 204.48 
 
 bot. angles 10.62 x 8^87 2 + 19.26 x 2 = 874.07 
 
 2,148.31 
 
 Then for the radius of gyration about this same axis x-x we have 
 
 r = 
 
 which is within 0.08 of 0.4 of the depth of the web. 
 
 As the section is symmetrical about the vertical plane through the 
 center of the cover plate, the vertical gravity axis y-y will pass through 
 the center of the cover plate and for the moment of inertia of the section 
 about this axis y-y we have the following: 
 
 cover plate + 506.96 = 506.96 
 
 webs 13.50 x7J56 a +0 = 771.57 
 
 top angles 4.96x8.76* +2.87x8 = 386.36 
 
 bot. angles 10.62x8.78 +6.91x2= 832.50 
 
 2,497.39 
 
354 STRUCTURAL ENGINEERING 
 
 Then for the radius of gyration about this axis y-y we have 
 
 
 40.58 
 
 = 7.84 ins., 
 
 which is a little larger than the radius about the x-x axis. 
 
 Now substituting the value of r (the least radius) in the column 
 formula, we have 
 
 16,000- 70 = 13,116 Ibs. 
 
 7.6o 
 
 for the actual allowable unit stress on the chord. Dividing this into the 
 stress, we have 
 
 521,000 ,- 
 
 = 39.72 ,.,,., 
 
 which is 0.86 n " less than the section assumed above, but as this is about 
 as close as we can obtain we will use the assumed section for chord be. 
 Member cd. There will be such a small difference between the allow- 
 able unit stress for the other top chord members and that found above 
 for chord be (as can be verified by actual calculations) that the allowable 
 unit stress found for that member will be used in designing the others. 
 So, dividing the maximum stress in chord cd, as given in Fig. 262, by 
 that unit stress, we have 
 
 589,000 
 
 = 44.9 sq. ins. 
 
 for the required area of cross-section of the member cd. 
 
 Then using the same size angles and cover plate as for chord be 
 and increasing the thickness of the web to -J" ', we have 
 
 1 cover pi. 23" x \" =11.50" 
 
 2 web pis. 18" x " - 18.00 n " 
 2 Ls 3i"x3i"xf" = 4.96 n " 
 2 Ls 6"x4"x T V =10.62" 
 
 45.08 n " 
 
 This section is quite close to the required section, being only 0.1 8 n " 
 larger, arid hence will be used for chord cd. 
 
 This completes the design of the main truss members,, as the bridge 
 is symmetrical about the center of the span, except the end posts, which 
 will be designed later, after the bending moment on them due to the 
 wind pressure is determined. 
 
 177. Designing of the Bottom Lateral System. The bottom 
 lateral system is really a horizontal truss in the plane of the bottom chord. 
 It is for the purpose of resisting the wind pressure on the lower portion 
 of the structure, and, in the case of through bridges (such as the one we 
 are now designing), the wind pressure on the train as well as the vibra- 
 tion due to the swaying of the train. A double system of diagonals, or 
 laterals, is practically always used, which is as indicated in Fig. 266, 
 where the bottom lateral system is shown in plain view, one system of 
 laterals being dotted to avoid confusion. 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 
 
 355 
 
 The bottom chords of the main trusses act as the chords in this lat- 
 eral system and the floor beams as posts. The stress in the chords due 
 to the wind and vibration is, as a rule, ignored unless it reverses the dead- 
 load stress, taking the wind pressure in that case on the unloaded struc- 
 ture as 350# per foot of span or where the maximum exceeds 25 per cent 
 of the maximum combined dead, live, and impact stresses (see specifica- 
 tions), in which case it is combined with the dead, live, and impact 
 stresses as in the case of viaduct columns (Art. 162). This combination, 
 however, is very rarely necessary in ordinary bridges. The stress in the 
 floor beams (acting as columns), due to wind and vibration, is ignored 
 entirely, as the laterals are connected to the bottom flange of the beams, 
 
 e Bo f. Chord. 
 
 
 6 Panefs <&, 2S- o"= ts 
 Fig. 266 
 
 which are in tension due to dead and live load, and hence the compres- 
 sion in these same flanges, due to wind and vibration, only tends to 
 reverse the dead- and live-load stress in them. 
 
 There are two ways of considering the bottom lateral system: One 
 is to consider the double system, in which case the two diagonals in any 
 panel resist equally the shear in the panel, that is, one diagonal is con- 
 sidered as having the same intensity of stress as the other, one being in 
 compression and the other in tension; the other way is to consider only 
 a single system, one system being in action when the pressure comes 
 from one direction and the other system acting when the pressure comes 
 from the opposite direction. We will consider the single system, in which 
 case the diagonals carry only tension, and then investigate for rigidity 
 after the sections are determined for the single system. 
 
 According to the specifications, the pressure or horizontal load on 
 the bottom lateral system, here considered, is 200 + 0.10 x 5,000 = 700# 
 per foot of span. So, for a panel load we have 
 
 W= 700 x 25 = 17,500 Ibs. 
 
 This is, according to the specifications, to be considered a moving load 
 (live load). Let us assume that this load acts in the direction indicated 
 by the arrows (Fig. 266) and that it moves over the structure from right 
 to left, that is, from G to A. Now, from Art. 90, we have 
 
 =^Fl + 2.... (n-l)l 
 
 for the maximum shear in the different panels, and hence for the maximum 
 stress in the corresponding diagonal (considering single system) we have 
 
 sec W. 
 
356 STRUCTURAL ENGINEERING 
 
 Tan a) = 25/16 = 1.56, and sec <o = 1.85, which is most readily obtained 
 from a table of natural functions after the tangent is computed. 
 Then for the maximum stress in diagonal cD we have 
 
 S=(I + 2 + 3) sec eo = 17 ^ x 6 x 1.85 = 32,375, 
 n o 
 
 say 32,000 Ibs. (tension), 
 and for the maximum stress in bC we have 
 
 S'= (1 + 2 + 3 + 4) seco)=i^| x 10x1.85 = 53,958, 
 n o 
 
 say 54,000 Ibs. (tension), 
 and for the maximum stress in aB we have 
 
 S" = (1 + 2 + 3 + 4+5 )secco = '- x 15 x 1.85 = 80,937, 
 n o 
 
 say 81,000 Ibs. (tension). 
 
 This completes the necessary determination of stress in the diagonals or 
 laterals (as they are called) due to wind and vibration, as the structure 
 is symmetrical about the center of the span, and we will next design the 
 sections for these members. 
 
 Taking the first lateral aB, we have 
 
 81,000 
 
 for the net area of cross-section required. Using 1 L 5"x3|"xf" = 
 5.81-0.75 = 5.06 n// , we have exactly the correct net area for that diag- 
 onal. For lateral bC, we have 
 
 54,000 
 
 for the net area of cross-section required and by using 1 L 5" x 3 J" x J" 
 = 4-0.5 = 3.5 n ", we have about the correct area, at least about as near 
 the correct area as is possible to obtain. For lateral cD, we have 
 
 32,000 
 
 for the net area of cross-section required, and by using 1 L 5" x 3|" x f " 
 = 3.05-0.37 = 2.68 n " we have 0.68 n " more net area than required, but 
 we will use this angle so as to have the bottom laterals made of 5" x 3J" 
 angles throughout. 
 
 As the structure is symmetrical about the center of the span, we 
 now have all of the laterals designed, considering the single system, and 
 we will now investigate the double system. The laterals are connected 
 to the bottom of the stringers at the points of intersection and hence 
 the longest unsupported length of lateral is between the stringer and 
 truss. This maximum unsupported length is about 8'-0", or 96". Now 
 taking the end lateral aB, which is composed of 1 L 5" x 3|" x J", we 
 have L/r = 96/0.96 = 100, which is quite satisfying as the maximum value 
 
DESIGN OF SIMPLE KAILROAD BEIDGES 
 
 357 
 
 for L/r is 120, and substituting the value of L and r in the column 
 formula we have 
 
 16,000 -70-^- = 9,000 Ibs. 
 
 \) \) O 
 
 for the allowable compressive unit stress, and dividing this into one-half 
 of the stress found above (the stress that it would be considered to resist 
 if acting in the double system) we have 
 
 40,500 
 
 for the required area of cross-section, which is considerably less than the 
 cross-section of the angle. So this lateral is sufficient, considering either 
 a single or double system, and the same is true for the other laterals, as 
 will be found upon investigation. So the bottom laterals as designed are 
 capable of resisting the forces coming on them from wind and vibration 
 when considered either as a single or double system. 
 
 The bottom laterals are subjected to stress from traction in addition 
 to the stress resulting from wind and vibration. As an illustration, sup- 
 pose a rapidly moving train to come onto the structure and the brakes to 
 
 floor Beam*. Bottom Chord-* Stringer-* 
 
 
 ( 
 
 
 I 
 
 A I 
 
 
 ~^ 
 
 
 / 
 
 ^ / 1 
 
 
 
 
 / 
 
 1 
 
 Fig. 267 
 
 be suddenly applied the wheels would skid or tend to skid along the 
 rails, whereby a longitudinal thrust would be exerted along the stringers 
 which would tend to bend the floor beams transversely as indicated in 
 Fig. 267 (where the train is assumed to be moving from right to left). 
 Now this bending of the floor beams, which would cause serious stresses 
 in them, is prevented by attaching the laterals to the bottom of the string- 
 ers so that the longitudinal thrust exerted by the stringers is transmitted 
 directly to the laterals, and from there to the bottom chords of the trusses 
 and on out to the end supports. As the stringers are connected to one 
 another rigidly, end to end, throughout the length of the structure, it is 
 evident that the traction from any loading would be transferred more or 
 less from stringer to stringer, and hence would be distributed to some 
 extent to all of the laterals; so in computing the stress in the laterals 
 due to traction we will consider that the laterals in any panel resist 
 the traction from an average panel load of fully loaded bridge. Then 
 placing wheel 1 (see Table A) at one end of the bridge, we have 
 
 for the weight of the full live load (Cooper's E50 loading) on the bridge. 
 Dividing this by the number of panels, we have 
 
 6 
 
 = 153,500 Ibs. 
 
358 
 
 STRUCTURAL ENGINEERING 
 
 Stringer 
 
 I52BO* 
 
 for the average panel load. Two-tenths (0.2) of this vertical load is 
 considered as traction and is exerted along the track, that is, the coeffi- 
 cient of friction of the wheels on the rails is taken as 0.2. So we have 
 
 152,500 x 0.2 = 30,500 Ibs. 
 
 as the traction force exerted 
 on the two stringers in each 
 panel, or 15,250* to each 
 stringer. This force on each 
 stringer is transmitted to 
 the laterals at two points, 
 where the laterals and 
 stringers connect, so we can 
 consider the case as indi- 
 cated in Fig. 268. One 
 component of the traction 
 force, 15,250# -f 2 applied at 
 each of the points c will, in 
 each case, act along the lat- 
 eral, causing compression in 
 the case of the part Be and tension in the part DC, and the other com- 
 ponent at each point c will act along member cc. 
 
 Without the members cc, there would be transverse bending on the 
 stringers from traction, and hence cc is an essential member in the 
 system.* 
 
 Then for the stress in a lateral, due to traction, we have 
 
 do. 
 
 Fig. 268 
 
 15,250 
 
 xsec</> = 
 
 15,250 
 
 x 1.18 = 8,997, say 9,000 Ibs., 
 
 which is compression in the part cB and tension in part cD, and just the 
 reverse if the train were moving in the opposite direction. For the stress 
 in member cc, which can be called a tie or strut, we have 
 
 15,250 
 
 x tan <j> = 
 
 15,250 
 
 x 0.64 = 4,880 Ibs., 
 
 which is compression in one and tension in the other. 
 
 These stresses are not very great in the first place, as is seen, and 
 they are not very likely to occur at the same time that the maximum 
 wind stress occurs and hence, as a rule, are ignored in the designing of 
 the sections of the laterals. However, the traction should be provided 
 for in so far as making the laterals capable of taking compression 
 (L/r<120) and the inserting of the ties cc. The tie cc does not neces- 
 sarily have to be a very heavy section. They are usually made of 1 L 
 3J"x3"x", in which case we have 
 
 7fi 
 16,000 -70^ = 9,934 Ibs. 
 
 * As far as the author knows, Dr. J. A. L. Waddell, M.A.Soc.C.E., was the first to 
 point out the necessity of these struts. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 359 
 
 for the allowable unit stress for compression. Then we have 
 
 4,880 
 
 for the required area in the case of compression, and less is required for 
 tension, as is obvious. So it is seen that the theoretical required area of 
 these struts (cc) is not' the governing feature, but that it is a matter of 
 obtaining sufficient rigidity. That means that L/r should not exceed 120 
 in any case. The L/r for a 3^"x3"xf" angle is about 0.87, which is 
 reasonably low. This angle is used as it is a very common angle in this 
 class of work, and it is about the minimum size permitted and it fits the 
 case wherein common judgment decides. 
 
 For the wind stress in the bottom chord ab or BC (Fig. 266) we have 
 
 2J W tan (o = 2J x 17,500 x 1.56 = 68,250 Ibs., 
 
 which is compression in ab and tension in BC, in this case. Now, as 
 shown in Fig. 262, the maximum combined stress, dead, live and impact, 
 is 335,000* tension. Twenty-five per cent of this is 83,750*, so the maxi- 
 mum wind stress in these chords can be ignored. (See specifications.) 
 In case of combined dead, live, impact, and wind stresses, the allowable 
 unit stress is raised 25 per cent, as the probability of all of these stresses 
 being a maximum at the same time is very remote. The same is true 
 of the other bottom chords, as will be found upon investigation. Now, 
 so far, the wind stresses in the bottom chords can be ignored ; but further, 
 suppose the bridge to be unloaded, that is, no live load on it, the wind 
 pressure of course will be less. Let us then assume a pressure of 350* 
 per foot of span, which is about correct, instead of 700*. Then the 
 stress in the chord ab or BC is 
 
 68,250 x = 34,125 Ibs., 
 
 which here, in the case of ab, is compression. But, as this does not 
 reverse the dead-load tension in the member, which is 55,000*, the wind- 
 load stress need not be considered. The same will be found to be true 
 for the other bottom chords, so the stresses in the bottom chords of this 
 bridge, due to wind or lateral pressure, as the specifications call it can 
 be ignored. The same is true for most all ordinary railroad bridges. 
 
 178. Designing of the Top Lateral System. The top lateral 
 system is really a horizontal truss in the plane of the top chords. In the 
 case of through bridges (such as we are designing) it is for the purpose 
 of holding the top chords transversely and resisting the wind pressure 
 coming on the top portion of the structure, but in the case of deck bridges, 
 where the track is on the top of the structure, the wind pressure on the 
 train and vibration due to the train must be provided for also, in which 
 case the designing of the top lateral system would be the same as shown 
 in the last article for the bottom lateral system. The top lateral system, 
 as a rule, has a double system of diagonals, the same as the bottom lat- 
 erals, except in the case of through bridges a portal is placed in each 
 end panel in the plane of the inclined end posts which takes the place 
 of diagonals in those panels. So, following the usual practice, the top 
 lateral system in this case will be made as shown in Fig. 269, where one 
 system is dotted to avoid confusion. 
 
360 
 
 STRUCTURAL ENGINEERING 
 
 The lateral pressure or load which this system must be designed 
 to carry, according to the specifications, is 200# per foot of span. This 
 horizontal load should be considered as a moving load, that is, as a uni- 
 form live load. Then for a panel load we have 
 
 P = 25 x 200- 5,000 Ibs., 
 
 one-half of which is to be considered as applied to each truss, that is, 
 wherever it is necessary to do so. 
 
 6 Pane/s @ 2S L O"=. /SO- O 
 
 Fig. 269 
 
 Now, assuming this load as moving onto the bridge from the right 
 end, and acting in the direction indicated by the arrows (Fig. 269), we 
 obtain 
 
 = 9,250, sa y 9,200 Ibs. 
 for the maximum tensile stress in diagonal cD, and 
 
 10 x 
 
 1.85 = 15,416, say 15,400 Ibs. 
 
 for the maximum tensile stress in diagonal bC. 
 
 Assuming a full panel load at each of the points F and E and a half 
 panel load at d (none at Z)), we have 
 
 5,000 5,000 
 3 x -^- + -~ = 5,000 Ibs. 
 
 O u 
 
 for the maximum compressive stress in strut dD, and assuming a full 
 panel load at each of the points F, E, and Z), and a half panel load at c 
 (none at C), we have 
 
 a 5,000 5,000 ,_,, 
 6 x - + ~ = 7,550 Ibs. 
 
 D 6 
 
 for the maximum compression in strut cC. 
 
 This really completes the necessary calculation for the stresses in 
 the top lateral system, as will be seen upon investigation. For example, 
 when panel point F alone is loaded, and the forces acting as indicated, 
 the stress in diagonal eF will be tension, as is obvious, but this diagonal 
 has maximum tension when the panel points are loaded from the left 
 end up to and including point E with forces acting in the opposite 
 direction, in which case the stress in the diagonal would be 15,400*, the 
 same as found for diagonal bC and loading points F and E, and the 
 forces acting as indicated, diagonal dE would be in tension; but the 
 maximum tension will occur in this diagonal when the span is loaded 
 
DESIGN OF SIMPLE RAILROAD BKIDGES 361 
 
 from the left end up to and including point D and forces acting in the 
 opposite direction, in which case the tension in diagonal dE would be 
 9,200*, the same as found for diagonal cD. So it is seen that one sys- 
 tem can be considered in action when the span is loaded from one direc- 
 tion and the other system when loaded from the opposite direction, and 
 as the lateral system is symmetrical about the center of span it is obvious 
 that the stresses determined above are sufficient for designing all the 
 members in the system. 
 
 The members (diagonals and struts) should be as deep as the top 
 chords, in order to hold the top chords rigidly. The diagonals, which 
 are usually designed as tension members, are as a rule made of two 
 angles latticed together, one angle being in the plane of the top of the 
 chord and the other being in the plane of the bottom of the chord. The 
 struts (members cC, dD, and eE) are compression members and are, as 
 a rule, made of four angles, two at the top of the chord and two at the 
 bottom. These are latticed together in the vertical plane so as to form 
 an I-section. 
 
 Now, beginning with diagonal bC (Fig. 269), we have 
 
 15,400 
 
 for the required net area of cross-section of the member, considered as a 
 tension member. This, as is seen, (theoretically) calls for two very 
 small angles, but (practically) the angles should be such that L/r be not 
 less than 120, to insure against vibration. The length of L can be taken 
 as one-fourth of the total length of the member. 
 
 (The diagonals are connected to each other at their intersection; 
 this alone reduces the length one-half, and each half can be considered 
 as a fixed column, thus reducing the length to one-fourth of the total 
 length.) The total length of the diagonal center to center of end con- 
 nections is about 30 ft., or 360 ins. Then using one-fourth of this, 
 we have 
 
 for the required radius of gyration. 
 
 According to this and to the required area of cross-section, found 
 above, 2" x 2J" or 3"x3" angles could be used, but as much better 
 details can be obtained by using 3J" x 3J" angles and the weight of the 
 top lateral system is a small item we will use 2 Ls 3J" x 3 J" x f " for 
 each of the diagonals throughout, as cB, eF, and fE require the same 
 section as bC, and the others, theoretically, require less. 
 
 The designing of the struts cC, dD, and eE is very similar to the 
 designing of the diagonals; it is mostly a matter of obtaining rigidity 
 and the selecting of sections that are satisfactory as to details. But in 
 all such cases the theoretical requirement should always be determined 
 before the selection of section is made. 
 
 These struts are more or less fixed at their ends, but as they are 
 wholly compression members we will take the distance center to center 
 of trusses, as their length, which is 16 ft., or 192 ins. In order to insure 
 
362 
 
 STRUCTURAL ENGINEERING 
 
 rigidity, L/r should not exceed 120. 
 
 _193 
 r ~120 
 
 Then, taking L as 192, we have 
 = 1.6 
 
 for the allowable maximum radius of gyration. The most satisfactory 
 section for these struts is four unequal leg angles, arranged in reference 
 to one another as shown in Fig. 270. The long legs of the angles are 
 turned out so as to make the radius of gyration about axis y-y as large 
 as is possible, for, as is obvious, the least radius is about that axis. The 
 lattice bars hold the angles about J" apart, and, as the radius of gyration 
 about axis y-y is the same for one pair of angles as it is for the two 
 pairs, we can select the angles from Table 5 by using the angles having 
 r3 equal to 1.6 the required radius. As is seen in this table, the 3" x 2^" 
 angles are too small, as r3 for them is less than 1.6; the 3J"x2J" are 
 large enough, but to use a 2|" leg would necessitate the using of J" 
 rivets, while J" would be used in all the other members; so we will use 
 the 3J"x3"xf" angles, in which case r3 is a little more than 1.71, as 
 seen in Table 5. Now substituting this value for r in the column formula, 
 we have 
 
 1 Q2 
 16,000 - 70^7 = 8,140 Ibs. (about) 
 
 for the allowable unit stress on a strut made of these angles. Dividing 
 this into the greatest stress found in these struts, which is 7,250*, we 
 obtain less than a square inch of metal for the required section. So it 
 is seen that the stress does not really influence the design and that each 
 
 -f 
 
 
 
 Fig. 270 
 
 Fig. 271 
 
 strut will be made of 4 Ls 3^"x 3"x f" = 9.2 n ", as this fits the case 
 most satisfactorily as regards rigidity and details. 
 
 179. Design of Portals. There are several types of portals in 
 use. The type shown in Figs. 269 and 271 will be used in this case, as 
 it is a type commonly used; it is quite simple, very rigid, and in fact 
 quite satisfactory for ordinary bridges. 
 
 In determining the stresses in portals, the first thing to do, the same 
 as in the case of all frames, is to locate all of the applied forces affecting 
 it. As is obvious, the maximum stresses in the portals will occur when 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 363 
 
 all of the panel points B to F, inclusive (Fig. 269), are loaded. As a 
 specific case, let us consider the portal at the left end of the bridge 
 (at bB) and let us assume that all of the panel points are loaded and 
 that the loads act as indicated in Fig. 269. Then assuming that the 
 diagonals carry only tension, the diagonal bC (the diagonal connecting 
 to the portal at 6) will be in tension and there will be no stress in 
 diagonal cB. One component of the stress in bC will be taken by the 
 portal and the other by the top chord be. The component taken by the 
 portal is applied (as is obvious) at b and is equal to the shear in panel 
 BC, the panel next to the portal. This force we will designate as R. 
 In addition, there will be one-half of a panel load at b and also one-half 
 of a panel load at B. Designating the panel load as P, we then have the 
 loading on the portal' as shown in Fig. 271: (.R + P/2)at point b and 
 P/2 at point B. If the pressure, or load, acted in the opposite direction, 
 the loading of the portal would be just the reverse, that is, the force 
 (tf + P/2) would be applied at B and P/2 at b. These forces applied 
 to the portal are held horizontally by the two equal horizontal reactions, 
 one at the bottom of each end post. Let each of these reactions be repre- 
 sented by // and we have 
 
 These two horizontal reactions at the bottom of the end posts and the 
 horizontal forces, (.R + P/2) and P/2, at the top of the portal form a 
 couple which must be balanced by two equal and opposite reactions, V 
 applied along and at the bottom of the end posts. Now, as the forces 
 (# + P/2), P/2, the two H's, and the two F's balance, evidently, the 
 frame formed by the end posts and portal can be treated as an inde- 
 pendent structure in equilibrium under the action of these forces. So, 
 taking moments about u (Fig. 271), and considering the end posts and 
 portal combined as a frame, we have 
 
 (3). 
 
 or taking moments about w, we have 
 - 
 
 In the designing of portals for bridges, there are two cases: one 
 when the end posts are considered as hinged at the bottom ends, and 
 the other is when the end posts are considered as fixed at the bottom ends. 
 The top ends of the end posts are practically always considered to be fixed 
 by the portal. In the case of light bridges, and especially when end 
 floor beams are not used, the end posts are not very rigidly fixed at the 
 bottom ends and the actual condition is most closely met by considering 
 the bottom ends of the end posts hinged, while in the case of heavy 
 bridges, with end floor beams and with wide shoes, there is no question 
 but that the bottom ends of the end posts are fixed. 
 
 In the case of the bridge we are designing, there is no question but 
 that the end posts will be quite rigidly fixed at the bottom ends, but for 
 the sake of illustration we will first determine the stresses in the portals. 
 
364 
 
 STRUCTURAL ENGINEERING 
 
 assuming the bottom ends of the end posts to be hinged, and then deter- 
 mine them, assuming the bottom ends of the end posts to be fixed. 
 
 By imagining the part to the left of section n-n (Fig. 271), including 
 the end post bu and a portion of the portal, moved bodily (forces and all) 
 to a new position we obtain the structure shown in Fig. 272 which is 
 acted upon by the known forces (R + P/2), H, -V , and by the unknown 
 forces S and SI, S being the stress in member bo and SI the stress in po. 
 The dotted members shown in the portal are assumed to have no stress 
 in them as they are redundant members intended only to brace the main 
 members of the portal, which are shown in full lines, and hence the dotted 
 members are ignored in the calculations. By taking moments about p 
 (Fig. 272) we eliminate 1 and V from the equation of moments, and 
 as the moments of H and (R + P/2) (about p) have the same sign, both 
 tending to produce clock-wise rotation about p, it is evident that the sum 
 of the moments of the two forces about p must be equal but of opposite 
 sign to the moment of S about p. So we have 
 
 from which we obtain 
 
 for the stress in portal member bo. Now it is readily seen from Fig. 272 
 that this stress S is compression, as H and (R + P/2) both tend to produce 
 clock-wise rotation about p, and as S alone prevents this rotation it must 
 
 Fig. 272 
 
 Fig. 273 
 
 evidently act to the left toward b, and hence is compression. In other 
 words, the member bo pushes to the left against the end post and un- 
 doubtedly is in compression. 
 
 To determine the stress, SI, in the portal member po, let us con- 
 sider it resolved at point p into two components, one (/) along the end 
 post and the other (h) perpendicular to the end post. Then taking 
 moments about b (Fig. 272), we eliminate -V, f, S, (R + P/2), and P/2 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 365 
 
 from the equation of moments and we have 
 
 from which we obtain 
 
 for the component of SI perpendicular to the end post. Then we have 
 SI = h sec B = 7 sec (4) 
 
 for the stress in the portal member (knee brace) po. Stress S\ is ten- 
 sion, as is seen by considering point b as the center of rotation; the force 
 H acting to the left would necessitate h acting in the opposite direction 
 or away from the post at point p, which indicates the component h to be 
 tension, and hence the stress SI is tension. 
 
 The stress $1 can also be determined from the summation of the 
 vertical forces. In which case we have 
 
 (-J 7 and / being the only vertical forces). From this we have 
 
 f=r, 
 
 and multiplying by sec </> we obtain 
 
 sec< ................... (5). 
 
 / equals V , and as these two forces must balance each other it is seen 
 that / acts as indicated in Fig. 272, which shows $1 to be tension. 
 
 By imagining the part to the right of section n'-n f (Fig. 271), in- 
 cluding end post wB and a portion of the portal, moved bodily (forces 
 and all) to a new position we obtain the independent structure shown in 
 Fig. 273 acted upon by the known forces H, V , and P/2 and by the 
 unknown forces (stresses) S2 and S3. Taking t as the center of moments, 
 it is seen that P/2 and H both tend to produce clock-wise rotation, and as 
 this rotation is prevented by S2 alone, we have 
 
 from which we obtain 
 P Hm 
 
 for the stress in portal member Bo. This stress is tension, as is readily 
 seen, as it acts away from the end post at point B, that is, the member 
 Bo pulls on the end post at point B and hence is in tension. 
 
 Resolving (at point t) the stress S3 into two components h' and /', 
 nnd taking moments about B, we have 
 
366 STRUCTURAL ENGINEERING 
 
 from which we obtain 
 
 and multiplying by sec we have 
 
 S3 = h' sec = ^-^sec .............................. (7) 
 
 for the stress in portal member ot. 
 
 By considering point B (Fig. 273) as the center of moments, the 
 stress S3 is seen to be compressive, as // acting to the left requires the 
 member ot to act to the right against the end post, and hence the member 
 ot is in compression. 
 
 As is seen by comparison, (4) and (7) are exactly alike. That 
 means that the stress in po is equal to the stress in ot, but, as stated 
 above, one is tension and the other is compression. 
 
 Now, having derived a formula for the stress in each member of the 
 portal, we will proceed with the determination of the stress in the mem- 
 bers. To do this it is first necessary to determine the value of P, R, H, 
 L, m, k, and sec 0. P is given in the last article as 5,000*. R is equal 
 to the shear in the panel BC (Fig. 269) when all the panels are loaded. 
 Beginning at the center of the span, we have one-half of a panel load at 
 D and a full panel load at C, making in all one and one-half panel loads 
 as the shear in the panel BC, hence we have 
 
 R = ^9. + 5,000 = 7,500 Ibs. 
 
 
 for this shear. Then from (1) we have 
 
 H= 7,500 + 5,000 =6)2501bs- 
 4 
 
 L is about 39 ft., as the height of the span is 30 ft. and the panel length 
 is 25 ft. It will be close enough, for designing, to take B as 45 degrees. 
 Then, A: = 8 ft., m = 31, and sec 0=1.4. Now, using the above values, 
 from (3) we obtain 
 
 = 7,500+^^ + 6,250x^ = 34,218, say 34,000 Ibs. 
 
 A O 
 
 (compression) for the stress in portal member bo. From (4) or (7) 
 we obtain 
 
 Sl = S3= 6 ^ X39 x 1.4 = 42,655, say 43,000 Ibs. 
 o 
 
 for the tensile stress in portal member po and compressive stress in ot. 
 From (6) we obtain 
 
 6,850x31 
 
 j _ 
 
 a O 
 
 for the tensile stress in the portal member Bo. By comparing (3) and 
 (6) it is seen that the stress in member Bo differs from the stress in 
 member bo bv the value of R. 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 
 
 367 
 
 h 
 
 (OJ 
 
 F 
 
 (b) 
 
 This completes the calculations for the stresses in the portals under 
 the assumption that the bottom ends of the end posts are hinged, and 
 we will now determine the stress in the portal, assuming the bottom ends 
 of the end posts fixed. To illustrate conditions, let e and g at (a) (Fig. 
 274) represent two thin wooden strips connected rigidly by another piece h. 
 If this frame be supported as shown at (6) and a force F be applied, 
 the pieces e and g would bend as indicated at (6), and we thus have 
 an illustration of the case just considered 
 where the end posts were assumed hinged at 
 the bottom ends. Now, suppose the bottom 
 ends of the pieces to be buried in concrete as 
 indicated at (c) and the concrete be allowed 
 to set and the force F be then applied. The 
 pieces e and g would then bend as indicated at 
 (c), and we thus have an illustration of the 
 case to be considered where the bottom ends 
 of the end posts are assumed to be fixed. The 
 points o' are the points of contra-flexure, where 
 the bending moment on the pieces e and g is 
 zero, as is readily seen from the sketch. The 
 points o f are at mid-distance from the con- 
 crete to the piece h, or m/2 from h, neglecting 
 the portion buried in the concrete as being too 
 small to materially change the length of the 
 part marked m. 
 
 Now, as there is no bending at the points 
 o' (points of contra-flexure), the frame can be 
 assumed as cut off at those points and we have 
 the independent structure shown at (d). From 
 this it is seen that the problem of determin- 
 ing the stresses in portals, when the bottom 
 ends of the end posts are considered fixed, is 
 the same as in the case of hinged ends, except 
 the distance m (Figs. 271 and 273) is one- 
 half as great, and hence the above formulas for 
 determining the stresses in the portals are ap- 
 plicable if m/2 be substituted for m and 
 (L-m/2) or (& + ra/2) for L. In an imaginary sense, we simply move 
 the bottom supports of the end posts half way up their unsupported 
 length, and proceed with the determination of the stress in the portals in 
 the same manner as if the bottom ends of the end posts were hinged. 
 Thus substituting in (3), we have 
 
 = 7,500 + M^1 + 6,250x^ = 22,109, say 22,000 Ibs. (compression) 
 
 F 
 
 'V 
 ^ 
 ^ 
 
 
 -^=-=3. 
 
 o' 
 
 
 0' 
 
 
 
 
 \\ 
 
 \\\ 
 
 \N\\\\\\\\\V 
 
 VW 1 
 
 (C) 
 
 Fig. 274 
 
 for the stress in portal member bo (Fig. 271), assuming the bottom ends 
 of the end posts fixed; and substituting in (4), we have 
 
 6180 * (30 -V) 
 
 8 
 
 SI = 
 
 = 25,703, say 26,000 Ibs. 
 
 for the tensile stress in portal member po and also for the compressive 
 
368 STRUCTURAL ENGINEERING 
 
 stress in portal member oi y and likewise substituting in (6) we have 
 
 S2 = P *- a Mgi + m; - V = 14 , 609 , say 15 ,ooo lbs . 
 
 6 K O 
 
 for the tensile stress in portal member Bo. Thus we have all the stresses 
 in the portals determined when assuming the bottom ends of the end posts 
 to be fixed. These stresses will be used in designing the sections of the 
 portal members as the bridge considered is designed for a heavy loading; 
 and we have provided end floor beams, consequently the bottom ends of 
 the end posts will be fixed quite securely. 
 
 Each member of the portals will be as deep as the end posts and 
 composed of two angles, latticed together vertically, similar to the top 
 laterals. 
 
 As seen above, the members are subjected to one kind of stress 
 when the wind load comes from one direction and just the reverse when 
 it comes from the opposite direction, that is, each portal member is sub- 
 jected to reversal of stress, and hence the members must be designed to 
 carry both compression and tension. 
 
 The members po and ot (known as knee braces) are each subjected 
 to 26,000* stress, considered to be either tension or compression. Taking 
 the case of tension first, we have 
 
 26,000 
 
 for the required net area of cross-section of each, which shows that two 
 very small angles could be used (theoretically) as far as tension is con- 
 cerned. As for compression, the length of each member is about 11.3 ft. 
 (0 = 45), but as the redundant members (those dotted, in Fig. 271) sup- 
 port each at mid-point, the length L can be taken as 11.3/2 = 5.6 ft., say 
 66 ins. As the members are subjected to reversal of stress, L/r should 
 be reasonably low, say 80. Then we have 
 
 , 
 
 for the required value of the radius of gyration. 
 
 Now, glancing over the table of angles (Tables 5 and 6) it is seen 
 that this calls for very small angles. So the case here is the same as 
 that of the top lateral struts very much a matter of judgment. 3" x 
 3J"xf" angles are as small as are used in this class of work (as it is 
 desirable to use J" rivets throughout), so we will try that section. Sub- 
 stituting the radius, found in Table 6, for this angle in the column 
 formula, we have 
 
 16,000-70^ =11,682 lbs. 
 
 for the allowable unit compressive stress in each of the members po and 
 ot. Dividing this into the stress, we have 
 
 26,000 -r 11,682 = 2.22 sq. ins. 
 
 for the required area, while 2 Ls 3" x 3" x f " have 4.96 n ", more than 
 twice the amount required for compression, and, as found above, only 
 1.62 n " is required for tension. So it is seen that 3" x 3" x f " angle/? 
 
DESIGN OF SIMPLE RAILEOAD BKIDGES 
 
 369 
 
 are much heavier than required, theoretically, yet smaller angles would 
 give the portals a flimsy appearance even if satisfactory details were 
 obtained by their use. So we are now to the point where the selection 
 must be made mostly upon appearance. As the stress in the members 
 bo and Bo is less than in members po and ot, and their lengths are also 
 less, it is not necessary to compute the required sections for those mem- 
 bers, but use the same as used for members po and ot. As the 3J" x 3 J" 
 angles are our minimum size we could use two of these for each portal 
 member, but for the sake of details and appearance we will use 2 Ls 
 4"x4"xf" for each of the main members and 2 Ls 3J" x 3$" x f" for 
 each of the redundant members, thus giving the main members a dis- 
 tinctive outline. 
 
 In the case of large bridges these members can be designed theoretic- 
 ally correct, as the required area of cross-section in those cases is such 
 that sections can be found which are both theoretically and practically 
 correct, but in the case of small bridges it is very much a matter of 
 judgment, as is seen from the above calculations. 
 
 180. Design of End Posts. The designing of the section of the 
 end posts is here taken as a special problem for the reason that, unlike 
 the other main truss members, it is subjected to bending stress in addi- 
 tion to direct stress. To correspond with the top chord, it will be made 
 of 18" webs, 23" cover plate and four angles. The length of the end 
 
 post, considering the x-x axis (see Fig. 265) is the total length L (shown 
 in Fig. 275 to equal AB) unless a collision strut OM be inserted, in which 
 case the length is L/2 and the length, considering axis y-y, is equal to m. 
 The collision strut should be used as a safeguard in case a derailed train 
 should strike the end post, and therefore we will use one in this case. 
 Then the length of the post in reference to the x-x axis is 1 9.5 ft. or 234 
 ins., and 31 ft. (=ra) or 372 ins. in reference to axis y-y. 
 
 Using the radii found for top chord be (Art. 176), which will be 
 about the same in this case, we obtain 
 
 . 
 16,000 - 70 ^- = 1 3,750 Ibs. 
 
 i .^O 
 
 for the allowable unit direct compressive stress in the end post, consider- 
 
 ing axis x-x, and 
 
 379 
 16,000- 70 = 12,678 Ibs. 
 
370 STRUCTURAL ENGINEERING 
 
 for the allowable unit direct compressive stress considering axis y-y. 
 Now dividing the latter, as it is the smaller, into the totnl direct stress 
 in the end post, as given in Fig. 262, we obtain 
 " 523,000 . 
 
 for the required area of cross-section of each end post, providing there 
 were no cross bending, due to the wind load, on the posts. But as the 
 bending stress on the compression side of the posts adds to the direct 
 stress, the two unit stresses must be combined. However, in case these 
 stresses be combined, the allowable unit stress can be increased 25 per 
 cent, so the above area may be about correct. As preliminary let us try 
 the section used for top chord be, which contains 40.5S n ". 
 
 In determining the bending moment on the end posts, which is due 
 to wind and consequently a transverse moment we can consider each 
 post as composed of two cantilevers as indicated at (a) in Fig. 275. 
 Then the moment on each post at any point either above or below the 
 point of contra-flexure (o') is Hx. Now it is seen that this moment is a 
 maximum at the points p and t and at the bottom of each post, and that 
 the maximum in each case is equal to Hm/2. As the points p and t are 
 the farthest out from the ends of the member, where the column formula 
 really applies, we will consider the bending at those points only. From 
 the last article, we have H = 6,250 Ibs. and m = 31ft. or 372 ins. Then 
 for the moment at p or t (Fig. 275 (a)) we have 
 
 M = 6,250 X~^= 1,162,500 inch Ibs. 
 
 a 
 
 Now using the moment of inertia found in Art. 176 for chord be in 
 reference to axis y-y and taking the horizontal distance to the extreme 
 fiber as 7.75 + 4=11.75" (see Fig. 265), we obtain 
 
 for the unit stress due to cross bending, assuming that the same section 
 as used for chord be is used for the end post. 
 
 Now dividing the area of the section into the total direct stress, 
 we have 
 
 . (about) :'- 
 
 for the actual direct unit stress, and adding this to the unit stress due 
 to cross bending we have 
 
 5,470 + 12,880 = 18,350 Ibs. 
 
 for the combined unit stress. The allowable unit stress as given above is 
 12,678. But this can be increased 25 per cent in the case of combined 
 stress, as per specifications. So we have 
 
 (12,678) 1.25 = 15,847 Ibs. 
 
 for the actual allowable combined unit stress, and as it is less than the 
 above (18,350#) a larger section will have to be used for each end post 
 than was just considered. 
 
DESIGN OF SIMPLE EAILEOAD BKIDGES 
 
 371 
 
 Let us assume the following sections: 
 
 1 cov. pi. 23"x|" = 11.50 n " 
 2 web pis. 18" x4" = 18.00" 
 2 Ls 3J" x 3V x |" = 4.96 n// 
 2 Ls 6"x4"xf" =11.72 n " 
 
 46.18" 
 
 As the post is stronger in reference to the x-x axis than in reference to 
 the y-y axis, owing to the length being greater in the last case (using 
 the collision strut), we need consider the above section only in reference 
 to the y-y axis. 
 
 O J 
 
 For the moment of inertia about the y-y axis (see Fig. 276), we have 
 cov. pi. 507 + =507 (about) 
 
 web pis. + 7^62 2 x 9 x 2 =1,045 
 
 top angles 2.87 x 2 + 88 2 x 2.48 x 2 = 397 
 bot. angles 7.52 x 2 + SlKJ^x 5.86 x 2 = 943 
 
 2,892 
 Then for the radius of gyration about axis y-y we have 
 
 Now substituting this radius in the column formula, we have 
 p = 16,000 - 70 JEU 12,710 (about) 
 
 for the allowable unit stress for direct compression, 
 and increasing this 25 per cent we obtain 15,890* 
 for the allowable unit stress in the case of direct 
 compression and bending stress combined. Then, 
 using the last assumed section, we have 
 
 11,320 Ib, (about) 
 
 for the actual direct unit compressive stress, and 
 
 1,162,500x11.75 
 - 
 
 Fig. 276 
 
 for the unit stress due to bending. Now, adding these two stresses 
 together, we have 11,320 + 4,730 = 16,050* for the combined unit stress, 
 which is only 160* (=16,050-15,890) greater than the 15,890* allowed, 
 and as this is about as close as we can obtain we will use the above 
 section for each end post. 
 
 181. Designing of Collision Struts. These members should at 
 least have sufficient section to resist the thrust that an ordinary loaded 
 car moving at an ordinary rate would exert upon them in case the car 
 were derailed. The problem involved is of such a nature that the entire 
 
372 
 
 STRUCTURAL ENGINEERING 
 
 premises must rest upon practical assumptions. We will assume that the 
 car and its load weigh 100,000* and that it has a velocity of 30 ft. per 
 second, which is a little over 20 miles per hour, and, further, we will 
 assume the car brought to rest in one second. Then substituting in 
 Formula A (Art. 23), we have 
 
 =93,750, say 94,000 Ibs., 
 
 for the constant force required to stop the car in one second. But at 
 the beginning of contact the force would be zero and a maximum when the 
 car was brought to rest, so the maximum force exerted by the car would 
 really be twice the constant force required to stop it, or 94,000x2 = 
 188,000*. This force would be exerted on the end post at a point about 
 9.5 ft. above the bottom chord (see Fig. 277), so that the horizontal force 
 exerted against the collision strut would be about 9.5/15 of the 188,000, 
 or 119,000*. 
 
 Then multiplying one-half of this by sec /?, which we will take as 
 1.56, (3 being the complement of 0, we have 
 
 119,000 
 
 Xlt56 = 92^820, say 93,000 Ibs. 
 
 for the stress in the collision strut. It may appear that the collision strut 
 should connect to the end post at the point where the 188,000* is applied, 
 
 but this is not the case, for if the blow 
 of 188,000* were struck at the point 
 where the strut connects to the end 
 post there would be but little resilience 
 and, consequently, the structure would 
 be subjected to severe shock. 
 
 It is seen from the amount of stress 
 that the section of the collision struts 
 need not be large as far as direct stress 
 is concerned, but the value of L/r should 
 not exceed 80 in order to obtain rigidity. 
 The best section for these struts is two channels latticed together, and 
 as 8" channels are about the smallest used in this class of work we will 
 tr y 2 [s 8" x 16.25*. The length of each strut, center to center of 
 end connections, is about 19.5 ft., or 234 ins., and, as seen from Table 
 3, the radius of gyration of the channels is 2.89, so we have 
 
 Fig. 277 
 
 234 
 
 2.89 
 
 = 81. 
 
 Then for the allowable unit compressive stress on each strut we have 
 
 p = 16,000 - 70 x 81 = 10,330 Ibs. 
 Now, dividing this into the stress, we have 
 
 93,000 
 
 for the section required. The area of the section of the two channels 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 373 
 
 assumed, as seen from Table 3, is 9.56 n ", which is 0.56 n " more than the 
 required area,, but we will use this section as it is about as close to the 
 required area as we can come and at the same time obtain good details. 
 
 The above analysis is only an attempt to provide for reasonable 
 accidents, as we might term it. The actual intensity of the blow struck 
 by a derailed car is distinctively problematic for the reason that there 
 could be so many different conditions assumed. As an illustration, sup- 
 pose a derailed car in the middle of a train should strike the end post 
 of a bridge. How much would the cars in the rear of the derailed car 
 increase the blow? And how much would the engine pulling on the 
 train increase it? And how much would the resistance of the wheels of 
 the derailed car bumping over the ties diminish it? And, further, sup- 
 pose a train traveling 50 miles per hour should jump the track just as 
 it gets to the bridge and the engine should strike the end post. There is 
 little doubt but that the bridge would be destroyed, yet it might glance 
 off and do but little damage. It is obvious that it would be absolutely 
 impracticable to provide for resisting the blow in the last case ; and prac- 
 tically the same is true in several cases that could be assumed, yet it 
 would not be good engineering to make no provision for reasonable acci- 
 dents, as past experience has taught us, and that is what we have 
 endeavored to do above. 
 
 182. Maximum Reaction on Shoe. For the dead-load reaction on 
 each shoe we have one-half of a panel load from D (Fig. 249), a full 
 panel load from B and C each, and one-half (so considered) of a panel 
 load from A, making in all three panel loads on each shoe. Then taking 
 the dead-load panel load as found in Art. 173, we have 
 
 12-3x26,375 = 79,125, say 80,000 Ibs. 
 
 for the dead-load reaction on each shoe. 
 
 As is evident, the maximum live-load reaction on a shoe will occur 
 when the span is fully loaded and the heaviest wheels are near the shoe 
 
 Fig. 278 
 
 and one directly over it that is, over the end floor beam. So, placing 
 the live load as shown in Fig. 278 with wheel (2) at A and taking 
 moments about G, using Table A, we have 
 
 15,274 + 274 x 49 + 49^ 50 
 
 R' = 
 
 150 
 
 for the maximum live-load reaction on the shoe at A, which is the same 
 as for the others, and, for the impact, we have 
 
 259,000 x 
 
 , say 173,000 Ibs. 
 
1J: 
 
 
 :" 
 
 I: 
 
 ilf*3S J 33 > ^ 
 
 -!-i 1^ 1^^?^ .v 
 
 l.S l ^ nrd3 I? 
 
 374 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 375 
 
 Now adding the above dead, live, and impact reactions together, we have 
 
 80,000 + 259,000 + 173,000 = 512,000 Ibs. 
 for the maximum reaction on each shoe. 
 
 183. Stress Sheet. After having completed the calculations as 
 above for the span the "stress sheet," Fig. 279, can be made. This draw- 
 ing contains all the information resulting from the calculations; in fact, 
 the drawing is really a summary of the calculations. After the stress 
 sheet (Fig. 279) is completed the detail drawings can be made. 
 
 184. Calculations of Details. In starting the detail drawings, a 
 large scale pencil sketch (1J", 2" or 3" scale) of each joint should be 
 made first. Such a drawing for joint LO is shown in Fig. 282. The 
 calculations of the details are made as the sketches are being drawn. 
 
 Joint LO. Taking first the case of joint LO (Fig. 282), the first 
 thing to do is to locate the center line of the pin in the end post. The 
 section of the end post is given on the stress sheet (Fig. 279), also in 
 Art. 180. 
 
 Taking moments about the center of the cover plate of the section 
 (see Fig. 280) we have 
 
 - 1.26 x 4.96 + 18 x 9.37 -f 11.72 x 16.47 '367.9 
 
 x = 
 
 46.18 
 
 46.18 
 
 = 7.96 ins. 
 
 for the distance from the center of the cover plate down to the gravity axis 
 x-x of the end post. Now, as far as the direct stress in the end post is 
 concerned, the pin could be properly placed on this gravity axis x-x; but 
 the member tends to bend downward as shown in Fig. 281, owing to its 
 own weight, and it is necessary to place the pin a short distance below 
 the gravity axis so as to counterbalance this bending. The pin being 
 placed below the gravity axis x-x, the direct stress tends to bend the 
 member upward as indicated by the dotted line, and, as is evident, if the 
 pin be placed just the correct distance below the gravity axis the end post 
 will be straight when the maximum stress in it occurs. For the weight of 
 the end post per foot of length we have 
 
 a 
 
 NJ 
 
 v..^ -t 
 
 If 
 
 " 
 
 
 0) 
 
 '* 
 
 
 ^VVV.V'A , 
 
 Fig. 28U 
 
 Fig. 281 
 
 W = 46.18 n " x 3.4# = 157 Ibs., 
 
 and increasing this one-third to provide for details we have 210* for the 
 total weight of the member per foot of length. This weight acts vertically 
 and, as the member is inclined, only the component perpendicular to the 
 member causes transverse bending. So, for the bending moment, we have 
 
Fig. 282 
 
 376 
 
DESIGN OF SIMPLE KAILEOAD BRIDGES 377 
 
 M = J x 210 x sin<9 x 19J? x 12 = 76,658 inch Ibs. (Sin0 = 0.64.) 
 Now by placing the pin a distance z below the gravity axis x-x, we 
 have 2 x 523,000 for the moment caused by the direct stress which tends 
 to bend the end post upward, and hence the end post will be straight 
 when 2 x 523,000 = 76,658, from which we obtain 
 
 . 
 
 which is the distance that the pin should be placed below the gravity axis 
 x-x. Now adding this to izT (given above) we have 7.96 + 0.1466 = 8.1066" 
 for the distance from the center of the cover plate down to the center of 
 the pin, and substracting one-half of the thickness of the cover plate from 
 this we have 8.1066 - 0.25 = 7.8566" for the distance from the under side 
 of the cover plate to the pin. The distance from the under side of the 
 cover plate to the working line in the top chord, which would be the cen- 
 ter line of the pins if the top chord were pin-connected, should be the 
 same as for the end post in order to obtain uniform construction of these 
 members, and hence at this juncture it is necessary to examine the top 
 chord sections and select a distance suitable for all top chord members 
 and end posts. 
 
 Considering the case of the top chord section U2-U3 (Fig. 279), and 
 taking moments about the cover plate as was shown above in the case of 
 the end post we obtain 
 
 - 1.26 x 4.96 + 18 x 9.37 + 10.62 x 16.49 
 ' = - 451)8 -- = 7 ' 76 ^ 
 
 for the distance from the center of the cover plate down to the horizontal 
 gravity axis. For the weight of the section we have 
 
 45.08 x 3.4 = 153 Ibs. 
 
 per foot and adding one-third for details we have 204# for the weight of 
 the member per foot of length. Then we have 
 
 M = J x 204 x 25 2 x 12 = 191,000 Ibs. (about) 
 
 for the bending on the chord due to its own weight, and for the distance 
 z' below the horizontal gravity axis, where pins would have to be placed 
 to balance the above moment, we have 
 
 191,000 
 
 Adding this to the above 7.76" and subtracting one-half of the thickness 
 of the cover plate we have 
 
 7.76 + 0.325 - 0.25 = 7.84 ins. 
 
 for the distance down from the under side of the cover plate to the work- 
 ing line. In the same manner we find that the corresponding distance in 
 the case of chord 71-172 is 7.75". So it is seen that 7" is about the cor- 
 rect value to use for the distance from the under side of the cover plate to 
 the pin center or working line, as the case may be, and hence this distance 
 will be taken in the case of all top chord sections and end posts throughout 
 the structure. Thus we have the pin LO located and the outline of the 
 portion of the end post shown (Fig. 282) can be drawn and the rivets 
 
378 STRUCTURAL ENGINEERING 
 
 spaced transversely in it as shown and then the drawing of the other 
 details can proceed. The end post should extend far enough beyond the 
 pin so that the bottom chord and end floor beam connections balance fairly 
 well about the pin. This part is at first determined merely by appearance. 
 After this the shoe can be sketched so as to clear the end post, the number 
 of rollers required is determined by sketching roughly their length, and at 
 the same time the length of the pedestal can be determined. After this is 
 done the required pin bearing on the end post can be figured. This bear- 
 ing must be sufficient to carry the maximum vertical reaction, which is 
 512,000* (see stress sheet, Fig. 279). Let us assume a 5" pin. Then 
 we have 
 
 t = 512,000 -r (5J x 24,000) = 3.87 ins. 
 
 for the required thickness of bearing on the two sides of the end post or 
 1.93" (= 1-Jf ) for each side. As shown, we have a f " gusset plate, \" 
 web, f " filler, and a T V' plate, making in all, f + i + f + iV = ^iV* which 
 is \" thicker than required. But this is about as close as we can come, 
 as the gusset plate should be f " thick (as will be seen later) and the f" 
 filler can not be reduced, as the bottom angles on the end post have that 
 thickness and the iV plate is as thin as should be used, owing to the coun- 
 tersunk rivets ( J" rivets should not be countersunk' in metal less than 
 f$" thick). So we will use the metal shown. There should be enough 
 rivets above the pin to hold these plates. First, there should be enough 
 rivets connecting the f " filler and the $" plate to the end post and gus- 
 set plate to transmit the pin pressure coming on the two. For this pres- 
 sure we have (f + T ^) 5J x 24,000 = 140,000* (about). The rivets are f" 
 shop rivets in single shear, each of which is good for 12,000 x 0.6 = 7,200*. 
 So we have 140,000 -=-7,200 = 19.5, say 20 rivets. We have just about 
 20 rivets above the pin, so the detail is satisfactory so far. The rivets in 
 the y filler are not included, as they simply hold that narrow filler. 
 There should be enough rivets in the -fa" plate to take the pin pressure 
 upon it, which is & x 5J x 24,000 = 57,750*. Dividing this by 7,200 we 
 have 57,750 -r 7,200 = 8 rivets whereas we have over twice that number, 
 and that part is amply strong. To be on the safe side there should 
 be enough rivets connecting the end post to the gusset plate to trans- 
 mit, in single shear, one-half of the stress in the end post, which is 
 523,000 -=- 2 = 261,500 (see Fig. 279). Dividing this by 7,200 we have 
 261,500 -=-7,200 = 36.3, say 37 rivets. We have about 38 above the pin, 
 so that part of the detail is satisfactory. We will next consider the bot- 
 tom chord connection. The net area of the bottom chord is 20. 9 n " (see 
 Fig. 279), and for its strength we have 20.9 x 16,000 = 334,400*. Then 
 as the rivets are I" field rivets, we have 334,400 -=- 6,000 = 55.7, say, 56 
 rivets, or 28 on each side. We have 30, so that detail is correct. 
 
 The end floor beam is cut, as shown (Fig. 282), to clear the shoe, pin, 
 and the details on the end post. Dividing the end shear on the end floor 
 beam (see Fig. 279) by 6,000 we have 143,200 -=-6,000 = 23.8, say 24 
 rivets for the number of J" field rivets required in each end connection. 
 As is seen, 24 rivets are used. The thickness of pin bearing on the shoe 
 need be (theoretically) but 1-Jf " on a side, as found above for the end 
 post, but there should be a point of bearing on each side of each web of 
 the end post (as shown) so as to distribute the pressure well over the 
 
DESIGN OF SIMPLE KAILEOAD BRIDGES 379 
 
 rollers and at the same time lighten the stress on the pin, and as every 
 part of such castings should be at least 1J" in thickness, we will make 
 each of these bearings 1J" thick, although we obtain an excess of bearing. 
 However, as the shoe is subjected to both bending and shear from the 
 direct load (see Art. 141) and wind, the metal is not so overly excessive 
 as at first appears. 
 
 Assuming one-fourth of the maximum reaction on the shoe, which is 
 equal to 512,000 -f 4 = 128,000*, as coming on each bearing of the shoe, 
 and taking moments about the center of bearing of the end post, we have 
 
 128,000 x 2f = 304,000 inch Ibs. 
 
 for the maximum bending moment on the 5^" pin. The lever arm 2f" 
 can be either computed or scaled from the drawing. For the bending 
 stress on the pin, due to this moment, we have 
 
 / =/-/r^/^4 = 18 ; 600 lbs - P er S <1- in - (about) 
 
 which shows the pin to be amply strong, as 25,000* per square inch is 
 allowed. It requires a moment of 408,350''* to stress the 5J" pin 25,000* 
 per square inch. The maximum bending moment allowed on the different 
 sizes of pins is to be found in tables of practically all structural hand- 
 books, and, in designing pins, usually these tables are consulted instead 
 of computing the fiber stress as is done above. 
 
 For the maximum shearing stress per square inch on the pin we have 
 
 128,000 Trr 2 = 128,000 -=- 23.75 = 5,390 Ibs. per sq. in., 
 
 which is quite low, as 12,000* is allowed. As seen from above, the pin is 
 really larger than required and, theoretically, should be made smaller; 
 but if reduced in size the bearing would increase, little would be saved, 
 and, judging from general appearance, a 5J" pin is as small as should 
 be used in this class of work; hence the assumed 5J" pin will be used. 
 
 Six-inch rollers are the minimum size permitted by the specifications, 
 and as that size appears to be about the correct size for this case, they 
 will be used. Then, for the linear inches of roller required, we have 
 
 512,000 
 
 600x6 
 
 = 142.2 ins. 
 
 Now, making each roller 30" long and deducting for the 7 " slots in 
 pedestals, we have (30 - 7 X f ) x 6 = 148.5", which is a few inches more 
 than is required, but is about as close as we can come and at the same time 
 obtain good details all around. So 6 6" rollers, each 30" long, as shown, 
 will be used. The rollers are notched J" into the cast shoe and pedestal, 
 so as to prevent their sliding transversely (due to wind), and the side 
 bars on their two ends control their relative longitudinal motion. It is 
 necessary that the pedestal and the base of the shoe project to clear the 
 side bars and bolt heads. In some cases it is necessary for the pedestal 
 and the base to project out farther to obtain sufficient bearing on the 
 masonry. In this case we have 
 
380 STRUCTURAL ENGINEERING 
 
 for the required area of bearing and we have, as shown, (3' 4J") x 
 (V - 4J") = l,154 n ", which is about 300 n " more than required, but the 
 roller nest requires about the same size of pedestal as shown and there 
 would be but little saving, if any, in reducing it to exactly the theoretical 
 size. So the size shown will be used. 
 
 In drawing the lateral connection at LO, we first determine the num- 
 ber of J" field rivets necessary to develop the lateral in tension. The 
 lateral is a 5" x 3^" x f" angle, which has a net area of 5.06 n " (allowing 
 for 1 V rivet hole). Multiplying this area by 16,000 and dividing by 
 6,000 we obtain 13.5, say 14 rivets. As is seen, 14 are used. The bolts 
 connecting the ^" lateral plate to the shoe should be sufficient to carry 
 the component (along the end floor beam) of the stress in the lateral. 
 This component is equal to 81,000 -f secco = 81,000 -r 1.85 = 43,800* (about) 
 (see Art. 177). Dividing this by 6,000 we obtain 7.3 for the number of 
 J" turned bolts required. As is seen, 7 are used. Part of the component 
 (43,800*) may be transferred along the floor beam to the other shoe, in 
 which case less than 7 bolts would be required ; but by using the 7 we are 
 sure that the detail is sufficiently strong. 
 
 For the component of the stress in the lateral along the bottom chord 
 we have 
 
 81,000 xsin<o = 81,000x0.8418 = 68,200 Ibs. (about). 
 
 Dividing this by 6,000 (the allowable shear on a J" field rivet) we obtain 
 11.3, say 12 rivets. We have, as shown, 14. However, some of these 
 are also needed to transmit the direct stress of the bottom chord into the 
 gusset plate, but as the maximum stress in the bottom chord and lateral 
 are not likely to occur at the same time, the detail as shown will be 
 considered sufficient. The shop rivets connecting the 2 4" x 4" angles 
 to the T V' lateral plate should be sufficient to transmit the 68,200* com- 
 ponent in bearing on the iV' plate. 
 
 One J" rivet at 24,000* bearing on a T V' plate will transmit 
 x T 7 F x 24,000 = 9,188*. Then we have 68,200 -f- 9,188 = 7.4 for the num- 
 ber of rivets required, and, as is seen, 8 are used. The number of rivets 
 required to connect the %" lateral plate to the bottom of the end floor 
 beam should be sufficient to take at least one-half of the component along 
 the floor beam of the stress in the lateral. This requires about 4 rivets, 
 but to obtain a good, rigid connection, more are needed. So 8 are used, 
 wherein we are governed very much by appearance and sense of fitness. 
 
 The 3J" longitudinal holes in the cast-steel pedestal should be cored 
 in casting and the J" slots cut at the machine shop. If the slots were to 
 be made in casting, the pedestal would be likely to warp in cooling. These 
 slots and holes in the pedestal are for two purposes : one is to save metal ; 
 and the other one is to permit dust and dirt that would accumulate between 
 the rollers to fall down into the holes. In other words, this arrangement 
 permits the dirt that is blown into the roller nest to be blown out, and thus 
 the roller nest is not likely to become clogged with dirt and rust. The 
 side bars at each end of the rollers should be at least ff" thick to be of 
 much service. 
 
 Joint LI. Taking the next lower chord joint LI (Fig. 283), the 
 cross-section of the floor beam and end connection angles of the same can 
 be drawn as shown, and the outline of the bottom chord collision strut and 
 
381 
 
382 STRUCTURAL ENGINEERING 
 
 the hanger can be sketched care being taken to provide about \" clear- 
 ance between these members. Then the rivet lines can be drawn and the 
 rivets spaced as shown. The stress in the collision strut as given in Art. 
 181 is 93,000*. Dividing this by 6,000* we obtain 15.5, say, 16 " 
 field rivets or bolts, as bolts should be used in this particular connection. 
 As is seen, 16 are used. Dividing the maximum end shear on the floor 
 beam, which is given on the stress sheet (Fig. 279) as 187,300, by 6,000 
 we obtain about 31 rivets for the number of J" field rivets required to con- 
 nect the floor beam to the truss. All of these should be above the bottom 
 chord, as there is no diaphragm in the bottom chord to transmit the 
 pressure across to the other side of the hanger and, consequently, the 
 rivets shown connecting the floor beam to the bottom chord must not be 
 considered to carry any of the end shear on the floor beam. As is seen, 
 there are 32 rivets connecting the floor beam directly to the hanger, 
 which is really one more than required theoretically. 
 
 After spacing the rivets in the collision strut and floor beam, the 
 gusset plate can be drawn to suit this spacing as shown. The rivets 
 connecting the gusset plate to the bottom chord should at least be suffi- 
 cient to transmit the component of the stress in the collision strut which 
 is equal to 93,000 x sin0 = 59,500*. Dividing this by 7,200 we obtain a 
 little over 8 J" shop rivets, or 4 on a side, whereas 12 (not including 
 the field rivets passing through the floor beam) are used, but the detail 
 is satisfactory, as this number is necessary to obtain a well-balanced 
 joint. Next, the bottom lateral connections can be drawn. The number of 
 rivets required in the lateral to the left of LI is 14 J" field rivets, as 
 determined at LO, and the detail can be drawn as shown. The lateral on 
 the right-hand side of LI (Fig. 283) is made of 1 5" x 3J" x J" angle, 
 which will have a net area of 3.5 n ". Then for the strength of this lat- 
 eral we have 3.5x16,000 = 56,000*. Dividing this by 6,000 we obtain 
 about 9 J" field rivets. This, as is seen, is the number used. The num- 
 ber of rivets connecting the lateral plate to the floor beam should be suffi- 
 cient to transmit the component of the lateral to the left of LI along the 
 floor beam. This component is equal to 5.06 x 16,000 x cosw = 80,960 x 
 0.5397 = 43,700*. Dividing this by 6,000 we obtain about 7 I" field 
 rivets, whereas 10 are used. After the rivets are spaced in the laterals 
 the lateral plate can be drawn to suit; that is, so the edges at no point 
 will be less than 1-J" from the rivets. Then the rivets connecting the lat- 
 eral plate to the floor beam can be spaced as shown. 
 
 Joint L2. We will next consider lower chord joint L2 (Fig. 283). 
 The floor beam connection at this joint must be exactly the same as at 
 LI, as the floor beams should be the same throughout. However, if the 
 spacing established at LI does not suit at L2, it must be changed so it 
 will be satisfactory for the two joints and, likewise, for the other joints. 
 In each case there must be 32 rivets above the chord as was determined 
 at LI. After drawing the cross-section of the floor beam and the end 
 connection, as shown (Fig. 283), the outlines of the bottom chord and 
 diagonal can be drawn. The net area of the main diagonal, as given on 
 the stress sheet (Fig. 279), is 23.9 n ". Multiplying this by 16,000 and 
 dividing by 6,000 we obtain about 64 J" field rivets for the end con- 
 nection, or 32 on a side, which, as is seen, is the number used. There 
 should be a sufficient number of rivets connecting the gusset plates to 
 
DESIGN OF SIMPLE KAILKOAD BEIDGES 383 
 
 the post to transmit the vertical component of the stress in the diagonal, 
 which is equal to 23.9 x 16,000 -=-sec0 = 294,000*. Dividing this by 6,000 
 we obtain 49 J" field rivets or, say, 24 on a side. As is seen, 24 on a 
 side are used. 
 
 There should be a sufficient number of rivets to the left of the floor 
 beam connecting the gusset plate to the chord (at L2) to transmit the 
 horizontal component of the stress in the diagonal. For this component 
 we have 23.9 x 16,000 x sin(9 = 244,700*, and dividing by 7,200 we obtain 
 34 y rivets (shop), or 17 rivets on each side of the chord, whereas 16 
 are used, but the rivets passing through the floor beam and those to the 
 right of the floor beam can be counted on for more than making up for the 
 one rivet missing. 
 
 The splice in the bottom chord just to the left of L2 should be suffi- 
 cient to develop the strength of the bottom chord member Z/l-Z/2. The 
 chord is spliced, as is seen, by the 12" x T V inside plates, by the 12" X J" 
 outside splice plates and the f " tie plates. The rivets through the web 
 of the channels to the left of the splice, which are field rivets, are in double 
 shear and bearing on the web of the channels. The strength of the chan- 
 nels is equal to 20.9 x 16,000 = 334,000*, or 167,000* for each channel. 
 Then considering one channel, the 16 field rivets in the web are good for 
 192,000* (=2x6,000x16) in double shear, and 145,600* (=0.52 xjx 
 16 x 20,000) in bearing on the web of the channel; and hence the latter 
 governs. The 6 rivets in the flanges of the channel are good for 
 6x6,000 = 36,000*. Adding this to the 145,600* we have 181,600*, 
 which is 14,600* more than is necessary to provide for ; or, in other words, 
 there is at least one rivet too many; but the rivets in the tie plates are none 
 too numerous to hold those plates and to omit one rivet in the web would 
 leave the spacing unbalanced; that is, unsymmetrical, and to omit two 
 would be too much of a reduction, so the rivets as shown will be used. 
 There should be enough rivets in the 12" x \" splice plate on each side 
 of the splice to develop the strength of that plate. The net section of the 
 plate is equal to 12 x \ - 1 = 5 n ". Multiplying this by 16,000 we have 
 80,000* for the strength of the plate. Dividing this by 7,200 we obtain 
 11 for the number of required shop rivets, and dividing it by 6,000 we 
 obtain 13.3, say 14, for the number of required field rivets. As is seen, 
 we have 12 shop rivets on one side of the splice and 16 field rivets on 
 the other side, so that the riveting as far as the outside splice plates are 
 concerned is quite satisfactory. 
 
 The net area of the 12" x^" plate is 5.25- 0.87 = 4.38 n ", and the 
 net area of the 12" xj" plate is 5 n ", making a total of 4.38 + 5 = 9.38 n " 
 in the splice plates, while the net area of the 15"x40* channel is 
 10.45 n " ; but the area of the tie plates can be counted to the extent of 6 
 field rivets. This is equal to 6,000 x 6 -r 16,000 = 2.25 n ". Adding this to 
 the area of the splice plates we obtain 11.63 n " for the total net area of 
 the plates splicing each channel which is quite sufficient. 
 
 The number of rivets in the laterals, shown at L2 (Fig. 283), is 
 obtained by developing the laterals. For example, the lateral to the left 
 of the floor beam is a 5" x 3J" x \" angle which has a net section of 
 4-0.5 = 3.5 n ". Multiplying this by 16,000 we obtain 56,000* for the 
 strength of the lateral, and dividing this by 6,000 we obtain 9 J" field 
 
384 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 385 
 
 rivets, whereas 9 are used. The number of rivets in the other laterals is 
 obtained in the same manner. 
 
 Joint L3. Next we will make a sketch of the details at the joint 
 L3, as shown in Fig. 284. In drawing this sketch, we can draw the 
 outline of the bottom chord, diagonals and post. Then we can draw the 
 end connections of the floor beam onto the sketch and draw the rivet 
 lines in all the members, and then we are ready to determine the number 
 of rivets and spacing of same for each member connecting at that joint. 
 
 The rivets in the connection of the two diagonals will be the same 
 for each. These diagonals are subjected to both tension and compres- 
 sion, and, according to the specifications, the sum of the two stresses 
 should be used in determining the required number of rivets in the end 
 connections. So we have (201,000 + 78,000) =279,000* for the stress to 
 be considered (see Fig. 279). Dividing this by 6,000 we obtain 46.5 J" 
 field rivets or, say, 24 on each side of each diagonal, arid, as is seen, 24 
 are used. The rivets on each side of the floor beam connecting the 
 gusset plate to the bottom chord should be sufficient to transmit the com- 
 ponent of the stress in the diagonal along the bottom chord. This com- 
 ponent is equal to 279,000* x sin = 279,000x0.64 = 178,000* (about). 
 Dividing this by 7,200* (the allowable single shear on a J" shop rivet) 
 we obtain 24.8 rivets or, say, 13 on a side and, as is seen, 14 are used. 
 The rivets in the diagonals should be spaced first and the edge of the 
 gusset plate located and drawn down to the bottom chord, and the rivets 
 arranged in the bottom chord to suit the gusset plate. Then the next 
 thing to do is to determine the rivets connecting the floor beam to the 
 truss, as was explained above in the case of panel points IA and L2. 
 
 After having the rivets spaced in the floor beam connection, the 
 gusset plate can be drawn completely; however, the spacing in the floor 
 beam connection selected for this joint must be the same as for the other 
 intermediate lower chord joints. The bottom lateral connections can 
 next be drawn. The rivets in each lateral are determined by developing 
 the strength of each lateral as was shown above for the other joints. 
 After the rivets are spaced in the laterals the lateral plate can be drawn 
 to suit the spacing; however, the rivets in the bottom of the floor beam 
 should be the same as required at LI, in order to have all of the inter- 
 mediate floor beams alike. 
 
 Referring to the sketch of the intermediate floor beam, drawn just 
 to the right of L3 (Fig. 284), the rivets connecting the end connection 
 angles to the floor beam must be sufficient to transmit the maximum end 
 shear on the floor beam in double shear or bearing, whichever requires 
 the greater number. These end angles are placed upon \" fillers (the 
 fillers being just as thick as the flange angles). These fillers are held 
 by rivets placed outside of the connection angles, and hence the thickness 
 of bearing on the rivets through the angles can be taken as the total 
 thickness of the 2 |" fillers and web, making lyV'j or * ne combined 
 thickness of the two angles, which is the least thickness as it is 1". So 
 for the allowable bearing on each J" shop rivet in this connection we 
 have Jx 1x24,000 = 2 1,000*, and for the allowable double shear on the 
 same we have 2x0.6x12,000 = 14,400*, which is less than the bearing 
 and hence governs. Then dividing the maximum end shear by the last 
 
386 STRUCTURAL ENGINEERING 
 
 figure we have 187,300 + 14,400 = 13 rivets, and, as is seen, 13 rivets are 
 used, not counting those passing through the flange angles. 
 
 There should be enough rivets connecting the stringer to the floor 
 beam to transmit the maximum end shear on a stringer. So we have 
 142,000 + 6,000 = 23.6 rivets in single shear, and, as is seen, 24 are used. 
 The bearing of these rivets is upon the 2 \" fillers and the T V web, so 
 that the shear governs. There should be enough rivets through these 
 fillers, outside of the stringer connection, to carry the total end shear on 
 a stringer and also the concentration on the floor beam. In the first 
 case, the rivets are in single shear and for the number required we have 
 142,000 4- 7,200 = about 20 shop rivets, whereas we have 20. In the sec- 
 ond case, these rivets are in bearing on the -%" web and for the number 
 required we have 187,300 -=-9,180 = about 20 rivets, so the correct number 
 is used. As a matter of fact, the field rivets in the stringer connection 
 (theoretically) bear on the web also and hence we have an excess of 24 
 field rivets in bearing on the web in the connection shown, but it is a 
 question in the case of such long rivets, and especially when field driven, 
 just how much bearing they will exert on the web, and there is some 
 question as to the bending, and to be on the safe side it is advisable to 
 compute the number on this important connection as is here shown. 
 
 It will be seen that in the case of the end connection of the floor 
 beam there is an excess of 5 rivets in bearing on the web (not considering 
 the ones passing through the flange they resist the flange increment) 
 and consequently 5 rivets could (theoretically) be omitted from the fillers, 
 but this would make the rivets look rather sparing, so the number shown 
 will be used. 
 
 The rivets through the fillers of the stringer connection are counter- 
 sunk on one side of the connection to permit the swinging of the stringers 
 in position during the erecting of the structure. The small angles under 
 each stringer (connected to the floor beam) are for erection purposes 
 and are not assumed to carry any load from the stringers. 
 
 Joint 71. We will next consider the drawing of the hip joint Ul 
 shown in Fig. 285. The first thing to do in making this sketch is to 
 select the center point of the joint and draw the center lines of the end 
 post and top chord. Then bisect the angle between these lines, and this 
 bisector will be the joint line where the two members meet. Now, using 
 the same transverse spacing as was used in the end post at LO (Fig. 
 282), the outlines of the end post and top chord can be drawn and the 
 rivets spaced. There should be enough rivets connecting the end post 
 to the gusset plates to transmit the total stress in the member. So 
 we have 
 
 523,000 + 6,000 = 87, say 88, J" field rivets, 
 
 or 44 on a side. As is seen, there are 38 in single shear, which provides 
 for 6,000x38 = 228,000* of the stress, and 4 passing through the small 
 outside splice plate which can be considered for bearing on the \" web 
 and hence provide for 8,750 x 4 = 35,000* of the stress in the end post. 
 Adding these two stresses we have 228,000 + 35,000 = 263,000*, and mul- 
 tiplying this by 2 we have 526,000, which is 3,000* more than the stress 
 in the end post, and hence the riveting as shown in the end post is suffi- 
 cient. There should be enough rivets connecting the top chord to the 
 

 387 
 
STRUCTURAL ENGINEERING 
 
 gusset plates to transmit the total stress in that member. As is shown, 
 there are 33 shop rivets (J" diameter) which provide for 33x7,200 = 
 237,600* of the stress in the top chord, and the 4 field rivets in bearing 
 on the f" web which provide for 6,563 x 4 = 26,250* stress. Adding these 
 two quantities together and multiplying by 2 we have 263,850x2 = 
 527,700*, which shows this riveting to be satisfactory, as the stress in 
 the top chord UI-L2 is 521,000* (see Fig. 279). Next, the hanger 
 Ul-Ll and the diagonal U1-L2 can be drawn and the rivets connecting 
 them to the gusset plates spaced as shown. The rivets connecting the 
 hanger to the gusset plates should be sufficient to develop the strength 
 of the hanger, which is 12.94x16,000 = 207,000* (see Fig. 279). Then 
 we have 
 
 207,000-6,000 = 34.5, say 36, J" field rivets, 
 
 or 18 on a side. As is seen, 18 are used. The number of rivets con- 
 necting the diagonal U1-L2 to the gusset plates should be sufficient to 
 develop the strength of the diagonal, which is 23.9x16,000 = 382,000*. 
 Dividing this by 6,000* we obtain 64 " field rivets, or 32 on a side; 
 whereas 32 are used, as is seen. 
 
 There should be enough metal in the gusset plates to properly 
 transmit all of the forces acting in the plane of the truss at the joint. 
 
 Fig. 286 
 
 These forces are as indicated in Fig. 286. As is readily seen, these 
 forces will not all be a maximum at the same time. About the most 
 severe stress will occur on the plates when the end post has a maximum 
 stress, as the top chord U1-U2 has about the maximum stress at that time 
 and the hanger Ul-Ll has quite a large stress, and the diagonal U1-L2 
 has a low stress. Combining S and S3 (Fig. 286) we obtain the resultant 
 R2, and combining SI and S2 we obtain the resultant Rl, which is equal 
 and opposite to R2. So, evidently, the maximum tension on the plates 
 will be on same section as cd, and combining S and S2 we obtain the 
 resultant R3 ; and, likewise, combining SI and S3 we obtain the resultant 
 7?4, which will be equal and opposite to R3. So, evidently, the maximum 
 
DESIGN OF SIMPLE EAILKOAD BEIDGES 389 
 
 compression on the gusset plates will be on some section as ab perpen- 
 dicular to these resultants (R3 and .R4), due mostly to S and 3. A 
 very satisfactory approximation can be obtained by making the plates 
 thick enough so that two-thirds of the metal along the section ab is suffi- 
 cient to transmit the maximum stress in the end post. By making the 
 plate f " thick, two-thirds of the section along section ab is about 48 x f x 
 f x2 = 40 n ". Dividing this into the maximum stress in the end post, we 
 have 523,000 -h40 n " = 13,100# (about) for the maximum compressive 
 unit stress in the plates, which is about the correct value, as it is about 
 the same as allowed in the chord U1-U2. So the gusset plates will be 
 made f" thick. The same thickness will be used at all of the other 
 joints, whether required or not, in order to have the sides of the truss 
 members in the same plane throughout without the use of fillers. 
 
 After the connections of the main members at joint 71 are drawn 
 as shown (Fig. 285), the portal can be drawn. In drawing the portal, 
 the first thing to do is to draw a top plan of the end post and locate 
 the center of the portal. Then locate the clearance line as shown and 
 draw the cross-section of the portal on the elevation of the end post (as 
 shown), and then draw the plan of the portal, keeping inside the clear- 
 ance line at every point. 
 
 The required number of rivets in the end connections of the portal 
 members is obtained by developing the strength of each of the members. 
 After the portal is drawn the top lateral and the bent lateral plates can 
 be drawn as shown, and thus the sketch of joint 71 is completed. 
 
 Joint U2. The details of this joint are shown in Fig. 287. The 
 outlines of the top chord as seen in elevation should be drawn first, and 
 then the outlines of the diagonal (C72-L3) and the post (C72-Z/2) can be 
 drawn as shown. The next thing after this is to locate the rivets in the 
 end connection of the diagonal. As this diagonal is subjected to both 
 tension and compression, the number of rivets in each end connection 
 must, according to the specifications, be sufficient to transmit the sum of 
 the two stresses. So we have 
 
 (78,000 + 201,000) -r 6,000 = 46.5 I" rivets 
 
 for the required number, or say 24 on a side, and, as is seen, 24 are 
 used, the same as in the end connection of this same diagonal at L3 (see 
 Fig. 284). 
 
 Next, the cross-section of the transverse strut can be drawn on to 
 the chord and then the transverse view, shown to the right, can be 
 drawn. In drawing this view it is best to draw first the cross-section 
 of the top chord, as shown, and then the elevation of the transverse strut. 
 Next the clearance line should be located and dotted in, as shown, and 
 then the knee brace and sub-strut can be drawn, at which time the con- 
 nections of the knee brace and sub-strut to the post are determined and 
 can be projected over to the other view of the post as shown. The next 
 thing to do is to determine the number of rivets required to connect the 
 post (C72-L2) to the gusset plates. 
 
 This member is subjected to 162,000* compression and 54,000* 
 tension. So, according to the specifications, there should be a sufficient 
 number of rivets connecting the post to the gusset plates at U2 to trans- 
 mit the sum of these two stresses. So we have (162,000 + 54,000) -5- 
 
J 
 
 Fig. 287 
 
 390 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 391 
 
 6,000 = 36 J" field rivets, or 18 on each side of the post, whereas 20 
 are used. 
 
 There should be enough rivets (at least) connecting the gusset 
 plates to the top chord to transmit the horizontal component of diagonal 
 U2-L3. This component, considering the sum of the stresses in the diag- 
 onal, is equal to (201,000 + 78,000) x sin (9 = 279,000 x 0.64 = 178,560*. 
 Dividing this by 7,200 we obtain about 25 shop rivets, say 13 on a side, 
 whereas there are almost three times this number shown (Fig. 287). 
 But the number cannot well be reduced, as the spacing of the rivets in 
 the chord angles is about what the specifications require, and the rivets 
 between the chord angles cannot be more sparingly spaced. So the rivet- 
 ing of the gusset plates to the chord will be considered satisfactory as 
 shown. 
 
 The next thing, the plan of the top chord, strut and laterals can be 
 drawn. There should be enough rivets in each lateral to develop the 
 strength of the lateral in tension. Each lateral is composed of 2 Ls 
 3J" x 3J" x f". So for the strength of each we have (4.96- 0.75) x 
 16,000 = 67,360*. Dividing this by 6,000 we obtain about 11 field rivets, 
 say 5 in each angle, whereas 5 are used 5 in the angle shown and 5 in 
 the angle at the bottom of the chord. 
 
 There should be enough rivets in each transverse strut to develop 
 the strength of the strut in compression. Each of these struts is com- 
 posed of 4 Ls 3|-" x 3" x f " arranged so as to form an I-section. So, 
 considering the 3J" legs turned horizontally and the vertical legs -J", 
 back to back, we have 
 
 1 Q9 
 p = 16,000 - 70 j~ = 8,140 Ibs. 
 
 for the allowable unit stress. Then multiplying this by the area of the 
 4 angles in each strut we obtain 8,140x9.2 = 74,888. Dividing this by 
 6,000 we obtain about 13 field rivets, say 7 in the top angles and 7 in 
 the bottom angles, whereas 8 are used so as to have symmetrical spacing. 
 
 The chord splice just to the left of joint U2 (Fig. 287) is what is 
 known as a butt joint. The chord sections are planed so they fit per- 
 fectly against each other and consequently the stress can be considered 
 as being transmitted from one chord segment to the other without the aid 
 of the splice plates, and hence the splice plates are considered as merely 
 holding the chord segments in position ; and, as is evident, the size of these 
 plates and the number of rivets connecting them to the chord is mostly 
 a matter of judgment. 
 
 Joint U3. The details at this joint will be practically the same as 
 at C72, except the connection of the diagonal and chord splice are 
 omitted, and consequently no larger scale sketch is necessary. 
 
 This completes the necessary large scale sketches and next the gen- 
 eral drawing, Fig. 288, and the shop drawings, Figs. 289 to 303, can be 
 made, wherein the details are, as we may say, simply transferred from 
 the above sketches to these drawings. 
 
 185. Camber. To prevent bridge trusses from sagging they are 
 "cambered," that is, they are built so that they curve upward. In the 
 case of an ordinary truss bridge the cambering of the trusses is accom- 
 
Fig. 288 
 
 392 
 
AHYKR. 
 
 /SO'-O"S.T Thro. ffivetedSpari 
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 Fig. 288 
 
 393 
 

 
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 394 
 
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 HI "S 
 
 II1MI 
 
 400 
 
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 JIfP 
 
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 mill! 
 
 402 
 
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 L> 
 
 r&*rt**sM***j? I I) 
 
 V < L *>-/./" 
 
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 S^OJSJi9AWJ t > > ^J 15*' 
 
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 403 
 
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 404 
 
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 408 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 409 
 
 plished by simply making the top chord longer than the bottom chord 
 and increasing the length of the diagonals to correspond. 
 
 The top chords are usually increased " for each 10 feet of length 
 (horizontal). The length of each diagonal is computed by taking the 
 mean of the top and bottom chord lengths (in the corresponding panel) 
 as the base of a right-angled triangle, and the height of the adjoining 
 post as the altitude, and the diagonal as the hypotenuse. This method 
 of cambering is satisfactory for trusses up to 300 feet in length, and 
 hence is used in the case of the above bridge. The panel length here is 
 25'-0", so by increasing the top chord J" for each 10 feet of length we 
 obtain S5'-0-^" for the cambered length of the top chord in each panel 
 as is shown for the chord sections in Figs. 292 and 294. The end posts 
 are not increased in length. For the mean lengths of the top and bottom 
 chords in each panel we have 
 
 Then for the cambered length of each diagonal we have 
 
 V (30) a + (25' - O^y) 2 = 39 ' - lf " 
 
 This length is used for the diagonals, as is seen in Fig. 297. 
 
 The camber affects the lengths of only the top chords, diagonals, 
 and top laterals. 
 
 186. Hints Regarding Shop Drawings. Drawing El (Fig. 289) 
 is the erecting diagram. This drawing is intended principally for the 
 use of the party erecting the structure. The mark (as is seen) of each 
 piece, or member, and the general dimensions of the structure are shown 
 and in addition a list of the shop drawings is given. 
 
 Drawing 1 (Fig. 290) shows the complete details of the stringers 
 and stringer bracing. The work of detailing in this case is practically 
 the same as for deck plate girders (Art. 136). There should be enough 
 rivets in the flanges to transmit the flange increment and at the same 
 time support the vertical load transmitted to them from the ties. To 
 obtain the spacing of the flange rivets at the ends of the stringers we have 
 
 R = 7,880#, which is the allowable bearing of a J" rivet on the 
 t" web; 
 
 v = 1,200# = + 100% impact (see Art. 136) ; 
 
 S = 142,000* ; 
 h = 41.5. 
 
 Then substituting these values in Formula (4), Art. 116 (the formula 
 required by the specifications), we obtain 
 
 7,880 
 
 p =IWW + - =2.18 (about), say 2J, ins. 
 
410 STRUCTURAL ENGINEERING 
 
 for the spacing of the flange rivets near the ends of the stringers, and, 
 as is seen, this spacing is used. The spacing at intermediate points is 
 obtained as explained in Art. 136. 
 
 The rivets connecting each pair of end stiffeners to the stringer 
 should be sufficient to transmit the maximum end shear on the stringer. 
 The two angles have a combined thickness of 1" in bearing on the rivets, 
 and the two fillers and web combined have a thickness of If" in bearing. 
 Then, using J" rivets, we have 
 
 24,000x1 xf = 21,000 Ibs. 
 
 for the allowable bearing on each rivet. The rivets are in double shear, 
 so for the allowable shear on each rivet we have 12,000 x 0.6 x 2 = 14,400. 
 As is seen, double shear governs. Then for the number of rivets required 
 in the angles we have 142,000 -~ 14,400 = 9.8 (about), whereas 11 are used, 
 but the rivets in the flanges should not be counted for very much, as they 
 take the flange increment. So the riveting, as shown, is about correct. 
 There should be a sufficient number of rivets connecting the \ n fillers to 
 the web to transmit the maximum end shear in bearing on the f " web. 
 Hence, for the number required we have 142,000-^7,880 = 18 (about), 
 and, as is seen, 18 are used. 
 
 The intermediate stiffeners are spaced in accordance with Art. 118 
 (see Art. 170). The spacing of the rivets in the intermediate stiffeners, 
 as previously stated, is mostly a matter of judgment (see Art. 136). 
 
 There should be a sufficient number of rivets in the end of each 
 lateral to develop the strength of the lateral in compression. The length 
 of each lateral is about 98". Then for the allowable unit stress on each 
 lateral we have 
 
 no 
 
 p = 16,000 - 70 ^ = 8,400 Ibs. (about), 
 y \) 
 
 Then for the strength of each we have 8,400x2.48 = 20,800*. Dividing 
 this by 6,000 we obtain a little over 3 J" field rivets, and, as is seen, 
 3 are used. There should be enough rivets connecting each lateral plate 
 to the stringer to transmit the sum of the components of the two laterals 
 along the stringers, as the two act in the same direction, one being in 
 compression and the other in tension. Each field rivet passing through 
 the flange will take the component exerted on it directly and hence it is 
 a matter of taking care of the component of the rivets in the lateral 
 plate outside of the flange angle. Two of these are good for 12,000#. 
 Resolving this along the stringer we have 12,000x0.69 = 8,280*. This 
 requires about 1 shop rivet, and, as is seen, 1 shop rivet is used. 
 
 The connection of the bottom laterals and struts to the bottom 
 flanges of the stringers is mostly a matter of judgment, as the stresses 
 due to traction are quite low (see Art. 177). However, there should be 
 enough rivets in these connections to insure rigidity. 
 
 Drawing 2 (Fig. 291) shows the complete details of the floor beams. 
 The end connections and the stringer connections, in the case of the 
 intermediate floor beams, were previously designed (see Art. 184) and 
 will not be considered here. The number of flange rivets in the inter- 
 mediate floor beams should be sufficient to transmit the flange increment. 
 To obtain the spacing of these rivets between the truss and the stringer 
 
DESIGN OF SIMPLE KAILKOAD BEIDGES 411 
 
 we have 
 
 r = 9,190#, which is the allowable bearing of a J" rivet on the 
 
 S = 187,300#; 
 h = 49.5, 
 
 Substituting these values in Formula (2) (Art. 116) we obtain 
 
 9,190x49.5 . . , . 
 P= !87,3QQ = 3- 4 "(<*<*) 
 
 for the required spacing, whereas 2J" spacing is used. 
 
 Apparently this spacing could be a little more than 2J", but as the 
 flange stress is zero at the end of the beam and practically a maximum 
 at the stringer connection, it is obvious that there should be a sufficient 
 number of rivets in each flange, between the end of the beam and the 
 stringer connection, to transmit the flange stress. Then dividing the 
 flange stress (see stress sheet, Fig. 279) by 9,190 we have 201,000 -r 
 9,190 = 22 (about) for the number of rivets required, whereas 20 are 
 used. From this it is seen that the number shown is about the correct 
 number. In all such cases, where the distance is short, the rivet spacing 
 should be verified in this manner. In most girders, however, the spacing 
 required by the flange increment governs. 
 
 The spacing of the flange rivets between the stringer connections 
 can be 6", the maximum spacing allowed, as the shear between the string- 
 ers is (theoretically) zero except for the small amount due to the weight 
 of the part of the floor beam intervening. 
 
 To obtain the spacing of the flange rivets in the end floor beam to 
 transmit the flange increment, between the end of the beam and stringer 
 connection, we have 
 
 r= 7,880*, which is the allowable bearing of a J-" rivet on the 
 
 2L web; 
 
 S = 143,200* (see stress sheet, Fig. 279); 
 fc = 49.5. 
 
 Then substituting these values in Formula (2) (Art. 116) we obtain 
 
 7,880x49.5 
 *= 143,200 =^ 
 
 for the required spacing. For the number required to transmit the flange 
 stress we have 153,000 -f 7,880 = 20 (about), and, as is seen, 20 rivets are 
 used in the top flange and hence the spacing shown is correct, although 
 it is less than required by the flange increment. 
 
 In the bottom flange there are 14 rivets between the end of the 
 beam and stringer connection, all of which can be considered as bearing 
 on the f " web, and those passing through the plates extending over the 
 flange angles can be considered as being in double shear in addition to 
 their bearing on the f " web. So we have 
 
 (14 x 7,880) + (4 x 14,400) = 167,900 Ibs. 
 
 for the value of the rivets transmitting the flange stress, whereas this 
 stress is 153,000*, and hence the rivets shown in the bottom flange, 
 between the end of the beam and stringer connection, are quite satis- 
 factory. 
 
412 
 
 STRUCTURAL ENGINEERING 
 
 The spacing of the flange rivets between the stringer connections 
 can be 6", as the shear between the stringers is practically zero. 
 
 There should be sufficient metal through the reduced portion, near 
 the end of the end floor beam, to resist the cross bending at that point. 
 Taking a section c-c (Fig. 282) we have the metal shown in Fig. 304. 
 
 Taking moments about the back of the top flange angles we have 
 - (23.62 x 19.75) + (12.37 x 16.62) + (8.36 x 1.96) _ 
 X= 44.35 
 
 for the distance from the back of the angles to the center of gravity of 
 the section. 
 
 V 
 
 * 
 
 -S'ffs. 
 
 Figr. 304 
 
 Then for the moment of inertia about the gravity axis x-x we have 
 the following: 
 
 8.36x13.56 +30.9 =1,568 for the angles; 
 12.37 xllV 1,122 =1,13 7 for the f" web; 
 
 23.62 x 4lJ3 2 + 1,435 = 4411 f or the A" P lates ; 
 4>ooo 
 
 295 
 Total moment of inertia = - for rivet holes. 
 
 4,/&oo 
 
 Then for the maximum stress on the outer element we have 
 
 This shows that the metal at the reduced portion of the end floor beams 
 is quite sufficient, as the allowable stress is 16,000#. 
 
 Drawings 3 to 14 (Figs. 292 to 303) are self-explanatory. The 
 making of these drawings is mostly a matter of copying the details from 
 the large scale sketches (Figs. 282 to 285 and 287). 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 413 
 
 187. Summary of Weight of Metal and Cost of the Above 
 150-Ft. Span. 
 
 12 Stringers @ 4,300*. . 51,600 Ibs. 
 
 Stringer bracing 2,120 Ibs. 
 
 2 End floor beams @ 3,010*. . 6,020 Ibs. 
 
 5 Intermediate floor beams @ 3,470*. . 17,350 Ibs. 
 
 4 End posts Ul-LO @ 8,850*. . 35,400 Ibs. 
 
 4 Top chords U1-U2 @ 4,830*. . 19,320 Ibs. 
 
 2 Top chords U2-U2. . @ 10,890*. . 61,780 Ibs. 
 
 4 Bottom chords LO-L2 @ 4,730*. . 18,920 Ibs. 
 
 2 Bottom chords L2-L2 @ 10,740*. . 21,480 Ibs. 
 
 4 Hangers U1-L1 @ 1,780*.. 7,120 Ibs. 
 
 6 Int. posts U2-L2 and U3-L3 @ 2,840*.. 17,040 Ibs. 
 
 4 Diagonals U1-L2 @ 3,760*.. 15,040 Ibs. 
 
 4 Diagonals U2-L3 @ 2,880*. . 11,520 Ibs. 
 
 4 Collision struts MI-LI @ 950*. . 3,800 Ibs. 
 
 2 Portal struts @ 2,270*. . 4,540 Ibs. 
 
 Top laterals 5,630 Ibs. 
 
 Transverse struts 2,770 Ibs. 
 
 Bottom laterals 6,460 Ibs. 
 
 4 Shoes @ 940*. . 3,760 Ibs. 
 
 2 Roller nests @ 1,180*. . 2,360 Ibs. 
 
 2 Cast pedestals @ 1,130*. . 2,260 Ibs. 
 
 2 Pedestals @ 1,090*. . 2,180 Ibs. 
 
 Pins 640 Ibs. 
 
 Anchor bolts 130 Ibs. 
 
 Total weight of bridge 279,240 Ibs. 
 
 Cost of span @ 3Jtf P^ pod 279,240 x 0.0325 = $9,075.30. This 
 price (3J^) per pound is only a fair average price for this class of work. 
 The price will vary from 2f $ to 4^ per pound. 
 
 The effective dead load from the metal is equal to the total weight 
 of metal in the span minus the weight of the end floor beams, shoes, 
 roller nests, pedestals, pins, and anchor bolts. So we have 379,240- 
 17,350 = 261,890* for the effective dead weight of metal in the span. 
 Dividing this by 150, the length of the span, we have 261,89o -=- 150 = 
 1,745* per foot of span, which is only 35* more than assumed above (see 
 stress sheet, Fig. 279). So the dead load assumed is quite satisfactory, 
 as 10 per cent variation would not materially affect the design. 
 
 DRAWING ROOM EXERCISE NO. 7 
 
 Design a single-track through riveted railroad bridge and make a 
 stress sheet for same the stress sheet to be upon tracing cloth. 
 Data: 
 
 Length of span = 6 panels at 26'-0" = 156'-0". 
 Height of trusses = 31'-0" c.c. of chords. 
 Live load, Cooper's JE50. 
 Dead load, to be assumed. 
 Specifications, A. R. E. Ass'n. 
 
5-f?ol/ers4l2-4.'tg. 
 
 '/'/^s. 
 
 '7&/>o//* 
 
 ^-oi" % Turned. 
 
 .ZTteP/s. /3"*$"*/-e" _, .< 
 
 /^^^^^ ^Z % 8 
 
 ' "" " . / \ >' '. / X ,' "*'/' 
 
 ft*****. 
 
 "rCastSket Mas. Plate. 
 
 \ t 'tt/tSi'** 
 
 X * /*/Jvr*.' f 
 
 Fig. 305 
 
 414 
 
DESIGN OF SIMPLE BAILKOAD BEIDGES 415 
 
 DRAWING ROOM EXERCISE NO. 8 
 
 Make a general drawing of the above 156-ft. bridge the finished 
 drawing to be upon tracing cloth and similar to the one shown in 
 Fig. 288. 
 
 188. Remarks. End floor beams are sometimes omitted in through 
 truss bridges in which case the end stringer rests directly upon the piers 
 as shown in Fig. 305. The ends of the stringers are usually constructed 
 as shown at (a). The stringers, as is seen, are braced to each other by 
 a cross-frame and they are connected to the shoes and trusses by a trans- 
 verse strut which forms the lower part of the cross-frame. 
 
 Circular rollers are sometimes used. The roller nests in that case 
 are usually constructed as shown in Fig. 305. The A. R. E. Association 
 Specifications practically prohibit the use of circular rollers as the mini- 
 mum diameter is specified as 6" and this size of rollers would place them 
 so far apart that the shoes would be unduly large. It is not considered 
 good practice in any case to use rollers less than 4" in diameter. 
 
 There is no question but that it is better practice to use end floor 
 beams., as the bottom ends of the end posts are held better than when they 
 are omitted and the thermal expansion and contraction are better pro- 
 vided for, as the entire structure is supported upon the shoes and will 
 move as a whole. In fact, there is no excuse for omitting the end floor 
 beams other than the slight saving in cost. 
 
 189. Dead-Load Stresses in Curved Chord Pratt Trusses. 
 As previously stated, the dead load is considered as uniformly distributed 
 over the span. The approximate dead weight per foot of span can be 
 obtained from (4), Art. 124, and adding 400 Ibs. to this to provide for 
 the weight of the deck (track) the total dead weight per foot of span 
 for single-track bridges is obtained, and by dividing this by 2 and mul- 
 tiplying by the panel length the panel load per truss is obtained. One- 
 third of this is considered applied at the top chord joints and two-thirds 
 at the bottom chord joints. 
 
 Let it be required to determine the dead-load stresses in the truss 
 shown at (a), Fig. 306, where P represents the panel load per truss. 
 
 Members bA and AB. Considering the part of the truss to the left 
 of section 1-1, as shown at (b), and resolving the forces vertically we 
 have 
 
 from which we obtain 
 
 Sl=Rsec6 
 
 for the stress in the end post bA, and resolving the forces horizontally 
 we have 
 
 SI sine -S2 = R sec0 x sin0 - S2 = 0, 
 
 from which we obtain 
 
 for the stress in the bottom chord AB. 
 
 Members BC, be and bC. Considering the part of the truss to the 
 left of section 2-2, as shown at (c), and taking moments about joint b, 
 
416 
 
 we obtain 
 
 STRUCTURAL ENGINEERING 
 
 h h 
 
 for the stress in bottom chord BC. Next, resolving the stress S3 (in top 
 chord be) into vertical and horizontal components at c and taking moments 
 about C we have 
 
 from which we obtain 
 
 Pd 
 
 for the horizontal component of the stress S3 in the top chord be, and 
 multiplying this by sec</> we obtain 
 
 for the stress in top chord be. 
 
 " ^ '? M 
 
 Fig. 306 
 
 Resolving the forces vertically and horizontally and summing up the 
 vertical we have 
 
 fl-P-F3-F4 = 0, 
 from which we obtain 
 
 for the vertical component of the stress /S4 in diagonal bC and multiply- 
 ing this by sec0 we have 
 
 for the stress in that member. As (R-P) is the shear in panel BC, it 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 417 
 
 is seen that the vertical component of the stress in diagonal bC is equal 
 to the shear in panel BC minus the vertical component of the stress in 
 the top chord be. Another way to obtain the stress in diagonal bC is 
 to prolong the top chord be until it intersects the bottom chord at [see 
 the diagram at (c)] and take moments about O. Thus, taking moments 
 about O we have 
 
 Rs-P (s + d) = S4:Xt, 
 from which we obtain 
 
 S4: = R S --P 
 
 for the stress in the diagonal bC. 
 
 In case the last method be used, the distances s and t can be deter- 
 mined in the following manner: Triangles cOC and cbm being similar 
 we have 
 
 s + 2d hi 
 
 d ~ hl-h 
 from which we obtain 
 
 /9*_*1\ ^ 
 
 hl-h 
 Similarly, since triangles OnC and bBC are similar we have 
 
 _ _ 
 
 h~~bC~ 
 from which we obtain 
 
 _/. + 2<A 
 
 - (-be- ) h ~ 
 
 Member cC. Considering the part of the truss to the left of section 
 3-3, as shown at (d), and resolving the forces vertically we obtain 
 
 for the stress in the post cC. Also, taking moments about [at (d)] 
 we have 
 
 from which we obtain 
 
 for the stress in post cC. 
 
 Members CD, cd and cD. Considering the part of the truss to the 
 left of section 4-4, as shown at (e), and taking moments about c we have 
 
 S7xhl=Rx2d-Pd, 
 from which we obtain 
 
 Rx2d Pd d 
 
418 STRUCTURAL ENGINEERING 
 
 (as seen above) for the stress in the bottom chord CD. Next, resolving 
 the stress in the top chord cd into horizontal and vertical components at 
 d and taking moments about D, we obtain 
 
 Rx3d-Px2d-Pxd d 
 
 ~ :6P 
 
 for the horizontal component and multiplying this by sec^l we obtain 
 
 for the stress in the top chord cd. 
 
 Resolving the forces horizontally and vertically and summing up 
 the vertical we have 
 
 R-2P-F8-V9 = 0, 
 from which we obtain 
 
 F9 = R-2P-F8 
 
 for the vertical component of the stress in diagonal cD and hence mul- 
 tiplying this by sec31 we obtain 
 
 for the stress in that member. Also, taking moments about 0' we obtain 
 
 ./ 
 
 Members dE and eD. Adding up from either end of the truss we 
 have 
 
 R-3P = 
 
 for the shear in panel DE and as the top chord de is parallel to the 
 bottom DE it is evident that the diagonals dE and eD have no stress 
 from dead load and hence in determining the dead-load stresses in the 
 truss we can ignore these members altogether. 
 
 Members dD, DE and de. Considering the part of the truss to the 
 left of section 5-5, as shown at (f), (ignoring diagonal eD) and summing- 
 up the vertical components and forces we have 
 
 2 
 
 R ~ 2 3 
 from which we obtain 
 
 for the stress in post dD. 
 
 If R -2(2/3 P) - F8 is minus 10 will be tension, as it would have to 
 act in the same direction as R in that case in order that the part of the 
 truss to the left of section 5-5 be in equilibrium, while if R -2(2/3 P) - FS 
 is plus 10 will be compression, as it would be acting in the opposite 
 direction to that of R. 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 419 
 
 The stress in dD can also be obtained by taking moments about O f 
 (see diagram at (f) ). Thus, taking moments about 0' we obtain 
 
 Rs' -P(s' + d) + P(s' + 2d) + f P(V 4- 3d) 
 (,' + 30 
 
 Taking moments about d we obtain 
 
 for the stress in the bottom chord DE, which, as seen above, is equal 
 to H8 y the horizontal component of the stress in the top chord cd. 
 
 Considering the diagonals eD and dE omitted from panel DE, as 
 there is no dead-load stress in them, it is obvious that the stress in top 
 chord de must be equal and opposite to the stress in the bottom chord 
 DE as these are the only horizontal forces in the panel and hence must 
 balance each other. So we have 
 
 S12 = S11=SS 
 
 for the stress in the top chord de. 
 
 Member bB. The only load on the bridge that could affect the 
 hanger bB is the one applied at B. This load, as is obvious, is supported 
 directly by the hanger and hence the dead-load stress in it is equal to fP. 
 
 As the truss is symmetrical about the center of the span and sym- 
 metrically loaded we need not consider further the dead-load stresses, as 
 we have now fully considered one-half of the span. 
 
 It is usual practice to determine graphically the dead-load stresses 
 in curved chord Pratt trusses. 
 
 190. Live-Load Stresses in Curved Chord Pratt Trusses. 
 
 Chords. The criterion for the placing of wheel loads for maximum 
 
 stress in the chords is exactly the same as given in Art. 91 for simple 
 
 parallel chord trusses. That is, the maximum moment about any panel 
 
 point will occur when the average 
 
 unit load to the left of the point is 
 
 equal to the average unit load on the 
 
 bridge. For example, to determine 
 
 the maximum live-load stress in chord 
 
 CD (Fig. 307) we would place a Fig. 307 
 
 wheel at C such that the average unit 
 
 load to the left of joint C would be equal (as nearly as possible) to the 
 
 average unit load on the bridge. Then by taking moments about c of 
 
 the forces to the left and dividing this moment by h we would obtain 
 
 the maximum live-load stress in the bottom chord CD and by multiplying 
 
 this stress by sec< we would obtain the maximum live-load stress in the 
 
 top chord be. The live-load stresses in the other chord members are 
 
 obtained in a similar manner. 
 
 Web Members. Let it be required to place the wheel loads so that 
 maximum stress will occur in diagonal cD (Fig. 308). If the top chord 
 cd and bottom chord CD were parallel maximum stress would occur in 
 diagonal cD when the shear in panel CD was a maximum, as the diagonal 
 in that case would carry all of the shear in the panel and hence the 
 
420 STRUCTURAL ENGINEERING 
 
 criterion for maximum shear given in Art. 90 would apply; but as the top 
 chord cd is inclined, and consequently carries some of the shear in panel 
 CD, the shear criterion will not apply exactly and hence we shall pro- 
 ceed to determine a criterion for the placing of the wheels for maximum 
 stress in the diagonal cD. Suppose the span is loaded from H (Fig. 
 308) up to panel CD as shown, which is about the same position as for 
 maximum shear in panel CD. Let P be the total load in panel CD, the 
 center of gravity of which is z distance from D, and let W be the total 
 load on the bridge, the center of gravity of which is x distance from H. 
 Considering the part of the truss to the left of section 1-1 and taking 
 moments about we obtain 
 
 for the stress in diagonal cD. 
 
 _ Wx 
 But R = ~L~ 
 
 Pz 
 
 and r - ~j~ ' 
 
 Substituting these values of R and r we obtain 
 Ws Ps Pa 
 
 Now suppose that there be a slight movement of the loads to the right 
 or to the left, say, to the left (W and P remaining constant), then SI, 
 x, and z will each receive an increment, A*S1, A.r, and A#, respectively. 
 Now adding these increments in (1) we have 
 
 Ws Ws \ Ps Ps \ /Pa 
 
 -Li* + Ti 
 
 Subtracting (1) we have 
 
 for the increment of the stress in the diagonal cD due to the slight move- 
 ment of the loads. But &z = A,r, as is obvious. So substituting A# for As 
 in (2) we have 
 
 Now Sly the stress in the diagonal cD, will be a maximum when 
 A$l = 0. So placing (3) equals and reducing and transposing we 
 obtain 
 
 TX7 T> / ~\ 
 
 ff-?(l+) < 4 >- 
 
 Expressing this equation in words we have: The average unit load 
 on the bridge is equal to the average unit load in panel CD multiplied by 
 
DESIGN OF SIMPLE KAILEOAD BEIDGES 
 
 421 
 
 (1 + a/s) when the maximum stress in diagonal cD occurs. This can be 
 taken as the criterion for placing the loads. 
 
 The increment of the stress in diagonal cD (W remaining constant) 
 can change signs only when a wheel passes the panel point D, so it is seen 
 that a load will be at that point when the maximum stress in the diagonal 
 occurs. One-half of this load should be considered in panel CD in com- 
 puting the value of P. 
 
 Next, let it be required to place the wheel loads so that maximum 
 stress will occur in post cC (Fig. 308). Considering the part of the 
 
 Fig. 308 
 
 truss to the left of section 2-2 with the wheel loads in about the position 
 shown and taking moments about 0' we have 
 
 from which we obtain 
 
 Rs' 
 
 - r 
 
 for the stress in the post cC. Substituting (W/L}x, and (P/d)z for 
 R and r, respectively, we have 
 
 (5). 
 
 Now, S2, x, and s will each receive a small increment if the loads 
 move slightly to the right or left. Let A$2, A,r, and A# be the increment, 
 respectively, that each receives due to a slight movement to the left. 
 Then substituting these increments in (5) we have 
 
 
 d d 
 
 and subtracting (5) from this we have 
 
 Ws' 
 
 P 
 
 - -j As 
 d 
 
 ('+*) 
 
 and substituting A,r for As, since A,r = As, we obtain 
 
 Ws ' - 
 
 '~L(s' + a) d 
 
 for the increment of the stress S2 in the post cC due to any slight move- 
 ment of the loads. Now, S2 will be a maximum when the increment 
 
422 
 
 STRUCTURAL ENGINEERING 
 
 = (see discussion of Art. 90). So placing (6) equal to zero and 
 reducing we obtain 
 
 JF_P 
 L ~ d 
 
 which is exactly the same as (4) except s' appears instead of s. Again, 
 suppose it be required to place the wheel loads so that maximum stress 
 will occur in diagonals bC. By loading panel BC, very much the same 
 as we did CD (above), and taking moments about 0' and adding incre- 
 ments, in the equation of moments we would obtain 
 
 JF_P/ o^\ 
 L ~ d\ s' )' 
 
 which expresses the criterion for the placing of the wheel loads for 
 maximum stress in the diagonal bC. This last equation is of exactly 
 the same form as (4) ; the only difference is the symbols have not the 
 same value as they have in (4). From this it is seen that equation (4) 
 expresses the general criterion for the placing of the wheel loads for 
 maximum stress in diagonals and intermediate posts. 
 
 In case the chords are parallel, a/s = Q as * in that case is equal to 
 infinity. The equation (4) then becomes 
 
 W P 
 
 Reducing and substituting nd for L we obtain 
 
 which is equation (5) of Art. 90. 
 
 It is evident that a live load moving onto the bridge from the right 
 would cause compression in diagonal fE (Fig. 308) and that this coin- 
 
 Fig. 309 
 
 pression would continue to increase until the load extended from H to a 
 short distance beyond F. Now if this compression were greater than 
 the dead-load tension in the diagonal we would have what is known as a 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 423 
 
 reversal of stress and the diagonal fE in that case would have to be 
 designed to carry both tension and compression or the member eF , which 
 would be known as a counter, would have to be inserted to carry the 
 reverse stress. In case the diagonal fE be composed of eye-bars, which 
 are often used in the case of long span bridges, the counter eF would be 
 used, as the bars would not carry compression; but if the diagonal fE be 
 a rigid member that is, capable of carrying compression the counter 
 would be omitted and the diagonal would be made sufficient to carry 
 both the tensile and compressive stresses in it, previously explained in 
 Art. 176 for diagonal cD of the 150-ft. span. 
 
 To determine the maximum live-load tension in the counter eF let 
 the bridge be loaded as shown at (a), Fig. 309. Let P represent the 
 load in the panel EF and W the total load on the bridge, and z the dis- 
 tance from F to the center of gravity of P and x the distance from H to 
 the center of gravity of W. Now, considering the part of the truss to 
 the left of section 6-6 as shown at (b) and taking moments about we 
 have 
 
 R (L + *) - r (a + *) - St' - Slt = 0, 
 
 from which we obtain 
 
 ^K .......................... (8) 
 
 for the stress in the counter eF. Substituting (W/L)x and (P/d)z for R 
 and r, respectively, we obtain 
 
 . 
 
 1, as is readily seen, is equal and opposite to the dead-load stress in 
 diagonal fE and hence is a known quantity. 
 
 Now, adding increments to S, x, and z in (9) and proceeding in the 
 same manner as shown above in the case of equations (1) and (5) we 
 obtain 
 
 <> 
 
 for the increment of the stress in the counter eF. The stress in the coun- 
 ter will be a maximum when this increment is equal to zero, as previously 
 explained. So placing (10) equal to zero and transposing we obtain 
 
 
 Expressing this in words we have: The average unit load on the 
 bridge is equal to the average unit load in the panel EF multiplied by 
 (a + s) -r (L + s) when the maximum stress in the counter occurs. This 
 can be taken as the criterion for placing the loads. 
 
 To determine the maximum stress in the counter we would first 
 determine the value of s, also of t and ', which can be determined suf- 
 ficiently accurately by scale in case of counters. Then we would place 
 the loading so as to satisfy equation (11) (as nearly as possible) and 
 
424 
 
 STRUCTURAL ENGINEERING 
 
 determine R and r by taking moments about H and F. Then the stress S 
 in the counter is readily found by substituting in equation (8). It can 
 also be determined by summing up the vertical forces and components 
 to the left of section 6-6. To determine it in this way, we would first 
 take moments about F (Fig. 309) and determine the horizontal com- 
 ponent of the stress in the top chord ef. Then multiplying this by tan</> 
 we would have the vertical component of the stress in chord ef, which we 
 will designate as V\. Then by determining the vertical component of 
 the dead-load stress in diagonal fE, which we will designate as V2, we 
 would have all of the vertical forces to the left of section 6-6 determined 
 except the vertical component of the stress in the counter eF. Let F3 
 represent this vertical component. Then summing up the vertical forces 
 and components to the left of section 6-6 we have 
 
 from which we obtain 
 
 for the vertical component of the stress in the counter eF and multiply- 
 ing this by sec# we would obtain the desired stress S in the counter. The 
 stress in other counters is determined in a like manner. 
 
 The maximum live-load stress in hanger bB (Fig. 310) is equal to 
 the maximum floor beam concentration at B. This, as explained in Arts. 
 148 and 171, is obtained by placing a wheel at B, such that the load in 
 panel AB will be equal (as nearly as possible) to the load in panel EC. 
 After having the loads thus placed the concentration at B, which is equal 
 to the stress in hanger bB, is obtained by taking moments about both 
 A and C. The maximum live-load stress in hanger gG is obtained in 
 the same manner as for bB. 
 
 Owing to the top chord segments having different slopes at the joints, 
 the intermediate posts of curved chord bridges are subjected to relatively 
 greater tensile stress from live load than the intermediate posts of parallel 
 chord bridges. For example, let us consider the case of post cC (Fig. 
 
 \ w \ w L.ve Load 
 
 IP \\9DeodLoad 
 \ d ' 3 
 
 Fig. 310 
 
 310). If panel points B and C alone were loaded with live load (ignor- 
 ing counter dC) it is readily seen that post cC would be in tension very 
 much the same as in the case of parallel chord bridges and it is obvious 
 that this tensile stress would be greater than it would be if the chords 
 be and cd had the same slope at c, as the chords, owing to their slopes 
 being different, pull upward (so to speak) on the post cC and also on 
 diagonal cD. Now, it is evident that the tensile stress in post cC will 
 increase as the stress in the top chords be and cd increases, and that it 
 
DESIGN OF SIMPLE KAILEOAD BRIDGES 425 
 
 will be a maximum when the stress in diagonal cD is zero, for in that 
 case the post alone would resist the upward pull from the chords be 
 and cd. So the problem in placing the live load for maximum tension 
 in post cC, is to place it so as to obtain zero stress in diagonal cD ana 
 at the same time as great a stress as possible in the top chords be and cd. 
 It is customary in practice to use an equivalent uniform live load (see 
 Art. 123) in determining the tension in intermediate posts of curved 
 chord bridges, as the work is very tedious if wfyeel loads be used. So 
 we will consider a uniform live load in this case. Suppose that this live 
 load moves onto the bridge from the left and loads it from A to m, just 
 so the dead-load tension in diagonal cD is reversed. Then the stress in 
 diagonal cD and also in counter dC would be zero. Now, as the load 
 continues to move on to the right past m, the counter dC will be in tension 
 and this tension will increase steadily, and likewise the stress in the top 
 chords be and cd (the stress in diagonal cD remaining zero) until some 
 point n is reached when the counter dC will have maximum tension and 
 the stress in diagonal cD will still be zero. Then, as the load continues 
 to move on to the right (past n), while the stress in the top chords be and 
 cd will steadily increase, the tension in counter dC will steadily decrease 
 until some point o is reached when the stress in the counter dC is zero. 
 This position is the one for maximum tension in the post cC as any 
 further movement of the load to the right would produce tension in diago- 
 nal cD (which has zero stress when the load extends from A to o) and 
 compression in post cC and hence the tension in the post would be rapidly 
 reduced. 
 
 The determination, as to location, of point o (the head of the uniform 
 load) in any case is very much a matter of trial. We could first load 
 from A to D, for instance. Then compute the stress in the counter dC 
 (or diagonal cD in case no counter is used) and if a stress occurs with 
 the load in that position we could move the load to the right or left, as 
 the case may require, until the stress in the counter is found to be zero 
 and this position of the load is the one required. 
 
 After the position of the load for maximum tension in the post is 
 found, as Ao, we can take moments about H and obtain R, the live-load 
 reaction at A. Then we can readily obtain the maximum live-load tension 
 in the post cC by taking moments about and considering all of the forces 
 to the left of section 2-2. Thus, taking moments about 0, we have 
 
 from which we obtain 
 
 W(s + 
 
 o = 
 
 for the maximum live-load tension in post cC, where W panel of live 
 load. 
 
 To this tension should be added the dead-load tension that occurs in 
 the post at the same time. The dead-load tension in the post is readily 
 obtained in the same manner as shown above for the live-load tension. 
 Thus, taking moments about we have 
 
STRUCTURAL ENGINEERING 
 from which we obtain 
 
 for the dead-load tension that occurs in the post at the same time the 
 maximum live-load tension occurs in it. P = panel of dead load and R' = 
 total dead-load reaction at A the same as occurs at H. 
 
 The maximum live-load tension in any other intermediate post is 
 obtained in a manner similar to that shown above for post cC. For 
 example, to obtain the maximum live-load tension in post dD, we would 
 load the bridge from A on to the right, until we obtained zero stress in 
 eD and then we would determine the maximum tension in the post by 
 taking moments about 0' . In the case of post dD the maximum tension 
 would occur in the member when the span was fully loaded, for in that 
 case both diagonals, eD and dE, would have zero stress. 
 
 Complete Design of a 225-Ft. Single-Track Through Pin-Connected 
 Curved Chord Pratt Truss Span 
 
 191. Data. 
 
 Length = 9 panels @ 25'-0" = 225'-0" c.c. end pins. 
 
 Width = 17'-0" c.c. trusses. 
 
 Height = 45'-0" at center and 31'-0" at hip. 
 
 Stringers spaced 6'6" c.c. 
 
 Live load, Cooper's 50 loading. 
 
 Specifications, A. R. E. Association. 
 
 192. Design of 25-Ft. Stringers. Proceeding in the manner out- 
 lined in Art. 170 we obtain the following for the stringers: 
 
 Maximum End Shear : Maximum Moment : 
 
 Z> = 5,000 Ibs. D= 375,000 in. Ibs. 
 
 L= 71,000 Ibs. L = 4,575,000 in. Ibs. 
 
 1= 66,000 Ibs. 7 = 4,223,000 in. Ibs. 
 
 142,000 Ibs. 9,173,000 in. Ibs. 
 
 142,000 -r 7,888 = 18 n " 9,173,000 -f 45 = 204,000# 
 
 1 web48"xf" = 18 n " 204,000 -f 16,000 = 12.75 n " 
 
 End stiff Ls3"x34x" 2 Ls 6" x 6" x 4" = 10.50" 
 
 Int. stiff Ls 3|" x 3| x |" J area of web = 2.25 n " 
 
 12.75 n " 
 
 193. Design of the Intermediate Floor Beams. Proceeding in 
 the manner outlined in Art. 171 we obtain the following for the inter- 
 mediate floor beams: 
 
 Maximum End Shear Maximum Moment 
 
 D~ 12,000 Ibs. D= 721,000 in. Ibs. 
 
 7,= 94,500 Ibs. L= 5,953,000 in. Ibs. 
 
 7 = 81,000 Ibs. 7= 5,1 04,000 in. Ib8. 
 
 187,500 Ibs. 11,778,000 in. Ibs. 
 
DESIGN OF SIMPLE KAILEOAD BRIDGES 427 
 
 187,500 -r 7,644 = 24.5 n " 11,778,000 + 53 = 222,000* 
 
 1 web56"x T y = 24.5 n " 222,000 -r 16,000= 13.87 n " 
 
 2 Ls 6" x 6" x " = 10.50" 
 4 area of web = 3.06 n " 
 
 13.56 n " 
 
 194. Design of End Floor Beams. Proceeding in the manner 
 outlined in Art. 172 and assuming the length of each beam to be 14'-6". 
 as they will rest upon the shoes in this case, we obtain the following for 
 the end floor beams: 
 
 Maximum End Shear : Maximum Moments : 
 
 D= 6,500 Ibs. D= 305,000 in. Ibs. 
 
 L= 71,000 Ibs. L = 3,408,000 in. Ibs. 
 
 1= 66,000 Ibs. I = 3,168,000 in. Ibs. 
 
 143,500 Ibs. 6,881,000 in. Ibs. 
 
 143,500 -r 8,266 = 17.37 n " 6,881,000 -r 52.37 = 131,000# 
 
 1 web 56" x f" = 21.00" 131,000 * 16,000 = 8.19 n " 
 
 " = 7.22-0.75= 6.47 n " 
 
 _ 
 
 - --(12,00 
 
 9.09 n " 
 
 195. Determination of Dead-Load Stresses in Trusses. From 
 (4), Art. 124, we have 
 
 p = 7x225 + 660 = 2,235 Ibs. 
 
 for the approximate weight of metal per ft. of span and adding 400 Ibs. 
 for the weight of the deck, we have 
 
 2,235 + 400 = 2,635 Ibs. 
 
 for the total assumed dead load per ft. of span or 2,635 4- 2 = 1,318 Ibs. per 
 ft. of truss. 
 
 Multiplying this by 25, the panel length in feet, we have 
 
 W = 1,318 x 25 = 32,950, say, 33,000 Ibs. 
 
 for the panel load on each truss. One-third of this will be considered at 
 the top joints and two-thirds at the bottom joints. The dead-load stresses 
 can be determined most readily by graphics. 
 
 Before we can proceed farther it is necessary that we determine the 
 exact outline of the truss. The panel points of the top chord will be on 
 the arc of a parabola. The height of the truss at the hip is 31 ft. and 
 45 ft. at the center of the span, making a drop of 14 ft. from the center of 
 the span to the hip. Laying off the panel lengths along the bottom chord 
 LO-D (Fig. 311) and taking C as the vertex of the parabola, CD as the 
 ,r-axis and considering half panel lengths we have 
 
 i *!. 
 
 14 ~ 7 2 
 from which we obtain 
 
 x = ~x 1 = 0.2857 ft. 
 
428 
 
 STRUCTURAL ENGINEERING 
 
 for the vertical distance from the horizontal line through C down to 
 panel point t/4. 
 
 Similarly, we have 
 
 fl 1 
 
 14" 7 2 
 from which we obtain 
 
 *'=j|x 9 = 2.5713 ft. 
 for the vertical distance down to panel point U3, and similarly we have 
 
 from which we obtain 
 
 14 
 
 = 7.1428 ft. 
 
 for the vertical distance down to panel point U2, and thus we have all of 
 the panel points or joints of the top chord located and the outline of the 
 
 Fig. 311 
 
 truss can be drawn as shown (Fig. 311). After the outline of the truss is 
 carefully drawn (say j% scale), the dead-load stress in each member 
 can be graphically determined as shown in Fig. 311; one-third of a panel 
 load of dead load, which is 33,000 xj = ll,000#, being considered as 
 applied at each top joint, and two-thirds, which is 33,000 xf = 22,000#, 
 at each bottom joint. 
 
 The reaction = 4 x 33,000 = 132,000#, which is applied at LO. We 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 429 
 
 obtain the diagram of the stresses shown (Fig. 311) by first laying off 1-2 
 (to, say, a ^ scale) equal to the reaction and passing around joint LO 
 counter clock- wise we obtain the polygon 1-2-3-1 for joint LO. Then 
 starting with LO-L1, at LI, and passing around joint LI counter clock- 
 wise we obtain the polygon 3-2-4-5-3. Considering 71 and passing around 
 counter clock- wise, beginning with the 11,000# load, we obtain the dia- 
 gram 6-1-3-5-7-6, and so on for the other joints as fully explained in Ex- 
 ample 2, Art. 96. 
 
 196. Determination of Live-Load Stresses and Impact in the 
 Trusses. In beginning the work of determining the live-load stresses 
 in the truss, a diagram showing the entire truss should be drawn carefully 
 to scale (say, to a -fa scale) and upon this diagram the important lengths 
 and the calculated values of the different angles should be given as shown 
 in Fig. 312. 
 
 End Post bA. The maximum stress will occur in this member when 
 the loading is placed for maximum shear in panel AB. By placing wheel 
 
 Fig. 312 
 
 4 at B (using Table A) we obtain 2,231 Ibs. for the average unit load on 
 the bridge and, considering one-half of the load at B as being in panel 
 AB, we obtain 2,400 Ibs. for the average unit load in the panel AB. This 
 position comes nearest to satisfying the criterion of Art. 90. So taking 
 moments about J (Fig. 312) we obtain (using Table A) 
 
 R = [16,364 + (284 x 109) + 109 2 ] ~|~ = 263,100 Ibs. 
 for the reaction at A, and taking moments about B we obtain 
 
 = 19,200 Ibs. 
 
 for the stringer reaction at A. Then for the maximum shear in panel 
 AB we have 
 
 S = 263,100 - 19,200 = 243,900 Ibs. 
 
 for Cooper's .E40 loading and 
 
 243,900 x 55 = 304,875, say, 305,000 Ibs. 
 
430 STRUCTURAL ENGINEERING 
 
 for Cooper's ESQ. Now multiplying this by sec# we have 
 305,000 x 1.28 = 390,400, say, 390,000 Ibs. 
 
 for the maximum live-load stress in end post bA. 
 
 The load extends over practically the entire span when this maximum 
 stress occurs, and hence the L in the impact formula (Art. 125) will be 
 taken as 225. So we have 
 
 = (3wS25) 390 ' 00 = 233 ' 0001bs - 
 
 for the maximum impact stress in bA. 
 
 Now adding the above live-load stress and impact and the dead-load 
 stress, given in Fig. 311, together we have 
 
 390,000 + 223,000 + 169,000 - 782,000 Ibs. 
 for the total maximum stress in the end post bA. 
 
 225' 
 
 Fig. 313 
 
 Diagonal bC. By placing wheel 3 at C (Fig. 313) and applying 
 Formula 4 of Art. 190 we obtain 
 
 By placing wheel 2 at C we obtain 
 432 
 
 225 
 
 From this it is seen that the criterion for placing the wheels for maxi- 
 mum stress in diagonal bC is nearest satisfied when wheel 3 is at C. Then 
 placing wheel 3 at C as shown in Fig. 313 and taking moments about J we 
 obtain ^ **. 
 
 R = (16,364 + 22,436 + 79*) - = 200,180 Ibs. 
 
 for the reaction at A. Then taking moments about C of the forces to left 
 we obtain 
 
 HI = [ (200,180 x 50) - 230,000] -r 37.86 = 258,290 Ibs. 
 
 for the horizontal component of the stress in the top chord be and multi- 
 plying this by tan< we have 
 
 V\ = 258,290 x 0.2744 = 70,875 Ibs. 
 
 for the vertical component of the stress in the top chord be. 
 Taking moments about C of wheels 1 and 2 we obtain 
 
 r = 9,200 Ibs. 
 
DESIGN OF SIMPLE BAILKOAD BRIDGES 
 
 431 
 
 Now adding up all of the vertical forces and components to the left 
 of sections 2-2 we have 
 
 from which we obtain 
 
 V = R-r-Vl 
 
 and substituting in the numerical values given above we have 
 
 V = 200,180 - 9,200 - 70,875 = 120,105 Ibs. 
 
 for the vertical component of the stress in diagonal bC and multiplying 
 this by sec0 we obtain 
 
 S = 120,105 x 1.28 = 153,734 Ibs. 
 
 for the stress in the diagonal due to Cooper's J240 and multiplying this 
 by 50/40 we obtain 192,167, say 192,000 Ibs. for the maximum live-load 
 stress in diagonal bC. 
 
 When this maximum stress occurs the span is loaded practically from 
 C to J and hence the L in the impact formula will be taken as 175 ft. So 
 we have 
 
 1 = 
 
 for the impact stress in diagonal bC. 
 
 Now adding together the above live-load stress and impact and the 
 dead-load stress, given in Fig. 311, we have 
 
 192,000 + 121,000 + 73,000 = 386,000 Ibs. 
 
 for the total maximum tension in diagonal bC. 
 
 Intermediate Post cC. Placing wheel 3 at D and applying Formula 
 4 of Art. 190 we obtain 
 
 and placing wheel 2 at D we obtain 
 
 382 _ 
 225" 
 
 Fig. 314 
 
 From this it is seen that the criterion for placing the wheel loads for 
 maximum stress in post cC is nearest satisfied when wheel 2 is at D. 
 Then placing wheel 2 at D, as shown in Fig. 314, and taking moments 
 about J we obtain 
 
 R = 145,200 Ibs. 
 
432 STRUCTURAL ENGINEERING 
 
 for the reaction at A and taking moments about D of wheel 1 we obtain 
 
 r = (10,000 x 8) -f 25 = 3,200 Ibs. 
 
 for the floor beam concentration at C. Then taking moments abouc C 
 of all the forces and components to the left of section 3-3 we obtain 
 
 H=(Rx5Q)+h, 
 
 and substituting in the numerical values given above we have 
 H = (145,200 x 50) - 37.86 = 191,700 Ibs. 
 
 for the horizontal component of the stress in the top chord be. Then 
 multiplying this component by tan< we obtain 
 
 F= 191,700 x 0.2744 = 52,600 Ibs., 
 
 for the vertical component of the stress in the top chord be. Now adding 
 up (algebraically) all of the vertical forces and components to the left of 
 section 3-3 we have 
 
 R-r-F-S = Q, 
 
 and substituting in the numerical values given above and transposing 
 we obtain 
 
 S = 145,200 - 3,200 - 52,600 = 89,400 Ibs. 
 
 for the stress in post cC due to Cooper's 40, and multiplying this by 
 50/40 we obtain 111,700, say, 112,000 Ibs. for the maximum live-load 
 compression in post cC. 
 
 When this maximum stress occurs the span is loaded practically from 
 D to J and hence the L in the impact formula will be taken as 150. So 
 we have 
 
 /= 112 > = >0 QQ lbs - 
 
 for the impact stress in post cC. Now adding together the above live-load 
 stress and impact and the dead-load stress, given in Fig. 311, we have 
 
 112,000 + 75,000 + 34,000 = 221,000 lbs. 
 
 for the total maximum compression in post cC. 
 
 Diagonal cD. Placing wheel 3 at D (Fig. 312) and applying equa- 
 tion 4 of Art. 190 we obtain 
 
 392 = 
 225 ~ 
 
 This position, as can be verified by trial, comes nearest to satisfying 
 the criterion for maximum stress in diagonal cD. So placing wheel 3 at 
 D, as shown in Fig. 315, and taking moments about J we obtain 
 
 R= (16,364 + 15,336 - f 54 2 )- 153,700 lbs. 
 
 for the reaction at A. Taking moments about D of the wheels to the left 
 we obtain 
 
DESIGN OF SIMPLE EAILKOAD BRIDGES 
 
 433 
 
 for the floor beam concentration at C. Next, taking moments about D of 
 the forces and components to the left of section 4-4 we obtain 
 
 Hl (153,700x50) (9,200 x25) 
 
 r 4 
 
 I 
 54' 
 
 L = 
 
 Figr. 315 
 
 for the horizontal component of the stress in the top chord cd. Multiply- 
 ing this by tan<^l we obtain 
 
 Vl^Hlx tan<l = 266,200 x 0.1828 = 48.660 Ibs. 
 
 for the vertical component of the stress in top chord cd. Now summing up 
 the vertical forces and components to the left of section 4-4 we have 
 
 Substituting the numerical values given above and transposing we obtain 
 V= 153,700 - 9,200 - 48,660 = 95,840 Ibs. 
 
 for the vertical component of the stress in diagonal cD and multiplying 
 this by secfll we have 
 
 S = V sec(91 = 95,840 x 1.199 = 114,912 Ibs. 
 
 for the stress in diagonal cD due to Cooper's 40 loading. Then by mul- 
 tiplying this by 50/40 we obtain 143,635, say, 144,000 Ibs. for the* maxi- 
 mum tension in diagonal cD due to the -E50 loading. 
 
 When this maximum stress occurs the span is practically loaded from 
 D to J and hence the L in the impact formula will be taken as 150 ft. So 
 we have 
 
 for the impact stress in diagonal cD. 
 
 Now adding together the above live-load stress and impact and the 
 dead-load stress, given in Fig. 311, we obtain 
 
 144,000 + 96,000 + 40,000 = 280,000 Ibs. 
 
 for the total maximum tension in diagonal cD. 
 
 Intermediate Post dD. Placing wheel 2 at E (Fig. 312) and apply- 
 ing equation 4 of Art. 190 we obtain 
 
 332 = __, ,20 A 75 
 225" 
 
434 
 
 STRUCTURAL ENGINEERING 
 
 This position of the wheels practically satisfies the criterion for maxi- 
 mum stress in post dD. So placing wheel 2 at E, as shown in Fig. 316, 
 and taking moments about J we obtain (using Table A) 
 
 R = 105,500 Ibs. 
 
 Fig. 316 
 
 for the reaction at A y and taking moments about E of wheel 1 we obtain 
 
 r = 3,200 Ibs. 
 
 for the floor beam concentration at D. Then taking moments about D 
 of the forces and components to the left of section 5-5 we obtain 
 
 HI = 186,500 Ibs. 
 
 for the horizontal component of the stress in top chord cd and multiply- 
 ing this by tan</>! we obtain 
 
 VI = 186,500 x 0.1828 = 34,092 Ibs. 
 
 for the vertical components of the stress in top chord cd. 
 
 Now, by summing up (algebraically) the vertical forces to the left 
 of section 5-5 we obtain 
 
 S = 105,500 - 3,200 - 34,092 = 68,208 Ibs. 
 
 for the stress in post dD due to Cooper's 40 and multiplying this by 
 50/40 we obtain 85,260, say, 85,000 Ibs. for the maximum compression in 
 post dD due to the E50 loading. 
 
 When this maximum stress occurs the span is practically loaded from 
 E to J and hence the L in the impact formula will be taken as 125 ft. So 
 we have 
 
 for the impact stress in post dD. 
 
 Now, adding together the above live-load stress and impact, and the 
 dead-load stress, given in Fig. 311, we obtain 
 
 85,000 + 60,000 + 12,000 = 157,000 Ibs. 
 
 for the total maximum compression in post dD. 
 
 Diagonal dE. Placing wheel 3 at E (Fig. 312) we obtain 
 
 342 ,40 
 
 _ =1.52 and _ 
 
 75 
 
DESIGN OF SIMPLE RAILROAD BEIDGES 
 
 435 
 
 This position of the load comes nearest to satisfying the criterion for 
 maximum stress in diagonal dE. So placing wheel 3 at E, as shown in 
 Fig. 317, and taking moments about J we obtain 
 
 R = 113,200 Ibs. 
 
 for the reaction at A, and taking moments about E of the wheels to the 
 left we obtain 
 
 r = 9,200 Ibs. 
 
 for the floor beam concentration at D. Then taking moments about E 
 of the forces and components to the left we obtain 
 
 HI = 248,100 Ibs. 
 
 for the horizontal component of the stress in the top chord de and mul- 
 tiplying this by tan<2 we obtain 
 
 VI = 248,100 x 0.0912 = 22,626 Ibs. 
 
 for the vertical component of the stress in top chord de. 
 
 Now, summing up the vertical forces and components to the left of 
 section 6-6 we obtain 
 
 F= 113,200 - 9,200 - 22,626 = 81,374 Ibs. 
 
 for the vertical component of the stress in diagonal dE, and multiplying 
 this by sec$2 we obtain 94,393 Ibs. for the stress in diagonal dE due to 
 
 Fig. 317 
 
 40 loading; and multiplying this by 50/40 we obtain 117,991, say, 118,- 
 000 Ibs. for the maximum live-load stress in the member. When this maxi- 
 mum stress occurs, the span is loaded practically from E to J and hence 
 the L in the impact formula will be taken as 125 ft. So we have 
 
 /= 
 
 118 > 000 = 83 > soo > sa y> 83 > 000 lbs - 
 
 for the impact stress in diagonal dE. 
 
 Now, adding together the above live-load stress and impact and the 
 dead-load stress given in Fig. 311, we obtain 
 
 118,000 + 83,000 + 21,000 = 222,000 lbs. 
 for the total maximum stress in diagonal dE. , 
 
4 36 STRUCTURAL ENGINEERING 
 
 Intermediate Post eE. Placing wheel 2 at F (Fig. 312) we obtain 
 
 and placing wheel 3 at F we obtain 
 
 292 1 Q , 40 /, 
 = 1.3 and -^1 
 
 From this it is seen that the criterion (Art. 190) for maximum stress 
 in post eE is the nearest satisfied with wheel 2 at F. So placing wheel 2 
 at F, as shown in Fig. 318, and taking moments about J we obtain 
 (using Table A) 
 
 R = 71,466 Ibs. 
 
 Fig. 318 
 
 for the reaction at A and taking moments about F of the wheel to the 
 left we obtain 
 
 r= 3,200 Ibs. 
 
 for the floor beam concentration at E. Then taking moments about E 
 of the forces and components to the left of section 7-7 we obtain 
 
 HI = 159,843 Ibs. 
 
 for the horizontal component of the stress in the top chord de and multi- 
 plying this by tan<2 we obtain 
 
 Fl = 14,577 Ibs. 
 
 for the vertical component of the stress in top chord de. Then adding 
 algebraically the vertical forces and components to the left of section 
 7-7 we obtain 
 
 S = 71,466 - 3,200 - 14,577 = 53,689 Ibs. 
 
 for the maximum compression in post eE due to the E40 loading and mul- 
 tiplying this by 50/40 we obtain 67,111, say, 67,000 Ibs. for the maximum 
 live-load compression in post eE. 
 
 When this maximum stress occurs the span is loaded practically 
 from F to J and hence the L in the impact formula will be taken as 100. 
 So we have 
 
 I = ( 
 
 67,000 = 50,250, say, 50,000 Ibs. 
 
 for the impact stress in post eE. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 437 
 
 Now, adding the above live-load stress and impact, and the dead- 
 load stress given in Fig. 311 we obtain 
 
 67,000 + 50,000 - 5,000 = 112,000 Ibs. 
 
 for the total maximum compression in post eE. 
 
 Diagonal eF. As the top chord ef (Fig. 312) is parallel to bottom 
 chord EF the diagonal eF will carry all of the maximum shear in panel 
 EF when the load moves onto the bridge from the right and hence the 
 ordinary criterion of Art. 90 for the maximum shear in panel EF applies. 
 According to this criterion, the maximum shear will occur in panel EF 
 when the average unit load on the bridge is equal to the average unit load 
 in the panel. 
 
 This is the nearest satisfied when wheel 3 is at F. So placing wheel 
 3 at F as shown in Fig. 319 and taking moments about J we obtain 
 
 R = 77,851 Ibs. 
 
 Fig. 319 
 
 for the reaction at A and taking moments about F of the wheels to the 
 left we obtain 
 
 r = 9,200 Ibs. 
 
 for the floor beam concentration at E. Then for the maximum shear in 
 panel EF we have 
 
 (R - r) = (77,851 - 9,200) = 68,651 Ibs. 
 
 This is assumed to be carried altogether by the diagonal eF, so, evi- 
 dently the vertical component of the stress in that member must be equal 
 to this shear; that is, 
 
 F = 68,651 Ibs. 
 
 Then multiplying this by sec03 we obtain 
 
 S = 68,652 x 1.145 = 78,605 Ibs. 
 
 for the stress in the diagonal eF due to 40 loading and multiplying this 
 by 50/40 we obtain 98,256, say 98,000 Ibs. for the maximum live-load 
 stress in diagonal eF. 
 
 When this maximum stress occurs the span is loaded practically from 
 F to J and hence the L in the impact formula will be taken as 100. 
 
 Then we have 
 
 /=! 
 
 \100 + 300 
 for the impact in diagonal eF. 
 
 ^ \ 98,000 = 73,500, say 73,000 Ibs. 
 J -f 300 / 
 
438 
 
 STRUCTURAL ENGINEERING 
 
 Now, adding the above live-load stress and impact, and the dead-load 
 stress given in Fig. 311 we obtain 
 
 98,000 + 73,000 + 00,000 = 171,000 Ibs. 
 for the total maximum stress in diagonal eF. 
 
 Fig. 320 
 
 Counter eD. Placing wheel 3 at D as shown at (a), Fig. 320, and 
 applying Formula 11 of Art. 190 we have (not considering wheel 15 on 
 the bridge) 
 
 232 inq n d40 - ?fi 
 
 1.03 and - = 1.26. 
 
 This position of the wheels comes nearest to satisfying the criterion 
 for maximum stress in counter eD. So placing wheel 3 at D and taking 
 moments about A we obtain 
 
 R = (10,816) = 48,071 Ibs. 
 
 for the reaction at J and taking moments about D of the wheels to the 
 right we obtain 
 
 r = 9,200 Ibs. 
 
 for the floor beam concentration at E. 
 
 SI is equal and opposite to the dead-load stress in diagonal dE which 
 is given in Fig. 311 as 21,000 Ibs. Then, resolving SI into horizontal 
 and vertical components at E, we obtain 
 
 VI = SI -f sec02 = 21,000 -=- 1.16 = 18,103 Ibs. 
 
 for the vertical component of the stress in diagonal dE. 
 
 Taking moments about D of the forces and components to the right 
 we obtain 
 
 (R x 150) - (VI x 25) - (r x 25) - (H2 x 42.43) = 0. 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 439 
 
 Substituting in the numerical values given above and transposing and 
 reducing we obtain 
 
 H2 = 153,853 Ibs. 
 
 for the horizontal component of the stress in top chord de and multiply 
 ing this by tan</>2 we obtain 
 
 F2 = 153,853 x 0.0912 = 14,031 Ibs. 
 
 for the vertical component of the stress in top chord de. 
 
 Now, summing up all the vertical forces and components to the 
 right of section 9-9 we obtain 
 
 and substituting in the numerical values given above we have 
 V = 48,071 - 9,200 - 18,103 + 14,031 = 34,799 Ibs. 
 
 for the vertical component of the stress in counter eD and multiplying 
 this by sec#3 we have 
 
 S = 34,799 x 1.145 = 39,844 Ibs. 
 
 for the maximum stress in counter eD due to the 40 loading and mul- 
 tiplying this by 50/40 we obtain 49,805, say, 50,000 Ibs. for the maximum 
 live-load stress in counter eD due to J550 loading. 
 
 When this maximum stress occurs the span is loaded practically from 
 D to A and hence the L in the impact formula will be taken as 75. 
 
 Then we have 
 
 1 = 
 
 for the impact stress in counter eD. 
 
 Now, adding the above live-load stress and impact together we have 
 (counters carry no dead load) 
 
 50,000 + 40,000 = 90,000 Ibs. 
 
 for the total maximum stress in counter eD. 
 
 No counter is needed in panel CD as the live-load shear is not suf- 
 ficient to reverse the dead-load shear; or, in other words, the dead-load 
 tension in diagonal cD is greater than the live-load compression in it. 
 
 Tension in Post dD. The maximum shear in end panel AB was 
 found above (in determining the stress in end post bA) to be 305,000 Ibs. 
 
 Substituting this shear in the formula given in Art. 123 we obtain 
 
 for the equivalent uniform live load per foot of truss for determining the 
 stress in the web members and hence we can use this load for determin- 
 ing the tension in intermediate posts. Then for the panel load we have 
 
 W = 3,050 x 25 = 76,250, say 76,000 Ibs. 
 
 The first part of our problem now is to place this uniform load so 
 as to obtain zero stress in counter eD and also in diagonal dE and at the 
 
440 
 
 STRUCTURAL ENGINEERING 
 
 same time as great a stress in chords cd and de as possible (see Art. 190). 
 The equivalent uniform live load will produce practically the same 
 stress in the counter eD (Fig. 320) as the wheel loads. So, if panel 
 points B, C, and D are loaded with the above uniform load the counter 
 eD will have a maximum tensile stress of 50,000 Ibs., the same as pre- 
 viously found for wheel loads. Dividing this by sec03 we obtain 
 
 50,000 4- 1.145 = 44,000 Ibs. (about) 
 
 for the vertical component of the maximum tensile stress in counter eD. 
 Now, any load at E, F, or any panel point to the right of panel DE will 
 tend to reduce the tension in counter eD. What we desire is to place the 
 load just so that the 50,000 Ibs. tension in the counter is exactly counter- 
 acted, for then the counter eD and diagonal dE will have zero stress and 
 hence the post dE will have maximum live-load tension. 
 
 By loading panel point E alone we obtain a reaction at A of 76,- 
 000 x 5/9 = 42,000 Ibs. Now, as the vertical component of the 50,000 Ibs. 
 
 Fig. 321 
 
 tensile stress in the counter eD alone is about 44,000 Ibs., it is obvious 
 that a panel load at E will not reverse the 50,000 Ibs. tension in the 
 counter. So next, let us place a panel load at both E and F. Then for 
 the reaction at A, which is equal to the negative shear in panel DE due 
 to the loads at E and F, we obtain 
 
 76,000 x^ + 76,000 x | = 76,000 Ibs. (about), 
 y y 
 
 Taking moments about D (panel points E and F alone being loaded) we 
 obtain 
 
 HI = (76,000 x 75) -r 42.43 = 135,000 Ibs. (about) 
 
 for the horizontal component of the stress in top chord de. Then mul- 
 tiplying this by tan</>2 we obtain 
 
 VI = 135,000 x 0.0912 = 12,300 Ibs. (about) 
 
 for the vertical component of the stress in top chord de due to the panel 
 loads at E and F. 
 
DESIGN OF SIMPLE KAILKOAD BRIDGES 441 
 
 Then subtracting this from the reaction we obtain 
 76,000 - 12,300 = 63,700 Ibs. 
 
 for the vertical component in counter eD. This is too great, as the coun- 
 ter would be more than reversed and hence the diagonal dE would then 
 be in tension. 
 
 Next, suppose the load extends just up to panel point F so that the 
 load at F will be only a half panel load or 38,000 Ibs. Then for the 
 reaction at A we have 
 
 76,000 x | + 38,000 x |= 59,000 Ibs. (about). 
 y y 
 
 Now, taking moments about D we obtain 
 
 0.0912 = 9,500 Ibs. (about) 
 
 for the vertical component of the stress in the top chord de. 
 Then subtracting this from the reaction at A we have 
 
 59,000 -9,500 = 49,500 Ibs. 
 
 for the vertical component of the compressive stress in counter eD. As is 
 seen, this is too large. So let us extend the load just 20 ft. beyond panel 
 point E as shown. Then the load at E will be 75,000 Ibs. and at F it 
 will be 24,000 Ibs. The reaction at A due to these two loads will be 
 
 ( 75,000 x^\ + ^24,000 x|U 52,300 Ibs. (about). 
 
 btain 
 8^400 ibs. (about) 
 
 Then taking moments about D we obtain 
 /52,300x75\ 
 
 \ 42.43 / 
 
 for the vertical component of the stress in the top chord de. Now, sub- 
 tracting this from the reaction at A we obtain 
 
 52,300 -8,400 = 43,900 Ibs. 
 
 for the vertical component of the compressive stress in the counter eD 
 due to the loads at E and F which just reverses the maximum tension 
 in the counter. So this is the position of the load being sought for. That 
 is, by placing the load so that it extends from A to 20 ft. beyond E the 
 stress in both the counter eD and diagonal dE will be zero and the chords 
 cd and de will have the greatest stress possible with that condition and 
 hence the tension in post dD will be a maximum. 
 
 So placing the load in that position and taking moments about A 
 we obtain 
 
 R' = (3,050 x 120)~= 97,644 Ibs. 
 
 . (about) 
 
 for the reaction at J. 
 
 Then taking moments about D we obtain 
 
442 STRUCTURAL ENGINEERING 
 
 for the horizontal component of the stress in top chord de and multiplying 
 this by sec<2 we obtain 
 
 418,000 x 1.004 = 420,000 Ibs. (about) 
 
 for the stress in top chord de. Now as the stress in diagonal dE is zero, 
 the tensile stress in the post dD can be obtained very quickly by drawing 
 the diagram shown at (b). From this diagram the tension in post dD 
 is found to be about 38,000 Ibs. 
 
 As is seen, when this maximum tension occurs in the post dD there 
 is 120 ft. of load on the bridge, so we have 
 
 1 = lon^Qnn 38 > 000 = 26 > 000 lbs ' ( about ) 
 
 for the impact stress. 
 
 Now, assuming diagonal dE as having zero stress we can obtain 
 the dead-load tension in post dD by drawing the diagram shown at (d) 
 
 The stress in top chord cd is given in Fig. 311 as 177,000 lbs. Then 
 drawing 1-3 equal (by scale) to 177,000 and parallel to cd, and 1-2 and 
 3-2 parallel to de and dD, respectively, we have the dead-load tension 
 in the post given by the line 3-2. This tension is found in this manner 
 to be 16,000 lbs. 
 
 Now adding together the above live- and dead-load stresses and 
 impact we obtain 
 
 38,000 + 26,000 + 16,000 = 80,000 lbs. 
 
 for the total maximum tension in post dD. 
 
 Tension in Post eE. As is obvious, the maximum tension will occur 
 in post eE when the span is fully loaded, for in that case both of the 
 diagonals eD and fE have zero stress. The counter eD also has zero 
 stress at the same time. 
 
 Now, as we are using a uniform live load we can readily obtain the 
 live-load tension in post eE by direct proportion as it will be propor- 
 tional to the dead-load tension given in Fig. 311. 
 
 So, letting S represent the live-load tension, we have 
 
 S 3,050 
 5,000 " 1,318' 
 from which we obtain 
 
 S = 11,570, say, 12,000 lbs. 
 
 for the maximum live-load tension in post eE where 3,050 is the uniform 
 live load and 1,318 the dead load per ft. of truss. 
 For the impact we have 
 
 Now, adding together the above live-load stress and impact, and the 
 dead-load stress given in Fig. 311, we obtain 
 
 12,000 + 7,000 + 5,000 = 24,000 lbs. 
 for the total maximum tension in post eE. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 443 
 
 Bottom Chords AB and BC. The maximum stress can be obtained 
 in these members by placing a load at B such that the average unit load 
 to the left is equal to the average unit load on the bridge (see Art. 90) 
 and taking moments about b; but the same can be found by simply mul- 
 tiplying the maximum shear in panel AB by tan#. The maximum shear 
 was found above (in determining the stress in the end post bA) to be 
 305,000 Ibs. So we have 
 
 305,000 x 0.8066 = 246,013, say, 246,000 Ibs. 
 
 for the live-load stress in each of the bottom chords AB and BC. 
 
 When this maximum stress occurs, the load extends practically over 
 the entire span, so for the impact stress in each of these chords we have 
 
 7 = 246 ' 00 = 140 ' 571 > sa ^ 141 > 000 lbs ' 
 
 Now, adding together the above live-load stress and impact, and 
 the dead-load stress, given in Fig. 311, we obtain 
 
 246,000 + 141,000 + 106,000 = 493,000 lbs. 
 
 for the total maximum stress in each of the bottom chords AB and BC. 
 
 Bottom Chord CD. The stress in this member (see Fig. 312) is 
 found by taking moments about panel point c. Then, evidently, the 
 stress will be a maximum when the load is placed so that this moment 
 is a maximum. According to Art. 90, the moment will be a maximum 
 when a load is at C such that the average unit load to the left is equal 
 
 >03' 
 
 Fig. 322 
 
 to the average unit load on the bridge. Placing wheel 7 at C, as shown 
 in Fig. 322, we have 
 
 (103 + 6.5)^.5? = 2,190 lbs. 
 ot) 
 
 for the average unit load to the left of C (considering one-half of wheel 
 7 as being to the left of C) and 
 
 [284 +(103x2)]^? =2,177 lbs. 
 
 for the average unit load on the bridge. This position of the load comes 
 nearest to satisfying the criterion for maximum moment about c (or C). 
 Then by taking moments about J (using Table A) we obtain 
 
 R = [16,364 + (284 x 103) + 103 *] ^^ = 249,880 lbs. (about) 
 
 /v/vO 
 
 for the reaction at A. Next, taking moments about c of the forces to the 
 left and dividing by the height of the truss at that point we obtain 
 
444 STRUCTURAL ENGINEERING 
 
 S = [ (249,880 x 50) - (2,155 x 1,000) ] ~ = 273,000 Ibs. (about) 
 
 OY.OO 
 
 for the maximum stress in the bottom chord CD due to the 40 loading 
 and multiplying this by 50/40 we obtain 341,250, say 341,000 Ibs. due to 
 the 50 loading which is the maximum live-load stress desired. 
 
 When this maximum stress occurs the live load extends over practi- 
 cally the entire span and hence the L in the impact formula will be taken 
 as 225. So for the impact stress we have 
 
 Now, adding together the above live-load stress and impact, and the 
 dead-load stress given in Fig. 311, we obtain 
 
 341,000 + 194,000 + 152,000 = 687,000 Ibs. 
 
 for the total maximum stress in bottom chord CD. 
 
 Top Chord be. Considering the forces to the left of section 1-1 (Fig. 
 322) it is readily seen that the horizontal component of the maximum 
 stress in top chord be is equal to the maximum stress in bottom chord CD. 
 So by multiplying the stress in CD (found above) by sec< (see Art. 190) 
 we obtain 
 
 SI = 341,000 x 1.037 - 353,617, say, 354,000 Ibs. 
 
 for the maximum live-load stress in top chord be. For the impact we 
 have 
 
 / - (aJ 30Q ) 35 ^ 000 = 20 ^ 000 lbs - ( about ) 
 
 Now, adding together the above live-load stress and impact, and 
 the dead-load stress given in Fig. 311, we obtain 
 
 354,000 + 202,000 + 157,000 = 713,000 lbs. 
 
 for the total maximum stress in top chord be. 
 
 Bottom Chord DE. The stress in this member is found by taking 
 moments about panel point d and hence the stress in the member will be 
 a maximum when the moment about d is a maximum. Placing wheel 11 
 at D, as shown in Fig. 323, we have 
 
 (152 + 10) = 2,160 
 
 o 
 
 for the average unit load to the left of D (or d) and 
 (284 + 210)^=2,195 
 
 for the average unit load on the bridge. This position of the load come* 
 the nearest to satisfying the criterion for maximum moment about d 
 (or D). 
 
 Then by taking moments about J we obtain 
 
 R = [ (16,364 x 75) + (284 x 105) + 105 ] = 254,260 lbs. (about) 
 
DESIGN OF SIMPLE RATLEOAD BRIDGES 
 
 445 
 
 for the reaction at A. Next, taking moments about d of the forces to the 
 left and dividing by the height of the truss at that point we obtain 
 
 S= [(254,260 x 75) - (5,848 x 1,000)] 
 
 1 
 
 42.43 
 
 = 311,600 Ibs. (about) 
 
 for the maximum stress in chord DE due to the -E40 loading and multiply- 
 ing this by 50/40 we obtain 389,500, say 390,000 Ibs. for the maximum 
 stress due to the E50 loading. 
 For the impact we have 
 
 390,000 = 222,856, say, 223,000 Ibs. 
 
 Now, adding together the above live-load stress and impact, and the 
 dead-load stress given in Fig. 311, we obtain 
 
 390,000 + 223,000 + 174,000 = 787,000 Ibs. 
 
 for the total maximum stress in bottom chord DE. 
 
 Top Chord cd. Considering the forces to the left of section 2-2 (Fig. 
 323) it is readily seen that the horizontal component of the maximum 
 
 Fig. 323 
 
 stress in top chord cd is equal to the maximum stress in bottom chord DE. 
 So by multiplying the maximum stress in DE by sec<l we obtain 
 
 390,000 x 1.016 = 396,240, say, 396,000 Ibs. 
 
 for the maximum live-load stress in top chord cd. 
 For the impact we have 
 
 396,000 = 226,000 Ibs. (about). 
 
 Now, adding together the above live-load stress and impact, and the 
 dead-load stress given in Fig. 311, we obtain 
 
 396,000 + 226,000 + 177,000 = 799,000 Ibs. 
 
 for the total maximum stress in top chord cd. 
 
 Bottom Chord EF. The stress is found in this member by taking 
 moments about panel point e and hence the stress in the member will 
 be a maximum when the moment about e is a maximum. 
 
 Placing wheel 13 at E, as shown in Fig. 324, we have 2,020 Ibs. for 
 average unit load to the left of E (or e) and 2,062 Ibs. for the average 
 unit load on the bridge. This position of the load comes the nearest to 
 satisfying the criterion for maximum moment about e (or E). 
 
446 STRUCTURAL ENGINEERING 
 
 Then taking moments about J we obtain 
 
 R = [16,364 + (284 x 90) x 90*] M^ = 222,320 Ibs. (about) 
 
 for the reaction at A. Next, taking moments about e of the forces to the 
 left and dividing by the height of the truss at that point we obtain 
 
 S= [(222,320 x 100) - (7,668 x 1,000) ]j~= 325,743 Ibs. 
 
 for the maximum stress in bottom chord EF due to the 40 loading and 
 
 325,743 x |9 = 407,178, say, 407,000 Ibs. 
 
 for the maximum stress due to the -E50 loading. 
 For the impact we have 
 
 1= ( 225 3 +3 OQ ) 407,000 -232,571, say 233,000 Ibs. 
 
 Now adding together the above live-load stress and impact, and the 
 dead-load stress given in Fig. 311 we obtain 
 
 407,000 + 233,000 + 184,000 = 824,000 
 
 for the total maximum stress in bottom chord EF. 
 
 Top Chord de. Considering the forces to the left of section 3-3 (Fig. 
 324) it is readily seen that the horizontal component of the maximum 
 
 stress in top chord de is equal to the maximum stress in bottom chord EF. 
 So by multiplying the maximum stress in the bottom chord EF, which, 
 as found above, is 407,000 Ibs., by sec<2 we obtain 
 
 407,000 x 1.004 = 408,628, say, 409,000 Ibs. 
 
 for the maximum live-load stress in top chord de. 
 For the impact we have 
 
 /= (a a5 3 + 300 ) 409,000 = 234,000 Ibs. (about). 
 
 Now, adding together the above live-load stress and impact, and the 
 dead-load stress given in Fig. 311, we obtain 
 
 409,000 + 234,000 + 185,000 = 828,000 Ibs. 
 
 for the total maximum stress in top chord de. 
 
 Top Chord ef. In accordance with usual practice we will consider 
 thfi stress in top chord ef to be equal and opposite to the stress in bottom 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 447 
 
 chord EF. This would be absolutely true if the shear in panel EF were 
 zero. If a uniform load be used the shear would be zero and in case 
 wheel loads are used the shear is slight and hence the usual practice is 
 not far wrong in any case. 
 
 We have now determined all of the stresses in one-half of the truss 
 and as the structure is symmetrical about the center of span these are all 
 the stresses necessary for designing the trusses. We can next write the 
 above stresses on the stress sheet, Fig. 328. 
 
 197. Designing of Members in Trusses. As the intermediate 
 posts are the governing members (see Art. 176) we will consider them 
 first. We will first ascertain as to whether channels can be used. The 
 longest post is J74-L4 (Fig. 328), which has a length of about 536 ins. 
 The average radius of gyration of 15" channels is about 5.4 (see Table 
 3). Then for the maximum L/r we have 
 
 536-5.4 = 99.2. 
 
 The maximum allowed for L/r for main members by the specifications 
 is 100. So, as far as L/r is concerned, 15" channels can be used. We 
 will next ascertain as to whether the area of the channels is sufficient. 
 Post U2-L2 has the greatest stress and hence will require the great- 
 est area. This post is about 454 ins. long. Now substituting in the 
 column formula, using the average radius, we have 
 
 p = 16,000 - 70 = 10,115 Ibs. 
 
 o.4 
 
 for the allowable unit stress. Dividing this into the stress in the column 
 we obtain 
 
 221,000 -f 10,115 = 21.74 sq. ins. 
 
 for the required area. From this it is seen that channels can be used. 
 
 Post U2-L2. The area just found for this post is 21.74 n ". Use 
 2 [sl5"x40#=23.52 n ". (The 15"x35# channels have 1.16 n " less 
 area than required.) 
 
 Post U3-L3. This member has a maximum compressive stress of 
 157,000 Ibs. and a maximum tensile stress of 80,000 Ibs. Combining, 
 according to the specifications, we have 
 
 157,000 + 197,000 Ibs. 
 
 & 
 
 for the maximum compression and 
 
 8( y)00 + 80 f = 120,000 Ibs. 
 
 2 
 
 for the maximum tension to be considered in designing the member. 
 
 The length of the member is about 509 ins. So substituting in the 
 column formula, using the average radius (as preliminary), we have 
 
 p= 16,000 -70 - = 9,400 Ibs. (about) 
 
448 STRUCTURAL ENGINEERING 
 
 for the allowable compressive unit stress. Then we have 
 197,000 4-9,40.0 = 20.96 sq. ins. 
 
 for the area of cross-section required for compression, and for the area 
 required for tension we have 
 
 120,000 -^ 16,000 = 7.5 sq. ins. 
 
 As is seen, the compression governs. Referring to Table 3, it is seen 
 that 2 [s 15" x 35* =20 58 n " have the nearest to the required area for 
 compression. So substituting the radius of these channels in the column 
 formula we have 
 
 16,000 -70 - = 9,615 Ibs. 
 
 o.oo 
 
 for the required unit stress. Dividing this into the maximum compres- 
 sion we obtain 
 
 197,000 -r 9,615 = 20.49 sq. ins. 
 
 for the required area. This is very close to the area of the 
 2 [s!5"x35# and hence these channels will be used. 
 
 Post U4-L4- This member has a maximum compressive stress of 
 112,000 Ibs. and a maximum tensile stress of 24,000 Ibs. Combining, 
 we have 124,000 Ibs. compression and 36,000 Ibs. tension to be consid- 
 ered in designing the member. The length of the member is about 536 ins. 
 Then, using the radius of the lightest 15" channel (as the stresses are 
 low), we have 
 
 16,000- 70 - = 9,324 Ibs. 
 0.0,0 
 
 for the allowable unit compressive stress on the member. Dividing this 
 into the above combined stress we have 
 
 124,000 -r 9,324 = 13.3 sq. ins. (about) 
 
 for the required area for compression. For the area required for tension 
 we have 
 
 36,000 ^ 16,000 = 2.25 sq. ins. 
 
 As is seen, the compression governs. 2 [s 15" x 33 # = 19.8 n " (which 
 are the lightest 15" channels) will be used for this member. As is 
 seen, there is an excess of 6.5 n " of metal, yet we cannot do better if 
 channels are used. The L/r would be too great if 12" channels were 
 used instead of the 15" channels. 
 
 Hanger U1-L1. This is entirely a tension member. For the re- 
 quired net area of cross-section we have 
 
 198,000 -r 16,000 = 13.37 sq. ins. 
 We will use 2 [s 15" x 33* = 19.8- (21/32 x4) =17.18 n "net. 
 
DESIGN OF SIMPLE KAILKOAD BEIDGES 449 
 
 (In this case the metal cut-out of the flanges is considered.) This is more 
 section than needed but 15" channels are used for the intermediate posts 
 and for the sake of appearance they will be used for this member. 
 
 Diagonal U1-L2. For the required net area we have 386,000- 
 16,000 = 24.12" net. Use 2 bars S" x 1 J" = 24.0". 
 
 Diagonal U2-L3. For the required net area we have 280,000 + 
 16,000 = 17.5 n " net. Use 2 bars 7" x 1 J" = 17.5 n ". 
 
 Diagonal U3-L4. For the required net area we have 222,000 -r 
 16,000 = 13.87" net. Use 2 bars 7" x 1 = 14 n " net. 
 
 Diagonal U4-L4. For the required net area we have 171,000 -r 
 16,000 = 10.68 n " net. Use 4 Ls 5" x 3J x | = 12.20 - 1.50 = 10.7 n " net. 
 
 Counter U4~L3. For the required area of cross-section we have 
 90.000-7- 16,000 = 5.62 n ". Use 1 bar 2-f" x2f" = 5.64 n " (standard bar). 
 
 Bottom Chords LO-Ll and L1-L2. For the required net area we 
 have 493,000 + 16,000 = 30.8 n ". Use the following section : 
 
 2 pis. 20" xf " = 25.0 -3.75 =21.25" net 
 
 4 Ls 3 J" x 3J" x T y = 11.48 - 1.75 = 9.73 n " net 
 
 30.98 n " net 
 
 Bottom Chord L2-L3. For the required area of cross-section we 
 have 687,000 -r 16,000 = 42.92". Use the following section : 
 
 2 bars 8" x If" = 22.00" 
 2 bars 8" x l T y = 21.00" 
 
 43.00" 
 
 Bottom Chord L3-L4- For the required area of cross-section we 
 have 787,000 ~ 16,000 = 49.18". Use the following section: 
 
 2 bars 8" x 1" = 24.00" 
 2 bars 8" x l T y = 25.00" 
 
 49.00" 
 
 Bottom Chord L4-L4. For the required area of cross-section we 
 have 824,000 + 16,000 = 51.5". Use 4 bars 8" x If = 52.0". 
 
 In designing the top chords, it is best to design the lightest section 
 first, using minimum thicknesses of web (if sufficient) so that the area of 
 cross-section can be increased for the others by merely increasing the 
 thickness of the webs. So in this case we will first consider top chord 
 C71-Z72, as it has the least stress and consequently will have the lightest 
 section. 
 
 Top Chord U1-U2. The first thing to do is to determine the gen- 
 eral dimensions of the section. The width of the posts will be 12J" 
 (see Art. 176) and the maximum thickness of eye-bars is 3" (2 8"x if" 
 bars at [71 ) and allowing, say, 2" for pin plates and clearance we have 
 12^ + 3 + 2 = 174-" for the required distance between the webs. Then if 
 the webs be f" thick (each) and the top angles be 4"x4" we obtain 
 17J + li + 8 = 26f, say, 27" for the width of the cover plate. The radius 
 of gyration in reference to the y-y axis (see Art. 176) is approximately 
 
450 STRUCTURAL ENGINEERING 
 
 equal to the distance from the center of the cover plate out to the outer 
 face of the web, which in this case is (17^ + li) -r 2 = 9.37. Now, in 
 order that the section be of economic depth, that is, so that the radius 
 of gyration about the x-x axis is the same as about the y-y axis, the webs 
 should have a depth 0.4 of which should equal the above radius. So for 
 the economic depth of the web we have 
 
 So we will make the webs 24" deep. According to the specifications, 
 the thickness of the cover plate should not be less than ^ of the dis- 
 tance between the rivet lines. The distance between the rivet lines in 
 this case will be about 17J + 1J + 4^ = 23 J". ^ of this distance is 
 about -j 9 6 of an inch. So we will make the cover plate -%" thick. The 
 thickness of the webs, according to the specifications, should not be less 
 than ^ of the distance between the flanges. If the top angles be 
 4"x4" and the bottom angles 6"x4" (the 6" leg along the web), we 
 have 24- (4 + 6) =14" for the distance between the flanges. Then for 
 the minimum thickness of the webs we have 14 x 1/30 = 0.46, which is 
 about -Q of an inch. 
 
 The top angles should be of minimum thickness and the bottom 
 angles should be of maximum thickness in order that the center of gravity 
 of the section be as near the center of the web as possible. 
 
 The length of the member (U1-U2) is about 312 ins. Now substi- 
 tuting this length and the approximate radius given above in the column 
 formula we obtain 
 
 p = 16,000 - 70 = = 13,670 Ibs. 
 9.o7 
 
 for the approximate allowable unit stress. Dividing this into the maxi- 
 mum stress in the member we obtain 
 
 713,000 -r 13,670 = 52.15 sq. ins. 
 
 for the approximate required area of cross-section. 
 
 In accordance with the above we obtain the following section: 
 
 l_ C o V er pi. 27"x T V' = 1 
 2 web pis. 24"x T y = 21.00 n " 
 2 Ls4"x4"x" = 5.72 n " 
 2 Ls 6" x 4" x T y = 10.62 n// 
 
 52.52 n " 
 
 This section is really the minimum and is very near the approxi- 
 mate section found above. 
 
 Now taking moments about the cover plate, in the manner shown 
 in Art. 176 we obtain 9.38" for the distance from the cover plate down 
 to the horizontal gravity axis of the section. 
 
 The next thing is to see if the eye-bars will fit into the top chord. 
 The largest pin, which will be at Ul, will be about 7" in diameter. From 
 the table of eye-bars, to be found in practically all structural hand- 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 451 
 
 books, it is seen that the 8" bars require 17J" head for a 7" diameter 
 pin. Half of this is 8|". So it is seen that the eye-bars fit into the 
 chord satisfactorily, as it is 9.38" from the gravity axis to the cover 
 plate and hence the bars will not interfere with the plate. 
 
 In the same mariner as shown in Art. 176 the radius of gyration 
 about the horizontal gravity axis, or x-x axis, is found to be about 9.5 and 
 9.3 about the y-y axis. Now using the least radius we obtain 
 
 q-i o 
 
 16,000 - 70 -^- = 13,652 Ibs. 
 
 for the actual allowable unit stress. 
 
 Then dividing this into the stress we obtain 
 
 713,000^13,652 = 52.22 
 
 for the required area of cross-section which is practically equal to the 
 above assumed section and hence that section will be used. 
 
 Top Chord U2-U3. Dividing the maximum stress in this member 
 by the allowable unit stress found for C71-J72 (which will be practically 
 the same as for this member) we obtain 
 
 799,000 ^ 13,652 = 58.53 sq. ins. 
 
 for the approximate area of cross-section required. The following section 
 has about this area: 
 
 1 cov. pi. 2T"x T y = 15.18 n " 
 2 web pis. 24"x 1 V' = 27.00 n " 
 2 Ls4"x4"xf" = 5.72 n// 
 2 Ls 6" x 4" x A" = 10.62 n " 
 
 58.52 n " 
 
 By taking moments about the cover plate we obtain 9.66" for the 
 distance from the cover plate down to the horizontal gravity, or x-x axis. 
 The radius of gyrations (found as previously explained) in reference 
 to the x-x axis is 9.3 and the radius in reference to the y-y axis is 9.2. 
 
 The length of the member is about 305 ins. Now, using the least 
 radius we obtain 
 
 16,000 -70 - = 13,680 Ibs. 
 y.,o 
 
 for the allowable unit stress. Dividing this into the maximum stress 
 in the member we obtain 
 
 799,000 * 13,680 = 58.40 sq. ins. 
 
 for the required area of cross-section. This is practically the same 
 as the approximate section found above and hence that section will be 
 used. 
 
 Top Chord U3-U4. Dividing the maximum stress in the member by 
 the allowable unit stress found for U2-U3 we obtain 
 
 828,000 -r 13,680 = 60.52 sq. ins. 
 
452 
 
 STRUCTURAL ENGINEERING 
 
 for the approximate area of cross-section required. The following section 
 has about this area: 
 
 1 cov. pi. 27"x T V' = 1 
 2_web pis. 24" xf" = 30.00 n " 
 
 2 Ls4"x4"xf" = 5.72 n " 
 2_Ls 6" x 4" x T V = 10.62 n// 
 
 61.52 n " 
 
 The allowable unit stress for this member is practically the same 
 as for U2-U3 and the above section is as near the required section as we 
 can obtain without changing the thickness of the bottom angles which 
 would be an objectionable thing to do, as the center of gravity would be 
 shifted, and hence the above section will be used. 
 
 Top Chord U4-U4* As the allowable unit stress for this member 
 is practically the same as for [73-174 and the stress is practically the same 
 the same section will be used as for [73-174. 
 
 198. Designing of the Bottom Lateral System. The load 
 
 specified by the specifications for the bottom laterals is 700 Ibs. per foot 
 of span. This load, according to the specifications, is to be considered 
 as a live load. 
 
 For the panel load we have 700x25 = 17,500 Ibs., and for determin- 
 ing the stresses we have the following: 
 
 17,500 -=-9 = 1,944.44. 
 Secw 1.78. 
 
 1,944.44x1.78= 3,461.1. 
 
 fe-z 
 
 V ^' 
 
 X 
 
 X 
 
 Fig. 325 
 
 Lateral Ab. For the maximum tensile stress in this member (see 
 Fig. 325, also Art. 177) we have 
 
 36 x 3,461.1 = 124,600, say 125,000 Ibs. 
 For the area of cross-section required we have 
 
 125,000 -f 16,000 = 7.81 sq. ins. 
 
 Use 2 Ls 6" x 3J" x " = 8.0" net. 
 
 Lateral Be. For the maximum tensile stress in this member we have 
 
 28 x 3,461.1 = 96,910, say 97,000 Ibs. 
 For the area of cross-section required we have 
 
 97,000 + 16,000 = 6.06 sq. ins. 
 Use 2 Ls 6" x 3 J" x |" = 6.09" net. 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 
 
 453 
 
 Lateral Cd. For the maximum tensile stress in this member we 
 have 
 
 21 x 3,461.1 = 72,683, say 73,000 Ibs. 
 
 For the area of cross-section required we have 
 
 73,000 -f 16,000 = 4.56 sq. ins. 
 
 Use 2 Ls 34" x 34" x //' = 4.87 n " net. 
 
 Lateral De. For the maximum tensile stress in this member we have 
 
 15 x 3,461.1 = 51,916, say 52,000 Ibs. 
 For the area of cross-section required we have 
 
 52,000 + 16,000 = 3.25 sq. ins. 
 
 Use 2 Ls 34" x 34" x f " = 4.21 n " net. 
 
 Lateral Ef, For the maximum tensile stress in this member we have 
 
 10 x 3,461.1 = 34,611, say 35,000 Ibs. 
 For the area of cross-section required we have 
 
 35,000 -=- 16,000 = 2.18 sq. ins. 
 
 Use 2 Ls 34" x 3^" x f" = 4.21 n " net. 
 
 This completes the designing of the bottom laterals as the structure 
 is symmetrical about the center of span. 
 
 The stringer bracing is designed as previously explained for deck 
 plate girder bridges and the transverse struts connecting to the stringers 
 iit the points of intersection of the laterals are designed as explained in 
 Art. 177 for the 150-ft. riveted bridge. 
 
 199. Designing of Top Lateral System. The load specified by 
 the specifications for the top laterals is 200 Ibs. per foot of span. This 
 load, according to the specifications, is to be considered as a live load. 
 
 For the panel load we have P - 25 x 200 = 5,000 Ibs., and for deter- 
 mining the stresses we have the following: 
 
 5,000 -=-9 = 555.55. 
 
 Seco>3 = 1.78. 
 Secw =1.78. 
 
 <D 
 
 Fig. 326 
 
 Lateral Be. For maximum tensile stress in this member (see Fig. 
 326) we have (see Art. 178) 
 
 - secwl x 28 = 555.55 x 1.82 x 28 = 28,310, say, 28,000 Ibs. 
 
454 STRUCTURAL ENGINEERING 
 
 For the area of cross-section required we have 
 
 28,000 ~ 16,000 = 1.75 sq. ins. 
 
 Use 2 Ls 3J" x 3 J" x f" = 4.22*" net. These are about the smallest an- 
 gles used in railroad work and hence are used in this case. 
 
 Lateral Cd. For the maximum tensile stress in this member we have 
 
 z> 
 
 - seco>2 x 21 = 555.55 x 1.79 x 21 = 20,883, say 21,000 Ibs. 
 
 u 
 
 For the area of cross-section required we have 
 
 21,000 -f 16,000 = 1.31 sq. ins. 
 
 Use 2 Ls3J"x3J"xf" = 4.22 n "net same as for Be. 
 
 Lateral De. For the maximum tensile stress in this member we have 
 
 secw3 x 15 = 555.55 x 1.78 x 15 = 14,883, say 15,000 Ibs. 
 y 
 
 Use the same section as used for. Be 2 Ls 3J" x 3" x f". 
 
 Lateral Ef. For the maximum tensile stress in this member we have 
 
 secco x 10 = 555.55 x 1.78 x 10 = 9,888, say 10,000 Ibs. 
 
 y 
 
 Use the same section as used for Be 2 Ls 3 \" x 3 J" x f ". 
 
 Strut Cc. For the maximum compression in this member we have 
 (see Art. 178) 
 
 21 x 555.55 + ( ~^- \ = 14,166, say, 14,000 Ibs. 
 \ X / 
 
 The length of member (c.c. of chords) is 17 ft., or 204 ins. If the strut 
 be composed of 4 Ls 3-J-" x 3" x " the least radius of gyration will be 
 about 1.7. Then L/r = 204 -f 1.7 = 120, which is just the maximum limit 
 permitted by the specifications and hence the above angles will be used. 
 It is seen that the stress in this strut is so low that it does not really 
 influence the design, and, as the stress in the other struts Dd, Ee, etc. 
 are less yet, there is really no need of computing it. We will simply 
 make each of the other struts of 4 Ls 3" x 3" x f " and let it go at that. 
 200. Design of Portal. Assuming the ends of the end posts fixed 
 and using the same symbols as used in Art. 179 (see Fig. 271) we have 
 in this case 
 
 P = 5,000 Ibs., 
 
 R = 5,000 x 28 = 15,555 Ibs., 
 
 t/ 
 
 H = 10,277 Ibs., 
 V= (20,555 x 24.16) -s- 17 = 29,212, say 29,000 Ibs. 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 
 Taking moments about p (Fig. 327) we have 
 
 455 
 
 and substituting the numerical values given above and reducing we obtain 
 S = 18,055 + (10,277 x 15,665) -=- 8.5 = 36,994, say 37,000 Ibs. 
 
 for the compressive stress in the part Bo of the portal. 
 
 In a similar manner taking moments about t we obtain 
 
 SI = 2,500 + 18,939 = 21,439, say 22,000 Ibs. 
 
 for the tensile stress in the part ob of the portal. 
 
 For the stress in the parts po and ot of the portal we have 
 
 S2 = V secc/. = 29,000 x 1.42 = 41,180, say 41,000 Ibs. 
 
 This is compression in one and tension in the other. With the applied 
 forces acting as indicated in Fig. 327, op would be in tension and ot in 
 compression. 
 
 The designing of the sections of the members of the portal in this 
 case is mostly a matter of obtaining rigidity as the stresses are relatively 
 low. For each of the members 
 Bb f op and ot we will use 
 2 Ls 4" x 4" x f " and for eac?" 
 of the secondary members, the 
 ones shown dotted in Fig. 327, 
 we will use 2 Ls 3|"x3J"xf". 
 These sections are obtained in 
 the manner shown in Art. 179. 
 201. Design of End Post. 
 As stated in Art. 180, colli- 
 sion struts are desirable, but 
 they are sometimes omitted and 
 for the sake of variety we will 
 omit them in this bridge. The 
 cross-section of the end posts 
 will be the same in form as 
 that of the top chords. The Fig. 327 
 
 length of the member as re- 
 
 gards the x-x axis (see Fig. 265) will be taken as the full length, 
 which is 478 ins., and as regards the y-y axis the length will be taken as 
 the distance from the lower end of the end post to the portal (distance 
 pu, Fig. 327). This distance is about 376 ins. 
 
 Using the radius in reference to the x-x axis found for the chord 
 section 72-173 as preliminary we obtain 
 
 = 16,000 - 
 
 4-7fi 
 
 = 12,403 
 
 for the approximate allowable unit stress, provided there were no cross 
 bending on the member. Dividing this unit stress into the maximum 
 
456 STRUCTURAL ENGINEERING 
 
 stress in the member we obtain 782,000 -r 12,403 = 63.04". Now as the 
 member is subjected to cross bending, due to wind, the section should 
 very likely be larger than this. Let us assume the following section : 
 
 1 cov. pi. 27"x T V = 15.1S n " 
 2 web pis. 24" xlJ" = 33.00" 
 2 Ls4"x4"xi" = 7.50 n " 
 2 Ls 6" x 4" x f" = 11.72" 
 
 67.40 n " 
 
 Taking moments about the cover plate of this section it is found 
 that the distance from the cover plate to the gravity axis x-x is 10.14 ins. 
 
 The radius of gyration in reference to the x-x axis is 9.22. The 
 moment of inertia in reference to the y-y axis is 5,742 and the radius in 
 reference to the same axis is 9.23. 
 
 Then for the allowable unit stress, provided there were no cross bend- 
 ing we would have 
 
 16,000- 70 - = 12,371 Ibs. 
 
 &&& 
 
 But in case of direct and bending stresses combined this can be increased 
 25 per cent (see specifications). So we have 
 
 12,371x1.25 = 15,464 Ibs. 
 
 for the allowable unit stress. 
 
 The maximum bending moment on the end post due to wind, consid- 
 ering the posts fixed at the ends, will occur at the bottom of the portal; 
 that is, at points p and t (Fig. 327). For this moment we have 
 
 M = 10,277 x 15.665 x 12 = 1,931,870 inch-lbs. 
 
 (see Fig. 327). 
 
 Then for the maximum bending stress we have 
 
 1,931,870 
 /= ' x 13. 5 = 4,542 Ibs. persq. in. 
 
 O j I *X /V 
 
 For the actual direct unit stress we have 
 
 782,000 -=-67.4 = 11,602 Ibs. 
 Now adding the bending stress to the direct stress we have 
 
 4,542 + 11,602 = 16,144 Ibs. 
 
 for the actual unit stress on the member due to direct stress and cross 
 bending combined. As this stress is very near the allowed, being only 
 680 Ibs. more (16,144-15,464 = 680), the above assumed section will 
 be used. 
 
 202. Maximum Reaction on Shoe. The dead-load reaction on the 
 shoe is the same as for the truss except for the half panel load at the end. 
 
DESIGN OF SIMPLE EAILKOAD BEIDGES 457 
 
 The panel load of dead load is 33,000 Ibs. (see Art. 195). Then for 
 the dead-load reaction on the shoe we have 
 
 4J x 33,000 = 148,500 Ibs. 
 
 Placing wheel 2 at one end of the span, as shown in Fig. 278 (Art. 
 182) for the 150-ft. span, and taking moments about the other end we 
 obtain 365,000 Ibs. for the maximum live-load reaction on the shoe. Multi- 
 plying this by 300 ~ (225 + 300) we obtain 208,500 Ibs. for the impact. 
 Then adding together the above dead- and live-load reactions and impact 
 we obtain 
 
 148,500 + 365,000 + 208,500 = 722,000 Ibs. 
 
 for the total maximum reaction on the shoe. 
 
 The stress sheet, Fig. 328, can now be completed and then the work 
 of designing the details can proceed. 
 
 203. Details. The general details of the entire span are shown in 
 Figs. 329, 330 and 331. These details are drawn to a f" scale on f" 
 layout; that is, the outline of the truss is drawn to a f" scale and the 
 details are drawn on this layout to a f " scale. 
 
 Joint LO. The detail of this joint is shown in Fig. 329. In 
 drawing this joint, first the joint line should be located, which is done 
 by bisecting the angle between the end post and shoe. Then the distance 
 from the cover plate of the end post to the center line of pin should be 
 determined as explained in Art. 184. Then the outline of the lower 
 end of the end post can be drawn and next the outline of the shoe can 
 be sketched. The shoe should be deep enough to resist the cross bend- 
 ing on it, as explained in Art. 142, and it should be long enough to cover 
 the rollers. The general dimensions of the shoe and number of rollers 
 are first assumed and then modified, if necessary, to agree with the 
 calculated requirements. 
 
 By drawing the end view and part cross-section of the joint, as 
 shown, to the left of I/O, about the desired length of the rollers is obtained. 
 The size of the rails in the pedestal is selected, the rails are spaced and 
 then a suitable diameter of roller can be determined. In this case, as is 
 seen, 70-lb. (per yd.) rails are used. Allowing for one-eighth of an inch 
 to be planed off the top of these rails, the width of bearing of each 
 against a roller is 2J" (see Carnegie and the Illinois Steel Company 
 catalogue) . 
 
 Resolving the stress in the end post vertically (which can be done 
 very rapidly by graphics) we obtain 612,000 Ibs. (about). One-half of 
 this, which is 306,000 Ibs., should be supported by the rails to the left 
 of the center line of the end post. Assuming 7" rollers (as shown), the 
 allowable bearing per inch of length of roller is 7 x 600 = 4,200 Ibs. Then 
 as there are five rails (not considering the 90-lb. guide rail at the center 
 line) to the left of the center line of the end post and there being seven 
 rollers, we have 
 
 (5 x 2i) 7 x 4,200 = 330,700 Ibs. 
 
 for the allowable pressure on the part of the rollers to the left of the 
 center line of the end post, which, as is seen, is about 24,000 Ibs. more 
 than necessary. The first five rails to the right of the center line of the 
 
458 STRUCTURAL ENGINEERING 
 
 end post will support the other half of the vertical component of the end 
 post and the other two rails, directly under the end floor beam, should 
 be about sufficient to support the maximum reaction of the beam. This 
 reaction (see Fig. 328) is 143,500 Ibs. For the allowable pressure on 
 the two rails we have 
 
 (2 x 2 J) 7 x 4,200 = 132,300 Ibs., 
 
 which is 11,200 Ibs. less than required, but as the maximum stress in 
 the end post and the maximum reaction on the end floor beam do not occur 
 at the same time (and the other rails are understressed), the design of 
 the roller pedestal, as shown, is satisfactory. 
 
 The shoe proper must be made to conform with the details of the 
 end post. Assuming a 1" pin, we obtain 
 
 782,000 -=- (24,000 x 7) = 4.65 ins. 
 
 for the required thickness of pin bearing on the end post, or 2.32" on 
 each side. We have on each side of the post, as shown, a &", \" and " 
 plate and an \\" web, making in all 2" (2.375) of bearing, which is the 
 thickness required. 
 
 For the required thickness of bearing on the shoe we have 
 
 612,000 -=- (24,000 x 7) = 3.64 ins., 
 
 or 1.82" at each side. We have at each side, as shown, 3 f" plates, 
 making 1J" (1.875) bearing which is about the required bearing. It 
 will be seen that in addition to this bearing there is a " filler and a 
 A" J aw plate, but as these extend only a short distance below the pin, 
 they will not be considered for bearing. 
 
 The bottom edges of the vertical bearing plates of the shoe are 
 planed so that they bear firmly against the sole plate and hence the rivets 
 passing through the plates simply hold them together. However, there 
 should be as many rivets in the vertical legs of the angles connecting 
 these plates to the sole plate as is possible to put in so that the pressure 
 will be well distributed over the sole plate by the outstanding legs of 
 the angles. 
 
 The pin plates on the lower end of the end post should be arranged 
 so that they firmly grip the post. For the allowable bearing on the %" 
 inside plate against the 7" pin we have 
 
 A x 7 x 24,000 = 94,500 Ibs. 
 
 Then the number of J" shop rivets required in single shear to hold this 
 plate is 
 
 94,500 -r 7,200 = 13. 
 
 As is seen, there are 21. For the allowable bearing on the J" outside 
 plate, we have 
 
 x 7x24,000 = 84,000 Ibs. 
 
 Then for the number of J" shop rivets required in single shear to hold 
 this plate we have 
 
 84,000-7,200 = 12. 
 
459 
 
L"" 
 
 Fig. 329 
 
 460 
 

 22Sfl Thro. S. Ypin 'ConnectedSpan 
 General Dran'ng 
 
 '(//a. -r 
 Spedrtcofions, A ft. Asm, 
 
 Fig. 831 
 
 462 
 
DESIGN OF SIMPLE RAILKOAD BEIDGES 463 
 
 This plate should have a sufficient number of rivets connecting it directly 
 to the top and bottom angles of the end post to transmit the total pin 
 pressure on it to these angles. There are 14 passing through it and the 
 angles,, so the riveting is satisfactory, although counting the other rivets 
 passing through the plate there is an excess of 15 rivets. For the allow- 
 able bearing on the f " filler we have 
 
 f x 7x24,000 = 105,000 Ibs. 
 
 Then for the number of J" shop rivets required in single shear to hold 
 the filler we have 
 
 105,000-^7,200 = 15. 
 
 As is seen, there are 23 passing through the filler, which is an excess of 
 8 rivets. The ^" outside plate and the f" filler acting together tend to 
 shear the rivets off against the web and angles of the end post. The 
 combined allowable bearing of these two plates against the 7" pin is 
 
 84,000 + 105,000 = 189,000 Ibs. 
 
 Then for the number of J" shop rivets required to hold these two plates 
 when considered as acting together we have 
 
 189,000 -r 7,200 = 27. 
 
 As is seen, there are 37 an excess of 10 rivets. 
 
 Although from the above there appears to be an excess of rivets in 
 these pin plates, nevertheless, the rivets are spaced about as sparingly 
 as possible, while yet obtaining good details, so the riveting is satisfactory. 
 
 According to the specifications, the net area of cross-section through 
 the pin hole of the bottom chord (at Z/0) should be 25 per cent more 
 than the net section of the member. The net section of the member (see 
 Fig. 328) is 30.98 n ". Then the area of section through the pin hole 
 should be 38.72 n ". 
 
 As is seen, the T V' plates have a net area through the pin hole of 
 5.25 n ", the i" plates have 13.00 n ", and the web has 16.25 n ", making a 
 total of 34.51 n " for the plates. In addition, each of the angles can be 
 considered to the extent of four rivets in shear. Each angle then is 
 equivalent to (4 x 7,200) -^-16,000 = 1.8 n ", or 7.2 n " for the four angles. 
 Adding this to the net area of the plates we obtain 34.51 + 7.2 = 41.71", 
 which is 2.99 n " more than required. 
 
 According to the specifications (A. R. E. Ass'n), the net area along 
 the center line between the pin and the end of the member must be equal 
 to the net area of cross-section of the member. As is seen, we have 
 
 for the net area along the center line between the pin hole and the end 
 of the member. As this is practically equal to the net area of cross-sec- 
 tion of the member, which is 30.98 n ", the detail of the end of the bottom 
 chord, as shown, is satisfactory as far as sections are concerned. 
 
 Considering the net area of cross-section through the pin hole, the 
 strength of the T y outside plate in tension is Gx^ x 16,000 = 42,000 
 
464 
 
 STRUCTURAL ENGINEERING 
 
 Ibs. Then for the number of J" shop rivets required to the right of the 
 pin hole to transmit this stress in single shear we have 
 
 42,000 -=-7,200 = 5.8, say 6. 
 
 As is seen, 6 rivets are used. 
 
 The strength of the \" inside plate in tension is 13 x-|x 16,000 
 = 104,000 Ibs. Then for the number of J" shop rivets required to the 
 right of the pin hole to transmit this stress in single shear, we have 
 
 104,000*7,200 = 15. 
 
 As is seen, 16 rivets are used. From the above, it is seen that the detail 
 of the end of the bottom chord at LO is correct both as to section and 
 riveting and hence the detail is satisfactory. 
 
 Resolving the forces in the 
 end post horizontally and verti- 
 cally, we have the loading on the 
 pin at LO shown at (a), Fig. 332. 
 Considering first the bending mo- 
 ment on the pin due to the hori- 
 zontal forces, shown at (b), the 
 bending moment at d is zero and 
 at e it is equal to + 246,500 x 2 = 
 493,000"*, which is the maximum 
 horizontal moment, the forces 
 being symmetrical in reference to 
 the center line cc. 
 
 The maximum bending mo- 
 ment on the pin due to the vertical 
 forces shown at (c), as is readily 
 
 (Hoc) 
 
 Fi 332 seen, occurs at both g and h and is 
 
 equal to 306,000 x J = 229,500"*. 
 Then for the resultant maximum bending moment we have 
 
 M = 1493,000 * + 229,500*= 543,000 inch Ibs. 
 
 Applying Formula (1), Art. 83, taking / as 25,000 Ibs., we find 
 that a 6^" pin is required for bending. For the maximum shear on this 
 pin we have 
 
 306,000 
 
 (?'= 
 
 10,385 Ibs. per sq. in. 
 
 So the 6J" is large enough as far as shear is concerned, as 12,000 Ibs. 
 per square inch is allowed. The bearing as computed above is for the 7" 
 pin and to use a 6J" pin would necessitate increasing the bearing by 
 about one-fourth, which would increase the weight of metal twice as much 
 as would be saved on the pin by reducing it to 6^" diameter. So there 
 is really no economy in using the smaller pin and therefore the 7" pin 
 will be used. 
 
 The other details at LO are considered to be self-explanatory, espe- 
 cially after Arts. 184 and 186 are carefully read. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 465 
 
 Joint Ul. In drawing this joint the first thing to do is to locate 
 the joint line by bisecting the angle between the end post and top chord 
 and then the outlines of the members can be drawn and the portal located 
 as explained in the case given in Art. 184; and next the required bearing 
 on the pin can be calculated. 
 
 We will assume a 1" pin the size used at LO; then for the required 
 thickness of pin bearing on the end post we have 
 
 782,000 + (24,000 x 7) = 4.65 ins., 
 
 which is about 4JJ". As is seen, we have on each side of the post a 
 T y and a f" plate and a f" filler. This with the H" w eb makes 2f" 
 bearing on a side or 4^" for the member, which is practically the required 
 bearing. 
 
 For the required thickness of pin bearing on the top chord we have 
 
 713,000 -r 168,000 = 4.24 ins. 
 
 As is seen, we have on each side of the chord 1 \" and 2 -jV plates 
 and 1 -f-g" filler. This, with the ^V' web, makes 2f " bearing on a side 
 or 4f " for the member. This is about V more than required, but it is 
 necessary to use this thickness in order to pack the joint. 
 
 At least one pin plate in each case should extend at least 6 ins. 
 beyond the end of the tie plate. 
 
 The number of rivets required to connect the pin plates to the end 
 post is determined in the same manner as shown above for joint LO. 
 For the allowable pin pressure on the T V' plate we have 
 
 7 x&x 24,000 = 73,500 Ibs. 
 
 Then for the number of f " shop rivets required to transmit this in single 
 shear, we have 
 
 73,500 -r7,200 = ll (about). 
 
 It is obvious that most of the pin pressure on this T V outside plate will 
 be first transmitted to the f " plate, then to the f " filler, and on to the 
 web and angles of the end post instead of being transmitted directly to 
 the web and angles by the rivets, as the rivets are quite long and conse- 
 quently will bend to some extent. Let us assume that the total pin 
 pressure on the -f$" plate is transmitted to the f " plate. Then for the 
 total pressure on the f " plate we have 
 
 73,500 + 7 x f x 24,000 = 178,500 Ibs. 
 
 For the number of J" shop rivets required to transmit this force, we 
 have 
 
 178,500 -7,200 = 25 (about). 
 
 There should be about enough rivets connecting this f " plate to the angles 
 of the end post to transmit this 178,500 Ibs. directly to the angles in 
 order to distribute the pin pressure well .over the cross-section of the 
 member. As is seen, there are 22 rivets, which is about the correct num- 
 ber. For the allowable pin pressure (bearing) on the f " filler we h&ve 
 
 7 xf x 24,000 = 105,000 Ibs. 
 
_/f.G6 STRUCTURAL ENGINEERING 
 
 Then for the number of |" shop rivets required to transmit this pressure 
 in single shear, we have 
 
 105,000 -5-7,200 = 14.5, say 15. 
 
 As is seen, there are 16 rivets between the lower end of the filler and 
 the -jV' plate, which we can consider as holding the filler. 
 
 As is seen from the above, if the rivets just below the pin be consid- 
 ered only to take the pressure on the T V' outside plate, the riveting of 
 the pin plates to the end post at 71 is about correct. 
 
 The allowable pin pressure on the T V' inside pin plate of the top 
 chord is 7 x T 7 F x 24,000 = 73,500 Ibs. It requires about 10 J" shop 
 rivets to transmit this in singular shear. As is evident, practically all 
 of the 73,500 Ibs. will be transmitted first to the " inside pin plate 
 and from there on to the web, and hence there should be a sufficient 
 number of rivets connecting the -J" plate to the web to transmit the 
 combined pin pressure exerted on both the $" and -J" plate. 
 
 For the allowable combined pin pressure of the two, we have 
 
 73,500 + (7 x \ x 24,000) = 157,500 Ibs. 
 
 It requires about 22 J" shop rivets to transmit this in single shear. As 
 is seen, there are 38 passing through the -J" inside plate. But 10 of 
 these should be considered as holding the T V' inside plate, thus really 
 leaving 28, which is 6 more than required. 
 
 The allowable pin pressure on the T V' outside pin plate is 73,500 
 Ibs. To transmit this in single shear requires about 10 J" shop rivets. 
 As this plate connects directly to the angles of the chord, there should 
 be enough rivets connecting it to the angles to transmit the entire 73,500 
 Ibs. As is seen, there are 14 rivets connecting the plate to the angles. 
 So the riveting is satisfactory as far as the T V' outside plate is con- 
 cerned. The allowable pin pressure on the -f^" filler is 
 
 7 x T % x 24,000 = 94,500 Ibs. 
 
 To transmit this in single shear requires about 13 J" shop rivets. 
 As is seen there are 26 twice as many as needed. As the web of the 
 top chord is only T V' thick, it is a question as to whether the rivets 
 connecting the pin plates to the member are not over-stressed in bearing 
 on the web. 
 
 The \" inside plate has a total pressure, as given above, of 157,500 
 Ibs. There are 38-10 (the 10 being considered to hold only the T V 
 inside plate) to transmit this, which stresses each rivet 157,500 -r 28 
 = 5,625 Ibs. The pin pressure on the T V' outside filler is 94,500, as 
 given above, and this stresses each rivet 94,500 -r 26 = 3,634 Ibs. Adding 
 these values we have 5,625 + 3,634 = 9,259 Ibs. for the maximum rivet 
 bearing on the T V' we b- For the maximum allowable bearing of a J" 
 shop rivet on %" metal, we have 
 
 T VxJx24,000 = 9,180 Ibs., 
 
 which is about the bearing exerted. So, taking all in all, the riveting of 
 the pin plates of the top chord at 71 is about correct. 
 
 The calculations for the details at the end of the hanger at 71 are 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 
 
 467 
 
 made in the manner as shown above for the end of the bottom chord 
 at LO. 
 
 The net section of the member is 17,18 n ". Then the net section 
 through the pin hole should be 17.18 (1 + 0.25) =21.57". 
 
 The net area of the two channels through the pin hole is 
 19.9 - (7 x 0.4 x 2) = 14.3". The net area" of the 2 J" plates through 
 the pin hole is (12.5 x \ x 2) - (7 xj x 2) =5.5 n// and of the f" plates 
 it is 6.87", making in all 14.3 + 5.5 + 6.87 = 26.67", which is about 
 5" more than necessary. , 
 
 The net section along the center line between the pin hole and the 
 end of the hanger should be equal to the net section of the member, or 
 17.18", according to the specifications. We have (3.05x9.25) 
 - (3.05x3.5) =17.54", which is about the section required, and as 
 the distance shown as 9^" can not be increased, on account of clearance, 
 the detail of the end of the hanger at Ul is satisfactory as far as the 
 
 Fig. 333 
 
 section is concerned. The strength of the J" plate in tension is 
 16,000 x 5. 5 = 88,000 Ibs. To transmit this in single shear requires about 
 12 J" shop rivets, or 6 on a side, which should be below the pin. As 
 seen, there are 8. The strength of the f " plate is 16,000 x 6.87 = 109,900 
 Ibs. This requires about 15 J" shop rivets, or 7.5 on a side. As 
 seen, there are 8 passing through this plate outside of the \" plate. 
 So, taking all in all, the detail of the end of the hanger, as shown at Ul, 
 is satisfactory. 
 
 In determining the bending moment on the pin at LO, there is no 
 question concerning the live-load stress in the members as the maximum 
 in the end post and bottom chord occurs simultaneously, but at Ul the 
 case is different, for at that joint the maximum stress in the members 
 does not occur simultaneously, and, consequently, it is necessary to ascer- 
 tain the position of the live load for maximum bending moment on the 
 pin at that joint. This can be done only by trial, as the bending on the 
 
468 STRUCTURAL ENGINEERING 
 
 pin due to any member not only depends upon the stress in the member, 
 but upon its lever arm as well. 
 
 It is seen from Fig. 329 that the hanger and the diagonal (at C71) 
 have the longest lever arms and hence most likely the pin will be sub- 
 jected to the maximum moment when these members have about the 
 maximum simultaneous stress. As this can be ascertained only by trial, 
 let us place the loads as shown at (a), Fig. 333, in which case wheel 4 
 is at B. Taking moments about J (using Table A) we obtain the reaction 
 263,100 Ibs. at A, and taking moments about B we obtain the concentra- 
 tion 19,200 Ibs. at A. Further, taking moments about C and A we 
 obtain the concentration 75, GOO Ibs. at B all due to Cooper's 40 load- 
 ing. Then beginning at joint A and drawing the stress diagram shown 
 at (b) we obtain the stresses in all the members at joint b (Ul) due to 
 Cooper's 40 loading and multiplying these by f $ we obtain the live- 
 load stresses shown at (a), which are due to the 50 loading. Next, 
 multiplying each of these stresses given at (a) (except the stress in the 
 hanger) by 300+ (225 + 300) we obtain the impact stress in each 
 member. Then adding together these live-load stresses and the impact 
 and the dead-load stresses (given in Fig. 328), we obtain the stresses 
 shown at (c). 
 
 There is some question as to how much impact should be added 
 to the hanger. Should L in the impact formula be taken as 50 or 225? 
 As is seen, the impact really added is just sufficient to balance up the 
 vertical components on the pin. This is a fair average for the impact 
 and seems to be a rational value. 
 
 The stresses given at (c) are resolved graphically into horizontal 
 and vertical components as shown. The horizontal components for half 
 of the pin are shown at (d) and the vertical components for half of 
 the pin are shown at (e). 
 
 Proceeding as outlined in Art. 83, for the bending moment on the 
 pin due to the horizontal components, we have 
 
 M e = 
 
 M t = + 245,000 x {J = +168,400 inch Ibs. 
 
 M g = +168,400+ (-98,500x2^) =-34,700 inch Ibs. 
 
 Similarly, for the bending moment, due to the vertical components, 
 we have 
 
 M e = 
 
 M t = + 305,000 x -1 J = +209,700 inch Ibs. 
 
 M g = 209,700 + (212,000 x 2^) = +646,900 inch Ibs. 
 
 M b = 646,900 + (90,500 x If) =+793,900 inch Ibs. 
 
 As the members are symmetrically arranged in reference to the cen- 
 ter line of the chord, the moment at g, due to the horizontal forces, is 
 constant between the two diagonals and hence for the resultant maxi- 
 mum moment at h, we have 
 
 M = x /34,700 2 + 793,900 2 = 797,200 inch Ibs. 
 
 This, as is readily seen, is the maximum resultant bending moment 
 on the pin when the live load is in the position shown at (a), Fig. 333. 
 
DESIGN OF SIMPLE BAILEOAD BEIDGES 469 
 
 In fact, it is about the maximum moment that can occur on the pin. 
 Applying (1), Art. 83, taking / as 25,000 Ibs., we find that the above 
 moment (79 7,200" #) requires a 6J" pin and hence the 7" pin assumed 
 is about the correct size as far as the bending moment is concerned. 
 
 In case of doubt, in any case, as to the position of the live load, the 
 loading can be placed in different positions and the moment on the pin cal- 
 culated for each position in the same manner as shown above for that one 
 position. This method of procedure, as is evident, is tedious. As a rule, 
 the position of the live load can be ascertained near enough (as the impact 
 is questionable) by mere inspection. 
 
 The maximum shear on the pin at each side of the chord is about 
 330,000 Ibs., which is the resultant of the stresses in the hanger and 
 diagonal. Then for the maximum shearing stress on the 7" pin we have 
 330,000 -r 38.4 = 8,590 Ibs. per sq. in. 
 
 From this it is seen that the 7" pin is amply large as far as shear 
 is concerned, 12,000 Ibs. being permissible, and as it is the correct size 
 for bending it will be used. 
 
 The details of the portal and laterals at [71 are considered to be 
 self-explanatory. The number of rivets required in each connection is 
 obtained by developing the section of the member connected, as explained 
 in Art. 184. 
 
 Joint LI. All of the details at this joint are practically self-ex- 
 planatory. The pin supports only the weight of the bottom chord. The 
 pin plates on the bottom chord are intended solely to replace the metal 
 cut from the chord by the pin hole. The rivets connecting the bottom 
 of the hanger to the lateral plate should be sufficient to transmit the 
 component of the lateral in panel LO-L~L along the bottom chord. This 
 component is about 105,000 Ibs. This requires 
 
 105,000 -- 6,000 = 17 J " field rivets. 
 16 are used. 
 
 The component of the stress in the same lateral along the floor 
 beam is about 70,000 Ibs. This requires about 12 J" field rivets. 12 
 are used. 
 
 The pin plates on the hanger are for the purpose of reinforcing the 
 hanger for bending. The bending on the hanger about the pin (at LI), 
 which is due to the longitudinal component of the stress in the bottom 
 lateral, is 105,000 x 14.25 = 1,496,250"#. The moment of inertia of the 
 two channels is 625.2, and the moment of inertia of the 2 -f-%" plates 
 is 246.1, making a total moment of inertia of 871.3. Then for the 
 maximum fiber stress in the hanger at LI, due to bending, we have 
 
 / = (1,496,250 x 7) - 871.3 = 12,020 Ibs. per sq. in. 
 
 From this it is seen that the hanger is amply strong to resist the bending 
 at LI. 
 
 The rivets connecting the lug angles to the bottom of the hanger 
 should be sufficient to transmit the longitudinal component of the stress 
 in the lateral. These rivets are in double shear and bearing on about 
 Jf" of metal. For the number of J" shop rivets required in double 
 shear, we have 
 
 105,000^ (7,200x2) = 7.3, say 8. 
 
470 
 
 STRUCTURAL ENGINEERING 
 
 -486OOO* 
 
 Co) 
 
 -6370OO' 
 
 J 
 
 .(c) 
 
 For the number required in bearing on the |f " metal, we have 
 
 105,000 -=-17,060 = 6. 
 
 As is seen, 8 are used the number required for shear. 
 
 Joint L2. The details of this joint are shown in Fig. 330. The 
 calculations for the end of the bottom chord are just the same as given 
 above for the other end of that member (at LO). 
 
 The pin bearing on the post should be sufficient to transmit the ver- 
 tical component of the maximum stress in the diagonal U1-L2. This 
 component is equal to about 300,000 #. Then for the required bearing 
 on the post, assuming a 7" pin, we have 
 
 300,000 -=- (7 x 24,000) = 1.78 ins. 
 
 or 0.89" on a side. The thickness of the web of the 40* channel is 
 0.52" (about ||"). This with the T y pin plate makes H" or 0.96", 
 which is about ^\ more than needed, but as counter-sinking (as a rule) 
 
 is not permitted in metal less 
 than T y thick the &" plates 
 shown will be used. 
 
 About the maximum bend- 
 ing moment on the pin will occur 
 when the bottom chord L2-L3 
 has maximum stress. (This can 
 be determined in the manner 
 shown above for joint 71.) The 
 wheels are in the position shown 
 in Fig. 322 when this stress 
 occurs. Taking moments about 
 J (Fig. 322) with wheel 7 at C 
 we can obtain the reaction R at 
 A and taking moments about B 
 we can obtain the concentration 
 at A. Then the stress in the 
 bottom chord AB can be obtained quickly by analyzing graphically the 
 joint A. Then the stress in chord BC is known, as it is equal to the stress 
 in chord AB. Next the stress in the diagonal bC (U1-L2) is readily de- 
 termined, as the horizontal component of its stress is equal to the difference 
 between the stress in chord BC and CD and as the vertical component of 
 the stress in the diagonal is equal to the stress in the post, we can readily 
 determine the stress in the post and thus we would have all the live-load 
 stresses in the members at L2 determined. 
 
 After determining these live-load stresses, the impact is determined 
 in each member by multiplying the live-load stress in it by 
 300 -=-(225 + 300). Then by adding together the live-load stress and 
 impact and the dead-load stress, we obtain the total stress in each 
 member as given at (a), Fig. 334. Two-thirds of a panel load of dead 
 load is added to the post to balance up the vertical component of the 
 diagonal. However, this should really be added to the post as the pin 
 is below the floor beam connection and hence most of the weight at the 
 joint is transmitted to the lower end of the post. 
 
 Fig . 334 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 471 
 
 The horizontal components of the stresses on one-half of the pin 
 are shown at (b) (Fig. 334), and the vertical components on one-half 
 of the pin are shown at (c). 
 
 For the bending moment on the pin, due to the horizontal com- 
 ponents, we have 
 
 M e = 
 
 M t = + (175,000 x If) = 306,250 inch Ibs. 
 
 M g = +306,250 + (-68,000 x 2| ) = +144,750 inch Ibs. 
 
 M h = +144,750 + ( 100,500 x 1 j) = +295,500 inch Ibs. 
 
 For the bending moment on the pin, due to the vertical component, 
 we have 
 
 M k = + (124,000 x 1 T V ) = 178,200 inch Ibs. 
 Then for the maximum resultant moment on the pin, we have 
 
 M = J295,500 2 + 178,200 2 = 345,000 inch Ibs. 
 
 Then by applying (1), Art. 83, taking / as 25,000, we find that the 
 above moment ( 345,000" #) calls for a 5J" pin (about). 
 
 The maximum shear on the pin is about 175,000 Ibs., which requires 
 that the pin be only about 4 j^'' diameter. 
 
 From the above it is seen that the assumed 7" pin is larger than 
 necessary, but the 7" will be used, for little would be saved by reducing 
 the size, as the thickness of pin bearing on the bottom chord (L2-LO) 
 and also on the post U2-L2 would have to be increased if a smaller pin 
 were used. 
 
 The other details at L2 are readily understood. The calculation 
 of the details of the bottom laterals, shown at this joint, is mostly a 
 matter of developing the members. The rivets connecting the bottom 
 of the post to the lateral plate should be sufficient to transmit the longi- 
 tudinal component of the stress in the lateral in panel L1-L2. 
 
 Joint U2. The top chord has a butt joint at this point; that is, 
 the ends of the members are planed so that they bear tightly against 
 each other and hence the stress is transmitted from one chord member 
 to the other chord member without passing through the pin. The splice 
 plates on the chord are intended only to hold the chords in line. How- 
 ever, there must be sufficient pin bearing on the chord to take the com- 
 ponent (along the chord U2-U3) of the stress in diagonal U2-L3. The 
 maximum stress in the diagonal (see Fig. 328) is 280,000 Ibs. Resolv- 
 ing this along the chord U2-U3 (which can be done graphically) we 
 obtain a component of 159,000 Ibs. To transmit this component, assum- 
 ing a 6" pin, requires 159,000+ (24,000 x 6) =1.1 // thickness of bearing, 
 or about \ n on each side of the chord. As is seen, there are 2 f" 
 plates and a T V' we b> making in all 1-&" bearing on each side of the 
 chord, which is quite excessive. 
 
 The pin bearing on the post should be sufficient to transmit the 
 maximum stress in the post, which is 221,000 Ibs. (See Fig. 328.) 
 This requires a bearing (assuming a 6" pin) of 221,000-=- (24,000x6) 
 
472 STRUCTURAL ENGINEERING 
 
 = 1.54" or 0.72" on a side. The web of the 40# channel is ft" thick. 
 This with the {%" plate makes j-j-" thickness of bearing on each side of 
 the post, which is excessive. 
 
 The maximum bending moment on the pin will occur when the 
 diagonal (J72-Z/3) has maximum stress, as the post (U2-L2) has about 
 the maximum stress at the same time. By placing the live load for 
 maximum stress in the diagonal (as shown in Fig. 315) and determin- 
 ing the reaction at A and the concentration at C (these are given on 
 pages 432 and 433) the live-load stress in each of the members at U2 
 can be readily determined by graphics. Then determining the impact 
 by multiplying each stress by 300 -r (150 + 300) and adding together 
 this impact and the live- and dead-load stresses and resolving the 
 resulting stresses horizontally and vertically, the maximum bending 
 moment on the pin can be determined in the same manner as shown above 
 for the other joints. This maximum moment is about 190,000 inch Ibs. 
 This requires about a 4J" pin. From the above, it is seen that the 6" 
 pin is larger than theoretically required. But it is usual practice to 
 limit the minimum diameter of a pin to -f^ of the width of the smallest 
 eye-bar connected. The eye-bars at U2 (as seen) are 7" wide. So, 
 according to this requirement, the pin should be 5.6" in diameter and 
 hence the assumed 6" pin is about the correct size, and will be used. 
 The other details at U2 are readily understood. 
 
 The details at 73 are similar to those at U2 and the details at L3 
 are very similar to those at L2 and hence the calculations of the details 
 at these joints are quite similar to those given above and consequently 
 are omitted. The same can be said regarding joints 74 and 1/4 (Fig. 
 331). The details at these joints are readily understood. 
 
 The calculations of the details of the intermediate floor beams, 
 shown in Fig. 331, are almost exactly as given in Art. 186 (pages 410 
 to 412) for the floor beams in the 150- ft. riveted span. 
 
 After the general drawings (Figs. 329, 330 and 331) are completed, 
 the shop drawings and shop bills for the structure can be made by 
 proceeding in the same manner as outlined in Art. 186 for the 150- 
 ft. span. 
 
 204. Camber. The camber of the trusses in this case is obtained 
 by lengthening the top chord y for each 10 feet of horizontal length, 
 as explained in Art. 185. The corresponding lengths of the diagonals 
 are calculated by using the mean lengths of the top and bottom chords 
 in each case. For example, the horizontal uncambered length of top 
 chord U2-U3 (Fig. 330) is 25 ft. So the cambered horizontal length 
 would be 25' - 0-jffr". Taking this as the base of a right angle triangle, 
 and the difference between the lengths of posts U3-L3 and U2-L2 as the 
 altitude, the cambered length of the top chord U2-U3 (given as 25' -5J") 
 is computed as the hypotenuse of the triangle. Then taking 25' - O^" 
 as the base of a right angle triangle and the length of post U2-L2 as the 
 altitude, the length of diagonal U2-L3 (given as 45' - 4") is calculated 
 as the hypotenuse of that triangle. The length of the end post is not 
 increased. 
 
 The determination of actual amount of camber of bridge trusses and 
 also the determination of the theoretical camber of the same will be 
 given later. 
 
DESIGN OF SIMPLE KAILKOAD BRIDGES 
 
 473 
 
 205. Graphical Determination of Live-Load Stresses in Curved 
 Chord Pratt Trusses. The influence line method, outlined in Art. 103,, 
 is the most convenient graphical method to use in this case, as the stresses 
 in the truss members are thus determined directly without considering 
 either shears or moments. 
 
 Chord Members. Let it be required to determine the live-load stress 
 in the top chord cd of the truss shown in Fig. 335, due to Cooper's 
 -E40 loading. 
 
 It is obvious that a single load moving from J to the left over the span 
 will cause a stress in chord cd which will be zero when the load is at J and 
 
 Fig. 335 
 
 increase constantly until the load reaches point D and that this stress will 
 then decrease constantly as the load moves on to the left from D, becoming 
 zero when the load reaches A. Then evidently the influence line for 
 the stress in chord cd is of the form c'o'd' , shown at (c). Now, know- 
 ing the form of the influence line, the next thing is to construct it. This, 
 as is readily seen, is a simple matter after the ordinate o'g' is known. 
 This ordinate o' 'g' ' , as we know, is the stress in the top chord cd, due 
 to a unit load at D. Then, of course, the first thing to do in construct- 
 ing the desired influence line is to determine the stress in the top chord 
 cd, due to a unit load at D. This stress can be determined very readily 
 
474 STRUCTURAL ENGINEERING 
 
 either analytically or graphically. The graphical determination is the 
 more convenient and hence that method will be used. By drawing, at 
 (a), the lines 00', OD and O'D we obtain, as we may say, the truss 
 ODO' , which will have end reactions exactly equal to those of the bridge 
 truss, due to a unit load at D, and as the member 00' coincides with 
 top chord cd, the stress, due to this unit load at D, will be the same 
 in the two. This is readily seen, for, taking moments about D, we 
 obtain 
 
 for the stress in each, where R represents the reaction at either A or 
 due to a unit load at D. 
 
 By drawing mp (at (b) ) and laying off nm (equal to one inch) 
 equal to one pound and drawing np, we have the influence line for the 
 reactions at A. Then gh is equal to the reaction at A and also at 0, due 
 to a unit load at D. 
 
 Then by drawing from h a line parallel to OD and from g a line 
 parallel to 00' , we obtain the line og, which is equal (in inches) to the 
 stress in chord cd, due to the unit load at D. In other words, by con- 
 structing a diagram of the forces at 0, we obtain the stress in 00', 
 which is equal to the stress in the chord cd, due to a unit load at D. 
 
 Then by drawing c'd' (at (c) ) and laying off o'g' og and draw- 
 ing c'o' and o'd' t we have the desired influence line for the stress in top 
 chord cd. 
 
 Then placing the loading for the maximum moment about D, accord- 
 ing to Art. 91 (see Fig. 323), and drawing the ordinates 1, 2, .... 7, as 
 shown, and multiplying each by the load or group of loads at each, and 
 adding together the results, the desired maximum stress in top chord cd 
 will be obtained. 
 
 Ordinates 1 and 2 are drawn at the center of gravity of equal 
 groups. Then by adding together the ordinates 1 and 2 and multiply- 
 ing by the weight of one of the groups, the stress in top chord cd, due 
 to the two groups, will be obtained. Similarly, by adding together the 
 ordinates 3 and 4 and multiplying by the weight of one of the groups 
 (at either of the ordinates), the stress in the top chord cd, due to those 
 two groups, will be obtained. Next, adding together ordinates 5 and 6 
 and multiplying by the weight of one of the wheels, the stress in top 
 chord cd, due to the two wheels (1 and 10), will be obtained. Multiply- 
 ing the ordinate 7 (which is equal to one-half of ordinate 8) by the total 
 uniform load the stress in chord cd, due to the uniform load, is obtained. 
 Then adding together all of these results, the total maximum live-load 
 stress in top chord cd is obtained. 
 
 The combined length of ordinates 1 and 2 (which can be quickly 
 combined by the use of a pair of dividers, as shown in Example 1 of 
 Art. 100) scales 1.54 ins. This means that if (as wm = l" = l#) one 
 unit load (1 Ib.) be placed on the truss exactly over ordinate 1 and 
 another exactly over ordinate 2, these two unit loads would produce a 
 stress of 1.54 Ibs. in top chord cd. Then evidently the two groups of 
 wheels at these ordinates, each weighing 80,000 Ibs., will produce a stress 
 
DESIGN OF SIMPLE EAILKOAD BEIDGES . 475 
 
 in the top chord cd of 
 
 1.54 x 80,000 = 123,000 Ibs. 
 
 The combined length of ordinates 3 and 4 scales 1.73 ins. Then 
 for the stress in top chord cd due to the two groups of wheels at these 
 ordinates, we have 
 
 1.74 x 52,000 = 90,480, say 90,500 Ibs. 
 
 Similarly for the stress due to wheels 1 and 10, the combined length 
 of ordinates 5 and 6 being 1.23 ins., we have 
 
 1.23x10,000 = 12,300 Ibs. 
 
 Ordinate 7 scales 0.41 ins. Then for the stress in top chord cd, 
 due to the uniform load, we have 
 
 0.41 x (2,000 x 105) = 86,100 Ibs. 
 Now, adding together the above, we obtain 
 123,000 + 90,500 + 12,300 + 86,100 = 311,900, say, 312,000 Ibs. 
 
 for the maximum stress in top chord cd due to Cooper's 40 loading, 
 and multiplying this by f$ we obtain 390,000 Ibs. for the stress due to 
 the 50 loading. As is seen from Fig. 328, this is 6,000 Ibs. less than 
 the actual stress, but as this difference is less than 2 per cent, the result 
 is accurate enough. That is, the change of section of the top chord 
 resulting from the error would be negligible. 
 
 As the horizontal component of the stress in top chord cd is equal 
 to the stress in bottom chord DE, as previously shown, and the position 
 of the loading for maximum stress in the two members being the same, 
 the influence line for the stress in bottom chord DE can be quickly 
 constructed after the one for the top chord cd is completed. Thus: By 
 drawing or we obtain the stress in bottom chord DE, due to a unit 
 load at D. Then drawing D'E' (at (d) ) and laying off r'o" ' or and 
 drawing D'r' and r'E', we obtain the influence line for the stress in 
 chord DE. The stresses can then be determined in this member in the 
 same manner as shown above for top chord cd. However, the stress 
 in bottom chord DE can be obtained more readily as the horizontal 
 component of the stress in top chord cd. The method, of course, can 
 be reversed; that is, the influence line for the stress in bottom chord 
 DE could be constructed and the stress in DE determined and then the 
 stress in top chord cd could be obtained by multiplying the stress found 
 in DE by the secant of the slope angle of chord cd. 
 
 The construction of the influence lines and determination of the 
 stresses in the other chord members are practically the same as shown 
 above for chords cd and DE. 
 
 For example, by drawing 0"0 f ", 0"E and E0" f and constructing 
 a diagram at ordinate z of the forces at 0" , the stress in top chord de, 
 due to a unit load at E, is obtained. Then the influence line for the 
 stress in top chord de, as well as for bottom chord EF, is readily 
 constructed. 
 
 As another example, by drawing tv (at (b) ) parallel to the end 
 
476 STRUCTURAL ENGINEERING 
 
 post, we obtain vu, which is equal to the stress in bottom chord AB (also 
 BC}, due to a unit load at B. Then drawing A'B' (at (e) ) and laying 
 off v'u* ' = vu (which can be done by using a pair of dividers) and drawing 
 A'v' and v'B' } we obtain the influence line for the stress in bottom chords 
 AB and EC, shown at (e). 
 
 Web Members. As an example, let it be required to determine the 
 stress in diagonal cD. The influence line for the stress in this member,, 
 as explained in Art. 103,, will be of the form A's'o'B' ', shown at (c) in 
 Fig. 336. To construct this influence line, first draw the influence lines 
 for reactions, as shown at (b). Then gh represents the reaction at A, 
 due to a unit load at D. Next, draw 00', OD and O'D. Then, drawing 
 og and oh (at (b) ) parallel, respectively, to 00' and OD, and ko 
 parallel to diagonal cD, we have the stress in the diagonal cD, due to a 
 unit load at D, represented by this line ko. Then, by drawing A'B' 
 (at (c) ) and laying off h'o' equal to ko, the part o'B' of the influence 
 line can be drawn. 
 
 Next, draw 01 and /DO', as shown at (a). Then, drawing vs and 
 su parallel, respectively, to 10' and 00' and is parallel to the diagonal 
 cD, we have the stress in the diagonal cD, due to a unit load at C, repre- 
 sented by this line ts. Then, by laying off t's' (at (c) ) equal to is, 
 the parts s' 'A' and s'o' of the influence line can be drawn, which will 
 complete the construction of the influence line A's'o'B'. 
 
 Any load at any point to the right of N will produce tension in 
 diagonal cD, and any load to the left will produce compression in the 
 same. Then, evidently, to obtain the maximum tension in the diagonal, 
 all the loads should be placed to the right of N. According to Art. 
 190, a wheel will be at D when the maximum tension in cD occurs. Then, 
 evidently, the loading will be in the position for maximum stress in 
 diagonal cD if the wheel be placed at &' that brings wheel 1 the closest 
 to N, the limit being that wheel 1 should not be to the left of N. From 
 this it is seen that the position of the loading for maximum stress in 
 the diagonal can be determined directly from the influence line instead 
 of applying the criterion of Art. 190. 
 
 The position of the loading for maximum tension in diagonal cD is 
 shown at (c). By multiplying the ordinates by the loads, as explained 
 above, the maximum tension in diagonal cD will be obtained. 
 
 In any case when a group of wheels comes at a point where the 
 influence line changes direction, as wheels 2 ... 5 do in this case, the ordi- 
 nate at the center of gravity of the group can not be used. In such 
 cases the weight of each wheel can be multiplied by the ordinate at it, 
 or the ordinates at the center of gravity of the wheels to either side of 
 the change of slope of the influence line can be used, or the sum of all 
 of the ordinates can be obtained by the use of dividers (see Example 
 
 1, Art. 100), and this sum multiplied by the weight of one wheel. Thus, 
 in the case shown at (c) the tensile stress in diagonal cD, due to wheels 
 2 ... 5, can be obtained by multiplying ordinate 1 by the weight of wheel 
 
 2, and ordinate 3 by the weight of wheels 3, 4 and 5, or by multiplying 
 the sum of all the ordinates 1, 2, 3 and 4 by the weight of one wheel. 
 
 The maximum compression in diagonal cD can be obtained by revers- 
 ing the loading; that is, bringing it on from the left, and placing the 
 wheel at t' that brings wheel 1 closest to N without passing that point. 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 
 
 477 
 
 The actual stress Is then obtained by multiplying the loads by the ordi- 
 nates, as previously explained. 
 
 The influence line for the maximum compression in post dD is 
 shown at (d). To obtain this influence line 00', OD, O'D, ODI' and 
 O'l f should be drawn. Then, by drawing fy parallel to 00' , yl parallel 
 to 01', and yz parallel to post dD, we have the stress in post dD, due 
 to a unit load at E, represented by yz. Then, drawing C'D' at (d), 
 and laying off y' z' equal to yz, we can draw the part y'D' of the influ- 
 ence line. Next, drawing qw parallel to 00' , wh parallel to O'D, and 
 ew parallel to post dD, we have the stress in post dD, due to a unit load 
 at D, represented by line ew. Then, laying off e'w f ', at (d), equal to ew, 
 we can complete the influence line C'w'y'D' by drawing C'w' and w' y' . 
 
 Fig. 336 
 
 The maximum compression in post dD can then be determined by 
 placing the loading on PD' with the wheel at z' that brings wheel 1 
 the closest to P, and proceeding as explained above in the case of the 
 diagonal dD. Influence lines for maximum tension in the posts will 
 be considered later. 
 
 The influence lines for maximum stress in the other diagonals and 
 posts and determination of the stresses are similar to the above. 
 
 The influence lines can be constructed more readily in the following 
 manner, taking first the one shown at (c) in Fig. 336: 
 
 The ordinate o'k' can be obtained in the manner explained above, 
 
478 
 
 STRUCTURAL ENGINEERING 
 
 and point N can be located by drawing OCx and O'Dx, as x is directly 
 over N. Having the ordinate o'k' determined and point N located, the 
 influence line A's'o'B' can be drawn. The ordinate #V, shown at (d), 
 can be obtained in the manner explained above, and point P can be 
 located by drawing ODx' and O'Ex* ', as x' is directly over P. Having 
 the ordinate y'z' determined and the point P located, the influence line 
 C'w'y'D' can be drawn. 
 
 A better way to locate points N and P is to draw the line OJ. 
 Then point F, where OJ intersects the diagonal cD, is directly over 
 point N; and point G, where the line OJ intersects diagonal dE, is 
 directly over point P. 
 
 It can be readily shown that this construction is correct. Referring 
 to Fig. 337, the structure OCDO'cO, supporting the loads PI and P2, 
 is a rigid structure. It is evident that the loads PI and P2 could have 
 
 Fig. 337 
 
 such relative weights that the member cD (also cC) would have zero 
 stress. In that case, OCDO' would be an equilibrium polygon. Then 
 by prolonging the segments O'D and OC to intersection, the point a? 
 is obtained, which would be at the resultant, or center of gravity, of 
 the loads PI and P2 (see Art. 45). Then both of the loads' could 
 be placed at x and the stress in member cD (also cC) would be 
 zero. Now, it is evident that as long as PI and P2 have this same 
 relative value, OCDO' would remain an equilibrium polygon, regardless 
 of the actual weights of the loads PI and P2; so, evidently, the stress 
 in cD due to any load at x will be zero, and hence x is the point of 
 zero stress for diagonal cD. 
 
 From similar triangles Dmx and DJO', we have* 
 
 from which we obtain 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 479 
 
 From similar triangles OAC and xmC, we have 
 
 = -, 
 iTom which we obtain 
 
 Now equating (1) and (2) equal, we have 
 
 6 A = a ' 
 from which we obtain 
 
 _ ^ f*\ 
 
 ~~ 77 10 ) . 
 
 c o/c 
 From similar triangles cFT and DFU, we have 
 
 (4). 
 
 But from similar triangles O'JO and cOT, we have 
 
 y_a 
 from which we obtain 
 
 h~L' 
 
 And from similar triangles JAO and JDU, we have 
 
 from which we obtain 
 
 Now substituting these values of ?/ and z in (4), we obtain 
 
 -- 
 t ~bk' 
 
 Equating (3) and (5) equal, we obtain 
 
 &r. ................ ............ 
 
 But c = d-g and t = d-s. 
 
 So, substituting these values in (6) and reducing, we obtain 
 
 which proves that points F and a are on the same vertical line. 
 
 Any of the other cases can be proved in a similar manner. Refer- 
 ring to Fig. 336, point 1 is the point of zero stress for diagonal bC and 
 
480 
 
 STRUCTURAL ENGINEERING 
 
 point 2 is the point of zero stress for post cC. Point 3 is the point of 
 zero stress for diagonal dE and point 4 is the point of zero stress for 
 post eE. The point of zero stress for diagonal eF is at 5. 
 
 Tension in Posts. Let it be required to determine the maximum live- 
 load tensile stress in post dD (Fig. 338). First construct the influence 
 line, shown at (c), for stress in diagonal dE. As explained on pages 
 424 and 425 (Art. 190), the maximum live-load tension will occur in 
 post dD when the load is so placed that the stress in diagonal dE is 
 zero. The stress in the diagonal dE due to dead load is tension. A 
 live load moving onto the bridge from the left would begin reversing 
 this dead-load tension from the start and continue reversing it until a 
 
 Fig. 338 
 
 point was reached where the stress in the diagonal would be zero and 
 from there on, as the load moved on to the right, the stress in the diagonal 
 would be compression and this compression would continue to increase 
 until the live-load extended from M to N (see the influence line at (c) ) 
 at which time the live-load compression would be a maximum. Then as 
 the load moved on past N the compression in the diagonal would decrease 
 until a point was reached, as the load moved on to the right, where the 
 stress in the diagonal dE would be zero for the second time. This last 
 position is the one desired, as the maximum tension in post dD will 
 occur when the load is in that position. The first thing is to find that 
 position. 
 
 Let S represent the dead-load tension in diagonal dE and S' the 
 live-load compression in the same when the load extends from M to N. 
 Then the load to the right of N must be just sufficient to produce a ten- 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 481 
 
 sion of S' S pounds in diagonal dE. An equivalent uniform live load 
 of w pounds per foot of truss will be used. Let y be the distance from 
 H to the head of the uniform live load and let A represent the required 
 shaded area. Then the required tension in the diagonal is Aw, which 
 must be equal to *S" - S, so we have 
 
 Aw = S'-S, 
 from which we obtain 
 
 S'-S 
 
 for the required shaded area in square feet. The next thing is to find 
 the value of y when the shaded area is equal to (S' S)/w. 
 Referring to the influence line at (c), we have 
 
 (7). 
 
 Cut ..' ,=K. 
 
 So, substituting this value of z, and ($'-) /w for A in (7), and reduc- 
 ing, we obtain 
 
 -s) ............................ (8). 
 
 From this equation (8) the head of the equivalent uniform live 
 load can be located. Then, having the load located, the influence line 
 for the maximum live-load tension in the post dD can be constructed 
 as shown at (d), and by multiplying the shaded area by w the desired 
 live-load tension in the post will be obtained. 
 
 To construct the influence line shown at (d) : First assume zero 
 stress in diagonal dE. Placing a unit load at D and drawing kh parallel 
 to O'D and kg parallel to O'd, we have the stress in top chord cd given 
 by the line kg; and as the horizontal components of top chords cd 
 and de must be equal (since the stress in dE is zero) by drawing from 
 g a line parallel to top chord de intersecting the vertical mk at m, we 
 have the stress in chord de represented by the line gm and the stress in 
 the post dD represented by the line mk. Then laying off m'k' at (d) 
 equal to mk the influence line as shown can be constructed. 
 
 In case the uniform live load does not extend to ordinate e, as shown 
 at (e), it is better to locate the head of the load with reference to P. 
 Let x represent the distance from P to the head of the uniform live load. 
 Then we have 
 
 . ce 2 
 A: ::x 2 :c 2 , 
 
 2 
 from which we obtain 
 
 (9). 
 
 206. Determination of Dead-Load Stresses in Pettit Trusses. 
 Let it be required to determine the dead-load stresses in the truss shown 
 
482 
 
 STRUCTURAL ENGINEERING 
 
 at (a), Fig. 339. Let JF1, W2, and so on (all of which are equal) 
 represent the loads at the lower panel points; PI, P2, and so on (all 
 of which are equal) represent the loads at the upper panel points; and 
 let R represent the reaction due to these loads. 
 
 Members Ul-LO, LO-L1, L1-L2 and U1-L1. As previously shown 
 (Art. 189), the dead-load stress in Ul-LO is equal to Rsec6, RtanB in 
 both LO-Ll and L1-L2, and Wl in Ul-Ll. 
 
 Member U1-U2. The stress in this member is readily obtained by 
 taking moments about L2. Thus, we obtain 
 
 for the horizontal component of the stress in U1-U2. The stress in the 
 member is then obtained by multiplying H by sec</>. 
 
 Fig. 339 
 
 Member U1-L2. Having determined H, the horizontal component 
 of the stress in top chord U1-U2, the vertical component of the same 
 is equal to 
 
 Now, considering the part of the structure to the left of section 1-1 and 
 summing up the vertical forces, we obtain 
 
 for the vertical component of the stress in U1-L2. Then we have 
 for the stress in U1-L2. 
 
 The stress in this diagonal can also be determined by taking moments 
 about the intersection of the chords as explained in Art. 189. 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 483 
 
 Members M3-L2 and M3-L3. In determining the stresses in these 
 two members the triangle L2-M3-L4: can be considered as a separate 
 truss, as shown at (b). The stress in M3-L2 is equal to ^(P3 + W3)secO', 
 and the stress in the hanger M3-L3 is simply equal to W3. 
 
 Members U2-L2. Considering the part of the structure to the left 
 of section 2-2 shown at (c) and resolving the stress in L2-M3 both 
 vertically and horizontally at L2 and resolving the stress in U1-U2 and 
 adding up the vertical forces, we have 
 
 from which we obtain 
 
 for the dead-load stress in post U2-L2. This stress can also be deter- 
 mined by taking moments about 0. Thus, taking moments about 0, we 
 have 
 
 from which we obtain 
 
 s + Zd 
 
 for the stress in U2-L2. 
 
 Member M2-M3. This member is not subjected to any direct stress. 
 It is for the purpose of stiffening post U2-L2. 
 
 Member U2-U3-U4- Considering the part of the structure to the 
 left of section 4-4, as shown at (d), and by taking moments about Z/4, 
 we obtain 
 
 for the horizontal component of the stress in top chord U2-U3-U4:. Then 
 by multiplying J/l by sec</>' the stress in the member is obtained. 
 
 Member M3-L4. Having HI, the horizontal component of the stress 
 in the top chord U2-U3-U4: determined, the vertical component F2 is 
 equal to H\ tan<'. Then resolving the stress in M3-L4 vertically and 
 horizontally and summing up the vertical forces shown at (d) we obtain 
 
 F3 = R - (Wl + W2 + W3) - (PI + P2 + P3) - 72 
 
 for the vertical component of the stress in diagonal M3-L4. Then for 
 the stress in the member we have 
 
 Member L2-L3-LJ^. Considering the part of the structure shown 
 at (d), and considering S2 and S3 instead of their components (see Art. 
 41), and taking moments about U2 of the forces shown at d, we obtain 
 
 for the stress in bottom chord L2-L3-L4. 
 
484 STRUCTURAL ENGINEERING 
 
 Member U2-M3. The vertical component of M3-L2 is equal to 
 J(P3 + ?F3); the vertical component of the top chord U2-U3-U4, as 
 found above, is equal to F2. Then considering the part of the structure 
 to the left of section 3-3 and summing up the vertical forces and 
 components on same, we obtain 
 
 F4 - R - ( Wl + W2 + PI + P2 ) - J (P3 + W3 ) - F2 
 
 for the vertical component of the stress in diagonal U2-M3. Then the 
 stress is equal to F4 x sec#'. 
 
 Member U3-M3. This member simply supports the load P3 and 
 hence the stress in it is equal to P3. 
 
 Members M5-L5 and M5-L4* The stress in hanger M5-L5 is equal 
 to W5. The vertical component of the stress in M5-L4 is equal to 
 ^(P5 + JF5). Then the stress in M5-L4 is equal to |(P5 + JF5)sec<9". 
 
 Member U4-L4. Considering the part of the structure to the left 
 of section 5-5 and resolving the stress in 74- 73- 72 and M5-L4 vertically 
 and horizontally and summing up the forces and components, we obtain 
 
 for the stress in post 74-L4. 
 
 Member U4-U5-U6. Considering the part of the structure to the 
 left of section 7-7 and taking moments about Z/6 we can obtain H2, the 
 horizontal component of the stress in top chord 74- 75- 76. Then the 
 stress in the member is equal to H2 x sec<". 
 
 Member L4-L5-L6. Considering again the part of the structure 
 to the left of section 7-7 and taking moments about 74, we obtain 
 
 for the stress in bottom chord L4-L5-L6. 
 
 Member M5-L6. Considering again the part of the structure to 
 the left of section 7-7 and letting F6 represent the vertical component of 
 the stress in top chord 74-75-76, we obtain 
 
 for the vertical component of the stress in diagonal M5-L6. Then mul- 
 tiplying VI by sec#" the stress in the member is obtained. 
 
 Members M4-M5 and M5-M6. These members have no direct stress 
 in them. They are for the purpose of stiffening the posts 74-L4 and 
 C76-L6. 
 
 Member U4-M5. Considering the part of the structure to the left 
 of section 6-6 and letting ^(P5 + ?F5) and FQ represent the vertical com- 
 ponent of the stress in M5-L4 and 74- 75- 76, respectively, and summing 
 up the vertical forces and components to the left of section 6-6, we 
 obtain 
 
 F8 = R - (Wl + W'2 . . . JF4) - (PI + P2 . . .P4) - J(P5 + 7F5) - F6 
 
 for the vertical component of the stress in diagonal 74-M5. Then the 
 stress in the member is equal to F8 x sec#". 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 485 
 
 Member U5-M5. This member supports the load Po and hence 
 *hr? stress in it is equal to Po. 
 
 Members U7-M7 and M7-L7. As is evident, the stress in U7-M7 is 
 equal to PI and the stress in M7-L7 is equal to Wl. 
 
 Members U6-M7 and M7-L6. The dead-load stress in these mem- 
 bers is due to the loads W7 and P7. We will assume that each is equally 
 stressed. Then the stress in each will be equal to J(?F7 + P7)sec0'". 
 
 Member U6-L6. Considering the part of the structure to the left of 
 section 8-8 and letting VQ and F9 represent the vertical component of the 
 stirss in members U4-U5-U6 and M7-L6, respectively, and summing up 
 ali the forces and components to the left of section 8-8, we obtain 
 
 6 = R - (Wl + JV2 . . . JF6) - (PI +P2 . . .P5) -F6-F9 
 
 for the stress in C76-L6. 
 
 Members U6-U7 and L6-L7. Considering the part of the structure 
 to the left of section 9-9 and taking moments about LI , we obtain 
 
 for the stress in top chord U6-U7. It will be seen readily that the 
 moments of the stresses in U6-M7 and M7-L6 about L7 are equal and of 
 opposite signs and annul each ' other, and consequently are not given 
 in the equation of moments. 
 
 The stress in L6-L7 is equal to the stress in top chord C76-C77. How- 
 ever, the stress in Z/6-L7 can be obtained by taking moments about 77. 
 
 As the truss is symmetrical in reference to the center of span and 
 symmetrically loaded, we have now sufficiently considered the analytical 
 determination of the dead-load stresses in the truss. 
 
 As a matter of fact, the dead-load stresses in such trusses are 
 graphically determined. The diagram of the dead-load stresses for the 
 right half of the truss is shown at (e). This diagram is obtained by 
 beginning at N and passing around the joints clock- wise. Thus, begin- 
 ning at point N, we draw 1-2 equal to R and then 1-3 and 2-3 parallel, 
 respectively, to bN and BN. Then passing on to joint B we have the 
 diagram 3-2-4-5-3 for that joint. Then passing on to joint b we have the 
 diagram 6-1-3-5-7-6 for that joint. Now, at C there are three unknown 
 forces, the same is true of joint c and, consequently, before we can go 
 farther we must determine one of the unknown forces at one of these 
 joints. By considering E-M3-C as a separate truss, as shown at (f), and 
 constructing the diagram at (g), we have the stress in M3-C given by 
 the line vu. Then considering joint C (at (a)) and passing around the 
 joint clock- wise, we have the part (starting at 7) 7-5-4-8 of the stress 
 diagram for that joint. We know that the stress in post cC will be rep- 
 resented by a vertical line from 7. So we can draw 7-y. We can also 
 draw 8-,r. Then laying off 0-10 equal to !(P3 + ^3) and drawing 9-10 
 parallel to M3-C, we have the stress in M3-C represented by this line 
 9-10 (see diagram at (g)) and thus we have the complete stress diagram 
 7-5-4-8-9-10-7 for joint C. For joint c we obtain the stress diagram 
 
486 
 
 STRUCTURAL ENGINEERING 
 
 (beginning at 11) 11-6-7-10-12-11. For joint d we obtain the stress 
 diagram (beginning at 13) 13-11-12-14-13. For joint D we obtain the 
 diagram 9-8-15-16-9. For joint M3 we obtain the stress diagram 14-12- 
 10-9-16-14. The line 16-14, which represents the stress in member 
 A/3-2?, must be parallel to 12-10,, which represents the stress in member 
 c-M3. This is the first check on the work. 
 
 By considering G-M5-E as a separate truss, the stress in M5-E can 
 be determined. Now,, considering joint E and passing around clock-wise, 
 we obtain the part 14-16-15-17 of the stress diagram for joint E. 
 
 Then drawing 14-r and laying off r-19 equal to ^(P5 + W5) and 
 drawing 19-18 parallel to the member M5-E, we have the complete 
 
 Fig. 340 
 
 stress diagram 14-16-15-17-18-19-14 for joint E. For joint e we 
 obtain the stress diagram 20-13-14-19-21-20. For joint / we obtain 
 the stress diagram 22-20-21-23-22. For joint F we obtain the diagram 
 18-17-24-25-18. For joint M5 we obtain the diagram 23-21-19-18-25-23. 
 The line 23-25 must be parallel to line 21-19. This is the second check 
 ^on the work. 
 
 The dead-load stress is assumed to be the same in the four members, 
 M7-t/6, M7-L6, M7-g and M7-G. This stress is due to the loads Wl 
 and PI. Then, by considering LQ-M 7-G as a separate truss, having a 
 reaction at each end of }(W7 + P7), the stress in M7-G can be graphically 
 determined. 
 
 Starting with 23, we have the part 23-25-24-26 of the stress diagram 
 for joint G. Then drawing a vertical line from 23 and a horizontal 
 line from 26 and laying off t-28 equal to J(7F7 + P7) and drawing 28-27 
 parallel to M7-G, we obtain the complete stress diagram 23-25-24-26-27- 
 
DESIGN OF SIMPLE EAILEOAt) BKIDGES 487 
 
 28-23 for joint G. For joint g we have the stress diagram 29-22- 
 23-28-30-29. 
 
 This completes the graphical determination of the dead-load stresses, 
 as the truss is symmetrical in reference to the center of span and the 
 load is symmetrical in reference to same. The distance 26-29 should 
 equal J(P7 + W7). This is the third and final check on the work. 
 
 207. Determination of Live-Load Stresses in Pettit Trusses. 
 
 Let it be required to determine the live-load stresses in the truss shown 
 at (a), Fig. 340, due to Cooper's E50 loading. 
 
 Members bA, AB and EC. The maximum live-load stress will occur 
 in these members when the wheel loads are placed for maximum shear in 
 panel AB. The placing of the loading will be in accordance with (5), 
 Art. 90. After the loads are thus placed, the next thing to do is to deter- 
 mine the reaction at A by taking moments about N. Let R represent this 
 reaction. The next thing to do is to determine the concentration at A 
 due to the loads in panel AB, which can be done by taking moments about 
 B. Let r represent this concentration. Then we have R-r for the shear 
 in panel AB. Then we obtain (R-r)secO for the maximum live-load stress 
 in end post bA and (R-r)tanO for the maximum live-load stress in each 
 of the members AB and BC (see Art. 174). 
 
 Member bC. The maximum live-load stress will occur in diagonal 
 bC when the loads are placed according to the requirements of (4) of 
 Art. 190. After the loads are thus placed, the next thing to do is to 
 determine the reaction at A, which can be done by taking moments about 
 N. Let R represent this reaction. Next, by taking moments about C, 
 the horizontal component of the stress in top chord be is readily obtained. 
 Let H represent this component. Then the vertical component of the 
 stress in be is equal to Htan(f>. Taking moments about C of the loads 
 in panel BC, the concentration at B is readily obtained. Let r repre- 
 sent this concentration. Now, for the vertical component of the stress 
 in diagonal bC, we have V = R - r - J/tan<, and multiplying this by sec0, 
 we obtain (R r Htan^secO for the stress in the member bC. 
 
 Member bB. The maximum live-load stress in this hanger, as is 
 obvious, is equal to the maximum live-load floor beam concentration, which 
 is determined as explained in Art. 148. 
 
 Member be. The maximum live-load stress will occur in this member 
 when the loads are placed for maximum moment about C. The placing 
 of the loads will be in accordance with (5), Art. 91. That is, a wheel 
 must be at C and the average unit load to the left of C must equal 
 (approximately) the unit load on the span. After the loads are thus placed 
 the next thing to do is to determine the reaction at A, which can be done 
 by taking moments about N. Then, by taking moments about C of the 
 forces to the left, the horizontal component of the stress in be is readily 
 obtained, and multiplying this component by sec<, the maximum live-load 
 stress in be is obtained. 
 
 Members cC and c-Ml. It is obvious if C-M\-E be considered a 
 separate truss, shown as C-o-E at (b), that the stress in cC and c-M\ will 
 not be affected in the least. The truss C-o-E really acts as a stringer 
 extending from C to E. Then, as is readily seen, the maximum live-load 
 stress in cC can be obtained by loading the span from the right, placing 
 
488 STRUCTURAL- ENGINEERING 
 
 a wheel at E and loading panel CE (no loads to the left of C), all in 
 accordance with (4) of Art. 190. 
 
 In applying equation (4) s should be taken equal to the distance 
 I A, shown at (b), and a should be taken equal to distance AC. 
 
 After the loading is placed,, the next thing to do is to determine the 
 reaction at A, which can be done by taking moments about N. Let R 
 represent this reaction. The next thing to do is to determine the con- 
 centration at C due to the loads in panel CE, which can be done by 
 taking moments about E, ignoring panel -point D altogether ; that is, CE 
 would be considered as a stringer. Let r represent this concentration 
 atC. 
 
 Next, by taking moments about C, the horizontal component of the 
 stress in top chord be is readily obtained, as previously explained. Let 
 H represent this component. Then the vertical component of the stress 
 in the top chord be is equal to HtaiKf>. Now, considering the part of the 
 structure to the left of section 1-1, and adding up the vertical components 
 and forces, we obtain R-r-Htan<f> for the maximum live-load stress in 
 member cC. The member oC is entirely ignored, as the vertical com- 
 ponent of its stress is included in the concentration r. 
 
 To obtain the maximum live-load stress in c-Ml, the loads would be 
 placed very much the same as for cC. In applying equation (4) of Art. 
 190 * would be taken equal to the distance I' A and a equal to AC. After 
 the loads are properly placed, the next thing to do is to determine the 
 reaction at A, which can be done by taking moments about N. Then 
 the next thing to do is to determine the concentration at C due to the 
 loads in panel CE, ignoring panel D. By taking moments about E, the 
 horizontal component of the stress in top chord ce is readily obtained, and 
 multiplying this component by tan</>', the vertical component of the stress 
 in top chord ce is obtained. Then considering the part of the structure to 
 the left of section 2-2 and summing up the vertical components and forces 
 to the left which consist of the reaction at A, concentration at C, the 
 vertical component of the stress in ce, and the vertical component of the 
 stress in cMl, the vertical component of the stress in cMl cstn be 
 obtained; and multiplying this component by sec0', the stress in cM\ 
 is obtained. 
 
 Members D-M1 and C-M1. As is obvious, the maximum live-load 
 stress in hanger D-M 1 is equal to the maximum live-load floor beam concen- 
 tration, which is determined as explained in Art. 148. The maximum live- 
 load stress in sub-diagonal C-M1, as is readily seen, is equal to one-half 
 of the maximum floor beam concentration multiplied by sec#'. 
 
 Member ce. The maximum live-load stress in this member occurs 
 when a load is at E and the average unit load to the left of E is equal 
 to the average unit load on the span. This is in accordance with Art. 
 91. After the loads are properly placed the next thing to do is to 
 determine the reaction at A, which can be done by taking moments about 
 N. Then considering the part of the structure to the left of section 3-3 
 and taking moments about E, the horizontal component of the stress in 
 top chord ce is readily obtained, and then multiplying this component 
 by sec<', the maximum live-load stress in ce is obtained. 
 
 Member E-M1. It is readily seen that if oE (shown at (b)) were 
 combined with E-M1, any load at D would cause compression in E-M1 
 
DESIGN OF SIMPLE EAILEOAD BEIDGES 489 
 
 and hence the member E-M\, as far as maximum stress is concerned, 
 acts just the same as an ordinary diagonal where DE is the panel 
 length. Then the maximum stress in E-M1 is obtained by loading the 
 span from the right with a wheel at E f and loading panel DE (no loads 
 to the left of D) according to (4) of Art. 190. In applying equation 
 (4) s should be taken equal to the distance I' A and a equal to the dis- 
 tance AD. After the loads are properly placed, the next thing to do is 
 to determine the reaction at A, which can be done by taking moments 
 about N. Then the next thing to do is to determine the concentration 
 at D, due to the loads in panel DE. Then, considering the forces to the 
 left of section 3-3 (oE and E-M\ being considered as combined, that is, 
 one member) and taking moments about E, the horizontal component of 
 the stress in top chord ce is readily obtained; and multiplying this 
 horizontal component by tanc// the vertical component of the stress in the 
 top chord ce is obtained. Then summing up the vertical components and 
 forces to the left of section 3-3 the vertical component of the stress in 
 diagonal E-Ml is readily determined ; and multiplying this component by 
 secQ' the maximum live-load stress E-Ml is obtained. 
 
 Member CE. It is readily seen that any load in panel CD or DE 
 will cause tension in CE, owing to the member CE acting as the bottom 
 chord of truss C-Ml-E. This stress, as we may say, is in addition to 
 the stress produced in the member acting as a main bottom chord sec- 
 tion of the structure. Then it appears that the maximum live-load stress 
 in bottom chord CE will occur when the loads in panels CD and DE 
 have some certain value as compared to the other loads on the structure. 
 This can be investigated most satisfactorily by the use of influence lines. 
 
 The stress in CE is finally obtained by considering the part of the 
 structure to the left of section 3-3 and taking moments about c. Then 
 evidently the stress in CE will vary as the moments about c and hence 
 the stress in the member can be investigated by constructing an influence 
 line for moments about c. 
 
 Let P represent a load at any point x distance from N. Then when 
 P is at any point to the right of panel point E we have M(Px/L~)a 
 for the bending moment about c. If x=b, the last equation becomes 
 M=(Pb/L)a, and if P = l we have M=(b/L)a. So if we draw 00' 
 at (c) and lay off uv equal to (b/L)a we can draw uO' ', which is the 
 influence line for moments about c for loads to the right of E. 
 
 If the load P moves to any point in panel DE, the moment 
 about c is 
 
 If x-b we have the same as given above for ordinate uv. If 
 and P = l the equation reduces to 
 
 , a + d 
 M = -L- a + d = ~L 
 
 Then if the ordinate nm be laid off equal to this last value of M' the 
 line nu can be drawn, which is the influence line for moments about 
 c for loads in panel DE. 
 
490 STRUCTURAL ENGINEERING 
 
 If the load P moves to any point in panel CD, the moment about c 
 will be 
 
 This is the equation to the line in, which is the influence line for moments 
 about c for loads in panel CD. 
 
 If the load P moves to any point to the left of C, the moment about 
 c will be 
 
 M"' = ( P j\ a-(x-b-2d)P = 
 
 which is the equation to line tO. But this value of M"' is exactly the 
 same as found above for M". Therefore, the lines tn and tO are in the 
 same straight line On and hence On is the influence line for moments 
 about c for loads to the left of D. Then evidently OnuO' is the complete 
 influence line for moments about c. As is seen, this influence line has 
 three different slopes and consequently there will be three different 
 moment increments to consider. 
 
 Let Wl represent the resultant of the loads to the left of D, W2 
 the resultant of the loads in panel DE, W3 the resultant of the loads to 
 the right of E and let W represent the total load on the span; that is, 
 W=W1 + W2 + W3. Then for the moment about c, we have 
 
 (1). 
 
 Now if all loads move to the left the distance dx, we have 
 
 AM = - Wldxtanfil + W2dxianp2 + W3dxtanf33 (2) . 
 
 for the increment of the moment about c. If the loads are in the posi- 
 tion for maximum moment about c this increment will be equal to and, 
 hence, (2) would become 
 
 = - Wltanpl + W2t&n/32 + JF3tan3 (3). 
 
 But, tan/?l = f^' 
 
 L + a 
 
 la + 
 = - 
 
 tan/33 = - 
 Li 
 
 Substituting these values of the tangents in (3), we obtain 
 
 a 
 L' 
 
 from which we obtain 
 
 (W2 - Wl)L + a(Wl + W2 + 7P3) =0, 
 from which we obtain 
 
 W _W1-W2 
 L = a 
 
DESIGN OF SIMPLE EAILEOAD BKIDGES 491 
 
 Expressing equation (4) in words, the average unit load on the span 
 is equal to the load to the left of D, minus the load in panel DE, divided 
 by a. This can be taken as the criterion for placing the load for maxi- 
 mum stress in bottom chord CE. 
 
 After the loads are placed in accordance with equation (4), the next 
 thing to do is to determine the reaction at A, which can be done by taking 
 moments about N. Then the next thing is to determine the concentra- 
 tion at D, which can be done by taking moments about C and E. 
 
 Let R the reaction at A, r = the concentration at D (due to the 
 loads in panels CD and DE), m = moment of load to the left of C about 
 c, and h = height of post cC. Then considering the part of the structure 
 to the left of section 5-3 and taking moments about c, we obtain 
 
 S=(Ra-m + rd}- 
 n 
 
 for the maximum live-load stress in bottom chord CE. 
 
 The moment about c of course could be obtained by the use of influ- 
 ence line OnuO' ' . 
 
 If W remains constant, the increment of the moment would change 
 signs only as a load passed joint D, so there will be a load at that joint 
 when the maximum moment occurs at joint c. Then in placing the load- 
 ing for maximum moment about c, that is, satisfying (4), a wheel must 
 be at joint D. 
 
 Member eg. The maximum live-load stress in top chord eg is 
 obtained by placing a load at G such that the unit load to the left of G 
 is equal to the unit load on the span. Then by taking moments about G 
 the horizontal component of the stress in eg is readily determined and 
 multiplying this component by sec<" the stress in the member is obtained. 
 
 Member eE. The maximum live-load stress in this member is 
 determined in the same manner as shown above for cC. The sub-diag- 
 onal E-M3 and hanger F-M3 are assumed to be omitted. The span is 
 loaded from the right, a wheel at G and panel EG loaded in accordance 
 with (4) of Art. 190. 
 
 The value of s (in equation (4)) is obtained by prolonging top chord 
 ce and a is taken equal to the distance AE. After the loads are prop- 
 erly placed, the next thing to do is to determine the reaction at A, which 
 can be done by taking moments about N. Next, the concentration at E 
 can be determined by taking moments about G. Then, the next thing 
 is to determine the vertical component in top chord ce, which can be done 
 by first taking moments about E, thereby obtaining the horizontal com- 
 ponent of the stress in the member and multiplying this component by 
 tan<', we obtain the vertical component of the stress in ce. 
 
 Let R represent the reaction at A, r the concentration at E, and V 
 the vertical component of the stress in top chord ce. Then considering 
 the part of the structure to the left of section 4-4 and summing up the 
 vertical components and forces, we have 
 
 S = R-r-F, 
 
 for the maximum live-load stress in post eE. 
 
 Member e-M3. The maximum live-load stress in this member is 
 determined in the same manner as shown above for member c-Ml. The 
 
492 STEUCTUEAL ENGINEEEING 
 
 members E-M3 and F-M3 are assumed to be omitted. The span is loaded 
 from the right with a wheel at G and panel EG loaded in accordance 
 with (4) of Art. 190. Top chord eg in that case would be prolonged to 
 determine the value of s, and a would be the distance AE. 
 
 After the loads are properly placed for maximum stress in e-M3, the 
 next thing to do is to determine the reaction at A, which can be done by 
 taking moments about N. The next thing to do is to determine the con- 
 centration at E, which can be done by taking moments about G. Then 
 the next thing in order is to determine the vertical component of the 
 stress in top chord eg, which can be done by first taking moments about 
 G, thus obtaining the horizontal component of the stress in top chord eg 
 and multiplying this component by tan</>", we obtain the vertical com- 
 ponent of top chord eg. Then considering the part of the structure to 
 the left of section 5-5, and summing up the vertical components and 
 forces, we obtain 
 
 Vl=R-r-V 
 
 for the vertical component of the stress in diagonal e-M3, and multiply- 
 ing this component by sec0", we obtain the maximum live-load stress in 
 member e-M3. 
 
 Members F-M3 and E-M3. As is evident, the maximum live-load 
 stress in hanger F-M3 is equal to the maximum floor beam concentration, 
 which is determined as explained in Art. 148. The maximum live-load 
 stress in sub-diagonal E-M3 is equal to one-half of this concentration 
 multiplied by sec0". 
 
 Member G-M3. The maximum live-load stress in this member is 
 determined in the same manner as explained above for diagonal E-M1. 
 The span is loaded from the right, a wheel is placed at G and panel 
 GF is loaded in accordance with (4) of Art. 190. In this case the value s 
 is obtained by prolonging top chord eg (see Art. 190) and a is equal to 
 the distance AF. The reaction at A is obtained by taking moments about 
 N. The concentration at F is obtained by taking moments about G, and 
 the horizontal component of the stress in top chord eg is obtained by 
 taking moments about G, and the vertical component of same is then 
 obtained by multiplying the horizontal component by tan</>". After these 
 are determined the vertical component of the stress in diagonal G-M3 is 
 obtained by considering the part of the structure to the left of section 6-6 
 and summing up the vertical component and forces, and then the stress 
 in G-M3 is obtained by multiplying this component by sec0". 
 
 Member EG. The maximum live-load stress in this member is deter- 
 mined in the same manner as explained above for bottom chord CE. In 
 applying equation (4) (given above) a would be taken equal to distance 
 AE. W\ would be the load between A and F, W2 the load in panel FG, 
 W3 the load between G and N t and W the total load on the span. 
 
 After the loads are properly placed, that is, in accordance with 
 equation (4), the stress in EG is readily obtained by considering the 
 part of the structure to the left of section 6-6 and taking moments 
 about e. 
 
 Member gG. The maximum live-load stress in this member is ob- 
 tained by considering members Gk and Hh as being omitted. The 
 span is then loaded from the right, a wheel at K and panel KG loaded 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 493 
 
 in accordance with (4) of Art. 190. The value of s in equation (4) of 
 Art. 190 is obtained by prolonging top chord eg and a is equal to the dis- 
 tance AG. After the loads are properly placed the reaction at A is 
 obtained by taking moments about N. The concentration at G is obtained 
 by taking moments about K. The horizontal component of the stress in 
 top chord eg is obtained by taking moments about G and the vertical 
 component of same is obtained by multiplying the horizontal component 
 by tan<". Having the above determined., the stress in gG is obtained 
 by summing up the vertical forces to the left of section 7-7 (ignoring 
 member G-M5). 
 
 Member GK. The maximum live-load stress in this member is de- 
 termined in the same manner as explained above for bottom chord CE. 
 In applying equation (4) (given above) a would be taken equal to dis- 
 tance AG, WL would be the load between A and H, W2 the load in panel 
 HK, W3 the load between K and 2V. After the loads are properly placed, 
 that is,, in accordance with equation (4), the stress in GK is readily ob- 
 tained by considering the part of the structure to the- left of section 8-8 
 (ignoring member Ar-M5) and taking moments about g. 
 
 Member g-Mo. To obtain the maximum live-load stress in this 
 member, we assume the members Gk and Hh to be omitted, and load 
 the span according to (5) of Art. 90. That is, the span would be loaded 
 from the right with a wheel at K arid the load in panel GK equal to the 
 load on the bridge divided by the number of panels. The panels con- 
 sidered in this case would be double panels that is, the number would 
 be 7 instead of 14. 
 
 After having the loads thus placed, the next thing to do is to deter- 
 mine the reaction at A and the concentration at G. These can be ob- 
 tained by taking moments about 2V and K, respectively. Let R repre- 
 sent the reaction at A and r the concentration at G. Then we have 
 
 for the maximum live-load stress in diagonal g-M5. 
 
 Member K-M5. To obtain the maximum live-load stress in this 
 member we assume member k-M5 to be omitted and load the span accord- 
 ing to (5) of Art. 90. That is, the span would be loaded from the right 
 with a wheel at K, and the load in panel HK equal to the total load on 
 the span divided by the number of panels. In this case the number of 
 panels would be 14. After the loads are thus placed, the next thing 
 to do is to determine the reaction at A and the concentration at H. Let 
 R' represent the reaction at A, and r' the concentration at H. Then we 
 have 
 
 S'=(R'-r')secO f " 
 
 for the maximum live-load stress in diagonal K-M5. 
 
 Members g-M5 and k-M5 would be assumed to have equal tensile 
 stress and likewise members K-M5 and G-M5. The latter members would 
 be subjected to compression from the floor beam concentration at H. 
 Assuming the concentration at H to be transmitted to panel points G and 
 K by sub-truss K-M5-G the maximum live-load compression in each of 
 the members G-M5 and K-M5 would be equal to one-half of the maximum 
 floor beam concentration at H multiplied by sec#"'. 
 
494 
 
 STRUCTURAL ENGINEERING 
 
 Member gk. The maximum live-load stress in this member is ob- 
 tained by placing the load for maximum moments about K. Then taking 
 moments about K (ignoring member &-M5) and dividing this moment by 
 the height of post kK, we would obtain the greater part of the maximum 
 stress in gk. Next the concentration at H, due to the loads in panels 
 GH and HK, can be determined by taking moments about G and K. 
 Then, multiplying one-fourth of this concentration at H by tan#"', and 
 adding the result to the stress found in gk, by taking moments about K, 
 the total maximum live-load stress in the member gk is obtained. 
 
 Maximum Tension in Post gG. The maximum live-load tension will 
 occur in this post when the live load extends from A to some point beyond 
 G, so that the stress in diagonal g-M5 is zero. The position of the load 
 
 Fig. 341 
 
 can be determined by trial as explained in Art. 205 for curved chord 
 Pratt trusses. After the position of the loading is found, the tension 
 in the member is readily determined by considering the part of the struc- 
 ture to the left of section 7-7. The use of an equivalent uniform live- 
 load will simplify the work very much and the result thus obtained 
 will be sufficiently accurate. 
 
 Stress in Counters. In case a member g-M3 were inserted, it would 
 be known as a counter. This member would not be stressed at all 
 unless the dead-load tension in either or both of the members e--M3 and 
 G-M3 were reversed. In case either, or both, of these members be sub- 
 jected to reversal, the counter g-M3 would be used or members e-M3 and 
 G-M3 would be designed to carry both tension and compression. In 
 
DESIGN OF SIMPLE KAILROAD BRIDGES 495 
 
 case the counter g-M3 is required, the maximum live-load stress (tension) 
 in it can be determined by assuming members eG and F-M3 to be omitted 
 and loading the span from the left with a load at E and loading panel 
 EG in accordance with (4) of Art. 190. After the loading is thus placed, 
 the next thing to do is to determine the reaction at N and the concentra- 
 tion at G, which can be done by taking moments about A and E, re- 
 spectively. Then by considering the part of the structure to the right 
 of section 6-6, the stress is readily determined. If counters are found 
 necessary at other points the stress in them can be determined in a 
 similar manner. 
 
 The determination of the stresses in trusses having the sub-paneling 
 at the top of the truss, as shown at (a) in Fig. 341, is very similar to 
 that given above for the truss shown in Fig. 339 wherein the sub-paneling 
 is at the bottom of the truss. 
 
 Dead-Load Stresses. The graphical determination of the dead-load 
 stresses in the type of truss shown in Fig. 341 is simpler than in the 
 case of the type shown in Fig. 339, as the sub-paneling does not bother 
 at all in the case of the truss shown in Fig. 341. 
 
 The graphical diagram of the dead-load stresses for the right half 
 of the truss is shown at (b). This diagram is readily followed throughout. 
 
 The analytical determination of the dead-load stresses is just the 
 same in some cases and similar in other cases to that given above for the 
 truss shown in Fig. 339. To save space, just the formulas for the dead- 
 load str.esses in part of the members in the right half of the truss shown 
 in Fig. 341 will be given: 
 
 Stress in sT equals RsecO. 
 
 Stress in TS and OS equals RtanO. 
 
 Stress in sS equals W\. 
 
 Stress in os equals [(R x 2d) - (Wl + Pl)d]sec<f>/hl (d = panel Igth.). 
 
 Stress in Os equals [-R - (Wl + PI) - F]sec#, where V represents the 
 vertical component of the stress in top chord os. 
 
 The stress in member oO equals R - (Wl + W2 + P1) - V. In this 
 case the part of the structure to the right of section 1-1 is considered. 
 
 The stress in member MO equals [Rx2d- (Wl+Pl)d]l/hl. This 
 is obtained by taking moments about o. 
 
 The stress in member mo equals [(R x4d)- (3W1 + 3P1+2W2 
 + 2P2)d]seccf)l/h2. This is obtained by considering the part of the 
 structure to the right of section 2-2 and taking moments about M. 
 
 The stress in member o-M3 equals [R- (W.1 + W2 + P1 + P2) 
 -Fl]sec#l, where V\ represents the vertical component of the stress in 
 top chord mo. In this last case the part of the structure to the right of 
 section 2-2 was considered. 
 
 The stress in sub-post n-M3 is equal to P3 and in N-M3 it is W3. 
 
 The stress in sub-diagonal m-M3 is determined by considering 
 n-M3-o as an independent truss supporting the load (W3+P3) at M3, as 
 shown at (c) in Fig. 341. 
 
 The stress in the member m-M3 equals (JF3 + P3) -f (cos02 
 + sin02cos<91). 
 
 The stress in member m-M3 can be determined most readily by 
 graphics (see the diagram at (d)). 
 
 The stress in M-M3 equals [12- (W1 + W2 + JF3 
 
496 STRUCTURAL ENGINEERING 
 
 Fl + F2]sec#l, where V\ and J T 2 represent, respectively., the vertical com- 
 ponent of the stress in mo and m-M 3. In this case the part of the structure 
 to the right of section 3-3 is considered. 
 
 The stress in mM equals R - (Wl + W2 + W3 + W4= + P1 + P2 + P3) 
 - V\ + F2. In this case the part of the structure to the right of section 
 4-4 is considered. 
 
 Following this method the dead-load stresses in the other members 
 of the truss are readily determined. 
 
 Live-Load Stresses. The determination of the live-load stresses in 
 the type of truss shown in Fig. 341 is very much the same as given above 
 for the truss shown in Fig. 340. The method of analysis will be suffi- 
 ciently shown by considering the members in panel EG. 
 
 Member eE. The maximum live-load stress will occur in this mem- 
 ber when the span is loaded from the right, a wheel at G and panel EG 
 loaded in accordance with (4) of Art. 190. In this case the top chord ce 
 would be prolonged to determine the value of s to be used in equation 
 (4) of Art. 190 and a would be taken equal to distance AE. After the 
 loads are properly placed for maximum stress in eE, the next thing to 
 do is to determine the reaction at A and the concentration at E, which 
 can be done by taking moments about T and G, respectively. The next 
 thing to do is to determine the vertical component of the stress in top 
 chord ce. This can be done by taking moments about E, thus obtaining 
 the horizontal component of the stress in ce (ignoring member e-M3, as 
 it is not subjected to any live-load stress at this time, the live load not 
 extending beyond ), and then multiplying this horizontal component 
 by tan<l, the vertical component of the stress in ce is obtained. Let R 
 represent the reaction at A, r the concentration at E, and V the vertical 
 component of the stress in top chord ce. Then considering the part of 
 the structure to the left of section 5-5 and summing up the vertical com- 
 ponents and forces, we have 
 
 S=R-r-V 
 
 for the maximum live-load stress in post eE. 
 
 Member EG. The stress in this member is obtained, as is obvious, 
 by taking moments about joint e. Then by placing the loads so that 
 there is a wheel at E with the unit load to the left of E equal to the unit 
 load on the span, which is in accordance with Art. 91, and taking 
 moments about e of the forces to the left and dividing this moment by 
 the height of post eE, the maximum live-load stress in EG is readily 
 determined. 
 
 Member F-M5. The maximum live-load stress in this hanger is 
 equal to the maximum live-load floor beam concentration at F, which 
 is determined as explained in Art. 148. 
 
 Members M4*M5, M5-M6 and f-M5. These members are not sub- 
 jected to live-load stress at all. /-M5 is subjected to dead-load stress 
 from the load at / only. The other two members are not subjected to 
 any direct stress whatever. They are for the purpose of stiffening the 
 posts eE and cjG. 
 
 Member e-M5. The maximum live-load stress in this member will 
 occur when the span is loaded from the right, a wheel at F and panel 
 EF loaded in accordance with (4) of Art. 190. 
 
DESIGN OF SIMPLE EAILROAD BEIDGES 497 
 
 The top chord eg would be prolonged to determine the value of s to 
 use in (4) of Art. 190, and a would be taken equal to distance AE. 
 After the loads are properly placed for maximum live-load stress in diag- 
 onal e-M 5, the next thing to do is to determine the reaction at A and the 
 concentration at E, which can be done by taking moments about T and F, 
 respectively. Let R represent the reaction at A and r the concentration 
 at E. Then considering the part of the structure to the left of section 
 6-6 and summing up the vertical components and forces, we have 
 
 S=(R-r-F)sec63, 
 
 for the maximum live-load stress in diagonal e-M5. V here represents 
 the vertical component of the stress in top chord eg, which can be deter- 
 mined by considering the part of the structure to the left of section 6-6, 
 and taking moments about G and dividing this moment by the height of 
 post gG, thus obtaining the horizontal component of the stress in top 
 chord eg, then multiplying this component by tan$2, the vertical com- 
 ponent V is obtained. 
 
 Member G-Mo. In determining the maximum live-load stress in 
 this member,, the members F-M5, g-M5 and f-M 5 are assumed to be 
 omitted. The maximum stress will occur when the span is loaded from 
 the right, a wheel at G, and panel GE loaded according to (4) of Art. 
 190. After the loads are properly placed for maximum stress in diagonal 
 G-M5, the stress is obtained by considering the part of the structure 
 to the left of section 7-7 (ignoring member g-M5) and summing up the 
 vertical components and forces. Thus let R represent the reaction at A, 
 r the concentration at E and V\ the vertical component of the stress in 
 top chord eg, then we have 
 
 for the maximum live-load stress in diagonal G-M5. 
 
 Member g-M5. The maximum live-load stress will occur in this 
 member when the live-load concentration at F is a maximum. Let r 
 represent this concentration. Then, considering e-M5-g as an independent 
 truss, we obtain 
 
 S = r -r (cos04 + sin04cos03) 
 
 for the maximum live-load stress in sub-diagonal g-M5. This stress in 
 g-M5 can be determined most readily by graphics, considering e-M5-g 
 as an independent truss. 
 
 Member eg. The live-load stress in this member can be investigated 
 most satisfactorily by the use of the influence line. 
 
 The stress in eg is finally obtained by considering the part of the 
 structure to the left of section 6-6 and taking moments about G. Then 
 evidently the stress in eg will vary directly as the moments about G and, 
 hence, the stress in the member can be investigated by constructing an 
 influence line for moments about G. 
 
 Let P represent a load at any point x distance from T. Then 
 when P is at any point to the right of panel point G, we have 
 
498 STRUCTURAL ENGINEERING 
 
 for the bending moment about G. If x-\j, the above equation becomes 
 
 and if P = l, 
 
 So if we draw 00' at (e) and lay off ts ba/L, we can draw tO' 
 which is the influence line for moments about G for loads to the right of G. 
 If the load P moves to any point in panel FG, the moment about G, 
 considering the part of the structure to the left of section G-6, is 
 
 f P4 
 
 m a, 
 
 w 
 
 which is the same as found above for M, so the influence line vt for 
 loads in panel FG is simply a continuation of the line tO' '. If x = 
 and P = l, we obtain 
 
 which is equal to the ordinate vw. 
 
 If the load moves to any point in panel FE, the moment about G is 
 
 M"= a-P(x- b-d)2. 
 
 If x - b + 2d and P = 1, we obtain 
 
 which is equal to ordinate MO, so the influence line vu for loads in panel 
 EF can be drawn. 
 
 If the load P moves to any point to the left of E, the moment about 
 G is M'"=(P*/L)a-P(a-b). If x = L, we obtain Af'" = (PL/L)a - 
 P(L-b)=Pa-Pa = Q. If x = b + 2d and P=l, we obtain M"' = (b + 
 2d)a/L 2d = b/L(a-2d), which is the same as found above for ordi- 
 nate uo. So the line Ou can be drawn, which is the influence line for 
 loads to the left of E. Then we have the complete influence line OuvtO' 
 for moments about G. As is seen, this influence line has three different 
 slopes and consequently there will be three different moment increments 
 to consider. Let W\ represent the resultant of the loads to the left of 
 E, W2 the resultant of the loads in panel EF, W3 the resultant of the 
 loads to the right of F, and let W represent the total load on the span ; 
 that is, W=W1 + W2+ W3. Then for the moment about G, we have 
 
 M = yWl + ylW2 + y2W3 ........................... . . (1). 
 
 Now if all loads move to the right a distance dx, we have 
 
 AM= Wldxtanpl + W2tan/32 - W3tan(33 ................ (2), 
 
 for the increment of the bending moment about G. If the loads are in the 
 position for maximum moment about G, this increment will be equal to 
 zero and, hence, (2) would become 
 
DESIGN OF SIMPLE RAILED AD BRIDGES 499 
 
 = Wltanpl + W2tan/32 - W3tanp3 .................... (3). 
 
 But, 
 
 b + d\ a + 2b L + b 
 
 Substituting these values of the tangents in (3), we have 
 
 Ll Li ft 
 
 Substituting L-a for b, we have 
 
 from which we obtain 
 
 W1 + 2W2 _W 
 a ~ L" 
 
 That is, the load to the left of E plus twice the load in the panel EF 
 divided by the distance a must equal the total load on the span divided 
 by the length of the span when the maximum stress in top chord eg 
 occurs. 
 
 W remaining constant, the increment of the bending moment about 
 G can change signs only as a load passes joint F, hence there will be a 
 load at F when the maximum live-load stress in eg occurs. Then by 
 placing the live load according to the above equation (4) ' with a load at 
 
 F, and taking moments about G (considering the part of the structure 
 to the left of section 6-6) the maximum live-load stress in eg can be 
 readily determined. 
 
 The reaction at A would be determined first, then the concentration 
 at E, due to the loads in panel EF, which can be done by taking moments 
 about F. Let R reaction at A, r = the concentration at E, due to the 
 loads in panel EF, and m = the moment of all loads to the left of E about 
 
 G. Then taking moments about G, we have 
 
 Dividing this moment by h, the height of post gG, we obtain the hori- 
 zontal component of the stress in top chord eg and multiplying this com- 
 ponent by sec<2, we obtain the maximum live-load stress in top chord eg. 
 
 The stress in top chord ce is determined in a similar manner. The 
 above equation (4)' would be applied to determine the position of the 
 loading. The value of a would be taken in that case equal to the dis- 
 tance AE. 
 
 208. Graphical Determination of Live-Load Stresses in Pettit 
 Trusses. The influence line method outlined in Art. 103 is the most con- 
 venient graphical method to use. The work as a whole is practically 
 the same as shown in Art. 205 for curved chord Pratt trusses. 
 
 As an illustration, let it be required to determine the live-load stress 
 in top chord eg of the truss shown at (a) in Fig. 342. The first thing 
 to do is to draw the influence line for reaction, as shown at (b). The 
 
500 
 
 STRUCTURAL ENGINEERING 
 
 stress in eg can be determined by taking moments about G and consider- 
 ing the part of the structure to the left of section 2-2. Then evidently 
 the influence line for the stress in eg will be of the form shown at (c). 
 By placing a unit load at G and drawing ec (at (b)) and ed parallel, 
 respectively, to SS' and SG, we have the stress in top chord eg, due to 
 the unit load at G, represented by the line ec. Then drawing AB, at 
 
 Fig. 342 
 
 (c), and laying off e f c r ec and drawing e'A and e'B, we obtain the 
 influence line A e'B for the stress in top chord eg. Then placing the live 
 ;oad on the span so that there is a load at G and the unit load to the left 
 of G equal to the unit load on the span, and multiplying the loads PT 
 their respective ordinates, as previously explained, the maximum live- 
 load stress in eg is obtained. In case an equivalent uniform live load be 
 'ised, the maximum stress in eg will be obtained by multiplying the area 
 
DESIGN OF SIMPLE EAILEOAD BRIDGES 
 
 501 
 
 of triangle Ae'B by the uniform load per foot, as previously explained. 
 The stress in bottom chord EG can be determined by taking moments 
 about e. Then evidently the influence line for the stress in EG will be 
 partly of the form Ck'D, shown at (d). But we found in Art. 207 (see 
 Fig. 340) that for moments the line Ck'n was a straight line and undoubt- 
 edly Ck'n will be a straight line if the influence line be constructed for 
 
 U3 
 
 Fig. 343 
 
 the stress in EG. So if k'm' were known, Jc'D could be drawn, thus 
 locating point o, and next Cn could be drawn and then the line on could 
 be drawn and thus we would have the complete influence line CnoD for 
 the stress in bottom chord EG. As is seen, the only thing needed to con- 
 struct this influence line is the ordinate k'm'. By placing a unit load at 
 E and taking moments about e (considering the part of the structure to 
 the left of section 2-2) and dividing by the height of post eE the stress 
 
502 STRUCTURAL ENGINEERING 
 
 in bottom chord EG, due to the unit load at E, is obtained, which is the 
 desired value of k'm'. By drawing Ae and eN we have the truss AeNEA. 
 Now it is readily seen that a unit load at E will produce the same stress 
 in AN as would be found in EG by taking moments about e. By draw- 
 ing km at (b), parallel to Ae, we have the stress in AN represented by 
 the line km. Then by laying off k'm' = km, the influence line CnoD can 
 be drawn as explained above. Then placing the live load on the span 
 in accordance with (4) of the last article, and multiplying the loads by 
 the proper ordinates to the influence line at (d), the maximum live-load 
 stress in bottom chord EG will be obtained. The other influence lines 
 shown in Fig. 342 are considered self-explanatory, provided Art. 205 is 
 thoroughly understood. 
 
 209. Remarks Concerning Pettit Trusses. Pettit trusses, as pre- 
 viously stated, are used for long span bridges. These trusses should 
 have economic heights, in which case, if single paneling were used, the 
 floor system would be very heavy, or the diagonals would have an un- 
 economic and awkward slope. These undesirable features are eliminated 
 by the sub-paneling. 
 
 All diagonals and sub-diagonals can be made out-and-out tension 
 members (eye-bars) if the sub-paneling be at the top chord. This will 
 result in a lighter truss than if the paneling be at the bottom chord. But 
 experience seems to indicate that trusses with sub-paneling at the bot- 
 tom chord are more rigid than trusses with sub-paneling at the top chord. 
 It is a question whether the lack of rigidity, however, is not due more 
 to poor details than to the form of the truss. 
 
 The stress sheets and the detail drawings for Pettit trusses are 
 gotten out in the same manner as previously explained for parallel and 
 curved chord bridges and the calculations of the details are similar to 
 those previously shown for those bridges. 
 
 A fair idea of the details of a Pettit truss can be obtained from 
 Fig. 343, where the details of a panel of the bridge indicated in Fig. 339 
 are shown. These details are of the Bismarck bridge designed by Ralph 
 Modjeski and built by the American Bridge Company. It is a Northern 
 Pacific Railway bridge over the Missouri River at Bismarck, North 
 Dakota. 
 
 The designing of the floor and lateral systems and end bearings of 
 Pettit truss bridges is exactly the same as for the curved chord Pratt 
 truss bridges, previously given. 
 
 MISCELLANEOUS TRUSSES 
 
 210. Warren Trusses. The truss shown in Fig. 344 is known as 
 the Warren truss. In case of a through bridge, there would be a floor 
 beam at each of the joints B, C, and D. 
 
 Dead-Load Stresses. Let P represent the dead load per panel at 
 each of the upper joints and W the dead load per panel at each of the 
 lower joints and let R represent the end reaction due to these loads. 
 The dead-load stresses in the web members are indicated at (a) in Fig. 
 344. These expressions can be written from mere inspection, As is 
 readily seen, 
 
DESIGN OF SIMPLE RAILROAD BEIDGES 
 
 503 
 
 Taking moments about b of the forces to the left of that joint and 
 dividing this moment by h, we obtain 
 
 for the stress in bottom chord AB. Taking moments about B of the 
 forces to the left of that joint, we obtain 
 
 for the stress in top chord be. 
 
 ,P ,P IP 
 
 \h +3(yr+ f>)fon0 I c + ^/,y PJ fone \cf 
 / ^T~ ~~W%. 
 
 L. 
 
 In a similar manner, taking moments about c, we obtain 
 
 for the stress in bottom chord BC and by taking moments about C, we 
 obtain 
 
 for the stress in top chord cd. 
 
 From this it is seen that the dead-load stress in any Warren truss 
 is easily determined. 
 
 Live-Load Stresses. In case the live load be a uniform load, the 
 stresses are very easily determined. Suppose the live load to be w 
 pounds per foot of truss. Then we have 
 
 2dxw = W 
 
 for the live-load panel load. 
 
 Loading joint D (alone), we have 
 
 W 
 
 for the maximum live-load compression in diagonal dC (as indicated at 
 (b)). Loading joints 7) and C, we have 
 
 , 3-fsecO ". ..;,:," 
 
 for the maximum live-load stress in each of the diagonals Cc and cB, 
 tension in cC and compression in cB. 
 
504 STRUCTURAL ENGINEERING 
 
 Loading joints D, C, and B, we have 
 
 for the maximum live-load stress in diagonal bB and end post bA, tension 
 in bB and compression in bA. 
 
 The chord stresses will be a maximum when the span is fully loaded ; 
 that is, when joints D, C, and B are loaded. The reaction at A is then 
 equal to \\W . Then taking moments about b we have 
 
 for the maximum live-load stress in bottom chord AB. Taking moments 
 about B, we have 
 
 for the maximum live-load stress in top chord be. Taking moments about 
 c, we have 
 
 for the maximum live-load stress in bottom chord BC. Taking moments 
 about joint C, we have 
 
 for the maximum live-load stress in top chord cd. 
 
 b c d 
 
 Fig. 345 
 
 From this it is seen that the stresses in any Warren truss due to a 
 uniform live load are easily determined. 
 
 The stresses in Warren trusses, due to wheel loads, are very readily 
 determined. As an example, let it be required to determine the maximum 
 live-load tensile stress in diagonal cC. The span would be loaded as 
 shown in Fig. 345. Let P represent the load in panel BC, the center of 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 5Q5 
 
 gravity of which is z distance from C; and let W represent the total load 
 on the bridge, the center of gravity of which is x distance from E. Then 
 by taking moments about E, we obtain 
 
 Wx 
 
 R= ^7 
 
 for the reaction at A and taking moments about C, we obtain 
 
 _P 
 
 ~2d 
 
 for the concentration at B. 
 
 Then for the live-load tensile stress in cC, we have 
 
 tWx Pz\ 
 S= (R-r) SG cO = (-J- - ^jsecfl. . .................... (1). 
 
 Now suppose the loads move a very short distance to the right or 
 left, say to the left, we have 
 
 Wx 
 
 secO- + - sec* 
 
 for the stress in cC. 
 
 Subtracting (1) from this last equation, we obtain 
 
 for the increment of the stress in cC. Now this increment would be zero 
 if the stress in cC were a maximum and hence the last equation would 
 become 
 
 But A,r = A#, as is readily seen, so we have 
 W P W P 
 
 when the stress in diagonal cC is a maximum. Expressing this in words, 
 the unit load on the bridge is equal to the unit load in panel BC when 
 the maximum live-load tensile stress in diagonal cC occurs. A load will 
 be at C. The maximum compression in diagonal cB, which is equal to 
 the tension in cC, occurs at the same time. To obtain maximum tension 
 in bB and maximum compression in bA, the span would be loaded from 
 the right with a wheel at B and panel AB loaded according to the above 
 equation (2). 
 
 Let it be required to determine the maximum live-load stress in bot- 
 tom chord BC. The span would be loaded as shown in Fig. 346. Let 
 W represent the total load on the bridge, the center of gravity of which 
 is x distance from E. Let PI represent the load to the left of B, the cen- 
 ter of gravity of which is z distance from B, and let P2 represent the 
 
506 
 
 STRUCTURAL ENGINEERING 
 
 load in panel BC, the center of gravity of which is y distance from C. 
 Then we have 
 
 Taking moments about c, we obtain 
 
 for the stress in Z?T. 
 
 c/ c/ 
 
 have 
 
 Fig. 346 
 
 Now, suppose the loads all move a short distance A.r, then we would 
 
 fcr the increment of the stress in chord BC. But if the stress in BC 
 were a maximum, this increment would be zero and we would have 
 
 But A,r = A2 = A2/, so this last equation reduces to 
 
 from which we obtain 
 
 W 
 L 
 
 (3) 
 
 This last equation shows clearly how the loads should be placed 
 for maximum stress in bottom chord BC. The case of the other bottom 
 chords is similar. The stress in the top chords is obtained by taking 
 moments about the bottom chord joints. The placing of the loads in that 
 case is simply a matter of applying (5) of Art. 91. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 507 
 
 The influence line for stress in diagonal cC is shown in Fig. 345 and 
 the influence line for stress in bottom chord BC is shown in Fig. 340. 
 These are readily understood; in fact, the drawing of the influence line 
 
 Fig. 347 
 
 for any of the members in a Warren truss is a simple problem,, provided 
 the work previously given on influence lines is understood. 
 
 There are very few out-and-out Warren trusses (such as shown in 
 Fig. 344) built. They usually have vertical posts and hangers, as shown 
 in Fig. 347. In that case the loading for maximum stress and the determi- 
 nation of the stresses are the same as for an ordinary Pratt truss. 
 
 Referring to Fig. 347, the stress in hangers dD and fF is the same as 
 in hangers bB and hH. 
 
 The posts cC, eE and gG in the case of through bridges have only 
 dead-load stress, which is due to the load at the top chord joints. To obtain 
 the maximum live-load stress in bottom chord CE, due to wheel loads, a 
 wheel would be placed at D, such that the average unit load to the left of 
 D would equal the average unit load on the bridge. The stress would then 
 be obtained by taking moments about d. For maximum stress in top chord 
 df the span would be loaded in reference to joint E. 
 
 That is, a wheel would be placed at E, such that the average unit 
 load to the left of E would be equal to the average unit load on the 
 bridge. The stress in df would then be determined by taking moments 
 about E. 
 
 212. Double-System Warren Truss. The truss shown at (a), 
 Fig. 348, is known as a "double-system" Warren truss. The dead-load 
 stresses and stresses due to a uniform live load can be determined very 
 readily by considering the truss composed of two independent trusses, 
 one of which is shown at (b) and the other one at (c). 
 
 The stresses are then determined in each of these trusses and com- 
 bined ; thus the stress in the structure as a whole is obtained. 
 
 Dead-Load Stresses. Let P represent the dead load per panel at 
 each top joint and let W represent the dead load per panel at each 
 bottom joint. The end reaction on the truss shown at (b) is 
 
 Then the dead-load stress in the web members is as indicated. 
 Taking moments about b, we have 
 
 d 
 h 
 
 for the stress in bottom chord AC. Taking moments about C, we have 
 
508 STRUCTURAL ENGINEERING 
 
 for the stress in top chord bd. Taking moments about d, we have 
 
 S2 = 
 
 The dead-load stresses indicated on the 
 
 for the stress in bottom chord CE. Taking moments about E, we have 
 
 for the stress in top chord de. 
 truss shown at (c) are 
 obtained in a similar 
 manner. By combining 
 the stresses given on 
 the top and bottom 
 chords and end posts 
 at (b) and (c) the 
 dead - load stresses in 
 the truss as a whole 
 are obtained. 
 
 Live-Load Stresses. 
 Let w represent a 
 uniform live load per 
 foot of truss. Then 
 we have 
 
 wd=W 
 
 for the live load per 
 panel. Referring to 
 the truss shown at (d) 
 and considering ^W at 
 // and W at G, we ob- 
 tain 
 
 sec0 
 
 for the maximum live- 
 load compressive stress 
 in diagonal Ef, and by 
 considering \W at H 
 and W at each of the 
 joints G and E, we ob- 
 tain 6(JF/8)sec0 for the maximum live stress in each of the diagonals dE 
 and dC, tension in dE, and compression in dC. Considering \W at H 
 and W at each of the joints G, E, and C, we obtain 12(JF/8)sec0 for 
 the maximum live-load tensile stress in diagonal bC. Considering \W 
 at H and W at each of the joints G, E, and C and \W at B, we obtain 
 2Wec6 for the live-load stress in end post bA. When the truss at (d) is 
 fully loaded, that is, W at each of the joints C, E, and G, and \W at 
 each of the joints H and B, we obtain, by taking moments about 6, 
 
DESIGN OF SIMPLE EAILKOAD BEIDGES 5Q9 
 
 for the stress in bottom chords ABC. Taking moments about C, we obtain 
 
 for the stress in top chord bd. Taking moments about d, we obtain 
 
 (2W)3 ~ (JJF)2 | - (W) Y =4JFtan0 
 h n n 
 
 for the live-load stress in bottom chord CE and taking moments about E, 
 we obtain 4^WtanO for the live-load stress in top chord df. The stresses 
 shown at (e) are obtained in a similar manner. Then combining the 
 stresses given on the top and bottom chords and end post at (d) and (e) 
 the live-load stresses in the truss as a whole are obtained. 
 
 In case wheel load be used, the stresses can be determined most 
 readily by the use of influence lines. As an illustration,, let it be required 
 to determine the stress in chord de (Fig. 349) due to wheel loads. First 
 construct the influence lines for shear as shown at (b). E is the center 
 of moments when the system drawn in full is considered. Then draw- 
 ing ms parallel to OE, the distance ns is obtained, which gives the stress 
 in de, due to a unit load at E. Then, laying off at (c) the ordinate b = ns, 
 the influence line C-4-J5 is obtained. D is the center of moments when the 
 dotted system is considered. Then drawing ut parallel to OD the dis- 
 tance vt is obtained, which gives the stress in de due to a unit load at D. 
 Then laying off the ordinate a vt, the influence line C-3-B is obtained. 
 Let us consider a single load passing over the span starting from K. 
 When it reaches joint H half of it is transmitted to the two systems. 
 Then, evidently, the point half way between the two influence lines at 
 that point will be on the influence line for stress in de. When the load 
 reaches joint G, the load will be supported by the system shown in full 
 and, hence, the point 6 will be on the influence line for stress in de. When 
 the load reaches joint F, the load will be supported by the dotted system 
 and, hence, the point 5 will be on the influence line for stress in de. 
 Tracing out in this manner, we obtain the influence line B-7-6-5-4-3-2-1-C 
 for stress in de. Then placing the wheel loads (mostly by trial) for 
 maximum stress in de and multiplying each load by its respective ordi- 
 nate to this zigzag influence line the maximum stress in de, due to wheel 
 loads is obtained. The maximum stress in the other top chord members 
 is obtained in the same manner. 
 
 The influence line for the stress in bottom chord DE is shown at (d). 
 Considering the system drawn in full, the center of moments in deter- 
 mining the stress in DE is at d. Drawing mx parallel to Ad, the dis- 
 tance nx is obtained, which gives the stress in DE, due to a unit load at 
 E and drawing or parallel to SA, the distance ko is obtained, which gives 
 the stress in DE, due to a unit load at C. Then laying off at (d) the 
 ordinates d and c equal, respectively, to nx and ko, the influence line 
 .E-9-11-F is obtained. Considering the dotted system, e would be the 
 center of moments in determining the stress in DE. Drawing wy paral- 
 
510 
 
 STKUCTUEAL ENGINEERING 
 
 lei to Ae, the distance ey is obtained, which gives the stress in DE 
 due to a unit load at F, and drawing zc parallel to Ke, the distance vz is 
 obtained, which gives the stress in DE, due to a unit load at D. Now, 
 laying off at (d) the ordinates / and e equal, respectively, to ey and vz, 
 the influence line E-W-12-F is obtained. Then by considering a single 
 load to move over the span as in the above case, the influence line 
 F-14-13-12-11-10-9-8-JB for the stress in bottom chord DE is obtained. 
 By placing the wheel loads (by trial) for maximum stress in DE and 
 
 multiplying each load by its respective ordinate to this influence line, the 
 maximum live-load stress in bottom chord DE will be obtained. The 
 stress in any of the other bottom chord members can be determined in the 
 same manner. 
 
 Let it be required to determine the stress in diagonal dE. Draw- 
 ing am parallel to diagonal dE, we have the stress in the diagonal due 
 to a unit load at E, and drawing bh parallel to dC, we have the stress 
 in diagonal dC, also diagonal dE, due to a unit load at C. Then laying 
 off at (e) the ordinates a'm' and b'h' equal, respectively, to am and 
 
DESIGN OF SIMPLE RAILROAD BEIDGES 
 
 511 
 
 bh, the influence line H-m'-h'-G is obtained. Then by considering a 
 single load to move over the span, the influence line //-20-19-18-m'-16- 
 h'-15-G for the stress in diagonal dE is readily drawn. Then by placing 
 the wheel loads (by trial) on the part of this influence line to the right 
 of 16 and multiplying each load by its respective ordinate the maximum 
 
 Fig. 350 
 
 tensile stress in diagonal dE is obtained. At the same time the maximum 
 compression in diagonal dC is obtained, as the stress in dE and dC are 
 equal and opposite. To obtain the maximum compression in dE (also 
 maximum tension in dC) the wheel loads would be placed to the left of 
 16 and each load multiplied by its respective ordinate. The stress in 
 any of the other web members due to wheel loads can be determined in 
 the same manner. 
 
512 STBUCTURAL ENGINEEEING 
 
 In order to obtain short panels, double-system Warren trusses are 
 sometimes sub-paneled as shown at (a), Fig. 350. The stresses are 
 determined in very much the same manner as shown above for the trusses 
 without sub-paneling. 
 
 Considering the case of dead load, let P represent the panel load at 
 each upper panel point and W the load at each lower panel point. We 
 can consider the sub-paneling as being independent sub-trusses or trussed 
 stringers as indicated at (b) and that the load from these sub-trusses 
 is transmitted directly to the main lower panel points, in which case 
 there would be 2W on each. 
 
 Then by considering the truss to be composed of two separate trusses 
 as shown in Fig. 348 the stress in all the main members can be readily 
 determined and then considering the additional stress in each bottom 
 chord and in the lower half of each diagonal due to the sub-trusses the 
 dead-load stresses throughout the truss are obtained. 
 
 Let it be required to determine the dead-load stress in diagonal eo. 
 The truss being symmetrical about I one-half of the load at that point 
 can be considered as transmitted by eo and also the load P at e. Then 
 we have (JF + P)sec0 for the dead-load stress in eo. This same stress 
 will be in oE and, in addition, the load W at F will produce a compressive 
 stress of ^WsecO in the member which should be added to the above. So 
 we have (W + P)sec6 + ^WsecO for the total dead-load stress in diag- 
 onal oE. 
 
 For the dead-load stress in diagonal od, we have ^PsecO from panel 
 point / and 2Wsec6 from panel point G. So for the total dead-load 
 stress in od, we have 2Wsec6 + ^PsecO. This same stress would be in oG 
 (so to speak) but the load W at F would produce a compressive stress of 
 ^WsecO, which should be subtracted from the above and hence we obtain 
 2WsecQ + ^PsecO-^WsecO for the total dead-load stress in diagonal oG. 
 
 The dead-load stress in each of the hangers bB, mD, oF, etc., is equal 
 to W. The stress in the bottom chord of each sub-truss, as CmE, is equal 
 to ^WtanO. This must be added to the stress found in each bottom chord 
 when considering the truss as composed of two independent trusses. 
 That is, the stress in each bottom chord is determined by considering the 
 truss to be composed of two independent trusses (2W at each lower joint), 
 just the same as shown above in the case of trusses without sub-paneling, 
 and then the stress QJFtanfl) in the bottom chord of the sub-truss is 
 added to this. 
 
 The stress in the top chords and upper half of the diagonals is just 
 the same as if there were no sub-paneling. 
 
 The stress in be (the upper half of the end post) is equal to the 
 shear in panel AC multiplied by sec0. So we have (7JF + 3^P) sec0 for 
 the dead-load stress in be. This same stress occurs in Ab and an addi- 
 tional amount of ^WsecO, due to the load at B. So for the stress in Ab, 
 we have (7W+ 3iP)sec0 + ^JFsec0. For the dead-load stress in bottom 
 chord AC, we have (7JF + 3P)tan0 + i*Ftan0. The part, $WtanO, is 
 due to the load W at B. 
 
 In determining the stress in the hanger cC the load at c, E, e, and 
 7 can be considered as not affecting that member. Then we have \P 
 from /, 2W from G, P from d and 2W from C, making in all (4fF f 1JP), 
 which can be considered to be the stress in cC. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 513 
 
 In determining the live-load stresses the work is greatly simplified 
 by using an equivalent uniform load. 
 
 Let w represent the live load per foot of span. Then we have wd W 
 for live load per panel. As an example, let it be required to deter- 
 mine the maximum live-load tension in diagonal od (shown at (a),, Fig. 
 350). We would load all panels from P to F (inclusive). This would be 
 equivalent to placing 2W at each of the joints 0, M, K, I, and G and \W 
 at E. The load at 0, K, and G are the only ones that produce stress in 
 od. So we have [(W)l/8 + (2W)3/8 + (2W)5/8]sccO for the maximum 
 live-load tension in diagonal od. 
 
 In determining the maximum live-load tensile stress in diagonal oG 
 the load at F would be omitted as it would cause compression in that 
 member. That is, the panels from P to G (inclusive) would be loaded 
 which would be equivalent to placing 2W at each of the joints 0, M, K, 
 and 7 and \^W at G. Then ignoring the load at I and M, we have 
 [(W)l/8+ (2W)3/8+ (l|JF)5/8]sec0 for the maximum live-load tensile 
 stress in diagonal oG. 
 
 To determine the maximum live-load compression in od, panel points 
 B, C, and D would be loaded, in which case the stress in od would 
 be l/8(2JF)sec0, which is due to the 2W at C. 
 
 In the case of diagonal oG, the maximum compression in it would 
 occur when panels B to F (inclusive) were loaded. 
 
 The load at E would produce no stress in oG. One-half of the load 
 at F would be transmitted to G. This would produce a compressive 
 stress of ^WsecO in oG but five-eighths of this would be transmitted back, 
 so we would have (J?Fsec#)f for the compressive stress in oG due to the 
 load W at F. The load 2W at C is the only load, except the one at F, 
 that affects diagonal oG. So we have [1/8W + (JJF)3/8] sec0 for the 
 maximum live-load compressive stress in diagonal oG. 
 
 The maximum live-load tensile and compressive stresses in any of 
 the diagonals due to a uniform live load can be determined in the same 
 manner as shown above for diagonals oG, od, eo, and oE. The maximum 
 stress in the chord members due to a uniform live load occurs when all the 
 joints from B to P (inclusive) are loaded. The sub-paneling is first 
 ignored and a load of 2W is assumed to be at each of the joints C, E, G, 
 I, K, M, and 0. Then the stresses in the chords are determined by con- 
 sidering the truss to be composed of two independent trusses as previously 
 explained. But to the stress in each bottom chord thus obtained the 
 stress ^WtanO must be added. ^Wtan.6 is the stress in the bottom chord 
 of each sub-truss as CmE, EoG, etc. 
 
 In case wheel loads be used, the stresses can be determined most 
 readily by the use of influence lines. The influence lines are easily drawn, 
 some of which are shown in Fig. 350. The influence lines for the top 
 chords are the same as if the truss were not sub-paneled. The influence 
 line E-8-9-10-15-11-12-13-14-F, shown at (d), is for bottom chord GI. 
 The influence line G-lS-^-lG-m-lS-lO-SO-H, shown at (e), is for diagonal 
 en, and G-15-fc-16-r-ra-18-19-20-# is for diagonal nl. The influence 
 line X-3-4-5-7-8-9-10-M, shown at (f), is for diagonal nf and #-3-4-5-6- 
 7-8-9-10-M is for diagonal Gn. The construction of these lines will be 
 readily understood by referring to the diagrams at (c) and (g). 
 
514 
 
 STRUCTURAL ENGINEERING 
 
 213. Whipple Trusses. The truss shown at (a), Fig. 351, is 
 known as a Whipple truss. Dead-load stresses and stresses due to uni- 
 form live load in this type of truss are readily determined by considering 
 the truss as composed of two independent trusses. In determining the 
 dead-load stresses in the truss shown at (a) the truss can be considered 
 as composed of two independent trusses, one of which is shown at (b) 
 and the other one at (c). Let W represent the dead load per panel, one- 
 
 third applied at each top joint and two-thirds at each bottom joint. Tak 
 ing moments about b and considering the truss at (b), we obtain 
 
 for the stress in CE. Likewise taking moments about a and considering 
 the truss at (c), we obtain 
 
 -- 
 
DESIGN OF SIMPLE EAILROAD BRIDGES 
 
 51 fi 
 
 for the stress in BD. Then the stress in bottom chord CD is equal to 
 
 U^W \ tan< + 2JFtan< = G| 
 Again taking moments about E and considering the truss at (b) we obtain 
 
 (- 
 
 V ^ 
 
 for the stress in bd and taking moments about D and considering the 
 truss at (c), we obtain 
 
 for the stress in ac. Then we have 
 
 ^ W ) 
 
 & I 
 
 = 11 Wtan<j> 
 
 for the dead-load stress in top chord be. The dead-load stress in any 
 chord member can be readily determined in this manner. 
 
 Fig. 352 
 
 Considering the trusses at (b) and (c) the dead-load stresses in 
 the web members, as indicated, are readily determined. 
 
 Let P represent the uniform live load per panel. The live-load 
 stresses in the chord members will then be as indicated on the trusses 
 
516 
 
 STRUCTURAL ENGINEERING 
 
 at (d) and (e). These are determined in exactly the same manner as 
 explained above for the dead-load stresses in the chords. The live-load 
 stresses in the web members are indicated at (d) and (e). These are 
 determined as previously explained. For example, considering the truss 
 at (d) and loading the panels from J to E (inclusive),, we obtain 
 
 ^ ( ~ + 2 + 4 + G ) sec<9 = 1 &|jPsec0 
 
 -Lu y (V i ti 
 
 for the maximum live-load tensile stress in diagonal bE and also 
 12 J x (P/10) for the maximum live-load stress in post bC. Likewise, 
 
 Fig. 353 
 
 loading the panels from J to F (inclusive) and referring to the truss at 
 (e), we obtain 
 
 1 
 
 for the maximum live-load stress in diagonal cF and also 8 J x (P/10) for 
 the maximum live-load stress in post cD. 
 
 If wheel loads are used the stresses can be determined most readily 
 by use of influence lines. Influence lines for stresses in Whipple trusses 
 are constructed practically in the same manner as shown above for the 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 517 
 
 double-system Warren trusses. Some of these influence lines are shown 
 in Fig. 352. The one at (c) is for top chord cd and the one at (d) is 
 for diagonal cF. The construction and use of influence lines in the 
 case of the Whipple truss is quite simple provided the work on influence 
 lines previously given is understood. 
 
 Fig. 354 
 
 214. Lattice Truss. The truss shown at (a), Fig. 353, is known 
 as a lattice truss. It can be considered composed of four independent 
 trusses, one of which is shown at (b) and another at (c) and by turning 
 these two end for end, we obtain the other two trusses. By determining 
 the dead and maximum live-load stress in each member of the trusses 
 shown at (b) and (c) and combining the stresses, the stress in the 
 
518 STRUCTURAL ENGINEERING 
 
 truss shown at (a) is obtained. The determination of the dead-load 
 stress is a simple problem and the determination of the live-load stresses 
 is practically as simple provided a uniform live load be used. Let P 
 represent the uniform live load per panel. Considering the truss at (b) 
 and loading joint C (only) we obtain f Psec# for the live-load tension 
 in eC and compression in eG. Loading joint G (only) we obtain 
 f Psecfl for the live-load tension in eG and compression in eC. Loading 
 both joint C and G we obtain -^g - Psec< for the maximum live-load ten- 
 sion in bC and fPsec# for the maximum live-load tension in kG, also 
 ^ Psee( and f Psec< for the live-load compression in b A and kK, re- 
 spectively. Loading joints C and G and taking moments about B, we 
 obtain 
 
 for the live-load stress in AC. Taking moments about C, we obtain 
 
 for the live-load stress in be. Taking moments about k (considering 
 reaction at K), we obtain 
 
 for the live-load stress in GK. Taking moments about e (considering 
 the forces to the left), we obtain 
 
 for the live-load stress in CG. The stresses due to a uniform live load 
 in the trusses shown at (b) and (c) are readily determined in this man- 
 ner. Then by combining these carefully for the chords and end posts, 
 the stresses throughout the truss are obtained. 
 
 In case ^yheel loads are used, the stress therefrom can be most 
 readily determined by the use of influence lines. These are constructed 
 very much the same as previously shown for double-system Warren and 
 Whipple trusses. 
 
 Several railroad companies use the lattice truss shown at (a). The 
 details of part of one of these trusses are shown in Fig. 354. These 
 details are taken from the standard plans of the Chicago & North West- 
 ern Railway Company (W. C. Armstrong, engineer of bridges). 
 
 DEFLECTION AND CAMBER OF TRUSSES 
 
 215. Analytical Determination of Deflection. Let it be required 
 to determine the deflection of joint C of the truss shown in Fig. 355, 
 due to any loads. In the case of pin-connected trusses the deflection 
 is due to the distortion of the members and to a small clearance between 
 the pins and pin holes, while in the case of riveted trusses the deflection 
 is due wholly to the distortion of the members. We will first consider 
 the deflection of joint C, due entirely to the distortion of the members. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 519 
 
 Let u represent the stress in any member as aE, due to a unit load 
 placed at C, and let S represent the stress in the same member due to 
 any loads placed at any panel points. 
 
 Fig. 355 
 
 Let L represent the length of member aB and A its gross area of 
 cross-section. Then for the work done on the member aB by the stress u, 
 we have ux^(uL/AE) which must undoubtedly be equal to J(Ad)l#, 
 where Ad represents the deflection of joint C, due to the stress u. So 
 we have 
 
 from which we obtain 
 
 Ad = 
 
 
 for the deflection of joint C, due to stress u. Now, evidently, the deflec- 
 tion of joint C, due to any stress S in member aB, will be directly pro- 
 portional to the deflection Ad, due to stress u. Then let Ay represent 
 the deflection of j oint C, due to stress S and we have 
 
 Ay _S 
 Ad = w' 
 
 and substituting (uL/AE)u for Ad and reducing, we obtain 
 
 SuL 
 ^ = ~AE 
 
 for the deflection of joint C, due to any stress S in member aB. 
 
 Now, evidently, if w', M", etc., represent the stress in the other mem- 
 bers of the truss, due to a unit load at C, and S', S" , etc., the stress 
 in these members, due to any loads placed at any joints, and Ay', Ay", 
 etc., the deflection of joint C, due to the stresses ', S" , etc., we can 
 write the formula 
 
 for the total deflection of joint C, due to any loads producing the simul- 
 taneous stresses S, S', S", etc. 
 
 By letting S = stress in any member due to any loading, 
 
 u = stress in any member due to a unit load at any point 
 
 considered, 
 
 L = length of any member in inches, 
 
 A = gross area of cross-section of any member in square 
 inches, 
 
520 
 
 STKUCTUEAL ENGINEERING 
 
 the general formula for deflection can be written as 
 SaZ 
 
 Example 1. Let it be required to determine the deflection of joint 
 L3 of the 150-ft. truss shown in Fig. 279, due to the given dead and live 
 loads. 
 
 The live-load stresses given for the truss members in Fig. 279 should 
 not be used in determining the deflection as they do not occur simul- 
 taneously. The maximum deflection will occur when the span is fully 
 loaded and a sufficiently accurate result will be obtained by using an 
 equivalent uniform live load, in which case the live-load stresses will be 
 directly proportional to the dead-load stresses, and hence they can 
 readily be determined directly from the dead-load stresses by the use of 
 a slide rule. 
 
 The maximum live-load stress in the end post (see Fig. 279) is 
 263,000 Ibs. Dividing this by the secant of the slope of the post, we 
 obtain 
 
 263,000 -1.3 -203,000 Ibs. 
 
 for the maximum live-load shear in the end panel LO-L1 (see Art. 174). 
 Then substituting this value for S in the formula of Art. 123, we obtain 
 
 P'= (203,000) 
 
 -d) =203,000-62.5 = 3,250 Ibs. 
 
 for the equivalent uniform live load per foot of truss for determining 
 the live-load stress in the web members, but as the chords contribute 
 most to the deflection we will use 
 
 3,250 - 3,250 (flft + 2-5) % = 3,120 Ibs. 
 
 per foot of truss for the equivalent uniform live load, which is really 
 the correct equivalent uniform live load for determining the maximum 
 chord stresses. (See Art. 123.) 
 
 LI 
 
 t> * ssooo # 
 
 i. + I&3OOO* 
 
 +^ 18000* 
 
 L2 L3 
 
 D+ 88000* 
 
 L f- 26OOOO * 
 
 +3^8000* 
 
 Fig. 356 
 
 As seen from Fig. 279, the dead load per foot of truss is 2,110-7-2 = 
 1,055 Ibs. Then multiplying each of the dead-load stresses given for the 
 truss members in Fig. 279 by f;-J|J the live-load stresses for determin- 
 ing the deflection of joint L3 are obtained. Then adding these to the 
 dead-load stresses, the total stress in each truss member, as shown in 
 Fig. 356, is obtained. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 521 
 
 Then from Figs. 279 and 356 the following table can be computed: 
 
 L S A SL u SuL 
 Member Ins. Lbs. Sq. Ins. EA Lbs. EA 
 
 Ul-LO 
 LQ-L2 
 L2-L3 
 U1-U2 
 U2-U3 
 U1-L2 
 U2-L3 
 U1-L1 
 U2-L2 
 U3-L3 
 
 ( = 30,000,000) 
 
 468 
 
 -337,000 
 
 46.18 
 
 -0.1138 
 
 -0.6500 
 
 0.0739 
 
 600 
 
 +218,000 
 
 23.52 
 
 +0.1853 
 
 +0.4166 
 
 0.0772 
 
 300 
 
 +348,000 
 
 36.98 
 
 +0.0941 
 
 +0.8332 
 
 0.0784 
 
 300 
 
 -348,000 
 
 40.58 
 
 -0.0857 
 
 -0.8332 
 
 0.0714 
 
 300 
 
 -392,000 
 
 45.08 
 
 -0.0869 
 
 -1.2498 
 
 0.1086 
 
 468 
 
 +203,000 
 
 26.48 
 
 +0.1196 
 
 +0.6500 
 
 0.0777 
 
 468 
 
 +67,000 
 
 19.80 
 
 +0.0527 
 
 +0.6500 
 
 0.0343 
 
 360 
 
 +71,000 
 
 14.44 
 
 +0.0590 
 
 
 
 
 
 360 
 
 -87,000 
 
 19.80 
 
 -0.0527 
 
 -0.5000 
 
 0.0263 
 
 360 
 
 -9,000 
 
 19.80 
 
 -0.0054 
 
 
 
 
 
 Total deflection of joint L3 = 1.0996 ins. 
 
 As L3 is at the center of the span and the bridge symmetrically 
 loaded, the deflection of L3 can be obtained, as shown by the above table, 
 by determining the deflection, due to just half of the truss members, and 
 then multiplying this deflection by two. In case of the determination of 
 the deflection of any other joint (except 73) all of the truss members 
 would have to be considered. For example, let it be required to deter- 
 mine the deflection of joint L2 (see Fig. 279). The table in that case 
 would be just the same as the above, except that the last two columns 
 on the right would have two values in each case as the u for the members 
 on one side of the center of span would be different from the u for the 
 corresponding members on the other side of the center of the span. For 
 example, placing a unit load at L2, we have four-sixths of a pound for 
 the reaction at one end of the truss, and two-sixths for the reaction at 
 the other end. Then for the u in one end post we would have $ x 1.3 = 
 0.8333, and | x 1.3 = 0.4166 for the u in the other end post; and likewise 
 | x 0.8333 = 0.5555 for the u in one bottom chord LO-L2, and f x 0.8333 = 
 0.2777 for the u in the other bottom chord LO-L2. Thus it is throughout 
 the span. The last two right-hand columns in the table would simply 
 have double numbers and the summation of the SuL/AE would not.be 
 multiplied by two. Otherwise, the table for each of the other joints 
 would be just the same as shown above for joint L3. 
 
 In case the loading causing the stresses S were placed unsym- 
 metrically, each member would be listed separately in the tables. In that 
 case it is advisable not to have two or more members with the same mark, 
 as in the case of the truss shown in Fig. 279. 
 
 It is customary to consider the distortion of compression members as 
 negative, and that of the tension members as positive. In that case the 
 tensile stresses, both u and S, should be indicated as plus, and the com- 
 pressive stress as negative, as shown in the above table. 
 
 In case the longitudinal displacement of any joint be desired, it 
 can be determined in the same manner as shown above for deflection, 
 except that the unit load applied at the joint in question would be con- 
 
522 
 
 STRUCTURAL ENGINEERING 
 
 sidered as acting horizontally instead of vertically. In that case it is 
 necessary to pay strict attention to the signs of the u's and S's, as they 
 may not have like signs, which they ordinarily have in the case of 
 deflection. 
 
 As an example, let it be required to determine the horizontal dis- 
 placement of joint L3 of the truss shown in Fig. 279, due to the stresses. 
 The unit load would be applied at L3, as shown in Fig. 357. The end 
 Z/6 being fixed and the end LO being on rollers, the only members that 
 would really be stressed by the unit load would be L3-L4 and L4-L6, 
 and hence we need consider only these members in determining the hori- 
 zontal displacement of L3. The stress in each of these members, due to 
 the unit force, as is obvious, is 1 Ib. Then making the following table, 
 the horizontal displacement of L3 is obtained: 
 
 L S A SL u SuL 
 
 Member Ins. Lbs. Sq. Ins. EA Lbs. EA 
 
 L3-L4 ... 300 -348,000 36.98 -0.0941 -1.0000 0.0941 
 L4-L6 ... 600 -218,000 23.52 -0.1853 -1.0000 0.1853 
 
 Total horizontal displacement of joint Z/3 = 0.2794 ins. 
 
 This shows that joint L3 will move about \" to the left. It will be 
 seen that this table is really compiled from the table given above in 
 determining the deflection of L3. 
 
 The horizontal displacement of any of the other lower chord joints 
 can be determined in a similar manner, and just as readily after the 
 tables for the deflection of those joints are made. 
 
 The determination of the horizontal displacement of the top chord 
 joints, while similar to that shown for the bottom chord joints, is not so 
 simple, owing to the fact that more members are involved. As an illus- 
 
 /' U4 
 
 Fig. 357 
 
 tration, let it be required to determine the horizontal displacement of 
 joint 74 (Fig. 357). The unit load would be applied at 74 as indicated. 
 The horizontal reaction of this load would be at L6, as indicated (the 
 end L6 being fixed), but in addition, the unit load at 74 would produce 
 a positive reaction of r at LO and an equal negative reaction r at L6. 
 These vertical reactions and the 1 Ib. horizontal reaction at L6 would 
 have to be considered in determining the u's for the members throughout 
 the truss. For each of the vertical reactions we have r=h/L. After the 
 u's are determined for the members throughout the truss, the horizontal 
 displacement of the top chord joints is determined by making a table for 
 each as explained above. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 523 
 
 tf 
 
 (a) 
 
 --'C' 
 
 In determining the deflection of pin-connected bridges, the distor- 
 tion (SL/AE) of each member should be increased by -^ in. to allow 
 for the clearance between the pins and pin holes. Otherwise the work 
 of determining the deflection of pin-connected bridges is exactly the same 
 as shown above for the riveted bridge. 
 
 216. Graphical Determination of Deflection. Let ACS at (a), 
 Fig. 358, represent a cantilever frame and suppose a vertical load P be 
 applied at C. This load will cause tension in member AC and compres- 
 sion in member BC and hence the distortion in AC will increase its 
 length while the distortion in BC will decrease the length of that member. 
 Let Ca represent the distortion of AC and Ce the distortion of BC and 
 let us assume that the joints A, C, and B are pin connected so that the 
 members AC and BC are free to turn about their ends. Now, undoubt- 
 edly, the distortion of the members AC and BC will cause the frame to 
 deflect to some position as 
 AC'B. As the joints A ^ I? 
 
 and B are fixed in posi- ^^ ^ a 
 
 tion the deflection of the 
 frame really consists of 
 joint C moving to C" and, 
 hence, the determination 
 of the deflection is simply 
 a matter of locating C' in 
 reference to joints A and 
 B, which remain fixed. By 
 taking A as a center and 
 A a as a radius, the arc C" 
 could be described, and 
 taking B as a center and 
 Be as a radius the arc 
 eC' could be described 
 and thus the point C' 
 would be located. 
 
 As another example, let D-F-G-E at (d) represent a cantilever frame 
 supporting a vertical load P at G. Let Fd, Ff, G'k, and Gm represent 
 the distortions of the members DF, FE, FG, and GE f respectively. Now, 
 owing to these distortions, the frame will deflect to some position as 
 D-F'-G"-E. As is obvious, the problem of determining the deflection of 
 the frame consists of locating F' and G" in reference to the fixed joints 
 D and E. First, taking D as a center and Dd as a radius, the arc dF' 
 could be described ; and taking E as a center and Ef as a radius the arc 
 fF' could be described; and thus the position of F' would be determined, 
 which is the deflected position of joint F. Now, having joint F located, 
 the deflected position of joint G can be determined in reference to E and 
 F. Suppose, for the time being, that joint G be disconnected and that 
 member FG moves to the parallel position F'G', while EG remains in 
 position: Then, taking F' as a center and F'k as a radius, the arc kG" 
 could be described ; and taking E as a center and Em as a radius, the arc 
 mG" could be described; and thus the position of G" would be deter- 
 mined, which is the deflected position of joint G; and hence the deflected 
 position of the entire frame would be known. 
 
 ''(e) 
 
 Fig. 358 
 
524 STRUCTURAL ENGINEERING 
 
 Now it is evident that, theoretically,, the deflection of any cantilever 
 frame could be determined in the same manner as shown for the above 
 two cases, and as any truss can be considered as being composed of one 
 or more cantilevers, the same is true of any truss. But, in practice, the 
 application of the method is practically impossible, as the distortions 
 of the members are so small compared with their lengths that if the 
 members be laid off to a reasonable scale it would be practically impos- 
 sible to measure the distortions and deflections. This very fact, how- 
 ever, makes possible the graphical method in general use, in which it is 
 assumed that the distortions are so small in comparison with the lengths 
 of the members that all such arcs as aC' , eC' ', dF' , mG" , etc., without 
 appreciable error can be considered straight lines; that is, the arc and 
 its tangent, in each case, are assumed to coincide. One can satisfy him- 
 self as to the accuracy of this assumption by trial. As a suggestion, sup- 
 pose the distance AB (Fig. 358 (a)) be laid off equal to 10 ft.; AC and 
 BC equal to 15 ft. and 18.5 ft., respectively; and Ca and Ce each equal 
 to -J of an in. and the arcs AC' and eC' described, as explained above, 
 and the difference between these arcs and their tangents noted: By 
 assuming that all such arcs as AC' and eC' , etc., coincide with their 
 tangents, the deflection of any truss can be determined by simply laying 
 off the distortions and drawing perpendiculars to these. As an example, 
 let us consider the case shown at (a), Fig. 358. The joints A and B 
 are considered fixed. Let us take any point A' at (b), as origin, to 
 determine the deflection of the joints A, C, and B. Now, as both A 
 and B are fixed, they will, as regards deflection, be at A' and hence we 
 have A and B located. Joint C in reference to A moves to the right the 
 distance Ca. Then at (b) lay off Ca from A' (the location of joint A) 
 to the right as shown. Joint C in reference to B moves toward B the 
 distance Ce. Then at (b) lay off Ce from A f (the location of joint Z?), 
 as shown. Then, by drawing from e and a the perpendiculars aC f and 
 eC' we have C' located, which is the deflected position of joint C. That 
 is, as A f is the origin (point of zero deflection), V represents the deflec- 
 tion of joint C, and h represents the horizontal movement of that joint. 
 
 Next let us consider the case shown at (d), Fig. 358. Here joints 
 D and E are considered fixed. Let D f at (e) be the origin to determine 
 the deflection of joints D, E, F, and G. As D and E are fixed, they will 
 both be at D', the origin. In reference to joint D joint F moves to the 
 right the distance Fd. So from D' (the location of joint D) lay off Fd 
 to the right as shown. In reference to E joint F moves toward E the 
 distance Ff. So from D' (the location of -E) lay off Ff as shown. Then 
 by drawing the perpendiculars dF' and fF f the point F' is located, which 
 is the deflected position of joint F. Now having F located, joint G can 
 be located in reference to joints F and E. In reference to joint F joint G 
 moves away from F the distance G'k, as shown. In reference to joint E, 
 joint G moves toward E, or to the left, the distance mG. So from D' 
 (the location of joint E) lay off mG to the left as shown. Then by draw- 
 ing the perpendiculars kG" and mG" ', the point G" is located, which is 
 the deflected position of joint G. We then have the deflection of all 
 the joints determined. The distances V and V" represent the deflection 
 of joints F and G, respectively, and h' and h" represent respectively the 
 horizontal movement of these joints. 
 
DESIGN OF SIMPLE RAILROAD BRIDGES 
 
 525 
 
 As another example, let it be required to determine the deflection 
 of the joints of the cantilever frame shown at (a), Fig. 359, where 
 
 Al, A2 AID represent the distortions of the members. The plus 
 
 sign signifies that the length of the member is increased by the distortion, 
 and the minus sign signifies just the opposite. First, select any point A' 
 at (b) as the origin. The joints A and B are fixed so they will both be at 
 A' '. In reference to A, joint C moves to the right the distance Al. So 
 from A' (at (b)) draw Al to the right. In reference to B, joint C moves 
 toward B the distance A4. So from B' (at (b)) draw A4 as shown, and 
 then drawing perpendiculars to these distortions the point C' is located, 
 which shows the deflected position of joint C. That is, joint C has moved 
 from A' to C' . Now, having C located, we can proceed to locate joint D 
 in reference to joints B and C. In reference to C, joint D moves down 
 the distance A5. So from C' lay off A5 downward as shown. In refer- 
 ence to B, joint D will move toward B the distance A9. So from B' 
 (at (b)) lay off A9 to the left. Then, drawing perpendiculars to these 
 distortions (A5 and A9), we obtain D' which shows the deflected position 
 of joint D. That is, 
 joint D has moved from 
 A' to D'. Now, having 
 both C and D located, 
 we can next locate the 
 joint E. In reference to 
 C, joint E moves to 
 the right the distance 
 A2. So from C" draw 
 A2 to the right. In 
 reference to D, joint E 
 moves toward D the 
 distance A6. So from 
 D', draw A6 as shown. 
 Then by drawing per- 
 pendiculars to these 
 distortions (A2 and A6), we obtain E' , which shows the deflected posi- 
 tion of E. That is, joint E has moved from A f to E'. Now having 
 joints E and D located, we can next locate joint F. In reference to E, 
 joint F moves downward the distance A7. So from E' draw A 7 down- 
 ward. In reference to joint D, joint F moves toward D the distance A10. 
 So from D' draw A10 to the left. Then by drawing perpendiculars to 
 these distortions (A7 and A10), we obtain point F', which shows the 
 deflected position of joint F. That is, joint F has moved from A' to F'. 
 Now, having joints E and F located, we can locate joint G. In reference 
 to E, joint G moves to the right the distance A3. So from E' draw A3 
 to the right. In reference to F, joint G moves toward F the distance A8. 
 So from F' draw A8 as shown. Then by drawing perpendiculars to these 
 distortions (A8 and A3), we obtain the point G' ', which shows the de- 
 flected position of joint G. That is, joint G has moved from A' to G'. 
 
 Referring to the diagram at (b), we can consider all of the joints 
 as being at A' before deflection, and that they deflect to the posi- 
 tions indicated. That is, C moves from A' to C' ', D from A' to D' , 
 E from A' to F', F from A' to F', and G from A' to G'. By measuring 
 
 (a) 
 
 Fig. 359 
 
526 
 
 STRUCTURAL ENGINEERING 
 
 the vertical distance down from A' in each case, the deflections are ob- 
 tained and the horizontal movements are obtained by measuring the hori- 
 zontal distance in each case from a vertical line through A' . 
 
 As all trusses can be considered as being composed of one or more 
 cantilever frames, it is obvious that the deflection of any truss can be 
 determined by constructing diagrams similar to those shown at (b) and 
 (e), Fig. 358, and at (b), Fig. 359. 
 
 These diagrams are known as "Williot diagrams," being named after 
 the French engineer Williot, who proposed them. 
 
 Example 1. Let it be required to determine the deflection of the 
 truss shown in Fig. 279, due to the stresses given in Fig. 356. The 
 
 distortions of the mem- 
 bers, due to these stresses 
 are given in Example 1 of 
 Art. 316 (see table). 
 
 First, draw the out- 
 line of the truss as shown 
 at (a), Fig. 360. The dis- 
 tortions shown there for 
 one-half of the members 
 (they are the same for the 
 other half of the truss) 
 are expressed in fractional 
 form for convenience. 
 
 As the truss is sym- 
 metrical about U3-L3 and 
 symmetrically loaded the 
 deflection will be fully 
 known if it be determined 
 for one-half of the truss. 
 By considering joint L3 
 as fixed in position and 
 member U3-L3 as fixed in 
 direction either half of the 
 truss can be considered as 
 a cantilever and the deflection determined as explained above. Let us take 
 L3 at (c) as a starting point. Then the position of joint U3 is obtained 
 by laying off the distortion of U3-L3 (which is j^") downward, using a 
 scale of iV^^V'. Then having joint U3 located, the position of U2 
 can be determined in reference to U3 and L3. Joint U2 moves towards 
 U3 and away from L3. Then laying off the distortion of member 
 U2-U3 to the right from U3 (at (c)) as shown, and the distortion of 
 member U2-L3 upward and parallel to U2-L3 from L3 and drawing 
 perpendiculars to these distortions, we obtain point U2, which shows the 
 position of joint U2. That is, joint U2 has moved from L3 to U2. Joint 
 L2 moves toward U2 and away from L3. Then laying off the distortion 
 of member U2-L2 upward from U2 and the distortion of L2-L3 to 
 the left from L3, and drawing perpendiculars to these distortions, we 
 obtain point L2, which shows the position of joint L2. Joint Ul moves 
 toward U2 and away from L2. Then laying off the distortion of member 
 U1-U2 to the right from U2, as shown, and the distortion of member U1-L2 
 
 Fig. 360 
 
DESIGN OF SIMPLE KAILKOAD BKIDGES 
 
 527 
 
 upward and parallel to U1-L2 from L'2, as shown, and drawing perpen- 
 diculars to these distortions, we obtain point 71, which shows the loca- 
 tion of joint 71. Joint LI moves downward from Ul and to the left from 
 L2. So laying off the distortion in 71-L1 downward from 71 (at (c) ) 
 and the distortion of L1-L2 to the left from L2 as shown and drawing per- 
 pendiculars to these distortions we obtain point LI, which shows the loca- 
 tion of joint LI. Joint LO moves toward 71 and away from LI. Then 
 laying off the distortion of member Ul-LO upward and parallel to Ul-LO 
 from 71 and the distortion of member LO-L1 to the left from LI, and 
 drawing perpendiculars to these distortions, we obtain the point LO, 
 which shows the location of joint LO. We now have the position of the 
 joints of the left half of the truss determined at (c) in reference to L3. 
 As is obvious, in reference to that point, the left half of the truss would 
 be bent upward, as indicated by the dotted outline, and of course the 
 right half would be bent up the same amount. Now, it is evident that 
 the deflection of the bottom chord joints is the distance they are below a 
 horizontal line through LO' . These distances are obtained for each joint 
 by measuring down from LO (at (c)). We thus obtain -" , J", and 
 l^V for the deflection of joints LI, L2, and L3, respectively. Then a 
 diagram of these deflections can be drawn as shown at (b). The deflec- 
 
 U2 -* 
 
 b U4 
 
 L2 
 
 L3 
 
 (aj 
 
 ; & . 3d 
 
 tion of any ordinary bridge truss can be determined in this manner. In 
 case of odd panels, it is best to begin at the middle of the center panel 
 in constructing the Williot diagram. For example, point a (Fig. 361 (a)) 
 would be considered fixed in position and line ab fixed in direction. The 
 diagonal 73-L4 and 74-L3 would be ignored, as they would have no 
 stress in them assuming that a uniform live- and dead-load extended 
 over the entire span. The point b would move downward the distance 
 A2, which is the distortion of each of the posts 73-L3 and 74-L4. Then 
 taking a as the starting point (at (b)), the point b is located by laying 
 off A2 downward from a, as shown. Then joint L3 is located by laying 
 off Al, which is one-half of the distortion of L3-L4, to the left from a 
 and joint 73 is located by laving off A3, which is one-half of the distor- 
 tion of 73-74, to the right "from b. Then, having joints L3 and 73 
 located, the work of determining the deflection proceeds as previously 
 explained. 
 
528 
 
 STRUCTURAL ENGINEERING 
 
 The above manner of determining deflection of trusses is quite 
 accurate as far as vertical movement is concerned, which is usually all 
 that is desired, but if the horizontal movement be desired, the work must 
 be carried out in a somewhat different manner. It will be seen that in 
 the above case we assumed the center of the truss to remain fixed and that 
 each end of the truss was free to move. This assumption is not true, as 
 we know, as one end of the truss is fixed and the other end is free to move. 
 
 Fig. 362 
 
 Now, evidently the only way to locate correctly the deflected posi- 
 tion of the truss would be to begin at the fixed end. 
 
 So let the diagram at (a), Fig. 362, represent the same truss as 
 considered above, and suppose the end L6 fixed and the end LO free (sup- 
 ported upon rollers). Then considering LC> fixed in position and member 
 Z/6-L5 as fixed in direction and taking L6 (at (b)) as the starting point 
 
DESIGN OF SIMPLE KAILROAD BRIDGES 529 
 
 and constructing the Williot diagram as shown, we obtain the relative 
 position of the joints, but the truss will be bent upward (so to speak) as 
 indicated by the dotted outline. So, to obtain the correct position of the 
 joints, the truss must be rotated downward about L6, until the joint LO is 
 again in the same horizontal line as L6. This means that joint LO will be 
 rotated through the arc ed. This arc is so nearly equal to the vertical 
 distance A that it can be considered equal to it. The distance A is equal 
 to the vertical distance LO-LO' ' , shown at (b). Now, evidently, all of the 
 other joints of the truss will be revolved through the same angle as LO 
 and the arc they describe in each case will be directly proportional to 
 their radius. These arcs are so small that the distance from L6 to the 
 undeflected position of any joint can be taken as its radius. Then the 
 radius for any joint of the bottom chord is the horizontal distance from 
 L6 out to the joint. . For example, the radius for joint LI is L1-L6 ; for 
 L2 it is L2-L6; and so on. The radius for joint 71 is J71-L6; for U2 
 it is 72-L6 ; and so on. The arcs can be considered as straight lines 
 perpendicular to these radii. 
 
 Then the arcs through which the joints LO, LI, L2, L3, L4, and L5 
 are rotated are represented, respectively, by the lines A, 1, 2, 3, 4, and 
 5 and the arcs through which joints 75, 74, U3, 72, and Ul are rotated 
 are represented, respectively, by the lines 6, 7/8, 9, and 10. Now, having 
 the lengths of the arcs through which the joints are rotated, it is an 
 easy matter to determine the true deflected position of any joint. As an 
 example, let us consider joint L4. By considering the member L5-L6 
 fixed in direction the joints would all move upward to the positions shown 
 on the Williot diagram at (b). That is, each would move from L6 to 
 the position indicated. Then laying off from L4 downward the arc 4, 
 we obtain the point L4', which is the true deflected position of joint L4. 
 That is, joint L4 really deflects from L6 to L4' and, hence, its vertical 
 movement is represented by the distance 0-L4' and its horizontal move- 
 ment by the distance L6-O. As another case, let us consider joint 75. 
 The line 6, which is perpendicular to the end post 75-L6, represents the 
 arc through which this joint is rotated. So, by laying off this arc from 
 75 the point 75' is obtained, which is the true deflected position of 
 joint 75. That is, joint 75 really moves from L6 to 75'. The true 
 deflected position of each of the other joints can be determined in this 
 manner, but time can be saved by laying off all of the arcs upward from 
 L6. In applying this method let us first consider joint L4. Laying off 
 arc 4 upward from L6, we obtain point L4". Now, the deflected posi- 
 tion of joint L4 is shown by the line L4"-L4. This is readily seen to be 
 true as L6-L4" is the arc through which the joint is rotated, and as the 
 joint was already deflected up to L4, the difference between the two posi- 
 tions would undoubtedly be the actual deflected distance. In the same 
 manner, laying off arc 7, from L6, we obtain point 74". Then the actual 
 movement of joint 74 is from 74" to 74. By laying off all the arcs 
 described by the joints from L6 we will obtain the points LO", L\ n ', Ul" , 
 72", etc. If these be connected by lines, we obtain the truss shown, which 
 is perpendicular to the truss shown at (a). We then have the deflected dis- 
 tance of each joint given. For example, the joint LO moves from LO" to 
 LO, joint LI from LI" to LI, joint 71 from 71" to 71, joint 72 from 
 72" to 72, and so on. By measuring the vertical and horizontal com- 
 
530 
 
 STEUCTUEAL ENGINEEEING 
 
 ponents of these distances the true deflected position of each joint can be 
 
 plotted as shown at (c), which in that case is for the bottom chord joints. 
 
 Now, it will be seen that all we need do to determine the deflection 
 
 of the truss is to construct the Williot diagram at (b) and draw the truss 
 
 LQ"-U\" L6. This truss can be drawn by dividing the line LQ>-LO" 
 
 into as many equal divisions as there are panels in the bridge and then 
 locating one of the points 71" ', 72" ', etc., which is easily done. For in- 
 stance, 71" is located by drawing from LO" a line perpendicular to end 
 post LO-U1 and drawing from LI" a line perpendicular to LG-LO" '. The 
 entire truss L0"-71" . . .L6 of course can be constructed in the .manner 
 indicated at (b). That is, by projecting points -LI", L2", etc., over from 
 the line C-L6 and then locating the points 71", 72", etc., by laying off 
 from LQ the arcs 10, 9, etc. 
 
 That the extremities of the arcs laid off from LG will form the truss 
 shown at (b) is readily seen. In the first place, there is no question as 
 to the location of the points LO", Z/l", J/2" . . .L6 and it is seen from 
 the construction that the two points L4" and 74", L3" and 73", and 
 so on, are in each case on the same horizontal line. So that in reality, 
 the only thing in question is as to whether the points 71", 72", 73", etc., 
 are in the same vertical line. This, however, is readily shown to be true 
 by comparing the construction at (b) with the diagram at (a). The arc 
 L6-74" is perpendicular to the radius L6-74, and, as is seen from the 
 construction Z/4" and 74", is on the same horizontal line and, hence, the 
 triangles L6-74"-L4" and L6-74-Z/4 are similar, the corresponding sides 
 being perpendicular each to each. In the same manner it can be shown 
 that triangle L6-73"-L3" is similar to L6-73-L3, and triangle L6- 
 72"-L2" is similar to triangle L6-U2-L2. Then, evidently, as 74-L4, 
 73-L3, 72-L2, etc., are of equal length at (a), the lines 74"-L4", 
 73"-L3" will be of equal length at (b) and, hence, the points 75", 74", 
 etc., will be in the same vertical line. 
 
 The determination of deflection or displacement by the use of the 
 Williot diagram and the rotation diagram combined, as shown at (b), 
 
 was first proposed by 
 Professor Mohr,* and the 
 diagram LO"-U\" . . . L6 
 is known as Mohr's rota- 
 tion diagram." 
 
 The construction of 
 the Mohr rotation dia- 
 gram is based upon the 
 fact that the lines joining 
 the extremities of the arcs 
 described by the joints of 
 a truss when rotated 
 through a small angle 
 
 about any point will, when the arcs are laid off from a point, form a truss 
 perpendicular to the first truss. For example, suppose the truss shown at 
 (a), Fig. 363, be revolved about any point through the small angle 6. 
 Let Al, A2. . . .A5 represent the arcs described by the joints. Then, by 
 
 Fig. 363 
 
 * See MoHtor's Kinetic Theory of Engineering Structures. 
 
DESIGN OF SIMPLE BAILEOAD BRIDGES 
 
 531 
 
 laying off these arcs from any point P, as shown at (b), and connecting 
 their extremities, we obtain the truss shown dotted. This truss is per- 
 pendicular to the truss at (a). All of this is readily shown to be true by 
 simple geometrical analysis. 
 
 217. Determination of Camber. It is obvious that, if each com- 
 pression member in a truss were lengthened an amount equal to its dis- 
 tortion and each tension member were shortened an amount equal to its 
 distortion, that the truss when supporting no load would be cambered 
 (curved upward) to an extent equal to the deflection and that the truss 
 would be straight when supporting the maximum full load. Camber ob- 
 tained in this manner is known as "exact camber." 
 
 In the case of trusses up to 300 ft. in length sufficient camber seems 
 to be obtained by merely increasing the length of the top chord about ^ 
 of an inch for each 10 feet of length, as explained in Art. 185, but longer 
 spans should have exact camber. 
 
 The position of the joints due to camber, in the case of exact camber, 
 is determined exactly in the same manner as the deflection; in fact, the 
 movement of the joints is exactly equal but opposite in direction. In case 
 the camber is obtained by merely lengthening the top chord, the posi- 
 tion of the joints is most readily obtained by the graphical method, and 
 the work is practically the same as for the deflection. As an example, 
 let it be required to locate the cambered position of the joints of the truss 
 shown in Fig. 279, due to the cambered length specified for that struc- 
 ture. (See Art. '185.) 
 
 First draw the diagram of the truss shown at (a), Fig. 364, and 
 write on each of the members the amount its length is changed to obtain 
 the camber. As is seen, 
 
 only the lengths of the ui +1" u? +,i" us *-%" U4 +" us 
 
 top chords and diagonals 
 are changed, and as the 
 lengths are increased in 
 every case, the plus sign 
 is used throughout. 
 
 We draw the dia- 
 grams at (c) by taking 
 L3 as the starting point 
 and assuming joint Z/3 
 as fixed in position, and 
 member C73-L3 fixed in 
 direction. Joint 73 does 
 not move in reference to 
 L3 (as the length of 73- 
 Z/3 is not changed), so 
 73 will be at the same 
 point as Z/3. Joint 72 in 
 reference to Z/3 moves to 
 
 the left -jffc- in. So from 73 (marked Z/3 also) lay off f$ in. to the left, 
 as shown. Joint 72 in reference to L3 moves away from Z/3 the dis- 
 tance -fo in. So from 73 lay off ^\ * n - upward and to the left as shown. 
 Then drawing perpendiculars to these, we obtain point 72, which shows 
 the position of joint 72. Joint Z/2 does not move in reference to either 
 
 *'-'v^LZ 
 
 N LO 
 
 Fig. 364 
 
532 STEUCTUEAL ENGINEEEING 
 
 joint U2 or L3. So drawing from point U2 a perpendicular to member 
 U2-L2 and from L3 a perpendicular to member L3-L2, we obtain point 
 L2, which shows the location of joint L2. To aid in understanding this 
 step, we can imagine the change in length of each of the members 
 U2-L2 and L2-L3 as being infinitesimal and the lines U2-L2 and L2-L3 
 as being perpendicular,, respectively, to these changes. 
 
 Joint 71 moves -f^ in. away from U2 and -/% in. from L2. Then 
 from U2 (at (c)) lay off -f^ in. to the left and from L2 lay off ^ i n ' 
 upward and to the left, and then drawing perpendiculars we obtain point 
 71, which shows the position of joint Ul. Joint LI does not move in 
 reference to either joint 71 or L2. Then by drawing from 71 a line 
 perpendicular to member 71-L1 and from L2 a line perpendicular to 
 member L1-L2, we obtain the point LI, which shows the location of joint 
 LI. Likewise, joint LO does not move in reference to either joint 71 
 or LI. Then by drawing from 71 a line perpendicular to member 71-LO 
 and from LI a line perpendicular to member LO-L1, we obtain point 
 LO, which shows the location of joint LO. Now, we have the position of 
 the joints shown at (c) for the left half of the truss, and as the truss 
 is symmetrical about the center of the span the diagram at (c) really 
 shows the position of all the joints in the truss. The diagram at (b) 
 shows the positions of the joints of the lower chord which are obtained 
 from the diagram at (c) by measuring upward from LO. 
 
 It is interesting to note how near the camber shown at (b) comes 
 to being equal to the deflection found for this same truss in Art. 215. 
 
 The cambered position of the joints, especially the lower chord 
 joints, of long spans is required mostly for locating the tops of the camber 
 blocks upon which the truss is supported while the structure is being 
 erected. These supports or blocks, placed under each panel point, should 
 have about the correct height, for otherwise the truss members will not 
 fit into place. 
 
CHAPTER XII 
 
 DESIGN OF SIMPLE HIGHWAY BRIDGES 
 
 218. Types. Steel highway bridges can be divided into the fol- 
 lowing types : beam bridges, plate girders, viaducts, pony truss, and high 
 truss bridges. Beam bridges and viaducts are usually deck structures, 
 while pony truss bridges are always through structures. Plate girders 
 and high truss bridges may be either through or deck structures. 
 
 219. Live Load. The maximum live load depends, as a rule, upon 
 the locality of the structure. It varies from street cars, heavy traction 
 engines, trucks, and dense crowds of people in and near cities and towns, 
 to light traction engines and droves of live stock in the country districts. 
 So, in respect to the live load, highway bridges can be classed, except for 
 isolated cases, as city bridges and country bridges. As each city bridge, 
 as a rule, is a special problem, we shall consider only country bridges, 
 although the general principles involved in the design of the two classes 
 are identical. 
 
 The live load for country bridges is specified as a uniform load of 
 so many pounds per square foot of floor the longer the span, the less 
 the load or a load of so many tons concentrated on two axles from 10 
 to 12 ft. apart, the transverse distance between the centers of the wheels 
 being from 5 to 7 ft. The concentrated load is used where it produces 
 a greater stress than the uniform load. The intensities of the loads used 
 vary for different localities depending, of course, in each case, upon the 
 actual loads found to exist and upon the probable future loads. As a rule, 
 the influence of the latter would, as is obvious, depend upon the financial 
 status of the locality paying for the structure. The live loads that the 
 author considers satisfactory for most country bridges are specified farther 
 on in the "Specifications for Steel Highway Bridges." 
 
 220. Impact. From actual tests made on highway bridges over the 
 State of Iowa, under the joint auspices of the State Highway Commis- 
 sion and the Iowa State College Engineering Experiment Station, it was 
 found that impact due to maximum loads is negligible. In some cases the 
 impact due to some light loads for instance, a trotting horse was 
 found to be quite high, especially for bridges having wood floors, but in 
 all such cases the stresses were so low that the impact in no way could 
 influence the design of such structures. In the case of bridges having 
 concrete floors no impact of any consequence was found in any case. 
 
 These tests were conducted by the author, assisted by Mr. C. S. 
 Nichols, Assistant Director of the Experiment Station, and others. The 
 results were verified by tests made in the State of Illinois by Prof. 
 F. O. Dufour and State Engineer A. N. Johnson. 
 
 Judging from the above, no impact should be considered in designing 
 highway bridges. 
 
 533 
 
534 STRUCTURAL ENGINEERING 
 
 221. Dead Load. The dead load, as in the case of railroad 
 bridges, consists of the weight of metal in the structure and the weight 
 of the floor. 
 
 Owing to the wide variation of the weight of metal in highway 
 bridges due to the design, details, width of roadway, loading, and so on, 
 it is practically impossible to give formulas for the weight of metal in 
 general. 
 
 In the case of beam and plate girder bridges, the weights of metal 
 in the different parts are readily assumed and verified as the calculations 
 for the designs are made. The approximate weight of metal in the 
 structures designated can be obtained from the following formulas where 
 p = weight of metal in pounds per foot of bridge, and L length of span 
 in feet, c.c. end bearings : 
 
 For pony trusses with concrete floors (without joists) 
 
 (1). 
 For through trusses with concrete floors (with joists) 
 
 (2). 
 
 The above formulas are for 16-ft. roadways. To obtain the weight 
 of metal for wider roadways add 15 Ibs. for each additional foot of width. 
 
 The weights obtained from the above formulas will, as a rule, be 
 within 10 per cent of the actual weight. The weight of reinforcing steel 
 in the concrete floors is not included. The weight of the floor in each 
 case must be computed and added to the weight of metal. 
 
 222. Specifications. Quite a number of very good specifications 
 for steel highway bridges have been written and published of late years, 
 but unfortunately these do not agree on some important points and for 
 that reason none of them will be designated here as a standard, but instead 
 the following specifications by the author, mostly a compilation, will be 
 used in this work: 
 
 SPECIFICATIONS FOR STEEL HIGHWAY BRIDGES 
 TYPES OF BRIDGES 
 
 1. The following types are recommended : 
 
 Spans up to 30 feet Rolled I-beams, 
 Spans 30 to 50 feet Plate girders, 
 Spans 40 to 90 feet Pony trusses, 
 
 Spans 90 to 150 feet Riveted high trusses, 
 
 Spans over 150 feet Pin-connected trusses. 
 
 LOADING 
 
 2. Dead Load shall consist of the entire weight of structure, including the 
 metal, concrete floor or wood floor (when allowed), and the fill on concrete floors. 
 
 In estimating the dead load concrete shall be considered as weighing 150 
 Ibs. and fill 120 Ibs. per cu. ft., and timber as 4.25 Ibs. per ft. B. M. 
 
 The dead load used in figuring the dead-load stresses must be within 10 per 
 cent of the actual dead weight. 
 
DESIGN OF SIMPLE HIGHWAY BRIDGES 
 
 535 
 
 3. The Live Load shall be as follows : 
 
 The live load on the roadway shall be taken as a uniform load of 
 
 100 Ibs. per sq. ft. for all spans up to 50 ft., 
 
 90 Ibs. per sq. ft. for spans 50 to 100 ft., 
 
 80 Ibs. per sq. ft. for spans 100 to 150 ft., 
 
 70 Ibs. per sq. ft. for spans 150 to 200 ft., 
 
 60 Ibs. per sq. ft. for spans over 200 ft., 
 
 or a 15-ton engine, as per Fig. 365, using the engine whenever greater stresses are 
 obtained than by using the uniform load. 
 
 The live load on sidewalks shall be taken as a uniform load of 50 Ibs. per 
 sq. ft. 
 
 4. Wind Load for high truss bridges shall be 300 Ibs. per lineal ft. of span 
 on the loaded chord and 100 Ibs. per lineal ft. on the unloaded chord. The wind 
 
 II- 
 
 Fig. 365 
 
 Fig. 366 
 
 load on all other structures shall be taken as 30 Ibs. per sq. ft. on one and one- 
 half times the vertical projection of the structure, all of these loads to be consid- 
 ered as moving loads. 
 
 CLEARANCE 
 
 5. The Clearance shall not be less than shown on the diagram in Fig. 366. 
 
 MATERIAL 
 
 6. All material shall be of medium steel except rivets, bolts, rockers, shoes, 
 and pedestals. Rivets and bolts shall be of rivet steel. Kockers, shoes, and 
 pedestals shall be of cast iron or cast steel. 
 
 7. Medium Steel shall have an ultimate strength of 60,000 to 70,000 Ibs. 
 per sq. in. ; an elastic limit of not less than one-half the ultimate strength ; elonga- 
 tion twenty-two per cent (22%} ; bending test through 180 degrees to a diameter 
 equal to the thickness of the piece tested without fracture on outside of bent por- 
 tion. It shall not contain more than 0.05 per cent of sulphur and if basic shall 
 not contain more than 0.04 per cent of phosphorus or 0.06 per cent if acid. 
 
 8. Eivet Steel shall have an ultimate strength of 48,000 to 58,000 Ibs. per 
 sq. in. ; elastic limit of not less than one-half of the ultimate strength ; elongation 
 twenty-six per cent (26%) ; bending test 180 degrees flat on itself without fracture 
 on outside of bent portion. It shall not contain more than 0.04 per cent of either 
 sulphur or phosphorus. 
 
 9. Cast Steel shall have an ultimate strength of not less than 65,000 Ibs. per 
 sq. in.; elongation of fifteen per cent (15%) ; bending test through 90 degrees to 
 a diameter of three times the thickness of the piece tested without fracture on 
 outside of bent portion. It shall not contain more than 0.08 per cent phosphorus 
 nor more than 0.05 per cent of sulphur. 
 
 10. All of the above steel shall be made by the open hearth process. 
 
 11. Cast Iron shall be what is known as grey iron. A test piece one inch 
 square in cross-section, loaded in the middle between supports 12 ins. apart, shall 
 support at least 2,500 Ibs. and deflect at least 0.15 in. before rupture. 
 
536 STEUCTUEAL ENGINEEEING 
 
 ALLOWABLE UNIT STRESSES 
 
 12. The Unit Stress in any part of any structure due to dead and live load 
 combined shall not exceed the following: 
 
 For Metal 
 
 Axial tension on net section 16,000 Ibs. per sq. in. 
 
 Axial compression on gross section where L is length of 
 
 member in ins. and r the least radius of gyration 
 
 in ins 16,000 - 70 (L/r) Ibs. per sq. in. 
 
 Shear on shop rivets and pins 12,000 Ibs. per sq. in. 
 
 Shear on field rivets, webs of girders and rolled beams. . . . 10,000 Ibs. per sq. in. 
 
 Shear on bolts 9,000 Ibs. per sq. in. 
 
 Bearing on shop rivets and pins 24,000 Ibs. per sq. in. 
 
 Bearing on field rivets 20,000 Ibs. per sq. in. 
 
 Bearing on bolts 18,000 Ibs. per sq. in. 
 
 Bending on pins, rivets and bolts ._. 25,000 Ibs. per sq. in. 
 
 Bending on beams and girders ". 16,000 Ibs. per sq. in. 
 
 Bearing on rockers or rollers where d is the diameter of 
 
 rocker or roller in ins (600 X d) Ibs. per lin. in. 
 
 Pin-bearing on rockers 12,000 Ibs. per sq. in. 
 
 For Timber 
 (Douglas fir, white oak, and long leaf yellow pine) 
 
 Bending 1,500 Ibs. per sq. in. 
 
 Tension with grain 1,500 Ibs. per sq. in. 
 
 Compression with grain 1,500 Ibs. per sq. in. 
 
 Shear across grain 800 Ibs. per sq. in. 
 
 Concrete 
 
 Compression due to cross bending 700 Ibs. per sq. in. 
 
 Direct compression and bearing of shoes and pedestals on 
 
 masonry 600 Ibs. per sq. in. 
 
 Tension 000 Ibs. per sq. in. 
 
 PROPORTION OF PARTS 
 
 13. The lengths of compression members shall not exceed 120 times their 
 least radius of gyration in case of pin-connected members nor more than 125 times 
 in case of members having riveted end connections. 
 
 14. Members subjected to alternate stresses of tension and compression shall 
 be proportioned for each kind of stress. 
 
 15. Whenever the live- and dead-load stresses are of opposite signs only two- 
 thirds of the dead-load stress shall be considered as counteracting the live-load 
 stress. 
 
 16. Members subjected to both axial and bending stresses shall be propor- 
 tioned so that the combined stresses will not exceed the allowed axial stress, 
 although if the bending is due to wind load the axial stress can be increased 25 
 per cent over the allowable specified above. 
 
 17. In proportioning tension members, the net area of cross-section shall 
 be considered as the effective section and in deducting for rivet holes the diameter 
 of each hole shall be considered as being & in. larger than the nominal diameter of 
 the rivet. 
 
 18. In determining the number of rivets the nominal diameter shall be 
 used. 
 
 19. Pin-connected riveted tension members shall have a net section through 
 the pin hole at least 25 per cent in excess of the net section of the body of the 
 member, and the net section back of the pin hole, parallel with the axis of the 
 member, shall be not less than the net section of the body of the member. 
 
 20. Plate girders shall be proportioned either by the moment of inertia of 
 their net section, or by assuming that the areas of the flanges are concentrated 
 at their centers of gravity, in which case one-eighth of the gross section of the 
 web, if properly spliced, may be used as flange section. 
 
 21. The gross section of compression flanges of plate girders shall be not 
 less than the gross section of the tension flange and the unsupported distance of 
 any compression flange shall be not greater than 15 times the width of the flange. 
 
DESIGN OF SIMPLE HIGHWAY BRIDGES 537 
 
 22. The flanges of plate girders shall be connected to the web with a suf- 
 ficient number of rivets to transmit the flange increment combined with any load 
 applied directly on the flange and, in case cover plates are used, the cross-section 
 of these plates must not exceed the cross-section of the flange angles. 
 
 DETAILS OF DESIGN 
 
 23. The strength of end connections shall be sufficient to develop the full 
 strength of the member, even though the computed stress is less, the kind of stress 
 to which the member is subjected being considered. 
 
 24. The minimum thickness of metal shall be in. except for fillers and 
 webs of channels. 
 
 The minimum distance between centers of rivet holes shall be three diameters 
 of the rivet; but preferably the distance shall be not less than 3 ins. for |-in. 
 rivets, 2 ins. for f-in. rivets, and 2 ins. for f-in. rivets. The maximum pitch shall 
 be 6 ins. for f-in. rivets, 5 ins. for f-in. rivets, and 4 ins. for f-in. and *-in. rivets. 
 Where two or more plates are used in contact, rivets not more than 12 ins. apart 
 in either direction shall be considered sufficient to hold the plates well together. 
 
 25. The minimum distance from the center of any rivet hole to a sheared 
 edge shall be 1^ ins. for |-in. rivets, 11 ins. for f-in. rivets, and 1 ins. for f-in. 
 and i-in. rivets. The maximum distance from any edge shall be eight times the 
 thickness of the plate, but shall not exceed 6 ins. in any case. 
 
 26. The pitch of rivets at the ends of built compression members shall not 
 exceed four diameters for a distance equal to one and one-half times the maximum 
 width of the member. In compression members the metal shall be concentrated 
 as much as possible in webs and flanges. The thickness of each web shall not be 
 less than one-fortieth of the distance between its connections to the flanges. Cover 
 plates shall have a thickness not less than one-fiftieth of the distance between 
 rivet lines. 
 
 27. Flanges of girders without cover plates shall have a minimum thickness 
 of one-twelfth of the width of the outstanding legs. 
 
 28. The open sides of compression members shall be provided with lattice 
 and shall have tie plates as near each end as practicable. Tie plates shall be 
 provided at intermediate points where the lattice is interrupted. In main members 
 the end tie plates shall have a length not less than their width and intermediate 
 ones not less than one-half their width. Tie plates shall have a thickness of not 
 less than one-fiftieth of the distance between rivet lines. 
 
 29. The minimum width of lattice bars shall not be less than 2 ins. for 
 f-in. rivets, 2 ins. for f-in. rivets, 2 ins. for f-in. rivets, and If ins. for ^-in. 
 rivets. The thickness of lattice bars shall not be less than one-fiftieth of the dis- 
 tance between end rivets for single lattice and one-sixtieth for double lattice. 
 
 30. The inclination of lattice bars with the axis of the member shall be 
 not less than 45 degrees, and when the distance between rivet lines in the flanges is 
 more than 15 ins., double lattice shall be used which shall be riveted at inter- 
 sections. 
 
 31. Abutting joints in top chords of high trusses when faced for bearing 
 shall be spliced on four sides sufficiently to hold the connecting members accurately 
 in place. All other joints in riveted work, whether in tension or compression, shall 
 be fully spliced. 
 
 32. Pin-holes shall be reinforced by plates where necessary, and at least one 
 plate shall be as wide as the flanges will allow and be on the same side as the 
 flange angles. They shall contain sufficient rivets to distribute their portion of the 
 pin pressure to the full cross-section of the member. 
 
 33. In case special permission be given, the field connections of small bridges 
 may be bolted. This refers to the connections of laterals, transverse bracing, joists, 
 floor beams, hand rails, and main splices in trusses. The holes in the connections 
 of the floor beams and joists (when the joists connect directly to the web of 
 the floor beams) must, if bolted, be sub-punched and reamed to iron templets and 
 the open holes in the main splices of trusses must be sub-punched and reamed to 
 size at the shop while the trusses are assembled. The bolts used must have hexa- 
 gonal heads and nuts and standard washers and must fit tightly into the holes. 
 If a desired fit can not be obtained without, turned bolts must be used. In no 
 case will the bolting of the connections of trusses shipped " knocked down" be 
 permitted. Eivets are preferable to bolts. 
 
538 
 
 STRUCTURAL ENGINEERING 
 
 34. Rivets carrying stress and passing through fillers shall be increased 50 
 per cent in number; and the excess rivets, where possible, shall be outside of the 
 connected member. 
 
 35. Provision for expansion and contraction to the extent of in. for each 
 10 ft. of length shall be made for all bridges. 
 
 Spans under 70 ft. in length shall be fixed at one end and allowed to expand 
 and contract upon planed surfaces at the other end. Both ends shall be anchored. 
 Slotted holes for the anchor bolts shall be provided at the sliding end. 
 
 The masonry plates or pedestals shall be cast iron in all cases. 
 
 36. The ends of 65-ft. spans and over shall be pin-bearing and have cast 
 rockers at one end and fixed cast shoes at the other. The rockers shall be of the 
 style shown in Fig. 367. 
 
 The fixed shoes shall be securely anchored to the masonry and the rockers 
 shall be anchored as indicated in Fig. 367. 
 
 Fig. 367 
 
 37. All laterals and transverse bracing in all structures shall be composed 
 of rigid members. 
 
 38. The webs of plate girders shall be stiffened by stiffening angles gen- 
 erally in pairs over bearings at points of concentrated loads and at other points 
 where the thickness of the web is less than one-sixtieth of the unsupported distance 
 between flange angles. The distance between stiff eners shall not exceed that given 
 by the following formula, with a maximum limit of 6 ft., but in no case exceeding 
 the depth of the girder: 
 
 d= (12,000-*) 
 
 where d = clear distance between stiffeners or flange angles, 
 t = thickness of web, 
 s = shear per square inch. 
 
 39. The stiffeners at ends and at points of concentrated loads shall be pro- 
 portioned by the formula of Paragraph 12 (16,000-70 (L/r) ) the effective 
 length being taken as one-half the depth of the girder. End stiffeners and those 
 under concentrated loads shall be on fillers and have their outstanding legs as wide 
 as the flange angles will allow. Intermediate stiffeners may be on fillers or be 
 crimped and their outstanding legs shall be not less than one-thirtieth of the depth 
 of the girder plus 2 ins. All stiffeners shall fit tightly against the flange angles. 
 
 40. Through plate girders shall have their top flanges stayed at each end 
 of every floor beam, or in the case of solid floors at a distance not to exceed 12 ft., 
 by brackets or gusset plates. 
 
 41. Pony trusses shall be riveted structures, with double-webbed chords and 
 shall have web members latticed or otherwise effectively stiffened. 
 
 42. The top chords of pony trusses under 70 ft. in length shall be securely 
 held in position at the panel points by gusset plates, knee braces, or solid webbed 
 vertical posts connected rigidly to the floor beams. The top chords of pony trusses 
 70 ft. in length and over shall be stiffened by knee braces in addition to having 
 solid webbed posts rigidly connected to the floor beams. 
 
DESIGN OF SIMPLE HIGHWAY BKIDGES 539 
 
 43. Trusses supporting wooden floors shall be given a camber by increasing 
 the length of the top chord J in. for each 10 ft. of length and trusses supporting 
 concrete floors shall be given a camber by increasing the length of the top chord 
 & in. for each 10 ft. of length. 
 
 Hip verticals and similar members and the two end panels of the bottom 
 chords of pin-connected bridges shall be rigid members. 
 
 44. All web splices in plate girders shall be sufficient to transmit the shear 
 and bending on the web at the point of splice. 
 
 45. The eye-bars composing a member shall be so arranged that adjacent 
 bars shall not have their surfaces in contact; they shall be as nearly parallel to 
 the axis of the truss as possible. The maximum inclination of any bar is limited 
 to 1 in. in 16 ft., and the bars composing a member must all have about the same 
 inclination. 
 
 WORKMANSHIP 
 
 46. All parts forming a structure shall be built in accordance with approved 
 drawings. The workmanship shall be equal to the best practice in modern bridge 
 works. 
 
 47. Material shall be thoroughly straightened in the shop, by methods that 
 will not injure it before being laid off or worked in any way. 
 
 48. The size of rivets called for on the plans shall be understood to mean 
 the actual size of the cold rivet. 
 
 49. The diameter of the punch shall not be more than ^ in. greater than 
 the diameter of the rivet ; nor the diameter of the die more than & in. greater than 
 the diameter of the punch. 
 
 50. Punching shall be accurately done. Drifting to enlarge unfair holes 
 will not be allowed. If the holes must be enlarged to admit the rivet they shall be 
 reamed. Poor matching of holes will be cause for rejection. 
 
 51. Where sub-punching and reaming are required, the punch used shall 
 have a diameter of & in. smaller than the nominal diameter of the rivet. The 
 holes shall then be reamed to a diameter ^ in. larger than the nominal diameter 
 of the rivet used. 
 
 If necessary to take the pieces apart for shipping and handling, the re- 
 spective pieces reamed together shall be match-marked so that they may be re- 
 assembled in the same position in the final setting up. That is, no interchanging 
 of reamed parts will be permitted. 
 
 52. Eiveted members shall have all parts well assembled and firmly drawn 
 together with bolts before riveting is commenced. 
 
 53. Web plates of girders, which have no cover plates, shall be flush with 
 the backs of the flange angles. 
 
 54. Eivets shall look neat and finished with heads of approved shape, full 
 and of equal size. They shall be central on shank and grip the assembled pieces 
 firmly. Eecupping and calking will not be allowed. Loose, burned, or otherwise 
 defective rivets shall be cut out and replaced. All rivets shall be driven by 
 pressure tools wherever possible. Pneumatic hammers shall be used in preference 
 to hand-driving. 
 
 55. Eye-bars shall be straight and true to size and shall be free from twists, 
 folds in the neck or head, or any other defects. The heads shall be made by up- 
 setting, rolling, or forging. Welding will not be allowed. The heads shall be so 
 proportioned that the bars will break in the body when tested to rupture. 
 
 56. Before boring the pin-holes each eye-bar shall be properly annealed and 
 carefully straightened. The pin-holes shall be in the center line of the bars and 
 in the center of the heads. Bars of the same length shall be bored so accurately 
 that, when placed together, pins jfe in. smaller in diameter than the pin-holes can 
 be passed through the holes at both ends of the bars at the same time without 
 forcing. 
 
 57. The distance center to center of pin-holes shall be correct within -^ 
 in., and the diameter of the holes not more than ^ in. larger than that of the pin, 
 for pins up to 5 ins. in diameter, and 3*3 in. for larger pins. 
 
 58. Pins shall be accurately turned and shall be straight and smooth and 
 entirely free from flaws. 
 
540 
 
 STRUCTURAL ENGINEERING 
 
 PAINTING 
 
 59. All metal work before leaving the shops shall be thoroughly cleaned from 
 all loose scale and rust and be given one coat of pure, boiled linseed oil, well 
 worked into all joints and open spaces, except pin-holes and finished surfaces, 
 which shall be given one coat of white lead and tallow. 
 
 All surfaces of riveted work coming into contact shall be painted before 
 being riveted together with one good coat of pure, boiled linseed oil and red lead. 
 
 After the structure is erected the metal work shall be thoroughly and evenly 
 painted (cleaning parts when necessary) with two coats of either red lead and 
 linseed oil or sublimed blue lead and linseed oil, tinted as directed. 
 
 BEAM BRIDGES 
 Complete Design of an 18-Ft. Beam Span with Wood Floor 
 
 223. Data. 
 
 Length of span = 18'0" c.c. of end bearings, 
 
 Roadway = 16'0" clear, 
 
 Live load, 15-ton traction engine as per Fig. 365, Art. 222. 
 
 Dead load as hereafter determined. 
 
 224. Design of Floor. The beams will 
 
 to" 
 
 
 ii 
 
 
 ^~^~^_ 
 
 
 3'^ 
 
 
 
 / 
 
 
 
 
 \ 
 
 y 
 
 
 a b 
 
 
 ti 
 
 
 ~7< 
 
 ,3' 
 
 7" 
 
 
 
 27" 
 
 
 * 
 
 (b) 
 Fig. 368 
 
 be spaced about 27" center to center. The 
 structure must be designed so that planks 16' 
 long can be used, as odd lengths are usually ex- 
 pensive and difficult to obtain. Let us assume 
 that 3"xl2" yellow pine plank will be used. 
 The large wheels on a traction engine are 
 about 20" wide. Owing to the plank deflect- 
 ing, the wheel of a traction engine will bear 
 mostly at its edges and consequently the pres- 
 sure on the plank will be somewhat as indi- 
 cated at (o), Fig. 368. From this sketch it is 
 seen that we can justly assume the load on 
 the wheel to be applied at two points, a and b, 
 one-half of the total load at each point. As- 
 suming the plank as forming a series of fixed 
 beams, we have at (a) the case of a fixed beam 
 supporting two equal loads of 5,000 # each. 
 In applying the formulas of Art. 69, L equals 
 27", and with reference to support A, kL 
 equals 7" for the load at a and 20" for the 
 load at b. Then we have k equals -fa in the 
 first case and -| ? in the second case. 
 
 Now applying (7) of Art. 69, we have 
 
 AT, 5,000 x 2 -[ 2 (1)' _ (i)'-i 
 
 for the moment at A. Substituting in a similar manner in (6), Art. 69, 
 
 we obtain 5,000 for the shear at A. Now, having the bending moment 
 
 and shear at A determined, the bending moment on the plank at any 
 
DESIGN OF SIMPLE HIGHWAY BEIDGES 541 
 
 point between the beams is readily determined. The maximum will be 
 at the loads, being the same at each load. Then we have 
 
 M = -25,920 + 5,000x7 = 9,080 inch Ibs. 
 
 for the moment at a. In the case of heavy wagons the loading on the 
 plank would be as shown at (6), Fig. 368. Considering the plank to be a 
 fixed beam, as before, and assuming the load on the wheel to be 2,000*, 
 we have 
 
 (2,000 x 27) -r 8 = 6,750 inch Ibs. 
 
 for the maximum bending moment at A and also at the center of the 
 span, it being a negative moment at A and a positive moment at the cen- 
 ter of the span. 
 
 As seen from above, the moment - 25,920"* due to the traction 
 engine is the greatest bending moment on the plank. Assuming each 
 plank to be 3" x 12", we have 
 
 for the moment of inertia. Then for the maximum fiber stress we have 
 /= ^25,920 xli^-5- 27 = 1,440 Ibs. 
 
 (see Art. 53). This stress is for live load alone. The dead load the 
 weight of the plank will raise this slightly. The 1,440* stress given 
 above is about the allowable intensity, but owing to the fact that lumber 
 under-runs in thickness and that the thickness is reduced also from 
 wear, we will use 4" x 12" plank. The weight of the plank can be taken 
 as 4.25* per ft. B. M. Then for the weight per ft. of a 4" x 12" plank 
 we have 4.25 x 4 = 17*. Considering a fixed beam, as before, we obtain 
 
 (17x2J ) + 12 = 7.17 foot Ibs. or 86 inch Ibs. 
 
 for the dead-load bending moment at the supports (see Art. 69). 
 
 Then adding this to the above maximum live-load moment we ob- 
 tain 
 
 25,920 + 86 = 26,006 inch Ibs. 
 
 for the total maximum bending moment on the floor plank. Now, apply- 
 ing Formula D of Art. 53, we have 
 
 /' = (26,006x2) +64 = 812 Ibs. 
 
 for the maximum bending stress on the 4"xl2" floor plank. This is 
 rather a low stress, but when we consider the wear from traffic and that 
 some plank may be only 8" wide and the thickness in some cases may 
 really be only 3^" it is evident that 4" is none too thick and hence we 
 will use that thickness. 
 
 It is true that, by planking for heavy loads, planks 3" thick could 
 be used. However, there is no question but that it would have to be 
 replaced sooner than if 4" plank were used and consequently the saving 
 is not as great as it appears. 
 
 225. Design of Beams. The beams will be about 2 7" = 2^ apart. 
 The 4" floor will weigh 4x4^ = 17* per sq. ft. Then we have 
 17x2^ = 38.25* for the dead load on a beam from the floor planks and 
 
542 
 
 STRUCTURAL ENGINEERING 
 
 assuming the beam to weigh 25* per ft. we have 38.25 + 25 = 63.25, say 
 64# for the total dead load per ft. of beam. Then we have 
 
 M=i-x.64xl8 2 x 12 = 31,104, say 31,000 inch Ibs. 
 8 
 
 for the maximum dead-load bending moment. 
 
 Prof. F. O. Dufour and the author, working independently of each 
 
 other, found that the wheel concentra- 
 tion on joists (spaced about 2' -3" 
 to 2' - 6" apart) supporting wood 
 
 floors was about 43 per cent of the 
 
 j I wheel load. To be on the safe side 
 
 i we will assume 50 per cent. The 
 
 ~^ I maximum live load moment will occur 
 
 Fig. 369 when one large wheel is in the middle 
 
 of the span as shown in Fig. 369. 
 Then we have 
 
 M" = ^-- x 9 x 12 = 270,000 inch Ibs. 
 
 18 
 
 for the maximum bending moment. Now, adding the maximum dead 
 and live load moments together we have 
 
 M'" = 31,000 + 270,000 = 301,000 inch Ibs. 
 
 for the total maximum bending moment on a beam. Then we obtain 
 301,000*16,000 = 18.8 for the section modulus. This calls for a 9"x21# 
 I-beam. The design of the bridge is shown in Fig. 370. 
 
 Complete Design of an 18-Ft. Beam Span with Concrete Floor 
 
 226. Data. 
 
 Length of span = 18' 0" c. to c. of end bearings. 
 Roadway = 16'-0". 
 
 Specifications, as given in Art. 222. 
 
 227. Designing of the Concrete Floor Slab. The I-beams sup- 
 porting concrete floor slabs are usually spaced from 2' 3" to 2'-6" 
 apart and it is usual to limit the minimum thickness of the slab to about 
 6" and the fill to 3" of gravel. So in this case we will assume the slab 
 to be 6" thick, the fill 3", and space the beams as shown at (a), Fig. 
 371. The weight of the slab per sq. ft. of floor is 150x = 75#, and 
 the weight of fill per sq. ft. is 120 x J = 30#, making in all 75 + 30 = 105# 
 per sq. ft. of floor. Now considering a transverse strip 1 ft. wide as a 
 continuous beam over eight supports we have 
 
 M = -J^xl05x2^5 2 x 12 = -673.7 inch Ibs. 
 
 for maximum bending moment on slab due to dead load. 
 
 As is seen, the large wheel of a traction engine will practically 
 extend from beam to beam and hence the maximum live load on the 
 transverse strip considered above will really be uniformly distributed. 
 We know that this load will be distributed along the joists to some 
 extent. Let us assume that it is distributed for 2 ft. along the joists. 
 Then we have the live load as shown at (6), Fig. 371. Now consider- 
 
X 
 
 ll 
 
 fin 
 
 543 
 
544 
 
 STBUCTUEAL ENGINEEEING 
 
 ing the transverse strip to be a fixed beam (which is a safe assumption) 
 we obtain 
 
 M' = --^x5,000-x 2.25x12 = 11,250 inch Ibs. 
 
 for the maximum bending moment on the slab due to live load. Then 
 adding the dead- and live-load moments together we have 11,923"* for 
 the maximum moment on the slab. This moment requires that the con- 
 crete slab (6" thick) be reinforced with |" square rods spaced about 18" 
 
 (a) 
 
 sooo 
 
 Vigaft^rg^ifa*. 
 
 (b) 
 
 Fig. 371 
 
 centers top and bottom. But this spacing would not be satisfactory as 
 the rods would be too far apart to properly distribute the stress through 
 the slab and consequently we will reinforce the slab with \" square rods 
 spaced 12" apart top and bottom in the transverse direction and two J" 
 square rods at bottom of slab between beams in the longitudinal direc- 
 tion. The maximum shear on the slab will be about 2,500 + 110 = 2,610*. 
 The cross-section of the 1-ft. transverse strip including the reinforcing 
 is about 79 n ". So the maximum unit shearing stress on the slab is 
 2,610 -5- 79 = 33* (about), which is quite satisfactory. 
 
 It will be found by calculation that the above slab is safe in case 
 the beams be spaced as far apart as 3' - 3". 
 
 228. Design of Beams. Assuming a beam to weigh 25* per ft., 
 the total dead weight on each intermediate beam will be (75x2.25) + 
 (30x2.25) +25 = 261.25, say 262* per ft. of beam. Then we have 
 
 M = |- x 262 x 18 2 x 12 = 127,333, say 127,000 inch Ibs. 
 
 o 
 
 for the maximum dead-load bending moment. 
 
 Now, assuming that each intermediate beam supports one-third of 
 a wheel instead of one-half as in the case of wood floors (see Fig. 369), 
 we obtain 
 
 x 9 x 12 = 180,000 inch Ibs. 
 
 for the maximum moment on each intermediate beam due to live load. 
 Adding the dead- and live-load moments together we have 
 127,000 + 180,000 = 307,000 inch Ibs. 
 
I 
 
 i I 
 I <* 
 
 j 
 
 545 
 
546 STEUCTUBAL ENGINEEEING 
 
 for the total maximum bending moment. Then we have 307,000 + 16,000 
 = 19.2 for the section modulus. This calls for a 9"x21# I-beam, which 
 is the same as found above for the span with wood floor. The outside 
 beams will very likely be subjected to about two-thirds of the live load 
 and one-half of the dead load carried by the intermediate beams. This 
 requires that the outside beams be 9"xl5# channels. The design of 
 the bridge would be the same as shown in Fig. 370, except the floor 
 would be of concrete. Fig. 372 shows the design of beam bridges sup- 
 porting concrete floors as used by the Iowa Highway Commission. It 
 will be seen that the spans given in Fig. 372 are the clear span in each 
 case. 
 
 PONY TRUSS BRIDGES 
 
 Complete Design of a 50-Ft. Pony Truss Bridge with Concrete Floor 
 
 and without Joists 
 
 229. Data. 
 
 Length =6 panels @ 8' -4" = 50'- 0" c.c. of end bearings. 
 
 Height = 6' - 6" c.c. chords. 
 
 Width = 19' - 3" c.c. trusses. 
 
 Roadway = 18' -0". 
 
 Specifications, as given in Art. 222. 
 
 230. Design of Concrete Slab. As the concrete slab is continuous 
 from one end of the bridge to the other and is supported directly upon 
 the floor beams (no joists) it is really a continuous beam over seven 
 supports and there is no objection to considering the slab as an out and 
 out continuous beam in determining the maximum moments and shears; 
 but sufficiently accurate results will be obtained by considering the slab 
 in each end panel as a beam fixed at one end and supported at the other, 
 and the slab in each of the intermediate panels as fixed beams. In 
 determining the moments and shears on the slab under either assumption, 
 a longitudinal strip 1 ft. wide is considered. In the case of live load 
 only one-fourth of a wheel is considered to be carried by this strip. It 
 is found from calculation that a slab 7J" thick reinforced top and bot- 
 tom longitudinally with f" square rods spaced 9" centers is sufficient 
 for all panels from 7' - 0" to 8' - 6" in length and a slab 8" thick with 
 the same reinforcing is sufficient for all panels from 8' 6" to 9"-0' in 
 length. 
 
 231. Design of Intermediate Floor Beams. Taking the slab as 
 7J" thick and the fill as 3" the dead load on the floor beam from the 
 slab is 124x8.33 = 1,033* per ft. of beam and assuming the beam to 
 weigh 55#, we have 1,033 + 55 = 1,088* for the total dead load on the 
 floor beam per ft. of length. Then we have 
 
 M =^ x 1,088 x 19.25 2 x 12 = 604,753, say 605,000 inch Ibs. 
 o 
 
 for the maximum bending moment on the floor beam due to dead load. 
 Placing the wheel loads as shown in Fig. 373, we obtain 
 
 M' = 8,437x8.12x12 = 822,000 inch Ibs. 
 for the maximum bending moment on the floor beam due to live load. 
 
DESIGN OF SIMPLE HIGHWAY BEIDGES 
 
 547 
 
 Now adding the above moments together we have 
 
 605,000 + 822,000 = 1,427,000 inch Ibs. 
 
 for the total maximum bending moment on the floor beam. Then we 
 have 1,427,000 fl6,00Q = 89.2 for the section modulus. This calls for an 
 I-18"x55*. 
 
 3' \I.5I.5' 
 
 
 
 T 
 
 n 
 
 
 
 
 I 
 
 
 ; j 
 
 T 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 R=S437. 
 
 3.G2' 
 
 
 3.62' 
 
 /9-3" 
 
 
 Fig. 373 
 
 232. Design of End Floor Beams. The maximum bending mo- 
 ment due to dead load is about one-half of the dead-load moment on the 
 intermediate beam and the live-load moment is exactly the same. So 
 for the total maximum bending moment on an end floor beam we have 
 
 318,000 + 822,000 = 1,140,000 inch Ibs. 
 
 Then we have 1,140,000 + 16,000 = 71.2 for the section modulus. 
 This also calls for an I - 18" x 55*. 
 
 233. Dead-Load Stresses in Trusses. From Formula (1) of 
 Art. 221 we obtain 
 
 Ibs. 
 
 for the weight of metal per ft. of span and for the dead load from the 
 floor we have 124 x 18 = 2,232* per ft. of span. Then for the total dead 
 load we have 420 + 2,232 = 2,652* per ft. of span or 1,326* per ft. of 
 truss. 
 
 Then for a panel load we have 1,326x8.33 = 11,045, say 11,- 
 000*. Practically all of this is at the bottom panel points and hence 
 we will so consider it. 
 
 Ul -1-57000 U 2+ 5 7000 U3 
 
 LO 1 
 
 LI 
 
 L2 (-4/ 
 > PQnels 
 
 L3 
 
 = 5Oo 
 
 LO 
 
 Fig. 374 
 
 Referring to Fig. 374 it is seen that the stress in the hangers Ul-Ll 
 and U3-L3 is a panel load, or 11,000*. The stress in U3-L2 is equal 
 to one-half a panel load multiplied by sec0 or (11,000 + 2) 1.63 = 8,965, 
 say 9,000* compression. 
 
 The stress in diagonal U1-L2 is equal to one and one-half of a panel 
 
548 
 
 STRUCTURAL ENGINEERING 
 
 load multiplied by sec0 or (11,000) 1 X ) .63 = 26,895, say 27,000* ten- 
 sion. 
 
 The stress in the end post LO-Ul is equal to two and one-half of a 
 panel load multiplied by sec0 or (11,000) 2^x1.63 = 44,825, say 45,000* 
 compression. The post U2-L2 does nothing but stiffen the top chord and 
 hence the only load coming on it is the weight of the metal at joint 72, 
 which is about 600*. The stress in bottom chord LO-L2 is equal to 
 Rtan6= (11,000) 2^x1.28 = 35,200, say 35,000* tension (see Art. 173). 
 
 The stress in top chord U1-U3 is equal to (11,000) 4 x 1.28 = 56,320, 
 say 57,000* compression. The stress in bottom chord L2-L3 is equal to 
 (11,000) 4|x 1.28 = 63,360, say 64,000* tension. This completes the de- 
 termination of the dead-load stresses in the trusses and we will next 
 take up the determination of the live-load stresses. 
 
 234. Determination of Live-Load Stresses in Trusses. As per 
 specifications (see 3 of Art. 222) the uniform live load is 100* per sq. 
 ft. of roadway. Then for the panel load we have 
 
 W= 100x9x8.33 = 7,497 Ibs. 
 We have the following for determining the stresses in the trusses: 
 
 Tan0 = 1.28, 
 Sec0 = 1.63, 
 
 | Wsec6 =2,036 Ibs., 
 WtanO =9,596 Ibs. 
 
 Loading panel points L5 and L4 (Fig. 375) we have 
 2,036x3 = 6,108, say 6,000 Ibs. 
 
 Ul +38000* U2 +38000* U3 
 
 LO 
 
 U4 
 
 US 
 
 L3 
 3 
 
 8-4 
 
 L4 
 Z 
 
 CD 
 
 L5 
 I 
 
 SO'-O 
 
 Fig. 375 
 
 for the tensile stress in diagonal U3-L4. Loading panel points 1/5, L4 
 and L3 we have 
 
 2,036x6 = 12,216, say 12,000 Ibs. 
 
 for the maximum compression in diagonal U3-L2. Loading panel points 
 L5 to L2 we have 
 
 2,036x10 = 20,360, say 20,000 Ibs. 
 
 for the maximum tensile stress in diagonal U1-L2. Loading panel points 
 L5 to LI we have 
 
 2,036x15 = 30,540, say 30,000 Ibs. 
 
 for the maximum compression in end post Ul-LO. 
 
 As is readily seen, the stress in hangers C71-L1, C73-L3, and U5-L5 
 is tension and equal to a panel load of 7,497 Ibs. There is no live-load 
 
DESIGN OF SIMPLE HIGHWAY BKIDGES 
 
 549 
 
 stress in the posts U2-L2 and Z74-L4. Loading all points, L5 to LI, 
 we have 
 
 9,596x2^ = 23,990, say 24,000 Ibs. 
 a 
 
 for the stress in bottom chord LO-L2 and 9,596 x 4^ = 43,182, say 43,000* 
 for the stress in bottom chord L2-L4. In a similar manner we obtain 
 
 9,596x4 = 38,383, say 38,000 Ibs. 
 
 for the stress in top chord U1-U3. These chord stresses can be deter- 
 mined more readily by 
 direct proportion. For 
 example, the dead-load 
 panel load is 11,000* 
 and the live-load panel 
 load is 7,49 7*, and 
 hence multiply- 
 ing (7,497 -=-11,000) 
 by the dead-load stress 
 in any chord member 
 
 we would obtain the R= 14-800' 
 
 live-load stress in it. 
 
 Thus, for the live-load Fig. 376 
 
 stress in L2-L3 we have 
 
 / 7,49 7 
 
 1 13.25' 
 
 74.25' 3' 
 
 2' 
 
 r 
 
 * 
 
 ^ 
 
 N. 
 
 ~ 
 
 3\ 
 
 
 
 T~ 
 
 
 S 
 
 J 
 
 
 11,000 
 
 x 64,000 = 43,650, say 43,000 Ibs. 
 
 We now have all of the stress due to the uniform live load deter- 
 mined and next we will consider the stress caused by the engine. Placing 
 the large wheels transversely over a floor beam as shown in Fig. 376 and 
 taking moments about A we obtain 
 
 (20,000 x 14.25) -r 19.25 = 14,800, say 15,000 Ibs. 
 
 for the reaction at J5, which is the maximum stress in hangers C71-L1, 
 [73-L3, and Z75-L5. This is greater than the stress caused by the uni- 
 form load and consequently it will be considered in designing these 
 members. 
 
 By placing the wheels as shown in Fig. 377 we obtain a stress of 
 [ (22,500 x 21.34) -r 50] 1.63 = 15,600, say 16,000* compression in diag- 
 onal t/3-L2, which is greater than the stress caused by the uniform load. 
 In a similar manner, by placing a large axle at L4 and the smaller one to 
 the right, we obtain a tensile stress of [ (22,500 x 13) -i- 50] 1.63 = 9,500, 
 say 9,000* in diagonal C73-L4. This stress is also larger than that ob- 
 tained by the use of the uniform live load and hence will be considered 
 in designing the diagonals U3-L2 and 75-L4. 
 
 The hangers 71-L1, etc., and the diagonals U3-L2 and 173-L4 are 
 the only members that are subjected (as ascertained by trial) to greater 
 stress from the engine than from the uniform live load and consequently 
 the stress in the other members due to the engine will not be considered. 
 In the case of shorter spans the stresses due to the engine will influence 
 the design of more members. In fact it is always advisable to determine 
 
550 
 
 STRUCTURAL ENGINEERING 
 
 the stresses due to the engine in most all the members of a truss to make 
 sure that the absolute maximum stress is obtained in each case. Influence 
 lines can be used to an advantage in determining the stress in the truss 
 and especially in the case of the engine load. 
 
 The maximum dead-load stresses in the truss are given in Fig. 374 
 
 3.66* 
 
 LO 
 
 U3 
 
 LI 
 
 L6 
 
 /4--O 
 
 - O 
 
 Fig. 377 
 
 and the maximum live-load stresses in Fig. 375. Combining these we 
 obtain the stress diagram shown at the top of Fig. 378. 
 
 235. Designing of the Sections of Truss Members. The work 
 of designing the truss members is practically the same as previously 
 given for railroad bridges. The vertical posts and hangers as far as 
 stresses are concerned could be very light section, but 4 Ls 3" x 2J" x J" 
 and a plate 7" x \" is about as light a section as can be used for these 
 members and obtain good details. 
 
 Diagonals U3-L2 must be designed to carry 25,000* compression or 
 12,500* tension. The effective length of this member is about 126". 
 Then using 2 Ls 3J"x2J" xi" = 3.76 n " (3J" legs vertically), we have 
 
 L 126 
 
 = 113 
 
 r 1.12 
 
 and hence we obtain 16,000-70x113 = 8,090* for the allowable unit 
 compression stress. Then 25,000 -=- 8,090 = 3.09 n " is the required area 
 for compression and 12,500 -r 16,000 = 0.78 n " is area required for ten- 
 sion. If less than 3^"x 2^" angles be used L/r would be too great, so we 
 will use 2 Ls 3" x 2J" x J" for diagonal U3-L2. The designing of the 
 other diagonals and the bottom chord is simply a matter of determining 
 the correct net section in each case as previously explained in the design 
 of railroad bridges. 
 
 The maximum stress in the top chord is 95,000*. Dividing this by, 
 say 12,000 Ibs., we obtain about S n " for the approximate area of cross- 
 section of the chord. Then assume the top chord to be composed of the 
 following : 
 
 2[ s 6"x8* =4.76" 
 
 1 Cov. i: 
 
 7.76 n " 
 
 The radius of gyration of this section with reference to the hori- 
 zontal axis is 2.38" and with reference to the vertical axis it is 4.08". 
 
551 
 
552 
 
553 
 
554 STKUCTUEAL ENGINEEEING 
 
 Considering the chord to fail vertically the length would be taken as 
 one panel length, or 100". Then we have 
 
 16,000- 70 = 13,060 Ibs. 
 
 for the allowable unit stress. Dividing this into the stress we have 
 95,0004-13,060 = 7.27 sq. ins. 
 
 If the chord be considered with reference to the vertical axis the length 
 should be taken as two panel lengths, or 200". Then we obtain 
 
 16,000-70 =12,570 Ibs. 
 
 \4.0S/ 
 
 for the allowable unit stress. Dividing this into the stress we obtain 
 95,000 + 12,570 = 7.56 sq. ins. 
 
 This shows that the assumed chord section is about correct and hence 
 will be used. The end posts are designed in a similar manner and the 
 laterals are designed to transmit a wind load of 300* per ft. of span. 
 Complete details of the span are shown in Fig. 378. 
 
 236. Long Pony Truss Spans. These spans are designed in the 
 same manner as shown above for the 50-ft. span. The ends of all spans 
 65 ft. and over in length should be pin-bearing and have a cast rocker at 
 one end and a cast fixed shoe at the other and have brackets stiffening 
 the top chord, as shown in Figs. 379 and 380. 
 
 THROUGH TRUSS BRIDGES 
 Complete Design of a 112-Ft. Through Pratt Truss Span 
 
 237. Data. 
 
 Length = 7 panels @ 16'-0" = 112'-0" c.c. end pins. 
 
 Height = 20' - 0" c.c. chords. 
 
 Width =17'- 0" c.c. trusses. 
 
 Roadway=16'-0". 
 
 Specifications, as per Art. 222. 
 
 238. Designing of the Joists and Concrete Slab is the same as 
 previously shown for beam bridges. The joists will be 9"x21* Is and 
 9" x 13.25* [s. The slab will be 6" thick reinforced with \" square rods 
 spaced 12" centers. The fill will be 3" thick. 
 
 239. Designing of the Intermediate Floor Beams. The weight 
 of the slab and fill is 105 Ibs. per sq. ft. of roadway. The joists can be 
 taken as weighing 21+2^ = 8.4 Ibs. per sq. ft. of roadway. Then from 
 the slab, fill, and joists we have (105 + 8.4) 16 = 1,814* for the load per 
 ft. of floor beam and assuming the floor beam to weigh 65* per ft. we have 
 1,814 + 65 = 1,879* for the total dead load on the floor beam per ft. of 
 length. Then for the maximum bending moment on the floor beam due to 
 dead load we have 
 
 M = i x 1,879 x rf x 12 = 814,500 inch Ibs. 
 o 
 
DESIGN OF SIMPLE HIGHWAY BEIDGES 
 
 555 
 
 Placing the wheel loads as indicated in Fig. 381 and assuming that 
 -f$ of the smaller wheels is transmitted to the floor beam under the larger 
 wheels, we have 
 
 M' = 9,520 x 7 x 12 = 800,000 inch Ibs. 
 for the maximum live-load bending moment on the beam. 
 
 4*0" 3 L 0' 
 
 5-0' 
 
 //-O 
 
 ie L o 
 
 Fig. 381 
 
 Adding the above dead- and live-load moments together we have 
 
 814,500 + 800,000 = 1,614,500 inch Ibs. 
 
 for the total maximum bending moment on the beam. Then we have 
 1,614,500 + 16,000 = 101 for the section modulus. This calls for a 
 20"x65* I. 
 
 The designing of the end floor beams is practically the same as for 
 the intermediate beams. The live load is exactly the same and the dead 
 load is about one-half as much as for the intermediate beams. 
 
 240. Determination of the Dead-Load Stresses in Trusses. 
 
 From (2) of Art. 221 we obtain 2.8x112 + 280 = 594* for the weight of 
 metal per ft. of span and for the weight of the concrete slab and the 
 fill we have (75 + 30) 16 = 1,680* per ft. of span, making in all 594 + 
 1,680 = 2,274* per ft. of span or 1,137, say 1,130 Ibs. per ft. of truss. 
 Then we have 
 
 W= 16x1,130 = 18,080 Ibs. 
 
 for the panel load. The expressions for the dead-load stresses are given 
 on the right half of the truss, Fig. 382, and the actual stresses are ffiven 
 
 Fig. 382 
 
 on the left half. About one-third of the weight of metal is assumed to be 
 at the top chord joints. This is about one-twelfth of the total dead load 
 and hence the distribution shown is assumed. For determining the stresses 
 we have the following values : 
 
STRUCTURAL ENGINEERING 
 Tan0=0.8. 
 
 556 
 
 WtanO= 14,464 Ibs. 
 Wsec0= 23,142 Ibs. 
 
 The stresses are determined as explained in Art. 173. 
 
 241. Determination of Live-Load Stresses in Trusses. We will 
 first consider the 80-lb. load specified in (3) of Art. 222. For the panel 
 load we have 
 
 P = 80x8xl6 = 10,2401bs. 
 
 For determining the stresses we have the following values: 
 P tan0 = 10,240 x 0.8 = 8,192 Ibs. 
 
 ip = 1,463 Ibs. 
 i-Psec0 = 1,872 Ibs. 
 Loading panel points G and F (Fig. 383) we obtain 
 
 Fig. 383 
 
 1,872 x 3 = 5,616, say 5,600 Ibs. 
 
 for the maximum compressive stress in diagonal fE. Loading panel points 
 G, F, and E we obtain 
 
 1,872 x 6 = 11,232, say 11,000 Ibs. 
 
 for the maximum tensile stress in diagonal dE. Loading panel points 
 G to D we obtain 
 
 1,872 x 10 = 18,720, say 18,800 Ibs. 
 
 for the maximum tensile stress in diagonal cD. Loading panel points 
 G to C we obtain 
 
 1,872 x 15 = 28,080, say 28,000 Ibs. 
 
 for the maximum tensile stress in diagonal bC. Loading panel points 
 G to B we obtain 
 
 1,872 x 21 = 39,312, say 39,000 Ibs. 
 
 for the maximum compressive stress in end post bA. Loading panel 
 points G to E we obtain 
 
 1,463 x 6 = 8,778, say 8,800 Ibs. 
 
DESIGN OF SIMPLE HIGHWAY BRIDGES 
 
 557 
 
 for the maximum compressive stress in post dD and loading panel points 
 G to D we obtain 
 
 1,463 x 10 = 14,630, say 14,000 Ibs. 
 
 for the maximum compressive stress in post cC. The stress in hanger bB, 
 as is evident, is equal to a panel load which is given above as 10,240*. 
 We now have the stresses in the web members determined and we will 
 next determine the stresses in the chords. The maximum stress in the 
 chords occurs when the span is fully loaded. For maximum stress in 
 bottom chord A-C we have 
 
 8,192 x 3 = 24,576, say 24,500 Ibs. 
 For maximum stress in top chord be and bottom CD we have 
 
 8,192 x 5 = 40,960, say 41,000 Ibs. 
 For the maximum stress in top chord c-f and bottom chord DE we have 
 
 8,192 x 6 = 49,152, say 49,000 Ibs. 
 
 We now have the stresses due to the uniform load determined and we 
 will now consider the stresses due to the engine. Placing the larger wheels 
 over a floor beam as shown in Fig. 384 
 
 we obtain (using the concentration 
 
 shown in Fig. 381) 
 
 1 9 
 
 (11,560 x $)|| = 16,320, say 16,300 Ibs. 
 
 for the maximum end reaction on a floor 
 beam which is also the maximum live- 
 load stress in each hanger and hence 
 this will be used in designing the hangers. 
 
 Placing the wheel loads longitudinally, as shown in Fig. 385, and 
 transversely, as shown in Fig. 384, we obtain the concentrations indicated 
 
 Fig. 384 
 
 Fig. 385 
 
 at E and F. For the shear in panel DE due to these concentrations we 
 have 
 
 This is equal to the maximum stress in post dD due to the engine, and 
 multiplying the same by sec# we obtain 
 
 8,370 x 1.28 = 10,714, say 10,700 Ibs. 
 for the maximum stress in diagonal dE due to the engine. But as these 
 
558 
 
 STRUCTURAL ENGINEERING 
 
 stresses (in dD and dE) due to the engine are less than produced by the 
 80-lb. uniform load we need not consider further the stresses produced 
 by the engine. 
 
 As is seen, the hangers are the only truss members receiving maxi- 
 mum stress from the engine load. 
 
 242. Determination of Wind Stresses in the Laterals. Let us 
 first consider the bottom laterals. The load as given in 4 of Art. 222 
 is 300* per ft. of span. Then for the panel load we have 
 
 P = 300 x 16 = 4,800 Ibs. 
 For determining the stresses we have the following values : 
 
 = 4,800,seco> = 1. 
 
 and Pseco> = 
 
 H 
 
 Fig. 386 
 
 Loading panels from h to e (Fig. 386) we obtain 
 939 x 6 = 5,634, say 5,600 Ibs. 
 
 for the maximum stress in lateral dE. Loading panel points h to d we 
 obtain 
 
 939 x 10 = 9,390, say 9,400 Ibs. 
 
 for the stress in lateral cD. Loading panel points h to c we obtain 
 
 939 x 15 = 14,085, say 14,000 Ibs. 
 for the stress in lateral bC. Loading panel points h to b we obtain 
 
 939x21 = 19,719, say 20,000 Ibs. 
 
 for the stress in diagonal aB. We now have all the stresses necessary 
 for designing the bottom laterals determined (see Art. 177) and we will 
 next consider the top laterals. The load as given in 4 of Art. 222 is 100* 
 per ft. of span. Then for the panel load we have 
 
 P' = 100 x 16 = 1,600 Ibs. and } P' secco = 313. 
 Loading panel points g to e (Fig. 387) we obtain 
 
 6 x 313 = 1,878, say 1,900 Ibs. 
 
 for the stress in top lateral dE. Loading panel points g to d we obtain 
 10 x 313 = 3,130, say 3,100 Ibs. 
 
DESIGN OF SIMPLE HIGHWAY BEIDGES 
 
 559 
 
 for the stress in top lateral cD. Loading panel points g to c we obtain 
 15 x 313 = 4,695, say 4,700 Ibs. 
 
 for the stress in top lateral bC. The maximum stress will occur in strut 
 dD when panel points g to d are loaded. The half of the panel load 
 
 D 
 Fig. 387 
 
 applied at D does not affect the strut and hence the maximum stress in 
 it is equal to 
 
 (f P' + J P') = f x 1,600 + 4 x 1,600 = 2,172, say 2,200 Ibs. 
 Loading panel points g to c we obtain 
 
 ^ P' + i p' \ = 3^)85, say 3, 
 
 000 Ibs. 
 
 for the maximum stress in strut cC. 
 243. Stress in Portal. The 
 
 maximum stress in the portal mem- 
 bers will occur when all of the panel 
 points are loaded. The forces on the 
 portal will be as shown in Fig. 388 : 
 
 P' = 1,600 Ibs. 
 
 R = 2P' = shear in panel be 
 when all panel points are loaded. 
 
 (*%>)= 4000* 
 
 c^-Qoo* 
 
 Fig. 388 
 
 Assuming the end posts fixed and taking moments about we obtain 
 
 V= (4,800 x 17.05) ^ 17 = 4,814, say 4,800 Ibs. 
 Now assuming = 45 we obtain 
 
 4,800 x 1.41 = 6,768, say 7,000 Ibs. 
 
 for the stress in each of the members bD and bE, the same being tension 
 in one and compression in the other. 
 
 Taking moments about D we obtain 
 
 (2,400 x 8.55 + 4,000 x 8.5) -r 8.5 = 6,414, say 6,400 Ibs. 
 
 for the compressive stress in member ab. Taking moments about E we 
 obtain 
 
d 
 
 n* *% 
 
 >|i vf J!M|\ 
 
 
 c S* 3 ^5 - f - 
 
 \ _ 
 
 
 
 
 IQ 
 
 is - 
 
 
 
 . . 
 
 T F 
 
 
 - 
 
 Q 
 
 *c 
 
 V; 
 
 i ?r 
 
 ji- 
 
 __ 
 
 * 
 
 - 
 
 
 * *in 
 
 ^ 
 
 
 
 M ^j 
 
 ! v^ 
 
 
 
 
 
 i s l 
 
 
 
 
 
 
 
 
 KJ 
 
 658^5 
 
 it 
 
 *?.. /' ; / 
 
 / 
 
 -N 
 
 X 
 
 \ 
 
 / 
 
 v . 
 "^ 
 
 7*f??*&/?j 
 
 560 
 
DESIGN OF SIMPLE HIGHWAY BRIDGES 561 
 
 (2,400 x 8.55 + 800 x 8.5) -h 8.5 = 3,214, say 3,200 Ibs. 
 
 for the tensile stress in member be. This completes the calculations of 
 the stresses in the structure and the stresses can be written on the stress 
 sheet Fig. 389. 
 
 244. Designing of Sections of the Truss Members. Intermedi- 
 ate Post. The length of each of these posts is 20 ft. or 240 ins. L/r 
 should not exceed 125. Then we have 240 -h 125 = 1.92 for the minimum 
 allowable value of r. Using this value of r we obtain 
 
 16,000 - 70 x 125 = 7,250 Ibs. 
 
 for the allowable unit stress. Dividing this into the stress in post U2-L2 
 (Fig. 389), we obtain 
 
 34,000-^7,250 = 4.69 sq. ins. 
 
 for the required area. It is seen from this that two 6" channels will 
 be satisfactory for section. The 5" channels can not be used as the r is 
 too small except in the case of the lightest weight which have not enough 
 section. Considering 2 [s 6"x8# = 4.76 n ", we obtain 
 
 16,000-70x^ = 8,821 Ibs. 
 
 for the allowable unit stress, and dividing this into the stress in post 
 U2-L2, we obtain 
 
 34,000 + 8,821 = 3.85 sq. ins. 
 
 for the required area of cross-section, which is 0.91 n/ ' less than contained 
 in the two channels ; but these channels will be used as they are nearest 
 to the theoretical requirement, and for the same reason these channels 
 will be used for the other intermediate posts. 
 
 Top Chord. Assuming 12,000* as the unit stress and dividing this 
 into the maximum stress in the top chord, we obtain 136,000^12,000 = 
 11. 3 n " for the approximate area of cross-section. 
 
 Let us assume the following section: 
 
 1 cov. 12"xJ" = 3.00 n " 
 2 [s 9" x 13.25* = 7.78 n " 
 
 10.78 n " 
 Taking r as 0.4 of the depth of the chord we obtain 
 
 p = 16,000 - 70 ( ^- ) = 12,267 
 
 for the allowable unit stress. Dividing this into the maximum stress in 
 the top chord we obtain 
 
 136,000 *12,267 = 11.1 sq. ins. 
 
 for the required area of cross-section, which is only 0.32 n " more than 
 the above assumed section. The actual r for the section is practically 
 3.6, the same as assumed, so the above assumed section will be used for 
 
562 STRUCTURAL ENGINEERING 
 
 the top chord throughout. The area required for U1-U2 is less than 
 this section, but as minimum thickness is used the area can not be changed 
 without changing the depth of the chord and hence the same section 
 throughout is used. 
 
 End Post. Let us assume the following section for the end post: 
 
 1 Cov. 12"xJ" = 3.00 n " 
 2 [s 9"x20# =11.76*" 
 
 14.76 n " 
 
 The radius of gyration about the vertical axis is about 3.8 and 
 about the horizontal axis it is about 3.5. As a collision strut is used we 
 need consider the strength only with regard to the vertical axis. Sub- 
 tracting the depth of the portal we have 
 
 25.6-8.5 = 17.1 ft., or 205 ins. 
 
 for the effective length of the end post. 
 Then we hr,ve 
 
 p = 16,000 - 70 ( y^ ) = 12,220 Ibs. 
 
 for the allowable unit stress for direct compression. 
 
 Referring to Fig. 387, it is seen that the maximum moment on the 
 post occurs at the bottom of the portal and is equal to 
 
 2,400x8.55x12 = 246,240 inch Ibs. 
 
 The moment of inertia about the vertical axis is about 220 and the dis- 
 tance to the extreme fiber is 6 ins. Using these values we obtain 
 
 246,240 
 
 220 
 
 x 6 = 6,715 Ibs. 
 
 for the unit stress due to the cross bending. 
 For the direct unit stress we have 
 
 108,000 -r 14.76 = 7,317 Ibs. 
 Adding this to the bending stress we have 
 
 6,715 + 7,317 = 14,032 Ibs. 
 
 for the total combined unit stress. The allowable unit stress given above 
 (12,220 Ibs.) can, according to the specifications (see 16 of Art. 222), 
 be increased 25 per cent in the case of combined stress. So for the 
 allowable unit stress we have 
 
 12,220x1.25 = 15,275 Ibs. 
 
 which is 1,243* greater than the actual combined stress, so the assumed 
 section is a little larger than required, but it will be used as the section 
 cannot be easily reduced. 
 
 The designing of the other members of the structure is mostly a 
 matter of determining required net areas, which is fully explained pre- 
 
563 
 
564 
 
 STRUCTURAL ENGINEERING 
 
 viously in the design of railroad bridges. After the sections are designed 
 the stress sheet, Fig. 389, can be completed as shown. 
 
 The general details of the span are shown in Fig. 390. These 
 details will be readily understood if the work on railroad bridges, pre- 
 viously given, is thoroughly studied. 
 
 Complete Design of a 162-Ft. Through Pin-Connected Curved 
 Chord Pratt Truss Highway Bridge 
 
 245. Data. 
 
 Length = 9 panels @ 18' - 0" = 162' - 0" c.c. end pins. 
 
 Height = 20' - 0" at hip and 27' - 10^" at center (see Fig. 392) 
 
 Width = 17' 6" c.c. of trusses. 
 
 Roadway = 16'-0". 
 
 Specifications, as per Art. 222. 
 
 246. The Designing of the Floor System is the same as pre- 
 viously shown (see Arts. 227, 228, and 239). 
 
 247. Determination of Dead-Load Stresses in Trusses. From 
 (2) of Art. 221 we have 
 
 2.8x162 + 280 = 734 Ibs. 
 
 for weight of metal per ft. of span or 367 Ibs. per ft. of truss. For the 
 weight of slab we have 
 
 75x8 = 600 Ibs. 
 
 per ft. of truss, and for the weight of fill we have 
 
 30x8 = 240 Ibs. 
 
 per ft. of truss. Now, considering one-third of the weight of the metal 
 at the top chord joints and two-thirds at the bottom chord joints, we have 
 
 f^pl 18 = 2,202, say 2,200 Ibs. 
 
 for the panel load on each top chord joint and 
 
 [(367) + 600 + 240] 18 = 19,524, say 19,500 Ibs. 
 
 for the panel load on each bottom chord joint. Now having the panel 
 loads computed the stresses are readily determined either analytically or 
 graphically in the same manner as previously shown for railroad bridges 
 (see Arts. 189 and 195). 
 
DESIGN OF SIMPLE HIGHWAY BRIDGES 565 
 
 248. Determination of Live-Load Stresses in Trusses. The 
 uniform live load produces maximum stress in all members except the 
 hangers, in which case the engine produces the maximum stress. The 
 uniform live load as per 3 of Art. 222 is 70 Ibs. per sq. ft. of roadway. 
 Then we have 
 
 P = 70x8x18- 10,080 Ibs. 
 
 for the panel load. 
 
 The maximum live-load stresses in the chords occur when the span 
 is fully loaded and hence are directly proportional to the dead-load 
 stresses in same and can be determined very quickly by multiplying each 
 dead-load stress by the live-load panel divided by the dead-load panel, or 
 the live-load stresses in the chords can be determined as explained in 
 Art. 189. 
 
 The maximum stresses in the web members due to the uniform load 
 are determined in very much the same manner as previously shown for 
 railroad bridges except that in the latter case wheel loads were used. 
 
 The maximum tensile stress in diagonal eF occurs when panel points 
 K, H, G, and F are loaded. (See Fig. 391.) The shear in panel E-F 
 iw then equal to ^-P and hence the stress in the diagonal is equal to 
 1 f-Psec63. The maximum compressive stress in post eE will occur at the 
 same time. This stress is equal to- 1 ^ - P-F2 where F2 represents the 
 vertical component of the stress in top chord ed, which is equal to 
 (l<l>-Px4d/h3) tan</>2 (moments about E). 
 
 The maximum tensile stress in diagonal dE will occur when panel 
 points K to E are loaded and is equal to (-^P F2) sec 62. The maxi- 
 mum compressive stress in post dl) will occur at the same time and is 
 equal to-^-P Fl, where V\ represents the vertical component of the 
 stress in top chord cd, which is equal to (^ 5 -Px3d/h2) tan</>l. 
 
 The maximum tensile stress in diagonal cD occurs when panel points 
 K to D are loaded and is equal to (-g^P VV) sec. 01. The maximum 
 compressive stress in post cC occurs at the same time and is equal to 
 -V P V, where V represents the vertical component of the stress in top 
 chord cb, which is equal to (\ l -Px2d/hl) tan 6. The maximum tensile 
 stress in diagonal bC occurs when panel points K to C are loaded and is 
 equal to (-% 8 -P-F) sec. 9. 
 
 The maximum stress in hanger bB, which is due to the engine load, 
 is determined as previously explained (see Art. 241). 
 
 The maximum tensile stress in counter fG will occur when panel 
 points K, H, and G are loaded, and is equal to (JP-F3) sec 03, where 
 F3 represents the vertical component of the stress in top chord fg. 
 
 The maximum tension in the posts can be determined as previously 
 explained for railroad bridges (see Art. 190, pages 422-426). 
 
 249. Stresses in Portals and Lateral System. From (4) of Art. 
 222 the wind load on the bottom laterals is 300* per ft. of bridge and 
 100* per ft. of bridge on the top laterals. Then we have 300x18 = 
 5,400# for the panel load on the bottom laterals and 100 x 18 = 1,800* 
 for the panel load on the top laterals. The stresses are determined as 
 explained in Art. 242 (also see Arts. 177-179). 
 
566 STRUCTURAL ENGINEERING 
 
 250. Designing of the Sections. The sections of the truss mem- 
 bers and lateral bracing are determined as previously explained (see 
 Arts. 197, 198, 199, 200, and 244). 
 
 After the stresses and sections are determined the stress sheet, Fig. 
 392, can be made. 
 
 251. Details. After the stress sheet is completed, the general 
 drawing, Fig. 393, can be made. The calculations for the details are 
 practically the same as previously shown for railroad bridges (see Art. 
 203). The student should verify the details shown in Fig. 393. 
 
 252. Truss Bridges with Wood Floors are designed in the same 
 manner as bridges with concrete floors. The total dead load is less than 
 for bridges with concrete floors, but the live load is the same. The first 
 cost is less, but in most cases bridges with concrete floors will prove to 
 be the cheaper in the end. 
 
 It is the opinion cf the author that all highway bridges should be 
 designed heavy enough to carry concrete floors even if wood floors are 
 used, as the county authorities usually insist on putting on the concrete 
 sooner or later, and if the bridge is not heavy enough to carry a concrete 
 floor it will have to be replaced by a new bridge, which practically means 
 the loss of the first bridge. 
 

 ki 
 
 c: 
 I 
 
 
 
 \ 
 
 / 
 
 ' 
 
 
 ,t 
 
 V 
 
 / 
 
 \ 
 
 ^ 
 
 , 
 
 \ 
 
 
 
 
 
 X 
 
 
 \ 
 
 
 
 X 
 
 
 
 
 \ 
 
 x 
 
 
 
 
 
 X 
 
 -^ 
 
 
 
 7 
 
 * 
 
 
 
 S 
 
 fe, 
 
 \ 
 
 \ 
 
 
 
 x 
 
 x 
 
 i( 
 
 
 \ 
 
 / 
 
 
 V 
 
 ?i 
 
 C<N 
 
 i 
 
 ; 
 
 ^ 
 
 -g 
 
 5 
 
 J! 
 
 \ 
 
 \ 
 
 
 
 x 
 
 / 
 
 
 
 \ 
 
 ^ 
 
 
 
 A 
 
 y 
 
 
 7 
 
 ^ 
 
 
 / 
 
 
 
 
 
 ^ 
 
 \ 
 
 
 
 
 
 / 
 
 
 \ 
 
 
 
 x 
 
 
 
 
 \ 
 
 / 
 
 
 
 ; 
 
 f 
 
 / 
 
 '; 
 
 ^ 
 
 ^ 
 
 
 t '- 9 ' m - '* 
 
 I 
 
 |<Qi 
 
 t 
 
 III 
 
 X < 
 
 iai 
 
 ?n 
 
 567 
 
568 
 
//"-> L3 
 
 3-Pone/s @ /8-O"= /62-O" 
 
 Genera/ f/ofe.s : 
 
 A II material 'fled. O. M s/sef except nvefs 
 
 andcashngrs. 
 
 All 'rivets '' 
 
 JrANDARD /6?ft. THRO.P/HCONft. TfrUSS H/GHWAY BRIDGE. 
 Concrete floor -/6&'/ 
 
 Fig. 393 
 
 569 
 
CHAPTER XIII 
 
 SKEW BRIDGES, BRIDGES ON CURVES, ECONOMIC HEIGHT 
 AND LENGTH OF TRUSSES, AND STRESSES IN PORTALS 
 
 253. Skew Bridges. Bridges crossing over streams, railroads, 
 streets, and roads at an angle are usually built on the skew. This is 
 
 done in the case of crossings 
 over streams in order to ob- 
 tain piers parallel to the flow 
 of the stream and in the 
 other cases to obtain mini- 
 mum span lengths. 
 
 Skew deck plate girder 
 railroad bridges are usually 
 constructed as shown at (a), 
 Fig. 394, through girder 
 spans as shown at (6), truss 
 spans as shown at (c) 
 and (d). The case shown 
 at (c) is where the skew is 
 a full panel length and the 
 case shown at (d) is where 
 the skew is less than a panel 
 length. 
 
 Skew highway bridges 
 are constructed very similar 
 to skew railroad bridges, as 
 far as the skew is concerned. 
 End floor beams are, as a 
 rule, used in the case of 
 highway bridges, while in 
 the case of railroad bridges 
 as a rule they are omitted 
 and details similar to those 
 shown in Fig. 305 are used, 
 except they are constructed 
 to fit the skew. 
 
 The outer ends of the 
 stringers in railroad bridges 
 should be square with the 
 track, as indicated in Fig. 
 394, so as to avoid having 
 ties resting upon masonry at 
 one end and upon a stringer 
 at the other end. By this 
 
 Fig. 304 construction we also obtain 
 
 570 
 
SKEW BRIDGES, BRIDGES ON CURVES, TRUSSES AND PORTALS 571 
 
 equal stringer concentrations on the first intermediate floor beams, as is 
 readily seen. 
 
 The end posts in each end panel of skew truss bridges should always 
 be parallel. The stresses in skew bridges are determined practically in 
 the same manner as previously shown for square spans; the only differ- 
 ence being that the skew must be taken into account, which is done by 
 assuming the loads to be applied along the center line of the bridge. Let 
 the diagram at (m) in Fig. 395 represent the plan of a skew truss bridge. 
 We would assume the live load 
 applied along the center line gh. 
 The influence line for the reac- 
 tion at A would be as shown at 
 (o). The influence lines for the 
 stress in chord cd and diagonal 
 cD would' be as shown at () 
 and (v), respectively. To sat- 
 isfy the criterion for maximum 
 live-load moment about, say, 
 point D, the distance sg would 
 be taken as A: in (5) of Art. 91 
 and as gw in case the maximum 
 moment at B is desired, and so 
 on. To satisfy the criterion (see 
 Art. 90) for maximum live-load 
 shear in, say, panel EC, the load 
 would extend from h to and into 
 panel BC so that the unit load in 
 that panel (with a load at C) 
 would be equal (or as nearly 
 equal as is possible) to the total 
 load on the span divided by the 
 distance gh. In determining the Fig. 395 
 
 dead-load stress the panel loads 
 
 are readily computed for all panel points. This load will be the same for 
 all panels except for points a, /, and F, in which case a slight correction 
 on account of the skew must be made. After the panel loads are computed 
 the dead-load stresses are readily determined either analytically or graph- 
 ically. However, it must be borne in mind that the truss is unsym- 
 metrical with reference to the center of span. 
 
 The maximum moment and shears on skew through plate girder 
 bridges are determined in exactly the same manner as shown above for 
 truss spans. Skew deck plate girder bridges can be treated the same as 
 square spans without any appreciable error. 
 
 254. Bridges on Curves. Bridges supporting curved tracks are 
 subjected to stresses due to centrifugal force and to the loads being 
 eccentrically applied. These stresses are in addition to dead, live, and 
 impact stresses. 
 
 Stresses Due to Centrifugal Force occur in the laterals system at 
 the loaded chord. The laterals and chords constitute the system which 
 is really a horizontal truss. For centrifugal force in general we have 
 
572 
 
 STRUCTURAL ENGINEERING 
 
 F= 
 
 Wv 2 
 gr 
 
 (1) 
 
 where W represents the weight of the moving body, in pounds, v the 
 velocity of the body in feet per second, g the acceleration due to gravity, 
 and r the radius of curvature in feet. 
 
 Let D equal degree of curvature. Then we have 
 
 50 
 
 r 
 
 and hence for a 1 curve we have 
 
 50 
 
 r = 
 
 -5,730ft. 
 
 0.00873 
 Let V equal velocity in miles per hour. Then we have 
 
 = 
 
 22 
 3,600 15 
 
 Now substituting these values in (1) we have 
 22" 2 
 
 F = 
 
 15 2 x 32.2x5,730 
 
 2 = . 0000117 JFF 2 (2) 
 
 for the centrifugal force for a 1 curve where W equals weight of 
 the moving body and V equals the velocity in miles per hour. Therefore, 
 for a 4 curve and 50-mile velocity, we have 
 
 Then if W be the uniform live load per ft. of span, 0.117 W would 
 be the centrifugal force per ft. of span which would be used to deter- 
 mine the stress in the chords and laterals. As an illustration, let Fig. 
 
 Fig. 396 
 
 396 represent the plan of the lateral system at the loaded chord of a 
 7-panel bridge supporting a track on a 4 curve. Let W represent the 
 live load per ft. of span. Then for a velocity of 50 miles per hour we 
 have 0.11 7W for the centrifugal force per ft. of span, and for the panel 
 load we have 
 
 Then the maximum stress in the laterals and chords due to this horizontal 
 
SKEW BRIDGES, BEIDGES ON CURVES, TRUSSES AND PORTALS 573 
 
 load will be as indicated in Fig. 396. These stresses are determined in 
 the same manner as previously shown for wind stresses. For example, 
 to obtain the maximum stress in diagonal eE panel points B, C, and D 
 would be loaded and in the case of diagonal fF panel points E, C, D, 
 
 Fig. 397 
 
 and E would be loaded, and so on. The 
 maximum chord stress would be obtained by 
 loading all panel points from B to G. The 
 compression in the floor beams due to centrif- 
 ugal force is not considered, as the laterals 
 are practically always connected to the ten- 
 sion flange of the beams and consequently the 
 stress due to centrifugal force only tends to 
 reverse the dead and live-load tensile stress 
 in those flanges. r 
 
 The stress in the chords due to centrif- 
 ugal force is always added to the dead, live, 
 and impact stresses in summing up the total 
 maximum stress. 
 
 Stress Due to Eccentricity occurs in the trusses, stringers, and floor 
 beams. Let the diagram shown in Fig. 397 represent the plan of a 
 bridge supporting a curved track, where the curve acb represents the 
 center line of track and the line oo represents the center line or axis of 
 the bridge which always bisects the middle ordinate (d) of the curve as 
 indicated. It is readily seen that loads near the center of the span will 
 load the outer truss and stringer more than the inner truss and stringer 
 and that just the reverse is true for loads hear the ends of the span. 
 
 The diagram in Fig. 398 shows the conditions at or near the center 
 of the span where g represents the center of gravity of the load. Taking 
 moments above A we obtain 
 
 Fig. 398 
 
 r= W^ Fy 
 
 (3) 
 
 for the vertical concentration on the truss at B where W represents the 
 weight of the load and F the centrifugal force. If the super-elevation 
 of the outer rail be in accord with the speed or velocity of the load the 
 resultant of W and F will pass through the center of the track, in which 
 case, taking moments about A, we have 
 
 for concentration on the outer truss. As is seen, the part We/b is due 
 to the eccentricity, and as is evident this is greater for a moving load 
 than for a static load, for in the latter case F would be zero and e would 
 be less, if not in reality shifted to the other side of the center line of the 
 
574 
 
 STRUCTURAL ENGINEERING 
 
 floor beam. If the velocity of the load be greater than the super-eleva- 
 tion of the outer rail provides, e will be increased. However, such a 
 velocity is not likely to occur for the heavy freight trains which produce 
 the maximum live-load stress in the trusses. If it were possible to locate 
 accurately the center of gravity (#) of the load, the concentration could 
 be determined in any case on the outer truss by taking moments about 
 A which would be simply the application of equation (3) and like- 
 wise the concentration on the inner truss could be determined by taking 
 moments about B. But as it is practically impossible to locate the center 
 of gravity of the load, as it varies so widely for the different engines and 
 cars, it is more practical to consider the velocity to accord with the super- 
 elevation of the outer rail and simply consider the concentration on the 
 outer truss to be as expressed by equation (4). 
 
 As is seen from Fig. 397, e is equal to one-half of the middle ordinate 
 at the center and ends of the span, and fair results will be obtained by 
 assuming e equal to one-half of the middle ordinate throughout the entire 
 length of the span, as the loads near the center of span contribute the 
 greater part of the chord stresses and the loads near the end contribute 
 the greater part of the web stresses and hence the actual discrepancy is 
 slight. Then, to obtain the maximum live-load stresses in the trusses, 
 we would first calculate the maximum stress in each member the same 
 as for ordinary bridges, assuming the load applied along the center line 
 of the bridge, and then increase each of these stresses by the amount 
 obtained by multiplying the stress in each case by e/b when e is taken 
 as half the middle ordinate and b as the distance between trusses. That 
 is, if S be the stress in a member due the load considered applied along 
 the center line of the bridge, the stress due to the eccentricity e would be 
 S x e/b, and hence the total stress in the member due to live load would 
 beS + Se/b. 
 
 In case more accurate results be desired than are obtained by taking 
 the eccentricity e equal to one-half the middle ordinate, the value of e 
 at each floor beam can be computed and the concentration on the trusses 
 at each of those points determined and the stresses in the trusses due to 
 same computed. However, in that case an equivalent uniform live load 
 should be used, as the work would be quite tedious if wheel loads be used. 
 When the stringers are symmetrically placed with reference to the 
 center line of the bridge, the maximum moment and shear on the outer 
 
 stringers are obtained by first 
 
 Truss or Mam Girder-? determining the moment and 
 
 shear for symmetrical load 
 and then increasing each of 
 these by the amount obtained 
 by multiplying each by e/b' 
 where b' is taken as the dis- 
 Fig - 309 tance between the stringers. 
 
 The moment and shear on the 
 
 inner stringers would really be decreased by the same amount; however, 
 it is usual practice to make the outer and inner stringers of equal 
 section. 
 
 In case the curve is quite sharp the stringers are usually off-set, as 
 shown in Fig. 399, in which case the stringers are about equally loaded 
 
 
 
 
 
 
 .---- 
 
 _ 
 
 
 
 
 
 
 
 51-ringerJ 
 
 
 Truss or Main 
 
SKEW BKIDGES, BRIDGES ON CURVES, TRUSSES AND PORTALS 575 
 
 and they are designed as in ordinary cases. The stringer bracing should 
 be designed sufficiently heavy to transmit the centrifugal force. The 
 stress in this bracing is determined in the same manner as shown above 
 for the lateral system of the structure. 
 
 In determining the maximum moment and shear on the floor beams, 
 the eccentricity of the loading must be taken into account. Otherwise 
 the work is the same as previously shown. 
 
 In the case of double-track bridges the determination of the stresses 
 due to centrifugal force is the same as for single-track structures, except 
 that both tracks must be considered. 
 
 According to the A.R.E. Specifications, "structures located on curves 
 shall be designed for the centrifugal force of the live load applied at the 
 top of the high rail. The centrifugal force shall be considered as live 
 load and derived from the speed in miles per hour given by the expres- 
 sion 60 - 2^Z), where D equals degree of curve." 
 
 The velocity given by this expression (60 2JZ)) for the different 
 degrees of curvature 'and the corresponding centrifugal force are as 
 follows : 
 
 D V F D V F 
 
 1 57.5. 0.039JF 4 50.0 0.117JF 
 
 2 55.0 0.071JF 5 47.5 0.132JF 
 
 3 52,5 0.097JF 6. 45.0 0.142ZF 
 
 where D represents the degree of curvature, V the velocity (speed) in 
 miles per hour, F the centrifugal force in pounds, and W the weight of 
 the load in pounds, which would be the weight of load per ft. of span if 
 a uniform live load be used. 
 
 255. Economic Depth of Trusses. As is evident, a, truss is of 
 economic depth as regards weight and stiffeners when the weight of the 
 web members is equal to the weight of the chord members. Of course 
 the weight of the floor system, which varies with the length of panels, is 
 part of the total weight of the structure, and hence the panel length is 
 involved. There are so many variables involved that the theoretical 
 determination of economic heights and panel lengths is practically impos- 
 sible. The most satisfactory values for heights and panel lengths are 
 obtained by actual calculations. The following values are recommended: 
 
 Length of Span 
 in Feet 
 
 -A 1 or tvaiiroau jjriuges 
 
 Length of Panels 
 in Feet 
 
 Height of Truss 
 in Feet 
 
 100 
 
 25 
 
 30 
 
 125 
 
 25 
 
 30 
 
 150 
 
 25 
 
 31 
 
 175 
 
 25 
 
 33 
 
 200 
 
 25 
 
 39 
 
 225 
 
 25 
 
 45 
 
 250 
 
 25 
 
 50 
 
 300 
 
 25 
 
 55 
 
 400 (14 panels) 
 500 (16 Panels) 
 
 28.58 (about) 
 31.25 
 
 62 
 
 72 
 
 Heights for intermediate lengths can be obtained by interpolating. 
 
576 
 
 STEUCTUEAL ENGINEEEING 
 
 Length of Span 
 in Feet 
 105 
 135 
 150 
 162 
 180 
 200 
 
 -For Highway Bridges- 
 Length of Panel 
 in Feet 
 15 
 15 
 16.6 
 18 
 20 
 20 
 
 Height of Truss 
 in Feet 
 20 
 22 
 24 
 28 
 32 
 36 
 
 (a) 
 
 ^r 3 ^, /?* 
 
 f^ 
 
 In order to obtain the required overhead clearance and a satisfac- 
 tory depth of portal the height of truss is limited to about 30 ft. in the 
 case of railroad bridges and to about 20 ft. in the case of highway 
 bridges. The minimum economic panel length for railroad bridges is 
 about 25 ft. and the maximum about 33 ft., while in the case of highway 
 bridges the minimum panel length is about 15 ft. and the maximum 
 about 20 ft. 
 
 256. Economic Span Length. The shortest bridge in the case 
 of a single span, as a rule, is the cheapest, as the cost of the two abut- 
 ments in most cases is practically independent of the span length. How- 
 ever, in a few cases the profile of the crossing is such 
 that the cost of the abutments is materially affected 
 by their location. In such cases care must be taken 
 in locating the abutments so that their cost is a 
 minimum, the usual limit being when the cost of the 
 span equals the cost of the two abutments. Often the 
 required water way or local conditions govern the 
 length of the span, in which cases economy of span 
 length can not be considered. In case a bridge is 
 composed of several spans the spans are practically 
 of economic length when the cost of superstructure 
 minus the floor system is equal to the cost of the sub- 
 structure minus the cost of the end abutments. This 
 is, of course, assuming that the total length of the 
 bridge is fixed either by the required water way or 
 local conditions. The problem can be solved only by 
 trial. First a profile of the crossing should be drawn 
 carefully to scale and the base of rail or roadway 
 located thereon, and next the cost of an average pier 
 computed, and then by using the formulae of Art. 
 124 (Art. 221 in the case of highway bridges) the 
 weight. of metal in the spans of different lengths can 
 be obtained (the weight of the floor system must be 
 subtracted in each case) and by careful comparison 
 of the different bridges possible we can readily as- 
 certain the economic length of spans. 
 
 257. Stresses in Portals. Portals other than the one previously 
 considered (Art. 179) are sometimes used. The plate girder portal 
 shown at (a), Fig. 400, is occasionally used in case of bridges having 
 shallow trusses. 
 
 el P+ 
 
 (0 
 
 Fig. 400 
 
SKEW BRIDGES, BEIDGES ON CURVES, TRUSSES AND PORTALS 577 
 
 Assuming the bottom ends of the end posts as hinged, the wind 
 pressure tends to distort the end posts and portal as indicated at (6). 
 
 Considering the forces to the right of section SS and taking moments 
 about d [see sketch at (c) | we obtain 
 
 F=2M ........ ......... ..(1) 
 
 e e 
 
 for the stress in the bottom flange of the portal. 
 
 But F=-R + P and H=R + P. 
 
 Substituting these values in (1) we obtain 
 
 When x = 
 
 /v c; 
 
 F = ............................................. (3). 
 
 When x w 
 
 F = -^(fl + P) .......................... " .......... (4). 
 
 Now it is seen from the above equations that the stress in the bot- 
 tom flange of the portal is zero at the center of the portal and a maximum 
 at each end, being tension at one end and compression at the other. 
 
 Taking moments about b we obtain 
 
 F = 
 
 for the stress in the top flange of the portal. Substituting 
 for V and ^(J2+P) for // in equation (5) we obtain 
 
 Whenr = 
 
 When# = w/2 
 
 F'-* 
 
 ~2 
 
 When x w 
 
 (R + P)+ (9). 
 
578 
 
 STEUCTUBAL ENGINEERING 
 
 From these equations it is seen that the maximum stress in the top 
 flange of the portal occurs at the ends, being compression at one end and 
 tension at the other,, and that the minimum stress in the top flange occurs 
 at the center of the portal where it is equal to R/2. 
 
 The shear on the portal is constant throughout its length and is 
 equal to V. In case the bottom ends of the end posts be fixed the above 
 will apply by substituting ^(h e) for (h e) and e + h/2 for h (see 
 Art. 179). 
 
 After the stresses in the portal are determined the required sections 
 are readily computed. Each of the flanges should be such that L/r does 
 not exceed 120. The portal shown at (a), Fig. 401, is used in case of 
 limited clearance. 
 
 Assuming the bottom ends of the end posts to be hinged, the wind 
 pressure tends to distort the end posts and the portal as is indicated 
 at (fc). The shear between g and / is constant and is equal to V. The 
 
 Fig. 402 
 
 moment at g and also at / is equal to Fx b/2 which is obtained by taking 
 moments about either g or /. The moment at o is zero and at any point 
 between o and / or between o and g it is Vs. Assuming the moment to be 
 zero at a and k the strut afgk can be considered as a continuous beam. 
 
 Then taking moments at either / or g we have (Vxb/2) ~-d for the 
 reaction at A: or a which is the shear in gk or af. The moment at any 
 point in gk or af is equal to [(Fx b/2) -r d~\x. If x-d we have Vxb/2 
 for the moment at g or / which is the same as found above. 
 
 The horizontal component of the stress in knee brace gm is obtained 
 by considering the forces to the right of section 2-2 and taking moments 
 about k. Then the stress in gm is obtained by multiplying this com- 
 ponent by sec</>. The stress in knee brace cf is obtained in the same man- 
 nerconsidering the forces to the left of section 1-1. 
 
 In case the bottom ends of the end posts are fixed the above will 
 apply by substituting $(h-e) for (h-e) and e + \(h e) for h. 
 
 The portal shown at (a), Fig. 402, is known as a lattice portal. It 
 is distorted by the wind pressure very much the same as the plate girder 
 portal. 
 
 As the shear is constant and equal to V throughout the length of the 
 portal the diagonals are assumed to be equally stressed. As one system 
 is in compression and the other in tension the moment of the stresses in 
 the diagonals about any point on a vertical line through the intersection of 
 
SKEW BEIDGES, BBLDGES ON CUEVES, TEUSSES AND POETALS 579 
 
 the diagonals is equal to zero as the moments balance. Then by taking 
 moments about c and considering the forces to the left of any section 
 as 2-2, shown at (6), we have 
 
 T^ 
 
 Ilh-Vx 
 
 for the stress in the bottom flange at b and taking moments about b we 
 have 
 
 
 H(h-e)-Vx P 
 
 for the stress in the top flange at c. The flange stress throughout the 
 portal can be obtained in this manner. 
 
 The stress in the diagonals is obtained by dividing the shear (F) by 
 the number of diagonals cut by a vertical plane, and multiplying this by 
 the secant of the slope of the diagonals. 
 
 The portal shown in Fig. 403 is used to 
 some extent, especially for long span bridges. 
 Let us first assume that the diagonals take ten- 
 sion only. Then diagonal db would have zero 
 stress when the wind pressure is applied as 
 indicated and diagonal ac would have zero stress 
 if the wind pressure were applied from the 
 opposite direction. 
 
 Assuming the bottom ends of the end posts 
 hinged, and taking moments about a (ignoring 
 diagonal bd) and considering the forces to the 
 left of section 2-2 we obtain 
 
 Fig. 403 
 
 (i) 
 
 for stress (compression) in be, and taking moments about c (ignoring 
 diagonal bd) and considering the forces to the right of section 3-3 we 
 obtain 
 
 S > = *L, h - e )+(R+? 
 
 6 \ 2 
 
 for the stress (tension) in ad. 
 
 The stress in the diagonal ac is equal to Fsec<. If the wind 
 pressure were applied from the opposite direction to that indicated 
 diagonal bd would be stressed, and the stress in be and ad would be 
 determined by taking moments about d and b, respectively. The stress 
 in diagonal bd would be Fsec</> and the stress in diagonal ac would be 
 
 zero. 
 
 In case the diagonal were considered to take compression as well as 
 tension we would assume the two equally stressed and then by taking 
 moments about o and m the stress in be and ad could be determined, as 
 the sum of the moments of the stresses in the diagonals about these points 
 would be zero as they would balance. 
 
580 
 
 STRUCTURAL ENGINEERING 
 
 If the lower ends of the end posts be considered fixed, the above 
 
 would apply if ^(h e) be substituted for (h-e) and e + ^(h-e) for h. 
 
 The depth of the portal, as a rule, is so shallow, as compared with 
 
 the length of the end posts, that the bending of the posts along the ends 
 
 of the portal can be ignored, and in case the lower ends of the posts are 
 
 fixed the points of contra- 
 
 lt t / flexure in the posts can be 
 
 N ^*i considered as being mid- 
 
 way between the portal 
 and the lower end of the 
 posts, as has been done in 
 all previous examples. But 
 in a few cases the required 
 depth of the portal is so 
 great that the bending of 
 the end posts along the 
 ends of the portal must be 
 taken into account in de- 
 termining the points of 
 contra-flexure in the posts. 
 Let the diagram shown, 
 
 Fig- 404 ( a ), Fig. 404, represent 
 
 such a case. The distor- 
 tion of the portal members is so small as compared with the flexure of the 
 posts that the points C and B can be assumed to remain in a vertical line 
 and likewise the points D and E. The problem is to determine the value 
 of z, which is the distance from the lower ends of the end posts up to the 
 points of contra-flexure. 
 
 Let R represent the horizontal component of the forces acting upon 
 the post at C, and R' the horizontal component of the forces at B. Then 
 the post ABC can be considered as a beam fixed at A and acted upon by 
 the forces R and R ' as indicated at (6). 
 
 Then, for the bending moment at any point on this beam between 
 A and B we have 
 
 (1). 
 
 Integrating, we have 
 
 dy 
 
 But -j*- = when x = 0, therefore C = 0, and we have 
 ax 
 
 (2), 
 
 Integrating again, we obtain 
 
SKEW BRIDGES, BRIDGES ON CURVES, TRUSSES AND PORTALS 581 
 But y when # = 0, therefore C' 0, and we have 
 
 (r 2 r 3 \ / r 2 r 3 \ 
 
 "y-y)- K '( c |-T; ................ 
 
 For the bending moment at any point between C and B we have 
 
 M' = EI^=R(h-x} ............................... (4). 
 
 dx~ 
 
 Integrating, we have 
 
 As is obvious (2) and (5) are equal when x-c in each. So by 
 substituting c for x in both (2) and (5), equating and reducing, we 
 obtain 
 
 C"- R' '' 
 C R - 
 
 Substituting this value of C" in (5) we obtain 
 
 2 \ 2 
 
 Of O 
 
 /C / 76 
 
 Integrating again, we have 
 
 ^+C"' (6). 
 
 As is obvious, (3) and (6) are equal when x = c in each. So by substi- 
 tuting c for x in both (3) and (G), equating and reducing, we obtain 
 
 C'"=+B'-j- 
 
 Substituting this value of C' " in (6) we obtain 
 
 fL.T4__7?' fy\ 
 
 X+ It (( ). 
 
 As points C and B are in the same vertical line (3) and (7) are 
 equal for x = c in (3) and x = h in (7). So substituting these values, 
 equating and reducing, we obtain 
 
 R_ 3c 8 - 3c*fe _ 3c 2 
 
 ^' " 3/ic- 2 - c- 3 - 2A 3 ~ 2hc + 2h 2 - c 2 " 
 
 When x z we have 
 
 (9). 
 
 From this equation we obtain 
 R (c-z) 
 
 R' (h-z) 
 
582 STRUCTURAL ENGINEERING 
 
 Now equating (8) and (9) and reducing we obtain 
 
 By applying equation (10) the points of contra-flexure in the posts 
 can be determined and then the stresses in the portal can be computed 
 as previously explained. 
 
 DRAWING ROOM EXERCISE NO. 9 
 
 (a) Design a single-track through pin-connected curve chord Pratt 
 truss railroad bridge and make a complete stress sheet of same upon 
 tracing cloth. 
 
 Data: 
 
 Length of span = 9 panels at 26'-0" = 234'-0" c.c. end pins. 
 
 Height of truss at center = 46'-0". 
 
 Height of truss at hip = 31'-0". 
 
 Top chord joints to be on the arc of a parabola. 
 
 Live load, Cooper's E50. 
 
 Dead load, to be assumed. 
 
 Specifications, A. R. E. Ass'n. 
 
 (b) Make a general detail drawing of the above bridge. 
 
 (c) Design a single-track through Pettit truss railroad bridge and 
 make a stress sheet for same. 
 
 Data: 
 
 Length of span = 14 panels at 28'-0" = 392'-0" c.c. end pins. 
 Live load, Cooper's 50. 
 Dead load, to be assumed. 
 Specifications, A. R. E. Ass'n. 
 
 (d) Design a pony truss highway bridge and make a general draw- 
 ing of same on tracing cloth. 
 
 Data: 
 
 Length of span = 6 panels at 8'-0" = 48'-0". 
 
 Height of truss = 6'-6". 
 
 Width of roadway = 16'-0". 
 
 Concrete floor 7" slab and 3" gravel fill. 
 
 Specifications, as per Art. 222. 
 
 (e) Design a through pin-connected curve chord highway bridge. 
 
 Data: 
 
 Length of span = 9 panels at 18'-6" = 166'-6" c.c. end pins. 
 
 Height of truss at center = 28'-0". 
 
 Height of truss at hip = 20'-0". 
 
 Roadway, 16'-0". 
 
 Concrete floor 6" slab and 3" gravel fill. 
 
 Specifications, as per Art. 222. 
 
CHAPTER XIV 
 
 DESIGN OF BUILDINGS 
 MILL BUILDINGS 
 
 258. Preliminary. The simplest type of mill building consists of 
 a steel roof supported upon brick or concrete walls as shown at (a), 
 Fig. 405. This type of building is satisfactory for power houses, pumping 
 stations, and small manufacturing plants where light material is handled. 
 
 The type shown at (b) consists of a steel roof supported upon steel 
 columns. The side and end walls for this type may be brick, hollow 
 tile, concrete, metal lathing plastered, or corrugated iron. In case the 
 material handled is heavy, an overhead crane would be installed. This 
 crane would run lengthwise of the building and be supported upon girders 
 riveted to the columns. 
 
 The type shown at (c) is practically the same as the one shown at 
 (6) except a shed, known as a lean-to, is attached to each side and a 
 monitor or ventilator is built onto the top of the roof for the purpose of 
 ventilating and lighting the building. 
 
 The type shown at (d) is simply a double building of the same 
 type as the one shown at (c) except, as a rule, it would be proportionately 
 larger. 
 
 The type shown at (e) is known as the saw-tooth construction. This 
 type of construction is used in the case of a very wide building in order 
 to obtain light in the interior of the building. 
 
 All of the types shown above are for steep roof construction where 
 the roof covering is corrugated iron, slate, concrete, or tile. In case of 
 flat roof construction, where the roof covering is felt covered with tar 
 and gravel, the types shown in Fig. 406 are used. 
 
 The type shown at (/) consists of a steel roof supported upon brick 
 or concrete walls. 
 
 The type shown at (g) consists of a steel roof supported upon steel 
 columns. The end and side walls may be corrugated iron, brick, con- 
 crete, hollow tile, or metal lathing plastered. 
 
 The type shown at (h) is the same as the one shown at (g) except 
 a lean-to is added to each side. 
 
 The type shown at (Ar) is practically the same as the one shown at 
 (h) except each lean-to is wider and the building is proportionately 
 larger in every way. 
 
 The types of buildings shown in Figs. 405 and 406 are general 
 types. These are often modified to suit certain special requirements; 
 in fact it is impossible to indicate types suitable for all cases. 
 
 As a rule the nature of the material to be handled and the required 
 location of the machinery govern the type of building used. 
 
 583 
 
Purlin 
 
 Monitor or l/enti lator^^. 
 
 Windows or Louvresy/r^. 
 
 Mf?C/OW5-~ 
 
 x-^v 
 
 |Z7/7-/O 
 
 fer-f 
 
 g ^ x 
 
 / , \ 
 
 vv^ 
 
 
 I 
 "^Column 
 
 1 
 
 I 
 "^-Column 
 
 
 
 584 
 
(9) 
 
 Clear 
 
 Lean-fo 
 
 Glass 
 
 
 (jlass 
 
 
 ^AlA 
 
 7_ 
 
 \ 
 
 7^^^ 
 
 ^ G/ass 
 
 
 
 -ffi3=F*q 
 
 
 7 
 
 
 
 \ 
 
 cr^r^rrr-rr, 
 
 
 
 ^j/\y\ 
 
 /\ 
 
 i 
 
 ^i't* 
 
 
 C- { 
 
 7\P\l7^^ 
 
 
 
 dCS> 
 
 
 
 
 
 
 cf&> 
 
 
 
 ' I 
 
 I 
 
 
 
 
 
 \ ' 
 
 
 Fig. 406 
 
 585 
 
586 
 
DESIGN OF BUILDINGS 
 
 587 
 
 The maximum length of span for the trusses as a rule is limited by 
 the length of the cranes which have a maximum length of about 80 ft. 
 more often 40 or 50 ft. is used. 
 
 The names of the different parts of an ordinary steel mill building 
 are given in Fig. 407. These same names, however, apply in general 
 to steel mill buildings. 
 
 The most common forms of roof trusses are shown in Fig. 408. The 
 steep pitch roofs are used when the roof covering is corrugated iron, 
 slate, or tile, and the flat roofs are used when the roof covering is felt 
 
 (d) Fink Truss 
 
 (b) Fan Truss. 
 
 ( CJ Warren Truss . 
 
 (d) Pratt Truss. 
 
 (e) Flat- Warren Truss. 
 
 (f) Flat Praff Truss. 
 
 Ho we Truss . 
 
 (h) Flat HoweTruss. 
 
 Fig. 408 
 
 covered with tar and gravel. Concrete roof covering may be used in 
 either case. 
 
 The details shown in Fig. 409 are those of a truss supported upon 
 columns. The purlins are channels which are supported on the roof 
 truss at panel points. The roof covering is corrugated iron connected to 
 the purlins by steel bands. 
 
 The details shown in Fig. 410 are those of a roof truss supported 
 upon walls. The roof covering is slate laid on 2" sheathing. The 
 purlins are channels which have 6" x 3" nailing strips bolted to them. 
 In this case the purlins are not at panel points and the top chord must 
 be designed heavy enough to resist, in addition to the direct stress, the 
 cross bending caused by the purlin loads. Wooden purlins as shown 
 at (a) are sometimes used, which is very good construction. 
 
 The details shown in Fig. 411 are those of a roof without purlins. 
 The sheathing consists of 2"x4" timbers laid on edge, acting as beams 
 extending from truss to truss. The bay length in the case of such con- 
 struction should be limited to about 14 or 15 ft. In case the roof load 
 
588 
 
 STRUCTURAL ENGINEERING 
 
 on the trusses is quite heavy and not supported at panel points, the top 
 chords of the trusses are constructed of two angles and a plate as shown 
 in Fig. 411. The plate is principally for transmitting shear. 
 
 Fig. 409 
 
 The details shown in Fig. 412 are those of an ordinary monitor 
 where louvre bars are used. The details shown in Fig. 413 are those of 
 an ordinary monitor where pivoted windows are used. 
 
 Pitch of a roof truss is the ratio of its height to its length. Thus: 
 A roof truss 60 ft. long and 20 ft. high would have J pitch, and if its 
 height were 30 ft. the pitch would be J. The terms "pitch" and "slope" 
 
Fig. 410 
 
590 
 
 {STRUCTURAL ENGINEERING 
 
 Fig. 412 
 
 are somewhat confusing. The pitch of an ordinary truss is obtained by 
 dividing its rise or height by its total length, while the slope is obtained 
 by dividing its height by one-half of its length. 
 
 Fig. 413 
 
 A lean-to truss, as shown at (d), Fig. 405, can be considered as a 
 half truss. 
 
 The pitch of a roof depends upon the roof covering. The following 
 is recommended: 
 
DESIGN OF BUILDINGS 591 
 
 Roof Covering Pitch 
 
 Corrugated Iron not less than J. 
 
 Slate not less than J, preferably J or J. 
 
 Tile not less than . 
 
 Felt, tar, and gravel. . .not more than ^5, preferably^ 
 
 259. Dead Load. Except for the roof trusses, the dead load for 
 mill buildings is readily estimated. In making these estimates the fol- 
 lowing units of weights may be used : 
 
 Timber 4.0 Ibs. per sq. ft. B. M. 
 
 Concrete 150.0 Ibs. per cu. ft. 
 
 Corrugated iron., #16 3. 5 Ibs. per sq. ft. 
 
 Corrugated iron,, #18 2.7 Ibs. per sq. ft. 
 
 Corrugated iron, #20 1.9 Ibs. per sq. ft. 
 
 Corrugated iron, #22 1.5 Ibs. per sq. ft. 
 
 Slate, J" thick, 3" double lap... 4.5 Ibs. per sq. ft. 
 
 Slate, fV thick, 3" double lap. . . 6.5 Ibs. per sq. ft. 
 
 Tile (plain) 18.0 Ibs. per sq. ft. 
 
 Tile (Spanish) 8.5 Ibs. per sq. ft. 
 
 Glass, y thick 2.0 Ibs. per sq. ft. 
 
 Gravel, felt, and tar roof covering. 5.0 Ibs. per sq. ft. 
 
 Hollow tile (gross) 40.0 Ibs. per cu. ft. 
 
 The approximate weight of roof trusses in pounds per sq. ft. of roof 
 surface is as follows: 
 
 Pitch- 
 Flat i i J i 
 
 20-ft. span 2.0 1.9 1.8 1.7 1.5 
 
 30-ft. span 2.6 2.5 2.3 2.1 1.8 
 
 40-ft. span 3.3 3.2 2.9 2.8 2.4 
 
 50-ft. span 3.8 3.6 3.4 3.2 2.7 
 
 60-ft. span 4.6 4.4 4.1 3.8 3.2 
 
 70-ft. span 5.2 4.9 4.6 4.3 3.7 
 
 80-ft. span 5.7 5.4 5.1 4.8 4.1 
 
 90-ft. span 6.4 6.1 5.7 5.3 4.6 
 
 100- ft. span 7.0 6.7 6.2 5.8 5.0 
 
 120-ft. span 8.0 7.6 7.1 6.7 5.7 
 
 260. Snow Load. The snow load on roofs varies with the locality, 
 slope of roof, and kind of roof covering. According to Trautwine, fresh 
 fallen snow weighs from 5 to 12 Ibs. per cu. ft., while moistened snow 
 compacted by rain will weigh from 15 to 50 Ibs. per cu. ft. 
 
 The following loads in pounds per sq. ft. of roof surface are con- 
 sidered to be sufficient allowance for snow: 
 
 Pitch 
 
 Locality Flat J J i * 
 
 Central States 30 30 25 20 15 10 
 
 Northwestern States 40 40 38 30 20 12 
 
 New England States 36 36 32 27 18 12 
 
 Rocky Mountain States 35 35 30 26 15 10 
 
592 STRUCTURAL ENGINEERING 
 
 These loads should be reduced 20 per cent in case the roof covering 
 be slate or corrugated iron. 
 
 No snow load need be considered south of latitude 35. However, 
 in some localities south of this latitude a load of 10 Ibs. per sq. ft. of 
 roof surface should be included in roof loads to allow for sleet. In fact, 
 load due to sleet, in all cases, should be taken as 10 Ibs. per sq. .ft. of roof 
 surface. 
 
 261. Wind Load. The wind pressure against vertical surfaces 
 varies with the velocity of the wind and to some extent with the area of 
 the exposed surface. The wind load (pressure) on vertical surfaces can 
 be safely considered as 30 Ibs. per sq. ft. except for steel towers, where 
 the exposed surfaces are small, in which case the load should be con- 
 sidered as 40 Ibs. per sq. ft. This 30 Ibs. pressure, in the case of mill 
 buildings, is considered to be applied to the vertical sides and ends of 
 the buildings. 
 
 In the case of wind load on sloping roofs the pressure is assumed to 
 act perpendicularly or normally to the roof surface. This pressure can be 
 determined only experimentally. Hutton derived, from his experiments 
 in 1788, the following formula: 
 
 r> T> s - \ 1 - 842 cosct ~ * 
 P n = P (sma) 
 
 for the normal pressure due to wind upon inclined surfaces, where P n 
 represents the normal pressure, P the corresponding horizontal pressure, 
 and a the slope of the surface. 
 
 Duchemin, in 1829, derived in the same manner the following 
 formula : 
 
 = p / 2 sina \ 
 \l + sin 2 a/ 
 
 for the normal pressure on inclined surfaces, where the letters signify the 
 same as in Hutton's formula. 
 
 These two formulas are in general use and the values for the normal 
 pressures when P equals 30 Ibs. are as follows: 
 
 Normal Pressure 
 Slope of Roof Hutton Duchemin 
 
 5 3.9 5.1 
 
 10 7.2 10.2 
 
 15 10.5 14.2 
 
 18-26' (i pitch) 13.0 17.4 
 
 21-48' ( J pitch) 15.0 19.8 
 
 26-34' (i pitch) 18.0 22.4 
 
 30 19.9 24.0 
 
 33-41' ( pitch) 22.0 25.5 
 
 40 25.1 26.7 
 
 45 (i pitch) 27.1 28.2 
 
 60 30.0 30.0 
 
 262. Cranes. In designing a mill building, it is absolutely neces- 
 sary that the designer have complete data as regards the type, weight, 
 
DESIGN OF BUILDINGS 
 
 593 
 
 and capacity of the cranes to be used, and also the clearance required 
 for their operation. 
 
 The data given in Fig. 414 will suffice in most cases for overhead 
 cranes. However, the general requirements in each case should be 
 
 Span 
 
 Q 
 
 Truck 
 
 Capacity 
 
 Span 
 
 ft 
 
 <q' 
 
 B 
 
 c 
 
 EdchlVhee 
 
 Total Net W 
 of Crane 
 
 5 Tons 
 
 4-0 Ft 
 
 A-'-ll' 
 
 5- 7" 
 
 0'-9" 
 
 e'-e" 
 
 IZ800 
 
 22400 
 
 5 
 
 (SO .- 
 
 5-3" 
 
 5' ll" 
 
 o^ 7 ^ 
 
 6'- S" 
 
 15500 
 
 31300 
 
 1 O 
 
 AO - 
 
 5-10 
 
 6- 6 
 
 o- /o' 
 
 9- 2" 
 
 19800 
 
 264QO 
 
 1 O 
 
 60 - 
 
 e'-o' 
 
 6'-S" 
 
 o'-/o" 
 
 9- 2" 
 
 22700 
 
 37600 
 
 1 5 - 
 
 -4O 
 
 6-0" 
 
 6-9' 
 
 o- 10' 
 
 9 - 2" 
 
 26500 
 
 33900 
 
 1 5 
 
 6O - 
 
 6-2" 
 
 6-n" 
 
 0- 10" 
 
 9-2" 
 
 29600 
 
 44000 
 
 20 
 
 ^4O 
 
 6-2' 
 
 7- 3" 
 
 o-n" 
 
 9-8" 
 
 32700 
 
 37600 
 
 2O 
 
 60 
 
 6'- 5" 
 
 7-6" 
 
 O- ll" 
 
 9-8" 
 
 37000 
 
 50700 
 
 25 
 
 ^.o 
 
 6-5" 
 
 7 1 7" 
 
 o-n" 
 
 9-6" 
 
 39300 
 
 4340O 
 
 25 
 
 6O - 
 
 6-9" 
 
 a'-o" 
 
 O'-M" 
 
 10-10" 
 
 44500 
 
 58700 
 
 25 - 
 
 SO 
 
 7-2" 
 
 S- 5" 
 
 O'-M" 
 
 13-4." 
 
 50800 
 
 7990O 
 
 3O .. 
 
 -40 
 
 e'-a" 
 
 S'- 0" 
 
 -0" 
 
 \O'-A" 
 
 46200 
 
 495OO 
 
 30 - 
 
 6O 
 
 v-o" 
 
 S L 6' 
 
 '-o" 
 
 II- 2' 
 
 51700 
 
 66600 
 
 3O 
 
 ao - 
 
 7 L 5" 
 
 8- /J" 
 
 '-O" 
 
 13-4' 
 
 56800 
 
 9O7OO 
 
 4-O 
 
 -40 - 
 
 r 1 o" 
 
 6 L 9" 
 
 -o" 
 
 ll'-4" 
 
 601 00 
 
 64200 
 
 4-O 
 
 60 - 
 
 -?-* 
 
 9 - l" 
 
 '-o" 
 
 n'-io" 
 
 67000 
 
 64300 
 
 4-O - 
 
 SO 
 
 7-9" 
 
 9-6' 
 
 -o" 
 
 13-4" 
 
 7560O 
 
 II 29OO 
 
 50 
 
 ^.0 
 
 7-7" 
 
 9'- 5" 
 
 -o" 
 
 II '-4" 
 
 74000 
 
 76100 
 
 5O 
 
 <ZG 
 
 7-9" 
 
 9-7" 
 
 -o" 
 
 II L 0" J : /0' 
 
 ^IZOO 
 
 9850O 
 
 50 
 
 &0 - 
 
 a-^" 
 
 .0'-3" 
 
 '-o" 
 
 O 1 ?' 5-/' 
 
 >*590O 
 
 131 200 
 
 Fig. 414 
 
 carefully studied; and often it is advisable to consult the company that 
 manufactures the crane to be used. 
 
 Complete Design of a Roof Supported upon Brick Walls 
 
 263. Data. 
 
 Size of building = 80 ft. long by 40 ft. wide. 
 Length of roof trusses = 40'-0" c.c. end bearings. 
 Height of roof trusses = lO'-O" (J pitch). 
 Length of bay = 20'-0". 
 
594 
 
 STEUOTUEAL ENGINEEEING 
 
 Roof covering, slate laid on 2" sheathing. 
 Purlins, steel channels. 
 
 264. Design of Purlins. As 2" sheathing is specified, the first 
 thing to do is to determine the distance that the purlins can be apart, so 
 that the sheathing will not be overstressed or the deflection be too great. 
 
 The load on the sheathing includes snow, wind, and the weight of 
 the slate and sheathing. 
 
 The maximum snow and wind load is not likely to occur at the same 
 time as the snow would be blown off. Let us assume that the wind pres- 
 sure is 18 Ibs. (Hutton see Art. 261) and the snow 10 Ibs. per sq. ft. 
 of roof surface. The 10 Ibs. snow load provides for sleet, ice, or frozen 
 snow. Using the weights given in Art. 259, we have the following dead 
 load per sq. ft. of roof: 
 
 iV' slate = 6.5 Ibs. per sq. ft. of roof surface 
 
 2" sheathing = 8.0 Ibs. per sq. ft. of roof surface 
 
 14.5 Ibs. per sq. ft. of roof surface 
 
 Adding this to the snow load we obtain 24.5 Ibs. This load acts 
 vertically while the wind load acts normally to the roof. 
 
 Let <f> be the slope of the roof. Then for the normal pressure due 
 to dead and snow load we have 
 
 24.5 x cos0 = 24.5 x 0.895 = 21.9, say 22 Ibs. 
 Now adding this to the wind load we have 
 
 18 + 22 = 40 Ibs. 
 
 for the total maximum normal 
 pressure on the roof per sq. ft. 
 of roof surface. Let x be the 
 distance between purlins, as in- 
 dicated in Fig. 415, and consider- 
 ing the sheathing to be a simple 
 beam 1 ft. wide, we have 
 
 \ 
 
 M = i f x 40 x x 2 x 12 = 60# 2 inch Ibs. 
 
 for the maximum bending mo- 
 ment. Taking 1,000 Ibs. as the 
 allowable unit stress on the 
 sheathing and applying Formula 
 (C) of Art. 53, we have 
 
 Fig. 418 
 
 from which we obtain 
 
 = 1,000 x 8, 
 
 *=11.5 ft. 
 
 for the distance that the purlins could be apart and the sheathing would 
 not be overstressed from the 40 Ibs. load. The sheathing should be 
 capable of supporting a 200-lb. man in addition to the 12.5 Ibs. dead 
 load. Considering the man to be midway between purlins (center of 
 
DESIGN OF BUILDINGS 595 
 
 span) and resolving the loads normally to the roof, we have 
 
 M' = J (14.5 x 0.895 x x 2 . x 12) + x 0.895 x - x 12 = 19.4o? 2 + 537,r 
 
 for the bending moment. Then we. have 
 
 19.4* 2 + 537^ = 1,000x8, 
 from which we obtain 
 
 * = 10.8 ft. 
 
 for the distance between the purlins which is only 0.7 ft. less than found 
 above for the 40 Ibs. load. But we know from experience that such 
 spacing would not. be practical as the deflection of the sheathing would 
 be so great that the slate would be broken. From this it is seen that 
 the spacing of the purlins depends entirely upon the maximum allowable 
 deflection of the sheathing. This deflection should not exceed ^J-^- of 
 the distance between th'e purlins. For the deflection due to the 200-lb. 
 man at mid span (see Art. 65) we have the following expression: 
 
 Placing this equal to ,r/800 and taking E as 1,000,000 and JP as 200, we 
 have 
 
 200 x 0.895 x ,r 3 _ x 
 
 48 x l,000,00(Xx 8 ~ 800' 
 from which we obtain 
 
 * = 51.8" or about 4'-4". 
 
 If the deflection due to the dead load of 14.5 Ibs. per sq. ft. of roof 
 surface be taken into account the value of * will be a little less than 4'-4", 
 so the correct distance will 
 be about 4'-0". The actual 
 spacing of the purlins de- 
 pends to some extent upon ^*^*\/ ' ^ 
 the design of the roof truss. ^sj\ si^ / i 
 In this case a Fink truss <^/\^ / 
 will be used and the most 
 suitable spacing obtainable 
 of the purlins is shown in 
 Fig. 416. This gives a dis- 
 tance of about 3 '-8" be- 
 tween purlins. Fig. 410 
 
 The dead and snow 
 
 load as given above is 24.5 Ibs. per sq. ft. of roof surface. Adding to 
 this the wind load of 18 Ibs., which we will consider to act the same 
 as the dead load on the purlins, we have 
 
 24.5 + 18 = 42.5, say 43 Ibs. 
 
 for the total load per sq. ft. of roof surface. 
 
 Now, as the purlins are 3'-8" apart, we have 
 
 43x3.66 = 157.38. say 158 Ibs, 
 
596 
 
 STRUCTURAL ENGINEERING 
 
 for the roof load per ft. of purlin. Assuming the purlin to weigh 12 Ibs. 
 per ft. we have 
 
 158 + 12 = 170 Ibs. 
 
 for the total load per ft. of purlin, 
 moment on the purlin we have 
 
 Then for the maximum bending 
 
 M' = Jx 170x20 x 12 = 102,000 inch Ibs. 
 
 Dividing this by 16,000 we obtain a section modulus of 6.3 which 
 calls for a 1" x 12J# [ or an 8" x 11J* [. The 8" x 11 J# [ will be used 
 for purlins throughout. 
 
 Theoretically the slope of a purlin should be taken into account 
 in determining the maximum stress on it, but as the above result would 
 not be changed by so doing (especially if the stiffening effect of the 
 nailing strips and sheathing be taken into account) we are really justified 
 in ignoring the slope, as was done in the above calculations. 
 
 265. Determination of Stresses in the Trusses Due to Dead 
 and Snow Load. There are really two loads to consider: The dead 
 and snow loads that can be combined with the maximum wind load, and 
 the dead and maximum snow load that could be combined with one-third 
 of the maximum wind load. The first is given above as 24.5 Ibs. per sq. 
 ft. of roof surface where the snow (sleet) is taken as 10 Ibs. Combining 
 this with the maximum wind load (Hutton) we obtained the, 43 Ibs. given 
 above. For the other load, taking the maximum snow for the Central 
 States, we have 
 
 Snow 
 20 
 
 Dead Load 
 14.5 + 
 
 Wind 
 6 =40.5 Ibs. 
 
 which is less than the combination first considered. So we will use the 
 24.5 Ibs. dead and snow load. The end reaction on each purlin, except 
 the eave and ridge purlins, including the weight of truss and purlins, is 
 [(24.5 + 2.9) 3.66+11.25] 10 = i;il6 Ibs. about, and hence the concentra- 
 tion on the truss at each of these purlins is 1,116x2 = 2,232 Ibs. Then 
 the load on the truss will be as shown in Fig. 417. 
 
 4-0-0' 
 
 Fig. 417 
 
 The concentration on the panel points, considering the top chord in 
 each panel as a simple beam, is shown at (a), Fig. 418. The diagram of 
 the stresses is shown at (b). This diagram is constructed by beginning 
 at joint A and passing around each. joint clock-wise. Thus, laying off 1-2 
 
DESIGN OF BUILDINGS 
 
 597 
 
 equal to the reaction and 2-3 equal to the 1,480 Ibs. load and closing on 
 1 and 3 we obtain 1-2-3-4-1 for the stress diagram for joint A. Starting 
 at 4 and passing around joint a clock- wise, we obtain 4-3-5-6-4 for the 
 
 -/8500 * V* -16500 * \yf -ft 000 * 
 
 (b) 
 
 Fig. 418 
 
 stress diagram for that joint. Starting at 1 and passing around joint B 
 clock- wise we obtain 1-4-6-7-1 for the stress diagram for that joint. 
 The stress diagrams for joints b and C can not be drawn as the structure 
 stands as there are three unknown forces at each joint. We get around 
 this difficulty by inserting the temporary member Cc and omitting mem- 
 bers be and ce. Having made this modification we obtain the stress 
 diagram 7-6-5-9-8-7 for joint b. Beginning at 1 and passing around 
 joint C clock- wise we obtain 1-7-8-12-*-! for the stress diagram for that 
 joint. The only correct stress obtained by drawing the last two diagrams 
 is the stress in CD which is represented by the line *-l. This stress, 
 as is readily seen, is not affected by the changing of the web members. 
 Having determined the stress in CD we can consider members be and ce 
 in place and Cc omitted, and proceed with the analysis. Considering 
 joint C and passing around clock- wise we obtain s-l-7-13-s for the stress 
 diagram for that joint. Having the stress in bC determined, which is 
 represented by the line 7-13, we can draw the stress diagram for joint b 
 which is 13-7-6-5-9-10-13. 
 
 The remainder of the diagram at (b) is readily constructed. This 
 diagram represents the stresses in the left half of the truss. The stresses 
 in the right half are the same, for corresponding members, and hence 
 the diagram at (6) is sufficient for determining the stress in all the truss 
 members. By scaling the diagram at (6) the desired stresses in the 
 truss members are obtained. These are shown on the members at (a). 
 
598 
 
 STRUCTURAL ENGINEERING 
 
 266. Determination of the Stresses in the Trusses Due to Wind 
 Load. The wind load (see table of Art. 2G1) is 18 Ibs. (Hutton) per 
 sq. ft. of roof surface. The combined dead and snow load considered 
 in the last article is 30.5 Ibs. per sq. ft. of roof surface. So by multiply- 
 ing the loads shown in Fig. 418 by ^ 8 5 we obtain the wind loads 
 shown at (a), Fig. 419, which, of course, act normally to the roof and 
 upon one side only. 
 
 The two ends of the trusses are considered to be equally fixed. Then 
 the reactions will be parallel to the loads. Laying off the load line MN, 
 
 (b) 
 
 Fig. 419 
 
 shown at (b), and drawing the ray diagram MON, and constructing the 
 corresponding equilibrium polygon ab' . . .0, and drawing the line OE 
 parallel to the closing line ag, we obtain ME,' which represents the reaction 
 R at A, and EN, which represents the reaction R\ at F. 
 
 Then beginning at joint A and passing around each joint clock-wise, 
 as explained in the last article, the stress* diagram shown at (6) is readily 
 drawn. By scaling this diagram the desired stresses, which are given at 
 (a), are obtained. 
 
 267. Designing of the Truss Members. Combining the stresses 
 shown in Figs. 418 and 419 we obtain the stresses shown in Fig. 420, 
 which are maximum stresses, and which the truss members must be 
 designed to transmit. 
 
 The members will be designed so that }" rivets can be used through- 
 out, which requires that the flanges of the angles through which rivets 
 pass be not less than 2". 
 
DESIGN OF BUILDINGS 
 
 599 
 
 Bottom Chord AB. This is a tension member and the required net 
 area of cross-section is equal to 28,340 -f- 16,000 = 1.77 n ". Two angles 
 will make the most satisfactory section for this member. The least size 
 permissible is 2|" x 2|", as }" rivets are to be used, and besides they 
 
 Fig. 420 
 
 should be at least this size to secure the necessary rigidity. So we will 
 use 2 Ls 2J" x 2" xj" = 2.38- 0.42 = 1.96 n ", which has 0.19 n " more 
 section than required, but the thickness is a minimum and hence this 
 section is as near the required as is possible to obtain. This same section 
 will be used for the bottom chord throughout, as it is the minimum. 
 
 Members cJc and cB. These are tension members and the required 
 section is only about 0.25 n ", so we will use 1 L 2J" x 2" xj" = 1.06- 
 0.21 = 0.85", which is the smallest angle permissible if f" rivets are 
 used. 
 
 Members ek and kC. These are tension members and the required 
 section for ek is 0.9 n ", and for JcC it is 0.61 n ". We will use 2 Ls 2" x 
 2" x^ r/ = 1.7 n// net, which are the smallest angles permissible. 
 
 Member cC. This is a compression member which has a length of 
 67". The value of L/r should not exceed 120. Then for the allow- 
 able r we have 67/120 = 0.55. Let us try 2 Ls 2"x 2"xi" = 2.12 n ". 
 If the 2" flanges are turned out the least radius of gyration will be 0.79. 
 Then we have 
 
 16,000 - 70 x -^- = 10,060 Ibs. 
 \j i y 
 
 for the allowable unit stress. Dividing this into the stress we obtain 
 8,750 -r 10,060 = 0.87 sq. ins. 
 
 for the required area of cross-section, which is about one-third of the 
 area contained in the above angles, but these angles will be used as they 
 are as small as we can use. The L/r = 67/0.79 = 0.85, which is compara- 
 tively low. 
 
600 STRUCTURAL ENGINEERING 
 
 Members dk and bB. We will make each of these members of 2 Ls 
 2J" x 2" x J", which are minimum size angles, the same as used for cC. 
 The actual requirement, however, can be ascertained in the same manner 
 as shown above for cC but, as is obvious, this work is not necessary. 
 
 Top Chord. This is a compression member. Considered in the 
 vertical direction the length would be a panel length, and considered in 
 the transverse direction the length should be taken as two panel lengths, 
 owing to there being some question as to the actual support given to the 
 chord by the purlins. 
 
 The length, considering the transverse direction, is from A to c, 
 which is about ll'-2" or 134". Owing to the purlin concentration being 
 applied between panel points there will be cross bending on the chord 
 which it must resist in addition to the direct stress. Let us assume 
 2 Ls 5"x3J"x-&" = 7.06 n ". The 5" legs will be placed vertically to 
 resist cross bending. Assuming that the angles are f" apart, (&-&), the 
 least radius of gyration, is 1.50. Then we have 
 
 16,000 - 70 x ~r = 9,750 Ibs. 
 1.50 
 
 for the allowable unit stress in compression. The concentrations on the 
 top chord due to dead and snow load are given in Fig. 417. The chord 
 
 in each panel can be considered as a fixed 
 beam. The maximum moment will occur 
 at supports b and d. The loads in panel 
 Ab (the same as in the other panels) are 
 shown in Fig. 421. This load (at g) is 
 obtained by resolving the 1,800 Ibs. purlin 
 concentration due to dead and snow load 
 (see Fig. 417) normally to the roof and 
 adding the concentration due to the 18 Ibs. 
 wind load, which is equal to 18x3.66x 
 Fig. 421 20 = 1,317 Ibs. Hence, for the concentra- 
 
 tion at g we have 
 
 1,800x0.895 + 18x3.66x20 = 2,928, say 2,900 Ibs. 
 Applying (7) of Art. 69, we have 
 
 M' = 2,900x67 (2P-P-fc) ............................. (1) 
 
 for the maximum moment which occurs at 6. 
 
 Substituting this value of k in (1), we obtain 
 M' = 2,900 x 67 (2 x 0.102 - 0.033 - 0.32) = 28,215, say 28,000 inch Ibs. 
 
 for the maximum bending moment on the top chord. 
 
 The moment of inertia of the 2 Ls 5" x 3" x T y about the gravity 
 axis perpendicular to the long legs is 8.9 x 2 = 17.8, and the distance from 
 the top of the chord down to this same axis is 1.63. Then applying 
 
601 
 
602 
 
 STRUCTURAL ENGINEERING 
 
 C of Art 53, we have 
 
 M' = 31,000 = 
 
 /x20 
 
 from which we obtain 
 
 / = 5,mibs. 
 
 for the maximum compressive unit stress due to cross bending For the 
 maximum unit stress due to direct compression, we have 
 
 35,000 -r8 n " = 4,375 Ibs. 
 
 Now adding this to the stress due to cross bending, found above, we have 
 
 1,375 = Ibs. 9,550 (about). 
 
 for the actual maximum unit stress on the top chord. The allowable, 
 as found above, is 9,750 Ibs. So the 2Ls 5" x 3J" x J" = 8 n ", assumed 
 above, are about the correct size, and hence will be used. 
 
 This completes the necessary calculations and next the details can 
 be drawn. Complete shop drawings for the building are shown in Fig. 
 422. The details shown here are considered to be self-explanatory. 
 However, the student should verify the riveting. 
 
 Fig. 423 
 
 268. Determination of Stresses in Trusses Supported upon 
 Columns. Stresses due to dead and snow load are determined just the 
 
DESIGN OF BUILDINGS 
 
 603 
 
 same as if the trusses were supported upon walls, and are fully treated 
 in Art. 265. The knee braces are ignored in the analysis. 
 
 Stresses Due to Wind Load. The columns are fixed to the masonry 
 by anchor bolts, which should extend well into the masonry, and by the 
 
 Fig. 424 
 
 addition of knee braces the columns are quite firmly connected to the 
 trusses, so that the columns can really be considered as fixed, in which 
 case the wind pressure will cause them to bend as indicated in Fig. 423, 
 where and 0' are the points of contra-flexure. 
 
 In determining the stresses in the truss due to wind pressure, we 
 
604 STRUCTURAL ENGINEERING 
 
 first locate the points of contra-flexure in the columns by applying (10) 
 of Art. 257. After the points of contra-flexure are located, we next 
 consider the ground moved up to that level as indicated at (#) in Fig. 424. 
 
 We next compute the value of the wind loads, PI, P2, etc. In com- 
 puting the horizontal loads (PI and P2) we take the wind pressure at 
 30 Ibs. per sq. ft. of vertical surface, and in computing the value of the 
 loads (P3 . . . P7) normal to the roof we use the pressure per sq. ft. 
 of roof surface given in Art. 261, the actual intensity of the pressure 
 used, of course, being in accord with the slope of the roof. 
 
 After the intensities of the loads (PI . . . P7) are computed, we 
 next determine the horizontal and vertical components of the reactions 
 on the columns at O and 0'. If the columns are of equal moment of 
 inertia we assume that the horizontal components are equal and that each 
 is equal to one-half of the horizontal component of all of the loads. 
 These horizontal components can be graphically determined by laying 
 off the loads as shown at (fo), drawing the vertical AP, and the horizontal 
 PB. Then the horizontal component in each case is represented by one- 
 half of the line PB. Let H represent this component. To determine 
 the vertical components of the reactions at O and 0' we complete the ray 
 diagram at (fr), then construct the corresponding equilibrium polygon 
 z-n-o . . . y as shown at (a). Then by prolonging the segments zn and 
 uy to intersect at I and from I drawing a line parallel to the line AB, 
 which represents the resultant F of all of the loads, we obtain the line 
 of action of the resultant, and prolonging this resultant to intersection at 
 N with line 00', and then laying off O'M equal to the vertical component 
 (=AP) of all of the loads, and drawing the line MO and the ordinates 
 NT and TU, we have the vertical component F2 of the reaction at O' 
 represented by the line NT and the vertical component V\ of the reaction 
 at by the line TU. 
 
 The line MO is really an influence line for the vertical component 
 of reaction at O' '. To show that this is true, let us assume that the re- 
 sultant F passes through point 0' . Then, as is readily seen, the total 
 vertical component of all of the loads would be taken by column kO' ', in 
 which case the vertical component on column Oc would be zero, and if 
 the resultant F intersected the line OO' midway between the two columns 
 the vertical components of the reactions at and O f would be equal and 
 each equal to one-half of the ordinate O'M. So, evidently, the line MO 
 is the influence line for the vertical component of the reaction at 0' '. 
 That is, if the resultant F intersects the line 00' at any point C between 
 and O' the vertical component of the reaction at O' and are repre- 
 sented by the ordinate CD and DG, respectively. 
 
 If the resultant F intersects the line 00' at any point between O 
 and 0' the vertical components of the reactions will be positive; but if it 
 intersects at any point E to the right of 0' the vertical component at 
 would be negative, and the vertical component at 0' positive. In that 
 case the ordinate EK (MO prolonged to K) would represent the positive 
 vertical component of the reaction at 0', while MS would represent the 
 negative vertical component of the reaction at 0. This last construction 
 is readily proven: Taking moments about we have 
 
DESIGN OF BUILDINGS 605 
 
 from which we obtain 
 
 But from similar triangles we have 
 
 OE__EK_ KK 
 
 00' ~ O'M ~" V ' 
 
 from which we obtain 
 
 EK=(FxOE) + 00', 
 and therefore 
 
 EK=F2. 
 
 After the horizontal and vertical components of the reactions at 
 and O f are thus determined, the actual resultant reaction at each of the 
 points is readily determined, but it is really just as convenient to use the 
 components in determining the stresses in the truss. 
 
 Having determined the horizontal and vertical components of the 
 reactions at and 0' in the above manner, we can readily determine 
 graphically the stresses in the truss due to the wind loads by first drawing 
 the auxiliary frames Oac and O'wk and beginning at either or 0' and 
 passing around each joint in the same direction. The stress diagram at 
 (c) is obtained by starting at and passing around each joint clock- 
 wise. The stresses obtained in the auxiliary trusses, including the 
 stresses in the columns, are to be ignored, but the stresses in the knee 
 braces and truss proper are not affected by adding the auxiliary trusses 
 and hence they are correct stresses. 
 
 Buildings with Lean-tos. In case the lean-tos are of simple con- 
 struction, as indicated at (/), Fig. 425, each lean-to column should be 
 considered as acting only as a simple beam in resisting the wind pressure, 
 in which case the wind pressure on the side of a lean-to would be trans- 
 mitted equally to the top and bottom of the columns. Then, considering 
 the case shown in Fig. 425, the pressure on the side AB would be trans- 
 mitted equally to points A and B, producing the loads PI and P2 (each = 
 3,000 Ibs.)- The horizontal pressure of the loads from B to C would be 
 transmitted directly to the column at C. The horizontal pressure repre- 
 sented as F (=8,600 Ibs.) is obtained by constructing the force diagram 
 shown at (6). Part of the force F is transmitted to point D and part 
 to E. The results obtained will be accurate enough if we consider the 
 part DE of the main column as a simple beam. Then taking moments 
 about E we have 
 
 for the part of F transmitted to point D. Then adding this force / to 
 the 3,200 Ibs. at D, we obtain the 9,650 Ibs. force shown at (d). The 
 part of F transmitted to D can also be determined by constructing the 
 influence line shown at (c). 
 
 After the force or load at D (all other loads are assumed as known) 
 is determined, the ray diagram shown at (e) can be drawn and the cor- 
 responding equilibrium polygon a-b-c-d-e-f-g-h ... w at (d) constructed. 
 Then prolonging Im and ab to intersection at I and drawing from I a 
 
Fig. 425 
 
 606 
 
DESIGN OF BUILDINGS 
 
 607 
 
 line parallel to the resultant GK we obtain 10 " for the line of action of 
 the resultant of all the wind loads affecting the roof. 
 
 By laying off O'P (=F = 8,400) equal to the vertical component of 
 all of these loads, and drawing ON and NO", we have the vertical com- 
 ponent F'2 of the reaction on the column at 0' represented by the dis- 
 tance NO" , and the vertical component V\ of the reaction on the column 
 at represented by the distance MP. F2 is positive and V\ negative. 
 Each of the horizontal components H on each column is equal to one-half 
 of the distance SK. 
 
 Now, having determined all of the loads and the horizontal and 
 vertical components of the reactions at O and 0' we can determine the 
 stresses in the roof truss in exactly the same manner as shown above for 
 the building without lean-tos. 
 
 In case the lean-tos are of the construction shown in Fig. 426, the 
 wind pressure will cause the columns to bend as indicated. If the shear, 
 indicated as HI, H2, etc., and the direct stress on the columns at the 
 
 Fig. 426 
 
 points of contra-flexure that is, the horizontal and vertical components 
 of the reactions at these points were determined, there would be no 
 difficulty in determining the stresses in the main and lean-to trusses. 
 The actual values of the horizontal and vertical components of the reac- 
 tions at the points of contra-flexure depend not only upon the relative 
 stiffness of the columns but upon the stiffness of the trusses as well. But 
 as this relative stiffness cannot be determined at all until the structure 
 is fully designed, and then only to a limited extent, it is evident that the 
 method used for determining the wind stresses must necessarily be based 
 to some extent upon practical assumptions. 
 
 By considering the main truss and the parts of the columns above 
 the points of contra-flexure, and O', as an independent structure, shown 
 at (a), Fig. 427, we can determine the horizontal and vertical components 
 of the reactions at and O' by drawing the ray diagram at (b) and the 
 corresponding equilibrium polygon at (a) and the influence line OJ, all 
 of which has been previously explained. 
 
 Considering the lean-to and the part of the main column below the 
 
608 
 
 STRUCTURAL ENGINEERING 
 
 point of contra-flexure O as an independent structure,, shown at (c), we 
 know H' and V which are applied at O, as they are equal and opposite, 
 respectively, to the //' and V found at (a), and we also know the wind 
 loads, but the components of the reactions at the points of contra-flexure 
 
 Fig. 427 
 
 S and ' and likewise the force H" are unknown. The force H" is the 
 reaction from the part of the structure above due to the wind loads on 
 the lean-to, and hence is equal to the part of these loads that is trans- 
 mitted through the main truss to the other side of the building. The 
 actual value of H" depends upon the rigidity of the columns and trusses. 
 This rigidity can be ascertained only by tedious analysis of the structure 
 after it is designed. However, a sufficiently accurate value of H" can 
 be obtained from the following empirical formula: 
 
 (i) 
 
 where 77 represents the horizontal component of the wind loads on the 
 lean-to and L the length of the main roof truss and e the distance from 
 
DESIGN OF BUILDINGS 
 
 609 
 
 the ground to the top of the lean-to and b the distance from the top of 
 the lean-to to the knee brace (see Fig. 426). 
 
 The values of e, b, and L are known (always) and H can be deter- 
 mined by constructing the force polygon CDE shown at (e). By sub- 
 stituting in (1) H" can be determined and then constructing the ray 
 diagram at (e), and drawing the corresponding equilibrium polygon at 
 (c) and the influence line ST, we obtain the vertical components V\ and 
 V% of the reactions at S and S', and then the only unknown forces re- 
 maining are the horizontal components of these reactions which are rep- 
 resented at H3 and H4. These forces (H3 and H4) would be consid- 
 ered to be equal if the two columns were of equal moment of inertia, but 
 as they never are we must assume relative values. We will assume the 
 lean-to column to be one-third as rigid as the main column. Then as 
 , we have 
 
 where H represents the total horizontal component of all the forces on 
 the structure shown at (c), including those at 0. 
 
 Considering the lean-to and part of the main column shown at (d) 
 as an independent structure, the only applied forces are H" , V" , and H' 
 at 0' '. These forces having been previously determined, we can deter- 
 mine the vertical component of the reactions at the points of contra- 
 
 V4 
 
 Fig. 428 
 
 flexure, S" and S" ', by constructing the force polygon at (/) and draw- 
 ing from 0' the line O'Y parallel to the resultant LM and then the 
 influence line S"Z. 
 
 Assuming the rigidity of the lean-to column to be one-third of that 
 of the main column, we have 
 
 H6 = -H f andH5 = ^H f . 
 
 4 4 
 
 We now have all of the components of the reactions at the points 
 
610 STRUCTURAL ENGINEERING 
 
 of contra-flexure determined, and we can now proceed with the determi- 
 nation of the stresses in the structure. 
 
 By considering the part of the structure above the points of contra- 
 flexure and 0' and adding the auxiliary trusses to the sides as indicated 
 at (tz), Fig. 428, the stresses in knee braces and main truss can be graph- 
 ically determined very readily by beginning at either or O f and passing 
 in the same direction around each joint. 
 
 By considering the part of the structure at (c), Fig. 428, as an 
 independent structure and adding the auxiliary frames, shown dotted, 
 the stresses in the knee brace and lean-to truss can be graphically deter- 
 mined very readily by beginning at S and passing in the same direction 
 around each joint. Likewise the stresses in the knee brace and the 
 lean-to truss shown at (d), Fig. 428, can be determined. 
 
 269. Determination of the Stresses in the Columns. The col- 
 umns are subjected to direct stress and cross bending similar to the case 
 of the end posts in bridges. (See Art. 180.) 
 
 After the horizontal components of the reactions at the points of 
 contra-flexure are determined, the bending moment at any point in a 
 column is readily computed. For example, the bending moment at any 
 point in column DE (Fig. 425) is equal to Hx, being a maximum at 
 D and E. 
 
 Complete Design of a Mill Building 
 
 270. Data. 
 
 Nature of building, machine shop. 
 
 Length of building=ll bays @ 20'-0" = 220'-0". 
 
 Width of main shop = 60'-0" c.c. of main columns. 
 
 Width of lean-tos = 24'-0" (lean-to on each side of building). 
 
 Roof covering, No. 20 corrugated iron. 
 
 Allowable intensities on metal 
 
 Tension 16,000 Ibs. per sq. in. 
 
 Compression. .16,000-70 L/r where L is the length of 
 member in ins. and r the least radius of gyration of 
 the cross-section in ins. 
 Crane 20-ton overhead in main shop. 
 
 271. Preliminary. In starting the designing the very first thing 
 to do is to draw a sketch of the cross-section of the building as shown in 
 Fig. 429, showing the general requirements as to type and pitch of roof, 
 spacing of purlins, height of lean-to, clearance for crane, height of clear 
 story, etc. The roof covering is to be of corrugated iron. The pitch of 
 the main roof will be ^ and the lean-tos about (see Art. 258). The 
 maximum spacing of the purlins is governed by the kind of roof covering. 
 In this case No. 20 corrugated iron will be used and, as stated in the 
 Carnegie handbook, the maximum span must not be more than 6 ft., and 
 less is preferable. In this case the purlins will be spaced about 4'-6" 
 centers. The distance from the knee brace to the crane girder, as specified 
 in Fig. 414, is 6'-2". 
 
 272. Designing of the Purlins. The weight of the corrugated 
 iron, as per Art. 259, is 1.9 Ibs. per sq. ft. of roof surface. The weight 
 
DESIGN OF BUILDINGS 
 
 611 
 
 of the purlin will be assumed to be 12 Ibs. per ft. of length. The snow 
 load (frozen snow and ice) will be taken as 10 Ibs. per sq. ft. of roof 
 surface and the wind load as 22 Ibs. per sq. ft. of roof surface as per 
 
 Fig. 429 
 
 Art. 261 for roof having ^ pitch. Then for the total load on each inter- 
 mediate purlin per ft. of length we have the following: 
 
 Dead load = 1.9x4.5 + 12= 20.5\ r ~~ 
 Snow load = 10 x4.5 = 45.0 J D ' D 
 Wind load = 22 x4.5 = 99.0 
 
 164.5, say 165 Ibs. 
 
 Then for the maximum moment on the purlin we have 
 
 M = J x 165 x 20 2 x 12 = 99,000 inch Ibs. 
 
 Dividing this moment by 16,000 we obtain 99,000 + 16,000 = 6.2 for the 
 section modulus, which calls for a 7"xl2.25# channel or an 8"xll.25# 
 channel. Use 1 [ 8" x 11.25* for each purlin. 
 
 273. Designing of the Lean-to Rafters. The wind load acts 
 perpendicularly to the rafter while the dead and snow load acts vertically. 
 So, for the concentration on the rafter at each intermediate purlin, using 
 the same unit loads as used in the last article, except the wind load which 
 is taken as 18 Ibs. (the lean-to roof has i pitch), we have: 
 
 Dead load = 20.5 x 20 x 0.908= 372 Ibs. 
 Snow load = 45.0x20x0.908= 817 Ibs. 
 Wind load = (18 x 4.5) x 20 =1,620 Ibs. 
 
 2,809 Ibs., say 2,800 Ibs. 
 
 Then applying this concentration at each purlin and considering 
 the rafter as a simple beam we have the case fully represented in Fig. 
 430. The maximum moment on the rafter due to these loads, as is 
 obvious, will occur at C, the center of span. So taking moments about C 
 we have 
 
612 
 
 STRUCTURAL ENGINEERING 
 
 M = [7,000 x 13.2 - (2,800 x 8.8) - 2,800 x 4.4] 12 = 665,280 inch Ibs. 
 
 for the maximum moment on the rafter due to concentrated loads. As- 
 suming the rafter to weigh 42 Ibs. per ft. of length we have 
 
 M' = Jx42x26.4 x 12 = 43,900 inch Ibs. 
 
 i 
 o 
 
 | 
 
 fc 
 
 * 
 
 c ^J 3| 
 
 
 <3.d 
 
 44' 
 
 i 
 
 ? ' 
 
 13.2' 
 
 264' 
 
 Adding 
 
 Fig. 430 
 
 for maximum moment on the rafter due to its own weight, 
 together the above moments we have 
 
 665,280 + 43,900-709,180 inch Ibs. 
 
 for the total maximum moment on the lean-to rafter. Dividing this by 
 16,000 we obtain 44.3 for the section modulus which calls for a 12"x40# 
 or 15"x42# I. We will use the 15"x42# I for each lean-to rafter. 
 
 274. Determination of Stresses in Trusses Due to Snow and 
 Dead Load. The snow and dead load per ft. of purlin as given in Art. 
 
 272 is 65.5 Ibs. Then for the 
 concentration on the truss at 
 each intermediate purlin we 
 have 65.5x20 = 1,310 Ibs. 
 Then, as two of these con- 
 centrations are transmitted 
 to each panel point of the top 
 chord of the truss (see Fig. 
 429), we obtain 1,310x2 = 
 2,620 Ibs. for the part of the 
 panel load due to snow, roof 
 covering, and purlins. The 
 weight of the roof truss as 
 per Art. 259 is 3.8 Ibs. per 
 sq. ft. of roof surface. Then, 
 for the part of the panel load 
 due to the weight of the roof 
 truss we have 3.8x9x20 = 
 684 Ibs. Now, adding this 
 to the part due to the roof 
 load we have 2,620 + 684 = 
 3,304, say 3,300 Ibs. (1,800 
 Ibs. of this is due to snow) 
 for the panel load due to 
 snow and dead load, and hence the snow and dead load on the truss will 
 be as shown at (a), Fig. 431. By beginning at A and constructing the 
 
 Fig. 431 
 
DESIGN OF BUILDINGS 
 
 613 
 
 stress diagram shown at (6) the stresses shown on the truss are obtained. 
 Multiplying each of these stresses by 1,500/3,300 the stresses marked D 
 which are due to dead load alone are obtained. 
 
 275. Determination of Stresses in Trusses Due to Wind Load. 
 The wind load per sq. ft. of roof surface is 22 Ibs. (Hutton), as per 
 Art. 261. Then for the panel load on the roof we have 22 x 9 x 20 = 3,960 
 Ibs. and hence the wind loads on the truss will be as shown in Fig. 432. 
 
 Fig. 432 
 
 The horizontal load at a is equal to 30 x 5 x 20 = 3,000 Ibs. The load at b 
 will be the same as at a plus the horizontal component of the wind load 
 from the lean-to. The horizontal load at the top and bottom of the 
 lean-to column is equal to 30x7x20 = 4,200 Ibs., as shown at (6). The 
 pitch of the lean-to roof is about J (i|), so the wind pressure per 
 sq. ft. of roof is 18 Ibs., as per Art. 261. Then the concentration on the 
 rafter at each intermediate purlin is equal to 18x4.5x20 = 1,620 Ibs. 
 and hence the wind loads on the lean-to roof will be as shown at (6). 
 
614 STRUCTURAL ENGINEERING 
 
 Then by constructing the force polygon at (c) we obtain 
 
 F = 8,250 Ibs. 
 
 for the horizontal component of the forces on the lean-to. Then adding 
 this to the 30x5x20 = 3,000 Ibs. from the clear story we obtain 3,000 + 
 8,250 = 11,250 Ibs. for the total wind load at b. Now having all the 
 wind loads determined we can proceed with the determination of the 
 stresses in the knee braces and truss due to same. 
 
 The points of contra-flexure in each column will be considered as 
 being midway between the bottom of the column and knee brace. By 
 constructing the ray diagram shown at (<^), and the corresponding equi- 
 librium polygon c-d-e . . . n-m at (a), and prolonging the segments cd 
 and nm to intersection at /, and drawing from I a line parallel to the 
 resultant AB, we obtain the line of action of the resultant of all of the 
 forces, which intersects the horizontal line 00' at N. 
 
 Then laying off 0' S equal to the vertical component of all of the 
 forces and drawing the influence line OS and dropping a vertical from 
 N we have the vertical component of the reaction at 0' given by the 
 ordinate NT and the vertical component of the reaction at by the ordi- 
 nate TU. The horizontal components J/l and H2 of the reactions at 0' 
 and are equal and each is equal to one-half of the horizontal component 
 of all the forces, which is represented at (e?) by the line CB. 
 
 Now having the components of the reactions at the points of contra- 
 flexure, O and 0', determined, the stresses in the knee braces and truss 
 due to wind loads, which are given at (a), are determined by adding the 
 auxiliary trusses, shown dotted, and constructing the stress diagram 
 shown at (e). The diagram shown at (e) is obtained by beginning at O 
 and passing around each joint clock-wise. The diagram at (e) should 
 be verified by the student. 
 
 276. Designing of the Truss Members. The maximum stresses 
 in each truss member (as given in Figs. 431 and 432), and also the ap- 
 proximate lengths of the members, are shown on the one diagram in Fig. 
 433. Having this data we can proceed with the designing of the members. 
 
 Member KF. This member is subjected to 35,200 Ibs. compression 
 and 20,000 Ibs. tension. Let us assume 2 Ls 5" x 3i"x T V' = 5.12 n " 
 as section. Assuming the 5" legs vertical and \" apart we have L/r = 
 177/1.50 = 118, and hence for the allowable compressive stress we have 
 16,000-70x118 = 7,740 Ibs. per sq. in. Then we have 35,200 -r 7,740 = 
 4.5 n/ ' for the area required for compression, which is 0.61 n// less than 
 the assumed section. 
 
 For tension the net area of cross-section of the assumed section is 
 5.12-0.55 = 4.57 n ". The actual area required for tension is equal to 
 20,000 -r 16,000 = 1.25 n ". From the above it is seen that the assumed 
 section is somewhat larger than required, but owing to the fact that it 
 is about as satisfactory as can be obtained (the value of L/r being about 
 correct) it will be used. 
 
 Members AF and FG. The member AF is subjected to 19,300 Ibs. 
 tension while member FG is subjected to 22,250 Ibs. tension and 13,300 
 Ibs. compression. The truss is held transversely at G and A by struts, 
 and as member FG is in compression it is necessary to consider (hori- 
 
DESIGN OF BUILDINGS 
 
 615 
 
 zontally) AG as one member subjected to the 13,300 Ibs. compression. 
 As the stresses are small the designing is simply a matter of obtaining 
 the proper value of L/r. Let us assume 2 Ls 5" x ?>\" x yV' = 5.12 n "; 
 the 5" legs horizontal and the vertical legs -J" apart. Then L/r = 
 260/2.44 = 106 in the horizontal plane and 130/1.03 = 126 in the vertical 
 plane. The first value (106) of L/r is quite satisfactory, but the last 
 
 M 
 
 *\1 J \ J I' ^ < H*V<*^ VV 
 
 *V-*%W 
 
 ^ - 24 7$0&*S? frC 
 
 G//g'-(o -3dow s >^jT ^ 
 
 Fig. 433 
 
 value is a little high (125 being the limit), but as the angles are fixed 
 more or less at the joints in the vertical plane the above assumed section 
 will be used. There should be enough rivets in these angles at points A 
 and G to develop the angles in tension in order to obtain rigidity. 
 
 Members BF and DH. Each of these members is subjected to 6,710 
 Ibs. compression. This stress is so small that the designing is simply a 
 matter of obtaining the proper value of L/r. Let us assume 2 Ls 2J" x 
 24"xi" = 2.38 n " for section. Then L/r = 72/0.78 = 92, which is quite 
 satisfactory, and hence the assumed section will be used. 
 
 Member CF. This member is subjected to 20,775 Ibs. tension and 
 24,700 Ibs. compression. Let us assume 2 Ls 3J"x3i"x T y = 4.18 n " 
 as section L/r - 130/1.08 = 120. Then for the allowable compressive 
 stress we have 16,000- 70 x 120 = 7,600 Ibs. Dividing this into the com- 
 pressive stress we obtain 24,700 -f 7,600 = 3.25 n " for the area required 
 for compression. For the net area of cross-section required for tension 
 we have 20,775 -f 16,000 = 1.30 n ". From the above it is seen that the 
 assumed section is larger for section than required, but as the value of 
 L/r is about as great as is permissible the assumed section will be used. 
 
616 STRUCTURAL ENGINEERING 
 
 Member CG. This member is subjected to 11,800 Ibs. tension and 
 21,600 Ibs. compression. Let us assume 2 Ls 4" x 4" x T y = 4.80 n ". 
 Then L/r = 144/1. 25 = 115, and hence for the allowable compressive stress 
 we have 16,000-70x115 = 7,950 Ibs. per sq. in. Dividing this into the 
 compressive stress we obtain 21,600 4- 7,950 = 2. 7 n// for the area of cross- 
 section required for compression. For the net area of cross-section 
 required for tension we have 11,800 -f- 16,000 = 0.73 n ". From the above 
 it is seen that the assumed section, as far as area is concerned, is larger 
 than required, but as L/r is of about the correct value the assumed section 
 will be used. 
 
 Member CH. This member is subjected to 5,975 Ibs. tension. The 
 area of cross-section required is about 0.37 n ". We will use 1 L 2^" x 
 2J"xJ" = 2.3S-0.44 = 1.94 n " net for section. This is about as small an 
 angle as can be used and the member be in harmony with the other 
 members of the truss. 
 
 Members GH and HE. The member GH is subjected to 19,450 Ibs. 
 tension and 10,650 Ibs. compression, while member HE is subjected to 
 25,525 Ibs. tension and 9,520 Ibs. compression. As the truss is not sup- 
 ported transversely at H it is necessary to design EG as one compression 
 member. Let us assume 2 Ls 5"x 3J"x T y = 5.12 n " for section. The 
 maximum radius of gyration is 2.44 and the least is 1.03. Then L/r 
 260/2.44 = 106 in one case and 130/1.03 = 126 in the other. For the 
 maximum allowable compressive unit stress we have 16,000 70x126 = 
 7,180. Dividing this into the maximum compressive stress we obtain 
 10,650 -f 7,180 = 1.48 n " for the required area of cross-section. For the 
 required area of cross-section for tension we have 25,525 -r 16,000 = 1.59 n " 
 net. From the above it is seen that the assumed section is larger than it 
 need be, as far as section is concerned, but that the value of L/r is about 
 as large as allowed, so the assumed section will be used. 
 
 Member GO. This member is subjected to 9,900 Ibs. tension and 
 2,600 Ibs. compression. These stresses are so low that the designing of 
 the member is simply a matter of obtaining the correct value for L/r. 
 The length of the member is 200 ins. and hence the radius of gyration of 
 the member must not be less than 200/125 = 1.60. By using 2 Ls 5"x 
 3J"x&" and 2 Ls 3J" x 3 J" x \" riveted together as shown at (a), 
 Fig. 433, we obtain about the correct radius and hence this section will 
 be used. 
 
 Top Chord AB . . . E. This member should be of the same section 
 throughout. It is subjected to a maximum compression of 43,600 Ibs. 
 and to a maximum tension of 22,100-8,600 = 13,500 Ibs. The purlins 
 hold the truss transversely more or less, but as there is some question as 
 to the rigidity thus obtained the top chord will be considered unsupported 
 transversely for half of its length, which is about 18 ft. or 216 ins. The 
 length as regards the vertical direction is 9 ft. or 108 ins. Then, as 
 regards the transverse direction, the radius of gyration must not be less 
 than 216/125 = 1.73, and as regards the vertical direction the radius of 
 gyration must not be less than 108/125 = 0.86. 
 
 There is a purlin concentration midway between panel points of 
 (3,960 + 3,300x0.83) = 6,700 Ibs. (see Figs. 431 and 432) which causes 
 a bending moment on the chord (considered as a fixed beam) of X 
 6,700x108 = 90,450 inch Ibs. 
 
DESIGN OF BUILDINGS 617 
 
 Now, it is seen that the section of the top chord must be sufficient to 
 transmit the direct stress and the cross bending given above and that 
 the radii must not be less than indicated above. 
 
 Let us assume 2 Ls 6" x4" x4" = 9.50 n " for section; the 6" legs 
 vertical and J" apart. The radii are about 1.71 and 1.91, which are 
 satisfactory values. For the allowable compression we have 16,000 70 x 
 216/1.70 = 7,100 Ibs. per sq. in. The actual direct stress is 43,600 + 9.5 = 
 4,600 Ibs. per sq. in. and the stress due to cross bending is (90,450 x 1.99) 
 + 34.8 = 5,180 Ibs. per sq. in. Now adding these two stresses together 
 we obtain 9,780, which is larger than the allowed, so the assumed section is 
 too small. Let us try 2 Ls 6" x 6" x f" = 8.72 n ". The radii are about 
 2.6 and 1.8. Then L/r in one case is equal to 216/2.6 = 83, and in the 
 other 108/1.8 = 60. Then for the allowable unit stress we have 16,000- 
 70x83 = 10,190. 
 
 For the stress due to direct compression we have 43,600 + 8.72 = 
 5,000 Ibs.' per sq. in., and for the stress due to cross bending we have 
 (90,450x1.64)4-30.78=4,820. Now adding these two stresses together 
 we have 5,000 + 4,820 = 9,820 Ibs., which is only 370 Ibs. less than the 
 allowable, so the 2 Ls 6" x 6" x f " will be used as section for the top 
 chord. We now have all of the truss members designed and the sections 
 can be written on the diagram as shown in Fig. 433. 
 
 277. Designing of the Lean-to Columns. We assume that these 
 columns are not fixed. The load on each purlin due to snow and dead 
 load, as given in Art. 272, is 65.5 Ibs. per ft. of purlin. The vertical 
 component of the wind load is equal to (18 x 4.5) x 0.908 = 73.5 Ibs. per 
 ft. of purlin. Then for the vertical load on the rafter at each interme- 
 diate purlin we have (65.5 + 73.5) x 20 = 2,780 Ibs. Then the concentra- 
 tion on the lean-to column due to the roof load is equal to 2,780 x 3 = 8,340 
 Ibs. Adding the weight of one-half of the rafter we have 8,340+ (13.2 X 
 42) =8,894, say 8,900 Ibs., for the total direct load on the column. 
 
 The wind load acting perpendicularly to the column is (considering 
 it uniformly distributed) equal to 30x20 = 600 Ibs. per ft. of column. 
 Then for the moment on the column due to this load we have % X 600 X 14 2 
 x 12 = 176,400 inch Ibs. Let us assume an 8"x34# H-beam (Carnegie) 
 for section, in which case L/r =168/1.87 = 89.8. Then for the allowable 
 unit stress we have 16,000-70x89.8 = 9.714 Ibs. and for the actual direct 
 unit stress we have 8,900 + 10 = 890 Ibs. For the maximum unit stress 
 due to cross bending we have 
 
 /= (176,400 x 4) + 115.4 = 6,110 Ibs. 
 
 Now adding the direct unit stress to this bending stress we have 6,110 + 
 890= 7,000 Ibs. for the maximum unit stress on the column, which shows 
 that the assumed H-beam is larger than required, but as this beam seems 
 to be the most satisfactory section obtainable it will be used throughout 
 for the lean-to columns. 
 
 278. Designing of the Crane Girders. The spacing and load on 
 
 the wheels of a 20-ton crane having a 60-ft. span are given in Fig. 414. 
 The maximum moment on the crane girder will occur when the wheels 
 are in the position shown at (o), Fig. 434. Taking moments about B 
 
618 
 
 STRUCTURAL ENGINEERING 
 
 9-8' 
 
 the reaction shown at A is obtained. Then taking moments about wheel 
 
 C we have 
 
 28,046x7.58x12 = 2,551,000 inch Ibs. 
 
 for the maximum moment on the crane girder due to the crane. Assum- 
 ing the crane girder to weigh 80 Ibs. per ft. and the rail J3_0 Ibs. per yd., 
 making in all 100 Ibs. per ft. of girder, we have J X 100 X 20 2 x 12 = 60,000 
 inch Ibs. for the maximum bending moment on the girder due to the weight 
 
 of the girder and rail. Adding 
 this moment to the moment due 
 to the crane, we have 2,551,000 + 
 60,000 = 2,611,000 inch Ibs. for 
 the total maximum bending mo- 
 ment on the crane girder. Di- 
 viding this by 16,000 we obtain 
 163 for the section modulus 
 which calls for a 24"xSO# I, 
 which will be used for each 
 girder. 
 
 The maximum reaction on the 
 crane girder will occur when the 
 crane wheels are in the position 
 shown at (6), Fig. 434. Tak- 
 ing moments about E we obtain 
 the reaction 56,100 Ibs. shown 
 at D, which is the maximum end 
 shear on the crane girder due 
 to the crane wheels. The end 
 shear due to the weight of the 
 crane girder and rail is equal to 
 100 + 20/2 = 1,000 Ibs. Then 
 
 AI 
 
 1 
 
 Ki 
 
 (P) O 
 
 7.56' 
 
 i 
 
 2.42 
 
 2-42. 
 
 7.58' 
 
 4 C 3- 
 20' 
 
 (a) 
 
 r *<-s>' 
 
 /o- 4-" 
 
 \ 20 '- " 
 
 (b) 
 
 Fig. 434 
 
 for the total end shear or reaction on the crane girder we have 56,100 + 
 1,000 = 57,100 Ibs. 
 
 279. Designing of the Main Columns. The load applied to the 
 top of each intermediate column due to snow and dead load is 13,200 Ibs., 
 as given in Fig. 431. The load from the lean-to rafter due to snow and 
 dead load is equal to 65.5 x 20 x 3 = 3,930, say 4,000 Ibs. The direct stress 
 due to the wind load is equal to the vertical component of the reaction at 
 the point of contra-flexure, the maximum of which is given in Fig. 432 as 
 11,500 Ibs. Then for maximum direct stress on the column above the 
 crane girder we have 13,200 + 4,000 + 11,500 = 28,700 Ibs. This load 
 comes on the leeward column. As is seen from Fig. 432, the maximum 
 moment on the column due to wind load occurs at the knee brace and is 
 equal to (//2 x 12.5) 12 = 11,500 x 12.5 x 12 = 1,725,000 inch Ibs. Let us 
 assume the part of the column above the crane girders to be a built 
 I-section composed of 4 Ls 6" x 3 \" x T V and a web 18" xf". The 
 distance from the truss down to the crane girder, which is 16 ft. or 192 
 ins. (see Fig. 429), will be taken as the length of this column. The least 
 radius of gyration of the section is 2.5. Then we have L/r = 192/2.5 = 
 76.8, and hence for the allowable unit stress we have 16,000- (70 x 76.8) 
 = 10,624 Ibs. The moment of inertia of the section with reference to the 
 
DESIGN OF BUILDINGS 
 
 619 
 
 H 
 
 axis perpendicular to the web is 1,576. Then for the maximum unit 
 stress due to cross bending we have 
 
 / = (1,725,000 x 9) + 1,576 = 9,850 Ibs. 
 
 The area of the cross-section is 26.87 n ". For the direct stress on the 
 column we have 28,700 + 26.87 = 1,068 Ibs. per sq. in. Adding this to the 
 bending stress we obtain 9,850 + 1,068 = 10,918 Ibs. for the total maximum 
 unit stress on the column (above crane 
 girder), which is only 294 Ibs. greater 
 than the 10,624 Ibs. allowable, so the 
 above assumed section will be used. 
 
 The part of the column below the 
 crane girder will be at least 11 ins. wider 
 than the part above (see Fig. 414), so 
 that the crane girder can be connected 
 directly to the column. This requires 
 that the part below the crane girder be 
 30" wide. So let us assume the part of 
 the column below the crane girder to 
 be composed of 4 Ls 6" x 3" x T V = 
 20.12 n " and a web 30" x A" = 9.38, 
 making in all 29. 5 n " of cross-section. 
 The least radius of gyration is 2.55. 
 The distance from the crane girder to 
 the base of the column is 19'-0" or 228". 
 So we have L/r = 228/2.55 = 89.4, and 
 hence for the allowable compression we 
 have 16,000-70x89.4 = 9,742 Ibs. per 
 sq. in. The direct load, considering the 
 windward column, applied from the part 
 of the column above the crane girder, is 
 equal to 13,200 + 4,000 + 1,700 = 18,900 
 Ibs. This causes a direct unit stress of 
 18,900 + 29.5 = 641 Ibs. The two angles 
 of the column directly under the crane 
 girders should be sufficient to carry the 
 maximum concentration from these gird- 
 ers. As the crane wheels are equal in 
 weight the maximum concentration due 
 to the wheels is equal to the maximum 
 end shear on the crane girder, which is 
 given in Art. 272 as 56,100 Ibs. The 
 concentration due to the weight of the 
 crane girders and struts is equal to 
 (80 + 20) 20 = 2,000 Ibs. Then for the 
 total concentration on the column from 
 the crane girders we have 56,100 + 
 2,000 = 58,100 Ibs. The area of cross- 
 section of the two angles is 10.06 n ". 
 Then we have 58,100 + 10.06 = 5,776 Ibs. for the direct stress on the two 
 angles due to the concentration from the crane girders. Adding to this 
 
 (a) 
 
 Fig. 435 
 
620 STRUCTUKAL ENGINEERING 
 
 the direct stress transmitted from the part of the column above the crane 
 girder we obtain 5,776 + 641 = 6,417 Ibs. for the maximum direct stress 
 on the two column angles directly under the crane girders. 
 
 As is seen from Fig. 4:32, the maximum bending moment on the part 
 of the column below the crane girder due to wind is equal to 12.5 X 
 11,500x12 = 1,725,000 inch Ibs., which occurs at the base of the column. 
 The moment of inertia of the column in reference to the gravity axis 
 perpendicular to the web is about 4,800. Then we have 
 
 / = ( 1,725,000 x 15) -r 4,800 = 5,390 Ibs. per sq. in. 
 
 for the maximum stress due to cross bending. Adding this to the above 
 direct stress we have 6,417 + 5,390 = 11,807 Ibs. for the total maximum 
 unit stress on the two angles directly under the crane girders. The 
 allowable stress is given above as 9,742 Ibs. per sq. in. ; but as the maxi- 
 mum crane load and maximum wind load are not likely to occur at the 
 same time this stress could safely be increased 25 per cent. So we have 
 1.25x9,742 = 12,179 Ibs. for the unit stress permissible for the combined 
 loading. This shows that the assumed section for the part of the column 
 below the crane girder is about correct and hence will be used. The two 
 outside angles and web of the column are stressed less than the angles 
 directly under the crane girders, but it is desirable to have a symmetrical 
 section and hence the main section of the column as given above is satis- 
 factory throughout. 
 
 280. Designing of Column Base. Case I. When column is sub- 
 jected to direct load only. In that case the pressure on the base will be 
 uniformly distributed as indicated at (a), Fig. 435. 
 
 Let P = direct load on column, 
 
 p = pressure per square in., 
 
 6 = width of base in ins., 
 
 d length of base in ins. 
 Then we have 
 
 CD 
 
 The value of p should not exceed 600 Ibs., which is the allowable 
 pressure on concrete masonry. 
 
 Case II. When the column is subjected to both direct load and cross 
 bending but the bending not sufficient to reverse the direct pressure. In 
 that case the pressure will vary as indicated at (6), Fig. 435. 
 
 LetP= direct load on column, 
 
 p= maximum pressure per sq. in., 
 p' minimum pressure per sq. in., 
 M bending moment in in. Ibs., 
 6 = width of base in ins., 
 d length of base in ins. 
 
 Then we have 
 
DESIGN OF BUILDINGS 621 
 
 = P 6M 
 and 
 
 , , , -j, (2). 
 
 bd bd~ 
 
 Case III. When the column is subjected to both direct load and 
 cross bending and the bending reverses the direct pressure. In that case, 
 assuming linear variation, the forces will be as indicated at (c), Fig. 435. 
 
 Let P = direct load on column, 
 
 p maximum pressure per sq. in., 
 
 / = stress per sq. in. on the anchor bolt =10,000 Ibs., 
 
 A = area of cross-section of anchor bolt or bolts on each side of 
 
 column, 
 
 M = bending moment in in. Ibs., 
 
 n ratio of the modulus of elasticity of steel to masonry = 15, 
 b = width of column base in ins., 
 d length of column base in ins. 
 
 The problem is to determine p and A. As is seen from the diagram 
 at (c) these could readily be determined if x were known. So we will 
 first derive an expression for the value of x. 
 
 It is obvious that the moments of the forces about o (center of the 
 column) must be equal to M. 
 Then we have 
 
 As the summation of the vertical forces must equal 0, we have 
 
 fx6-/*=P ................................... . ..... (5) 
 
 From the stress diagram at (c) we have 
 
 P x 
 
 ' d-x' 
 
 v; 
 
 from which we obtain 
 
 From (5) we obtain 
 
 From (4), (5), and (6) the following expression for x can be obtained: 
 
 = ............ (8) 
 
 jo of 
 
622 STEUCTURAL ENGINEERING 
 
 The value of x for any case can be determined from (8) and then 
 the value of p can be obtained from (6) and A from (7). For conven- 
 ience the anchor bolts are assumed to be exactly at the edge of the base 
 plate, which of course is not absolutely true, but the error resulting from 
 the assumption is small. 
 
 Now, considering the windward column, the direct load above the 
 crane girder as previously given is 28,700 Ibs. and the load from the 
 crane girder is 58,100 Ibs., and assuming the column to weigh 2,000 Ibs. 
 we have 
 
 P = 28,700 + 58,100 + 2,000 = 88,800 Ibs. 
 
 for the direct load on the column. As previously found the bending 
 moment 
 
 M = 1,725,000 inch Ibs. 
 
 Let us assume d equals 40" and b equals 16". Substituting these values in 
 (8) and assuming / equals 10,000 we obtain 
 
 x 20 ins. (about). 
 Substituting in (6) we obtain 
 
 p = 6661bs., 
 
 which is quite satisfactory, as the pressure is due to direct load and cross 
 bending which are not likely to occur simultaneously. In fact p could 
 safely be 800 or 900 Ibs. 
 
 Substituting in (7) we obtain 
 
 A = (^ x 20 x 16 - 88,800 \ * 10,000 = 2.77 sq. ins. 
 
 for the area of cross-section of anchor bolts. Use two If" anchor bolts 
 on a side. 
 
 281. Drawings. The data given in Articles 270 to 280 is suffi- 
 cient for making the stress sheet, Fig. 436, except for the gable fram- 
 ing, girts, and bracing, which can be considered more conveniently 
 as the work on the stress sheet progresses. The gable framing is de- 
 signed to resist a wind pressure of 30 Ibs. per sq. ft. on the end of the 
 building. Each gable column can be considered as a vertical beam 35 ft. 
 long and, as they are about 11 ft. apart, the load on each is 11x30 = 330 
 Ibs. per ft. of length. Then the bending moment on each is x330x 
 35 2 x 12 = 606,370"*. Dividing this by 16,000 we obtain 37.9 for the 
 section modulus. This calls for a 12"x35# I which will be used for 
 each gable column. 
 
 The bracing between the trusses and main columns in the end bays 
 should be sufficient to resist the wind pressure on the end of the building. 
 The stresses in this bracing are determined as indicated in Fig. 437. The 
 diagram at (a) represents the bracing between the main columns and the 
 diagram at (6) represents the stresses in the same. The diagram at (c) 
 represents the bracing in the plane of the roof and the diagram at (d) 
 represents the bracing in the plane of the bottom chord of trusses. The 
 designing of the bracing in intermediate bays is mostly a matter of judg- 
 ment as it is intended mostly for rigidity. 
 
It'! 
 f-ftH 
 
 
 rrM 
 
 623 
 
624 
 
 STRUCTURAL ENGINEERING 
 
 The girts are designed to resist a horizontal wind pressure of 30 Ibs. 
 per sq. ft. and in some cases they should act as struts, that is, L/r should 
 not exceed 125. After the stress sheet, Fig. 436, is completed the shop 
 
 20-0' 
 
 (a) 
 
 drawings and bills for the structure can be made. A fair idea of the 
 details can be obtained from the general drawing, Fig. 438. 
 
 HIGH BUILDINGS 
 
 282. Preliminary. In the case of modern high buildings such as 
 office buildings, hotels, etc., all loads including exterior walls, partitions, 
 floors, and live loads are practically in every case supported upon a steel 
 frame which consists of columns and beams. While the designing of 
 
626 STRUCTURAL ENGINEERING 
 
 these buildings as a whole is architectural work the designing of the steel 
 frame is purely a structural engineering problem and hence will be herein 
 treated as such. 
 
 283. Dead Load. The dead load includes all permanent parts of 
 the building and also all permanent fixtures and mechanisms supported 
 by the building. In estimating dead loads the following units of weights 
 may be used: 
 
 Steel 490 Ibs. per cu. ft. 
 
 Concrete 150 Ibs. per cu. ft. 
 
 Brick walls 140 Ibs. per cu. ft. 
 
 Hollow tile walls 40 Ibs. per cu. ft. 
 
 Hollow tile floors (tile alone).. 50 Ibs. per cu. ft. 
 
 Wood 4.5 Ibs. per sq. ft. B. M. 
 
 Plaster 5 Ibs. per sq. ft. 
 
 The weights of the other materials can be obtained from handbooks. 
 
 284. Floor Load. The floor load consists of the weight of the 
 floor proper and the load supported on the floor. The former is usually 
 referred to as the dead load and the latter as live load. The dead load 
 is always first assumed and then verified by calculations. The live load, 
 as a rule, can only be estimated. The building ordinances of the differ- 
 ent cities specify what these loads shall be in each case. In the case of 
 office buildings the live load for the first floor is usually taken as 100 to 
 150 Ibs. per sq. ft. of floor; the other floors from 40 to 75 Ibs. per sq. ft. 
 
 Designing of a Ten-Story Office Building 
 
 285. Data: 
 
 Length = 7 bays @ 15'-0" = 105'-0" 
 Width =4 bays @ 15'-0" = 60'-0" 
 
 f Basement = lO'-O" 1 
 
 Height =1 1st story = 18'-0" Il36'-0" 
 
 1 9 stories @ 12'-0" = 108'-0" J 
 Dead load to be computed. 
 
 f First floor 100 Ibs. per sq. ft. of floor. 
 Live load < All other floors 50 Ibs. per sq. ft. of floor. 
 
 I Roof 25 Ibs. per sq. ft. of roof. 
 
 Wind load 20 Ibs. per sq. ft. of vertical exposed surface. 
 Type of floor solid concrete. 
 All metal to be incased in concrete. 
 
 In order to resist the wind pressure on the building the floor 
 beams will be rigidly connected to the columns. These beams will 
 be considered as fixed beams for all loads. 
 
 The floor framing in general will be as shown in Fig. 439. 
 Allowable unit stresses will be the same as for highway bridges 
 (see 12 of Art. 222). 
 
 286. Wind Stresses. As seen from Fig. 439, it is necessary to 
 compute the wind stresses in bents BE, CC, and EE, at least. The wind 
 stresses in bents AA will be one-half as much as in bents BE, and like- 
 wise the stresses in bent DD will be half as much as in bents EE and FF. 
 
DESIGN OF BUILDINGS 
 
 627 
 
 Let us first consider bents BB, the elevation of which is shown in 
 Fig. 440. The wind load at the roof is equal to 
 
 20 x 15 x (6 + 4) =3,000 Ibs. 
 
 At each floor from the tenth down to the third (inclusive) the wind load 
 is equal to 20x15x12 = 3,600 Ibs. as shown, and at the second floor it is 
 
 60-0" 
 
 
 
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 General Plan of F/oor Framing . 
 
 Fig. 439 
 
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 20x15x9 = 2,700 Ibs. Having the wind loads determined we can pro- 
 ceed with the determination of the stresses due to same by beginning at 
 the roof and working downward story after story. 
 
 The columns and the floor beams connecting to them will be con- 
 sidered fixed and the point of contra-flexure in each column of each story 
 will be considered to be at mid-story, while the point of contra-flexure in 
 each floor beam will be considered to be at mid-span. Any part of the 
 structure between points of contra-flexure can be considered as an inde- 
 pendent structure. 
 
 In determining the stresses the building as a whole will be considered 
 
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 Bents - BB 
 
 Fig. 440 
 
 628 
 
DESIGN OF BUILDINGS 
 
 629 
 
 to act as a vertical cantilever beam, in which case the direct stress on 
 the columns will vary directly as their distance from the neutral axis. 
 
 Considering the part of the structure above the points of contra- 
 flexure of the columns in the tenth story we have the independent struc- 
 ture shown at (a), Fig. 441. As is obvious, the neutral axis will be at 
 
 
 7.5' 
 
 /S' 
 
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 I vi 
 
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 *' ' 7SO* 
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 (a) 
 
 
 
 ? 
 
 
 
 
 1 
 
 ._-J 
 
 Fig. 441 
 
 O x , the center of the building. Then the vertical reaction on the columns 
 will vary directly as their distance out from as indicated at (6). 
 For the moment of the wind load about O x we have 
 
 M = 3,000x6-18,000 ft. Ibs. 
 
 This moment must be balanced by the moment of the vertical reactions on 
 the columns. Let R represent the reaction at F. Then the reaction at 
 a point one foot (unit) out from O l would be 12/30, and hence the reac- 
 tion at G and H will be (12/30) 15 and similarly at F and K it will be 
 (12/30) 30. Then taking moments about we have 
 
 from which we obtain 
 
 < 5 
 
 The number 75 is a constant that can be used in determining the 12 In all 
 the other stories. 
 
630 STRUCTURAL ENGINEERING 
 
 Now having the vertical reaction R at F determined, we can readily 
 determine the reactions on the other columns as the intensity in each case 
 is directly proportionate to the distance of the column from 0^ Then, 
 for the reaction at G and H, we have (15/30) x 240 = 120 Ibs. and for 
 the reaction at K we have (30/30) x 240 = 240 Ibs., which of course is 
 the same as at F. 
 
 Having the vertical reactions on the columns determined we can next 
 determine the vertical shear on the floor beams by simply adding up the 
 vertical forces beginning at either A or E. For example, beginning at A 
 we have 
 
 Fl = 240 
 
 for the shear on girder AB. For the shear on girder EC we have 
 
 F2 = 240 + 120 = 360. 
 
 The structure being symmetrical about O i the shears and reactions on 
 one side of Oj will be equal and opposite to those on the other side, and 
 hence only one-half of the structure need be considered in determining 
 these stresses. 
 
 Having the vertical reactions on the columns and the vertical shear 
 on the floor beams determined, the horizontal shears H\, H2, and H3 on 
 the columns can be computed. Considering the part of the structure 
 shown at (c) as an independent structure and taking moments about A 
 we have 
 
 240x7.5-7/1x6 = 0, 
 from which we obtain 
 
 #1 = 300 Ibs. 
 
 Then summing up the horizontal forces, we have 
 
 3,000- 300 -F = 0, 
 
 from which we obtain 
 
 F = 2,700 Ibs. 
 
 for the direct compression in beam AB. Considering the part of the 
 structure shown at (d) as an independent structure and taking moments 
 about JB, we have 
 
 240 x 7.5 f 360 x 7.5 - H2 x 6 =6, 
 
 from which we obtain 
 
 #2 = 750 Ibs. 
 
 Then summing up the horizontal forces we have 
 
 2,700- 750 -Fl = 0, 
 from which we obtain 
 
 Fl = 1,950 Ibs. 
 
 for the direct compression in floor beam BC. Likewise, considering the 
 part of the structure shown at (e) as an independent structure and taking 
 moments about C we have 
 
 360x15-^3x6 = 0, 
 
DESIGN OF BUILDINGS 
 
 631 
 
 from which we obtain 
 
 H3 = 900 Ibs. 
 
 Then adding up the horizontal forces we have 
 
 1,950- 900 -F2 = 0, 
 from which we obtain 
 
 F2 = 1,050 Ibs. 
 
 for the direct stress in beam CD. The direct stress in beam DE is equal 
 to 1,050-750 = 300 Ibs., which is and should be equal to HI. 
 
 These shears and stresses can now be written on the diagram in 
 Fig. 440. 
 
 Next considering the part of the structure between the points of 
 contra-flexure in the columns of the tenth and ninth stories we obtain the 
 independent structure shown at (a), Fig. 442. Taking moments about 
 
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 1980*0* 
 
 (a) 
 
 I65O* 
 
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 32^-0 
 
 Fig. 442 
 
 O 2 of the wind forces on the building above this point (see Fig. 440), we 
 have 
 
 M = 3,000 x 18 + 3,600 x 6 = 75,600 ft. Ibs. 
 
 Letting R represent the vertical reaction on the column at F we have 
 
 from which we obtain 
 
632 STRUCTURAL ENGINEERING 
 
 Then the reaction on the column at G is equal to 1,008 (15/30) =504 Ibs. 
 Beginning at A and adding up the vertical forces (algebraically) 
 we have 1,008-240 = 768 Ibs. for the shear on beam AB, 1,008-240 + 
 504-120 = 1,152 Ibs. for the shear on beam EC. Now considering the 
 part of the structure shown at (c) as an independent structure and taking 
 moments about A, we have 
 
 HI x 6 + 300 x 6 - 768 x 7.5 = 0, 
 
 from which we obtain 
 
 HI = 660 Ibs. 
 
 for the horizontal shear on the column at F. 
 
 Similarly, considering the part of the structure shown at (d) as an 
 independent structure and taking moments about B, we have 
 
 H2 x 6 + 750 x 6 - 768 x 7.5 - 1,152 x 7.5 = 0, 
 
 from which we obtain 
 
 H2 = 1,650 Ibs. 
 
 Considering the part of the structure shown at (b) as an independent 
 structure and taking moments about C, we have 
 
 H3 x 6 + 900 x 6 - 1,152 x 15 = 0, 
 
 from which we obtain 
 
 H 3 = 1,980 Ibs. 
 
 for the shear on the column at 2 . Now adding up the horizontal forces 
 beginning at A we have 3,600 + 300 - 660 = 3,240 Ibs. for the direct com- 
 pression in beam AB, 3,600 + 300-660-1,650 + 750 = 2,340 Ibs. for the 
 direct compression in beam BC, 2,340 + 900-1,980 = 1,260 Ibs. for the 
 direct compression in beam CD, and 1,260 + 750-1,650 = 360 Ibs. for 
 the direct compression in beam DE. 
 
 Continuing in the manner shown above, considering the parts of the 
 structure between points of contra-flexure in the column of consecutive 
 stories on down to the base of the building, the direct stresses and shears 
 on the columns and floor beams, as shown in Fig. 440, can be determined. 
 
 It will be seen from Fig. 440 that the horizontal shear on each 
 column varies from the ninth down to the second story by a constant and 
 chat the vertical shear on the floor beams in each bay varies from the 
 tenth down to the third floor by a constant, and hence these stresses can 
 be quickly computed by the use of the constants, which can be determined 
 after the stresses are computed in the three top stories. Also, it will be 
 seen (see Fig. 440) that the direct stress in the floor beams in each bay 
 is the same from the tenth down to the third floor. After the shears on 
 the columns and floor beams are computed by the use of the constants, 
 the direct stress on the columns can be determined by beginning at the 
 ninth story and adding up the vertical forces included between the points 
 of contra-flexure. As an example, for the direct stress in the outer left- 
 hand column of the sixth story, we have 4,272 + 2,496 = 6,768 Ibs. For 
 the next column to the right we have 2,496 + 2,136 + 3,744 = 3,384 Ibs., 
 and so on. To obtain the stresses in the second story floor beams and 
 
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 Fig. 443 
 
 633 
 
634 
 
 STRUCTURAL ENGINEERING 
 
 first story columns we first take moments about O 10 of the wind forces 
 above that point. For this moment (see Fig. 440), we have 
 
 M = 3,000 x 117 + 28,800 x 63 + 4,500 x 9 = 2,205,900 ft. Ibs. 
 
 Then dividing this by the constant 75 (previously determined) we obtain 
 29,412 Ibs. for the reaction or direct stress in the outer left-hand column. 
 The reactions on the other columns can then be quickly determined by 
 direct proportion, and the remainder of the work of determining the stress 
 is the same as previously explained. 
 
 To obtain the stresses in the floor beams of the first floor and in the 
 basement columns we would first take moments about 0^ (see Fig. 440) 
 of the wind forces above this point. For this moment, we have 
 
 M = 3,000 x 131 + 28,800 x 77 + 4,500 x 23 + 2,700 x 5 - 2,727,600 ft. Ibs. 
 
 Then dividing this by the constant 75 we obtain 36,368 Ibs. for the reac- 
 tion or direct stress on the outer left-hand column. The reactions on the 
 other columns are readily determined by direct proportion, and the re- 
 mainder of the work of de- 
 
 he termining the stresses is 
 
 the same as previously ex- 
 plained. 
 
 From the above it is 
 seen that all of the direct 
 stresses and shears in the 
 columns and floor beams 
 of the bent are obtained by 
 fully analyzing only the 
 three top stories and the 
 first floor and basement. 
 
 We will next consider 
 the bents CC (through the 
 light court), the elevation 
 of which is shown in Fig. 
 443. Considering the part 
 of the structure above the 
 points of contra-flexure of 
 the columns in the tenth 
 story we obtain the inde- 
 pendent structure which is 
 shown at (a), Fig. 444. 
 As these bents are unsym- 
 metrical we have to deter- 
 mine the location of the 
 neutral axis 00. The re- 
 actions on the columns will vary directly as their distance from this 
 neutral axis, as indicated at (6), and the sum of the reactions on one 
 side must equal the sum of those on the other. 
 
 Let 2 = the distance from A to the neutral axis, and for the sake of 
 simplicity let a, b, and c represent the bay lengths as indicated, and let R 
 represent the reaction at E. 
 
 
 z- 
 
 
 
 
 
 21.25' 
 
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 Fig. 444 
 
DESIGN OF BUILDINGS 635 
 
 Then we have R for the reaction at E, (R/z) (2 - a) for the reaction 
 at G, (R/z) (a + b-z) for the reaction at H, and (R/z) (a + b + c-z) 
 for the reaction at K. Then, as the sum of the reactions on one side of 
 the neutral axis must be equal to the sum of those on the other side of the 
 neutral axis, we have 
 
 from which we obtain 
 
 
 for the distance from A to the neutral axis. Now, substituting the 
 numerical values of the bay lengths in (1) we obtain 
 
 which gives the location of the neutral axis. 
 Taking moments about O i we have 
 
 3,000x6 = 0, 
 from which we obtain 
 
 for the reaction on the column at E. The number 43.23 is the constant 
 for determining the reaction R for all the other stories. Having the 
 reaction at E determined the reaction on the other columns is readily 
 determined by direct proportion. For example, the reaction at H is 
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 mined the shears on the columns and the shears and direct stresses on the 
 floor beams can be determined, as previously explained, by considering 
 the independent portions shown at (c), (d), (e), and (/). In fact, the 
 remainder of the work of determining the direct stresses and shears on 
 the columns and floor beams shown in Fig. 443 is the same as previously 
 explained for bents BB. 
 
 The shears and direct stresses in the columns and floor beams for 
 bents EE and FF are given in Fig. 445. These are determined in the 
 same manner as previously shown for bents BB. It will be seen that the 
 neutral axis for bents EE and FF is midway between the central columns. 
 After the shears and direct stresses on the columns and floor beams, as 
 given in Figs. 440, 443, and 445, are determined the stresses at the column 
 connections are readily obtained, as will be seen later. 
 
 287. Designing of the Roof Framing. The columns and beams 
 will be spaced as shown in Fig. 446. 
 
 The roof covering will be a 4" solid concrete slab. 
 
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 Fig. 445 
 
 636 
 
DESIGN OF BUILDINGS 
 
 637 
 
 Live load (Chicago) ............ 25 Ibs. per sq. ft. of roof. 
 
 Dead loadT Concrete slab ....... 50 lbs * per sq< ft> of roof - 
 
 \Suspended ceiling. ... 1^0 Ibs. per sq. ft. of roof. 
 
 60 Ibs. per sq. ft. of roof. 
 
 C2 
 
 C3 
 
 C3 
 
 C3 
 
 C2 
 
 r- 7 "x /s < 
 
 / 
 
 Cl 
 
 'cj- 
 
 c/ 
 
 C/ 2 . .C/ 
 
 C4 
 
 C4- 
 
 C/ 
 
 Roof 
 
 Fig. 44G 
 
 Beams 1. The dead load on each of these beams is as follows 
 
 Concrete slab and ceiling = 60x5 = 300 Ibs. per ft. of beam. 
 Concrete around beam 75 Ibs. per ft. of beam. 
 
 Beam (assumed) = 15 Ibs. per ft. of beam. 
 
 390 Ibs. per ft. of beam. 
 Then, for the maximum bending moment due to dead load we have 
 
(538 STRUCTURAL ENGINEERING 
 
 J x 390 x 1? x 12 = 131,625 inch Ibs. 
 
 The live load is 25 x 5 = 125 Ibs. per ft. of beam. Then, for the live load 
 moment on each beam we have 
 
 Jxl25xl5 2 x 12=42,187 inch Ibs. 
 
 Adding the two moments together we obtain 131,625 + 42,187 = 173,812 
 inch Ibs. for the total maximum moment on the beam. Dividing this by 
 16,000 we obtain 10.8 for the section modulus which calls for a 7"xl5# 
 I-beam, which will be used. 
 
 T-, , , /Dead load = 390x7.5 = 2,925 Ibs. 
 End shear 
 
 3,862 Ibs. 
 
 Beams 2. These beams will be connected rigidly to the columns to 
 resist wind pressure and consequently will be fixed beams for all loads. 
 The dead load on each of these beams is as follows: 
 
 Fire wall [see sketch at (6) ] = 1 x 140 x 4 = 560 Ibs. per ft. of beam. 
 Concrete slab and ceiling = 60 x 2.5 150 Ibs. per ft. of beam. 
 Concrete around beam = 120 Ibs. per ft. of beam. 
 
 Beam (2 [s) = 30 Ibs. per ft. of beam. 
 
 860 Ibs. per ft. of beam. 
 
 Then for the maximum bending moment at the ends of the beam due 
 to dead load, considering each bracket to be 18 ins. long, we have 
 A x 860 x!2 2 x 12 = 123,840 inch Ibs., and one-half of this, or 61,920 inch 
 Ibs., will be the moment at the center of span (see Art. 69). The live 
 load on the beam is 2.5x25 = 62.5 Ibs. per ft. of beam. Then for the 
 bending moment at the ends of the beam due to live load we have y 1 ^ x 
 62.5 x!2" 2 x 12 = 9,000 inch Ibs., and one-half of this, or 4,500 inch Ibs., 
 will be the moment at the center of span. 
 
 The maximum shear on these girders due to wind pressure is one-half 
 of the maximum given in Fig. 440, or 360 + 2 = 180 Ibs. Then for the 
 moment at the end of the beam (at end of bracket) we have 180 x 72 = 
 12,960 inch Ibs. 
 
 Now combining the maximum dead and live moments we have 
 123,840 + 9,000 = 132,840 inch Ibs. Dividing this by 16,000 we obtain 
 8.3 for section modulus, which calls for 2 [s 6"x8#, but to obtain 
 rigidity 2 [s 7"x9.75# will be used. The stress due to wind load is 
 not considered as it does not exceed 50 per cent of the live- and dead-load 
 stresses. This is in accordance with practice. 
 
 For the maximum end shear on the beam we have 
 
 Dead load = 860 x 7.5 = 6^450 Ibs. 
 
 Live load= 62.5x7.5= 469 Ibs. 
 
 6,9 19 Ibs. 
 
 Beams 8. For the load on each of these beams we have the fol- 
 lowing: 
 
DESIGN OF BUILDINGS 
 
 63$ 
 
 Concrete slab = 50x5 = 250 Ibs. per ft. of beam. 
 Concrete around beam = 120 Ibs. per ft. of beam. 
 Beam = 21 Ibs. per ft. of beam. 
 
 Ceiling =10x5 =50 Ibs. per ft. of beam. 
 
 Live load = 25x5 =125 Ibs. per ft. of beam. 
 
 Total = 566 Ibs. per ft. of beam. 
 
 Considering the beams fixed to the columns by brackets we have T V X 566 X 
 12 2 x 12 = 81,504 inch Ibs. for the maximum moment at the ends of the 
 beam due to dead and live load. The moment due to wind load is 539 X 
 72 = 38,808 inch Ibs., which is less than 50 per cent of the live- and dead- 
 load moment and hence will not be considered. 
 
 Dividing the dead- and live-load moment by 16,000, we obtain 
 81,504 -r 16,000 = 5.1 for the section modulus, which calls for a 6" x 12.25* 
 I, but for the sake of rigidity and also to allow for rivet holes at the 
 brackets we will use 1 I 7"xl5# for each beam. 
 
 For the maximum end shear on the beam we have 
 
 Dead load = 441 x 7.5 = 3,307 Ibs. 
 
 Live load = 125 x 7.5 = _93^1bs. 
 
 4,244 Ibs. 
 
 Beams 4- There will be two concentrations on each of these beams, 
 each of which we will assume is equal to twice the maximum end shear on 
 beams 1. The concrete around the beam will be assumed to weigh 120 
 Ibs. per ft. of beam and the beam itself will be assumed to weigh 21 Ibs. 
 The loading will then be as indicated in Fig. 447. These loads include 
 both live and dead load. 
 
 Considering the beam to be fixed we have T V X 141 x 12 2 X 12 = 20,304 
 inch Ibs. for the maximum moment due to the uniform load. A; = 3. 5/12 = 
 0.29 for one load and 8.5/12 = 
 0.71 for the other. Then substi- 
 tuting in (7) of Art. 69 we obtain 
 229,120 inch Ibs. for the maxi- 
 mum moment due to the concen- 
 trations. This moment occurs at 
 the end of the beam. 
 
 Adding together the two mo- 
 ments we obtain 20,304 + 229,- Fig 447 
 120 = 249,424 inch Ibs. for the 
 
 total maximum bending moment on the beam due to dead and live load. 
 Dividing this by 16,000 Ibs. we obtain 15.5 for the section modulus which 
 calls for a 9"x21# I, which will be used for each of these beams. 
 
 For the maximum end shear on these beams we have 7,724 + 141 X 
 7.5 = 8,781 Ibs. The wind stress is so small that it is negligible. 
 Beams 5. The dead load on each of these beams is as follows : 
 
 Fire wall (same as 2) =560 Ibs. per ft. of beam. 
 
 Concrete around beam = 150 Ibs. per ft. of beam. 
 
 Beam (2 channels) = 30 Ibs. per ft. of beam. 
 
 740 Ibs. per ft. of beam. 
 
 
 \ 
 
 s 
 
 ^ 
 
 I 
 
 N 
 
 
 
 
 35' 
 
 5' 
 
 3.5' 
 
 
 15' 
 
 
 12' 
 
 
 1.5' 
 
 
 
 is' 
 
 
 
640 STRUCTURAL ENGINEERING 
 
 For the maximum moment on the beam due to this load, considering 
 the beam fixed, we have T Vx740xl2 2 x 12 = 106,560 inch Ibs. 
 
 The moment due to the two concentrated loads will be one-half of 
 that found for beams 4 or 229,120 H- 2 = 114,560 inch Ibs. Adding the two 
 above moments together we have 
 
 106,560 + 114,560 = 221,120 inch Ibs. 
 
 for the total maximum moment on the beam. Dividing this by 16,000 we 
 obtain 13.8, which calls for 2 8"x 11.25* [s, which will be used. 
 
 Beams 6 to 12. For the sake of uniformity beams 6 will be the same 
 size as beams 4, although the moment does not require such a large sec- 
 tion. Likewise, beams 7 will be the same size as beams 3 ; beams 8, 9, 
 and 10 will be the same size as beams 1 ; and beams 11 and 12 will be the 
 same size as beams 5. 
 
 288. Designing of the Tenth-Story Columns 
 Load on Column Cl 
 Dead load 
 
 From beams 2 6,450x2 =12,900 Ibs. 
 
 From beams 1 2,925x2 = 5,850 Ibs. 
 
 From beams 4 141 x 7 = 1,058 Ibs. 
 
 19^708" Ibs. 
 
 Live load = 25 x 15 x 7J = 2,812 Ibs. 
 
 Total = 22,520 Ibs. 
 Load on Column C2 
 Dead load 
 
 From beams 2 = 6,450 Ibs. 
 
 From beams 1 = 2,925 Ibs. 
 
 From beams 5 = 740x7J = 6,450 Ibs. 
 
 15,825 Ibs. 
 
 Live load = 25 x 7| x 7J = 1,406 Ibs. 
 
 Total = 17,231 Ibs. 
 
 Load on Column C3 
 Dead load 
 
 From beams 5 = 6,450x2 =12,900 Ibs. 
 
 From beams 1 = 2,925x2 = 5,850 Ibs. 
 
 From beams 3 = 3,307 Ibs. 
 
 22,057 Ibs. 
 
 Live load = 25 x 15 x 7J = 2,812 Ibs. 
 
 Total = 24,869 Ibs. 
 
 Load on Column C4 
 Dead load 
 
 From beams 1 = 2,925x4 =11,700 Ibs. 
 
 From beams 4 = 1,058x2 = 2,116 Ibs. 
 
 From beams 3 = 3.307x2 = 6,614 Ibs. 
 
 20,430 Ibs. 
 
 Live load = 25 x 15 x 15 = 5,625 Ibs. 
 
 Total = 26,055 Ibs. 
 
DESIGN OF BUILDINGS 641 
 
 Load on Column C5 
 Dead load 
 
 From beams 1 = 5,850 Ibs. 
 
 From beams 3 = 3,307 Ibs. 
 
 From beams 7 = 2,200 Ibs. 
 
 From beams 8, 9 and 10 = 3,900 Ibs. 
 
 From beams 6 = 1,000 Ibs. 
 
 16,257 Ibs. 
 
 Live load = 25 X 12| X 15 = 4,687 Ibs. 
 
 Total = 20,944 Ibs. 
 
 Load on Columns C6. Total = about f of load on column C3 = 16,600 
 Ibs. 
 
 Load on Column C7 
 Dead load 
 
 From beams 2 = 6,450 Ibs. 
 
 From beams 11 = 6,020 Ibs. 
 
 From beams 3 = 3,307 Ibs. 
 
 From beams 1 = 5,850 Ibs. 
 
 From beams 8 and 9 = 2,600 Ibs. 
 
 24,227 Ibs. 
 
 Live load = 4,200 Ibs. 
 
 Total = 28,427 Ibs. 
 
 Load on Column C7'. The load on this column is about equal to the load 
 on column C7 plus 10,000 Ibs. for elevator, which makes a total load of about 
 38,000 Ibs. 
 
 Load on Column C2'. This load on this column is about equal to the 
 load on column C2 plus 10,000 Ibs. for elevator, which makes a total load of 
 about 27,000 Ibs. 
 
 The direct wind loads on these columns are negligible, the maximum 
 being only about 367 Ibs. 
 
 The maximum shear on these columns due to wind is 1,194 Ibs. in one 
 direction and 554 Ibs. in the other direction (see Figs. 443 and 445). Then 
 for the maximum bending moments on the columns, we have 1,194X6X12 
 = 85,968 inch Ibs. and 554X6X12 = 39,888 inch Ibs., respectively. 
 
 The columns should be designed to resist this bending in addition to the 
 direct load on them. However, the unit stress for the combined loading can 
 be safely increased 50 per cent over that obtained by the formula, 16,000 
 -707,/r. 
 
 Let us assume the following section: 
 
 4-Ls 4"x3"x T V' = 8.36" 
 1-web 12" x' = 3.75 n " 
 
 The Max. 7 = 294.6 and the Min. 7 = 30.2 
 The Max. r= 4.91 and the Min. r= 1.58 
 
 Columns C4 in the center of the building will be subjected to the 
 
642 STRUCTURAL ENGINEERING 
 
 greatest bending in the two directions. For the fiber stresses we have 
 
 /= 39,888 x 4.1 + 30.2 = 5,415 Ibs. 
 and 
 
 f = 85,968 x 6.2 + 294.6 = 1,809 Ibs. 
 
 For the allowable direct stress, we have 16,000- 70 x 144-=- 1.58 = 
 9,620 Ibs. per sq. in. 
 
 Dividing the area of cross-section of the column into the load on the 
 column we have 26,055 -r 12.11 = 2,150 Ibs. per sq. in. for the direct stress. 
 Adding this to the maximum wind stress we have 5,415 + 2,150 = 7,565 Ibs. 
 per sq. in., which is quite low, so we will assume a lighter section. 
 
 Let us try 
 
 4 Ls 3i"x2J"xi" = 5.76 n " 
 1 web " 12"x" = 
 
 8.76 n " 
 
 Max. 7 = 213.5, Min. 7=15.9 
 Max. r= 4.94, Min. r= 1.35 
 
 For the allowable direct stress we have 16,000- 70 x 144-=- 1.35 = 
 8,530 Ibs. per sq. in. For the unit stress due to cross bending we have 
 39,888 x 3.62 -r 15.9 = 9,080 Ibs. 
 
 For the stress due to direct load we have 26,055 -r 8.76 = 2,970 Ibs. 
 per sq. in., which is low. Adding together the stresses due to direct load 
 and cross bending we have 2,970 + 9,080 = 12,050 Ibs. per sq. in. The 
 allowable is 8,530x1.5 = 12,795 Ibs. per sq. in. for combined loading, 
 which is very near the actual stress and hence the last assumed section 
 can be used. This section is about the minimum that would be used, so 
 all columns in the tenth story will have this section. 
 
 289. Designing of the Tenth-Floor Framing. The framing in 
 this floor will be as indicated in Fig. 448. 
 
 Beams 1. For the dead load on each of these beams we have the 
 following : 
 
 4" concrete slab = 50 x 5 = 250 Ibs. per ft. of beam. 
 Plaster =5x5= 25 Ibs. per ft. of beam. 
 
 Concrete around beam = 75 Ibs. per ft. of beam. 
 Beam = 18 Ibs. per ft. of beam. 
 
 Total = 368 Ibs. per ft. of beam. 
 
 _ 2 
 
 For the maximum moment due to dead load we have ^ x 368 x 15 x!2 = 
 124,200 inch_lbs. For the maximum moment due to live load we have 
 -| (50x5) x!5 2 x 12 = 84,375 inch Ibs. Then for the total maximum mo- 
 ment we have 124,200+84,375 = 208,575 inch Ibs. Dividing this by 16,000 
 we obtain 13.04, which calls for an 8" x 18# I, which will be used. 
 For maximum end shear on the beam we have 
 
 Dead load = 368 x 7^ = 2,760 Ibs. 
 
 Live load = 250 x 7 = 1,875 Ibs. 
 
 Total = 4,635 Ibs. 
 
 Beams 2. The dead load on each of these beams consists oi a story 
 
DESIGN OF BUILDINGS 
 
 643 
 
 of 12-in. curtain wall, a strip of concrete floor 2.5 ft. wide, and the weight 
 of the beam and concrete encasing it. 
 For the weight of the wall we have 
 
 8" hollow tile= 40 x 0.66 = 26.4 Ibs. per sq. ft. of wall. 
 
 4" brick face = 140 x 0.33 = 46.2 Ibs. per sq. ft. of wall. 
 
 Total = 72.6 Ibs. per sq. ft. of wall. 
 
 The curtain wall in each bay will have two openings for windows, 
 as shown at (a), Fig. 449. The load on the beam from this wall will be 
 
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 Fig. 448 
 
 as shown at (6), but, as the load near the ends of the beam is supported 
 directly upon the brackets, the load for computing the stress on the beam 
 can be taken as indicated at (c) . 
 
644 
 
 STRUCTUKAL ENGINEERING 
 
 Then for the maximum moment on the beam due to the curtain wall, 
 we have 
 
 A+ 182 x II 2 x 12 = 22,022 inch Ibs. 
 
 -x 2,504x11 x 12 = 41,318 inch Ibs. 
 Total =63,340 inch Ibs. 
 
 2' 
 
 2' 
 
 (a) 
 
 Tig. 449 
 
 For the remainder of the dead load we have 
 
 Concrete slab = 50 x 2.5 = 125 Ibs. per ft. of beam. 
 
 Beam (2 [s) = 30 Ibs. per ft. of beam. 
 
 Concrete around beam =150 Ibs. per ft. of beam. 
 
 Total ="305 Ibs. per ft. of beam. 
 
DESIGN OF BUILDINGS 645 
 
 For the moment on the beam due to this load we have 
 
 ~ x 305 x IT* x 12 = 36,905 inch Ibs. 
 12 
 
 For the live load on the beam we have 50 x 2.5 125 Ibs. per ft. of 
 beam. Then for the live-load moment on the beam we have 
 
 -i x 125 x IT' x 12 = 15,125 inch Ibs. 
 12 
 
 Adding together all of the above moments we have 
 
 63,340 + 36,905 + 15,125 = 115,370 inch Ibs. 
 
 for the total maximum bending moment on the beam due to dead and live 
 load. 
 
 The maximum shear on the beam due to wind load is 1,162 -r 2 = 576 
 Ibs. (see Pig. 440). Then for the moment on the beam due to wind load 
 we have 576x66 = 38,016 inch Ibs., which is less than 50 per cent of the 
 live- and dead-load moment and hence it will be ignored. Dividing the 
 dead- and live-load moment by 16,000 we have 115,370 -r 16,000 = 7.2 for 
 the section modulus, which calls for 2 [s 6" x 8*, but for the sake of 
 rigidity we will use 2 [s 7"x9.75#. 
 
 ,, fDead load = 6,504 Ibs. 
 
 Maximum end shear ^ T . -, j ^ orv ,, 
 
 l^Live load = 937 Ibs. 
 
 7,441 Ibs. 
 
 Beams 3. For the dead load on each of these beams we have the 
 following : 
 
 Partition = 40 x 0.33 x 12 = 160 Ibs. per ft. of beam. 
 Concrete slab = 50x5 = 250 Ibs. per ft. of beam. 
 Beam = 21 Ibs. per ft. of beam. 
 
 Concrete around beam = 150 Ibs. per ft. of beam. 
 Plaster = 5x5 = 25 Ibs. per ft. of beam. 
 
 636 Ibs. per ft. of beam. 
 
 Then for the maximum moment on the beam due to dead load, we 
 have T V x 636 xll 2 x 12 = 76,956 inch Ibs. The live load is 50x5 = 250 
 Ibs. per ft. of beam. Then for the moment on the beam due to live load, 
 we have ^ x 250 xll* x 12 = 30,250 inch Ibs. Adding together the dead- 
 and live-load moments we obtain 107,206 inch Ibs. for the total maximum 
 moment due to dead and live load. The maximum shear on these beams 
 due to wind load, as given in Fig. 440, is 1,152 Ibs. So, for the maximum 
 bending moment on the beams due to wind, we have 1,152x66 = 76,032 
 inch Ibs. Now subtracting from this 50 per cent of the live- and dead- 
 load moment we obtain 76,032 - 53,603 = 22,429 inch Ibs., which must be 
 added to the maximum dead- and live-load moment. So we have 
 
 107,205 + 22,429 = 129,635 inch Ibs. 
 bending moment which the beam must be designed to resist. Dividing 
 
646 
 
 STRUCTUKAL ENGINEERING 
 
 this by 16,000 we obtain 8.1 for the section modulus which calls for a 
 7"xl5# I, but as beams 1 are 8-in. beams we will make these the same 
 size; that is, for each we will use 1 I 8" x 18*. This excess size also 
 provides for rivet holes cut out at the bracket. 
 
 Maximum end shear 
 
 
 \ 
 
 S 
 
 
 
 
 3' J 
 
 ' <* 
 
 3' . 
 
 
 
 
 
 
 
 
 
 33//^er^ 
 
 nnn 
 
 
 2' 
 
 
 II ' 
 
 
 z' 
 
 
 
 15' 
 
 
 
 Dead load = 636 x 7.5 = 4,770 Ibs. 
 Live load =1,875 Ibs. 
 
 6,645 Ibs. 
 
 Beams 4- The maximum dead load consists of a 4-in. partition, 
 two concentrations from beams 1, the weight of the beam, and the concrete 
 encasing it. For the uniform dead load on the beam we have the 
 following: 
 
 Partition (same as 3) = 160 Ibs. per ft. of beam. 
 
 Beam = 25 Ibs. per ft. of beam. 
 
 Concrete around beam = 150 Ibs. per ft. of beam. 
 
 Total = 335 Ibs. per ft. of beam. 
 
 For each concentration due to dead load we have 2,760x2 = 5,520 
 
 Ibs. For each live-load con- 
 centration we have 1,875 x 2 = 
 3,750 Ibs. Then for the to- 
 tal of each concentration we 
 have 5,520 + 3,750 = 9,270 Ibs. 
 The loading will then be as 
 indicated in Fig. 450. For 
 the maximum moment (con- 
 sidering the beam fixed) due 
 Fig. 450 to the uniform load we have 
 
 x 335 x IT 2 x 12 = 40,535 inch Ibs. 
 
 For the maximum moment due to the concentrated loads we have 
 
 [9,270 x 11 (2 x O27 2 - (m* - 0.27) + 9,270 x 11 (2 xOT? - (XTS* 
 
 -0.73)] 12 = 240,800 inch Ibs. 
 
 Now adding together the above moments we obtain 
 40,535 + 240,800 = 281,335 inch Ibs. 
 
 for the total maximum dead- and live-load moment. The maximum shear 
 on these beams due to wind load, as seen from Fig. 445, is 731 Ibs. So, 
 for the maximum moment on the beams due to wind load we have 731 x 
 66 = 48,246 inch Ibs., which is less than 50 per cent of the combined dead- 
 and live-load moment and hence will be ignored. 
 
 Then for the required section modulus we have 281,335 + 16,000 = 
 17.57, which calls for 1 I 9" x 21 #, which will be used. 
 
 For the maximum end shear on the beam we have 
 
 Dead load = 8,032 Ibs. 
 
 Live load= 3,750 Ibs. 
 
 Total =11,782 Ibs. 
 
DESIGN OF BUILDINGS 647 
 
 Beams 5. The dead load supported by each of these beams consists 
 of a story of 12-in. curtain wall, two concentrations from beams 1, and a 
 uniform load which consists of the weight of the beam and the concrete 
 encasing it. 
 
 The bending moment due to the curtain wall = 63,340 inch Ibs., the 
 same as for beams 2. 
 
 The bending moment due to the concentrations = 120,400 inch Ibs., 
 one-half of that for beams 4. 
 
 For the uniform dead load we have the following: 
 
 Beam (2 [s) = 25 Ibs. per ft. of beam. 
 
 Concrete around beam = 150 Ibs. per ft. of beam. 
 
 Total = 175 Ibs. per ft. of beam. 
 
 For the maximum bending moment due to this load we have 
 
 ~ x 175 x IT* x 12 = 21,175 inch Ibs. 
 2 
 
 Adding together all of the above moments we obtain 
 
 63,340 + 120,400 + 21,175 = 204,915 inch Ibs. 
 
 for the total maximum moment due to dead and live load. Dividing this 
 by 16,000 we obtain 12.8 for the section modulus which calls for 2 [s 
 8"x 11.25*, which will be used. 
 
 The wind stress is ignored as it is less than 50 per cent of the live- 
 and dead-load stresses combined. 
 
 For the maximum end shear on the beam we have the following: 
 
 Dead load = 8,290 Ibs. 
 
 Live load = 1,875 Ibs. 
 
 10,165 Ibs. 
 
 Beams 5' . The loading on these beams is exactly the same as on 
 beams 5, except for the wall, which is solid, no openings for windows at 
 all. This wall, as determined for beams 2, weighs 72.6 Ibs. per sq. ft. 
 Then, ignoring arch action, we have 72.6x12 = 871 Ibs., per ft. of beam. 
 For the moment due to this load we have 
 
 ^r x 871 x II 2 x 12 = 105,391 inch Ibs. 
 12 
 
 Now, adding this to the other moments found for beams 5, we obtain 
 105,391 + 120,400 + 21,175 = 246,966 inch Ibs. 
 
 for the total maximum bending moment on the beam. Dividing this by 
 16,000 we obtain 15.4 for the section modulus, which calls for 2 [s 8" x 
 11.25* the same as used for beams 5. 
 
 For the maximum end shear we have the following: 
 
 Dead load = 10,604 Ibs. 
 
 Live load = J^^ Ibs. 
 
 12,479 Ibs. 
 
648 STEUCTURAL ENGINEERING 
 
 Beams 6, 7, 8, and 9. For the sake of uniformity beams 6 will be 
 the same size as beams 3, although the moment does not require such a 
 heavy section. Likewise, beams 7, 8, and 9 will be the same size as 
 beams 1. 
 
 Beams 10. For the dead load on these beams we have the following: 
 
 Partition = 160 Ibs. per ft. of beam. 
 
 Plaster = 25 Ibs. per ft. of beam. 
 
 Concrete slab = 250 Ibs. per ft. of beam. 
 
 Beam = 25 Ibs. per ft. of beam. 
 
 Concrete around beam- 75 Ibs. per ft. of beam. 
 535 Ibs. per ft. of beam. 
 
 Then for the dead-load moment we have 
 
 i x 535 x IS* x 12 = 180,562 inch Ibs. 
 8 
 
 The maximum moment due to live load is 84,375 inch Ibs. the same as 
 for beams 1. Adding, we have 
 
 180,562 + 84,375 = 264,937 inch Ibs. 
 
 for the total maximum bending moment on the beam. Dividing this by 
 16,000 we obtain 16.56 for the required section modulus, which calls for 
 1 I 8"x23#. Metal would be saved by using the 9"x21# I but it is 
 desirable to have beams 10 of the same depth as beams 1. 
 Maximum end shear is as follows: 
 
 Dead load = 3,960 Ibs. 
 
 Live load = 1,875 Ibs. 
 
 5,835 Ibs. 
 
 Beams 1'. The loading on these beams is exactly the same as that 
 on beams 1, except a partition is supported at mid-span. For this con- 
 centration we have 160x5 = 800 Ibs. This produces a moment of 400 x 
 7.5 x 12 = 36,000 inch Ibs. Adding this to the maximum moment on beams 
 1 we have 
 
 208,575 + 36,000 = 244,575 inch Ibs. 
 
 for the total maximum moment on beams V '. Dividing this by 16,000 we 
 obtain 15.3 for the required section modulus, which calls for an 8" x 20.5* 
 I, which will be used. 
 
 Beams 3' and 4' ' Beams 3' support less load from partitions than 
 beams 3, but the difference is not sufficient to change the section, so beams 
 3' will be the same size as beams 3. Beams 4' support no partitions and 
 consequently could be lighter than beams 4, but for the sake of uniformity 
 they will be the same size as beams 4. 
 
 The floor framing for the other floors down to the first floor would 
 be the same as the tenth floor if it were not for the wind stresses. How- 
 ever, the work of designing the framing in these floors is exactly the same 
 as shown above for the tenth floor, and hence the method of procedure is 
 fully shown above. 
 
 As an example, let us consider beams 4 in the fourth floor. The 
 
DESIGN OF BUILDINGS 649 
 
 sum of the dead- and live-load moments is 231,335 inch Ibs., as found 
 above. The maximum shear on the beams due to wind, as given in Fig. 
 445, is 4,023 Ibs. Then for the maximum moment due to wind load we 
 have 
 
 4,023x5.5x12 = 265,518 inch Ibs. 
 
 Subtracting 50 per cent of the dead- and live-load moment from this, we 
 have 265,518-140,667 = 124,851 inch Ibs. So we have 
 
 281,335 + 124,851 = 406,186 inch Ibs. 
 
 for the moment which the beam must be designed to resist. Dividing this 
 by 16,000 we obtain 25.4 for the section modulus, which calls for a 10" x 
 30# I, which will be used. 
 
 The floor framing on the first floor is designed to support a live load 
 of 100 Ibs. per sq. ft. of floor. The work of designing this framing is 
 the same as for the other floors. 
 
 290. Designing of the Ninth-Story Columns. As an example, 
 let us consider column C4. (See Fig. 448.) 
 
 For the maximum direct load on this column we have the following: 
 
 From roof (constant) =26,055 Ibs. 
 
 From beams 3 = 4,770x2 = 9,540 Ibs. 
 
 From beams 4 = 8,032x2 =16,064 Ibs. |*36,854 Ibs. 
 
 Live load = 50 x 15 x 15 = 11,250 Ibs. J 
 
 Weight of column and concrete 
 
 encasing it = 2,400 Ibs. 
 
 Total = 65,309 Ibs. 
 
 The maximum direct wind load equals 514 Ibs. (negligible) (see Fig. 
 443). The maximum shears on the column equal 2,626 Ibs. and 1,218 
 Ibs. (see Figs. 443 and 445). Then for the maximum moments on the 
 column at the brackets we have 2,626 x (6-2) 12 = 126,048 inch Ibs., and 
 1,218 x 48 = 58,464 inch Ibs. 
 
 Let us assume the following section: 
 
 4Ls 4x3x T 5 e = 8.36" 
 
 1 web 12 xA = 3.75 n " 
 
 Total = 12.11 n " 
 
 Max. 7 = 294.6 and Min. 7 = 30.2 
 Max. r= 4.91 and Min. r- 1.58 
 
 For the allowable unit stress for dead and live loads we have 16,000 - 
 70 x 144 -r 1.58 = 9,620 Ibs. Dividing this into the load we have 65,309 + 
 9,620 = 6.79 n// for the required area of cross-section. For the actual 
 direct stress we have 65,309 + 12.11 = 5,390 Ibs. per sq. in. 
 
 For the maximum unit stresses due to cross bending (from the wind 
 load) we have 
 
 / = 126,048 x 6 + 294.6 = 2,568 Ibs. 
 
 and 
 
 /' = 58,464 x 4.15 -r 30.2 = 8,038 Ibs. 
 
650 STRUCTURAL ENGINEERING 
 
 Adding the maximum of these to the direct stress we obtain 
 8,038 + 5,390 = 13,428 Ibs. 
 
 for the maximum combined unit stress on the column. For the allowable 
 unit stress for the combined dead, live, and wind loads we have 9,620 x 
 1.5 = 14,430 Ibs. From this it is seen that the above assumed section is 
 about the correct size and hence could be used. 
 
 The other columns in the ninth story are designed in the same 
 manner as shown for C4. Unless the loads are quite different the columns 
 throughout the story would be of the same section using the maximum 
 section required. 
 
 After the columns in the tenth and ninth stories are designed, as ex- 
 plained above, the columns in the eighth can be designed in the same 
 manner, and then the columns in the seventh story ; and so on down to the 
 basement columns, using in each case the loads found at the stories above. 
 As an example, let us consider column C4 in the first story. 
 
 For the direct load on the column (see loads on C4 given above), we 
 have the following: 
 
 From the roof = 26,055 Ibs. 
 
 From 10th, 9th ... 2nd floors = 3 6,854 x 9 =331,686 Ibs. 
 
 Weight of column above and concrete encasing it = 37,800 Ibs. 
 
 395,541 Ibs. 
 
 For the direct wind load on the column we will take 15,006 Ibs. (see 
 Fig. 443). 
 
 For the maximum moments due to wind loads we have 
 
 14,442 x (9 - 3) 12 = 1,039,824 inch Ibs. 
 and 
 
 6,700x72-482,400 inch Ibs. 
 
 Let us assume the section shown in Fig. 451 which 
 is composed of the following section: 
 
 2 [s 15"x40# = 23.52 n " 
 
 2 [s 12"x25# = 14.70 n " 
 Total = 38.22" 
 
 Max. I = 1,389, Min.' 7=711 
 Fi&. 451 Max. r- 6.02, Min. r= 4.3 
 
 For the allowable unit stress for live and dead load we have 
 16,000-70x216-^-4.3 = 12,470 Ibs. 
 
 Dividing this into the sum of the dead and live-load stresses we obtain 
 395,541 -r 12,470 = 31.7 n// for the required section for dead and live load 
 alone. 
 
 For the total direct stress dead, live, and wind loads combined we 
 have 
 
 395,541 + 15,006 = 410,547 Ibs. 
 
652 STRUCTURAL ENGINEERING 
 
 Dividing this by the area of the cross-section we obtain 410,547 -5- 38.22 = 
 10,740 Ibs. for the actual- direct stress" on the column due to dead, live, and 
 wind loads combined. For the unit stresses due to cross bending we have 
 
 /= 1,039,824X9.4-:- 1,389 = 7,035 Ibs. 
 and 
 
 /' = 482,400X7.5-J-71 1 = 5,090 Ibs. 
 
 Then for the maximum unit stress we have 
 
 10,740+7,035 = 17,775 Ibs. 
 
 For the allowable stress we have 12,470X1.5 = 18,705 Ibs. for combined 
 dead, live, and wind loads. From this it is seen that the assumed section is 
 about correct and hence could be used. 
 
 The other columns throughout the building are designed in the same 
 manner. 
 
 291. Calculation of Details. A fair idea of the details of the 
 building can be obtained from Fig. 452. As is seen, knee braces are used 
 throughout instead of the ordinary brackets. The connections of the 
 beams to the columns in all cases are sufficient to develop the bending 
 moment. 
 
 As an example, let us consider the details connecting beam 3 to column 
 4, as shown at (a), Fig. 452. The moment at 0, as given in Art. 289, is 129,635 
 inch Ibs. and the end shear is 6,645 Ibs., which we will assume applied at 0. 
 Then, for the moment about the center of the column (see Art. 66) , we have 
 129,635+6,645X18 = 249,245 inch Ibs. Then dividing this by the arm 
 b (15") we obtain 249,245 -r- 15 = 16,616 Ibs. for the stress in the knee brace, 
 provided there were but one brace, but, as there are two holding the beam, 
 stress in each knee brace will be 16,616 -r- 2 = 8,308 Ibs. As is seen, the stress 
 in the knee brace causes shear and tension on the rivets connecting it to the 
 beam and column. So, as a compromise, we will compute the rivets to 
 take the total stress in each knee brace. Then we have 8,308 -i- 4,420 
 = 1.88, say 2 -f" field rivets required to connect each knee brace to 
 the beam; also to the column. As is seen, 4 are used in order to obtain good 
 connections. We will next determine the number of rivets required to 
 connect the end of the beam to the column. Taking moments about we 
 have 
 
 129,635 
 
 from which we obtain 
 
 R= 7,200 Ibs. 
 
 for the force exerted along the column by the beam. So we have required 
 7,200 H- 4,420 = 1.6, say 2- f" field rivets. As is seen, 4 are used. All other 
 connections of the beams to the columns throughout the building can be cal- 
 culated in the same manner. 
 
 The other details throughout the building are readily calculated, provided 
 the work previously given in this book is understood. 
 
For one re// on/y. 
 
 s Fo//owed ' by4OOO Ibs.per ft.. 
 CLASS -_-4O 
 
 TABLE A. 
 
 /a*. 
 
 ff- 
 
 /03 
 
 J S- 
 
 232 
 
 %- 
 
 e 
 
 7734 6793 5857 4999 
 
 65_ 
 
 68X5888 50 
 
 30/6 
 
 22/6/5/6 
 
 208 
 
 '299107999091 7439 
 
 59/9 5O484242 
 
 35/4 
 
 2851 
 
 2421 
 
 78. 
 
 4JI8 339C 
 
 2/55 
 
 78 
 
 V936 8336 6836 5436 
 
 320 840 460 
 
 8728 
 
 7936. 65/8 
 
 28582247 
 
 /233 
 
 559 884 
 
 7668 
 
 692856034398 3268 
 
 2248 17O2 
 
 '22 
 
 8/8 
 
 480 
 
 300 
 
 282 
 
 529 
 
 6708 
 
 6018 4798 3678 2658 
 
 /738 
 
 '257841 
 
 503 
 
 230 
 
 /OO 
 
 30O 
 
 547 859 
 
 '249J704 
 
 5648 
 
 5208 4O8B 3068 2/48 
 
 '3289/2 
 
 56f 
 
 esa 
 
 80 
 
 289 
 
 436Z 
 
 m 
 
 i&4_ 
 
 T3JC 
 
 3496 
 
 30/b 
 
 ?480 3065 37Z&WS6 
 
 zest. 
 
 >475 3O6C.37/O 443t523i 
 
 345 
 
 65. 
 
 78 
 
 3834 4562 5366 &3 
 
 /80_ 
 
 303t 2876 38/6 
 
 33762087/44 
 
 030 
 
 559 884 
 
 2034 ?<& 3174 
 
 5094 
 
 78378&K> 
 
 ceo 
 
 287 529 854 
 
 /244 
 
 /624 
 
 2544 3564 4684 59O4 
 
 7789 884i c r)60 
 
 /OO 30O 
 
 547 859 
 
 7O4 
 
 2/34 
 
 3/54 
 
 7745494 66/4 
 
 78? ea^9 994', 
 
 80 
 
 30O 60O 
 Tad 
 
 9/2 
 
 2744 
 HM 
 
 386Q5084 6404 7824 
 SW 
 
 is? 
 
 m 
 
 Moments /n lOOOft Ibs. for One Truss 
 
4*11 U 
 
 snjnpou 
 uoi-p-ag 
 
 fifi-sixv 
 UOUOUAA 
 
 .o snioy 
 
 X3C-SIXV 
 
 J.U2LUO14 
 
 DI+JKJfJO 
 
 spuncy ui 
 joey JJ 
 
 4 s Si 
 
 C* 
 
 *** 
 
 *?fj*g 
 
 0000 
 
 dot QO oo 
 
 $- 
 
 oooooooooo 
 
 ^0 r- 06 VQ CD cri <n 
 
 &* 
 
 ^^3 
 
 t^^o 
 vsluiro crir^vsioco 
 
 qm 
 
 <ji<yocarrO 
 
 <j)0iy> 
 
 ji 
 
 
 OCOvB* 
 
 OCD 
 
 oooooo 
 
 co 
 
 *3 
 
 c^ 
 
 Jill 
 
 rfioo 
 in col 06 
 
 CCCQCO 
 
 ^-^^ 
 
 654 
 
B 
 
 -2 
 
 c 
 
 o 
 
 O 
 
 8 
 
 
 
 en 
 
 S6|OUJ 
 
 dug' 
 
 fifi-sixy 
 
 "XX-sixy 
 oi.|Jz>uj.p 
 
 rQ'oaSssojQ 
 
 
 Uw 
 
 ^wooO 
 
 (BCD 
 
 < cuoq 
 
 > tf) ^ 
 
 oo 
 
 <fr oo fi) o 
 
 in 
 
 oo 
 
 fi 
 
 <n^ 
 
 CO 
 
 655 
 
656 
 
<? 
 
 O 
 IL U- 
 
 "o 
 
 CO O- 
 UJ 
 
 p 
 
 Qu 
 
 w 
 
 8 jf%! 
 
 t4r 
 
 i * 
 
 snjnpoycjoipas 
 
 snippy 
 
 fifi-sixv/ 
 
 JOi i 
 
 N 
 
 jovpog Oicuj 
 
 FJflAPJQJO J4C 
 
 zis 
 
 o? ? 
 
 s y> o vs r- 
 
 
 _Nco^cn^in^vsi-oo<PO 
 
 657 
 
U) 
 
 ui 
 _J 
 CD 
 
 < 
 
 10 g- 
 U) ^ 
 
 Sifi 
 
 -5 S 
 
 *f 
 
 oo-stxy 
 jbsni 
 
 XX-eixy 
 
 uoi^aiAQjosnipoy 
 
 jo^u2>aJO W 
 
 .jovjoog cuojj. 
 ni IAOJO 40 ^40 
 
 6ao-| 
 
 6SOjgjooajy 
 
 spunoj ui 
 ' 
 
 ffS7>u^oii^ 
 
 
 2S 
 
 ^ 
 
 i* 
 
 cad 
 
 d6 
 
 666 
 
 66666 
 
 6 d o o 
 
 d6 
 
 vv vei 
 
 658 
 
of 
 
 DC 
 OL. 
 
 sn | n PH 
 
 oo-sixy 
 
 uouaiAcuo smpoy 
 
 XX jofifi-eixy 
 
 jOJ-aj.U9QlOa3UD.jSig 
 
 32 IS 
 
 00 
 
 n?? 
 
 l^ 
 
 ^|S|S?^R 
 606 
 
 SS3C58 
 
 ^S 
 
 ^! 
 
 66d6 
 
 vmt 
 
 ooo 1 
 
 tt 
 
 ll 
 
 j 
 
 <-*: 
 
 0) 
 
 if 
 
 Q_ 
 
 8- L 
 
 !?.- 
 
 S-S-s^S 
 t gt E 
 
 m 
 
 -fe o ? o 
 
 U'i s 
 
 o.> v J: 
 E oca ,/>' 
 
 N H 'I II 
 
 H 3^5: +. 
 
 M= O from wh'ich we obftJin 
 >U i= Scc-fi'ooModulus-(m 
 
 ss/Buy 
 
 XX uofifisixy 
 
 uoi{CuA)^o smpoy 
 
 3C3C ao'fifi Sixy 
 
 sixy 
 
 Vogutc 
 
 spuooj ui 
 -j.ooj. 
 
 &2U.OUJ UI 
 
 t- t* t 
 
 a! 
 
 S? 
 
 00 
 
 sassa^ 
 
 y>v9vj)y>U)i 
 
 *;** 
 
 SS?P 
 
 oStrico 
 
 cSS 
 
 odd 
 
 ?;*** J**f***3?|*?* 
 
 t*t* 
 
 do 
 
 sa 
 
 i 
 
 ^iss^s 
 
 as? 
 
 222S 
 
 ^SSS^c^^ 
 
 SSSi 
 
 riri^c' 
 
 2g^5S 
 
Areas and Weights of Bars and P/afes 
 
 A = area of cross -sect/on in sq. inches 
 T Qul6- / W= weight in pounds per /inea/ff. at 490 pounds per cuff: 
 
 WWh 
 
 /n 
 Mes 
 
 
 Thickness 
 
 Extreme 
 
 Length 
 /n 
 Feef 
 
 
 
 / " 
 
 2 
 
 " 
 /6 
 
 i" 
 
 %' 
 
 r 
 
 /f 
 
 ^ 
 
 w 
 
 A 
 
 W 
 
 A 
 
 W 
 
 A 
 
 W 
 
 A 
 
 W 
 
 A 
 
 W 
 
 / 
 
 O.2& 
 
 O.S5 
 
 O.3/3 
 
 W6 
 
 0.375 
 
 /.28 
 
 0.438 
 
 /49 
 
 V5W 
 
 170 
 
 C.563 
 
 /.9? 
 
 70 
 
 1 
 
 0.375 
 
 /.28 
 
 0469 
 
 /.9 
 
 0563 
 
 /92 
 
 0.656 
 
 Z?3 
 
 O.750 
 
 2.55 
 
 0.844 
 
 2#7 
 
 it 
 
 2 
 
 O.438 
 
 /.49 
 
 054'/ 
 
 /.86 
 
 0656 
 
 223 
 
 O766 
 
 260 
 
 3875 
 
 298 
 
 0.934 
 
 335 
 
 if 
 
 z 
 
 0.500 
 
 /.70 
 
 0.625 
 
 2/2 
 
 0750 
 
 255 
 
 0875 
 
 293 
 
 /.OO 
 
 40 
 
 //3 
 
 383 
 
 /t 
 
 % 
 
 0563 
 
 /.9/ 
 
 0703 
 
 2.39 
 
 0.&44 
 
 237 
 
 0.934 
 
 3.35 
 
 /./3 
 
 3S3 
 
 /27 
 
 4.30 
 
 it 
 
 ?4 
 
 0625 
 
 2./2 
 
 O.7SI 
 
 Z65 
 
 0.933 
 
 J./9 
 
 /.09 
 
 372 
 
 /.25 
 
 4.25 
 
 /4/ 
 
 47<$ 
 
 ///, 
 
 3 
 
 0.750 
 
 Z.55 
 
 0.938 
 
 3./9 
 
 //3 
 
 333 
 
 /.3I 
 
 4.46 
 
 /.50 
 
 J/O 
 
 /.69 
 
 574 
 
 n 
 
 3 
 
 O.87S 
 
 2.98 
 
 /.09 
 
 372 
 
 /3/ 
 
 447 
 
 /53 
 
 5.?0 
 
 /.75 
 
 5.95 
 
 /.97 
 
 670 
 
 n 
 
 4 
 
 /OO 
 
 3.40 
 
 /?5 
 
 42S 
 
 /.50 
 
 S/0 
 
 /75 
 
 5.9S 
 
 ZOO 
 
 6.80 
 
 ?.?5 
 
 765 
 
 n 
 
 5 
 
 /25 
 
 4.25 
 
 /.56 
 
 S.3/ 
 
 f.M 
 
 6.30 
 
 2/9 
 
 744 
 
 250 
 
 8.5O 
 
 ZS/ 
 
 9.57 
 
 it 
 
 6 
 
 /50 
 
 S./O 
 
 /.8ff 
 
 6.38 
 
 Z.Z5 
 
 7.65" 
 
 263 
 
 8.93 
 
 300 
 
 /0.20 
 
 338 
 
 //4 
 
 If 
 
 7 
 
 /75 
 
 95 
 
 2/9 
 
 7.44 
 
 263 
 
 8.93 
 
 306 
 
 /O4/ 
 
 3.50 
 
 //.9O 
 
 3.94 
 
 /3.39 
 
 ti 
 
 8 
 
 2OO 
 
 6O 
 
 25O 
 
 3.5-0 
 
 3.OO 
 
 /020 
 
 350 
 
 //.90 
 
 4OO 
 
 /30 
 
 4.50 
 
 /53O 
 
 n 
 
 9 
 
 225 
 
 7.6S 
 
 ?.0/ 
 
 9.56 
 
 338 
 
 i/48 
 
 394 
 
 /3.40 
 
 450 
 
 /5.3O 
 
 S06 
 
 /722 
 
 n 
 
 /O 
 
 250 
 
 8.50 
 
 3./J 
 
 /a 62 
 
 375 
 
 /?7 
 
 4.38 
 
 /4S8 
 
 00 
 
 /7OO 
 
 563 
 
 /9/4 
 
 n 
 
 // 
 
 Z7S 
 
 9.14 
 
 344 
 
 //.6 
 
 4/3 
 
 /4O3 
 
 4.8/ 
 
 /6.36 
 
 5S0 
 
 /87O 
 
 6./9 
 
 2/02 
 
 ii> 
 
 
 
 300 
 
 /0.20 
 
 3.75 
 
 /2.7S 
 
 4.50 
 
 /5.30 
 
 5.25 
 
 /7.05 
 
 600 
 
 ZO40 
 
 6.75 
 
 2295 
 
 M 
 
 /s 
 
 325 
 
 //.06 
 
 4.06 
 
 /3.8/ 
 
 438 
 
 /6.58 
 
 69 
 
 /934 
 
 650 
 
 22./0 
 
 7.3/ 
 
 24J6 
 
 n 
 
 /4 
 
 3.50 
 
 //.90 
 
 4.38 
 
 /4S3 
 
 J.Z5 
 
 /7<36 
 
 6./3 
 
 20.82 
 
 700 
 
 2380 
 
 7.88 
 
 26.78 
 
 H 
 
 /5 
 
 37S 
 
 f?75 
 
 4.69 
 
 /S.94 
 
 563 
 
 &/<4 
 
 6.56 
 
 22.32 
 
 7.50 
 
 25.50 
 
 844 
 
 2870 
 
 it 
 
 /6 
 
 4.00 
 
 /3.60 
 
 S.oo 
 
 /7.0O 
 
 6.00 
 
 20.40 
 
 700 
 
 23.80 
 
 .00 
 
 2770 
 
 9.00 
 
 30.60 
 
 IV 
 
 /7 
 
 425 
 
 /444 
 
 3/ 
 
 /#06 
 
 6.38 
 
 2/6S 
 
 744 
 
 25.28 
 
 850 
 
 28.89 
 
 9.56 
 
 3252 
 
 II' 
 
 Iff 
 
 450 
 
 /5SO 
 
 5.63 
 
 /9./2 
 
 6.7S 
 
 2296 
 
 7S8 
 
 2679 
 
 900 
 
 3O.60 
 
 /0./3 
 
 34.44 
 
 n 
 
 20 
 
 5.OO 
 
 /700 
 
 6.25 
 
 2/24 
 
 750 
 
 25.50 
 
 &7S 
 
 2975 
 
 /000 
 
 3400 
 
 //.25 
 
 38.27 
 
 /' 
 
 22 
 
 S.50 
 
 /8.69 
 
 6.68 
 
 2336 
 
 8.2S 
 
 2806 
 
 9.63, 
 
 3272 
 
 //OO 
 
 3740 
 
 /238 
 
 42O4 
 
 ii 
 
 24 
 
 6.OO 
 
 20.40 
 
 7.50 
 
 25.52 
 
 9.00 
 
 3O.60 
 
 /0.50 
 
 3572 
 
 /200 
 
 40.80 
 
 /3.50 
 
 45.92 
 
 n 
 
 26 
 
 6.50 
 
 2Z./2 
 
 S./3 
 
 27.62 
 
 9.75 
 
 33/6 
 
 //.3 
 
 3868 
 
 /300 
 
 4420 
 
 M.63 
 
 4973 
 
 n 
 
 28 
 
 7OO 
 
 Z3JO 
 
 6.75 
 
 2976 
 
 /0.50 
 
 35.72 
 
 /2.2S 
 
 4/65 
 
 /4.0C 
 
 47.60 
 
 /5.7S 
 
 5356 
 
 // 
 
 30 
 
 750 
 
 25.50 
 
 9.36 
 
 3/.68 
 
 //.2~ 
 
 38.78 
 
 /3/J 
 
 4464 
 
 /5.00 
 
 5/.00 
 
 /68 
 
 3740 
 
 n 
 
 32 
 
 BOO 
 
 27.20 
 
 /O.OO 
 
 34.00 
 
 /?00 
 
 40& 
 
 /4.OO 
 
 47.60 
 
 /6.OO 
 
 54.40 
 
 /&O0 
 
 6122 
 
 // 
 
 34 
 
 8.50 
 
 28.85 
 
 /0.63 
 
 36./2 
 
 /275 
 
 43.36 
 
 /4.S8 
 
 505', 
 
 /7.OO 
 
 7.7<3 
 
 /9I3 
 
 6504 
 
 // 
 
 36 
 
 900 
 
 30.59 
 
 //.2 
 
 38.24 
 
 /350 
 
 45.92 
 
 /575 
 
 S3.58 
 
 /800 
 
 6/20 
 
 2025 
 
 6868 
 
 ii 
 
 38 
 
 9.5O 
 
 32.32 
 
 //& 
 
 40.39 
 
 /4.2 
 
 4S4S 
 
 /6.63 
 
 56.56 
 
 /900 
 
 64.62 
 
 2/38 
 
 7268 
 
 if 
 
 40 
 
 /0.00 
 
 34.00 
 
 {250 
 
 4?4S 
 
 /5.00 
 
 5/.OO 
 
 /7.5O 
 
 5950 
 
 ?0.0 
 
 6000 
 
 2250 
 
 7654 
 
 n 
 
 660 
 
Areas and Weights of Bors and Plates 
 
 A = area of cross- sect/on in sq. inches 
 IQ D I6~S W* weight in pounds per //nee/ ft. at 4 9O pounds per cu. ft 
 
 Width 
 in 
 Inches 
 
 Thickness 
 
 Exirfme 
 
 Length 
 in 
 reef. 
 
 
 f" 
 
 2L" 
 /6 
 
 i" 
 
 J3_" 
 16 
 
 7 ff 
 
 J5_" 
 16 
 
 A 
 
 W 
 
 A 
 
 W 
 
 A 
 
 W 
 
 A 
 
 W 
 
 A 
 
 W 
 
 A 
 
 W 
 
 / 
 
 0.625 
 
 2.13 
 
 O.G88 
 
 2.34 
 
 0.75 
 
 2.55 
 
 O.QI3 
 
 2.76 
 
 0.88 
 
 2.98 
 
 0.94 
 
 3./9 
 
 7O 
 
 2 
 
 0.93$ 
 
 3.19 
 
 1.021 
 
 3.51 
 
 /J25 
 
 3.83 
 
 f.2/9 
 
 4./4 
 
 /.3/ 
 
 4-.4t> 
 
 /.<?-/ 
 
 4.7ft 
 
 // 
 
 2S 
 
 1.094 
 
 3.72 
 
 /.203 
 
 4.09 
 
 /.3/3 
 
 4-.46 
 
 /.4-2Z 
 
 4.83 
 
 /.53 
 
 S.2/ 
 
 /.64 
 
 5.58 
 
 // 
 
 2 
 
 /.Z5 
 
 4,25 
 
 1.375 
 
 4.68 
 
 50 
 
 5./0 
 
 /.625 
 
 5.53 
 
 1.75 
 
 -.35 
 
 /.88 
 
 6-38 
 
 // 
 
 H 
 
 J.40& 
 
 4 t 78 
 
 1.541 
 
 5".2 
 
 /.68ft 
 
 5.74 
 
 /.828 
 
 6.72. 
 
 /.97 
 
 6.69 
 
 2.// 
 
 7-/7 
 
 f/ 
 
 z-k 
 
 1.563 
 
 5.31 
 
 1.7/9 
 
 5.84 
 
 /.87S 
 
 6,38 
 
 2.03' 
 
 6.5/ 
 
 2./9 
 
 7.4-4 
 
 2.34 
 
 7.97 
 
 f/ 
 
 3 
 
 1.875 
 
 6.38 
 
 2.063 
 
 7.O/ 
 
 2.25 
 
 7.65 
 
 2.4-3 
 
 829 
 
 2.63 
 
 ft.93 
 
 2.81 
 
 9.56 
 
 // 
 
 3^ 
 
 2.188 
 
 7.44 
 
 2-4U 
 
 8>J8 
 
 2.615 
 
 833 
 
 2-B44 
 
 9.67 
 
 3.O6 
 
 fO.4t 
 
 3.28 
 
 S/./6 
 
 // 
 
 4 
 
 2.5 
 
 $.0 
 
 2.75 
 
 9.35 
 
 3.00 
 
 /a 20 
 
 3.2.5 
 
 I/.O5 
 
 3.5-0 
 
 //.96 
 
 3.YS 
 
 /2.7S 
 
 n 
 
 
 
 3.125 
 
 10.63 
 
 3438 
 
 I/.69 
 
 3.7S 
 
 12.75 
 
 4.063 
 
 13.81 
 
 4r38 
 
 /4.88 
 
 ^.69 
 
 /S.94 
 
 if 
 
 6 
 
 3.75 
 
 12-75 
 
 4.125 
 
 /4.03 
 
 4-.5O 
 
 15.30 
 
 4.875 
 
 16.58 
 
 S.2S 
 
 /7.8S 
 
 S.63 
 
 /9</3 
 
 // 
 
 7 
 
 4.375 
 
 14.88 
 
 4.8 1 3 
 
 16-36 
 
 5.250 
 
 17.85 
 
 5.688 
 
 1934 
 
 6./3 
 
 20.83 
 
 6.^6 
 
 22.31 
 
 // 
 
 g 
 
 5.00 
 
 I7.O 
 
 5.50 
 
 18.70 
 
 6.00 
 
 20.40 
 
 6.50 
 
 22.10 
 
 7.00 
 
 2330 
 
 7.W 
 
 25. 5 
 
 n 
 
 3 
 
 5.62S 
 
 19.13 
 
 6.188 
 
 21.04 
 
 6. 75 
 
 2235 
 
 7.3/3 
 
 2436 
 
 7.88 
 
 26.78 
 
 8.4+ 
 
 28.69 
 
 n 
 
 /O 
 
 6.25 
 
 21.25 
 
 6S75 
 
 23.38 
 
 7.50 
 
 25.50 
 
 S-/25 
 
 27.G3 
 
 8.7S 
 
 29.75 
 
 9.38 
 
 3/.S8 
 
 // 
 
 // 
 
 6JB7S 
 
 23.28 
 
 7563 
 
 25.71 
 
 8.25 
 
 28.05 
 
 8338 
 
 30.39 
 
 9.63 
 
 32.73 
 
 /0.3/ 
 
 35.06 
 
 tt 
 
 /'Z 
 
 7.50 
 
 25.50 
 
 8.25 
 
 28.05 
 
 9.00 
 
 30.60 
 
 9.7S 
 
 33.15 
 
 'O.50 
 
 35.70 
 
 //25 
 
 38.25 
 
 fi 
 
 /3 
 
 8J3 
 
 27.63 
 
 8.94 
 
 30.4 
 
 9.75 
 
 33.2 
 
 10.56 
 
 35.3 
 
 /I38 
 
 38.7 
 
 /2./9 
 
 41.4 
 
 n 
 
 t 
 
 8-75 
 
 29.75 
 
 9.63 
 
 32.7 
 
 IO.SO 
 
 3^.7 
 
 11.38 
 
 38.7 
 
 12.25 
 
 41.7 
 
 /3J3 
 
 44.6 
 
 // 
 
 IS 
 
 9.3$ 
 
 31.88 
 
 10.31 
 
 35. / 
 
 II.2S 
 
 38.3 
 
 /Z./9 
 
 41.4 
 
 /3./3 
 
 44.6 
 
 /4.06 
 
 47.8 
 
 // 
 
 7? 
 
 10.00 
 
 34.0 
 
 11.00 
 
 37.4 
 
 12.00 
 
 4O.$ 
 
 /3.00 
 
 44.2 
 
 /4.00 
 
 47.6 
 
 /5.0-0 
 
 51.0 
 
 // 
 
 17 
 
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 11.69 
 
 39.7 
 
 1275 
 
 43.4 
 
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 4-7. 
 
 '4.88 
 
 SO. 6 
 
 /S.94 
 
 S4-.2 
 
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 /8 
 
 /f.2S 
 
 28-M 
 
 /2.3$ 
 
 4-2.1 
 
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 45.3 
 
 /4.63 
 
 49.7 
 
 /5.75 
 
 S3. 6 
 
 /6.8B 
 
 57. + 
 
 // 
 
 20 
 
 J2JO 
 
 41.50 
 
 13.75 
 
 4-6.8 
 
 /s.oo 
 
 s/.o 
 
 I6.2S 
 
 55.3 
 
 '7.50 
 
 59.5 
 
 /8.7S 
 
 63.8 
 
 // 
 
 22 
 
 (3.75 
 
 46-75 
 
 /5.I3 
 
 S/.4 
 
 16.50 
 
 5-6,1 
 
 '7.8ft 
 
 6O.8 
 
 /9.2S 
 
 65. S 
 
 20.63 
 
 70./ 
 
 // * 
 
 24- 
 
 15.00 
 
 51.0 
 
 16.50 
 
 S6J 
 
 18.00 
 
 6/,2 
 
 19.50 
 
 66.3 
 
 21.00 
 
 7/.4- 
 
 22.50 
 
 76.S 
 
 // 
 
 -26 
 
 16.25 
 
 5 $.25 
 
 17.8$ 
 
 60.8 
 
 /9.50 
 
 66.3 
 
 Z/./3 
 
 7/.8 
 
 2?. 7 5 
 
 77.^- 
 
 24.38 
 
 ft 2. 9 
 
 if 
 
 28 
 
 /7.50 
 
 53.50 
 
 /9.25 
 
 65.5 
 
 21.00 
 
 7/.4 
 
 22.75 
 
 77.4 
 
 24.50 
 
 83.3 
 
 26.25 
 
 89.3 
 
 if 
 
 30 
 
 18.75 
 
 63.75 
 
 20.63 
 
 7OJ 
 
 22.50 
 
 76. S 
 
 2438 
 
 82-9 
 
 26.25 
 
 89.3 
 
 2$./3 
 
 95.6 
 
 if 
 
 3Z 
 
 20.00 
 
 6$.0 
 
 22.00 
 
 74.8 
 
 24.00 
 
 81.6 
 
 26.00 
 
 88.4 
 
 28.00 
 
 9S.2 
 
 30.00 
 
 /oz.o 
 
 n 
 
 34 
 
 21.25 
 
 7.1.25 
 
 23.38 
 
 79.5 
 
 25.50 
 
 86.7 
 
 27.63 
 
 93.9 
 
 29.75 
 
 IO/.2 
 
 31.88 
 
 /08.4- 
 
 ti 
 
 36 
 
 22.50 
 
 76.50 
 
 24.75 
 
 84.2 
 
 27.00 
 
 31.8 
 
 29.25 
 
 99.5 
 
 3 /.SO 
 
 /O7J 
 
 33. 7S 
 
 //4.Q 
 
 it 
 
 38 
 
 23.75" 
 
 30.75 
 
 26.13 
 
 88.8 
 
 28.50 
 
 963 
 
 30.83 
 
 105.0 
 
 33.25 
 
 II 3.1 
 
 3S.63 
 
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 25.00 
 
 85.00 
 
 27.50 
 
 93.5 
 
 30.00 
 
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 32.50 
 
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 119.0 
 
 37.SO 
 
 /27.S 
 
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 661 
 
Areas; Weights and Max. Lengths ofW/cfeff&fes, 
 
 A = ore.a of cross secf/or? /'/? sg./h3. 
 /y= we/ghf/r? pounds per /5f 
 iable-y L = moximum/eng/-/? fr? /7. ohfoinoh/e. 
 
 mm 
 
 m 
 
 Inches 
 
 Thickness 
 
 **_ 
 
 3 /8" 
 
 7 //6" 
 
 z.// 
 
 2. 
 
 A 
 
 W 
 
 L 
 
 A 
 
 W 
 
 L 
 
 A 
 
 tr 
 
 L 
 
 A 
 
 W 
 
 L 
 
 42 
 
 /3J3 
 
 44.6? 
 
 37.5 
 
 /S.7S 
 
 53.55 
 
 41*6 
 
 /8.38 
 
 62.48 
 
 45.8 
 
 21.00 
 
 7/.4O 
 
 45.8 
 
 46 
 
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 48. 88 
 
 37.5 
 
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 58-65 
 
 41-6 
 
 20. t '3 
 
 68.43 
 
 45.8 
 
 23.00 
 
 78.20 
 
 4^.8 
 
 48 
 
 15.00 
 
 5/.00 
 
 4-1. & 
 
 /8.00 
 
 61. 2O 
 
 45.0 
 
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 7/.40 
 
 4-6.6 
 
 24.00 
 
 8/.60 
 
 46.6 
 
 SO 
 
 15.63 
 
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 37.5 
 
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 63.80 
 
 45.0 
 
 21.88 
 
 74.W 
 
 46.6 
 
 25.00 
 
 8S.OO 
 
 46.6 
 
 52 
 
 /6.2S 
 
 5536 
 
 37.5 
 
 /9/SO 
 
 66.30 
 
 45.0 
 
 22.75 
 
 77.40 
 
 46.6 
 
 26.00 
 
 88,40 
 
 46.6 
 
 56 
 
 17-50 
 
 53.50 
 
 37.5 
 
 21.00 
 
 7/.4O 
 
 45.0 
 
 24.50 
 
 83.30 
 
 46- 
 
 2$. 00 
 
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 46.6 
 
 60 
 
 18.75 
 
 63.80 
 
 37.5 
 
 22.50 
 
 76.50 
 
 45. 
 
 26.25 
 
 89.30 
 
 46,6 
 
 30.00 
 
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 46.6 
 
 66 
 
 20.63 
 
 70.10 
 
 33.3 
 
 24.75 
 
 84,20 
 
 4-1. & 
 
 28.88 
 
 98.20 
 
 46.6 
 
 33.00 
 
 112.2. 
 
 46.6 
 
 68 
 
 21.25 
 
 7230 
 
 33.3 
 
 25.50 
 
 86.70 
 
 4/>6 
 
 29.75 
 
 /o/.z 
 
 46.6 
 
 34.00 
 
 //S.6 
 
 46.6 
 
 72 
 
 22.50 
 
 76-50 
 
 3/6 
 
 27.00 
 
 9/.80 
 
 32.5 
 
 31.50 
 
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 43.3 
 
 36.00 
 
 /22-4 
 
 43.6 
 
 76 
 
 23.15 
 
 80.80 
 
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 28.50 
 
 96.^0 
 
 38.3 
 
 33.7.5 
 
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 39.6 
 
 38-00 
 
 /29.2 
 
 43.6 
 
 80 
 
 25.00 
 
 85.00 
 
 3/.6 
 
 30.OO 
 
 /O2.0 
 
 38.3 
 
 35.00 
 
 S/9.0 
 
 39.6 
 
 40.00 
 
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 4/.6 
 
 84 
 
 26.25 
 
 89.30 
 
 27. / 
 
 31.50 
 
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 36.6 
 
 36.75 
 
 /2S.O 
 
 38.7 
 
 42,00 
 
 /4Z.8 
 
 39.6 
 
 SO 
 
 28.13 
 
 9S.60 
 
 25,0 
 
 33.75 
 
 //4.8 
 
 33.3 
 
 39.38 
 
 /3S.9 
 
 36.6 
 
 45.00 
 
 /53.0 
 
 37-5 
 
 96 
 
 30.00 
 
 /O2.0 
 
 22.3 
 
 36.00 
 
 /22.4 
 
 29. / 
 
 4-2-00 
 
 /42.8 
 
 3/.6 
 
 48.00 
 
 /63.Z 
 
 33.3 
 
 JOO 
 
 31.25 
 
 106.3 
 
 Z/.6 
 
 37.50 
 
 /27.S 
 
 25.8 
 
 43.7S 
 
 /48.8 
 
 29. / 
 
 50.00 
 
 /70.O 
 
 30.0 
 
 662 
 
Tab/e 70 
 
 ffad/^ of Gi/rahbn ofCo/umn Sechons. 
 
 T b. 
 
 
 
 
 y\ 
 
 
 ^^ 
 
 
 =.3G/7 
 
 X S X. 
 
 y\ 
 
 - ^* 
 
 \--3c 
 
 /? 
 
 I 3C 
 
 L 
 
 
 
 oc 
 
 5_.^-.;'_^ 
 
 J 
 
 =.3eh r y =-.55b 
 
 
 jj . y- 
 
 a: 
 
 663 
 
Tbb/e 11 
 
 Shearing and Bearing l/a/ues offffvefe 
 
 D/amefer of 
 R/vefin tnches 
 
 Area 
 of 
 fti'vet 
 
 Sing/e 
 Shear 
 /2OOO* 
 oersg.in.. 
 
 Bearing l/a/ues for Dtfferenf Thicknesses ofP/o/es @24ooo/bs. 
 
 frac/io/i 
 
 Dec/ma/ 
 
 %' 
 
 % 
 
 %" 
 
 7,6 
 
 ^" 
 
 */ 
 
 1C 
 
 /// " 
 //. 
 
 3 /4" 
 
 /3/ " 
 -V6 
 
 7 /8" 
 
 3 /8 
 
 0.375 
 
 0. 1104- 
 
 /3ZO 
 
 225O 
 
 2820 
 
 338O 
 
 
 
 
 
 
 
 
 
 '/Z 
 
 0.500 
 
 O.IK3 
 
 23&0 
 
 3000 
 
 375O 
 
 4500 
 
 5Z50 
 
 6000 
 
 
 
 
 
 
 
 S /B 
 
 0.625 
 
 0.306B 
 
 3680 
 
 37SO 
 
 4690 
 
 563O 
 
 6560 
 
 7SOO 
 
 8440 
 
 
 
 
 
 
 %t 
 
 O.750 
 
 0.44-18 
 
 S300 
 
 4SOO 
 
 S&30 
 
 675-0 
 
 7880 
 
 9000 
 
 /0/3O 
 
 //2SO 
 
 
 
 
 
 7 /8 
 
 0.875 
 
 0.60/3 
 
 7Z20 
 
 5250 
 
 &5O 
 
 7880 
 
 9/9O 
 
 /O5OO 
 
 //820 
 
 /3/20 
 
 /4430 
 
 /f7SO 
 
 /70SO 
 
 
 / 
 
 t.OOO 
 
 OJ8S+ 
 
 9430 
 
 60OO 
 
 7SOO 
 
 SOOO 
 
 /0S00 
 
 /ZOOO 
 
 /3WO 
 
 /5OOO 
 
 /650O 
 
 /SOOO 
 
 /9S00 
 
 z/ooo 
 
 
 Diome+ej- of 
 Rivefin inche.* 
 
 Area 
 of 
 Kivet 
 
 Single 
 She?r 
 /OOOO* 
 perfgjn. 
 
 Bearing Values forDi'fferenJ-rt)icf(nessesofF'Jofes@20OOO/bs. 
 
 Frochon 
 
 Decimal! 
 
 i 
 
 */,6 
 
 % 
 
 7 /,6 
 
 ^ 
 
 9 /K 
 
 % 
 
 % 
 
 % 
 
 '& 
 
 % 
 
 ft 
 
 0.375 
 
 0. H04 
 
 IIOO 
 
 1880 
 
 2340 
 
 28/0 
 
 
 
 
 
 
 
 
 
 '/Z 
 
 O.SOO 
 
 0. 063 
 
 I960 
 
 2.500 
 
 3130 
 
 37 SO 
 
 4360 
 
 000 
 
 
 
 
 
 
 
 */8 
 
 0.42S 
 
 0.3068 
 
 3O70 
 
 3I3O 
 
 39/0 
 
 4690 
 
 S470 
 
 6250 
 
 7O3O 
 
 78/O 
 
 
 
 
 
 3 /4 
 
 O.7SO 
 
 0.4418 
 
 -4420 
 
 3150 
 
 4690 
 
 5630 
 
 6560 
 
 70O 
 
 84*0 
 
 9380 
 
 /O3/0 
 
 //2*O 
 
 
 
 7 /8 
 
 0.275 
 
 0.60/3 
 
 6010 
 
 4380 
 
 547O 
 
 S7O 
 
 7660 
 
 8750 
 
 9840 
 
 /O940 
 
 W30 
 
 /3/30 
 
 /4220 
 
 
 / 
 
 f. OOO 
 
 0. 73fr 
 
 7850 
 
 5000 
 
 6250 
 
 750O 
 
 Q7SO 
 
 IOOOO 
 
 i/ZSO 
 
 /2S00 
 
 /37SO 
 
 /SOOO 
 
 /6250 
 
 I7SOO 
 
 664 
 
Allowab/e Bearing ar?d ' A?o/77e/7fs 0/7 Pins 
 Tab/e /2 
 
 Diameter 
 
 of 
 Pin 
 
 Bearing 
 of 140 00^ on 
 /"ofmeM 
 
 Allowob/e 
 
 Moment^ 
 4+25000* 
 
 Diomefer 
 
 of 
 
 Pin 
 
 Bearing 
 of240Q0*W 
 
 /"ofrtefa/. 
 
 Allowab/e 
 Moment 
 
 atzsooo* 
 
 2" 
 
 4SOOO& 
 
 I9600" 
 
 7" 
 
 I&8000* 
 
 84/800"* 
 
 3 
 
 54-000 
 
 Z8000 
 
 7% 
 
 174000 
 
 935300 
 
 2^ 
 
 60000 
 
 38300 
 
 7^z 
 
 180000 
 
 1035400 
 
 2^ 
 
 66000 
 
 51000 
 
 7 3 /+ 
 
 186000 
 
 1142500 
 
 3 
 
 72000 
 
 66300 
 
 8 
 
 192000 
 
 1256600 
 
 3% 
 
 78000 
 
 84300 
 
 Q'/4- 
 
 198000 
 
 1378200 
 
 3'/z 
 
 84000 
 
 105200 
 
 Q'/z 
 
 204000 
 
 / 5 07 300 
 
 3% 
 
 30000 
 
 / 29 400 
 
 Q 3 /4- 
 
 2/0000 
 
 /644200 
 
 4- 
 
 96000 
 
 I5UOO 
 
 9 
 
 2/6000 
 
 1783200 
 
 4& 
 
 102000 
 
 I8S400 
 
 9'/4- 
 
 222000 
 
 1942500 
 
 4& 
 
 /oeooo 
 
 Z23100 
 
 m 
 
 228000 
 
 2/04-300 
 
 4 3 /4- 
 
 1 14000 
 
 Z63000 
 
 9^ 
 
 234000 
 
 Z274900 
 
 S 
 
 120000 
 
 306800 
 
 /O 
 
 240000 
 
 24-54400 
 
 H 
 
 126000 
 
 355200 
 
 (Oji 
 
 246000 
 
 2643/00 
 
 s& 
 
 I3ZOOO 
 
 4-08300 
 
 /0/2 
 
 252000 
 
 284J200 
 
 s& 
 
 138000 
 
 4-66600 
 
 /0% 
 
 258000 
 
 3043/00 
 
 e 
 
 14-4000 
 
 S30/00 
 
 // 
 
 264000 
 
 3266800 
 
 6%. 
 
 160000 
 
 599200 
 
 //x^ 
 
 270OOO 
 
 34-946OO 
 
 6'/z 
 
 /56000 
 
 674000 
 
 //J/z 
 
 276000 
 
 3732800 
 
 6% 
 
 /6ZOOO 
 
 7S4800 
 
 i/a 
 
 28200 O 
 
 398/600 
 
 665 
 
INDEX 
 
 PAGES 
 
 Applied forces, definition of 17 
 
 Beams, theoretical treatment of 51 to 105 
 
 Beams, shearing stresses on 51 to 53 
 
 Beams, bending stresses on 53 to 56 
 
 Beams, reaction on 56 
 
 Beams, deflection of 72 to 106 
 
 Beams, cantilever 75 to 80 
 
 Beams, simple 80 to 84 
 
 Beams, fixed at one end and supported at other 86 to 90 
 
 Beams, fixed 90 to 94 
 
 Beams, overhanging 84 to 86 
 
 Beams, continuous 94 to 105 
 
 Beam bridges, railroad 198 to 214 
 
 Beam bridges, highways 540 to 546 
 
 Bending stress, increment of 64 to 65 
 
 Bending moment, maximum on simple beams 128 to 132 
 
 Bending moment, maximum on simple trusses 138 to 139 
 
 Bridges on curves 571 to 575 
 
 Buildings, mill 583 to 624 
 
 Buildings, office 624 to 652 
 
 Camber of trusses 518 to 532 
 
 Cantilever beams 75 to 80 
 
 Castings 2 
 
 Centrifugal force, stresses due to 571 to 575 
 
 Center of gravity, determination of 44 to 47 
 
 Columns, theoretical treatment of 106 to 114 
 
 Column, formulas 106 to 109 
 
 Continuous beams, theoretical treatment 94 to 105 
 
 Conventional signs 11 
 
 Cover plates, theoretical length 181 to 183 
 
 Dead load, railroad bridges 197 to 198 
 
 Dead load, highway bridges 534 
 
 Deflection of beams 72 to 106 
 
 Deflection of trusses 518 to 532 
 
 Design of I-beams and plate girders, theoretical 172 to 194 
 
 Design of railroad bridges : 
 
 Beam bridges 198 to 214 
 
 Deck plate girders 215 to 259 
 
 Through plate girders 259 to 295 
 
 Viaducts 295 to 324 
 
 Truss bridges 324 to 532 
 
 Design of highway bridges : 
 
 Beam bridges 540 to 546 
 
 Pony trusses 546 to 552 
 
 Through Pratt trusses 554 to 564 
 
 Curved chord through trusses 564 to 589 
 
 Design of a 10 ft. beam bridge, railroad 199 to 201 
 
 Design of a 15 ft. beam bridge, railroad 201 to 214 
 
 Design of a 50 ft. deck plate girder, railroad 216 to 244 
 
 Design of a 75 ft. deck plate girder, railroad 244 to 248 
 
 Design of a 90 ft. deck plate girder, railroad 249 to 253 
 
 Design of a 60 ft. through plate girder, railroad 260 to 286 
 
 Design of a 150 ft. Pratt truss bridge, railroad 326 to 413 
 
 667 
 
668 INDEX 
 
 PAGES 
 
 Design of a 225 ft. curved chord truss bridge, railroad 426 to 473 
 
 Design of buildings 583 to 652 
 
 Distortion, definition of 19 
 
 Distortion, determination of 20 to 22 
 
 Drawing instruments 9 
 
 Drawings, size of 11 
 
 Drawings, shop 11 
 
 Drawing Room Exercise No. 1 13 
 
 Drawing Room Exercise No. 2 214 
 
 Drawing Room Exercise No. 3 259 
 
 Drawing Room Exercise No. 4 259 
 
 Drawing Room Exercise No. 5 295 
 
 Drawing Room Exercise No. 6 324 
 
 Drawing Room Exercise No. 7 413 
 
 Drawing Room Exercise No. 8 415 
 
 Drawing Room Exercise No. 9 582 
 
 Economic depth of plate girders 180 to 181 
 
 Economic depth of trusses 575 
 
 Economic length of spans ' 576 
 
 Eccentrically loaded columns 112 to 114 
 
 Elasticity, definition of 19 
 
 Elastic limit 19 
 
 Equilibrium, definition of 31 to 35 
 
 Equilibrium of couples 35 
 
 Equilibrium polygon 38 
 
 Equilibrium polygon through two and three points 159 to 161 
 
 Equation of the elastic curve 72 to 75 
 
 Fabrication 7 
 
 Fixed beams 90 to 94 
 
 Flange stress in plate girders 178 to 180 
 
 Flange rivets, spacing 185 to 187 
 
 Flange splice 254 
 
 Flange increment 184 to 185 
 
 Force, definition of 14 
 
 Force, measure of 14 
 
 Force, direction of action 16 
 
 Force, line of action '. 16 
 
 Force polygon 26 
 
 Graphical statics, theoretical 144 to 161 
 
 Graphical determination of resultants of parallel forces 35 to 43 
 
 Graphical determination of reactions and moments 147 
 
 Graphical analysis of deck plate girders 255 to 259 
 
 Graphical analysis of through plate girders 286 to 290 
 
 Graphical analysis of curved chord bridges , 473 to 481 
 
 Horizontal shearing stress in beams 65 to 68 
 
 Highway bridges 533 to 569 
 
 Impact, railroad bridges 198 
 
 Impact, highway bridges 533 
 
 Increment of flange stress 184 to 185 
 
 Increment of bending stress 64 to 65 
 
 Inertia, definition of 47 
 
 Inertia, moment of 48 to 50 
 
 Influence lines 162 to 171 
 
 Length of cover plates (plate girders) 181 to 184 
 
 Lettering 9 to 10 
 
 Live load for railroad bridges 195 to 197 
 
 Live load for highway bridges 533 to 535 
 
 Length of span (economic) 576 
 
 Manufacture of steel 1 
 
 Maximum shear on trusses 134 to 138 
 
INDEX 669 
 
 PAGES 
 
 Maximum moment on trusses 138 to 139 
 
 Mill buildings, design of 583 to 624 
 
 Modulus of elasticity 20 
 
 Moment of a force 23 
 
 Moment of a couple 23 
 
 Motion ^ 22 
 
 Office buildings, design of 624 to 652 
 
 Parabola, graphical construction of , 61 
 
 Pettit trusses, determination of stresses in 481 to 502 
 
 Pins, determination of stress in 118 to 120 
 
 Plate girders, economic depth 180 to 181 
 
 Pony trusses (highway) 546 to 554 
 
 Portals, stresses in 576 to 582 
 
 Radius of gyration 48 to 50 
 
 Railroad bridges 195 to 533 
 
 Rankine's column formula 106 to 107 
 
 Reactions, definition of 17 
 
 Reactions, on simple beams 124 
 
 Reactions, on simple trusses 132 
 
 Relation of stress and distortion 19 to 20 
 
 Relation of shear to cross bending 61 to 64 
 
 Resultant of parallel forces 43 to 44 
 
 Rivets and riveting 3 to 5 
 
 Rivets, size of 12 
 
 Rivets, conventional signs 11 
 
 Rivets, stress on 115 to 117' 
 
 Rivets, in flanges of plate girders 185 to 187 
 
 Rollers, theoretical treatment 120 to 121 
 
 Shafting, theoretical treatment 121 to 123 
 
 Shear on beams 51 to 53 
 
 Shear, maximum on beams 124 to 128 
 
 Shear, maximum simple trusses 134 to 138 
 
 Skew bridges 570 to 571 
 
 Specifications, railroad bridges 195 
 
 Specifications, highway bridges 534 to 540 
 
 Straight line column formula 108 to 109 
 
 Stiffeners 190 to 191 
 
 Stress, definition of 17 to 19 
 
 Stress, maximum 68 to 72 
 
 Stress on beams 51 
 
 Stress on rivets 115 to 117 
 
 Stress on trusses 139 to 143 
 
 Stresses in curved chord bridges 415 to 426 
 
 Structural material 1 to 2 
 
 Structural shapes 2 
 
 Structural draughting 9 to 13 
 
 Tracings 11 
 
 Truss bridges (railroad) 324 to 532 
 
 Types of railroad bridges 195 
 
 Types of bridge trusses 324 
 
 Viaducts, railroad 295 to 324 
 
 Warren trusses 502 to 513 
 
 Web splice 189 to 190 
 
 Whipple trusses 514 
 
 Wind load, railroad bridges 198 
 
 Wind load, highway bridges 535 
 
THIS BOOK IS DUE ON THE LAST DATE 
 STAMPED BELOW 
 
 AN INITIAL FINE OF 25 CENTS 
 
 WILL BE ASSESSED FOR FAILURE TO RETURN 
 
 HIS BOOK ON THE DATE DUE. THE PENALTY 
 
 WILL INCREASE TO SO CENTS ON THE FOURTH 
 
 DAY AND TO $1.OO ON THE SEVENTH DAY 
 
 MAR 4 
 MAR 19 1934 
 
 FEB 141936 
 
 MAY 7 1937 
 
 OCT 291940 
 
 1943 
 
 OCT 3 1934 
 APR 271935 
 
 LD 21-50m-l,'3l 
 
Y.C 40403 
 
 UNIVERSITY OF CALIFORNIA LIBRARY