' r \ I I: t I. THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES GIFT OF Jolm S.Prell ■ £'^.:^> ■ * *tf tit^ti-m^- 1 «v ^*,' •# ' J;?- '^ ^'is ^i \..:. '•/ WORKS OF PROF. W. H. BURR PUBLISHED BY JOHN WILEY & SONS. A Course on the Stresses in Bridges and Roof- trusses, Arched Ribs, and Suspension Bridges, with an Appendix on Cantilevers. 8vo, 500 pages, 13 folding plates, cloth, $3.50. Elasticity and Resistance of Materials of Engin- eering. For the use of Engineers and Students. Contain ing the latest engineering experience and tests. 6th Edition, Rewritten and Enlarged. 8vo, iioo pages, cloth, $7.50. Ancient and Modem Engineering and the isth- mian Canal. 8vo, XV -|- 473 1 ages, profusely illustrated, includ- ing many half-tones. Cloth, $3.50, net. Postage, 27 cents additional. A COURSE THE STRESSES BRIDGE AND ROOF TRUSSES, ARCHED RIBS AND SUSPENSION BRIDGES, PREPARED FOR THE DEPARTMENT OF CIVIL ENGINEERING AT THE RENSSELAER POLYTECHNIC INSTITUTE. JOHri S. PRELL Gvil & Mechanical Engineer. SAN FRANCISCO, CAL. By WILLIAM H. BURR, C.E., raoPKSSOR of civil engineering in Columbia college in the city of new york ; MRMBER OF THE AMERICAN SOCIETY OF CIVIL KNGINEERS. NINTH EDITION REVISED AND ENLARGED. FOURTH THOUSAND, NEW YORK JOHN WILEY & SONS LONDON CHAPMAN & HALL, Ltd. 1904 Copyright, 1886, By VVm. H. Burk, PRESS OF BRAUNWORTH & CO. eOOKBINDERS AND PRINTERS BROOKLYN, N. Y. ^ Library Id 1^04- PREFACE TO THE EIGHTH EDITION. The eighth edition of this work has been considerably enlarged by a very material extension of Arts. 35. 36, 37, and ^ya, on Swing Bridges, which have been entirely rewritten. In this revision, I am under much indebtedness to Mr, Henry W. Hodge, C.E., for the computations involved in the numerical work of the two actual cases of swing bridge spans in Arts. 36 and T,y. The general theory of this class of structures, as first completely given in the first edition of this work, has proved to be so well adapted to the rather complicated requirements of the case that it is now almo:~t universally adopted in this country. The essential improve- ments and simplifications presented in this new matter make the labor of computation but little more than that required for non-continuous spans. W. H. B. New York City, /ufy, 1893. 733584 PREFACE TO THIRD EDITION. Since the publication of the first edition of this book •engineering practice in iron and steel construction, especially in the department of bridge building, has made very material progress. The distribution of metal in pin structures has been considerably modified so as to produce concentrations in larger members ; but chiefly the treatment of moving loads has experienced such a radical transformation as to bring it to a thoroughly rational basis. Hence, portions of the book as originally written have been cancelled and re- placed by entirely new matter, so amplified and extended as to bring the work in all its details abreast of the best practice of the present day. My indebtedness to the published papers of Prof. H. T. Eddy, of the University of Cincinnati, on the arched rib, will be evident to any reader even slightly acquainted with his valuable work entitled " Researches in Graphical Statics." Certain matters are of such common occurrence in the fol- lowing pages that it may conduce to clearness to mention them here. The word " ton " signifies a ton of 2,000 pounds, unless it is otherwise specifically stated. The word " stress " means the force acting in any member of a structure, while " strain " is the distortion which accom- panies the stress. The sign + indicates a tensile stress, and the sign — , a compressive one. Unless otherwise stated, the stress in any member of a structure will be represented by inclosing with a parenthesis the letter or letters which belong to it in the diagrams or plates. Thus (A B), or (a), signifies " stress in the member A B," or " stress in the member a." VI PREFACE. As a matter of convenience to those who may be familiar ■with the first and second editions, it is well to state that pages 19 to 60, Arts. 72, 74 and 85 are entirely new. Portions of pages at several other places in the book have also been re-written, but it is unnecessary to name them here. For convenience in swing bridge computations, Appen- dix IV. has been inserted. Formulae for moments and re- actions are there collected in the simplest form for applica. tion. W. H. B. Phcenixville, Pa., Feb. ii,th, 1886. 'r; I. CONTENTS. CHAPTER 1. General Consideration of the Laws Governing the Action of Stresses io Trusses. PAG? Art. I. — The Truss Element i " 2. — General Case 3 " 3. — Web and Chord Stresses in General . 4 " 4. — Overhanging Truss — Parallel Chords — Bracing with two Inclina- tions 8 " 5. — Overhanging Truss — Parallel Chords — Uniform Bracing — Vertical and Diagonal Bracing II CHAPTER n. Special Non-continuous Trusses vrith Parallel Chords. Art. 6. — Distribution of Fixed and Moving Loads 18 6a. — Trusses with Uniform Panel Loads 20 " 7. — Position of Moving Load for Greatest Shear and Greatest Bending. 20z ' ' 8. — Fixed Weight 36 " 9. — Single System of Bracing with two Inclinations 38 " 10. — Single System of Vertical and Diagonal Bracing — Verticals in Tension 50 " II. — Single System of Vertical and Diagonal Bracing — Verticals in Compression 54 " 12. — Two Systems of Vertical and Diagonal Bracing — Verticals in Compression 59 " 13. — Truss with Uniform Diagonal Bracing — Two Systems of Triangu- lations .... 66 " 14. — Compound Triangular Truss 6g " 15. — Methods of Obtaining Stresses — Stress Sheets 74 " 16. — Ambiguity caused by Counterbraces 74 Viii COh^TENTS. CHAPTER III. Non>continuous Trusses with Chords not ParalleL PAGE Art. 17. — General Methods 75 " 18. — Curved Upper Chord — Two Systems of Vertical and Diagonal Bracing 78 " 19. — General Considerations 88 *' 20. — Position of Moving Load for Greatest Stress in any Web Member. 89 *• 21. — Position of Moving Load for Greatest Stresses in Chords 92/ " 22. — Horizontal Component of Greatest Stress in any Web Member — Constant Value of the Same for Vertical and Diagonal Bracing with Parabolic Chord : : . 93 " 23. — Bowstring Truss — Diagonal Bracing — Example 93 *' 24. — Bowstring Truss — Vertical and Diagonal Bracing with Counters. . 102 *' 25. — Bowstring Truss — Vertical and Diagonal Bracing without Coun- ters 105 " 26. — Deck Truss with Curved Lower Chord, Concave Downward — Ex- ample loS " 27. — Crane Trusses iii " 28. — Preliminary to the Treatment of Roof Trusses — Wind Pressure — Notation 115 " 29. — First Example 117 " 30. — Second Example ir8 " 31. — Third Example 121 " 32. — Fourth Example 122 " 33. — General Considerations t . . 123 CHAPTER IV. Swing Bridges. Ends Simply Supported. Art. 34. — General Considerations .... 125 " 35. — General Formulae for the Case of Ends Simply Supported — Two Points of Support at Pivot Pier — One Point of Support at Pivot Pier 126 *' 36. — Ends Simply Supported — Two Points of Support at Centre — Partial Continuity — Example 133 " 37. — Ends Simply Supported — Two Points of Support at Centre — Complete Continuity — Example 159 " yja. — Tables and Diagrams — Turntables and Engines 179 " 38. — Ends Simply Supported — One Support at Centre — Example .... 187 CHAPTER V. Swing Bridges. Ends Latched to Supports. Art. 39. — General Considerations 203 " 40. — Ends Latched Down — One Point of Support between Extremities of Truss — Example 203 CONTENTS. ix CHAPTER VI. Swing Bridges. Ends Lifted. PACK Art. 41. — General Considerations 212 " 42. — Ends Lifted — One Point of Support between Extremities — Ex- ample 212 " 43. — Final Observations on the preceding Methods 222 CHAPTER VII. Continuous Trusses other than Souring Bridges. Art. 44. — Formulae for Ordinary Cases — Reactions — Methods of Procedure . 224 CHAPTER VIII. Arched Ribs. Equilibrium Polygons 229 Bending Moments 234 General Formulce 236 Arched Rib with Ends Fixed 237 Arched Rib with Free Ends 247 Thermal Stresses in the Arched Rib with Ends Fixed 250 Thermal Stresses in the Arched Ribwith Ends Free 256 Arched Rib with the Fixed Ends — / and n Variable 259 Determination of the Stresses in the Members of an Arched Rib — • Example — Fixed Ends — Consideration of Details 267 " 54. — Arched Rib Free at Ends and Jointed at the Crown 278 CHAPTER IX. Suspension Bridges. Art. 55. — Curve of Cable for Uniform Load per Unit of Span — Suspension Rods Vertical — Heights of Towers Equal or Unequal — Gen- eralization 271) " 56. — Parameter of Curve — Distance of Lowest Point of Cable from either Extremity of Span — Inclination of Cable at any Point. . . 281 " 57- — Resultant Tension at any Point of Cable 282 " 58- — Length of Curve between Vertex and any Point whose Co-ordi- nates are x and y, or at which the Inclination to a Horizontal Line is ? • 283 " 59- — Deflection of Cable for Change in Length, the Span remaining the same 285 60. — Suspension Canti-Levers 287 Art. 45. ' 46. ' 47- ' 48. ■' 49- ' 50. •< 51. ' 52. ' 53- X CONTENTS. PACK Art. 6i. — Suspension Bridge with Inclined Suspension Rods — Inclination of Cable to a Horizontal Line — Cable Tension — Direct Stress on Platform— Length of Cable 288 " 62. — Suspension Rods; Lengths and Stresses 292 " 63. — Pressure on the Tower — Stability of the Latter — Anchorage 294 " 64. — Theory of the Stiffening Truss — Ends Anchored — Continuous Load — Single Weight 296 " 65. — Theory of the Stiffening Truss — Ends Free — Continuous Load — Single Weight 307 *' 66. — Approximate Character of the Preceding Investigations — Deflec- tion of the Truss 311 CHAPTER X. Details of Construction. A.RT. 67. — Classes of Bridges — Forms of Compression Members — Chords Continuous or Non-Continuous 313 -Cumulative Stresses 315 - Direct Stress Combined with Bending in Chords • 316 -Riveted Joints and Pressure on Rivets 323 -Riveted Connections between Web Members and Chords 325 -Floor-Beams and Stringers — Plate Girders 326 -Eye-Bars or Links 343 -Size of Pins 345 -Camber 350 -Economic Depth of Trusses with Parallel Chords 353 -Fixed and Moving Loads 354 -Safety Factors and Working Stresses " 358 -General Observations 365 CHAPTER XI. Wind Stresses and Braced Piers. Art. 80. — Wind Pressure 366 ' ' 81. — Sway Bracing 370 " 82. — Transverse Bracing for Transferring Wind Stresses from One Chord to Another — Concentrated Reaction 379 " 83. — Transverse Bracing with Distributed Reactions 384 84. — Stresses in Braced Piers 387 " 85. — Complete Design of a Railway Bridge 395 APPENDIX I. The Theorem of Three Moments 431 APPENDIX IL The Resistance of Solid Metallic Rollers 443 APPENDIX III. The Schwedler Truss 447 -APPENDIX IV. Reactions and Moments for Continuous Beams 453 APPENDIX V Csntilevevs 455 CONTENTS. X APPENDIX V. Cantilevers. PAGE Art. I. — Cantilever Structures — Positions of Loading for Greatest Stresses in the Cantilever Arm 455 " 2. — Positions of Moving Load for Greatest Stresses in the Anchor Arm, and the Greatest Stresses Themselves 464 " 3. — Positions of Moving Load for Greatest Stresses in Anchor Spans and the Greatest Stresses Themselves 467 " 4. — Wind Stresses 469 ' 5. — Economic Lengths of Spans and Arms 472 *' 6. — Wind Pressure 474 CHAPTER I. GENERAL CONSIDERATION OF THE LAWS GOVERNING THE ACTION OF STRESSES IN TRUSSES. Art. 1.— The Truss Element. A TRUSS may be defined to be a structure so composed of individual pieces that, if all the externally applied forces called loading are parallel in direction, the other external forces called reactions will be parallel both to each other and the loading. The simplest of all trusses is a triangle, and all trusses, however complicated, containing no superfluous members, are, and may be considered, assemblages of triangles simply. That the triangle is the truss element arises from the fact that it is the only geometrical figure whose form may not be changed without .varying the lengths of its sides. In the elementary truss of the figure, let any force act ver- tically downwards at B, and consider the two triangles ABD and BDC having the common side BD. Since all external forces are parallel, the reaction at A is to the reaction at C as DC x?. to AD. For if BD be taken to represent the vertical force acting at B, and DF be drawn parallel to ^^ as well as EF parallel to AC, then will DF represent the stress in AB, and BF that in BC. But by the construction ED : EB = DC : AD, but ED is the vertical component of the stress in AB as well as the re- action at A, while BE is the same component of the stress in BC, and, similarly, the reaction at C. It is to be particularly noticed that EF is the common horizontal component in each of the members AB and BC, and also the resultant stress in AC I I 2 THE ACTION OF STRESSES IN TRUSSES. When, therefore, the truss is horizontal, as is supposed in the figure, the vertical component in each of the members AB and BC is equal to the reaction at its foot ; also the hori- zontal component of stress in each of these members is equal to the horizontal component in the other, as well as to the resultant stress in the third horizontal member. These simple principles constitute the foundation of all stress analyses in trusses. Fig. I. Art. 2. — General Case. Again, consider any truss whatever, as that in Fig. i, in which the supports are not on the same level, nor are any two of the triangles of which it is composed similar. Suppose a vertical load to act at any apex, as A, the reactions will be vertical. Let the truss be cut by any plane which divides the line GH {AH is the trace of such a plane), then the part of the truss which is found on the left of AH is held in equilib- rium by a component of the vertical force at A, the vertical reaction at C, and the induced stress in GH. Since there is equilibrium, the lines of action of those forces must intersect in a point ; and since the forces acting through C and GH have lines of action intersecting at /?, the line of action of the component of the vertical force at A must pass through the same point. Thus the line of action DA for one component is established. In precisely the same manner BE is erected and produced until it intersects GH, produced, in E and the line of action AE of the other component established. Connect D and E, then, so far as the reactions are concerned, the case will not GENERAL CASE. 3 be changed if the actual truss be supposed displaced by the simple truss DEA. Let AN represent the vertical load at A, then make N^O, parallel to AE, and DA, produced, intersect at O. If MO be c^rawn parallel to DE, AM will evidently represent the reaction at C, and MN the reaction at B. Pro- duce AN until it intersects GH in G, and draw DK and EF in a horizontal direction, then, from similar triangles : OM DG MA = GA^ ^"^ OM EG MN ~ GA ' ividing one by the other. MN DG DK MA ~ EG ~ EF' But DK + EF is equal to the span, hence the reactions are inversely as the segments of the span, or any load or system of loading, vertical in direction, to which any truss whatever is subjected, is divided into reactions according to the law of the lever. Farther, whatever the internal stresses of the truss may be, at the ends the sum of the vertical components must be equal to the reactions. THE ACTION OF STRESSES IN TRUSSES. Art. 3. — Web and Chord Stresses in General. In the preceding case no account was taken of the stresses to which the individual members of the truss were subjected, but it will now be necessary to consider them. For this pur- pose PI. I., Fig. I, will be used, in which the points of support will be taken in the same level, the loading, vertical ; and the truss will be considered as made up of similar triangles. The first and last suppositions in nowise affect the generality of the conclusions, but the operations are thereby simplified and given a character approaching more nearly to that of the ordinary operations of the engineer. In PI. I., Fig. I, is the representation of a truss placed upon two supports A and L in the same horizontal line. CH and AL are parallel, and the oblique members included between them may have any inclinations whatever, only they are made symmetrical in reference to a vertical centre line through O. Let any weight W act at any point, as N, and let t and /' represent the tangents of the greater and less inclinations, respectively, of the oblique members to a vertical. Erect ver- ticals at A and L which will intersect the prolongations of CH in B and K\ then, as has been shown, BR and UK drawn through A" will be the lines of action of the components of W which act on the two parts of the truss. The force parallelo- gram WRNU can then be drawn, in which R T and UV are to be drawn parallel to AL. NT= VW is the reaction at A, and vVFthat at L. Resolve NR in the direction of /TV and NM' by drawing M'R parallel to FN, then will M'R represent the stress in FN. The stress in FN will induce the stresses Fb and Fb"y in FG and FO, and in the same manner the stresses shown in the figure will be induced at all the points on the left of FN. It is to be noticed that all the inclined stresses at the points A, C, Q, D, P, E, O, F, as well as M'R, have the same vertical component, NT; also, that all the horizontal stresses at C, Z>, F, and F are equal to each other and act in the same direc- tion, each one having the value NT x {t + t'). Let n be the WEB AND CHORD STRESSES IN GENERAL. 5 number of the points C, D, E, and F, then the total horizon- tal stress acting along CH from left to right will be NT X (/ + /') X n. At N there is the horizontal force NM' acting from left to right. Let d equal the depth of the truss, or AB, and / and /' the segments ^iVand NL respectively of the span AL or s\ then from the figure it is seen that NM' = NT(^-t\ At the points O, P, Q, A, there are horizontal stresses act- ing from right to left. From the diagram it is seen that the stress at 6> is (2 / X NT) ; at P, {t + /') x NT\ at Q, {t + /') x NT; at A, t' x NT\ hence the total stress on the left of N acting from right to left is Hence there is deduced the important result that NM' is just equal to the total horizontal stress on the left of N, and possesses the same line of action but is opposite in direction, therefore the two forces balance each other. Next prolong GN until it cuts UV in F, then will NY represent the stress induced in GN by the component UN. The stress at iV, acting from right to left, is therefore UY, or UY^NVy. {^-t Although the diagrams are not drawn, it is plain that the horizontal stresses at M and L are NV x (/ -)- /') and NV x /' respectively ; also, that their directions are from left to right ; hence the total horizontal stresses, on the right of N, which act from left to right, are 6 THE ACTION OF STRESSES IN TRUSSES. Hence the stress UY is balanced by the horizontal stresses on the right of ^V. All the internal horizontal stresses acting along AL are, therefore, balanced. According to the two force parallelograms drawn from G and H, it is seen that all the horizontal stresses on the right of N, which act from right to left along CH, are n' X NVx{t + t') = NVx^-^, in which n' is the number of apices G and H B\xtNV=NTj- hence /' / a a But it has already been shown that all the stresses which act along CH from left to right are NT y^{t ^t')y.n = NT ^. Hence all the horizontal internal stresses of the truss are perfectly balanced among themselves. This important characteristic belongs only to the " truss " proper, and distinguishes it from all other bridge superstruc- tures. If an irregular truss, like that in Fig. i above, were treated, precisely the same result would be reached, but the resultant horizontal stress would be expressed by 2Pt or 2'Pt,m which Pis a variable portion of W; and ^ would be a "mean " depth, such that 2 1 d = I, and 2' td= /'. The portion FG is subjected to all the stress induced at the points C, D, E, and F; EF to all that induced at C, D, and E', etc. It is important to notice this accumulation of stresses from panel to panel in the horizontal lines CH 3ind AL, for IVEB AND CHORD STRESSES IN DETAIL. y it shows that the stresses in those portions are not uniform from end to end. A stress induced at one point may be felt at any distance from that point. The upper and lower portions of the truss, CDEFGH and AQPONML, are called the top and bottom " chords," and all members included between the chords, whether inclined or vertical, are called " braces " or web members. The various portions into which the chords are divided, usually equal to each other, are called panels. From the figure it is seen that the vertical components of the stresses in the braces or web members on one side of N are equal to each other ; also, that the chord stresses have no vertical component, being horizontal. Farther, the vertical component in any brace or web member is equal to the reac- tion found on the same side of the load as itself. In a truss provided with horizontal chords, therefore, the ofifice of the web members is solely to transfer, so to speak, the load from its point of application to the abutments or piers of the bridge ; their duty, therefore, is precisely the same as that of the web in a flanged girder, hence their name " web mem- bers." In other words, the braces or web members take up the shearing stress at any section. Let sec i and sec i ' be the secants of the angles of inclina- tion of the web members, corresponding to the tangents / and t\ and let 5" and S' be the shearing stresses in the two segments of the span ; then the web stresses in the left-hand segment will be 5 sec i and 5 sec i', and those for the right- hand segment 5 ' sec i and 5 ' sec i ' . The general principles brought out in the preceding results, therefore, are these : With horizontal chords the iveb stresses are products of the shears by the secants of the inclinations, and the chord stresses are functions of the tangents of the inclina- tions of the braces or web members from vertical lines. From an inspection of the force parallelogram at D, for instance, it is seen that the increment of chord stress at any panel point is equal to the algebraic sum of the horizontal com- ponents of the stresses in the web members intersecting at that point. The sum is numerical when, as in the figure, the 8 THE ACTION OF STRESSES IN TRUSSES. bfc ces slope on different sides of the vertical line passing through the panel point, but the numerical difference is to be taken when both braces are found on the same side. Two web members intersecting at any panel point have stresses of opposite kindo induced by the same shearing stress. The stress in CH is of course compressive, while that in AL is tensile. The preceding general results have been deduced on the supposition of the application of but one weight, but they are equally true for any system of loading. For the effect of any system of loading is simply the summation of the effects of the individual loads of which it is composed, hence only those principles which are true for the individuals can be true for the system, and those at least must hold, for the action of each load is independent of all the others. Art. 4. — Overhanging Truss. — Parallel Chords. — Bracing with Two In- clinations. Probably the simplest case of a truss subjected to the action of external loading occurring in the practice of the engineer is a simple truss fixed at one end, and is the case with one arm of a swing-bridge when open and subjected to its own weight as load. Now, in all cases of actual trusses, the load will be sup- posed divided in its application to the truss between the upper and lower chords. It will not, however, be equally divided, because the floor system of the bridge will rest wholly on one chord or the other. h c d e f T Fig. I. In the figure, let the truss be fixed at AB, and let W and OVERHANGING TRUSS. W be the panel loads on the upper and lower chords respec- tively, except at the extremity P of the upper chord, where h W will rest. Let a represent the angle QPR, and a' the angle SQT\ the line PR is vertical, and y^/^ horizontal. A simple and direct application of the principles and for- mulae of Art. 3 gives the following results : Stress in " M " K " G " E " D " N '' L " H " F - V " C ec a. {ihW + (2i VV + 2W') {3vy + 1>^V') « (4yV + aW) isvv + SlV) ( W + IV')- sec a! {qw + 2 IV) {2lW + 3in (3U^ + 4 IF') {Ah^V-^ SW) {ShW + 6W') It will be observed that in every instance the stress in any brace is the " shearing stress " in the section to which the brace belongs, multiplied by the secant of the inclination to a vertical line of the brace in question. For instance, if the brace L be divided by a vertical plane, the weights on the right of it, or the shearing stress, are {^\IV -^ 2 IF'), and this multiplied by the secant of a is the stress desired. The chord stresses are also determined by a direct and sim- ple application of the principles and formulae of the preceding article. Stress in / is \ IV tan o: " " e " i^lVtan a-\-(lV' + ^W) (tan a + fan cr') " " ^ " 2,\ IV tan a + {2 IV+ 3 IV) (tan a + tan a') " " c " zWtan r*' + (4.i IF-I-6IF') (tan a^tan a) " " b " 4^ Wtan n' + (8 W^ 10 W ') {tan a + tan a) " " ^ " 5 ^ W tan a + {\2\ IV+ 1 5 IV) (tan a + tan a') , (0- In the lower chord the stresses are as follows : 10 THE ACTION OF STRESSES IN TRUSSES. Stressing is W tana' ■^\W {tan a-'rtana'^ ~1 " " h'' 2W' tana' + {2lV+W'){tana+tana') | u u j^u ^Wfana' + {4lW+T,lV'){tana + tana') - " " / '' 4W' tana' + {SW+6W'){tana + tan a') u "f^*' ^W tana +{i2lW+ioW') {tana + tana) \ u a ^u eW tana' + {iSlV+isW'){tana + tana')) K2). In determining these quantities, it is to be remembered that the stresses cumulate from the free end to the fixed ; i.e., the stress developed at any panel point is felt throughout those portions of the chords included between that point and the fixed end of the truss. General formulae for the Eqs. (i) and (2) may easily be found. Let n be the number of the panel, from the free end, in the chord AP {/is number i ; r, 2 ; ^, 3 ; etc.), then the formula expressing the results in Eq. (i) is the following: Stress in any panel = {n — ^) W tan a -\- \ - — — W + -^^ ' W \ \tan a + tan a\ .... (3). This expression gives the stress in any panel of AP. The formula which expresses the results shown in Eq. (2) is the following : Stress in any panel = nW tan a' + { — W + -^ W ^ \tana + tana'\ . . . (4). In which n denotes the number of the panel in the chord BQ starting from the free end ; i.e., g is number i, h number 2, etc. The weight at P has been taken at half that applied at other panel points in the same chord. In the case of a swing- bridge, however, it is greater than that, since some of the details of the locking apparatus, etc., are hung from that point. Yet the Equations (i) to (4) may still be used, only a OVERHANGING TRUSS. \\ simple term is to be added to each of those equations. Let / be the panel length, d the depth of the truss, and IF, the actual weight hung from P. Also, let Wy—\W=w'. In order to find the additional stress produced in any panel d of the chord AP, let the moment of w be taken about the in- tersection of H and K in the lower chord ; this moment is w\{n — \)p -{- d tail a\. Consequently the additional stress desired is I p \ s— w^An— \)^-\- tana r . . . . (5). The stress s is to be added to each of equations (i) and (3) if Wi> ^ W, otherwise it is to be subtracted. In precisely the same manner, the additional stress for the lower chord BQ is s' = wn ^ (6). d The stress s is to be added to equations (2) and (4) if fFj > \ W, otherwise it is to be subtracted. If, as in Fig. i, AP is the upper chord, the stress in QP and all members parallel to it will be compressive ; while the stress in QS and all braces parallel to it will be tensile. Like- wise the stress in AP is tensile, and that in BQ compressive. If the truss were turned over so that ^^ would become the top chord, the expressions for the stresses in equations (i) to (6) would remain exactly as they are, only the signs of the stresses would change. The condition of stress would be ex- actly represented in the preceding paragraph by simply chang- ing "compressive " to " tensile," and " tensile " to " compres- sive." Art. 5. — Overhanging Truss — Parallel Chords — Uniform Bracing — Ver- tical and Diagonal Bracing. The two most frequent cases of Fig. i, Art. 4, are, first, that in which a — a', and, second, that in which a' — o. The first of these cases is represented in Fig. i, and the second in Fig. 2. 12 THE ACTION OF STRESSES IN TRUSSES. The web stresses for this case will be precisely the same in general form as those given in Art. 4, but sec a will be written for sec a. Very simple general formulae can be written for these web stresses. Let n' denote the number of any brace starting from O, which is called i ; then observing the general values in Art. 4, the stress in any brace 71' parallel to O will be + d = \ ^ W + ^^ ~ — - IV i sec a + w sec a. . . (i). Fig. I. The expression (+ d), of course, denotes tensile stress in any brace parallel to O. In precisely the same manner, the compressive stress in any brace parallel to A^ {n possessing the same signification as before ; i. e., n for A^ is 2 ; for L, 4, etc.) is , ( («' — i) „. n „., ) , , — b — \ IV + — W > sec a + w sec a . . (2). In determining the chord stresses, it is to be remembered that the weights IV rest on the lower chord AP. Making tan a — tan a' in Eq. (3) of Art. 4, the stress in any lower- chord panel is C— {n-h)lVtana-\- \ {n—\yW+n{n—\)W' \ tana-\-s . (3). Making the same change in Eq. (4) of the previous Article, the upper-chord tensile stresses will be found to be T—71W' tan (x + \ n^ f F+ ;/(;/— i ) f f ' \ tan a + s (4). VEKHA NGING TR USS. 13 Some of the results given by the formulae should always be checked by the method of moments. Let it be desired to determine the stress in k by the method of moments. Let the origin of moments be taken at the intersection of G and H. The moments which balance each other about that point are that of the stress in k acting with the lever-arm d, the depth of the truss, and those' of the weights applied to the truss on the right of the panel point in question ; these latter act against the former. Calling the panel length/, and taking the moments mentioned: 7^ = 3 W • 1/ + 2 W • \p ^ \W • IP ^- w ' ip. :. T=4VV'^ + 4W^+3^^ • • • (5). The result of Eq. (5) ought to be the same as that of Eq. (4). Two or three panels in each chord ought to be treated in the same manner. Fig. 2. The cantilever truss represented in Fig. 2 shows the rase in which m' of Fig. i, Art. 4, is equal to zero. The note-fion is precisely the same as that used before. The general expression for the stress in any inclined brace is simply Eq. (i) repeated — that is: + d i 2 I) JT' [ sec a + IV sec a (6). Making sec a' = i, there results for the compressive stress in any of the verticals 2, 4. 6, etc. : 14 THE ACTION OF STRESSES IN TRUSSES. -I) b = i ^— 2— ^ ^¥+-^'1 + ^ (7). Making faft a = o, in Eq. (3) of Art. 4, gives the com- pressive chord stress in any panel of the lower chord AP. Hence, for that chord : C i ,,- (n — i)^ „. n (n — i) „., , = j (« - i) ^V + ^- — ■— W + — ^^ ^ W \ tmi a + wn P (8). In a similar manner, from Eq. (4) of the previous article, for the tensile stress in any panel of the upper chord, there results the equation : T = IV + ti {n — i) W \ tan a + wn (9). If Wi should be less than \W, the term which expresses the additional stress, whether in braces or chords, will be sub- tractive, as will be indicated by the sign of w. Fig. 3. Again, applying the moment test to any panel, as c, by taking the origin of moments at R, the notation remaining the same as before, there results : .'.C^{4W+slV')^^+4^^ (10). OVERHANGING TRUSS. 15 This result ought to agree with that shown by Equation (8), and several panels in each chord should be tested. In the great majority of cases it is not convenient to apply a general formula, but the numerical values are usually deter- mined directly from the diagram, and the stress in each mem- ber written along it, as shown in Fig. 3. Fig. 3 shows a truss which frequently occurs in the practice of the American engineer ; it is in reality one arm of an open swing-bridge. Let the panel length = p = \2 ft., and the depth of the truss = d = 20 ft. The tangent of CQP = 12 ^ 20 = 0.6, and the secant of CQP — 1.166. The tangent of DEP = 1.2, and the secant of DEP— 1.562. The panel loads at E, F, H, etc. = W^' = 3-00 tons; at C, D, G, K, etc., W= 5.00 tons, at P, IVi = 4.00 tons. No load is taken at Q. In the figure there are two pystems of right-angled triangu- lation ; P, E, D, H, K, etc., is one system, and C, F, G, L, M, etc., is the other. This does not, however, complicate the matter in the least, for each system of triangulation is regarded as ail individual truss carrymg its own weights only. Calcula- tions are therefore made for each system of triangulation as if they were independent trusses, and then the two are added. The weight of the portion EQP is supposed divided be- tween E and P, thus showing W^ > i W. The vertical braces are evidently in compression, while the inclined ones are in tension. The figures in the diagram denote tons (2000 lbs.) of stress; -I- indicates tension, while — indicates compression. Stress in PE = IV^ see DEP — 4 x 1.562 = 6.248 tons. " " CF =W sec DEP ^t^ ^ 1.^62^ 7.810 " " " DH={U\ + W + lV)x 1.562 = 18.744 " " " GL= 2PV+lV'x 1.562 = 20.306 " The other brace or web stresses are found in precisely the same manner. l6 THE ACTION OF STRESSES IN TRUSSES. Stress in CP — W^ tan DEP =4 x 1.2 = 4.8 tons " " DC = Wta?i DEP + 4.8 = 6.0 + 4.8 = 10.8 tons. " " GD= { W\ + IV + W) X 1.2 + 10.8 = 14.4 + 10.8 = 25.2 tons. " '' KG = {W^- W + W)x 1.2 + 25.2 = 40.8 tons. Other lower-chord stresses are found in exactly the same manner. By an inspection of the diagram it is seen that the general expression for the stress in FE is precisely the same as that for CP\ the same can be said oi HE \n reference to DC; LH in reference to DG ; etc. The explanation of this is simple. If the truss be divided by a plane normal to the paper and parallel to the inclined braces, only vertical and horizontal members will be cut. But the truss is in equilibrium, and since the loading is wholly vertical, the sum of the horizontal stresses must be zero ; or, the stress in the lower-chord panel cut must be equal and opposite to the stress in the upper- chord panel cut. Let the moment test be applied to the stress in the panel MK. The origin of moments for the loads applied to the system PEDH, etc., is N, and the origin for the other system is L. Taking moments about those points : C'd + C"d = 16 X 36 + 4 X 72 + 8 X 24 -h 5 X 48 .-. C — C -\- C" = 64.8 tons. Again, for the lower-chord panel adjacent to A, B is the moment origin for the whole load. C= {48 X 42 + 5 X 84 + 4 X 96) ^ 20 = 141 tons. Thus the numerical results are verified. According to one of the principles of Art. 3, the horizontal component of the stress in any inclined web member ought to be equal to the increment of chord stress at either of its VERHA XG/.VG TR USS. 17 extremities, and such will be found to be the case. If, for example, 20.3 be multiplied by the cosine of the angle GLH, the result will be 15.6 = 40.8 — 25.2. This last is a verification of the web stresses, and both methods of checking are perfectly general and may be applied to all trusses, as should be done in actual cases. CHAPTER II. SPECIAL NON-CONTINUOUS TRUSSES WITH PARALLEL CHORDS. Art. 6. — iDistribution of Fixed and Moving Loads. The trusses treated heretofore have been of rather an ele- mentary character, and general principles have been con- sidered instead of special and practical applications. Before taking up the technical treatment of trusses it will be neces- sary to consider some preliminary matters. The total load on a bridge-truss always consists of two parts, the fixed load and the moving load. The fixed load consists of the entire weight of the bridge, including tracks, flooring, etc. The moving load, as its name indicates, con- sists of that load (whether single or continuous) which moves over the bridge. The truss is, of course, always subjected to the action of the fixed load. If the truss is of uniform depth the panel-fixed loads will be uniform in amount for one chord ; but if the depth is vari- able it may be necessary to make a varying distribution of the weight of the trusses and lateral bracing. The amount and rate of this variation can only be determined by the cir- cumstances of each particular case. The moving load on a railway bridge may be taken as con- tinuous or as a series of single weights as actually applied at the wheels under the locomotives and cars. The assumption of continuity of moving load was formerly always made, a larger amount per lineal foot being taken to represent the extra locomotive w^eight. In such a case, if the moving load extends from the end of the bridge to the centre of any panel, or to the end of that panel, the panel point immedi- i8 RAILWAY BRIDGES. 19 ately in front of the train will not sustain a full panel load ; but if it be assumed that this panel point does sustain the full load, then a small error on the side of safety will be committed. Such an assumption was formerly made, and the consequent method of computation will be given in some of the Arts, which follow in this chapter. At the present time (1885), however, the demands of the best practice require the moving load to be taken at the actual points of application of locomotive and car wheels. This method of computation will be given in several of the first cases taken. If the span is short, or less than 125 feet, the moving load should be taken entirely of locomotives, as two will nearly cover the structure. The amount and character of the moving load, however, is usually indicated by specifica- tions. The moving load of a bridge may pass along the upper chord or the lower chord. In the first case the bridge is called a " deck " bridge, and in the latter case a " through " bridge. The methods employed in the determination of stresses in the various truss members are exactly the same in both cases. " Pony " trusses are through trusses not sufficiently high or -deep to need overhead cross-bracing. Every truss-bridge is composed of the following parts: Upper and lower chords. Upper sway-bracing, Web members, Floor system, including beams, stringers, ties, floor-hangers, lower sway-bracing, and rails. The sum of the w'eights of the parts is the " fixed " load of the bridge. In the case of highway bridges the calculations are pre- cisely the same as for railway bridges, except that the mov- ing load is assumed to be uniform per lineal foot of bridge. The greatest load that can ordinarily pass on a bridge is a dense crowd of people, the greatest w'eight of which can be taken at eighty-five pounds per square foot. The late Mr. 20 SPECIAL NON-CONTINUOUS TRUSSES. Hatfield, of New York City, found by experiment that it was scarcely possible to exceed seventy pounds per square foot. The moving panel load of a highway bridge may then be found by multiplying the width of the clear way, including sidewalks, by the product of the panel length with the load per square foot. If the span is not over 125 feet, or about that value, the moving load for the truss members may be taken at eighty- five pounds per square foot, or sixty pounds for greater lengths. In all cases, however, the floor beams and joists should be designed for a moving load of 100 pounds per square foot, in order to provide for the increased fatigue of those members due to shocks and sudden application of loads. In some cases highway bridges are subjected to enormous, concentrated loads of a special character. Such loads can only be known from local considerations, and the bridges must be built with a view to sustaining such special weights. All the methods or principles used, then, in the following cases, which will be those of railway bridges, are equally ap- plicable to highway structures, and no further special atten- tion will be given to the latter. Art. 6 a. — Trusses with Uniform Panel Loads. Fig. I represents a simple truss of the Pratt type, sup- ported at each end on an abutment, and supposed to carry a system of uniform moving panel loads in addition to its own weight or fixed load. The moving load will be taken as traversing the lower chord. Although the truss has but nine panels, the formulae will be written so as to apply to any number of panels. 7RUSSES WITH UNIFORM PANEL LOADS. 20a The inclined web members, except the end-posts, are in tension. The end-posts, vertical web members (except the two Zi's), and upper chord are in compression, while the lower chord is in tension. The two vertical tension mem- bers, Tx, are really not members of the truss, but they are virtually hangers, to carry the loads at their feet to the upper chord panel points at their tops. The moving load is sup- posed to pass on the bridge from right to left, and the nota- tion is so written as to apply to any number of panels. li\ — uniform lower fixed panel load. w — uniform lower moving panel load. W — uniform upper fixed panel load. / = panel length. / = length of span. a = uniform angle of inclination of inclined mem- bers to a vertical line. It will be convenient and more simple to treat the fixed and moving loads separately. Case I, — Fixed Loads. The inclined members marked c^, c^, etc., and shown by broken lines, will be considered omitted, as they are not stressed by the fixed or dead loads. If iV = i is the total P number of upper chord panel points, or the total number of lower chord points at which loads are applied, the reaction R due to the fixed loads only will be : K^^SK^J^ ^^^ Then, if n be any of the subscript numbers of c^, c^, Ci, . . . T-;, the shear in any of the members T will be : S = R -{N- n) {W + W,) = n{W, + W) (2). 20b SPECIAL NON-CONTINUOUS TRUSSES. Both the second and third members of Eq. (2) show that the fixed load shear in any main web member Z„ is equal to the total fixed load between the centre of the span and the member T,^. If the number of panels in the span is odd, this princi- ple shows that the shear in the inclined web members, or diagonals, of the centre panel will be zero ; and it also shows that if the number of panels is even, the shear in each of the diagonals running upward from the centre panel point of the lower chord will be | {W + Wy). The stress in any inclined main web member T^ will there- fore be : (7;,) = S sec a ^ {W + W,) (n - ^\ sec a . (3). The stress in any vertical post P^ will be : (/'„):=(^^'+^0(«-f) + ^' . . (4). N In these formulae n will never be less than — . 2 The stress in the inclined end-post EP will be : (^EP) = ^ -^ seca, , . . (5). The stress in Zi will evidently be : (^1)= ^1 (6). Inasmuch as the arrangement of the truss members is symmetrical about a vertical line through the centre of the span, it is necessary to find the stresses in the members of one-half the truss only. When loads are so symmetrically disposed about the centre of the truss that the inclined web members c,, c^, etc., which are shown as broken lines in Fig. i, and are called " counters " or " counterbraces," are not stressed, they (the loads) are said to be " balanced." All uniform fixed or dead loads in sym- metrical trusses are balanced. TRUSSES WITH UNIFORM PANEL LOADS. 20C The fixed-load chord stresses remain to be determined. In accordance with the principles already established, the lower chord stress in panels i will be the shear or vertical component in £/* multiplied by tan a. Hence : {l) = ^^ {W + l]^,)tan a . . . . (7). Since the stress in 71 has no rectangular horizontal com- ponent, the two stresses in the lower chord panels i will be identical. The stress in lower chord panel 2 will be equal to the shear in T^f-i multiplied by ta7i a, added to (i). Hence, by the aid of Eq. (3), in which ji — N — i : (2) = ^ ( W + W,) tancY + (^ - i) ( ^V' + W,) tan a = -(iV - I) ( ^F' + W^ tan a . . . (8). Similarly, the stress in lower chord panel 3 is equal to the shear in 7v_2 multiplied by tan a, added to (2). Hence, by aid of Eq. (3), as before, in which ;/ is now equal to iV— 2 : (3) = (A^- I) {W + W,) tana + (~- 2) {W + W,) tan a = ^(jV- 2}{IV' + W,)tana . . . (9). In a similar manner: (4) =^{N-s){W' + W,) tana . . (10). The second member of Eq. (7) and the third members of Eqs. (8), (9), and (10) show that if n' indicate the number of any lower chord panel, as exhibited in Fig. i, the general expression for any lower chord stress will be : n' =z - {N - (n' - i)\ (IV + W,) tan a . (11). 20^!" SPECIAL NON-CONTINUOUS TRUSSES. If the number of panels in the truss is even, the greatest — ^— "^l but if that number is odd, the value of ri will be greatest value of ■n K ^ will be — . 2 The upper chord will evidently be subjected to compression, and the stresses in that member can be written immediately from the lower chord stresses. If the truss be divided through any main panel, as a2,bi, c4, etc., it is at once clear that the only horizontal stresses in the section are those in the two chord panels cut, since the posts that are cut are vertical, and have no horizontal components of stress. The upper and lower chord stresses are, therefore, equal in pairs, but of opposite signs. Hence : («) - - (2) (^) = - (3) (^) = - (4) etc. = — etc. !^ (12). In the case of an even nurpber of panels in the truss. Fig. 2, the stress in the centre pair of panels of the upper chord will be greater than the result given by Eq. (ii)with the greatest value of n', by the amount ^ {W' + W^) tan a. This is for the reason that two inclined main web members intersect at the middle point of the lower chord, while their upper ends include the centre pair of panels of the upper chord. The horizontal component of the stress in each of these inclined main web members \s h {W -{- W^ tan (y. If the end-posts of the truss are vertical, as shown in Fig. 2, the stress in the end lower chord panel, o, is zero ; but the Fig. 2. stresses in the panels 1,2, 3, etc., are given by Eq.(i i) precisely as before. Then, also, in the same general way as before : TRUSSES WITH UNIFORM PANEL LOADS. 2(V (^)--(2) V. .... . (13). etc. = — etc. ) The expressions for the web stresses will remain unchanged •and as given by Eqs. (3) and (4). As the end-post, EP, is now vertical : {EP) - - {W' + Wy) + W . . . (14). The central vertical member, P^, is, strictly speaking, not a truss member for the fixed load ; hence its stress due to fixed load is simply the amount of that load applied at its end. No formula is to be applied. Case II. — Moving- Loads. . The moving load will be taken as passing on the span, panel by panel, from A towards B, Fig. i. When the first panel load is at the foot of c, the second will be at the foot of Ti. Hence the stress in c, will be the shear in that mem- ber, /. e., w ~ + w -J-, multiplied by sec oi\ ox\ P (c,) = {i + 2)w J sec a . . . (15). When the head of the train is at the foot of c^ there will be panel loads at the foot of /\ and 71, and the shear in c-i will be (zv ~ + w ~ + w ^j ; hence : (cs) = {l + 2 + ^) wj sec a . . . (16). Similarly, for any other inclined web member (except the ^end-post) with the subscript n : {T„) = (i + 2 + . . . + n)w y sec a . . (17). 20/ SPECIAL NON-CONTINUOUS TRUSSES. The general expression for the sum of the series in the parenthesis gives: {T,) = n{^yvo^jseca . . . (i8). As the shear in any incHned web member is equal to the compression in the post whose upper end it meets in the upper chord : (/-„) = n (^) «. f (.9). As Zi is simply a hanger : {T,) = w , , . , , , (20). The stress in EP is : N {EP) = — w sec oc (21). The stress in any lower chord panel whose number is w'can be taken directly from Eq. (11) by placing w for {W' + Wi). Hence : («') = — \N — {n' — i)] zu tan a . . (22). The upper chord stresses wiU be found from the lower, as indicated by Eq. (12); and the greatest value of ;/' will be N N — I -when the number of panels in the span is odd and when that number is even. If the end-posts are vertical, as shown in Fig. 2, the relation shown by Eq. (13) still holds, while the web stresses remain unchanged. The vertical end-post stress becomes: {EP)='^w (23> TJiUSSES WITH UNIFORM PANEL LOADS. 20g Case W.a. — Moving Loads — Deck Span. If the moving load traverses the upper chords, as indicated in Fig. 3, the post {P) subscripts shown in Figs, i and 2 are to be increased by one, so that the post and tie meeting in the lower chord will have the same subscript. Fig. 3 ex- FiG. 3. hibits the notation required. The expressions for the stresses in the main ties and posts, T and P, will be precisely as given in Eqs. (18) and (19) : ( 7;j = « ( ) w -J sec a {Pn) = n n 4- I w (24). In the cases where a counter member, c, acts the part of a post, there is a little ambiguity in the post stress, and it may be advisable to make it equal to the shear in T^-x in- creased by w. In that case: (i^„)=|(..- 1)^^+1}^. No other change is needed in the formulae for the through truss to meet the " deck" condition, except that. 1\ will have no moving load stress, and if the end-post is vertical Eq. (23) will take the form : (^-)=(f-:) w. (25). As there is more or less concentration at a panel point with 2Qk SPECIAL NON-CONTINUOUS TRUSSES. an actual moving load, it is usually advisable, in the case of vertical end-posts, to take : {EP) = (^ +i)w.... {2Sa). Case III. — Wed Members zvith Two Inclinations — Fixed Loads. This case is represented by Fig. 4, in which the end-posts and the main web members parallel to them, i. e., the web members in compression, have the inclination a' ; while the main web members in tension, such as 1 -^ to T^, have the inclination a' to a vertical line. Inasmuch as the truss should be symmetrical, however, the members P^ and 7^ of the centre panel must have the same inclination, a, to a vertical line. Strictly speaking, therefore, the web members have three inclinations. For the reasons explained in the preceding cases, the fixed loads will be balanced, and, hence, there will be no stresses corresponding to those loads for the members Zi to 7^, inclu- sive, and Px to P^, inclusive, or for members corresponding to them for a different number of panels from that shown. Inclination of EP to vertical — a. Inclination of T» to vertical = a" . Inclination of Pt or Ti to vertical = a. As before, W will be the upper chord panel fixed load, and W^i the lower chord panel fixed load. Also, / = panel length, /= length of span, and N = i = total number of loads sustained by the truss on the lower chord. The number TRUSSES WITH UNIFORM PANEL LOADS. 20i of loads sustained in the upper chord will be {N + \). Hence, the reaction at either A or B will be : jV A^ 4- I R, = - U\ + '—^^^ W . . . . (26). By using the same process employed in establishing Eq. (2), the shear in any member, T, with the subscript, n, will be: S=n{W' + IV,) - f^lVr + '^^^— -^ W'^ = (^n-'-^y_lV'+lV,)-'^ .... (27). The second parenthesis in the second member will b^- recognized as the reaction R . Similarly, the shear in any member P with the subscript n will be : IIV N + I \ S' = n{W' + W,) + W - r^ W, 4- — ^ W'\ = {n-^){^' + m+^-^ ^^^)- Hence, the stress in any member T between the centre panel and the end of the truss, i. e., in Fig. 4, from T^ to T^, inclusive, is : i^n)= \{n-^y,W' +W,)-^lseca" . . . (29), Similarly, for any member P between the centre panel and the end of the truss : iPn)=- ][n-^){lV' + W,) + ^\seca'. . . (30). N And for the centre panel, for which n = — : ^ 2 w {Pd = m=--^seca .... (31). lOJ SPECIAL NON-CONTINUOUS TRUSSES. If the number of panels in the lower chord is even, so that a panel point is at the centre of the span, there will be no members with the inclination a, and Eq. (31) will not be needed ; Eqs. (29) and (30) will then give all the fixed load web stresses. Eq. (30) will give the end-post stress (^P), by N giving ;/ its proper value, — . The equation then becomes : fN W'\ {EP) = ('J {W + W,) + -^ j sec a . . (32). The chord stresses due to the fixed loads can best be written by the method of moments. If the number of any lower chord panel be indicated by ;/, the origin of moments for the stress in that panel will be the panel point over it in the upper chord, and the distance of the moment origin from the nearest abutment (the lever arm of the reaction R^ will be (;/' — \) p + d tan a', ^^ being the depth of the truss. The distance from the centre of the {ri! — i) loads in the lower chord to the same origin will be \ [n' — 2) p + d tan a ; and the distance from the centre of the (>/ — i) loads in the upper chord to the same origin will be \ n'p. The lever arm of («') will be d. The equation of moments about the origin described will then be : {n')d= R, I («' - i)p + dtan a \ - (;/' - i)W, |i(V - 2)p + dtana']^ - {n' - i)W' "^ p . . (33). By substituting the value of R^ from Eq. (26) and then dividing both members by d, remembering that/ = d {tan a + tan a"), the following expression for («') will result: («') = ^ ^iN-n' + i){lV' + IV,) + IV' (tan a' + tan a") + I («' - i)W, - ^{W, + W) - ^ I tan a" . (34). 7-A' (/SSES WITH UNIFORM PANEL LOADS. 20k The moment origin for any upper chord panel, U^, is at the panel point underneath it in the lower chord, distant ?ip from the abutment. The lever arm of the {n — i) lower chord ft'f) panel loads will be — , and that of the n upper chord panel loads will have the value ^ ~ + d tan «". The equation of moments will then be : {U^)d = R,np - {n - i)W, f - nW [(« - i)^ + dtan a-" I (35)- By inserting the value of R, from Eq. (26) and remembering that p = d {tan a' + tan a") : Un = - \{N- n + \) {W + W,) + W {tan a' + tan a") — nW tan a" (36). Eqs. (34) and (36) will give all the chord stresses in the truss except that in the centre lower chord panel, which will be called {m). The fact that the two web members over it are equally inclined to a vertical makes a separate expression necessary for that panel. By taking moments about the centre panel point of the upper chord there will result; (m) d = Ri '-^ — ~Wi—p -W ^ ^ 2 24^2 4 Hence : l^^dtana'^ (37). (^) = [^(^+^) ( PT' + FTO + ^^ ^'] {tan a' + tan a") - ^I'V tan a" (38). 2 20/ SPECIAL NON-CONTINUOUS TRUSSES. Eq. (38) will give the stress in any centre lower chord panel, when there is one. It will evidently not be needed when the number of panels in the lower chord is even. In the application of Eq. (34) the greatest value of n will N . N + \ be — for an odd number of panels in the span, and for 2 ^ ^ 2 an even number. If the end posts, EP, are vertical, tan a becomes equal to zero, but no other changes are required in any of the preced- ing formulae. Case IV. — Case III., with Moving Loads. As heretofore, w will represent the uniform panel moving load. The latter will be taken as passing from A to ^ in Fig. 4. All notation will remain as in Case III. With the progress of the moving load from A to B the stresses which it produces in all the members marked 7i, T.^, etc., will be ten- sile, but those in P^, P-i, etc., will be compressive. Between A and the centre of the bridge, therefore, the moving load will produce stresses in the web members of a kind opposite to those caused by the fixed load. If, then, the moving load stress of one kind in any member is greater than the fixed load stress of the other kind in the same member, that mem- ber must be so designed as to sustain both tension and com- pression, i.e., it must be countcrbraccd. The resultant moving load stress in any web member of the half of the truss first traversed by that load will be, then, the excess of the great- est latter stress over that produced by the fixed load. The point at which the counterbracing is to begin will depend upon the numerical data for each case, but it is to be found by the preceding method. The moving load stress in any web member, T^, of Fig. 4, between the end A and the centre panel, is given at once by Eq. (18): (7;)= +«(--+ ^)tc.^..r.^" . . . (39«\ TRUSSES WITH UNIFORM PANEL LOADS. 20m Or between the end B and the centre panel : (T;) = + n \^-^-) w^jsec a" . . . (39). Or for the centre web member corresponding to T^ : ■{Tm) = + m i^L——^ li'j-sec a . . . (40). Similarly, Eq. (19) will give the post stresses between the end A and the centre panel : {P^ = — n { — ^ — \ w Y sec a" . . . (41a). Or between the end B and the centre panel : {P^)^-n{^-^)zvtseca' . . . (41). And, for the centre panel of an odd number: {P^) = - VI (^^^) IV ^ sec a . . . (42). In case of an even number of panels in the span, Eqs. (40) and (42J will not be needed. The greatest chord stresses will exist with the moving load over the entire span, and they can be at once written from Eqs. (34) and (35) by making W equal to zero and then writing iv for ]\\. By thus adapting Eq. (34) for the lower chord : Vn («') = -\- w\ — {N — n' + i) {ta7i a + tan «") + (^«' - - - l) tan a" (43). The same process of adaptation applied to Eq. (35) will give: (Un) — — w - {N— n + i) {tan a + tan a") . . . (44). 20;/ SPECIAL NO/V-CONTINUOUS TA'USSES. Again, applying the same process to Eq. (38) : (m) = + w — ^^-5 ' {tan a + tan a ) . (45). o Eq. (45) will give the stress in the centre panel of the lower chord when the number of panels in the span is odd, but that equation will not be needed when the number is even. These complete all the expressions for stresses required for this truss used as a through structure. Case V. — The Truss of Fig. 4 used as a Deck Structure, Fixed Loads. The formulae of Case III. are to be applied for the stresses in this case precisely as they stand, with the notation as shown i?i Fig. 4. Moving Loads. When the moving loads traverse the upper chord, the notation of the web members will be represented by Fig. 5, which is like that in Fig. 4, except that each pair of braces ^ Fig. 5. intersecting in the lower, or unloaded, chord will be affected by the same subscript. The remaining notation will be the same as that used in connection with Fig. 4, including the uniform moving panel load w, and depth of truss, d. Let p — d tan (x" = p" , Let p — d tan a = p' . 7RUSSES WITH UNIFORM PAN EI LOADS. 200 As in Case IV., the moving load will be taken as passing across the bridge from A to B. The shear in any web mem- ber, T,^, between A and the centre panel of an odd number in the lower chord will be the same as that indicated in £9.(39^:) diminished by 7iu' In other words, that shear is the saniv as if the upper chord panel points, with their moving loads, were first placed over the corresponding panel points in the lower chord and then moved back to their actual positions ; in the first of these positions the shear would be indicated by Eq. (39^), and the movement back to the actual positions p" would require the diminution mv y. Hence, the stress in any Z,, between A and the centre panel will be : {T„) = + ?i n + I + niv P P ' P' . w J sec a — nw j- sec a 1 2 7 j sec a (46). The stress in T^ [in = | JV) at the middle panel, corre- sponding to Ti in Fig. 5, will be: {Tm) = +mw^^^ —^ "j]^ sec a . . (47). The greatest shear in the P„^i at the middle panel, corre- sponding to /*5 in Fig. 5, will exceed that in T^n by i zu. Hence : {Pn.^l) = -'^ P (m + I p"\ I sec a (48). Between the middle panel and the end of the span B, with an odd number of panels in the span, T„^^x^ corresponding to 7^5 in Fig. 5, will have the same shear as {Prn + \)\ hence: ( 7;„ + 1) = + w p m + I / V 2 I + - 2 sec a" (49). 20/ SPECIAL NON-CONTINUOUS TRUSSES. For the members T^ with subscripts greater than {in + i), i.e., where n'>{m + i) : (r„) = + w [^ j ;« (^— ^ ^ j + (« - ?« - I) c^ + ;;/ + 2 /' + j^"^ a .-.(7;)= + w 4 1 — '^ ^ — {in + i) — in i— / 2 ^ / -("-■)|V7] sec a ■ (5o> With an even number of panels in the lower chord, Eqs. (47), (48) and (49) will not be needed. In that case the P^ and T„, {in — ^) N + i) ) which intersect the lower chord at its centre panel point will have the same shear given by Eqs. (46) and (47). Hence : (51). (52). / -r \ , p \ m + I p" {Tra) = + mw ^ -| — ^ ^ ]^sec a /r> \ p \ in + I p" I „ For values of « > »« : fm + I /"\ , fn + m-\- I p'\~] m [—^ -r_^^in- m) (^ ^ ^JJ sec a ~n{n + i) /" -/' /n „ , , — ^^ ' - ni n ^— \sec a' (53), L 2 / pA ^^^' .•.(r„)= + The greatest shear in any /^,„ for which n < \ N, or less than \{N -\- i), according as the number of panels in the TKUSSES WITH UNIFORM PANEL LOADS. 20q span is odd or even, respcctivel)-, i.e., between the end A and the centre panel, will be that indicated by Eq. (46). Hence : / ( « + I /" ) ^ ^ / ( 2 / ) sec a (54)- Similarly, when the number of panels in the span is odd, and where « > (w + i), Eq. (50) gives: i^n) W ' p \ n {11 + i) .7 '( 2 P" — p' P' ) i~1 — {m 4- i) — m . — {n — i) - } + - \ sec a' . (55). The end-post stress is {hP) = zu sec 01 (56). EQ- (55) ^^''1 give Eq. (56) by making n = N+ i and N m = When the number of panels in the span is even and n > ni, Eq- (53) gives: (P.) = _ -/ [i(^) - « ^' - „ t^ ,,, „. . (57). The end-post stress is iV+ I {EP) = — z^ sec a' (58). Eq. (57) .will give Eq. (58) by making n = JV + i and m = i(iV+ i). These expressions complete all that are required for the movinof load stresses. 20r SPECIAL NON-CONTINUOUS TRUSSES. Case VI. — Truss with all Web Members Equally Inclined to a Vertical Line — Warren Girder. I. — Through Span — Fixed Load. This truss is illustrated by Fig. 6, in which the notation is precisely like that in Fig. 4. The inclination of all web members to a vertical line = a. / = length of span, and/ — panel length. d= depth of truss, and N = i. P tan ar = — , ; and sec a = '^ \ -\- tan^. 2 d W' = uniform upper chord panel, fixed load. Wi = uniform lower chord panel, fixed load. The reaction at either end of the span due to the fixed load alone is given by Eq. (26) : TV" N + I R, = fw,+'^W .... (59). ^^■y" The stress in any tensile web member Tn, between that end o( the truss for which n is the greatest and the centre of the span, is given by Eq. (29) by making a" = a, whether the numb.^ of panels in the span is even or odd, or whether the %uss is deck or through. Hence : {T„) = + -^^ W' (IV' + W,)--^} sec a (60). T/? LOSSES WITH UNIFORM PANEL LOADS. 20S The equally general value of (/*„) is given by a similar change in Eq, (30). Hence: W = -\^{n-^{W' +W,)+^\seca , (61). If n' represents the number of any lower chord panel, then Eq. (34) will give the expression for the stress in it by making tan a = tan a" = tan a. Hence : («') = + [(«' - I) {N- n' + I) {W + W^ ^\{N{W^^ W) + W')-] tan a . . , (62). In precisely the same manner Eq. (36) will give, for the upper chord stresses: {(7,,)= -n{N-n + i){W' + W,) tana . (63). The end-post stress {£P) will always be found by making n = N"m Eq. (61), which will then be reduced to : {EP) =-f^{W,+ W')+ ^) sec a = R,seca . (64). //. — Through Span — Moving Load. Fig. 6 will again be used, and the uniform panel moving load w will be taken as moving from A to B. The stress in any member T^ will be at once given by Eq. (18): (r„)= +n{^-^)w^^seca . . . (65). Eq. (19) will then give the stress in any post (/*„) : (P„)= -«(^)z^|..rr.. . . (66). Counterbracing must begin when the moving load stress in 20/ SPECIAL NON-CONTINUOUS TRUSSES. Tn or P^ becomes equal to that of an opposite kind produced by the fixed load, as explained in Case IV. The end-post stress {EP) is found by placing n =^ N in Eq. (66) : N {EP) = — — zv sec a i^7) The stress in any lower chord panel ;/' can at once be writ- ten by making W = o in Eq. (62), and then writing w for W^ : («') = + (;/' - i){N - n' + i) + N- w tan a (68). Precisely the same operation applied to Eq. (63) will give any upper chord stress U^ : {Un) = — n {N — n + \)w ta^i a . . (69). Eq. (69) completes the expressions for the stresses in all the members of the truss. Case Mil.— Case VI. as a Deck Span. I. — Deck Span — Fixed Load. The formulae for web and chord stresses remain exactly as found in Case VL, I., ivitk the notation as given in Fig. 6. II. — Deck span — Moving Load. When the moving load passes along the upper chord from A to B, the notation of the web members must be as shown Fig. 7. in Fig. 7. It will be observed that the only change from Fig. 6 consists in giving the same subscript to any two web mem- bers which intersect in the lower chord. TRUSSES WITH UNIFORM PANEL LOADS. 2011 The required formulae for the web stresses can be at once written from those of Case V., in which p' — p" = )^ p. As usual, the uniform panel moving load will be w. By making P" •V = i in Eq. (46): Similarly, Eq. (54) gives : n^ p + — w -. sec a . 2 / «- p — w ~ sec oc . 2 / (70). (71). The end-post stress {^EP) will result from Eq.(7i) by mak- ing ft = N + I : tV 4- I {hF) = — — w sec a . (72). Any lower chord stress {ti) can be written from Eq. (62) by making W^ = o and then writing w for ]\" : [n) — [{n — \){N — n ^ 1)+ ^ (A^ -r- i)] w tan a . (73). In the same manner Eq. (63) gives, for any upper chord stress : [Un) = — n {N — n + i) u> tan a . . . (74). All formulae for the deck truss are thus made complete. Case VIII. — Vertical Ties and Inclined Posts. — Hoive Truss. I. — Through Span — Fixed Loads. The truss to be treated in this case is shown in Fig. 8, in which all inclined web members are in compression. If the a 6 c d \ ' \ T '-' Fig. compression members and chords are built of timber, it is the common Howe truss. The notation will be that already used. 20Z/ SPECIAL NON-CONTINUOUS TRUSSES. W = uniform upper chord panel fixed load. W^ = uniform lower chord panel fixed load. / = length of span. / = panel length. d = depth of truss. Tan a — ^. Sec a = y i ^-cL N=L u P The posts P.y to /'j, inclusive, in Fig. 8, are counterbraces, and will not be stressed by the fixed load ; hence they are to be neglected in finding the fixed load stresses. The reaction at each end of the truss due to the fixed load only, will be : Rr=-^i}V ^-W,) .... (75). It is agreeable to the notation of the figure to consider the main web members in the right half of the truss. The shear in any inclined web member P^ between the centre panel and the end B of the span will be the reaction at A^ less the loads n (W + W^), exactly as indicated by Eq. (3) of Case I. Hence: {P„)= - {ri - j^ {W + W,) sec a .... (76). Similarly : (zj- + [(«-^)([^'4- fro- w^] . . . {77)' In Eqs. {76) and {77), n '^( — -f i j for an odd number of panels in the span; and n > ( ■ ) for an even number of panels. In the latter case, however, the fixed load stress 1 • 1 11 • 1 . • T A^ + I m the middle vertical tie is, it ?/? = : 2 {Tm)= + W, (78). tr'usses with uniform panel loads. 2(ym The end-post stress is found by making « = iV in Eq. (76): {EP) = -~{W' -^ IV,) sec a . . (79). The lower chord stresses will be found by taking mo- ments about the upper chord panel points. If n' is the num- ber of any lower chord panel between the centre and end panels, inclusive, the equation of moments will be: («') d=-{lV' + IV,) . 71' p - n' {IV + W,). ^^-^^p ... [n'^^^ ^'l^i^N-n' ^ i){W' + W,)tana . (80). which is identical with Eq. (11). The upper chord stresses at once follow: (d) =-(i) 1 {c) = - (3) etc.= — («') 1^ (81). A vertical end-post at each end of the span will carry sim- ply the load at its top, and none of the preceding expressions will be changed. //. — Through Span — Aloving Load. The uniform panel moving load will be w. The greatest moving load stress in any counterbrace (/*„), for which n < — for an odd number of panels in the span, or N — I n <_ for an even number of panels, will be exactly sim- ilar to that given by Eq. (18) : (/*„)= — n (^ j zv^. sec a . . . (82). 20X SPECIAL NON-CONTINUOUS TRUSSES. If n = N, {P„) = {EP) = - — w sec ex . (82«). Eq. (82) also gives the stress in any main post P^ as it stands, by giving n the proper value, which will be greater than the limits indicated above. Any main member Z'„ will have the stress: (7;)= +n(^^)^^^~i .... (83). The lower chord stresses can be written at once from Eq. (80): {n') = -I {N — n' ■{- i)w tan a . . (84). As before : W=-(3) f etc.= - («') J (85). Expressions for all the stresses are thus completed. ///. — Deck Span — Fixed Load — ]\[o7'ing Load. Fig. 9 shows this truss as a deck structure. The notation is like that shown in Fig. 8, except that the subscripts of the vertical braces {T^ are so written as to make each pair of web members intersecting in the lower chord take the same subscript. TRUSSES WITH UNIFORM PANEL LOADS. 20/ Fixed Load. The expressions for all the stresses due to the fixed load are precisely the same as those in Case VIII., I., Eqs. (75) to (81), inclusive, but the subscripts for the vertical members Tn must be used, as shown in Fig. 8. Moving Load. All the chord stresses due to the moving load will be given by Eqs. (84) and (85) exactly as they stand, as the notation for the chords is the same in Figs. 8 and 9. All the web stresses due to the moving load will be given by Eqs. (82), (82^), and (83) just as they stand, but the iveb notation shown in Fig. 9 must be used. Vertical end-posts will change nothing, as they will carry that load only which rests at their tops. Tables for Computations. The following Tables, which explain themselves, will enable the preceding equations or formulae to be applied to actual cases with considerable economy of labor. The numerical value of the expression {n — i ) {N — ;?' + i) can readily be taken from the first of the Tables in the following manner. Clearly [ji — i) (.V — n -\- \) —{ri — i) {N — («' — i) — I + i) = {n — i) [A^— {n — i)]. Hence, if {n — i) is placed for n in the Table, the desired result will be the tabular result diminished by {n' — i). As an ex- ample, let iV = 17 and ;/ = 7. Then take ;/ = ;/'— i = 6. The tabular result is 72, which, diminished by 6, is 72—6 — 66 = in' — i) {N — n' + iV Indeed, the desired result is always found in the Table in the column with heading {N — I) and opposite n = («' — i). 202 SPECIAL NON-CONTINUOUS TRUSSES. Values of n {N— ti + i). n I. .. 3... 4... 5--- 6... 3 3 4 3 4 4 6 6 4 5 5 8 g 8 5 6 6 10 12 12 10 6 7 7 12 15 16 15 12 7 8 8 14 18 20 20 18 14 8 9 9 16 21 24 25 24 21 16 10 10 18 24 28 30 30 28 24 18 10 II II 20 27 32 35 36 35 32 27 20 II 12 12 22 30 36 40 42 42 40 36 30 22 12 13 13 24 33 40 45 48 49 48 45 40 33 24 13 14 M 26 36 44 50 54 56 56 54 50 36 26 14 15 '5 28 39 48 55 60 63 64 63 60 55 48 39 •28 15 16 16 30 42 52 60 66 70 72 72 70 66 60 52 42 30 16 17 17 32 45 S6 65 72 77 80 81 80 77 72 65 56 45 32 17 18 18 48 60 7? 78 84 88 90 90 88 84 78 70 60 48 34 18 19 36 51 64 8^^ 91 c6 99 ICO 99 96 9> 84 75 f4 51 36 '9 20 20 38 54 68 80 90 98 104 Toa 8 10. . . no Tn8 104 98 90 8n i5--- 16 68 18 :::: .;;; 54 38 j 1 j 1 Values r /« + I \. n = I I 2 3 3 6 4 10 5 15 6 21 7 28 8 36 9 45 10 55 II 66 12 78 13 9' 14 105 15 120 16 17 18 171 19 20 « (« + i) 2 136 153 190 210 Values ■H^y n = I 2 I 3 3 4 6 5 10 6 15 7 21 8 28 9 10 36 45 II 55 12 66 13 78 14 15 105 16 17 136 18 19 20 « (« — i) 91 120 153 171 190 2 Art. 7. — Position of Moving Load for Greatest Shear and Greatest Bending. That method of computation which treats the moving load as composed of a system of isolated weights requires some simple method of finding the greatest possible shear in a given panel for a given system of loading. Among the first to use such a method was Theodore Cooper, C. E. ; and the results of the following investigation are the same as those determined by him. The method and formula first developed apply to any single system of triangulation so far as the web stresses are concerned, but for the chord stresses they only apply to such a system when composed of alternately vertical and inclined members. Subsequent modifications for web members all inclined will be made for the chord stresses. GREATEST STRESSES. Case I. 21 Let the moving load consist of the advancing weights Wy, IVi, W3, .... IV,^ separated by the distances a, b, c, d, etc. ; then let the weights W^i, W^, .... Wn' be Fig. I. found in the panel DC, in which it is desired to find the greatest possible shear, and at the distances b' , b" , . . from C ; it being understood that the moving load advances from F toward G. The last load W^ is found at the distance X from F. The length of span FG is /, while the length of panel Z>C is/. With the assumed position of loading, the reaction R dit G will be : R = \ W, a + b + c + + W. + W, b + c + d -{- c + d ^ + ^„f + X + X + ;r . . . (I) The parts of the weights W^, W^, . . resting on 22 SPECIAL NON-CONTINUOUS TRUSSES. h' h" DC, which pass to D, are W^ -, W^ —, . . . . Hence the shear in the panel DC will be S = R-(W,^ + W,^'+ ) .... (2) If the train advance by the amount A x, the new reaction R', at G, takes the value : R' = R + {IV,+ IV,+ W,+ . . . .+W,)^;. (3) and the new shear will become : S'=--R'-{m^-+W,~+ . . . )-{W^+W,+ . . .)— (4) V p ' P -{W,+ W,+ . . . )^ (5) Or, S' -S = ~\^{W,+ W^^- W,+ . . . + W^) -{W,+ W,. ..){/-= n)]^ (6) Whenever S' — S becomes equal to zero, S'vvill be either a maximum or a minimum. If the difference is positive just before it becomes equal to zero, S' will be a maximum, and that is the only case of interest in the present connection. Hence, by placing 5' — 5 equal to zero, the following condi- tion is obtained : n{W, + W., + . . .)=W,+ W,+ W, + . . . + Wn ..{7) The shear in the panel in question will therefore take its greatest value when n times the moving load which it contains is equal to, or most nearly equal to, the entire moving load on the bridsre. GREATEST STRESSES. 23 That equality will seldom or never exist unless one of the weights W is placed on a panel point, since IVi + IV2 + . . . is seldom or never an exact divisor of the entire load on the bridge. If a weight rests on a panel point, any part of such a weight may be taken as acting in one adjacent panel and the remainder in the other ; the desired equality may thus be obtained. In case Eq. (7) should hold, the position of the moving load is a matter of indifference so long as the panel in ques- tion contains the same number of loads JVi + IV^ + ■ • •> as there is no trace of d', d", etc., in that equation. A load may then always be taken as resting at the rear extremity of the panel where the greatest shear in it exists. These considerations frequently essentially simplify com- putations. When the value of x has been found for the position of the greatest shear, the latter being determined by the preced- ing method, Eq. (2) may be put in the following convenient shape by the aid of Eq. (i) : S='^[IV^ a +{Wi+W,)d + {IV,+W, + W,)c+ . . . . P -f . . . +{IV, + W, + . . . +IV„._,)?] . (8) The sign (?) stands for the distance between the wheel con- centrations Wn'^i and Wn', since the latter rests directly at the panel point in question. It is thus seen that all the parts of Eq. (8) may be taken at once from tables, except that term involving x. Case II. In the preceding case it has been supposed that for the greatest shear in DC, the front weight W^ is found between /? and C; but let JV^ + IV2 + etc., be supposed between D and G. 24 SPECIAL NON-CONTINUOUS TRUSSES. With the notation remaining the same as before, the shear ^ will become: s = R-{w, + w, + Qtc.)-{wA + w,j+ . . .);(9). while S' takes the value : -{W,^W,+ . . . )~. Hence for a maximum, the following expression must never become negative : -{W, + W,+ . . . )(^-^ = n^^=o. . .(10). But Eq. (lo) is identical with Eq. (7). Hence, the same conditions for a maximum obtain wherever may be the head of the moving load. The second member of Eq. (8), how- ever, must contain the negative sum of all the weights be- tween D and G. EXAMPLE. If each one of the weights W^, IV^, etc., is equal to any- other, i.e., if they are all uniform, and if a — b — c — d = . . . = p, Eq. (7) shows that the front weight W^ must be taken at the first extremity of the panel in question. The same result holds if the first weight VVi is not exceeded in amount by any that follows it, provided that a, b, c, etc., still equal /. In cases where the same system of concentrated loads is to be used for a number of spans, it will be shown that the tabulation of the products of the sums of the weights Wi, W^, etc., by the distances a, b, c, etc., can be advantageously used GREATEST STRESSES. 25 to shorten and simplify computation. In other cases, how- •ever, the quickest and simplest method is partly graphical ; it is as follows : ( ■% F ^ ^ 3 B ' f "^^^^^A 1 \ VI W Fig. 2. 1 In Fig. 2 let AB be the length of span, and W any weight resting anywhere in the span. Erect a vertical at B and let BC represent W by any convenient scale ; then draw the straight line^C The vertical intercept WD w'lW represent the reaction at B due to W, by the same scale on which BC represents that weight. In the same manner if 5/^ repre- sents W^, then W^E will represent the reaction at B due to W^. Thus there must be as many verticals BF, BC, etc., as there are different weights resting in the span, and the total reaction at B, for any given position of the moving load will be the sum of the vertical intercepts WD, W^E, etc., erected at each load ]V for that position. h' h" The negative shears Wx -, W^ —, etc., appearing in Eq. (2) are most readily found in the same manner. If GH, Fig. 3, is a panel length and [Fany weight represented by the verti- cal line GK, drawn to any convenient scale, while WH is equal to b' , b" , etc., then the vertical intercept WE between G"// and A'// will represent W -It G due to W, , /k— , etc., I.e., the reaction 26 SPECIAL NON-COi\ TINUO U S TRUSSES. If the reaction at B, Fig. 2, is then given by ^IVB, and /' A" W\ — \-lV« 1- etc. = ^ WE, Fig. 3, the shear (see Eq. (2) ) / ' P will be : S=2WD- 2W£. If Figs. 2 and 3 are drawn on profile, or cross-section paper, the shears for any span can be found with great ease and rapidity. Position of Moving Load for Greatest Bending Mojuent. The Fig. and notation of the preceding cases will be used in connection with this. Moments will be taken about the panel point C, horizontally distant /' from G. There will be supposed to be n weights in front of C {i. e., between C and G), and the weight W^„. will be taken at the distance x' from C towards G. The bending moment M will then take the value : M=Rl' - Wi(a + I? + c + . . . + x) + W^{ b + c + . . . -\-x') + W,,x'. Or, after taking the value of R from the preceding cases: M=j[W,a + {W^^ W,\ d + (W, + M^, + Ws)e+ . . . + {w, + iv, + w,+ . . . ^iVn)x]-w,a-{w^ + jj:;)d -{lV, + 1V., + W,)e- . . . -{W,+ W^ + W^+ ... ^ W^.)x' (Ill If the train advances by the amount Ax, the moment be- comes : M' = M^^-j{W, + W^ + W^+ . . . ^W^) Ax-{W, + W^ +-...+ W,,) AX, (12). GREATEST STRESSES. 27 Hence, for a maximum, the following value must never be negative : -f . '. . +W„.)\=o (13). Or, the desired condition for a maximum takes the form : I ~ W,+W^ + W^+ . . . -\-l^V„ (H). It will seldom or never occur that this ratio will exactly exist if Wn' is supposed to be a w/io/e weight ; hence, IV,,. will usually be that part of a whole weight at C which is neces- sary to be taken in order that the equality (14) may hold. It is to be observed that if the moving load is very irregu- lar, so that there is great and arbitrary diversity among the weights W, there may be a number of positions of the train which will fulfil Eq. (14), some one of which will give a value greater than any other ; this is the absolute maximum desired. Since IV^' will always rest at a panel point for the greatest bending moment, x' in Eq. (11) may always be put equal to zero when that equation expresses the greatest value of the moment. The latter then becomes: M = j[lV,a + {IV,+W,)d + . . . + {W,+W,+ . . . + W^)x]-lV,a-{U\+JV,)b- . . . -iU\+lV, + .. . . +^„.-i)? (15)- In this equation, of course x corresponds to the position of maximum bending, while the sign (?) represents the distance between the wheel concentrations r^,j_i and JV„. It is known that for any given condition of loading the greatest bending moment in the beam or truss, will occur at that section for which the shear is zero. But if the shear is zero at that section, the reaction R must be equal to the sum of the weights {li\ + /f 2 + ^^ 3 + • • • + ^^«) between G and C; the latter now being that section at which the greatest 28 SPECIAL NON-CONTINUOUS TRUSSES. moment in the span exists. Hence for that section Eq. (14). will take the form : /' R I ~ w^ + w. + w,+ . . . + w,: or, the centre of gravity of the load is at the same distance from one end of the truss as the section or point of greatest bending is from the other. In other words, the distance between the point of greatest betiding for any given system of loading, and the centre of gravity of that loading is bisected by the cen- tre of span. If the load is uniform, therefore, it must cover the whole span. It will be observed that Eq. (15) is composed of the sums of Wi, Wi + W^, etc., multiplied by the distances a, b, c, etc., precisely as in Eq. (8), hence the same tabulation as there indicated may be used to advantage. Limitations of the preceding methods. The preceding methods are limited to a single system of triangulation. By the use of certain assumptions in refer- ence to the distribution of the loading between two or more systems of triangulations in the same truss, a somewhat simi- lar investigation might be made for such cases, but such analysis would not be rigorously exact. Hence it is as well to pursue the usual method and assume that each system acts as an independent truss, then place the moving load in such a position for each system that the front panel load for that system will be the greatest possible. This panel con- centration will then be the forward panel-moving load, and the succeeding ones may either be those concentrations which actually correspond to the forward one, or may be supposed to be composed of a uniform load equivalent to the concentrated one. The latter plan will be employed hereafter. Application of the preceding method to an all-inclined tveb system. As was observed at the beginning of this Art. the analy- sis for chord stresses, as already given, is directly applicable GREATEST STRESSES. 29 to a single system of triangulation in which a vertical web member is found in each panel. The general demonstration, however, is easy. If the web members are all inclined, the formulae, as al- ready given, are directly applicable in any case to the deter- mination of stress in that chord which does not carry the moving load, since moments are taken about the panel points of the chord traversed by the moving load. But let it be required to determine the position of the moving load for the maximum stress in DC, Fig. i, and the expression for the corresponding moment. As before, let the load move from F toward G. Let q represent the hori- zontal distance of D from H, i. e., q = KH ; evidently q is constant for the same span. Let x^ represent the horizontal distance from H of the first load to the left of D, and let that load be represented by W^„. That portion of the loads rest- inij in DC, which is transferred to D, is 2W-. I' will now f ' p represent GD 4- q. By taking moments about H, Eq. (11) will take the form : M ^j\W^a -^ {\\\ -V W^b + . . . +(^Fi + IF2 + . . . + W,^x-\ - lV,a-{lV,+ W,)d -... -{IV,+ W, + ... + m„)x, ~^2Wd' (16). By advancing the train Ax, since Ax = Ax^= Ab\ Eq. (13) will become : M' -M= AxY-^{W,^W.^ ... +^„)-(I^,+ I^2+...+ ^n) -^:s^l=o (17). The condition for a maximum or minimum then takes the shape : L _ ^1 + ^ 2 + . . . + m ^ + f^w (18). 30 SPECIAL NON-COA'TINUOUS TRUSSES. Eqs. (i6) and (i8) are the general expressions of which Eqs. (15) and (14) are special forms. After X and, hence, x-^ have been determined by the aid of Eq. (18), the maximum moment will be given by Eq. (16), in which the tabulations already indicated can be advantage- ously employed. Application of preceding methods to a system of concentra- tions followed by a uniform load. If the uniform load does not reach to the panel under con- sideration, which is usually the case, W^ in Eqs. (7), (14) and (18) represents the total uniform load on the bridge, but the formulae are in no wise changed. In Eqs. (8), (15) and (16), however, it is to be observed that while W^^ again represents the total uniform load, x will represent the distance from its centre of gravity to the end of the span {i. e., half the length covered by the uniform load), also that the distance between IVn and W^-i will be equal to x plus the space which sepa- rates Wn-x from the front of the uniform load. In the case of the existence of this uniform load it will happen that IVn^ will not rest at a panel point. The last term in the negative expressions of the second members of Eqs. (8) and (15) will then h& {W^ + W^ + . . . + IV„.) x' ; x' being the distance of W^' in front of the panel point C. Eq. (16) is general, and needs no change on this account. If the concentrations are so few that the uniform load ex- tends over a portion of the panel in question, the observa- tions made above still hold. But in addition to them, ^„. or W^n will represent the amount of uniform load in the panel, and x or x^ will represent the distance from its centre of gravity to the panel point. The interval or space between Wn_x or ^n-1 and W^' or W\ will then be the distance from either of the former to the centre of gravity of the uniform load. Finally, in Eq. (17) or (18) '^W will be either wholly or partly composed of uniform load. Modifications for Skew Spans. If a skew bridge is under treatment, the preceding methods GREATEST STRESSES. 31 apply in all respects, so far as the general principles are con- cerned. It will be sufficiently accurate in all cases to treat the Fig. 4. Fig. 5. moving load as if it were passing along the centre line LL L'L of each track. In the case of the single track bridge, Fig. 4, if the load is passing from right to left, the moving load does not rest on the truss CD until it passes the point C , and continues to act on that truss until it passes to the same distance to the left of D, it being borne in mind that all moving load is trans- ferred to the trusses by transverse floor beams placed nor- mal to the axis of the bridge. It results from these con- siderations that if the load passes from right to left in Fig. 4 and along LL, the reactions at D will be greater than a half of those at K by the amount of the half products 32 SPECIAL NON-CONTINUOUS TRUSSES. of the total load corresponding to those reactions by the ratio CC CD Hence, if R is any reaction at K and '^W the total load, the corresponding reaction at D will be : 2 ^ CD ' On the other hand, with the load moving in the same direction, the reaction at A will be : 2 -^CD In Eqs. (8), (15), and (16), then, there is to be written for {IV, + W,+ . . . +lVn)x the expression, {W, + W,+ . . . +W,){x±CC\ according to the direction in which the train is moving ; dut the negative portions are to remain ujtchanged. It is to be remembered that the quantity x is to be measured on the centre line. In the case of the double track skew bridge of Fig. 5, in which there are the two trusses AB and CD only, precisely the same observations hold. For one track, however, CC" is to be used and CC for the other. Separate computations, are to be made for each track for each truss. If there are three trusses in Fig. 5, each pair of trusses con- stitutes a single track bridge for the track between, and is ta be treated precisely as Fig. 4. If the skew is so great that one or more floor beams have their end or ends resting on the masonry, obvious modifica- GREATEST STRESSES. 33 tions must be made according to the preceding general prin- ciples. TJie G rap J deal Method. With convenient means for constructing an accurate equi- librium polygon with a large number of loads, this method is a very rapid one for either shears or moments. A perfect familiarity with the principles and operations of Art. 45 is here supposed. Let the entire moving load for a given truss be represented by the system of forces i, 2, 3, 4, and 5, in Fig. 6. They are given in actual position under the polygon PKO. P is the B 5 r^ pole, EF the load line, and PO an indefinite horizontal line normal to EF. As usual, the moving load is supposed to move from O toward P. The polygon PQK .... NO is formed in the ordinary manner by making its sides parallel to the lines radiating from P. For reasons that will pres- ently be evident, QF should be continued considerably beyond C, while NO should be carried somewhat below the horizontal line through P. After the constructions indicated have been made to any 3 34 SPECIAL NON-CONTIA'UOUS TRUSSES. convenient scale, let the span under consideration be laid off to the same scale by which the horizontal separations between the loads, i, 2, 3 and 4, etc., are laid down, and let indefinite vertical lines be drawn through the panel points, but let all this latter construction be made on tracing cloth. In order to avoid confusion, neither the panel points nor the vertical lines through them will be shown in the Fig. In the first place, let the position of the moving load for the maximum shear in some web member be determined by Eq. (7;, and in this position let the loads, i, 2, 3 and 4, be supposed to rest on the truss ; and let B\ represent the dis- tance between the last concentration and the right hand of the span {j. c, B4 is x of Eq. (8) ). Now let the tracing cloth be superimposed on the equilibrium polygon in such a manner that AB shall represent the span ; then erect the verticals BD and AC, and draw CD, to the latter of which PH is drawn parallel. FH will then represent the reaction at the left end of the span, /. e., at A. If from this reaction the negative shear shown in Eq. (8), or found by the method of Fig. 3 be subtracted, the result will be the shear desired in the web member under consideration. In this manner all the maximum shears may be found. Again, let it be required to find the greatest moment at a given panel point, for which the moving load has been found by Eqs. (14) or (18), to occupy such a position that the dis- tance from the right end of the span to the last concentra- tion (/. e., X in Eqs. (15) and (16) ) is represented by B' 5 in Fig. 6. Also with the position of moving load thus deter- mined, let it be supposed that the load 2 rests at the panel point considered. Now, let the tracing cloth be so super- imposed on the equilibrium polygon that A' B will represent the span, then erect the vertical lines B' D' and A'C\ and join CD' . Since load 2 was found at the panel point, at which the moment is to be determined, KL will represent the max- imum moment in question. Each linear unit in KL, meas- ured by the same scale to which the span and distances, t-2, 2-3, etc.. are laid down, will represent as many moment units as there are force units in the pole distance PT. If PT is GREATEST STRESSES. 35 in pounds and KL in feet, the FT X KL will be the mo- ment desired in foot-pounds. In the same manner all maxi- mum moments may be determined. In every position of moving load required, the vertical intercept between the closing line and equilibrium polygon, drawn through the panel point considered, will represent the maximum moment at that point. PH' drawn parallel to CD' will give FH' as the reaction at F, though it is of no special value in this con- nection. The application of this method to the different panels of a truss will give all the greatest chord stresses. This method is, of course, subject to all the modifications that have been outlined for the various special cases and conditions. It cannot, however, be applied to the mov'.ig load on skew bridges. The reaction for the centre line of the track must be reduced to the trusses, while the negative moments of the loads in advance of the panel point which serves as the moment origin remain unchanged. The gener- ality of this method is not, therefore, comple;te. The Maximum Floor-bcairi Reaction. The moving load iscarried to ea;ch transverse floor beam by the adjacent stringers. Hence each floor beam is a pier for two adjacent spans of string-trs, and it becomes necessary to determine that position of the moving load on those two spans which will subjec'c the floor beam to its greatest load. In the Fig. let a sect.ion of the beam be shown at R, while / and /' are the two adjacent stringer spans traversed by the moving load ; then let the ^s be measured from the right and left ends of /, and /', while W, IV\ etc, Jl\. W„_, etc., rep- resent the weigh ts or wheel concentrations resting in the two spans, the reaction R will then have the value: IV, xi + W^x.-, -h etc, Wx -^ VV^ x' -t- etc, . . R^. ^_1_. + ^- . (19). If the whole system of loading move to the left by the distance ^x, th e new reaction will be : 36 SPECIAL NON-CONTINUOUS TRUSSES. (JFi + W^ + etc.) A,r (^F+ ^i + etc.) A;»r ^ ~ ^ I, ^ I In that position which gives a maximum or minimum,, J^i — R = 0; hence : {Wi + JV2 + ^s + etc.) - ^{IV+ W + W" + etc.) (20). It will seldom happen that Eq. (20) will be satisfied unless ^5-■ -->U-- |<- 1 av — >i k- h ^ & x--^ _l r L. W, W2 etc. "" etc. w w J L nr _IL Fig. 7. a concentration rest on the poii?t R, so that the proper por- tion of it may be taken for one span or the other, precisely as in the problems of maximum shear and maximum mo- ments. Ordinarily the two adjacent spans afe equal, or /=/, W, + W, + etc. = W + W' + etc. . . (21). Eq. (21) shows that when the two span.'5 are equal, the amounts of load each side of R must also be eUual. After the proper position of loading has bee^n determined^ Eq. (19) will give the maximum reaction desirec^* Article 8.— Fixed Weight. » As the weight of a structure forms a very considerable portion of its total load, it becomes a matter -of importance / FIXED WEIGHT. 3/ to find at least an approximate value for it, when the moving load has been once assumed. The weight of ties, guard timbers or rails, rails, spikes, etc., may be taken at 350 to 475 pounds per lineal foot of single track, and forms an invariable part of the fixed weight, (/. r., weight of structure), since it is independent of span, length of panel, or depth of truss. With ordinarily heavy traffic and standard gauge, 400 pounds per lineal foot is usually taken. After having fixed the weight of track, the stringers are to be designed. A very little experience will enable their weight per lineal foot to be assigned in advance, so that the total load resting upon them may be used in making compu- tations. These track stringers are almost invariably plate girders, and the main object of the computation is to deter- mine the area of flange section. If that area, as computed, makes the weight of the stringer very different from that assumed, it will be necessary to again assume a weight, guided by the results of the first computation, and re-calcu- late for the flange area ; and then repeat the operation until sufficient accuracy is obtained. The weight of the track stringer thus obtained becomes a part of the load sustained by the floor beam. The weight per lineal foot of the latter is then to be assumed, and calcu- lations made and repeated, if necessary, precisely as in the •case of the stringers. With a very little practice the weights of the stringers and floor beams may be so accurately assigned in advance, that a re-calculation is seldom or never necessary. The lateral and transverse systems of bracing should next be subject to computation, and as the wind pressure is their only load, their own weight does not affect the operations ; hence no re-calculation will ever be required. The remaining calculations are those of the truss proper, and the accurate assignment of their own weights presents more difficulty than any other part of the operation. Ex- perience, however, enables even this weight to be quite closely taken in advance. For each single track of standard 38 SPECIAL KON-CONriNUOUS TRUSSES. gauge the weight of two through trusses, designed for the moving load taken in Art. 9, together with the lateral sys- tem, in pounds per lineal foot, may be approximately taken at five times the length of the span in feet, or double that amount for a two truss double track structure. If this is not sufficiently close, one or more re-calculations will be required. For spans over two hundred and fifty feet in length, this rule gives too small results. By thus designing the floor system, and lateral and trans- verse bracing before the computations are made for the trusses, it is only necessary to assign before each step, the weight of that part immediately under consideration, and, hence, enables the actual weight to agree very closely with the assumed, without re-calculation. If the bridge is of the deck variet}' and carries the ties di- rectly on the upper chord, so that they act as a transverse load on the latter, thus obviating a system of track stringers and floor beams, the total fixed weight will be reduced about 125 pounds per lineal foot. If, on the other hand, such a bridge carries a regular system of stringers and floor beams, the truss weight may be the same as if it were a through bridge. If the span is less than about eighty or more than about two hundred and fifty feet, the weight per lineal foot will some- what exceed the value given by the preceding rule. If the bridge is a highway structure, the same general method of operations (and taken in the same order) is to be followed as for a railway bridge. A rule for truss weights can- not, however, be so easily given, because the moving load is so very variable ; they may equal or exceed those of railway structures of the same span, or may not exceed a third of that value. The weight of a highway floor, if of plank and timber joists, will be from twenty to twenty-five pounds per square foot. Art. 9 — Single System of Bracing with Two Inclinations. The first case taken will be that shown in Fig. 2 of PI. I. The span j is 120 feet; depth d 20 feet ; panel length / 20 RAILWAY BRIDGES. 39* feet; angle NAR, i8° 30', and angle NAM, 33° 40'. The moving load will be taken as two coupled consolidation loco- motives, each with weights distributed as shown in Fig. i, Art. "]"], followed by a uniform load of 1.5 tons per lineal foot. The bridge will be supposed to be a single track " through " structure. Each truss will be taken to weigh approximately 240 pounds per lineal foot. The upper lateral bracing will be taken at thirty pounds per lineal foot for each truss. The total weight concentrated in each of the upper chord panel points will then be 10 x 270 = 2,700 pounds. The fixed weight concentrated in the lower panel points will be taken at 9,300 pounds. Hence, if W is the upper chord fixed panel load and W the same for the lower : W = 2700 pounds = 1.35 tons, for one truss. W ■= 9300 " = 4.65 " tan NAR — 0.333 t^n NAM - 0.666 sec " = 1.054 sec " — 1.202 In the above fixed weight, the ties, rails, guard timbers, etc., were taken at 400 pounds per lineal foot. The stresses due to the fixed load only, in each of the truss members will first be found, and it will be convenient to begin by determining those in the web members. On page 7 it is shown that the stress in any web member of a truss with horizontal chords is the vertical shear, multi- plied by the secant of its inclination to a vertical line. But the vertical shear in any web member of such a truss is simply the algebraic sum of all the vertical forces or weights {includ- ing the end reaction^ between the end and the zveb Member in question. In web member 5, for example, the fixed load shear will be the difference between the reaction R and the weights at A, B, M and Z; or, since the truss is symmetrical with the centre, the shear will be the weight at C added to half the weight at K. In fact, as a general principle, when the truss and its load are symmetrical with the centre, the shear in any web member will be the load between that member and the 40 SPECIAL NON-CONTINUOUS TRUSSES. (I). centre. Hence, the shear in the web member of half the truss will possess the following values : Shear in 6 = i^ = 2.325 tons '< 5 = 1 J/F + W = 3.675 " " 4 = illV + W =^ 8.325 " '' ;^ = l^W + 2W' = 9.675 '' " " 2 = 2hJ'V + 2IV' = 14.325 " " " I = 2hW + 3IV' = 15.675 " If the plus sign indicates tension and the minus sign com- pression, the web stresses will take the following values : Stress in 6 = + 2.325 x 1.202 = -f 2.79 tons " " 4 = + 8.325 X " = + 10.00 " " " 2 = + 14.325 X " = + 17.19 " " " 5 = - 3-675 X I-054 = - 3-87 " a u 2 =z — 9.675 X " = — 10. 18 " " " I = — 15.675 X " z= — 16.52 " ^ (2). On the same page, 7, it was shown that the increment of chord stress at any panel point is equal to the algebraic sum of the horizontal components of the web stresses intersecting at that point ; /;?// t/iosr horizontal components are the vertical shears multiplied by the tangents of the respective inclinations to a vertical line. Hence : The chord increment at : A = 15.675 X ^ + 14.325 X I B = 9.675 X ^ + 8.325 X I ^ = 3-675 X ^ + 2.325 X I 14.79 ^^^^ 8.79 2-79 5-24 ^(3)- R = 15.675 X 1 + o 3/= 9.675 X i + 14.325 X I - 12.79 Z = 3.675 X 1 + 8.325 X I = 6.79 K= o + 2.325 X I = 1.55 Hence the following are the upper chord stresses : (i) = - 14.79 = - 14-79 tons ) (2) = - (14.79 + 8-79) = - 23-58 " [ . (4)- (3) = - (23-58 + 2.79) = - 26.37 " ' The lower chord stresses will take the values : J? AIL IV AY BRIDGES. 41 (i) = 5.24 tons (2) = 5.24 + 12.79 = 18.03 " (3) = 18.03 + 6.79 = 24.72 " (5). When the position of the moving load is once determined for a required maximum stress, the latter is found from the re-ac- tion and panel loads by precisely the same general methods used with the fixed loads. Hence the proper position of the moving load is first to be found for each of the web stresses. Eq. (7) of Art. 7 is an expression of the condition which obtains with the greatest shear in any web member. The number of panels, n, is 6. W^, W», IV^, etc., are the single locomotive weights given in Art. yy ; IV^ has the value 3.75 tons ; W2, W3, Wi and W 5, 6 tons, etc., etc. If the moving load passes on the bridge from the right, and W^ rests at the foot of web member 10, U\ . . . \V^ will be on the bridge, and )i times 3.75 {i.e. ,6 x 3.75 = 22.50 tons) will be less than IVi + IV^ + . . . + IV^ = 27.75 tons. But ^ ^ (3-75 + 6) tons is much greater than the total moving load on the bridge. Hence, Ifo at G is the position desired for web member 10 ; W^ will then be 8.083 ^^et from G toward H. At this point the tabulation mentioned in connection with Eqs. (8) and (15) of Art. 7, may be used. The tabulation for the two locomotives is P'iven herewith. I 2 3 4 I 15,000 X 8.o3 121,200 121,200 2 39,000 X 5.75 224,250 345.450 3 63,000 X 4.50 283,500 628,950 4 87,000 X 4.50 391,500 1,020,450 5 111,000 X 7.08 785,880 1,806,330 6 126,000 X 4.83 608,580 2,414,910 7 141,000 X 5.67 799.470 3,214,380 S 156,000 X 4.83 753.480 3,967,860 9 171,000 X 9.00 1,539,000 5,506,860 10 186,000 X 8.o3 1,502,880 7,009,740 II 210,000 X 5.75 1,207,500 8,217,240 12 234,000 X 4.50 1,053,000 9,270,240 13 258,000 X 4.50 1,161,000 10,431,240 14 282.000 X 7.08 1,996,560 12,427,800 15 297,000 X 4.83 1.434.510 13,862,310 16 312,000 X 5.67 1,769,040 15.631.350 17 327,000 X 4.83 1,579.410 17,210,760 18 342,000 X 4.00 1,368,000 18,578,760 42 SPECIAL NON-CONTINUOUS TRUSSES. A little consideration of the table will make its composition evident. It is reproduced from a table in actual use, and is given in pounds and foot pounds. When [F2 rests at G, the x of Eq. (8) will be 5.25 feet. It should be explained that each quantity in column 4 is the sum of all the preceding and opposite numbers in column 3. As an example, 5,506,860 is the sum of all the numbers in column 3 from the top down to and including the ninth. Column 4, then, represents the positive parenthesis in Eq. (8) of Art. 7, less the term multiplied by x. When W^ rests at G", the x of Eq. (8), Art. 7, will be 5.25 feet. Hence that equation in connection with the preceding table gives : f,, I /i 1 1000 i02045o\ I 121200 Shear in 10 = ( x 5.25 ^ =^^^ ) x . I20\2 2/20 2 = 3650 pounds = 1.825 tons (6). This quantity will shortly be needed again. If the same locomotive weight W.^ rest at H, W^ . . . W^, or one complete locomotive, will be found on the bridge, and n times (3.75 + 6) or6 x 9.75 = 58.50 tons is still in excess of 42.75 tons, t. c, half the entire locomotive weight. Hence, W^ at H is the position of moving load, which gives the greatest shear to web member 8, and x, in Eq. (8) of Art. "jy becomes 2.83 feet. That equation then gives : „, . „ I /171000 „ 3q6786o\ I 121200 Shear in '^ — - ^ x 2.83 + -^-^ — — x . 120 \ 2 2/20 2 = 15520 pounds = 7.76 tons (7). If the same weight, W.^, be placed at K, it will be found that W-Q, will rest at the right extremity of the span, there will therefore be eleven weights on the bridge. Since 6 x (1200a + 7500) = 1 17000 > 105000, this position of the moving load gives the greatest shear in brace 6. Eq. (8) of Art. 7 thea gives, since x = 5.75 feet : „, . - 8217240 I 121200 Shear m o = -—^ x . 2 X 120 20 2 = 31400 pounds = 15.7 tons (8). RAIL IV A V BRIDGES. 43 If W-i, be placed at Z, it will be found that six times {W^ + W.^ is less than W^ . . . . W^r,, which will then rest on the bridge ; but 6 ( W^ -v W^ ^ W^ > {W^ . . . . H^ie), hence W^ must rest at L for the greatest shear in 4. The value of x will then be 4.83 feet. Hence : I A 1 2000 „ i38623io\ I 345450 S/iear m 4 = ( ^ x 483 + ^ ^^ j - — x ^^^^ ^I20\2 2/20 2 = 55480 pounds = 27.74 tons (9). In the case of brace 2, the condition of maximum shear obtains with JFg at M. This position of moving load places 10.5 feet of the uniform load of 1.5 tons per lineal foot on the bridge. The value of x in Eq. (8) of Art. 7, conse- quently, becomes 5.25 feet. The tabulation given above is not quite sufficient to completely cover this case ; although the result in line 18 forms by far the greater portion of the moment product which must be divided by the span, in order to obtain the shear. The application of Eq. (8) of Art. 7, to this particular case becomes, then : I / 342000 3000 — [^ X 10.5 + ^^ — 120 \ 2 •'2 i857876o\ 345450 Sheaf tn 2 = — ^ y^ — ^ iO-5 + "^T ^ ^°-5 ^ 5-^5 +- 2 X 20 = 84430 pounds = 42.2 tons (10). The moving load shears in web members i, 3, 5, 7 and 9, will be precisely the same as those in 2, 4, 6, 8 and 10 re- spectively, because each pair, as 10 and 9, 8 and 7, etc., inter- sect in a chord which carries no moving load. The web stresses due to the moving load will then take the following values : Stress in 10 = + 1.825 x 1.054 = ~ 1.92 tons. - " " 8 = + 7.760 X *' = + 8.1 " " 6 = + 15.7 X 1.202= -^ 18.87 " r • (lO- " " 4 = + 27.74 X " = + 33.35 " * " 2 = + 42.20 X '' = + 50.72 •' 44 SPECIAL NON-CONTINUOUS TRUSSES. Stress in 9 = — 1.825 x 1.202 = — 2.19 tons. 1 u <. 7= — 7.762 X " = — 9.33 " I " " 5 = -157 X 1.054 = -16.49 " r • (^2). " " 3 = — 27.74 X " =—29.13 " i " " I = —42.20 X " =—44.30 " j By comparing (11) and (2), it is seen that the moving load stress in web member (10) is of a kind opposite to that caused by the fixed load in web member 3, while those two members are, in reality, identical ; the latter is compression and the former tension. Since it is evident that no piece of ma- terial can be compressed and extended at the same time, it is clear that the resultant stress will be the numerical differ- ence or algebraic sum of the induced stresses. In other words : 1/ the action of forces external to a piece of material tend to subject that portion of material to stresses of opposite kinds, the resiiltant stress will be equal to the numerical differejice of the opposite stresses, and will be of the same kind as the greater. By comparing (11) and (12) v/ith (2), it will be seen that in all the web members of that half of the truss first traversed by the train, the fixed and moving load produce stresses of opposite kinds. The moving load stresses predominate in the mem- bers near the centre of the span, but near the ends the fixed load stresses are the greatest. Ir. web member 8, the resul- tant stress is 8. 1 — 3.87 = 4.23 tons of tension ; but if the bridge carries no moving load, it is subjected to 3.87 tons of compression. Now, that member may be so designed and constructed that it can resist tension or compression, accord- ing to the demands upon it ; in such a case it is said to be co2in- terbraced. If, however, it is formed to resist compression only, the member 14 must be introduced between /Tand (7 to take the moving load shear in tension. In order to provide for the movement of the load in the opposite direction, 13 must be introduced between L and D. Such web members as 13 and 14 are called counterbraccs. Other web members than counterbraces are called main braces or main web members. The duty of counterbraces, then, or of count erbraced main RAILWAY BRIDGES. 45 web members, is to transfer moving load shear to the farther abutment or pier. It is now clear that the extent to which a main web mem- ber must be counterbraced is found by taking the excess of the moving load stress over that caused by the fixed load. At first sight it would appear that the same method should hold in determining counterbrace stresses ; since it may be supposed that the main brace will carry shear until its fixed load stress is neutralized. This presupposes, however, that there is such an exact adjustment of members that each will perform just the amount of duty assigned to it. As that is an end that can never be confidently realized, it is only pru- dent to suppose that the counterbrace takes all the moving load shear, and this will be assumed in all that follows. This procedure appears the more advisable when one reflects that counterbraces are subject to the greatest fatigue of all truss members, and that the amount of metal concerned is trifling. It now becomes necessary to determine where the coun- terbraces are to begin. Since the fixed and moving load stresses, or shears, neutralize each other in equal amounts, it at once results that counterbraces or counterbraced zveb mem- bers must begin at that point, or with that web member, in which the stress or shear produced by the moving load is greater than that of the opposite kind produced by the fixed load " Stress " or " shear " is used indifferently, as the former is simply the product of the latter by the secant of the inclina- tion to a vertical. In order, therefore, to find the stress in a counterbrace, it is only necessary to ascertain the moving load shear, and multiply it by the secant of the inclination. The secant of the angle between counterbrace 13 or 14, and a vertical line is 1.944 ; and since the shear it has to carry is by Eq. (i i) 'j.'j6 tons: Stress in counterbrace {i/^)='j.y62 x 1.944= + 1 5.09 tons.(i 3^ Again, by comparing (11) with (2) it is seen that the fixed load shear in 10 (or 3) is — 9.675 tons, while the moving load 46 SPECIAL NON-CONTINUOUS TRUSSES. shear is +1.825 tons. Hence 13 and 14 are the only counter, braces needed. It is farther seen that in the half of the truss traversed last in order by the train, the stresses produced by the fixed and moving loads are the same in kind. Hence, in the main braces the fixed and moving loads induce stresses of the same kind and the resultant is si^nply the numerical sum. A tabulation of all the resultant web stresses, then, gives the following values : Web mem ber I = —60.82 tons. ~ n " 3 =-39-31 " t< .^g^ir3>967.86o^i7i^ gH^^ 121 981 2 2 J 14 ; 200 EE= - •' DC = BA- 7,009,740 210,000 2 2 10,431,240 282,000 X 1.8 I : 14 X 6.8 I — X 14 2 121,200 2 345.450 10.05 16.68 - 25.33 "15,631,350 327,000 •^ -^^ + —^ X 2.5 I 345,450 - X — — —= 35-79 14 2 The shears in the inclined web members only have been given, because each pair of braces that intersect in the chord not traversed by the moving load take their greatest stresses with the same position of moving load. Braces 2 and 3, 4 and 5, etc., thus go together in pairs. The resultant web stresse.s, by the aid of the preceding results, will then take the following values: Brace i " 3 " 5 " 7 -[3(^F+ W^) + 35.79] X 1.22 = — 59.04 tons. •[2(W^ 4- M^^) + 25.33] X 1.22 =— 41.15 " -[ W+ W + 16.68] X 1.22= -25.48 '' — 10.05 X 1.22 = — 12.26 " - 4.46 X 1.22 = - 5.44 " RAILWAY BRIDGES. 53 Brace 2 .. . 2^F^ + 3^F+25.33 = + 35.23 tons. " 4 . . . W^ ^ 2IV+ 16.68 =^ + 22.38 " '' 6 . . . jr+ 10.05 = + 11.55 " The positions of the moving load for the greatest chord stresses are found by the aid of Eq. (14) of Art. 7, and result in the following quantities : For upper chord 3 • . W- 2iiF . . • w; = w,, . . x= 2h. . . . ^.-1 = ^K 2 . . . w, " n .. . w^ - w,, . . X = 4 " . ■ . w,._, = VV, I . . W, " B The greatest stress in upper chord i occurs with the maxi- mum shear in BA, and is found by taking the product of that shear by the tangent of its inclination to a vertical line. The shear has already been determined to be 35.79 tons, and the tangent is 0.7. Hence ; Stress in upper chord i = — 35.79 x 0.7 = — 25.05 tons. Eq. (15) of Art. 7 then gives by the aid of the tabulation on page 41, the following bending moments: 3 Upper chord 3 =_ 2 2 =- 7 "13.862,310 312,000 2 2 "13,862,310 312.000 2,414,010 '^ ^'^ ^1,896.754 ft.lbs. 2 1,020,450 1,648,391 " " As the depth of the truss is 20 feet, the moving load chord stresses become : Upper (3) = — 1,896,754 -^ 20 =: — 94,837 lbs. = — 47.42 tons. " (2) = — 1,648,391 -=- 20 =— 82,420 '' =—41.21 " " (i)= =-25.05 " By combining these results with those due to the fixed load, the following resultant chord stresses are found : Upper (i) = — [25.05 + 3(1^+ W^) X 0.7] = — 33.87 tons. " (2) = - [41.21 ■\- S{W ■\- W) X 0.7] = - 55.91 " " (3) = - [47.42 + 6{W + W) X 0.7] = - 65.06 " 54 SPECIAL NOA'-COA'TIiVrOUS TRUSSED: It is to be observed that the upper and lower chord stresses are the same in pairs, i. c, in the same obhque panel. The reason is obvious. If a panel, as DFEC, be divided by any line cutting DF, DE and E.C, it will be evident that no hor- izontal forces whatever exist except the stresses in DF and EC \ hence, by the first principles of statics, those stresses must be equal in amount and opposite in kind. The same result holds for any oblique panel, and, indeed, in all vertical and diagonal bracing with horizontal chords and vertical load- ing. The stresses in one chord can then always be written from those in the other, as is done here : Lower (i) = + 33.87 tons. " (2) =-- + 55.91 " '' (3) = + 65.06 " In reality the moments remain the same whether the upper or the lower extremity of any vertical brace is taken for the moment origin ; hence the equality of upper aud lower chord stresses in the same oblique panel. If the truss becomes a through one (/. r., with the load on the lov/er chord), the same pairs of web members do not inter- sect in the unloaded chord, hence a different position of the moving load must be taken for the greatest shears in half the braces. A simple inspection of the diagram will show at once that the position of the moving load for the greatest shears in the oblique or compression braces must be the same whether that load traverses the upper or lower chord. For the vertical braces, however, the moving load must be ad- vanced in the lower chord at least one panel (more in some cases) beyond its position on the upper. Hence the vertical- brace stresses will be greater in a through truss than those found with the moving load on the upper chord, while the stresses in the oblique braces remain the same. Conse- quently this type of truss is better adapted to the deck than the through form. Art. 11. — Single System of Vertical and Diagonal Bracing. — Verticals in Compression. This type of truss is very common in American bridge RAILWAY BRIDGES. !); practice. Its compression members are the shortest possible, and its details are simple in character, and both those fea- tures are conducive to economy. The skeleton diagram ta ■which reference is to be made is given by Fig. i, of Plate II. The bridge is supposed to be a "through" structure, hence the floor system will rest on the lower chord. The loads and stresses in this Art. will be given in pounds. The principal dimensions and fixed load data are as follows : Span = 184' iig" Panel length = 20' 6g" Depth = 27 o" Number of panels = 9 Upper chord fixed load = 230 lbs. per lin. ft. per truss. Lower " " — 533 " " " Upper " " per truss panel — IV — 4726 lbs.. Lower " " '' " =IV = 10953 " 1 5679 lbs. ta?i ABL — 0.761 sec ABL ~ 1.26 The moving load will consist of a train of two coupled con- solidation locomotives, each with concentrated weignts as shown in the diagram below, followed by a uniform train of 2,240 pounds per lineal foot. It will be taken to move from right to left. The moving load is comparatively light, hence the fixed load is also assumed rather low. As usual, Eq. (7) of Art. 7 will be used in determining 4^ 2:M0l>s. pt-rLiii. ft. the positions of moving loads for the greatest shears, and, hence, for the greatest stresses. For a reason that will here- after appear, the greatest shear for every panel in the truss will be found. The application of Eq. (7), Art. 7, will giv*' 56 SPECIAL NON-CONTINUOUS TRUSSES. the following quantities for use in Eq. (8) of the same Art.: Max. sJiear OP . MN. FG . EH . DI . CJ . BK . AB . in W^at P . . . IV„=W, ... ^=4.1 W. " i\^ . . '' =IV, . . . x= 2.2 " " 6^ . . • " = ^12 • • • ;r = I .o " '' H . . " = J'Fig . . . ;r = 2.0 w, - / . . . *' =IV] 2x= 9.0 U U CY '' —W Uniform 2x = 29.5 " " A' . . " = W ' load, 2^ =: 50.2 " " z . . " = IV ] 2x = 70.75 ft. , The tabulation to be used in connection with Eq. (8) and (15) of Art. 7, is given below and is formed precisely like that in Art. 9. I 12,000 X 7.666 92,000 2 32,000 X 4 5S3 146,666 238,666 3 52,000 X 4 25 221, coo 459,666 4 72,000 X 4 583 329.976 789,642 5 92,000 X 10 666 981,332 1.770,974 6 106,000 X 4 666 459.333 2,230,307 7 120,000 X 5 583 670,000 2,900,307 8 134,000 X 4 666 625,333 3.525.640 9 148,000 X 9 333 1,381,333 4,906,973 10 160,000 X 7 666 1,226,666 6,133.639 II 180,000 X 4 583 825,000 6,958,639 12 200,000 X 4 25 850,000 7,808,639 13 220,000 X 4 5S3 1,008.333 8,816,972 14 240,000 X 10 666 2,560,000 11,376,972 15 254,000 X 4 666 1.185,333 12,562,305 16 268,000 X 5 583 1.496.333 14,058,638 17 282,000 X 4 666 1,316,000 15,374.638 18 296,000 X 3 25 962,000 16,336,638 Eq. (8) of Art. 7 gives by the introduction of the quanti. ties found above, and by the aid of the tabulation : Shear in OP = 185 459,666 72,000 -^^-^ 4--^-! X 4.1 = 2,040 lbs. 3,525,640 148,000 J.J J. ^ +-±^ X 2.2 " MN=-— 185 FG ^M 6,958.639^200,000^ ^^ 185 '|_ 2 2 92,000 2 X 20.55 = 8,171 17,110 RAIL WA Y BRIDGES. 57 Shear in EH - -r- 185 DI " CJ BK AB 185 I I I 11,376, 2 972 254,000 + ^^ X 2.0 2 9^°^ = 29.884 '• 2x20.55 16,336,638 2 612,160 + 2 "" '^•^ •^3«.^«'= 45.783" 2X20.55 1 1 658,080 + % X 14.75 - " = 63,735 " " 2 — " = 86,132 " " 750,480 - + ^ ; X 35.38 - " =110,105 " Since the fixed load shear in HM is 15,679 lbs. it is seen that counterbrace 10 is the first counter needed. Combining the fixed load shears with those above due to the moving load, the following resultant stresses in the inclined web mem- bers are found : .Stress (10) = + ( 17,110 " (9) = + ( 29,884 ) X (7)= +( 45,788 + PF+ W) X (5) - + ( 63,735 + 2W+2W') X (3)= +( 86,i32 + 3?'r+3J^0 X (i)z=: -(110,105 + 4?F+4[r) X ) X 1.26 = + 2\, ^60 lbs. = + 37.654 " = + 77,445 '' = + 119,817 " = + 167,800 " = - 217,750 " Eq. (21) of Art. 7 shows that if [^4 be placed at the feet of the vertical brace 2, that member will take its maximum moving load stress, which, by Eq. (19) of the same Art. takes the value : (2) 4.05 X 6,000 4- (11.72 + 16.32 4- 15.97) 10,000 + 2.92 X 14,000 20.55 + 10,000 = + 34,600 lbs. The stresses in the vertical braces become : Stress (8) = - (29,884 + H'*) = - 34,61 1 lbs. '' (6) = -(45,788 + f r + 2 fF) =-66,193 " " (4) = - (63,735 + 2?f^ 4- 3^) =-99,819 " " (2) = + (34,600 + + ^)= + 45,553 " These complete the greatest stresses in the braces. 58 SPECIAL NON-CONTINUOUS TRUSSES. Eq. (14) of Art. 7 gives the following values for the posi- tions of moving load for the greatest stresses in the chords: Upper land \ . . . W^„_i= IV^ ■ . . lVn = 72,Soo . . . 2x =65. //f. 2 ..." = iVg ..." =82,300 ..." =73.5 " I . . . " = H^5 . . . " =78,000 ..." =69.7 " The "2.f " shows the distance covered by the uniform load. Eq. (15) of Art. 7 then gives the following maximum bend- ing moments by the aid of the tabulation already given : For upper 3 and 4 J/=^ 16,336,638 737>6oo - \ - X 32.5 9 6,958,639 5,478,153//. lbs. M = - 16,336,638 , 756,640 + 4,906,973 36.8] = 4,910,013//. lbs. M=- 16,336,63 8 ^ 748,130 ^ 1, 770,97 A = 3,830,813//. lbs. The moving load stresses in lower chord i and 2 are found by taking the product of the greatest shear in brace i by its. vertical tangent : Lower (i) and {2) — 1 10,105 x 0.761 = 83,790 lbs. Since the depth is 27 ft. the greatest moving load upper chord stresses are the following: Upper (3) and (4) =- 5,478,153 -=- 27 = - 202,900 lbs. (2) =-4,910,013 H- 27 =- 181,850 " (i) =- 3,830,813 -^ 27 =- 141,900 " Since [W +W') tan ABL = 11,932, the resultant upper chord stresses become : RAIL WA Y BRIDGES. 59 Upper (i) =— (141,900 + 7 X 11,932),= - 225,424/^5. (2) ==— (181,850 + 9x 11,932; =— 289,238 " " (3)^««'(4) =— (202,900 + 10 X 11,932) =— 322,220 " In the lower chord, the resultant stresses are : Z,'].2 tons. (i i) = + 4.0 tons. (19) = + 28.8 " Only the diagrams for the next three positions of the mov- ing load will be given, for they will sufificiently illustrate the rest ; they all embody exactly the same operations. Figs. 2, 3, and 4, of PI. IV., are all drawn to the same scale of 20 tons to the inch. With the head of the train at the foot of 20, R =21.56 tons, and /^= 32.10 feet ; hence, (/) = {R X 137.5 - a'» -^ 32-10 = -85.75 tons, since '2,P — o at the foot of 19, and/ is the panel length. Divide the truss through/, 19, and m, then in Fig. 2, PL IV., make 1 — 6 vertical, and f parallel to that panel and 84 NON-CONTINUOUS TRUSSES IN GENERAL. equal to 85.75 tons; it acts toward 2 as shown. Make 2—3 equal to 21.56 — ■ze'' = 4.56 tons, and then draw in and 19. These latter must act in the direction shown. Now suppose g, 19, 13, and ni severed. From the upper extremity of 19 draw ^ parallel to that panel (referring to Fig. 2) until it in- tersects I — 6 in I ; I — 2, acting downward, is the tension in 13. It should be stated that the shear 2 — 3 = 4.56 tons acts downward, as shown. Next make 1—6= 21.56 tons, acting down, and draw / and 20, acting in the directions shown ; finally, draw h and/z'. 1 — 4, which acts upward (not indicated), is the compression in 15 ; and, in like manner, 4 — 5 is the compression in 17. Scaling from the diagram : (11) = + 5.2 tons. (20) = +• 16.0 tons. (13) = + 3-9 " (15) = - lo.i " (17)=— 9.0 " (19) = + 17.6 " With the head of the train at the foot of 22, R' — 28.25 tons, and 2P= o at the foot of 20 ; hence {g) is the stress to be found by moments. Ig = 33.75 feet. {g) = {R' X 125 - w'p) ^ 33.75 = - 98.3 tons. Fig. 4, PI. IV., is the diagram used for this position of the load, and is constructed in precisely the same manner as Fig. 2. 2 — 5 is equal to R' — w' = 11.25 tons, the shear for^, /, and 20, and it acts downward, i — 7 is the shearing stress, 28.25 tons, for k, k, 22, and 20, also acting down, as shown. I — 2 is the stress in 15 ; 2—3, that in 13 ; 1—4, that in 17 ; and 4 — 6, that in 23. Below are the numerical values : (15)=+ 3.8 tons. (13)= + 6.0 tons. (22) = + 16.0 " (23) = - 10.8 " (20) = + 20.8 " (17) = — 12.2 " Fig. 3 is the diagram for the head of the train at the foot of 24, and is constructed in precisely the same manner as CURVED UPPER CHORD. 85 Figs. I, 2, or 4. R' — 35.75, and '2P — o for the toot of 20. h - 3375 feet. Hence, {g) = {R' X 125 -3w'/)^3375 = - II3-5 tons. (15) = + 4.8 tons. (24) = + 25.2 tons. (22)= + 15.0 " (23) = - 9.4 " (17) = - 3.2 " (25)= - 15.7 " With the head of the train at the foot of 26, ^P — o for the foot of 22, since R' = 35.75 tons. 4 = 34.65 feet. Hence, (//) = {R' X 1 12.5 - 3w» -^ 34.65 = - 124.7 tons. (22) = + 16. 1 tons. (26) = + 34.3 tons. (17)= + 7.0 " (25)= - 2.5 " (24) = + 12.5 " (27) = — 28.0 " (23) = — 10.2 " With the head of the train at the foot of 28, R' = 53.1875 tons, and 2P = o at the foot of 22. 4 = 34-65 feet. Hence, (//) = {R' X 112. 5 — 6w'J>) -^ 34.65 = — 136.1 tons. (17) = 4- 6.9 tons. (25) = — 1.5 tons. (24)= + 12.8 " (28)= +25.1 " (23) = — i.o " (27) = — 16.0 " (26) =+21.5 " (29)= — 15.5 " With the head of the train at the foot of 30, R' = 63.125 tons, and ^P = o for the foot of 24. 4' = 34-95 feet. .-. (//') z= (R' X 100 — 6w'p) -^ 4' = — 144- 1 tons. (17) = + lo.o tons. {27) = — 9.3 tons. (23) = + 4.5 '' (30) = + 35-6 '* (26) = + 15.7 " (29) = — 10.7 " (25)=+ 1.0 '' (^31) =-27.5 " (28) = + 20.8 ♦* With the head of the train at the foot of 32, R' = 73-875 tons, and 2P = o for the foot of 24. 4' = 34-95 feet. 86 NOX-CONTIXUOUS TRUSSES /.V GENERAL. {h') = {R' X loo X low'p) ^ 4' = — 150.6 tons. (27) = — 9.9 tons. (30) = + 36.0 " (29) = — I.O " (32) = + 24.8 tons. (31) =-26.8 " (33) =-16.7 " With the head of the train at the foot of 34, R' = 85.4375 tons, and 2P = o for the foot of 24. 4- = 34.95 feet. .-. (k') = {R' X 100 — i4zv'p — wp) -^ 34.95 = — 154.7 tons. (17) = + 8.8 tons. (32) = + 24.6 tons. (25)= + 8.9 " (31)= - 17.4 " (23) = + 5-5 " (34) = + 55-2 " (30)^+26.5 " (33) =-15.4 " (29) = - 0.8 " (35) = - 46.8 " With the head of the train at the foot of 36, R' = 97.8125 tons, and ^P= o for the foot of 24. 4' = 34-95 feet. Hence, {/i) = {R' X 100 — iSw'p — yup) ^ 34-95 = — 156.5 tons. (17) = + 1 1.0 tons. (24) = — 2.0 " (23) = + 4.5 " (25) = + 12.0 " (27) = - 5-0 " (30) = + 29.8 " (29) = + 8.4 " (32) = + 12.0 tons. (31) = -20.3 (34) = + 60.0 (33)= - 2.8 (36) = + 40.0 (35) = - 50.8 With the head of the train at the foot of 37, R' = iii.oo tons, and 2P= o for the foot of 24. 4 = 34-95- Hence, (/i') = {R' X icx) — 22w'p — 6wp) -^ 34.95 = — 156.CXD tons. (17) r= + 10.9 tons. (32) = + 10.5 tons. (24) = ~ 2.2 " (31) = - 16.7 " (23) = + 4.4 " (34) = + 54-2 " (25) = + II.9 " (36) = + 37-0 " (28) = + 1.5 " (35) = - 43-5 " (27) = - 5-5 " (37) = + 78.0 " (30) = + 26.0 " (38)= - 1 1 1.0 " (29) = + 8.7 " CURVED UPPER CHORD. The chord stresses with this loading arc : 87 (a) — — 87.7 tons. (^k) = + 157.6 tons {f) - - 131.2 " (/') = + 150.5 " {c') = - 137-0 " (;«')= + 149-7 " (^^') = - 153-5 " («•) •-= + 132.0 " (r') = - 152.6 " {0') = + 124.5 '' (/')= - 158.7 '' (/) = + 81.8 " {g')= - 156-6 " (g) = + 50.0 " {h') = - 156.0 " (r) = The same checks apply as with the fixed load. The great- est stresses are found by combining the fixed and moving load stresses. The greatest chord stresses are thus found to be the following : (^') = — 140.4 1 :ons (^') = — 200.8 (^ = - 217.5 (''") = - 239-3 in = -- 243.0 (/') = - 250.8 W) = - 249.85 (/O = -249.1 (r) =0 (^') = + 70.0 tons. (/') = + 131-5 (o') = + 190-5 (;/') = + 210.0 (;«•) = + 233.4 (/') = + 239.7 (k) = + 249.1 In the case of the web stresses the combination is effected by taking the algebraic sum. It will be seen that a number of the braces near the centre need counterbracing, t. e., acting consistently with the assumptions that were made. It is uncertain to what extent the existence of the counters renders this counterbracing unnecessary; their influence is therefore neglected. It is to be borne in mind that the greatest result of a given sign is to be selected from all the preceding moving-load stresses, and added, algebraically, to the stress in the same- member caused by the fixed load. 88 NON-CONTINUOUS TRUSSES IN GENERAL. (i8')= + (18) = + 19.0 24.8 tons. i^7)={- 1 1.2 tons. 12.0 " (19) = + (20) = + 28.8 20.8 (23) = { - 8.9 ' 7-4 ' (22) = + (24) = + 16.1 27.8 ii •(25) = 1; 17.2 ' 10.5 ' (26) = + 38.3 i< (27)= - 29.4 ' (28) - + (30) = + 33-6 44.6 (29) = 1 ; 21.8 ' 2.4 ' (32) = + 41.3 a (31)= - 35-8 ' (34) = + 81.0 i(. (33)= - 29.8 ' (36) - + 74.8 n (35)- - 71.0 ' (37) = + 109.8 a (38)= - 175.0625 tons In actual practice it perhaps would hardly be worth while to counterbrace the web member 29. These results, then, are the greatest values of the stresses to which the different members of the truss are subjected. Art. 19. — General Considerations. It is clear that in the graphical treatment of such a problem the stress diagrams should be as large as possible. The scale used for all the results obtained in Art. 18 was ten tons to the inch, and it is not usually best to use more tons to the inch than that. Another method, but a far more tedious one, of applying precisely the same general principles is to determine the stresses produced in every member of the truss by each indi- vidual panel load, and then combine the results. The steps of the different operations in such a case are precisely the same as those gone through above. Although the truss taken consists of but two systems of triangulation, precisely the same method is applicable to any number of any kind of systems, or to the ordinary "bow- string" truss of one system. In the latter case there is no ambiguity unless counter- braces are used. GENERAL CONSIDER A TIONS. 89 It is to be particularly noticed, also, that the method is per- fectly independent of the character of the curve of the upper chord ; it may equally well be applied to trusses with both chords curved, or to trusses with parallel chords ; in fact, it is perfectly general, though usually not desirable. Art. 20. — Position of Moving Load for Greatest Stress in any Web Member. Two principal cases occur in connection with types of structures ordinarily used in engineering practice. That one to be treated first is the case of the intersection of the chord sections, in any panel, lying below the inclined web member in the same panel ; the other is the case of the intersection lying above the inclined web member. Applications of these principal cases to special features can easily be made after the general results are obtained. Case 1. — The Intersection of Chord Sections below the Inclined Web Member. Let / be the length of span ; i the distance from end of span to the point of intersection, H, of the chord sections in the panel in question ; m the distance from the end of the span to the same panel, whose length is// 5 the stress in the web member under consideration, and h its lever arm about H ; a, b, c, etc., the distances separating W^ from Wo, W2 from W^, etc., etc. ; W^, IV^, etc., the weights resting be- tween G and D, and PV^, JV^, etc., the weights resting in the panel/, while Wn, distant x from £, is the last weight rest- 90 NON-CONTINUOVS TRUSSES IN GENERAL. ing on the span from C to toward E. b' is the distance front. D to the nearest weight, W^. The reaction at 6^ is : • Ji = WA ^ )+ W^\^ )+ etc...+ W„~. (I). By taking moments about H : Sh = Ri- W,{/ + t-a-b x)- JV,{/ + i-d-c x\ - etc. - {w/-^ + ^ / ~ ^ ~ ^ + etc.) {vi + i). . (2). By rrioving the entire load the distance Ax toward G, re- membering that the change in the value of R will be A X A K = {Wi + IV2 + • • • + IV„) ~j , the change in S/i becomes : {W,+ W, + fU.)^{m + t) . . . (3). For a maximum or minimum a S/i = o, hence : lV^ + lV, + ... + irn=--( IV, + IV, + etc. )^y^W., + W,^ etc.) pi (4). Bearing in mind that the first parenthesis in the second member of Eq, 1^4) represents the load between the panel / and the left end of the span, and that the second represents the load in panel / itself, it will be at once seen that when the load extends from E to W^, S is the maximum main stress; and that when it extends from G to lV^ {i. e., to the weight farthest toward E), S is the maximum counter stress. Eq. (4), therefore, as it stands, gives the condition for maxi- mum main or counter stress. GREATEST WEB STRESSES. 9 1 Eq. (4) is perfectly general in character and covers all sys- terns of loading whatever, but it may be put in special forms for convenient application in special cases. Example I. — Uniform Load. If the load is continuous, or only partially so, and w is its intensity {i.e., its amount per lineal unit) at any point distant X from E, then the various concentrations W^, W^, W^, etc., will be represented by wdx. If, further, the load is uniform and continuous, w is constant, and W^ ■¥ W^ -\- . . . + W„ = wxi, Xi representing the length of uniform load on the bridge. In the same manner, if x^ represents the length of uniform load from D towards G,{W]^ -\- W„ -\- etc.) = wx^, ; and (W^8 + Wi + etc.) = wrp\ r being the fractional part of the panel/ covered by the uniform load w. Eq. (4) then becomes : / ^ /(?« + i) wx, = — ". zvx; + ivrp — : — -. ' I ^ ^ pi Or, x^^ -jx.i + rl{^^ ^ \\ .... (5). As with the general case so with Eq. (5), it is so written as to give both maxima, main and counter stresses. If the load is placed for the greatest main stress, for which x,^ — o, while Xx — np -\- rp; in which np is the length of load- ing from C towards E : 'b np + rp -/(t-) ••• -77;;r^^ • ■ i^\ \i / " If. m If the load is placed for the greatest counter stress , ^2 = - 7 +ir('^+i) . . . . (7). 92 NON-CONTINUOUS TRUSSES IN GENERAL. Or, as there is i.o load between C and E, and if np = x^ is the length of load from D towards G, Eq. (7) will become : \ I / p \t J I (m Eqs. (6) and (8) will enable the position of moving load to be at once computed without trial. Example II. —Loads at Panel. Points Only. If loads are located at the panel points only, then JFj, W^^, Ws, etc., will be the panel loads, and a, d, c, etc., the panel lengths, and equal to each other in case those, lengths are uni- form ; the parenthesis in Eq. (2) multiplied by {in + /) will also disappear. Substituting R from Eq. (i) in Eq. (2) with the last parenthesis dropped, there will result : Sh — Wxia ^ b + c + • " ^ x) (^ ^- i j + \\\ {b + c + • • • ^- x) (-^+i)4-.-.etc. + W^3(r + ...+;r)^+...+ff^„y' -(^Fi + ^F2 + .-.etc.)(/ + /) (9). The last term of the second member represents the loads between G and D. The position of loading for a maximum of 5 will, in the general case, be determined by trial, by ascertaining at what position the second group of positive quantities in the second member ceases to increase more rapidly (as the load pro- gresses) than the negative difference between the first posi- tive group and the negative last member. This can only happen if the panel weights toward, or in the vicinity of W^ are very heavy relatively to those toward, or in the vicinity of W^. If the heavy panel loads are W-^ and those near it, GREATEST WEB STRESSES. g2a I. e., if the heaviest panel loads are at the head of the train, the following analysis shows the positions for maxima stresses, in which, it is to be observed, Wi is the rear panel load for counter stresses. Since {a + b + c -h - - • + x) {i + I) <{i + 1)1, and, hence : (a + d + e + ' • • -h ;ir) U + I j < (z + /), it is clear that for maxima main stresses the loads must extend from the farther ejid of the span to the main tnember in question. Since the counter shear is negative, i. r., opposed in sign to the main shear, the negative portion of the second member of Eq. (9) must be as large as possible for the maximum counter shear, and the positive portion as small as possible. t xt Hence the portion W^ {c +'•■-{• x)- + ■'• + IV„~ must be omitted, and the load W^ placed at the panel point nearest the end G ; i. e., the load must cover that portion of the span between the eounter and the nearest end of the span, for the maxima counter stresses. Hence, for main web stresses under the assumed con- ditions : S =i^\ W-i^ {a + b + c + • ' • + x) + W^ {b + c ■\- ' • ■ -\- x) + . . . + W„x [ ^ . . . (10). And for counter web stresses : S=j\Wi{a + b-\-c+'-'^-x) + W^{b+c-\---- + x)-ir-'- etc. [ (^+i)-(^i+ ^^. + ---etc.)(^'") . . . (II). ^2b NON-CONTIXUOUS TRUSSES IN GENERAL. The conditions on which Eqs. (lo) and (ii) are based are precisely the same as if the chords are parallel. In the latter t see (X \ I -\- t case d— oo , /^ = co r^j ar, — , = — ; — - = o, and — j-- = sec a. Id In h Case II. — The Intersection of Chord Sections Above the Inclined Web Member. This case is illustrated by Fig. 2, in which let the stress S in the member DC be under consideration. The moving load is supposed to pass on the bridge from E towards G for the main web stress. H is the point of intersection of the chord sections, while GH and GD are the distances i and m respect- ively. All other notation remains precisely as in the previous case. The reaction, R, under G, is given by Eq. (i). Bear- ing in mind that the distance DH is now {i — ni), and taking moments about H, there will result : Sh = Ri - \\\{i-l +^ + ^ + "- + ;r)- W^{i-l + <^ -h ^ + •• ■ +-r) — etc. [wjL ^' + W,^- ^- "" + etc.) ii-m) . . (12). Precisely the same operation which follows Eq. (2) shows that the desired condition for a maximum is given by the following equation : PF, + 1^2 -(- W^3 4- .. . + W„ = -.( W^ + \\\ -h etc.) + (^F3+H/,-hetc.)^^'^^ . . . (13). GREATEST WEB STRESSES. g2C The different portions of Eq. (13) evidently represent ex- actly the same loads as the same portions of Eq. (4). It is also clear, from the same considerations, that if the loads extend from E to W^, Eq. (13) gives the position for a maxi- mum main stress, and a maximum counter stress if they reach from G to IV4 (/. e\, to the weight farthest towards £). Eq. (13), like Eq. (4), applies to any system of loads what- ever, and can be applied to special cases in the same manner. Example III. — Uniform Load. By using the same notation and the same process of reason- ing as in Ex. i, Eq. (13) takes the form for the greatest main .stress : Or, for the greatest counter stress : -e-o EXAiMPLE IV. — Loads at Panel Points Only. Let it be first supposed that / is less than /; i. c, i < /. The same general considerations that were given in Ex. II., applied to Eq. (12), will cause it to take the form : Sh^-W^{a-¥b^-c^-.'-^x){\-'^-W.Xb ^-c + .'• ^ x) \} —^ij ■ ' etc. + W^{c ^ • ' ■ ^ x)^- -^ etc. + (^^1 -H W^a -h etc. )(/-/) . . . (16). Since {a -{■ b -^ c -{■ • • • ^ x){l — i) < l{l — i)\ hence : {a + b + c + ' ■ • -\- x) (\ -yj <{l-t). <^2d NON-CONTINUOUS TJiUSSES IN GENERAL. Therefore, in order that the second member of Eq. (i6) shall have its greatest positive value, the loads must be at all the panel points. Eq. (i6) also shows that in the case now under consideration there can be no reversal of stress in any web member. In the inclined member, S, the stress will always be tension, and always compression in the vertical member passing through its upper extremity. The positions for the actual maxima stresses can be found by trial only, as they will depend on the amounts of the panel loads and their locations relatively to each other. Let it next be supposed that i is greater than /; i. e., i > /. In this case (/ — z) becomes negative, or (/ — /) positive, and the conditions for maxima values are the same as those fixed for Eq. (9) ; hence they need no further attention. A remaining example with the intersection of chord sec- tions beloiv the inclined web member, and between it and the end of the span, can be treated in precisely the same general manner as the preceding. The moving load between G and D would lie, in the general case, partially on one side of the point of intersection and partially on the other. This form of truss, however, has little or no technical interest, and needs no further attention. The preceding treatment applies to any forms of trusses, whether deck or through, with one or either chord horizontal. In the application of any particular formula it is only neces- sary that the point of intersection of the chord sections shall be located according to the conditions on which the formula is based. The treatment also applies to any system of web members, whether they are all inclined at different angles to a vertical line, or at equal angles ; or, again, if, as in the Figures, a part of them are vertical. It is only to be borne in mind that two web members intersecting in the unloaded chord take their greatest stresses together only when that chord does not change its direction at that point of intersection. In case that direc- tion does change, the value of "z" will be different for the two members, although " ;;/ " will remain the same. The tabulations mentioned in Art. 7 and given in Arts. 9 GREATEST WEB STRESSES. g2e and 1 1 can be used in the application of the preceding formulae precisely as with parallel chords. Short methods of compu- tation, well known to every engineering ofifice, make their practical applications easy and rapid. In designing trusses with variable depth, special care must be taken in determining counter web stresses. It frequently happens, in cases similar to Fig. i, that counters must be car- ried at least one panel nearer the end of the span than paral- lel chords would require. Again, the vertical web members are frequently subjected to heavy tension, with special con- ditions of moving load, at panel points where the chords change direction. Numerical Example. The truss to be taken in this example is exactly similar to that shown in Fig. i, except the upper chord will be straight from the hip to the top of the inclined end posts. Let /= 297 ft. ; / = 27 ft. .-. -=ii. " m = ip = ^\ ft. " the centre depth of truss be constant for three panels and equal to 42 feet. " the depth of truss at the first panel point from the end be 24 feet. As the web member whose stress, 5, is now desired is \n the fourth panel, the distance, /, is 81 feet; /. e., 3/. The moving load to be used is that shown in Fig. i of Art» y"], and tabulated on page 41. If this load passes on the bridge from right to left, and if the first driving wheel rests at the panel point at the right extremity of the panel in question — /. r., the fourth panel from the left end — 93.8 feet of the uniform load of 3,000 pounds per lineal foot will rest on the bridge in addition to the two locomotives. In Eq. (4) the parenthesis (W^i + W^2 + etc.) will, with the above position of loading, be equal to zero, while {W^ + W^^ + etc.) will equal 92/ NON-CONTINUOUS TRUSSES IN GENERAL. either 15,000 or 39,000, or some value between. With the values taken in this example, Eq. (4) will become : W^i + W^2 + • • • + ^. = 22 (W^3 + H^4 + . . • + etc.) ; and with the position of loading assumed, this equation will "take the following numerical form : 623,400 > (22 X 15,000 = 330,000), and < (22 X 39,000 = 858,000). Hence the position of loading assumed satisfies Eq. (4). The resulting reaction at the left end {G, Fig. i) of the span is easily and quickly found by Eq. (i), with the aid of the tabulation on page 41, to be: 7*? = 215,000 pounds. The lever arm, //, of the stress desired is found from the truss dimensions already given, and is 15 1.2 feet. The distance b' in Eq. (2) is 18.917 feet. The preceding data substituted in Eq- (2), gives : Sh = 16,687,458 . • . 5=1 10,370 lbs. The elements of this example have designedly been so ehosen as to show how greatly the computations may be shortened and simplified by a proper choice of the depth, chord inclination, and intersection distance, /. Art. 21. — Position of Moving Load for Greatest Chord Stresses. Since the chord stresses under any given system of loading depend only on the truss depths at the various panel points, measured in a direction normal to the opposite panels, the positions of loading for their maxima values with inclined •chords will be identical with those determined in Art. 7 for parallel chords. No new conditions, therefore, are to be developed here. The equations of Art. 7 are to be applied exactly as they stand. It is only to be remembered that the lever arm for any panel is the normal erected on it to the opposite panel point. HORIZONTAL COMPONENT OF GREATEST STRESS. 93 Art. 22. — Horizontal Component of Greatest Stress in any Web Member — Constant Value of the Same for Vertical and Diagonal Bracing with Parabolic Chord — All Loads at Panel Points. Of more interest, perhaps, than real value to the engineer, is the expression for the horizontal component of the greatest stress in any web member, though it may very easily be written. It may be useful at times as a numerical check. Let the figure represent a truss of one system of triangula- tion, subjected to the action of vertical loads passing along the lower chord AB. It is desired to find the horizontal component of the greatest tensile stress in GH. Let Hh and Gg be verticals passing through // and G. The following notation will be used : n = panel length (uniform) in AB. = number of any panel from B ; 2, and 4 for OH. for PQ, n has the value rp d d' = Mg = Gg = Hh. N = number of panels in AB. I w R = AB=:Np. = moving panel load. = reaction at B. nip = BM. For the greatest tension in GH the moving load must extend from B to H. 94 NON-CONTINUOUS TRUSSES IN GENERAL. The distance of the centre of such a load from B is — i^. 2 Hence, ^=(/--f-)7(«.-.)- = -|('^.-.).-2i^>^| Now let moments be taken about G. Hence, (J///) = {r(„, - r)p - («, - ,) a. I («, _ ^)^ - M I ^^ ^ Or, (MH) = ^ [^-^ I «, („, - .) . (3). If a is the angle of inclination of GH to a horizontal line^ then : {GH) = H^sec a (4). Eqs. (3) and (4) apply to all tensile web stresses. For com- pressive web stresses as typified by {GM) there would be found the Hon Comp. {GK), instead of Hon Comp. (GF), by taking moments about A/; d' would then represent Mm. By making r = o in Eq. i : hORIZOXl-AL COMPONENT OF GREATEST STRESS. 95 Hor. Comp. {GK) = ^|; ^^ - '^^P^ n, [u, - i) . (5). Hence, for the horizontal component of the greatest com- pressive stress in GM : And, {GM) = H,' sec a' .... (7). By means of the Eqs. (3), (4), (6), and (7), every web stress in the truss may be determined by formula. If GM is vertical, r = o and d = d' m Eq. (6), and /// = o, as it should. If GH is vertical, Eq. (3) shows H^ to be zero in the same manner. It is to be borne in mind, in the application of these formu- las, that n is counted along the loaded segment ; also that d', for tension, is taken at the head of the train, and one panel in front of it for compression. If the moving load passes along the upper chord, exactly the same formulae hold true, but d' taken at the head of the train will give compression, and tension when taken a panel length in front. If the curve KFC is a parabola, with vertex at the centre of the span, if K and C coincide with A and B, respectively, and if GM and all corresponding web members are vertical, Eq. (3) becomes : H, = i|^ I '^.TLll/ _ LlA^l } „, („,-.). (8). From the ordinary equation to the parabola : j^ = ax and — = adi ; 4 96 NON-COXTIXUOUS TRUSSES IX GEXERAL. in which d-^ is the depth of the truss at the midle of the span. Hence, ^ = ^37"- In this equation put y — r^p and x = d^ — d, then j = ('^1 — 0/ and X = d^ — d', successively. There will re- sult : 4 4"i 4 4«i . Remembering that / = Np : 4 («i TV — «i^) ^=4 JV^ Putting these values in Eq. (8), also Np = /, there will re> suit: ^ _ wfN^ ( TV- (;?i- I) N-n^ { ' - ~Mjr \ K- I) N-{n,-i)~ n^N-n^ \ "' ^''' '^ As this is the horizontal component of the greatest tension in any diagonal web member and constant, t/iaf greatest stress itself is the hypothejmse, parallel to the brace in question, of a right-angled triangle of wJiich the base is H-^ — — — . 8^4 B01VS7'R/.VG 7 A' CSS. gj This furnishes a very short method of finding the stress in any inclined web member. The similarity between //j and the total stress in the hori- zontal chord, with the truss wholly loaded, is interesting. If the trussing is so designed that the diagonal or inclined braces sustain compression, Eq. (6) gives precisely the same general result, but with the sign changed. In such a case there would be substituted in the parabolic / / equation y = «,/ and x — d-^ — d' also, r = {n^ — i)/ and X — d^ - d\ <■/ and rt'' having changed places. If no web members are vertical, y will have for one value / in the equation to the parabola, - — [n^ — r) p instead of ^hP^ the other values to be substituted remaining the same. This new value gives, ^i{n,-r)N-{_n,-rn Now making the substitutions in Eq. (3) instead of Eq. (8) 8^1 \«i — I n^ — rj ^^ ^ ' "^ = Td, (^) "• Art. 23. — Bowstring Truss — Diagonal Bracing — IjKample. The first form of bowstring truss to be treated is that shown in Fig. I. All braces are inclined, and each apex in the upper chord is vertically over the centre of the panel below. The truss is supposed to be designed for a highway bridge^ There is a sidewalk on either side. 98 NON-CON TINUOU S TRUSSES IN GENERAL. The greatest moving load will be assumed to be that of an advancing crowd of people, from the left end of the span, weighing eighty-five pounds per square foot. As the span is a short one, and the roadway heavy, the whole of the fixed load will be put upon the lower chord. The following are the data required : Span = 72 feet. Depth of truss at centre =11.7 feet. Radius of circumference of circle passing through apices in upper chord = 60 feet. Number of panels = 6. Panel length = 12 feet. Width of roadway, from centre to centre of trusses = 20 feet. Width of each sidewalk = 6 feet. \V = 900 pounds per foot = 5.4 tons per panel. %v = 32 X 85 X 12 = 16.32 tons per panel. ■As usual, W^and w refer to fixed an3 moving loads respect- ively. C -101.0 2> Fig. I. In all the diagrams that follow, the lines indicated by any two letters are parallel to the members of the truss at the extremities of which the same letters are found. In this truss and in the two which follow, the upper and lower chord sections found in any panel intersect outside of the span RR , hence the positions of the moving load, for the greatest web stresses, are precisely the same as those which would be taken for a truss with parallel chords. With the head of the moving load at J/, the truss is first to be considered as divided through the members AB, BM, and ML ; then through BC, BL, and ML. BOWSTRING TRUSS. Fig. 2 is the complete diagram for this position. 99 R (reaction) = 27.1 tons. {ML) ={R X 18 — 21.72 X 6) -^ 9.3 = 38.4 tons. 9.3 feet is the depth of truss through B. « ore; Fig. 2. The shear is, 5 = 27.1 — 21.72 = 5.38 tons. The diagram needs no explanation. It gives: {BM) = + 18.5 tons. {BL) = - 2.1 tons. With the head of the moving load at L: R = 37.98 tons. Fig. 3 is the complete diagram. (CD) Fig. 3. The truss is first supposed to be divided through BC, CL^ and LK\ then, through CD, CK, and LK. {LK) = {R X 30 — 2 X 21,72 X 12) ^- 1 1.7 = 52.8 tons. The shear is, -S" = 37-98 - 43-44 = - 546 tons. lOO NON-CONTINUOUS TRUSSES IN GENERAL. The diagram gives: {LC) — + 20.0 tons. {CK) 0.60 tons. With the head of the moving load at AT: R — 46.14 tons. Hence, {KH) = (7? X 42 — 3 X 21.72 X 18) H- 1 1.7 = 65.4 tons. Fig. 4 is the complete diagram. (CD) W» Fig. 4. The shear is, 5 = 46.14 — 65.16 = — 19.02 tons. The diagram gives: (KD) = + 21.7 tons. (^//) = - 7.4 tons. With the head of the moving load at //: R' = 40.7 tons, and S = - (40.7 - 5.4) = - 35.3 tons. {BG) = {R' X 18 - 5.4 X 6) -^ 9.3 = 75.3 tons. mE) Fig. 5. Fig. 5 is the diagram, and it gives : (//£) = + 20.9 tons. i-^G) = — 4.0 tons. ROW STRING TRUSS. lOI With the head of the moving load at G, or with the moving load over the whole bridge : 7? = 2.5 x {W + w^^ 54.3 tons. In this case no chord stress is found by moments, but the diagram, Fig. 6, is worked up from the end of the truss. (KT^) Fig. 6. It gives: {RA) = — 100. 1 tons. (^^) =-109.1 " \bC) = - 103.0 " \CD) = - 102.0 " {RM) = + 84.6 tons. \mL) = + 93.6 " {LK) = + 97.0 - {AM) = + 19.5 " The results of these diagrams are collected and written in Fig. I. Both web and chord stresses may be checked by moments as follows. Moments about K give : {CD) = — (54.3 X 36—2 X 21.72 X 18) -f- 1 1.7 = — 100.2 tons. The diagram gave 102.0 tons. The agreement is close enough for the purpose, but in an actual truss the difference ought not to be greater than one per cent, of the smallest result. Again, BC and LK intersect in a point about 30.8 feet to the left of R, and the normal distance from that point to CL 102 nojv-coa't/xl'ous ti^usses in general. produced is about 48.5 feet. Hence, with the head of the moving load at L, and, by taking moments ^bout the point of intersection : {CL) = (2 X 21.72 X 48.8 - 7? X 30.8) ^ 48.5 = + 19.6 tons. The diagram gave + 20.0 tons. The agreement is suf- ficiently close. Numbers of checks like the two above should be applied. The Fig. i shows that the web members MB, BL, CL, CK, DK, DH, HE, and EG must be counterbraced. Art. 24. — Bowstring Truss — Vertical and Diagonal Bracing with Counters. The case next to be taken is that of an ordinary " bow- string" truss with vertical and diagonal bracing, and is rep- resented in Fig. I. The inclined braces are not supposed capable of resistance to a compressive stress. Inasmuch as counters are almost invariably introduced in such a truss in ordinary engineering practice, they will be supposed to exist in this case. Fig. I. The truss is supposed to be designed for a highway bridge, furnished with sidewalk on each side. The greatest moving load will be taken as a crowd of peo- ple weighing eighty-five pounds per square foot, covering roadway and sidewalks, advancing panel by panel from R until the truss is entirely covered. Since the span is a short one, and the roadway very heavy, the whole of the fixed load will be taken as applied to the lower chord. BOirs TRIXG 1 'R USS. 103 The following are the data required ; they are taken from the preceding Article; Span = 72 feet. Depth of truss, 12 feet. Radius of circumference of circle passing through upper ex- tremities of verticals — 60 feet. Number of panels = 6. Panel length = 12 feet. Width of roadway, from centre to centre of trusses = 20 feet. Width of each sidewalk = 6 feet. W = 900 pounds per foot = 5.4 tons per panel, w = 32 X 85 X 12 = 32,640 pounds = 16.32 tons per panel. As in Art. 18, if the plane dividing the truss cut more than three members, some one of these members must be neg- lected or assumed not to exist. For the sake of brevity, two letters inclosed by a paren- thesis will denote the stress in the member indicated by these letters. The placing the load for the greatest web stresses, is done in accordance with the general principles established in Art. 20. When the head of the moving load is at Z, the existence of AK must be ig- «^ (j^jq nored ; for if BL be then omitted, y^ A' will suffer compression. With the head of the train at Z, R—Zj.i tons ; hence Fig, 2. {LK) = (/? X 24 - 21.72 X 12) ^ {BK= 10.8) = 36.1 tons. Fig. 2 is the complete diagram for this case, and explains itself, ac is the shear '^P — 27.1 — 21.72 = 5.38 tons. Scaling from Fig. 2 : {BL)— + 19.2 tons. {BK) — + 2.0 tons. BH is also omitted for this loading. 104 NO.V-CONTINUOUS TRUSSES IN GENERAL. With the head of the moving load at K, R = 37.98 tons, and CG and BH are omitted. {KH) — [Ry. 36—2 x 2 1 .72 x 18) -^{CH= 12) = 48.78 tons. Fig. 3 is the diagram for this case, ce is the shear, acting downwards, 2P^= 37-98 — 2 x 21.72 = — 5.46 ton. This diagram gives the results : iCH) i.o ton. (AT) = + 16.0 tons. With the head of the moving load at H, R = 46.14 tons. DF and CG will be omitted. It is unnecessary to give the diagram for this case, drawn precisely as the two preceding ones have been. The diagram gives : It is (DG) = -1.0 ton. (HD) = + 1 7. 1 tons. Neither will the diagram for the head of the load at G be given, as it is constructed exactly like the others. It gives, omitting GC and DF: {GE)— + 15.00 tons. For the greatest chord stresses the moving load covers the entire truss, and Fig. 4 of the next Article is the complete diagram for the case. All the explanation which the diagram needs is there given. BL, KC, GC and FD are supposed to be omitted. Taking moments about B, with uniform load {iv + W) • {KH) = (54.3 X 24 — 21.72 X 12) ^ 10.8 = 96.5 tons. Others may be checked in the same way. The most rational circumstances under which the greatest tensile stresses can be supposed to occur in the verticals, are BOWSTRING TRUSS. 105 those under which they are found in the next Article ; and the results there obtained are used in this case. They are introduced, without more explanation, in Fig. i. Their dia- grams will be found in the next Article. These results are by no means satisfactory, but nothing better can be done with such a form of truss. Some of the web stresses should be checked by moments, by the general method. Fig. I shows the greatest web stresses selected from all the preceding results. It is far more convenient to treat the fixed and moving loads together, as has been done in this case, than to treat them separately, as, of course, may be done. Again, the stresses caused by each panel load on all the members of the truss may be found, and their effects com- bined, but this also requires far more labor than the method followed. The stress diagram for each position of the moving load might have been worked up from the end of the truss, as was that for the chord stresses, but it saves considerable labor to find one chord stress by moments, and begin the diagram with that. Art. 25. — Bowstring Truss — Vertical and Diagonal Bracing without Counters. It is evident, from what has preceded, that the existence of the counters causes considerable ambiguity in the web stresses, Fig. I. and it is much more satisfactory from a strictly technical point of view to leave them out, as shown in Fig. i. io6 NON-CONTINUOUS TRUSSES IN GENERAL. Fig. I is exactly the same as Fig. i of the preceding Arti- cle, with the counters omitted, and in this Article will be found the stresses existing in it with precisely the same data as were used above. The moving load is brought on panel by panel from R, ac- cording to the general principles established in a preceding Article. With the head of the moving load 2X L, R = 27.1 tons. As before, {LK ) = 46.9 tons. (HP Fig. 2. Fig. 2 is the complete diagram for this loading. {RL) — {LK ) = 46.9 tons, and ^<^ = 27. i — 2 1 .72 = 5.38 tons. Hence, {^^) = ii-9 tons. With the head of the moving load at K, R = 37.98 tons. {KH) = (R X 24 — 12 X 21.72) -i- 10.8 = 60.27 tons. Fig. 3 is the diagram for this loading, and it explains itself. Hence, (BK) = + 24.9 tons. {BH)= - 15.8 tons. Fig. 3. It is unnecessary to show the diagrams for the two cases of the head of the moving load at H and G. They give respect- ively : BOWSTRING TRUSS. 10/ {DH) = + 17.1 tons, {DG)= — i.o ton, and {GE) = + 15.0 tons. Fig. 4 is the diagram for the moving load over the whole bridge, bd is the reaction R = R' = 54.30 tons ; dc = 21.^2 tons ; dixxd fg is equal to \ (21,72) tons. The greatest chord stresses, taken from Fig. 4, are written in Fig. I. The greatest web stresses are selected from all the preced- ing results, and also written in Fig. i. The stress {BC) may easily be checked by moments, as follows, by taking the origin at H. The normal distance of H from {BC) is 1 1;9 feet. Hence, {BC) = ;2^(zt'+ ^F)x36-3x \2{zu+W)\ -m 1.9=98.56 tons. Other and similar checks should also be applied. It is seen in Fig. i that the diagonals need counterbracing, but there is no ambiguity, and the superiority of the design over that in the preceding Article is evident. (RI^) 2fh = (EC) - -naeton^i Fig. 4. It is to be carefully borne in mind that the diagrams must always be drawn as large and as accurately as possible. Those of the present Article were constructed to a scale of ten tons to the inch. The figures do not show the scale. io8 NON-CONTINUOUS TRUSSES IN GENERAL. Art. 26. — Deck Truss with Curved Lower Chord, Concave Downward — Loads at Panel Points — Exsunple. The truss shown in Fig. i is, in some respects, a peculiar one. It has one prominent characteristic which distinguishes it from the bowstring trusses which have been treated in the three preceding articles, in that the chord sections (upper and lower) in any panel, excepting those two at the centre, inter- sect in the upper chord within the limits of the span. All the web stresses, except those in Z?iV and DM, will have their greatest values when the moving load covers the whole truss, as was shown in Art. 20. NJM is horizontal, consequently the JL -3».» X - sa.9 C Fig. I. intersections of NM with CD and DE are found at an infinite distance from the truss, but ON and CD intersect between E and 7^ (near the latter point) : all other intersections are found between A and G. The positions of moving load for the greatest stresses in DN and DM are the same, therefore, as for a truss with parallel chords. Observations relating to the positions of the points of inter- section of the chord sections in the figure apply to a truss for which the following are the data: Radius of circumference of circle passing through the apices of the lower chord = 60 feet. Vertical distance of centre of circle below D = 66 feet. Depth of truss chrough D = 6.3 feet. Span = AG = y2 feet. AR = GR = 18 " DECK TRUSS. Uniform upper chord panel length — 12 feet. NM = 12 feet. ON = ML = IT, " OP = LK = 1 1.7 " Uniform panel fixed load = 5.4 tons = W. " '' moving load = 16.32 '^ = w. 109 (CS) (BT) CDir)/,\ / \ / -. / j\ (SMI -T \ — y f * — \IP) / \/ • \ 4 ^ fsc; _V c \. «lX =2? =i'«.5 ivrx^ jl,M«j_V Ic ^cd ^ 21. jz d-e = (C2); fc ^ r.MTv-; Fig. 2. The loading is the same as that used in the preceding bow- string trusses. Let the angle which DN or DM makes with a vertical line be denoted by ot. Then, tan ix. — 0.952 ; sec (v — 1.38. For greatest tensile stress in DN. With the moving load extending from A to C\ Reaction R — 37.98 tons. The shear 5 = 2 x 21.72 — 37.98 = 5.46 tons. Hence, {^^) — S sec a = + 7.53 tons. For greatest compressive stress in DM. With the moving load extending from A to D: Reaction R = 46.14 tons. The shear 5 = 46. 14 — 3 x 21.72 = — 19.02 tons. Hence, (DM) = S sec a = — 26.25 tons. no NQN-COA'TIN UO U S TRUSSES IN GENERAL. For the other web stresses the moving load must cover the whole truss, and Fig. 2 is the complete diagram for that con- dition of loading. With the data given below the figure, no explanation is needed. The results of the diagram will be found in Fig. i, together with those determined above by the trigonometrical method. One advantage inherent in this form of truss, as in all in which the chord intersections are found within the limits of the span, is the little counterbracing required. In the truss taken, the two web members DN and DM dive all that require such treatment. Indeed, with a sufificiently small radius of lower chord and centre depth, together with an odd number of upper chord panels, a truss may readily be designed which will require no counterbracing at all. A dis- advantage, however, is the small depth at centre, just where a great one is needed, with the resulting heavy chord stresses. As the Fig. i shows, no stress exists in PR or KR ; never- theless those members would ordinarily be inserted for the purpose of stiffening the whole structure. The truss may be supported directly, as at G, or there may be an end post, as AR. The greatest stress in AR, will be the reaction R added to ^{w -\- W). Hence, {AR) = — (54.3 + 10.86) = — 65.16 tons. As checks, moments about D give : (NAI) = (54.3 X 36 — 2 X 21.72 X 18) -^ 6.3 = + 186.0 tons. The normal distance from C to ON is 7.5 feet (nearly). Hence, by moments about C: (ON) — (54.3 X 24 — 21.72 X 12) -^ 7.5 = + 139.0 tons. OP, prolonged, cuts the upper chord at a point about three feet from D toward £, and the normal distance from that point of intersection to OB, prolonged, is about 23.5 feet. Hence, (OB) = (54.3 X 39 - 21.72 X 27) ^ 23.5 = +65.1 tons. CRANE TRUSSES. Ill The agreement of the last result with that obtained by diagram is very close, but the other results, by the two methods, show a difference of about two per cent. This is close enough for the present purpose, but in practice the figure and diagrams should be drawn large enough to make this difference, at most, one per cent, of the smallest result. A truss of this character, with more than one system of triangulation, gives indeterminate stresses, but the approxi- mate method of Art. i8 may be used. Approximate deter- minations may also be made by treating each system, with its weights, by the methods just given, and combining the results for the chords. Art. 27. — Crane Trusses. A form of truss which has been used for powerful cranes, A Jl.^ Fig. I. 1 1 2 ^^0A^- CO A' TIN UO US TR I \SSE S IN GENERA I. under circumstances requiring much head room, is that shown m Fig. I. It revolves about a vertical axis midway between E and F. CRANE TRUSSES. H^ In the example taken, the weight, W, hanging from D, the peak, is supposed to be ten tons. Each chord of the truss ;«, /, k, etc., or a, b, c, etc., is made up of chords of quadrants of two circumferences of circles. The radius for the chord mlk, etc., is 25 feet, and that for the other chord is 22.6 feet. EF is 5 feet. Denoting the chord panels by single letters: rt' = 5 feet. m = b = c = 6 " / = ./=; " /i = g " Fig. 2 is the complete diagram for the stresses in the truss, supposing the on/j/ load to be the ten tons hanging from the peak. If it should be necessary to take into account the weight of the truss, it would be done precisely as the fixed weights of trusses have been treated in the preceding Articles. The lines in Fig. 2, denoted by letters and figures, are par- allel to lines denoted by the same letters and figures in Fig. i. The diagram gives the following results : (0 = + 12.5 tons. ( 2) = — 14 tons. (3) = + 6.00 " (4) = - 13.7 " (5) = + 1.50 " (6) = - 134 " (7) = - 2.60 " ( 8) = — 14.0 " (9) = — 7.20 " (10) = - 13-2 " (II)- - 5-' 8 tons. {a) = — 8.7 tons. (.«) = + 17.8 tons. {b)=^ — 24.0 " (/) = + 30.4 " i^) = - 35-9 " {k) = + 38.7 " {d) = -44.0 " {h) = + 44-2 " (^) = -48.2 " {g) = + 47-2 " (/) = - 54 . I tons. These results may easily be checked by moments. The 114 NO^r-CONTINUOUS TRUSSES IN GENERAL. different lever arms, with two exceptions, to be used, are shown in Fig. i. The normal distance from ^ to C is about 4.6, and from e to B about 5.3 feet. These lever arms were scaled from the drawings, and may not be exactly right, but near enough for the purpose. By moments about C\ {S) = + (10 X 22.25) ^ 4-6 = + 48.4 tons. By moments about B\ {e) ~ — (10 X 26.0) H- 53 = — 49.0 tons. The chord sections d and k, prolonged, meet at ^, and mo- ments about that point give : (7)= - (10 X 3.7) 14.7 2.5 tons. These results agree sufficiently well with those obtained by the diagram. If the chain, rope, or cable pass along either chord, the ten- sion in it will tend to produce an equal amount of compres- sion in the panels of that chord. The resultant stress, there- fore, in any panel will be the algebraic sum of this amount of compression, and the stress due to the weight W. Fig. 3 is a skeleton diagram of the ordinary crane, which revolves about the centre line of ED as an axis. AB is the --^ weight hung at the peak, A. BC is parallel to DA, and represents the amount of tension in that member. AC \?> the compression in AE due to the weight W. As before, the tension in the rope or cable tends to produce an equal amount of compression in any member along which it lies. Let / denote the normal dis- tance from DA to any point in the centre line of DE\ then any Fio. 3. section of DE will be subjected to the bending moment 3/= {DA) X /. ROOF TRUSSES. II.S DE will also be subjected to a direct stress (tension in Pig. 3) equal to the vertical component of the stress in DA. The greatest resultant intensity of stress in any section will be the combination of the intensities due to these two causes. Art. 28. — Preliminary to the Treatment of Roof Trusses — Wind Pressure — Notation. Four, only, of the principal types of roof trusses with straight rafters will be treated, since the method used for any one is precisely the same in character as that to be used for any other. The wind will be assumed to act on one side of the roof, and its resultant action will be assumed to be normal in direc- tion to the rafters. If such is not the case, f and i> will represent empirical determinations of the horizontal and vertical components, respectively, in the Articles which fol- low, and the methods will remain precisely the same. Let/ be the intensity of this normal wind pressure, and let / and /i be the lengths of two adjacent panels of the rafter, while ^ is, the horizontal distance between two adjacent and parallel rafters. The total normal wind pressure supposed exerted or concentrated at the point between the two panels, will then be — — ^^ ~ ; or if / = /i , as is usually the case, pdl. If ^ is the angle which the rafter makes with a horizontal line, the horizontal component of this normal pressure will be : /= ^"^^^^ ^^) sin 6 ■ or pdl sin 6 • and the vertical component will be : pd[/ + /i) ^ ,, „ V = ~ ^ cos 6 ; or pdl cos 0. Finally, let the total fixed weight of the roof be supposed concentrated at the panel points of the rafters ; and let the weight of such load at any panel point be represented by W^. The total vertical load at any panel point will then be : The vertical reactions due to the vertical component of the Il6 A'ON-CO.VTIXUOUS TRUSSES IN GENERAL. wind pressure and the fixed load are found by the principle of the lever in the usual manner ; that at the left of the span will be called R, and that at the right, R' . The vertical reactions due to the horizontal forces /, will, however, be called R\ and they will be found for the different cases by taking moments about any point in the horizontal line joining the feet of the rafters. If b is the vertical pro- jection of a rafter, and 2c the span, or distance between the feet of two rafters meeting at the ridge, there will result : ^ 2 2C 4C At the foot of the rafter pressed by the zvind, this reaction will be downward in direction ; at the foot of the other rafter it will be uptvard. The total horizontal reaction at the points of support will be equal to bp, the total horizontal force of the wind, and its direction of action will be opposite to that of the wind. The horizontal component of the wind pressure and the horizontal reaction produce a couple, equal and opposite to that whose force is R", and whose lever arm is 2c. R" must be numerically less than R, or the wind will turn over the roof bodily. If the foot of one rafter is supported on rollers, the hori- zontal reaction will be wholly exerted at the foot of the other. If the foot of neither rafter is supported on rollers, the horizontal reaction will be assumed to be equally divided between the points of support. The stresses for the vertical and horizontal loads are found by separate diagrams, although they might be found by one only, because the slope of the roof may, in some cases, be so small as to make it needless to consider the forces/". If rollers are used at the foot of one rafter, the wind may press that one or the other. In treating a large roof it may, then, be necessar\^ to take the wind first in one direction and then in the other. These two, with the case of no rollers, make three possible cases, and an example will be taken in each one. ROOF TRUSSES. 117 Art. 29.— First Example. The truss represented in Fig. i is a roof truss, applicable to short spans. There is no " moving load " in such a case. The wind pressure, however, may act on one side and not on the other, and for that reason VV, \\\, and W^^ are taken as differing from each other, as was explained in the preceding Article. tiM^ Fig. I. There is no essential error in this case in assuming all the load concentrated at the points indicated. The wind press- ure is assumed to act on the left side of the truss, so that W>W^> W^. Also, JFi = 1 ( JF+ W^). A and B are equal in length, and E is horizontal. Figs. 2 and 3 are the stress diagrams for Fig. i, and the lines in them, indicated by letters primed, are parallel to, and repre- sent stresses in the members marked with the same letters in Fig. 2. Fig. I. The kinds of stresses in the different members are shown by the signs + or — , in both figures, signifying tension or compression, respectively. Il8 NOA'-CONTINUOUS TRUSSES IN GENERAL. Fig. 2 is the diagram for the vertical loads, and Fig. 3 that for the horizontal loads/". {^ c^__^ -7^ ^ -S- ^ ^ yi ^ " < ^-i^ -^ ^ ^^ ^--/, -X -;^ -X-;S Fig. 3. Rollers are supposed to be under the foot of neither rafter. Consequently the horizontal reaction at each end is /, as shown, /i is equal to hf- In Fig. I, R" acts downward at the left of the span, and upward at the right. The resultant stress in any member is the algebraic sum of those given by the two diagrams. Fig. 3 does not show all its lines parallel to the members of Fig. I. There^ is, of course, a diagram similar to Fig. 2, for the right half of the truss, but it is not needed. The stresses may also be found by the method of moments, by locating the origin of moments according to the general principle stated in Art. 17. Art. 30. — Second Example. Fig. I of this Article represents a very common roof truss. As before, the total load is supposed concentrated on the rafters at the points indicated. Wind pressure is taken as acting on the left of the roof, making JV > JJ\ > JV.2. Jl\ is simply the panel weight of the roof, and U\ = }^{IV+ W^. Each rafter is divided into four equal parts, and W is taken equal to 2W^_. Hence the reaction R — 3 fF. This does not at all affect the generality of the diagram. The lower ex- KOOF TRUSSES. 119 tremity, however, of D', in Fig. 2, will not usually be found at a'. Fig. I. Fig. 2 is the stress diagram for the vertical loading taken, and the notation has precisely the same meaning as before. It is seen from Fig. i, that there is some ambiguity in regard to the stresses C, Q, F' , and M'. It may be assumed, however, that E' = H' and P' = Q', which makes F' = Fig. 2. The kinds of stresses are shown on the diagrams. I20 NOJV-COA^TIXUOUS T/^USSES IX GENERAL. Fig. 3 is the diagram for the horizontal stresses f and Rollers are supposed to be placed at the foot of A. Hence all horizontal reaction will be found at the foot of A^. As shown, that reaction will be equal to ^f. The vertical reac- tion R" will be directed downward at the foot of A, and upward at the foot of A-^. Considering the horizontal forces only : Ai =Bi = c = n,' = -A'. K,'=L,'=0' = -4/- A". The resultant stress in any member is found by combining the results of the two diagrams in the usual manner. This style of truss is so frequently used that formulae for the stresses due to the vertical loading are given below, in which a is the length of the rafter, c the half span RR\ and d the height of W^ above O. These expressions may be readily derived from Fig. i. A'=.jR, B' =A'-W C = A' - 2lV~, a D' = A' -iW H'=E' =-W. a F' = 2H' = 2-W, a ROOF TRUSSES. 121 ~ '00 O' =L' -Jlf'=j{R-^lK), N' = M' + S' = |^ ^' Art. 31.— Third Example. Figs. I, 2, and 3 are roof-truss and stress diagrams, respect- ively. The wind pressure is supposed to act on RIV^, and the total load is taken to be concentrated as shown. -/^ v3 sp Fig. I. cte ^ /^' act, R dc Z1C6 zw Fig. 2. The rafters are each divided into three equal parts. ^22 NON-CONTINUOUS TRUSSES IN GENERAL. The notation has the same signification as that used be- fore, and it is unnecessary to explain the diagrams. Fig. 2 is the diagram for the vertical loading, and Fig. 3, that for the horizontal forces/" and /i = \f. Rollers are supposed to be placed at the foot of A-^ in Fig. I. Hence the total horizontal reaction will exist at the foot of A, and its value will be 3/", as shown. As before, the direction of R" will be upward at the foot of A-^, and down- ward at the foot of A. The resultant stress in any member will be found by com- bining the results given in Figs. 2 and 3. \--^'^-~\\ ^'>^ '.C',' ^ ^ f < f H-~. 'x'^-+ r Fig. 3. Fig. 3 does not show K' equal to ///, as it is not a scale diagram. The members FH and K are in tension, while D and E are in compression. Wy Wx> W.^, the last being the fixed panel weight of the roof. Art. 32. — Fourth Example. In this Article, as before, Fig. i is the truss, and Fig. 2 is its stress diagram for the vertical loading. Wind pressure is Fig. I. GENERAL CONSIDER A TIONS. 123 taken as acting on RW]^\ the reaction R, therefore, is \ (3 W + \\\ + W.^ and W> \\\> W^. The rafters are each divided into three equal parts, and F and D are vertical. IV2 is the fixed panel weight of the roof, and W^ = ( IV + W.2) -i- 2. Fig. 2. The kinds of stresses in the various members are shown in Fig. 2. It is unnecessary to give the diagram for the horizontal components of the wind pressure. The method of drawing it will fall under some one of the three cases already given. Art. 33. — General Considerations. In all the preceding cases of roof trusses the stresses may be obtained by the method of moments, and therefore the graphical results may be checked by that method. The operations are precisely similar if a part of the load is hung from points in the tie RR', in all the cases, or if the loading is even more eccentric than that assumed. In many cases the sides of large buildings are braced to 124 NON-CONTINUOUS TRUSSES IN GENERAL. the roof by oblique members extending from that panel point of RR' adjacent to the side, to some point in that side. In such cases the wind will cause stresses, in the different mem- bers of the roof truss, which must be determined independ- ently of those already found and added, algebraically, to them. CHAPTER IV. SWING BRIDGES. ENDS SIMPLY RESTING ON SUPPORTS. Art. 34. — General Considerations. Without regarding the nature of the supports or attach- ments at the extremities of the two arms, swing bridges are divided into two classes : those with center-bearing turn- tables, and those with rim-bearing turn-tables. In the first class the entire reaction at the pivot pier is exerted through a central pivot, or a nest of one or more series of solid, con- ical rollers ; usually the latter. In such a case there may be a circular drum or framework, supported on wheels running on a circular track, but they are used solely for the purpose of steadying the bridge while open. In the case of a rim-bearing turn-table, however, the reac- tion at the pivot pier is exerted through the circular track on which the wheels supporting the drum or framework of the table turn. The object of a pivot, in such a case, is simply to enable the bridge to turn truly about a center. In reference to the truss, there is evidently only one point ■of support, at the pivot pier, with a center-bearing turn-table, i. c, at the center. With a rim-bearing turn-table, however, there may be two or more points of support at the pivot pier, though it will be shown hereafter that, by separating the different systems of triangulation, it will never be necessary to consider more than two at once. Again, with either turn-table there may be three different methods of supporting or securing the extremities of the two arms of the bridge. These extremities may simply rest on .supports, so that the reaction will always be zero, or upward ; 125 126 SWING BRIDGES. this is the only case which will receive more than a passing notice in this chapter. Again, those extremities may be fastened down or latched to the piers when the bridge is not open. The reaction may then be nothing, upward or downward. In these two cases the reactions at the extremities of the arms will be zero when the bridge is simply closed, and sup- porting no moving load. In the first, when the moving load is on one arm, the ex- tremity of the other may be slightly raised from its support ; in the second case, however, that extremity will be held down by the latching apparatus, /. e., the reaction will be downward. The object of the latching apparatus is thus seen to be the prevention of thchammering of a truss-end on its support. Finally, the third method is to raise, by proper machinery, the truss ends, when the bridge is closed, any diesired amount. The object of this arrangement is to insure a reaction at the extremities, which will always be nothing or upward, and thus obviate the hammering before mentioned. In this case the whole of the bridge weight does not rest at the pivot pier, as the lifting of the ends takes up a part of it; in fact, may take up the whole of it. Recapitulating, then, the ends of a swing bridge may be : (l.) Simply supported, (2.) Latched down, (3.) Lifted up. The detailed consideration of the first case will next be taken up. Art. 35. — General Formula for the Case of Ends Simply Supported— Two Points of Support at Pivot Pier — One Point of Support at Pivot Pier. With two points of support at the pivot pier there usually arises the case of a continuous beam resting on four points of v4 ^^ :b l^ C £5 j> ^ — i^ ^ a 2d, M^ M-^ M^ GENERAL FORMULA. 12/ support, as shown in Fig. i. The notations of the spans and bending moments at the different points of support are suf- ficiently well shown in the figure. The points of support will all be taken in the same horizontal line, as the formulae will then also apply to any configuration belonging to a state of no stress, provided the truss may be considered straight between any two points of support (see Appendix). Any truss may be considered straight when an equivalent solid beam has a neutral surface which is plane before flexure ; a straight solid beam is " equivalent " to a straight truss when equal moments of inertia and resistance are found at the same section in the two structures. The theorem of three moments, in the ordinary form, does not apply, then, to a continuous truss with one chord curved, and none of the following investigations apply to such a case. Again, in the span /, there will be supposed no load, as such is usually the case. The load on /, ought always to be supported on short girders or beams resting at B and C, for there is the less complication of stresses in the trusses, and consequently less liability to uncertainties ; besides, such an arrangement is probably more economical in material. In the present case M^ and M^ will each be equal to zero. Let z denote the distance of the point of application of any force, P, from the left-hand end of the left-hand span, or right-hand end of right-hand span. In /j, z would be meas- ured from A, while in 4 it would be measured from D. The formula expressing the theorem of three moments for all supporting points in the same level becomes, by using the notation of Fig. i : M4x + 2Mlk + 4) + M4. + yi/^/i' - ^)s + \hp{li - s')z = o. 1 2 The symbols 2 and 2 indicate summations for the spans /i and 4. Applying the above equation to spans 4 and 4, and then 128 SWING BRIDGES. to 4 and 4> there will result, bearing in mind the circumstances^ of the present case : 2J/2(/, + 4) + J/3/2 + 7i/'(/i'-^)^ = . (i). M^ k + 2J/3 (4 + 4) -f yhp{ii -z')z = o . (2). If Equation (i) be multiplied by /„ and Equation (2) by 2 (4 + 4) ; and if the results so obtained be subtracted, and if the following notation be used : 4,4 2 s '2 '2 *i '3 there will result : ,^ c'l^hPii -n')n- 2(^4 (r + I) ^P{i - ;«2) ;« , , Equation (i) then gives : _ M^ c %^P{i -n')n ,. ' 2(^+1) ... V4;- Equation (2) will evidently give another expression for- M2, but it is not necessary to write it. Let Rx, R2, Rsj and R^ be the reactions at the points of support A, B, C, and D respectively, Figure i. Then, adapt- ing the formulae for reactions from the theorem of three mo- ments (see Appendix) to the notation of the present case,, there may at once be written R,= :SP(i-n)+^ (5). R,^2P»-^-^'-^' . . . (6). 4 4 ^8= + ^-^ +^^^--/- • . (7)- R,= 2P{i -;«) +-^ .... (8).. 4 As should be the case, there is found : R, + R, + R, + R, = 2P + 2P. GENERAL FORMULA. 1 29 In ordinary swing bridges where 1^ = 1^ = /, and a single weight P rests on /, : R.=p{^-ri)\'~^'^''^''~- — r~^l • (5-). R, = c\R,-R,-P{i-7i)\-R, . . {7a). R,:=P-Ri- Ri + R, . . . . (Sa). It may sometimes be convenient to use the following equa- tion derived from Equation (i): J/3 = - ; 2 J/2 (r + I) + r/2^P{i - ;/2) n\ . (9). These are all the equations necessary for the solution of the case of two supports at the pivot pier, frequently exist- ing if the turn-table is rim-bearing. If there is only cme point of support at the pivot pier, the -A I, S z^ c e. zs- ^ M, :m:^ M^ case reduces to that of a continuous beam of two spans only, as shown in Figure 2. As A and C are points of support only, M^ and J/3 are each zero; hence if m now stands for y, the equation immediately preceding Equation (i) gives: ^. = -//i^'-'^)"^^f^' -'-')"'■ . (.0). -» ({ > .^8l^j^:iO^{»^^ Fig. I. From what has preceded, it is clear that the greatest coun- ter stresses are to be found by carrying the moving load for- ward from the center of the bridge toward the end, and on one arm only, at the same time considering it a simple non- continuous span. The greatest main web stresses, on the other hand, are to be determined by moving the system of balanced panel weights, already described, from the two extremities of the bridge toward the center, as will presently be illustrated. It is to be remembered that verticals in compression will sustain their greatest stresses at the same time with the diagonal tension members which cut their upper extremities, if the moving load traverses the lower chord. As one arm of the bridge is a simple truss supported at I40 SWING BRIDGES. each end, for the unbalanced loads on it, it is evident that the greatest compression in the upper chord and tension in the lower will exist, near the ends, for the moving' load over the whole of one arm. Since moving loads on both arms at the same time balance each other, it results that the greatest tetision in the upper chord nnd corApression in the lower, at the center, ivill exist with the moving load over the wJiole of both arms. This is true for the center only. For other panels adjacent to the center, // zvill be necessary to take single-balanced pafiel moving weights, and f^id for each, all panels in which the stress is of the same kind as that caused by the fixed load alone, and the amount of that stress in those panels. Having obtained the results for each pair of balanced weights, they are to be combined in the manner already shown. The greatest compression in the lower chord and tension in the upper, near the ends, however, will exist ivith the bridge open or closed and subjected to its own weight only. From these results the greatest chord stresses are to be found. The maximum moving load compression in EP w'xW. next be found, and then the counter stresses and that in Tig, for all of which the arm of the draw will be treated as a non-con- tinuous span, with the moving load passing over it from the center toward the end. The positions of the moving load for the greatest value of these stresses will, hence, be found by the principles of Articles 7, 20, and 21. Equation (7) of Article 7 shows that the third driver of the front locomotive must be placed £t the foot of Tloin order to give the greatest shear in EP, or, what is the same thing, the greatest reaction at its foot. Such a position of the moving load will make the shear in question 45.16 tons. Hence: {EP) = — 45.16 X 1.4 = — 63.23 tons. In order to determine whether the counter Ci is necessary, let it be supposed omitted, and the resulting greatest moving ENDS SIMPLY SUPPORTED. I4I load compression found in T^. Equation (^4) of Article 20 shows that the latter will exist with the third driver of the front locomotive at the foot of P^, whence will result a reac- tion at the foot of ^5"/* and a load at the foot of Tio of 32.1 tons and 5.72 tons, respectively. By taking moments about the point of intersection of the chords, and using the same lever arm for 7^4 as was employed for the fixed load stresses : = — 34.08 tons. But the fixed load tension in T^ has already been found to be 22.57 tons, which is much smaller than the 34.08 tons of moving load compression just found. Hence, as Ti can really sustain no compression, the counter c^ must be introduced, and its greatest tension found with the same position of moving load, and, also, under the assumption that T^ does not exist. Remembering that the lever arm of c^ is 22 panel lengths divided by the secant for c, : , , \2.\ X 20 — 5.72 X 21 (ci) = ^-^ sec c,, = + 23.72 X 1.4 = + 33.21 tons. By a precisely similar use of Equation (4) of Article 20, it will be found that the greatest moving load compression will exist in 7^3 with the first driver of the front locomotive at the foot of Pg. This position will give a reaction of 13.05 tons at the foot of £P, and an advance load of 1 1.02 tons at the foot of P^. Again taking moments about the point of intersec- tion of the chords, the greatest moving load compression in Ts will be found to be : -^ 13.05 X 20— 11.02 X 22 ^ ( io) = — -^ — sec Ts = — 14.45 tons. 22 But the fixed load tension in T3 has already been found to be 36.815 tons, which is over two and a half times the 14.45 tons of moving load compression. Hence no compression 142 SWING BRIDGES. can ever exist in T^, and no counter will be needed in the same panel with it, nor, indeed, any other counter than c^. The same condition of loading will give the greatest mov- ing load tension in P^, but computations will show that it is only about one-third the fixed load compression (as would be inferred from the results for T^ ; hence no counterbrac- ing of that member is required. These operations show the method always to be pursued in determining the requisite counters, and, also, that it is essentially identical with that used for non-continuous spans. As T'lo is a simple hanger, it will take its greatest tension with the greatest floor beam reaction at its foot. This latter will exist with the same position of moving load as that which gives the greatest stress in EP — i.e., with the third driver of the front locomotive at the foot of Tiq, and its value is 25.57 tons. Now the shear existing at the same time in EP has been found to be 40.16 tons, and as the co-existing vertical component of the stress in a is between three and four tons only, the counter Ci must act in tension to carry sufficient shear over to EP to make up its shear of 40.16 tons. In other words, the vertical components of (^i) and (a) added to the stress in Zlo must equal 40.16 tons. Hence the stress in Zio is simply the greatest possible load at its foot. .-. (r,o) = + 25.57 tons. The member d sustains no stress, under the condition of partial continuity, with the moving load on one arm only; hence it is to be ignored in finding the stress in P^. The greatest compression in this latter member will therefore exist with precisely the same position of moving load as was used for EP, but changed end for end on the span, so as to be headed toward the center rather than toward the end of the arm. By using the same shear as for EP, therefore : (/*i) = — 45. t6 X sec P■^ — — 59.57 tons. The tension member 7\ is a simple hanger, with the maxi- ENDS SIMPLY SUPPORTED. 143 mum floor beam reaction at its foot, as was found for T^a- Hence: Ty^= + 25.57 tons. The moving load will pass on the arm from the end toward the center, for the remaining web members. The position of moving load for 7*2, in accordance with Equation (4) of Article 20, requires the third driver to be at the foot of /g, vvith both locomotives and tenders, but no train load, on the arm. This position gives a reaction of 32.1 tons at the foot of /'i, and an advance weight of 5.72 tons at the foot of Zi. Hence : _ 32.1 X 25 — 5.72 X 24 ^ o ^ . r, = ^— — ^ ^-^ ~ sec T^ = + 38.16 tons. 2 23 ~ o In the same manner it is found that the greatest tension in 7^3 will exist with the second driver of the first locomotive at the foot of P^, and with the corresponding driver of the second locomotive on the arm and nine inches from its end. The reaction at the foot of /\ will be 15.24 tons, and the weight at the foot of /*2, 3.14 tons. Hence, by moments, as usual, about the point of intersection of the chords : Since Pc, sustains its greatest compression under the same condition of moving load as T^ : 15.24 X 25 Pi = ^-^ ~ - — 16.57 tons. I 23 ^' Again, Equation (4) of Article 20 shows that the first driver of the front locomotive must be at the foot of Tlo in order to give 7^4 its greatest tension. The second wheel of the tender will then be on the arm and one foot from its end. Hence the reaction at the foot of P-^ will be 4.62 tons, and the weight at the foot of P-^, i.i tons, and : _ 4.62 X 25 — I.I X 22 ^ 7; - — —^ sec 7; = ^^ 5.96 tons. 144 SWING BRIDGES. The same position of moving load gives the greatest com- pression in P^ ; hence : _, 4.62 X 25 Pz= — = — 5.25 tons. ^ 22 ^ ^ These results complete the moving load web stresses for the moving load on one arm only, leaving the chord stresses yet to be determined. The lower chord panels i and 2 receive their greatest stresses with £P, and they are to be found by simply multi- plying the shear in that member by the tangent for £P. Hence : (i) = (2) T= 45.16 X 0.982 = + 44-345 tons. Similarly, the panels 4 and 5 sustain their greatest stresses with Pi. Hence : (4) = (5) = 45.16 X 0.859 = + 38.79 tons. In order to determine the greatest stress in lower chord panel 3, recourse must be had to Equation (14) of Article 7. That equation shows that, with the moving load passing on the arm from right to left, the third driver of the second locomotive must be at the foot of P^, with 24 feet of the uniform train load resting on the arm adjacent to the center. This position gives a bending moment about the top of P^ (the center of moments for panel 3) of 3,672,530 foot- pounds. Hence : 3,672 5^0 ^^^ "^ ^ 30.666 " "^ 121,070 lbs. = -f 60.535 tons. The upper chord stresses at once result from those in the lower, but it is first to be observed that since T^ is a tension member only, it will not be stressed with the moving load on essentially the whole of the arm, since the counter c^ will then come into action. The upper chord stresses in a and d will therefore be equal, and as the foot of P^ will be the center ENDS SIMPLY SUPPORTED. 1 45 of moments for those stresses, the position of the moving load for their greatest value will be precisely the same as for (3), with the exception that the direction of motion is to be reversed, thus placing the third driver of the second locomo- tive at the foot of Pg. Hence : 5 672 Ej ■^o {a) = {b)= - ~^~j- seca=- 125,350 lbs. = - 62.675 tons. "y* J J J Finally: [c) = — (3) sec a = — 60.61 tons. These computations determine all the greatest moving load stresses in one arm considered as a simple non-continuous span. In finding the greatest stresses due to the moving load on both arms, the condition of partial continuity requires the use of balanced loads — t.e., loads simultaneously on each arm which, if the trusses were continuous over the center, would produce equal and opposite center bending moments. The most obvious and simple balanced loads are those of equal magnitude placed at symmetrical panel points in each arm, and such balanced loads will be used in the following com- putations. As has already been observed, the use of the locomotive concentrations and the ordinary moment diagram for the greatest stresses in this case, would lead to excessive compli- cation and great labor without any corresponding advantage. Essential accuracy can be attained by computing a system of panel concentrations or weights from a position of the moving load which will give results differing in no sensible degree from those determined by the most refined calcula- tions. The desired position is largely a matter of judgment, but a heavy concentration should be found at the panel point covered by the head of the train. In the present instance the third driving wheel of the front locomotive will be placed over a panel point which will be called panel i. A small concentration will, of course, exist at the panel point 10 14^ SWING BRIDGES. in front of panel i ; this will be called the " advance load.** The desired concentrations will then be as follows: Advance load, 5.70 tons. At panel i, 25.55 " " - 2, 17.75 " " " 3, 25.55 " •' " 4, 19-35 " The advance load is sometimes neglected in finding the web stresses, as the resulting computations are somewhat simplified, and the small error committed is on the side of safety. All reactions in this case must be found by the equations of the preceding Article for perfect continuity, as all loads are to be balanced. If a panel load, /*, rests on the left-hand arm, or span, /j, the reaction R^ will be by Equation (5«) of Article 35 : 7?. = /'(.-»)|'-(' + ")"- — - — r\ ■ (I). ( 4(^+.)-,-^f And by Equation {6d) of Article 35, the reaction R^ will be: R, = P{\-7i^)n . J .^ . . . (2). In this case, c = 137.5 -^ 20.5 = 6.7073. If the panel points be numbered from i to 4 from the end of the arm to the center, panel i being at the foot of Zlo, and panel 4 at the foot of Zi ; and if P be the general value of a panel load, n will have the values 0.2, 0.4, 0.6, and 0.8 in the formulae for Ri and R^, and the expressions for those reactions will be : Panel load Pat panel i, R^ = 0.716102 P; R^ = 0.005443 P. " 2, i?i =0.453178/'; 7?4 = 0.009526/'. " " " " 3, i?i = 0.232204 P; /?4 = 0.010886 P. " " " " 4, 7?i = 0.074153 P; i^4 = 0.008165 P ENDS SIMPLY SUPPORTED. 1 47 Since balanced loads only are to be employed in this case, It will be necessary to determine the reactions R^ and Ri with a panel concentration on one arm balanced by an exactly equal concentration symmetrically placed on the other. The reactions for each such concentration will be those given above, but R^ and Ri will be interchanged with the exchange of one arm for the other. When the two concentrations are simultaneously placed on the two arms — i.e., when they are balanced — the reaction at each end of the bridge will equal that at the other, and will be represented by the sum of R^ and -^4, as given above for a single panel load. If R is that reaction : For panel i, R = Ry + R^ = 0.7215 P. " 2, R = R, + R^ = 0.4627 P. " 3, R= R^ + R, = 0.2431 P. " 4, R = R^ + R^ = 0.0823 P. The preceding numerical work can be thoroughly checked by finding M2 for both arms fully loaded, from Equation (4) of the preceding Article (having previously found J/g), and using it in Equation (5) of the same Article. The result- ing value of Ri = Ri should equal the sum of the preceding values of R, because both results are for both arms fully loaded. The sum of the four values of R is 1.5096 P, while Equation (5) of Article 35 gives R^ = R = 1.5097/*. Hence the check is satisfactory. In this verification all panel loads P are assumed equal, but this does not in any way affect the numerical work which it was desired to verify. As the reactions for each of the panel loads, and the panel loads themselves, are now known, all stresses can readily be found. In all the cases except those of simple hangers, or where a trigonometric multiplier only is needed, the method of moments will be employed, precisely as with the fixed load and with the moving load on one arm only. The greatest stress in EP will evidently exist with the great- est reaction at its foot — i.e., with the moving load over the whole of both arms. Hence the reaction is : 148 SWING BRIDGES. 07215 X 25.55 = 18.435 0.4627 X 17.75 = 8.210 0.2431 X 25.55 = 6.210 0.0823 X 19.35 = 1.595 .-. R = 34.450 tons. Hence: {EP) = — 34.45 x sec EP — — 48.23 tons. The same condition of loading gives the greatest stress in lower chord panels i and 2 ; therefore : (i) = (2) = 34.45 x tan EP^ + 33.83 tons. The hanger T^q will be treated precisely as in the case of moving load on one arm only, but with the balanced concen- trations. Hence : (7;o)= + 25.55 tons. The position of loading for the greatest stress in c-^ is with the advance weight 5.7 tons at the foot of /"lo, as the resulting reaction is then greater in comparison with the load between the end of the arm and foot of c^ than with any other position. If the advance load were very large it would have to be placed at the foot of c^ for the greatest stress in that mem- ber. The reaction then becomes : 0.7215 X 5.7 = 4.112 tons. 0.4627 X 25.55 — J 182 0.2431 X 17.75 ^ 4.315 " 0.0823 X 25.55 = 2.103 " R = 22.350 " Remembering that T^ is now to be neglected : , , 22.35 X 20 - 5.7 X 21 , _ ^ {c^ = — -^^ ■^-^ sec c^—+ 20.83 tons. Since the stress in c^ is much less than was found with the moving load on one arm, as was to be anticipated, it is clear that no other counter stresses need be found. Indeed, it is evident from the general conditions of the two cases that ENDS SIMPLY SUPPORTED. 1 49 the greatest counter web stresses must be found with the moving load on one arm only, as has already been observed. The conditions of loading for the greatest stresses in the main web tension members 7^, To,, and T^ will be found by bringing the moving load on the bridge from the end of the arm toward the center, every panel concentration at the same time being balanced. In the present case, the advance load of 5.7 tons is to be placed one panel in front of the foot of the mem- ber in question, as it is small in comparison with that which follows it. This advance load might be so large as to require it to be placed at the foot of the member whose stress is sought. The choice between these two positions can only be deter- mined by trial. One of these positions of the moving load will give the greatest stress desired, for the reason that the end reaction will then be the least in comparison with the moving load between it and the member considered, thus insuring the greatest possible shear in the latter. It is to be observed that the design of the truss is such that the inclined web members are subject to pure tension, and the vertical posts P^ and P^ to pure compression, necessitating the introduction of the counter c-^. If it is desired to omit the latter member, making a rather more excellent design from a purely engineering point of view, T^, and perhaps other adjacent inclined web members, will be subjected to compression, with possibly some of the vertical posts in tension, and all such members must be counterbraced. The requisite computations for this case will be considered at the end of this Article. In order to find the greatest stress in T^, the advance load of 5.7 tons is to be placed at the foot of P^\ hence the reaction R will be : 0.7215 X 25.55 = 18.435 0.4627 X 5.7 = 2.637 K. = 21.072 tons. ,_. 25.55 X 21 - 21.072 X 20 ,.. .-. (T;) = -^^^ ^^^ '— se< /4 = + 7-5 1 tons. 150 SWING BRIDGES. The advance load must be at the foot of P^ for the greatest stress in /"g, and the resulting reaction will be : 0.7215 X 17.75 = 12.808 0.4627 X 25.55 = 11.820 0.2431 X 5.7 = 1.385 R = 26.013 tons. 25.55 X 22 + 1775 X 21 - 26.013 X 20 _ ^ — + 25.315 tons. For the greatest stress in T^, the advance load must be at the foot of Tx ; whence : 0.7215 X 25.55 ^ 18.435 0.4627 X 17.75 — 8.212 0.2431 X 25.55 = 6.213 0.0823 X 5.7 = 0.470 ^ = 33-330 tons. 25 . 5 5 X 2 3 + 17. 75 X 22 + 25.55 X 21 - 33.33 X 20 sec T.2 = + 48.64 tons. ••• (^^^ 23 In illustration of the effect on the stress in T, of placing the advance load at its foot, rather than at the foot of 7",, the following would be the value of that stress, remembering that the reaction R would be 26.013 tons, as found for T^: 5.7 X 23 + 25.55 X 22 + 17.75 X 21 - 26.013 X 20 (A) - 23 sec T.i= + 31.3 tons, which is seen to be about two-thirds of the greatest stress. The result would have been very different, however, if the 5.7 tons were displaced by a large concentration. ENDS SIMPLY SUPPORTED. 151 As Tx is a simple hanger, it receives only the greatest load at its foot : (Ti) = + 25.57 tons. The greatest reaction at the foot of /\ will exist when the advance load of 5.7 tons is placed at that point, and its value will be : 0.7215 X 19.35 = 13.962 0.4627 X 25.55 = 11.820 0.2431 X 17.75 = 4.315 0.0825 X 25.55 = 2.103 R — 32.200 tons. Since the upper chord d cuts the lower chord panel i at a point 0.8 panel length from the foot of EP, that point of intersection must be taken as the center of moments for /^j ; hence : (/'O = 32.2x0.8+19.35 X 0.2 + 25.5 5 X 1.2 + 17.75 X 2.2 + 25.55 x3'2 4,2 sec P^ = — 56,875 tons. The posts P2 and P^ take their greatest stresses with the web members T^g and T^ respectively ; hence the reactions for the latter members are to be used for the former. There- fore : iD\ 25.55x22 + 17.75x21—26.013x20 o {Pi)= '^-^ —^-^ = — 18.025 tons. Also: fr,\ 25.55 X 21 — 21.072 X 20 (^3) = - -^^ Yi = ~ 5.235 tons. These complete the web stresses, and leave only those for the chords [except (i) and (2) already found] to be deter- mined. As the counter c^ comes into action with the moving load over the whole of the bridge, the two upper chord panels 152 SWING BRIDGES. a and b will sustain the same stress, and panel point 2 at the foot of P2, will be the center of moments for their stress. Now if each panel load be called unity, the reactions Rx ot the loads at panel points i, 2, 3, and 4 will be 0.7215, 0.4627, 0.2431, and 0.0823 respectively ; and if each of these reactions multiplied by its distance from the center of moments be greater than the product of the panel load that produces it multiplied by its lever arm, the effect of placing that load on the span will be the production of a stress of the same kind as that due to the reaction alone. If the moments of these unit reactions and that of the unit load at panel point i be taken about panel point 2, there will result : For panel load I, (0.7215 x 2 = 1-443) — i = -I- 0.4430 " " " 2, 0.4627 X 2 .= + 0.9254 " " " 3, 0.2431 X 2 = + 0.4862 " " " 4, 0.0823 X 2 = + 0.1646 Since all these unit moments are positive, all loads will produce compression in a and b. It is clear, also, that the heaviest load must be placed at the foot of P^\ this will locate the two heaviest concentrations at the feet of /s and Zi, with the head of the train toward P^. The actual moments of the panel loads about the foot of P^ can now readily be found by multiplying those loads by the unit moments given above, and then by multiplying the sum of the results by the panel length, as follows : 19.35 X 0.4430 = 8.572 25.55 X 0.9254= 23.644 17.75 X 0.4862 = 8.630 25.55 X 0.1646= 4.206 Total moment . . . 45.052 x 27.5. Hence the stress in a and b will be : (^) = (^)=-45-Q52x27,5 29.333 Y. sec a =^ — 45.052 X tan T^ a sec a — — 42.31 tons. ENDS SIMPLY SUPPORTED. 153 In order to find the greatest stress in lower panel 3, Tnoments must be taken about the top of Pj. By taking unit moments, as before, about that center: For panel point i, (0.7215 x 3 = 2.1645) — 2 =: + 0.1645 " 2, (0.4627 X 3 = 1. 3881) — I = + 0.3881 " 3, 0.2431 X 3 = + 0.7293 *' " " 4, 0.0823 X 3 = + 0.2469 Since all these results are positive, all loads on the arm will produce tension in lower panel 3, or, what is the same, com- pression in upper panel c. For the greatest stress in panel 3, one of the heaviest concentrations must be placed at the foot of P^_, so that the two heaviest concentrations will be located at the feet of P^ and r,o with the head of the train toward EP. Hence : 25.55 X 0.1645 =. 4.203 17.75 X 0.3881 = 6.889 25.55 X 0.7293 = 18.633 19.35 X 0.2469 = 4.778 Total moment . . . 34.503 x 27.5. (3) = 34-503 X ^^ = 34.503 X tan 7; = + 30.95 tons. Also : ic) = — (3) X sec a — — 30.985 tons. Unit moments about the top of P^ for the stress in lower panels 4 and 5 will result as follows : For panel point i, (0.7215 x 4 = 2,8860) — 3 = — 0.1140 " " " 2, (0.4627 X 4 = 1.8508) — 2 = — 0.1492 " " " 3, (0.2431 X 4 = 0.9724) — I = — 0.0276 " " " 4, 0.0823 X 4 = + 0.3292 These results show that the panel load at the foot of 7i is the only one which produces tension in panels 4 and 5 ; while it is evident that one of the heaviest concentrations should be placed at the foot of 7i. Hence, by placing the 154 SWING BRIDGES. advance load of 5.7 tons at the foot of T^, and the heavy load of 25.55 tons at the foot of Zl : For panel 3, 57 x (— 0.0276) —— 0.1575 " " 4, 25.55 X 0.3292 = + 8.41 1 Total moment . . . 8.2535 ^ 27.5. Hence (4) = (5) = 8.2535 X ^ = 8.2535 X tan /'i = + 7.09 tons. 32 In order to find the greatest compression in panels 4 and 5,. the advance load of 5.7 tons is to be placed at the foot of 7*2 with the head of the train toward /\. Hence: For panel i, '17.75 x (—0.1140) = — 2.024 " 2, 25.55 X {- 0.1492) =- 3.812 " 3, 5.7 X (- 0.0276)= -0.157 Total moment . . . 5-993 x 27.5. Hence: (4) = (5)= — 5-993 x tan P^ = — 5.15 tons. The advance load might be so small that it would be neces- sary to place it at the foot of 7i for the greatest compression in 4 and 5. In order to determine the greatest stresses in panels e and 6, let unit moments be taken about the foot of P-^ : For panel i, (0.7215 x 5 = 3.6075) — 4 = — 0.3925 " 2, (0.4627 X 5 := 2.3135) - 3 = - 0.6865 " 3, (0.2431 X 5 = 1.2 155)- 2 = -0.7845 '' 4, (0.0823 X 5 = 0.41 1 5) - I = - 0.5885 These results show that all loads produce tension in d and e and compression in 6. Hence, by placing the two heaviest concentrations at the feet of T^ and P^ with the head of the train toward P^ : ENDS SIMPLY SUPPORTED. 155 - 19-35 X 0.3925 - 25.55 X 0.6865 - 17-75 X 0.7845 -25.55 X 0.5885 /. (6) =-{e) = ~ 54.096 X ^ = 42 7-595 17.540 13-925 15-036 ^•096 - 35.42 tons. (^= {e) sec ft = +35.41 X 1.064 = + 37.685 tons. {cP) — — {e) tan ft — — 35.41 x 0.364 = — 12.895 tons. These complete all the moving load stresses, and, with those due to the fixed load, they enable all the resultant maxima stresses to be at once written. The following tabu- lation shows the results of all the computations : TABLE I. Fixed Load. Moving Load on One Arm. Both Arms. T, T^ T, T, + 8.325 + 50.02 + 36-815 + 22.57 - 8.14 + 25-57 + 38.16 \ +18.85 / -14-45 ) + 5-96 { -34-oS + 25-57 + 33-21 — 63.23 -59-57 -16.57 - 5-25 + 25-57 + 48 . 64 + 25-315 + 7-51 + 25-55 + 20.830 -48.23 -56.875 — 18.025 - 5-235 -12.895 -42.31 -42-31 -30.985 + 37685 + 35-42 + 33-83 + 33-83 + 30.95 i + 7-09 \- 5-15 i + 7-09 I- 5-15 -35-41 EP Px Pi P3 cP + 7.00 -23-73 -29-595 -19.09 -36.785 + 4-92 + 20.37 + 44.98 + 98.93 + 92.98 - 4-91 -20.35 -44-93 -77-52 -77-52 -92.98 a b c d -62.675 — 62.675 — 60.61 I 2 3 4 5 6 + 44-345 +44-345 + 60.535 + 38-79 + 38-79 156 SWING BRIDGES. The stresses in Table I show that the tension member T^m must be counterbraced so as to sustain the tension of 19.48 tons due to the moving load on one arm and the fixed load compression of 8.14 tons. T^, is shown under a compression of 14.45 tons, due to moving load on one arm, but the fixed load tension of 36.815 tons is more than two and one-half times as great ; hence no counterbracing would be necessary in any case. Although 34.08 tons compression is shown opposite 7*4, it belongs to the case in which c^ does not exist, and which will be treated at the end of this Article ; it there- fore needs no further consideration here. The entire lower chord must be counterbraced, as must also the upper chord panels a, b^ and c. These latter members are always built with sections for compression. Table II shows the final resultant stresses desired. TABLE II. Member. Tension. Compression. Member. Tension. Compression. + 33-895 + 98 66 + 62.13 + 30.08 + 17-43 + 33-21 + 7.00 a b !/ e \ I i 2 3 4 5 6 + 4.92 + 20.37 + 44-98 + 136.615 + 128.39 + 39-435 + 23.995 + 15.605 - 57-755 - 42.305 - 15-63 Ti Ti /"lo — 8.14 EP - 4-91 - 20.35 - 44-93 - 82.67 - 82.67 -128.39 - 56.23 - 83.30 - 47-62 - 24.34 - 49-68 Pi Pi cP It is noticeable in Table I that P^ receives greater stress with moving load on one arm only, than on both arms. This is in consequence of the inclination of the upper chord panel d, by which it takes a considerable shear that would other- wise exist in /'j, and which would correspondingly augment its stress. The same Table also shows that the greatest stresses exist in the various portions of the arm under the conditions of loading indicated in a former portion of this Article: The preceding constitutes all that pertains to the computa- ENDS SIMPLY SUPPORTED. 157 tion of stresses in the trusses, but it is still necessary to com- pute the reactions R^ and R^, at the extremity of the arm and over the center pier respectively. The supports at the latching and locking points must be designed to resist the reaction ^,, and the drum must be designed to support the reaction R.,. R^ will manifestly have its greatest value with the moving load on one arm only. The determination by trial of the maximum reaction at the foot of EP, including the proportionate part of those wheel concentrations in the panel adjacent, is somewhat laborious, and as each locomo- tive and tender weight divided by its total length averages 3,152 pounds, or 1.576 tons, per lineal foot, it will be essen- tially accurate and much simpler to assume that the total maximum reaction to be supported at the locking and latch- ing point at the foot of EP'\s\ ^^ ^ ^37-5 X 0.775 ^ 6 ^ 5^ ,8 ,^„3 The concentration of 6 tons is added to provide for the driving-wheel concentration that must be supported at the instant of entering on or leaving the arm while the whole of the latter is otherwise loaded. The reaction for both arms covered with moving load is best determined in precisely the same manner. Each panel load will be 27.5 x 0.775 = 21.312 tons. Now the reaction i?, for four such panel loads will be : (.7215 + .4627 + .2431 + .0823) X 21.312 = 1.5096 X 21.312. .'. R^ = 32.173 tons. Hence : 7?,' — 4 X 21.312 — 32.173 + — ~ X 21.312 -\- 6 — 77.6^^ tons. The panel load added is for the half panel adjacent to the center and the central panel, while the 6 tons is for the driv- ing-wheel concentration, as just explained in connection with R.. To R' must be added the total fixed load carried 158 SWING £ JUDGES. to the center. From the data given at the beginning of this Article, that total fixed load will be: 24 X 0.85 , 4 X 11.7 + 5 + -^ ^ = 62.0 tons. Hence the total reaction R^ to be supported at the drum will be : R^ = 77 '^74 + 62 = 139.674 tons. As the ends of the span are simply supported, no fixed load reaction is to be added to the value of R^ as determined by the moving load only. The omission of counters. It has already been remarked that the best design, from a purely engineering point of view, is secured by omitting the counters and counterbracing the main web members. In the present instance the counter <:- would be omitted and the main members 7^ and P^ counterbraced. The member T^^ will also receive much greater tension than in the preceding case. In considering these counter stresses, the arm of the bridge is to be treated as a simple non-continuous span, as was done in the earlier part of the Article, with the moving load passing from the center toward the end. Equation (4) of Article 20 shows that the third driver of the front locomo- tive should be at the foot of 7"]o in order to give that mem- ber its greatest tension. The first parenthesis of the second member of that equation disappears in this case because in is zero. The resulting reaction at the foot of EP is 50.86 tons, of which 5.70 tons is due to the wheel concentrations between the foot of EP and the foot of Zio- Hence, by moments about the chord intersection : ^ (50.86 — 5.70) 20 Tio = — ±-L-L — = + 43.01 tons. The moving load compression in T^ has already been found to be 34.08 tons, in deciding upon the necessity for c^ in the ENDS SIMPLY SUPPORTED. 159 preceding case. The same position of the moving load (the third driver at the foot of P.^ will give the greatest tension in /g, and using the data already employed for Ti, there will result : „ 32.1 X 20 — 5.72 X 21 P, = ^ ^^ = + 23.73 tons. Again, it was found that the greatest moving load compres- sion in Tg would be 14.45 tons, while the fixed load tension is 36.815 tons ; hence T^ can never suffer compression. The re- sults for 7^3 show also that the fixed load compression in P^ will always largely overbalance any possible moving load tension. The only counterbracing needed, therefore, will be that for the members Ti and P3. These computations, in combina- tion with the results given in Table I, show that Zio, Ti, and Pg will sustain the following resultant stresses : 7^10 ; + 34.87 tons. — 8.14 tons. Ti ; + 30.08 " — 1 1.5 1 " P, ; + 4.64 " - 24.34 " All the other stresses remain unchanged. Art. 37. — Ends Simply Supported — Two Points of Support at Center — Complete Continuity — Example. The case of complete continuity to be considered in this Article involves the existence of web members in the panel over the drum (?>., the span 4 of Article 35), designed by actual computation to take the shear which may pass the center when the structure carries unbalanced loads. It has been shown, in Article 35, that when the condition of continuity is fulfilled, which is the only condition there contemplated, either of the reactions R^ or .^3 may be nega- tive, i.e., downward, while both the reactions i?, and P^ are always positive, or upward. It will be found that provision of the nature of heavy anchorage must be made in order to meet the requirements of P2 and ^3 with some conditions of loading of draw spans. This indicates that the web members in the center panel will be very heavy, and such will be found to be the case. l6o SWING BRIDGES. The moving load stresses in all cases will be determined by means of the formulae of Article 35. The truss to be con- sidered is the same as treated in the preceding Article, and is shown by Figure i of Plate X. The data, reproduced from Article 36, are as follows : Length between centers of end pins . . 295.5 feet. Panel length in arms 27,5 " Length of center panel or span ... 20 " Center depth at CP — 42 feet. Dep*-h at Ti = 32 feet. Depth at Pg = 30| " " " P^ = 2g\ " " " Zio — 28 " Truss centers 16 ft. apart. Weight of rails, ties, and guards, with bolts and connec- tions, 0.2 ton per lineal foot. Total lower panel fixed load .... 8.32; tons. " upper " " " . . . . 3.375 " Total 11.700 " Lower panel load at end of arm ... 5 tons. Upper " " " top " CP .... 2.94 " The trigonometric quantities required will not be re- peated. As the ends of the span are simply supported, the fixed load stresses are those existing when the draw is open, and, hence, are identical with those found for the fixed load in Table Ij =^1 -I -' ^1 S' 5' fil s Y^ (mm) r^ht^(^ H i ' 'i^ ^r- i -r - i — -, — -' —14^9— -54:3:: Fig. I. I of the preceding Article. As they are there arranged in convenient shape for combination with the moving load .stresses about to be found, they will not be reproduced here. ENDS SIMPLY SUPPORTED. 1 6 1. The moving load is identical with that used in Article 36, and is shown by the diagram on preceding page. For the same reasons given in that Article, the system of panel concentrations which was used there will be employed here, instead of the wheel weights shown in the diagram. These panel concentrations are as follows : Advance load , 5.7 tons. At panel i . 25.55 tons. At panel 2 . . 1775 " " " 3 • 25.55 " " " 4 • . 19-35 " On account of the complete continuity at the center, there will be no condition of a simple, non-continuous span for the moving load on one arm only, but all moving load stresses will be found by the use of the preceding panel concentra- tions in connection with the formulae of Article 35. The reactions R-^ and R^ will be precisely the same as those found on page 146 of Article 36. They are: Panel load Pat panel i, R^ — o.y \6\02 P ; R^ =. 0.005443 P. " 2, i?i = 0.453178 P; 7^4 = 0.009526/1 " 3, R^ = 0.232204 P\ Ri = 0.010886 P. " 4, Ri = 0.074153 P; Ri = 0.008165 P As unbalanced loads will be considered in this case, these reactions will frequently be used separately in the succeeding computations; but in those special circumstances which cause the loads to be balanced, they will be united by addition, as was done in the preceding Article. As all loads produce positive or upward reactions at each end of the bridge, the post BPwiW receive its greatest stress when the moving load completely covers both arms, and in such a manner that the reaction at its foot will be the great- est possible, as shown by the values of the panel reactions given above. The reaction will therefore be : 25.55 X 0.7161 = 18.297 tons. 17.75 X 0.4532 == 8.044 " 25.55 X 0.2322 = 5.932 " 19.35 X 0.0741 = 1.434 " Left arm total = 33.707 " II -l62 WING BRIDGES. 19.35 X 0.0054 = 0.104 ton. 25.55 X 0.0095 = 0.243 " 17.75 X 0.0109 = 0.194 " 25.55 X 0.0082 = 0.209 " Right arm total = 0.750 " v. {EP) = - (33707 + 0.75) sec EP= - 34.457 X 1.4 = — 48.24 tons. .*. (i) = (2) = 34-457 X tanEP= + 33.835 tons. As Ti cannot sustain any compression, the hanger T^q will be stressed by the entire load at its foot, unless that load added to the coexistent vertical shear in the upper chord panel a exceeds the reaction at the foot of EP. Since the latter has been shown to be 34.457 tons, while the maximum load at the foot of T^q is only 25.55 tons, that excess cannot exist, but ci will be required to carry over shear to make up the reaction found. Hence: (7,o) -= + 25.55 tons. Since the function of c^ is to carry over shear to make up the reaction at the foot of EP, its stress will be the greatest when that reaction exceeds by the greatest amount possible the moving load between it and c^, added to the coexistent shear in upper panel a. As the advance load is small in com- parison to that which follows it, this condition will exist when the moving load covers both arms, with the advance load at the foot of Zio- The reaction at the foot of EP will then be : 5.7 X 0.7161 = 4.082 tons. 25.55 X 0.4532 = 11.579 17.75 X 0.2322 — 4.122 25.55 X 0.0741 = 1.893 Total from left arm .... 21.676 Right arm reaction as above. 0.75 Total reaction .... 22.426 , , 22.426 X 20 — 5.7 X 21 .'. (^i) — — -"^ ■ sec f 1 = + 20 925 tons. ENDS SIMPLY SUPPORTED. 1 63 The advance load would have to be placed at the foot of Pg if it were sufficiently large. At the end of the Article the case will be considered in which the counter c-^ is omitted and the members 7*4 and P^ counterbraced. It will then be shown that 7^ and Pj can never sustain any other kinds of stresses than those induced by the fixed load. Hence, no other counter than c, will be required in the present instance, and the main web stresses will next be found. Since the function of the main web members is to carry load or shear over to the center, one condition to be fulfilled in order that they may receive their greatest stresses is, in gen- eral, to make the reaction at the end of the arm in which they are located as small as possible. As the main web stresses in the left arm are now sought, therefore, no moving load must be placed on the right arm ; because all such load increases the upward reaction at the foot of EP. This condition holds in general because the chords of draw spans are usually either parallel or so inclined to each other that their intersections lie witJioiit the span. If, however, their intersections lie within the span, the reactions at the free end of the span will have moments of the same sign as those loads between the moment origin and the web members in question. Hence in those cases, the reactions (with moving loads omitted between moment origin and the free end) should be as large as possible, and the moving load should cover the whole of the other arm. The intersection of the panels d and 5 is 0.8 panel length from the foot oi EP, and 0.2 panel length from the foot of Tiq; hence the entire bridge is to be covered with moving load for the greatest stress in P^, precisely as was done in the pre- ceding Article for the same member. Therefore, from that Article : (/\) = - 56.875 tons. The member Ji is again a simple hanger, and its greatest stress is the greatest reaction at its foot. Hence; 7i =: + 25.57 tons. 1 64 SIVING BRIDGES. For the remaining web members the moment origin is twenty panel lengths from the end of the span, and the mo- ments of the reactions will have signs opposite to those of the loads. No moving load must, therefore, be placed on the right arm, in order that the reactions may be as small as possible. The advance load will be placed at the foot of T^ for the greatest stress in T^, thus making the reaction at the foot oiEP: For panel i, 25.55 x 0.7161 = 18.295 tons. " " 2, 17.75 ^ 0.4532 = 8.04 " " 3. 25.55 ^ 0.2322 = 5.933 " " " 4, 5.7 X 0.0741 = 0.422 " Total reaction = 32.690 " ^ 25.55 X 21 + 17.75 X 22 + 25.55 X 23 - 32.69 X 20 ^^^ ^ 23 ^ = + 49.37 tons. The greatest stress in T^ requires the advance load at the foot of T2, giving the reaction : For panel i, 17.75 ^ 0.7161 = 12.71 tons. " 2, 25.55 X 0.4532 = 11.58 " " " 3, 5.7 X 0.2322 = 1.325 " Total reaction = 25.615 " 17.75 X 21 + 25.55 X 22 - 25.615 X 20 • • W 3; — — — — ^^ ^^^ ^ 3 = + 2579» Since P^ takes its greatest stress with T^ : 17.75 X 21 + 25.55 X 22 - 25.615 X 20 ^^^)- 23 = — 18.37 tons.. ENDS SIMPLY SUPPORTED. 1 65 The greatest stress in 7!, requires the advance load to be placed at the foot of T-^. Hence the reaction will be; For panel i, 25.55 ^ 0.7 161 = 18.295 tons. " 2, 5.7 X 0.4532 = 2.585 " Total reaction = 20.880 " /-T^N 25.55 X 21 — 2O.8S X 20 ™ , „ ^fL, ^ .'. (Ti) = -^-^^ sec Ti= + 7.765 tons. Since P^ takes it greatest stress with the same position of moving load : /ox 25.55 X 21 — 20.88 X 20 (Ps) = - -^^ = - 5.405 tons. In order to illustrate the effect of moving the load so as to bring the advance concentration of 5.7 tons at the foot of the diagonal whose stress is sought, let the stress in T^ be found in that manner. The reaction as found for Z^ will be 20.88 tons. Hence : (7.)^ = 25.55x21+57^x22-20.88x20^^^ ^^ = + 14.91 tons. This result is only a little more than one-half the greatest stress in T^ as found above. If the advance load had been large, the result would have been different. The greatest stresses in the members cc and LS in the center panel are next to be found. These members carry the shear past the center for all unbalanced loads on either arm ; hence the stresses in them will occur when there exists the greatest amount of unbalanced load — i.e., when the moving load covers one arm while the other arm is free of it. This also follows clearly from the fact, demonstrated in Article 35, that all load in the arm, or span, /, causes ^3 to be negative and K2 positive ; and, hence, since the stresses in a" and LS result from the downward pull of R^, any load in the arm, or 1 66 SWING BRIDGES. span, 4 would reduce the negative R^ by balancing some load in /i, and thus correspondingly reduce the stresses in question. Those considei^tions are concomitant with complete conti- nuity over the center. On page i6i are found values of R\ and R^ for a unit panel load at each panel point of either arm, and by using these in Equation {ya) of Article 35 the following values of R^ result, remembering that P is unity, that c = 6.707, and that n has the values 0.2, 0.4, 0.6, and 0.8 respectively: For panel i, R^— — 0.6044 " 2, 7^3= - 1.0579 " 3, ^3 = - 1-2095 " 4, Rs= - 0.9076 The shear in the center panel will be R^ + R^ (a numerical difference) ; hence, by using actual panel loads : (— 0.6044 4- 0.0054) X 19.35 = — 11.59 tons. (— 1.0579 + 0.0095) X 25.55 = — 26.785 " (— 1.2095 + 0.0109) X 17.75 = ~ 21.275 " (— 0.9076 + 0.0082) X 25.55 = — 22.98 " .-. R^ + R,= - 82.630 '* Since the two rods cc pull against opposite ends of the strut LS, the preceding shear of 82.63 tons will be equally divided between them. Hence: (cc) = 41.315 X sec cc = + 57-715 tons. The stress in LS is of course the horizontal component of that in cc ; therefore : (LS) = — 41.315 X fan cc — — 40.33 tons. The preceding calculations show, therefore, that the total downward reaction to be supplied, or upward pull to be re- sisted, at the foot of each cP is 82.63 + 0.75 =83.38 tons; and there must be sufficient weight of drum, cross beams, etc., as well as strength of connections, to fulfill the require- ments of that condition. ENDS SIMPLY SUPPORTED. 1 6/ Since the center panel, or span, is divided into two stories, the upper half of cP will sustain a stress different from that in the lower half in consequence of the action of the two parallel tension braces cc. Again, unbalanced moving loads will produce stresses in one of the cP posts that are not only different in amount from those in the other, but, also, different in kind. It has just been shown that the cc tension braces inclined in the same direction as /\ receive their greatest stresses when the left arm is entirely covered with moving load, the right arm being free from it. It necessarily follows that the left post cP receives at the same time its greatest compres- sion. The same condition of loading, it has also been shown, induces the greatest possible downward reaction, or pull, at the foot of the right cP. In fact, this pull produces a heavy tension in the lower story of that cP at the foot of which the reaction R^ exists ;*but in the upper story of the same member a small tension only will be found, due to the small moving load compression in the right-hand member d induced by the reaction R^. It will be further apparent that the moving load on one arm only will cause the greatest compression in that cP nearest the loaded arm, if it be observed that the member d takes its greatest tension with the same condition of load- ing ; because if the other arm be loaded, the upward reaction at the end of the arm will be increased by the amount of i?^, which will correspondingly relieve the tension in d. Let the greatest compression in the lower story of cP be found with the same condition of moving load used for cc, and in doing this let the truss be supposed to be divided through d, cP, and 6 (the cc which is divided does not act and is there- fore neglected). Moments will be taken about the intersec- tion of d and 6 in the left arm, by the reactions R^ and R^. Hence, remembering that 20.5 ~ 27.5 = 0.7455 : Lower half {cP) = °-75 >< 9-9455 - 83.38 x 4-9455 ^ ' 4.2 = — 96.4 tons. l68 SWING BRIDGES. This is, of course, equal to the vertical component of the tensile stress in d added to the vertical components of the two stresses {cc). It will presently be shown that the tension in d is 37.875 tons; hence: Lower half {cP^ — 37-875 x 0.364 + 2 x 41.315 = — 96.4 tons. The compression in the upper story of cP will be simply that in the lower story diminished by the vertical component of (cc). Hence : Upper half {cP) — — (96.4 — 41.315) = — 55.085 tons. This can, again, be checked by moments by supposing the truss divided through d, upper half cP, LS, cc, and 6, and by including the moments of {LS) and {cc) now known ; but it is quite unnecessary. Continuing the same position of loading, the tension in the lower half of the other ^/* will now be found. Let the truss now be supposed to be divided through the right arm d, lower story of right-hand cP, cc, and 6; then let mo- ments be taken about the intersection of 6 and d in the right arm. Remembering that ^4 = 0.75 ton, and R^ = 83.38 tons, while the lever arm of {cc) = 57-715 tons is 4.2 panels divided by sec for cc : T u If ^ • / D\ 0-75 X 0.8 + 83.38 X 4.2 Lower half tension (cP) = —^-^ ^^-^ -^— 1^7.715 X 4.2 _ 3/ / J ±_ __ ^ -2.21 tons. 4.2 X sec cc This result can be checked by adding the vertical compo- nent of the upper {cc) to the vertical component of the com- pression now existing in d (in consequence of the upward reaction ^^4), whose horizontal component will presently be shown to be 2.455. Hence, this checking operation gives : 41.315 + 2 455 X tan f5 — 42.21 tons. ENDS SIMPLY SUPPORTED. 1 69 The upper half of r/* receives only the vertical component t)f the compression in d ; hence : Upper half tension {cP') = 2.455 ^ ^^'^ /^ = + 0.895 tons. This result can also be checked by moments in the same general manner indicated for the upper half of the other cP, but it is not necessary. The computations for the web stresses are thus completed, leaving those for the chord stresses yet to be shown. Unit panel moments, as described on page 152 of Article 36, will first be taken about the foot of P.^ for all the panel points of the two arms, those on the right arm being represented by Rjs! i'ld the following will be the results : For right arm, R^ x 2 = + zR^ " panel i, 0.7161 x 2 — i = + 0.4322 " " 2, 0.4532 X 2 =4- 0.9064 " " 3, 0.2322 X 2 = + 0.4644 " " 4, 0.0741 X 2 = + 0.1482 All ihese results are positive, which shows that a// loads on both arms produce compression in d; but with the entire left arm loaded, the counter Ci comes into action and causes th'^ stress in a to be the same as that in d. Therefore, for the greatest compression in a and d, moving load must be so placed on the right arm as to give ^4 its greatest value of 0.75 ton (as already shown), while the left arm is entirely Tovered, with the heavy concentrations at panels 2 and 4. The desired moment will then be: 0.75 X 2 =1.5 19.35 X 0.4322 = 8.363 25.55 X 0.90641^ 23.159 17.75 X 0.4644= 8.243 25.55 X 0.1482 = 3.786 Total . . . 45.051 . . ,,. 45-051 X 27.5 .,\ (a) 1= (^) = — tt£-^ <-^ sec a = — 45.05 1 V / V / 29.333 ^^ 5 X ^an T^ X sec a ^ — 42.31 tons. 170 SWING BRIDGES. Again, for the chords c and 3, unit moments about the foot of P^ give : For R,, ^4x3 + 3^4 " panel i, 07161 x 3 - 2 = + 0.1483 " " 2, 0.4532 + 3 - I = + 0.3596 *' " 3, 0.2322 X 3 = + 0.6966 " " 4, 0.0741 X 3 = + 0.2223 As all these results are positive, it follows that all panel loads produce compression in c and tension in 3. It is also evident that the heavy concentrations must be placed at panels i and 3. Introducing the panel concentrations: 0.75 X 3 = 2.25 25.55 X 0.1483 = 3.789 1775 X 0.3596 = 6.383 25.55 X 0.6966 = 17.798 19.35 X 0.2223 = 4.301 Total . . . 34.521 ••• (3) = ^"^'^^0666^^^ = 34.521 tail T,= + 30.965 tons. Also: (<:)= — (3) sec a = — 31.005 tons. For the stresses in 4 and 5 (equal to each other), the unit moments are to be taken about the top of /\ ; they are as^ follows : For Ri, R4 X 4 = + 4R4 "■ panel i, 0.7161 x 4 — 3 = — 0.1356 " " 2, 0.4532 X 4 — 2 =: — 0.1872 " " 3, 0.2322 X 4 — I = — 0.0712 " " 4, 0.0741 X 4 = + 0.2964 These results show that all loads on the right arm with that at panel 4, only, produce tension in 4 and 5. Hence, the moving load on the right arm is to remain as before, while the heavy concentration is to be placed at the foot of ENDS SIMPLY SUPPORTED. 17 1 Ti, with the advance load of 5.7 tons at the foot of P^. By the introduction of these moving loads : For R^, 0.75 X 4 = + 3 " panel 4, 25.55x0.2964 =+ 7.573 + 10.573 3» — 5'7 X 0.0712 = — ,406 Total . . . +10.167 (4) = (5) = 10.167 X tan T.2= -\- 8.735 tons. Since the loads at panel points i, 2, and 3 produce com- pression in 4 and 5, the greatest compression possible must be found. It will result from placing the heavy concentra- tion at panel point 2, with the advance load of 5.7 tons at panel point 3. The actual concentrations then give : For panel point i, — 17.75 x 0.1356 = — 2.407 " " " 2, — 25.55 X 0.1872 =— 4.783 " •' " 3, — 5.7 X 0.0712 = — .406 Total ... = — 7.596 .•. — (4) = — (5) = — 7.596 X tan T2= — 6.525 tons. It is thus seen that the panels 4 and 5 are the only por- tions of the lower chord that can ever be subjected to com- pression by the moving load. In order to determine the stress in d, the unit moments must be taken about the foot of cP, and they are : For right arm, ^4x5 = + 5^4 " panel point i, 0.7161 x 5 — 4 = — 0.4195 '* " " 2, 0.4532 X 5 - 3 := - 0.7340 " " " 3, 0.2322 X 5 — 2 = — 0.8390 " " " 4, 0.0741 X 5 — I = — 0.6295 Whence, all loads on the right arm give compression to d, and they must be removed when the greatest tension in d 1/2 SWING BRIDGES. is sought. The greatest moving load compression in d will therefore be : .{d) = — 0.75 X 5 X -^-^ X sec ft = — 2.455 X ^^^ ft 42 = — 2.61 tons; and since under this loading (on the right arm only) the rods cc sloping upward to the left do not act, the compression in e will be the horizontal component in d. Hence: {e) = — 2.455 tons. If all the load on the right arm is removed, and all that on the left arm retained, the preceding show that d will be sub- jected to its greatest tension. By placing the heavy con- centrations at panel points 2 and 4, and introducing all the panel concentrations on the left arm in the unit moments, there will result : For panel point i, 19.35 x 0.4195 = 8.1 18 " " " 2, 25.55 X 0.7340 = 18.753 " " " 3, 17.75 X 0.8390= 14.892 " " " 4, 25.55 X 0.6295 = 16.084 Total . . . 57.847 .-. {d) = 57.847 X ^^ -A sec ft = 37.875 X sec ft 42 = +40.3 tons. Under this condition of loading, the rods cc sloping up- ward toward the right do not come into action ; therefore the compression in 6 must equal the horizontal component of the tension in d\ or: (6) = — 37.875 tons. It has been shown that moving load on one arm, only, pro- duces compression in e ; hence, to obtain the greatest tension ENDS SIMPLY SUPPORTED. ^71 in that member, the whole of both arms must be loaded with balanced moving loads, which is precisely the condition used in Article 36, on page 155, where there was found: {e) — + 35.42 tons. This completes the computations for all the moving load stresses, and Table I shows them grouped so that they may be combined with those due to the fixed load as found in Article 36. TABLE I. Member. Stress. Member. Stress. T, + 25. 57 tons. c — 31.005 tons. T, + 49 37 d \ + 40 3 T, + 25 79 " "(- 2 61 T, + 7 765 " ' \ +35 42 " r.o + 25 55 " "/- 2 455 " Ci + 20 925 " I + 33 S35 " EP -48 24 " 2 + 33 835 •' P^ -56 875 '• 3 + 30 965 " P^ - 18 37 " S + 8 735 " P. - 5 405 " 4 '(- 6 525 " Upper cP ( + / - 55 895 " 085 •' 5 \ + 8 /- 6 735 " 525 •' Lower cP i + 42 21 6 - 37 875 '• /-96 40 " cc + 57 715 ;] a - 42 31 LS - 40 33 b — 42 31 By comparison of these results with those of Table I on page 155, it is observed that the assumption of complete continuity produces web stresses either slightly greater in every instance (except those due to the moving load on one arm only) than the hypothesis of partial continuity, or identical with it. The chord stresses are also usually slightly greater, or equal, ex- cepting, also, those due to moving load on one arm only in the case of partial continuity, which are materially greater. Toward and at the ends of the span, therefore, the assump- tion of a simple non-continous span gives much larger stresses, both in chords and web, than are found under the supposition of continuity, as would be anticipated. With these exceptions, however, the differences are unimportant, and either method 174 SWING BRIDGES. could be used with indifference, and that selected which would result in the least labor of computation. An enormous difference is seen, however, in the magnitude of the stresses both for the center posts and central diago- nals, and in the labor of their computation, cc and LS suffer no stress under partial continuity, but are subjected to heavy stresses if the continuity is complete. The posts cP also suf- fer very heavy compression in the latter case, and a compar- itively light compression in the former, due only to the fixed load. The lower stories of the same members are also sub- jected to heavy tension when the continuity is complete. In the present instance this tension, added to the vertical com- ponent of the stress in the lower cc, is 42.21 + 41.315 — 83-525 tons; and since the fixed load compression in <:/^ is, by Table I of Article 36, 36.785 tons, the total maximum upward pull to be resisted at the foot of cP'xs: 83.525 - 36.785 = 4^-74 tons. Now the lower chord fixed load at the foot of cP is 24 X 0.3025 = 7.26 tons, and the weight of the drum, etc., which may be assumed to be concentrated at the same point, is not more than 9 or 10 tons ; or, say, 9.74 tons. Hence the amount of anchorage which would have to be provided under the assumption of complete continuity is : 46.74 — (7.26 + 9.74) = 29.74 tons ; and unless this were supplied, each foot of r/' would, in turn, rise and fall with varying conditions of moving load, so that not only very destructive hammering would take place, but, also, the condition of those spans continuous over two sup- ports would be displaced by that of two unequal spans con- tinuous over one support, thus vitiating the computations underh'ing the design of the trusses. The realization of the conditions requisite for the case of complete continuity, therefore, involves considerably in- creased difficulty and expense in connection with the design of the center portion. ENDS SIMPLY SUPPORTED. 175 Inasmuch as the main truss members have been found to sustain essentially the same stresses in either case, except where those under partial continuity largely exceed the others, and inasmuch as all draw-bridge formulae drawn from the theory of the continuous beam involve the uniformity of the moment of inertia of all normal sections, thereby incur- ring a very considerable error in at least some of the resulting computations, it is much more rational and in harmony with better judgment to use the methods of partial continuity in all cases. These and similar considerations have led engi- neers to almost universally adopt the partial-continuity as- sumption in the construction of swing bridges. It is to be observed that the unbalanced uplift at the foot of cP and the stresses in cc and LS will increase rapidly as the length of the center panel or span decreases ; and, hence, that it is advisable to make that center panel as long as possible. The method of computation will be in no way changed if the center panel is designed with one storj^ only, instead of two. By combining the fixed load stresses in Table I of Article 36 with those due to the moving load given in Table I of this Article, Table II, giving the resultant stresses in all the mem- bers, at once results. TABLE II Member. Tension. Compression. Member. Tension. Compression. Ti Tons + 33-895 + 99-39 -\- 62 605 + 30.335 + 17-41 + 20.925 4- 7.0 Tons b c d e I 2 3 4 5 6 cc LS Tons •f 20.37 + 44-98 + 13Q.23 -f- 128.40 + 28.925 + 13-485 Tons - 21.94 7'o Ti Ti Tio — 8.14 - 4.91 - 20.35 - 44-93 - 84.045 - 84 045 - 130.855 EP - 41.24 - 80.605 - 47-965 - 24.495 - 91-87 - 133-85 - 37-39 Pi Pi Upper cP Lower cP + 5-425 + 4-92 + 57-715 - 40-33 a 1/6 SWING BRIDGES. These resultant stresses are not very different from those in Table II of page 156, except in those members already indi- cated. In this case, upper chord c and lower chord 3 do not need counterbracing, while they required such treatment in the case of partial continuity. In actual practice, however, the entire lower chord, and the upper chord panels a, b, and c, would all be designed with compression cross sections — i.e., they would be counterbraced. The lower story of ^P must also be counterbraced, and ordinary construction would make it so even if it were not required. The existence of the central diagonals cc designed to trans- fer shear, results in some ambiguity in the upper chord stresses at the center. As the moving load passes on the bridge, one pair of those diagonals receive their greatest stresses, and then they may either be supposed to be relieved with the further progress of the train, or they may be supposed to hold essentially their greatest stresses while the other pair gradu- ally take the same condition, as the moving load covers the entire structure. Either supposition fulfills the requisites for equilibrium, but the former will give the greatest possible stress in e and was used in the preceding computations, be- cause the stress in the same upper chord member will be decreased under the' latter supposition by an amount corre- sponding to the horizontal components of the stresses in the diagonals cc. This condition of ambiguity cannot be avoided under the assumption of complete continuity, but disappears if the continuity is assumed to be partial only. The proper method, therefore, is to compute the greatest possible stress in each member, and use it in the design, and this has been done in the present case. The reaction R^ and R2 at the ends and adjacent to the center, respectively, remain to be written, but they can be taken directly from results already found in the preceding Article. For this purpose, and for the reasons fully explained in Article 36, the moving load will be taken as uniform and at 0.775 to'"" P^i" lineal foot for each truss. The greatest end reaction will exist with the moving load over the entire struct- ure, and it has been shown on page 1 57 that its amount will be : ENDS SIMPLY SUPPORTED. 177 21 ■^12 ^1 = 32.173 + — +6= 48.829 tons. The half panel load is that adjacent to the end, whicK does not affect^ the trusses but forms a part of the reaction to be supported by the truss ends, while the 6 tons is the wheel concentration. Since the moving load on each arm produces a negative or downward reaction at the opposite side of the drum, the greatest reaction R2 will be produced with the moving load on the adjacent arm only. The reac- tion R^ due to the four panel loads at the panel points 1,2, 3, and 4 will be : i?,' = (0.7 161 4- 0.4532 + 0.2322 + 0.0741) = 31.448 tons. Hence: 7^2' = 4 X 21.312 — 31.448 ■] — X 21.312 + 6 = 78.399 tons. The third term in the second member is for the half panel adjacent to the center and the central panel, while the 6 tons is for the driving-wheel concentration, as already ex- plained. As shown on page 158, the total fixed load at the foot of Pi is 62 tons; hence the total reaction desired is: 7?2 = 78.399 + 62 = 140.399 tons. Tkc omission of counters. If all counters are omitted, it will usually be necessary to counterbrace some of the main web members. In the present case, T^ will be most in need of such treatment. The position of moving load required to give its maximum com- pression to 7^4 is the same as that used in finding the great- est tension in c^ on page 162, and the corresponding reactioa at the foot of EP was there found to be 22.426 tons. Hence : ,^. 22.426 X 20— 5.7 X 21 ™ — (7^) = ^-^ —sec 7^ = — 21.47 tons. 178 SWING BRIDGES. But it has been shown that the fixed load tension in T^ is 22.57 tons. No counterbracing, therefore, according to these results, is required. But the tension excess is so small that the member should be made with a compression section, otherwise the moving load shock might overcome the excess and cause the member to buckle. A rule of good practice is to counterbrace if at least one-fourth of the computed mov- ing load stress added to itself overcomes that due to the fixed load, and perhaps a better rule is to add one-third. With the same position of loading there will result: ,„. 22.426 X 20 — 5.7 X 21 (P3) = -^ ^^ ^ ' == + 14.95 tons. If one-fourth of 14.95 tons be added to itself, the result will be less than 19.09 tons, the fixed load compression in P^\ hence no reversion of stress can take place, although the character of the pin connection would enable very consider- able tension to be resisted. For the greatest compression in Z3, and tension in /*2» the advance load of 5.7 tons must be placed at the foot of /g. The reaction at the foot of EP will then be : 5.7 X 0.4532 = 2.583 25.55 X 0.2322 = 5.932 17.75 X 0.0741 ~ 1.316 Total from left arm . . . 9.831 tons. Right arm reaction . . . 0.75 " Total . . . 10.581 '\ Hence : ,„., 10.581 X 20 — 5.7 X 22 ^ ^ ^^^ ^ — (T;) = 2 ^-J. sec 2^3 = — 5.265 tons. 22 And ,„v 10.581 X 20 — 5.7 X 22 (^2) = — ^ ^, ^^ = + 3.75 tons. TABLES AND DIAGRAMS. 179 Both of these results are so small in comparison with the ■opposite fixed load stresses that no counterbracing is re- quired. These computations show that even if the counter c^ is omitted, no counterbracing will be required, except in the case of 7"^, as explained. Article 37a. — Tables and Diagrams. — Turntables and Engines. The labor of stress computations by the methods given in the three preceding Articles can be reduced to a very small amount by the aid of the tables and diagrams which follow. They are devised for the purpose of showing the reactions at TABLE I. either extremity of a three-span, rim-bearing drawbridge for a square crossing and for arms of equal length. The ratios, €, of either arm to the center span or panel, as shown i8o SWING BRIDGES. immediately preceding Equation (3) of Article 35, are given- in Table I, with the greatest, least, and mean values, taken from fifteen drawbridges as they have actually been designed. They cover, therefore, a range that will include nearly all practical cases of pin-connected structures. The columns of the table show the reactions R^ and R^ for a unit panel load placed successively at distances from the free end of the arm /[ which vary by .05/]. These reactions due to unit panel loads are computed from Equations (5^) and (6«) of Article \ \ V \ \ > V '^ \ M ^? * y-. 03 ^d ^ ' « *^ '*' ^ N \ \ k 1 1 ^, S; k V ^ <^ Ss 1 J "^ 55=1 .4 .5 Fig. I. 35, with the values of c given at the heads of the columns of the table. As an example, if the unit load be placed at the distance 0.2/1 from the free end of the arm, the reaction 7?, will have the values 0.71355 and 0.72112 for the values of I ^ U = c = 8.822 and 4.395 respectively. The corresponding reactions, ^4, will be .004401 and .007311. The actual reac- tions for any panel load will be found by simply multiplying those given in the table by the actual panel load in pounds or tons, as the case may be. It will be observed that the reactions vary very little TABLES AND DIAGRAMS. I8l between the limits of range of the values of c. Hence, for values materially outside of those limits, the corresponding reactions may be assigned by the aid of Figures i and 2 with sufficient accuracy. Those figures exhibit the results given in Table I. Figure i gives the reactions 7?,, and Figure 2 the ibO / '' N. 140 / / s y 130 / \ / \ 120 1 / 110 ¥ f / y s, s. y 100 V / s V \ / / \ \ 90 / 1°/ \ \ "0 / / ^ — \ \ 80 / / / ^' s Ik ^ \ a / / #, / \ \ 70 « 1 / %V" \ \ / / i V 60 // 1 // 50 \ 1 / I / \ 40 I u \\ , 7 / 30 / // A / 20 // f i II ul 10 1 \l \ J i .5 Fig. 2. reactions R^. The three curves so nearly coincide in Figure i that only those for the two values of r = 8.822 and c = 4.395 are shown. For practical working, that figure should be drawn to at least double the size shown. In Figure i the tabular values of the reactions R^ have been taken full size, while those for R^, in Figure 2, have been multiplied 1 82 SWING BRIDGES. by io,ooa The figures will enable either of the reactions to be read at a glance for any length of panel in any length of arm. They, as well as Table I., are, therefore, perfectly gen- eral for a wide range of the value of c. After the reactions are read from the table or the diagrams, moments for either the web or chord stresses can readily be written ; and from these moments the stresses themselves will at once result in accordance with the methods of the preceding Articles. If the chords are parallel, moments will not be required for the web stresses, as the latter can be at once written from the shears. In many cases, particularly if the chords are parallel, the values given in Tables II. to VIII. will be found very conven- ient for seven- to nineteen-panel drawbridges, in which the center panel is equal in length to the others. They have been computed by Mr. Frank C. Osborn, consulting engineer, of Cleveland, Ohio, for his own practice, who has kindly given the author the privilege of using them in this connection. They express the shears and moments on the basis of each panel load being unity, and of each panel being one unit in length. Each actual shear will therefore be found by multi- plying each tabular shear by the actual panel load, and each actual moment will be found by multiplying each tabular moment by the product of the actual panel load by the actual panel length. These tables also show the greatest shears and moments at the various panel points. The reactions due to the various panel loads, by the aid of which the shears and moments are obtained, are computed from Equations (5^) and (6«) of Article 35, except those which belong to the simple spans, which, of course, follow the law of the lever. If the chords are not parallel, the web stresses must be determined by the method of moments, as illustrated in the two preceding Articles. Even if the length of the center panel should differ to some extent from that of the others, Table I. and Figs, i and 2 show that the values in Tables II. to VIII. will not be sensibly changed. TABLES AND DIAGRAMS. 183 TABLE II. (BCD E .F G i J ^ A ALL EQUAL A f^ 7 PANELS 1^ Note. — Shear in panel ab =■ reaction at a, and Shear cd = reaction at d. Loads at : Shear in Panel : Moment at ab he cd b c d b and g + 568 + 0.210 -0.432 + 0.210 -0.432 -0.790 + 0.568 + 0.210 + 0.136 + 0.420 —0.296 —0.370 c and f. Maximum ■ As a Simple Span - + 0.778 + I .000 + 0.210 -0.432 + 0-333 -0-333 — 1.222 — 1. 000 + 0.778 + I. 000 + 0.556 + I. 000 -o!666 TABLE III. (B C D E F G b 2 rf 9 PANELS tL r f A S A 9 PANELS l'^ A ALL EQUAL Note. — Shear in panel ah = reaction at a, and Shear de = reaction at g. Loads at; Shear in Panel Moment at : ab be cd de b c d e i and i + 0.665 + 0.364 + 0. 131 -0-335 + 0.364 + 0.131 -0.335 —0.636 + 0.131 -0-335 —0.636 —0.869 + 0.665 + 0.364 + 0.131 +0.330 + 0.728 +0.262 —0.005 + 0.092 +0.393 -0.340 -0.544 -0 476 Maximum As a Simple Span - + 1.160 + 1.500 + 0.495 -0-335 + 0.750 -0.250 + 0.131 -0.971 + 0.250 -0.750 -1.840 — 1.500 + 1.160 + 1.500 + I . 320 + 2.000 + 0.485 —0.005 + 1.500 — 1.360 184 SWING BRIDGES. TABLE IV. A A Ji t k T ALL EQUAL A I I PANELS Note. — Shear in panel ab = reaction at a, and Shear cf = reaction aty. Loads at : Shear in Panel : Moment at : ab be cd de e/ b c d e / + 0.726 + 0.471 + 0.252 + 0.089 -0.274 + 0.471 + 0.252 + 0.089 -0.274 -0.529 + 0.252 + 0.089 -0.274 -0.529 -0.748 + 0.089 -0.274 -0.529 -0.748 — 0.911 + 0.726 + 0.471 + 0.252 + 0.089 + 0.452 + 0.942 + 0.505 + 0.178 + 0.178 + 0.412 + 0.757 •+0.268 + 1.615 -0.095 — C.117 + o.oog + 0-357 -0.369 —0.646 -0.738 -0-554 c and k Maximum . . . . - + 1.538 + 2.000 + 0.812 -0.274 + 1.200 —0.200 + 0.341 -0.803 + 0.600 —0.600 + 0.089 -1. 551 + 0.200 -1.200 — 2.462 —2.000 + 1.538 + 2.077 + 0.366 —0.212 + 2.000 -2.307 As a Simple j Span 1 + 2.000 + 3.000 + 3.000 TABLE V. B C D E F G H I K L M N b c 3" e / 13 PANELS m n ALL EQUAL A 13 PANELS A A ALL EOUAL A Note. — Shear in panel ab — reaction at a, and Shear ^ = reaction at^. Loads at : Shear in Panel : ab be cd de 'f fS + 0.768 + 0.548 + 0.350 + 0.185 + 0.065 —0.232 + 0.548 + 0.350 + 0.185 + . 065 —0.232 -0.452 + 0.350 +0.185 + 0.065 —0.232 —0.452 —0.650 + 0.185 + 0.065 —0.232 -0.452 —0.650 —0.815 + 0.065 -o.23« -0.452 -0.650 -0.815 -0-93S j — I Maximum As a Simple Span + 1.916 + 2.500 + 1.148 — 0.232 + I . 667 -0.167 + 0.600 -0.684 + 1.000 —0.500 + 0.250 -'•334 + 0.500 — 1. 000 + 0.065 -2.149 + 0.167 -1.667 -3 084 — 2.500 TABLES AND DIAGRAMS. loi Loads at : Moment at : b c d e / S b and n +0.768 +0.548 +0.350 +0.185 +0.065 + 0-537 + 1.097 + 0.700 + 0.370 + 0.129 + 0.305 + 0.645 + 1.050 + 0.556 + 0.194 + 0.074 + 0.193 + 0.400 + 0.741 + 0.259 —0.158 -0.259 -0.250 -0.074 + 0.323 -0.390 —0.710 Maximum \ + I. 916 +2.500 + 2.833 + 4.000 + 2.750 + 4.500 + 1.667 + 4.000 + 0.323 —0.741 + 2.500 -3.501 TABLE VI. B CDEFGHIKLMNOP A y J /! I III )i op q A A 15 PANELS A A ALL EQUAL Note. — Shear in panel ab = reaction at • Or, for double-track spans : r-^ (4 Hence the total moment of inertia will be : f=/' + r (5). The nominal horsepowers of engines fitted to a number of railroad and heavy city draw spans which have proved to iSje SWING BRIDGES. be very satisfactorily operated are given in the tabulation below : Bridge. 500 feet double track railway 400 feet city bridge 270 feet city bridge , 400 feet double track railway 300 feet double track railway 362 feet single track railway 217 feet double track railway Weight of moving parts. Engines. LBS. H.P. 2,400.000 4,200, COO I, 800,000 40 50 40 2,000,000 35 1.250.000 684,000 600,000 to to to 00c The tendency has been toward an increase of engine power, in consequence of some of the earlier and smaller engines having shown insufificient capacity' in winds and other contingencies of drawbridge operation. The preceding considerations apply to rim-bearing turn- tables only, which are now exclusively used for all draw spans over a length sufficiently great to require the structure to be of the " through " type. The center- or pin-bearing type, in which the entire moving weight of the structure is carried on a center pin or pivot, is used for short spans only. This center pin or pivot may vary from eight to twelve or more inches in diameter, so that the pressure per square inch of bear- ing surface will not take greater values than two thousand pounds to twenty-five hundred pounds. The center pin, of wrought iron, or, preferably, steel, frequently rests upon a flat disk of phosphor-bronze in order to reduce friction and wear. The bearing faces of the pin and disks should be channeled or grooved to allow entrance for a heavy lubricant. In the Transactions of the American Society of Civil En- gineers for 1874, Mr. C. Shaler Smith gave, as the result of a number of tests of center-bearing draw spans, the fric- tional resistance, if exerted at the circumference of the center pin, at i-fo of the total weight turned. If W^ is that weight DRUMS AND ENGINES. 1 87/ in pounds, and F the force of friction supposed exerted in the circumference of the pin, then : F=.og W (6). If, again, / is time in minutes required by one man to open the draw, or close it, and if ^ is the diameter of the center pin in feet, he gave : / z= .09 H^^ "^ (7). 4 X 10,000 ^ ^ The time given by Equation (7) will usually be insufficient for the requirements of one man, even if no other considera- tion than that of friction be involved. In the contingency of an unbalanced wind pressure on one arm of the draw, one man may not be able to put the bridge in motion, nor to hold it against the wind. Hence none but the smallest draws in unimportant situations are made de- pendent on one man power. On account of the intermittent character of the operations of the motive power of a draw-bridge, gas or hot air en- gines are admirably adapted to that purpose. Their increased economy and essential cessation of expenses when the bridge is not being turned meet the requirements of the situation in a very satisfactory manner. As far as they have been tried they leave little to be desired. Art. 38. — Ends Simply Resting on Supports— One Support at Centre — Example. The general principles fundamentally involved in this case are not different from those of the two preceding ones, except in the number of points of support at the first pier. All the fixed load of the bridge is carried to the central point of sup- 188 SWING BRIDGES. port, whether the bridge is open or closed ; the end supports furnish reactions for the moving load only. The truss to be taken as an example is the one shown in the accompanying figure, in which the arms are of equal length. The general formulae to be used for the reactions at A, B, and C, and for the bending moment at the centre, are equa- tions (ii), (12), (13), and (10) respectively of Art. 35. These equations may be written as follows, remembering that /^ = /a = /, and M.^ = M: ^=-ip\-^''^'' ^)z+ip{i'-z')z\ (I). R, = j^hp{l-a)+M\ ........ (2). R,= j\^hPa+hp,~2M'^ (3). R,= 'j^^hp{l-z)+M'^ (4). It is to be remembered that c is measured from A or (7, according as the left or right arm is considered. The following are the data to be used : Total length = AC=2l=2AB= 2BC = 144 feet. Uniform depth = dD — bB — 16 feet. Panel length — AD = DE = etc. = 13 feet. BH = BH' = 7 feet. EXDS SIMPLY RESTING ON SUPPORTS. 189 Total fixed weight per foot = 1,200 pounds (nearly). Upper chord panel fixed weight = W — 2.73 tons. Lower " " " " = W = 5.00 " Uniform panel moving load = w = 19.50 " The moving load traverses the lower chord, and the weight of the floor system is taken at nearly 350 pounds per foot. On account of the extra weight of the locking apparatus, the fixed weight at ^, or C, will be taken at 3 tons, and will be denoted by u\. As is clear from the figure, all inclined web members, ex- cept the end posts, are for tension only, while the verticals are compression members. As the ends A and C are neither latched down nor lifted up, either arm is a single truss simply supported at each end, for all moving loads which rest upon it, so long as there are no movmg loads on the other arjn. For exactly the same reasons, therefore, as those given in the preceding Article, any counter, as dE, will sustain its greatest stress when the moving load extends from its foot to the centre, if no other moving load rests on the bridge. It must still be borne in mind that in connection with any counter stress, the stress in the vertical which cuts its upper extremity is to be found, for such a one may be the greatest stress in the vertical. Again, resume the general expression for the shear in any web member : s— S — n(W ^- W) — n'w ; in which S is the shear at one extremity of the arm, and n and n' the numbers of fixed and moving panel weights re- spectively between the same end of the arm and the web member in question. In considering the main web members, 5 will be taken adjacent to the centre, and, in the present example, at an indefinitely short distance from B in the arm AB. For a given condition of loading in AB, it is evident that the smaller is ^1 the greater will be 6". But Eq. (i) shows I no SWING BRIDGES. 1 that M \s always negative. Hence so long as '2P{1 — z) re- mains the same, Eq. (2) shows that R^ decreases as M in- creases (numerically). Again, Eq. (i) shows that J/ will have its greatest numerical 2 value, other things remaining the same, when ^P{P — z^)z has the greatest value possible ; i. c, when the moving load covers the whole of the arm BC. With a given value, there- 1 fore, for ^P{1 — z), R^ will be the least possible when the moving load covers the whole of the other arm, or the whole of BC; consequently 5 will be the greatest under the same conditions. Now having found under what circumstances S is the greatest, precisely the same reasoning used in the pre- ceding Articles shows that s will be the greatest, under the same circumstances, when ;/ is zero. Any inclined main wch vieinber, then, will sustain its greatest tensile stress when the moving load extends from its foot to the free extremity of the arm in which it is found, and covers at the same time the zvJiole of the other arm. A few main web stresses in all trusses of this case are a little singular in character, but are no exceptions to this rule. Those for the example taken will be noticed in the proper place. Any vertical web member, unless acting as a counter, will sustain its greatest compression in connection with the great- est tension in the inclined main web member which cuts its upper extremity. In seeking the greatest main web stresses, it may happen that the reaction Ry becomes zero ; this, however, changes nothing in the method. Since either arm is a simple truss for all moving loads rest- ing on it (supposing none on the other), every such load tends to cause the same kind of stress throughout the same chord. Consequently, as in the previous Article, the greatest tension in the lower chord, and compression in the upper, will exist when the moving load covers one arm only. These stresses will be found in that portion of the arm adjacent to its free extremity. The greatest chord stresses of the same kind as those E.VDS SIMPLY RESTING ON SUPPORTS. 191 caused by the fixed load, can be found with the least labor by first determining all the stresses due to t/ie fixed load alone, and tabulating them. The stresses caused by the moving load alone are then to be determined by the aid of the following considerations. Let that moment be considered negative which causes tension in the upper chord and compression in the lower. All moments, then, caused by the fixed or moving loads are negative, and all those produced by the upward reactions R-^ are positive. Now the compression in any lower chord panel may be found by taking moments (such will be negative if compression exists) about the panel point vertically over that extremity nearest the centre. The general expression for such compression will be : R{np — n'wt in which n is the number of the panel from the free extremity of the arm, n the number of the moving panel loads on the arm, / the distance of the centre of gravity of the moving loads 71 w from the origin of moments, / the panel length, and d the depth of the truss. The numerator of this expression must be negative, and it is desired to find what value of n will give it its greatest negative value. Now since every panel moving load on the arm AB increases R^ (as a positive quantity), it appears from the figure that 71 must not be greater than (« — i), and, farther, it must belong to loads between the panel considered and the free extremity of the arm. Since, however, / varies with 71 the above expression may have its greatest negative value when 71 is less than (;/ — i). These considerations are independent of the general charac- ter of R^\ it has already been seen, however, that, with a given loading on AB, R^ will be the least when the moving load covers the whole of BC also. Hence in order to find the greatest tension in the upper chord and compression in the 192 SWING BRIDGES. lower due to the moving load, the method of procedure is as follows : Throughout the whole operation the moving load is to entirely cover one arm, as BC. The moving load is then to cover the other arm from the free extremity to any panel point, and the stresses in the panels situated between the end of the train and the centre are to be computed. This operation is to be repeated for every panel in that arm not wholly covered by the moving load. From these results the greatest stresses may be selected and then added to the fixed load stresses. The character of these operations and the reasons for them will be much more evident after the example is treated. The following values will be needed, and depend only on the data already given. ^= 13 z — 26 ^=39 ^=52 ^ = 65 /-^= 59 / - ^ = 46 ^-- = 33 I — z — 20 I — z = 7 (/2 — ^)^ = 65,195.00 (/■^ — s^)z = 1 17,208.00 {P - z')z = 142,857.00 (/'^ — z^)z = 128,960.00 {P - z^)z = 62,335.00 Angle AdD = a. " HbB = fi. tan a tan fi 0.8125. 043 75- sec a = 1.29. sec yS = 1 .09. 4/2 20736 P= 19.5 tons = w ; = 0.00004823 P I 0.271 The following values of M are found by substituting the proper numerical values in Eq. (i). The moving load is taken to cover the whole of BC and so much oi AC d^^ is indicated by the values of z. ENDS SIMPLY NESTING ON SUPPORTS. 193 In arm AB \ 2 = 13 M= — 547.00 (( <( <( (( (( « « << (t ; ^ = r3 [26 • . M= — 657.00 (13 ; ^r = ■<> 26 . 39 . . M= — 792.00 ; ^= - 13 26 39 • ' 152 . M= —913.00 ; ^= " 13 26 39 • • 52 . M= — 972.00 165 The following values may now be written. Those of 7?i are found by simply substituting the proper numerical quanti- ties in Eq. (2) ; the moving load, as the values of M show, is taken to cover the whole of BC. 2= 13 p i - :2 {I— z) = 15.99 tons . . R^— 8.39 tons. = 28.46 " . . R,= 19.33 - 13 26 39 3740 R^ - 26.40 " 13 26 39 152 42.82 " ^1 = 30.14 " 13 26 39 52 .65 13 = 44.72 " . . i?i = 31.22 " IQ4 SWING BRIDGES. The stresses in the counters will first be sought, i.e., those in the arm AB. As a trial let the moving load cover the points F, G, and H, and, as before, let R^ be the general expression for the re- action at A. Hence: y?i = 3 X 19.5 X -^ '- = 16.25 tons. 72 Since 3 -}- 2 x 7.73 > R^, 2P = o at the panel point E, and no counter is needed between e and F. Moving load over EH. 26 1; 7?i = 4 X 19.5 X ^^ = 28.71 tons. 72 As 3 + 2 X 7.73 + 19.5 > Rxy '2P= o at E, and dE is the first and only counter needed. The vertical component of the stress in dE is s = R^-T, + 7.73 == 17.98 tons. Hence, {dE) = 17.98 x sec a = + 23.19 tons. The greatest compression in the end post Ad, and tension in the vertical d'D will exist when the moving load covers the whole of the arm AB, the other carrying none, and with such loading : •3 •? y?l zz: 5 X 19.5 X -^ = 44-69 tons. 72 Hence, (Ad) = - {R^ - t,) x see a = - 53.78 tons. At the same time : (d'D) = + {19.5 + 5.00) = + 24.5 tons. The stresses in the main web members are next to be de- termined. Those in Ad and d'D will occur under circumstances to be indicated hereafter. E.VDS SIMPLY RESTIXG OX SUPPORTS. 19! The following operations are in accordance with the princi- ples already shown. Moving load on BC and at D. R^ = 8.39 tons ; hence, for the shear in De: J= - /?i + (3 + 7.73 + 19-5) = 21.84 tons. Hence, (^^) = s x seca = + 2S.iy tons. Also, {e£) = — [s + 2.73) = — 24.57 tons. Moving load on BC and DE. 7?i = 19-33 tons ; hence, for the shear in Ef: -y ^ - /?i + (3 + 2 X 7.73 + 2 X 19.5) = 38.13 tons. Hence, {Ef) =-s x sec a = +49.19 tons. Also, {fF) = — (i- + 2.73) = — 40.86 tons. Moving load on BC and DP. Rx = 26.40 tons ; hence, for the shear in Fg: ^=-^i + (3 + 3x 773 + 3 X i9-5) = 58.29 tons. Hence, (Fg) = s x sec a = + 75.19 tons. Also, (^6^) = — (s + 2.73) = — 61.02 tons. Moving load on BC and DG. Ri = 30.14 tons; hence, for the shear in G/i : s=— Ri + (t,+4x 7.73 + 4 X 19.5) = 81.78 tons. Hence, (G/i) = s x sec a = + 105.5 tons. Also, (/^^) = — (s + 2.73) = — 84.51 tons. Moving load over BC and AB. Rx = 31.22 tons ; hence, for the shear in Hd : s= -Rx + {s + S ^ 7-73 + 5 X 19-5) = 107.93 tons. jq6 , " SWING BRIDGES. Hence, {Hb) = s x sec ^ = + 117.64 tons. The same panel weights have been taken for //and H' as for D, E, F, etc., though, strictly speaking, they would be a little smaller. At d, however, the fixed weight will be taken as 2.73 X 7-^13 =147 tons. Hence, (dB) = — (2 x 107.93 + 1.47) = —217.33 tons. Thus the web stresses, with the exceptions noticed, are completed. It has been shown that the greatest compression in the up- per chord and tension in the lower will exist when the moving load covers the whole of one arm, as AB, for which condition of loading, as has already been seen : i?i = 44.69 tons. Now, /?! — (3 + 7.73 + 19.5) = 14.46 tons is that part of the total panel load (fixed and moving) at E, which may be con- sidered as passing directly to A ; while (19.5 + 7.73) — 14.46 = 12.77 tons is the remainder, which maybe taken as passing directly to B. The following values will now be needed : 14.46 X tan a ~ 1 1-75 tons. 12.77 ^ tana— 10.37 " (7.73 + 19.5) X tan a = 22.12 " The chord stresses then follow : {AD) = {DE) = {R, - 3) X fan ^ = + 33.87 tons. {de) = {ef) = — (2 X 14.46 + 27.23) X tan a= - 45.62 tons. ifg) = { V^i X 65 — 4 X 19.5 X 32.5 {GH) = - {lib)' = -1 5 — ^^-^^ ^-^ = - 35-99 tons. qg^ SH'/NG BRIDGES. Moving load over BC and AF. Ri = 26.40 tons. Moments about g give : /Z7^\' / )\' ^1 X 52 — 3 X 19.5 X 26 ^ ^ i^FG) = - {gh) = -^ -J^ ^>* = - 9.26 tons. Moving load over BC and AE. R^ = 19.33 tons. Moments about /"give: ,r-r-\' / /- ^, 7^11 X ^9 — 2 X I9.5 X 1 9. ^ {EF) = - {fg)'= -i ^ -^-^-5 ^A = _ 0.4 tons. Moving load over BC and at D. R^ — 8.39 tons. Moments about e give: / r^ r-^ . r^, -/?1 X 26 — I9.5 X I 3 {DE) = - (r/) = ^- g-^^ ^ = - 2.21 tons. •{EF)' = —{fg)' = —2.21— (19.5-8.39) x/«;z «'= —11,24 tons. {EG)' = — (gli)' - — 11.24 — 9-03 — — 20.27 tons. {GH)' = — (//Z')' = — 20.27 — 9.03 = — 29.30 tons. Other chord stresses, with the different conditions of load- ing taken, are not indicated, as they were found to be less, for the same panels, than those that are given. They might be needed, however, in some cases. A very important result occurs, which has not before been noticed, when the moving load covers ^C and one panel load rests at A. In such a case Eq. (i) gives : M= - -,^2(P-s')z= - 486.00. 4/ And Eq. (2) /?i = -- = - 6.75 tons. Under the circumstances just named, therefore, the condi- ENDS SIMPLY RESTING ON SUPPORTS. 199 tion of things at A is equivalent to hanging a weight of 6.75 tons at that point, as the end of the overhanging arm AB. With such a weight, the following stresses result : 6.75 X tana — 5.48 tons. {AD)" = - {de)" = - 5.48 tons. (BE)" = - {efY = - 10.97 " {EFY = - (/-)" = - 16.45 " {FCy = -(^//)"= -21.94 " {GH)"= - {bh)" = - 27.42 " (BH)" = — (27.42 + 6.75 fan (i) = — 30.37 tons. From these results are to be selected the greatest chord stresses. For examples : Resultant {GH) — — (35.99 + 75.00) = — 1 10.99 tons. (^^) = -(21.94 +47-44)= - 69.38 " In short, precisely as the operation has been done before. The resultant web stresses caused by this negative reaction, are: {Ad) = (3 + 6.75) y. sec 01 — ^ 12.58 tons. {cW) = - (3 + 2.73 + 6.75) = — 12.48 tons. These are the " singular " stresses already mentioned. Collecting and arranging the results, the following resultant stresses are obtained : {dE) = + 23.19 tons. {Ad)= + 12.58 " {Ad) = - 53.78 tons {dD) = + 24.50 " {dD) = - 12.48 " {De) = + 28.17 " {eE) = - 24.57 " {Ef) = + 49-19 " {fF) = - 40.86 " {Fg) = + 75.19 " {gG) = — 61.02 " {Gh) = + 105.50 " (/.//)= - 84.51 " {Hb) = + 117.64 *" {bB) = - 217.33 " {AD) = - 7-92 tons; + 33.87 tons. {DE) = - 22.13 " ; + 33-87 " 200 SWING BRIDGES. {EF) = - 42.61 \ tons ; + 35.25 tons, {FG) = - 69.38 " ) + 2.76 " {GH) = — 1 10.99 (/?//)= -153-82 (de) = + 7.92 — 45.62 tons. (, the reaction at A will be negative, in the search for main web stresses. In such a case the method of operation is simply an extension of that used in the example. If the numerical value of this negative reaction is equal to, or less than, a moving panel load (which may rest at A), a weight equal to this reaction is to be taken as hung from A, and the panel load (moving) at D is to be taken as hung from that point, while the arm AB is to be considered as an overhanging one. If the reaction, however, is greater than a moving panel load, then two such loads are to be taken as hanging from A and D with the overhanging condition of the arm, A whole panel moving load is taken at A for prudential reasons. If the load were of uniform density, then a half panel moving load would be taken at A. Negative reactions by the formula for any number of mov- ing panel loads near the end are to be treated in exactly the same way ; for it is to be remembered that negative reactions in an actual truss, in this case, cannot exist. It may be urged that the case of partial continuity, taken in the preceding Article, should be treated according to the principles developed in this, by taking the middle span equal to zero. ENDS SIMPLY RESTING ON SUPPORTS. 20 1 Making such an assumption, however, would be a depart- ure from the real state of the truss. The safe way would be to determine the greatest stresses by both methods, and select the greatest of the two sets of results. Differences would be found only in the upper chord ten- sion, lower chord compression, and main web stresses. In the present case, if there are two or more systems of triangulation, each is to be treated precisely as the example has been. This case really includes that of a centre-bearing turn-table with two points of support at the centre, as shown in the h. Ti- figure. HH' is free to " rock " on the central point B, and as the motion is always very small, BH (horizontal distance) is essentially equal to BH' (also horizontal) at all times. From this it results that the reaction at H will always be equal to that at //', consequently the diagonals Hh and H'h must be introduced. Now as HH' is really a part of the truss, attached to and moving with it, the whole bridge, AC, is simply a continuous truss of two spans supported on the fixed point B. All the conclusions and formulae, therefore, of this Article, apply to it directly. R^ will be the reaction at B, and M will be the moment over the same point. According to the principles established, HJi will receive its greatest stress when AB, only, carries moving load. Since the pressure on H is always equal to that on H' also to a half of the reaction at B, there results : {Hh) = R'9 X sec hHh in which i^'g is the reaction at B due to the moving load on 202 SWING BRIDGES. AB only, considered as a simple truss. The greatest stress in H'h is, of course, equal to {Hh). The greatest stress in Hh (equal to [H'h')) is found, as before, by putting the moving load on BC and AG. No locomotive excess has been taken, but precisely the same conditions of loading hold whether such excess is taken or not. It will only be necessary to remember that the locomotive may be at either end of the train, and that the greatest results arising from the two positions are to be selected. CHAPTER V. SWING BRIDGES. ENDS LATCHED TO SUPPORTS. Art. 39. — General Considerations. It has already been stated that the object of fitting the ends of a swing bridge with a latching apparatus is to enable those ends to resist a negative reaction, or in other words, to prevent their rising from the points of support. All "ham- mering " of the ends will thus be prevented. It has further been shown in the preceding Chapter that if there are always two points of support at the center, for each system of triangulation, the ends will never tend to rise. It was also observed in the preceding Article that with a pivot, or centre-bearing turn-table, the bridge always presents the case of continuity with two spans only, whatever may be the number of apparent points of support at the centre. In this chapter, then, it will only be necessary to consider the one case of continuity with a single point of support between the extremities of the bridge. Art. 40.— Ends Latched Down— One Point of Support Between Extremi- ties of Bridge — Example. The general formulae required in this case are Eqs. (i), (2), (3), and (4), of Article 38, and they are here reproduced. M ie, = i|i^p(/-.i + 7i/| (2). 203 204 SWING BRIDGES. /?2-^|i/'^ + i^/^^-2J/| (3). i?3 = y|i^/^(/-^i + ^/| (4). These involve the condition 1^ = 1^ — /, which will appear in the example. If this condition does not exist in any case, the formulae to be used are Eqs. (lo), (ii), (12), and (13) of Article 35, but they are to be used in precisely the same manner as will be Eqs. (I), (2), (3), ajid (4). This case was essentially treated in the preceding Article, insomuch that with ordinary moving loads precisely the same conditions of loading, for the greatest stresses, are required in the two cases. The results themselves, however, will be dif- ferent for the upper chord compression, lower chord tension, and counter stresses. It will probably be as expeditious and labor saving, never- theless to find the reactions and chord stresses due to each moving panel load, and then combine the results thus found with those due to the fixed load alone, in the usual manner. Such is the method to be used in the example. Since Eq. (i) shows that M is always negative, Eq. (2) 1 shows that with a given value of 2P{1 — z), R^ will have its greatest positive value when no moving load is upon the span /g. The expression for the shear in any counter: s= R,-n'u'-n{lV+ IV'); (in which n' is the number of moving loads between R^ and the counter, and n the number of fixed loads similarly lo- cated), will have its greatest value for ;/' = o. Hence, /^ sec a = + 15.62 tons. With the moving load covering DH, dD acting as a counter will sustain a tensile stress equal to [dD) = + (19.5 + 5.00) = 4- 24.5 tons. With the same condition of loading. Ad receives its great- est compressive stress : {Ad) = —Ri X sec a = — 37.98 sec a = — 48.99 tons. Main Web Stresses. The main web stresses are found precisely as in Article 38, and there is no need of repeating the operation here. The values of the stresses will be reproduced in the proper place. Chord Stresses. The chord stresses due to the fixed load alone are the same as those determined on page 195 of Article 38. They will 2o8 SWING BRIDGES. not De reproduced, but references will be made to them as they are. Those caused by the moving load alone will be determmed by placing a panel load at each panel point successively, and finding all the chord stress in both arms due to it, then tabulating the results, and, finally, combining them in the manner already shown in several instances. The counter dE will be supposed to come into action for the weights E, F, G, and H. The panel loads at F, G, and H will cause apparent com- pression in some, or all, of the inclined members Ef, Eg, and Gh. The resultant action of fixed and moving loads in those members, however, will in all cases be tension. The detailed expressions for the chord stresses due to one moving panel load only will be given, as all the others are like it. For this purpose take w ^X E. 7?i = + 7.07 tons ; R^= — \.2>y tons, w — Ry^— 19.5 — 7.07 = 12.43 tons. {AD) — {DE) — + Ry tan « = + 5.74 tons. {de) = {ef) = — 2 X i?i X tan a — — 11.49 tons. {EE) = - (fg) = + 1 1.49 + 5.74 = + 17.23 tons. {EG) — — {gh) = 17-23 — 12.43 X tana = + 7.13 tons. {GH)= — {hb) = 7-13 — 12.43 X i<^^ a = — 2.97 tons. {HE) = — 2.97 — 12.43 X t^n ft = — 8.41 tons. {CD') = — {d'e') — R3 X tana— — 1.52 tons. {DE') = - {e'f) = 2 X " " = - 3.04 " {E'F) = - yfg) = 3 X " " = - 4-56 " {F'G') = - {gh) = 4 X " " = - 6.08 " {G'H') = - {kb) = 5 X " '' = - 7.60 " {H'B) = - 7.60 - RsX tanft = - 8.42 " The following checks by moments should be observed. Moments about b give : /zj-zjN Ri X 72 ~ 19.5 X 33 o . {HB) — — ^ '- ^ ^ ^ = — 8.40 tons ; {H'B) = ^^4^ ^ 16 = — 8.415 tons. ENDS LATCHED DOWN. 209 Moments about f give : {EF) — ^ ^^ = + 17.23 tons. The following tables are found by following the same operation for all the weights. {de) {ef) (/^) igk) {hb) w ^t D — 12.3 — 8.76 — 5-22 - 1.68 + 1.86 " " E - 17-78 - 17-78 - 10.82 - 3-87 + 3-09 " " F - 11-49 - 11-49 - 17-23 - 7-13 + 2.97 " " G - 6.08 - 6.08 — 9.12 — 12.16 + 0.65 " " JI - 1-77 - 1-77 - 2.66 - 3-55 -4.42 {AD) {DE) {EF) {FG) {GH) {HB) W 3l.\. D + 12.3 + 8.76 + 5-22 + 1. 68 - 1.86 -3-77 " " E + 8.89 + 8.89 + 10.82 + 3-87 - 3-09 -6.94 " " F + 5-74 + 5-74 + 17-23 + 7-13 - 2.97 — 8.41 " " G + 3-04 + 3-04 + 9.12 + 12.16 — 0.65 - 7-54 " " H + 0.89 -f 0.89 4- 2.66 + 3-55 + 4 42 - 3-63 {de) (e'f) i/Y) (g'A') {A'b) w at D + 0.69 + 1-38 4-2.07 4-2.76 + 3-45 " " E + 1.24 + 2.49 + 3-73 + 4-97 + 6.22 " " F 4-1.52 + 3-04 + 4-56 -1- 6.08 4-7.60 " " G 4- 1-37 4-2.70 4- 4.10 4- 5-46 -f 6.83 " " H 4- 0.66 4-1.32 4- 1.97 4- 2.63 4- 3-29 14 210 SWING BRIDGES. (CD) (DE) (£F') (FG) iC/l) {HB) 7f/al D — 0.69 -1.38 — 2.07 — 2.76 - 3-45 — 3-82 " " E -1.24 - 2 49 - 3-73 - 4-97 — 6.22 -6.88 " " /r -1.52 ! - 304 -4-56 - 6.08 — 7.60 — 8.42 '■ " G -1-37 i -2.73 — 4.10 - 5-46 - 6.83 -756 " " J/ -0.66 — 1.32 - 1-97 — 2.63 - 329 - 364 Using the main web stresses and the fixed weight chord stresses found in Article 38, the following greatest stresses at once result : idE) = + 15.62 tons. {Ad) = + 12.58 (< {Ad) = — 48.99 tons {dD) = + 24.50 It {dD) = - 12.48 " {De) = + 28.17 u {eE) = - 24.57 " im = + 49-19 << i/E) = - 40.86 " {Fg) = + . 75-19 << {gG)= - 61.02 " {Gh) = + 105.50 << {/iB)=- 84.51 " {Hb) = + 117.64 (( {bB) = - 217.33 " {AD) = - 7.92 ton. 5 ; 4- 28.42 tons. {DE) = - 22.12 << ; + 16.16 " {EF) = - 42.59 (( ; + 18.89 " {EG) =- 69-34 (( {GH)=- I 10.96 li {BH)=- 153.83 a {de) = + 7.92 a ; — 46.98 tons. (?i = + 15.14 . . Rs = — 0.851 tons. 2 = 39 " ■ . .']/= - 134-29 . . i?, = + 7.07 . . ^3 = -1.87 " 2 = 65 '■ . . M- - 58.59 . . i?i = + 1.09 . . ^. = — 0.81 21 3-531 System ADeFg, etc. = 26 feet . . M — — 110.18 . . i?i = + 10.94 . - R^ = — 1.53 tons. = 52 " . . M = — 121.22 . . ^1 = + 3-74 - . i?^ = - 1.68 " 3.21 Each of these results, it is to be obsen^ed, is for a single panel moving load placed at the panel point denoted by the value of z. They have been used in the two preceding cases. The chord stresses due to each panel moving load alone will first be found. As these are all found by exactly the same method, the detailed expressions for two only (one in each system) will be given. System ADdEf, etc. Panel moving load at/": ^ = 39 feet ; Rx— -^r 7.07 tons ; i^g = — 1.87 tons. {Ad) — — R^ tan a = — 5.74 " {df) = — 2 Ri tan a = — 1 1.49 " {fh) = — 11.49 + (12.43 — 7.07) tan n — — 7.20 " {hh') = — 7.20+ i2.43(/(2«a' + /<:^;^/S)= + 8.30 " {DE) = -{Ad) {EG) = 5.74 + 2 i?i tan a {GB) = 17.23 — 2 X 12.43 ^ tana (Cd) = — R^ tan a {df) = - 2 " " {fJi) = - 4 " " {hh') — 6.08 — ^3 {tan a + tan ft) {DE') = - {Cd') {E'G') = -f- 3 i'^g tan a {G'B) = + S '' " = + 5-74 = + 17-23 = — 2.96 = + 1.52 = + 3 -04 =: + 6.08 = + 8.42 = — 1.52 = — 4.56 = — 7.60 2i6 SWING BRIDGES. As numerical checks, the moment method gives the fol- lowing results : , „„- 7^1 X 65 — 19.5 X 26 ^ ^ ifiB) = —^ — ^-^^ = — 2.96 tons. (M') = ^-^^ = +8.41 tons. (G'B) = ^^^ = - 7.60 tons. System ADeFg, etc. Panel moving load at e: ^=26 feet; 7?i = + 10.94 tons ; R^= — 1.53 tons, (Ae) = — Rx tan a = — 8.89 " [eg) = {Ae) — 2.38 X taji a = — 10.82 " igg') = (^ All the results for the two systems give the four tables below : W ZX d {Ad) W) ifh) {hh) {DE) {EG) {GB) — 12.30 - 8.76 - 1.68 + 3.77 + 12.30 + 5-22 - 1.86 w zX. f - 5-74 -11.49 — 7.20 + 8.30 + 5-74 + 17-23 — 2.96 w 3.t h - 0.89 - 1.77 + 3.63 4- 0.89 + 2.66 + 4-43 {Cd) {df) (/VO {hh') {DE) {EG) {G'B) w 2X d + 0.69 + 1.38 + 2.77 + 3.83 — 0.69 — 2.07 -'3-46 W 2X f + 1.52 + 3-04 + 6.08 + 8.42 - 1-52 -4.56 — 7.60 ■w zX h + 0.66 + 1.32 + 2.63 + 3.64 -0.66 + 1-97 - 329 {Ae) -8.89 - 3 04 {eg) — 10.82 — 9.12 igg) {DF) {FB) + 3-87 + 12.16 w aX e " " E + 6.84 + 7-54 + 17-78 + 6.08 {Ce) {eg) ig'g) {DF) {FB) TO at ^ + 1.24 + 1-37 + 3-73 + 4.10 + 6.88 + 756 -2.49 -4-97 -2.73 -546 The open draw stresses due to the fixed weight alone are the followine : 3 X tan a = 2.44 tons. W X " =4.06 " JV X " = 2.22 " 3 X tand = 3.75 tons. IV X " = 6.25 " W X " = 3.41 " 2i8 SWING BRIDGES. W X /an /3 = 2. ig tons. W X " = 1. 19 " (Ad) = + 3 X tan a = + 2.44 tons. (de) ={Ad)+ JV tan a = + 6.50 " (e/) = (de) + |2 (3 + W) + W\ tana = + 19.88 " {fg) = (V) + ]2 {W + W) + W\ tan a = + 36.50 " ig^i) = ifg) + (3 + f'^ + 2 W) {tan a + /«« 6) + /F Z^?/ »~ 4/3/4/5 • 1 6 (/, + 4) (4 + 4) (4 + 4) (4 + 4) + 44= (4 + 4) (4 + 4) + ) -iT/, ^ 4 4'-(4 + 4) (4 + 4) + 4 4- (4 + 4) (4 + 4) - 4= 4' > . (14). Any bending moment may thus be found. It is seen that all moments are given in terms of M-^, which is still unknown. However, the bending moment at the other free end of the truss, from o. Fig. i, will be zero; conse- quently its general expression, put equal to zero, will give ^J/i in terms of A,B,C, etc., and the lengths of the different spans, /. e., in terms of known quantities. When AI^ is known, all the other bending moments are at once given by Eqs. (i i), (12), (13), (14), etc. As an illustration, if there are five spans, M^ — o and Eq. (14) will at once give M^. Eqs. (11), (12) and (13) then give the other moments desired. Another method may be followed by which a less number of equations will suffice for a greater number of spans. For example, the Eqs. (11), (12), (13) and (14), v/ith a similar value for J/g are aufficient for the solution of a case of ten spans. Let A' be the quantity corresponding to A, which would appear in the equation involving J/9 and J/g, in a continuous FORMULA. FOR ORDINARY CASES. 227 truss of ten spans, and corresponding to Eq. (i). Let B', C\ D' represent similar quantities in equations corresponding^ to Eqs. (2), (3) and (4). The following five equations may then be written by the aid of Eqs. ( i ) to (5): 2M, (/lo + 4) + Vl/8/9 + A ^o . . (15) J/9/9+ 2i^/8(/9 + 4) + J/7/8 + i5' = o . . (16) J/8/8+ 271/, (4 + /;) + J/eA + 6" = . . (17) M^l^^ 2.\U{1., + U) + AhU^ D- ^o . . (18) Ms/e + 2 J/5 (4 + 4) + J/4 4'+ £ =0 . . (19) A value for J/5 may be written by changing, in Eq. (14), J/i to J/9, 4 to /,o, 4 to 4, 4 to 4, 4 to 4, 4 to 4, A to A', B to B\ C to C\ and D to D'. A value of J/g in terms of J/j would be equal to the value of M^, given by Eq. (13), with exactly the same changes made, in so far as the same quan- tities appear. These pairs of values of the two quantities J/5 and J/g, equated, would give two equations from which Afi and J/y could be immediately deduced. All the other mo- ments would then follow. The Eqs. (ii), (12), (13) and (14), are sufficient in them- selves for the solution of a case of nine spans, in the manner just indicated. The preceding operations represent the most direct method of finding the bending moments over the points of support. All things considered, it is probably as short as anything that can be derived. Prof. Merriman has, however, given a more elegant method by the use of so-called " Clapyronian numbers." Any method involves sufficient tedium. The preceding formulae will be very much simplified if a single weight, only, rests upon some one span, since all the quantities. A, B, C, etc., except two, will then disappear. The various reactions may be immediately determined by Eqs. (2i)-(27) of Appendix I., after the bending moments are found. And when the reactions are known, the stresses in the individual members, for a given condition of loading, are found precisely as for a simple truss supported at each end. 228 CONTINUOUS TRUSSES. If the ends of the truss are not simply supported, the end moments must be known, else the problem will be indeter- minate. In such a case the preceding methods are in nowise changed, but the end moments, instead of being zero, will appear as known quantities. CHAPTER VIII. ARCHED RIBS, Art. 45. — Equilibrium Polygons. Preliminary to the specific treatment of arched ribs it will be necessary, first to consider some general principles re- garding equilibrium polygons for any given system of vertical forces, and then those involved in the theory of flexure. In the figure below, let AB K h& any straight, simple beam, subjected to the action of the vertical forces B, C, D, etc. Let x be measured from any section positive and horizontal toward A, and letP signify any external force such as the reaction at A or any of the forces applied to the beam ; Fig. I. then will 2Px represent the bending moment to which the beam is subjected at the section denoted by x. Now let there be imagined any force /^acting parallel to AEK, and let the moments 2Px be taken at each of the points B, C, D, E, F, G, H. Then if the quotients of those moments divided by the horizontal component of T be supposed represented by the vertical lines bB, cC, dD, etc., respectively, will the poly- gon AbcdefghK he one equilibrium polygon for the given sys- tem of loads ; so that if the beam AK were displaced by a tie in which exists the stress T, and the given loads hung from the joints b, c, d, etc., the whole system would be in equilib rium. 229 230 ARCHED A'IBS. In order to establish this, it is only necessary to show that no piece of the polygon is subjected to bending; for if that is the case, the line of action of the resultant stress must co- incide with its centre line. Consider any portion of the system, as that lying on the left of the vertical line dD. Those forces which have moments about the point d are the external forces to the left oidD and the stress T in the tie AK ; the latter has a lever-arm «, equal to the normal distance from ^ to AK, and its moment is op- posite in sign to 2Px. Consequently the resultant moment about d will h& M = 2Px — Tn. Btit by construction 2Px = Tn, hence M = o; and the same is, of course, true of every other joint. If T^ is the horizontal component of 7\ then evidently Tn = 7; (dD) = 2:Px. If t' be the general representative of the vertical ordinates dB, cC, etc., then, in general, 2Px ^ h but 7"^ is a constant quantity. From these considerations follows this important principle : The vertical ordinates of the cqiiilibritim polygon of any sys- tem of vertical loads arc proportional to, and may represent, the bending moments found at the various sections of a beam sub- jected to the action of the same system of loads, and having the same span. Since the stress T was taken arbitrarily, it is evident that there may be an indefinite number of . equilibrium polygons for any given system of loads ; the principle stated above, however, is perfectly general, and is true for all. Since vTj^— ^Px — constant for any given section, it fol- lows that any variation of T, and therefore Tf,, produces an opposite kind of variation in 7'. Hence the height of an equi- librium polygon is proportioned to the reciprocal of T or Tj,. The method of constructing the equilibrium polygon given above is not the most convenient, nor the one commonly used. The method ordinarily used is the usual one for con- EQUILIBRIUM POLYGONS. 231 structing the equivalent polygonal frame, and is the follow- ing : Let AK, Fig. 2, represent any span, inclined in this case but ordinarily horizontal, and i, 2, 3, 4, etc., the vertical loads acting along their respective lines of action. In Fig. 3 let the portions i, 2, 3, 4, 5, 6, 7, 8, and 9 of the vertical line 1-9 rep- resent those loads taken by any assumed scale. Since BC Fig. 2, represents the sum of all the applied loads, it is also equal to the sum of the two reactions or shearing stresses at A and A. In the case of the simple beam taken, those quantities will of course be determined by the law of the lever only. Suppose A' C and A' B to represent the shearing stresses or reactions at ^ and A" respectively. Then draw y^'P parallel to AK, and on it take any point P. From P draw the radial lines a, b, c, d, /, as shown, and starting from A or K in Fig. 2, draw the lines a, b, c, d, /, parallel to the lines denoted by the same letters in Fig. 3. Then will Fig. 2 represent the equi- librium polygon for the given span and system of loading. The line AK ox PA' is called \}!\q. closing line of the polygon. The reaction at A is 232 ARCHED RIBS. evidently composed of the numerical sum of the vertical com- ponents in / and AK, while that at K is equal to the numer- ical difference of the vertical components in a and AK. The point P, from which the radial lines are drawn, is called the pole, and the normal distance from the pole to the load line BC, t\\Q pole distance. The pole distance evidently rep- resents the horizontal component of stress common to all the members of the polygon. In order that the equilibrium polygon, constructed accord- ing to the principles given above, shall exactly fit the span, it is only necessary that a proper observance be paid to the scales used. 2Px From the equation 7^ = — = — it is seen that the scale for the ■I h forces does not affect the height of any joint of the polygon ; it depends only on the scale according to which x or the hor- izontal span is drawn. Let the line PP' be drawn parallel to EC, and let P' be the pole of a new equilibrium polygon ; the pole distance will, of course, remain the same as before. But the pole dis- tance represents the horizontal component of the stress in the closing line, and it has already been shown that if 7)^ remains. the same, 7> cannot vary. Hence, any movement of the pole parallel to the load line does not chafige the vertical dimensions EQUILIBRIUM POLYGONS. 23: ■of the equilibrium polygon. But if the pole distance is changed, the vertical di>nensions are changed in the inverse ratio. The determination of the deflection polygon of an arched rib with ends fixed, involves the use of an equilibrium poly- gon, similar to that required for a system of forces whose resultant is a couple. Its method of construc- tion is not at all different from that just given. In Fig. 4, let the forces 1,2, 3, 4. 5.6, 7, 8, and 9, act vertically, BC being horizontal, and let the sum of I, 2, 3, 8, and 9 be numeri- cally equal to the sum of 4, 5, 6, and 7. The double line, DE, in Fig. 5, represents the forces shown in Fig. 4. In Fig. 5, draw the line rt: in a horizontal direction through the upper extremity of force i, and take any point on it for the pole P. lines in the usual manner as shown. From C, in Fig. 4, draw b' parallel to b in Fig. 5, until it in- tersects the line of action of force 2. Then draw the other lines, ^', d\ etc., parallel to r, d, etc., until the lines of action of the other forces are intersected, b' , c\ . . . . h' , k', will then be the equilibrium polygon for the system of forces assumed. It is seen that the polygon does not close. This simply shows that the resultant of the system is a couple, whose mo- ment is the force a, in Fig. 5, multiplied by AB (vertical) in Fig. 4. The following general principle then results : The equilib- rium polygon for any system of parallel forces zvhose resultant is a couple, is not a closed o?te. This principle is indeed true for any system of forces. P might have been taken at any other point, as P' in OP. In that case, however, the equal forces a, acting at A and C, Fig. 4, would be parallel to a line drawn from the Fig. 5. From P draw the radial n t 4 -o4 ARCHED RIBS. upper extremity of force i to P' . The vertical dimensions of the polygon, measured from either of the forces a (in gen- eral inclined), will always be the same if the pole remains in the linePP'. Art. 46. — Bending IVIoments. An arched rib is any truss curved in a vertical plane, both of whose chords are convex or concave in the same direction, neither being horizontal ; the ends may be fixed or free. In Fig. 2 of Art. 45, let ADEK represent an arched rib sustaining the loads i, 2, 3, 4, .. . . . 9. Now it has already been seen that, so far as equilibrium is concerned, any given system of loading may be sustained by any one of a set of equilibrium polygons consisting of an indefinite number. On the other hand, it is evident that no polygon or arched rib can be drawn, which is not an equilibrium polygon for some system of vertical loading; but if that arched rib sustains some other system of loads than that which, it may be said, properly belongs to it, and if its joints be prevented from turning, it will be subjected to bending, which will vary from one section to another. The arched rib ADEK sustains a system of loading for which AfK is the equilibrium polygon, hence the former will be subjected to varying degrees of bending at various sec- tions. When the rib ADK is subjected to the action of its load, stresses are developed in its different parts, whose hori- zontal components are all the same because the load is wholly vertical. Now if an equilibrium polygon can be found in which the horizontal component of stress Tj, is the same as that developed in ADEK, then all the circumstances of stress and bending in the latter can be determined, as will be seen hereafter. Suppose AfK to be that polygon, then let v' denote the portion of a vertical line intercepted between it and the arched rib, as DD' . The moment about any point D will then be M ^^Px- T„ (7' 4^ v). But since AfK is the equilibrium polygon, '^ Px — Tf^v — O. RENPIXG MOMEXTS. 235 .'. M = — T^ v'. When the polygon lies above the rib, v' is negative, and, hence, M positive. Let the polygon which has the same value of 7\ as the arched rib be called the true equilibrium polygon ; then, since Th, is a constant quantity for the same rib, there is established the following important principle: J lie bending moj/ients to zvJiich the different parts of an arched rib are subjected are proportional to, and may be represented by, the vertical intercepts included between the rib and the triie eqiiilibriuni polygon. This principle has been demonstrated for a beam with free ends only, but it is true also for a beam with fixed ends, as will now be shown. In order to fix the end of the rib it is only necessary to im- press upon the rib at A, the point of fixedness, the proper couple whose moment is in ; in the fixed rib, as in the free, let Tj^ represent the horizontal thrust. The true equilibrium polygon for the fixed rib will be that found by increasing the vertical dimensions of a polygon for a free-end rib, formed by vising Tf^ and reactions for fixed ends, by a constant amount m equal to -7fr . Again, taking moments about any point of the centre line of the rib, there will result M—^ Px + ;// — 7; (f + ™- + t^) = — Tj,v', ^ h as before. This shows that the principle stated above is true fpr both fixed and free-end ribs. In truth this equation might have been written first, and the special case of the free-end rib deduced by making m = o. An arched rib, then, when subjected to the action of a load, suffers bending in the same manner as a straight beam, but to a different degree. In so far as it plays the part of a beam, it must be governed by the general laws of bending or flexure. The formulae to be given in this connection are those approximate ones based on the common theory of flexure, and found in the ordinary works on that subject. :236 ARCHED RIBS. Art. 47. — General Formulae. Let ^^ denote the total shearing stress at any section, P an/ applied load or external force, M the bending moment at any section, and D the deflection found above ; then the six gen- eral equations of flexure demonstrated in the Appendix on the Theorem of Three Moments, some of which are made use of in the graphical treatment of arched ribs, are the following: 5 = ^P, M =:2Px, p, _M^ EI ' nM D =2 nP X = 2. r-T ■ • EI D„ =^nPy. If the beam is originally straight and parallel to the axis of X, n becomes dx and y = o. As usual, E and / represent the coefficient of elasticity and moment of inertia of the cross-section, respectively. The limits of the summations are the section considered and any section of reference. The quantities S, M, P' and D then refer simply to that portion of the beam over which the summation extends. One very important deduction is to be drawn from the above equations, or rather from the second and fifth of them. It is seen from these two equations that -^^^ stands in the same relation to D that P does to M. Consequently if -ry be put ARCHED RIB IVITH ENDS FIXED. 237 /n the place of P in any graphical construction, D will be represented in the place of M. If, therefore, an " equilibrium polygon " be constructed for any span by taking -^^ as loads instead of P, the vertical ordinates of the polygon will repre- sent the deflections at the sections denoted by the corresponding values X. This polygon may be called the " deflection polygon," and its construction plays a very important part in the determina- tion of the true equilibrium polygon for an arched rib, in the majority of cases. Art. 48. — Arched Rib with Ends Fixed. The ends of an arched rib or any girder are considered fixed when the angles formed by their centre lines with any assumed line, at the fixed sections, do not vary under any applied load. Let BAD, PI. v., be the centre line of any arched rib ; it will, of course, be considered fixed at the points B and D. This line may be any curve, though for convenience the arc of a circle has been drawn. In the demonstration no atten- tion whatever has been given to the character of the curve, so that it is equally applicable to any other curve. In the present case the centre line of the rib will be divided into equal parts for the application of the load, and each of those parts will be n ; consequently that quantity will have a finite value, and the results will not be strictly accurate, though near enough for all technical purposes. The piers or points of fixedness are supposed to be immov- able, whatever may be the character of the load ; but if that is the case, the summation of the strains at any given dis- tance from the neutral axis of the rib, considered as a truss, taken throughout the whole length of the rib BAD, must be equal to zero. Take that distance as unity, then there re- sults, for one condition, since 71 is constant, the equation : 258 ARCHED RIBS. ^ nP' = :^ ^^ = —-:e m = o. "EI EI " ^..M=o. The quantity -=-= is brought outside of the sign ^\ because El the moment of inertia of the cross-section of the rib is s7ipposed to be the same throughout its entire length. More will be said on this point hereafter. It has already been shown that the area included between the equilibrium polygon and the curve BAD is made up of vertical strips, whose lengths (the vertical intercepts) repre- sent the actual bending moments at the different sections of the rib. Hence 2M represents the sum of those vertical lengths or intercepts drawn at the points to ivhich the inonients M belong, and the equation 2 M =^ o shows that the sum on one side of BA D must be equal to that 07i the other. But, as will be seen, there may be an indefinite number of equilibrium polygons which will fulfill this condition ; conse- quently at least one other condition must be obtained. Since the points B and D are fixed, the sum of all the de- flections, both horizontal and vertical, taken between those two points, must be equal to zero. It has been shown that the vertical deflection at any point, when n and / are con- n sidered constant, is Z> = — - '2Mx ; also, from the reasoning • El applied to the curved girder, that the horizontal deflection is n Dh — -jrj 2My. Now, when these summations extend from ^to D, since those points are fixed, both D and D,^ must equal zero. The three equations of condition, then, which must be fulfilled for the rib, are : 2lMy= o. It has already been stated, and it is evident without much ARCHED RIB WITH ENDS FIXED. 230 thought, that any polygon whatever is an equihbrium polygon for some load. Hence consider i^^'^Z?, PL V., an equilibrium polygon for its proper load, and consider it subjected to that load ; denote its moments by M^,. Again, suppose the polygon a, a , a", etc., to be the true equilibrium polygon for the given load, and denote its mo- ments, represented by the vertical ordinates drawn from its closing line, by M^. Then, from the principle which precedes the equation M = — Tf{v' in the general discussion of equilibrium polygons, there follow the equations : .'. ^]Mx = ^]M,x - ^]M,x = o. Or, ^"Jlax = ^"Jhx. In the same manner, ^'^^Maj^ = '^'^^M^.y. This last equation will be used in fixing the pole distance of the true equilibrium polygon. It must be remembered that M represents the actual mo- ment to which the rib is subjected at any point. The application of these two conditions will be shown in the course of the construction of the true equilibrium poly- gon, as they are needed. In the figure of PI. V., let the scale for linear measurements be 10 feet to the inch, and the force scale 15 tons per inch. The curved centre line of the rib is divided into ten equal parts of 10.95 feet each, and that is the constant value of n. The load is not therefore uniformly distributed. The panel length, or horizontal distance between the points of appli- cation of the loads, is thus a variable quantity. If the versed sine of the centre line of the rib is small, n may be taken equal to the span divided by the number of panels. But if the versed sine is not larger, even, than in the present case, n 240 ARCHED RIBS. cannot be so taken without sensible error, as will be seen. The other data are as follows : Span = loo feet. Radius = 75 feet. Angular length of curve = 83^ 'i^'j. Panel fixed load = 4 tons. Panel moving load = 10 tons. Centre rise of rib = 19. i feet. In the figure BD is the span, and C the centre ; b\ b", b"\ etc., are the panel points equidistant in the curve, and through which the loads are supposed to be applied. Now the actual moment area for the arched rib is supposed, really, to be the difference between the moment area of the true equilibrium polygon for the applied loads and that of the rib itself considered as an equilibrium polygon for the proper load ; both systems of loading being supposed applied to a straight beam fixed at each end. The first portion of the problem which presents itself, then, is to determine the true equilibrium polygon for the given load. The construction will first be made, and it will then be shown that the two conditions given above are satisfied. Let the moving load cover the left half, BC, of the span, and suppose the half panel loads at B and D to rest directly on the abutments. According to the scale taken, lay off B6 equal to 7 tons, half the total load on the panel Ab'^"' , and B^ equal to half the fixed panel load on Ab"^; then lay off 6— 10, divided into four equal parts, equal to the four equal panel loads on b\ b", b'" , and b^''. In the same manner lay off 5 — I, equal to the four fixed panel loads on the right half of the span. Assume 6" as a convenient pole, and draw the radial lines from it to i, 2, 3,4, 10. Starting from C, draw C — 6 until it intersects a vertical through ^'^ at Ui; from the latter point, a^ a^ parallel to Cy until it intersects the ver- tical through b'" ; proceed in the same manner until the poly- gon ECF is drawn. If the ends were free EF^owld be the closing line, but one must now be found that will satisfy the ARCHED RIB WITH ENDS FIXED. 24 1 condition ^ Ma = o, or, in other words, the sum of the verti- cal intercepts drawn from the closing line downwards must be just equal to the sum of those drawn from the same line upwards. The proper closing line is easily located by trial. If 6V= 0.5 inch and vv' = 0.96 inch, there results: .B Til 2^ Mf, = ^vv' + a-^c — U'f" — a^c'" — a^c^'' — Cc^ — a^f^ — a-f + ^8*^^" + ^9^*^ + i^'V" = 1.88 - 1.89 = - o.oi inch. This sum is sufficiently near zero. The lines vv' and v"v"' are drawn vertically through points n b and h^, distant — from B and D on the curve BAD, and 4 their halves are taken because in the summation 2 iiMa there 1 , , n ,11 ,, ,,, vv' would appear terms — x vv and ~ x v v , or n x — and v"v"' n X 2 Similar terms will hereafter appear in similar summa- tions. The closing line HK then satisfies the condition 2 ^AI^ — o for the equilibrium polygon ECF. There still remains the condition 2 J/^x = 2 M,,x. This equation will be satisfied by making each of its members equal to zero. The closing line HK must, then, also make 2^MaX = o. This simply means that the vertical ordinates of the polygon a, measured from HK, multiplied by their horizontal distance, from D, will form a sum equal to zero when their products are added. If the ordinates below HK are taken positive, as v"v"' , agC^^. a^c", etc., and those above, as a^c"^, negative, and if the ordinates and distances be taken by scale from the drawing, there will result, nearly, v"v"' X De + a^c^ x De' + a-^c' x De^ + etc. = + 99.7 Orfi^ X De" + a^c^ xDe"+ etc. = — 102.0 16 242 ARCHED RIBS. The numerical values are nearly enough equal, and the line HK will be taken as the proper closing line. The next step is to find the closing line for the curve BAD of the rib, considered as an equilibrium polygon, which will satisfy the same general conditions. Using precisely the same method of procedure as for the pohgon ECF, the line H" K ' is found to be the one desired, for that line makes the sum of the intercepts above it just equal to the sum of those below it. Ac^ is about 0.62 of an inch. For this curve the summation may be written : 2^J/j = bH' -f- 2<5'Vi +• 2b"C'i, — 2c^b"' — ic^b^" — Ac^ = o. Since the curve is symmetrical in reference to A, the static moments of the ordinates on one side of H" K'\ about DK" , will evidently be equal to the same moment of those on the other side. The second condition may now be applied. That condition is 2^May = -S^il/ftj. It has already been shown that if M^ and Jlfi, be considered loads applied at distances y from the assumed origin, the ordinates of the equilibrium polygon so constructed will represent the quantities 2 Maj = -D^, or Through A, therefore, draw the horizontal line RAS. As- sume any line, as AC, as the closing line of the deflection polygon, and lay ofi At equal to a half of vv'. Also make tt equal c'a-^; ft" equal d'a^^; t"t"' equal a^c", etc. /"/'" is measured in a direction opposite to that of the preceding, because it represents a moment of an opposite sign. In the same way At measured to the right of A is equal to a half of v"v"' ; tty equal to ^^^9 ; t^t,, equal to «8<^^"\ etc.* Draw bb'^ and note its intersection C" with AC. From C" draw C"d^ parallel to Ct until it intersects a horizontal drawn through b' in d' \ draw d'^d'' parallel to Ct' until it intersects a horizontal through b" in d" ; draw d''d'" parallel to Ct" * If the drawing had been made accurately, /'"/iy would have been exactly equal to Cc. ARCHED RFB WITH ENDS FIXED. 243 until it intersects a horizontal through //", etc. The polygon C'd'd'', etc., will intersect the horizontal RAS at a point distant from A, on the left of it, from which / is drawn to a point on the right of A at the intersection of the deflection polygon formed by using /, / , /, , / , as before, and which is shown in the figure. The sides of the deflection polygon on the right of CA are parallel to radial lines drawn from C to the points /, /, / , /, . The distance (/ — / ) represents, in an exaggerated manner, the horizontal deflection of the end of a vertical beam fixed at C, whose length is CA, and which is subjected to the bending moments vv , ca, c"a^, etc., at verti- cal distances from C equal to the heights of b, b\ b", etc., above BD. In the same manner, / represents the same quantity for the same beam when subjected to the corresponding mo- ments on the right-hand side of A C. The line / can be determined with less work and more sim- ply when the meaning of the construction is once clearly seen, by laying ofT, on the left of y^, as before, loads repre- sented by the algebraic sums (At + At), (//' + //,), {t't" -\-t t }, etc., and then drawing the equilibrium polygon as usual. The distance from A to the intersection of the polygon with AR will then be equal to /. The deflection polygon d'd"d"'d^^ is constructed in pre- cisely the same manner as the preceding. Make As equal to a half of bH' ; ss equal to cb' ; s s" equal to c^^b" ; s's" equal to c^b"\ etc.;* then draw radial lines from those points of division to C. The point d' is at the intersection of C"d' , drawn parallel to Cs, with a horizontal line drawn through b' ; d'd" is drawn parallel to Cs until it intersects a horizontal line drawn through b" , and the other sides of the polygon are constructed in the same way. The polygon cuts the horizontal line RAS in a point distant i/ from A. There will, of course, be another deflection poly- gon, precisely the same as the last, on the right-hand side of AC, found by taking ^K'D, c^b^^, etc., and laying them off from A towards S. * With a sufficient accuracy of construction, As^^ would equal a half of Ac^ 244 ARCHED RIBS. If the intercepts used in the deflection polygons for Ma and >/ft, represent those moments by the proper scale, then by the same scale CA will not, in general, represent the true pole distance. But this fact has the same proportional effect on both /and /. Consequently any result depending on the equality of / and /, will not be affected. Instead of using Acr^ and Cc^ in the manner shown, greater accuracy might have been attained by taking half intercepts n at the distance — on both sides of A and C. Such an oper- 4 ation, however, is unnecessary in all ordinary cases, since moments in the vicinity of A and C have very little effect on the horizontal dimensions of the deflection polygon. Mo- ments, on the contrary, in the vicinity of HH" , have great effect. Now /, represents 2 ^ M^y ; and /, -2 ^ Mj,y ; and in order that the second condition may be satisfied they should be equal. Since /^ is less than /, it shows that the quantities M^ are too small, or, in other words, the pole distance BC is too large. This last statement is evidently true, if it be remem- bered that the pole distance is inversely proportional to the vertical ordinates which represent the moments. Lay off, therefore, on yi (7 produced, the distance 6" J/ equal to /i, and draw through M the horizontal line MN. With a radius CN equal to /, draw the arc of a circle cutting MN in N, then produce the line CP. All moments represented in the lower equilibrium polygon a will have to be increased in the ratio of 67V to CM. To make this reduction, draw a horizontal line, for instance, through c^ until it cuts CN mp, then make c^a^ equal to Cp\ a^ will be one point in the true equilibrium polygon. All other points might be found in the same way, but having found one point, as a^ , a much shorter method may be used. It has already been shown that the vertical dimensions of two equilibrium polygons for the same loading and span are inversely proportional to the pole distances, the vertical dimensions being measured from the closing lines. In the ARCHED RIB WITH ENDS FIXED. 245 figure the vertical dimensions of the polygon ECF must be increased in the ratio of Z, to /, or in the ratio of Cc^ to Cp. Hence, on CN produced make CP equal to BC, and draw the horizontal line OP cutting CM produced in O, then will CO be the pole distance for the true equilibrium polygon. In order to find the true pole, make CL parallel to HK, then draw LC parallel to BC^ and equal to C0\ the point C will be the true pole. According to previous principles, the reactions or vertical shearing stresses at B and D will be Z — 10 and L — i respect- ively, and, since the closing line must be parallel to BD in the true equilibrium polygon, LC must be parallel to BC. From C draw the radial lines shown to i, 2, 3, and 10; these lines will be parallel to the sides of the true polygon ; /. e., draw rtV^ parallel to C 6 until it cuts a vertical through b'^'^ \ a^^a" parallel to 6" 7 until it cuts a vertical line drawn through d'" ; a^a^ parallel to C 5, etc. The polygon aa'a" a^^a^ so formed will be the true equilibrium polygon. As a check some of the points as a' or a" should also be found by the previous method. Thus make 6^^ equal to c'a, and draw gp' parallel to CD ; Cp' should then be equal to ca'. Other points may be treated in the same manner. It may now be seen that the polygon a a a" a^ satisfies the three conditions 2^M=o, 2 Mx = o, and 'S^My — o. The first conditions are evidently satisfied by the method of locating the closing lines in the lower polygon UiO^ rt'g, and the 'curve BAD, for any intercept be- tween the curve and the upper polygon a is the difference between two intercepts each of which belongs to a sum equal to zero, hence the sum of those intercepts is zero. The second condition is satisfied by the location of the point C. Another check on the degree of accuracy attained in the construction is found, in what has just been said, i. e., the * The reason for the true closing line being horizontal is given, though for soother purpose, on page 273. 24^3 ARCHED RIBS. moment area lying above the curve BAD must be just equal to that lying below it. By. actual measurement LC is equal to 3.65 inches, hence the constant horizontal component of stress in any portion of the rib is 54.8 tons. The resultant stress at any point of the arched rib is equal to 54.8 tons multiplied by the secant of the inclinations at that point. The bending moment at any point to which the rib is subjected is found by multiplying the vertical intercept between the equilibrium polygon and the curve by (54.8 tons = Tj). For example, the actual mo- ment to which the rib is subjected at h' is {b' ci x 54.8). The line of action of Tf, (54.8 tons in this case) is, of course, along the true closing line H"K". This is an important matter, as will hereafter be seen. A line drawn through C parallel to RF would cut off on the load line the reactions which would exist were the ends free. The reaction at B, in the present case, is thus seen to be much greater than would be found in the case of free ends. A point in the vertical line passing through the centre of gravity of the load is found at the intersection of the sides a a and a^ a^ in G. G' , at the intersection of Ea^, and Fa^, prolonged, is in the same vertical line. The pole of the deflection polygons might have been at A, and the moments laid off from C, in which case the half of Acr,, and the same of Cc", would have been the moment dis- tances adjacent to C. Precisely the same value iov LC would probably not be found, because the sum ^ My is not con- tinuous, and consequently not exact. For this reason it would be better in an actual case to divide the real panel lengths into two or more equal parts in order to find the true pole distance, LC\ and consequently the true equilibrium polygon. Th can then be used to find the stresses in the members of the actual rib in a manner that will hereafter be shown. The diagram should of course be drawn to as large a scale as possible, and it may often be advisable to exaggerate the vertical scale so as to make the intersections of the true poly- gon and curve well defined. ARCHED RIB WITH FREE ENDS. 247 The effect of such an exaggeration may easily be shown since the various steps of the construction remain precisely the same. Suppose AC to be m times as large as it would be made by the scale according to which the span BD is laid off; w denotes the degree of exaggeration. The distance / will be m times as great as it ought to be, and consequently the height of the equilibrium polygon will be increased beyond its true value in the same ratio. But if the true height is only — of that found, the true pole distance will be ;« x CO, in and LC must be made equal to that. The span has been supposed half covered by the moving load, but any other portion might have been taken as well. It is to be noticed that the method is perfectly general, and entirely independent of the character of the curve BAD, or of the loading. Art. 49.— Arched Rib with Free Ends. The treatment of the arched rib with free ends is not dif- ferent in any respect, except one, from that given in the pre- vious case. The exception is this, that the condition 2 M =0 must be omitted, since the bending moments at the free ends must disappear. In this case, again, the centre line is divided into equal parts. Observations made under this head in the preceding Article apply here also. The expression '2nP ' = 2 -^, sometimes called the bend- ing, is of course based on the common theory of flexure, and denotes simply the difference of inclination of the neutral surface of a straight beam at the two sections indicated by the limits of the summation ; with a constant moment of inertia it is usually written j^- j Mdx. The condition of fix- edness of the two ends of the ribs requires the position of the neutral surface to remain unchanged at those two sections, consequently 2nP' must be equal to zero between those 248 ARCHED RIBS. limits. In the case of free ends, the position of the neutral surface may be any whatever consistent with the elastic prop- erties of the material at those sections. The middle points of the free-end sections must, however, retain their primitive positions, or the summation of the deflections, either horizon- tal or vertical, between those points must be equal to zero. The only remaining conditions, therefore, are ^^^nMy = o = But the ends of the rib may have any relative vertical movements, whatever, without changing the circumstances of bending. Consequently the only condition to be fulfilled is, '2. j^nMy— o. The same amount and proportion of loading will be taken as in the previous case; the same radius, span, notation, and scale will also be taken. The figure of PL VI. represents the construction. The moving load is assumed to cover the half span BC. As before, take C as the pole for the trial polygon, then make B 5 equal to a half panel fixed load, supposed applied at yJ, and ^5 6 a half panel (fixed + moving) load, supposed applied at the same point ; also make 5—4 equal to load at b^, and 6 — 7 the load at b"^^ , etc. Draw radial lines from C to the points of division, i, 2, 3, 4, 5 • • • • 10, and construct the polygon E, a-^, a2, a^, . . . . F, precisely as before ; in truth it is exactly the same polygon that was used in the preceding case. Since there can be no bending moments at B and D, the equilibrium polygon must pass through those points, hence EF is the closing line of the trial polygon, ECF. As has been seen, the only condition to which the equilib- rium polygon is subject is 2^ nAIy = o ; or, as before, as n is constant. :ElMy = ^iMay -^^M.y = O. Or, 2lA/,y=:Eyf,y. The method of constructing the deflection polygon is pre- ARCHED RIB WITH FREE ENDS. 249 cisely the same as that followed in the previous case, only in the present one a half of the ordinates representing M^ and J/ft will be laid off from A in order to keep all the points s and t within the limits of the diagram. As the half intercepts at the distance - from B and E are very small, and as their omission will lead to simplicity in the diagram, and not cause much of an error, they will be neg- lected. In an actual case, however, the omission should be made with caution. Since the moments at B and D are zero, make As equal to \cxb\ ss equal to \c.^b" , s's' equal to \cjj"\ s"s" equal to )^cj)'"' ; then draw horizontal lines through d'd", b"b"', etc., cutting AC 2,% before. As the end moments are zero, d' will be on AC, then d'd" will be parallel to Cs, d"d"' will be parallel to Cs', and so on until the point d^' on the line RS is reached. The two portions of the deflection polygons for the mo- ments Ma will not be similar, yet it is only necessary to con- struct a single deflection polygon, as was shown in the pre- ceding Article. The sums .^"^ J/^j' and ^^ J/^j' may each be divided into pairs of terms, each member of the pair having the same value oi y ; this may be done for any case in which the moments may be taken in pairs. The moments in each pair of terms will of course be located equidistant from AC. Make, therefore. At equal to \ (a^d + rt'/*), /'/, = to h {^-d' + ^8<^^"'), t,t„ equal to hagc'" + a-^c"^^), and tj„, equal to ^ {a^c^ + a^c"^^). Draw radial lines as shown, and make d/ d/' parallel to Ct, d/'d/" parallel to Ct,, etc., until the point ^/ is reached. Lay off on /iC produced, CM equal to Ad,'^, draw J/A^ par- allel to BD, and with a radius CV equal to 2{Ad^) find the point TV, and produce CN through that point. In order to find the true equilibrium polygon it is only necessary to in- crease the ordinates of the trial polygon in the ratio of CM to CN. Hence draw d'p parallel to CD, then make Ca^ equal to Cp ; d' will be one point in the true equilibrium polygon. In the same manner make Cg equal a-j/c'", and determine/' as before, 2 CO ARCHED RIBS. then make c^'" equal to Cp' ; o!" will be another point in the true polygon. A shorter way to proceed, however, is the one indicated in the previous case. Draw CL parallel to EF, then a line parallel to BC through L. Make CP equal to BC, and draw OP parallel to BD ; OC is the true pole distance. Make, therefore, LC equal to OC, and 6" will be the true pole from which radial lines are to be drawn to i, 2, 3, 4, 5, . . . . lO. Starting from any given point, as B, make Ba' parallel to C 10, a'a" parallel to 6"' 9, etc., etc. ; the side parallel to C i should pass through the point D. The two methods should be used to check each other, as was indicated in the previous case. Since LC is 3* inches, TJ^ — 3.^ x 15 = 52.5 tons, and its line of action is evidently BD. A II those directions of a general character zvhich accompany and follow the construction in the preceding case apply with equal force to the present one and those which follow. It is particularly important in the graphical treatment of all arched ribs to make the polygons approach as nearly their ultimate limits, i. e., curves, as possible ; for that reason it will be advisable in most cases to divide the actual panels into two or more equal parts in the search for the true equilibrium polygon. Art. 50. — Thermal Stresses in the Arched Rib with Ends Fixed.* Thermal stresses are those stresses which are co-existent with any variation of temperature, in the structure consid- * In reality the deflection which produces stress, in the case of variation of temperature, is not the whole deflection. If the points B and Z>, in PI. VII., were free to move, there would be no thermal stresses, but there would be deflec- tion. This deflection, in the present case, would be (if loo units become 100.12204 for an increase of 180" F.) (100 -i — -—0.12204 I -^ 100 X 19.1 — 19.1 = .021 ft. Strictly speaking, the deflection to be used, then, for a change of 165° F. is 0.25 — 0.021 = 0.23 ft. The difference is so small, however, that it may be neglected, especially since the error is a small one on the side of safety. These observations are general and apply to all cases. THERMAL STRESSES IN ARCHED RIB. 25 I ered, and whose values depend upon that variation. Any variation of temperature in the material of which an arched rib is composed will cause a variation in its length, and con- sequently a deflection at any given point. Although the temperature is supposed to change, yet the ends of the rib are supposed to remain in their normal positions, so that the general conditions ^>/J/=:oand ^)iMy — o = .^'«J/;ir, hold for thermal stresses as well as for any other, remembering that 11 is constant. Any change of form, such as that arising from the applica- tion of loading, will cause extra stresses, which are to be de- termined in precisely the same manner as that used for ther- mal stresses ; in fact, they belong to the same class of stresses. Every kind of material has its own coefficient of linear ex- pansion ; wrought iron, for instance, expanding .12204 units in every 100 for a change of temperature from 32° to 212° F., while tempered steel gives the empirical quantity . 1 2396 for the same conditions. (D. K. Clark, " Rules, Tables, and Data.") The arched rib shown in PI. V., and already treated for ordinary stresses, will be supposed to be of such a material and to be subjected to such a change of temperature that the point A will suffer a vertical deflection of 3 inches. It is a matter of indifference in which direction the deflection takes place ; it will be supposed upward in the present case. The effect of the thermal variation is to cause bending moments at the various sections of the rib from which the deflection results. Also, to keep the ends in their original positions requires the existence of a horizontal force, such as the stress which may be supposed to exist in a horizontal tie. The stresses and bending, then, caused by thermal variations are the same as those which would be caused by a horizontal force having the proper line of action ; the problem then resolves itself into finding the proper value and line of action of this horizontal force. The figure to be used, and which will be referred to, is that of PI. VII., and represents the same rib precisely as PI. \'^. For the sake of greater accuracy, ;/ will be taken half as great as in Pis. V. and VI. ; its value will then be 5.47 feet. 252 ARCHED RIBS. It will first be assumed that the points b, b^, b^, etc., are at a uniform horizontal distance apart of 5 feet, and at the same time equidistant on the curve ; that which might be allowable in a very flat curve. The conditions '2^ nM = o and 2^?iMx — o show H" K" to be the line of action of the horizontal force, which will be called T^. The line H" K" is the same line as N"K" in PI. v., since it is located by exactly the same condition, the vertical intercepts between it and the curve BAD represent- ing the moments M. Since the only force acting on the rib is 7)i, the bending moment M will be equal to T,, multiplied by the proper vertical intercept. Thus the moment at b, will be T^ X a^b/, and that at bg, T^ x a^bg. The value of Tf^ is determined by either of the conditions 2^n3fy=nf,E/, or 2^^nMx=I)£/; by way of variety the lat- ter will be taken. Assume any point as C for the pole, and make ^(T the pole distance. Makei5i equal to i^^^g; i —2 equal to agbg ; 2 — 3 equal to a^b^ ; . . . . and 9— 10 equal to ab, and construct the polygon Caa" . . . . a^^ in the usual way. B — 10 is one half //'b', and a^N' is a vertical line through //, while Bb' is one quarter of Bb. Draw the horizontal line a^E, then the vertical intercepts included between a^£ and the deflec- tion polygon a^d^C, of the kind iaJi), represents in an exag- gerated manner the deflections (vertical) of the points in the rib vertically above them. Now since the actual moment in an equilibrium polygon is equal to the vertical ordinate multiplied by the pole distance, the actual vertical deflection at A is CE multiplied by its pole distance ; however, since the real deflections are proportional to the vertical ordinates of the kind [aJi), those deflections may be at once -found by multiplying those ordinates by a proper ratio. That ratio is a known quantity, because the real deflection at A is known to be three inches. The ordi- nate CE measures 1.61 inches on the drawing, and represents 16. 1 feet on the actual rib; the ratio desired is therefore ^ = — - — = ^ . To illustrate, the deflection at the 12 X 16.1 193.2 64.4 THERMAL STRESSES IN ARCHED RIB. 253 ^section / is 13.9 -^- 64.4 = 0.2 16 feet. Similarly, the deflection at ^3 is 6.2 ^ 64.4 = 0.096 feet. These quantities will be used farther on. It will now be necessary to give a little consideration to the general equation '^hjSIx — DEI. If the ordinates of the kind {ah), measured from H" K'\ are denoted by y\ M, from what has already been said, can be written as T^^y . Hence the general equation may be written, 2nMx = T^^ny'x = DEI. Or, since n is constant, n 2j/ x ' This last is the equation from which 7\ is to be found. Now since j' is positive or negative according as it is meas- ured on one side or other of the line H" K" , and since n is assumed to be uniform and horizontal, the quantity n'^y x is the difference between the statical moments of the moment areas on the different sides of that line in reference to the sec- tion considered. Written as an integral expression, it would be fy'xdx. The area of the surface Ab(,ina^ is 1.12 sq. in., or 112 sq. ft. full size. The distance of the centre of gravity of the same area from AC is very simply found by construction. Take any point «, for a pole, and a-jm for the pole distance. Make rniH]^ =i?im^= ^a-ib-i, nixiUi equal to a(,b^, and so on, making, however, nif^nii equal to a half of Aa^,. Then construct the equilibrium polygon r, e-^, e^, . . . . c-. The sides ee^ and e^e-f produced will cut each other in the vertical line GH passing through the centre of gravity of the area Ama^. The area BH in is to be treated in precisely the same man- ner, taking a^ as pole, and a^H ' as pole distance, and making Bn equal to a half of BH" . The vertical line KL passing through the centre of gravity of the area is found as before at the intersection of the sides ^i and/s/g prolonged. The area BH'ni is, of course, equal to the area Ama^, con- 2 54 ARCHED RIBS. sequently the value of 7i'2, y'x for the section A will be the product of the common area by the horizontal distance be- tween their centres of gravity, i. e., 3.26 inches in the figure, but 32.6 ft. full size. The cross-section of the rib will be assumed to be of such form that £/has the value of 2,000,000 foot-tons. Hence for the point .'^,Z>= 3 inches or 0.25 ft., 7t'2y'x = 112x32.6 = 3651.2; ^ 2000000 X 0.25 •■• ^A= z ^ = 137 tons. 3651.2 -^^ For any other section, as /, the deflection is 13.9 -^ 64.4 == 0.216 ft. The vertical line passing through the centre of gravity of the area included between ;// and a vertical line through / passes through the intersection of the sides e^a-j and ec^, prolonged, and the distance between KL and it is, as shown, 3.04 inches in the drawing. For this section n2yx = 1 12 X 30.4 — 49.82 X 5.8 = 31 16; ™ 2000000 X 0.216 •■■ ^•= ^776 ='39 tons. In precisely the same manner, for the point h^, ™ 2000000 X 0.C96 7fc = ^— = nS tons. 1421 -^^ The values should have been the same, except for the errors incident to a small scale and the fact that polygons were used where curves really belonged ; yet the difference between the extreme values is only 2^ per cent, of the larger, which is not very much of an error. If, however, the true value of n (5.47 feet, nearly) be taken along the centre line of the rib, a decidedly different result will be found, since n'2 y'x will have a different value. The quantity x, as before, will be measured from a^ to the intersections of the dotted lines drawn through the points {b), while j' will represent any vertical ordinate (belonging to any THERMAL STRESSES IN ARCHED RIB. 25c point {bf) from the line H" K", taken positive downward. The point m will be assumed midway between a^ and a^ . The following values are measured from the original draw- ing : = — 6.2 I {xy) = — 4-22 = -6.1 xy' = - 33-25 = — 5.5 xy' = — 60.00 = - 4.6 xy' = - 74.52 = - 3.2 xy' = - 69.44 = - 1-35 ^J = - 36.18 x= 1.36 j' x= 5.45 7 .r = 10.9 y x= 16.2 y jr = 21.7 y X = 26.8 y - - 277.61 X = 32.0 y= 0.7 ;rj/ = 22.4 X = 36.8 y= 3-2 ;rj' = 117.76 X = 41.6 j' = 6.0 xy = 249.6 A- = 46.0 y = 9.2 .rjj/ = 423.2 X — 49.0 j' = 11.98 n^j) = 293-5 1084.06 The first values of x and y' belong to that portion of the moment area adjacent to Aac,. The last values of x and y' belong to that portion of the moment area adjacent to BH ' \ and half of each product is taken, so that it may be multiplied by the full value of n = 5.47 feet. The formula then gives: „ DEI 0.25 X 2000000 ^ , 1 s « 7\ = — ^-^— - = = I n-3 tons (nearly).* y X 4411.30 * The method by deflection polygon given in the next Article, produces a re- sult essentially the same as the one above. £C is the pole distance laid down to a scale of 400,000 foot-tons to the inch. n = 4 X 16. 1 X 5.47 0.62 X 400000 6^2 — 113.6 tons. The agreement is much closer than can ordinarily be expected with the scales used. 256 ARCHED RIBS. The difference between the results of the two methods is 137.00 - 113-3 = 237 tons. This difference is by no means small, and shows how care- fully the approximate method ought to be used. With a wrought-iron rib, the deflection taken, 3 inches at the middle of the span, belongs to a change of temperature of about 165° F., a very extreme case, which accounts for the large values of Tj^. This shows, however, in a very marked manner, the impor- tance of putting together an arched rib at about the mean temperature, for then the variation of temperature to be taken in the calculation of thermal stresses will only be about half the variation between the extreme limits. The effect of T^ is the same as if that force were applied at m and acting toward H" for the portion Bm fixed at B, but applied at m and acting toward vi for the free-end portion (so considered) mA. The change of temperature, 165°, changes the radius from 75 feet to 74.536 feet, and increases the length of the curve 0.122 feet. Art. 51. — Thermal Stresses in the Arched Rib with Ends Free. The method to be used in the present case is somewhat shorter and simpler than the one used in the preceding, but will probably not give as nearly correct results when the scale used is small. The figure to be used is that shown in PI. VIII., and the curve BAD is precisely the same as that shown in Pis. V., VI., and VII. For the sake of greater accuracy, the curve BA will be di- vided into ten equal parts. There will then result, n = bby = bxb% = etc., = 5.47 (nearly) feet. Since BD is the true position of the closing line for free ends, the effect of the variation in the temperature will be the same as that of a horizontal stress, T^ , whose line of action is BD, and which will produce a deflection equal to the thermal deflection. What maybe called the "thermal THERMAL STRESSES TV ARCHED RIB. 257 moment," therefore, at any point will be equal to 7\ multi- plied by the vertical ordinate of the curve at that point, or M— Tf^y'. The moment at b^, for instance, is M = T^ >^b^n^. Supposing the rib to be of wrought iron, a change of tem- perature of 137° F. will cause the length to change from 109.454 feet to 109.556 feet, and the radius from 75 feet to 74.67 feet ; the corresponding upward vertical deflection at crown A will be i.^ inches, or ^ of a foot. Let the two general equations be compared : 2Px = T^y. 2nMx = 2M'x = £1.1). From these two equations it is seen that if M' be taken as vertical loading, and £1 as pole distance, then the ordinates of the resulting equilibrium polygon will represent the deilec- tions according to the same scale by whicJi x is measured. It is important also to notice that M' and £1 are of the same de- nomination (foot-pounds or foot-tons, as the case may be), consequently M' is to be measured in the same scale as that ac- cording to which £1 is laid dozvn. Since nM = n Tj,y' {n being constant), the vertical ordinates of the kind ((^«) are proportional to the moments J/ or J/', and they may be taken to represent those moments ; since, however, that would carry the lower extremity of the load line B — 10 off the diagram, one third only of those ordinates will be taken in the plate. As before, £/ will be assumed to be 2,000,000 foot-tons, and the scale according to which it is to be laid down for the pole distance at 200,000 foot-tons to the inch. Hence make £7//^ parallel and equal to BCD. Since BD is ten inches in the diagram, ££ is the pole distance, and £ will be taken as the pole. As in the case of the same rib with external loading, the half intercept at the distance - from B will be neglected. 4 Make £1 equal to ^{\AC), i-2 equal to ^b^ns, etc., and construct the polygon a a^ai //" in the usual man- '7 258 ARCHED RIBS. ner. BA is divided into ten equal parts for the sake of greater accuracy. The polygon thus constructed will represent the actual de- flections to a scale of lo feet to the inch. Now Ea is equal to 1.12 (3.32 inches on the original draw- ing) inches, or 11.2 feet full size, whereas it ought to be but 0.125 foot ; and since the pole distance is to remain the same, the moments must be reduced in the ratio of \ to 13.2 ; or the moments as actually taken must be multiplied by - — = ^ ^ 8 X II. 2 ^. The reduction for the bending moment at ^-J, there- r . , . ^ „ ^ 0.636666 X 200000 . ^ fore, IS \ AC y. 200000 -=- 8q.6 = — ^ ^r — ^ , smce AL "* ^ 89.6 is equal to 1.91 inches. Hence, „, M' 0.636666x200000 ^ ,^ Tn = — - = — ^-TT^ = 7446 tons. * y 89.6x19.1 But since Af = ?zM: Ti, = — — — - = — = 13.6 tons. y ny n This operation may be considerably shortened by remem- bering that 0.636666 -^ 19.1 = I -^ 30, and that this ratio is constant for all points. If, therefore, the moment at any other point, as b^, be taken, precisely the same result will be obtained. The method used in the preceding Article gives, at least approximately, the same result. Taking Bn, Bn^, etc., for the different values of x, and bn, b^n^, etc., for r, there will result : X = 4.0 feet, y' = 3.7 feet, .-. xy' = 14.8 X = 8.4 " y' = 6.8 " .-. xy' — 49.6 X = 13.2 " y — 9-5 '' •'■ 'V = 1 01.6 x= 18.0 " y = 12.0 " .-. xy' = 183.6 ARCHED KfBS WITH FIXED ENDS. 259 X = 23.1 feet, }>' = 14.2 feet, X = 28.2 " y = 16.0 " x= 33.6 " y = 17.4 " X = 39.0 " y = 18.3 " X — 44.4 " y = 18.9 " ^y = 307-2 xy =z 425.8 -I'j' = 554-4 xy = 643.5 xy = 799.2 With these values the partial summation is : 2:^iyx = n^j'x = 16846.00. A product for the moments adjacent to B is nearly : 0.92 X I X 2.73 = 2.51. Another for those adjacent to AC is nearly : 19.1 X 48.64 X 2.73 = 2536.5. Taking the sum of these results for the complete summa- tion : „ DEI 0.125 X 2000000 7* = s:^ — - = 5 =13.0 tons (near y). " 2.nrx 19385.00 -^ \ Jj The polygon a, a^, a^^, a^ // is an exaggerated repre- sentation of the movements of the points b, b^, b^, etc., when the temperature is changed 137° F. In all cases, as large a diagram as possible must be used, in order to reduce the scale for the pole distance, so that that distance may be the largest possible. The use to be made of T^ will be shown farther on. Art. 52. — Arched Rib with Fixed Ends — / and n Variable. The rib to be taken in this case, and the loading, are pre- cisely the same as taken in the preceding cases. As before, the centre line is divided into ten equal parts of 10.95 feet each. The data are therefore the following : Span = 100.00 feet. Radius = 75.00 " Panel fixed load = 4.00 tons. Panel moving load = 10.00 " Centre rise of rib = 19.10 feet. 26o ARCHED RIBS.^ The moving load is supposed to cover the half span BC. The figure to be referred to is that shown, in Pi. IX. The point h is midway between b' and b'\ while k is mid-^ way between b" and b" . The moment of inertia /of the cross-section of the rib will be taken as 2,000,000 foot-tons throughout Bh, 1,777,778 foot- tons throughout hk, and 1,600,000 foot-tons throughout kA. The same values hold for similar portions of AD. I, 2, 3, etc., .... 10, is the load line, and ^C the pole distance of the equilibrium polygon Ea^ Ca-fF. That polygon is drawn in precisely the same manner, in fact, is precisely the same one as that shown in PI. V. ; the radiating lines drawn from C are, therefore, omitted. As the ends are fixed, EF cannot B nM be the closing line; but the condition ^ — — = o must first be imposed. It has already been shown that for any point in the arched rib the moment M = M^ — M^. Hence if /j be the moment of inertia for the portion 2Akoi the rib, there may be written : For Bh, i has the value j = 0.8 ; for M, the value ^ = 0.9 ; and ior Ak, the value - = i.o. -'I The closing line HK must be so located that : But for the equilibrium polygon Ea-^a^, etc., the summation '2 ^ niMa has the value : 2 j^ niMa = O.Sn' x vv' + o.8n x a^c' + O.gn x rto"'l/' series by writing — and instead of the whole quantities 22 themselves. If vv be taken at 1.04 inches and v"'v" at 0.66 inches, the above summation (after dropping 11 in the manner shown) gives a result of — 0.02 of an inch only. This agreement is sufficiently close. The vertical intercepts and their products by /, are the fol- lowing : + 0.8 X — = + 0.8 X --^ = + 0.42 2 2 + 0.8 X a^c' = + 0.8 X 0.56 = + 0.45 + 0.9 X a./' = + 0.9 X 0.06 = + 0.05 — I X a^c'" = — I X 0.30 = — 0.30 — I X a^c^" = — I X 0.50 = — 0.50 — I X Cc"' = — I X 0.46 = — 0.46 — I X a^c"^ = — I X 0.31 = — 0.31 — I X rt-^r^' = — I X 0. 10 = — 0. 10 + 0.9 X ^gC^'^zz: + 0.9 X 0.15 = +0.13 + 0.8 X rtgc^ = + 0.8 X 0.42 = + 0.34 o V'v"' _ 0.66 ^ + 0.8 X = + 0.8 X = + 0.26 2 2 The other condition for the closing line HK'\s 2 nMatx = O. 262 ARCHED RIBS. Taking the products of the ordinates (nMi) of different signs, by their horizontal distances from DK", as was done in Art. 48, there will result the sums + 91.8 and —90.5. The alge- braic sum is only + 1.3, which is near enough to zero. HK will therefore be taken as the proper closing line. The condition which locates the closing line H"K'' is similar to that which placed HK\ it is the following: 2 tiiMi, = 0.8 X 2n' y. H'b -\- 0.8 x 2n x c^b' + 0.9 x 2n Y-c^b' — 2n X b"'c^ — 2n x b^c^ — fi x Acr, = o. Since the curve BAD is symmetrical in reference to AC, the line H"K" will evidently be horizontal. For the same reason, 2n' and 2n are written in the summation in all its terms except the last. As before, Jt is one half «, and by so writing it, n may be dropped from the series. By making Ac^ — 0.58 inch, the summation (after dropping n) gives 2.18 — 2.16 = 0.02 only; H"K" will therefore be assumed to be the proper closing line for BAD. The vertical intercepts and their products by i are the fol- lowing : o bH' o 1. 1 3 2 X 0.8 X = 2 X 0.8 X — 0.90 2 2 2 X 0.8 X /;>! = 2 X 0.8 X 0.66 = 1.06 2 X 0.9 X b"c.i — 2 X 0.9 X 0.12 — 0.22 2 X b"'c^ = 2 X 0.27 = 0.54 2 X //^r4 = 2 X 0.52 = 1.04 * Ac5= 0.58 = 0.58 As before, since the curve BAD is symmetrical in reference to A, the two conditions -^^/z/J/^ = o, and 2: j^niM^x = o, are equivalent. The equation expressing the condition that the horizontal deflection of D in reference to B is nothing, is the general one already given : ARCHED RIB WITH FIXED EA'DS. 263 ^B nMy _ Hi Or, as in preceding cases : KriM^y = 2lrtM,y. The quantity n^ is any standard value of n, just as /j is a standard value of /, and r is such a variable ratio that for any n section ^ — r/i, or r = — . In the present case n^ will have the value 10.95 feet ; consequently r will be unity for all sections except d, and for that one it will be ^. The ratio r might have been used in the previous summa- tions of the present Article in exactly the same manner and with exactly the same values as in the present one. The ratio t has the same value as before. The principles on which the remaining constructions are based are precisely the same as those shown in the preceding Articles ; the difference in the construction itself is simply this, that riM^ and rtil/h are taken instead of J/„ and Mf,. In other words, r and /, in general, in this case, have values dif- ferent from unity, while in the preceding cases r = i — i. The following are the values of {riM^ : h. X 0.8 vv + v"v") = At', X 0.8 ttxC + ^9'-''') = tt' \ X 0.9 V' -a^^''')^ ft" ; X [ - a^c" — a-f^'''') — - /"/'" ; X I I X I X CV == - ft'" ; Draw radial lines from C to the points / ; then draw the horizontal lines bC" , b'dxd', b"dx'd\ etc. C"d^ is parallel to Ct\ d^di' is parallel to Ct \ dx"d^" is parallel to 67", etc. CC'd-^d-l'dx'dx'dx is then the deflection polygon for the mo- ments M^. 264 ARCHED RIBS. The following are the values for (riM^ : ^ X 0.8 X 2H'b = As; I X 0.8 X 2b' c^ — ss' ; I X 0.9 X 2b" C2 — ss" ; IX I X (- 2b"'c^= — s"s"' ; IX IX (—2/^%)= — ^"j-i^; IX IX {- Ac^) = - j*M. The point A belongs to the curve BAD. The sides of the deflection polygon C"d'd"d"'d^''d^ are parallel to radiating lines drawn from C to the points s. Ad^^ represents ^^j^riM^y, and Ad^ represents ^^j^riM^y. Since the first is less than the second, the moments M^ must be increased in the ratio of Ad-^ to Ad". Hence on AC pro- longed, make CM equal to Ad-^" ; draw T^/iV- parallel to BD and with a radius 67V equal to Ad" find the point N. Pro- long the line 67V; this line will enable the true moments Ma to be determined in the manner already shown in the other cases. Draw c"p parallel to BD, c^a" equal to Cp will give a point d' in the true equilibrium polygon. Again, Ck " equals KF, hence K"a^, equal to Ck'", gives the true point a^. Also, 6^ is equal to HE ; and H"a, equal to Cp' , gives the true point a. All points in the true polygon might be thus determined, but it is advisable to check by the other method already shown. For this purpose draw CL parallel to HK, and LC parallel to BD. Then make CP equal to BC and draw OP parallel to BD. Take LC equal to CO. C is the pole and LC the pole distance of the true equilibrium polygon. Finally draw radiating lines from C to the load points i, 2, 3, 4, 5, 6, etc. Starting from any point already determined, as a, draw ad parallel to C\o; dd' parallel to C'(^; d'd' parallel to C'?>, etc. The different points found by the two methods ought to coincide. The polygon add'd"a^"a"d'^a'''^^d'^^^d^a^ is the true equili- brium polygon, which was to be found. ARCHED RIB WITH FIXED ENDS. 265 C L is 3.86 inches ; hence 7"^, whose line of action is H" K" is equal to 3.86 x 15 = 57.90 tons. In the determination of the thermal stresses the same fig- ures will be used, and there will be supposed such a change of temperature that the point A will suffer a vertical deflec- tion of 1.5 inches or 0.125 of a foot ; the same, in fact, as was supposed in a previous case. As the ends of the rib are fixed, the general conditions, 2j^riM = o; and 2^ri3fj/=2^riAfx = o hold as well for thermal stresses as others. Consequently H" K" will be the line of action of the horizontal stress 7^, induced by the variation of temperature. As before, let y denote the vertical ordinate of any point in the curve BAD from the line H"K", then there may be written : ^1 >r,^ •7,7- J^ ^h^h ^A ■ , DEh Tn = ^A ■ , Hi^. ^rty X In order to save confusion in the figure, ?Zi will be taken at its previous value 10.95 feet. Also, EI^ = 1600000 foot-tons ; D = 0.125 foot. A half of the moment at A will be supposed applied at a point e distant —^ from A on the curve BAB, and none at 4 all at A. The co-ordinate x will be measured from c^ towards K'\ The following values then result: X = 2.72 ft. riy' = ^ X I X 5.7 = 2.85 ri'y'x = 7.752 x= 10.90 " riy' = I X I x 5.2 = 5.2 riy'x = 56.68 X = 21.7 " rty' =1x1 X 2.7 = 2.7 riy'x = 58.59 123.022 266 ARCHED RIBS. x= 3.2 ft. x = 4i.6 " x=4y.g " Hence, r?jj''= — I X0.9 X 1.2 = — 1.08 r{y= — ixo.Sx 6.6 = — 5.28 r«y=:— ^xo.8x 1 1.3 =—4.52 rty,Y = - 34.56 rz)/> = — 2 1 9.648 rtyx = — 206. ^oS —460.716 ^^rz/x = 123.022 — 460.716 = — 337.694. The negative sign will be dropped hereafter, as it refers simply to the direction in which j was measured. Making the substitutions : T, 0.125 X 1600000 10.95 X 337-694 54.1 tons. The method by the deflection polygon gives nearly the same result, as will now be shown. Comparing the two equations : 2Rv= T^.y, 2niriMx — EI-^ . D, it is seen that if £/i be taken as the pole distance in the deflection polygon, UiriAI must be the general expression for the load at any point. Since EI^ is i,6oo,ooo foot-tons, CD will represent it at 320,000 foot-tons per inch ; and that will be taken as the pole distance. Z> — 3, measured downwards, will be the load line. If the loads were taken at 10.95 x riM, or lo.g^riy', the lower limit, 3, of the load line would not be on the diagram. The loads will therefore be taken as ^riy' and a proper reduc- tion will be made afterwards. Hence, make D- I = 4 ''('' ' I — 2 — 4 riy' 2 -3 = 4 riy' 3 -4 = - 4 W' ' 4 - 5 = - 4 fiy ' 5 -D = - 4 riy ' 4 X 0.285 inches. 4 X 0.52 4 X 0.27 — 4 X 0.108 — 4 X 0.528 — 4 X 0.452 STRESSES IX THE MEMBERS. 26 J The values of riy' are taken from the tabic immediately above. By drawing radial lines from C to the points i, 2, 3, 4, 5, the polygon Cd-^d,,d'ididr,d^d'i is formed in the usual manner. The deflection Dd-i measures 2.67 inches or 26.7 feet, full size. Since — = 2.74, the deflection with the true moment- 4 loads would be 26.7 x 2.74 = 73.16 feet, whereas it should be but one-eighth of a foot. Hence, measured by the same scale, the quantities ;/ir/J/must be ^ of those taken in the ^ 8 X 73.16 figure, the pole distance CD remaining the same. Since M — T^y', there may be written for the point A : 10.95 X riT^y' = 10.95 X 0.285 x 320000-^ 8 x 73.16. As r = ^, i = I and J'' = 5.7 feet there results: ^k = 547 tons. The deflection at other points might be used in the man- ner already shown in a preceding Article. It is thus seen that the constructions are equally simple in principle whether r and / are constant or variable. If the ends had not been fixed, it would only have been necessary to use the condition -^^riMy^O. Art. 53. — Determination of Stresses in the Members of an Arched Rib — Example — Fixed Ends — Consideration of Details. It has been shown, in the preceding Articles, how to deter- mine the horizontal tension 7\ in the various cases which may arise ; the method of using this horizontal tension in the determination of the stresses in the individual members of an arched rib remains to be shown. For this purpose there will be taken the rib shown in PI. 268 ARCHED RIBS. X., Fig. 3, having ends free, /. e., free to turn about the points J/ and L. The curve which has hitherto been used, and called the '■ centre line " of the rib, is the centre line of the neutral sur- face of the arched rib considered as a beam ; consequently the centres of gravity of the various cross sections of the rib must be found in this " centre line," m all cases. In other words, the " centre line " which has been used in the preceding articles is the locus of the centres of gravity of the normal cross sec- tions of the actual rib. In the rib taken as the example, PI. X., Fig. 3, the apices in the upper and lower chords lie in the concentric circum- ferences of circles having radii of 78 and 72 feet, respectively ; and the centres of gravity of the normal cross sections will be supposed to lie on the circumference of a circle having the same centre, whose radius is 75 feet. Thus the centre line is precisely the same as has been used in the preceding Articles. The same span and loading will also be taken. The extremities of the span, or points M and L, at which the horizontal tension or force is applied, imist lie in the centre line. All the loading will be assumed to be applied at the apices of the upper chord, although the operations would be exactly the same if the fixed load were divided in any proportion be- tween the two chords. The apices of the triangles of the web system were located as follows : As was done in finding Z^, the centre line was divided into ten equal parts. The upper chord panel points are vertically over those points of division. The upper chord panels were then bisected, and radii were drawn through these points of bisection. The lower chord panel points were taken at the intersections of those radii with the circumference of the circle whose radius was 72 feet. There is only one point to be observed in forming the chord panels, i. e., the panel points must be so located that the load will act exactly as was supposed in determining T^. The following loads will then be assumed to act through the upper chord panel points : ST/C£SS£S IN THE MEMBERS. 269 7 tons at the intersection of i and 3 ; 14 " " " intersections of 4 and 5, 6 and 7, 8 and 9, 10 and 1 1 ; 9 " " " intersection of 12 and 13; 4 " " " intersections of 14 and 15, 16 and 17, 18 and 19, 20 and 21 ; 2 " " " intersection of 22 and 24. As the ends are free, L — \o — 50.7 tons in PI. VI., gives the reaction R at the left end of the span, or at L in the ex- ample. Also Z — I in PI. VI., gives the reaction as 30.3 tons at M in the example. In Art. 49 it is found that Tf^ — 52.5 tons for this case, and the two methods in Art. 51 give the thermal stresses in the horizontal tie as 13.6 and 16.7 tons. The thermal tension will be taken at 15 tons. The total ten- sion in the tie will then be 52.5 + 15.00 = 67.5 tons, as shown. It is a matter of no consequence whether the tie exist or not. If it does not exist, the abutments at M and L must then supply the horizontal force of 67.5 tons. Fig. 4 of PI. X. is the complete diagram for the stresses with the load taken ; it is drawn to a scale of 20 tons to the inch, nearly. The lines indicated by letters or figures in the diagram are parallel to the members of the rib indicated by the same letters or figures, though the parallelism is not ex- actly shown in the plate for all the lines. With the following explanations relating to the diagram, little more is needed : db — b'c = c'd' r^d'f— 14 tons. on = ti m = in k = k Ji =4 " ///■= 9 " q'd = ( l) + 7. pd = (24) + 2. PS ^{£). ae is the reaction R, and c'd is the reaction at M, while 270 ARCHED RIBS. e'P'xs the horizontal tension 67.5 tons. Although the diagram appears very complicated, yet, it is really composed of very simple yf^'^-sided figures, as may easily be seen. Let the rib be divided through c, 7, B, and T; then the portion of the rib between that surface of division and L is held in equilibrium by the action of the stresses in the members divided (considered as forces external to that portion), the applied loads, {T), and the reaction R. The resultant of the loads and i? is a verti- cal shear represented by e'c' in the diagram. The forces act- ing upon the portion of the rib in question are then repre- sented by the lines cc', (T), (B), (7), and (c) in Fig. 4, and these constitute a simple five-sided figure. The arrow heads show the direction of action of these forces, and enable the kifid of stress to be recognized at a glance. The whole diagram is thus composed of just such pen- tagons. As a check on the accuracy of the construction of the diagram, if it is worked continuously from L to J/, the four- sided figure involving (7^), (23) and (24) should exactly close. It is far more conducive to accuracy, however, to work up the diagram from both ends, and if the work has been accu- rately done, the diagrams will give the same stress in that member which becomes common to both where they meet. The stresses, as determined by the original diagram from which Fig. 4 was constructed, are written in Fig. 3. If an arched rib is subjected to a load, advancing panel by panel, the stresses due to the fixed load alone may first be de- termined and then tabulated. The stresses in all the mem- bers of the rib due to each panel moving load may then be found and tabulated also. The greatest stress of either kind in any member may then be determined by a combination of these results in the usual manner. Some of the stresses found by diagram should be checked by moments in the following manner. The horizontal dis- tances of the panel points, in the left half of the rib holding- 14 tons each, from a vertical line bisecting the span and pass- ing through the intersection of 12 and 13, are 10.9 feet, 21.7 feet, 32 feet, and 41.5 feet. The normal distance from sy/^ESSES IN THE MEMBERS. 2/1 the intersection of (12) and (13) to E, is 6.15 feet (by scale). Hence, taking moments about that intersection, / c. 507 X 50 — 14 (10.9 + 21.7+ 32 +41.5) — 67.5 X 22.1 (^J = -^^ = — 72 tons. The diagram gave 72.5 tons, and the agreement is suffi- ciently close. On account of the ill-defined intersections of the prolonged chord sections in any panel, the method of moments for the web stresses is not satisfactory unless one chord stress in the panel is known. The web stress can then be found by mo- ments in a manner to be presently illustrated. Again, let (C)be determined by taking moments about the intersection of 16 and 17. Draw a vertical line through that point. The horizontal distances of the three upper chord panel points on the right of that line, from the same, are 10.3 feet, 19.8 feet, and 28.5 feet (by scale). In the same manner the vertical distance of the point above 7" is 19.25 feet. Hence, 30.3 X 28.5 -4(iO-3 + 19-8) -67-5 X 19-25 _ (^)_ __ _ — 90.5 tons. The diagram gave 95 tons, and the agreement is not close. This illustrates in a marked manner the great fault of the graphical method. In constructing the original diagram, shown by Fig. 4 of PI. X., the rib was drawn to a scale of 5 feet per inch, and the diagram itself to a scale of 10 tons per inch, and although the greatest care was taken, yet the stresses found for the right half of the rib may, in some members, be wrong to the extent of even twenty per cent. The method requires the largest and most accurate figures possible, and the very nicest instruments, for extended diagrams. By far the most accurate, and, all things considered, the most satisfactory method, is the combination of moments and 272 ARCHED RIBS. diagram, so freely used in the treatment of bowstring trusses In this method a stress in either chord is found by moments, the other two stresses (one a chord and the other a web) in the same panel are then immediately found by a simple five- sided figure or diagram. The chord stresses {E) and [G) have just been found. In Fig. I take bd equal to 67.5 tons and parallel to T, and make dc (parallel to E as well as T) equal to (£") = — 72.00 tons. A section is supposed to be taken through the panel in which /", 13 and E are found; consequently the vertical shear .S = 50.7 — 65.00 = — 14.3 tons. The remainder of the diagram needs no explanation. It gives: (f) ^ ah ^ S -14:3. tons dc =(E) \3) ioL^CT) V;r (T) cU 'i Fig. I. (13) = — 20.3 tons ; (/) = + 17.4 tons. Fig. 2 is drawn in precisely the same manner by using \G) = — 90.5 tons, which has already been determined by the moment method. The section is taken through G, 17 and /i. Fig. 2 gives : (17) = — 5.5 tons ; (/i) = + 23.0 tons. All the other stresses may be found in the same manner. The difference in the case of (//) between the results of the ST/i£SS£S IN THE MEMBERS. 273 two methods is four tons, or about 17 per cent, of the smallest result. By the method of Figs, i and 2 accurate results may be obtained by taking a scale of even twenty tons to the inch, but a larger diagram is preferable. If the ends of the rib are fixed, the shortest method of find- ing the stresses is in no way different from that given in con- nection with Figs. I and 2, excepting this : the chord stress which is found by moments will have a different value. If the ends are fixed, the reaction R, at the left end of the span, will be L — \o of PL V. ; and the horizontal tension ( T) tvill be taken as 54.8 + 113.3 = 168. i tons, from Arts. 48 and 50. The line of action of T^ for both external load and thermal stresses is H"K" of either PI. V. or PI. VII.; let PI. V. be considered. The action of T^ through H" (taken as acting toward K"), so far as the rib BAD is concerned, is equivalent to 7"^ aeting through B toward D, combined zvith a right hand couple luhose force is Tft and ivJiose lever arm is BH" . Let the moment ot this couple be called M. This moment will cause compression throughout the upper chord of the rib and tension throughout the lower. Let J/j represent the moment (of the external forces and T^) about any panel point of the rib, as F, Fig. 3 (the reaction and /"ft being taken for the particular case, as just indicated). Then any chord stress, as {BE), will be «i being the normal depth of the rib, as shown. Particular care is to be taken in regard to the signs of J/ and J/i, i. e., it is to be noticed whether they tend to produce the same or dif- ferent kinds of stress in BE. After {BE) is found, the diagrams are to be drawn precisely like Figs, i and 2, and the resulting stresses, scaled from the diagrams, are the ones desired. The following, but longer methods, may also be used • lb 274 ARCHED RIBS. The first portion of the operation is simply the application of the method by diagram, or the combination of moments and diagram, already given in connection with the case of ends free. All the individual stresses in the rib are to be found in this manner ; tJiose for the iveb members are the true web stresses desired if the rib is of uniform normal depth. The chord stresses thus found are, however, in all cases, to be modified. Let the normal depth of the rib at any section be «i ; the distance BH'\ PI. V., h " ; and the general expression for any chord stress due to the moment M — T^h', c. Then will result : _M_T\h^ Then let {c) be the general expression for any chord stress already found without considering the moment M \ the nu- merical value may be either positive or negative. Finally, the resultant chord stress desired will be, for the upper chord, (c) - c=(c)- ^f, and for the lower. If the rib is of uniform normal depth, i.e., if Wj is constant, the web stresses will not be affected by the moment M, for it (the moment M) will cause uniform chord stresses throughout the rib. If the normal depth, however, is not constant, the moment J/ will cause web stresses which may be determined very ac- curately in the following manner ; STRESSES IN THE MEMBERS. 275 Let it be desired to determine the stress in the web mem- ber DF of a portion of an arched rib, shown in Fig 3, and let w denote that stress. The stress in DE is c, found by the method just given. In Fig. 4, take {DE ) =c and parallel to DE in Fig. 3. The lines {GF) and (DF) in Fig. 4, are then parallel to GF amd (jos:) - c rojp) - ^ Fig. 4. DF in Fig. 3, and they are the stresses in those members. All the web stresses and the chord stresses in one chord may be thus found. This operation is simply the method of Figs. I and 2 applied to this case. The web stresses may be found by using moments, only, in the following manner : Take ^ as any convenient point in GF. Let/] represent AB, and 4, AC; these lines are normal to DE and DF re- spectively. Let M be still considered right-handed and positive, and let it first be assumed that c is compression. Moments about A give : _ _ -r/i + T„h' If c is tension, or belongs to a panel in the lower chord, then {w) will be the stress in a member like DG. There will then result : — cli + TJt" w — — — - — . 4 In either of these formulae, (w) will represent tension or compression according as the result is positive or negative. Let {w) represent any web stress already found by neglect- ing M, then will any resultant web stress desired be : {w) + w\ the signs of both these quantities being implicit. 2^6 ARCHED RIBS. As the lever arms /, , A , and Wj are scaled from the draw- ing, the rib should be laid down as accurately and to as large scale as possible. In important cases these different methods should be used as checks. A very common system of bracing for arched ribs, although a very unsatisfactory one, is that shown in Fig. 5. The web Fig. 5. members, bB, cC, dD, etc., are normal to the centre line of the rib, and are designed for tension only. The other web members are for compression only Let the ends be supposed free, and take AOP for the true equilibrium polygon for a given load. For any given loading the stresses in the different members are indeterminate, unless about half of the compression web members are neglected. With the assumed position of the equilibrium polygon AOP, for instance, it is seen that compression will increase in the upper chord from b io d (nearly) ; from that point to g (nearly) it will decrease. The compression in the lower chord will increase from G to L (nearly) and then decrease from that point to N. The points of greatest chord stresses of either kind are those at which the polygon and centre line of rib are parallel. From these considerations it results that the web members bC, cD, De, Ef, Fg, Gh, Hk, Kl, IM, and mN may be omitted ; they must be omitted, in fact, if the stresses are to be deter- minate. Having made these omissions, the stresses are to be found by the methods already given, as those methods. . are perfectly general. STRESSES LV THE MEMBERS. 277 Precisely the same observations apply to the case of fixed ends, or to that in which the normal members are in com- pression and the others in tension. It is by no means certain that the stresses thus found will really exist in the rib, but the assumption is the best that can be made. This system of bracing is, at best, very un- satisfactory. The free ends of an arched rib are sometimes arranged, in regard to support, as shown in Fig. 5, PI. X. There are two ties or sets of ties, T' and T'\ instead of one, 7),. A and B are the points at which these ties take hold of the rib. E is the intersection of AB and the centre line, EF, of the rib. The span to be used in finding T,, for either external loads or thermal variation is the horizontal distance between E and the corresponding point at the other end of the rib. T\ T^, and T" are parallel to each other, and ac is normal to the three. Now if T' and T" are determinate, there may be written : T' --T ' ^ - ac ^'^• In such a case the systems of triangulation in the rib may be separated, T' will belong to one and T" to the other. The stresses may then be found in each system separately, and the results combined for the resultant stresses of the rib. If the web members may be counterbraced, the resultant stresses are thus determinate. It is not certain, however, that the tensions T' and T" will have the values given above. For the determination of the stresses in the rib, however, it is not necessary to resolve Tj^ into T' and 7"", except for the panel ABDC. In the case of a design, if the dimensions required by the calculations of this Article give a value to the moment of inertia / very different from that assumed in the determina- tion of 7"/^, for either thermal variations or external load, it will be necessary to make an entirely new set of calculations- 2y8 ARCHED RIBS. with another value of /. This must be done until the agree- ment between the assumed and required values of / is suf- ficiently close. Art. 54. — Arched Rib Free at Ends and Jointed at the Crown, Suppose the rib to be represented in the figure. Since there is a joint at A, the bending moments must be zero at that point ; consequently the equilibrium polygon for any load must pass through that point. This fact furnishes a Fig. 6. very simple method of determining 7^. Denote by "2 Px the moment of all the external forces about the joint A ; then, since J/ must be equal to zero, ^Px- T^{AC) T,= ^Px AC In this rib variations of temperature produce no variations of stress in BD, except that due to the slight change of A C, as shown by the formula above. This, however, is a very small quantity, and would ordinarily be neglected. If neces- sary, it would be allowed for by taking the value oi ACat the lowest temperature to which the rib would be subjected. Other arched ribs are seldom constructed, but they are to be treated by the same general methods, precisely, as those used in the preceding cases. CHAPTER IX. SUSPENSION BRIDGES. Art. 55. — Curve of Cable for Uniform Load per Unit of Span — Suspen« sion Rods Vertical — Heights of Towers, Equal or Unequal — Gen- eralization. Fig. I. In the figure, let EH' C represent the cable of a suspension bridge carrying a load extending over the whole span. In the ordinary experience of an engineer, the load carried by a suspension bridge cable is nearly uniform in intensity in reference to a horizontal line ; so nearly uniform per foot of span, in fact, that it is assumed to be exactly so, and such an assumption will be made in the present instance. The use of the stiffening truss, to be presently noticed, makes this assumption essentially true. Let {ED + BC) = I = span ; BH' = h^ ; DH' = Jh \ w = load for horizontal foot, and let x be measured hori- 279 28o suspensiojv bridges. zontally from H', the lowest point of the cable. The height of the highest tower is, of course, h^, and that of the other //j. The ordinate of any point P is x, the load on H'M is, con- sequently, W= wx. Draw PK tangent to the curve at P, then by the first principles of statics, it is known that the direction of the cable tensions at /"and H' and the direction of W must intersect in one point N. Since, however, %v is uniform along x, the resultant direction of W passes through N, half way between H' and M. Hence FH' = H'K; or, since FK \s, the snhidingent, the abscissa, FH', of the curve is equal to half the subtangent, consequently the curve is the or- dinary parabola. Again, it is known that the horizontal component of the' tension of a cable will be a constant quantity if the loading (as in the present case) be wholly vertical ; let that compo- nent be denoted by H. Let GNP be taken for the triangle (right angled) of forces at P, in which NP represents the cable tension at P, GN the load, W^ wx and C/'the constant horizontal component H. Then let AP be drawn normal to the curve at P\ the triangles AFP and GNPv^WX be similar. There can now be at once written the relation : AF FP x I GP~ GN wx w ' H .' TT . AF — — = constant w but GP = H .'. 747^=— = constant . . (i). w Now AF is the subnormal of the curve of the cable, and since it is constant, the curve is the ordinary parabola. The preceding results may be generalized in a very simple and easy manner. If any two points, as P and Q, be considered fixed, and if the portion PQ of the cable carry the same intensity of load w as before, there will at once result the general case of a flexible cable carrying a load whose intensity, along a straight line, and direction are uniform. There may then be stated PARAMETER OF CURVE. 28 1 the general principle : If a perfectly flexible cable carry a load 2iniforin in directio?i and intensity in reference to a straight lifie, the cable will assume the form of an ordinary parabola whose axis zvill be parallel to the direction of the loading. This principle finds its application in the case of a suspen- sion bridge with inclined, but parallel, suspension rods. Art. 56. — Parameter of Curve — Distance of Lowest Point of Cable from either Extremity of Span — Inclination of Cable at any Point. Attending to the figure and notation of the previous Ar- ticle, the equation of the curve, the origin of co-ordinates being taken at //', is: x^ — 2py ; in which 2/ is the parameter. Let BC = Xi and ED — Xo, then there may be written Xi^ = 2phi, x-i = 2ph2 and ix^x.^ = ^p\/h^h<^. Hence {x^^x^f=l^=2p{Vh',+Vh:,f=2p{h, + 2VT,h^ + h.;) . . . (I). .'. P = ^ = = ==:= . . (2). 2 ( Vhi + V/h f 2(//i -I- 2 V/h fh + /^2) If the towers are of the same height, then //j — li^ = h and ^=£ (3). Now Xx is the horizontal distance from the lowest point of the cable to that end of the span at which //^ is found, /. e., BC in the figure, while x^ is the other segment of the span, and by the equations immediately preceding : ^1= /T-^ /7- (4)- 232 SUSPENSION BRIDGES. / If //i = //2 ; -n = jTa = - • • . . (6). Referring to the figure, since KH' = H'F = y, if i is the incHnation to a horizontal Hne, of the curve at any point Py. then X tan i = 2y ; 21/ . . / 4y hence, tan i — -^ .'. sec i — \/ i + -~V At the summits of the towers t/' (7). tan i\ = ~—^ and tan i^ = — - .... (8). Xi x^ \i hi = /^2, ta)i ii = tan 1^= . . . (9). Art. 57. — Resultant Tension at any Point of the Cable. In the first Article of this Chapter there was recognized the general principle that if the loading on a cable is uniform in direction, the component of cable tension normal to that direction will be constant at all points of the cable. In the present case the resultant tension at the lowest point of the cable will be this constant component H. From Eq. i, Art. 55, //= zuAF. But AF is the subnormal of the curve, and, from Analytical Geometry, it is known to be equal to one half of the parameter, or equal to/ (using the same notation as before). Hence, after taking the value of p from the previous Article: H=wp = -.= 7^— = , • (0- 2 ( V //i + V //2 f 2 (h■^ + 2 V //i /t^ + A,) LENGTH OF CURVE. 283 Let R denote the resultant tension at any point, then by the triangle of forces GNP, in the figure : R= H seci= HA/ 1 + ^ . . . . (2). Eq. (2) gives the tension at any point. At the summits of the towers there arc found : ^i = //j/i + ^i- (3). /' R, = Hi/i^^ (4). If hi = //j, consequently Xx = X2 = — , then : H='il,R,^R, = H^,^'-^ . . . (5). S& Art. 58. — Length of Curve between Vertex and any Point whose Co- ordinates are x and y, or at which the Inclination to a Horizontal Line is /. The usual expression for the length of a part of one branch of a parabola, beginning at the vertex, as determined by the integral calculus, may easily be put in the following form, denoting by c the length in question : x" c = A}' :,l?'(/-^-Hyp.,Og.(^/y,.^)|(.). Or, using the values for tan i, sec i, and /, determined in the preceding Articles : c = -<- \tan i sec i 4- hyp. log. {tan i -\- sec i)\ . . . (2). 284 SUSPENSION BRIDGES. The total length of the cable will, of course, be found by putting x\ and hy for x and y in Eq. (i), or Zj for i in Eq. (2) ; then X2 and fi^ for x and j/, or ^2 for /, and adding the results. Denoting those results by fj and c^ the total length will then be : Ci + c>. An approximate formula sometimes used is determined as follows. In the figure of the first Article of this chapter con- sider H'Pto be the arc of a circle, and let x and y be taken as heretofore ; also let R be the radius of the circle. The ordinary expression for the length of a circular arc, in the integral calculus, is : (nearly), if X is small compared with R. Again, performing the divi- sion indicated and omitting all terms in the quotient after the second, there will result : / ^-(i+2^'^-^) =^0 +6^,^ • • (3^- If j/^ be omitted in the expression x^ = 2Ry — jj^, and the resulting value of R be inserted in Eq. (3), there will at once be found : 2y^\ As before, to find the total length of the curve, x^ and /i^, and X2 and /12 must be inserted in succession in Eq. (4), and the results added. If the heights of the towers are equal to each other and to k, the total length will be <..f) ... ...(s). DEFLECTION OF CABLE. 28; The expressions (4) and (5) are evidently not close approxi- Tnations except for very flat curves, in which case the nature of the curve is a matter of indifference. Art. 59. — Deflection of Cable for Cheinge in Length, the Span Remaining the Same. The approximate formula (4) of the preceding Article is usually used in determining the deflection. The total length of the cable is : Ci+ C2 = x\ 4- x^ +-[— + -r ) , Differentiating : 'i — y^_ = h, there results : 2/2 ( 4^ "^-xWi { ^|ATLf^.h3,p.,o,(^./77^)[ -2^1 * (7)- In Eq. (7) /z is then to be found by trial, as before, if v is given ; or if Ji is given, v at once results. The deflection of the middle point of the truss will be : // - //i. It is to be noticed that in Eqs. (5) and (7) all the quantities, ji, j'2, and //, increase in the same direction with v. This ma- terially simplifies the approximation by trial. The determination of v in Eq. (5) might be made without assuming ^1 and x^ to remain constant, for there are two other equations of condition : .f 1 + X2 ^= /, and These, with Eqs. (5) and (6) would be sufficient in order to find the four unknown quantities, j'^, j.^, Xi and ^2 . Such a degree of extreme accuracy, however, is unneces- sary. Art. 60. — Suspension Canti-Levers. In the figure, ABD represents a suspension canti-lever. The cable BC goes over either to another span or to an anchorage, while A is the end of the canti-lever. The cable AB is in precisely the same condition as the half of a cable belonging 288 SUSPENSION BRIDGES. to a span equal to 2AD; consequently its tension R at any point and its inclination at the same point are to be found by the formulas already given. In fact, all the circumstances are pre- cisely the same except this, the platform is subjected to a thrust, uniform throughout its whole length, and equal to the constant hori- FlG. I. zontal component, //, of the tension R. Art. 61. — Suspension Bridge with Inclined Suspension Rods — Inclination of Cable to a Horizontal Line — Cable Tension — Direct Stress on Plat- form — Length of Cable. In this case the suspension rods, or suspenders, are all sup- posed to be equally inclined to a vertical or horizontal line, and, consequently, are parallel to each other ; they are also supposed to take hold of the platform or stiffening truss at points equidistant from each other. These conditions cause the cable to be subjected, in each of its parts, to the action of parallel loading of uniform intensity in reference to the span. As was shown in the first Article of this Chapter, the curve of the cable will be composed of common parabolas having axes parallel to the suspension rods. In the figure, A is the lowest point of the cable, while BD and FG represent suspension rods on either side of A. The angle a is the common inclination of all suspenders to a ver- IXCLIXED SUSPENSION RODS. 289 tical line, it also represents the inclination of the axis of either of the parabolas AC ox AE to the same line. The vertex of the parabola y^C is on the left of A, and the vertex of AE is on the right of the same point. Let X be horizontal in direction and measured from J, and let the length of any suspender, as BD be denoted by j, but let CP be designated by )\. Either parabola, as A 6, will then be referred to oblique co-ordinates in the usual manner. If OB be drawn tangent to the curve at B, AO v^\SS. be X equal to OD, or — . If i represents the inclination of the curve at any point, as B, to a horizontal line, the triangle OBD will give: BD _2y _ sin i sin i OD X cos (a + z) cos a cos i — sin a sin V 2y — ^^—cos a .'. tan i= . . . (i). 2y . ^ ^ I + ~^^ sin a X In the usual manner, sec i = Vi + tan^ i, or, a/ I + ^^ sin a ■ 4/ sec I = ' . . (2). 2y . ^ ^ I H ^^ stfi a X At the point (7, if CM = /i^, AP= x^, and AM= a, in Eqs. (l) and (2), hi sec a is to be put ior y, and {a — h^tan a) for;jr. Exactly similar equations apply to the other portion of the span. For the point A, Eqs. (i) and (2) apparently become inde- terminate, but only apparently, for the relation. y hi sec a . l?"'\a- hi tan af ' * ' * • U> 19 290 SUSPENSIOA' BRIDGES. gives, 2y _ 2^1 sec a X (« — h]^ t<^n of ' and when x = o, consequently, - becomes zero, making tan i X equal to zero also. If OBD be taken as a triangle of forces, OB will be the cable tension R dX B \ while OD will be the horizontal com- ponent H, and BD will represent wx ser a. w is the total load per unit of span on AD. From the triangle in question, H _ cos {a -'r i) _ X ^ tJ — "^^^'^ ^^^ ^ wx sec oc sin t 2y 2y rr w{a — lu tan a^- , . ••• ^=^ — - <4). As was to be expected, Eq. (4) shows // to be a constant quantity, but it is not a rectangular component in this case. The same triangle gives for the resultant tension at any point : R= sjiywx sec of -\-H^ + 2wx H tan a. . . (5). For the point C, x becomes (a — h^ tan ex). If / is the span, these equations apply to the other portion of it, by taking //^ for //j, and (/ — a) for a. If the towers are of equal heights, //j becomes equal to h^, and a = I — a = —. 2 'Let p be the horizontal distance between any two suspend- ers, then the tension, /, in the suspender will be : / = wp sec a (6). The direct stress in the platform is caused by the horizon- tal component of the tension in the suspension rods. This stress may exist as tension in the platform, in which case it INCLINED SUSPENSION RODS. 29 1 ^vill exert no action on the towers. Remembering that all the suspension rods must, at any instant, be subjected to a uniform stress, it is evident that the direct tension in the plat- form will have its greatest value at the centre, and will be equal to 7it sin a — mvp tan a ; in which ;/ is the number of suspension rods in each half of the span, supposing towers to be of equal heights. If «' be the number of suspenders between the end of the span and any point, the tension in the platform at that point will be fit sin a = nwp tan a. If the towers are of unequal heights, there will be a greater number of suspenders on one side of the lowest point of the cable than on the other. Let n^ be the number in that por- tion of the span adjacent to the highest tower, and n^ the number in the other portion ; n^ will be greater than n.^. In this case, then, the platform at the foot of the highest tower will sustain a thrust given by the expression (Wi — n-^ t sin a = {n^ — n^) ivp tan a. If the platform is to sustain a direct thrust only, at the feet of the two towers it will have to sustain thrusts given by the -expressions Tilt sin a = n^ivp tan a n^t sin a = n-fcvp tan a. If tt represents the number of suspension rods between the centre and any point, the thrust at that point will be n't sin a = liwp tan a. In the case of a suspension cantilever, in addition to the thrust given above there will be one denoted by H, uniform throughout its length. Other calculations for a suspension cantilever are precisely the same as those already given. 292 SUSPENSIOJ^ BRIDGES. The length of the cable from the lowest point to any other point at vvhicli the inclination to a horizontal is i, is readily found by means of the formula used for the cable with verti- cal rods. In the present case the inclination of the cable at any point to a line perpendicular to the axis of the parabola is {i^- a) ; consequently there is simply to be found the length of the parabolic arc between the points at which the incli- nations to the axis are (90 — {i+a) ) and (90 — a). The formula mentioned then gives c 2\ — {tan {i+oi) sec {i-{-a) — tan a sec a -f , , tan (i + «) + sec (i +a)\ , , hyp. log. ^^ ' ^^ ) . . (7). ^^ ^ tan a + sec a J ^' ' It is known from analytical geometry that p takes the fol- lowing form in terms of the oblique co-ordinates used in this case: " ~ 2y ~ 2hx Eq. (7) is, of course, to be applied to both branches of the curve to obtain the total length. From what was said in the demonstration of the approxi- mate formula, it may be seen that it can be applied to the present case by changing x \.o {x -^ y sin a) and, y to y cos a. The formula then becomes : 2f co^ a c = X + y sin a -{ -. , -. r . . (o). 3 (;r + J sm a) Art. 62. — Suspension Rods; Lengths, and Stresses. In the following calculations it is virtually assumed that the cable lies in a vertical plane, and that the suspension rods are vertical. This, however, does not affect the generality of the results obtained, for in all cases the suspension rods are sup- posed parallel to each other, and the lengths found by the SUSPENSION RODS. 293 formulae of this Article are to be taken as the vertical pro- jections of the true or actual lengths. The true lengths are therefore to be found by multiplying the values of /io, //i, //^j, etc., by the secant of the common inclination to a vertical line, of the suspension rods. Since a flat parabola nearly coincides with a circle, the camber may be supposed to be formed by a parabolic arc. Let the co-ordinate x be measured from A toward B, in Fig. 14, PL XII., and y perpendicular to it ; also let AB = x\ = half span. Then since the curve of the cable is supposed to be a parabola in a vertical plane : In the same manner for the camber: Then the total length of any suspender is: h = y + y + c. When the suspenders are separated by a constant distance, d, simpler formulae may be found. Each suspender is composed of the sum of two variable lengths (y and j") and a constant length, <:. Now if {j^ + 6) be written for y^ : J= (j'l + <^)— 2, will evidently be the sum of the two variable lengths referred to. Hence, if //q, /h, Jh, 7^3 * * * //„ represent the lengths of the suspenders as shown in the figure : h = ^, h = C + -i (Ji + «^)» -^1 294 * SUSPENSIOA'^ BRIDGES. h = c + —2- ( Ji + 6), hz = c + ^ (ji + S), {n-ifd\ Having computed the lengths of the suspenders for one half the span, the results may be used for the other half if the piers are of the same height ; otherwise the lengths must be computed separately. The stress in any suspension rod is the vertical load which it carries, multiplied by the secant of its inclination to a vertical line. Art. 63. — Pressiire on the Tower — Stability of the Latter — Anchorage. Let P^ — vertical component of pressure on tower head. '' A = horizontal " 7? = resultant " Tj, and 7^', Fig. 13, PI. XII., = tensions of the cable on different sides of the pier head. a, a', and d represent inclinations to the vertical as shown. When friction on the saddle is considered : Pj, = Tp cos a + Tp'cos a' ; P^= Tp sin a — Tpsin a'; R=VPv' + Pk'; cos 6 = ^. When friction on the saddle is not considered, Tp = Tp ; .'. P^= Tp {cos a + cos a') ; Pj^= Tp {sin a — sin a') . (i). j^=VpJTW; cose = ^. PRESSURE ON THE TOWER. 295 In the same case \{ a = a ; P^^2W, /\ = o, R^2W, 8 = 0. {W=^ weight of load and structure.) There are two cases in which the resultant pressure on the tower, caused by the tension in the cables, may be vertical in direction. Both, however, are founded on the single condi- tion that the horizontal components of the cable tension, on each side of the tower head are equal to each other. This condition will exist i{ a = a' in Eq. ( i ), making /\ = o ; or if the saddle be supported on rollers and roller friction be omitted. In the latter case P^ = o because Tp sin a = Tp sin a , and not because a necessarily equals a'. In discussing the stability of position of masonry towers, let the distance of the centre of pressure from the centre of figure of the section of the pier be denoted by q' . If this latter does not exceed g (the limit of safety for q'), which may be ascertained by determining the line of resistance for the pier, stability of position will be secured. It is supposed, of course, that 6 has some value greater than zero ; otherwise q'^= O. The stability of friction for masonry towers will be secured, at any joint, if the obliquity of the resultant pressure be less than the angle of repose. Iron and timber towers are to be treated, each as a whole, as long columns, by Gordon's formula. If the anchorage is a mass of masonry, the stabilities of po- sition and friction are to be considered. Let IV = weight of mass and q' the normal distance from a vertical line through its centre of gravity to the centre of fig- ure of its base. Let Tp equal the tension in anchor chains and / the normal distance from the centre of pressure to its line of direction ; and q the distance from the centre of press- ure of the base of the foundation to its centre of figure. Then, in order that stability of position may be secured : TpPL^ lV{q + q) Stability of friction is secured if the greatest obliquity of 296 SUSPEA^SIOiV BRIDGES. the resultant pressure on any section (including the base) is less than the angle of repose for the surfaces in contact. If it were not for friction between the anchor chain and its supports, on the circular part of the chain (see Fig. 15, PI. XII.), the tension would be the same throughout its whole length ; but on account of friction, the tension diminishes on the circular part, from link to link downward, according to the law of friction between cords and cylinders, and is, therefore, the least at the bottom. The diminution of tension of the anchor chain is computed by this formula : T TpEf' in which Tp — tension of anchor chain before friction takes effect ; Tp = tension of any point below the first point of sup- port ; E = base of Napierian system of logarithms ; f— coef- ficient of friction ; d = length of arc considered. The anchor- age can be ruptured only by the breaking of chain, or bolt, or plate, or pulling out the whole masonry. The probability of the latter can be determined by comparing the tension of the chain, at the upper surface of the masonry, with the weight of the whole masonry. Art. 64.* — Theory of the Stiflfening Tmss — Ends Anchored — Continuous liOad — Single Weight. It has been seen that when a suspension bridge cable carries a load covering the entire span, of uniform intensity Fig. I. per horizontal unit, its centre line forms a parabolic curve. When, however, such a cable carries an isolated weight, or a partially uniform load, it is evident that the centre line of the * This and the following two Arts, form the substance of a paper presented to the Pi Eta Scientific Society, in June, 1879. THEORY OF THE STIFFENING TRUSS. 29/ cable will assume a form different from that of the preceding parabola, unless such a change is prevented by some special device. Such a special device is the stiffening truss. The objections to a change of form in the cable of a sus- pension bridge are of two kinds. Not only would destructive undulations result, but, also, the determination of stresses would become exceedingly complicated and uncertain. Two cases may arise : the stiffening truss may be securely anchored at its ends ; or its ends may simply rest upon sup- ports and be free to rise, in which case there can be no nega- tive or dozvnward reaction. The former case will be considered first. It is desired to have the cable retain, for all positions of the moving load, the same parabolic form. Now, it has already been seen that such a result can be attained only by assuming a uniform pull on the suspension rods from end to end of the span. Let T be the general expression for this uniform pull for any suspension rod, and let / be its intensity per unit of span, so that if/ be the panel length of the stiffening truss. T = //. Let zv be the weight per unit of span of the fixed load sustained by the cables. This will, of course, be composed of the weights of the truss, suspension rods and cable or cables. Let w be the moving load per unit of span ; / the span ; R the reaction at i? ; R the reaction at A, and let the moving load pass on the bridge from B. Also let x-^ be the distance from B to the head of the mov- ing load ; the latter being supposed continuous from B. Since all the forces acting on the stiffening truss are vertical in direction, there are only two general conditional equations of equilibrium, and those simply indicate that the sum of all the external vertical forces, as well as the sum of the moments of the same, about any point, must be zero. Those two equations are the following : wl -{■ w' x^—tl — R —R = o . . . . (i). (-W + w')^-t^- Rx,+ U- w) ^^~ '""'^^ + '2 2 ^2 /?'(/- .0 = 0. . . (2). 298 SUSPE.VSWiV BRIDGES. Eq. (2) can be at once written by taking moments about the point x^ at the head of the moving load. Eqs. (i) and (2) are the only equations of condition neces- sary for equilibrium, but they hold three unknown quantities, i. e., t, R, and R ; hence, any one of those three quantities may be assumed at pleasure, and the other two determined from Eqs. (i) and (2). This indetermination simply means that unless another condition be imposed, it cannot be ascer- tained how much the truss will carry as a simple truss, and how much in connection with the cable. This other condition is virtually the following : the stiffening truss must act wholly in co7inection with the cable, and carry no load zuhatever as an ordinary truss. The direct consequence from this condition of the problem is, that the sum of all the uniform upward forces, T ^^ pt, must be equal in amount to the sum of all the loads of the kinds ^' and w . But the line of action of the resultant of the latter is not, for a partial moving load, the line of action of the resultant of the former: consequent!}, the truss will be subjected to the action of a couple. In order that equilib- rium may be assured, therefore, another couple of equal mo- ment, but opposite sign, must be applied to the truss ; the forces of this couple must act at the extremities^ and B, and they are nothing more than the reactions R and R . From this there at once results : R^ - R. This condition, in Eq. (i), gives: t=XV+ 7C''^ (3). Eq. (2) gives : R = -R = '■^{^-■f). . . . (4). Eq. (4) shows that both reactions, R and R , are zero for Xi — /. or for ic' = o. THEORY OF THE STIFFENING TRUSS. 299 It is also seen that R and R are always iiuincrical/)' equal, but liave opposite directions, hence R is a doxvmvard reaction, and its maximum value ivill indicate the amount of anchorage required at each end of the truss. Using Eq. (4) : dR IV xu'x-L IV Xx I dx-x ~ 2 2/ 2/ ~" ' ' * ^ 2 ■ Putting Xi = " in Eq. (4) : R = ^^ (5). Eq. (5) shows the maximum value of (— R) and gives the amount of anchorage required at either end of the truss ; it aiso shows the greatest shear to be provided for at either end of the truss. The general value for the shear at any section of the por- tion of the truss covered by the moving load is : Or, S = R + tx — tvx ~ w'x .... (6). w'xi w'xi Ixi \ , , . This value of 5 shows it to be positive near the end of the bridge ; it then decreases as x increases, passes through the value zero, and then increases as a negative quantity. As a negative quantity it attains its maximum value for x = x^; it then becomes : 5.= -'-^'(i--^)=^' . . . (8). Hence the two reactions at the ends of the truss and the shear at the head of the moving load arc always numerically equal. 300 SUSPENSION BRIDGES. I Eq. (8), consequently, takes its maximum value for x-^ = ~ , and that value is given in Eq. (5) ; this last equation, therefore, gives the maximum shear which is to be provided for at the head of the moving load. Now since this maximum shear is to be provided for at both ends and at the middle of the truss, it would probably be advisable in all ordinary cases to design all the web members of the truss to be of uniform size, and capable of carrying this maximum shear, although there would then be a little waste of material in the vicinity of the quarter points of the span on each side of the centre. This supposes, of course, that the chords of the stiffening truss are parallel and horizontal. If the chords are not parallel the amount of shear carried by the web members will depend on the inclination of one or both the chords. For all values of x and x^, for the portion of the span cov- ered by the moving load, the total shear will be given by Eq. (7). For the portion of the span not covered by the moving load the general value for the shear is {measuring x from A) : ^ _,, w'xi w' Xi w'xiX , . S=- R' + wx-tx=~'-^^ -^ . . (9). This expression attains its greatest values for x =^ o and X = I — x^. In the first case the shear becomes — R', and in the second R. These results show nothing new. Since the maximum shear in a simple truss, of the span I and uniform loading of intensity (w + w'), is i(w + 'w')l, it is seen that the maximum shear in the stiffening truss of same span is only one-fourth of that due to the moving load alone in the case of the simple truss. The general value of the bending moment to which the truss is subjected, for the portion covered by the moving load; is: x^ M= Rx -{w + w' -t)— . . . ( 10). THEORY OF THE STIFFENING TRUSS. 30 1 Eq. (10) shows that if x = -n, the bending moment is equal to zero. Hence, at the head of the moving load, for all its positions, there is a section of contraflexure or no bending, and consequently the loaded and unloaded portiotis of the stiffening truss are each in the condition of a simple beam supported only at each end, and loaded uniformly throughout its length. If the values of R and / from Eqs. (3) and (4) be inserted in Eq. (10), and if -^ be put equal to zero, it will be found that the bending moment has its maximum value for x ~ ■ — ; as might have been anticipated. Putting X — — in Eq. (10) : 2 ^^^ .... ("). By differentiating in respect to .I'l it will be found that M has its maximum value for x-^ = \l. Denoting this value of M by M^, there results : J/i = (12). Eq. (12) shows the maximum bending moment to which any loaded portion of the truss can be subjected. If a simple truss supported at each end be subjected to the action of a uniform load, of the intensity {%v -f- w'), through- out its entire length, the greatest bending moment will be : M' = g-^ = -^ (if zv = o) ; .-. J/o' = ^J/o'>^vW . . • (13)-* From what has alread}' been shown it is evident that the * The subscript o indicates that 7z/ = o in M'. 302 SUSPE.VSION BRIDGES. greatest bending moment for the portion of the truss not covered by the moving load will occur at the distance ( '—) from the reaction R . The general value, therefore, for the greatest moment for that portion will be : M=lR'{l-x,) + {i-zv)^-^^^ . . (14). Putting the differential coefificient of M in respect to .Vj (after inserting the values of R' and /) equal to zero, there results : ^j — -^l — L- V 9^^ 9^ .-. ^i = |/±- = /ori/ . ._' . . (15). The latter value (^/) gives a maximum, and inserted in Eq. (14): M,' = - %'^' =-M,= - \M^ (nearly) . . (16). 54 Eqs. (12) and (16) show that the greatest bending moments, to which the stiffening truss is subjected, are equal, but of oppo- site kinds ; and it is seen that they occur when one-third or two-thirds of the span are covered by the moving load. The chords, therefore, of the stiffening truss must be designed to resist both tension and compression. Eqs. (10) and (14) are general expressions for all the bend- ing moments to which any portion of the truss can possibly be subjected, but in all ordinary cases it would probably be best to make the chords uniform in section (supposing the depth of the truss to be constant) from end to end, and capable of resisting the moments given by Eqs. (12) and (16). Eq. (13) shows that the greatest bending moment to which a stiffening truss can be subjected is only \ of that found in a simple truss supported at each end and loaded with a uni- THEORY OF THE STIFFENING TRUSS. 303 form load equal in intensity to that of the moving load on the stiffening truss. It may be interesting to notice that the resultant load on the portion -r^ of the truss is downward^ since {zu + w') > {t=.w-\- '' -^\ : but that that on the portion (/ — .r,) is upward, since zv < /. If the bridge is traversed by a single concentrated load or weight, W, the general method of procedure is precisely the same as before. Let the weight W pass on the bridge from the end B, in the figure, and let x\ denote its distance from that point ; also measure x from the same point. The general equations of condition are : Rjr R + tl-wl-W=^o . . . (17). - R'{l-x,) + {w - t) ^"-"^^ - {w - t) '^ +Rx, = o . (18). Eq. (18) is written at once by taking moments about the point of application of W. By precisely the same method as before, there may be found the result, R = - R' . In Eqs. (17) and (18) let there be put R = — R', then there results : W x^ R=-R'^ ^^U-'y) • • • (20) When Xi = - , R = o and R' = o Eq. (20) shows that //-r reactioji nearest the weight W will always be positive, or upward ; and that the other will be nega- tive, or downward. 304 SUSPENSION BRIDGES. The maximum value of R or R' is found (by making x\ — o W in Eq. (20) ) to be — , and that ts the amount of anchorage required for the weight W, alone, at eaeh end of the truss. The point of appHcation of the weight IF divides the span into two segments. The general value of the shear for the shorter segment is: S = tx-wx + R = ~x-\- wU -7-) • • (21)- This has its greatest value for x — x-^\ it then becomes : 2 Hence S ' is the tmiform maximum upward shear that must be provided for, throughout the whole length of the truss. The general value of the shear in the longer segment of the span is : S=R' -w{l- x)+t{l- x)^wU+Y -^-^ . . (22). This expression attains a positive maximum for x = x^, that being the least positive value of x admissible ; the re- sulting value of 5 is ^ W, which shows nothing new. The negative maximum for ;r = / is simply the reaction R'. Putting 5 = o in Eq. (22) there results : X = Xi -\ — . ^ 2 Hence for all points of the span between - + x^ and / there will be negative or downward shear. The maximum negative shear, however, to be provided for, is shown by Eq. (22) to exist in one-half of the truss when IF rests on the opposite end ; or, when in that equation x^ — O and x > -. The negative THEORY OF THE STIFFENING TRUSS. 305 or downward shear to be provided for has, then, for its gen- eral expression : S,= -w{^-^-\) .... (23). Eq. (23) is to be appHed to each half of the truss, and it is also seen that the web members which take up S^ should increase, in ultimate resistance, uniformly from the centre to the ends of the truss ; supposing the chords to be parallel and horizontal. In many cases, however, it may be best to design them of uniform dimensions belonging to those at the ends. The bending moment for any point of the smaller segment of the truss is : M=Rx+{t-w)-^W{l--j-)x+ ^^ . . . (24). Since both terms of this moment are positive, it will attain its greatest value for x — x^ ; it then becomes : J/, = -(.r,-^) .... (25). Eq. (25) gives the general value of the greatest positive bending moment to which any point of the truss will be sub- jected, for which case x^ must 7iever be made greater than \l. The bending moment J/i will evidently cause compression in the upper chord, and tension in the lower. • For the longer segment of the truss the general value of the bending moment is : Or, M ao M=R'{l-x)^- {t- w) ^ ^ = _ j^(^i _-ij(/_,^) + ___v__i ^ (26). 3o6 SUSPENSION BRIDGES. There is evidently a point of contra-flexure for the longer segment of the truss, for if the second member of Eq. (26) be put equal to zero, there results x = 2^1. Hence all the por- tion I — 2,Vi. of the truss, will be subjected to a 7iegative bending moment causing tension in the upper chord and compression in the lower. In Eq. (26), putting -rj-, \ == 0> there results : ^ = ^1 + - {^7)' This value of x, in Eq. (26) gives : Wl Wx. Wx? Eq. (28) gives the general value for the maximum negative bending moment at any point in the entire truss, for any position of the weight W. In Eq. (28), it is to be remembered, x^ must always be less than - ; also, that the point at which M' exists will be given by the value of x in Eq. (27). The formulae for a continuous load, taken in connection with those for a single weight, will give all the circumstances of bending or shearing which can exist with any condition or position of loading. When the " shear " has been determined for any section, the stress in the web member which is to carry it (if the chords are parallel and horizontal) will be found at once by multiplying that " shear " by the secant of inclination of the web member to a vertical line. If there are two or more systems of triangulation in the truss, then each system is to be treated as a single truss in the usual manner. If desirable, after the reactions R and R and the upward load T = pt are known, the stresses in the individual members of the stiffening truss can be traced as in the case of an ordi- nary truss supported at each end. THEORY OF THE STIFFENING TRUSS. 307 Art. 65. — Theory of the Stiflfening Truss — Ends Free— Continuous Load — Single Weight. In this Article the notation of the previous one will be con- tinued, and the same figure will be referred to. The case of a continuous load will first be treated, and, as before, it will be supposed to pass on the bridge from B. Since the ends are not anchored, in this case there can be no negative or downward reaction, consequently R' will be zero. As before, putting the sum of all the vertical forces acting on the truss equal to zero, and taking moments about the head of the moving load, there result the two general Equa- tions of condition : wl -V wx\ — tl — R^o (i). (^ + ^^,^ _ /) ''^ - (zf - t) -^^^ - Rx^ = o . . (2). Since there are now but two unknown quantities, R and /, the problem is perfectly determinate. Eqs. (i) and (2) give : ^ = «'''Vi(i-7-) (3) t = w+ —3 1^ (4) The general value for the shear at any section of the truss, for the portion covered by the moving load, is S = R^[^t-w- w) X = IV {x, - x) - w' -Y[i - '^j . (5). Evidently >S has its maximum positive value for a: = o ; its greatest negative value for x = Xi, and the value zero for x = Jx^ 3o8 SUSPENSION BRIDGES. In order to find the head of the moving load for that po- sition which makes R a maximum, let -y— be put equal to zero. There is then found x\ — — . Hence the moving load cover- ing half the span gives the maximum reaction R. I Placing Xx^= — in Eq. (3), the greatest value of R becomes i?i = u'' / -r- 4. In order to determine the greatest web stresses it is neces- sary to find the greatest shear at any point whose abscissa is X. This maximum shear at once results by placing the first derivative of 5 in respect to x^, from Eq. (5), equal to zero. That operation gives : 2X^ ( X\ /2 • I r- I - -7 I^ O .-. Xi = — , / \ // ' 2(/- ;r)* By the introduction of this value of x-^ in Eq. (5), the great- est shear for any section located by x becomes : i /2 max. S — w' \ —-, r - U (^ - ^) It is clear that this value is a maximum for the reason that d^ S -j-^ is a negative quantity in which x^ does not appear. It is further evident that max. 5 is a positive, or upward shear, from the fact which was observed at the bottom of the preceding page, that the greatest negative shear occurs at the head of the moving load. By making x = x-^xn Eq. (5) that greatest negative shear becomes : , x^ ( Xi s, = -^ rV-7 It will be necessary to apply max. S and S ^ to 3. half of the span, regarding the shear for a positive direction on one side of the centre as negative for the other. These two values of the shear will enable all the greatest web stresses to be determined. Since R' — o, the whole truss will be subjected to bending moments of the same sign ; such bending moments, in fact, as THEORY OF THE STIFFENING TRUSS. 309 will put the upper chord in compression and the lower one in tension. The general value of the bending moment for that portion of the truss covered by the moving load is, J/= Rx - {w + w - /) — , T,^ , W'X?X 7U'X^ W'X^X^ ,^. Ix Since the shear S is zero for x = , " ^ — , that value of x in / + Xy Eq. (6) will give the maximum value of M. This latter is : M^ = wx^ f ~ ^^ , (7). Putting -j — = o, there is found, dxx ;ri = -(- I ± VS) = + 0.618/. The absolute maximum bending moment exists, therefore, when the moving load covers 0.618 of the span. That mo- ment has for its value : 0.045 1 w7^ (nearly) (8). Ix If ;ri = o.6i8/be put in the expression ^ = - — ? — , there will / "T" -t J result X = 0.382/, nearly. Eq. (6) gives the general value of the bending moment for any position of the load, but it would probably be the most convenient to make the moving load cover 0.618/, and design the chords for the distance 0.382/ from each end to resist the bending due to that position of the load, and then design all of the middle 2 (0.5 —0.382)/= 0.236/ to resist the moment rjio SUSPENSION- BRIDGES. ^iven by the expression (8). By this arrangement there would be a little surplus of material at the middle of the truss. If a single weight rests upon the bridge, the two general equations of equilibrium, obtained in precisely the same man- ner as heretofore, are : wl^W-tl-R=o (9). (w - /) ^ - (w - t) i!—^ -Rx, = o . . (10). These equations then give : 2Wx, f . R= w[\ -~^ ..... (12). It Xi = - , i = w + ^-, and R = o. 2 I The general values of the shear S, and moment M, are the following : S = R + {t - %v)x, ... 5 = ^(i-^--)+^-i^. . (13). M= Rx -\- U -w) — 2 ... M=w{.-^)x^^ . . (14). Eqs. (13) and (14) show, since Xi must not be taken less than X, that if the maximum shear and bending moment are desired for any section, the weight, W, must be placed at that section, and x be made x^ in those equations. APPROXIAfATE CHARACTER OF INVESTIGATIONS. 311 That section at which the bending moment will attain its absolute maximum value is found by putting x — x^ in Eq. (14), then taking the first differential coefficient of M in re- spect to ^'1, equating to zero and solving. There results: x^^\l±\l= \L This value in Eq. (14). when x — x^, gives: M = 3^ Wl. The equations for the continuous and single moving loads, used in combination, will give moments and shears for any character and position of loading whatever. The general observations m regard to finding web and chord stresses, at the close of the last Article, apply equally well to this case. As was to be anticipated, R and R' in the two preceding Ar- ticles, are independent of the fixed load w. Art. 66. — ^Approximate Character of the Preceding Investigations — De- flection of the Tnass. In the two preceding Articles it has been virtually assumed that the deflection of the cable, due to its lengthening under stress, is just sufificient to allow the truss to take the deflec- tion due to the loads T — pt, w', and W xn the different cases. Such, however, is not really the case. In all ordinary cases, the cable does not deflect to that ex- tent. The result is, as is evident, that the truss is not sub- jected to the amount of bending assumed. The error, however, is only a small one, and on the safe side, as should be the case. It has been found by experiment that if the ends of the truss are anchored, the stiffening truss will be subjected to a maximum moment equal to that existing in an ordinary truss -supported at each end, with the same span, and carrying load 312 SUSPENSION BRIDGES. over its entire length of about one-eighth the intensity of the moving load on the suspension bridge. This same case, treated analytically, as was seen, gave about \ instead of \. Approximate values of the deflection of the stiffening truss can be found by the ordinary formulae used for solid beams in the subject of resistance of materials, in the differ- ent cases, where /, R, R', and the moving load are known. CHAPTER X. DETAILS OF CONSTRUCTION. Art. 67. — Classes of Bridges — Forms of Compression Members— Chorda Continuous or Non-continuous. Regarding the systems of construction, truss-bridge struct- ures are divided into two classes at the present time, i. e., bridges with " pin connections " and bridges with " riveted connections." In the former class the connection of web members with the chords or with each other is made by a single pin only, as in Fig, 3, PL III, (the figure shows simply the upper chord and tension web members together witn one end post). The pins are shown at 1,2, 3, 4, and 5, where the tension members join the chord. In the latter class the con- nections mentioned are made by means of rivets, as shown in Fig. 6, PI. XI. In that figure ^ is a tension, and B a com- pression web member, while C is a portion of the lower chord. Screw connections for tension web members and simple abutting connections for compression ends have been used, but are not usually employed at present. A screw connection is formed by passing a tension member through the chord at one of the joints, and placing upon the end of it a nut ; this method, therefore, can only be conveni- ently used when the tension members are of circular cross- section. An abutting connection for a compression member is formed by simply abutting either end against the chord, which is properly formed for the purpose at the joint. The end of the post or strut is inserted in the chord, or else a projection of the chord passes into the end of the post ; or, again, some simple device is employed for the purpose of 313 314 DETAILS OF CONSTRUCTION. keeping the ends of the post in position, and for nothing else, as the entire compressive stress in the post or strut is trans- mitted through the abutting surfaces. Occasionally screw, abutting, and pin connections are com- bined in a single bridge. Without re<.arding systems of construction, truss-bridge? are divided into : " Deck " bridges, /. e., the applied load is on the chord in compression. " Through " bridges, i. e., the applied load is on the chord in tension. " Pony" trusses are through truss-bridges when the trusses are not sufficiently high (or deep) to need overhead cross- bracing, and they are seldom put up for spans of over eighty feet, although there are examples of one hundred feet (and even more) in length. The lateral stability of such long pony trusses, however, i:. very precarious. Figs. I and 2 of PI. XI. show the ordinary forms of plate girder, stringers and floor beams, with plate hangers at the ends of the latter. The various forms of cross-sections of upper chords and posts, and compression members generally, are almost innu- merable, and subject only to the fancy of the engineer or builder. The principle which should always be kept in view is this : That the material should be as far as possible from the neutral axis. Figs. 3, 4, 5, and 6 of PL III. show methods of building up the upper chord. It consists of riveting plates to a pair of channel bars, or to a pair of channel bars and an X beam. The bars or beams are frequently built of plates and angles. The blackened portions represent sections. Chords are continuous or non-continuous according as they are built up in a continuous manner from end to end, or built up of panels abutting against each other at the panel points. The former are principally used at present. Tension members are always of rectangular or circular sec- CUMULATIVE STRESSES. 31$ tion. Fig. 5, PI. XII., is a lower chord "eye-bar." In writing of channel bars, I beams, angle-irons, iron bars and rods, they are indicated as follows: C, X, L) CD* O; that is, by skeletons of their sections. Art. 68. — Cumulative Stresses. Stresses are said to be cumulative in any part of a struct- ure when they are transmitted through that part to other parts, whose whole duty is to sustain them, the part in ques- tion being subject at the same time to its own stress. The member in which the stresses are cumulative is, therefore, overstrained to some extent in some one or more portions of it. The two channel bars in Fig. 3, PI. III., are the portions of that upper chord which are subjected to cumulative stresses. If C'C is supposed to be the centre line of the bridge, then the compressive stress in the chord increases as C'C is ap- proached from the end, in consequence of the components in the direction of the centre line of the chord of the stresses in the inclined ties. A, B, C, etc. This increase of stress is pro- vided for by riveting plates to the upper flanges of the two Cs, as shown in the figure. It is evident that the plates do not receive the stress which they are intended to bear, except indirectly through the Cs and the rivets which connect the latter with the plates. Now since the Es are supposed to have their own share of direct compressive stress to sustain, it is plain that the material in the vicinity of the front of the pin (looking from the centre of the pin toward the centre of the bridge) is subjected to a much greater intensity of com- pressive stress than should exist in the structure. This re- lates only to the material in front of the pin, and that is the only vicinity in which cumulative stresses would exist if the plates could be so securely riveted to the Cs that the whole chord could be depended upon to act as one piece. In prac- tice, however, no such riveted work exists. The C^ must in- evitably yield to some extent before they bear sufficiently on the rivets to give to the plates their proper share of the stress. The result is that not only the material in front of 3l6 DETAILS OF CONSTRUCTION. the pin but the whole of the Cs are overstrained by these cumulative stresses. The only remedy is to so proportion the chord that those parts which are designed to sustain stress shall receive it immediately, and not indirectly through some other part. The Fig. 5, PI. III., shows a method of accomplishing this object. The plate ab is a light one riveted to the top flanges of the Cs, and extends throughout the whole length of the chord. The increase of the areas of cross-sections are ob- tained by riveting plates to the flat sides of the Qs, and by adding an X> if necessary, as shown. The parts of the chord thus receive stresses immediately from the pins, and cumula- tive stresses are obviated. It is also evident that if the stresses are applied to the cen- tres of gravity of the cross-sections, or parts of the cross-sec- tion, no cumulative stresses will exist. Cumulative stresses are as liable to occur in riveted connec- tions as in pin connections ; in fact, more so. It may be said to be impracticable to so construct riveted work that cumula- tive stresses will not exist, and in this respect pin connections have the advantage of riveted connections. Fig. 6, PI. XI., illustrates the matter for a riveted chord. The plate C is common to the whole chord, and all web members are riveted to it, as shown by A and B, so that before the rivets can take their share of the stress in transferring it to the plate and Ls ED, it (the plate C) will necessarily yield to such an extent that cumulative stresses will exist throughout its whole length to a greater or less degree. Art. 69. — Direct Stress Combined with Bending in Chords. If direct stress is not applied to the centres of gravity of the ends of a piece subjected to compression, it is clear that bending must take place. In Figs. 3 and 4 of PI. III., the horizontal components of the oblique forces in the ties A, B, etc., do not act through the centres of gravity of the sections of the chord, hence there must be a bending in those chords. If the chords were DIRECT STRESS WITH BENDING IN CHORDS. 317 perfectly straight, and if the centres of the pin-holes were all at the same distance from the centres of gravity of the dif- ferent cross-sections, as well as in the same straight line, then the total direct stress to which the chord is subject at any section would produce bending at that section, and the lever, arm would be the same for all sections. Camber and deflec- tion from loading, however, so complicate the matter that it is quite impossible to make even a satisfactory approximate computation of the chord bending arising from this cause. All chords in compression, therefore, should be so designed that the axes of the pins may traverse the centres of gravity of their sections, even though the ties rest directly on the upper chord. It is clear that when this bending exists, the proper distribution of direct stress is greatly disturbed, though to an indeterminate extent ; and is, except in most rare cases, a very faulty construction. If it be supposed that the total direct stress in the chord acts as if the latter were perfectly straight, so that it all produces flexure ; and if it then be supposed that the in- crement only, at each pin or panel point produces flexure in the adjacent panel ; it is evident that the first supposition will make the flexure the greatest possible, while the second will make it the least possible. It is farther evident that if cumulative stresses occur, flex- ure must necessarily exist, for the simple reason that the di- rect stress is not uniformly distributed over the cross-section of the chord. Although this flexure is indeterminate in amount and shows a faulty design, the attempt to utilize it has sometimes been made, and the analysis on which the practice was based will now be given, it being premised that the ties are supposed to rest directly on the chords, as shown in Figs. 4 of Pis. III. and I. Suppose, in PL III., Fig. 6 to be an enlarged cross-section of the chord in Fig. 4, and let fg pass through the centre of gravity of the cross-section, being parallel to ab and cd' ; then, since the increment of the chord stress transmitted through the pin from the ties AA is applied to the cross-sec- 318 DETAILS OF CONSTRUCTION. tion of the chord at a distance h below the centre of gravity; there will be an excess over the uniform intensity of stress in the cross-section at the lower side EE, and a deficiency at the upper side ab. This excess or deficiency (the same in amount, of course, for certain sections only) is found in a very simple manner, as follows : Let P be the increment of direct compressive stress given to the chord by the ties AA, and let P^ be the total direct compressive stress in the section. Put .S" for the area of the: cross-section. Now the variation of the intensity of stress from the mean is due to the moment Ph, and since this moment is constant for all points between any two pins, as i — 2 or 2 — 3, Fig. 4, PI. III., the variation in intensity is also constant between^ these points. RT The moment Ph — —-t- , in which rt^ equals the distance "1 from /g- to ££, and R the intensity of stress at EE due to- bending, gives The intensity of stress at ad due to bending is, of course,, equal to Now the total intensity of stress at EE, Fig. 6, PI. III., is- P P d — d equal to -7^ + i?; and that at ab, -^ — R — -:, — ^; the inten- p sity at 7^=-^, whatever may be the figure of the cross-sec- tion. The variation of intensity at any point in the section may be easily found from 7^ by a simple proportion, and the p total intensity by adding that to ~. DIRECT STRESS WITH BENDING IN CHORDS. 319 As a first case, let the chord be a non-continuous one, so that each panel, so far as the panel moving load is concerned, is a simple beam supported at each end. If the load rests on the upper chord immediately, as shown in Fig. 4, PI. III., it will produce tension at the lower side of the chord and compression at the upper by simple flexure, an opposite tendency to that exerted by the moment Ph. The moments due to the moving load on any panel vary (that is, increase) from the joints to the middle point of the panel, where, of course, the moment is maximum. Denote by R' the greatest intensity of the tensile stress caused by the moment of the moving load, then ^ _ d^'2wx^ I in which 'S.WX'^ expresses the greatest moment of the applied load. For a uniform load : ^WXx - — g- , and it exists at the centre of the panel. Now, ordinarily, the chord would be required to resist the bending moment ex- pressed by '^wx, but h may be so chosen that for its max- imum value R — R', and then no extra metal will be required on account of the flexure produced by the moving load. This yalue of h is found as follows : put Phd]^ _ d{2wx-^ _ 2wxj In all ordinary cases of uniform load 2wx^ = —^ > 220 DETAILS OF CONSTRUCTION. in which w is the intensity of uniform load. ••• ^^ = ^- The value of h, therefore, is independent of the form of cross-sections. \i R <, R', additional material will be needed in order to prevent an excess of compressive stress at the upper part of the chord, and a deficiency at the lower side. When R > R' , there is an excess of compressive stress at EE, Fig. 6, PI. III. When the lower chord sustains the moving load directly, as in Fig. 4, PI. XL, the only change arises from this : That it is in tension instead of compression, and h is measured above the centre of gravity of the cross-section, instead of below it ; also, the section is rectangular. In the case of the lower chord, however, if the stress in one panel is given to the adjacent one through the medium of a pin, then the total stress in the panel under consideration must be put for P in the formulae above. This must also be done in every case where the total chord stress produces bending. The chord stress used may be taken as the max- imum (that which exists when the moving load covers the whole truss), for in all other cases there is a surplus of ma- terial with which to resist the bending. This method of neutralizing the flexure produced by the direct application of the moving load to the chords is very unsatisfactory in many ways and should never be used. At all places in the panel except the centre the metal is still over-strained by^ the flexure due to its own stress, and in the vicinity of the panel points this condition exists to a very serious extent. When this consideration is coupled with the great uncertainty attached to the hypothesis on which the analysis is based, the unsatisfactory character of the method is sufficiently evident to effect its exclusion from the best practice. If the chord is continuous, the objections to the method al- ready mentioned gather considerably increased force. At DIRECT STRESS WITH BENDING IN CHORDS. 321 and near the ends of the panels the fixed and moving load produce flexure in the same direction as the direct chord stress, and to twice the amount of that at the centre. It is not necessary, therefore, to consider this case farther. A considerable saving of material can be effected by plac- ing the ties directly on the upper chord of a deck bridge, and with a proper design it in no manner conflicts with the best practice. In all cases the axis of the pm should traverse the centre of gravity of the chord section, or as nearly so as practi- cable, in order, if possible, to eliminate all flexure due to the direct chord stress. The chord section should then be so formed that the combined stresses due to flexure in the ex- terior fibres, and the direct chord stress shall at no point ex- ceed a proper value per square unit. This value may be taken at 8,000 to 9,000 pounds per square inch for wrought- iron upper chords, or 10,000 to 11,000 for mild steel members of the same kind. These values may be taken comparatively high for the reason that an indefinitely small portion only of the material is subjected to these intensities, and that small portion is well supported against fatigue by the material about it, which is considerably understrained. If the notation previously used in this Art. be still main- tained, the maximum external moment '^wx-^ will develop in the most remote fibres at the distance d^ from the neutral axis the intensity. R^ d^^P ^^^^ If, on the other hand, P^ is the total direct stress in the chord and 5 the area of cross section, while/ is the greatest allowable combined stress, then there will result : di^zux. Pi , . ^ = ^T^+S • • ■ (^)- Eq. (2) shows that in the most ef^cient design the moment of inertia /of the section must be the greatest possible. At the same time considerations affecting the joint details render 322 DETAILS OF CONSTRUCTION. it advisable that the centre of gravity should lie not far from the mid-depth of the section. These two conditions are ful- filled by placing large quantities of the material, and as nearly as possible in equal amounts, at the top and bottom of the chords, as shown in Fig. i8 of PI. XII. The cover plate be and angles dd are made as light as the circumstances of proper design will permit, but the angles aa are made as heavy as possible. An unequal-legged angle is a very good one {ox aa with the longest leg horizontal, and it is sometimes necessary to rivet a narrow plate to those horizontal legs in order to properly balance the section. The centre of gravity XxViO. fg will usually lie a little above the centre of figure. As the centre of the span is approached from the end the chord section must be rapidly increased but in no case should that increase be made by tkickenifig the cover plates or increas- ing their number, as such an operation inevitably means cumulative stresses or flexure by direct stress. In rare cases it may be admissible to slightly thicken a cover plate, if there is but one, but, as a rule, Fig. 6 of PI. III., shows a design to be carefully avoided. All increase of section should be ob- tained by thickening the side or web plates, or increasing their number ; or, again, by increasing the angles, or, finally, by introducing an interior eye-beam, as shown in Fig. 5, PI. III. If the chord is non-continuous, '^wx^ is simply the bending for a span equal in length to a panel and due to the track load, own weight and superimposed moving load, and is easily determined. If the chord is continuous, on the contrary, the analysis for the moving load bending is not simple. The total bending for this case, however, may properly and safely be taken at three-fourths its value for a non-continuous chord. The upper chord section required in the case of combined bending and direct stress is readily found by the aid of Eq. (2). The radius of gyration r can be easily and with suf- ficient accuracy predetermined ; so that S-r can be put for / in that equation. After that substitution is made, there at once results : RIVETED JOIXTS. 323 S = i(^' + /',) (3). This is a very convenient formula for practical use. Art. 70.— Riveted Joints and Pressure on Rivets. In riveted bridge work the pitch of rivets {i. e., the dis- tance from centre to centre) should not be less than three diameters, although it sometimes is ; if possible it should be from four to eight diameters of the rivet, provided that value does not exceed about fourteen or sixteen times the plate thickness. The diameter of the rivet is determined by the amount of stress which the joint is to carry, so that the in- tensity of pressure against the surface of the rivet in contact Avith the plate shall not exceed a given value. If the rivet and hole were in ideally perfect contact, this intensity could easily be found, having given the amount of stress which the rivet is to carry. But such is never the case. The only re- sort left, therefore, is to assume that the rivet does fit per- fectly, and fix a low enough value for the intensity of pressure against its surface to make the joint safe. In Fig. I, PI. XII., suppose EF to be a part of a plate in which is drilled or punched the rivet-hole ADBK, and sup- pose the stress to be exerted on the plate in the direction of the arrow at K, then the surface of contact between the plate and rivet will be projected in ADB ; contact will not take place throughout the whole semi-circumference when the plate is not subjected to stress, unless the rivet fits the hole with absolute accuracy. Since all material is elastic to some degree, there will be a surface of contact when the plate is subject to stress, even when the rivet does not accurately fit the hole, and this sur- face will evidently increase with the stress in the plate. But suppose that the rivet fits the hole exactly, then the pressure on the surface of contact, ADB, will be of uniform intensity, and the case will be similar to that of fluid pressure on a cylinder. Let/ denote this intensity whose direction is normal to ADB at every point of it (friction is omitted from 324 DETAILS OF CONSTRUCTION. consideration), then the total pressure in the direction of the arrow at K exerted by the plate on the rivet for each unit of length of the latter is equal to pAB. Let t be the thickness of the plate, as shown, and put d for the diameter AB, then the total pressure against the rivet in the direction of the arrow is P^ptd. The quantity/ is the greatest mean value of the intensity of compressive stress which it is desirable to put upon the material under the given circumstances. It is usually taken as high as 12,000 lbs. per square inch, although 10,000 is a safer value. Of course, the actual maximum value of the intensity immediately in front of the centre, (T, is much greater than either 10,000 lbs. or 12,000 lbs. If Z' is the amount of stress which the joint is required to carry, then the number of rivets, so far as the previous consideration is concerned, is equal to T' ■ 7 = n. ptd The riveted joint itself, as shown in Fig. 5, PI. XL, may now be examined. The distance c should be at least 23^ diameters of the rivet. By the arrangement of the rivets shown, when the pitch is from four to eight diameters, the strength of the plate of the width w will only be decreased by about the amount of metal taken out in one rivet-hole, although experiments to settle this point definitely are wanting. After having determined the pitch and distance of r, as above, there are five methods of rupture of the joint only which need serious attention. These five are: (i) tearing of the plate though the rivet-hole E, (2) tearing of the cover- plates through the rivet-holes at the middle of the joint, two in the figure, (3) shearing of the rivets, (4) and (5) rupture by compression at the surface of the contact between the rivets and the plates. The safe shearing stress to which rivets are subjected in bridge structures is usually taken at 7,500 lbs. This gives a safety factor of from 5 to 6. RIVETED CONNECTIONS. 325 Put 5 for the intensity of the maximum safe shearing stress on rivets (7,500 lbs. for wrought iron),/ for the intensity of the maximum compressive stress (10,000 to 12,000 for wrought iron), and 7" for the maximum working tensile stress; also, ri for the number of rivets on the line through the middle of the joint (two in the figure). Let t and / ' represent the thickness of the plate and covers as shown. Then equal liability to rupture in the five ways mentioned is expressed as follows : Tt (zu — d) = 2 TV' {w—n'd) = . 2 . S = ntdp = 4 2nt'dp =^ T'. It almost always happens that these quantities are not each equal to T', but none of these should be less. If only one cover-plate is used, the 2Tt ' should be replaced by Tt ', 2S by S, and 27it' by nt'. The form of this Equation of condition may be somewhat changed by piling of the plates, etc., but it will remain essentially the same, and serves to illustrate the principle which must govern in all cases. We see, therefore, that in obtaining the amount of pressure which should be put upon a rivet,/ ought to be multiplied by its diameter, and not by the semi-circumference ADB, Fig. i. Art. 71. — Riveted Connections between Web Members and Chords. When web members, as A and B, Fig. 6, PI. XL, are riv- eted to the chord, the centre line (/. r., the line joining the centres of gravity of the sections of the members) should pass through the centre of gravity of a system of points situated at the centres of the rivet-holes ; otherwise the intensity of stress in any section of the member will not be uniform, and it (the web member) will be subjected to flexure. It is supposed, of course, that each rivet carries the same amount of stress, which, however, is probably seldom true, but it is the best assumption that can be made. Fig. 6 represents a proper dis- tribution of rivets in reference to the centre lines Ac and Be. It will be observed that the strut, composed of two un- equal legged angles, has its connection with the chord through both legs by means of the angle lugs. This should 326 DETAILS OF CONSTRUCTION. always be done in similar cases, for in no other way can an angle-brace develop its full strength. The practice of rivet- ing single legs, only, of angle-braces to chords is highly ob- jectionable, for the reason that the actual resistance of the brace is far below the nominal. When three or more pieces are riveted together at the same joint, all the centre lines of stress should intersect at one point, if flexure is to be avoided. It is frequently im- practicable to do this in riveted connections, and the imprac- ticability constitutes a serious objection to that character of work. If T is the total tension in the member A of Fig. 6, PI. XL, and C the compression in B, then there will be developed at a the bending moment : T ^ ac sin 6 ; and at b the bending moment : C ^ be sin d ; it being supposed that ac, be, and ab are the centre lines of stress of the two members and chord. It is very true that the metal is well supported in the vicinity of the joint, but unless provision is made for the flex- ure, as shown in Art. 69, which is seldom or never the case, some of the metal will be over-strained. Hence, this flexure should always be made a minimum, and reduced to zero if possible. In pin connections this bending at the joints is. of course, entirely obviated when all centre lines of stress intersect at the centre of the pin. Art. 72. — Floor-Beams and Stringers. — Plate Girders. The load applied to a bridge rests immediately on the floor- beams, generally speaking, and is transferred through them to the joints of the truss. In railway bridges the track and ties lie on stringers, which rest on the floor-beams. There are two or more stringers for each track. In highway bridges, the floor-beams support stringers, say FLOOR-BEAMS AND STRINGERS. 32/ two feet apart (sometimes less and sometimes a little more), running parallel to the centre line of the bridge, which carry the floor. In railway bridges, the moving load is applied to the beams at the ends of the stringers, but in highway bridges the greatest moving load, for which the beam is to be de- signed, may be taken as uniformly distributed over the entire length of the beam, or rather that part of it between the points of support. Floor-beams should always be supported at the ends by a single hanger, or by some equivalent arrangement, which rests at the centre of the pin, or centre of the chord in riveted con- nections. Double hangers may be made tolerable by some equalizing device, usually of an expansive character, but as a rule they cannot be too strongly condemned, for in such cases the deflection of the beam will throw the greater part or all of the weight of the beam and its load on the inner hangers. The result will be not only a great overstraining of the latter, but a prejudicial redistribution of stresses in both web mem- bers and chords. The excessive load on the inner hangers will cause an overstrain in the inner tension braces which will extend to the inner lower chord members, and even to the posts and upper chord. Floor-beams are frequently built into vertical posts. In such cases an essentially central bearing on the pin should be provided. Plate girder stringers for railway bridges arc either sup- ported in between the floor-beams, or partially so and par- tially above, or are supported wholly on the top of the floor- beams. With proper designing there is little difference in cost in the various methods. The first, however, is far prefer- able, for the reason that it gives the greatest stifl'ness to the floor system, other things being equal. The depths of plate girder stringers is usually found be- tween one-ninth and one-twelfth of their spans, i. e., the panel lengths. The floor-beam depth should be as great as economical considerations will permit, in order that the de- flections may be the smallest possible. 328 DETAILS OF CONSTRUCTION. The webs of stringers and floor-beams slightly aid resistance to flexure, but rivets in stiff'eners and splice plates, if such ex- ist, decrease this resistance to some extent. Hence, it is the best practice to disregard the resistance of the web to flexure, and to assume that it resists the shear only. This is the more advisable when it is remembered that the rivets, in giving stress to the flanges, produce a flexure in the latter which is always neglected. This flexure arises from the fact that the rivet holes never pass through the centre of gravity of the flange angle section. The true depth of a plate girder is the vertical depth be- tween the rivet hole centres, but by a curious confusion between rolled and built sections, it is commonly taken as the vertical distance between the centres of gravity of the flange angles. All stress is given to the flanges by, or through, the rivets, binding them to the web, hence their proper distribution be- comes a matter of importance ; it will be shown by two ex- amples. The exact analytical determination of the web thickness cannot be reached, but the following approximate analysis is frequently used. It is shown by the theory of elasticity that if two planes at right angles to each other and to a plane normal to the neutral surface of a bent beam, be so taken that their intersec- tion shall be found in that neutral surface while their common inclinations to it is45°, then there will exist at the neutral axis the same intensity of stress on the two planes, but one stress will be tension and the other compression. It is farther shown that the common intensity of the two stresses is the same as that of either the transverse or longitudinal shear at the same point, which is also known to be | the mean for the whole section in the case of a solid rectangular beam. Now, since the intensity of shear at the neutral surface of such a beam is a maximum and zero at the top and bottom surface, and since it has been assumed that the entire web takes the shear only, it follows that if the shear be assumed to be uniformly distributed tliroughout any transverse section FLOOR-BEAMS AND STRINGERS. 3^9 of the web, that the latter may be supposed to be composed of an indefinitely great number of columns, each of which is an indefinitely thin strip of the web, making an angle of 45° with the axis of the beam. In a direction normal to these columns an equal intensity of tension will, of course, exist. One of the preceding assumptions is an error on the side of danger, by making the shear at the neutral surface only two-thirds of its actual value ; while the other, by making the shear at the top and bottom surfaces equal to the mean, instead of zero, is an error on the side of safety, and its in- fluence largely predominates over the former. The elementary columns of the web may be assumed to have their ends fixed at, or by, the flange rivets of a built beam, and if d' is the vertical depth, between rivet hole centres of the two flanges, the length of the elementary columns will be : 1= d' sec Of^" = \./\\\ d' (i). If 5 is the greatest total shear at any transverse section, A the area of that section of the web ; then taking the depth as d\ and s the mean shear, or : 5 these elementary columns will be subjected to an intensity of compression equal to s. Hence if t, the thickness of the web, is sufficiently great, there may be taken by Gordon's formula : ^ = -^^ (^)- Or, ' = '♦''57731) (4). If, for wrought iron, a — 3000 and /= 8000, there will re- sult : / = 0.0183 /i/ i (4). 8000 — s 330 DETAILS OF CONSTRUCTION. The empirical constants for steel are yet to be determined. In applying Eq. (4) to wrought iron plate girders, it will be found that the resulting values of t are excessive for large beams. The approximations already indicated, and the ad- ditional fact that the elementary columns are held in place throughout their whole length by the tension in the web, equal in intensity to the compression, and at right angles to the latter, are suflficient to justify the anticipation of such results, Eq. (3), therefore, has its chief value as the basis of an empirical formula for the web thickness. Although the web resists " shear," it is evident, from the preceding analysis, that the method of failure of a web will be that of buckling, in which the corrugations will be at right angles to the elementary columns. Hence, if the web is so held that these corrugations are prevented, its resistance will be very materially increased. This is accomplished by rivet- ing angles, usually in pairs, on each side of the web at proper intervals, so that the web plate is securely held between them. The office of these " stiffeners," it is to be remem- bered, is simply to stiffen the web, and prevent its buckling. If they are assumed to act as struts, the transverse shearing strain in the web at the section considered, must in so much exceed the compressive strain in the stiffener-stiaits, that the rivets can transfer to it its proper load, at the same time pre- supposing a perfect condition of riveting. In reality, neither of those conditions can possibly exist. These stiffeners are ordinarily riveted to the web at right angles to the axis of the beam. If no elementary column is to be without the support of, at least, one of these stiffeners in some portion of its length, they must be placed at a dis- tance apart, measured along the axis of the beam not greater than the vertical depth between rivet hole centres ; and that limit is very commonly given in specifications, although it is sometimes placed at once and a half that depth. About the same amount of stiffening would be secured by placing the stiffeners at 45° with the axis of the beam, and at intervals of twice the depths, but the difficulties of con- struction would be increased. FLOOR-BEAMS AND STRINGERS. 331 A safe rule, and one frequently used, though purely con- ventional, is to introduce stiffeners when the mean shear, i. e., S -^A, exceeds 4000 pounds per square inch for wrought iron. The function of the rivets holding the flanges to the web is next to be considered. In reality, these rivets may have two offtces to perform. If the load of the girder rests on one of its flanges, the flange rivets will sustain it directly ; but their chief office is to give the flanges their proper stress. If then the stresses be determined for any two points between the end of a girder, and the point of greatest flange stress, the shearing or bearing resistance of all the rivets between those points must'be equal to, or not less than the resultant of the difference between the determined flange stresses and the load resting on the flange between the same points. If no load rests on the flange, but is carried directly by the web, the " resultant " is evidently the simple difference between the determined flange stresses. These elementary considerations constitute the entire method of finding the pitch and number of rivets in the flanges of built beams, and will be applied to two examples. The complete design of the truss, treated in Art. ii, is to be given in subsequent pages, and in the present connection the stringers and floor-beams will be discussed. The strin- gers will be placed 7 feet apart centres, and the rails will be laid on 8 inch by 8 inch ties 9 feet long, spaced 16 inches from centre to centre. The ties, rails, guard rails, splices, spikes, etc., will then weigh about 325 pounds per lin. ft. The depth of stringers will be taken at 27 inches throughout their lengths, and the iron of each stringer will be assumed to weigh 100 pounds per lineal foot. The total fixed load will then amount to 263 pounds per lineal foot of each strin- ger, while the moving load is one-half of the concentrations given in the engine diagram of Art. ii. The principles es- tablished in Art. 7 show that the four 10,000 pound driving wheel loads will produce the greatest bending moment when either wheel ^ or ^ is at the distance 1.062 feet from the centre of the panel, and will be found under the wheel in question. 332 DETAILS OF CONSTRUCTION: With that position, the following moving load bending moments will exist : « Maximum 1 19,200 ft. lbs. 7^ feet from the end 105,400 " " 5 " " " " 85,600 " " 2i " " " " 44,800 " " It is evident that the last three of these values are not the greatest moments for those points, but as the flanges are to be of uniform section, it is not necessary to seek them, as might easily be done by the aid of the principles of Art. 7. The vertical depth between the rivet hole centres of these stringers will be taken at 2 feet. The flange stresses at the various points will then be : . 26% X (20.55)- IIQ,200 ^^ ,, „„ At centre . . . ^ „ ^ ^^ + -^ = 66,540 lbs. CD. 8x2 2 •' Iff 1 263 X 7.5 X 13.05 105,400 ^ u -c-n 7*ft.fromend — ^ '-^ ^-^-+ ^^ = 59,144 • ^^• ' ■ 2x2 2 ^ ,. ., .. 263 X 5 X 15.55 85,600 ., ^rr c " " " — ^ 2 ^-^ + • — - — = 47,920 " . GH. ^ 2X2 2 263x2.5x18.05 ^ ^4^800 ^ ^^^^g^ ,. _ ^^^ ^3 2X2 The allowed working stresses in the flanges of the stringers will be taken at 7,000 pounds per sq. in. of gross section in compression and 8,000 pounds per sq. in. of net section in tension. The diameter of rivets in the stringers and floor beams of railway bridges is chiefly a matter of judgment ; it usually ranges three-quarters to seven-eighths of an inch. In the present instance, rivets of the latter diameter, before being driven, will be taken. The metal punched out for a rivet should leave a hole not more than one-sixteenth of an inch greater in diameter than that of the cold rivet. But the metal immediately about the edge of the hole is materi- FLOOR-BEAMS AND STRINGERS. 33^ ally injured for tensile purposes, and in the tension-chord angles the disc of metal rendered valueless should be taken one-eighth of an inch greater in diameter than that of the cold rivet, i.e., for the present case, one inch. In the com- pression flange no metal need be deducted for the rivet holes. The upper flange section at the centre will be : 66,540 -4- 7,cxx) =9.5 sq. in. The net lower flange section will be : 66,540 ^ 8,000 = '^.l sq. in. Tlic following flanges will satisfy the conditions : ttj.^ n i 2 — 5" X 4'' 30 lb. angles. Upper flans^e . . . . ■ „ „ , ( I — 10 X I cover plate. Lower flange .... 2 — 5" x ^' ^"j Va. angles. The 62 lb. angles have a thickness of three-quarters of an inch, so that the excess of gross section exactly covers the metal destroyed by the punch. It is the best practice to run a plate the whole length of the upper flange in order to make it act as a unit in resisting compression. The lower flange needs no cover plate, and, again, the additional rows of rivet holes would produce more dead metal. Although there is a little waste of metal near the ends in small plate girders, it is economy to save labor by making the flanges of uniform section throughout their lengths. This economy ceases when the flanges become so heavy that one or more cover plates become necessary in the tension flange and more than one in the compression flange. Cover plates should be carried at least a foot beyond the section at which the additional metal is required, and rivets should be closely pitched in that portion. If rivets pierce both legs of an angle in tension, metal should be deducted for all the rows in both less. 334 DETAILS OF CONSTRUCTION. The locatiori of the preceding sections is shown in Fig. I, of PI. XI. The increments of stresses for the different seg- ments of the flanges will be : EC or FD — 66,540 — 59,144 = 7'396 lbs. GE or HE = 59,144 — 47,920 = 1 1,224 " GK or HL = 47,920 — 25,367 = 22,553 " KA or BL ^ = 25,367 " The extent of flange over which the weight of each driving wheel weighing 10,000 pounds will be distributed is indeter- minate, but as the ties are 16 inches apart centres, it will be sufTficiently accurate to take that distance as 2i feet. As each driving wheel passes over the entire stringer, the result- ant stresses which the rivets in the various sections into which the beam is divided will be obliged to carry, are as follows: EC or ED = y'( 7,396)^ + (10,000)^ = 12,440 GE or HE = y'(i 1,224)^ + (10,000)- = 15,000. GK or HL — ^(22,553)^ + (10,000)- = 24,700. KA or BL — y/{2^,^(^'jj + (10,000)^ = 27,300. In order to determine the number of rivets in any of these sections, it will be necessary to fix the thickness of web plate. The minimum thickness permissible is, to some extent, a matter of judgment, but it is safe to say that no built beam for railroad purposes should have a less thickness of web than YF of an inch, and a limit of | is still better practice. The latter limit will be used here. With the position of moving load already determined in Art. II, the greatest shear at the end of the stringer is 24,830 + 2,700 = 27,530 pounds = S. The sectional area /i of a 27 by I inch plate is 10.125 sq. ins. = A. Hence s — S -^ A — 2,720 pounds ; also / = 24 x 1.414. = 33-94 inches. Hence Eq. (4) gives: / = 0.45 inch. Since this result is greater than the assumed value of /, the FLOOR-BEAMS AND STRINGERS. 335 hypothetical elementary columns are not capable of sustain- ing their loads without exceeding by a little the proper work- ing stress ; but as the hypothesis involves a considerable safety error, the assumed value of t is probably ample. How- ever, as the additional metal is very small in amount, a pair of 3 by 2\ 1 6-pound L stiffeners will be introduced at the distance of 27 inches from each end as shown. The working resistance to shearing offered by rivets in the truss will be taken at 7,500 pounds per sq. in., and the limit- ing pressure between rivets and walls of holes will be fixed at 12,000 pounds per sq. in. under the same circumstances. The floor of a bridge is subject to shocks due to track imperfec- tions, and the above values should be reduced by 25 per cent., making the working resistance to shearing 5,625 pounds, and to pressure 9,000 pounds per sq. in. As the web is embraced by an angle iron on each side, each rivet will be subjected to double shear, and the thickness of bearing surface of the web will be much less than that of the two flange angles. In the de- termination of the shearing and bearing resistances of rivets, the cold diameter, before being driven, should, as a margin of safety, be considered. Hence, those resistances for the seven-eighths rivets under treatment, are: 2.\. 71 (0.875)^ • 5,625 = 6,750 pounds; and, 0.875 ■ 0.375 • 9'000 = 3,000 " The latter quantity is much the smaller, and will govern the number of rivets in each section, as follows : EC or FD 12,440 ^ 3,000 = 4 rivets required. 6^£ or ///^. ... 15,000 -^ 3,000 = 5 " " GKox HL 24,700 -^ 3,000 = 8 " " KA or BL 27,300 ^ 3,000 = 9 " " The nearest whole number is taken in each case. These results give at and near the end about nine rivets to each two and a half feet. A uniform pitch of three inches, therefore, will be assumed throughout each flange. If the 33^ DETAILS OF CONSTRUCTION. load is uniform in intensity, it is well known that the varia- tion of flange stress is very little for a considerable distance either side of the centre. This example shows, however, that concentrated loads may require just as close centre riveting, i.e., just as small pitch, as at the ends. The number of sec- tions into which a beam must be divided is a matter of judg- ment in each particular case. The greatest shear at the end of the stringer has already been seen to be 27,530 pounds, hence the end stifTeners at A must transfer that amount to the floor-beam web. An inten- sity of 4,000 pounds per square inch of normal section is a safe and proper value for those members. The end stiffeners will then be assumed to be 2 — 4" x 4" 35 lb. angles, one being on each side of the web at each end of the beam and extend- ing the full depth between the legs of the flange angles of the latter. Fillers whose thickness is just a little less than that of the heaviest flange angle will be required under these end stiffeners. Light intermediate stiffeners may be bent to fit if the flange angles are not too heavy ; otherwise fillers must again be used. The number of rivets required to transfer the greatest re- action at the stringer ends to the floor-beam (see Fig. 2, PI. XT.) is found by taking the reaction of the locomotive load (found in Art. 11 to be 34,600 pounds) and adding to it the fixed load, or, 20.55 ^ 263 = 5,405 pounds, then dividing the result by the bearing capacity of seven-eighths rivet in the three-eighths web of the floor-beam, as that is less than the resistance of the same rivet in double shear. The number of rivets needed will then be (34,600 + 5,405) -^ 3,000 = 13. Seven rivets will be placed in each 4x4 end angle, as shown in Fig. 2, of PI. XL, making 14 for the end of each stifl"ener. If the loads were very great, it would be neces- sary to reinforce the web plate of the floor-beam where it receives the ends of the stiffeners, by riveting a plate on each side. In the present instance, however, it is un- necessary. The bill of material, with the weights of one stringer, may now be written as follows : FLOOR-BEAMS AND S7AVNGERS. 337 1 27 X I Plate 20.55 ft. long 695 lbs. 2 5x4 30 lb. angles " " " 411 2 5 X 4 47 " " " " " 644 I 10 X I Plate " " " 257 4 4 X 1^ Filling plates 19 ins. " 60 43x4 •' " 19 " " 32 4 4 X 4 35 lb. angles 27 " " 105 4 3 X 2| 16 " " 27 " " 48 5 Rivets 1 20 Total weight of stringer 2,372 lbs. The actual weight per foot is thus about 15 lbs. more than was assumed. Although this makes an insignificant differ- ence in the flange section, and is amply provided for in the present instance, in practice the actual weight should be a little under the assumed, and not over it. The arrangement of connecting the stringers to the floor- beams shown in Fig. 2 of PI. XI., has the merit of making a very stiff floor system. It is proper to say, however, that many other methods are used. The small angle brackets seen under the ends of the stringers and riveted to the lower flange angles of the floor-beam are simply for convenience in erection, and are not considered as essential to the resistance of the joint. According to the preceding bill of material, the actual maximum weight concentrated at the stringer ends will be: 34,6cx) + 20.55 ^ 300 = 40,800 nearly. The depth of the floor-beam will be taken at 36 inches throughout its entire length, so that the vertical distance be- tween rivet centres in the two flanges will be about 33 inches. If the weight of the floor-beam be assumed at 150 lbs. per lineal foot, the total flange stresses at the centre, at the stringer points, and at points 2^ feet distant from the ends, will be : 338 DETAILS OF CONSTRUCTION. 40,800 X 5 150 X (17)^ _ 2.75 8 X 2.75 ~ ^ '^ ' 40,800 X 5 iSOJiA ^j3 ^5840, 2.75 2 X 2.75 /:?' t » 40,800 X 2. 5 150 X 2.5 _ _ ^t_! 2 + ^ ± X 14.5 = 38,090. 2.75 2 X 2.75 ^^ -^ '^ Using the same working stresses in the flanges as were fixed for the stringers, there will be found for the upper flange section at the centre : 76,170 -f- 7,000 = 10.9 sq. ins. ; and for the net lower flange section : 76,170 ^ 8,000 — 9.5 sq. ins. The following flanges will satisfy these requirements : ( 2 — 5" X 4" 36 lb. angles, ^^ ■' 1 I — 12 X tV cover plate. Lower flange 2 — 5" x 4" 55 lb. angles. The thickness of the 55 lb. angle is about 0.7 inch, so that if seven-eighth cold rivets are used the metal destroyed by the punch is just about equal to the excess of the total section over the net. The total shear at the end of the floor-beam is 40,800 + 8.5 X 150 = 42,075 lbs. = 5. If the web thickness be assumed at three-eighths of an inch, /^ = 36 x | = 13,5 sq. ins. Hence, s =■ S ^ A — 42,075 -f- 13.5 = 3,120 lbs. / = 33 X 1,414 — 46.66. These quantities inserted in Eq. (4) give : t = 0.68 inch. This value shows that if the hypothetical elementary columns are to sustain 3,120 lbs to the sq. in., the web must FLOOR-BEAMS AND STRINGERS. 339 be 0.68 inch thick. But such a thickness is plainly excessive, and shows how the formula errs in the direction of safety. A web thickness of three-eighths of an inch will be assumed, and 3" X 2\" 16 lb. angle stifTeners placed half-way between the stringer supports and the ends, as shown at EF, Fig. 2, PI. XI. These stiffeners will require 3" x |" filling plates 28 inches long. If shearing and bearing resistances of 5,625 and 9,000 lbs. per sq. in., respectively, be taken, as was done in designing the stringers, the bearing value of one rivet in the three- eighths web will be, as before, 0.875 ^ 0.375 x 9,000 = 3,000 pounds. Hence the number of seven-eighths rivets required between the three sections of the beam will be : Centre and C (76,170 — 75,840) C and EF (75-840 — 38,090) EF'' AB 38,090 3,000= I, 3,000= 13, 3,000 = 13. These conditions will be sufificiently near fulfilled if a pitch of 2| inches be taken for a distance of 6 feet from each end, and six inches for the remaining 5 feet between the stringers. The flange stresses do not require a six-inch pitch at the centre, but that value should not be exceeded, in order that the flanges may be properly bonded. The pitch in the cover plate on the upper flange should be 2f inches for a distance of 15 inches from each end, and over the remaining portion of its length that pitch may be doubled. In no case, how- ever, should the pitch in a compression plate exceed about 16 times its thickness. The end floor-beams are suspended from the pins by the plates shown riveted to the heavy end stifTeners, as at AB. Those StifTeners transfer half the weight of the beam in addi- tion to the 40,800 pounds at the stringer ends, to the suspen- sion plates, or, 40,800 4- 1 50 x 8.5 = 42,100 lbs. As these end stiffeners do not in this instance sustain this entire weight at any section, the latter cannot be analytically determined. They should be heavy, however, and will be taken as 5 x 4 .36 lb. angles. 340 DETAILS OF CONSTRUCTION. The number of rivets required between the end stiffeners and the web is 42,100 h- 3,000 = 14, but it is convenient to take 15 as shown. The dimensions of the suspension plates cannot be fixed until the diameter of the pin is determined, and they will be found in a later Art. The bill of material for the floor-beam with its weight will now be : 1 36 X I Plate 17 ft long 17 x 45 — 765 2 5 X 4 36 lb. angles " " " 34 x 12 = 408 1 12 X ^ cover plate " " " 17 x 12.5= 213 2 5 x 4 55 lb. angles " " " 34 x i8|^ = 623 4 3 x f filling plates 2\ " " 9^ x ^\ =■ 58 4 3 X 2| 16 lb. stiffening angles 3 ft. long 4 x 16 = 64 4 5 X I end " plates 2i" " 10.4 x 81 = 87 4 5x4 36 lb. angles 3 " " 4 x 36 = 144 4 3 X 2^ 16 " " I " " i^ X 16 = 21 \ rivets =130 Total weight of beam .... 2,513 lbs. The weight per lineal foot is then 2,513 -^ 17 = 148 pounds, or less than that assumed, as it should be. In the cases of large plate girders it is necessary to make splices in the web plate, a splice plate being used on each side of the web. The combined thickness of these splice plates should be at least 50 per cent, in excess of that of the plates spliced. The number of rivets on each side of the joint should be such that the total bearing resistance in the web, or the total shearing resistance of the rivets, shall at least equal the greatest possible transverse shear at the joint considered. At least two rows of rivets should be found on either side of the joint. It is sometimes customary, if web plates have no splices, to take one-sixth of the web section as acting in either flange. If no rivet holes were punched for the stiffeners, this method would be allowable. But such rivet holes frequently take out FLOOR-BEAMS AND STRINGERS. 341 considerable metal, and as the tension side of the plate only is affected, one-sixth of the remaining metal ceases to be a proper proportion. On the whole, therefore, it is better to neglect the bending resistance of the web, and allow it to balance, so far as it may, the effect of the rivet holes being out of the centre of gravity of the flange angles. That depth of plate girder Avhich will give the least weight depends entirely upon the manner and amount of the load- ing. With very heavy concentrated loads, it may be half the span ; on the other hand, with very light loads it may be less than one-twentieth of the span. As all bending moments may be supposed to be caused by some uniform load, either fictitious or real, the following analytical discussion may be of some value as well as interest : Economic Depth of Plate Girders with Uniform Flanges. If a plate girder carries a uniform load, and is designed with flanges of uniform cross sectional area, the depth which will give the least weight of girder may easily be obtained. Let / = span in feet. d — depth " '• t — web thickness in inches. / = allowed working stress in lbs. per sq. in. for the flanges. /' = " " " '' " end stiffeners. a = sectional area of each intermediate stiffener in jr^. _/y. n = number of stiffeners (intermediate). lu = total load per lin. ft. of girder (in pounds). The flange stress at centre will then be : F= — %d' The volume of the web plate in cu. ft. will be : 12* 342 DETAILS OF CONSTRUCTION. If one-sixth of the latter is taken to be concentrated in the flange, the volume of the two flanges in cu. ft. will be: 2FI I Idt _ wP Idt 144/ 6 \2 ~ S7^P^ 36' The volume of the end stiffeners will be : wld 144/' .and that of the intermediate stiffeners: nad. The volume of the entire girder will then take the value: -, wP Idt wld , , , V— — -^—j + -- + , + nad . ....... (5). 576/(3^ 18 144/ ^^' By taking the first derivative ; dV _ _ wP It wl _ did) ~ S7^pd^ ^ 18 ^ 144/^ + na-0. Solving for d^: ^ , // wl S7^P { —o + Ti + ^^^ ^ m8 144/ .d='-A/ ^ . . . (6). ^ /(8// + ^ + I44««) If one-sixth of the web is not concentrated in each flange, 12// will take the place of 8// in Eq. (6). If all stiffeners, both end and intermediate, are omitted, Eq. (6) will take form : .= ^Y|. ..... (7). In reality / is seldom or never exactly the same for both EYE-BARS OR LINKS. 343 flanges, since it is the working stress in reference to the gross section. It will be sufficiently near, however, for all usual purposes to make it a mean of the two actual working stresses for the gross sections. It should be borne in mind that local circumstances fre- quently compel a different depth from that given by Eq. (6). It will also be found that a considerable variation from that depth will cause a comparatively small variation in weight. Again, the difficulties of handling a deep girder, and the shop cost per pound, may, and usually does, make the economic depth a little less than that found by the aid of Eq. (6). Art. 73. — Eye-Bars or Links. An "eye-bar" or "link" is a tension member of a pin- connection bridge, fitted at each end with an eye for the in- sertion of a pin. Two views of an eye-bar are shown in Fig. 5, PI. XII.; A is the body of the bar, D the neck, and C the eye. The head of an eye-bar is the enlarged portion in which the pin-hole is made. The eye-bar is one of the most im- portant members of a pin-connection bridge, and the deter- mination of the relative dimensions of the head has been the subject of much experimenting. A mathematical investiga- tion; however, with the same object in view is a matter of considerable complexity, although an approximate solution of the problem may be obtained, and its agreement with the results of experiment is quite close. Before taking a general view of the stresses which may arise in an eye-bar head, it must be premised that a difference of ^V" to ^V" between the diameter of the pin and that of the pin-hole is considered exceedingly good practice. Before the eye-bar is strained, therefore, there is a line of contact only between the pin and eye-bar head, but on account of the elasticity of the material, this line changes to a surface when the bar is under stress, and increases with the degree of stress to which the bar is subjected. This line and surface of contact is, of course, in the vicinity of K, Fig. 3, PI. XII., i.e., on that side of the pin toward the nearest end of the bar. The consequence of this is that, when the bar is strained, the 344 DETAILS OF CONSTRUCTION. portion about KA, Fig. 3, is subjected to direct compressiort and extension ; that about BL, DH, and FM to direct ten- sion and bending, while in the vicinity of CN (also CQ) there is a point of contra-flexure, and the stress in the direction of the circumference changes from compression to tension as H is approached from K. It should have been said before that if w represents the width of an eye-bar, as shown, then its thickness, /, is gener- ally included between the limits \w and \w. These limits of the relative values of the quantities are seldom exceeded. Fig. 2, PI. XII., represents a method of laying down an eye-bar head which has been determined by a very extensive system of experiments given by a member of the British In- stitution of Civil Engineers, and one that has stood the test of long American experience ; in short, there is probably no better method known. Let r represent the radius of the pin- hole, and w the width of the bar. Then take EN = o.66iu. The curve DRBK is a semicircle with a radius equal to r 4- 0.66w, with a centre, A, so taken on the centre line of the bar that QB = 0.87a'. GF is a por- tion of the same curve, with A' as the centre (yi'C= y^Q ; GH'xs any curve with a long radius joining 6^/^ gradually with the body of the bar. HG should be very gradual in order that there may be a large amount of metal in the vicinity of CG, for there the metal is subjected to flexure as well as direct tension. FD is a straight line parallel to the centre line of the bar. Fig. 3 shows another method founded on the results of a mathematical investigation. Take r and w as before. Then BC= AC = r + 0.S77V, DH— \w =■ o.66w, ED = EF= 2r + w. DF is described with ED until DCF= 45°. BAB is de- scribed with BC until BCA = 35°. BN is drawn from Z as a centre located in such a position as to cause that arc to be at the same time tangent to j97Vand AB. DN \s a straight line drawn parallel to the axis of the bar. PF'xs any easy curve which will appear the best. The dotted lines in both Fig. 2 and Fig. 3 show the slope that should be given in order to clear a die. SIZE OF PINS. 345 The outline of the head is now usually formed of a portion of the circumference of a circle whose centre is the centre of the pin-hole. In such a case no dimension of the head should be less than the corresponding one determined by either of the methods just given. Fig. 4 shows the head thickened in such a manner that the mean maximum intensity of pressure between pin and pin- hole shall not exceed a given amount, p. Let 7" represent the maximum intensity of tension in the body of the bar; then, as has been shown in discussing the pressure against the bodies of rivets : wtT wtT= 2rpt' .-. t' 2rp Art. 74.— Size of Pins. The exact analytical determination of the pin diameter in any particular case is, like many other matters, involving the elasticity of materials, an impossibility, although the problem in its simplest form was subject to a very able mathematical investigation by Charles Bender, C. E., in Van Nostrand's Magazine for October, 1873. One or two reasonable assump- tions, which, in a great majority of cases, must be very nearly accurate, give the problem a very simple character. The first of these assumptions is that tJic pressure applied to any pin has its centre at the eentre of the surface of contact. Fig 3 of PL XI. represents the lialf of a pin-connected joint, LL being the centre line, and by this assumption the centre of pressure be- tween each of the eye-bars A, B, C, D, etc., and post bearing P, and the pin is located half-way between the faces of those members normal to the axis of the pin. If, however, a pin is held by a compression member, such as an upper chord or post, then the centre of pressure in that member may be taken as the eentre of such a surface as zvill reduce the bearing intensity to its maximum limit. It is to be premised that the general considerations touch- ing the distribution of pressure between rivets and plates given in Art. 70 hold equally true for pins. The greatest al- 346 DETAILS OF CONSTRUCTION lowable bearing intensity between pins and eye-bars of wrought iron ranges from io,ooo to 12,500 pounds per square inch of the surface found by multiplying the diameter of pin by thickness of bar. The latter product is always considered the bearing surface. If two bars only, such as A and D, act on each end of a pin it is clear that the centre line of the latter will be convex toward D. The result will be a movement of the centres of pressure of those bare toward each other ; so that the lower arm of A will be less than half the thickness of that member plus half that of B. The second assumption given above seems thus very reasonable, and may be extended to the case of a pair of eye-bars, only, at the end of a pin. When, on the other hand, a number of eye-bars of various sizes take hold of a pin, particularly if the bending moments have dif- ferent directions at different sections of the pin, the axis of the latter may be essentially straight and the centres of press- ure should be taken according to the first assumption. This is, in reality, the best practice in all cases, for if the centre of pressure departs from the axis of the bar, the latter will be subjected to a bending moment equal to the tension in the bar multiplied by the distance of the centre of pressure from its axis. Hence the necessity of so fixing the diameter of pin that it shall be as stiff as possible. In Fig. 3 of PI. XL, let a be the distance between the centres of eye-bars A and D ; a, that between D and B ; a'' that between B and £, etc., etc. These distances a, a', a", etc., should always be taken as the thickness of the head plus one-eighth of an inch ; the latter amount representing about the proper clearance in the best work. Then let Ta, Ta, T^, etc., represent the total tensions in the bars A, D, B, etc. The bending moments about the centres of those bars will then be : About centre of D ... .a T^ '' B ... {a + a') Ta - T^a'. " " '' E . . . . {a + a + a") 7^ + T(,a" — Ta{a' + a"). Etc., etc., etc., etc., etc. SIZE OF PINS. 347 The rod 7? is a counter and does not usually act when the pin receives its greatest bending. The preceding moments are all similarly formed and are about vertical axes until the centre of the post bearing, P, is reached. The tension T of the main tension brace T pro- duces a moment about an axis normal to its own. Let it be supposed that the resultant moment of all the chord members A, B.^ C, D and E about the centre of P is right-handed look- ing vertically down, as shown by M' in Fig. i. Let M' represent that moment by any con- venient scale. The moment of T by a!°, or 7^", will be right-handed looking upward ; and let Mt represent that moment by the same scale as before. The latter line is drawn normal to the axis of the member T. The line M will now represent by the same scale the moment to which the pin is subjected at the centre of P, and its direction is that of the axis of the moment. ^^^- ^' The greatest pin bending in the lower chord will usually take place with the greatest chord stresses, but the upper chord pins will receive their greatest moments by the great- est web stresses. When a number of bars are coupled to the pin in such a joint as that shown in Fig. 3 of PI. XL, it is usually necessary to test a number of sections in order to find the greatest mo- ment; unless the bars are very nearly of the same size and placed alternately as shown when the greatest moment will be found at the centre of the pin. It is frequently advisable, however, to employ different sized bars in order to reduce the bending moments ; a small bar being placed at the end of the pin. The same reduction of bending moments is brought about even more effectually by the arrangement of lower chord bars shown in Fig. 2. In that figure it will be observed tJiat the lower chord eye- bars are so grouped on any one pin, that the stresses in them- for eaeh half of the pin, form couples which have opposite 348 DETAILS OF CONSTRUCT ION. signs and, thus, to a great extent, or wholly, neutralize each other. By varying the sizes or thicknesses of the bars and by re- sorting to the method of grouping shown in Fig. 2 (which Fig. 2. represents a portion of an actual lower chord) the bending moments in lower chord pins may easily be reduced to any desired extent in any case whatever. It is evident that the resultant moment shown in Fig. i could be obtained by resolving the stress T into its vertical and horizontal components and combining their moments with those of the lower chord stresses, making the components of M vertical and horizontal instead of vertical and inclined. The bending of pins is very much increased by thickened eye-bar heads, since the thickening increases the lever arm of the tensile stress in the eye-bar. The thickened eye is a most excellent thing for the bar, but necessitates an increased diameter of pin. The preceding operations illustrate the general method of finding the bending moment to which a pin is subjected in all cases ; the component moments are determined from the stresses in the individual truss members, and the resultant is then found by the moment triangle or polygon. The pin diameter is then readily found in the following manner. If M IS, the external bending moment, /the moment of in- ertia of the normal section of the pin about its diameter D and K the intensity of stress in the fibres most remote from D, then it is known from the theory of flexure, since 1 , that 64 M: 32 D 2.2 j/: M K (I)- SIZE OF PIXS. 349 If K is known, Eq. (i) gives D at once after M is found by the general method cxempHfied by Fig. i, or in any other manner. For wrought-iron pins in the trusses of railway bridges, K is usually taken at 15,000 pounds. This value in Eq. (i) gives for wrought-iron pins : D = 0.089 VJT (2). For steel pins, under similar conditions. A' may be taken at 20,000 pounds, for which : i; = 0.081 \jJi (3). There are numerous tables showing the bending moments of pins of all usual diameters with given values of A', so that in practice the computations expressed in Eqs. (i), (2) and (3) are seldom necessary. The value of M is determined for any particular case, after which, by the simple inspection of a table, the proper diameter may be chosen. It is seen by Eq. (i) that the diameter of a pin varies di- rectly as the cube root of M and inversely as the cube root of A'. It may sometimes happen that Mt, in Fig. i, is so small that it may be neglected ; in which case M — M' . No pin should possess a diameter less than eight-tenths the width of the widest bar coupled to it. When bending and bearing are properly provided for, a safe shearing resistance will be amply secured. If the apparent moment in the pin is sufficient to cause failure by flexure, it does not, by any means, follow that failure will actually take place; for the distortion of the pin beyond the elastic limit will relieve the outside eye-bars of a larger portion (in some cases perhaps all) of the stress in them. This result will pro- duce a redistribution of stress in the eye-bars, by which some will be understrained and the others correspondingly over- strained. Thus, although the pin may not wholly fail, the safety of the joint will be sacrificed by the overstrained metal in the eye-bars. 350 DETAILS OF CONSTRUCTION. Art. 75.— Camber. Camber is. the curve given to the chords of a bridge, caus- ing the centre to be higher than the ends, or rather it is the amount of rise of the centre above the ends. It is given to a truss so that the chords may not fall below a horizontal line when the load is applied. Fig. 8, PI. XII., represents a truss with exaggerated camber. The actual amount varies from ■^l^th to yaVoth of the span. Camber may be given to a truss either by lengthening the upper chord or shortening the lower one ; the latter method is preferable because the upper chord is sometimes not hori- zontal, and different panel lengths would have to be shortened by different amounts. On account of the unavoidable play at the joints of all work, the shortening of the lower chord, or lengthening of the upper, must be increased by about ^V^ of an inch per panel in order to secure the desired camber. The lower chord shortening is made uniformly throughout its length ; that is, each panel length is shortened by a con- stant quantity. The true chords will, therefore, become arcs of circles of very large radii, and vertical posts will become radial. By means of the equation of the circle, y^ = 2Rx — x^, R being the radius, the amount of shortening or lengthening of chord to produce a given camber may be determined if the play at the joints be omitted. In the equation above, y rep- resents the half span and x the camber desired, hence the radius R = __y' + x^ 2X This is the radius of the lower chord when cambered. Generally it will be near enough to put R — — 2X The angular length of the lower chord will be 1 J' -1 2xy a— 2 sin-^ -f^ — 2 stn~^ -s — ^ , R jr + x^ CAMBER. 351 and the length in feet : / = Ra. The length of the upper chord will then be : /■ = {R^- d) a. The difference in length of chords will be : D ^ i - I = da. This is the amount by which the lower chord is to be shortened, or the upper lengthened, in order to produce the required camber, if no play or strains exist. Since x is very small compared with y : a — 2 sur^ -^ — =^ = -^ p , ., = ^^— (nearly). If the span is l^ : S X , „ ^^^ a = —z— ; and, // = -j— . h n If r is such a ratio that d = r/j : D = 8rx. On account of the play at the joints, x should be taken a little larger than the camber desired. Frequently r is about one-eighth, and for such a value: D = x; or, neglecting the play at the joints, t/ic difference in lengths of the chords should equal the camber. If the chords are to be horizontal under the greatest loads, while T and C represent the supposed uniform intensities of tension and compression in the lower and upper chords respectively, E and E' representing the coefficients of elas- ticity ; T C 352 DETAILS OF CONSTRUCTION. This formula can only be approximate, for the chords are never exactly uniformly stressed, and the coefficient of elas- ticity is probably never the same throughout either chord. Since d, the depth of truss, does not vary, these formulae apply only to trusses of uniform depth. A "through" truss has been supposed, but the same for- mulae exactly apply to a deck bridge. It is to be borne in mind that one-half the horizontal dis- tance between the centres of end pins is to be taken iox y in determining R. If this distance is assumed in designing the truss, then the panel length is to be found by dividing / or /' by the number of panels. If the panel length is first assumed, and the camber pro- duced by shortening or lengthening it, then this horizontal distance is essentially equal to the assumed chord length diminished or increased by D — da. In order to hold the camber in a truss, the diagonals must be shortened, as shown in Fig. 9, PI. XII. The diagonal which was bd before cambering, becomes ed afterward, ad and be are supposed to be panels in the upper and lower chords respectively before putting in the camber ; afterward be becomes ef, while ad remains the same ; the lower chord is supposed to be shortened. Let x be the amount of shorten- ing of each panel of the lower chord == 2be = 2/e ; d, the depth of the truss ; and/ the original panel length equal to ad. Then ed = ^d200 75 " 100 3,100 100 " 1 50 3,000 1 50 " 200 2,goo 200 " 300 2,800 300 " 400 2,700 400 " 500 2,500 Floor-beams and stringers are really bridges of short spans equal to their lengths, consequently they must be designed for the heavy loads belonging to those short spans. Fig. i shows the locomotive weight specified by Fig. I. Mr. Theodore Cooper, C.E., in his " General Specifica- tions for Iron Bridges and Viaducts," while Fig. 2 shows the heavy passenger locomotive used by Mr. Jas. M. Wil- son, C.E., in his standard specifications for the Pennsyl- vania R. R. FIXED AND MOVING LOADS. 357 These represent the heaviest engines of their type now in use. There is, however, a heavy decapod engine shown by Fig. 3, beginning to make its appearance. K:-5-'0'-^^-5'6--^ — 9'0--^— 8'0-— ►]. 9'6-— *)— 5'0-4^5'6-4"5V4-3^ij ^-»J Fig. 2. The moving load usually specified consists of two of any of these types of locomotives followed by a uniform train of 3,000 pounds per lineal foot. Occasionally the locomotive concentrations are followed by those of the train. Besides the preceding heavy moving loads, there are in- Km. A si s' 81 «l SI I v.. .',• 1' .0. Li I ' ,r ' I ' ' ' I -M'S- FiG. 3. numerable lighter ones depending upon local circumstances of trafific. The actual concentrations play a very important part in bridge computations. The old method of a uniform load, even with an engine excess, no longer fulfils the requirements of the best engineering practice, particularly in the treatment of short spans. This is well illustrated by the following table which shows the uniform load per lineal foot which will produce the same 358 DETAILS OF CONSTRUCTION. chord stresses at the centre of the span as the actual concen- trations shown in Fig. i. span in feet. Equ iv. uniform : load Span '. in feet. Equiv. uniform load in lbs. per lin. ft- in lbs. per lin. .ft. 55 3750 25 4,838 50 3,866 20 5,137 45 4,004 15 5,760 40 4,242 12 6,000 35 4-336 10 5,766 30 4,572 5 9,600 Above 55 feet the equivalent uniform load per lineal foot will slowly decrease until it reaches a value of about 3,200 lbs. for 100 feet and over, /. r., supposing the moving load to consist of a train of such locomotives. Art. 78. — Safety Factors and Working Stresses. Although the subjects of safety factors and working stresses properly belong to the domain of the resistance of materials, they may here be touched upon in a general manner. The fixed weight of a long span bridge is much greater, per foot, than that of a short span. Again, it has been seen in the preceding article that the moving load for a long span is much less than that for a short span. For both these rea- sons, the variations of stress in passing from a loaded to an unloaded condition are much greater in the material of a short span than that of a long one. Consequently, the mate- rial will be much more fatigued in a short span than in a long one. Although the subject of the fatigue of metals is yet in an unsettled state, it is clearly established that these conditions of stress in short spans demand a larger safety factor, or smaller working stress, than those in the long spans. Again, in any bridge or truss whatever, carrying a moving load, some parts are subject to a much greater variation of SAFETY FACTORS AND WORKING STRESSES 359 stress in the process of first bein' closely to a suddenly applied load, although it is quite certain that it does not. For this reason some engineers have doubled the moving loads, in making their calculations, and then fixed the values of the safety factors as if all loads were gradually applied. But no track is in perfect condition, and all rough places, or lack of continuity, such as rail joints more or less open, produce shocks which cause greater stress than any suddenly applied loads. The amounts of these last stresses are inde- terminate, for the extent of their causes can scarcely be determined. Again, Mr. J. W. Cloud, C. E., at the Philadelphia meeting of the American Institute of Mining Engineers, February, 1881, pointed out the existence of certain unrecognized stresses ; such as those caused by the vertical component of the thrust of the connecting-rod of a locomotive, which alter- nates in direction twice in each revolution of the driving- wheels, thus producing a pulsating effect, as well as those which arise from the lack of balance of the driving-wheels in a vertical direction. All these causes produce stresses which it is impossible to measure, and the safety factor must cover all uncertainties. It is possible that a more highly perfected track and the production of more nearly uniform material in connection with an extended experience may justify the reduction of safety factors. The following " Table of Tubular and Truss Bridges for Single and Double Track Railways, constructed of Iron and SAFETY FACTORS AND WORKING STRESSES. 36 1 Steel and having Spans exceeding 300 feet," gives the work- ing stresses, loads, and other interesting data of some of the principal bridges of the world. It is taken (with the excep- tion of No. 18) from the " Proceedings of the Institution of Civil Engineers" of Great Britain, Vol. LIV. The greater portion of it is there given in connection with a paper by Mr. T. C. Clarke, on " Long Span Bridges." No. Date of Erection. ?; O u ^'i- n C C/i_ - o ^ >^ (/)_, >? - r 6= • u> ■ D : a. 2. IT Sr o ffic m ft 03 ft - n Span in Feet between Points of Bearing^ Tons of Iron. (2000.00 lbs.) •^ Width between Centres Trusses. of .''B 4, „ u. -P' » 4. -u n Total Dead Load of Iron Tia a\ 8 o\ ^ 8 § and Timber in Track. » K> » w Live Load of Engines 8 ^ 8 ■^ 8 8 4>. and Cars. a) 5 1 ^j 8, -^ ■^ S ^0 '^ 1 From Fixed Load only. H Ul M ' n 3- 8 oFo o'S n 1 R 8 R From Total Fixed and Moving Loads. s 5 tfl Q. to B 4- 8 en 8 "8 From Fixed Load only if 1 I ^ 0° 00 c^ >o From Total Fixed § § 8 § 8 and Moving Loads. m z » *»j £ ■u Test Load. Tons. (2000 lbs.) - - - " ^^ ^ ? ^j -vl P Centre Deflection. 00 CO H OJ i Dead Load of Iron, Tin Deflection of Span from iber, &c. 1^ ^ its own Ln p Weight on Removal of Scaffold'g. H > H ;? > H H > ^ s*" S c r* — ^ angular truss 3 cliords cast ailed iron, onnections ; angular truss il 3 :q J? 3 P 0T3 "■ s - 0. 3 1x5 Si ' c S3 wo P 2-33 7 "TO 3 g ETS Sag- 5 cr 3 a 3 3 g) J3 St r 3 S c Tl ? 1:1 • X) p t ^5 '!£ V-3 ^5 3 c Z 3 1 — o"'3uOM--:i ii o . — ■- 2 -^ _— o "1 ^ ' 3 3 ■» " '-•T ' Qli rt w ".^ ^ n; C ,„ 3 S , T3 rt ed ir ction lartr ca ctual = w 3 "O nj a w C C rt . = ii — u »i b.s < <5 8 R 8 o 8 o M OO o o o g OO VO S o S 8 8 8 vC NO ir> " I^ £? m Di ^ -i Oi ■£0 O 'rt V rt irt ^ 3 •3 c rt u .yi S J3ffi ' OvO OO s •^ On :r ::• lO c o in o in o OO o r» o t^ 8 « s Ov M "' '^ xn W^' ■ . £E «^ •'^ i B '.' 3 -l.s Q _< Q — M Q — ^^ — S '-' 00 no' r'N ON o r-^ 2 t^oo NO 1-^ 2S 00 00 vo r^ 13 -^ o o m f*^ m t^ 8 C4 ^ <4- ■«• ■a- ^ N — ~ in Ul m • 00* OO I^NO f-. c* CI « NO 00 c^ ■+ N m tN. r^ ON U^OO -* 0^ O r^ ^ r- Jl m rr> 2l (55 oS J ^"C 13 O 3 S U "a 'CC g > WW 2 s. ■^ 3n - I! J^ M :; NO No. 1 s » 00 OO OD 00 OO 00 s. Date of Erection. ^ 00 4. KJ> 2 1 < n X » JO ?3H O ■<" w - 3 2.0 2_ re n < o 3- 3 W : S < ■-t ■ u- - ■ C?'< ■ " ■ S • 2 ° N ■ T • ft ■ re' ?5 c o O O > rn 5 3) T^K Q t/i < n ^ 1 1 P W 2 o 2, 3 d 3 3 re 3) re 3 3 •s. 5' re 3 n 3 3 n j:. OJ l Vl ^ ^ OJ o-^ J" z d s z u „ .. « „ ^ re 3 8 ~ On 4. OJ 4. ^ CO P IW z 1) n , ? u\ -^i W -J - 4. M 4. OJ M 4^ ro M 4. N NO 4^ ON O OJ ^^ w -J OD 00 Ln-J OJ M OJ 3 3- ' ' Ln ^ <^ 4. ro K) OJ N 4. NO +■ ^ * Total Dead Load of Iron ?Fs N Jo NO - Moving Loads. ? z OJ 4». M It 'oJ Test Load. Tons. (2000 lbs.) „ „ « « ^ i__^ OJ Ui OJ '!n •^ 3 Centre Deflection. fj OO'jJ ON Ol VO UN , H 4^ ■^ ^ o Dead Load of Iron, Timber, &c. Deflection of Span from its own 3 Weight on Removal of Scaffold'g. u> O t/! D ■0 t/)3 ,3 o sriw C 'TQ sS;r s s- cr «! — " « re W 2 3 -1 ;p 2 S w KT ?r w T3 ■ ■0 3E GENERAL OBSERVATIONS. 365 Art. 79. — General Observations. All abutting surfaces in bridges, or similar structures, should be very carefully machine finished. Where pins bear against portions of the upper chord, as at c, d, and e of Fig. 5, PI. III., the amount of bearing surface should be determined as for rivets, and sufficient area given by riveting on thickening plates, if necessary. A thickening plate is shown at the joints of the same figure. The num- ber of rivets for the thickening plate is determined by the amount of pressure allowed on each one, as has already been shown. If a finished piece is to fit into a finished cavity, however well the work may be done, there must be at least ^^ inch ^' play." One end of a truss bridge, unless the span is very short, usually rests upon " expansion " rollers, from two to four inches in diameter. An approximate formula for the resist ance of such rollers is given in the Appendix. CHAPTER XI. WIND STRESSES AND BRACED PIERS. Art. 80.— Wind Pressure. In a paper presented to the American Society of Civk Engineers (Transactions, Vol. X.), Mr. C. Schaler Smith gives some very valuable information in regard to wind pressure. The highest observed pressure which has come within his knowledge is 93 pounds per square foot. This pressure derailed a locomotive at East St. Louis, Mo., in 1871. In his- own specification he says : — " The portal, vertical, and horizontal bracing shall be pro- portioned for a wind pressure of 30 pounds per square foot on the surface of a train averaging 10 square feet per lineal foot, and on twice the vertical surface of one truss." The wind pressure on a train is a moving load, and should be so considered, while the wind pressure on the trusses is a fixed load. His experiments on the Rock Island draw-bridge showed that the wind pressure against the two trusses was over 1.8 times that on the exposed surface of one. Again, quoting from his paper : — " The Erie specifications are as follow : Fixed load, roadway chord, 150 lbs. per lineal foot. other " 150 " Moving " roadway " 300 " " " " Iron in tension at 15,000 pounds. " " compression, factor 4. "The Pittsburg-, Cincinnati and St. Louis Railway requires 366 WIND PRESSURE. 367 300 pounds per foot for the train, and 30 pounds per square foot on one truss only. " For the bridge over the Missouri, at Glasgow, 50 pounds per square foot on one truss, and 300 pounds per lineal foot of train were used. " For the Eads bridge, at St. Louis, 50 pounds per square foot on the structure alone was the specified pressure. " For the Kentucky River bridge the wind pressure was assumed at 314 pounds per square foot on spans, train, and piers, and factor 4 was used in proportioning the bracing. " The Portage bridge, New York, was built to resist 30 pounds per square foot on structure and train, and 50 pounds per square foot on the structure alone. " The 520 feet span over the Ohio, at Cincinnati, was de- signed to withstand 50 pounds per square foot on structure alone, or 30 pounds per square foot on train and structure combined. " A fully loaded passenger train, and the heaviest possible freight train, will leave the track at the respective pressures of 31 4 and 56^ pounds per square foot." Engineers frequently specify 30 pounds per square foot of trusses and train combined, or 50 pounds per square foot of trusses alone. 300 pounds per linear foot of single track is also frequently used for moving wind pressure on train. The following refers to the single track bridge at Platts- mouth, Neb., and is from the Railroad Gazette, 17th Dec, 1880: "The structure is also designed to resist a lateral wind pressure of 500 pounds per lineal foot on the floor, and 200 pounds per lineal foot on the top chord of the through spans and the bottom chord of the deck spans; these quantities are about equivalent to a wind pressure of 30 pounds per square foot on the bridge when covered by a train, and to 50 pounds per square foot on the empty bridge." The following are a set of rules recommended for English practice, almost exactly in the words of the report : 368 WIND STRESSES AND BRACED PIERS. Report of the Committee appointed to eoisidcr the Question of Wind PressKre on Raihvay Striiet2ircs, by the Board of Trade of London, made on the 20th May, 1881. The following rules were recommended : ( i). For railway bridges and viaducts, a maximum pressure of 56 pounds per square foot should be assumed for purposes of calculation. (2). That when the bridge or viaduct is formed of close girders, and the tops of such girders are as high or higher than the tops of passing trains, the total wind pressure upon such bridge or vi-aduct should be ascertained by applying the full pressure of 56 pounds per square foot to the entire verti- cal surface of one main girder only. But if the top of a train passing over the bridge is higher than the tops of the main girders, the total wind pressure upon such bridge or viaduct should be ascertained by applying the full pressure of 56 pounds per square foot to the entire vertical surface, from the bottom of the main girders to the top of the train passing over the bridge. (3). That when the bridge is of the lattice form, or of open construction, the wind pressure upon the outward or windward girder should be ascertained by applying the full pressure of 56 pounds per square foot, as if the girder were a close one, from the level of rails to the top of the train passing the bridge or viaduct, and by applying, in addition, the full pressure of 56 pounds per square foot to the as- certained vertical area of surface of the iron work of the same girder, situated below the level of the rails or above the top of a train passing over such bridge or viaduct. The wind pressure upon the inward or leeward girder or girders should be ascertained by applying a pressure per square foot to the ascertained vertical area of the surface of the iron work of one girder only, situated below the level of the rails, or above the top of a train passing over the said bridge or viaduct, according to the following scale : {a). If the surface area of the open spaces does not exceed f of the whole area included within the outline of the WIND PRESSURE. 369 girder, the pressure should be taken at 28 pounds per square foot. {b). If the surface area of the open spaces lie between | and I of the whole area included within the outline of the girder, the pressure should be taken at 42 pounds per square foot. {c\ If the surface area of the open spaces be greater than I of the whole area included within the outline of the girder, the pressure should be taken at 56 pounds per square foot. (4). That the pressure upon arches and piers of bridges and viaducts should be ascertained, as nearly as possible, in conformity with the rules above stated. (5). That in order to insure a proper margin of safety for bridges and viaducts, in respect of the strains caused by wind pressure, they should be made of sufficient strength to with- stand a strain of 4 times the amount due to the pressure calculated by the foregoing rules. And that for cases where the tendency of the wind to overthrow structures is counter- balanced by gravity alone, a safety factor of 2 will be sufficient. John Hawkshaw. W. G. Armstrong. W. H. Barlow. G. G. Stokes. W. YOLLAND. The evidence before us does not enable us to judge of the lateral extent of the extreme high pressures occasionally re- corded by anemometers, and we think it desirable th:t experiments should be made to determine this question. If the lateral extent of exceptionally heavy gusts should prove to be very small, it would become a question whether some relaxation might not be permitted in the requirements of this report. W. G. Armstrong. G. G. Stokes." Up to the date of the above report, the highest pressure per square foot ever recorded at Glasgow was 47 pounds ; 24 370 WIND STRESSES AND BRACED PIERS. while the highest ever recorded at Bidston, near Liverpool, was 90 pounds per square foot. At another time, at Bidston, 80 pounds per square foot was recorded. The above committee also found that, if P is the maxi- mum pressure per square foot, Fthe maximum run of wind in miles per hour, both these quantities being observed by anemometers, the following equation very nearly held true : 100 Art. 81. — Sway Bracing. The construction of the upper and lower sway bracing of a truss must, so far as the jar and oscillation of a moving road are concerned, be a matter of judgment ; but the stresses due to the action of the wind may be determined with suffi- cient accuracy. Although a moving train will partially shelter one truss, it seems no more than prudent, with the ordinary open style of American bridge, to consider the action of the wind as exist- ing constantly, during the passing of a train, over the whole of the projection of each truss in the bridge on a plane nor- mal to the direction of the wind. If this is considered exces- sive, however, for low trusses, that portion of the windward truss sheltered by the train may be omitted. Let Fig. I and Fig. 2 represent a single concellation rail- way truss bridge, with vertical and diagonal bracing, and let the wind be supposed to blow in the direction shown by the arrow, which is normal to the planes of the trusses. Primed letters belong to the truss DC'N'O,' but all are not shown. As the truss is a " through " one, all wind pressure against the floor system will act in the lower chord. With the wind pressure between thirty and forty pounds per square foot, the following loads may be taken at the various panel points : SWAV BRA CING. 3 7 1 At C, G\ K', L, M\ N\ C, G, A', L, M, N. 0.35 tons. " n sindO' 0.18 " " Intermediate points 0-35 " " ^ and (9 0.18 " " Intermediate points 3.01 " \ ' /^ Fig. I. I Fig. 2. M N 'A \y V V / \ ob/ / e' E \ /\ /\ /\ / N°'^ ./>", pro- ducing a bending in DN which, has its greatest moment M at O. Hence : M=2{F-hF')a (6). If A' is the greatest intensity of compressive stress (due tc flexure) in the cross section of the post DN, at O, d^ the greatest distance of any compressed fibre from the neutral axis of the cross section, and /the moment of inertia of the section about an axis passing through its centre of gravity and lying in the plane of the truss ; then, by the well-known formula — K = '!^ (7). At O there will then exist the intensity of compression : — { — + A^) ; in which q is the area of cross section of 382 WIND STRESSES AND bRACED PIERS the column. The intensity of compression — ( — -I- K] is- \^ / in addition to the regular truss stresses arising from vertical and wind loads. If, for lack of head room, a flanged beam only is used, as shown in Fig. 2, instead of the lateral bracing of Fig, I, then that beam will be subjected to combined compres- sion and bending. Let + 7^ represent the total wind pressure, per panel, for both trusses, on \A />, and let w represent the release of weight in AB and in- crease in CD. Also let AB = a, and BD — b. \F is to be taken as dJMI M^ applied at yJ and zv, at the same point. Equal and opposite forces are also to be supposed to act at D. The moment M = Fa = ivb, exists at C and gives: BA Fa With the direction of wind shown by the arrow, the bend- ing caused by w will increase uniformly from nothing at A to M = ivb at C. The bending moment, therefore, to be resisted by this beam A C, and by the Joints between it and the chords A and C, is : M=Fa = wb (8). The direct compression in AC is: (AC)= -^F. (9). Hence, if q is the area of cross section of the beam, and if K, /, and di retain the same general signification as in Eq. (7), the greatest intensity of compression in the beam (at its ends) will be : TJiANSVERSE BRACING. g -(-;h-^^ (-)• The direct compression in CD is {CD) =^ -w (II). The bending moment in CD at C is : M=Fa (12). The greatest compressive intensity is found at once by Eq. (10), after writing 2U for IF, and giving to the remaining notation its general signification. The two preceding cases are those of through trusses. In the case of a deck truss the lateral bracing is of much more simple character ; it is shown in Fig. 3. At C and A are the two lower chords. CA is a lateral strut, while BC and cj|l DA are lateral ties. No parts are subjected to bending. If F is the panel wind pressure (for both trusses) acting along AC, there will result : Fig. {CA) = - IF {BA) = - zv. (13)- (14). {BC) = + VF^ + W' (15). The horizontal component of {BC) is equal to F, and acts at B. Thus all wind pressure is carried to the upper chord. The compression {BA) is in addition to the regular truss stresses induced by the vertical and wind loads. By these methods all the wind pressure may be carried to either chord. The truss stresses of the sway truss in the horizontal plane of that chord have already been found in the preceding Article, or rather, the methods for finding 384 WIND STRESSES AND BRACED PIERS. them and the effect of iv on the stresses in the vertical trusses have there been completely given. The wind has been taken in one direction only; with the other direction, opposite but symmetrically located parts would be stressed by the amounts found. Art. 83. — Transverse Bracing with Distributed Reactions. In the preceding articles it has been assumed that the hori- zontal reactions of the wind pressure were concentrated at the extremity (top or bottom, as the case may be) of one post in the tranve^se panel considered. This assumption, however may not be admitted ; or some other may be substituted in its place. Let Fig. I represent a transverse panel, with the wind blowing in the direction shown by the arrow; As before, the following notation will be used : AB = ON ^ d. BC= OD = a. DC = AN = b. F =total wind pressure, per panel, for one truss, on \{AB-\- BC). rp' , about an axis lying in the plane of the vertical truss, will be : M = 2fi{F +F')a (5). Since the windward truss is always relieved of a part of its weight the bending moment 2{i — n) {F + F')a, at ^, will seldom or never be needed. The value of M, from Eq. (5), put in Eq. (7) of the pre- ceding Article, and in the expression following that equation, will enable the greatest compressive intensity in the post to be found. If the transverse panel, Fig. i, represents the portal of a bridge, the distances AB and BC, or ^^ and a, represent tn- clincd distances in the plane of the portal. F ' will {or may) then include, also, the reaction of the horizontal sway truss in the plane of AN, while F will include the reaction of the horizontal sway truss in the plane of OB, if there is such a sway truss. 95 $86 WIND STRESSES AND BRACED PIERS, If « = I, as is sometimes assumed : {AN)=-\j' ^{F ■\- F')^ (6). i^OB) = - 1"/"+ (F + F')--! = (AN). . . . (7). If « = I in the formulae of this Article, those of the cor responding cases in the preceding Article at once follow. In Fig. 2 let the notation be as follows: AB=CD^ a. BD= CA = b. Total wind pressure for both trusses, per panel, along AC=F. Horizontal reaction at Z? = nF. (8), B={i-n)F. (9). If Fig, 2 represents a portal, F will or may include the reaction of a horizontal sway truss. The bending moment on both DC and CA, S* at C, also on the joint at the same point, is : m Ml = nFa. (lO). KiG. fJjjB This is the greatest bending in DC and CA of that kind which produces compression in the lower flange of the beam CA. The relief of panel load in the truss AB and increase of that in CD is ; w Fa_ b (10). Let X represent any variable portion of CA ; then the bending moment at any point of CA is: ST/(£SS£S IN BRACED PIERS. 387 Fa M = n Fa — wx = n Fa 7- x. . . (i i). For the point or joint A, x becomes equal to b, while the expression for the moment is : M\^ ~ {\ - n) Fa (12). This is the greatest bending of the kind opposite to M ^, in CA. It is also the greatest bending in AB. The con- nections at C must resist the moment M-^, while those at A must resist M\. The direct compression in CA is 4/^ " CD " w. \{n = \\ M,^-M\ = \Fa. .... (13). These various bending moments, substituted in Eq. (10) of the preceding Article for J/, will enable the greatest intensi- ties of stress in the members CA, CD, and y^^ to be at once found. Art. 84. — Stresses in Braced Piers. The general treatment of stresses in braced piers may be exemplified by that of a single " bent " represented by a skeleton diagram in Fig. i, in which the horizonal web mem- bers are compressive ones. The plane of the " bent " is verti- cal and normal to the centre line of the truss whose end rests upon it ; or if the track is curved, this plane is normal to it. The bent shown in Fig. i may be considered one of a pair, in parallel planes, which, being braced together, compose the complete braced pier. The dotted rectangle .iJ/A^.5 repre- sents a skeleton section of the truss supported by the piers, the upper chords of which rest upon the top of the pier at A and B. A skeleton section of the train is also shown. The direction of the wind is supposed to be shown by the arrows a, normal to the track at the top of the pier. If the 388 WIND STRESSES AND BRACED PIERS. trusses are loaded with a train, the wind pressure against them and the train will be carried to the top of the piers in the manner shown in Art, 8i. The wind will also act against the pier itself. Let the train be supposed to cover the whole of the two spans adjacent to the top of the bent (in all ordi- nary cases one of these spans will be the distance between two adjacent bents) ; then let H represent half the total pressure against trusses, and P{ half that on the train cover- ing the two spans. The height of the centre of action of /^i" above AB, Fig. i, is h. Also let b = AB. The pressure /\" will decrease the train reaction at A and increase that at B by the amount : Fig. 1. ^ K.= ^'. (I) Let k' represent the vertical distance of the centre of ac- tion of H from the horizontal line AB. The wind pressure on the truss AMNB will cause an in- crease of truss reaction at A, and an equal decrease of that at B, which will be denoted by F, and its value will be: V = Hh' consequently if /'= Fi- F = P;'h Hh' b the total horizofital force to be taken as acting at A, and with the wind, will be (// + /\" — 2/' tan a) added to the wind pressure acting directly at A. In Fig. 2, r^ represents this STJ?£SS£S IX BRACED PIERS. ;89 force, laid down to any desired scale, measured to the right of d represent the panel wind pressures against the pier at the points C, E, G, and A', while those shown on the left of c rep- resent the panel pressures at D, B, F, H, and L. The panel pressures at A, B, K, and L are half those at the other points. Let W represent the total weight of adjacent trusses and moving load resting at the top of the pier. Let Wi represent the panel weight of the pier itself resting at the points C, E, G, D, F, H: 4 n\ will be taken as applied at the points A, B, K, and L ; then the resultant reactions at A and B, with the wind blowing, will be, respectively, — — t and V t . 2 2 The small segments Fig. 2. (2). It has been implicitly supposed that two equal and opposite forces, equal in magnitude and parallel to /\'', act along AB. One of these forms, with Px' itself, the couple /*l'7^ ; the other is the wind pressure which, combined with the half panel press- ure at A, and {H—2t' tan a), is represented by cd in Fig. 2. The quantity /' is the force of a couple whose lever arm is b. One force /' is therefore supposed to act at A, and the other at B. Vi will be considered larger than V; hence /' will act upward at A and downward at B. If a is the angle between ^^or BL and a vertical line, the /' at B will cause a compression m AB equal to /' tan «', while the /' at A will pull to the left by the same amount. Consequently the force 2/' tan (n will act on the point A and toward the left. In the diagrams and in the equations which follow, positive and negative signs indicate tensile and compressive stresses, respectively. 390 WIND STRESSES AND BRACED PIERS. The stresses due to vertical loads at A and B, and the other panel points, will be the following : fW W\ iAC)"=-[-j~r+^)seca. {EGj'~-[ " - + 2 {GK)'^-( - " + '^-^) " 2 / (-)--(-. ^■.'^0 (^^)"=-(" "+-^ (FH)" = -( " " +5i^ 2 (i7Z)" = -( " - + Zi^ {AB)"=-[^ +-^) tana. (CD)" = -IV ta7i a. {EF)"= - " (GH)'. ' it «( {KL)"=+{^~-t +L^ytaKa. The difference between the horizontal component in HL and {KL) is 2t' tan ex, and it acts towards the right. The stresses caused by the horizontal wind pressure acting through A, /?, Cy D, etc., are shown in Fig. 2, as has already oeen noticed. The diagonals sloping similarly to AD are as- sumed not to be stressed. The diagram explains itself. The resultant stresses, finally, are to be found by combin- ST/?ESSES IN BRACED PIERS. 391 {~DF) = -(•• ,. , 3 ^^) <( {'FH) = -(" .. , ^^ " {'Hi) = -(" -^^) (( ing the results of the diagram in Fig. 2 with those expressed by the equations already written. They are the following: iAC) = -[^-^ + ^)-^'^-^-- (C£) = -(" "+-2-) " +(6^^)- (EG) = -(" " +^!") " +{£G). (GK) = -(" " +^2^) " +{GK). nv jv\ (BD) = _ (^^ + /' + - ' j ^^f « - (BD). - [DF). - (FH). - iHL). (AB) = -i{lV+ IV) tan a - {AB). (CD) = - VV tan n - (CD). (Rf) = - " " - (EF). (GH) = - " " - (GN). (KL) = + (t - ^' + -f) '^'' « - (^^)- It is not necessary to reproduce the stresses in the oblique web members, since they can be scaled directly from Fig. 2. All the stresses may be checked by the method of moments in the' usual manner, and such checks should always be applied. The two reactions R and R' are the following : R =. i{W - 2t' ^ S IV,) + Ri. R' = \{W + 2/' + 8 \\\) + /?i'. 392 WIND STRESSES AND BRACED PIERS. It is to be remembered that R( is to be taken as positivt in these expressions ; also that R-l— — R\i as shown in Fig. 2. The lateral force Fi to be resisted at the foot of the bent by friction or some special device, is the total wind pressure against the train, truss, and bent. If y is the coef^cient of friction at A' and L, Fig. i, the lateral resistance of friction offered at K is f'R, and that at L,f'R'. It is supposed that both the reactions R and R' are upward, also that both coefficients of friction are the same. The expression for {KL) has been written on the assump- tion that all frictional resistance is exerted at L. Strictly, however, the stress in KL may be taken as : {KL\ = {'kL) -f'R; . always supposing, numerically, {KL) > /R. The circumstances of particular cases frequently require calculations to be made with the structure free of moving load, as well as covered with it. In such a case it is only necessary to put for IV, in the preceding operations, the weight of trusses only. The wind has been taken in but one direction only, though the pier is to be designed for both directions, since it is only necessary in the resultant stresses to change the letters B, D, F, LL, L, to A, C, E, G, K, and vice versa. If MN, Fig. I, should coincide with AB, or if the truss should rest upon the top of the pier, it would only be necessary to take t' = Fj + V, remembering that h is the distance (vertical) from the centre of /-*/' to the top of the pier. It should be stated that 2t' tan a may be treated as a single force acting toward the left and along AB, Fig. i. It will then give rise to the diagram in Fig. 3, which shows all the stresses produced by its action. In that Fig. ad repre- sents 2/' tan a. In such a treatment of the question, cd. STRESSES IN BRACED PIERS. 393 Fig. 2, would represent H + A" added to the half panel pressure at A. The resultant stresses would then be found by combining the results of the two diagrams with those of the equations. All the results of these two methods will not agree ; the latter will give the greatest. This ambiguity cannot be av^oided, for it results from the fact that the pier cannot be so divided as to sever these members only. Mr. J. A. Powers, C. E., has called the attention of the writer to the fact that the web members of a braced pier carry- ing a double track railway, similar to that shown in Fig. 4, will receive their greatest stresses with the windward track only loaded. The vertical member GH vazy be sup- posed to carry its proper proportion of the load which rests on each track. This ^^^- 3- supposition, however, does not affect the statement made above. Let the wind have the direc- tion shown by the arrow, and let \\\ as before, represent the fixed and moving weight resting on GB, while iv is that part of W which is carried to B. If GH acts • QD Fig 4. If GH does not act : CF W. DF In the latter case, the beam AB will carry Wi'= ^ W to A. 394 IVJND STRESSES AND BRACED PIERS. If the angle FBN = CAM— a, the force h = zv' tan a — u>i tan a will act along /^i5 as an unbalanced horizontal one. If GH acts, Wx tan a becomes equal to zero. Then in the preceding investigation, there is to be put, (// + Ji) for //, while iv' is to be taken as acting vertically- down at B, and zi\ or o (as the case may be) at A. The preceding methods and diagrams remain exactly the same as before. In the formulae, however, w/ or o is to be put for the 2 at A, Fig. I, and w' for that at B in the same figure. Noth- ing else is changed. If W rests on AG and GB at the same time, a horizontal force equal and opposite to h is developed at A^ Fig. 4. Hence Ji will be balanced and disappear. If W rests on AG alone, with the direction of the wind remaijiing the same, Ji will change its direction, thus giving much smaller web stresses than those existing with W on GB alone. If, for any reason, the load on a single track pier does not rest over its centre, h will have a definite value, and the above considerations must govern the determination of the web stresses. This condition may exist if it becomes necessary to place braced piers under a single track railway curve. The stresses caused by the traction, or pull, of the loco- motive, in the members of a braced pier, are simple in char- acter and easily determined. In such a case, the pier is simply a cantilever with the traction, or pull, as a single force acting at its extremity. The traction acts along the line of the rails, and the length of the cantilever is the height of the pier. The stresses thus determined are to be combined with those already found. COMPLETE DESIGN OF A RAILWAY BRIDGE. 395 Art. 85. — Complete Design of a Railway Bridge. The main sections and details of this design shown on Pis. XI. and XII. arc based on the following general specifica- tions. The span length, depth of truss, panel division, moving load, fixed loads, and. stresses resulting from the preceding, shall be as determined in Art. ii. The clear width between trusses shall be 14 feet. A wind load of 150 pounds per lin. ft. of span shall be taken for the upper chord, and the same amount for the lower, and shall be treated as a fixed load in each chord. In addition to this fixed load, 300 pounds per lin. ft. of span shall be taken as a moving wind load for the lower chord, since the train passes along the latter. The greatest allowed tensile stresses under the preceding loads shall be : For lower chord eye-bars . . 10,000 lbs. per sq. in. " main brace eye-bars nearest end of span .... 10,000 " " " " " firsc counter-brace 7-333 " '' " " " vertical adjacent to end of span 7,333 " " " " Other tension braces to be proportioned according to loca- tion between 7-333 and 10,000 " " " " For plate hangers on floor-beams (net section) .... 8,000 " " '• " " bottom flanges of floor-beams and stringers (net section) 8,000 " " " " ■' lateral braces . . . . , 12,000 " " " " The greatest allowed compressive stresses shall be : For upper chord and end posts : Flat ends. Pin ends. 7,800 ^ 7,800 , . P= ^^72 . • . /= ^72 • • (i). 50,000^^ 3o,oooP^ For intermediate posts at centre of span, a reduction of 20 per cent, shall be made from the preceding values, and all other intermediate post stresses shall be proportioned accord- ing to location, between end and centre values. 39^ COMPLETE DESIGN OF A RAILWAY BRIDGE. In the preceding column formulae, "/" is pounds per square inch ; "/," length; and "r" the radius of gyration of normal section in direction of failure, and in the same unit as " /." For lateral compression braces the above values may be increased 20 per cent. For top flanges of stringers and floor-beams the greatest compressive stress shall be 7,000 pounds per square inch of gross section. The greatest mean bearing intensity of pressure between pins and pin-holes or rivets and rivet-holes, shall be 12,000 pounds per square inch. In stringers and floor-beams and their con- nections with each other or with the trusses, this value shall be reduced 25 per cent. The greatest shearing intensity in rivets or pins shall be 7,500 pounds per square inch, and in stringers and floor- beams and their connections with each other and the trusses, this amount shall be reduced 25 per cent. The greatest bending stress in the extreme fibres of wrought- iron pins shall be 15,000 pounds per square inch, and the cen- tres of pressure shall be taken at the centres of the bearing surfaces. No cast-iron whatever shall be permitted in any part of the structure, and all parts shall be accessible for inspection and painting. The unsupported width of any plate in compression shall not exceed thirty times its thickness. The pitch of rivets in compression members shall not ex- ceed sixteen times the thickness of the thinnest plate through which the rivets pass. An initial stress of 5,000 pounds shall be added to that produced by the vertical loading in all adjustable tension members. These meagre specifications are sufificient for the design when it is premised that all details of construction, such as eye-bar heads, connections, etc., shall be consistent with the best engineering practice. COMPLETE DESIGN OF A RAILWAY BRIDGE. 397 Although it is customary to add the total stress of adjust- ment to that caused by the vertical loading, in adjustable tension members, the practice is not strictly correct. Just what part of the initial stress should be added is not exactly determinate, but it is certainly not the whole. For this rea- son the apparently small value of 5,000 pounds has been taken. As the widths of the compression members depend, to some extent, on the thickness of the tension members, and as the design of the latter is of the greatest simplicity, it conduces to the greatest convenience to begin with them. All eye-bars should be as thin as considerations of resist- ance will permit, as pin bending will then be reduced to a minimum. By taking the stresses from Fig. i of PI. II., and subjecting them to the preceding specifications, the following sections are obtained : Bars. Brace. Total stress. A llowed stress. Sections. 2 45-553 lbs. 7,333 lbs. per sq. in. 2-4" X f" 3 167,800 a 10,000 a u u " 2—6 X if 5 119,817 a 9,340 Li u u (( 2—6 X l| 7 77,445 (< 8,670 (< ii it soo \ 1 CO _ Centra ^^-EE^S""'' "^ 45353 Fig. 2. diameter of pin must be assumed, for the moment at the centre of the pin will depend partially on the thickness of the pin plates, which bear against the pins. A diameter of 4I inches will be taken ; hence each inch in length of the pin will take 4,375 x 12,000 = 52,500 pounds. It will be seen hereafter that the floor-beams will be riveted into the posts below the pins in such a manner that the pin plates will not only carry the column pressures to the pin, but the floor- beam loads also. Hence, in determining the thicknesses of bearing areas in the pins, these two loads and the pressures fiBsm 2 — 3 due to initial stresses in the counter rods must be added. The vertical brace 8 is the only post subject to the initial stresses in the counters and the vertical component of each counter at its top is 4,000 pounds, or 8,000 for the tuo. 400 COMPLETE DESIGN OF A RAILWAY BRIDGE. The total lower chord load has already been found to be 45,. 553 pounds in the case of brace 2. This maximum lower chord panel load will not usually occur with the greatest post stress, but all possible cases are covered by combining the two. The bearing thickness at each side of each post is thus found to be: For brace 4 (99,819 + 45,553) "^ 52,500 x 2 = 1.4 inches. " " 6 (66,193 + 45,553) ^52,500 " =1.06 " " 8 (34,611 + 45,553 + 8,000)^-52,500 " =0.84 " By regarding the principles afTecting pin bending as de- veloped in Art. 74, it will be found that the arrangements of o — 4 482W I ■ I 3 4S200 -a \ ^J48aoQ 53166C j. =^ :^_ ^Co ___Centre yS?__ Fig. 4. lower chord eye-bars and braces shown in Figs. 2, 3, 4 and 5 will reduce the lower chord pin bending to the least amounts possible. The values of these least pin moments for the principal 4S200 ,4 — 5 '-— ^. ,^ — ^ 1 53700 ^) — -1 — - I 53700 _ ^^P-i"-: liRS. Fig. 5. sections are found to be as follows, and they can be verified by remembering the thicknesses of the eye-bars (already de- termined) and the fact that a play of one-eighth of an inch is allowed between each contiguous pair of heads. Joint 1 — 2. The section of plate hanger at the end of the floor-beam is shown shaded, and the distance between its centre and that COMPLETE DESIGN OF A RAILWAY BRIDGE. 40 1 of either of vertical braces 2 is 3 inches. Taking moments about the centre of the plate-hanger : Moment about Vert, axis 65,800 x 1.5 = 98,700 in. lbs. Hor. " 22,800x3.0= 68,400" " Herue resultant moment ^ ^ (98,700)" + (68,400)^ =120,000" " . (2). Joint 2 — 3. The bearing area of post 4 on the pin is shown shaded. In the remaining cases it will be assumed that the adjacent lower chord panel stresses take their greatest values together, in accordance with which assumpt'^n the eye-bars will be stressed for this joint as shown in Fig. 3. The force 78,400 pounds is the corresponding stress in one eye-bar of brace 3. The tangent of the inclination of the latter to a horizontal line is 1.32, hence the vertical component of brace 3 (one eye- bar) is (i 12,800 — 65,800) 1.32 = 62,040 pounds. The moments about vertical axes are : About a 65,800 X 2.625 — 56,400 X 1.25 = 102,225 in. lbs. . (3). " b 65,800 X 4.0 — 112,800 X 2.0 = 37,600 " " The moment about a horizontal axis through c is : 62,040 X 1.5 = 93,060 in. lbs. The vertical moment about c is the same as that about b, hence the resultant moment about c is: |/ (37,600)" -4- (93,060)- = 100,360 in. lbs. . . .... (4). Joint 3—4. By the preceding method the resultant moment of the in- clined brace was resolved into vertical and horizontal com- ponents ; in this and the following cases, however, the resultant moment itself will be taken. As before, the post bearing is shaded in Fig. 4. The head of the counter rod c^ is assumed to be one inch thick. The secant of the inclina- tion of the inclined bar to a horizontal line is 1.66. Hence the inclined stress is (3 x 48.200—2 x 56,400) 1.66= 53,166 pounds. The vertical moments are then as follows : 26 402 COMPLETE DESIGN OF A RAILWAY BRIDGE. About a . . . 48,200 X 1.2 = 57,840 in. lbs. " a . . (f 12,800 — 96,400) X 2.94 = 48,216 "■ " " b . . . . 48,200 X 3.5 — 16,400 X 6.44 = 63,100 " " The inclined moment about b is ; 53,166 X 2.31 = 122,800 in. lbs. In Fig. 6 ab is normal to the inclined brace, and represents 122,800 inch-pounds by scale, while be is vertical, and repre- sents 63,100 inch-pounds. Hence the resultant moment about b is represented by ac, and has the value : R = 97,500 inch pounds (5). S^otn^ 4 — 5. The thickness of the head of the centre rod c is 1.5 inches; and the inclined eye-bar stress of brace 7 is (3 X 53,700 — 3 X 48,200) 1.66 = 27,500 pounds. The vertical moments are as follows : About (z' 48,200 X 1,155 = 55,671 in. lbs. (6). " a 2(53,700 — 48,200) X 2.875 — 31,625" " " i 2 (53.700 — 48,200) X 4.03 — 48,200 X 1. 155 = — 11,340 " " " ^ 3 (53,700 - 48.200) X 3.7 -11,340 = 49»7oo " " The inclined moment about d is 27,500 X 2.6 = 71,500 in. lbs. Hence the resultant moment about d is, by Fig. 7: R = 57,000 inch-pounds ' (7). In addition to the preceding, the moments of the greatest vertical components of the inclined eye-bar stresses about the centres of the post bearings should be examined. It is here unnecessary to go into all these in detail. The greatest oc- curs at joint 3 — 4. The vertical component of the maximum COMPLETE DESIGN OF A RAILWAY BRIDGE. 403 •Stress is ( 1 19,817 ~ 2) 0.8 = 47,930 lbs. in question is : Hence the moment 47,930 X 2.31 — 1 10,700 in. lbs. (8). The preceding results show that while it is quite unneces- sary to take moments at the centre of all bearings, a thor- ough examination of the lower chord joints must be made in order to find the greatest moments. Eq. (2) gives the greatest resultant moment of 120,000 inch-pounds. Hence a wrought-iron pin 4,375 inches in diam- eter will be sufificient to meet the requirements of the speci- fications. But since the bars in braces 3 and 5 are 6 inches wide, and since the eye-bar heads are no thicker than the bodies of bars, the requirement of 12,000 pounds per square inch bearing pressure against pins cannot be met by a less diameter than 5 inches in the case of brace 3. For a reason that will appear hereafter, the diameter of the large pins will be taken at 5f inches. Fig. 6. Fig. 7. Fig. 8. In the cases of braces 7 and 9 a much smaller pin may be used, and while it is not economy in the shop to have a large number of pin diameters, two, or even three, are not too many for a span of this length. The cosine of the inclination of brace 7 to a horizontal line is 0.61, hence the horizontal component of the greatest stress in one of its eye-bars is (77,445 -^ 2) X 0.61 — 23,620 pounds. It will be seen here- after that the side plates of the upper chord panels 3 and 4 will be 0.5 inch thick, and since 4^\ x 12,000 -^ 2 = 25,000, 404 COMPLETE DESIGN OF A RAILWAY BRIDGE. it appears that with a pin diameter of a^-?^ inches no thicken- ing plates at pin-holes D, E, F, and M will be needed. It will be found that the thickness of bearing plates at the top of brace 6 must be \\ inch, and the clearance at each side of eye-bar head (between i-inch side plate of chord and pin plate of post) will be about y^ inch. The vertical component of eye-bar stress (for brace 7) will be (77,445 -h 2) x 0.8 = 30,980 pounds. Hence : Pin moment at centre of eye-bar head = 23,620 x 0.875 = 20,670 in. lbs. Vertical " " post pin plate = 23,620 x 1,875=44,290" " Inclined " " " = 38,720 x i.oo = 38,720 " " The resultant of the last two moments is shown by Fig. 8 to be : R = 37,000 inch-pounds (9). The moment of the greatest vertical component in brace 7 at the bottom of post 8 is : 30,980 X 2,6 = 80,550 inch-pounds. . (10). Eqs. (6), (7), (9), and (10) show moments far below the allowed resisting capacity of a 4^ wrought-iron pin, i. e., 108,000 inch pounds. Hence, at C, L, K, and J 5| inch pins will be used, while at D, E, and I, ^^^ pins will be taken. Before determining the diameters of the pins in the inclined end post, it will be necessary to fix the sections of that member, and it will be convenient to find those of the upper chord at the same time. As each upper chord panel is a beam of considerable span, carry- ing its own weight, the depth should not be small, and it will be taken at eighteen (18) inches. The radius of gyration of the normal section about a horizontal axis through its centre of gravity must first be found. % — ^ G — -^^--H ri- -.t- jL Fig. 9. COMPLETE DESIGN OF A RAILWAY BRIDGE. 405 As the upper chord stress in panels 3 and 4 is about 322,000 pounds, the area of that panel section will not be far from 44 square inches. Fig. 9 represents a trial section of that area. AB is a 21 X ^»g inch cover plate. C and D are 3x3 " 21 lb. angles. £ a P u 5x3 "50 6^ " // " 18 X 1 " side plates. The 5-inch legs of E and F are horizontal. The centres of gravity of C and D are 0.9 inch from lower surface of AB, and those of E and F are 0.9 inch from lower surface of the horizontal 5-inch legs. Static moments about a horizontal line through the centre of gravity of the section of AB give: 3x3 angles . . . 2 x 2. i x 1.2 = 5.04 5x3 " ... 2 X 5.0 X 17.4 = 17400 Side plates ... 18 x 9.3 = 167.4 Total 346.44 Hence, 346.44 -4- 44 = 7.9 inches ; or the centre of gravity g of the e?itire section is y.6 inches from the lower surface of AB. In such computations some dimensions are taken a little full because adjacent surfaces do not have mathemati- cal contact. The elements of the moment of inertia of the section about the horizontal axis GH through g take the values : Cover plate 11. 81 x 7.9- = 737-o6 , ( 2 X 2.0 = 4.0 3 X a angles \ ,^., „„ •^ -^ ^ ( 4.2 X 6.7- = 188.54 r y^ . u j 2 X 4^ = 9.0 • ^ ^ ( lO.O X 9.52 = 902.5 Side plates .... J 18 x Is^ - 1^=486.0 ( 18 X 1.42 = 35.28 Moment of inertia .... 2,362.38 The moment of inertia of AB about a horizontal axis through its own centre of gravity is so small that it has been neglected. The 3x3 and 5x3 angles each have a moment 40^ COMPLETE DESIGN OF A RAILWAY BRIDGE. of inertia of 2 about a horizontal axis through the centre of gravity of each respective section. The least radius of gyration about a horizontal axis for the entire section then becomes : V 2,362.38 -^- 44 = 7-33 inches. The panel length is 20.55 ^^- Hence l^r = 20.55 X 12 -f- 7.33 = 33.7. The preceding value in Eqs. (i) gives: Flat ends. Pin ends, p = 7,620 lbs. per sq. in. p — 7,5 10 lbs. per sq. in. For one pin and one fiat end / = (7,620 + 7,510) -^- 2 = 7,565 lbs. per sq. in. The upper chord will be continuous after the bridge is erected, but the extremities will be hinged at the upper ends of the inclined end posts in the manner shown in Fig. 4 of PI. XI. Hence upper chord panel i will have one pin end and one flat end ; all other panels will be flat end columns. The upper chord sections will now be as follows : Upper chord i. Required area = 225,424 -=- 7,565 = 30.0 sq. ins. 1 — 21 X y'L inch cover plate. 9.2 " " 2—3x3 " 18 lb. angles 3.6 " " 2—5x3 " 30 " " 6.0 " " 2 — 18 x I " side plates. 13.5 " " Total 32.3 " " Upper chord 2. Required area = 289,238 -i- 7,620 = 38.0 sq. ins. 1 — 21 X ^'g- inch cover plate . 9.2 " " 2 — 3x3 " 18 lb. angles. 3.6 " " 2 — 5 X 3 " 36 " " 7.2 " " 2 — 18 X i " side plates. .18.0 " " Total 38.0 " " COMPLETE DESIGN OF A RAILWAY BRIDGE. 407 Upper chord 3 and 4. Required area = 322,220 ^ 7,620 = 42.3 sq. ins. 1 — 21 X y"?^ inch cover plate = 9.2 " 2 — 3x3 " 18 lb. angles 3.6 " 2 - 5 X 3 " 47 " " 9-4 " 2 — 18 X y9^ " side plates. 20.25 " Total 42.45 " The end post bears on pins at top and bottom ; hence it is a pin-ended column. It is about 408 inches long, and it will be most convenient to take its depth identical with that of the upper chord, or 18 inches. The radius of gyration may then be taken, as before, at 7.33 inches. Hence, i ^ r = 408 -r- 7.33 = 55.7. The second formula of Eq. (i) then gives: / = 7,070 lbs. per sq. in. Inclined end post. Required area = 217,750 -^ 7,070 = 30.8 sq. ins. 1 — 21 X yig- cover plate .... 9.2 " " 2 — 3 X 3 18 lb. angles . . 3.6 " " 2 — 5 X 3 30 " " . . 6.0 " " 2 — 18 X I side plates. ... 13.5 " " Total 32.3 " " All these actual areas agree sufificiently near in character and amount with the trial section to make re-computations of the radius of gyration quite unnecessary. A very little ex- perience makes such a result possible in all ordinary cases. A very close but approximate rule for all box or semi-closed sections like those just considered, is to take the radius of gyration at four-tenths (0.4) the depth of the side plates. In the present case it would make r = 0.4 x 18 = 7.2 inches, while the exact value is 7.33 inches. In building a section such as these, the angles C and D, Fig. 9, should be made as light as possible, in order that the cover plate AB may be, to a considerable extent, balanced by the heavy angles £, F. In this manner the centre of gravity, 408 COMPLETE DESIGN OF A RAILWAY BRIDGE. g, of the section may be brought down sufficiently near to the mid depth to give all the space needed inside the chord for the eye-bar heads, if the pin axis should be made to pass through g, at the same time there is gained the incidental but important advantage of an increased moment of inertia. If the chord were subject to no bending from its own weight, the axis of every pin should pass through the centre of gravity of the section. It has been shown in Art. 69 that this flexure cannot be satisfactorily neutralized by the direct stress, par- ticularly if the chord is continuous, as in the present case. It is best, therefore, to reduce the bending stresses by making the chord depth as great as possible. For these reasons it was taken at eighteen (18) inches. If the panels were non- continuous the greatest stress per sq. in. in the exterior fibres of panels 3 and 4 would be : j^ Md 88,800 X 10.4 K = ~j— = — '-^ p = 300 lbs. / 2,362 ^^ Now, when it is remembered that the chord is continuous it is evident that flexure may be neglected in the sections found. This point, however, should always receive careful attention. In the present case, //le axes of pins in the upper chord and end posts zvill be placed eight (8) inches from the line AB, thus allowing a small counter moment from the direct stress due to a lever arm of 0.4 inch. A compression member with the same degree of end con- straint in all directions ought to have equal capacity for re- sistance in all directions. If the radius of gyration be taken in different directions about g. Fig. 9, for the different sec- tions as formed, it will be found that this condition is fulfilled. Finally, the unsupported width of any plate in compression, measured transversely between rivet heads, should not be more than about thirty times the thickness. An examination of the sections will show that this condition also has been fuffilled. The details about the pin bearings at the upper and lower ends of the inclined end post may now be considered. COMPLETE DESIGN OF A RAILWAY BRIDGE. 4O9 Fig. 4 of PL XI., shows the detail at the upper end of the inclined end post. The end panel of the upper chord does not rest its extremity immediately against the upper end of brace i, but they are separated along the line ab by about the distance of | inch, and each bears directly against the end pin. The diameter of the latter is taken by trial at 5^ inches. By the specifications the bearing value of this pin against a one-inch plate is 5| x 12,000 = 61,500 pounds. Hence for plates J, y^^, |, \ and ^ inch the bearing values will be as follows : 61,500 X ^ = 15,375 pounds. " X tS = 19,220 " " X I = 23,060 " X J = 30,750 " " X^=: 34,596 " The arrangement of thickening plates for upper chord i is clearly shown by Fig. 4 ; there is a half-inch plate inside and next the fths web and a | jaw plate inside the half-inch thick- ener. Against the web outside is a ^^g. inch thickener. The total bearing thickness is then | + i 4 | + ^^ = if| inches. Hence 61,500 x \\\ = 1 1 1,500 pounds. The half of the stress in upper chord i is 112,712 pounds and the two quantities are sufificiently near in amount. All rivets about the joint are f inch in diameter. The shearing resistance of one rivet at 7,500 pounds per sq. in. is 3,300 pounds, while the bearing values against the various plates are : I rivet against -^-^ plate = 2,800 pounds. " " " I " = 3,400 " " " " I " = 4,500 " The total bearing pressure against the jaw and thickening plates is 23,060 + 30,750 + 34,596 = 88,406 pounds, and there are 21 rivets through those plates, as shown in Fig. 4. Ap- plying the bearing and shearing values given above to the number and distribution of rivets located in that figure, it will be seen that there is a little excess of both those resistances. 4IO COMPLETE DESIGN OF A RAILWAY BRIDGE. The same figure shows the number and distribution of both rivets and thickening plates at the upper end of the incHned end post. There is a jj inch jaw plate outside, then a half- inch thickener and a f^ plate between that and the web. There is also a quarter-inch thickener inside. The amount of bearing thickness is thus the same as for the upper chord. An examination of the number and location of the rivets will show that there is again a little excess in the bearing and shearing resistances. The pitch of rivets in the immediate vicinity of the joint is three (3) inches, in all other parts of the upper chord and end posts it will be six (6) inches. The preceding arrangement is for one side of the chord and end post, since both sides, of course, are alike. As Fig. 4 shows, the pin passes through the jaw plates only. The four (4) jaw plates hold the post and upper chord securely to- gether in case of any derailment or other accident tending to knock the end post out of place. They are further reinforced by the light f inch cover plate shown at a. The latter also performs an important of^ce in transferring upper lateral loads to the end posts. One portion of it, or both, must, of course, be riveted in the field. It will be observed that each jaw plate has a "play" or clearance of \ inch, to provide for imperfections of workmanship and secure ready erection. The figure shows what rivets must be countersunk, both out- side and inside. The eye-bars of brace 3 lie adjacent to the interiors of the upper chord and end post, while those of brace 2 are inside of the first. Assuming that the greatest stresses in those braces occur together (which is a small error on the side of safety), the end pin will be subjected to the bending moments shown in Fig. 10. The component moments are as follows: „ , 167,800 . ,, For brace 3- • x 1.71 = 143.470 i"- lbs. " " 2- • • "^^'^^^ X 2.90 = 66,053 " " 2 217,750 X 0.62 — 67,500 " " COMPLETE DESIGN OF A RAILWAY BRIDGE. 4II The latter moment arises from the fact that the upper chord and end-post bearings have their centres separated by 0.62 inch. In the figure, be is normal to brace 3, and ac is horizontal^ while ad is normal to brace i. Hence db is the resultant moment of 220,- 000 inch pounds. A pin 51 inch in diameter will a little more than supply the required resistance with K— 15,000, as Eq. (i) of Art. 74 demon- strates, or as may more simply be found by reference to any reliable table of pin mo- ments. The thickening plates and rivets just found will now be a very little excessive, but they will be retained. These large rounds frequently vary in standard sizes by quarter-inches, and a 5J inch diameter may be turned to 5| with little waste. Fig. 7 of PI. XI. shows the lower end of the inclined end post with the number and location of the | rivets and thick- ening plates. The operation of designing them is precisely similar to those already employed, and they will not now be repeated. An examination of the plates and rivets in con- nection with the preceding values, will show that the shear- ing and bearing resistances of both the rivets, plates, and 5=^ inch (assumed) pin required by the specifications are secured. The line ab is 2\ inches below the centre of the pin-hole, giving about 2 inches of solid metal below the pin. Figs. 8, 9 and 10, of PI. XI., show two elevations and a plan of the pedestal at the lower extremity of each end post. The centre of the pin-hole is taken six (6) inches above the bottom I inch plate. The figures show with perfect clearness the arrangement of the various parts. The 4 and 19 inch spaces give ample clearance for the sides of the end post which enter them. The vertical component of the end-post stress is 172,820 4^2 COMPLETE DESIGN OF A RAfLWAY BRIDGE. pounds. The total bearing thickness under each half of the pin is I + ^ + 8 = if inches. But 64,500 x !« = 104,610; or greater than 172,820 -^ 2 ■= 86,410. Hence, ample bearing surface at 12,000 pounds per square inch is secured. Now since the half of the end-post bearing is at the centre of each of the 4 inch spaces, it may at first sight appear as if either equal bearing areas ought to be found each side of those spaces, or as if all ought to be on one side. But it is better to mass the metal as much as possible ; at the same time the weight should be distributed somewhat on the f inch plate. The arrangement shown accomplishes these results and gives a little excess of bearing area. The 5 by 3 inch angles are but 15 inches long, while the | inch bottom plate is 24 by 36 inches. The bearing thickness at a is 1.25 inches; hence the up- ward pressure at that surface is 64,500 x 1.25 = 80,625 pounds. The 4 inch space gives about \\ inch total clearance for the side of the end-post and the eye-bar (i| inches thick) of lower chord panel i, or ^ inch each for the three clearance spaces thus formed. The pin moment about the centre of the end-post bearing is: 80,625 X 1.875 inches = 151,171 m/; /<^j\ . . . (11). Again, taking moments about the centre of the angle bear- ing b, there are two moments with horizontal axes but with opposite signs formed by the upward pressure at a, and the half vertical component in the inclined end post, thus: -f- 80,625 X 4.81 = -^ 387,806 in. lbs. — 86,413 X 3.00= — 259,230 " " Resultant^ -I- 128,576 " " The stress in the 5 x i| inch eye-bar of lower chord i has the following moment about a vertical axis passing through the centre of b : 65.760 x 1.14 = 74,970 in. lbs. COMPLETE DESIGN OF A RAILWAY BRIDGE. 413 Hence the resultant moment about the centre of <^ is : -v/(i28,576)^ + (74,970)- = 150.000 ?/'/.//''i-. . . . (12). As the moment (11) is greater than (12) and far less than the resisting capacity of the assumed 5 J inch pin, the latter diameter will be retained. A smaller pin would give sufficient bending resistance, but would necessitate additional metal in the thickening plates, and would increase the variety in pins and pin-holes. It is frequently desirable to hang one pair of eye-bars (either braces 2 or 3) outside of chord and end post at the upper end of the latter. In such a case the angle flanges at i, Fig. 4, PI. XL, would be cut away, and more rivets would need to be countersunk about the pin-hole on the outside of the outer jaw plate. In the present instance, however, the pin necessary at the upper chord end is but little different from those required by braces 3 and 5. Hence, for the sake of uniformity, the pins at B, C, A, L, K and/, will be given a diameter of 5| inches, while the others are 4Y^g^ inches. The pin-plates at the upper and lower end of the intermediate posts will now be found. It will be assumed that the maximum post stress and the greatest panel load occur together. This is not possible, but it is difficult to determine the exact maximum load on the lower pin-plate, and the assumption involves a safe error. It will farther be assumed that the greatest panel load for the intermediate posts is the same as the greatest load on brace 2. This also involves a slight safe error. In consequence of these assumptions and the additional fact that the smaller part of the load in each lower pin-plate is the panel moving load, no addition for impact will be made in fixing the thickness of the pin-plates. The manner of supporting the ends of the floor-beams is clearly shown in Figs. 14, 15 and 16, PI. XI. They are built into the posts below the pin. The channels are continued 30 inches below the centres of the pin-holes, and a 4 by 4 inch 414 COMPLETE DESIGN OF A RAILWAY BRIDGE. 36 pound angle is riveted to each as shown 2X a a Fig. 14. The end stiffeners of the floor-beam (Fig, 2, PI. XI.) are brought against these latter and riveted fast to them in erec- tion. The number of rivets required for this connection will be found later on. In order to freely admit the end stiffeners of the floor-beam, the 10 inch channels of the post will be separated 10 inches. Vertical Brace 4. The top pin plates will carry 99,819 lbs. " bottom " " " " 99,816 + 45,553 = I45>372 lbs. Since 5| x 12,000 = 64,500 pounds, the total thickness of bottom pin plates will be 145,372 -^- 64,500 = 2.25 inches ; and since the shearing resistance of one f inch rivet is 3,300 pounds, the total number of rivets in the channel flanges will be 145,372 -=- 3,300 = 44. The lower part of Fig. 14, PL XL shows the required arrangement of pin plates and rivets There are three | inch outside pin plates on each side of the post. The rivets about the pin-hole on the outside will be countersunk in order that the eye-bar of brace 3 may lie close against the post. The total thickness of pin plates at the top of the post will be 99,819 -=- 64,500 = 1 1^6 inches, and the total number of rivets, 99,819 -^ 3,300 = 30. The upper part of Fig. 14, PI, XI. shows the required arrangement of pin plates and rivets. As the total number of the latter must be divided by 4, 32 rivets are used. There is one | inch outside pin plate and one yV inch inside plate riveted to the former. The object of placing the latter inside is to keep the upper chord as narrow as possible. Vertical Brace 6. The top pin plates will carry 66,193 lbs. " bottom" " " " 66,193 + 45,553 = 1 11,746 lbs. The total thickness of bottom pin plate will be 111,746^ 64,500= if inches, and that of the upper 66,193 — 50,250 COMPLETE DESIGN OF A RAILWAY BRIDGE. 415 = i/g inches, since the upper pin is 4y^e inches in diameter and 4y\ x 12,000= 50,250 pounds. The total numbers of \ rivets below and above, respectively, are 1 1 1,746 -r 3,300 = 34 and 66,193 -^ 3,300= 20. Fig. 15, PI. XL, shows the required pin plates and rivets. At the bottom there is a \ inch plate next to the channels and a half-inch plate outside. At the top there is a | inch plate outside and a ^g i^ich plate inside, as shown. Vertical Brace 8. Four adjustable ties meet the upper extremity of this post, and it has already been shown that each tie adds 4,000 pounds to the post stress. Hence, the top pin plates will carry 34,611 x 16,000 =50,611 lbs. bottom " " " 34,611 -f 16,000 + 45,553=96,164 " The total thickness of bottom pin plates will be 96,164 -=- 50,250= \\ inches; and that of the top plates 50,611-^ 50,250 = I inch. The total numbers of rivets required are 96,164^ 3,300 = 29 and 50,611 ^3,300= 16. Fig. 16, PL XI., clearly shows the arrangement of plates and rivets. There are more rivets show^n at the top than is necessary tor bearing or shearing alone, for the reason that the notch a must be cut out of one channel to let the counterbraces 9 take hold of the pin inside the post, as there is not room enough outside. Upper Chord Joints and Thickening Plates. It will readily be seen that the pin-hole at the joint point between upper chord i and 2 is the only one needing a thicken- ing plate. The greatest tension in an eye-bar of brace 5 is 119,817 -^ 2 = 59,909 pounds, and the sine of its inclination to a vertical line is 0.61. Hence, its horizontal component is 59,909 X 0.61 = 36,544 pounds. The thickness of the side plates of upper 2 is \ inch; hence. \ x I2,000 x 5| = 32,250 pounds. A little over 4,000 pounds, then, is all that need be resisted by a thickening plate. This might safely be 4i6 COMPLETE DESIGiX OF A RAILWAY BRIDGE. neglected, but the y^g joint plate shown by Fig. 13, PI. XL, will be extended to cover the pin-hole. Precisely the same operation shows that no thickening plates are needed at the other pin-holes. There will be joints in the upper chord as near as possible to, and on the left of the pin-holes at 6", D and E of Fig. i, PI. II., and at corresponding points in the other half of the truss. These joints are formed as shown at Fig. 13, PI. XI. At C the joint will be 12 inches from the centre of the pin hole. It is formed by riveting top and bottom and side plates to the chords, as shown. All the joints are formed precisely like this, except that in the other cases the y^^ plate between the angles extends each side of the outer one, as shown on the left only. Latticing and Batten Plates. The dimensions of latticing and batten plates are matters of judgment and experience. Evidently no segment of a column between lattice points ought to be less in resistance per square unit of section than the column as a whole, but experiments are yet lacking to give quantitative results. Single latticing with centre lines making angles of 60° with the axis of the member will be used here. On the under side of the upper chord and end post the lattice bars will be 4 inches by |, and each end will be held by two rivets. Fisr. 1 1 shows /oo/ \oo\ v /oo/ \p -/ / / / ^ \^ // IT /op/ VoN 7o=r _ Fig. II. this latticing. It will weigh about eight (8) pounds per lineal foot of member. The two battens (one at each end) on the under side of the end post and those at the ends of the upper chord (four in all) will be 21 inches by 21 inches by | inch. All other battens (one on that side of each vertical post opposite to the chord joint) will be 2i x 15 x | inches. The bottom plate of each joint forms, of course, a batten. On the intermediate posts, the latticing will be single and COMPLETE DESIG.V OF A RAILWAY BRIDGE. 4^7 60° as before, but the lattice bars will be 2 inches by i\ inch. This latticing (both sides) will weigh about 9 pounds per lineal foot of post. Floor-beam Supports. The method of suspending the floor-beam from the pin at the lower extremity of brace 2 is shown in Fig. 2, PI. XI. Two plates riveted to the end stiffeners of the beam take the 5| pin with its centre line six inches above the upper flange. The greatest moving load carried by the beam end has been already found to be 34,600 pounds. One-third of this will be added for impact, and as the fixed load is 8,000 pounds, the total load to be resisted by the plate hangers becomes : 4 X 34.600 + 8,000 =54..30 pounds. The greatest allowable load per square inch in these hang- ers is 8,000 pounds; hence the required net area is 54,130 -^ 8,000 = 6.8 square inches. These plates will be taken 12 inches wide. By deduction of the pin-hole the available net width becomes 12 — 5.375 = 6.625 inches. One plate 12 X i^g and another 12 x \ inch gives the required area. Rivets I inch in diameter will hold these plates to the end stiffeners. The shearing resistance in this case is less than the bearing, and the former for one rivet at 7,500 pounds per square inch, is 4,500 pounds. Hence the required number of rivets is 54,130 -f- 4,500 — 12 rivets. In consequence of the deflection of the beam some of the upper ones will be subjected to slight tension. Hence 16 rivets are shown. The figure shows the number and distribution of rivets and plates. The vertical pitch of rivets is 3 inches. The manner of attaching the floor-beams to the lower ex- tremities of the intermediate posts is shown by Fig. 14, PI. XI. a and « are 4 x 4 inch 36 pound angles 27 inches long riveted to the inner surfaces of the lo-inch channels, as shown. The vertical centre lines of the rivet rows in the channels 27 4i8 COMPLETE DESIGN OF A RAILWAY BRIDGE. are coincident with the central Hnes of the latter, thus insur- ing an equal division of the floor-beam load between the pin-plates. The number and distribution of the \ inch rivets in the angles o, a will, of course, be the same as those in the plate hangers of Fig, 2, PI. XL Both these methods of supporting floor-beams insure a central application to the pin, and the latter insures addi- tional stiffeners to the floor system and entire structure. The two lines of \ inch rivets take less than a square inch of section from the two :o inch 50 pound channels. Hence the remaining net section of our nine square inches is more than sufficient to carry the total floor-beam load at all points. Upper Lateral System. A wind pressure of 150 pounds per lineal foot will be taken as acting in the horizontal plane of the upper chord. The panel wind load will then be 20.55 ^ 150 = 3>o83 pounds. Fig. 12 shows a half plan of the upper lateral system. The diagonals are ties, and the other members are struts. Fig. 12. The secant of the inclination of T-^ to a horizontal line normal to the axis of the bridge is 1.57, and 3,083 x 1.57 — 4,848 pounds. The wind load in the upper chord is a fixed one. The line ah is the centre of the span. The following stresses may now be written, remembering that as the tension diagonals will each be adjustable, 5,000 pounds must be added for initial stress: C0MFLE7E DESIGN' OF A RAILWAY BRIDGE. 419 7^4 =: — + 5,000 = 5,000 pounds. 7; ^ 4,848 + " = 9,848 Zj = 9,696 + " = 14,696 " Tx = 14.544 + " = 19-544 Pi = 3,083 + 3,200 = 6,283 pounds /*3 = 6,166 + " = 9,366 " Pa =: 9,249 + " = 12,449 P, = 12,332 + " = 15,532 The greatest allowable stresses in the lareral systems may be taken as follows : For tension .... 14,000 pounds per sq. in X) 7^ 9,^00 , . compression. . g — *' " (13). 30,000 r^ The latter formula is for flat-end members of angle iron, as those will be used for lateral compression members. It will be observed that it gives less value than the formula (i) for columns of the box type, like those used in the trusses. Under these stresses the tension members become : Ti r - ii o. Ts I - iJ o. T2 I — 1} o. Ti I - i| o. It is not advisable to have any tension member less in sec- tional area than i square inch. Hence, T^ and T^ are a little larger than the stresses require. All the struts except P, will be formed of 3 — 3 x 3 inch angles. A section of this strut is shown at Fig. 1 1. PI. XI. The two angles c lie on the upper chord and are riveted to it, as shown. These two angles are designed to carry all the stress of the strut. The only office of the angle a is to keep the strut stiff in a vertical plane; it takes hold of the lower flange of the chord with two rivets, as shown at c, Fig. 4, PI. 420 COMPLETE DESIGN OF A RAILWAY BRIDGE. XL The strut is thus of the same depth as the chord, and' takes hold of both those members in such a manner as to give them great rigidity. At each end of the strut there is a 20 X 12 X y\ inch plate, and there is a set of single 60° lacing, 2 X y^6 inch. The radius of gyration of the section of the two top angles about a vertical line midway between the two, is 1.3 inches. The length of the strut is about 192 inches. Hence Eq. (13) gives 5,300 pounds as the greatest allowable stress per square inch. As P^ requires only the lightest angle that ought to be used, all these struts will be made alike. /s . . . . f 3 — 3" X 2|" 15 pound angles. P. As the detail for these struts would give some trouble at the end of the upper chord, P-^ will be a single 6" x 4' angle with the 6" leg horizontal, as shown at a. Fig. 4, PI. XI. The tie 7i is attached to the upper chord by the detail shown at b, Fig. 11, PI. XI. A piece of 6" by 4", 60 pound angle, 12 inches long, with the 6 inch leg lying on the cover plate of the chord, carries two pieces of 3 ' by 3", 21 pound angles about ^\ inches long, and with edges parallel to the axis of 7"i. One end of each of the latter angles rests squarely against the vertical 4-inch leg of the 6" by 4" angle. Six three- quarter inch rivets are then passed through the angles in the manner shown. Each such rivet will resist about 4,000 pounds in single shear. The tie T-^ passes through the 4-inch leg of the heavy angle (between the 3-inch angles), and carries a nut at its end, which gives the requisite adjustment. Ti and all the other lateral ties are held by the same detail, except that 4 rivets only are needed, as shown at b\ Transverse Bracing. A skeleton sketch of the intermediate transverse bracing for the vertical plane of any two opposite posts is shown in Fig. 16, Pi. XII. If the upper lateral system is designed to- COMPLE TE DESIGN OF A RAIL IV A V BRIDGE. 4^ I carry the whole wind load to the ends of the upper chord, as has been supposed, the duty of the intermediate transverse bracing is entirely indeterminate. As the sections must be determined in some manner, however, the method of Art. 82, Fig. I, will be applied. The slight analytical superabundance of stability thus secured is no more than is required by a rapidly moving load. F' of Art. 82 will here be taken as 3,083 pounds and F= o. Also, d = 1/ ft. and a = 6 ft. It will here be assumed that all the wind load is applied in the windward truss. This is the usual assumption in practice, although the conditions taken in Art. 82 are exactly true. Eq. (i) of that Art. then gives : w = (3,083 X 27) -f- 17 = 4,932 pounds. As the tangent of the inclination of 7^ to a vertical line is 2.833 ^n<^ the secant, 3.0, the stresses are: Z= 4,932 X 3 + 5,000= 19,796 pounds.i I o. P=4,932x 2,833 4- 5,000= 18,974 " 2-3" x3"24 lb. angles. The sizes are based on the same allowed working stresses as for the upper laterals. The strut P is shown at Fig. 19, PL XII. The two angles are held if inches apart by separa- tors, and present a horizontal upper surface. The ends are secured to a batten plate in proper position on the post and in the manner shown. The separation of the angles permits the tie 7" to pass between them and through the batten plate and take a nut inside the post. The washer a is formed from a piece of 3 by 2 inch angle, one-half inch thick, with the 2 inch leg sheared off until the proper angle is formed. The 3 by 2 inch angle b forms a check to keep the angle washer a in place. The length of the strut P is 192 inches, while the radius of gyration of a 3 by 3 inch angle about an axis through its centre of gravity and parallel to one leg is 0.92 inch. Hence the allowable stress per sq. in. by Eq. (13), is 4,000 pounds. 422 COMPLETE DESIGN OF A RAILWAY BRIDGE. Both ends of T are held by precisely the same detail. At the upper end of the post, however, the pin-plates take the place of the battens at the intermediate points. The rivets securing the batten plate to the post are seen to give an ex- cess of resistance. Portal Bracing. Fig. 17, PI. XII., shows a skeleton sketch of the portal bracing. The sketch is taken in the plane of the portal. The strut Pis placed 7 ft. 6 in. from the top of the post, while the length of the latter is 33.8 ft. The computations are made precisely as in connection with Fig. 16, PI. XII. The force acting at the upper extremity of the end post is 12,332 pounds. The tangent of the angle be- tween Zand the end post is 2.27, while the secant is 2.48. Hence, if w = 12,332 x 33.8 -^ 17 = 24,664, then : T= 24,664 X 2.48 = 61,170 pounds. . I — 6" X 4" 60 lb. angle. /•= 24,664 X 2.27 = 55,990 " ..I— 6" X 6" 75 " " As shown by the preceding results, this bracing is com- posed entirely of angles. This is done in order to secure the utmost stiffness or rigidity in the portal. Fig. 12, PI. XI., shows the method of securing the ends of the members 7" and P. The upper extremity of 7" is shown with the six-inch leg of the angle lying on the end post at a. In order that the proper number of three-quarter inch rivets may be brought into play, a | inch plate lies underneath the .angle, as shown. The method of securing the lower end of T, and each end of P xs clearly shown at b. Three-quarter inch rivets are used for all these connections. At the inter- section of the two Ts, one is cut and a firm joint is made by a centre plate a half-inch thick, aided by angle lugs. The length of the strut P is about 192 inches, and its radius of gyration, 1.9 inches. Hence Eq. (13) gives the working stress at 7,000 pounds per square inch. The tension allowed in this angle bracing is 12,000 pounds per square inch. An ornamental wrought-iron bracket may be placed in the angle between P and the end post. COMPLETE DESIGN OF A RAILWAY BRIDGE. 423 Lower Lateral Bracing. The lower lateral bracing is shown in skeleton plan by Fig. '6, PI. XII. It is designed to resist a uniform fixed wind load of 150 pounds per lineal foot in addition to a uniform moving wind load of 300 pounds per lineal foot. The fixed panel load, therefore, will be 20.55 x 150= 3,083 pounds, and the moving panel load, 20.55 x 300 = 6,166 pounds. The secant of the angle between T^ and P^ is 1.57. The line ab is the centre line of the span. Remembering that there are nine panels in the lower lateral system, and that the greatest al- lowable tension is 14,000 pounds, the following stresses and sizes may at once be written from the preceding data: 7'i stress 58,164 + 5,000 = 63,164 lbs. i — 2"| round. ^2 " 44,700+ " =49700 " I - 2"! " ^3 " 32,313 + " =37,313 " I - I ^ " T^ " 21,000 + " = 26,000 " I — I 7; " 10,770 + " = 15,770 " I - I X The initial stress is included in the total by the addition of 5,000 pounds, as has been done before. The floor-beams form the struts in the lower lateral system, except in the cases of the end struts I\, hence, only the latter need be provided. The friction on the wall-plate will evi- dently relieve P^ of some of its stress, but it is uncertain to what extent ; hence, initial tension only will be neglected. Consequently : Pi stress 41,620 lbs I — 6" x 4 " 50 lb. angle. The 6 inch leg of this angle is horizontal, and it is riveted to the top flanges of the stringers zvlicre it crosses the latter. By this means the stringer ends are held rigidly in position, and the general stiffness is increased. The rods of the lower lateral system will necessarily pass through the webs of the stringers, and will be secured to the webs of the floor-beams (as closely as possible to their upper flanges) by precisely the detail shown in Fig. 19, PI. XII. 424 COMPLETE DESIGN OF A RAILWAY BRIDGE. The web-plate now takes the place of the batten in that figure. The lower lateral 7^ takes hold of the pedestal by the clevis and 3 inch pin bolt shown in Fig. lo, PI. XI. A 9" X 7'' X I ' plate is riveted above and below the f inch base plate of the pedestal in order to give the proper bearing area. The method of securing the end of P, to the pedestal is shown by the same figure with perfect clearness. Expansion Rollers. A set of expansion rollers under one end of each truss must be provided. A diameter of 2^ inches will be assumed. According to Appendix II., the resistance per lineal inch of a roller is: 4 ^|/.^^' = |^|/4^(for£ = £> Since all metal is wrought-iron E = E' . The greatest al- lowable intensity of pressure on the roller will be taken at 12,000 pounds per square inch, or «; = 12,000. Also R = 1.47 and E — 26,000,000. Hence the allowable load per lineal inch, by the above formula, is 1,050 pounds. The maximum vertical component of the end-post stress is 172,820 pounds^ Hence the number of lineal inches of roller bearing required is 172,820 -=- 1,050 = 164 inches. The set of rollers shown by Fig. II, PI. XII., gives very closely the required amount. The clear space between each adjacent pair of rollers is ^ inch. The ends of each roller are turned down to | inch and pass through a 2i x | inch wrought-iron strap on the outside of which each roller end takes a nut. The rollers are thus held rigidly in their proper relative positions. A i x ^ inch collar is turned down at the centre of each roller to take the i x y'^ inch shoulder which is shown in the wall plate, and by means of which all lateral motion of the rollers is prevented. Wall Plates. The mean pressure per square inch on the total surface of COMPLETE DESIGN OF A RAILWAY BRIDGE. 4-5 the wall plate should not exceed 200 pounds. The total area of wall-plate surface shown in Fig. 12, PL XII., is 30 x 30 = 900; hence the total allowable weight is 900 x 200 = 180,000 pounds. The maximum total vertical pressure of 172,820 pounds is thus provided for. The plan of the wall plate at the roller end is shown in the figure ; the upper elevation also belongs to it. At the fixed end either the pedestal or the masonry must be sufficiently high to fill the roller space. Both those alterations, however, are unadvisable for obvious reasons. It is better to fill the roller space with the wall plate. The lower elevation of Fig. 12, PI. XII., shows the arrangement to be adopted. The same | inch thick wall plate is to be taken, 4 — 3i x 3^ x | inch angles then run in the direction of the rollers across the ■entire plate. At right angles to these /\ lines of 3 x 3 x 4 inch angles are placed. The latter are cut to fit in between the former and will, of course, require filling strips under- neath. The top of this gridiron arrangement is then planed ofl until the proper height of wall plate is reached. The rectangular spaces thus formed are then filled with Portland cement rammed hard and flush with the planed upper sur- faces of the angles. A solid wall plate is thus formed with the interior surfaces completely protected against corrosion. At diagonally opposite corners are seen the holes for the i| inch anchor bolts. Wind Pressure on Chords and End Posts. The effect of the wind load on both upper and lower chords has been shown in detail in Art. 81 ; the principles there established remain to be applied here. The chord stresses in the lateral trusses are the same in kind as those produced by the vertical loading in one lower chord and one upper chord. Just to what extent these wind stresses may be allowed to exist without necessitating any increased chord section is a matter of experience only ; but as the greatest wind stresses and those due to the vertical loading so rarely combine in jnost localities that with the working stresses specified in this 426 COMPLETE DESIGN OE A RAILWAY BRIDGE. case the 7vind load may he allowed to reach \tJis the value of the greatest vertical loading without requiring any increase in chord section, in all localities not ordinarily subject to cyclones or tornadoes. In the present instance the total fixed and moving vertical load is equivalent to about 2,040 pounds per lineal foot of each truss. The total wind load in the lower chord is 450- pounds per lineal foot, and its depth of' truss is only 17 feet. If reduced to the same truss depth ao that for the 27 vertical load, /. r., 27 feet, it would be 450 x — = 720 pounds. Again, the overturning effect of the wind on the train (dis- cussed at the close of Art. 81) throws on the leeward truss the additional weight of ^ = 140 pounds per lineal foot. It is assumed that the centre of wind pressure on the train is 8 feet above the end supports of the floor-beam. The total wind effect on the loading of the leeward truss is then 720 + 140 = 860 pounds per lineal foot, or a little in ex- cess of four-tenths the vertical loading. As three-eighths the vertical loading is 765 pounds per lineal foot, the chord sections should be increased for 95 pounds per lineal foot. As the increase in area, however, would be but one-sixth of an inch for one chord, and as the* transverse bracing will slightly relieve the leeward truss, no change will be made. If the lower chord needs nc revision, the upper need not be considered. The total pressure of wind against the upper extremities of the end posts, and, hence, against their lower extremities also, has already been seen to be 12,332 pounds. If this is assumed to be equally divided between the end~post feet, each of the latter will carry 6,166 pounds. Each end of P(in the portal. Fig. 17, PI. XII.) is 26.3 feet from the end- post foot. Hence at the former point the end post suffers the bending moment 6,166 x 12 x 26.3 = 1,945,990 in. lbs. The moment of inertia of the end-post section about the neutral axis normal to the cover plate is 1,886, and since the COMPLETE DESIGN OF A RAILWAY BRIDGE. 4-7 half total width of the chord is 12.5 inches, the stress per square inch in the extreme fibres of the 5x3 angles is K' = 1,945,990 X 12.5 -f- 1,886 = i2,go4 pounds per sq. in. The direct post stress will be about 7,200 more, or a total of 20,104 pounds per sq. in., whereas 15,000 should not be ex- ceeded. \{ K — limit of compressive bending stress per square inch, which must not be exceeded ; M =■ bending moment in inch pounds ; / the moment of inertia of the total post section about a neutral axis normal to the cover-plate ; 2d = total width of end post ; P = total direct stress of compression, due to vertical loading and overturning action of the wind against the train (/ of Art. 81 is its panel value), and A = total area of section ; then the sectional area must be increased until the following equation holds true : ' Md P , , ^=^^A ^^4). It has been seen above that the wind effect in the leeward truss is 140 lbs. per lin. ft., or 20.55 ^ HO = 2,877 lbs. per panel. Hence, 4 x 2,877 ^ ^-26 = 14,500 lbs. is that part of P due to the wind ; or : P— 14,500 + 217,750 = 232,250 lbs. Also, 2^ = 25 ; ov d = 12.5 inches ; And, M = 1,945, 990 /;/. lbs. If 2 — 3" X 3" 25 lb. angles are riveted on the outside of each side plate of each post in the manner shown in Fig. 13, the centres of gravity of those angles will be 8.3 inches from the neutral axis about which / is taken, and the moment of inertia of each angle section about a parallel axis through its own centre of gravity is 2.25; hence: /= 1,886 + 4 X 2.5 X (8.3)2 + 4 X 2.25 = 2,585. Finally : A = 32.3 + 4 X 2.5 = 42.3 sq. ins. 428 COMPLETE DESIGN OF A RAILWAY BRLDGE. These quantities placed in Eq. (14) give: K = 9,412 + 5,500 = 14,912 lbs. per sq. in. which shows that the desired section is obtained. Fig. 13 shows an elevation of parts of the end post, which is supposed to be intersected by the portal strut at ab. cc' , and dd' are the 3" x 3" 25 lb. angles. ^ is 9 ft. below ab and c' 2 ft. 9 in. above it. dd' is half the length of cc'. Below c and above c' , the preceding figures show that no increase of section is needed. If much increase is needed, unequal legged angles with id' Fig. 13. the larger legs normal to the side plates can be most advan- tageously used. It will ordinarily be suf^ciently accurate to increase the K' section in the ratio of K These computations show what an important factor the wind load may be in a country subject to tornadoes and cy- clones. In such exposed localities the chord-wind stresses should not be allowed to exceed 25 per cent, of those due to the vertical loading without providing correspondingly in- creased sections. The wind effect on the stringers mentioned in Art. 81 is such a small per centage of the vertical load that it need not be considered. Conclusion. It is not necessary here to produce in detail the complete list of weights of all the parts, although this must invariably COMPLETE DESIGN OF A RAILWAY BRIDGE. 429 6e done in practice. If the estimated weight comes out greater than the assumed, a revision of the computations must be made with a sufficiently increased fixed weight to exceed, at least by a little, the estimated weight. If, on the other hand, the estimated weight is considerably less than the assumed, the latter may be reduced in a recomputation, in order to reach a proper degree of economy. APPENDIX I. THE THEOREM OF THREE MOMENTS. Art. I. — The object of this theorem is the determination of the relation existing between the bending moments which are found in any continuous beam at any three adjacent points of support. In the most general case to which the theorem applies, the section of the beam is supposed to be variable, the points of support are not supposed to be in the same level, and at any point, or all points, of support there may be constraint applied to the beam, external to the load which it is to carry ; or, what is equivalent to the last condi- tion, the beam may not be straight at any point of support before flexure takes place. Before establishing the theorem itself, some preliminary matters must receive attention. In Fig. I, let ABC represent the centre line of any bent beam; AF, a vertical line through A ; CF, a horizontal line Fig I. through C, while A is the section of the beam at which the deflection (vertical or horizontal) in reference to C, the bend- ing moment, the shearing stress, etc., are to be determined. 431 43' APPENDIX I. As shown in the figure, let x be the horizontal co-ordinate measured from A, and y the vertical one measured from the same point ; then let z be the horizontal distance from the same point to the point of application of any external ver- tical force P. To complete the notation, let D be the deflec- tion desired ; M^, the moment of the external forces about A ; S, the shear at A ; P' , the strain (extension or compression) per unit of length of a fibre parallel to the neutral surface and situated at a normal distance of unity from it ; /, the general expression of the moment of inertia of a normal cross- section of the beam, taken in reference to the neutral axis of that section ; E the coefficient of elasticity for the material of the beam; and M the moment of the external forces for any section, as B. Again, let A be an indefinitely small portion of any normal cross-section of the beam, and let y be an ordinate normal to the neutral axis of the same section. By the " common theory " of flexure, the intensity of stress at the distance y from the neutral surface is {y'P'E). Consequently the stress developed in the portion A, of the section, is EP'y A, and the resisting moment of that stress is EP'y'^A. The resisting moment of the whole section will therefore be found by taking the sum of all such moments for its whole area. Hence : M=EP':Sy"-A = EP'L Hence, also, EI If n represents an indefinitely short portion of the neutral surface, the strain for such a length of fibre at unit's distance from that surface will be nP'. If the beam were originally straight and horizontal, n would be equal to dx. P' being supposed small, the effect of the strain 7iP' at any section, B, is to cause the end K, of the tangent BK, to move vertically through the distance nP'x. THEOREM OF THREE MOMENTS. 433 If BK 2iX\A BR (taken equal) are the positions of the tan- gents before and after flexure, 7iP' x will be the vertical dis- tance between TsTand R. By precisely the same kinematical principle, the expression nP'v will be the horizontal movement of A in reference to B. Let '2nP'x and ^nP'y represent summations extending from A to C, then will those expressions be the vertical and horizontal deflections, respectively, of A in reference to C. It is evident that these operations are perfectly general, and that X and y may be taken in any direction whatever. The following general, but, strictly, approximate equations, relating to the subject of flexure, may now be written : S =:2P (i). M^ = 2Ps (2). 2nP'=2n^ (4). D =2nP'x=^"^ .... (5). A = 2.P>=^^^ .... (6). Z>ft represents horizontal deflection. Art. 2. — Some elementary but general considerations in reference to that portion of a continuous beam included be- tween two adjacent points of support must next be noticed. If a beam is simply supported at each end, the reactions are found' by dividing the applied loads according to the simple principle of the lever. If, however, either or both ends are not simply supported, the reaction, in general, is greater at one end and less at the other, than would be found 28 434 APPENDIX I. by the law of the lever ; a portion of the reaction at one end is, as it were, transferred to the other. This transference can only be accomplished by the application of a couple to the beam, the forces of the couple being applied at the two adja- cent points of support ; the span, consequently, will be the lever arm of the couple. The existence of equilibrium re- quires the application to the beam of an equal and opposite couple. It is only necessary, however, to consider, in connec- tion with the span AB, the one shown in Fig. 2. Further, from what has immediately preceded, it appears that the force of this couple is equal to the difference between the actual reaction at either point of support and that found by the law of the lever. The bending caused by this couple will evidently be of an opposite kind to that existing in a beam simply supported at each end. These results are represented graphically in Fig. 2. A and B are points of support, and AB is the beam; AR and BR' are the reactions according to the law of the lever; RF=i Fig. 2. R'F is the force of the applied couple; consequently ^7^ = AR + RF and BF= BR' ~ {R'F= RF) are the reactions after the couple is applied. As is well known, lines parallel to CK, drawn in the triangle A CB, represent the bending mo- ments at the various sections of the beam, when the reac- tions are AR and BR'. Finally, vertical lines parallel to AG, THEOREM OF THREE MOMENTS. 4.- in the triangle QHG, will represent the bending moments caused by the force R'F. In the general case there may also be applied to the beam two equal and opposite couples, having axes passing through A and B respectively. The effect of such couples will be nothing so far as the reactions are concerned, but they will cause uniform bending between A and B. This uniform or constant moment may be represented by vertical lines drawn parallel to AH or LN (equal to each other) between the lines AB and HQ. The resultant moments to which the various sections of the beam are subjected will then be represented by the algebraic sum of the three vertical ordinates included between the lines ACB and GQ. Let that resultant be called iM. Let the moment GA be called Ma, and the moment BQ — LN — HA, Ml,. Also designate the moment caused by the load P, shown by lines parallel to CK in ACB, by J/j. Then let X be any horizontal distance measured from A towards B ; ./the horizontal distance AB\ and s the distance of the point of application, K, of the force /* from A. With this notation there can be at once written : M=Ma(^-^)+M,{^-^ + M,. . . (7). Eq. (7) is simply the general form of Eq. (2). It is to be noticed that Fig. 2 does not show all the mo- ments Ma, Mb and M^ to be of the same sign, but, for con- venience, they are so written in Eq. (7). Art. 3. — The formula which represents the theorem of three moments can now be written without difficulty. The method to be followed involves the improvements added by Prof. H. T. Eddy, and is the same as that given by him in the "American Journal of Mathematics," Vol. I., No. i. Fig. 3 shows a portion of a continuous beam, including two spans and three points of support. The deflections will be supposed measured from the horizontal line NQ. The spans are represented by 4 and 4 ; the vertical distances of 436 APPENDIX I. NQ from the points of support by Ca, Cj, and C(, ; the moments at the same points by Ma, M^ and M^, while the letters 5 and R represent shears and reactions respectively. 2V 2^ # o R Fig. 3. In order to make the case general, it will be supposed that the beam is curved in a vertical plane, and has an elbow at b^ before flexure, and that, at that point of support, the tangent of its inclination to a horizontal line, toward the span 4 is /, while /' represents the tangent on the other side of the same point of support ; also let d and d' be the vertical distances, before bending takes place, of the points « and r, respectively, below the tangents at the point b. A portion of the difference between r„ and r^ is due to the original inclination, whose tangent is /, and the original lack of straightness, and is not caused by the bending ; that por- tion which is due to the bending, however, is, remembering Eq. (5) : D = Ca — c^ — IJ - d= 2^ —^jr- By the aid of Eq. (7) this equation may be written : £{(^a~(^b-^a^-^) = ^l L) M„ I M, + (8). In this equation, it is to be remembered, both x and z (in- volved in J/i) are measured from support a toward support b. Now let a similar equation be written for the span 4, in THEOREM OF THREE MOMENTS. 437 which the variables x and z will be measured from c toward b. There will then result : E(c, -c- IJ' - «") = i'^ [ ] i/, (^) + Ah (7) + ^,}f] . . . (9). When the general sign of summation is displaced by the integral sign, 71 becomes the differential of the axis of the beam, or ds. But ds may be represented by udx, u being such a function of x as becomes unity if the axis of the beam is originally straight and parallel to the axis of x. The Eqs. (8) and (9) may then be reduced to simpler forms by the fol- lowing methods : In Eq. (8) put (10). a fl — x\ xn __ I /*" u {la — x) xdx _ fa r" U{l^-X)dx Also, Also, '^r U (4 -X)dx = '-^^f\la -X)dx = ^a ^a ^^a 'a 2 . . (12). In the same manner: ^.a x^n I /*" iix^dx 'laI~Ub I ^d P'^ uxdx ~ U b I . . (13). Also, Xa /*" Uxdx laJb I = '''''.'' f\xdx . . (14). 438 APPENDIX I. And, h^ f\xdx = ^^ f\dx = ^^^^^^"^^ . . (15). Again, in the same manner: ^l — Y^ =hathn^ M^^A X . . . (16). Using Eqs. (10) to (16), Eq. (8) may be written: E{c^-C^-lat - d) =^ {Ma Ua 4 X^ + M^ 11^ la^a ) + Uxaha^\MxA X . . . (17). Proceeding in precisely the same manner with the span 4, Eq. (9) becomes : £{cc-ci,- /,f-d') = ^ {M^u^ i^x, + M^,u;i;x; ) + ti,,i,,^\M,xAx . . . (18). The, quantities Xg and x^. are to be determined by applying Eq. (10) to the span indicated by the subscript ; while Ua, ia, Uc and Zc are to be determined by using Eqs. (11) and (12) in the same way. Similar observations apply to 2ia, 4 , x„, n^y ij and x^, taken in connection with Eqs. (13), (14) and (15). If / is not a continuous function of x, the various integra- tions of Eqs. (10), (i i), (13), and (14) must give place to sum- mations (-2") taken between the proper limits. Dividing Eqs. (17) and (18) by 4 and 4. respectively, and adding the results : \ la 4 4 4 / 4 I * McUcicXc+ Mi,Uci^'Xc) .... (19). in which T= t + f THEOREM OF THREE MOMEA'TS. 439 Eq. (19) is the most general form of the theorem of three moments if E, the coefificient of elasticity, is a constant quan- tity. Indeed, that equation expresses, as it stands, the " the- orem " for a variable coefficient of elasticity if {ie) be written instead of i \ e representing a quantity determined in a man- ner exactly similar to that used in connection with the quan- tity i. In the ordinary case of an engineer's experience, T= o, /i = d'=: o, /= co)istant, 11 = ii^ = ii^ = etc. = c =i secant of the inclination for ivhicJi t — — t' is the tangent; consequently •_•'_•_•'_• _■ —I ^a — ^a — ^c — ^ c — ^ ,a — ^,c — ~J' From Eq. (10), 2/„ 2/. Xa- ^ , -c- ^ From Eq. (13), « ~ 6 ' ' ~ 6 ' The summation ^M^xAx can be readily made by referring to Fig. 2. The moment represented by CK in that figure is, consequently the moment at any point between A and K, due to F, is, Between K and B, Using these quantities for the span 4 440 APPENDIX I. ^lul/.xJx =JlM^xdx +f'"M,xdx = \P{la^ - ^)z. For the span 4, the subscript a is to be changed to c. Introducing all these quantities, E(\. (19) becomes, after providing for any number of weights, P: C \ 1(1 if, / }-iP{l^-z')z^^kp{Q-£^)z. . . . (20). Eq. (20), with <:' equal to unity, is the form in which the theorem of three moments is usually given ; with c' equal to unity or not, // applies only to a beam whicJi is straight before flexure, since 7"= t-Vt'=o — d — d'. If such a beam rests on the supports a, b, and e, before bending takes place, ^'~ — - = ^— ^ — -, and the first mem- ta 'c ber of Eq. (20) becomes zero. If, in the general case to which Eq. (19) applies, the deflec- tions ea, Ci,, and e^ belong to the beam in a position of no bending, the first member of that equation disappears, since it is the sum of the deflections due to bending only, for the spans 4 and 4) divided by those spans, and each of those quantities is zero by the equation immediately preceding, Eq. (8). Also, if the beam or truss belonging to each span is straight between the points of support {such points being sup- i>o.sed in the same level or not), tia= i^n = ^ha — constant, and Uf.— Uc = Uic = another constant. If, finally, /be again taken as constant, Xa and x^, as well as ^M^xAx, will have the values found above. From these considerations it at once follows that the second member of Eq. (20), put equal to zero, expresses the theorem of three moments for a beam or truss straight be- tween points of support, when those points are not in the same level, but when they belong to a configuration of no bending in the beam. Such an equation, however, does not belong to a beam not straight between points of support. THEOREM OF THREE MOMENTS. 441 The shear at either end of any span, as l„, is next to be found, and it can be at once written by referring to the ob- servations made in connection with Fig. 2. It was there seen that the reaction found by the simple law of the lever is to be increased or decreased for the continuous beam, by an amount found by dividing the difference of the moments at the extremities of any span by the span itself. Referring, therefore, to Fig. 3, for the shears 5, there may at once be written : 5^ = i/>^L^_^!^^LZ_^ . . . (21). 5,'=:SPf +^^^-^* .... (22). *a 'a s, = :ep^+'^^^-^ .... (23). The negative sign is put before the fraction — ^ ~ , in 'a Eq. (21), because in Fig. 2 the moments J/^ and J/^ are rep- resented opposite in sign to that caused by P, while in Eq. (7) the three moments are given the same sign, as has already been noticed. Eqs. (21) to (24) are so written as to make an upward re- action positive, and they may, perhaps, be more simply found by taking moments about either end of a span. For example, taking moments about the right end of 4: SJa - ^P{la - =) + Ma = M^. From this, Eq. (21) at once results. Again, moments about the ieft end of the sam.e span give : 442 APPEiVDIX I. This equation gives Eq. (22), and the same process will give the others. If the loading over the different spans is of uniform inten- sity, then, in general, P — wdz; w being the intensity. Con- sequently : :SP(/2 - ^) ^ = /w(/2 - z')zdz = w -. ^ o 4 I " In all equations, therefore, for — '2 P{la — z^) z there is to /^ I ^ be placed the term w^-- ; and for — ^ P{l^f — ^) z, the term 4 4 43 Wg— . The letters a and c mean, of course, that reference is 4 made to the spans 4 and 4. From Fig. 3, there may at once be written : R =Sa + S, (25). R' =S, + S, (26). R" =67 + 5, (27).. ctc.= etc. + etc. APPENDIX II. THE RESISTANCE OF SOLID METALLIC ROLLERS. An approximate expression for the resistance of a rollei may easily be written, and although the approximation may be considered a loose one, it furnishes an excellent basis for an accurate empirical formula. The following investigation contains the improvements by Prof. J. B. Johnson and Prof. H. T. Eddy on the method originally given by the author. The roller will be assumed to be composed of indefi- nitely thin vertical slices par- allel to its axis. It will also be assumed that the layers or slices act independently of each other. Let E be the coefficient of elasticity of the metal over the roller. Let E be the coefificient of elasticity of the metal of the roller. y\g. i. Let R be the radius of the roller and R' the thickness of the metal above it. Let w = intensity of pressure at A. " / = " " any other point. " /• = total weight which the roller sustains per unit of length. " X be measured horizontally from A as the origin. " ^ = AC. " -f = DC. 444 APPENDIX II. From Fig. I : h h h h ,'. d = AC = AB + BC = w (^ + I') ; And (I). A'C' = A'B' + B'C:=^p(^ + ^)l. . . . (2). Dividing Eq. (2) by Eq. (i): p = A'C -^ But p-^r 'pdx = '^ r ' A'C'dx. If the curve DAH be assumed to be a parabola, as may be done without essential error, there will result : /'' A'Cdx = -ed. e 3 Hence : P=^we (3). But: e = V 2J^d - d- = V 2^ nearly. By inserting the value of d from Eq. (i) in the value of e, just determined, then placing the result in Eq. (7) : P=t^lJiJf4) (4). THE RESISTANCE OF SOLID METALLIC ROLLERS. 445 P=^ R V 2lif , E + E ' EE' (5). The preceding expressions are for one unit of length. If the length of the roller is /, its total resistance is i/' P = P' = \l\/ 2^R{^^§) ... (6). Or if ^ = K / ^'=f^V-^^^' . (7). In ordinary bridge practice Eq. (7) is sufficiently near for all cases. A simple expression for conical rollers may be obtained by using Eqs. (4) or (5). As shown in Fig. 2, let s be the distance, parallel to the .axis, of any section from the apex of the cone; then consider J, . -_-^, Fig. 2. ■a portion of the conical roller whose length is dz. Let Rx be the radius of the base. The radius of the section under con- -sideration will then be R^yR.: 44^ APPENDIX II. and the weight it will sustain, if R^ = R ; R, / E + E' dP' = y A/ 2w^ —EE^' ^^^' Hence : ^■=/>^'=^V"'' ^ + E' EE' (8). Eqs. (6), (7), and (8) give ultimate resistances if w is the ultimate intensity of resistance for the roller. It is to be observed that the main assumptions on which the investigation is based lead to an error on the sid*' cJ safety. If for wrought iron, w = 12,000 pounds per square inch, and E = E' = 28,000,000 pounds, Eq. (5) gives : APPENDIX III. THE SCHWEDLER TRUSS. The general principle applied in Chapter III. to bowstring trusses, enables the characteristics of the Schvvedler truss to be very simply shown. In fact, that truss is a special bow- string, having the least possible number of diagonal braces under the conditions assumed. Fig. I represents the elevation of such a truss, and the problem involved is the determination of such depths, near ^ C \ K . f « • / ^ '^^ V\ yv Xl "TTTPr/t- b c d c \ g 'I ) k Fig. I. the ends, that one diagonal only will be needed in each panel ; it being premised that the inclined web members or diagonals are to sustain tension only. Let W — total (upper and lower chord) panel fixed load. " R' = half the fixed load (or weight) of the bridge. " w = panel moving load. " / = length of span. " ^ = any vertical brace or truss depth, as Cc. '* dx ~ vertical brace or truss depth, as Dd, adjacent to d and toward centre. " / = panel length. " « = inclination of any diagonal, as Cd, to the hori- zontal lower chord, i. c, Cdc = a. 447 448 APPENDIX rii. Let X = distance from A to the intersection of the pro longation of any upper chord panel, in the left half of the truss, with the prolongation to the left of the lower chord. " J = the normal distance from that point of inter- section to the prolongation of the diagonal immediately under the upper chord panel prolonged. Let the moving load pass on the bridge from A toward L, and let ;/ be the number of panel-moving loads from A. The reaction at A, for any position of the moving load will be : R = R^+nzv(^i- ^'+J^^ y . . . (I). Then let the truss be imagined divided through the panel immediately in front of the train. If moments be taken about the point of intersection de- noted by Xy and if T represents the tension in the diagonal just in front of the train, whose lever arm is y, there will result : Ty = Rx - n{lV + w)fx + ^-^-^^^\ . (2). Eq. (2) is so written, it is important to notice, that if the second member is greater than zero, or positive, T will be tension. Hence, if T is tension : But, j/ = (x + (n + l)p) sin a (3). Also, from similar triangles ; ^-^^=—^-^.'.x + {n + i)p = -^^.. (4). By the aid of Eq. (3) : THE SCHWEDLEK TRUSS. ^g { n + I)/ — R{7i + i)p + n{lV + w) 2 R — n{W +w) ~ {x -f (;^ + I )p) sin a sin a — * Hence, by the aid of Eq. (4) : n{ W + wY (^ n{W ^ivf. [R- ^ 'Yn + i)p /< Using the last two members of this inequality : d R — n{PV + w) (6). ^ ,^ , / R- n{lV + zv) \ Or;d> dJi - j^ ^^ — '- ) The second member of (6) is the least value of the depth d which can exist without inducing compression in the diagonal under consideration. This diagonal is the one immediately in front of the train, and the principles given in Chapter III. show that if this position does not induce com- pression, no other will. Inequality (6) shows that if R is greater than n{lV + w), or R>n(JV + zc), d will always be less than d^. U R = n{W V lu), then d= d^. Again, \i R d-^\ hence, zvhen n becomes so great that R is less than ii{ IV + zv), or R < n{ IV + zv), T zvill be compression, and the diagonal imist be coiinterbraced or else intersecting diagonals mnst be placed in the panels, as shozvn in Fig. i, near the middle of the truss. The value of n given by R ii{ IV + w) — 52. Hence Cd is the only diagonal needed in the panel CDdc, and d = CV is to be computed from Eq. (6). The centre depth — Ee — Ff was taken at 16 feet ; let DdhQ taken at 15.5 feet. Since « = 2 ; R — ti {W ^ w) = \o, and i? - ^^ — ' = 36. Substituting these values, and di= 15.5, in Eq. (6) : d = o.gi dj = 14. 1 1 feet. Hence let d— 14.5 feet = Cc (Fig. i). Next, let the head of the train be at d, i.e., let « = i. Then by Eq. (8) : y?= 32 + 16 =48. fi{ M/ 4- iv^ Also; R - n{lV + w) ^ 22, and 7? - -^^ — ^I__' = 35. For this position of load, d^ = 14.5 feet. Hence, by Eq. (6): ^= 14.5(1 ) = 9-94 feet. Hence let d — 10.00 feet = Bd (Fig. i). 452 APPENDIX III. Fig. I represents the truss, drawn to scale, with the various depths given or computed as above, i. e., Ee — Ff — 16.0 feet. Cc = Hh — 14.5 " Bb = Kk^\o.o " In the three panels adjacent to each end of the truss only two main diagonals are thus seen to be necessary, and ir^ those no compression will ever exist. In each of the three middle panels, however, two intersecting diagonals will be necessary, since no diagonal must sustain compression. As is evident, the expression R — n{ \V + w) is the vertical shear at the head of the train. Hence the limiting case of the inequality (7) : R = n{W +w), gives the two points, in the two halves of the truss, at which the vertical shear at the head of the train is zero. Between these points intersecting diagonals or counterbraces are needed, and only between them After the truss depths are fixed by the preceding method, the stresses in the individual members are to be found in the usual manner — as in any other bowstring truss — as shown in_ Chapter III. APPENDIX IV. REACTIONS AND MOMENTS FOR CONTINUOUS BEAMS. Case I. — Tluee Spans with Two Intermediate Points of Support and Two End Supports. The notation of Art. 35 will be used and reference will be made to Fig. i of that Art. M, = [a%^: - J)| - 2/3' (A + 4) ip(. - g^ ^ !4 (4 + 4) (/, + /,) -4"! (i). M, = - I J/34 + A'2J'(^' - ^)f I - 2 (4 + 4)- • • (2)- ^■=M'-7>f (-'>■ Ji. = 2P'^-^-^^-^' (4). ^3 = 7 + 2iJ^j-~^ (5). In ordinary swing-bridges where /j = 4 = / and a single weight P rests on /i : -.=H(.-;)-(--";);\-#i^,4--<^)- 454 APPENDIX IV. These formulae are in no wise changed for a single weight P resting on 4, except that R^ and R^ are interchanged. Also : ^^^ w -^.) /-/>(/-.) _ ^^ ^^^^ ,'.R.^ = P-R^- R,^ R^ (lo). Case II. — Two Spans with One Intermediate Support. Reference will be made to the notation and Fig. 2 of Art. 35. ^■ = ^^^0-7.)-f -■ • • • C^)- ^, = 24-^ .i/'^.-f (,3). lf/, = /, = /.■ '2 — M These formulae are based on the supposition that there may be negative reactions at A and C o{ Fig. 2, Art. 35. If no negative reactions are possible, and if the load is on one arm only, that arm will be a non-continuous beam for such load, and the reactions will be found by the simple principle of the lever. APPENDIX V. CANTILEVERS. Art. L— Cantilever Structures.— Positions of Loading for Greatest Stresses in the Cantilever Arm. Cantilever structures are formed of continuous or semi- continuous trusses in which the reactions and stresses are equally determinate with those in non-continuous trusses. Fig. I, PI. XIII., typifies a cantilever structure with an anchor arm at each end. The anchorage at A holds the anchor arm ;/ in position under all conditions of loading, thus enabling the cantilever arm ;// to sustain the suspended span /. The latter is a simple, non-continuous truss with inclined end-posts, suspended at the lower extremities of the vertical tension members at C and D, thus readily allowing expansion and contraction to take place both in the suspended span itself and in the cantilever arms adjacent to it. As can- tilevers are erected without lower false-works between the piers Band E, each cantilever arm, with the adjacent half of the suspended span, is built out from each of those main piers ; hence the members FG sustain compression during erection, while HD take tension. These erection stresses are frequently very heavy, and the members themselves must be designed to take them. At the same time, proper details must be arranged so that after the structure is complete the requisite expansion and contraction can take place at each end of the suspended span. These results are usually accom- plished by folding wedges, or wedges and rollers, removed after erection, in connection with oblong pin holes. Similarly, nearly the whole of the upper chord of the sus- pended span must be designed to take the tensile erection stresses, and nearly the whole of the lower chord the corre- sponding compression erection stresses. The erection stresses in the cantilever arms arc, of course, the same in kind as 45^ APPENDIX V. * those caused by the moving and fixed loads, and no reversion takes place. The conditions of loading for the greatest stresses in the suspended span, in addition to the erection stresses, are pre- cisely the same as those for the ordinary non-continuous truss given in detail in Art. 7, and they will receive no fur- ther attention here. Greatest Web Stresses in Cantilever Arm. In seeking the greatest web stresses in the cantilever arm, reference will be made to Fig. i, PI. XIII., and a perfectly general system of loading will be assumed. The chords of the cantilever arm will also be assumed to be not parallel. Let the greatest stress 5, in MN, be desired, and let the system of weights or concentrated loads W{, W2, etc., extend over the entire suspended span /, and over the cantilever arm ;«, from C to some point to the left of M. Then let i rep- resent the distance from C to the point of intersection of the chord sections in the same panel with MN, while A is the length of the normal dropped from the intersection point to MjV prolonged. Also let ^1 represent the distance from D to the centre of gravity of the load on /; ^9 the distance from the same point to the centre of gravity of the load to the left of C; g^ the similar distance from M ior the load to the left of M) and ^, the similar distance from M of the loads Wn, W^, etc., in the panel in question. The loads on the left of M are W^, W2, etc. The distance from C to M is represented by k. R is the downward reaction or upward pull on the anchorage at the extremity of n ; while R^ is the upward reaction at the main pier. The weights over the various portions of the structure will be represented as follows : 2 W for total weights in /. CANTILEVERS, 457 The centre of gravity distances, gy, g.,, and g^, can be read- ily found by aid of tabulations like those given in Arts. 9 arid II, while ^4 will be found by taking moments of IV^, W^, etc., about M. Then, by moments in the ordinary manner: B ^'^klV^{^^i±±^^^lV (I). R., = reaction at C from load on suspended span =^ 2W (2). y?,^y?+i?,+5j^=^(^^i)in^^4-(^^'±^-i)SjF(3). If the length of the panel in question is/, then that portion of the loads l\\, JV^, etc., resting in it and carried to M, is: (w, ^ rr, + etc.) (i - ^) (4). If the cantilever arm be now supposed divided through the panel MN, and if the moments about intersection of chords be taken of all the forces external to that portion of the structure to the left of this line of division, including the reactions R and — R^, there will result : 5 = -, I ^' i^r + (^, - / 4- /) 1 JF - (f-Fi + W.^- etc. + W, + fr; + etc.) {k + i) -{W,^ W, + etc.)^-3 + (^^4 + ^^'5 + etc.)(/J' + ifj [ . . (5). In order that 5 may be a maximum or a minimum, AS must be zero when gx, gi, gs, and gi each varj' by the same small amount, as when the loads are all advanced slightly to the left. It is to be remembered, however, that the variation of ^4 will be opposite in sign to that of the others. By thus 458 APPENDIX V. making AS equal to zero, there will result as the desired con- dition : W,+ IV, + etc. + {W, + W, + etc.) ^-^^ = \2W + S:W. P . ^ .-. ( W, + Wr, + etc.) ^^-^^^ = -^ i IV + i IV. . (6). In case there may be a number of maxima indicated by Eq. (6), the greatest must be sought by trial. Eq. (6) shows, as might have been anticipated, that the loads IV^+ ff^a + etc, between the panel in question and the main pier, do not directly affect the conditions for greatest stress. If the load is of the uniform intensity w,-d,Vidi x is the por- tion of the panel/ covered by it, Eq. (6i is only satisfied by making x = /. This result might also have easily been anticipated, for it is clear that if the load is uniform it must at least reach to M, for the panel MN. The amount by which it may extend to the left of M is a matter of indiffer- ence. If the chords are parallel, i is infinitely great and Eq. (6) becomes : ^4+ IV, -^ etc. = j 2 IV. . . . . (7). In the general case, when Eq. (6) has fixed the proper position of loading, the corresponding values of ^i, £:,, g^, and g^ can be at once obtained by the methods already indicated, and Eq. (5) will then give the desired greatest value of 5. In the case of a uniform load, u>, that equation takes a much simplified form, since : W^ + W2+ etc. = o ; Wi + IV, + etc. = wp ; I 2W=wk; 2JV=wl; ^ , . ^ A P CANTILEVERS. 459 Substituting these values in Eq. (5) : If the chords are parallel and a is the angle between ^' and a vertical line,// = (00 = z) cos ix\ hence: S = zv i~ -\- k —^ \ sec IX (9). With the same uniform load, w, per lineal foot, and with kw on the cantilever arm in, Eqs. (i), (2), and (3) become : R =^"l(Uk-^\ (I0\ n \2 2mJ /?, = '^' (.1). 2 „ wl fm \ , fm k \ , . R^ = — [^+\]+wk[ — + I ) (12). 2 \n J \n 2n J ^ ' If the load, w, covers the entire arm ;;/, k = ;;/, and : R^w-{l+vi) . . . (13). 2n ' ^ wl , „ %vl (m ^ \ f m ^ \ I . R, = --, and ye, = ^ (- + l) + wm (^- + . ). (.4). If a single weight, W, rests on the suspended span, /, at the distance ^, from D : ^=,K'-M.;^,= ^,and^, = i^('^.). . (.5). If a single weight, W, rests on the arm m at the distance k from C: ji^KiaiJzJ); ji. = o: and R, = w('^i^+ i). (16). .460 APPENDIX V. Greatest Chord Stresses in the Cantilever Arm. Reference will again be made to Fig. i of PL XIII., in which the compression web members are vertical ; the nota- tion will remain as before. The greatest chord stress, S^., in any panel, as MN, will be sought. Let //' represent the normal dropped from M on the chord panel whose greatest stress is S^. ; then, if moments be taken about M, h^ will be the lever arm of Sc, and there will result : 5c = .-J I Rx {m — k) — R {in + n — k) -(W, + JV, + ctc.)s,^. . . . (17)- .-. 5, = i U^^W- ^j^2W- {/+ k)2lV . -{IV,+ W^, + etc.)^,| .... (18). For a maximum or minimum ^Sc = o when gx, g^, and g^ are the only variables in the second member of the equation ; hence : -'^w^ ^w-{w, + w, + t\.c.) = hiv. . . (19). This condition for a maximum is seen to be independent of i; hence it is precisely the same whether the chords are parallel or inclined. After the values of g, gg, and ^3, corre- sponding to the position of loading fixed by Eq. (19), are in- serted in Eq. (18), the desired chord stress will at once result. It is to be borne in mind, however, that there may be several maxima fixed by Eq. (19), and that the greatest of these, to be found by trial, is the " greatest stress " desired. If the load is of the uniform amount, zi>, per lineal unit of structure, Eq. (19) is satisfied only by making ^'rF= kw; hence, the moving load iniist cover the suspended span, and the cantilever arm from its end to the vertical line through the cen- CANTILEVERS. 461 tre of Diomcnts — i.e., the distance k. Eq. (18) then takes the value, for uniform loading : 5. wk 2h' {k + I). (20). For a panel in the horizontal chord, //' is simply the depth of truss at the origin of moments. In case all web tnembers arc inclined, as indicated in Fig. i, so that any lower chord panel point M^ is at the horizontal distance q from that in the upper chord, Eq. (18) takes the form : 5e = I \g!^W- (i±^ii'fr-(7+ k ^- q)'kw -(W^, + ^F, + etc.)(/-^)"^| . . . . (21). Again making ^5^= o, with g^, g.,, and g.^ the only vari- ables, there results for a maximum or minimum : f I^ = i±^il^+ {W, + W, + etc.)^^ ^ Or, / ^^2W=^^W-{IV,+ lV, + etc./-^-^, (22). (23). If ^ = o, Eq. (19) at once results. The observations fol- lowing Eq. (19), regarding /, gi,g2, and gs, obviously hold true for this case also. If the moving load is of the uniform in- 462 APPENDIX V. tensity w, Eq. (23) is satisfied only by making it cover k -\- p\ hence that equation takes the form : k + q=^k+^^ = k+q; . . . . (24). which shows that t/ie laiiform moving load imist cover the sus- pended span, and the cantilever arm from its end to the further extremity of the panel in which the chord stress is sought. For this uniform moving load Eq. (21) takes the form: -^k^q){k-^p + -y^-^p-q)^-^y (25)., h' \ 2 If ^ z= o in this equation, Eq. (20) immediately follows. If the panel in the lower chord is under consideration, M, Fig. I, becomes the origin of moment, and as the rhoving load is supposed to traverse the upper chord, the conditions. for a maximum and the corresponding formulae are precisely the same as for the case with vertical compression web mem- bers. If, with all inclined web members, the moving load tra- verses the lower chord, the preceding conditions and formulae apply precisely as they stand. It is only to be borne in mind that the formulae involving q are to be used for the chord carrying the moving load. If the tension members are vertical, and the compression members inclined, as in the Howe truss, or as exemplified by the inclined post, P, Fig. i, of PI. XIII., the positions of moving load for greatest chord stresses will be at once found by making q — p in Eq. (23), which gives: —j^^W=2W (26). for the general load, and the same condition for a uniform load as that described in italics immediately after Eq. (24). By making q — p, in Eq. (25), the expression for the greatest chord stress under a uniform load, w, becomes — CA A' riLE VERS. 4^3 " //' ( 2 2 ) 2ll'- ' ' If the truss is of uniform depth, /^' is that depth. The preceding investigations hold true for any system of triangulation or for any system of loading. They do not apply to multiple systems of bracing, as such systems do not admit of exact calculation of stresses. When they are used, each system must be assumed to act independently of all the others (although the assumption cannot be definitely estab- lished) and to Z2sry the system of maxima concentrations permitted by the moving load used, or else such a uniform load as will be practically equivalent to the real load. Although there is not at the present time (1890) any con- structive or other proper reason for the use of multiple sys- tems of bracing in any pin structure of proportions hitherto employed, the above assumptions are \\ithin the limits of safety, and may be used where necessary and unavoidable. Another constructive device is that shown in Fig. 2. It may be used advantageously in cases where the cantilevers are supported on high iron or steel piers or towers, when the latter would become too expensive if constructed of masonry of corresponding dimensions. It leads to appreciable econ- omy by the shortening of the clear cantilever opening, as well as the anchor arm. It consists in separating the feet of the two inclined posts, P and IB' , of Fig. r, PI. XIII., by some convenient distance, o, and omitting all bracing in the rect- angle, BB'B^B^, Fig. 2, so that no shear can be transferred past B' to the left, or past B^ to the right. Since this last condition exists, the reaction, R, and the positions of loading for all the maxima web and chord stresses in the cantilever arm, as given by the preceding formulae, will hold for this case without any change whatever. It is only to be carefully observed that /// and ;/ are to be measured from the extremi- ties of the cantilever and anchor arms, respectively, to the extremities of the open panel, 0, as shown in Fig. 2. The reaction, R^ at B\ will equal the sum of the vertical compo- nents of all the inclined stresses in the first panel of m — i.e., 464 APPENDIX V. the total shear in tliat panel, added to the panel loads acting at B and B . Similarly, the reaction R" at B^ will equal the sum of the vertical components of all the inclined stresses in the first panel of n — i.e., the total shear in that panel, added to the panel loads acting at B^ and B^. Finally, the reactions, 7?,' and 7?/', of Fig. 2, added together, will equal Fig. 3. the reaction, 7?„ of Fig. i, PI. XIII. Thus it is seen that the formulae already established for the latter case will meet the former without any change whatever. The case of the subdivided panel shown in Fig. 3 is also covered by the preceding formulae. The general and detailed considerations governing their applications are precisely the same as those given at length in Art. 7 ; it is not necessary, therefore, to give them further attention here, except to ob- serve that for the greatest stress in be, the panel length to be taken is ad, while ae is the panel for the greatest stress in ab. Art. 2. — Positions of Moving Load for Greatest Stresses in the Anchor Arm, and the Greatest Stresses Themselves. No load placed on the anchor arm of a cantilever structure will affect in any way the reaction or any shear in the canti- lever arm, except such as may result from its slight deflectional movement, being too small to be expressed in appreciable amounts. It at once follows from this fact that all stresses in the anchor arm, due to loads on that arm, may be compu- ted precisely as if it {AIB'L of Fig. i, PI. XIII.) were a simple, non-continuous truss supported at each end {i.e., A and B' CANTILEVERS. 465 of Fig. I, PI. XIII.). The greatest of these stresses so found are then to be combined with the greatest stresses of the same kind due to the load on the cantilever arm and result- ing from the reaction, R, at the anchorage ; this combination will give the resultant greatest stresses desired. Although the preceding general observations are compre- hensive and sufficient for the determination of all the greatest stresses desired, it will be well to extend them in some detail, without, however, establishing any formulae. Inasmuch as all upper chord stresses in the anchor arm, due to load on that arm only, will be compressive, it is clear that the greatest tension in the upper chord, so far as it may exist, will be found with no moving load on the anchor arm, and with the moving load so placed on the cantilever arm as to give the maximum bending moment (and, hence, chord stresses) over the main pier between the anchor and cantilever arms. This position of the moving load on the cantilever arm can be found for the various cases from Eqs. (18), (20\ (21 ), and (25) of the preceding Art. The resulting moment over the main pier, divided by n, will give the reaction, R, with which the greatest tension throughout the upper chord and the greatest compression throughout the lower chord are to be found — i.e., so far as this tension or compression exists. If the anchor arm is very long in comparison with the can- tilever arm, the fixed load compression in the upper chord and tension in the lower may be so great that there will be but little upper chord tension and lower chord compression, even for the conditions producing maxima. Inasmuch as all loads on the suspended span and cantilever arms produce tension in the upper chords of the anchor arms and compression in the lower, the positions of moving load for the greatest compressions in the upper chord of either anchor arm, or greatest tensions in the lower, are precisely the same as if it (the anchor arm) were a simple, non-continu- ous truss, the cantilever arms and suspended span, at the same time, carrying no moving load. The greatest compres- sive stresses in the upper chord and tensile stresses in the lower chord, thus found in the anchor arm, as for a simple, 466 APPENDIX V. non-continuous span, are to be combined with the upper chord tension and lower chord compression produced by the fixed load of the cantilever arms and suspended span, whence will result the maxima chord stresses desired. The greatest shears in the anchor arm will evidently be affected by the magnitude of the reaction /?, Fig. i, PI. XIII., at its extremity. If a downward shear, producing tension in members sloping similarly to OQ, or compression in IB\ is called a main shear, then any condition of loading on the structure which will increase the downward reaction, R, will increase the main shears and, hence, the stresses in the main web members, as AL, LV, UT, etc., and IB' will be called. But the maximum value of the downward moving load reac- tion or pull at the extremity of the anchor arm will be found with no moving load on the latter, and for the maxi- mum value of the bending moment over the main pier B', as given by Eqs. (i8), (20), (21), and (25), of the preceding Art., as already stated on page 465. Finally, that part of the maximum main shear in any panel of the anchor arm due to the moving load on that portion of the structure alone, is found precisely as if it were a simple, non-continuous truss with the given system of loading, advancing from the extrem- ity of the anchor arm to the main pier. The positions of moving load, therefore, for the main shears or main web stresses are, first, so much on the cantilever arms and sus- pended span as will produce the maximum downward pull at the extremity of the anchor arm, and, second, with this con- stant condition in the main span of the structure, an advanc- ing moving load from the extremity of the anchor arm, treated as a simple, non-continuous span, to the main pier. The maximum moving-load shears or stresses thus found, com- bined with the fixed-load shears or stresses in the anchor arm, will give the resultant greatest web member stresses desired. It may sometimes happen that the anchor arm will be so long that the greatest downward pull at its extremity due to the moving load will be less than an existing upward reac- tion due to the fixed load. In such a case no anchorage, of course, will be required, as the reaction at the extremity of CA N TIL E VERS. 4^7 the anchor arm will always be upward. The preceding con- ditions for greatest main web stresses, however, hold in all cases without exception. The greatest counter shears and, hence, stresses in counter web members such as QV, or VL if under compression, or AL in tension, Fig. i, PI. XIII., will, for the same general rea- sons just stated in connection with the greatest main shears, be found under the moving load advancing from the main pier B' to the extremity of the anchor arm, under the supposition that the latter is a simple, non-continuous truss, with no mov- ing load whatever on the cantilever arms or suspended span. If a portion of the web members are vertical posts, as shown in Fig. i of the plate, so that the kind of stress in them is the same whether acting as counters or main web members, those nearest the extremity of the anchor arm will probably take their greatest stresses as counters, but those more remote are doubtful and must be computed both as main and counter members in order to find the greatest values. Those near the main pier will take their greatest stresses as main web members. Art. 3. — Positions of Moving Load for Greatest Stresses in Anchor Spans and the Greatest Stresses Themselves. The anchor span is shown as the span ;/ in Fig. 2, of PI. XIII., on the left of which there is supposed to be a cantilever arm similar to vi on the right. The anchor span is always of such length and weight that the reactions, R and Ry, are invari- ably upward. Hence the anchor span is identical in character with the anchor arm, whose length and weight are so great that the reaction at its extremity over the anchorage is always upward. Hence the positions of moving load for the greatest stresses are precisely the same as those established for the anchor arm in the last Art. The greatest moving-load stresses in the span ;/, Fig. 2, PI. XIII., are to be determined precisely as if the trusses were of a simple, non-continuous character ; and the fixed-load stresses are to be found in the same trusses under exactly the same conditions. The greatest difference of moments at the two 468 APPENDIX V. sections, BR^ and AR, is then to be found and divided by the span length n, thus giving a reaction ivJiich is to be treated pre- ciscly as the downward reaction at the extremity of the anchor arm in the preceding article. This downward reaction will pro- duce stresses in the chord and web members of the anchor span, which are next to be determined in the usual manner for a load hanging at the extremity of the span n, with the other extremity fixedly held. If both cantilever arms flanking the span n are of the same length ni, and of the same weight similarly disposed, the fixed-load moments at the sections AR and BR^ will always be equal (but not otherwise), and the difTerence in moments mentioned above will be entirely due to the moving load on one of the adjoining cantilever spans. Any difference of moments at those two sections resulting from dissimilarity or inequality of fixed loads must be added to or subtracted from the unbalanced moving-load moment just described. Finally, the fixed-load balanced mo- ment at either of the sections AR or BR^ is then to be deter- mined and divided by the depth of the truss in order to find the uniform tension throughout the length of the upper chord, and the uniform compression throughout the length of the lower, if the truss has a uniform depth. The stresses resulting from these four sources — viz., the moving load and the fixed load on the span n\ the unbalanced moment over the piers due to possibly both moving and fixed loads, and the balanced moment due to fixed load only — are to be so combined as to produce maxima in each and all the truss members. The greatest web stresses will require the greatest unbal- anced moving-load moment alternately at each end of the anchor span, with the usual progressive movement of moving load for ordinary non-continuous trusses ; but the greatest upper chord compression and lower chord tension will be found with no moving load on the adjacent cantilever arms or suspended spans. The greatest upper chord tension and lower chord com- pression will be found by combining the fixed-load stresses, found as indicated above, with the moving-load stresses CANTILE VERS. 469 induced by the greatest possible moving-load moments simultaneously at each end of the anchor span, the latter, at the same time, being entirely free from moving load. If the anchor span is not of uniform depth, the balanced fixed-load moments at its extremities will induce stresses in the web members, which are to be determined in addition to those in the chords for combination with the other stresses described above in the search for the maxima. Art. 4. — Wind Stresses. The stresses due to the action of the wind on the various portions of a cantilever structure play a very important part in its design. The conditions of wind loading are much more varied than in ordinary non-continuous spans, and the result- ing stresses must be computed with great care and thorough- ness. The pressures against the different members of the structure constitute a fixed load, and all resulting stresses are to be found by precisely the same general methods as given in the preceding articles for the vertical fixed loads — i.e., own weights. Fig. 5 of PI. XIII. shows the horizontal system of lateral bracing between the horizontal upper chords, ABCDE of Fig. I, If the wind blows on the entire structure in a given direction, as shown by the arrow in Fig. 5, there will in gen- eral be a horizontal reaction induced at A, in direction and amount depending on the relative proportions of anchor and cantilever arms and suspended truss. This reaction must be provided for, at A, by a proper connection between the end of the anchor arm and masonry at that point, so designed that the requisite longitudinal expansion and contraction may at the same time take place. The amount of this reac- tion is to be determined precisely as the fixed-weight reac- tion, R, at the same point, in Fig. i. This reaction, R, will be composed of two parts, one of which is due to the wind pressures along the upper chord, ABCD, Fig. i, and the other to those along the lower chord, ALB'NG; and each is to be found as indicated above. By means of these horizontal reactions at A, and the panel wind pressures at the upper 4/0 APPENDIX V. and lower chord points, all the fixed-load wind stresses in the upper and lower lateral systems, both of which are pro* jected in Fig. 5, may be at once completely established. The wind pressures against the moving train form a mov- ing load in those chords carrying the train (the upper chords in Fig. I, PI. XIII., and the intermediate and lower chords in Fig. 3). The conditions under which these moving wind loads produce their horizontal reactions at A, and their great- est stresses in the web and chord members of the lateral sys- tem, are precisely the same as those described in detail for the vertical or train moving loads in the preceding Arts. The cantilever conditions, so to speak, including span lengths for the former loads, are identical with those for the latter. Hence the positions for the wind maxima must be deter- mined by exactly the same general methods as are used for the vertical loads. The computations for the moving wind stresses are simplified by the fact that that load is of uni- form intensity. The combination of the fixed and moving wind load stresses according to the usual methods will give the desired resultants in all the members. In the case of the anchor span of Fig. 2, PI. XIII., the same general observations as those already made are to be applied. The fixed wind loads on the span are to be treated precisely like the fixed vertical loads, and the resulting wind stresses in the upper and lower lateral systems are to be computed accordingly. Again, the moving wind loads accompany the moving vertical or train loads, and all the conditions deter- mining the greatest stresses for the latter determine those for the former also. The same set of greatest moving wind stresses must therefore be found as for the moving vertical loads, and followed by their combination with the fixed wind stresses, in order to reach the resultant wind stresses desired. Inasmuch as it is known that the highest wind pressures cover comparatively small areas, it will usually be necessary to supplement the preceding wind computations, which are based on the assumption that the wind presses equally over the entire structure, by others resulting from the application CA JV TILE VERS. 47 1 of the highest pressure to some particular portion, as the clear cantilever span or the anchor span, or, possibly, the anchor arm, and provide for the resulting stresses. The extent to which these supplementary and special computations are to be made will depend upon the judgment of the engineer acting on the circumstances of each case. The transverse bracing in the vertical and inclined planes of the various pairs of web members is to be designed under the same assumptions of wind transference, and in accord- ance with the same principles used for the same general purposes in ordinary non-continuous spans. The wind loads in the upper chords of the suspended trusses of the cantilever span should always be carried down to the lower chords of the cantilever arms at their extremities, and along those chords to the piers, in order that the overturning wand effect on the cantilever and anchor arms may be reduced to a mini- mum. In the same manner, the wind loads in the upper chords of the cantilever and anchor arms, which are resisted by the main piers, should be carried down to them by the transverse bracing between the inclined posts acting as por- tals at those places. In high cantilever trusses the overturn- ing wind effect is frequently a very serious matter, and should, indeed, be carefully computed in all cases, and its results com- bined with the fixed load and wind stresses already deter- mined. It will throw a considerable portion of the fixed load in the windward truss into the leeward. It will also consid- erably increase the downward pull on the windward side of the anchorage, and relieve that at the leeward side by the same amount. These changes in the downward pull of one side of the anchorage can be found by dividing the wind moment about the top of the pier between the anchor and cantilever arms by the transverse width between the trusses where they are attached to the anchorage. The same effects occur, of course, similarly in the anchor span. These overturning effects, resulting from the pressure of the wind against the structure, affect the fixed-load stresses only. It may also be necessary to consider the truss stresses arising from the over- turning effect of the wind on the train. This, however, will 472 APPENDIX V. depend upon the special circumstances affecting any particu. lar case. Art. 5. — Xiconomic Lengths of Spans and Arms. The problem of the relative lengths of suspended spans, cantilever arms, anchor arms, and anchor spans for the most economical amounts of material in them is not capable of exact mathematical treatment, but may be very simply solved in a manner sufficiently accurate for all practical purposes. In order to do this it is only necessary to observe that the weight per lineal foot of any single span of bridge structure, omitting the floor, is very closely found by multiplying the span length by a numerical factor, determined by actual com- putation of weights of similar spans. In the case of a canti- lever arm, the length of that arm would be used instead of span length. In the analysis which follows, there will be used : a — cantilever arm factor ; b = suspended span factor ; f = anchor " " ; rt^= anchor " " L = length of cantilever span — I + 2ni of Fig. i, PI. XIII. / = " suspended span. Economic Length of Suspended Span and Cantilever A rm. The total weight, W, of the suspended span and cantilever arm trusses will be : -{L-lf + bP=W, Therefore, by differentiation : a{L-l)- 2bl = o. .-. /= — ^ Z. . . . (i). ^ ^ a + 2b ^ ^ Hence, cantilever arm = w = ■ j L (2)» a + 2b If there be taken loosely, a — 20 and b= j, I will be nearly 0.6 L and in nearly 0.2 L. It has been found, however, that / may be advantageously taken about 0.5 Z to 0.55 Z, and m about 0.25 L. CANTILEVERS. 473 Economic Length of Anchor Arm, There is to be found in this case the entire weight of the trusses for two anchor arms and the cantilever opening, or the weight of / + 2n + 2m = 5 of Fig. i, PL XIII. By using the proportions established in the preceding case, there will result : Anchor arm length = n. Suspended span length = -j (S — 2n) = I, Cantilever arm length = — {S — 2n) = tn. Hence the entire weight desired will be : ad 2cn' a + -^{S-2ny=W. .... (3). By differentiation : ; r {S — 211) = O. .-. n = -7 7- ; S. (4). a-\-2b^ ' c{a + 2d} + 2ab ^^' en — ,;l = —. TT r S\ and, m — —. r- -. S. (5). c{a + 2b) -\- 2ab c {a ■\- 2b) -\- 2ab ^^' If, as before, there be taken loosely : a = 20\ b — 7 \ and c = 10, there will result : 7t = 0.226 S ; w = 0.1 13 S, and / = 0.32 S. It has been found in the best practice that the anchor arm should be 1.67 to 2 times the length. of the cantilever arm, with the suspended span about twice the length of th^ latter, or a very little more. Economic Length of the Anchor Span. The economic length of the anchor span will be found by considering the total weight of the trusses for an anchor span 474 APPENDIX V, and an adjoining cantilever span, or the weight of « + 2m. + / in Fig. 2, PI. XIII. The proportions established in the preceding cases give : Anchor span length = n. Cantilever arm " = r (S — n). a + 2d ' Suspended span " = — y {S — n). Hence the total weight desired is: dn^^-^^AS-tif^W, (6> a + 2b^ ' ^ ' Differentiation then gives : f , ab \ ab c^ ab o f^\ nid+ T ) = 7 S, ,', n = -j7 Tx— — T S. (7). a + 2b J a + 2b ' d{a + 2b) + ab Hence: 5; and m= ,, ^ .. , , 5. . (8). d{a + 2b) + ab ' d {a + 2b) + ab If the same rough factors as before be taken, viz., a = 20\ b = y, and d ■= 12, there will result : « =0.265; / = 0.438 5; and ?« = 0.153 5. These results, however, make n too small for the ordinary requirements of navigation where such exist. The strict economic requirements in such cases are, therefore, neglected, and n taken from 0.3 to 0.4 5. These results show the importance, however, of making the anchor span, compara- tively speaking, very short. Art. 6. — Wind Pressure. The great importance of wind stresses in cantilever struct- ures makes it of much interest to observe that the latest CANTILEVERS. 475 investigations indicate a somewhat lower maximum on large areas for the usual highest winds than has hitherto been contemplated. The highest pressure actually observed and measured at a very exposed point on the site of the Forth bridge, over a period of six years, on a surface of 300 (15 ft. by 20 ft.) square feet, was 27 pounds per square foot, although 41 pounds per square foot was at the same time observed on a small area of 1.5 square feet immediately adjacent to the large one. A pressure of 50 pounds per square foot, therefore, over the entire surfaces exposed in a 500 feet cantilever span is probably as rare at any given location as a cylone at the same place. Again, some later investigations of a most valuable char- acter, by Mr. O. T. Crosby, and given in his paper, " An Experimental Study of Atmospheric Resistance," read before the West Point Branch of the United States Military Service Institution, in 1890, appear to quite invalidate the formula, V- . . . /*=——, in which /* is the pressure in pounds per square foot, and Fthe wind velocity in miles per hour, given in Art. 80, page 370. Mr. Crosby's experiments with a surface one square foot in area show that with velocities varying from 30 to 130 miles per hour, very closely : 7 This is such a radical departure from the hitherto accepted values for P that further tests are desirable, although it is difificult to detect any ground of error in Mr. Crosby's results. PIJ PI. I Fig.i B c . J) jP -^ . i . . G^ ff" w h\- /\ "'" A ■:• /-V'^ / \ --A V \ /^T '?y' '\ --'"/ ^ / ) ^ 7^ -'-\a^ :3 yyi' a- ? -' ' j"-^ ^. V ,.'-''/> - -- ir :>R jr -55. ai F -gj oe H " ' / / 2 / J / \ \ - r + 2 / r / CI •4- / / »/\ \ \ / 1 3.1. 87 / tSS. 91 / tff J. 0!& / +6s.oe \ \ A 1 1 6 2 i 3 6 + /I p r t* FiQ4- ^ /.. B I 'in. ^ ri.m. r n o' p' q' t" o c f-^ o o v., o A a b ^ ;: "" T T' ^(1 li V ^ 1 j ^ ■f'ff ^•L' E Ua U «7^ ■^- E Fiff.C. "A Pi.m. V. Fi(). i a h W \\ — -J r \f^ ■5\ -^- —I X-Fiq.G. /y./i: PiJV. /v.\ P/.V. n.Yf. Pi.rr. pi,m. pi.m. /y. 17//. /'i.vm. b, J / h. J^\ /y./i. j_ c /'/.X. I^' G H K L O r Q S m. 5 /v.x. \ ( Fin. I. C /v. XI W /6 c o o o o o o o o o o ~^ 1 FitfJf 0, ;i >->' o 0. !o o' ' o "i — / /'///. /'T. Ol lO 0' ;o ^ K i^ ■6 i Ol |0 O, lO C, 10 Oi 10 1 1 / /;./. /^. 1- /2.' T'/. 4"ip tx/^ovf^^ fY/ul de^low Fiq. 9. ir. I'l. .\l „ ./ K ff JT f - •^■ ♦ «£?=- i. ^0j; oi *^ is iO:^ Fi Roc}: uJ I'l xu. /; / o !- — Fiff.f. T: /■■ B\\0 \ II- fiff-^ -' Fiff.J. .. ( o V .1 D O ) v_c.^ ' --i-' ftrf. 7. ®) c /><7./. 3,0tt3 •H ^^ "fe Fiff.W. H 1 II : />y./Z te:'oi o To; O'Ol -...I3"i--. Fig.n. 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Vol. II Small 4to, 10 00 Foster's Treatise on Wooden Trestle Bridges 4to, 5 00 Fowler's Ordinary Foundations Svo, 3 5p Greene's Roof Trusses Svo, i 25 Bridge Trusses Svo, 2 50 Arches in Wood, Iron, and Stone Svo, 2 50 Howe's Treatise on Arches Svo, 4 00 Design of fi.mple Roof-trusses in Wood and Steel Svo, 2 00 Johnson, Bryan, and Tumeaure's Theory and Practice in the Designing of Modern Framed Structures Small 4to, 10 00 Merriman and Jacoby's Text-book on Roofs and Bridges: Part I. Stresses in Simple Trusses Svo, 2 50 Part 11. Graphic Statics Svo, 2 50 Part in. Bridge Design Svo, 2 50 Part rV. Higher Structures Svo, 2 50 Morison's Memphis Bridge 4to, 10 00 Waddell's De Pontibus, a Pocket-book for Bridge Engineers. . i6mo, morocco, 3 00 Specifications for Steel Bridges i2mo, 1 25 Wood's Treatise on the Theory of the Construction of Bridges and Roofs . . Svo, 2 OO Wright's Designing of Draw-spans: Part I. Plate-girder Draws Svo, 2 50 Part n. 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Svo, 2 50 Elements of Analytical Mechanics Svo, 3 00 MATERIALS OF ENGINEERING. Baker's Treatise on Masonry Construction Svo, 5 00 Roads and Pavements Svo, 5 00 Black's United States Public Works Oblong 4to, 5 00 Bovey's Strength of Materials and Theory of Structures Svo, 7 50 Burr's Elasticity and Resistance of the Materials of Engineering Svo, 7 50 Byrne's Highway Construction Svo, 5 00 Inspection of the Materials and Workmanship Employed in Construction. i6mo, 3 00 Church's Mechanics of Engineering Svo, 6 00 Du Bois's Mechanics of Engineering. Vol. I Smt^ll 4to, 7 50 ^Eckel's Cements, Limes, and Plasters Svo, 6 00 Johnson's Materials of Construction Large Svo, 6 00 Fowler's Ordinary Foundations Svo, 3 50 Keep's Cast Iron Svo, 2 50 Lanza's Applied Mechanics Svo, 7 50 Marten's Handbook on Testing Materials. (Henning.) 2 vols Svo, 7 50 Merrill's Stones for Building and Decoration Svo, 5 00 Merriman's Mechanics of Materials. Svo, 5 00 Strength of Materials i2mo, i 00 Metcalf's Steel. A Manual for Steel-users i2mo, 2 00 Patton's Practical Treatise on Foundations Svo, 5 00 Richardson's Modern Asphalt Pavements Svo, 3 00 Richey's Handbook for Superintendents of Construction i6ino, mor., 4 00 Rockwell's Roads and Pavements in France i2mo, i 25 7 Sabin's Industrial and Artistic Tectinoiogy of Paints and Varnish 8vo, 3 oo Smith's Materials of Machines i2mo, i 00 Snow's Principal Species of Wood 8vo, 3 50 Spalding's Hydraulic Cement lamo, 2 00 Text-book on Roads and Pavements i2mo, 2 00 Taylor and Thompson's Treatise on Concrete, Plain and Reinforced 8vo, 5 00 Thurston's Materials of Engineering. 3 Parts Svo, 8 00 Part I. Non-metallic Materials of Engineering and Metallurgy 8vo, 2 00 Part n. Iron and Steel 8vo, 3 50 Part in. A Treatise on Brasses, Bronzes, and Other AUoys and their Constituents 8vo, 2 50 Thurston's Text-book of the Materials of Construction 8vo, 5 00 Tillson's Street Pavements and Paving Materials Svo, 4 00 Waddell's De Pontibus. (* Pocket-book for Bridge Engineers.). . i6mo, mor., 3 00 Specifications for StL i Bridges i2mo, i 25 Wood's (De V.) Treatise on the Resistance of Materials, and an Appendix on the Preservation of Timber Svo, 2 00 Wood's (De V.) Elements of Analytical Mechanics Svo, 3 00 Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and SteeL Svo, 4 00 RAILWAY ENGINEERING. Andrew's Handbook for Street Railway Engineers 3x5 inches, morocco, i 25 Berg's Buildings and Structures of American Railroads 4to, 5 00 Brook's Handbook of Street Railroad Location i6mo, morocco, i 50 Butt's Civil Engineer's Field-book i6mo, morocco, 2 50 Crandall's Transition Curve i6mo, morocco, i 50 Railway and Other Earthwork Tables Svo, i 50 Dawson's "Engineering" and Electric Traction Pocket-book. . i6mo, morocco, 5 00 Dredge's History of the Pennsylvania Rai'lroad: (1S79) Paper, 5 00 * Drinker's Tunnelling, Explosive Compounds, and Rock Drills. 4to, half mor., 25 00 Fisher's Table of Cubic Yards Cardboard, 25 Godwin's Railroad Engineers' Field-book and Explorers' Guide. . . i6mo, mor., 2 50 Howard's Transition Curve Field-book '. . i6mo, morocco, i 50 Hudson's Tables for Calculating the Cubic Contents of Excavations and Em- bankments Svo, I GO Mohtor and Beard's Manual for Resident Engineers i6mo, i 00 Kagle's Field Manual for Railroad Engineers i6mo, morocco, 3 00 Philbrick's Field Manual for Engineers i6mo, morocco, 3 00 Searles's Field Engineering i6mo, morocco, 3 00 Railroad SpiraL i6mo, morocco, i 50 Taylor's Prismoidal Formulae and Earthwork Svo, i 50 * Trautwine's Method of Calculating the Cube Contents of Excavations and Embankments by the Aid of Diagrams Svo, 2 00 The Field Practice of Laying Out Circular Curves for Railroads. i2mo, morocco, 2 50 Cross-section Sheet Paper, 25 Webb's Railroad Construction i6mo, morocco, 5 00 WeUington's Economic Theory of the Location of Railways Small Svo, 5 00 DRAWING. 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Clerk's Gas and Oil Engine Small Svo, 4 00 Coolidge's Manual of Drawing Svo, paper, i 00 Coolidge and Freeman's Elements of General Drafting for Mechanical En- gineers Oblong 4to, 2 50 11 Cromwell's Treatise on Toothed Gearing i2mo, i so Treatise on Belts and PuUeys i2mo, i 50 Durley's Kineniatics of Machines 8vo, 4 00 Flather's Dynamometers and the Measurement of Power i2mo, 3 00 Rope Driving i2mo, 2 00 Gill's Gas and Fuel Analysis for Engineers i2mo, i 25 HaU's Car Lubrication i2mo, i 00 Bering's Ready Reference Tables (Conversion Factors) i6mo, morocco, 2 50 Button's The Gas Engine 8vo, 5 00 Jamison's Mechanical Drawing 8vo, 2 50 Jones's Machine Design : Part I. Kinematics of Machinery 8vo, i 50 Part II. Form, Strength, and Proportions of Parts 8vo, 3 00 Kent's Mechanical Engineers' Pocket-book i6mo, morocco, 5 00 Kerr's Power and Power Transmission 8vo, 2 00 Leonard's Machine Shop, Tools, and Methods 8vo, 4 00 *Lorenz's Modern Refrigerating Machinery. 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(Berrmann — Klein.). .Svo, 5 00 Wolff's WindmiU as a Prime Mover Svo, 3 00 Wood's Turbines Svo, 2 50 MATERIALS OF ENGINEERING. Bovey's Strength of Materials and Theory of Structures Svo, 7 50 Burr's Elasticity and Resistance of the Materials of Engineering. 6th Edition. Reset Svo, 7 50 Church's Mechanics of Engineering Svo, 6 00 Johnson's Materials of Construction Svo, 6 00 Keep's Cast Iron Svo, 2 50 Lanza's Applied Mechanics Svo, 7 50 Martens 's Bandbook on Testing' Materials. (Benning.) Svo, 7 50 Merriman's Mechanics of Materials. Svo, 5 00 Strength of Materials i2mo, i 00 Metcalf's SteeL A manual for Steel-users i2mo. 2 00 Sabin's Industrial and Artistic Technology of Paints and Varnish Svo, 3 00 Smith's Materials of Machines i2mo, i 00 Thurston's Materials of Engineering 3 vols., Svo, S 00 Part II. Iron and Steel 8vo, 3 50 Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their Constituents 8vo, 2 50 Text-book of the Materials of Construction 8^0, 5 00 12 r Wood's (De V.) Treatise on the Resistance of Materials and an Appendix on the Preseivation of Timber 8vo, 2 00 Wood's (De V.) Elements of Analytical Mechanics 8vo, 3 00 Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and Ste»l. 8vo, 4 00 STEAM-ENGINES AND BOILERS. Berry's Temperature-entropy Diagram. i2mo, i 25 Carnot's Reflections on the Motive Power of Heat. (Thurston.) i2mo, i 50 Dawson's "Engineering" and Electric Traction Pocket-book. . . .i6mo, mor., 5 00 Ford's Boiler Making for Boiler Makers i8mo, 1 00 Goss's Locomotive Sparks 8vo, 2 00 Hemcnway's Indicator Practice and Steam-engine Economy i2mo, 2 00 Button's Mechanical Engineering of Power Plants 8vo, 5 00 Heat and Heat-engines 8vo, 5 00 Kent's Steam boiler Economy 8vo, 4 00 Kneass's Practice and Theory of the Injector 8vo, i 50 MacCord's Slide-valves 8vo, 2 00 Meyer's Modern Locomotive Construction •. 4to, 10 00 Peabody's Manual of the Steam-engine Indicator i2mo, i 50 Tables of the Properties of Saturated Steam and Other Vapors 8vo, i 00 Thermodynamics of the Steam-engine and Other Heat-engines 8vo, 5 00 Valve-gears for Steam-engines 8vo, 2 50 Peabody and Miller's Steam-boilers ' 8vo, 4 00 Pray's Twenty Years v/ith the Indicator Large 8vo, 2 50 Pupin's Thermodynamics of Reversible Cycles in Gases and Saturated Vapors. 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(Du Bois.) 8vo, 5 00 Whitham's Steam-engine Design 8vo, 5 00 Wilson's Treatise on Steam-boilers. (Flather.) i6mo, 2 50 Wood's Thermodynamics, Heat Motors, and Refrigerating Machines. . .8vo, 4 00 MECHANICS AND MACHINERY. Barr's Kinematics of Machinery 8vo, 2 50 Bovey's Strength of Materials and Theory of Structures 8vo, 7 50 Chase's The Art of Pattern-making ». . . . i2mo, 2 50 Churchts Mechanics of Engineering 8vo, 6 00 13 Church's Notes and Examples in Mechanics 8vo, 2 00 Compton's First Lessons in Metal-working i2mo, i 50 Compton and De Groodt's The Speed Lathe i2mo, 1 50 Cromwell's Treatise on Toothed Gearing i2mo, i 50 Treatise on Belts and Pulleys i2mo, i 50 Dana's Text-book of Elementary Mechanics for Colleges and Schools. .i2mo, 1 50 Dingey's Machinery Pattern Making i2mo, 2 00 Dredge's Record of the Transportation Exhibits Building of the World's Columbian Exposition of 1893 4to half mcrocco, 5 00 Du Bois's Elementary Principles of Mechanics: Vol. I. 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