THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES EAST ABUTMENT OF BLACKWELL'S ISLAND BRIDGE, NEW YORK Copyright, 1907, by Underwood & Underwood, New York. JOHfl S. PRELL ' Civil & Mechanical Engineer. SAN FRANCISCO, CAL. Cyclopedia tf Civil Engineering A General Reference Work ON SURVEYING, RAILROAD ENGINEERING, STRUCTURAL ENGINEERING, ROOFS AND BRIDGES, MASONRY AND REINFORCED CONCRETE, HIGHWAY CONSTRUCTION, HYDRAULIC ENGINEERING, IRRIGATION. RIVER AND HARBOR IMPROVEMENT, MUNICIPAL ENGINEERING, COST ANALYSIS, ETC. Editor-in- Chief FREDERICK E. TURNEAURE, C. E., Dr. Eng. DEAN, COLLEGE OF ENGINEERING, UNIVERSITY OF WISCONSIN Assisted by a Corps of CIVIL AND CONSULTING ENGINEERS AND TECHNICAL EXPERTS OF THE HIGHEST PROFESSIONAL STANDING Illustrated with over Three Thousand Engravings EIGHT VOLUMES CHICAGO AMERICAN TECHNICAL SOCIETY COPYRIGHT, 1908 BY AMERICAN SCHOOL OF CORRESPONDENCE COPYRIGHT, 1908 BY AMERICAN TECHNICAL SOCIETY Entered at Stationers' Hall, London All Rights Reserved, TA Editor-in-Chief FREDERICK E. TURNEAURE, C. E., Dr. Eng. Dean, ColJege of Engineering, University of Wisconsin Authors and Collaborators WALTER LORING WEBB, C. E. Consulting Civil Engineer American Society of Civil Engineers Author of "Railroad Construction, ' "Economics of Railroad Construction," etc. FRANK O. DUFOUR, C. E. Assistant Professor of Structural Engineering, University of Illinois American Society of Civil Engineers American Society for Testing Materials HALBERT P. GILLETTE, C. E. Consulting Engineer American Society of Civil Engineers Managing Editor "Engineering-Contracting" Author of "Handbook of Cost Data for Contractors and Engineers," "Earthwork and its Cost," "Rock Excavation Methods and Cost" ADOLPH BLACK, C. E. Adjunct Professor of Civil Engineering, Columbia University, N. Y. EDWARD R. MAURER, B. C. E. Professor of Mechanics, University of Wisconsin Joint Author of "Principles of Reinforced Concrete Construction W. HERBERT GIBSON, B. S., C. E. Civil Engineer Designer of Reinforced Concrete AUSTIN T. BYRNE Civil Engineer Author of "Highway Construction," "Materials and Workmanship 713592 Authors and Collaborators Continued FREDERICK E. TURNEAURE, C. E., Dr. Eng. Dean of the College of Engineering, and Professor of Engineering, University of Wisconsin American Society of Civil Engineers Joint Author of "Principles of Reinforced Concrete Construction," "Public Water Supplies," etc. THOMAS E. DIAL, B. S. Instructor in Civil Engineering, American School of Correspondence Formerly with Engineering Department, Atchison, Topeka & Santa Fe Railroad ALFRED E. PHILLIPS, C. E., Ph. D. Head of Department of Civil Engineering, Armour Institute of Technology ^ DARWIN S. HATCH, B. S. Instructor in Mechanical Engineering, American School of Correspondence CHARLES E. MORRISON, C. E., A. M. Instructor in Civil Engineering, Columbia University, N. Y. Author of "Highway Engineering." ERVIN KENISON, S. B. Instructor in Mechanical Drawing, Massachusetts Institute of Technology ** EDWARD B. WAITE Head of Instruction Department, American School of Correspondence American Society of Mechanical Engineers Western Society of Engineers EDWARD A. TUCKER, S. B. A rchitectural Engineer American Society of Civil Engineers ERNEST L. WALLACE, S. B. Instructor in Electrical Engineering, American School of Correspondence American Institute of Electrical Engineers A. MARSTON, C. E. Dean of Division of Engineering and Prof essor of Civil Engineering, Iowa State College American Society of Civil Engineers Western Society of Civil Engineers Authors and Collaborators Continued CHARLES B. BALL Civil and Sanitary Engineer Chief Sanitary Inspector. City of Chicago American Society of Civil Engineers ALFRED E. ZAPF, S. B. Secretary, American School of Correspondence SIDNEY T. STRICKLAND, S. B. Massachusetts Institute of Technology Ecole des Beaux Arts, Paris RICHARD T. DANA Consulting Engineer American Society of Civil Engineers Chief Engineer, Construction Service Co. ^ ALFRED S. JOHNSON, A. M., Ph. D. Textbook Department, American School of Correspondence Formerly Instructor, Cornell University Royal Astronomical Society of Canada WILLIAM BEALL GRAY Sanitary Engineer National Association of Master Plumbers United Association of Journeyman Plumbers R. T. MILLER, Jr., A. M., LL. B. President American School of Correspondence GEORGE R. METCALFE, M. E. Head of Technical Publication Department, Westinghouse Electric & Manufac- turing Co. Formerly Technical Editor, "Street- Rail way Review" Formerly Editor "The Technical World Magazine" MAURICE LE BOSQUET, S. B. Massachusetts Institute of Technology British Society of Chemical Industry, American Chemical Society, etc. HARRIS C. TROW, S. B., Managing Editor Editor of Textbook Department, American School of Correspondence American Institute of Electrical Engineers Authorities Consulted THE editors have freely consulted the standard technical literature of America and Europe in the preparation of these volumes. They desire to express their indebtedness, particularly, to the following eminent authorities, whose well-known treatises should be in the library of everyone interested in Civil Engineering. Grateful acknowledgment is here made also for the invaluable co-opera- tion of the foremost Civil, Structural, Railroad, Hydraulic, and Sanitary Engineers in making these volumes thoroughly representative of the very best and latest practice in every branch of the broad field of Civil Engineer- ing; also for the valuable drawings and data, illustrations, suggestions, criticisms, and other courtesies. WILLIAM G. RAYMOND, C. E. Dean of the School of Applied Science and Professor of Civil Engineering in the State University of Iowa; American Society of Civil Engineers. Author of "A Textbook of Plane Surveying," "The Elements of Railroad Engineering." V JOSEPH P. FRIZELL Hydraulic Engineer and Water-Power Expert; American Society of Civil Engineers. Author of " Water Power, the Development and Application of the Energy of Flowing Water." ^ FREDERICK E. TURNEAURE, C. E., Dr. Eng. Dean of the College of Engineering and Professor of Engineering, University of Wisconsin. Joint Author of "Public Water Supplies," "Theory and Practice of Modern Framed Structures," " Principles of Reinforced Concrete Construction." ^ H. N. OGDEN, C. E. Assistant Professor of Civil Engineering, Cornell University. Author of "Sewer Design." ^* DANIEL CARHART, C. E. Professor of Civil Engineering in the Western University of Pennsylvania. Author of "A Treatise on Plane Surveying." ** HALBERT P. GILLETTE Editor of Engineering -Contracting; American Society of Civil Engineers; Late Chief Engineer, Washington State Railroad Commission. Author of " Handbook of Cost Data for Contractors and Engineers." CHARLES E. GREENE, A. M., C. E. Late Professor of Civil Engineering, University of Michigan. Author of "Trusses and Arches, Graphic Method," "Structural Mechanics.' Authorities Consulted Continued A. PRESCOTT FOLWELL Editor of Municipal Journal and Engineer; Formerly Professor of Municipal Engineer- ing, Lafayette College. Author of " Water Supply Engineering, " "Sewerage." LEVESON FRANCIS VERNON-HARCOURT, M. A. Emeritus Professor of Civil Engineering and Surveying, University College, London; Institution of Civil Engineers. Author of "Rivers and Canals," "Harbors and Docks," "Achievements in Engineer- ing," " Civil Engineering as Applied in Construction." PAUL C. NUGENT, A. M., C. E. Professor of Civil Engineering, Syracuse University. Author of " Plane Surveying." *> FRANK W. SKINNER Consulting Engineer; Associate Editor of The Engineering Record; Non-Resident Lec- turer on Field Engineering in Cornell University. Author of " Types and Details of Bridge Construction." HANBURY BROWN, K. C. M. G. Member of the Institution of Civil Engineers. Author of " Irrigation, Its Principles and Practice." SANFORD E. THOMPSON, S. B., C. E. American Society of Civil Engineers. Joint Author of "A Treatise on Concrete, Plain and Reinforced.' JOSEPH KENDALL FREITAG, B. S., C. E. American Society of Civil Engineers. Author of "Architectural Engineering," " Fireproofing of Steel Buildings." *> AUSTIN T. BYRNE, C. E. Civil Engineer. Author of "Highway Construction," "Inspection of Materials and Workmanship Em- ployed in Construction." ^ JOHN F. HAYFORD, C. E. Inspector of Geodetic Work and Chief of Computing Division, Coast and Geodetic Survey; American Society of Civil Engineers. Author of "A Textbook of Geodetic Astronomy." WALTER LORING WEBB, C. E. Consulting Civil Engineer; American Society of Civil Engineers. Author of "Railroad Construction in Theory and Practice," "Economics of Railroad Construction," etc. Authorities Consulted Continued EDWARD R. MAURER, B. C. E. Professor of Mechanics, University of Wisconsin. Joint Author of " Principles of Reinforced Concrete Construction." *r* HERBERT M. WILSON, C. E. Geographer and Former Irrigation Engineer. United States Geological Survey; American Society of Civil Engineers. Author of " Topographic Surveying." " Irrigation Engineering," etc. MANSFIELD MERRIMAN, C. E., Ph. D. Professor of Civil Engineering, Lehigh University. Author of " The Elements of Precise Surveying and Geodesy," "A Treatise on Hydraul- ics," "Mechanics of Materials," "Retaining Walls and Masonry Dams," "Introduction to Geodetic Surveying," "A Textbook on Roofs and Bridges," "A Handbook for Surveyors,'' etc. ^* DAVID M. STAUFFER American Society of Civil Engineers; Institution of Civil Engineers; Vice- President, Engineering News Publishing Co. Author of ' Modern Tunnel Practice." ** CHARLES L. CRANDALL Professor of Railroad Engineering and Geodesy in Cornell University. Author of "A Textbook on Geodesy and Least Squares." N. CLIFFORD RICKER, M. Arch. Professor of Architecture, University of Illinois; Fellow of the American Institute of Architects and of the Western Association of Architects. Author of " Elementary Graphic Statics and the Construction of Trussed Roofs." ^ JOHN C. TRAUTWINE Civil Engineer. Author of "The Civil Engineer's Pocketbook." ^ HENRY T. BOVEY Professor of Civil Engineering and Applied Mechanics, McGill University, Montreal. Author of "A Treatise on Hydraulics." ^ WILLIAM H. BIRKMIRE, C. E. Author of "Planning and Construction of High Office Buildings," "Architectural Iron and Steel, and Its Application in the Construction of Buildings," " Compound Riv- eted Girders," 'Skeleton Structures,' etc. ^- IRA 0. BAKER, C. E. Professor of Civil Engineering, University of Illinois. Author of "A Treatise on Masonry Construction," " Engineers' Surveying Instruments, Their Construction, Adjustment, and Use," "Roads and Pavements." Authorities Consulted Continued JOHN CLAYTON TRACY, C. E. Assistant Professor of Structural Engineering, Sheffield Scientific School, Yale University. Author of " Plane Surveying: A Textbook and Pocket Manual." FREDERICK W. TAYLOR, M. E. Joint Author of "A Treatise on Concrete, Plain and Reinforced." ^ JAMES J. LAWLER Author of " Modern Plumbing, Steam and Hot- Water Heating." FRANK E. KIDDER, C. E., Ph. D. Consulting Architect and Structural Engineer; Fellow of the American Institute of Architects. Author of "Architect's and Builder's Pocketbook," " Building Construction and Super- intendence, Part I, Masons' Work; Part II, Carpenters' Work; Part III, Trussed Roofs and Roof Trusses," " Strength of Beams, Floors, and Roofs." ^ WILLIAM H. BURR, C. E. Professor of Civil Engineering, Columbia University; Consulting Engineer; American Society of Civil Engineers; Institution of Civil Engineers. Author of " Elasticity and Resistance of the Materials of Engineering;" Joint Author of " The Design and Construction of Metallic Bridges." WILLIAM M. GILLESPIE, LL. D. Formerly Professor of Civil Engineering in Union University. Author of " Land Surveying and Direct Leveling," " Higher Surveying." *r> GEORGE W. TILLSON, C. E. President of the Brooklyn Engineers' Club; American Society of Civil Engineers; Ameri- can Society of Municipal Improvements; Principal Assistant Engineer, Department of Highways, Brooklyn. Author of " Street Pavements and Street Paving Material." *** G. E. FOWLER Civil Engineer; President, The Pacific Northwestern Society of Engineers; American Society of Civil Engineers. Author of " Ordinary Foundations." ^ WILLIAM M. CAMP Editor of The Railway and Engineering Review; American Society of Civil Engineers. Author of " Notes on Track Construction and Maintenance." W. M. PATTON Late Professor of Engineering at the Virginia Military Institute. Author of "A Treatise on Civil Ensrineerinrr." Foreword HE marvelous developments of the present day in the field of Civil Engineering, as seen in the extension of railroad lines, the improvement of highways and waterways, the increasing application of steel and reinforced concrete to construction work, the development of water power and irrigation projects, etc., have created a distinct necessity for an authoritative work of general reference embodying the results and methods of the latest engineering achievement. The Cyclopedia of Civil Engineering is designed to fill this acknowledged need. C, The aim of the publishers has been to create a work which, while adequate to meet all demands of the technically trained expert, will appeal equally to the self-taught practical man, who, as a result of the unavoidable conditions of his environ- ment, may be denied the advantages of training at a resident technical school. The Cyclopedia covers not only the funda- mentals that underlie all civil engineering, but their application to all types of engineering problems; and, by placing the reader in direct contact with the experience of teachers fresh from practical work, furnishes him that adjustment to advanced modern needs and conditions which is a necessity even to the technical graduate. C. The Cyclopedia of Civil Engineering is a compilation of representative Instruction Books of the American School of Cor- respondence, and is based upon the method which this school has developed and effectively used for many years in teaching the principles and practice of engineering in its different branches. The success attained by this institution as a factor in the machinery of modern technical education is in itself the best possible guarantee for the present work. C, Therefore, while these volumes are a marked innovation in technical literature representing, as they do, the best ideas and methods of a large number of different authors, each an ac- knowledged authority in his work they are by no means an experiment, but are in fact based on what long experience has demonstrated to be the best method yet devised for the educa- tion of the busy workingman. They have been prepared only after the most careful study of modern needs as developed under conditions of actual practice at engineering headquarters and in the field. C. Grateful acknowledgment is due the corps of authors and collaborators engineers of wide practical experience, and teachers of well-recognized ability without whose co-opera- tion this work would have been impossible. Table of Contents VOLUME VI BRIDGE ENGINEERING By Frank O. Dufourf Page *11 Bridge Analysis Early Bridges Trusses Girders Truss and Girder Bridges Deck and Through Bridges Truss Members Lateral Bracing Portals Sway Bracing Knee-Braces Kinds of Trusses Chord and Web Character- istics Weights of Bridges Loads (Dead, Live, Wind) Principles of Analysis Resolution of Forces Method of Moments Stresses in Web and Chord Members Warren Truss under Live and Dead Loads Position of Live Load for Maximum Moments Counters Maximum and Minimum Stresses Pratt, Howe, Baltimore, Bowstring, Parabolic, and Other Trusses under Dead and Live Loads Engine Loads Position of Wheel Loads for Maximum Shear and Mo- ments Pratt Truss under Engine Loads Impact Stresses Snow- Load Stresses Wind-Load Effects Top and Bottom Lateral Bracing Overturning Pratt Truss under Wind Loads Girder Spans Floor- Beam Reactions Plate-Girder Reactions Bridge Design Economic Considerations Types of Bridges for Various Spans Economic Proportions of Members Clearance Diagram Weights and Loadings Specifications Stress Sheet Design of a Plate-Girder Railway Span Masonry Plan Determination of Span Ties and Guard -Rails Web and Flanges Rivet Spacing Lateral Systems and Cross- Frames Stiffeners Web-Splice Bearings Design of a Through Pratt Rail- way Span Stringers Floor- Beams Tension Members Intermediate Posts Lacing Bars End -Posts Pins Transverse Bracing Shoes and Roller Nests HIGHWAY CONSTRUCTION . By A. T. Byrne and A. E.Phillips Page 267 Country Roads Road Resistances to Traction Axle Friction Air Resistance Tractive Power and Gradients Effects of Springs on Vehicles Location of Roads Contour Lines Levels Cross-Levels Bridge Sites Mountain Roads Alignment Zigzags Construction Profile Width and Transverse Contour Drainage Ditches and Culverts Earthwork Roads on Rocky Slopes Earth and Sand Roads Grading Tools - Rollers Sprinkling Carts Road Coverings Gravel Roads Macadam Roads City Streets Catch- Basins Pavement Foundations Stone- Block Pavements Properties of Stones Cobblestone Pavement Belgian Block Pavement Brick Pavement Paving Brick Con- crete Mixers Gravel Heaters Melting Furnaces Wood Pavements Asphalt Pavements Mixing Formulae Footpaths Curbstones Artificial Stone Pavement Selection Safety Life of Pavements Cost Relative Economies REVIEW QUESTIONS Page 399 INDEX Page 405 *For page numbers, see foot of pages. TFor professional standing of authors, see list of Authors and Collaborators at BRIDGE ENGINEERING PART I BRIDGE ANALYSIS 1. Introduction. The following treatment of the subject of Bridge Analysis, while not exhaustive, is regarded as sufficiently elaborate to develop and instill the principal theoretical considera- tions, to illustrate the most convenient and practical methods of analyzing the common forms of trusses and girders, and also to lay a sufficient foundation for the analysis of such other trusses as are not specifically mentioned or treated herein. The necessary steps and operations required for a proper analysis of the several types of bridges are fully demonstrated by sketches and computations, the numerical values being mechanically obtained by the use of a slide rule, which is a handy instrument for quickly performing the operations of multiplication and division, and for squaring and extracting the roots of numbers. The values given may differ from the exact value by one unit in the second decimal place (seldom more) and are sufficiently accurate for the purpose of design. All bridge computers should be proficient in the use of the slide rule. The problems given in the back of this Instruction Paper, exemplifying the practical application of the subject-matter treated in the various articles, should be solved by the student as each article is mastered. HISTORY 2. Early Bridges. Early bridges were not bridges according to the present conception of the term. They were simple pile trestle bents placed at frequent intervals and connected by wooden beams on which the floor was placed. The Pans sublicius, built over the Tiber, at Rome, about 650 years before Christ was born, was of this trestle type. Also the famous bridge b'uilt by Csesar across the Rhine in 55 B. C. was of the same kind of construction. As civiliza- tion progressed, the arch type was developed; and in 1390 the great Copyright, 190S, by American School of Corespondence. 11 BRIDGE ENGINEERING bridge at Trezzo over the River Adda was built of one span of 251 feet, which has never been eclipsed or equaled. 3. Truss Bridge Development. The first truss bridge is sup- posed to have been originated by Palladio, an Italian, who used the king-post truss (Fig. 1) about 1570. Its importance was not recog- nized, and it became entirely for- gotten until it was rediscovered in 1798 by Theodore Burr, an Ameri- can, who used it in his construction. About the same time, Burr invented the truss that bears his name, which was in reality a series of king-post trusses (see Fig. 8). This was found to be unstable under moving loads, and was therefore stiffened by the use of an arch (Fig. 2), or was built somewhat as an arch, there being considerable rise at the center of the span (Fig. 3). By 1830 the principle of the double cross-bracing in the panel was understood; and in quick succession came the patents of Long, Howe, Pratt, and Whipple on forms of trusses which bear their respective names. It remained for Squire Whipple in 1847 to place the science of bridge building on a rational and exact mathematical basis such Fig. l. King-Post Truss. Fig. 2. King-Post Truss Bridge Stiffened by Arch. as is now used. Previous to this time, and indeed several years afterwards for Whipple's work did not become generally known until a much later date bridges were built, not from previously computed strains, but by "judgment." All parts of a bridge were made of the same size, and if one started to fail it was replaced by a larger one; or small models were made and loaded proportionally, broken members being replaced by larger ones. There is no doubt 12 BRIDGE ENGINEERING that many of the bridges built at this period were very weak as well as very strong. The failures are not remembered; but the sound judgment of many of our earlier bridge engineers is recorded in the wooden structures they left behind them, some of which have stood the demands of traffic for over a century. After 1850, bridges were built from computed stresses; wood was discarded; and the develop- ment became rapid, until about 1870, when the introduction of sub- diagonal systems brought the truss system to practically what it is to-day. DEFINITIONS AND DESCRIPTIONS 4. Trusses. A truss is a series of members taking stress in the direction of their length, placed together so as to form a triangle Fig. 3. Burr Tru^s Bridge, Arched. or system of triangles, which, when placed upon supports a certain distance apart, will, in addition to their own weight, sustain certain loads applied at the points where the members intersect. These points are called panel points. 5. Bridge Trusses. A bridge truss is one in w r hich the members that carry the superimposed loads are in the same plane. Usually this plane is vertical. 6. Truss Bridges. A truss bridge is a structure consisting of two or more usually two bridge trusses connected by a system of beams called the floor system, which transfer to panel points the load for which the trusses are designed. 7. Girders. These are beams consisting of a wide, thin plate, called a web plate, with shapes, usually angles and narrow, thin plates called flanges, at the top and bottom edges. All are firmly riveted together. (See Part IV, "Steel Construction.") 8. Girder Bridges. These consist of usually two, sometimes 13 BRIDGE ENGINEERING three, girders connected as in the case of truss bridges by a system of beams. 9. Deck Bridges. In cases where the floor system connects the trusses at their tops, the bridge is called a deck bridge, since the traffic moves on a deck, as it were (see Fig. 4). Fig. 4. Deck Bridge. Fig. 5. Through Bridge. 10. Through Bridges. In cases where the floor system con- nects the bottoms of the trusses, the bridge is called a through bridge, as the traffic moves through the space between the trusses (see Fig. 5). 11. Members of a Truss. Each truss consists of a top and bottom chord, end-posts, and web members. The web members are further divided into hip verticals, intermediate posts, and diagonals. Fig. 6 shows these various classes, A-A being top chord, B-B Fig. 6. Showing the Members of a Truss. bottom chord, A-B end-posts, vertical members C-b intermediate posts, A-a hip verticals, and A-b and C-b diagonals. 12. Pony=Truss Bridges. When the height of the trusses of a through bridge is less than the height of the loads that go over them, they are called pony trusses, and the bridge a pony-truss bridge. 14 BRIDGE ENGINEERING 13. Lateral Bracing. In all deck bridges, and in all through bridges except pony-truss bridges, the chords which are not con- nected by the floor system are connected by a horizontal truss system called the lateral bracing. In all bridges the chords which are con- nected by the floor system are connected by a horizontal truss system, also called the lateral bracing. One of these systems is called the Stringers Fig. 7. Through Bridge, Showing Top and Bottom Systems of Lateral Bracing, also Portal Bracing and Floor System. top lateral system, as it connects the top chords; and the other is called the bottom lateral system, as it connects the bottom chords (see Fig. 7). 14. Portals. In through bridges, the end-posts of the pair cf trusses are connected by a system of braces in order to preserve the rectangular cross-section of the bridge. This is called the portal bracing (see Fig. 7). 15. Sway Bracing and Knee=Braces. These serve the same purpose as the portal braces, and are either small struts or systems 15 BRIDGE ENGINEERING of cross-bracing placed at the intermediate posts. The former are called knee-braces, and the latter sway bracing. 16. Floor Systems. In both highway and railway bridges, there are beams running from the intermediate posts or hip ver- ticals across to the like members opposite. These are called floor- beams. In highway bridges, there are smaller beams running parallel to the trusses and resting at their ends upon the floor-beams. These are called floor-joists, and the plank or other floor rests directly upon them. In railway bridges, two beams or girders per track run parallel to the trusses and are connected at their ends to the floor-beams. These are called track stringers (or simply stringers). The ties rest directly upon them. The various members of the floor system of a railway bridge are shown in Fig. 7. The diagonals connecting the top chords, and those connecting the bottom chords, are the top and bottom laterals respectively. CLASSES OF TRUSSES 17. Names. Trusses may be classified according to their names, the character of their chords, and the system of webbing. Table I gives the classification of the more important of these accord- ing to name. TABLE I Chronological List of Trusses ORIGIN King-Post 1570 Palladio Italy Fig. 1 King-Post 1798 Theodore Burr America Fig. 1 Burr 1798 Theodore Burr America Fig. 8 Warren 1838 England Fig. 9 Howe 1840 William Howe America Fig. 10 Pratt 1844 Thos & Caleb Pratt America Fig. 11 W T hipple Bowstring Baltimore 1847 1847 1877 Squire Whipple Squire Whipple Penn. R. R. America America America Fig. 12 Fig. 13 Fig. 14 Of the types of trusses listed in Table I, the Warren, Howe, Pratt, Bowstring, and Baltimore are now built; and of these construc- tions probably 90 per cent are Pratt trusses. The Baltimore truss is used for long spans only. 18. Chord Characteristics. In most types of bridges the 16 BRIDGE ENGINEERING Fig. 8. Burr Truss. Fig. 9. Warren Truss. Fig. 10. Howe Truss. Fig. 11. Pratt Truss. Fig. 12. Whipple Truss. 17 BRIDGE ENGINEERING chords are parallel. When such is the case, the stresses increase from the end toward the center, and there is a considerable difference between any two adjacent panels of the same chord. This neces- sitates different areas for each section. When the chords are not parallel, as in the bowstring truss, the stresses in the chords are so nearly equal that the same area is used throughout or nearly through- out the entire chord. Also, the stresses in the diagonals are nearly equal. These conditions would seem to indicate that this was a very economical form of truss. Theoretically it is; but practical con- siderations such as the beveled joints and the posts w r hich must be constructed to withstand reversals of stress customarily limit this type to the longer spans. 19. Web Characteristics. The web systems of the Burr, Warren, Howe, Pratt, and Bowstring trusses are called single sys- tems; that of the Whipple truss is a multiple system; while those of the Baltimore trusses are examples of webbing with sub-systems. As the maximum economical panel length has been found to be about 25 feet, which makes the economical height of the truss about 30 feet, and as the length of the span should not be more than ten times the depth, the span for trusses with simple systems of webbing is limited to about 300 feet. In order to increase the limiting span, multiple systems like that of the Whipple or similar ones were intro- duced. Calculations of stresses in members of the Whipple truss are somewhat unreliable on account of the fact that we are unable to tell just how the effects of the loads are distributed. For this reason, that type has gone out of use, and the sub-systems are used instead. These allow spans of twice the above limit; and, indeed, trusses with this type of webbing have been built up to and over 600 feet. This style of webbing can be applied to the bowstring truss, almost all long-span bridges being of this type with sub-systems of webbing. WEIGHTS OF BRIDGES 20. Formulae. In order to obtain the stresses due to the weight of the structure, the latter quantity must be known. As this weight can be determined only after the structure has been designed, it is evident that an assumption as to the weight must be made. The best method is to use the actual weight of a similar structure of like span which has been built. As the necessary data 18 BRIDGE ENGINEERING 9 for this is not always available, it is customary to use formulae to derive an approximate weight of sufficient accuracy for purposes of computation. Table II gives some of the most reliable formulse. TABLE II Formulae for Weights of Bridges CLASS OF BRIDGE Heavy Interurban Riveted ,., _ Ann i i o/ i 27/1 4. hjf 1 -1- 1 I E S. Shaw W DUU ^ L.Ctl ~r Zi i U ~\- ^-CHil-p '.: r\r\fv 1 First-Class High- 1 / 1 ' \ way Riveted w = 300 + l + 22b-\ blfl -\ I j E. S. Shaw First-Class High- 15 \ 1,000 / way Pin w = 34 + 226 + O.lQbl + 0.71 J.A.L.Waddell Light Country Highway w = 250 + 2.51 Author Railroad Truss E 50 w = (650 + 70 F. E. Turneaure Railroad Truss E 40 w = | (650 + 70 F.E. Turneaure Railroad Truss . E 30 w~= \ (650 + 70 F E. Turneaure Railroad Deck Girder E 50 w = 124.0 + 12.0Z Author Railroad Deck Girder E 40 w = 123.5 + 10.0Z Author Railroad Deck Girder E 30 w = 111.0 + 8.81 Author In the above formulae, w = Weight of steel per linear foot of span; I = Length of span in feet; b = Breadth of roadway, including sidewalks. In using the formulae of Table II, remember that a span has two trusses. The weights for highway bridges do not include the weight of the wooden floor, which may be assumed as 10 pounds per square foot of floor surface All highway bridges have steel joists. The weights of railroad spans do not include the weight of the ties and rails, which may be assumed at 400 pounds per track per linear foot of span. If solid steel floors are to be used, 700 pounds per linear foot of span are to be added to the weights computed from the table. All the weights given for railroad spans are for single track. Double- track truss-spans are about 95 per cent'heavier; and double-track girder-spans are 100 per cent heavier. Through girder spans are about 25 per cent heavier than deck girder spans; and through spans are about 15 to 40 per cent heavier than deck spans. The spans on which Table II is based are of medium steel. Bridges built of soft steel or wrought iron will weigh 10 to 15 per cent more. * The author is indebted to the distinguished engineers whose names appear in Table II, for permission in this connection to make use of the formulae given opposite their names. 10 BRIDGE ENGINEERING In order to give an idea of the relative weights of steel in different classes of bridges, let it be required to compute the dead weight of a 100-foot span of each class. For heaviest highway bridges to carry heavy interurban cars: 00 ( i + Ym) = 1 358 lbs - per linear ft - 27 X 16 12 Fig. 13. Bowstring Truss. For heavy riveted highway bridges to carry heavy farm engines: u = 300 + 100 + 22 X 16 + 16 * ( 1 + y^ \ 870 Ibs. per linear foot. Fig. 14. Two Forms of Baltimore Trusses. For heavy pin-connected highway, bridges to carry heavy farm or traction engines: w = 34 +22 X 16 + 0.16 X 16 X 100 + 0.7 X 100 = 710 Ibs. per linear ft. For light country highway bridges to carry 100 pounds per square foot of floor surface: w = 250 + 2.5 X 100 = 500 Ibs. per linear foot. If the total weight is required, the weight of the wooden floor must be added. Take, for example, the last bridge: 20 BRIDGE ENGINEERING 11 Weight of steel = 500 X 100 = 50 000 pounds. " floor = 100 X 16 X 10 = 16000 pounds. Total dead load = 66 000 pounds. The weight per linear foot for a railroad truss bridge of 100-foot span is: w = 650 + 7 X 100 = 1 350 Ibs. per linear foot. This is about the same as that for a heavy interurban bridge The reason for this is that in addition to the heavy rolling stock of the electric road, the heavy highway traffic must be provided for. A deck girder of 100-foot span weighs: w = 124 + 12 X 100 = 1 324 Ibs. per linear foot. 21. Actual Weights of Railroad Spans. In case actual weights can be obtained, a more exact analysis can be made. The weights of bridges indicated in the accompanying tables and diagrams, are based on actual constructions recently erected. These bridges rep- resent the very best modern practice of engineers and manufacturers The weights of through truss-spans made of medium steel and designed for E 50 loading, are given in Fig. 15. The weights include the weight of the ordinary open steel floor, and they also include the weight of the ties and rails, which is taken at 400 pounds per linear foot per track. The weight of steel in medium steel deck plate-girder spans designed for E 50 loading, is given in Table III. TABLE III Weights of Deck Plate-Girders, Medium Steel Loading E 50 SPAN (in feet) W FIGHT (in pounds) SPAN (in feet) WEIGHT (in pounds) 15 5300 70 59500 20 7 800 75 67300 25 11 800 80 76300 30 14 500 85 94200 35 18800 90 105 500 40 23 300 95 114200 45 27400 100 123 600 50 32 400 105 146000 55 38 800 110 161 700 60 45 500 115 174 900 65 51 500 120 187000 The spans are the distance center to center of bearings; and the weights do not include the weight of the ties and rails, which is to be taken at 400 pounds per linear foot per track. Intermediate spans may be interpolated. 12 BRIDGE ENGINEERING 22. Actual Weights of Highway Spans. The actual weights of highway spans for heavy interurban trolley-cars and traffic, should preferably be computed from the formulae of Shaw or Waddell (Table II). 7 The weights of country bridges, including floor, may be taken from the diagram of Fig. 16. LOADS 23. Classes of Loads. Those weights just given constitute what is called the dead load of the bridge. The traffic which passes 130.000 / 1 / 1 I 0.000 / / 1 MOCOO / i IOQOOO / / | 90.000 / / 80.000 7 / 70000 7 / 6QOOO / 1 / | 50,000 / 4-0 OOO/ , P a n in K'c t Fig. 15. Weights of Through Truss-Spans. Fig. 16. Weights of Country Bridges, In Medium Steel, E 50 Loading. eluding Floor. over the bridge is called the live or moving load. In addition to the two classes mentioned, is the effect of the wind, which is designated as the wind load. These loads vary with the class of bridge, be it highway or railway, and with the purpose for which it is intended. 24. Live Loads for Highway Bridges. Highway bridges are usually divided into several classes according to the traffic, which may be that of heavy interurban cars, light trolley-cars, farm engines, road rollers, teams, human beings, or some combination of these loadings. The standard specifications of J. A. L. Waddell or of Theodore Cooper are obtainable for a very small sum. Their pur- BRIDGE ENGINEERING 13 chase is advised, and the reader is referred to them for further infor- mation. The trusses of country highway bridges are usually designed for a live load of 100 pounds per square foot of roadway. This may be considered good practice; and it is the law in some States. The floor system of these same bridges should be of sufficient strength to sustain 100 pounds per square foot of roadway, or a 12-ton farm engine having 4 tons on the two rear wheels, which are 12 inches wide and 6 feet apart, and 2 tons on each of the front wheels, which are 6 inches wide and 5 feet apart. The axles of this engine are spaced 8 feet center to center. 25. Live Loads for Railway Bridges. The loads for any par- ticular railroad bridge are not always the same, on account of the great variation in the weights and wheel spacings of engines and cars. It is customary to design the bridge for the heaviest in use at the time of construction, or for the heaviest that could reasonably be expected to be built thereafter. As the computations with engines were formerly somewhat laborious on account of the different weights and spacing of wheels, it has been proposed by some engineers to use a uniform load, called the equivalent load, which would give stresses the same, or very nearly the same, as those obtained by the use of engine loads. However, as these loads are different for each weight of engine, and also different for the chord members, the web members, and the floor-beam reaction of each different length of span, and as the labor of the computations, using engine-wheel loads, has been greatly reduced by means of diagrams, it does not seem as if this method would ever come into very general favor except for long-span bridges, where the live load is much smaller than the dead load. The equivalent loads for Cooper's E 40 (see Fig. 85) are given in Table IV. Most railways specify that their bridges shall be computed by using two engines and tenders followed by a train. The spacing of the wheels, and the load which comes on each wheel of the engines and tenders, are fixed by the railway company. The train is repre- sented by a uniform load. Formerly there was a great diversity of practice among the different roads in regard to the engine and. train loads specified ; but practice has of late years become quite uniform, It BRIDGE ENGINEERING TABLE IV Equivalent Uniform Loads Loading E 40 SPAN (in feet) EQUIVALENT UNIFORM LOAD SPAN (in feet) EQUIVALENT UNIFORM LOAD Chords Web Floor- Beam Chords Web Floor- Beam 10 9000 12 000 8200 46 6330 7240 5240 11 9310 11 640 7960 48 6220 7 140 5200 12 9 340 11 330 7830 50 6 110 7060 5 140 13 9340 11 080 7600 52 6040 6940 5 130 14 9210 10 860 7 460 54 5960 6820 5 120 15 9030 10 670 7 330 56 5880 6 720 5 110 16 8850 10 500 7 120 58 5800 6 620 5090 17 8650 10 350 6940 60 5 730 6530 5080 18 8430 10 240 6780 62 5690 6490 5080 19 8220 10 100 6630 64 5700 6450 5070 ,20 8000 10 000 6500 66 5620 6450 5070 21 8040 9 780 6390 68 5560 6380 5060 22 8040 9 580 6290 70 5510 6340 5060 23 8010 9 400 6200 72 5490 6320 5030 24 7960 9 230 6 120 74 5460 6 300 5010 25 7890 9080 6040 76 5440 6290 4990 26 7 780 8 930 5 970 78 5420 6 270 4970 27 7660 8 790 5900 80 5400 6250 4950 28 7 540 8660 5830 82 5 370 6230 4930 29 7420 8 540 5 770 84 5340 6200 4910 30 7300 8430 5720 86 5310 6 180 4890 31 7220 8 320 5680 88 5270 6 150 4870 32 7 140 8 190 5650 90 5250 6 130 4860 33 7 050 8080 5620 92 5250 6 110 4830 34 6 960 7980 5600 94 5210 6 090 4810 35 6870 7890 5570 96 5 170 6060 4 780 36 6820 7820 5 530 98 5 150 6 040 4 760 37 6760 7750 5 500 100 5 140 6 020 4 740 38 6700 7690 5 460 125 5 100 5 770 4 720 39 6630 7630 . 5 430 150 5010 5570 4700 40 6 560 7 570 5400 175 4890 5350 4686 42 6 530 j 7 450 5 340 200 4740 5240 4660 44 6 470 7340 5300 250 4510 5030 4 640 with an apparent tendency to standardize in accordance with the classes of loading specified by Cooper. Cooper's Class E 50, which represents the heaviest engines now in common use, was invented by Theodore Cooper, a consulting engineer of New York City. It is given in Fig. 17. Lighter loadings for light traffic on the same general plan are advocated by Mr. Cooper, and are given at length in his "General Specifications for Iron and Steel Railway Bridges and Viaducts" (1906 edition). 26. Wind Loads. Some designers require that the stresses due 24 BRIDGE ENGINEERING 15 A -GH- ~. In Fig. 24, the sum of the vertical forces is again used. Here the section cuts the member with the known tensile stress S r + R p p p p P + Si cos S X cos = S s = +7.5-S6C <. BRIDGE ENGINEERING 23 Sec is equal to VW + 25 2 -h 30, which is equal to 1.302; and therefore, S 3 = +7.5 X 1.302 = +9.765, which is a tensile stress of 9,765 pounds. 31. Stresses in Chord Members. The stresses in chord may be obtained by either the method of moments or the method of resolu- tion of forces, this latter being usually the resolution of horizontal forces. In accordance with the text of Article 29, the following rule may be stated with regard to the solution of stresses in chord members by the method of moments: Pass a plane section cutting the member whose stress is to be computed, and as few others as possible; then take the center of moments at such a point s< t ' t / / / / /' / \ 1 >' / ' Fig. 29. Calculation of Stresses iu Chord Members by "Tangent" or "Chord-Increment" Method. that the lines of action of as many forces as possible, the unknown one excepted, pass through that point; write an equation of the moments about this point of the known loads and forces to the left of the section, assuming the unknown force to act away from the section, and taking the known forces to act as given, the tensile stresses to act away from the section, and the compressive stresses to act towards the section; place the equation equal to zero, and solve. The stress will solve out with its correct characteristic sign. In the majority of cases a section can be made to cut three mem- bers only, one of the three being the one whose unknown stress is desired. In such cases, take the center of moments at the inter- section of the other two, and proceed as before. As examples of this latter case, note the centers of moments at U 2 , Fig. 26, and 0, Fig. 27, and also the equations resulting therefrom. When the method of resolution of forces is used, it is usually designated as the tangent method or the chord increment method. The simplest application of this method is to trusses with horizontal chords and vertical posts in the web members. Then the stress in any chord member is equal to the product of the sum of the shears 33 24 BRIDGE ENGINEERING in the panels up to that section, and the tangent .of the angle which the diagonals make with the vertical. This can readily be proved by reference to Fig. 29. Let it be re- quired to determine the stress in the chord member S K , Pass the section a a. The stresses S v S v S 3 , and S t are now computed, and are Sj= V l sec ; S 2 = + T 7 ", sec 9; S 3 = +T T 3 sec 9; and S 4 = + F 4 sec 9. Now noting the directions of the known stresses and Lo L,' L e L=> L. L 5 L 6 Fig. 30. Illustrating Method of Notation of Stresses and Members in a Through Bridge. L. L e L 3 L 4 L 5 Fig. 31. Illustrating Method of Notation of Stresses and Members in a Deck Bridge, assuming S, to act away from the section, the equation of the hori- zontal component is: 4- S l sin $ + Sj sin $ + S, sin + S t sin ij> + S, = 0. Now, substituting the values of S v S v etc., and remembering that sec < = COS '-; from which, - . the equation becomes : 7.= & = - (F, 4- V, + V s + VJ tan 0; S x = -2Ttan ^. From inspection of Fig. 29, it will be noticed that the stress in any section of the chord is equal to that in the section to the left of BRIDGE ENGINEERING 25 it, plus the increment (horizontal component) of the diagonal; hence the name chord increment method. 32. Notation. The practice hitherto used in designating stresses by S v S 2 , etc., will now be discontinued, as it is inconvenient in the extreme; moreover, it is not the method used in practical work. The notation to be used is that given in Figs. 30 and 31, the former being for a through and the latter for a deck truss. The practical advantages of this system are very great. When U l U 2 is noted, it is at once known to be the top chord of the second panel; U 2 L 2 is known to be the second vertical ; while U 2 L 3 is at once recognized as the diagonal in the third panel. A stress in a member, as well as the member itself, is designated by the subscript letters at its ends. Thus U l L 2 may mean the member U 5 u a Fig. 32. Calculation of Stresses in a Six-Panel Warren Truss Through Bridge. itself or the stress in the member. The text will clear this up. In analysis, the stress would be implied, while in design the member itself would be intended. 33. Warren Truss under Dead Loads. The Warren truss has its web members so built of angles and plates or of channels, that they can take either tension or compression. The top chord is of structural shapes, while the lower chord may be of built-up shapes or simply of bars. Let it be required to determine all of the stresses in the six- panel truss of a through Warren highway 120-foot span for country traffic. The height is to be 20 feet. The outline is given in Fig. 32. According to Fig. 16, the total weight of the span, including wooden floor, is 76 000 pounds. Each truss carries one-half of this, or 76 000 35 26 BRIDGE ENGINEERING -r- 2 = 38 000 pounds. As there are six panels, each panel load is 38 000 -=- 6 = 6 333 pounds. This means that we must compute the stresses in the above truss by considering that a load of 6 333 pounds is at points L v L 2 , Z/ 3 , L 4 , and L 5 . Of course there is some weight at L and L 6 ; but this does not stress the bridge, as it is directed over the abutments or supports. The reactions at L and L 6 are each equal to (5 X 6 333) -T- 2 = 15 833 pounds. The shears are next computed, and are: F, = + 15 833 - = +15 833 F 2 = +15 833 - 6 333 = +9 500 F 3 = + 15 833 - 2 X 6 333 = + 3 167 It is unnecessary to go past the center of the bridge, as it is symmetri- cal. The V l represents the shear on any section between Z/ and i t ; F 2 represents the shear on any section between L l and L 2 ; and V s represents the shear on any section between L 2 and L 3 . The secant of the angle is : (20% 10) 2 ) - 20= 1.12. The stresses in the web members are computed as follows: For L () U r Pass section a a. Assume stress acting away from the section, as shown. Then, F, + L U l cos < = 0; L U l = - F, sec ; L U, = - 15 833 X 1.12 = - 17 700 pounds, which shows that L U l has a compressive stress of 17 700 pounds. For U l L v Pass section 6 6. Assume stress acting away from the section, as shown. Then, F, - U t L, cos < = 0; U,L, = +'F, sec <; tfjL, = +15833 X 1.12 = + 17 700 pounds, which shows that U^ has a tensile stress of 17 700 pounds. For LJJ r Pass section c c. Then, F 2 - + L,U 2 cos = 0; L,U 2 = -F 2 sec $; L,E7, = - 9500 X 1.12 = - 10 640 pounds. For U 2 L r Pass section d-d. Then, + 9500 - U 2 L 2 cos = 0; UgLf = +9500 X 1.13'- +10640. ' BRIDGE ENGINEERING 27 Far L 2 U 3 . Pass section e e. Then, + 3 167 + L 2 U 3 cos = 0; L 2 U 3 = -3 167 X 1.12 = -3 540. For U 3 L 3 . Pass section / / . Then, + 3 167 - U 3 L 3 cos < = 0; U 3 L 3 = +3 167 X 1.12 = +3540. The computation of the stresses in the chords is made by the method of moments, and is as follows: For L U L V Section b b cuts U l L l and UJJ 2 , besides the mem- ber whose stress is desired, and therefore the center of moments will be taken at their intersection U^ The equation is: + 15 833 X 10 - L L, X 20 = 0, whence, L L, = ( + 15833 X 10) 4- 20; = + 7 917 = a tension of 7 917 pounds. For LJj r Either section c c or d d may be used, and each shows the center of moments to be at U y The equation is: + 15 833 X 30 - 6 333 X 10 - L,L 2 X 20 = 0; L^ = ( + 15 833 X 30 - 6 333 X 10) 4- 20; = +20 583 = a tensile stress of 20 583 pounds. For L 2 L 3 . Either section e e or / / may be used, and each shows the center of moments to be at U r The equation is: + 15 833 X 50 - 6 333 X 30 - 6 333 X 10 - L 2 L 3 X 20 = 0; whence, L 2 L 3 = +26917. The center of moments for UfJ^ is at L t ; for U 2 U 3 , it is at L 2 ; and for U a U 4 , it is at L 3 . The following equations can now be written : + 20 X U,U 2 + 20 X 15833 = 6; whence U,U 2 = -15833; + 20 X U 2 U 3 + 40 X 15 833 - 20 X 6 333 = 0; whence U 2 U' 3 = -25 333; + 20 X 17,17 4 + 60 X 15 833 - 40 X 6 333 - 20 X 6 333 = 0; whence U 3 U 4 = -28500. A diagram of half of the truss should now be made, and all the stresses placed upon it. The dimensions should also be put upon this diagram. The student should cultivate this habit, as it shows him at a glance the general relation of stresses and the general rules of their variations. Fig. 33 gives the half-truss, together with the stresses and dimensions. The stresses in the members of the right half of the truss are the same as those in the corresponding members of the left half. 37 28 BRIDGE ENGINEERING From inspection of the above diagram, it is seen that the chord stresses increase from the end toward the center; that the web stresses decrease from the end toward the center; and that all members slant- ing the same way as the end-post L^J l have stresses of that sign, Fig. 33. Dimensions and Stress Diagram of Half a Six-Panel Warren Through Truss. while all that slant a different way have an opposite sign. These relations are true of all trusses with parallel chords and simple systems of webbings. 34. Position of Live Load for Maximum Positive and Negative Shears. The dead load, by reason of its nature, is an unchangeable load. The stresses due to it are the same at any and at all times. "1 ^+y ) < -y > e W Ibs. per lin. Foot. Y * I r ft * Fig. 34. Calculating Maximum Positive and Negative Shears in Simple Beam under Live Load. Conventional Method. With the live load, the case is different. The live load represents the movement of traffic upon the bridge. At certain times there may be none on the bridge, while at other times it may fill the bridge partially or entirely. In such cases the shears due to live load will vary. 38 BRIDGE ENGINEERING 29 Conventional Method. It has been found that the maximum positive shear at any section of a simple beam occurs when the beam is loaded from that section to the right support, and that the maximum negative shear occurs at the same section when this beam is loaded from the section to the left support. This can be proved as follows : Let a beam be as in Fig. 34, and let a a be the section under consideration. The reaction R l is due to the load wy on the part y, and to the load wx on the part x. That is, Now the shear at the section a a is 7? t wx-, or, '2\J~ WX = Vc * w x 21 / X T' From inspection of this last equation, it is seen that wo; ^ -, is the amount that is added to the reaction by loading the part x. x ~2~ y Also., that -j is less than unity, is evident. The amount in brackets in the last equation represents the effect of the loading of the segment x of the beam. As this is negative and will only reduce the positive valued term -^ , it is therefore proved that to get the largest positive shear the beam should be loaded from the section to the right support. From further inspection of the equation, it will be seen that the term in brackets, which represents the effect of the load on the seg- ment x on the shear, is always negative; and that the term ,which represents the effect of the load on the segment y on the shear, is always positive. Hence, to get the largest negative shear at the section, the load should be on the segment x. That is, the loading should be from the section to the left support. 30 BRIDGE ENGINEERING In a truss, the loads are placed at the panel points; and the above rules in application, should be formulated as follows: To get the maximum positive shear at a section or in a panel, load all panel points to the right of it. To get the maximum negative shear at a section or in a panvl, load all panel points to the left of it. Example. Determine the maximum positive and the maximum negative shears in the panels of the 7-panel Pratt truss shown in Fig. 35, the Fig. 35. Calculation of Shears in Panels of 7- Panel Pratt Truss. live panel load being 40 000 pounds, the truss is not required.) (It will be noticed that the height of For maximum + V in 1st panel, load L u L 2 , L 3 , L 4 , L 5 and L 6 . + V + v + v + v + v + v 2d 3d 4th 5th 6th 7th ' ' L 2 , L 3 , L 4 , L 5 , and L 6 . ' ' L 3 , L 4 , L s , and L 6 . " L 4 , L 5 , and L 6 . ' ' L 5 and L 6 " L 6 . " no panel points at all. The reaction produced by each of the loadings is equal to the shear for that particular case, since the shear at any section or in any panel is equal to the reaction minus the loads to the left of the section or panel, and, according to the method of loading, there are no loads to the left of the section ; therefore the reaction is equal to the shear. For the first panel, the computation is made as follows, the center of moments being, of course, at L 7 : (72, = + V,) X 7 X 20 = 40 X 20 + 40 X 2 X 20 + 40 X 3 X 20 + 40 X 4 X 20 + 40 X 5 X 20 + 40 X 6 X 20. It will be seen that as 20 occurs in all terms of this equation, it can be factored out by dividing both sides by 20, and the result will be the same. The equation can now be written: 40 BRIDGE ENGINEERING 31 + V, X 7 = 40 + 40 X 2 + 40 X 3 + 40 X 4 + 40 X 5 + 40 X 6, and can still be simplified by writing: + V l =- (1+2 + 3 + 4 + 5+6)=+ 120.00, which is the form customarily used, the panel length being taken as a unit of measurement. The other shears are now easily computed in a similar manner: + v a = -^ (1 + 2 + 3 + 4+5)=+ 85.71 + F 3 = lP_(l + 2 + 3 + 4) = + 57.14 40 + V< = ~ (1 + 2 + 3) = +34.28 40 + F 5 = -- (1 + 2) = +17.14 + F B = 1 (1) = +5.71 + V 7 = ^.(0) -. +0 In computing the maximum negative shears, sometimes called the minimum shears, the reaction is not the same as the shear, as there are loads to the left of the section, and these must be sub- tracted. The loadings are: For maximum V in 1st panel, load no points. " L,. "Li anc!L 2 . " L,, L 2I andL 3 . " L lf L 2 , L 3 , andL 4 . " L,, L,, L 3 , L 4> and L 5 . " L 1( L,, L 3) L 4 , L s , and L c . It is evident that the maximum F 1 is equal to zero, there being no loads on the span. The maximum negative shear in the second panel is equal to the reaction produced by loading the panel point L v and the load at i 1 . Thus, 7R, = 40 X 6 - V 2 = R, - load at L, 41 32 BRIDGE ENGINEERING The other shears are next computed as follows: _ V 3 = ~ (6 + 5) - 2 X 40 = - 17.14 . 40 - V. = (6 + 5 + 4) - 3 X 40 = -34.28 -57.14 = - 85.71 40 - F 7 = - (6 + 5 + 4 + 3 + 2+ 1)-6X 40 = - 120.00 7 The maximum positive and the maximum negative live-load shears should now be written side by side, and inspected, in order to observe any existing relations which might help to lessen the labor of future computations. The values are given in thousands of pounds below : LOCATION MAX. + LIVE-LOAD SHEAR MAX. LIVE-LOAD SHEAR F, + 120.00 - 0.00 v, + 85.71 - 5.71 vl + 57.14 -17.14 v< + 34.28 -34.28 V, + 17.14 -57.14 V, + 5.71 -85.71 V 7 + 0.00 -120.00 It is at once seen that the negative shears are numerically equal in value to the positive ones, but that they occur in reverse order. This simplifies the labor required in the derivation of the negative shears; for, after computing the maximum positive shears, these may be written in reverse order, and the negative sign prefixed; the result will be the maximum negative shears. The above method for maximum live-load shears is called the conventional method. It is the one that is almost universally used, and its use will be continued throughout this text. Exact Method. On account of the fact that the floor stringers or joists transfer the loads to the panel points, it would be impossible to have a full panel live load at one panel point and no load at the panel point ahead or behind. In order to have a full panel load at one point, the stringers in the panels, on both sides of the point must 42 BRIDGE ENGINEERING 33 be full-loaded, and this would give a load at the panel point ahead, provided the bridge was fully loaded up to and not beyond the panel point ahead, equal in value to one-half of a full panel load (see Fig. Fig. 36. Illustrating "Exact" Method of Calculating Live-Load Shears in Panels. 36). The uniform live load, in order to produce full panel loads at L 2 , L 3 and L 4 , will also produce one-half a panel load at L r By the methods of differential calculus, it can be proved that the true maximum positive live-load shear occurs in a panel when the f Y//////&7//////// ///////////////////, -rn. panels Fig. 37. Calculating Maximum Positive Live-Load Shear in Panel. live load extends from the panel point to the right into that panel an amount (see Fig. 37) equal to m-1 ' in which, n = Number of the panel point to the left of the panel under considera- tion, counting from the right; m = Total number of panels in the bridge; p = Panel length. Let the truss of Fig. 35 be considered, the live load being 2 000 43 34 BRIDGE ENGINEERING pounds per linear foot of truss, and let it be required to determine the true maximum positive live-load shear in the 5th panel from the right end. x = y-^-j X 20 = 13.333 feet. There will now be (4 X 20 + 13.333) X 2 000 = 186 666 pounds on the truss; and the left reaction will be { 186 666 X (4 X 20+13.333) -r- 2} -=-140 = 62 200 pounds. From this must be subtracted the amount of the load on the 13.333 feet, which is transferred to the point L r This is equal to the reaction of a beam of a span equal to the panel length, loaded for a distance of 13.333 feet from the right support with a uniform load of 2 000 pounds per linear foot. This 1 "^ c \'\ f \ amounts to (13.333 X 2 000 X - ^~ ) - 20 = 8 890 pounds. The true shear is now: + V 3 = +62 200 - 8 890 = +53 310 pounds. The + V 3 , as computed by the conventional method, was -f 57 140, making a difference of 3 730 pounds between the two. If the true shears were computed and compared with the conventional shears, it would be found that the V l would be the same, and that the remainder of the conventional shears would be greater than the corresponding true shears. The difference between any two corre- sponding shears would increase from the left to the right end; that is, the difference between the conventional and exact shears would be greatest in the panel L 5 L 6 . To get the maximum negative shear in any panel, load from the left support and out into the panel under consideration an amount p x, and proceed in a manner similar to that above described. As this method of exact or true shears is seldom employed, problems illustrating its application will here be omitted. 35. Position of Live Load for Maximum Moments. In order to obtain the maximum moment at any point, the live load must cover the entire bridge. Let the beam of Fig. 34 be considered, and let it be required to obtain the maximum moment at the section a a. The reaction, as before computed, is: BRIDGE ENGINEERING 35 all terms of which are positive. The moment at the section is : M = R, X and substituting for R l its value, M=R l Xx-wx~; But y = I x; therefore, * - The first term of this equation represents the effect of the load on the portion x, and the second term represents the effect of the load on the portion y. The value of M will always be positive. The quantity x varies betw ? een and I. When x = 0, M is equal to 0. When x = I, the moment is equal to + wy 2 x -f- 2. For all values of x between and I, the first term is positive; and the second term being positive in all cases, it is therefore proved that for maximum live-load moments at any point, the entire span should be loaded, as loads on both segments add positive values to the moment value. 36. Warren Truss under Live Load. In order to analyze a truss intelligently, it is necessary to know its physical structure; that is, it must be known what character of stress can be withstood by the different members. The top chords of all trusses are built to take only compression, and the bottom chords are built to take only tension ; while some web members of some trusses are built for tension stresses, some for compression stresses, and some for both. The characteristic of the Warren truss is that the web members are built so as to be able to withstand either tension or compression. Let it be required to determine the live-load stresses in the Warren truss of Fig. 32. Let the live load per square foot of roadway, which is assumed to be 15 feet wide, be 100 pounds. The live panel load is then 100 X 15 X 20 -i- 2 - 15 000 pounds, and the live-load reaction under full load is 2 X 15 000 = 37 500 pounds. 45 36 BRIDGE ENGINEERING As the live load must cover the entire bridge to give maximum moments and therefore maximum chord stresses, as the -chord stress is equal to the moment divided by the height of the truss a simple method for the determination of live-load chord stresses presents itself. The live load and the dead load being applied at the same points, and being different in intensity, the stresses produced will be proportional to the panel loads. The maximum live-load chord stresses (see Fig. 33) will then be equal to the dead-load chord stresses multiplied by 15 000 -=- 6 333 = 2.371, and they are as follows: L U 1 = -2.371 X 17700 = -42000 U 1 U 2 = -2.371 X 15833 - -37530 U 2 U 3 = - 2.371 X 25 333 = -60 050 U 3 U 4 = -2.371 X 28 500 = -67 600 L L, = +2.371 X 7 917 =+18 770 L,L 2 = +2.371 X 20583 = +48800 L 2 L 3 = +2.371 X 26917 = +63850 The next step in order is to determine the maximum positive shears, and from these write the maximum negative shears. This is done as follows: + Live-Load V Live-Load V + 2 + 3 + 4+ 5) = +37 500 6 V 2 = ^-^(1 + 2 + 3 + 4) =+25000 -2500 y 16000 (1 + 2 + 3) = +15000 -7500 6 V 4 = 1500 (1 + 2) = + 7 500 - 15 000 + 2 500 - 25 000 + -37500 The stresses produced by the positive shears are called the maximum live-load stresses, and are: + L U, cos ( + 37 500 = .-. L U l = -37 500 X 1.12 = -42 000 - U,L, cos + 37 500 = .-. l^L, = +37 500 X 1 12 = +42 000 + L 1 U 2 cos + 25 000 = .'. L,C7 2 = -25 000 X 1.12 = -28 000 - C7 2 L 2 cos + 25 000 = /. C7 2 L 2 = +25 000 X 1.12 = +28 000 + L 2 C7 3 cos< + 15000 = .-. L 2 C7 3 = -15000 X 1.12 = -16800 - U 3 L 3 cos + 15 000 = .'. U 3 L 3 = + 15 000 X 1.12 = +16 800 The stresses produced by the negative shears are called the minimum live-load stresses, and are: 46 BRIDGE ENGINEERING 37 + LJJ l cos + =0 - f7,L, cos <}> + =0 + L 1 ?7 2 cos $ - 2500 = - U 2 L 2 cos - 2 500 = + L 2 U 3 cos = 0. In the first four panels from either end, the live-load shear, which is of a different sign from that of the dead-load shear, is smaller than the dead-load shear, and therefore will not cause a reversal of stress in the member in that panel. For example, take U 3 L t ; then, for dead-load stress, -C7 3 L 4 cos < + 66.0 = --.U 3 L 3 = +66.0 X 1.231 = +81.20 For live-load stress, -C/ 3 L 4 cos < - 27.0 = .-.U 3 L 3 = -27.0 X 1.231 = -33.25 The total stress = + 81.20 - 33.25 = + 47.95, which is still tension. Considering L 9 U 10 , the stress equations are : For dead-load stress, + L 9 f/ 10 cos < - 66.0 = /. L U }0 = +81.20 For live-load stress, + L 9 t/ 10 cos < + 27.0 = .:L U M = -33.25 The total stress, as before, is + 47.95, or a tension of 47 950 pounds. An inspection of the center panel and the two panels on each side of it, shows that the live-load shear is of a different sign from the dead-load shear, and is also greater in value than the dead-load shear. If the members shown in Fig. 39 were the only ones in the panels, then the dead-load stresses would be: - U 4 L 5 cos + 44.0 = / 4 L 5 = + 54.20 -t/ 5 L 6 cos + 22.0 = t/ 5 L 6 = +27.10 + L 7 U K cos - 22.0 = L 7 U S = +27.10 + LJT cos < - 44.0 = L K U 9 = +54.20 and the live-load stresses caused by the shear of opposite sign from that of the dead-load shear, are: -t/ 4 L 5 cos $ - 45.0 = t/ 4 L s = -55.40 - U S L cos $ - 67.5 = f/ 5 L 6 = -83.10 + L 7 C7 8 cos i + 67.5 = L 7 f/ 8 = -83.10 + L a U 9 cos + 45.0 = L 8 C7 fl = -55.40 As no diagonal acts under dead load in the center panel, we may assume that C/ 6 L 1 acts under live load. The stresses which occur in this are: + U ti L 7 cos $ + 94.5 = U K L 7 = +116.30 - U b L 7 cos - 94.5 = U e L 7 = - 116.30 40 BRIDGE ENGINEERING The above shows that compressive stresses will occur in the diagonals which were built for tension only. These stresses are : C7 4 L 5 = +54.20 - 55.40 = - 1 200 pounds C7 6 L 6 = +27.10 - 83.10= -56000 " L 7 U 8 = +27.10- 83.10= - 56000 " L S U 9 = +54.20 - 55.40 = - 1 200 " U 6 Lj = - 116.30 = -116 300 " If some provision were not made for these stresses, they would cause the members to crumple up and the truss to fail. In order to allow for them, diagonals are placed in the panels, as shown by the dotted and dashed lines. These members will take up the above stress; and moreover, as they slope the opposite way from the main members, they will be in tension. In order to prove this, assume L 5 U 9 to act when the live load is on points L 5 , L 4 , L 3 , L 2 , and L r Now, U 5 L 6 will not be regarded, as its stress will be zero. Then the stresses will be: For dead load, + L 5 f/ cos < + 22.0 = L 5 [/ 6 = -27.10. For live load, + L 5 7 U cos $ - 67.5 = LJJ 6 = +83 10; and the total stress in L,U 6 will be - 27.10 + 83.10 = + 56.00. In a similar manner, the stresses in the other members are: LJJ 5 - +1.2; L 6 U 7 =+ 116.30; U,L S =+ 56.00; and 7 8 L 9 =+1.2. These diagonals are called counters or counter-bracing. From a consideration of the foregoing, it is evident that: (a) // the live-load shear in any panel is of opposite sign and greater than the dead-load shear in the same panel, then a counter is required. (b) The stress in a counter is equal to the algebraic sum of the dead-load shear and the live-load shear of opposite sign times the secant of the angle it makes with the vertical. This is true for any truss with horizontal chords and a simple system of webbing with diagonals and verticals. 38. Maximum and Minimum Stresses. Some specifications require the member to be designed for the maximum stress, while others take into account the range of stress. In this latter case it is necessary to determine the minimum as well as the maximum stress. Except where a reversal of stress occurs and this does not happen in trusses with horizontal chords few specifications require any 50 BRIDGE ENGINEERING 11 but the maximum stresses to be com- puted. For that reason, little space will here be devoted to the minimum stresses, their computation in succeed- ing articles being thought to illustrate them sufficiently. (a) The maximum stress in a member is equal to the sum of the dead-load stress and the live-load stress of the same sign. (b) The minimum stress is equal to the sum of the dead-load stress and the live- load stress of the opposite sign, or to the dead- load stress alone, according to which gives the smallest value algebraically. By this latter statement it should be seen that if the maximum stress is 58.60, then or + 18.00 would be smaller than -3.00. (c) It is evident that the minimum in all counters and in all main members in panels where counters are employed will be zero, for when the counter is acting the main member is not, and therefore its stress is zero. The reverse is also true. (d) An exception to a is seen in the case of the counters. Here it is evident that the maximum stress is equal to the algebraic sum of the dead-load shear and the live-load shear of opposite sign times the secant of the angle which the counter makes with the vertical. While it is true that in trusses with horizontal chords the loading for maxi- mum shears will give the maximum live-load stress to be added to the dead load for the maximum stress, it is not always true that the loading for minimum live-load shears will give the stress to add to the dead-load stress to get the minimum stress. However, the loading for the minimum live-load shears will give the live-load stress to be added to the dead-load stress for the minimum stress, except in the case 51 -12 BRIDGE ENGINEERING U of verticals placed between panels each of which contains counters, and in that case it may or may not do so. In such cases a loading must be assumed preferably the one for minimum shears and the shears in the panels on each side of the vertical must be com- puted for the loading assumed. If the resultant shear is the same sign as the live load, then the main diagonal acts; if it is of different sign, then the counter acts. As an example, let it be re- quired to find the minimum stress in the vertical U 5 L 5 of the truss of Figs. 39 and 40. It is assumed that the loading for minimum shears will give the result. The section a a is then passed, and the live load placed on L. and all points to the left. The shears will then be as shown in Fig. 41. To obtain the shear in the panel L t L 5 , aiv. 4- 44.0 .+ 22.0 llv. 9.0 - 67.5 Totalv 4- Fig. 41. Stress Diagram for Vertical iii Truss of Fig. 40. under this loading, it must be re- membered that a load is at L 5 ; and so the shear is the shear in the panel I/ 5 L 6 with the panel load at I/ 5 added, or, 67.5 + 58.5 = 9.0. The diagonals now act as indicated by Fig. 41, and the total stress in C7 5 L 5 is determined by passing a circular section around L T 5 , and it is : -Load at f/ 5 - f/ 5 L 5 = 0. As there is no load at U 5 , the stress in C7 5 L 5 is = 0. The same result will occur if points L 4 or L 3 and to the left are loaded; but if points L 2 and to the left are loaded, the members C7 4 L 5 and ?7 5 L 6 will act, and the stress in U 5 L 5 will then be equal to the shear on the section a a. The stresses are: Dead-load, 22.0; and live-load, + 13.5, which gives a total of 8.5; but as the maximum stress is -22.0 - 126.0 = - 148.0, it is evident that and not -8.5 is the minimum. The computation of the maximum stress is as follows: Load points L 6 and to the right. The shear on a a is, for BRIDGE ENGINEERING 43 dead load, +22.0; and for + live load, +126.0; and the equations of the stresses are: + 22.0 + C/ 5 L 5 = C7 5 L 5 = - 22.0 + 126.0 + U 5 L 5 = C7 5 L S = - 126.0 Max. = - 148 TRUSSES UNDER DEAD AND LIVE LOADS 39. The Pratt Truss. The Pratt truss is used to perhaps a greater extent than any other form; probably 90 per cent of all simple truss spans are of this kind. Let it be desired to determine the stresses in the 8-panel 200-foot single-track span shown in Fig. 42, the height being 30 feet, the dead panel load being 30 000 pounds, and the live panel load 62 400 pounds. The secant is {^ +~3() 2 ) * + 30 = 1-302, and the cosine is 0.7685. The dead-load reaction is 3 X 30.0 = 105.0. The dead-load shears are : y, = +105.0 V, = + 75.0 V 3 = + 45.0 V 4 = + 15.0 F 5 = - 15.0 The dead-load chord stresses may be tabulated as follows (see Articles 27 and 29): Dead- Load Chord Stresses MEMBER SEC- TION CEN- TER OF MO- MENTS STRESS EQUATION STRESS L L, = L,L 2 a a u l + 105.0 X 25 - L,L 2 X 30 = + 87.5 T T ^2-^3 b-b U 2 + 105.0 X 50 - 30.0 X 25 - L 2 L 3 X 30 = + 150.0 L 3 L 4 c c U 3 + 105.0 X 75 - 30.0 (25 + 50) - L 3 L 4 X 30 = + 187.5 UJJ* a a L 2 + 105.0 X 75 - 30.0 X 25 + U,U 2 X 30 = -150.0 u a u, b-b L 3 + 105.0 X 75 - 30.0 (25 + 50) + U 2 U 3 X 30 = -187.5 U 3 U, c c L< + 105.6 X 100 - 30.0 (25 + 50 + 75) + U 3 U 4 X 30 = -200.0 In determining dead-load stresses in web members, it is cus- tomary to assume one-third of the dead panel loads as applied at the II BRIDGE ENGINEERING upper chord points. This, as will be seen, makes no difference in the stresses in the chords or in the diagonals, the stresses in the verticals only being different from what is the case when all the dead load is taken on the lower chord. The stresses in the diagonals (see Articles 27,28, and 30) are: Dead-Load Stresses in Diagonals MEM- BER SEC- TION SHEAR ON SECTION ST RESS EQUATION STRESS LU, + 105.0 + 105.0 + L n U , X 0.7685 = -136.70 UiL 2 a a + 75.0 + 75.0 - t/jZ/ 2 X 0.7685 = + 97.60 U L b-b + 45.0 + 45.0 - U L 3 X 0.7685 = + 58.60 U 3 L 4 c c + 15.0 + 15.0 - U 3 L 4 X 0.7685 = + 19.53 In determining the stresses in the verticals, it is to be remem- bered that one-third the dead panel load (or 10.0) is at the panel o a 'i b - 2 c Fig. 42. Outline Diagram of 8-Panel Single-Track Pratt Truss Span. points of the upper chord, and two-thirds (or 20.0) is at the lower chord. The stress in the hip vertical U l L l is determined by passing a circular section around L,. It is solved thus: -20.0 + f/.L, = t/A = +20.0 In a similar manner the stress in U 4 L 4 is found to be : -10 - U 4 L 4 '= U 4 L 4 = -10.0 In order to find the stress in the remaining verticals, sections 1 1 and 2 2 are passed, cutting them, and the shears on these sections computed. The shears are: F,_, = + 105.0 - 2 X 20 - 1 X 10 = +55.0 F 2 - 2 = + 105.0 -3X20-2X10= +25.0 The stress equations are written, remembering that as the verticals 54 BRIDGE ENGINEERING 45 make an angle of zero with the vertical, their cosine is equal to unity. These equations are: + U 2 L 2 + 55.0 = U 2 L 2 = -55.0 + U 3 L 3 + 25.0 = U 3 L 3 = -25.0 The live-load chord stresses will be proportional to the dead- load chord stresses, as both loads cover the entire truss in exactly the same manner. The ratio of the panel loads by which the dead- load chord stresses are multiplied in order to get the live-load chord 62 400 stresses, is = 2.08, and the chord stresses are: U l \ 30000 L U L, = L,L 2 = + 87.5 X 2.08 = + 182.0 L 2 L 3 = +150.0 X 2.08 = +312.0 L 3 L 4 = +187.5 X 2.08 = +390.0 U 1 U 2 = -150.0 X 2.08 = 312.0 U 2 U 3 = -1875 X 2.08 = -390.0 U 3 U 4 = -200.0 X 2.08 = -416.0 As the entire bridge is to be loaded to get the maximum stress in I/oC/j, it is therefore equal to the dead-load stress times the above ratio; or L U t = -136.70 X 2.08 - 284.20. The maximum live-load stress in U 1 L 1 is determined by passing a circular section around L v and is solved (see Fig. 43) from the equation : + U,L - 62.4 = .'. 17, L! = +62.4 For VJj v the section a a is passed, and the points L 2 and to the right are loaded. The maxi- mum shear is: + V 2 = + -^p (1+2 + 3 + 4 + 5 + 6)= + 163.8; and the stress equation is: + 163.8 - U,L 2 X 0.7685 = .'. U,L 2 = +213.2. In a similar manner, pass section b b, and load points L 3 and to the right, and the shear and the stress equations for U 2 L 3 are : + V 3 = + -^i (1+2 + 3 + 4 + 5)= +117.0 + 117.0 - U 2 L 3 X 0.7685 = .'. U 2 L 3 = +152.4 55 46 BRIDGE ENGINEERING For UJji, the section c c is passed, and the panel points to the right are loaded. The shear and stress equations are: + V 4 = + - 6 |^ (1 + 2 + 3 + 4) = +78.0 O + 78.0 - C7 3 L 4 X 0.7685 = .'. U 3 L 4 = +101.6 For the maximum stresses in the verticals, sections 1 1, 2 2, and 3 3 are passed, and in each case the panel points to the left of these loaded. The shears are Fj _ , = -^ (1+2 + 3 + 4 + 5)= +117.0 O 62.4 V,-, = - ~- (1 + 2 + 3 + 4) = +78.0 + 2 + 3) = +46.8 AO 4 F 8 _, = - U + 117.0 + f/ 2 L 2 = + 78.0 + C7 3 L 3 = u, \ The stress equations for U 2 L 2 and U 3 L 3 are simple, as only three members are cut. They are: .-. f/ 2 Lo= - 117.0 .-. U 3 L 3 = - 78.0 It is seen that the section 3 3 cuts the member LJ7 5 , and therefore the stress in this must be determined before the stress equation can be written, as its ver- tical component will enter into it. However, by comparing the dead- load shear in that panel, which is 15.0, and the live-load shear F 3 _ 3 , which is +46.8, it is seen that the resultant shear is + ; and, as this is of opposite sign from the dead- load shear, a counter is required and is acting. The stress in U 5 L 4 is zero, and the diagonals act as in Fig, 44, the section 3 3 then cutting three members. The stress equation is + 46.8 + U t L 4 = 0, from which U 4 L 4 = - 46.8. Care should be taken not to add to this -46.8 the -10.0 derived as dead-load stress on page 44, in order to get the maximum stress, as the 10.0 previously derived was the dead-load stress in UJj 4 when U 3 L 4 and LJJ b were acting. The dead-load stress which goes with the live-load stress of 46.8 acts simultaneously with it, \ \- Fig. 44. Calculation of Stress in Diag onal of Span of Fig. 42. 56 BRIDGE ENGINEERING 47 and is the dead-load stress in U^L^ when the members U,L 4 and U 4 L 5 are acting as in Fig. 44. The dead-load shear on the section 3 3 would then be the left reaction minus the loads at points U v U 2 , U 3 , L v L 2 , L 3 and L 4 ; or, F 3 _ 3 = +105.0 -3X10-4X20= -5.0; and -5.0 + U 4 L t = .-.U 4 L t = +5.0. Remember that this +5.0 can act only when the live load tends to produce a stress of 46.8; and thus the total stress in C7 4 L 4 with live load in that position is -46.8 -f 5.0 = -43.8, while with dead load only in the truss the stress is 10.0. The dead-load shears and the maximum + and live-load shears should now be written for inspection, in order to investigate for counters and then for the minimum stresses. Those whose derivation has not been given should be easily computed by the stu- dent at this time. The shears are DEAD- LOAD + LlVE-LOAD LlVE-LOAD V l + 105.0 + 218.4 F 2 + 75.0 + 163.8 - 7.8 V 3 + 45.0 + 117.0 -23.4 V 4 + 15.0 + 78.0 -46.8 From a study of these it is seen that a counter is required in the 4th panel according to rule a, Article 37; and according to rule b of the same article, the maximum stress is (-46.8 -f 15.0) X 1.302 = + 41.4, the minimum stress for it and also U,L t being zero according to the same article. A counter is also required in panel 5, as the truss is symmetrical. The minimum live-load stress in U^ is zero, and occurs when no live load is at the point L r The minimum live-load stresses in the diagonals U^ and ?7 2 L 3 occur when the truss is loaded successively to the left of the sections a a and b b, in which case the shears are 7.8 and 23.4 respectively. The stress equations are -U,L 2 - 7.8X0.7685 = .-. U t L 2 =- 10.16 -U 2 L 3 - 23.4 X 0.7685 = .'. U 2 L 3 = -29.15 The minimum live-load stress in U^ is obtained by passing 57 48 BRIDGE ENGINEERING section 1 1 and loading the panel points to the left. The live-load shear is the same at this section as it is at the section 6 6 namely, 23.4. The stress equation is + U 2 L 2 - 23.4 = .'. U 2 L 2 = +23.4 To determine the minimum live-load stress in U 3 L 3 , proceed as indicated on page 42. By loading points L 3 and to the left, the live- load shear in the 4th panel will be 46.8, and in the 3d panel under this same loading it will be -46.8 + 62.4 = + 15.6. The sign of the total shear in the two adjacent panels, and the members acting, are shown in Fig. 45. The stress in U 3 L 3 is then determined by using a 10 10 10 eo.o E4 10 U 5 L4- eo.o Ls eo.o d.lv. + 4-5.O + I5.O l.l.v. * 1 5.6 4-6.6 Totalv + d.l.v. + 15.0 - 15.0 l.l.v. - 15.6 -78.0 Total v Fig. 45. Fig. 46. Stress Diagrams for Verticals hi Span of Fig. 42. circular section around U 3 , and is simply the dead load at U 3 , there being no live-load stress in the member when the bridge is loaded as has been done. In finding the minimum live-load stress and also the minimum stress in U t L v the same method of procedure will be followed. Let L^ and to the left be loaded. Then the shear in the 5th panel is 78.0, and under this same loading the shear in the 4th panel is -78.0 + 62.4 - -15.6. The sign of the total shear in each of the adjacent panels is given in Fig. 46. It should be remembered that a resultant shear with the same sign as the dead-load shear causes the main diagonal to act, while a resultant shear of opposite sign to that of the dead-load shear causes the counter to act. The members 58 BRIDGE ENGINEERING 49 acting are shown, and a section 4 4 can be passed. The dead-load shear at this section is 105 - 3 X 20 - 4 X 10 - +5.0; and accordingly, - U 4 L 4 + 5.0 = 0. Therefore, U 4 L 4 = + 5.0 = Dead-load stress in this case. The live-load stress which acts at the same time is: -U 4 L 4 - 15.6 = .-. U 4 L 4 = -15.6, the term 15.6 representing the live-load shear on the section 4 4. This is not the minimum stress, as will next be shown, but it illus- trates the fact that the loading for minimum live-load shears does not always give the minimum live-load stress. By loading L v the live-load shear in the second panel, and likewise all others from this to the right support, will be 7.8. The total shears, together with their sign, and also the members they cause to act, are given in Fig. 47. The minimum live-load stress in U 4 L^ is found to be zero, and the dead-load stress is 10, as is derived by passing a circular section around U 4 , the equation 10 U 4 10 eo I EO Ls eo d.l.v. + 15.0 -15.0 U.v - 7.8 - 7.8 Total v + Fig. 47. Stress Diagram for Vertical in Span of Fig. 42. being as follows: -Live load at U t - U 4 L t = .'. U 4 L 4 = for live load. -Dead load at U 4 - U 4 L 4 = .'. U 4 L 4 = - 10.0 for dead load. A diagram of half the truss should now be made, and all dead and live load stresses placed upon it, and these should be combined so as to form the maximum and the minimum stresses. Such a dia- gram, together with all stresses, is given in Fig. 48. The stresses are written in the following order: Dead load, maximum live load, minimum live load, the maximum, and the minimum. In the chord and end-post stresses, there is no minimum live-load stress recorded, it being zero. Where pairs of stresses occur simultaneously, a bent arrow connects them. 40. The Howe Truss. The physical make-up of the Howe truss 59 50 BRIDGE ENGINEERING QpOO o ^ II 1 1 ! I I o0 O.Q n-XO 1000 'QirrU H- + + differs from that of the Pratt in that the diagonals are made to stand com- pression only, and the verticals can stand tension only. In the Pratt truss it was found that M g none of the inter- .2 mediate posts could 5] be brought into I tension by any ^ loading. In the Howe truss it will 3 be found that none E of the verticals can a be brought into . OQ compression. ^ Let it be required to determine the | stresses in a Howe 3 truss of the same g span, height, and K loading as the Pratt " truss of Article 39. tc E An outline diagram is given in Fig. 49. The dead-load shears and the maximum and min- imum live-load shears will be the same as for the Pratt truss, and they are : . BRIDGE ENGINEERING 51 DEAD-LOAD V + LIVE-LOAD V - LIVE-LOAD V V + 105.0 + 218.4 - r + 75.0 + 163.8 - 7.8 V + 45.0 + 117.0 -23.4 V + 15.0 + 78.0 -46.8 V - 15.0 + 46.8 -78.0 Inspection of these shows that counters are required in the 4th and 5th panels (see Article 37). The dead-load lower chord stresses will be computed by the U, / U E / U 3 / : U 4 / Pig. 49. Outline Diagram of 8-Panel Single-Track Howe Truss Span. tangent method (see Article 31), the section being y y { , etc. The tangent of < is 25 -=- 30 = 0.8333. The stresses may be conven- iently tabulated as follows: Dead-Load Chord Stresses (Lower Chord) MEM- BER SECTION STRESS EQUATION STRESS L n L, y - Vi - 105.0 X 0.8333 + L L, = + 87.5 L t L 2 y y-i -(105.0 + 75.0) 0.8333 + L,L 2 = + 150.0 L 9 L 3 y y* -(105.0 + 75.0 + 45) 0.8333 + L 2 L a = + 187.5 L 3 L 4 y - 2/4 -(105.0 + 75.0 +45.0 + 15.0) 0.8333 + L 3 L 4 = + 200.0 A simple method for the determination of the upper chord stresses, is to pass a section and to equate the sum of the horizontal forces. Pass section 1 1. The only horizontal forces are the stresses in I/^ and UJ7 2 ; and as these are parallel, one must be equal and opposite to the other. In a like manner the stresses in the other sections of the top chord are found. The stresses are : U,U 2 = -L L, = -(+ 87.5) = - 87.5 U 2 U 3 = -L,L 2 = -( + 150.0) = -150.0 U 3 U 4 = -L 2 L 3 = -( + 187.5) - -187.5 61 52 BRIDGE ENGINEERING A consideration of the Pratt truss shows that this method can be applied to it in determining the chord stresses. As it is known that the diagonal web members are in compression under the dead load which produces a positive shear in the left half of the truss, it is evident that positive live-load shears will produce compressive stresses, and negative live-load shears tensile stresses, in the diagonals in the left half of the truss. Also, from Article 30, the stress in a diagonal is V sec <. The stresses can now be written directly without the aid of the stress equation : L U l = -105.0 X 1.302 = -136.70 L,7 2 = - 75.0 X 1.302 = - 97.60 L,U 3 = - 45.0 X 1.302 = - 58.60 L~[7 4 = - 15.0 X 1.302 = - 19.53 Likewise the stresses in the verticals can be written directly, remem- bering that here the secant is unity, and that the shear at the section cutting the member is to be used, not forgetting that J of the dead panel load is applied at the top panel points. The shears and stresses are: V. - , = +105.0 - 10 = +95.0 [7,1/1 = +95.0 V 2 _ 2 = +105.0 - 20 - 2 X 10 = +65.0 U,L 2 = +65.0 V 3 _ 3 = +105.0 -2X20-3X10= +35.0 U 3 L 9 = +35.0 . The member C7 4 L 4 cannot be easily determined by passing a section 4 4, for this cuts four members. It is determined by passing a circular section about the point L 4 , the equation being + U t L 4 - 20.0 = 0, from which U 4 L t = + 20.0, which is equal to the dead panel load at the point Z/ 4 . The live-load chord stresses are determined by multiplying the dead-load chord stresses by the ratio of the live to the dead loads. This has been found to be equal to 12.08. The live-load chord stresses are found to be : L L, = +182.0 UJJ t = -182.0 L,L 2 = +312.0 UJJ 3 = -312.0 L 2 L 3 = +390.0 U 3 U 4 = -390.0 L 3 L 4 = +416.0 As the character of the stresses which can be taken by the diagonals and the verticals is known, the maximum and minimum live-load stresses can be written without first writing the stress equations. The maximum live-load stresses are: BRIDGE ENGINEERING 53 L U t = -218.4 X 1.302 = -284.36 tf.L, = +218.4 L,C7 2 = -163.8 X 1.302 = -213.27 U 2 L 2 = +163.8 L 2 U 3 = -117.0 X 1.302 = -152.33 U 3 L 3 = +117.0 L 3 [7 4 = - 78.0 X 1.302 = -101.56 C7 4 L 4 = + 78.0 It should be noted that when L 4 and all panel points to the right are loaded, the shears and the members acting are as shown in Fig. 50. The dead-load shear on the section 4 4 is +15.0, less the load at C7 4 , or + 15.0 10.0 = +5.0; and the equation of stress is J7 4 L 4 + 5.0 = 0, from which J7 4 L 4 = +5.0. Thus it is seen that in this 10 10 eo.o 10 U 5 10 J EO.O 6S.4. eo.p 6EA / / 10 d.l.v. + 15.0 - I5.O l.l.v. -- 7.8 + 15-6 Total v. + + d.l v. + ^-5.0 + 15.0 uUy 7.8 7.8 Total v. + + Fig. 50. Fig. 51. Stress Diagrams for Members of Howe Truss Span of Fig. 49. case the dead-load stress is +5.0 when the live-load stress is +78.0. The maximum stresses in the counters (see Article 37) are : (-46.8 + 15.0) 1.302 = -41.4. The minimum live-load stresses are now written as follows: L,[7, = + 7.8 X 1.302 = +10.16 LM 3 = +23.4 X 1.302 = +29.15 L a U 4 = U a L t = 17.L, - C7 2 L 2 - -7.8 U 3 L 3 ( See discussion / 4 L 4 j following. If live panel loads were placed at points L v L 2 , and L 3 the live- load shear in c c would be 46.8; and the dead-load shear being + 15, the counter would act, and the stress in U 3 L 3 would be tensile and equal to the s'um of the dead and live panel loads which are at its lower end L y If points L V and L 2 had live panel loads on them, 63 54 BRIDGE ENGINEERING the resultant shear in c c would be 23.4 + 15.0 = 8.4; the counter would act, and the stress .in U 4 L 4 would be tensile and equal to the dead panel load which is at L y There being no live panel load at L 3 , the live-load stress in U^ would be zero under this loading. If a live panel load be placed at L l only, then the shears and the mem- bers acting will be as shown in Fig. 51, and V 3 . 3 for dead load = + 45.0 - the load at U y or = 45 - 10 = +35.0. The F 3 . 3 for live load = 7.8, and the stress equation U 3 L 3 7.8 = 0, from which U 3 L 3 = 7.8. So this live-load compression stress of 7 800 pounds occurs at the same time as the dead-load tensile stress of 45 000 pounds. By loading various groups of panel points in succession and determining the resulting live-load stresses in U 4 L 4 , it will be found that under no loading can a negative live-load stress be produced. The minimum live-load stress is therefore zero, and occurs when there is no live load on the bridge. The stresses should now be placed on an outline diagram similar to that of Fig. 48, and the stresses in corresponding members com- pared with those in that figure. This is left for the student. 41. Bowstring and Parabolic Trusses. A bowstring truss is shown in Fig. 13, the full lines representing the main members, which are the members under stress by the dead load. The dotted members represent counters which may be stressed by the action of the live load. As before mentioned, the stresses in the chords and also in the webbing are quite uniform. When the end supports and the panel points lie on the arc of a certain curve, called a parabola, then, under full load, the stresses in all panels of the lower chord are equal; the stress in all verticals is tensile and is equal to the panel load at the lower end ; and the stress in all diagonals is zero. Under partial load, the stresses in the webbing are exceedingly small, and the chord stresses remain almost equal. If it is desired to have a parabolic truss, first decide upon the length of span, the number of panels, and the height at the center. The height of any vertical post is given by the formula : k.H- *. BRIDGE ENGINEERING which, All distances are in feet. Suppose, as an example, that it was desired to determine the heights of the vertical posts in an 8-panel parabolic truss of a height approximately equal to 24 feet. One-half H = Approximate height at center; d = Distance of vertical post from center; I = Span; h = Height of vertical Fig. 52. One-Half of 8-Panel Parabolic Truss. the truss is shown in Fig. 52. At the center, d = 0, and the equation reduces to h = H, which is 24 feet. For UgL v d = 20; then, from which, For L\L 2 , h = 24 - 4 X 3JX 20 t ~ 160 2 h *= 22.5 feet. d = 40 h = 24 - 4 X.24_X 40 2 160 For l\L v d = 60 h = 24 - 4 X j 4 X h = 18.0 feet. = 10.5 feet. Inspection of the above results shows that the span or the center height must become quite great before the clearance at U t L^ will be sufficient to allow the traffic to pass under a portal bracing at this point. For this reason these trusses are usually built as through trusses with bracing on the outside of the truss, which connects to the floor-beams extended. 65 56 BRIDGE ENGINEERING \ In the bowstring truss, the panel points of the top chord usually lie on the arc of a parabola which does not pass through the supports. For example, suppose that it was decided to have the span and pan- els the same as shown in Fig. 52, but the height at L 1 was to be '28 feet, and at L 4 36 feet. By substi tu ting these values in the equation just given, and solving for I, the place will be deter- mined where the para- bolic curve cuts the lower chord extended, and the lengths of the vertical posts may be computed as before. Substituting these re- sults : 28 = 36 - 4 X 36 X 60 2 (-36 + 28)Z 2 = - 4 X36 X 60 2 I = /4 X 36 X 60 5 "V 8 = 254.5, which shows that the arc cuts the lower chord extended at a point 254.5 -T- 2 = 127.25 feet from the center of the span (see Fig. 53). BRIDGE ENGINEERING 57 The other vertical posts are: U 3 L 3 h 4 X 36 X 20 2 36 - ., , = 35.11 feetj U 2 L 2 254. 5 2 4 X 36 X 40 2 254.S 2 32.44 feet; h = 36 - =~j = 28.00 feet, which checks. 254.5 The analysis of a bowstring truss will now be given. Both the maximum and minimum stresses will be determined, as reversal of Fig. 54. Outline Diagram of 5-Panel Bowstring Truss Span. stresses is liable to occur in the intermediate posts. The loading for minimum live-load stresses can be ascertained only by trial, care being taken to compute the dead-load stresses for the arrangement of web members caused by that particular live loading. Let it be required to determine the maximum stresses in the 5-panel 100-foot bowstring truss shown in Fig. 54, remembering U, that the diagonals take only ten- sion. The height of U^ is 20 feet, and of U^ 25 feet. The dead panel load is 17 200 pounds, and the live panel load is 50 000 pounds. The full lines show the main members which act under dead-load stress, and the dotted lines show the counters which may act under the action of the live load. One-third of the dead panel load, or 5 730 pounds, is taken as acting at the upper panel points, while the remainder, 11 470 pounds, acts at the lower L, 11.47 Fig. 55. Resolution of Forces around Panel Point in Bowstring Truss of Fig. 54. 67 BRIDGE ENGINEERING Fig: 61. Fig. 62. Analysis of Stresses in Various Members of the Bowstring Truss of Fig. 54. 68 BRIDGE ENGINEERING 59 Fig. 66. Analysis of Stresses in Various Members of the Bowstring Truss of Fig. 54. 60 BRIDGE ENGINEERING ones. Articles 27, 28, and 29 should be carefully, reviewed before going further. The shear times the secant method cannot be con- veniently employed for the live-load stresses in the members U^L 2 and LJJ 2 , as the section will cut the member U^Lf v and the vertical component of its stress must be reckoned with in the stress equation The method of moments as illustrated in Fig. 27, Article 29, will be used for these members. The dead-load reaction is 2 X 17.2 = +34.4. The dead-load chord stresses should first be computed. By resolving the horizontal forces around L v it is seen that L L t = LJj 2 (see Fig. 55). Passing the section a a, taking the center of moments, at U v and stating the equation of the moments to the left of the section, there results (see Fig. 56) : + 34.4 X 20 - L,L 2 X 20 = .'. L,L 2 = +34.4 For -LjL 3 , the section b b is passed ; the center of moments is at U 2 ; and the equation of the moments to the left of this section (see Fig. 57) is: + 34.4 X 2 X 20 - (11.47 + 5.73) 20 - L 2 L 3 X 25 = .'. L 2 L 3 = +41.26. By passing a vertical section cutting L^ and L U V the stress in I/o^can be determined by taking the sum of the vertical forces to the left and equating them to the vertical component of the stress (see Fig. 58). The equation is: + 34.4 + L U 1 X 0.707 = 0, from which L U 1 = -34.4 X 1.414 = - 48.7. A section a a (Fig. 59) shows that the center of moments for UJJ 2 is at U 2 ; and stating the moments of the stress, and the forces to the left of the section, there results an equation in which an unknown lever arm enters. This lever arm I is readily computed to be 24.28 feet, and the equation can now be written : + 34.4 X 2 X 20 - (11.47 + 5.73) 20 + U t U 3 X 24.28 = .'. U,U 2 = -42.51. The stress in U Z U 3 is determined by passing a vertical section in the 3d panel, and taking the sum of the horizontal forces. As there is no dead-load stress in the members L 2 U 3 and U^, their compo- nents will be zero. Therefore (see Fig. 60) it is evident that U 2 U 3 must be equal and opposite to L 2 L 3 and will be equal to 41.26. By reference to Fig. 55, the stress in LJJ l is seen to be tensile and equal to +11.47. BRIDGE ENGINEERING 61 Pass a circular section around U 2 and take the sum of the vertical components, assuming that the stress in U 2 L 2 acts away from the section. The length of UJJ 2 is 1/5 2 + 20 2 - 20.6, and therefore the vertical component of UJJ 2 will be (42.51 -- 20.6) X 5 - 10.32, which acts upward. The stress equation of UyL 2 (see Fig, 61) is: + 10.32 - 5.73 - U 2 L 2 = .'. U 2 L 2 = +4.59, showing that a tensile stress occurs in U 2 L 2 when all panel points are loaded. The simplest method of ascertaining the stress in U t L 2 is to pass a vertical section cutting members as shown in Fig. 62, and to equate the horizontal forces and stresses. The horizontal component of U l U 2 is: 45 * X 20 = 41.30, which acts to ward the left. The equation of stress is, then: -41.30 + 34.40 + U,L 2 sin = 0; but sin = 0.707; U,L 2 = +6.90 X 1.414 = +9.76 All the dead-load stresses being computed, the next operation will be to determine the live-load chord stresses. These are pro- portional to the dead-load stresses in the same ratio as the live panel load is to the dead panel load. This ratio is 50 H- 17.2 = 2.907, and the chord and end-post live-load stresses are: L U, = -48.71 X 2.907 = -141.7 U t U 2 = -42.51 X 2.907 = -123.6 U 2 U 3 = -42.26 X 2.907 = -123.0 L L 2 = +34.40 X 2.907 = +100.2 L 2 L 3 = +41.26 X 2.907 = +120.3 Also, the stress in U 2 L 2 when the live load covers the entire bridge is not 2.907 X 4.59, as it must be remembered that part of the dead load is at the panel points of the upper chord. Taking a circular section around U 2 (see Fig. 61), and noting that there is no load at U 2 , it is seen that the stress in U 2 L 2 due to live load is simply equal to the vertical component of the live-load stress of UjU 2 and wjll be tensile. It is: U 1 U 2 = (123.6 -f- 20.6) X 5 = +30.0. The maximum live-load stress in U 1 L l is tensile, and equal to the live panel load at L 1 (see Fig. 55). 71 25 + 30.0- t/jLs cos 0=0; but cos < = =0.782; 02 BRIDGE ENGINEERING To obtain the maximum stress in U^, load L 3 and L 4 . The erv shear V 3 will then be ^ (1-f 2) = +30.0. The section will cut o the members as shown in Fig. 63, and the equation of stress will be : J /. t/^3 = +38.4. If panel points L^ and L 2 were loaded, it is evident that the stress in L 2 U 3 would be +38.4. To obtain the maximum live-load stress in UJLi 2 , a section is passed cutting UJJ V L^, and U^ (Fig. 64). The center of moments will be at the intersection of UJJ 2 and L } L 2 , and this point lies some place to the left of the support L . The lever arm of U : L Z will be the perpendicular distance from this point to the line U V L Z extended. The panel points L 2 , L 3 , and L^ are loaded. The left 50 reaction is then (1 + 2 + 3) - + 60.0. The lever arms are 5 easily computed, and these, together with the members cut, are shown in Fig. 64. The equation of stress is: - 60.0 X GO.O+t/jLaX-TO.S = .-. U,L 2 = +50.80. If a load were put on L l only, then the reaction at L would be 4 X 50 = 40; and the equation of stress would then be: O -40.0 X 60.0 + 50 X (60.0 + 20.0) + t/,L 2 X 70.8 = .'. J7,L 2 = -22.6. As this is compression and greater than the dead-load stress, + 9.76, a counter is required in that panel. In order to get the stress in the counter, it must be inserted, U^ being left "out, and the dead and live load stresses computed and their difference taken. Fig. 65 gives the lever arms, center of moments, and the forces acting in this case. The dead-load stress is: -34.4 X 60.0 + (11.47 + 5.73) (60 + 20) - L,C7 2 X 62.5 = .'. L,U 2 = -11.02; and the live-load stress is: -40 X 60 + 50 (60 + 20) - L.O, X 62.5 = .-. L,U 2 = +25.60, and the stress in the counter is the algebraic sum of these two, or -11.02 + 25.60 = +14.58. 72 BRIDGE ENGINEERING 03 When a live panel load is at L v LJJ 2 is acting, as has just been proved. As this load at L t causes a negative shear in all panels to the right, this negative shear in the center panel will cause L 2 U 3 to act. A section may now be passed as shown in Fig. 66, and the stress equations for U 2 L 2 written: For dead load, +34.4 - 11.47 - 2 X 5.73 - U 2 L 2 = .'. U 2 L 2 =- ,11.47 For live load, +40.0 - 50.0 - U 2 L 2 = .'. U 2 L 2 = -10.00 Total = + 1.47 This is evidently not a maximum for U^L^ for when a full live load was on the span, the stress was +30.0 due to live load and +4.59 due to dead load. It might be well to consider what effect is produced by loading L 3 and L 4 . The loading of L 2 and L l need not be considered, since it is evident that, as this causes the total shear in panel 2 to be positive and the total shear in panel 3 to be negative, therefore U^ and L 2 U 3 Fig. 68. Stress Diagram of Half -Span of Parabolic Truss of Fig. 54. will act, and this causes a tensile stress in U 2 L 2 equal to the vertical components of the dead and live load stresses in UJJ 2 less the dead panel load at U 2 . With a live panel load at L 3 and L v the left reaction 50 is (1 + 2) = +30.0. The section, the live-load forces, the cen- ter of moments, and the members acting are shown in Fig. 67. The dead-load stress in U 2 L 2 will be the same as when the truss has no live load on it. The stress equation for the live load is: -60 X 30 - (60 + 20 + 20) X L 2 C7 2 = .'. L 2 U 2 = -18.0. 04 BRIDGE ENGINEERING The dead-load stress being +4.59, this stress of 18.0 causes a reversal of stress in the vertical. For this reason the ver- ticals of bowstring trusses are, like web members of Warren trusses, built so as to take either tension or compression. The minimum stresses in the diagonals will be zero, for when one diagonal in a panel is acting, the other is not. The diagram of half of the truss in Fig. 68 gives all the stresses. It is to be noted by the student, that in some cases one method for the deter- mination of s tresses is preferable to others in that it saves labor of computation. The analysis of the truss of Fig. 68 illus- trates this fact. 42. The Baltimore Truss. Baltimore trusses are of two classes those in which the half-diagonals, called sub-diagonals, are in compression, and those in which the sub-diagonals are in tension. The latter class is the one most usually built, as it is more economical on account of many of its members being in tension, in which case these members are cheaper and easier to build than if they were com- pression members. Fig. 14 shows boiii types of truss. The Baltimore truss does not have a simple system of webbing, and tbi that reason the analysis is here pre- sented. As the tension sub-diagonal truss is the type in most common use, its analy- sis will be given. Let it be required to compute the maximum stresses in the 14-panel 280- foot span of Fig. 69. The height is 40 feet, the dead panel load 24 000 pounds, d o * a 74 BRIDGE ENGINEERING 65 and the live panel load 40 000 pounds. One-third of the dead panel load is applied at the upper ends of the long verticals and also of the half-verticals. These half-verticals are designated as sub- verlicals. Attention is called to the system of notation used for the ends of the sub-verticals. The full lines in Fig. 69 represent the main members, being stressed by dead load only. The heavy lines indicate those members that take compression, the light lines those that take tension, and the broken lines the counter-braces. In this, as in nearly all Baltimore trusses, the diagonals make an angle of 45 degrees with the vertical. The dead and the positive live-load shears in the various panels should be computed. They are : DEAD-LOAD V + LIVE-LOAD V 7, +156.00 V l = (1 + 13) |j = +260.00 40 V 2 + 132.00 V 2 = (1 + .... 12) ^ = +223.00 V 3 +108.00 V 3 = (1 + 11)^- = +188.50 V t + 84.00 V t = (1 + 10) Y = + 157.20 7 5 + 60.00 V 5 = (1 + . . . . 9) -p = + 128.50 V 6 + 36.00 V = (1 + .... 8) YJ- = + 102.80 40 V 7 + 12.00 Vj = (1 + 7) ~ = + 80.00 It is only necessary to determine the negative live-load shear in panels 5 and 7, in order to ascertain if there is a counter required. These shears are : - V s = (10 + 11 + 12+ 13) ~ - 4 X 40 = -28.60 - F 7 = (8 + 9 + 10 + 11 + 12 + 13) ~ - 6 X 40 = -60.00 From a comparison of these with the dead-load shears, it is seen (see Article 37) that a counter is required in panel 7 only. The dead-load stresses are first to be computed. The stress in any sub-vertical is found by passing a circular section around its lower end, and equating the sum of the vertical forces, assuming in this, as in all cases, that the unknown stress acts away from the 75 BRIDGE ENGINEERING section. Take M l m v for example. Fig. 70 gives the section, the forces acting, and the members cut. Then, + M,w 1 - 16.0 = .'. M 1 m 1 = +16.0 As all sub-verticals have the same dead load at their lower end, it follows that the dead-load stress in all sub-verticals is the same, a tensile stress of 16 000 pounds. The dead-load stresses in the sub-diagonals are determined by resolving the forces around the joint at their lower end. The com- ponents perpendicular to the diagonal are taken (see Fig. 71). Take M, 16.0 Fig. 70. Diagram for Calculating Stress in Sub- Vertical of Baltimore Truss. m 2 U 2 . The known forces or stresses are the dead panel load of 8.0 and the stress in w 2 M 2 , which is 16.0 and which being tensile acts away from the sec- tion. The stress equation is : + m 2 U 2 - 8.0 sine (f> 16.0 sine $ = 0. < = 45, sine < = 0.707,and m 2 U 2 - 8.0 X 0.707 - 16.0 X 0.707 = .-. m 2 U 2 = +16.96. This equation may be put in another form by multiplying and dividing the numerical values bv 2: Fig. 71. Diagram for Calculating Stress in Sub-Diagonal of Baltimore Truss. (8.0 + 16.0) X 1.414 = 0; or, m,P,- + % which proves the well-known saying that the stress in the sub-diagonals is equal to one-half the panel load, times the secant of the angle (f>. It also shows that the vertical component of the sub-diagonal is equal to BRIDGE ENGINEERING 67 in one-half the panel load. This fact should be remembered, as it will be frequently used further on. In a similar manner, the stress in all the tension sub-diagonals will be found to be the same, + 16.96, and the stress in the compression sub-diagonal m^Li is -16.96. The stress in the member L m l and in the upper half of any main diagonal (i. e., Ujin^ U 2 m 3 , and Ugn^) is determined as in the diagonals of the Pratt or Howe truss, for the section passed cuts but one mem- ber, which has a vertical component. Take ^TO, (see Fig. 72). Then +156.0 + L m l 156.0 Fig. 72. Stress in Diagonal of Baltimore Truss. cos 45 = 0, from which i w 1 = 220.5. For Ujtn 2 the section is passed as in Fig. 73, and the equation of stress is + V 3 t/,m 2 cos 45 = 0, or +108.0 - U^ X 0.707 - 0, from which L r ,m 2 - + 152.9. In a similar manner, U 2 m 3 = +60.0 -f- 0.707 = +84.84; U 3 m t = + 12.0 -i- 0.707 = + 16.96. The stresses in mJJ v mJL v and w 3 L 3 may be determined by Fig. 73. Stress in Upper Half of Main Diagonal of Baltimore Truss. Fig. 74. Stress in Diagonal of Baltimore Truss. resolving the forces about TO,, m v and TO S ; but a neater solution is to pass a vertical section cutting the member whose stress is desired, and to equate to zero the shear and the vertical components of all 77 68 BRIDGE ENGINEERING the members cut (see Fig. 24, Article 28). The section for passed as in Fig. 74. The equation of stress is then: m 1 U 1 cos 45 + mA cos 45 + F 2 = 0; but the vertical component of is = 12; and therefore, m l U l X 0.707 + 12 + 132 = .-,m l U l = -203.6. For w 2 L 2 , the section is as shown in Fig. 75, and the stress equation is : -m. z L 2 X 0.707 + vert, component m 2 U 2 + K 4 = -m 2 L 2 X 0.707 + 12 + 84 = .-. m 2 L 2 = + 135.6. [Fig. 75. Calculating Stress in Lower Half-Diagonal of Baltimore Truss. In a similar manner, passing a section cutting U 2 U 3 , m 3 U 3 , m^, and MjLg, the stress equation may be written: - m 3 L 3 X 0.707 + 12 + 36 = .-. m 3 L 3 = +67.85. The stresses in the verticals are best determined by resolving the vertical forces at their lower end. Referring successively to diagrams a, b, and c of Fig. 76, the stress equations are: + U 1 L 1 - 16.0 - 12.0 = .'. t^L, = +28.0 + f/ 2 L 2 - 16.0 + 96.0 = .'. U 2 L 2 = -80.0 + C7 3 L 3 - 16.0 + 48.0 = .'. U 3 L 3 = -32.0 96 and 48 being the vertical components of mJL 2 and m 3 L 3 respec- tively. 78 BRIDGE ENGINEERING 69 The chord stresses are easiest computed by considering the resolution of horizontal forces at the panel points. As the diagonals make an angle of 45 with the vertical, their horizontal and vertical 16.0 16.0 Fig. 76. Calculating Stresses in Verticals of Baltimore Truss of Fig. 69. components are equal. For instance, the horizontal component of the members Ljn^ Ujm 2 , and U 2 m 3 are equal to the shear in that panel, which is their vertical component. At point L (see Fig. 77), there results: + L A/i horizontal component of M ^L = 0; or, + L M, - 156 = .-. L M, = +156.0; and from Fig. 70 it is evident that L M t = MJj r At point L l (see Fig. 78), L t M 2 is equal to M 1 L V less the horizontal component of MJj v and the equation is: - 156 + 12 + L,Af 2 = . L,Af 2 = + 144.0; and M 2 L 2 = + 144.0. At point L 2 (see Fig. 79), L 2 M 3 is equal to the sum of the horizon- tal components of M^L 2 and m 2 L 2 ; that is, + L 2 M 3 - 144.0 - 96.0 = .-. L 2 M 3 = M 3 L 3 = +240.0. Fig. 77. Chord Stress in Baltimore Truss. In a similar manner, at point L 3 , the stress equation is: + L 3 M 4 - 240.0 - 48.0 = .-. L 3 M 4 = M t L 4 = +288.0. there results the At the upper panel point U l (see Fig. equation: % + Z/il7 2 + hor. comp. C7,m 2 + hor. comp. m l U 1 = 0; U,U 2 + 108.0 + 144 = 0; or, U,U 2 = -252.0. For the member U 2 U 3 (see Fig. 81), the equation is: 79 70 BRIDGE ENGINEERING + UJJ Z + U 2 U 3 - hor. comp. m 2 C/ 2 + hor. comp. U 2 m a = + 252.0 + U 2 U 3 - 12 + 60 = /. U 2 U 3 = -300.0. In a similar manner, by resolving the horizontal forces at U 3 , it will be seen that the action of m 3 U 3 will neutralize that of U 3 m t , us M, L . Fig. 78. Fig. 79. Bottom Chord Stresses in Baltimore Truss. they are equal and pull in opposite directions, and U 3 U 4 is equal to The live-load stresses in the chords, the end-post, and the sub- diagonals are all proportional to the dead-load stresses in the same Fig. 80. Fig. 81. Top Chord Stresses in Baltimore Truss. ratio as the live panel load is to the dead panel load. This ratio is 40 = 1.667. By reference to Fig. 70, it will be seen that the live- load stress in the sub-verticals is +40.0 for each one. The following stresses can now be determined : L m l = -220.5 X 1.667 = -367.5 ,t/, = -203.6 X 1.667 = -339.5 U,U 2 = -252.0 X f/ 2 C7, = -300.0 X U 3 U t = -300.0 X L L, = +156.0 X 1.667 = +260.0 L,L 2 = +144.0 X L 2 L 3 = +240.0 X L 3 L 4 .067 = -420.0 .667 = -500.0 .667 = -500.0 .667 = +240.0 .667 = +405.0 +288.0 X 1.667 = +481.0 80 BRIDGE ENGINEERING 71 m.L, = - 16.96 X 1.667 = - 28.28 m,U s = m 3 U 3 = +16.96 X 1.667 = +28.28 The vertical U^ will have its maximum live-load stress when points J/ t and L 1 are loaded, for these are the only loads which cause a stress in that member (see Fig. 76a). The equation is: - ~ - 40 + = 0, from which, t/jL, = +60.0. The maximum live-load stresses in Ujm 2 , U 2 m 3 , and U 3 m 4 are obtained in a manner exact- ly like that used in obtaining dead- load stress, only the live-load posi- tive shear is used. The stresses are: Stress in Lower Half of Main Diagonal of Baltimore Truss. U } m 2 X 0.707 + 188.5 = U 2 m a X 0.707 + 128.5 = U 3 m t X 0.707 + 80.0 = .'. U,m 2 = +2665 . . U 2 m a = +181 5 .-. U.ni = +113.1 In the determination of the maximum live-load stress in the lower halves of the main diagonals, ra^, ra^, and m 4 Z 4 , one of the peculiarities of this truss becomes apparent. A section being passed as in Fig. 82, the panel point ahead of the section, and all between the section and the right support, must be loaded. This of course produces a stress in ra 2 ?7 2 , and the vertical component of this enters the stress equation. The shear in the section a a under this load- ing is: V a _ a = +188.5 - 40 = +148.5; and the stress equation is: -m 2 L 2 X 0.707 + ~ + 148.5 = .-. m 2 L 2 = +238.0. If the truss had been loaded from the section to the right, there being no load on M 2 , no stress would result in m 2 U 2 , and the stress in 81 72 BRIDGE ENGINEERING would have been 157.2 0.707 = +222.2. In a similar manner, Fig. 83. Stress in Main Vertical of Baltimore Truss. by loading successively points M 3 and to the right, and M 4 and to the right, the stress equations of m 3 L 3 and m 4 L 4 are: -m 3 L 3 X 0.707 + -y + 128.5 - 40 = .'. m a L 3 = +153.3 -w 4 L 4 X 0.707 + ~ + 80.0 - 40 = .'. m 4 L 4 = + 84.8 The maximum live-load stresses in the main verticals occur when the panel points to the right of the section which cuts the member under considera- tion are loaded. There being no load at the end of the sub-vertical just to the left of the section, there will be no stress in the sub-diag- onal which the sec- tion cuts. The chords, of course, do not exert a vertical com- ponent; and so the only unknown term of the stress equation is the stress in the member itself. Fig. 83 shows how the section should be passed when U^L 2 is considered. The stress equation is: + U 2 L 2 + F a _ a = 0; + U 2 L 2 + 128.5 = 0; .'. U 2 L 2 = -128.5. In a similar manner, by passing a section cutting U 2 U 3 , m 3 U 3 , C/gZj, L 3 m v and loading M 4 and to the right, it is seen that the stress equation for U 3 L 3 is: + U a L a + 80.0 = .'. U 3 L 3 = -80.0 The components of m 3 U 3 and L 3 m 4 are zero, as can readily be proved by solving for them under this loading. Fig. 84 gives all the stresses, and they are written in order of dead load, live load, and maximum. 43. Other Trusses. The analysis of the foregoing trusses will enable one to solve any of the trusses of modern times. For the solution of the Whipple (sometimes called the "double-intersection Pratt") and others which are not mentioned in this text, the student BRIDGE ENGINEERING 73 is referred to the text- books of F. E. Tur- neaure and Mans- field Merriman. ENGINE LOADS 44. Use of En= gine Loads. It was formerly the custom for railroads to spec- ify that the engine to be used in computing the stresses in their bridges should be one of their own which was in actual use. The engines of differ- ent roads were usual- ly different both in regard to the weight on the various wheels and in regard to the number and spacing of the wheels. Of late years, consider- able progress has been made towards the adoption of a typical engine load- ing as standard. These typical engines (see Fig. 17, Article 25) vary only in re- gard to the weights on the wheels, the number and spacing of wheels being the same in all engines. 71 BRIDGE ENGINEERING Ti vO ci, < 3' O~ & The distance between wheels is an even number of feet, instead of an odd number of feet and inches and frac- tions thereof. For examples of load- ings which are in almost universal use, consult the specifications of Cooper or Waddell. The labor of computation of stresses when engine loads are used is consid- erably lessened by the use of the so- called engine diagrams. Fig. 85 gives a diagram which has been found very convenient. The first line at the top represents the bending moment of all the loads about the point to the right of it. All the loads are given in thou- sands of pounds, and all the moments are in thousands of pound-feet. The practice of writing results in thousands of pounds or, in case of moments, in thousands of pound-feet or pound- inches is to be recommended, as it saves the unnecessary labor of writing ciphers. Throughout this text this practice has been extensively followed, the stresses being written to the near- est ten pounds or one-hundred pounds as the case may be. For example, 6 433 may be written 6.43 or 6.4, the few pounds which are neglected mak- ing no appreciable difference in the design. The distances are in feet. As an example of the use of the first line at the top, suppose that it is de- sired to find the moment of all the loads to the left of a certain point when wheel 6 (the numbers of the wheels are placed inside of the circles 84 BRIDGE ENGINEERING 75 representing the wheels) is just over the point. The moment will be 1 640 000 pound-feet, which is obtained by reading off the 1 640 just to the right of the line through wheel 6. When using the first line for values at sections in the uniform load, the values given represent the moment of all wheel and uniform loads about the points in the line or section to the left of the value given. For example, if it is desired to find the moment about a point in line 2, it will be 19 304 000 pound-feet, the value 19 304 appearing to the right of the line. The line of figures below the wheels indicates the distances between any two wheels. The third line of figures indicates the distance from the first wheel to the wheel to the right. For instance, 37 is the distance from wheel 1 to wheel 7, The values in the fourth line indicate the sum total of all the loads to the left of the value given. For example, 245 signifies that the loads 1 to 15 inclusive weigh 245 000 pounds. The values in lines 5 and 6 are similar to those of lines 3 and 4, except that the starting point is at the head of the uniform load. For example, 40 in line 5, and 112 in line 6, indicate that it is 40 feet from the head of the uniform load to the wheel 12, and that wheels 18 to 13 inclusive weigh 112000 pounds. The values in lines 7 to 16 indicate the value of the moment of all the wheels from the zigzag line up to and including the one to the left or the right, according as the value is to the left or the right of the zigzag line. For example, 2 745, line 11, indicates that the moments of wheels 8 to 14 inclusive about the zigzag line just under wheel 15, is 2 745 000 pound-feet; or the value 1 704, line 14, shows that the moments of wheels 13 to 18 about the zigzag line just under wheel 12 is 1 704 000 pound-feet. When line 4 of figures is under the uniform load, the values refer to the vertical line to the right; 'thus 324 is the value of all loads to the left of line 3 about that line. For values of moments at points which fall in between wheels, or at positions in the uniform load where the value of the moment is not given, a very important principle of applied mechanics is used. It is: M a = M' + Wx +. , 85 76 BRIDGE ENGINEERING in which, M, = Moment at section desired; M ' = Value of moment at preceding vertical line; W = Sum total of all loads to the left of and at the point where M' is taken; x Distance from section under consideration to vertical line to which M' is referred; w = Uniform load on the distance x. Let it be desired, for example, to determine the moment at a a point c, 3 feet to the right of wheel 13. The position of the loads is given in Fig. 86. The moment is: M... = 7 668 + 212 X 3 = 7 668 + 636 = 8 304 = 8 304 000 pound- feet, there being no uni- form load. To illustrate the method when applied to points in the uniform load, assume the point to be 7 feet to the right of line 2. The po- sition is illustrated in Fig. 87. The moment is: eo 6708 eo 74' 7668 Jft.- 79' eo ese Fig. 86. Calculation of Moment at a Point under Engine Load. Mb-b = 19 304 + 304 X 7 + 7 " X 2 . = 21 481 = 21 481 000 pound-feet. The use of the moment diagram is now apparent. Reactions due to any position of the engines may be determined by dividing the span into the value obtained for the moment at the right end of the span. Likewise, if the moment of the reaction about any panel point is determined and from it the moment of the wheel loads about that same panel point are subtracted, then the result, divided by the height of the truss, will give the chord stress. For example, if the right end of an 8-panel 196-foot span truss, height 25 feet, came 7 feet to the right of the vertical line 2, then the moment at this point (see Fig. 87) would be 21481000, and the reaction would be 21 481 000 - 196 - 109 600. This position of the loads would cause the panel point L 6 to come 3 feet to the right of wheel 13. The moment of the reaction about L R is 109 600 X 6 X 24.5 = 16 111 200; 88 BRIDGE ENGINEERING 77 and the chord stress U 6 L 6 for this position of the engine is: 16111200-8304000 25 -31 2 000 pounds. In using the engine to determine the shear in any particular panel, it must be remembered that the shear is not the left reaction less all the loads to the left of the panel point on the right of the section, as the loads in the panel under consideration are carried on stringers, and these stringers transfer a portion of the loads to the 163 64 19304 eo 7ft. eo 2000 lt>s. per L 5' 10' 10' 109' 119' 304 Fig. 87. Calculation of Moment at Point under Uniform Load. panel point on the left of the panel, and a portion to the panel point on the right of the panel. Only that portion of the loads in the panel which is transferred to the left panel point should be subtracted from the reaction, as should all of the loads to the left of the panel under consideration. If, in a 6-panei 120-foot span Pratt truss, the wheel 6 comes at L 2 , the left reaction will be : Rl = 120 284 = 143 - 6 ; and the loads in the first two panels will be in position as indicated by Fig. 88, the wheel 3 being 1 foot to the right of point L r Let it be required to determine the shear in the panel L t L 2 when the loads are in this position. It will be the reaction 143.6 minus loads 1 and 2 and also that portion of the loads 3, 4, and 5 which will be trans- ferred by the stringers to -the point L r As the stringers are simple 87 BRIDGE ENGINEERING G G G -6 Fig. 89. Shear Diagram for Panel under Engine Load. beams, the amount transferred to L t will be the reaction of the stringer L t L 2 . Re- ferring to Fig. 89, the reaction is: R Li = (20 X 9 + 20 X 14 + 20 X 19) - 20 = 42.0 The shear in the second panel is now found to be: F 2 = 143.6 -(10 + 20 + 42.0) = +71.6 In the majority of cases where it is necessary to determine the shear in a panel, none of the loads will be in the panel to fhe left of the one under consid- eration. In this case the operation is somewhat simplified, as the engine dia- gram can be used directly. If the engine be placed so that the third wheel is at L 2 , wheel 16 will be just over the right support, and the left reaction will be : Ri = 12041 *- 120 = 100.3. As there are no wheel loads in the first panel, the amount to be sub- tracted from the reaction will be that proportion of the loads 1 and 2 which is transferred to Z^ and this (see Fig. 90) is 230 -f- 20 - 11.5. The shear in the second panel is then 100.3 - 11.5 = +88.8. From inspection of the resulting shear in the second panel when wheel 6 is at L 2 and when wheel 3 is at Z/ 2 , it is seen that different wheels at L 2 will give different shears in the panel to the left. Evi- dently there is some wheel which will give the greatest shear possible. The same is true of the relation between wheels and moments- The 88 BRIDGE ENGINEERING 79 next two articles are devoted to subject-matter which will enable. one to tell which of several wheels is the correct wheel at the point, without the necessity of solving for the shear each time every wheel is at the point. 45. Position of Wheel Loads for Maximum Shear. By methods of differential calculus, it can be proved that, for any system, either of wheel loads or wheel loads followed by a uniform load (see Fig. 91), the correct wheel that should be at the panel point 6 in order to Fig. 90. Determination of Shear in Panel under Engine Load. give a very great or maximum shear in the panel a b, is such a W W wheel that the quantity Q = - G is positive when q = - (G + -P) is negative. In these equations, W = Total load on the truss; m = Number of panels in the truss; G = Load in panel under consideration; and P = Load at panel points on right of panel. If a load is directly over the panel point a, it is not to be included in the weight G; neither is P included in the weight G. If a wheel load should come directly over the right end of the truss, it should not be considered in the quantity W. The only way to determine which wheel is the correct one, is to try wheel 1, then wheel 2, and so on, until the wheel or wheels are reached that will give the Q and q signs of an opposite character. 89 80 BRIDGE ENGINEERING The process should not be stopped there, but the next succeeding wheels should be tried until Q and q again have the same sign. As an example, let it be required to determine the position of the wheel loads to produce the maximum positive shears in a 6-panel 120-foot Pratt truss. This work should be arranged in tabular form, and Table V is found to be convenient. TABLE V Determination of Position of Wheel Loads for Maximum Shear (m = 6) PANEL POINT WHEEL AT POINT i o ci r-t CO ! + (M ' 05 + _f p" 00 i- 23 d + + 1C j:' ' + 4 p" -dH CD CO OO CO GO CO;H + + 1 o co gs + + VERTICALS H? P" t" CO i-H O coco'-cd oo i- c^ * 1 1 J P" t~ 1C O ^ 6 1 1 Wheel 11 is 4 ft. from L 8 Wheel 9 at L . U,U 2 u t u, -168.2 -183.8 114 114 0.724 0.724 -121.8 -133.2 j Wheel 13 at L 4 . / 5 ft. of uniform load on bridge. / 15 ft. of uniform load on bridge. \ Wheel 1 off bridge. L A L 2 L 3 + 107.5 + 168.2 113 114 0.727 0.724 + 78.2 + 121.8 Same loading as for L U l Same loading as for U y U 3 In the case of U^, it should be noted that only the wheels 10 to 16 inclusive cause the stress (see Fig. 94), and that the loaded length is the distance from wheel 10 to wheel 16. Some specifications do not call for impact stresses. The unit- stresses in these specifications are made low, and the sections designed are large enough to withstand the additional stresses due to impact. In cases where the impact stresses are required, they must be con- sidered in computing the maximum and minimum stresses. 49. Snow=Load Stresses. In some localities the snowfall is considerable, and its weight should be taken into account in com- puting stresses. This should be done by considering it as an addi- tional dead load of 15 pounds per square foot of floor surface for every foot of snowfall. As it covers the. entire floor surface, the stresses will be proportional to the dead-load stresses. Also it is evident that the snow load should not be taken into account in railroad bridges unless they have solid floors, as most of it falls through the open spaces between ties and stringers. As an example, let it be required to determine the snow-load stresses in a member of a highway bridge, the dead-load stress in the member being +84.0, the dead panel load being 12 000 pounds, and 103 94 BRIDGE ENGINEERING the snow bemg l\ feet deep on the roadway, which is 14 feet wide. The snow panel load is: - (14 X 15 X 1 X 20) = 3 150 pounds. In the above equation, 14 is the width of roadway; 15 is the weight in pounds of one square foot of snow one foot deep; and 20 is the length of one panel. One-half of the weight of snow must be taken, as half is carried b each truss. The snow-load stress is then: In like manner, all snow-load stresses can be computed. Most of the standard specifications which have been published do not specify snow loads; and in fact it is not customary to include the snow load in any designs except those for bridges in extreme northern latitudes. It is hardly probable that the greatest load will come upon a country bridge when it is covered with snow. Also, in cities, the sidewalks are cleaned of snow; and so is the roadway if the city is of large size. WIND=LOAD EFFECTS 50. Top Lateral System Through=Bridges. The unit-loads for this system are given in Article 26. Common practice is to take 150 pounds per linear foot of top chord, the end-post being con- sidered part of the top chord in this computation. In many of the longer-span modern bridges, the diagonals of this system are designed to take either tension or compression ; but in the majority of the shorter spans, 200 feet and under, while generally consisting of angles or other stiff shapes, they are designed to take tension only. The verticals or top lateral struts take compression. This combination of tension diagonals and compression verticals makes the so-called Pratt system of webbing; and indeed the lateral systems, both top and bottom, are Pratt trusses in a horizontal posi- tion. Fig. 95 shows the side elevation of the truss of Article 47, and also the top and bottom laterals. The diagonals shown in full lines act when the wind is right, and those shown by dotted lines act when the wind is left. Wind right indicates that the wind is blow- ing from the right hand when a person stands facing the righ' end of 104 96 BRIDGE ENGINEERING the bridge. Wind left indicates that the wind blows from a person's left when standing as above described. The wind load of 150 pounds is di- vided between the two trusses, this being exact enough for practical pur- poses; for, by actual experiment, the difference between the readings of wind-pressure gauges placed at points opposite each other in the top chords of a through-bridge was only from 8 to 10 per cent. The problem, then, is one of a deck Pratt truss with a dead panel load of 150 X 20 = 3.0 divided between the two chords. Fig. 96 shows the distri- bution of loads and the reaction, it being considered that the portal brac- ings and the end -posts (see Fig. 95) are stiff enough to distribute the reaction equally between the bearing points L , L ', L 6 , L 9 '. Each panel load is indicated by an arrow, and is equal to 3.0 ^ 2 = 1.5. The re- action at each of the points L , L/, L , and L G ' is 10 X 1.5 + 4 = 3.75. The truss being symmetrical, the stresses in like members on each side of the center will be the same. The shears in the top system are: V 1 == +2 X 3.75 - 2 X 1.5 = +4.5 y a - a = +2 X 3.75 - 3 X 1.5 = +3.0 V 2 = +2 X 3.75 - 4 X 1.5 = +1.5 and the secant ^ is (17 2 + 20 2 )* -=- 17 = 1.544. The stresses in the diag- onals are: U 1 'U 2 = +1.544 X 4.5 = +6.95 U 2 'U Z = +1.544 X 1.5 = +2.32. 106 BRIDGE ENGINEERING 97 The vertical U 2 'U 2 = 3.0; and by passing a section 6 6 around U 9 f , the stress in U 3 'U 3 is found to be 1.5. In obtaining the chord stresses in this system, the case is the same as if the reactions were applied at U^ and U b r , as the portal and end-posts are not in the same plane as the lateral system. The tan- gent method is the simplest to use in this case. The tangent is 20 -f- 17 - 1.176, and the stresses (see Fig. 95) are: ff/CY = - 4.5 X 1.176 = -5.29 U*' U 3 r = -(4.5 + 1.5) X 1.176 = -7.06 17,17, A U 2 U 3 = -USU,' = +5.29 Fig. 97 is a diagram with the stresses caused by wind right and u l W.R.- 529 W.L.+ W.R.-7.06 W.L.+ 5E9 v U, U e W.R.* Fig. 97. Wind Stress Diagram of Pratt Truss of Fig. 95. W.R.+ 59 W.L. - 7.O6 wind left indicated thereon. The stresses for wind left can easily be written by inspection. 51. Bottom Lateral Bracing, Through=Bridges. Fig. 95 shows the lower lateral system with the panel points loaded with the fixed or dead wind load. In this case it is all taken as acting on one side, it being assumed that the floor system protects the leeward truss. The problem then becomes that of determining the stresses in a deck Pratt truss of 6 panels of 20 feet each, the height being 1 7 feet. When wind is right, the members shown by broken lines in Fig. 95 do not act. The fixed wind load (Article 26) is 150 pounds per linear foot of chord. The panel load will be the same as before, 3.0, but all will be on one chord. The shears are: V l = 2J X 3.0 = +7.5 Fa-a = +7.5 F 2 = +7.5 - 3.0 = +4.5 Fb-b = +4.5 F 3 = +7.5 - 2 X 3.0 = +1.5 107 98 BRIDGE ENGINEERING The secant being 1.544, as previously computed, the web stresses are: L 'L, = +7.5 X 1.544 = +11.60 L/L, = -7.5 L/L., = +4.5 X 1.544 = + 6.95 L/L, = -4.5 L 2 'L 3 = +1.5 X 1-544 = + 2.32 L/L, = -3.0 The stress in L 3 'L 3 is determined by passing section c c and resolving the vertical forces at L 3 f (see Fig. 95). By using the tangent method, the chord stresses are computed as follows: L 'L,' = -7.5 X 1.176 = -8.82 L/L/ = -(7.5 + 4.5) X 1.176 = -14.12" L,'L 3 ' = -(7.5 + 4.5 + 1.5) X 1.176 = -15.88 L,L 2 . -L 'L/ = -(-8.82) = +8.82 L,L 3 = - W = -(-14.12) = +14.12 The wind load acting on the train is 450 pounds per linear foot. It is evident that the train may cover the span either partially or entirely, and therefore its action on the lower lateral system is the same as if it were stressed by a live load of 450 pounds per linear foot of truss. The live panel load is 450 X 20 = 9.0. The maximum live- load reaction is 5 X 9.0 + 2 = 22.5, and the positive live-load shears are: V l = +22.5 q A F, = (1 + 2 + 3 + 4) ^ = + 15.0 Q O V,= (1+2 + 3)^ = +9.0 It is unnecessary to go further than the center, as only the maximum stresses are required in the members: The web stresses are com- puted as given below : L 'L, = +22.5 X 1.544 = +34.75 L/L, = -22.5 L/L 2 = +15.0 X 1.544 = +23.15 L/L 2 = -15.0 L 2 'L 3 = + 9.0 X 1.544 = +13.91 L 3 'L 3 = - 9.0 The maximum chord stresses due to this load of 450 pounds per linear foot of train, occur when the train covers the entire span; and they are directly proportional to the stresses produced by the fixed load, in the same ratio as the live panel load is to the fixed panel load. 9 This ratio is ^- = 3.0. The chord stresses, therefore, are: o.U 108 BRIDGE ENGINEERING 99 L/L ' = - 8.82 x 3 = Z-/Q *^\ -26.46 " ^ -~^ L/L/ = -14.12 X + -42.36 / \ ~*~~ L/L 3 ' = -15.88 X 3 = \\ -47.64 /\ L,L a = + 26.46 + 8.82 X 3 "| L 2 L 3 - + 14.12 X 3 = /^o + 42.36 Table X, Article 53, gives the stresses in the top and bottom lateral systems for wind right and wind left. 52. Overturning Ef= feet of Wind on Truss. When the wind blows on the top chord, it tends to overturn the truss. As the truss is held down by its own weight, the action of the wind does not overturn it, but causes the dead-load reaction on the windward side to be less and that on the lee- ward side to increase by a like amount. The amount is V = -^- X = the sum -=- , where b of all .the wind panel loads, h = the height of the truss, and b = the distance center to center of. trusses. The effect upon the leeward truss is the same as if two ver- 109 100 BRIDGE ENGINEERING tical loads, each equal to V and acting downward, were placed at the hips U l and U 5 (see Fig. 98). The effect on the windward truss is the same as if two vertical loads, each equal to V and acting upward, were placed at the hips C// and U 6 f . The stresses in the leeward truss will now be worked out. The stresses in the windward truss are the same, but with opposite signs. 10 X 1.5 25 2 X 17 The truss is that of Article 47. Here V = 11 000. Fig. 99 shows the truss with the loads in the correct position, the Fig. 99. Truss under Wind Loads. reactions each being 11.00. V 1 = +11.00, and V 2 = + 11.00- 11.00 = 0. The shears in the 2d, 3d, 4th, and 5th panels are also zero. As the shear in these panels is zero, the stress in the diagonals and vertical posts is zero X secant $ = zero. The stress in the hip verticals U^ and U b L. is zero, as there are no loads at L t and L y The stress in the end-post is -11.00 X 1.28 - -14.08. Taking the center of moments at TJ V the stress equation of LJL^ = LJL 2 is: -1^2 X 25 + 11.00 X 20 - 0; whence L t L 2 = +8.8. The stress in all the lower chord members will be found to be + 8.8. By summing the horizontal forces at the section a a, noting that, as UjL 2 is zero, its component is also zero, there results: +L^L 2 + UJJ^ = 0; whence U t U 2 = -i t L 2 - - ( + 8.80) = -8.80. This is also the stress in all members of the top chord. It is now seen that the overturning effect of the wind on the truss causes stresses only in the end-posts and chords. The wind on the lower chord causes no overturning effect, as it is transferred directly to the abutments. 53. Overturning Effect of Wind on Train. The wind blowing upon the train tends to overturn it, and in so doing the pressure on 110 BRIDGE ENGINEERING 101 the leeward stringer is increased and that on the windward stringer decreased by the same amount. This difference in pressures is transferred to the floor-beam and then to the panel points (see Fig. 100), where its value is: L = W X (8.5 + a) where = Panel load due to wind on train; 8.5= A constant established by the Specifications (see Article 26, p. 15); a = Distance from base of rail to center line of lower chord. It may be taken as 3 feet in most cases, as this is approximately the usual depth of floor). b = Distance center to cen- ter of trusses. 'For the case in hand, W = 20 X 450 = 9 000. Therefore, 9 000 X (8.5 + 3) L = 17 Illustrating Overturning Effect of Wind on Train. = 6 090 pounds. The action of the wind in tending to overturn the train is the same as if the truss were under a live panel loading of L, the panel load L acting upward on the windward and downward on the leeward truss. The chord stresses due to this will be proportional to the dead- load stresses in the same ratio as this panel load L is to the dead panel load. For the truss of Article 47, this ratio is ^ = 0.303, and ^U UUU the chord stresses caused by the overturning effect of the wind on the train (see Table VII) are: tf,C7 2 = -64.0 X 0.303 = -19.39 U 2 U 3 = -72.0 X 0.303 = -21.82 L L, = L,L 2 = +40.0 X 0.303 = +12.12 L 2 L 3 = +64.0 X 0.303 = +19.39 The stress in f/,L, is +6.09, the panel load at L r The maximum positive shears are : V l = ^^(1 +2 + 3 + 4 + 5)= +15.22 111 102 BRIDGE ENGINEERING .09 .09 34-4)= +10.15 (1 + 2 + 3) = +6.09 = i(i + 2) = +3.05 D and the maximum web stresses are found to be: L U t = -15.22 X 1.28 = -19.50 U,L 2 = +10.15 X 1.28 = +13.00 U 2 L 3 = + 6.09 X 1.28 = + 7.80 U 3 L t = + 3.05 X 1.28 = + 3.91 U 2 L 2 = - 6.09 U 3 L 3 = - 3.05 It is unnecessary to compute the shears further than one panel past the middle of the span, as only the maximum stresses are usually required. The wind stresses from various causes are grouped together and given in Table X. From Table X it is seen that large wind stresses occur in some of the members. Most specifications require that the stresses due to wind shall be neglected in the design unless they exceed 25 per cent of the sum of the dead-load and live-load stresses. The subject of wind stresses does not ordinarily receive the con- sideration it should have; in fact, it appears to be common practice, in the case of spans up to 200 feet, to neglect the action of the wind in all members of the bridge except the top and bottom lateral diagonals, the top struts, the portal, and the bending effect in the end-post. For the last two effects mentioned, see the next succeeding article. 54. Portals and Sway Bracing. One- half of the wind on the top chord is trans- ferred to the hips U l f U l and U e 'U 6 . From there it is carried to the abutments by means of the portal bracing and the end- posts. Various styles of portal bracing are in use, but few are so easily analyzed and constructed as that of Fig. 101. This form Fig. 101. Style of Portal Bracing in Common Use on Spans up to 250 Feet. 112 BRIDGE ENGINEERING 103 TABLE X Wind Stresses in Pratt Truss WEB MEMBERS OVERTURNING L a U, C/.L, [7,1,3 U,L, [/,, r/,L, U 3 L, Wind Right ! on Truss -14.08 o on Train -19.50 -13.00 + 7.80 + 3.91 + 6.09 -6.09 -3.05 Wind Left on Truss + 14.08 on Train + 19.50 -13.00 : -7.80 -3.91 -6.09 + 6.09 + 3.05 Maximum + Stress + 33 . 58 + 13.00 ; +7.80 + 3.91 + 6.09 + 6.09 + 3.05 Maximum - Stress -33.58 -13.00 : -7.80 -3.91 -6.09 -6.09 -3.05 CHORDS MEMBER L L t L,L, L,L 3 U,U 2 U,U 3 Direct Wind Right +8.82 + 14.12 + 5.29 , +26.46 + 42.36 Wind Left - 8.82 -14.12 -15.88 - 5.29 - 7.06 -26.46 -42.36 -47.64 Overturning Truss Wind Right + 8.80 + 8.80 + 8.80 - 8.80 - 8.80 Wind Left - 8.80 - 8.80 - 8.80 + 8.80 + 8.80 Overturning Train Wind Right + 12.12 + 12.12 + 19.39 -19.39 -21.82 Wind Left -12.12 -12.12 -19.39 + 19.39 + 21.82 Maximum + Stress + 20.92 + 56 . 20 + 84.67 + 28.19 + 35.91 Maximum - Stress -56.20 -77.40 -91.71 -33.48 -37.68 LATERAL SYSTEMS MEMBER Ui'U, U,'U, UjU, U 3 'U 3 L 'L l L t 'L. L,'L 3 Wind Right on Truss + 6.95 + 2.32 -3.0 -1.5 + 11.60 + 6.95 + 2.32 on Train + 34.75 + 23.15 + 13.91 Wind Left on Truss -3.0 -1.5 on Train Maximum + 6.95 + 2.32 -3.0 -1.5 + 46.35 + 30 ..10 + 16.23 The stresses -in L,'L,, L,'L,, and L 3 ' L 3 are not given in the above table. These members are the floor-beams, and the small stress due to wind is neglected in their design. 113 104 BRIDGE ENGINEERING of portal is at present being used almost universally on all spans up to 250 feet. Let it be required to analyze a portal of this form, all the dis- tances being as indicated in Fig. 101 ; and let : w = Wind panel load of upper chord; m' = Number of panels in upper chord; Then, P = (m r - l)w; and, V = (P + w) + w ; also, H l = H 2 = { (P + w) + w J- -=- 2. The stress in BC, the center of moments being at D, is: Q \ 2 a \ The stress in AB, the center of moments being at E, is: , wa + HJ . I SAB = + - = +w + H, a a For the stress in BD, the center of moments is taken at C, and the perpendicular distance c to BD is determined. The stress in BD, then, is: 2 c The stress in BE is: S BE = - //, A- It must be remembered that A, is not the height of the truss, but is the length of the end-post from L to U v For the truss of Article 47, w = 1.5; m' = 4; and P = 4.5. The value h, = (20 2 + 25 2 )* - 32.0 feet. The distance a must be so chosen that BD will not interfere with engines or other traffic which passes through the bridge. It will be assumed as 5 feet in this case. Then V = (4.5 + 1.5 + 1.5) ^ = 14.08; and H l - H 2 7 ' 5 0-7* = = 3.75; whence, SBC = - ( 6.0 + 3.75 X ^-\ = - 26.25 114 BRIDGE ENGINEERING 105 S AB = 1.5 + 3.75 X ~ o + 21.75 2 + 5.0 2 = 9.85. The distance BD = ^BC~ + CD = Then, from similar triangles DCB and DFC, is obtained the proportion : CF _ BC . CD~BD' C= = 4. 3 feet; and 32 SBD = +3.75 X -. = +27.90 4 .6 S BB = -3.75 X = -27.90 When the wind blows from the other side, the stresses in the diagonals are reversed, and those in the top are transposed. The members shown by broken lines take no stress. When the wind blows, the end-posts tend to bend as shown in Fig. 102. This is with- \ Fig. 102. Illustrating Tendency of End-Posts to Bend under Wind Load. Fig. 103. Bending Tendency when End-Posts are Fixed at Lower End. stood by the cross-section of the post at the points E and D. The bending moment caused at these points by the wind is H i X I and H 2 X /. For the truss under consideration, MD = ME = 3.75 X 27 X 12 = 1 215 000 Ib.-ins. If the posts are fixed at the lower end, then they will tend to bend as shown in Fig. 103, the post resisting the bending at two points D and d. The section at each point withstands in this case only half of the moment just computed, or 1 215 000 + 2 = 607 500 Ib.-ins. A further discussion of this will be given in Part II, on "Bridge Design." 115 106 BRIDGE ENGINEERING Various forms of sway bracing are used to connect the inter- mediate posts and thus stiffen the cross-section of the bridge at those points. The form of portal just given is often used, as is also the form shown in Fig. 104. Here h is the height of the truss. The braces BD are called knee-braces. Here w is the wind panel load of the top chord, and 2wh 2w , = _ (wa + H 2 l) -=- a , c , = + ( W + Hi ) a The stress in B'B is the direct compression due to wind right or Fig. 104. A Type of Portal and Sway left, and differs in accordance Bracing in Frequent Use. . , , ... c , with the position of the top strut. There is also a bending moment at B r and B, which is: Ms, = M K = -V'g + HJi. The bending moment at D and D' is equal to H 2 l or Hj, -r- 2, according to whether or not the lower ends of the posts are fixed. The determination of the stresses for the truss of Article 47 is left to the student. When the wind is from the other side of the truss, the signs of the stresses in the knee-braces and the members C'B' and CB are reversed. 55. Final Stresses. The class of stresses which go to make up the maximum or minimum for which the member is designed r is .determined by the specifications used. The dead-load and live-load stresses are always included, and then those due to impact and wind should be added if required. In computing the maximum stresses, the algebraic sum should always be used. In a large majority of cases, all stresses which go to make up the maximum have the 116 BRIDGE ENGINEERING 107 same sign, but some exceptions have been noted, as in the middle vertical of a Pratt or Howe truss. The minimum stresses ure, with rare exceptions, obtained by combining stresses with signs of opposite character. GIRDER SPANS 56. Moments and Shears in Floor=Beams. In any bridge the floor-beam acts as a support for either the joists or stringers, and the moments and shears occurring in it are due to the loads which come on the joists or stringers. In a highway bridge the joists are spaced so closely that the load which they transmit to the floor-beams may be considered as uniformly distributed, providing the live load is a uniform load, in which case, _ (2P L + Pn) X panel length in inches 8 V = (2P L + P D ) + 2, where M = Maximum moment in pound-inches; V = Maximum shear; PL = Live panel load; PD = Weight of stringers and floor material in one panel. It will be seen that these formulae are those for the maximum moment and shear in a uniformly loaded beam, the total load being 2P L +P D . As an example, let it be required to determine the maximum moment and shear in the floor-beam of a highway bridge whose panels are 20 feet long, and trusses 16 feet center to center, the live load being 100 pounds per square foot of floor surface, the flooring weighing 10 pounds per square foot, and there being 5 lines of joists weighing 15 pounds per linear foot, and 2 lines of joists weighing 8 pounds per linear foot. P L = 1JL x 20 X 100 = 16 000 pounds. PD = 5 X 20 X 15 + 2 X 20 X 8 + 16 X 20 X 10 = 5 020 pounds. Therefore, (2 X 16000 + 5 020 } 20 X 12 ~T~~ = 1 110 600 pound-inches at center of floor-beam. V = (2 X 16 000 + 5 020) -H 2 = 18 510 pounds at ends of floor-beam. 117 108 BRIDGE ENGINEERING In the case of a single-track railroad bridge, there are only two stringers upon which the weight of the track, the engine, and the train is supported. These join the floor-beam at points equally distant from the center of same. The weight of the ties, rails, and fastenings is usually taken at 400 pounds per linear foot of one track. As regards the live load, the proposition reduces itself to placing the wheel loads so that the sum of the reactions of stringers in the adjacent panels will be a maximum on the floor- beam under consideration. This is discussed in Article 47, page 87 (see Fig. 94). In determining the values of the maximum moment and shear in the floor-beam, the case is that of a beam symmetrically loaded with two equal concentrated loads. Each load is equal to the dead weight of one stringer, one-half the track weight in one panel, and the maxi- mum sum total of the reactions due to the wheel loads on the stringers in adjacent panels which meet at that point. This latter quantity is called the floor-beam reaction. For a general arrangement of the loads, see Fig. 105. The distance a has become standard for single-track spans, and is 6 feet 6 inches. Let it be required to determine the maximum shears and moments in the floor-beam of the truss of Article 47. The weight of the stringer may be obtained by the formula of Table II, and is: Stringer = 20 (123.5 + 10 X 20) -f- 2 = 3 200 pounds. The weight of one-half of the ties, rails, etc., in one panel is: i Track = (400 X 20) + 2 = 4 000 pounds. The weight that comes from the engine wheels is given in Article 47, page 87 (see Fig. 94), and is 65.55. Each load is therefore the sum of all the above weights, as follows: 3 200 + 4 000 + 65 550 = 73 750. The maximum shear (see Fig. 105) is seen to be 73 750 pounds; and the maximum moment occurs at C and D, and is: p 1 Kl r- C/ /-51m 1 qers' i D Floor 3eam C. a->. For the engine under consideration, a = 8 feet, and P = 10 000 pounds, and the equation reduces to : 131 122 BRIDGE ENGINEERING which is to say that when the load on the girder is greater than 1 times the span, then wheel 2 should be placed at the section in order to give the maximum shear. For loading E 40, the following is true: For all' sections up to and including the center of all spans, place wheel 2 at the section to give the maximum shear. In Fig. Ill it is immaterial whether or not any additional loads come on the span at the right end when the loads move forward the Fig. 111. Beam of Fig. 110 with Loads Moved Forward. distance a, as they would only tend to increase the left reaction and therefore the shear P 7/ ' b _ b . If the relation deduced is true for the case when no extra loads come on at the right end, it will be true when they do. The live-load shears at the left end and at the tenth-points, wheel 2 being at the section in all cases, are computed from the gen- eral formula, which is: V = R - 2 P, in which, R = Left reaction; ~ P = All loads to left of section, and is equal to 10 000 pounds for all sections except the end of the girder. The computations and results can be conveniently placed in tabular form, and are given in Table XVIII. In order to illustrate the use of the relation W> 1 ^ I, let point 3 in the above span be taken. Place wheel 2 at point 3; then, as wheels 1 to 13 are on the girder, the total weight W is 212. As I = 100, \\l = 125. Therefore, as 212 is greater than 125, wheel 2 is the correct wheel. 132 BRIDGE ENGINEERING 123 TABLE XVIII Maximum Shears in a Deck Plate-Oirder g REACTION EQUATION R !,- V REMARKS ^ 1 (13 904 + 4X274) -100 150.00 150.00 Wheel 18, 4 ft. from rt. end 1 2 (12041+5X258) + 100 10816 + 100 133.31 10 108.16! 10 123. 31 1 Wheel 98.16 Wheel 16, 5 ft. from rt. end 15 at right end 3 ( 7 668 + 4X212) -100 85.16; 10 75.16 Wheel 13, 4 ft. from rt. end 4 ( 5 848 + 4X172) -100 65.36 10 55.36 Wheel 1 1, 4 ft. from rt. end 5 ( 4 632 + 2X152) -100 49.36 10 39.36! Wheel 10, 2 ft. from rt. end The curyes of maximum live-load moments and shears are shown in Fig. 112. They should always be drawn. From them the shear or moment at any desired section can be determined. For exam- ple, let it be desired to determine the maximum live-load shear and moment at a point 24 feet from the left end of the girder. By drawing the ordinate, shown by a broken line in Fig. 112, and scaling, the following values are found : F 24 = 88 000 pounds; A/ 24 = 2 440 000 pound-feet. A similar set of curves for the dead-load shears and moments should be made. The set for the deck plate-girder in hand is shown in Fig. 113. These are easily constructed by laying off the max- imum values of the shear at the Fig. 112. Curves of Maximum Live-Load Moments and Shears. end, and the maximum value of the moment at the center. The shear curve is a straight line from the end to the center, while the moment curve is a parabola from the center to the end. The stresses in the lateral systems of plate-girders are computed in a manner the same as that employed for the lateral systems of trusses, the unit-load being taken according to the specifications used. 133 124 BRIDGE ENGINEERING 59. Stresses in Plate=Qirders. The stresses in plate-girders are treated in the Instruction Paper on Steel Construction, Part IV, pages 251 to 263, and the student is referred to this treatise for infor- mation regarding this subject. The stress in the flange is seen to depend upon the distance from center of gravity to center of gravity. This distance, in turn, depends Pounds Fig. 113. Curves of Dead-Load Shears and Moments in a 100-Foot Span Deck Plate-Girder. upon the depth of the girder. Certain approximate rules have been proposed in order to determine this, but the following formula will give the width of the web plate in accordance with best modern practice : 7 in which . 005 I + . 543 ' d = Width of the web plate, in the even inch; I = Span, in feet. For example, let it be required to determine the width of the web plate of a plate-girder of 80-foot span center to center of end bearings. d = 0.005X80 + 0.543 = 0^4 = 85 ' 2 (say 86) inches ' If the resultant value had been 85 inches, the width would have been taken as either 84 or 86. The reason for this is that the wide plates kept in stock at the mills are usually the even inch in width and can therefore -be procured more quickly than if odd -inch widths were ordered, in which case the purchaser would be forced to wait until 134 BRIDGE ENGINEERING 125 they were rolled often a period of several months. The distance back to back of flange angles, the so-called depth of girder, is one-half inch more than the width of the web. This is due to the fact that each pair of flange angles extend one-fourth inch beyond the edge of the web plate, so as to keep any small irregularities caused on the edge of the web plate by the rolling, from extending beyond the backs of the angles. EXERCISES AND PROBLEMS 1. Determine the maximum positive shears in the first six panels of a 9-panel 114-foot Pratt truss, the live panel load being 8.0. Use the exact and also the conventional method. ANSWER: x XACT HEARS N I RS F, 16 + 32.00 + 32.00 V 2 14 + 24.54 + 24.90 V 3 12 + 18.05 + 18.70 V 4 10 + 12.48 + 13.35 V, 8 + 7.85 + 8.90 vl 6 + 4.50 ' + 5.34 2. Find the maximum and minimum stresses in L,C/ 2 and U 3 L 3 of an 8-panel 160-foot through Warren truss. Height 20 ft.; dead panel load 10.00, all on lower chord; live panel load 12.00. ANSWER: In LJJ 2 : d. 1., -28.00; 1.1., - 35. 30 and +1.68; max., -63. 30; min., -26.32. In t/ 3 L 3 : d. 1., +16.80; l.l., + 25. 20 and -5. 04; max., +42. 00; min., + 11.76. 3. In the truss of Problem 2, determine the maximum stress in L 2 L 3 by the method of moments, and also by the tangent method. ANSWER: d. 1. = +67.50; 1.1. = +81.00; max. - +148.50. 4. Determine the dead-load stresses in the members U 2 L 2 and L 4 U 6 of a 9-panel 180-foot through Warren truss. Height is 24 feet; dead panel load is 10.0, one-third being at each panel point of the upper chord, and two- thirds being at each panel point of the lower chord. ANSWER: U^L, = +30.60; LfJ. = -1 .80. 5. Determine the stress in the counter of a through Howe truss of 8 panels and 160-foot span. Height is 30 ft.; dead panel load, 9.6; live panel load; 11.5. ANSWER: 4.59. 135 126 BRIDGE ENGINEERING 6. In the truss of Problem 5, determine the maximum and minimum stress in U 2 L 2 , L 2 U 3 , and L 3 U 4 . ANSWER: V,L, w. wr. d. 1. + 20 . 80 -17.30 - 5.76 1.1. + 30.30 -25.90 -17.30 1.1. -14.40 +5.18 0.00 Max. + 15.10 -43.20 -23.06 Min. + 19.36 -21.20 0.00 6 at IO'= 60' Fig. 114. Deck Parabolic Bowstring Truss. 7. In the deck parabolic bowstring truss of Fig. 114, determine the maximum stress in L,L 2 , L,E/ 2 , and U 3 L 3 . The dead panel load is 4.0, all on upper chord; and the live panel load, 20.0. ANSWER: L^ = +201 .9; LJj\ = +21 .8; U 3 L 3 = 33.6. L, 6 at 1 5= 90' Fig. 115. Through Bowstring Truss. 8. In the through bowstring truss of Fig. 115, determine the maximum stress in U^ and LjC7 2 , the dead panel load being 5.0, and the live panel load 15.0. ANSWER: U^ = +33.50; LJJ 2 = +38.0. 136 BRIDGE ENGINEERING 127 9. Determine the maximum and minimum stresses in the members _Z7,L,, t/,L 2 , t/,L 2 , and U 3 L 3 of a 7-panel 175-foot through Pratt truss 30 feet high. Dead panel load is 10.0, all on lower chord; live panel load is 15.0. ANSWER : V, ** W! U,L. d. 1. +10.0 + 26 . 00 -10.00 0.00 1. 1. + 15.0 + 41.70 -21.40 -12.85 1.1. 0.0 - 2.78 + 6.42 0.00 Max. + 25.0 + 67.70 -31.40 -12.85 Min. + 10.0 + 23.22 - 3.58 0.00 10. Determine the maximum and minimum stresses in the members Ujn v m 3 L 3 , U 2 L 2 , and m 2 U 2 of the deck Baltimore truss shown in Fig. 116. Dead panel load, 30 000 Ibs.; live panel load, 50 000 Ibs. One-third of dead panel load is applied at the lower ends of all the verticals. ANSWER : CT.m, m 3 L 3 L\L 3 m,U, d. 1. 1. 1. 1.1. + 19058 + 333 . 5 - 15.1 + 84.8 + 191.5 - 50.5 -110.0 -211.0 + 10.7 + 21.2 + 56.6 0.0 Max. Min. + 524 . 3 + 175.7 + 276.3 + 34.3 -321.0 - 99.3 + 56 . 6 + 21.2 11. In the truss of Problem 10, determine the maximum stress in M 2 U 2 and L 3 L 4 . , ANSWER: M 2 7 2 = -840.0; L 3 L 4 = +960.0. 12. Determine the position of the wheel loads of Cooper's E 40 loading to produce the maximum positive live-load shears in the panels of a 7-panel 175-foot Pratt truss. ANSWER : L v wheel 4 ; L 2 , wheels 3 and 4 ; L 3 , wheel 3 ; L 4 , wheel 3; L 5 , wheel 2; L v wheel 2. 13. Determine the maximum positive live-load shears for the truss of Problem 12. ANSWER: V l = 192.8; F 2 = 137.8; F 3 = 90.8; F 4 - 52.6; F 6 =25.0; F.= 6.8. 137 128 BRIDGE ENGINEERING 14. Determine the position of the wheel loads of Cooper's E 40 loading to produce max- imum moments at the panel points of the truss of Problem 12. ANSWER; L v wheel 4; L v wheel 7; L 3 , wheels 11 and 12; L 4 , wheels 13 and 14. 15. Determine the maximum moments at the panel points of the truss of Problem 12. Loading, Cooper's E 40. ANSWER: M t 4820000; M 2 = 7745000; M 3 = 9192000; M 4 = 9 082 000, all in pound-feet. 1G. Compute the maximum live-load web stresses in the truss of Problem 12, the height being 32 feet. Loading, E. 40. 17. Compute the maximum live-load chord stresses in the truss of Problem 12, the height being 32 feet. Loading, E 40. 18. Compute the impact stresses for all members of the truss of Problem 12. 19. Determine the maximum live-load shears at the tenth-points of a 65-foot span deck plate-girder. Loading, E 40. ANSWER: F - 103.0; V l = 86; F 2 = 69.7; F 3 = 54.5; F 4 - 40.8; F 5 = 28.4. 20. Compute the shear due to impact in the girder of Problem 19. ANSWER: F = 84.7; V\ = 71.5; F 2 -58.8; F 3 = 47.0; F 4 = 35.6; F 5 = 25.4. 21. Compute the maximum live-load moments at the tenth-points of the girder of Problem 19. Loading, Cooper's E 40. 138 BRIDGE ENGINEERING 129 ANSWER: POINT WHEEL MOMENT IMPACT MOMENT 1 2 6540 5520 2 2 11 320 9520 3 3 14860 12500 4 4 16850 14050 5 4 16860 14530 1 . 45' from center 4 16920 14650 All moments are in thousands of pound-inches. 139 BRIDGE ENGINEERING PART II BRIDGE DESIGN 60. General Economic Considerations. The prime considera- tion which influences the decision to build is cost. After the decision to build has been made, the problem is one of a purely engineering character, whereas in the first case it was one of either a political or an engineering character, or both. The engineering problem is an economic one, in which maximum benefits must be obtained at a minimum cost. A map of the proposed bridge site and the approaches, as well as of the country for a considerable distance up and down stream, should be made. This map should show the contours, the soundings, the borings, the high and low water-mark elevations, and the excep- tional flood line. On this map the bridge should be plotted in its proposed location and also in various others. In the case of each of these locations, various schemes taking into account different numbers of piers and spans should be considered. Several authors have attempted to present formulse having a more or less theoretical derivation and purporting to indicate the correct number of piers and spans for a minimum cost. The use of these formulse should not be encouraged, since they do not in any case give results close enough to serve for anything but a rough guide. The cost of abutments will vary somewhat with the location and the character of the approach. This variation is usually small, and ordinarily an approximate location of the abutments can be quickly made. As the number of abutments is in all cases constant, their effect upon the problem of the location of the bridge is small, the main proposition being that of the cost and the number of piers and spans. The cost of the piers will usually not be constant, those closer to the middle of the stream costing more on account of the depth of the water and the more difficult character of the foundation. Piers Copyright, 1908, by American School of Correspondence. 141 BRIDGE ENGINEERING should not be placed on a skew; neither should they be placed directly in the maximum line of action of the current. If a skew is unavoid- able, it should be as small as possible. The cost of piers should be ascertained by the most careful estimates. In the case of small bridges where there are only one or two piers, the matter is very simple, but with a considerable number of piers the problem becomes very complicated and requires weeks and sometimes months or years for its solution. The determination of the cost of the superstructure is a com- paratively simple matter. In certain instances the class of bridge is limited to some extent by the specifications. Cooper, in Article 2 of his "Specifications for Steel Railroad Bridges and Viaducts" (edition of 1906), gives the following: Types of Bridges for Various Spans SPANS KIND op BRIDGE Up to 20 feet 20 to 75 " 75 to 120 " 120 to 150 " Over 150 " Rolled beams Riveted plate-girders Riveted plate- or lattice-girders Lattice or pin-connected trusses Pin-connected trusses One railroad expresses a preference for plate-girders for all spans from 20 to 115 feet; and for spans from there to 150 feet, riveted trusses. The question as to whether the bridge will be deck or through is one which is decided by the controlling influences of water-way, false work, time of erection, and extra cost of masonry. If the clear height required for the water-way is sufficiently small, the deck bridge should be chosen, as in this class the cost of false work is less, the time of erection is less, and the cost of masonry is less by an amount equal to the cross-section of the piers times the depth of the truss. Deck bridges also cost less than through bridges of equal span. The conditions permitting, girders should be used in preference to trusses. While for equal spans girders are heavier and therefore cost more, the steel work alone being considered, little or no false work is required, and the time of erection is much less than in the 142 BRIDGE ENGINEERING 133 case of trusses. This makes the total cost of girder bridges less than those in which trusses are used. Another item in favor of girders is their great stiffness. While pin-connected bridges cost less and are easier to erect, their stiffness is not so great as that of riveted bridges, which cost more. The time required for the erection of riveted bridges is also greater than that for pin-connected bridges. This is on account of the great amount of time required to make the riveted connections. For long spans, say over 200 feet, it is necessary to use pin-connected bridges, as the extreme size of the connection plates prohibits the use of the riveted type. Also, it is unnecessary to use riveted long-span trusses to obtain stiffness, as the weight of the pin-connected bridges is so great when compared with the live load that sufficient stiffness is obtained. The cost of spans of different lengths and character may be obtained directly from the bridge companies; or their weights may be computed from the formulae given in Article 20, p. 9 (Part I, "Bridge Analysis"), and multiplied by the unit price which your experience indicates is correct, thus giving the total cost. Evidently the solution of problems of this nature cannot be made within the limits of this text, but the following example will tend to indicate somewhat the manner of procedure in a problem of this kind. For example, if the length between abutments is 1 400 ft., the cost of each abutment is $12 000, and the cost of each pier is $15 000, then, if we have fourteen 100-foot plate-girder spans, each costing $4 300, and thirteen piers, the total cost will be $279 200. On the other hand, if nine piers and ten 140-foot truss spans, each cost- ing $9 200, are used, the cost will be $251 000, showing a balance of $28 200 in favor of the truss scheme. The live loading is E 50. 61. Economic Proportions. The depth of girders is given in Article 59, Part I. In the case of trusses, the effect of an increase in the height is to increase the stresses in the web members and to decrease the stresses in the chord members. This variation does not affect the weights to any considerable extent; in fact, a variation of 20 per cent in the height will not affect the weight more than 2 or 3 per cent. The height of the bridge is usually fixed by some considerations which are in turn determined by the specifications. The height must 143 134 BRIDGE ENGINEERING be sufficient to clear whatever traffic will pass through. It should also be sufficient to prevent overturning on account of the wind pressure on the truss or on the traffic. In addition, the height of the bridge is influenced by the depth of the portal bracing. A deep portal bracing is desirable, in that it stiffens the trusses under the action of the wind and the vibration due to the passing traffic; but a deep portal bracing increases the height of the truss and therefore the bending in the Panel Length 18 toes feet. / end-posts due to the wind. Judg- ment on the part of the engineer should be used in order to determine the limiting height for securingamax- imum amount of benefit as regards stiffness and a minimum amount of bad effect due to the bending in the end-posts. Fig. 117, which gives the height for any given length of span, may be said Variations of a foot c Double Track y y _ I -Stnqle Track ^ Sf an in fet. 3 50 .100 ISO 117. Curves Showing Relation between Height of Truss and Length of Span in Double- and Single-Track Railway Bridges. represent the best modern practice (1908). Pig or more from those given do not affect the weight to any appreciable extent. The distance from center to center of trusses for highway bridges depends upon the width of the street or, if in the country, the width of the roadway. Streets, of course, vary in width in differ- ent localities, but country highway bridges usually have a roadway of from 14 to 16 feet in the clear. In the case of railroad bridges, the distance from center to center of trusses depends upon whether the track is straight or on a curve, and also upon whether the bridge is a deck or a through bridge. 144 BRIDGE ENGINEERING 135 The actual amount varies in most cases, and is fixed by specification. Some specifications require that when the track is straight, the dis- tance from center to center of trusses shall be 17 feet; or that, in case one-twentieth of the span exceeds the 17 feet, then one-twentieth of the span shall be used. For deck plate-girders the common practice appears to be to space them as given below : Width of Plate-Ciirder Bridges for Various Spans SPANS DISTANCE CENTER TO CENTER OF PLATE-GIRDERS Up to 65 feet 65 to 80 " 80 to 115 " 6 feet 6 inches 7 feet inches 7 feet 6 inches For through plate-girders the spacing should be such that no part of the clearance diagram will touch any part of the girder. In case of double-track plate-girders with one center girder, great care should be exercised in order that the center girder shall not be so deep nor have so wide a flange as to interfere with the clearance diagram (see Fig. 126). On account of the wind on a train which runs on track placed at the elevation of the top chord of deck bridges, the overturning effect is exceedingly great, and special care should be taken that the height and width are such as to prevent overturning. In through bridges the clearance must be such as to allow the clearance diagram to pass. Special attention should be paid to the knee-braces and also to the portal braces. When the bridge is on a tangent, the spacing of the trusses is a comparatively simple matter, being just sufficient for the clearance diagram; but on curves, allow- ance must be made for the tilt of the diagram due to the super- elevation of the outer rail, and also allowance must be made for the fact that the length of the cars between trucks forms a chord to the curve, and as such the middle ordinance must be taken into account. It is also necessary to allow for that part of the car which projects over the trucks, as this will extend beyond the outer rail by an amount greater than one-half the width of the clearance diagram. (See Figs. 119 and 120.) 145 "o ^^ '^Jl 1 "^-^ t2 T / n 2 !f h h ;; s: .00 O ^ M < 8 g3 -. ^ '& M* \ K I |! s ' VpTOll J_Q ^ u'n T>JL)li'2)'~} : ^ ) S K^ o "fu il 1 !^ ? 3 -ssi V "o U-) | 9o o O ^"S 0! 2*- r^ a 2 Vj v^-,* ,,-,11 ,,9 -,9 m B^-2 fe o|g "t NJ llfil S 2S^a 8 ss-S^ ,,o-,22 . "zSo liijs ,9 l^g 'Pi| Vj 1 / 1 * -^-t=^ & ^il /_ ^ L:: 13 / . c "' b ' ,,^-,11 k * M C i-,9 (/ > 5j;3i! ^ og^cs "CD "01 'o r- -i r -Sg S rtg- c bD-r3 S J ^2^8, * SHI /~VpVL\_ {0 ^ ^ui"~i 2jl}U 2>3 i i Ki "(0 , c "Ss-g To , fill! > & ^ o - V) H ; ^ S g*l cO m 5 |s^? ru "id _ < S ^S.2 'o ^ SS< g | tO . r^ ' i &$! <^ u 5Sg^ _j g .s "*^-^ cQ m ooo D feg J 1 S2^ "tD 1^ 2L-,-b- i BIS " ' r L 6 | S | ifl ? M 1 fe-a to 1 1 II M S2 i "o n" E 1 L ; oa l ^ ss -^ 1 l& s \ 1 ,x M ^ o ""^'* -," .9 -, J J \I.J ^ --^"' 1 BRIDGE ENGINEERING 137 62. The Clearance Diagram. The clearance diagram is not supposed to represent the outline of the largest engine or car which may run over the line, but repre- sents the maximum amount of space which may be taken up by objects which are to be shipped over the line. For instance, the lower part of the clearance dia- gram may allow for snow-plow or ballast distributors, and the upper part may take into account the passage of such material as carloads of lumber, piles, or tele- graph poles. The standard clear- ance diagram of the Lehigh Valley Railroad is given in Fig. 118. This diagram is for the clearance on straight track only. On curves, the diagram tilts as shown in Fig. 119, and to allow for this tilting the Lehigh Valley Railroad requires 2j inches additional clearance on the inside of curves for each inch of elevation of the outer rail. In addition to this tilting effect, the clearance should also be increased on account of the Fig. 119. Clearance Diagram on Curves, Showing Tilting. Fig. 120. Standard Car on Curve, Showing Necessity for Wider Spacing of Trusses. length of the cars and their projection over the outer and inner rails. Fig. 120 shows a standard car according to the specifications of the American Railway Engineering & Maintenance of Way Association, in such a position on a single-track span as to show the effect of the curve upon the widening of the spacing, center to center of trusses. 147 138 BRIDGE ENGINEERING This car is 80 feet long, 60 feet between centers of trucks, and is as wide as the clearance diagram, 14 feet for single track. It is evident that the trusses cannot be spaced so as to interfere with the clear- ance line of the body of the car and its projecting ends. These clearance lines are represented as broken lines in Fig. 120, and are marked c-c. Note that the center of the track is seldom in the center of the floor-beam. Also, it is evident that the sharper the curve, the greater the required distance between trusses, and accordingly the greater the floor-beams in length. This varies the moment in the different floor-beams and therefore makes them more costly. The stringers, also, are more costly, on account of the fact that their ends are skewed. On account of the eccentricity of the track, one truss takes more of the load than the other, and therefore the trusses are not the same a fact which further increases the cost. From the above it is seen that almost all conditions incident to the building of a bridge on a curve tend to increase the cost; and hence a fundamental principle of bridge engineering: Avoid build- ing bridges on curves. 63. Weights and Loadings. For the weight of steel in any particular span, and for the loading required for any particular class of bridge, see Articles 20 to 23, Part I. The weight of the ties and the rails and their fastenings is usually set by the specifications at 400 pounds per linear foot of track. For highway bridges the weight of the wooden floor is usually taken at 4^} pounds per square foot of roadway for every inch in thickness of floor. Highway bridges are divided into different classes according to their loadings (see Cooper's Specifications). The decision as to the class to be employed depends somewhat upon the distance to the nearest bridge across the same stream. In case the nearest bridge is only a few miles away and is of heavy construction, it is not actually necessary to construct a heavy bridge at the proposed site, the heavier traffic being required to pass over the other bridge. In case a heavy bridge is not in the neighborhood, then one should be constructed at the proposed site. If the proposed site is on a road connecting adja- cent towns of large size, then a heavy bridge should be constructed and provision made for future interurban traffic, even if none is at that time in view, since it will be more economical to do this than to erect a new bridge in the future. 148 K OOOZ/ I ^ R ^ o/ \ (0 S (V ^ X C .3 o r CL 140 BRIDGE ENGINEERING In the case of railroad Bridges, new ones are nearly always con- structed to carry the heaviest main line engines. These are usually of a class corresponding to Cooper's E 40 or E 50. In some localities branch-line bridges are built for the same live loadings; but in the majority of cases the branch-line bridges consist of the old bridges from the main line. 64. Specifications. For any particular bridge the specifica- tions are either written by the engineer in charge, or some of the very excellent general specifications which are on the market in printed form are used. Some railroads use these general specifications with the addition of certain clauses which are desired by the chief or bridge engineer. The principal differences in these general specifications are in regard to the allowance for impact. Whenever highway design is mentioned in this text, it is to be in accordance with Cooper's Highway Specifications (edition of 1001). Wherever plate-girder design is given, it is in accordance with Cooper's Railway Specifications (edition of 1906); and wherever truss design is given, it is in accordance with the general specifications of the American Railway Engineering & Maintenance of Way Association (second edition, 1906). 65. Stress Sheet. Before the sections are designed, the com- puter makes a skeleton outline of the truss, and on this places the dead-load and live-load stresses, and, in case the wind should be considered, the wind-load stresses. This is sent to the designer. The designer determines the various sections, and also the moments and shears in the stringers and floor-beams. These are placed on a sheet usually 17 by 23 inches. This is called a stress sheet. This sheet is now given to the draftsman, who makes a shop drawing. The stress sheets for railroad bridges are usually more elaborate than those for highway bridges. Plate I is the stress sheet of a highway bridge; and Plate II (Article 78) and Plate III (Article 93) are examples of the best modern practice in the making of plate-girder and truss- bridge stress sheets. 66. Floor System. Perhaps no part of bridge design is better standardized than the construction of the open steel floors for railroad bridges. The stringers are usually placed 6 feet 6 inches apart, and consist of small plate-girders, or, if the panel length is short, of one or more I-beams. I-beams are economical in regard to 150 BRIDGE ENGINEERING 141 TABLE XIX Safe Spans for I Beams (Based on unit-stress of 10 000 Ibs. per square inch in extreme fibre) fc, Z o < N^Q dS-i 8 2^7 K^ 6 25 30 35 30 35 40 31% 40 50 65 42 50 60 70 80 55 65 75 80 90 65 75 85 90 100 80 85 90 95 100 MOMENT OF INERTIA 1 1-Beam ENGINE CLASS E 40 ENGINE CLASS E 50 Safe span C to C of Bearings Safe span C to C of Bearings 1 Beam per rail 2 Beams per rail 3 Beams per rail 1 Beam per rail 2 Beams per rail 3 Beams per rail 9 in. 9 " 9 " 10 " 10 " 10 " 12 " 12 " 12 " 12 " 15 " 15 " 15 " 15 " 15 " 18 " 18 " 18 " 18 ' 18 20 ' 20 ' 20 ' 20 ' 20 ' 24 ' 24 " 24 " 24 " 24 " 92 102 112 135 163 175 218 274 332 403 443 515 619 718 774 809 .890 1023 1 063 1 188 1 180 1 277 1 453 1 502 ; 1 650 2112 2182 2356 2427 2497 5 ft. 3 in. 5 9 " G " 6 " 6 " 9 ' 8 " 6 ' 7 ft. 9 in. 8 " 9 " 9 " 6 " 10 " " 11 " " 4 ft. 9 in. 5 " 5 " 3 ' 5 " 9 ' 7 " ' 7 ft. in. 7 " " 7 " 9 " 8 " 6 " 10 " " 3 ft'. 6 in.' 4 ' 3 " 4 ' 6 " 4 ' 9 " 5 ' 9 " 7 ' 3 " 3 ft. in. 3 " 6 " 9 " 3 ' 11 " 6 ' 12 " 6 ' 12 " 9 " 13 " 9 " 15 " 3 ' 4 " 4 " 9 " 6 " " 7 " 9 ' 9 " 6 ' 11 " 3 11 " 3 " 12 " 3 " 13 " 6 " 8 ' 6 " 7 ' 6 " 12 " 9 ' 15 " 6 ' 6-3 " 11 " 6 ' 15 " " 14 " " 12 ' " 11 ' " 11 ' 9 " K " 3 ' 21 " 6 ' 10 " 9 " 15 " 3 ' 19 " 3 " 16 " ' 16 " 9 ' 20 " 9 " 10 " 6 " 15 " " 17 " 9 " 18 " 9 " 20 " " 12 ' 6 " 12 ' 9 " 13 ' 6 " 12 ' 9 " 13 ' 3 - 14 ' 3 " 19 " 6 ' 18 " 9 ' 19 " 6 ' 21 " ' 24 " 6 " 23 " " 24 " " 26 " " 12 " 3 " 11 " 6 " 12 " " 12 " 9 " 17 " 6 " 16 " 6 " 17 " 3 " 18 " 6 " 21 " 9 " 20 " 9 " 21 " 3 " 23 " " 14 ' 6 " 21 " 3 ' 15 ' 9 " 16 ' " 17 ' " 17 ' 3 " 17 ' 6 " 23 " ' 23 " 9 ' 24 " 6 ' 25 " " 25 " 3 " 29 ' 3 " 29 ' 9 " 30 ' 9 " 31 6 " 32 ' " 14 " " 14 " 6 " 14 " 9 " 15 " " 15 " 3 " 20 " 6 " 21 " " 21 " 6 " 22 " " 22 " 6 " 25 " 9 " 26 " 3 " 27 " 3 " 27 " 9 " 28 " " DIAGRAM OF ENGINES g g 8 8 o, o o o q o o o' o o' o o" tD \Q <0 ru =z=^- i ^\^p*^- T ing made heavy enough to act as girders as well as tension members. Figs. 128 to 130 show sections of solid floors. The ballast is laid di- ^ rectly upon these 4 Fig. 121. Sec Itailroac use r ^r r ;tional View Showing Open-Floor Construction Bridge of Short Span, Single Track. I-Beams i for Stringers. Lehigh Valley Standard. h ., "H h j*0. 1 4 '' 1 * . 1 8 I" }*- '^^_ 8 i. I 8 i ^y; f solid floors, w r hich are first covered with a good damp- proof paint. The floors should also be supplied with good drainage fa- cilities. er Concrete is sometimes laid di- J Fig. 122. Se and Truss - G'C" ^ L-tion of Open-Floor Construction of Deck-Gird Bridge, Single Track. Lehigh Valley Standard Plate-Girders used for Stringers. rectly upon the steel floor, and the ballast put upon this concrete, Fig. 123. Floor Construction of a Through-Girder Bridge, Single Track. Lehigh Valley Standard. 159 BRIDGE ENGINEERING 145 which has previously had a "layer of some good waterproof- ing applied on its upper sur- face. 67. Practical Considera= tions. The possibilities of the rolling mill and the various shops of a bridge company, such as the drafting room, forge, foundry, templet shop, assembling shop, and riveting and finishing shop, and also the shipping and erect'rg facil- ities, should be well known in order to make the most eco- nomical use of them. This requisite knowledge comes only from experience. The best way to obtain this experi- ence without being actually employed in the shops, is to go into the shops every chance that presents itself, keep your eyes and ears open, and ask all the questions you can. The use to be made of handbooks of the various steel manu- facturers is given in Part I of "Steel Construction," and should be thoroughly studied before going further. Some one of these handbooks is in- dispensable to persons design- ing steel structures. That of the Carnegie Steel Company (edition 1903) is one of the best, and will be frequently referred to in the present text. 155 146 BRIDGE ENGINEERING Copies may be procured from the Carnegie Steel Company, Frick Building, Pittsburg, Pa. The usual price to students is 50 cents, to others $2.00. DESIGN OF A PLATE=GIRDER RAILWAY-SPAN 68. The Masonry Plan. In some cases the general dimensions of the masonry are limited; such a case, for example, would occur in the crossing of a street or narrow waterway. Here the length of the span and the distance above the street or the surface of the water, Drain Pipe / Fig. 129. Solid-Floor Construction of Plates and Angles. are the limited dimensions. The span and under-clearance may be unlimited, as in the case of a country stream crossed by a roadway which is a considerable distance above the surface of the water. The term unlimited is not here used in its exact meaning, as the span in this case is really limited by the cost, which rapidly increases with the length of the span. In some cases, as when the engineer is in a bridge office, the masonry plans are sent in by the railroad. In such cases many of the limited dimensions are fixed. The most usual dimensions to be fixed are the elevation of base of rail, the elevation and size of the bridge seat, and the length of the span under coping. These limit the depth of the girder, or the depth of the floor if it be a through girder, and also limit the length of the bearing plates at the end. Fig. 131, the masonry plan of a road crossing, shows in general what can be expected. All the dimensions usually fixed are given, and those marked x and?/ may or may not be, but x should never be less than 3 feet. 69. Determination of the Class. As before mentioned, the deck plate-girder should be used if possible, since its cost is less. There are some cases, however such as track elevation in cities where the additional cost required to elevate the track so as to use a 156 BRIDGE ENGINEERING 147 deck plate-girder will more than balance the saving in its favor. In such cases the through plate-girder is used. The case whose design is under con- sideration will be taken similar to that of Fig. 131, and the span will therefore be a deck one. 70. Determination of the Span, Cen= ter to Center. Fig. 132 shows the various spans namely, under coping, center to cen- ter of end bearings, and over all. The span under coping is that span from under cop- ing to under coping lines of the abutments, and is so chosen as to give the required dis- tance between the abutments at their base. The span center to center is equal to the span under coping plus the length of one bearing plate. The span over all is the extreme length of the girder. The length of the bearing plate is influenced by the width of the bridge seat, and also by the maximum reaction of the girder. The length should seldom be greater than 18 inches and never greater than 2 feet, as the deflection of the girder will cause a great amount of the weight to come on the inner edge of the bearing plate and also on the masonry, in which case the masonry is liable to fail at that point and the bearing plates are over-stressed. Cast-steel bearings are now almost universally used. They decrease the height of the masonry, and distribute the pressure more evenly and for a greater distance over the bridge seat. When these castings are used, the bearing area between them and the girder may be made quite small, thus doing away to a great extent with the 157 148 BRIDGE ENGINEERING deteriorating effect due to the deflection of the girder as mentioned aboVe. Fig. 133 shows the end of a girder equipped with a cast- steel pedestal. Table XX gives the length of the bearing on the I/Center Line ot Rood Under Clearance Line? | * 6P-0' Elevation 7SfeOx.| Fig. 131. Masonry Plan of a Road Crossing. masonry for various spans, Cooper's E '40 loading being used and cast-steel pedestals being employed. TABLE XX Length of Masonry Bearings SPAN LENGTH OF BEARING 1.5 to 24 ft r 12 inches 25 to 44 16 ' 45 to 69 21 1 70 to 79 2.3 SO to 89 29 ' 90 to 115 36 As an example, let it be required to determine the span center to center of a deck plate-girder of 00-foot span under coping, the I i L j^ Masonry Plate M r Span Under Copinq Bearmq Area J ionry Bearing ^p^ i i A Span Center to Center Bearmq ^ Fig. 132. Diagram Showing Various Spans Considered in Bridge Construction. loading being E 40. From Table XX it is seen that the length of the masonry l>earing will l>e 21 inches, and therefore the span center 158 BRIDGE ENGINEERING 149 to center of bearings will be 60 + 2 X (I X 1 ft. 9 in.) - 61 feet 9 inches. In Articles 71 to 77 the above girder will be designed ; and also such information as is of importance regarding the subject-matter of each article will be treated. The dead- and live-load shears are computed by the methods of Part I, and are given in Fig. 134. 71. Ties and Guard=Rails. The length of ties for single-track bridges is 10 feet. For double-track bridges the length is in most cases the same. In some double- track bridges, however, either each tie or every third tie extends entirely across the bridge. In other cases every third tie on one track extends to the opposite track, thus act- ing as a support for the foot- walk which is laid upon them. It is the best prac- tice to limit the length of the ties on double-track bridges to 10 feet, since, if they extend into the opposite track in any way whatsoever, unnecessary expense is incurred whenever repairs or renewals are made, because both tracks must necessarily be disturbed to some extent. The size of the ties varies with the weight of the engines and the spacing of the stringers or girders on which they rest. They are usually sawed to size instead of hewn, and the following sizes may be purchased on the open market namely, 6 by 8, 7 by 9, 8 by 9, 9 by 10, a =5 rr^rr "-^A / i \ "- 1 '//^////^////////, f , Section A-A Span Undtr Coping 5paa Center to Center Span Over AU Fig. 133. End of a Girder Equipped with a Cast-Steel Pedestal. lit 150 BRIDGE ENGINEERING 1600 300 Fig. 134. Shear arid Moment Diagram. 160 BRIDGE ENGINEERING 151 and 10 by 12 inches. Larger sizes may be obtained on special order. The elevation blocks (see Fig. 127) should be of length to suit the width of the cover-plates and the spacing of the supports. They are usually made of the best quality of white oak, since the cost of renewal is great enough to demand that they be made of material as permanent as possible. The guard-rails should be placed in accordance with the specifica- tions (see Articles 13 and 14). Some railroads specify that the guard- rails shall be in 24-foot lengths unless the bridge is shorter than 24 feet, in which case one length of timber should be used. For method of connection and other details, consult Figs. 121 to 127. The guard-rails and the ties are usually made of Georgia long-leaf yellow pine, prime inspection. Other wood, such as chestnut, cedar, and oak, may be used. In addition to the wooden guard-rail, a steel guard-rail usually consisting of railroad rails is placed within about 8 inches of the track rail. In designing ties, the problem is that of a simple beam symmet- rically loaded with two equal concentrated loads, the weight of the rail and tie itself usually being neglected. For the case in hand, which is that of a deck plate- A -^ J^j girder, loading E 40, the con- t^--^^^--^-.^^----^ ^J i centra ted load for which the tie ^^-^ '""^ must be designed is, according Rr to Specifications (Article 23, 3d rai-^ Q Q'*'} T\r>nnr1o A rvr-rl i n , Fig. 135. Distribution of Loading on Ties Of part), S 666 pounds. According Deck plate-Girder Bridge. to Article 23, 100 000 pounds is on four wheels. This gives 25 000 pounds on one wheel, and ac- eording to Article 15, one-third of this, or 8 333 pounds, will come on one tie. Fig. 135 shows the condition of the loading, the space center to center of rail being taken as 4 feet 10 inches. Some designers take this distance as 5 feet; but as the standard rail head is about 2 inches, and the standard gauge 4 feet 8^ inches, the distance here taken seems to be the more logical one. The formula to be used in the design of this beam is that given OT in "Strength of Materials," and is M = In this case M=10 X 161 i_J. r BRIDGE ENGINEERING and c = d -r- 2, and therefore - = c 6 8 333 = 83 330 pound-inches. In the above formula, 7 = W-^ 12, Substituting the value of the moment in the above formula, and solving for S, there results' 499 980 bd a For a 6 bv 8-inch tie, the unit-stress would then be : S = - = 1 310 pounds. If a 7 by 9-inch tie is used, the unit-stress is found to be 880 pounds. Since according to Article 15 of the Specifications, the unit-stress cannot be greater than 1 000 per square inch, it is necessary to use a 7 by 9-inch tie. If the engine loading had been E 50, the mo- ment would have been 100000 pound-inches, and then the stress in a 7 by 9-inch tie would be 1 0(>0 pounds per square inch, and the stress in an 8 by 9-inch tie would be 930 pounds, which would neces- sitate the use of the latter. The guard-rails on this bridge will be placed according to the Lehigh Valley standard, and hence their inner face will be 4 feet 1^ inches from the center of the track. Elevation blocks will not be required, as the bridge is on a tangent. 72. The Web. The economic depth of the web, according to Article 59, Part I, will be: Fig. 136. End Rivets Transferring Shear to Web. Cl ft. 9 in. Depth = 0.543 rs = 72.5. The depth might be taken as 72 inches, but 74 inches will be decided upon, as this will decrease the area of the flange and also will not affect the total weight to any great extent- The unit-stress for shear 169 BRIDGE ENGINEERING 153 Is 9 000 pounds per square inch (see Specifications, Articles 40 and 41). The maximum shear in the girder occurs at the end, where it is 117 800 pounds. The area required for the web is then 117 800 -f- 9000 - 13.09 square inches, and the required thickness is 13.09 -r- 74 = 0.177 inch. This latter value cannot be used, since, on account of Ar- ticle 82 of the Specifications, no material less than f inch can be used. The web plate will therefore be taken as 74 in. by | in. in size. Some engineers insist that the net section of the web should be considered. Consider Fig. 136, the shear being trans- ferred to the web by the end rivets. The web will not tend to shear along the section B-B, in which case the rivet-holes should be subtracted; but it will shear along section A- A, a section which is unaffected by the rivet-holes. The web splice should come at one of the stiffeners, and will therefore be considered in Arti- cle 76. 73. The Flanges. This portion of the girder is usually built either of two angles or of two angles and one or more plates. In heavy girders where the flange areas are large, additional area is OD- tained by using side plates or side plates and four angles. Sometimes two chan- nels are used in the place of side plates and angles. Fig. 137 shows the different methods of constructing the top flanges of girders. The lower flanges are usually of the same construction. Fig. 137 b has the web extending beyond the upper sur- 163 154 BRIDGE ENGINEERING faces of the upper flange angles. This is done in order that the ties may be dapped over it, and thus prevent the labor usually required for cutting holes in the lower face of the tie in order to allow for the projecting rivet-heads. Fig. 137 g is usually uneconom- ical, since the thinness of the channel web requires a great many rivets to sufficiently transmit the shear from the web to the flange, and also since the cover-plates must be very narrow. Specifications usually state that the flanges shall Have at least one-half of the total flange area in the angles, or that the angles shall be the largest that are manufactured. The largest angles are not usually employed, since their thickness is greater than three-quarters of an inch and therefore the rivet-holes must be bored, not punched. The reason for this is that the depth of the rivet-hole is too great in proportion to its diameter, and on this account the dies used for punching frequently break. Also, the punching of such thick material injures the adjacent metal, which makes it undesirable. In reality the flange area of only the short- span girders is small enough to allow the flange area to be taken up by the angles. In choosing the thickness of the cover-plates, care should be taken so that the outer row of rivets will not come closer to the outer edge of the plate than eight times the thickness of the thinnest plate. In at orkss not over 5" c3Qt or less. Fig. 138. Diagram Showing Relation be- tween Thickness of Coyer-Plates and Position of Rivets. case eight times the thickness of the plate is greater than 5 inches, then 5 inches should be the limit. Also, the distance between the inner rows of rivets should not exceed thirty times the thickness of the thinnest plate. These limitations are placed by Article 77 of the Specifications, and Fig. 138 indicates their significance. The determination of the required flange area depends upon the distance between the centers of gravity of the flanges; and in order to determine this exactly, the area and composition of the flanges should be known. The above condition requires an approximate design to be made, the supposition being that the flanges consist of two angles and one or more plates as shown in Fig. 138. 164 BRIDGE ENGINEERING 155 The distance back to back of angles will be taken as 74 + 2 X | = 74^ inches. Article 74 of the Specifications requires ^ inch; but I ; inch is better practice, since the edges of the web plate are very liable to overrun more than -^ inch. Some specifications require 3- inch. In the computation of the approximate flange area, the center of gravity of each flange will be assumed as one inch from the back of the angles. The approximate effective depth is then 74j less 2 X 1 inch, which equals 72 j inches. The approximate stresses in the flange areas are : , 275 000 x 12 I or dead load, - - = 4o GOO pounds. For live load,] . 12 . - 222000 pounds. The approximate flange areas are now obtained by dividing these amounts by the allowable unit-stresses for dead and live load, which are (see Specifications, p. 8, Article 31): 20 000 and 10000 pounds per square inch respectively; and the resulting areas are: For dead load, - -- -- = 2 . 28 square inches. For live load, = 22. 20 square inches. These amounts give a total of 24.48 square inches as the approximate net flange area required. It will be assumed that one-half the total area, or 24.48 -r- 2 = 12.24 square inches, is to be taken up by angles. If 12.24 sq. in. is distributed* over two angles, then 12.24 -7-2 = 6.12 square inches is the net area for one angle. Of course it is not to be assumed that the area of the angle chosen must be exactly 6. 12, but that this 6. 12 square inches is the approximate area of the angle to be chosen, and the net area of the angle (see Specifications, Article 149) must not be 2 j per cent less than this, although it may be greater. From Steel Construction, Part I, Table VII, or from the Car- negie Handbook, p. 117, a 6 by 6 by f-inch angle gives a gross area of 8.44 square inches and a net area of 8.44 - 2 X (f + |) X f = 6 . 94 square inches, f-inch rivets being used and so spaced that two rivets are taken out of each angle (see Specifications, Article 63, and Fig. 139). A 6 by 4 by -j f-inch angle, giving a gross area of 7.47 and a net area of 6.66 square inches, one rivet-hole being out> 105 150 BRIDGE ENGINEERING Fig. 139. Calculation of Size of Angle and Cover-Plate. could have boon used, but { e inch is too thick to punch, and there- fore the above angle is chosen. The required net area of the cover-plate is now found to be 24.48 - 2 X 6.94 = 10. 60 square inches. Since the legs of the angles are 6 inches and the thickness of the web is | inch, the outer edges of the angles are 12f inches apart; and since the cover-plate must extend somewhat over the edges of the angle, and the width of the cover-plate should be in the even inch, the width of the cover- plates must be at least 14 inches, as shown in Fig. 139. On account of the 1-inch rivet- holes to be deducted, the real or net width of the cover-plate ' is: 2 X n + m = 14 2 X 1 = 12 inches. The thickness of all the cover-plates at the center is now: 10 Cii) t = r> - - = O.SS5 inch say, Jjj inch. A thickness of I of an inch is decided upon, for the reason that plates are rolled only to the nearest sixteenth of an inch. The approximate section at the center has now been determined, and is: 2 Angles (i by <> hy ^-inch = 1I5.8S sq. in. net. Cover-plates I inch thick = lO.oOsq. in. net. Total = 24.38 sq. in. net. This approximate section must now be examined, and, if it shows too great an excess or a de- ficiency, must be revised. In order to deter- mine the effective depth the distance between the centers of gravity of the flanges must first be com- puted, the gross areas be- ing used. Theoretically, perhaps, the net areas should be used; but this is an unnecessary refinement, since the effect on the final result is of no practical importance. 0.875" Fig. 140. Determination of Center of Gravity. 166 BRIDGE ENGINEERING 157 In computing the center of gravity (see Fig. 140), the axis is taken at the center of the cover-plates, as this reduces the moment of the cover-plates to zero. The distance of the center of gravity of the angles from their back (Carnegie Handbook, p. 117, column 6) is 1 .78 inches. The distance of this center of gravity from the center of the cover-plate, is 1 .78 + 0.875 -^ 2 = 2.22 inches. Gross aceaof the angles = 2 X 8.44 =10.88 sq. in. " cover-plates = I X 14 = 12.25 sq. in. Total = 29.13 sq. in. The center of gravity is now found to be 10.88 X 2.22 -=- 29.13 = 1.286 inches from the center of the cover-plate, and 1.286-0.875 -=-2=0:848 inch from the back of the angle. The effective depth h e is 74.25 - 2 X 0.848 = 72.554 inches, and the required flange areas are: 275 000 X 12 72 . 554 X 20 000 1 340 OOP X 12 "727554 X 10 000 = 2.272 sq. in. for dead load. = 22.200 sq. in. for live load. Total = 24.472 sq. in. The values of the moments, as taken from the curves, must be mul- tiplied by 12 in order to reduce them to pound-inches. A total of 24.86 square inches is given by the section approxi- mately designed, and the difference between that and the section as above determined is : (24 . 472 - 24 . 38) -r- 24 . 472 = . 38 per cent, and as this is less than 2^ per cent (see Specifications, Article 149), it may be used without any further change. If there should have been a deficiency or an excess greater than 1\ per cent, then it would have been necessary to revise. In case a revision of section is necessary, the size and thickness of the angles generally remain the same as those taken in the approximate design, the thickness of the cover-plates being decreased or increased as the case may be. The total thickness of the cover-plates, - inch, is too thick to be punched. In such cases as this, the section is made up of two or more plates whose total thickness is equal to that required. If plates of more than one thickness are decided upon, then their thickness should decrease from the flange angles, outward. For the case in hand, one plate \ inch thick and one plate \ mcn thick will be decided upon. The flange section at the center as finally designed is: 167 158 BRIDGE ENGINEERING SHAPE NET SECTION GROSS SECTION 2 Angles 6 by 6 by 1 Cover-plate 14 by 1 Cover-plate 14 by fin. fin. Jin. 13.88 sq. in. 4 . 50 ' ' 6 . 00 " 16.88 sq. in. 5 .25 " 7.00 " Total 24. 38 ' ' 29.13 " The above is the section required at the center of the girder; for any other point it will be less, decreasing toward the end, where it will be zero. Evidently, then, the cover-plates will not be required to extend the entire length. The following analysis will determine where they should be stopped. If the load were uniform, the moment Fig. 141. Diagram Showing Curve of Required Flange Areas. curve would be a parabola. Although under wheel loading the curve of moments is not a parabola, yet it is sufficient for practical purposes to consider it as such. The curve of flange areas, like that of moments, is to be considered a parabola (see Fig. 141). Let j = Net area, in square inches, of the outer cover- plate; a., = Net area, in square inches, of the next cover-plate; (i x , etc. = Net areas of the other cover-plates; A = Net area of all the cover-plates and the flange angle. Then, from the properties of the parabola, where L = Length of cover-plate in question; I = Length of span, center to center; a = Net area of that cover-plate and all above it; and A = Total net area of the flange, $ of the gross area of the web not being considered in this quantity. 168 BRIDGE ENGINEERING 150 The lengths of the cover-plates for the section above designed (see Fig. 141) are: L, - 61.75 - - = 26.45 feet. Lo = 61.75 J - ^p = 40.00 feet. \ J4.OS One foot is usually added on each end of the cover-plate as theoretically determined above. The results are also usually rounded off to the nearest half-foot. This is done in order to allow a safe margin because of the fact that the curve of flange areas is not a true parabola. The final measurements of the cover-plates are: 14 in. by f in. by 28 ft. 6 in. long. 14 in. by in. by 42 ft. 6 in. long. In most cases the cover-plate next to the angle on the top flange only is made to extend the entire length of the girder. Although this is not required for flange area, it is done in order to provide additional stiffness to the flange angles toward the ends of the span, and to prevent the action of the elements from deteriorating the angles and the web by attacking the joint at the top (see broken lines, Fig. 141, for length of first cover-plate extended). EXAMPLES FOR PRACTICE 1. The dead-load moment equals 469 000 pound-inches; and the live- load moment, 4 522 000 pound-inches. Design a flange section entirely of angles, if the distance back to back of angles is 45 inches. 2. The dead-load moment is 3 340 000 pound-inches, and the* live- load moment, 21 235 000 pound-inches. Design a flange section using 6 by 6-inch angles and three 14-inch cover- plates, the distance back to back of flange angles being 78} inches. 3. In each of the above cases, design the flange section considering that i of the web area is taken as effective flange area. (For demonstra- tion of the methods to be employed in the solution of this problem, see the succeeding text.) While the section of a plate-girder is composite that is, it con- sists of certain shapes joined together, and is not one solid piece- nevertheless these shapes are joined so securely that the section may be considered as a solid one and its moment of resistance computed accordingly. Let Fig. 142 be considered. The moment of resistance of the section is: in the derivation of which the moment of inertia of the flange about its own neutral axis is considered as zero, and A equals the net area of one flange. Now, as the values of h e and h seldom differ by more 169 160 BRIDGE EX( ] IX EE H ING than one inch, for all practical purposes they may be considered* as equal. The above expression then reduces to: M = S X h(A + - ~ ) = S X h (net area of flange + one-sixth gross area of web) Since the rivet-holes decrease the moment of resistance of the web, one-sixth of the gross area cannot be considered, as is theoreti- cally indicated in the above formu- la. It is common practice to take one-eighth, instead of one-sixth, of the gross web area. Substituting this value in the above equation, and transposing, there results: Area of flange -f -J- gross web area = ^.- O/l The flange section will now be designed for the moments previously given, considering of the gross web area as efficient in withstanding the moment. The gross area of the web is 74 x | = 27 . 75 square inches ; and | of this is 3.47 square-inches. The total approximate amount of flange area required is, as in the first case, 24.48 square inches. According to the above formula, | of the web area, or 3.47 square inches, may be considered as flange area, and therefore 24.48 3.47 = 21.01 square inches, is the approxi- mate area of the angles and cover-plates of the flange. The ap- proximate area of one angle is then 21 ,01 -^ (2 X 2) = 5.25 square inches. A 6 by 6 by f %-inch angle gives the gross area of 6.43 square inches and, two rivet-holes being deducted, a net area of 5.305 square inches (see "Steel Construction," Part I, Table VIII, or Carnegie Handbook, p. 117). As this is quite close to the approxi- mate area determined above, this angle will be taken. The ap- Fig. 142. Section of Plate-Girder. 170 BRIDGE ENGINEERING 101 proximate area of the cover-plates is 21.01 2 X 5.305 = 10.40 square inches. As before, the gross width of the cover-plate will 10.40 be taken as 14 inches. The thickness is then - 14 (-, = 0.867 inch say f inch. The gross area of the angles being 12.86 square inches, and that of the cover-plates 12.25 square inches, the center of gravity of the section is found, by a method similar to that previously employed, to be 1 .10 inches from the center of the cover-plate, or 1.10 0.438 = . 662 inch from the back of the flange angles. This makes the effective depth 72.93 inches. For this section, the true live-load flange stress is (1 340 000 X 12) -=- 72 . 93 = 221 000 pounds, and the actual dead-load flange stress is (275000 X 12) -=- 72.93 = 45400 pounds. The actual areas required for the live and dead load are 22. 10 and 2.27 square inches, which are obtained by dividing the above flange stresses by 10 000 and 20 000 pounds, respectively. The total required area is the sum of the tw r o areas above, and is equal to 24.37 square inches. The total area required in the flange angles and cover-plates is therefore 24.37, less - the gross area of the web, 3.47, which leaves 20.90 square inches. The same angles as decided upon before will be used. This gives a required area for the cover-plates of 20.90 10.61 = 10.29 square inches. The required thickness is then 10.29 -T- (14 2) = 0.857 say inch. The following section of the flange will therefore be decided upon: SHAP E NET SECTION GROSS SECTION 2 Angles 6 in. by 6 1 Cover-plate 14 in 1 Cover-plate 14 in in. by T 9 ff in. . by f in. . by in. 10. Gl sq. in. 4 . 50 ' ' 6.00 " 12.86 sq. in. 5.25 " 7.00 " Total = 21.11 " 25.11 " As the total net area above is within 1\ per cent of the required net area, that section will be taken (see Specifications, Article 149). Note that in this case, the thickness of the cover-plates in the final design is the same as that determined in the preliminary design. Also note that the total net area is about 4 square inches, or 20 per 171 162 BRIDGE ENGINEERING cent, less than in the flange as first designed, in which case none of the area of the web was considered as withstanding the bending moment. The 4-inch cover-plate on the top flange will extend the entire length of the grider, and is therefore 62 feet 9 inches long. The lengths of the other cover-plates are: For Wnch plate at the bottom, L = Gl .7.5 -y 21' 11 = 4 ' ? ' 5 fe, which is the formula for the rivet spacing in the vertical parts of the flanges of any girder, providing the flange is not subjected to localized loading. It is to be used for the rivet spacing in both the top and bottom flanges of through girders, but not in the top flanges of deck plate-girders for railroad service. It is to be used, how- ever, in the bottom flanges of deck plate- girders for railroad service. The dis- tance /i e is not ordinarily used, the distance between rivet lines being used instead (see Fig. 145). The rivet spacing in the cover-plates and horizontal legs of the angles is made to stagger with that in the vertical legs, and usually the staggering is with every other rivet in the vertical flange. The term stagger signifies that the rivets in the top flange are not placed opposite the rivets in the vertical legs of the flange angles or, that in case there are two lines of rivets in the vertical legs of the angle, a rivet near the outer edge of the cover-plate is placed in the same section w r here a rivet occurs near the lower edge of the vertical legs of the angle, and vice versa. EXAMPLES FOR PRACTICE 1. Determine the rivet spacing at a section where the shear is 147 200 pounds, the value of one rivet. 4 920 pounds, and the effective depth of the section 84 inches. ANS. 2.82 inches. 2. Determine the stress on a rivet at a section where the shear is 299 400 pounds, the spacing 2\ inches, and the effective depth of the girder 84} inches. ANS. 8 870 pounds. Fig. 145. Determination of Rivet Spacing. - 174 BRIDGE ENGINEERING 165 The rivet spacing is usually determined at the tenth-points; and a curve is plotted with the spacing as ordinates and the tenth- points as abscissae. The rivet spacing at any intermediate point can be determined from this curve. When one-eighth the gross section of the web is considered as flange area, then only that proportion of the shear which is transferred to flanges is to be considered in com- puting the rivet spacing, on account of the fact that some of the shear is transferred directly to the bending moment in the web. In order to determine the distance between rivet lines, the gauge, or distance out from the back of the angles to the place where the rivets must be placed, must be known for different lengths of legs. Table XXI gives the standard gauges, and also the diameter of the largest rivet or bolt which is allowed to be used in any sized leg. No gauges should be punched otherwise unless your large experience or instructions from one higher in authority demand it, and this should be so seldom that indeed it might be said never to be necessary. TABLE XXI Standard Gauges for Angles (All dimensions given in inches) MAXI- MAXI- MAXI- L a MUM RIVET ou BOLT L g. MUM RIVET OR BOLT L (/ MUM RIVET OR BOLT 8 7 4* 31 2 it j 2 l! H l ' 6 3* - 2| if | i| 1 5 3 1 2i it a J 4 21 21 H f i L Oi 02 L 02 8 3 3 6* 2i 2 7 2J. 5 2 1 G 21 2$ "When thickness is J inch or over. 175 166 BRIDGE ENGINEERING Fig. 146. Determination of Distance between The distance between rivet lines for the girder being designed (see Fig. 146), is, in the first case: h, - h - (2 9l + ^) = 74.25 - (2 X 1\ + 2J) = 67 . 00 inches. In the second case, where | of the web is considered, the above distance is 74.25 (2 X 2J + 24) = 67.25 inches. The compu- tations and the rivet spaces at the tenth- point, and at the ends of the cover-plates in the bottom chord of the plate-girder, are shown for each case in Table XXII. The value of v is the value of a |-inch rivet in bearing in a f-inch web (see Specifications, Article 40, and Carnegie Handbook, p. 195, second table). This value is 4 020 pounds. In the first column, 7.98 indicates that the end of the cover-plate next to the flange is 7.98 feet from the end of the girder, and that this section is taken just to one side of that point, the side being that nearest the end of the girder. In a similar manner, 7.98+ indicates that the section 'is taken to that side of the point which is nearest the center of the girder. A like interpretation should be placed on 15.55 and 15.55 + , the point under consideration in this -case being the end of the outer or top cover-plate. In the fifth column, values are given which indicate that portion of the shear which is transferred to the flanges. For example, O q ? n = 74 700, and the difference between 97 700 and 74 700 represents that portion of the shear which is taken up directly by the web in the form of bending moment. An inspection of the headings of the third and fourth columns will tend to make this matter clearer. Where there is local loading, as in the top flange, the rivets, in addition to the stress caused by the transferring of web stresses, are stressed by the vertical action of the angles being pressed downward by the ties and the consequent upward pressure of the web. Accord - 176 BRIDGE ENGINEERING 107 TABLE XXII Rivet Spacing in Bottom Flange Flange Taking All the Moment SECTIOX TOTAL SHEAR (Pounds) ft, (Inches) (Pounds) RIVET SPACING (Inches) REMARKS 117800 67 4290 2.80 1 97 700 67 4290 3.38 2 79 300 67 4290 4.16 3 61 300 67 4290 5.38 4 5 44200 28 600 67 67 4290 4290 7.46 11.52 I See Specifica- \ tions, Art. 54 One-Eighth of Web Area Considered hr = 67.25 inches; v - 4 920 pounds SECTION TOTAL SHEAR (Pounds) NET FLANGE AREA PLUS i WEB AREA (Sq. Inches) NET FLANGE AREA (Sq. Inches) REDUCED SHEAR (Pounds) RIVET SPACING (Inches) 117800 14.08 10.61 88800 3.75 1 97 700 14.08 10.61 74700 4.45 7.98- 90000 14.08 10.61 67 900 4.88 7.98 + 90000 20.08 16.61 74800 4.42 2 79 300 20.08 16.61 65800 5.02 15.55- 67500 20.08 16.61 56 000 5.81 15.55 + 67500 24.58 21.11 58 100 5.69 3 61 300 24.58 21.11 . 52 700 6.28 4 44200 24.58 21.11 38000 8.71 5 28600 24.58 21.11 24 600 13.42 ing to Article 15 of the Specifications, the weight of one driver is distributed over three ties (see Fig. 147). W Let j- , = iv, the load per linear inch caused by one wheel W, which load is assumed to be uniformly distributed over the dis- tance /; ws = i'i , the vertical load or stress that comes on one rivet in the space s; v = -T , the stress on a rivet due to the distribution of flange stresses when s is a space, and V the shear at that point. When these two stresses act on the rivet, the maximum stress will be v n , the ultimate amount that it is allowed to carry, and this will act as shown in Fig. 147 : Then, 177 168 BRIDGE ENGINEERING + (ws) 2 from which, which gives the spacing at any point in the girder flange under localized loading. Note that if w equals zero that is, if there is no localized loading there results: which is the same as previously deduced for flanges with- out localized load- ings. The rivet spacing for the top flange of the girder Which is being designed is given in Table XXIII. Here W = 20000; / = (3X 7 + 3 X 6) = 39 . , 20 000 inches; w = -~ = 513; h r = 67 inches; and v w = 4 920 pounds, which is the bearing of a | -inch rivet in the f-inch web. The top cover-plate is run theentire length of the span. Fig 147. Rivet Spacing Determined by Stresses Distributed under Localized Loading. 178 BRIDGE ENGINEERING 109 TABLE XXIII Rivet Spacing in Top Flange Flange Taking All the Moment SECTION w- TOTAL SHEAR (Pounds) or V(x)- RIVKT SPACE (Inches) 1 2 3 4 5 262600 - 117800 97700 79 300 61 300 44200 28600 3 080 000 2 140000 1 390 000 840000 435000 181500 1 825 1 550 1 285 1 050 835 660 2.70 3.17 3 . 83 4.68 5.88 7.45 One-Eighth of Web Area Considered w = 513; h t = 67i inches; r u = 4 920 pounds. SKCTIO.N . u? 2 REDUCED SHEAR (Pounds) (*) RIVET SPACE (Inches) var- 262600 97600 2 100 000 1 538 3.20 1 262 600 81 000 1 450 000 1315 3.74 2 262 600 65 800 ! 985 000 1 100 4.50 15.55- 262600 56000 695 000 980 5.02 15.55 + 262 600 58 100 765 000 1 014 4.85 3 262 600 52 700 616 000 938 5.24 4 262600 38000 320 000 763 6.44 o 262 600 24600 134 000 629 7.82 The points other than the tenth-points referred to in the first column are for sections taken just to the left and right of the top cover-plate. The values of the reduced shears given in the third column are ob- tained as has been previously explained. Although the rivet spacing in the lower flange is considerably greater than that in the upper flange, and accordingly a smaller number of rivets would be required, yet the spacing in the lower flange is made the same as that in the upper. Convenience in the preparing of plans and facility in manu- facture make this action economical. Theoretical spacing greater than 6 inches should be dealt with according to Article 54 of the Specifications. The values of the rivet spacing given in Tables XXII and XXIII are plotted in Fig. 148. Note that the effect of the localized loading is to decrease the rivet spacing, and also note that the effect increases from the ends toward the center. 179 170 BRIDGE ENGINEERING Note.-5ccond Cover Plate oi Top Flanoc Lxtends Lntire Length, of Oirdcr. Fig. 148. Plotted Values of Rivet Spacing Given in Tables XXII and XXIII. 180 BRIDGE ENGINEERING 171 The size of the flange angles and the width of the cover-plates for different spans, are a matter of choice. Once the size is deter- mined, the thickness can be computed. The sizes very generally adopted in practice are as follows: SPANS WIDTH OF COVER-PLATE 15 to 20 feet 5 x 3J inches none 20' 25 6x4 ' ' 25' 40 6x6 " 40' 75 6x6 14 inches 75' 100 6x6 16 " 100 ' 120 8x8 20 " For another method of the presentation of this subject, see "Steel Construction," Part IV, pp. 264 to 268. EXAMPLE FOR PRACTICE 1. Determine the rivet spacing for the top chord of a plate-girder, loading E 40, and 7 by 9-inch ties being used. The web is inch thick; distance from back to back of angles, 6 feet 6| inches; flange angles, 6 by 6 by ^-inch; and cover-plate, 14 by ^-inch, two f-inch rivets'being out of each. First, consider the flange as taking all the bending moment; and second, consider one-eighth the gross area of the web. The total unreduced shear is 80 000 pounds in both cases. ANS. 3.21 inches; 3. 76 inches. 74. Lateral Systems and Cross=Frames. There are two methods in use in common practice in determining where the panels of the lateral bracing shall fall namely, (1) To choose the number of panels so that the panel points come opposite the stiffness, and (2) to choose the number of panels so that the placing of the panel points is inde- pendent of the stiffener spacing. The lateral systems should be of the Warren type; and in both of the above cases the angles that the diagonals make with the girder should not be greater than 45 degrees. Also, it is best to have all panels the same length and to have an equal number of panels. This latter condition will simplify the drafting very much, since one-half of one girder can be drawn and the other half will be symmetrical, the opposite girder being similar to the one drawn, but being left-handed. The members of the lateral systems will take tension or com- pression according to the direction the wind blows. Cross-frames are placed at intermediate points to stiffen the girders. These are 181 ftfii?* 1"^ s *!* 1 ? "S si'vj 1 ^ l" ^>"^ leS'Sai ^ S ^>^ 1IJU II^IP^ fill BRIDGE ENGINEERING 173 diagonal bracings (see Plate II), and are placed at certain intervals according to the judgment of the engineer. Good practice demands that their number should be about as indicated below : SPAN OF GIRDEII NUMBER OF CROSS-FRAMES to 20 feet 20 to 35 " 3.5 to 70 " 70 to 85 " 85 to 110 " The above is not intended to serve as a hard and fixed rule. Varia- tions from the limits given are to be made as the case demands. In all cases they are put at the panel points of the bracing, the top and bottom parts acting as sub-verticals in the lateral system. Also, the cross-frames should divide the span into equal parts if possible. In cases where that is not possible, the shortest divisions should come near the ends of the spans. If the panel points are to be located at the stiffeners, the number of panels is a function of the depth of the girder (see Specifications, Articles 47 and 48) . In this case the number of panels is given by : ,. _ Span in inches. Depth of girder in inches' no fraction being considered. As an example, let it be required to determine the number of panels in a girder 85 feet center to center of bearings, the depth being 90| inches back to back of angles. Then, 85 X 12 N = -^ ------ = 1 1 . 3, or, say, 1 1 panels. yu . o Each panel will then be 92.8 inches long. This, according to Article 47 of the Specifications, being greater than 5 feet, would not be allowed as a space between two stiffeners; but one stiffener can be placed in between, and then the panel points will come at every other stiffener. The cross-frames should be five in number. The arrangement of panels and cross-frames is shown in Fig. 149. Here the cross-frames are marked C. F., and the broken lines represent the lower lateral system. In case an even number of panels were desired, then ten would be the number chosen and the general arrangement would be as shown in Fig. 150. The length of a panel would be 85 X 12 -s- 183 174 BRIDGE ENGINEERING 10 = 102 inches, or 8 feet 6 inches, which would allow of one stiffener in between and still keep the stiffener spacing within the limit of 5 feet. The cross-frames at the ends of the span are designated as end cross-frames, and those in between are designated as intermediate cross-frames. In case the spacing of the stiffeners is not required to be such as to coincide with the panel points of the lateral bracing, the panel length will depend upon the spacing of the girders, being equal to or Fig. 150. Arrangements of Panels and Cross- Frames. greater than the spacing in order to keep the angle which the diagonals make with the girder less than 45 degrees. In this case, , T Span in feet Distance center to center of girders in feet For the girder considered on page 1 74, the number of panels would be OK - = 11.3 or, say, 11 panels if odd numbers were to be used, 7.5 and 12 if even numbers were to be desired. For the case in hand, the panel points of the bracing will be taken at the stiffeners, and an even number of panels will be used. Then, W _ 511?.. 9. The arrangement of the panels and cross-frames, and also the maxi- mum stresses in the diagonals, are shown in Plate II, the stresses being determined according to Article 50, Part I, "Bridge Analysis," and Article 24 of the Specifications. All the wind is taken as acting on one side of the bridge; and no overturning effect, either on the girder 184 BRIDGE ENGINEERING . 175 or on the train, is considered. Also, note that the wind stresses in the flanges are not considered. Should the student determine these, he will find them too small to be considered according to Specifications, Article 39. Before designing the lateral diagonals which consist of one or two angles, Articles 31, 33, 34, 35 (last portion), 38, 40, 63, and 83 of the Specifications should be care- fully studied. The upper lateral bracing is to be designed first. Carnegie Handbook, pp. 109 to 119, is to be used. The member U^U l must be designed for a compressive stress equal to 23.20 + 0.8 X 20.6 = 39.68, and a tensile stress of 20.6 + 0.8 X 20.6 = + 37.08. The length of the diagonal measured from center to center of girders is 1/6.5'-' +6.2 2 - 9 feet, or 108 inches. In reality the length is not 108 inches, as the cover-plate takes off a certain amount, as shown in Fig. 151. The true length, which is to be taken as a column length in designing, is 108 2y, and y is readily computed to be 9.70 inches, thus making the true length 88.6 inches. The least allowable rectan- gular radius of gyration is obtained from the relation that the greatest value of - = 120, and therefore the least value of r = = 0.74. r 120 It will be assumed that a 6 by 4 by r 9 6 -inch angle with an area of 5.31 square inches will be sufficient. Here the length equals 88 . 60 inches, and the least rectangular radius of gyration is 1 . 14; hence, 88.60 Fig. 151. Determination of Length of Diagonal in Lateral Bracing. The required area is = 13 000 - 60 X = 8 330 pounds per square inch. 39 360 8330 4 . 73 square inches. As the angle assumed has an area of 5.31 square inches, which is considerably greater than the 4.73 square inches required, this angle cannot be used, and other assumptions must be made until the area of the angle 185 170 BRIDGE ENGINEER ING assumed and the required area as computed are equal or very nearly so. A 6 by 4 by ^-inch angle with an area of 4.75 square inches will now be assumed. The length is 88 . 60 inches as before, and the least rectangular radius of gyration is 1.15. The unit-toad P = 8340 pounds per square inch, and the required area is-^ = 4 . 72 square 8 340 inches. As the area of the angle assumed and the required area as computed are very close, this sized angle will be used. The section must now be examined for tension, and in order that both legs of the angle may be considered as effective section, both legs must be connected at the end. The area required will be ' - 2.04 square inches. Considering one rivet-hole is taken out of the angle, the net area is 4 . 75 1 X (I + i ) X i = 4 . 25 square inches, which is amply sufficient. If the 4-inch leg only were as- sumed to be connected, the grcss area would be 4 X % = 2. CO square inches, and the net area would then be 2.00 - 1 X (I + I ) X -^-inch = 1 .50 square inches, which is not sufficient. If the 6-inch leg were connected, the area would be sufficient. See Fig. 152 for method of connection and rivets. The number of rivets required (see Specifications, Article 38 and 40) is computed as follows: If the member were not subjected to both tension and compression, the number of rivets required in single shear would be: = 39360 (9000 + 50 per cent of 9 000) X 0.6013 = 39360 8100 = 4. 86 (say 5). But according to Article 38 of the Specifications, this number must be increased 50 per cent, and accordingly 4.86 X li .7. 29 (say 8) Fig. 152. Method of Connecting Angle Legs in Lateral Bracing and Cross-Frames. 186 w ^ "i M - 15 BRIDGE ENGINEERING 177 shop rivets are to be used. In the above formula, 0.6013 is the area of the cross-section in square inches of a |-inch rivet. In order that both legs should be connected, a clip angle as shown in Fig. 152 is used ; and the same number of rivets must go through both legs of the clip angle, since the stress in the vertical leg of the main angle is transferred to the clip angle and from there into the connecting plate. The above number of rivets makes the joint safe so that it will not shear off in the plane between the connection plate and the hori- zontal leg of the angle. The joint must also be designed so that there will be sufficient rivets in bearing to prevent them from tearing out of the connecting plate. The number required is: _ 39 360 " (15 000 + 50 per cent of 15 000) X $ X $ _ 39 300 7~380 = 5.34. This 5.34 must be increased 50 per cent, making a total of 5.34 X 1.5 = 8.01 = say, 8 shop rivets as before. The above rivets are shop rivets, since it is assumed that the span, being a small one, will be riveted complete in the shop and shipped to the bridge site ready to place in position without any further riveting. In case the girders are shipped separately, then the lateral bracing must be riveted up in the field ; and according to last part of Article 40 of the Specifications, the rivets, being field rivets, must have the allowed unit-stresses reduced one-third, which is equivalent to having the number of shop rivets increased 50 per cent. This will make the required number of field rivets 8 X 1.5 = 12. As a rule, the connection plates are |- inch thick, seldom more. Also, the members of the upper lateral system are connected on the lower side of the connection plate in order not to interfere with the ties. Note that the use of the clip angles requires a smaller connec- tion plate than would be necessary if these angles were not used, since in the latter case all the rivets must then be placed in one row in the horizontal leg of the angle. The number of rivets required in the connection plate and the flange of the girder must be sufficient to take up the component of that member parallel to the girder. For the case in hand, the num- ber (see Fig. 153) is: 187 178 BRIDGE ENGINEERING x _6\2 ~s ~ p = is ooo - oo x f = 8 42 1 . lo Uo 6.-SH Fig. 153. Calculation of Rivets in Connection Plate and Flange of Girder. pounds per square inch. , . . 33400 Ihe required area is - = 3 . 97 square inches. As this is very near the area assumed, and as trials with other angles do not give required areas which come any closer, this angle will be used. The rivets required in single shear are: 33 400 X 1.5 = 6.21 (say 7) shop rivets, and 8 100 G. 21 X 1.5 = 9.3 (say 10) field rivets. The rivets required in bearing in the f -inch connection plate are : oo 4-nn -'---- X 1 .5 = 6.78 (say 7) shop rivets, and 7 ooU * 6.78 X 1.5 = 10.17 (say 11) field rivets. The above computations show the joint to be weakest in bearing, and therefore 7 shop or 11 field rivets must be used. It is not neces- sary to investigate this member for tension, as the computations for the first diagonal indicate that the area will be sufficient, both legs being connected. The member UJJ 2 f must be designed for a maximum compres- sive stress of 16.0 -f- 0.8 X 14.1 = -27.28. A 6 by 4 by f-inch 188 BRIDGE ENGINEERING 179 angle with an area of 3.61 square inches and a least rectangular radius of gyration of 1.17 will be assumed. The unit-stress P, as computed from the formula in the Specifications, is 8 460 pounds per 27 9 SO square inch; and the required area is " = 3 . 23 square inches. This angle will be used, as the given and required areas are close together, and as the next smaller angle a 6 by 3 i by f-inch angle with an area of 3.42 square inches gives a required area of 3.58 square inches, thus being too small. The rivets required in single shear are : 27 280 X 1 5 = 5 . 04 (say 5) shop rivets, and 8 1UU 5.04 X 1.5 = 7.6 (say 8) field rivets. The rivets required in bearing in a f -inch web are : '' X 1 .5 = 5 . 54 (say 6) shop rivets, and 5 . 54 X 1.5 = 8.3 (say 9) field rivets. In order to make the joints safe, 6 shop or 9 field rivets should be used. The member U 2 'U 2 must be designed for a maximum compressive stress of 9.6 + 0.8 X 8.0 = -16.00. A 3i by 3 by f-inch angle with an area of 2.30 square inches and a least rectangular radius of gyration of 0.90 will be assumed. The unit-stress P is 7 090 pounds per square inch, and the required area is - - = 2 . 26 square inches. i UJU As the required and the 'actual areas are very close together, this angle will be used. The rivets required in single shear are : ^n * * *> = 2.96 (say 3) shop rivets, and o 1UU 2.96 X H = 4.44 (say 5) field rivets. By computation similar to the above, it is found that 4 shop or 5 field rivets are required in bearing. Since the bearing requires the most rivets to make the joint safe, 4 shop or 5 field rivets must be used. If the Specifications would have allowed a 3^ by 3i by j 5 6 -inch angle with an area of 2.09 square inches, this angle would have exactly fulfilled the requirements, the required area being 2.09 square inches. The member U 2 U 3 ' must be designed for a maximum compres- 189 180 BRIDGE ENGINEERING sive stress of 8.0 + 0.8 X 4.1 = -11.28. A3 by 3 by f-inch angle with an area of 2.11 square inches and a least radius of gyration of 0.91 will be assumed. In this case the unit-stress is 7 160, and the area required is 1.58 square inches. The required area is considerably less than the area of the angle assumed; but it must be used, since it is the smallest allowed by the Specifications, which require that the material shall not be less than f-inch, and from Table XXI it is seen that 3 inches is the smallest size leg in which a |-ineh rivet can be used. The stresses in all the members of the lower lateral system are less than the stresses in the member just designed, and therefore all members of the lower lateral system will be "made of one 3 by 3 by f-inch angle. For the last member designed in the upper lateral system, and for all members in the lower lateral system, 3 shop or 5 field rivets will be required at the ends. These are more than sufficient to take up the stress, but it has been found that less- than three rivets do not make a good joint. The stress sheet, Plate II, shows the general arrangement of the lateral system, the number of rivets in the connections and also in the connection plates where they join the flanges. The intermediate cross-frames do not lend themselves to a theo- retical design, since the stresses which come upon them are not easily ascertained. It is good practice to require that all members be of the sizes as given below: RIVETS SPAN OF GIRDKR ANGLES (in Feet) (in Inches) SHOP FIELD 30 to 65 34. -x 34. x f 3 4 65 to 110 4 x 4 x f 4 5 The angles in the intermediate cross-frames will therefore be 3^ by 3-1- by |-inch. The end cross-frames (see Fig. 154) act in a manner somewhat similar to the portal bracing in a bridge, since they transfer all the wind which comes on the top chord and on the train to the abutment. This load, which acts at the level of the ties, is in this case (see Article 24 of the Specifications): 190 BRIDGE ENGINEERING 181 X 61 ft. 9 in. = 18 525 pounds. It is usually assumed that half of this is' transferred to the point a by means of a-b, and from there down a-b' to the masonry. The other half goes directly down b-a' to the masonry. This causes stresses as shown in Fig. 154. Note that the stress in a-b will always be com- pression; but the stresses in the diagonal will be either tension or compression according to the direction the wind blows. The mem- ber a-b will be a 3-^by 3Hwf-inch angle. To form the connections at its end, 3 shop or 5 field rivets will be used. The maxi- mum compressive stress for which the diagonals are to be designed is 12.70 + 0.8X12.70 -22.86. Here the length is 108 inches if the angle tends to bend one way; but if it bends as shown by the broken lines in Fig. 154, the length will be one-half of this. For this reason, angles with unequal legs should be used, the longer leg extending outward. This allows the greatest rectan- gular radius of gyration to be used. A 4 by 3 by j 7 6 -inch angle with an area of 2.87 square inches and a radius of gyration of 1 .25 will be assumed. The unit-load P is computed to be 8 750 pounds, and the required area is therefore 22 860 -=- 8 750 = 2.61 square inches. This does not coincide very closely with the given area, but will be used since this angle comes nearer to fulfilling the condition than any of the other sizes rolled. The joint will require more rivets in bearing than in single shear. It is not necessary to perform the complete computations in order to determine this, since a comparison of the values of a rivet in single shear and in bearing shows that the value in bearing is less than that Fig. 154. Action of End Cross-Frames. 101 182 BRIDGE ENGINEERING in single shear, and therefore the number of rivets required in bearing will be greater than that number required in single shear. The number of rivets required in bearing is: ^K j p ~ sav > ^ shop rivets, and 4.00 X 1.5 = 6 field rivets. 75. The Stiffeners. According to Article 47 of the Specifica- tions, these should be placed at certain intervals whenever the unit- shear is greater than S = 10 000 - 75 X (~ = 10 000 - 14 800 = -4 800 pounds. This negative sign signifies that whenever the unit shearing stress is greater than zero, the stiffeners must be placed through- out the entire length of the span at distances not to exceed 5 feet. The interme- diate stiffeners should have the outstanding leg long enough to give good sup- port to the flange angle (see Fig. 155). The filler bars or fillers are put in so as to allow the stiffener angles to remain straight throughout their entire length; otherwise they will have to be bent as shown in Fig. 156. This bending is called crimping. Stiffeners must also conform to Article 48 of the Specifications. This would require a different sized stiffener at each point, and also a different number of rivets in each stiffener. This is not usually done in practice. In practice the stiffener for the first intermediate Section A-A Fig. 155. Use of Straight Stiffeners, with Filler Bars. 102 BRIDGE ENGINEERING 183 point is designed, and the remainder are made the same in size and have the same number of rivets. An exception to this is where a stiffener comes at a web splice. In this case the size is usually kept the same, but the number of rivets is changed somewhat to conform to the requirements of the splice design. The second intermediate stiffener comes at the first tenth-point, and is 6.175 (say 6.2) feet from the end, since it is at the first panel point, or opposite the first panel point, of the lateral system. The first stiffener will be 3 . 1 feet from the end ; and scaling off the value of the shear at this point (see Fig. 134), it is found to be 108000 Fig. 156. Crimping of Stiffener Aiiglt ' where No Filler Bars are Used. Fig. 157. Section of Intermediate Stiffener Construction. pounds. Here the length I to be used in the formula for the unit allowable compressive stress is 74 2 X f = 72.75 inches, the f being the thickness of the flange angle. The section of the material which according to Article 48 of the Specifications is to be considered as a column, is shown in Fig. 157. The assumed column cannot bend about the axis B-B,but about the axis A- A, and therefore the radius of gyration about the axis A - A must be computed. The moment of inertia of the fillers and the web plates about their own axes is considered as zero. A 4 by 4 by ^-inch angle with an area of 3 . 75 square inches will be assumed to be sufficient to withstand the stress. The moment of inertia of this and the filler bars and the web portion is /A-A = 2(5.55 + 3.75 X 2.12 2 + 3.00 X 0.563 2 ) = 46.70. 184 BRIDGE ENGINEERING & \ \ \ \ x-V \0icater /: Less thar The radius of gyration, then, is, */ ,?'' = 1.764, and the unit-stress \ 15. (JO computed with this value and a length of 72 . 75 inches is 8 140 pounds. The required area is now determined to be 108 000 + 8 140 = 13 . 27 square inches. The value 15.00 used in the above computation for the radius of gyration is the value of the area of the angles, the filler bars, and the web portion. A 5 by 3^-inch angle with the 5-inch leg out would have given better support to the flange, but would not make so good a job, as it would have extended about inch be- yond the curved part of the hori- zontal leg of the flange angle. The bearing determines the number of rivets in this case. The number is 108000 4- 4 920 = 22 shop rivets in the web. The angle must now be inves- tigated in order to see if these 22 rivets can go in one row without being closer together than 2| inches, which is three diameters of the f-inch rivet. The total length in which these rivets must be placed is 72.25 inches, and therefore we have 72.25 -j- 22 = 3.3 inches as a spacing. Since this is greater than 2| inches, 22 rivets can be placed in one row. If the spacing as determined above had been less than 2f inches, it would have been neces- sary to use two rows of rivets spaced as shown in Fig. 158; and then the distance center to center would be more than 2| inches, although the spacing in a vertical line would be less than that. The four angles at the ends of the girders are called the end stiffencrs. These are placed in pairs on opposite sides of the web (see Plate II, Article 74). The total end shear is 117 800 pounds, and this is assumed to be carried by the two pairs of end-stiffener angles, each carrying one- half. This amount would require lighter angles than the angles used for intermediate stiffeners. It is the customary practice to make them the same size and thickness as the intermediate stiffeners, Fig. 158. Rivets Placed in Two Rows to Give Necessary Number and Spacing. 194 BRIDGE ENGINEERING 185 additional strength being allowed in order to withstand the effects of the end cross-frame when in action. The bearing determines the number of rivets required in each pair of stiffeners. The number required is 117 800 sn P rivets. Some engineers arbitrarily choose the stiffeners regardless of the shear, enough rivets, however, being put in the end stiffencrs to take up all the shear; and the spacing in the intermediate stiffencrs is made the same. One noted engineering firm determines its stiffencrs according to the following: FLANOE ANGLE STIFFKXERS HORIZONTAL LEO THICKNESS END IXTEKMEI HATH 4 in. Any .3x3 x Jin. 3 x 3 x I! in o in. Any 4x4 x \ in. Six 3- 1 , x i in 6 in. Over | in. 4x4 x A in. 31 x 3.V x s in 6 in. Less than \ in. 5 x 3i x A in. 5 x 3.1 x j|- in 8 in. Any 6x6 x | in. 6 x 4 x s in No rational method has as yet been determined for ascertaining the stresses in the stiffeners of plate-girders. Results obtained by placing extensomcters on the stiffeners of actual plate-girders appear to indicate that the stresses are very small, in fact in most cases not being greater than 1 500 or 2 000 pounds per square inch. PROBLEMS FOR PRACTICE 1. Design, according to Cooper's Specifications, the end stiffeners :1 the shear is 150 000 pounds, the distance back to back of angles is 6 feet G^ inches, the web inch thick, and the flange angle 6 by 6 by ^-inch. Use fillers. 2. Design the intermediate stiffeners for the girder of Problem 1, above, where the shear is equal to 75 000 pounds. L'se crimped stiffener angles. Note that in this case the angles lie close against the web, no filler bars being used in between. 76. The Web Splice. Web splices are required because of the fact that wide plates cannot be rolled sufficiently long. Web splice s should be as few as possible, and good practice demands that they be placed at the same points as the stiffener angle. The tables on page 30 of the Carnegie Handbook give the extreme length of plates which can be procured for any given width. The 195 186 BRIDGE ENGINEERING length of plates for widths which are not given in these tables, should be taken equal to the length of the next plate given whose width is less than that of the desired plate. From the first table it is seen that a 74 by f -inch plate can be rolled up to 400 inches, or 33 feet 4 inches, in length. Therefore, if the girder under consideration is spliced at the center, the web plates will be required to be ~ = 30 feet 10 i inches, which value does not exceed the 33 feet 4 inches as given above. According to Articles 46 and 71 of the Specifications, a plate must be placed on each side of the web as shown in Fig. 159, and enough rivets placed in each side to take the total shear. The total thickness of both plates, and also their length, must be sufficient to stand the total shear, but must not be less than | inch. The total shear at the center of the girder under consideration (see Fig. 134, p. 150) is 28 600 pounds. 28 600 The area required in each of the two splice plates is * /\ y uuu ^1.59 square inches; and as their length is 62.25 inches, the thickness must be 1.59 -f- 62.25 = 0.0255 inch, but they must be made | inch thick according to the Specifications. The width should be somewhat greater than twice the width of the stift'ener angle leg. This would make the width in this case about 10 inches. The bearing governs the number of rivets required in this case, and they are 28 600 -r- 4 920 = 5.81, say 6, shop rivets. More rivets than this will be required by practical considerations, as indicated by A-* J Section A-A Fig. 159. Splice Plates Placed on Each Side of Web. 196 BRIDGE ENGINEERING 187 Article 54 of the Specifications or in order to make the spacing in the stiffener angle the same as that in the other stiffeners. This detail is to be left to the draftsman, the required number only being put on the stress sheet In case $ of the gross area of the web is considered as efficient flange area, then provision must be made in the splice for the bending moment which the web takes. A very economical and efficient splice is shown in Fig. 1GO. The horizontal plates take the stress due to the moment, and the vertical plates take the stress due to the shear. The web equivalent is 3.47 square inches and the total moment is 1 615 000 pound-feet, which is composed of 275 000 pound- o o I ' o o o o o o O o o o .0 o o o o o o o o t) (J , 1 1 (Sh ear Plat* 4 ^ 6 Shop: k e Shop | Lii Shop,' o) y- 5.^ \ 600^ b ro ooo^ o~o"o~ o 0000 1 o o o o o o / g o o / Fig. 160. Splice Consisting of Vertical and Horizontal Plat feet due to dead load and 1 340 000 pound-feet due to live load. Therefore that proportion of the 3.47 which is taken up by the dead load is: and that proportion taken up by the live load is: 3.47 = 2.88 square inches. The equivalent flange area is assumed to act at the center of gravity of the flange ; and the bending moments equivalent to the above areas are, for dead load; 0.59 X 20000 X 72.554 = S5G 000 pound-niches; and for live load : 2.88 X 10000 X 72.554 = 2 090 000 pound-inches. 107 188 BRIDGE ENGINEERING These bending moments must be taken up by the horizontal splice plates The stresses in these plates (see Fig. 160) are, for dead load: 856 000 -^v Q-- = 15 /SO pounds; and for live load, 2 090 000 = 38 500 pounds. o4.2o While the allowable unit-stresses are a maximum at the center of gravity of the flange and are those given by the Specifications, they decrease rapidly towards the center of the girder, being zero at the neutral axis of the entire section. The unit allowable stress at the center of the horizontal plates will not be so great as the maximum allowable, but will be proportional to the distance from center (see Fig. 160). The horizontal plates 'will be taken 8 inches in width. The unit-stresses are easily determined by means of the similar triangles oab and 006'. The dead-load stress is determined from the proportion : 207)00 and is 14 950 pounds. For live load, the unit-stress will be one-half of this amount, or 7 475 pounds. The area required for this plate is, for dead load, rr-TT = 1 ^ 38 ^00 square inches, and for live load ~ f ,^~. .- =F 5.16 square inches, making a total of 6.21 square inches for both plates. Assuming two rivet- holes out of the section, the net width is 8 2 Q + -J) = 6 inches; 6 ^1 and the required thickness for one plate is - = . 52, say , 6 i ~ X 6 inch. The joint will be weakest in bearing in the -jj-inch web. The number of rivets required is : 15 670+ 38 500 4 920 The design of the shear plate is as fpllows: The shear is 28 COO 28 600 pounds, and the required area is^nnn =3.18 square inches. As 010 the length of the plate is 46J inches, the required thickness is ^ 2. X 4o.2o 198 BRIDGE ENGINEERING 189 = 0.034 inch, but on account of the Specifications it cannot be less than | inch thick. It will, however, be made j 9 6 inch thick, since it will then fill out even with the horizontal tension plates and no filler will be required. Bearing in the web plate decides the number of rivets, which is: 28600 "4920 = sh P riVetS ' The width of this shear plate should be, as before, 10 inches. The same conditions limiting the spacing of the rivets apply here as in the case where the splice was designed for shear only. The length of the horizontal plates should be sufficient to get in all the rivets, and this is a detail which is left to the judgment of the draftsman. f 1 - ' Proportion,, <***** Bearing Plate, and PROBLEMS FOR PRACTICE 1. A plate- girder is 87 ft. 9 in. center to center of end bearings. The dead-load moment is 9 125 000 pound- _t _ inches, and the live- r r ' load moment is \ 38 205 000 pound- inches, the total shear at the sec- tion being 202 700 pounds. The web is 90 by -.v-inch, arid the flange angles are G by 6 by |-inch. Design the web splice when no part of the web is considered as taking bending moment. 2. For the girder of Problem 1, above, design the splice when J of the gross area of the web is considered as effective flange area. 77. The Bearings. Articles 113 to 119 of the Specifications should be carefully studied before proceeding; also Article 87. Article 70 should be referred to, and the remarks there made about bearings should be read. In case the length of bearing is such as to allow a simple -2-inch plate, care must be taken that the bearing plate does not extend past the flange angles more than 2 inches, or that the masonry plate does not extend past the bearing plate over 2 inches. Reference to "Steel Construction," Part II, p. 96, to Fig. 161, 109 100 BRIDGE ENGINEERING and to the discussion which follows, will explain the reason of this. I = ~ = 250 X I X _ " 12" 1X9 . 16 X 12' and as s = 10000, } 10000X from which, Z = 1 .94, say 2 inches. In case it is desirable to have a simple masonry plate instead of a cast-steel pedestal, and to have the plate extend over the sides of the angles a distance greater than 2 in- ches, then some arrangement must be made for sup- porting the pro- jecting portion. Fig. 162 shows one of the methods most commonly used. Notwith- standing the brac- ing of the gusset plates, the mason- ry plate is not ade- quately supported, the greater pro- portion of the stress coming upon the ends. The disadvan- tage of having the masonry plate too long is plainly shown by Fig. 163. Here the girder is shown deflected under a live load, the rear end of the plate being tilted up and the greater part Pig. 162. Arrangement where Masonry Plate is Used instead of Cast-Steel Pedestal. 200 BRIDGE ENGINEERING 191 o'f the pressure coming upon the forward end. The use of this style of plate is not to be recommended for spans over 40 feet. The design for the bearing of the girder under consideration will now be made. The total reaction of one girder must now lx> computed. This will be due to the weight of the steel in the girder, to the weight of the track, and to the reaction pro- duced by the E 40 loading when wheel 2 is directly over the end support. This reaction is: Btormq Plate Masonry Plate Fig. 163. Effect of Having Masonry Plate Too Long. Weight of Steel, (123.5 10 X 61 . 75) 61.75 4 400 Weight of Track,- ;; (61.75 + 1.75)$ Reaction Due to Engine Loading 1 1 430 pounds 6 350 ' ' = 99700 " Total 117 480 pounds. The square inches of bearing surface required is ^ 2oO 470; and, as the length is 1 foot 9 inches, or 21 inches, the total width of the 470 cast-steel pedestal will be = 22.4, say 23 inches, or 1 foot 11 inches. A bearing plate must be riveted to the lower flange where it rests upon the pedestal. The pedestal must be so constructed as to allow this bearing plate to set in it. Hand-holes should be provided in the casting in order to allow the bolts which connect the casting to the girder to be inserted. These bolts should be at least f inch in diameter. Anchor bolts f inch thick and at least 8 inches long should be provided and fox-bolted to the masonry. The thickness of the metal in all parts of the casting should be at least \\ inches. The details of the pedestal are given in Fig. 164, the length of the bearing being made 12 inches so as to allow one rivet to be driven in the flange angle in the space between the end stiffeners. Allowance should be made for a variation of 150 degrees in tem- 201 192 BRIDGE ENGINEERING perature. The coefficient of expansion for steel per unit of length is 0.0000065, and the amount of expansion for 150 degrees of tempera- ture will be : 0.0000065 X (61 ft. 9 in.) X 150 = 0.00 foot. This is about f inch, and therefore the holes in the flanges at one end of the girder should be made oblong and long enough to allow the Fig. 164. Sido and End Elevations Showing Construction of Pedestal and Connection to Bearing Plate. girder to move 4- inch, or | inch either backward or forward from a central position. In determining the length of this slotted hole (see Fig. If Jo), it must be noted that the J-inch bolt takes up part of this hole, and therefore its length should be | + $ = say If inches. The width of the hole should be sufficient to allow for any over-run in the diameter of the bolt. It should be at least 1 {- inches wide. PROBLEMS FOR PRACTICE 1. Determine the distance center to center of bearings, and the size of the masonry plate, for a plate- girder of 40-foot span under coping, the loading being E 40. ANS. 41 ft. 4 in.; 350 square inches. (NOTE Interpolate values in Ta- ble I, Cooper, p. 80.) 2. If the girder span is 78 feet under coping, and the loading K 40, determine the maximum end reaction and the width of the masonry plate. ANS. 147 130 pounds; 24 \ inches. 78. The Stress Sheet. Plate II (p. 1 72) shows the stress sheet for Fig. 1(55. Slotted Bolt-Hole in Flange at End of Girder to Allow for Contraction and Expansion Due to Temperature Changes. BRIDGE ENGINEERING 193 the girder which has just l>een designed. It represents the best modern practice in that it gives, in addition to the sizes of all the sections, the curves for the maximum shears and moments, the rivet-spacing curve, and the number of rivets required in the different parts of the structure. This general form has been adopted by one of the largest bridge corporations in this country, and is to be recommended since it gives the* draftsman all necessary data and thus prevents the loss of time by an inexperienced man in recomputing certain results. The results just referred to are the shears, the moments, the rivet spacing, and the number of rivets required in the various parts. Formerly it was not customary to give this information on the stress sheet, and the draftsman was therefore required to do all this computation which had previously been worked out by the designer but had not been placed on the stress sheet in available form, and thus unnecessary loss of time resulted. DESIGN OF A THROUGH PRATT RAILWAY=SPAN 79. The Masonry Plan. The same remarks which are made in Article 67 apply here. In this case the length of the masonry plate is usually determined by considerations relative to the number and length of the rollers in the bearing, and not by the bearing per square inch upon the masonry, the size of the plate as determined by the above considerations being usually much larger than if it had been determined by the unit bearing stress. A preliminary design of the masonry plate is usually made in a manner similar to that done in the case of the plate-girder; or the length of the masonry plate may be approximately determined from the following: LENGTH OF MASONRY PLATE FIXED END ROLLER END 100 feet 23 inches 23 inches 125 " 2G " 26 " 150 " 28 " 28 " 175 " 31 " 31 " 200 " 35 35 The above masonry plates are for single-track bridges, with or with- out end floor-beams, the length being the same in either case. 194 BRIDGE ENGINEERING 80. Determination of the Span. The determination of the span is made in exactly the same manner as described in Article 68. Care should be taken, in case end floor-beams are not used, to allow for the pedestal stones, which are square stones resting directly upon the bridge seat, and upon the top of which rest the masonry plates of the stringers. Their height must, of course, be such as to keep the stringers level. In case these stones are used, their size must be determined; and if it is greater than that of the bearing or masonry plates, then their size determines the width of the bridge seat and the span center to center of bearings. 81. The Ties. In the design of the ties, as well as in all the design which follows, the Specifications of the American Railway i Fig. 166. Spacing of Stringers and Rails, and Position of Loads. Engineering & Maintenance of Way Association will be followed. Whenever reference is made to these specifications, the number of the article will be enclosed in parentheses, as "(>)/' which signifies that Article 5 of the Specifications is to be referred to. The stringers in the bridge in question will be taken 6 ft. 6 in. center to center. The maximum loading (7) is such as to bring 8 333 pounds on one tie, and to this must be added 100 per cent for impact, making a total of 16 667 pounds. In order to illustrate the method of assuming the distance, center to center of rails, as 5 feet, that distance will be used in this case. The maximum moment will then be 9 X 16 667 = say, 150 000 pound-inches. The size of the tie will be determined as in Article 71, the allowable unit-stress being 2000 pounds per square inch (5). If a 7 by 9-inch tie is used, the unit-stress will be 1 590 pounds. If a 6 by 8-inch tie is used, the unit-stress will be 2 340 pounds. It is evident that a 7 by 9-inch tie must be used. See Fig. 166 for spacing of stringers and 204 BRIDGE ENGINEERING 195 rails, and for position of the loads. Note that, although impact is taken into account in this case, the size of the tie is the same as that designed for the plate-girder, although the unit allowable stress also differs. 82. The Stringers. The width, center to center of trusses, will be assumed as 17 feet, since this is sufficient to clear the clearance diagram in cases of single-track bridges of spans less than 250 feet. The span which is to be designed in the following articles is a through-Pratt with 7 panels of 21 feet each, making a total span, center to center of bearings, of 147 feet inches. See Plate HI (p. 251) . Rivets I inch in diameter will be used throughout, except in channel flanges. The length of the stringers end to end will be 21 feet, and accord- ing to Cooper's Specifications/ p. 32, the maximum moment for the live load will be 226000 pound-feet per rail. The coefficient of impact (9) will be ( 5x77) = 0.935, and therefore the moment > 21 -|- oOO ' due to impact will be 0.935 X 226000 X 12 = 2535000 pound- inches, making a total of 5 247 000 pound-inches due to live load. The section modulus for any particular beam is equal to the bending moment divided by the unit-stress, and this is equal to the moment of inertia divided by one-half the depth of the beam. This latter quantity is constant for any given beam, and for I-beams may be found in column 11, Carnegie Handbook, p. 98. On account of the cheapness of I-beams, they will be used for stringers in this bridge; and sufficiently heavy shelf angles will be used to take up any distorting influences due to the eccentric connections which are unavoidable in this case. In case an I-beam had not been decided upon, the stringers would have been small plate-girders with a span of 21 feet and depth according to formula? given. They would have been computed in exactly the same manner as a plate-girder span of 21 feet center to center of bearings. Since the dead-load moment cannot be determined until the size of the stringer is known, an approximate design must first be made by using the live-load bending moment alone; and then, with the size determined in this manner, the extra section modulus required for the dead-load moment due to the weight of the beam and the track must be computed. If this extra section modulus, added to the one previously determined, is greater than that given by the beam in 205 196 BRIDGE ENGINEERING question, a larger size beam must be used and a recomputation made. 5 247 000 The section modulus (17) required for live load only is 10 UUU = 327.9. As this is too large for one beam, two beams will be used, thus giving a required section modulus of 164 for one beam. Two 24-inch 80-pound I-beams will be used, giving a total section modulus of 2 X 174 - 348. Assuming the rails and ties to weigh 400 pounds per linear foot, 400 the dead load per linear foot per stringer is 80 + = 180 pounds. The dead-load moment is therefore - =119 000 pound-inches. This requires an additional section modulus of 119 000 = 7.45. This, added to the 164 as determined above, makes 16 000 a total required section modulus of 171.25, which, being less than 174 (which is that for one I-beam), indicates that the above chosen beam is sufficient in strength, and it will therefore be used. The number of rivets in the end connections will now be deter- mined. The total end reaction for one I-beam is equal to the weight of one-half the beam, one-eighth the track in the panel, and one-half the maximum live-load reaction for one rail. These quantities are: \ Live-Load Reaction = ^ = 25 700 pounds. Impact = 25 700 X 0.935 =24 030 " 400 1 Weight of Track = ^ X ~ 9 = 530 " Weight of Stringer = ~ X 80 = 840 " Total = 51 100 pounds. The coefficient of impact is that for a loaded length of 21 feet. From p. 177, Carnegie Handbook, sixth column, it is seen that the longest connection angle which can be used with a 24-inch I-beam is 20f , say 20 inches. In this case the thickness of the connection angles must be = 0.23 inch; but according to (36), f 10 000 X 22 inch will be used. The angles chosen will be 6 by 3 by f-inch. The 6-inch leg will be placed against the web of the floor-beam in order to allow for sufficient room for rivets to be driven. 2oe BRIDGE ENGINEERING 197 The rivets will tend to shear off at places between the webs of the stringer and floor-beam and the connection angles. They will also tend to tear out of the web of the stringer and out of the web of the floor-beam. As the thickness of this latter is not known, the determination of the rivets for this condition will be made under the next article. The bearing value of a |-inch rivet in a |-inch plate (19) is 1 X ^ X 24 000 - 10 500 pounds, and therefore ~] = 5 lu ouu shop rivets are required in bearing in the web of the stringer. The value of a |-inch shop rivet in single shear (18) is 0.6013 X 12 000 - 7 220 pounds, and the number of rivets required to prevent shearing , . 51 100 between the connection angles and the webs is -=~^. = 7 shop rivets. / __' ' The value of a f-inch field rivet in single shear (18) is . 6013 X 1 000 = 6 013 pounds, and therefore =9 field rivets are required to b OJ o connect the connection angle to the web of the floor-beam. As men- tioned above, the number of rivets in bearing in the web of the floor- beam will be determined in the K/9 Field next article; and if the number required for bearing is greater than 9, then that number must be used instead of 9. Fig. 167 shows the connection of the stringer to the floor-beam web, and also the number of rivets as determined above, in their proper 7 Shop /I-beam Wb Floor Beam Web /I-beam Web F5Kop Id positions. The distance between Fig. 1 67. Connection of Stringer to Floor- , , . Beam Web ; also Number of Rivets. the webs ot the stringers must be such as to prevent their flanges from touching at the top. The stringers should be connected at the bottom by a system of lateral bracing of the Warren type. The size of these angles cannot be determined by theoretical considerations, but is usually chosen to be 3| by 3^ by f-in. See Plate II (p. 172) for the general arrangement of this bracing. 83. The Floor=Beams. All floor-beams should be of sufficient depth to allow the use of small-legged connection angles at the ends 198 BRIDGE ENGINEERING where they join the end-posts. The thickness of the web should also be greater than that which is theoretically computed, in. order that sufficient bearing may be given so that the rivets for the stringer connections will not require the stringers to be of too great a depth. The depth of the floor-beam will, of course, vary somewhat with the length of the panel and with the loading, but should not be less than 36 inches in any case. A considerable variation in the depth will not affect the weight of the floor-beam or the bridge to an appreciable extent. A good plan is not to exceed a depth of 5 feet, with panel lengths of 25 feet and E 50 loading. In this bridge the depth of all intermediate floor-beams will be taken as 48 inches. It is good practice not to consider | the web area when designing flanges of floor-beams and stringers, and the design here given does not consider the web as taking any bending moment. The design of an intermediate floor-beam will now be made. The loads for which it is designed are the floor-beam reaction due to the live load (see Cooper,p. 32), the floor-beam reaction due to impact, the dead weight of the stringers and track, and the weight of the beam itself. The latter weight is distributed uniformly over the entire length of the beam, and the other loads act as concentrated loads spaced 6 feet 6 inches apart at equal distances from the center. The computation of the concentrated loads is as follows: Live Load = .................... ........ 68 000 pounds Impact = 68 000 X ( Q7 o t ~r oUU 60 500 Dead Load of Stringer = 2 ( 21 X ^ -- 8 ) . . 3 320 400 Dead Load of Track = - --- X 21 ......... 4 200- Total = 136 020 pounds. The moment at points under the loads (see Fig. 168) is 136 020 X 5.25 X 12 = 8 575 000 pound-inches. This is due . to the con- centrated loads only. The weight of one floor-beam may be approxi- mately determined by the same formula as used to determine the weight of plate-girder spans; only, in place of the length of the span, the length of the panel must be substituted. The total weight of the above floor-beam, then, is: TF = 0.45 X (123.5 +10 X 21) X 21 = 3 160 pounds. BRIDGE ENGINEERING 199 The dead-load moment at the center due to this weight will be: Wl 3 160 X 17 X 12 = 80 700 pound-inches, O O making a total moment at the center of the beam of 8 575 000 + 80 700 <0 2 i I i i ^ 5'- 3" c 6 "" 6 " j. 1 5'-3 M 17-0" Fig. 168. Diagram Showing Loads on Floor-Beam. = 8 655 700 pound-inches. Note that the dead-load moment at the center of the beam is added to the concentrated-load moment at the point where the concentrated load is applied. This will give the total moment at the center of the beam as shown by Fig. 169, since the concentrated-load moment, is constant between the points of application. The end shear -is readily computed to be 136020 + 1 580 = 137 600 pounds. The curves of moments and shears are shown in Fig. 169. The total depth of the floor- beam, back to back of angles, is .rf Center toCcnUr Fig. 169. Shear and Moment Diagram. 48 j inches ; and the effective depth will, for approximate computa- tion of the flange area, be taken as somewhat less, say 44^ inches, since the flange angles will probably be 6 by 6-inch and the center of gravity of most of these angles lies about If inches from 8 655 700 the back. The approximate flange stress is pounds, and the required net area (17) 44.5 will be = 194 500 194 500 16000 200 BRIDGE ENGINEERING = 12.2 square inches. In assuming the size of the angle, it is to be remembered that when, as in this case, no cover-plates are used, no rivet-holes will be taken out of the top flange, and only one rivet- hole will be taken out of the vertical flange. Two j6 by 6 by f -inch angles give a gross area of 7.11 square inches each, and a net section of 7 . 1 1 . 625 = 6 . 485 square inches each, or 12 . 97 square inches net for both. As this is near the required area, these angles will be taken; and a recomputation will now be made with the actual effective depth, in order to see if sufficient variation in the areas occurs to require another angle to be taken. The actual effective depth is now 48} 2 X 1 .84 = 44. 57 inches; and making computations with this, it will be found that a net area of 12.10 square inches is required. As this is practically the same as was determined at first, no 75KO I Li i Field |/7ShOp change will be made in the size of the angle. The web is to be designed for a total shear of 137600 pounds. The required area (18) is 137 600 lo 7b souare inches ?" Web of I-beam^ fl a* Web ot I-beam;, ^^RJ 7 Shop-' fc: I F5hop Ft eld /Floor Beam Web 10000 and the required thickness is Pis. 170. Calculation of Number of Rivets UIK> 05t Fig. 171. Position and Number of Rivet to Connect End Angles with Posts. Impact = 51 400 X 300 Dead Load of Stringers = 300, BO X 2 X 21 Dead Load of Track = 48 000 1 6SO 2 100 Total 103 ISO pounds. The maximum moment due to the above load is 103 180 X 5.25 X 12 = 6 500 000 pound-inches. The weight of the beam may be assumed as 3 160 pounds. This is the same as was computed for the intermediate floor-beam, but will be used for this beam, since the size of the web will be the same as in the others; and, although the flange area will be less, the end connections will be somewhat heavier owing to the connection of the beam to the end-post and the roller bearing, 212 BRIDGE ENGINEERING 203 and this additional weight will cause the total weight of the end floor-beam to be about the same as that of the intermediate ones. The total moment at the center will then be 6 500 000 + 80 700 = 6 580 700 pound-inches. The depth of the end floor-beam will be somewhat greater than the depth of the intermediate floor-beams. This is due to the fact that it extends downward a greater distance, resting upon the bearing plate, which comes directly upon the top of the rollers. The exact depth cannot, of course, be determined until after the roller bearings are designed; but it may be safely assumed as four or five inches deeper than the intermediate floor-beams, and in case this is not enough, the draftsman can easily fill in the remaining distance with filler plates, as this distance will not be very great. In case this depth is too great, the flange angles may be bent upward at the end, or a re-design may be made. The depth will be assumed as 52 inches in this case. The effective depth will be assumed as 48 inches, and this gives an approxi- mate flange stress of 658 70 =137 000 pounds, O and an approximate net flange area required of 137 000 ..- _ = 8.57 square inches. A 6 by 6 by -^ -inch angle gives a gross area of 5.06 square inches, and a net area of -5.06 (f + |) r 7 T = 4.62 square inches. A recomputation with the true effective depth requires 8.42 square inches net. Two of these angles give 9.24 square inches; and as this coincides very closely with the required area, it will be used: The size of the web plate is 52 by f-inch. The pitch or spacing of rivets in the flanges is: 7 880 X 48 IMW = 3 62lnches - The maximum end shear as above computed is taken by two stringers; and therefore the number of rivets required in bearing to form the connection between the stringers, connection angles, and the floor-beam web is, for each angle: 104 760 2,-^sTr = 8 field rivets. 2 X 6 560 213 204 BRIDGE ENGINEERING The value 6 560 in the above equation is the value of a f-inch field rivet in bearing in the f-inch web. The number of rivets required in the end angles on the floor- beam is : 104 760 -788F = 14sh P rlvets - These rivets go through the web of the floor-beam. The connection of the floor-beam to the end-post is made by means of field rivets and a large gusset plate. This gusset plate is usually f inch in thickness. The number of rivets through the end connection angles and this gusset plate is governed by single shear, since the rivets will shear off between the angles and the gusset plate before they will tear out of the gusset plate, as the value of a rivet is greater in bearing than in shear. The number required is: The general arrangement of the intermediate floor-beams is shown in Fig. 172. The ends of the lower flange are bent up as shown, in order to allow the I-bar heads or any other section of the lower chord to have clearance. This makes it necessary for the floor- beam web to be spliced at the ends, as shown. The distance which this plate should extend above the floor-beam proper depends upon the distance which the lower chord is bent up. In any case the length of the connection on the post should be at least equal to the depth of the floor-beam. Two splice plates, one on either side of the web, are placed here in a manner similar to that of a splice as designed in the plate-girder when shear only was considered. Here shear only is considered, and the number of rivets which must be on each side of the splice will be: 137 640 ~7~880~ = P " The 7 880 which occurs in the above equation is the value of a f-inch rivet in bearing in a f-inch plate (19). Inspection of Plate II (p. 172) will make this design clearer. Plate II also shows the shape of the end floor-beams. The small shelf angle shown in Fig. 172 should have sufficient rivets to prevent any twist of the stringers due to their being con- nected on one side of their web only. This number is a matter of 214 BRIDGE ENGINEERING 205 judgment. Experience seems to indicate that enough rivets to take up one-third of the total reaction of the stringers will be sufficient. This will require shop rivets, and the number will be : 103 180 ' 3X7 220 = P nve m sm S le shear. 84. The Tension Members. Tension members usually consist of long, thin, flat plates with circular heads forged upon their ends. Pig. 17S. General Arrangement of Rivets, Splices, Connections, etc., for Intermediate Floor-Beams. These circular heads have holes punched through their centers and then very carefully bored. Through these holes are run cylindrical bars of steel called pins. These pins connect them with other mem- bers of the truss. See Carnegie Handbook, p. 212, for table of I-bars. The I-bars given are standard I-bars ; and while departures from these widths and minimum thicknesses may be made, it may be done only at great cost to the purchaser. Note that there are no standard 9-inch I-bars. The thicknesses given are the minimum thicknesses 215 206 BRIDGE ENGINEERING for that width of bar, and do not indicate that thicker bars of that width cannot be obtained; but on the contrary thicker bars of that width can be obtained, and this should be done, the minimum thick- ness as given in the table being avoided if possible. It has been found that bars which have a ratio of thickness to width of about one-sixth give good service and are easy to forge. This relation gives us a rough guide which will enable us to determine the approximate width and thickness of any bar of a given area. Once the approximate dimensions are determined, the actual dimen- sions can be chosen from the market sizes of the material (see Car- negie Handbook, pp. 245 to 250). An expression for the approximate de,pth of the bar will now be derived by using the above relation. Let A = Area of bar, in square inches; d = Width of bar, in inches; t = Thickness of bar, in inches. Then, also, Substituting the value of t in the expression for A, there results: d = VGA. The stresses in all the members in the truss under consideration are computed by the method described in Part I, and are placed on the stress sheet, Plate III (p. 251). In the succeeding design, the student should obtain his stresses from Plate III without his attention being again called to the matter. Table XXIV gives the tension members and their dead-load, live-load, impact, total, and unit stresses (15), together with the required area, the number of bars, the approximate depth of bars, and the final sizes used. The first seven columns in Table XXIV are self-explanatory. The number of bars to be used in any particular case is a matter of judgment. One fast rule is that an even number of bars should always be used, except in the case of counters, where one is permis- sible. This is due to the fact that the placing of one of the main members in the center of the pin would create a large moment, and 216 BRIDGE ENGINEERING 207 ' * i s y 03 : :3S? : : oi ^-i o o t^ co i eS therefore an ex- cessively large pin would be required, 0* - ffconjoo H A and accordingly a is #> '. 1 X X X X X X || t^ ,-( CD CO O * 5 very large head on the I-bar in pro- ^ 1(0 i. ^3 Tf O "3 O O Tf< O O5 1C CO : : t>i ; o >o 5b 0) ^2 S portion to its width all of which are very undesirable AREA*; ONK BAH ^ CO ID S 1 - and costly. In gen- eral the number of s l i ! -(^ifOOOCO i I i-H i-H jg X E 2* H 5 - OOOtOCOCDOCD 5 '5 from the ends to- ward the center of ea 1 i the truss, and the < 2 . | H O H TJH^HfllCr-IOC02GO COTt<05(NC:i>Tj S o3 ^ h^- 02 for the member L 2 U 1; for example, the required area of one bar is: M * ^ 'o one IAA H 4 ts tS* S tT sq kf 3 ^^ ^3 = 9 22 H S Kf sq K5 K? Kf ^ 4" 4 1 square inches. 217 208 BRIDGE ENGINEERING The approximate depth of this bar is determined by taking the square root of 6 times the area as above determined. It is: d = V 6 X 9 . 22 = 7 . 44 inches. As this is nearer 7 than 8 inches, a 7-inch bar will be chosen; and looking in the first column, Carnegie Handbook, p. 248, for an area which will be equal to or in excess of 9.22, it is found that a If -inch bar satisfies this condition, and therefore the section of this member consists of two bars 7 by If inches. According to (80), the first two sections for the lower chord are to be made of built-up members. This requires that instead of I-bars they are to be made of angles and plates, or, in case the stress is light, of channels. The depth of the section is limited by the size of the greatest I-bar head. As the diameter of the I-bar head depends upon the size of the pin, it cannot of course be determined accurately before the pin is designed. It is customary to assume the largest head, and to design the section so as to clear this. The size of the largest head for bars of given width is given in the Carnegie Hand- book, p. 212. The design of the member L Z/ 2 will depend upon the size of the largest head of the 7-inch I-bar of the member U { L V This is 17^ inches; and in order that the head may have some clearance, it will be necessary to add \ inch to the top and the bottom, making a total of 18o inches. Since the flange angle, as in the case of plate-girders, will extend over the plate about \ inch, the plate itself may be 18 inches wide and still give sufficient clearance. The total stress is 234 200 pounds, and the allowable unit- stress (15) is 16000 pounds per square inch. The required net area, then, is: 234 200 16 000 = 14.64 square inches. According to the Specifications, the thickness of the plate cannot be less than f inch. The gross area of two 18 by f-inch plates is 13.5 square inches, and the gross area of four 3^ by 3^ by f-inch angles, which are assumed to be sufficient, is 9.92 square inches, thus making a total gross area of 23 . 42 square inches. If 5 rivet-holes are assumed to be taken out of each web, and one rivet-hole taken out of each angle, this will require a certain number of square inches to be 218 BRIDGE ENGINEERING 209 less e-Pb or less deducted from the section, and this is computed as follows: Out of webs, 2 X 5 ({ + i) X | = 3.75 sq. in. Out of angles, 4 ( + i) X | . . = 1 .50 " " Total = 5.25 sq. in. The net area of the section is now determined to be 23 .42 5 .25 = IS. 17 square inches. This is somewhat greater than the required net area, but must be used, for according to (39), these are the smallest and thinnest angles that may be used. Figs. 173 and 174 show the cross-section and the general detail at L 2 . The width of the member cannot be determined until after the section of the end-post is computed, since it must fit inside of the end -post, the horizontal legs of the angles being cut off to allow this. The end-post, Arti- cle 87, is 14^ inches inside. If it is assumed that all the pin-plates on the end-post are placed on the outside, and all those at L on L L 2 are on the inside, then the width of L L 2 , back to back of plates, must be 14 - (2 X V + 2 X |) = 12}- inches or less, J-inch clearance being allowed between the sides of the angles and the web plates of the end-post (see Fig. 173). The total net section through the pin-hole at L 2 (26) must be 1 X 18.17 = 22. 7 square inches, or 11.35 square inches for one side. The plate which is to increase the section must be on the out- side, since the intermediate post U 2 L 2 and the two I-bars of member UJ^ must go inside. The gross width of this plate is 1 1J inches (see Fig. 174), and the net width is 2w = 11| - 5 = 6 inches. The net area through the pin is: Two 3J by 3 by f-in. angles = 4 . 96 square inches. One 18 by $-'m. plate = 9 - 5 X i = 6.50 " " Total = 11 .46 square inches Since this is greater than the 11 .35 required, no plate will be necessary to fulfil (26) in this respect. Fig. 173. Cross-Section Showing Construc- tion of Lower Chord Member. 219 210 BRIDGE ENGINEERING BRIDGE ENGINEERING 211 Sufficient bearing area must be provided at this point. The total stress is 234 200 pounds, the total bearing area required is 234 200 = 9.76 square inches, and the total thickness for one side is 24 000 '-' = . 976 inches. Since the thickness of the web is i inch, the 2X5 pin-plates must be . 976 . 50 = . 476 inch (say \ inch) thick. A |-inch pin-plate must be used, and as the total thickness of the 50 234 200 bearing area is now 1 .00 inch, this pin-plate will take ' X ' = 58 550 pounds. The joint is weakest in shear, and will therefore 58550 require-. = 8 -t- (say 9) shop rivets. In case it is necessary to put the member U^ on the outer side of LyL^, then the outer legs of the upper angles must be cut off to allow U^L 2 to pass. This will decrease the section by an amount (3^ f ) X f = 1 20 square inches. Considering the pin-plate, which is (18i - 2 X 3*) - i = Hi inches, the | inch being allowed for clearance between the edges of its flange angles, the total net section through the pin-hole on one side will be: One Angle 3 by 3 by f-in. = 2.48 square inches One Cut Angle 3| by 3J by f-in. = 1.31 " One Web (18 - 5) J sq. in. =6.50 " One Pin-Plate (11J - 5) f sq. in. = 2.34 " Total = 12.63 square inches. This is greater than 11 .35 as required, and is therefore safe. The distance from the center of the pin to the end must now be determined (26). The total net section of the body of the member is 18.17 square inches, or 9.09 square inches for one side, and the thickness of the web and the pin-plate is 1 inch. The distance from the pin to the end of the member is then - =9^ inches, and the distance to the center of the pin is 9& -f -^ = Hf , say 12 inches (see Fig. 174). Rivets should be countersunk where necessary to prevent interference with I-bars. For signs, see "Steel Construction," Part III, p. 192, and Carnegie Handbook, p. 191. 221 212 BRIDGE ENGINEERING At point L of this member, the pin is 6^ inches in diameter, and, as previously mentioned, the legs of the angles are cut (see Fig. 175). 9.76 The total bearing area required for one side is '- = 4 . 88, and the 4 88 ' required thickness is . 781 inch. Subtracting the thickness of the ^-inch web from this gives 0.281 inch. A pin-plate f inch thick must be used. The net area through the pin (26) must be 11 .35 square inches. Fig. 175. Elevation and Section Showing Pin Connection at End of Truss. This net area, remembering that the angle legs are cut and therefore their area is that of a bar 3-2- by f-inch, computed for one side, is as follows : Two Angles, legs cut, 3| by 3 by f-in. = 2 . 62 square inches One Web 18 X i - 6i X i sq. in. = 5.88 " " One Pin-Plate Hi X $ - Gi X f sq. in. = 1 .87 " " Total 10 . 37 square inches. This shows the section to be deficient, and the thickness of the pin- plate must be made $ inch. This gives a net area through the pin of 11 .62 square inches. The distance between rivet lines (see Fig. 173) is 17 \ inches, and (44) the tie-plates must be 17J (say 18) inches long, and their 1 7 thickness 50 0.346 inch (say f inch). BRIDGE ENGINEERING 213 The lattice bars (45) must be 2 inches wide, and (47) must be double. From (45) and Table XXV, page 219, the thickness must be r 7 6 inch, the distance c being 17.25 X secant 45 - 2 ft. O t V in. The design of the hip vertical U^ is also made in accordance with (80) of the Specifications. It will be assumed that the section consists of one 8 by f-inch plate, and four 3 by 3 by f-inch angles, since this is the lightest section that may be used according to the Specifications, the 8-inch plate being chosen as it gives some clearance between the inner edges of the legs of the angles. The total stress in the member is 141 600 pounds, and the unit- stress is 16000 pounds per square inch, thus requiring a net area of 8.85 square inches. The plate gives a net area of 2 . 20 square inches, and the four angles give a net area of 6.88 square inches, making a total of 9.08 square inches, one rivet-hole being taken out of each angle, and two out of the web, at any particular section. The net area is somewhat greater than that required, but must be used, as this is the minimum sec- tion allowed by the Specifications. Fig. 176 shows a cross-section of this member as above determined. This member will be connected to the upper chord and end-post by means of a pin \vhich is 6| inches in diameter, the diameter of the pin being determined later. The total stress is 141 600 pounds, and this will be taken by two plates, one on either side of the member. The net section of the member is 9.08 square inches, and the section through the pin (26) must be 25 per cent in excess of this, making a total of 11.35 square inches, or 5.68 square inches for each plate. The total width of these plates will be taken as 12 inches, and this (see Fig. 177) will make the required thickness: 5.68 5.68 Cross-Section, of Hip Vertical. Fig. 177 shows the details of these pin-plates. Since the above thickness is too great to be punched in one single piece, the above thickness will be made up of two plates, each inch thick. The area at section A-A must be equal to that of the body of the bar. It 214 BRIDGE ENGINEERING is 12.00 X ;> = 6.00 square inches for one side, or 12.00 square inches for both sides. As this is greater than the 8 . 85 square inches as above computed, the area at A-A is sufficient, as is also the width of the plate, which was assumed as 12 inches. One of the plates will be riveted directly to the member, and the other will be riveted to it as a pin-plate. The section back of the pin (26) must be equal to the net section in the body of the member. The net section is 4.54 square inches for one side, and the total thickness Fig. 177. Connection of Hip Vertical to Upper Chord and End-Post. of the pin-plates is 1 . 125 inches, making the distance from the end 4 54 of the member to the pin - -' = 4|- inches, and the distance to 1 .125 the center of the pin 4| + -^- = 7| inches. The joint between the plates and the main member will be weak in shear, the rivets tending to shear off between the f-inch angles and the plate, and also between the two plates themselves. As each side takes one-half of the above stress, the number of rivets required to connect the plates to the main member will be : 141 600 - 2 7220 = 10 shop rivets, and the number of rivets required to connect the inner ^-inch plate to the outer one which is connected to the member itself will be: 141 600 -T- 4 7 220 = 5 shop rivets. 224 BRIDGE ENGINEERING 215 The distance from the center of the pin to the top of the main part must be greater than one-half the diameter of the largest I-bar head that is, 17^ -r- 2 = say, 9 inches. At the lower end, this member is connected to the bottom chord by means of a couple of clip angles and four or five rivets. Only sufficient rivets are required to prevent the sagging of the bottom chord, since the floor-beam is connected to the hip vertical above the lower chord, and hence no stress comes on the joint at the lower end (see Fig. 178). The width of the plate has been assumed as 8 inches. This width is liable to be changed after the design of the intermediate Fig. 178. Connection of Hip Vertical to Lower Chord. posts has been made, since it will be economical to have all the inter- mediate floor-beams of the same length; and therefore the width of this plate will be changed so as to make the width of the hip vertical the same as the width of the intermediate posts. 85. The Intermediate Posts. The post U 2 L 2 must be designed to stand a total stress of 163 GOO pounds. Where possible, it is economical to make the intermediate posts out of channels, as this saves a large amount of riveting. As seen by the stress sheet, the length of these posts is 30 . 1 feet center to center of end pins. It is usually required that must not be more than 100, and this con- dition requires that the least radius of gyration cannot be less than 30.1 X 12 100 3.62. 210 BRIDGE ENGINEERING From Carnegie Handbook, p. 101, it is seen that a 12-inch 30- pound channel has a radius of gyration of 4.28, and will fulfil the conditions. The area of two of these channels is 17 . 64 square inches. The unit allowable stress (16) is: 30.1 x 12 16 000 - 70 X 4.28 = 10 090 pounds per square inch. 1 63 600 The required area is then determined to be -7^7^ = 16.2 square V-5ShojD^-9 Shop he Field 1 go ^ , ,s Shop 9 Shop LizField^ Fig. 170. Cross-Section of Intermediate Post, Showing Diaphragm. inches; and as this coincides very closely with the area given, these channels are efficient and will be used. Fig. 179 shows the cross-section of this post. The radius of gyration which was used above was the radius of gyration of the chan- nels about an axis perpendicular to their web. The radius of gyra- tion of the entire section about an axis perpendicular to the web will be the same as that of one channel. In order to have the sections safe, the radius of gyration about the axis B-B must be equal to or greater than the other. The radillS of gvration 8 " about the axis B-B can be in- creased or decreased by spacing the channels. The exact distance which will make the two rectangular radii of gyration equal may be determined by the methods of "Strength of Materials," or it may be found in columns 14 and 15 of the Carnegie Handbook, p. 102. For any particular case it is equal to the value given in column 14, plus four times that given in column 15. For the channels under consideration, it is equal to 7.07 + 4 X 0.704 = 9. 89 inches. Any increase in this distance will only tend to increase the radius of gyration about the axis B-B, and will make the post safer about that axis. Fig. 179 shows a diaphragm. The web of this diaphragm cannot be less than f inch, and the size of the angles cannot be less than 3^ by 3^ by f -inch, as this is the least allowed by the Specifications. The function of this diaphragm is to transfer one-half of the floor-beam reaction to the outer side of the post. The rivets which connect the 226 BRIDGE ENGINEERING 217 angles to the diaphragm web are shop rivets, and (see design of floor- 1 O*7 P.f)f) beam) must be- -- ^^7: = 9 in number. The rivets which connect 2i X 7 ooO the diaphragm angle with the outer channel of the post are also shop 137 600 rivets, and are ^ oorT = ^ m numDer > 5 on each side. The same rivets which connect the floor-beam to the post go through the diaphragm angle on that side of the diaphragm next to the cen- ter of the bridge, and must therefore be field rivets and take the 137 600 entire floor-beam reaction. These must be^ -= 23 in number, 6 Olo 12 on each side. The exact distance, back to back of the channels of the post, cannot be determined until after the top chord has been de- signed, since the post must slide up in the top chord and also leave room on each side for the diagonal members of the truss. The width is determined by the packing of the members at joint L 2 (see Fig. 174), and is found to be 9| inches. Since this is less than that required above, the post must be examined for bending about an axis parallel to the web of the channels. According to the methods of "Mechanics" and "Strength of Materials," with the help of the Carnegie Handbook, p. 102, the moment of inertia about this axis is found to be 286.42, and the ra- dius of gyration 3 .96. The unit allowable compressive stress is then computed to be 9 580 pounds per square inch, and the required area 1 no p.r\f\ --- = 17.10 square inches, which, being less than 17.64, shows the section to be safe. This member is connected to the top chord at its upper end by a 5-inch pin. The total stress is 163 600 pounds, and the total bear- , . 163 600 ing area required is -^77^- = 6.8 square inches, or 3.4 square inches for each side (19). The total thickness of the bearing area for each side is ^ = 0.68 inch. The thickness of the web of a 12-inch o 30-pound channel is 0.513 inch, which leaves 0.68 - 0.513 - 0. 167 inch as the thickness of the pin-plate, but it must be made f inch according to the Specifications. Fig. 180 shows the arrangement of the plates and the rivets. 218 BRIDGE ENGINEERING The sum total of the pin-plates and the channel web is 0.888 inch, and therefore on one side the stress transferred to the pin by . 1 0.375 X 163 600 means of the pin-plate, which is 0.3/5 inch, is - X ^ U . ooo = 34 600 pounds. This plate will tend to shear off the rivets between 34 600 it and the channel web, and therefore = 5 shop rivets are , / ZiZ\} required. The stress that is shown on the stress sheet is the stress in the post above the floor-beam. The stress in that part below the floor- T v I'jrij l'l r l 1 ' / 1 \ \ | J !! i ft -V- J liiF y 1 1 M J ! 01 Li \ ) l| y r 1 ~-- N 1 ! |!'l 1 1 l^i 10 01 1 ^~-~-~ ( 1- t' b 1 1 , \ ;> , x ii 10 |o 10 1 0! o! 1 1 I v'" ^ L J Fig. 180. Arrangepient of Plates, Rivets, Pin, etc., at Connection of Intermediate Post to Top Chord. beam is equal to the vertical component of the diagonal in the panel ahead of the post in question. In this case it is the vertical com- ponent of the stress in V \L 2 , and is equal to 242 000 pounds, and this 242 000 requires a total bearing area ot -7^77^- = 10.1 square inches, and a total thickness of 24000 1.01 inches on each side, the pin being 2X5 5 inches in diameter. From this total thickness must be subtracted the thickness of the web of the channel, and this leaves 1.01 0.513 = 0.497 inch as the total thickness of the pin-plates required. This shows that we must use one ^-inch plate. The total thickness of the bear ing area is now 0.51 3 + 0.50 = 1.013 inches. Each plate takes a total stress of 1 .013 949 nnn X = 59 700 BRIDGE ENGINEERING 219 pounds; and the joint being weak in shear, the number of rivets 59 700 required will be = 9 rivets in single shear. The detail will be / _ V ) similar to that in Fig. 180. The distance, back to back of the channels in this post, will probably not be greater than 12 inches, and this will make the dis- tance between rivet lines about 9 inches. According to (44), the end tie-plates must be at least 9 inches long and of course 12 inches wide. g The thickness cannot be less than = . 18 inch, but they will be 50 made f inch (36). Between the tie-plates the channels will be con- nected by means of lattices. The Specifications (45) require that they should not be less than 2\ inches in width and (1 .414 X 9) = 0.318 (say f) inch in thickness. Table XXV gives the thickness of lacing bars for any distance between rivets. TABLE XXV Thickness of Lacing Bars xxv<> LJL SINGLE LACING ('=*>' *=>) DOUBLE LACING (*=^: = 45) t t c i in. Oft. 10 in. i in. 1 ft. 3 in. A in- 1 ft. in. A in- 1 ft. 6| in. t in. - 1 ft, 3 in. 1 in. 1 ft. lOJt in. I 7 in. 1 ft. Si in. T 7 s in- 2ft. 2i-in. Jt in. 1 ft, 8 in. i in. 2 ft. 6 in. T in- 1 ft. 10* in. T in. 2ft. 9|in. f in- 2ft. 1 in. t in. 3ft. liirf. A width of 2} inches is chosen above, since according to Carnegie Handbook, p. 183, a |-inch rivet is the largest which can be used in the channel flange. The post U 3 L 3 must be designed for a total stress of 87000 pounds. It will be assumed that two 10-inch 20-pound channels 220 BRIDGE ENGINEERING with a radius of gyration 3 .66 and an area of 5 .80 square inches each will be sufficient. The length, as before, is 30.1 feet, and the unit- stress is: P = 16 000 - 70 X - = 9 080 pounds. 3.66 The required area is =9.60 square inches. Since the y 080 total area of the two channels is 11 .76 square inches, and the required area is 9.6 square inches, it is seen that they do not coincide very closely. These channels, however, will be used, since the thickness of the web is the thinnest allowed by the Specifications, and the width of the channels is the smallest that can be used and still give sufficient room to make the connections with the end connection angles of the floor-beams. The lower end of this post also has a diaphragm which must transfer half of the stress to the outer channel of the post. The sides of the diaphragm are the same as in the posts previously de- signed ; and the number of rivets ^ ~)/ 5 P^~ 3 P | ^ p..^ required is computed in a simi- \ j"? ") -v lar manner and found to be as ^sSho 9^(5 I I' rie indicated in Fig. 181, which shows the cross-section of this post. At the upper end the bear- Fig. 181. Cross-Section of Intermediate . Li , Post. ing area required on one chan- cy i nn nel is- = 1.814 square inches, and the thickness required *- /\ ^4 UUU 1 814 is = 0.363 inch, a 5-inch pin being used. As the web of the 5 channel is 0.382 inch thick, it will give sufficient bearing area without pin-plates. At the lower end, the vertical component of 7,Z, 3 is 157 500 pounds. The bearing area required on each side of the post is 157 500 3 28 3.28 square inches, and the thickness is = 0.66 s\ ^i~L UUvJ inch. The thickness of the channel web being . 382 inch leaves . 660 0.382 = 0.278 inch as the required thickness of the pin-plate; BRIDGE ENGINEERING 221 but f inch must be used, making a total thickness of 0.382 + 0.375 = 0.757 inch. The plate will carry ~^. 5 X ~ = 39 000 39000 pounds, and this requires = 6 shop rivets in single shear. The distance, back to back of channels, will be the same as in T T r /s / \ / \ / \ / , ( \ / \ \ \ / / \ \ / / \ \ s / \ x / / \ \ -cT 1 '%' . / i "T s*. i^ 7' ? / | 1 Jx s Lfe |i i^ 1 P * Is 4 1 ( V 04, ) C ^ <^ ] Fig. 182. End and Side Elevations Showing Detail of Construction at Lower End of Intermediate Post. C/ 2 L 2 , and therefore the tie-plates and lacing bars will be the same. Fig. 182 gives a detail of the lower end of U 3 L y 86. The Top Chord. The top chords of small railway bridges may be made of two channels laced on their top and bottom sides. This is not very good practice, since it leaves the tops of the channels open and lets in the rain and snow, wjiich tends to deteriorate the joints. It is better to add a small cover-plate, even if this does give 222 BRIDGE ENGINEERING an excessive section. In case of stress such as is demanded, the chords may consist of two channels and a cover-plate. In this case it is necessary to place small pieces called flats upon the lower flanges of the channel, in order to lower the center of gravity of the section and to bring it near the center of the web. This section makes a very economical section in that it saves much riveting. On account of channels being made only up to 15 inches in depth, 'the use of this section is quite limited owing to the fact that it is not deep enough to allow the I-bar heads sufficient clearance, for the I-bar heads in bridges of even ordi- nary span will exceed this amount. The most common section is that which consists of two side plates, four angles, and one cover-plate. Sometimes this section has flats placed upon the lower angle in order to lower the center of gravity, as ex- plained above. According to (33), the sec- tion should be as symmetrical as possible, and the center of gravity should lie as near the center of the web as is consistent with economy. In case the stress is great enough to demand a heavier section than that above described, additional plates are added upon the sides of the original plates, and heavier and larger cover-plates and angles are used. Fig. 183 shows different types of chord sec- tions. Ir addition to the cover-plate being designed to withstand the total stress, close attention must be paid to (42). This clause has been inserted on account of practical considerations, since it has been found out BRIDGE ENGINEERING 223 that if plates are made much thinner than the proportions here required, they will crumple up and fail long before the allowable unit of stress as computed from the formula has been .reached. In some cases especially where the stress is light the proportions laid down in (42) and (36) will govern the design of the section, instead of the required net area as determined by the formula for the allowable unit compressive stress. The design of the first section of the top chord will now be made. Here, as in the case of the first sections of the lower chord, the diameter of the head of the greatest I-bar determines the width of the plates in the section The head of the 7-inch I-bar w T hich constitutes the member U t L 2 is 17^ inches, and, allowing a clearance of \ inch on either side of the head, the total depth inside the chord should be 18J inches. As in the case of the lower chord, plates 18 inches wide may be used. The size of the angles to be chosen is a matter of judgment. Usually any size should be chosen at first, and the preliminary design will indicate at once what size should have been taken. For this case, 3a by 3i by f-inch will be assumed at first. For sections of this character, the radius of gyration is approxi- mately equal to OAh, in which h is the height, or rather the width, of the side plate. The approximate radius of gyration is r = 0.4 X 18 = 7. 2 inches, and the length is equal to one panel length, or 21 feet. The allowable unit of stress (16) is: P = 16 000 - 70 X 21 X 2 12 = 13 550 pounds. 449 500 The required area is-r^^vr- = 33.2 square inches. The correct proportion for sections of this character is that . 4 of the total area should be taken up by the web. The area of the web would then be 0.4 X 33.2 = 13.28 square inches, and the thickness would be 1 *3 28 = 0.37 inch. According to this, a f-inch plate should be used, 2 X 18 14 ^ but (42) requires that it shall be.-^r- = 0.483 inch or thicker. Therefore an 18 by i-inch plate must be used for the web. The correct proportion for sections of this character is that the width between plates should be about f the width of the side plates. 224 BRIDGE ENGINEERING This will give the required width between plates equal to X 18 = 15.75 inches. The cover-plate (42) must not be thinner than the distance between the connecting rivet lines. The rivet lines are, in this case, 15.75 + 2 X 2 = 19.75 inches apart, and therefore the 19 75 thickness of the cover-plate cannot be less than 40 0.494 inch. The cover-plate will therefore be taken as \ inch thick. The width of the cover-plate (see Fig. 184) | t i 9 j?" d must be about 15.75 + 2 X 3 + \ = 231 inches (say 23 inches). The cover-plate will be taken 23 by i-inch. The center line of pins will be taken at the center line of the w 7 eb, and the center of gravity of the section will be assumed as ^ inch above this. In order that the center of gravity may be near that assumed, the moment of the cover-plate about the assumed center of gravity axis should be about equal to the moment of the flats about the same axis. The moment of the cover-plate about the assumed axis is: gPlS. 18"xgPls Neutral AMS^ Center Line of Pins? qpprox. Fig. 184. Section of Top Chord. 23 X 23 V Q (9.0 - 0.5 + 0.25 + 0.25) = * ; and the moment of the flats about the same axis is: A (9.0 + 0.5 + 0.25 + 0.5) = 10.25,4, in which A is the area in square inches of both of the flats. Equating these two expressions, and solving for A, there results: 23 X 9 A = .- 10.1 square inches. 2 X 10.25 Assuming the flats to be 4 inches wide, the thickness on each side will be 1 .25 inches. As this is too thick to punch, the flats on each side will be composed of two 4 by f-inch plates. The total area is : 234 BRIDGE ENGINEERING 225 One cover-plate = 2.3 X i = 11.5 sq. in. Two web plates = 2 X 18 X i =18.0 " " Two flats 4 X 1J = 10.0 " " Total 39 . 5 sq. in. But the required area is 32.2 square inches, which is considerably less than the area above given, and which does not include the angles and hence we can use the smallest size angles, which are those pre- viously assumed. The area of each of these angles is 2.48 square inches, thus making the total area of the section 39 . 5 + 4 X 2 . 48 = 49.42 square inches. This is considerably in excess of the area as required according to the formula for compression; but it is the least allowed by the Specifications. Note that this is the case where (42), instead of the formula for compressive stress, is the ruling factor in the determination of the section. The center of gravity of the approximate section must now be determined, the moment of inertia and the radius of gyration about the neutral axis must be computed, and the required area must be determined by using this radius of gyration as computed. If the required area as determined with the actual radius of gyration is less than the approximate area, then the thickness of the angles or the plates must be increased and the section then examined for its radius of gyration and required area. If the area is sufficient, the section is used; if not, another recomputation is in order. In the determination of the center of gravity of the section, the moment is taken about the top of {lie cover-plate. The moments are computed as follows: Cover- plate (23 X *) X \ 2 . 88 Webs 2 (18 X i) X (9 + |) 175.60 Top angles 2 (2.48) X (1.01 + i) 7.50 Lower angles 2 (2.48) X (J + i + 18 H- \ - - 1.01) 89.30 Flats" 2 (4 X H) X 19| 196.25 Total ...... 471.53 The center of gravity is newfound to be 7^-^. - =9.55 4 X 2.48 inches from the top of the cover-plate. The distance from the top of the cover-plate to the middle line of the web is9+j + ^ = 9.75 inches, and this leaves a distance of 9.75* 9.55 = 0.2 inch from the center line of the web to the neutral axis. This distance is gen- 226 BRIDGE ENGINEERING erally represented by the letter e, and it is known as the eccentricity of the section. The moment of inertia about this axis must now be computed. The relation used is that the moment of inertia about any axis is equal to the moment of inertia about some other axis, plus the product of the square of the distance between the two axes by the area of the section whose moment of inertia is desired. The moments of inertia of the various parts of the section (see "Steel Construction," Part IV, pp. 292 and 293) are computed and are as follows: Cover-plates 955 . 26 Webs 486 . 72 Top angles 325 . 74 Lower angles '. 359 . 74 Flats... ..-1017.37 Total. . .. 3 184.83 The radius of gyration is equal to the square root of the quotient obtained by dividing the moment of inertia by the area. It is r = J 3 184 ' 83 = 8.04. Af 49.42 Using this value of the radius of gyration in the formula for the com- pressive stress, there is obtained 13 800 pounds as the unit allowable , 449 500 stress in compression, and this requires an area or =32.5 13 800 square inches. Since this is considerably less than the actual area of the section, the section will not be changed but will be taken as first assumed. In order that the section should be safe about both axes, the moment of inertia about the axis perpendicular to the cover-plate should be equal to or greater than that as above computed. By com- puting the moment of inertia about the axis perpendicular to the cover-plate, it is found to be 3 256 .3, which gives a radius of gyration of 8 . 1 1 ; and since both of these are greater than those first computed, it is seen that the section is safer about the axis perpendicular to the cover-plate than it is about an axis perpendicular to the web plates. There are small stresses in this member due to its own weight and to the fact that the pins are not placed directly upon the neutral axis (see "Strength of Materials," p. 82). These stresses are seldom more than 1 000 pounds per square inch in the extreme fibre; and 236 BRIDGE ENGINEERING 227 since the section has such an excess of area, they will not be computed, as it is evident that there is sufficient strength in the member to with- stand them. The section just designed is that for the top chord having the greatest stress; and since this is the minimum section allowed by the Specifications, it must be used in all the sections of the top chord. The section as finally designed is: One cover plate, 23 by i inch; Two webs, 18 by \ inch; Four angles, 3 by 3^ by f -inch Four flats, 4 by f-inch. A pin 6j inches in diameter w r ill be used at the point U r The stress in the member UJJ 2 is 378 200 pounds, and the bearing area 378 ^00 required is - - = 15. 75 square inches, or 7.875 for each side. 7 875 This makes a total required thickness of ' = 1 . 265 inches for one side. Since the thickness of the web plate is ^ inch, it will be necessary to provide pin-plates whose total thickness must be 1.265 0.5 = . 765 inch. Two f -inch plates will give a thickness of . 75 inch ; and since this is less than the required thickness by an amount not over 2\ per cent, they may be used. The total thickness of the bearing area is now 1 . 265 inches. The stress transferred to the two f -inch plates is: s = JL X 189 100 = 113 500 pounds. The rivets required to keep the outer plate from shearing off the 113 500 other are - - - =8 shop rivets, and the rivets required to keep both ^ X / 2ZO of the | -inch plates from shearing off the web of the chord section are I 1 O !XAf) ~~7~99ff~ = ^ shop rivets in single shear. The bearing of a f-inch shop rivet on a |-inch plate is 10 500 pounds, and therefore the num- ber of rivets required to keep these pin-plates from tearing the rivets 113 500 out of the *>-inch web plates is =11 shop rivets in bearing. 10 500 Fig. 185 shows the detail of this end of the top chord section. The pin- plates should extend well back on the member, and at least one pin- 237 228 BRIDGE ENGINEERING plate should go over the angle, and enough rivets, as computed above, should go through the angles and this pin-plate. Experiments on full- sized bridge members go to show that unless the pin-plates cover the angles and extend well down on the member, the member will fail O before the unit-stress reaches that value computed by the formula for compression. Since the ends of the chord are milled at the splices, and therefore butt up against each other and allow the stress to be transmitted ,0000 O O gPlate 0..0 2 8 Plate p O \ O O O I Fig. Detail of Top Chord Section at Point U r directly, only sufficient rivets need be placed in the splice to keep the top chord sections in line (55). At the point U 2 , it is not necessary to put in a pin-plate to take the stress in the upper chord; but it is only necessary to provide a pin-plate to take up the difference in stress between the two chord sections. This difference in stress is equal to the horizontal com- ponent of the maximum stress in the member U 2 L 3 . This is 110 000 pounds, and the area required on each side for bearing is 2.3 square inches; and as a 5-inch pin is used here, the thickness of the bearing inch. As this thickness is less than the thick- 2 3 area is - = 0.46 5 ness of the web plate, no pin-plates will be required. At the point U 3 , a bearing area will be required to withstand the horizontal component of the member U 3 L 4 . This is 56 300, and the BRIDGE ENGINEERING 229 bearing area required on each side is 56300 24 000 X 2 1.18 = 1.18 inches. The required thickness of the bearing area is - J = 0.24 inch, as a 5- 5 inch pin is used here also. As this thickness is less than the thickness of the web plate, no pin-plate will be required. The under parts of these members must be stiffened by tie or batten plates, and these plates (44) must be equal in length to the distance between rivet lines. This is 19n inches. They will be made 20 inches long and 23 inches wide. The thickness of these plates (44) must be ' = 0.39 inch (say T V inch). The size of the tie-plates 50 will then be 20 in. by W in. by 1 ft. 11 in. Since the distance between the rivet lines is greater than 15 inches, double latticing must be used (47); and according to Table XXV the lacing must be \ inch thick; also, according to (45), it must be 1\ inches wide, as the rivets used are \ inch in diam- eter. The lattices will then be 2-fc by 4-in. 87. The End=Post. Since the minimum section as chosen for the top chord is about 50 per cent in excess of that required by the compression formula, it will be assumed to be sufficient for pig Calculation of End . Pos , the section of the end-post, and it will now be investigated to see if it is safe. In addition to the stress due to direct compression, the end-post is stressed by its own weight, by eccentric loading due to the pin being in the center of the web instead of at the center of gravity of the section, and to a bending moment at the place w T here the portal brace joins it. This is due to the bending action of the wind on the top chord. These different stresses will now be computed; and since the post is in all cases stressed by a combination of bending and compres-, sive stresses, this fact should be considered in the design. In deter- mining the stress in the end-post due to its own weight, the entire 239 230 BRIDGE ENGINEERING weight must not be used in computing the bending action, but only that component of it which is perpendicular to the end-post. The length of the end-post is readily computed, and is as shown in Fig. 186. The general formula for accurately computing stresses due to bending when the member is also subjected to compression, is: My, ~/T^T WE in which, S = Stress in pounds per square inch in the extreme upper fibre of the beam; M = Exterior moment causing the stress, and is considered positive if it bends the beam downward, and negative if it bends the beam up- ward; ?/,= Distance from the neutral axis to the extreme upper fibre; / = Moment of inertia of the section; P = Direct compressive stress, in pounds; I = Total length, in inches; E = Modulus of elasticity of steel, which is usually taken as 28 000 000 pounds per square inch. In this case the force causing the bending is that component of the weight perpendicular to the end-post. This is Wl sinf'>, in which W is the weight of the steel in the end-post; and this is computed and is as follows: Cover-plate 1 435 Ibs. Web plates '. 2245" Angles 1250" Flats... ..1245" 6 175 Ibs. Add 25 per cent for details 1 544 " Total. ... 7 719 Ibs. Substituting in the above formula the various values, there results: X 7 719 X 36.7 X 0.572 X 12 X 9.55 S = -- , 1C . 410500 X (36.7 X 12) 2 o loo 10 X 28 000 000 = 800 pounds per square inch compression in the tipper fibre due to bending. In the above equation, the stress in the member is 410 500 pounds; the distance y l is the distance from the neutral axis to the top of the cover-plate, and the coefficient of elasticity of steel is taken as 28 000 000. 240 BRIDGE ENGINEERING 231 In computing the stress due to the eccentric loading, the moment is equal to the product of the total stress in the member by the dis- tance~from the neutral axis to the center of gravity axis causing a negative moment. Substituting in the above formula for combined stresses, there results: - 410500 X 0.2 X 9.55 S = - 3 185 - 410500 X (36.7 X 12) 2 4 70 10 X 28 000 000 = 270 pounds per square inch tension in the upper fibre. In order to find the compression in the lower fibre, it is only necessary to notice that the stresses are proportional to the distances from the neutral axis. Accordingly (see Fig. 187), the stress in the lower fibre due to the weight is 895 pounds tension, and the stress in the lower fibre due to the eccen- tric loading is 302 pounds compression. Before computing the stress due to the bending moment caused by the wind on the upper chord, it is necessary to in- vestigate the post to see if it is fixed or hinged at its lower end. This is very important, since, if the post is found to be hinged, the bending moment will be one-half of that which will occur when the post is not hinged. An end-post is considered hinged when the product of one-half of the total stress times the distance between the web plates is greater than the product of the wind load acting at the hip, or joint U v times Fig. 187. Calculation of Stress in Chord. the length of the end -post. In this case the first value is 410 500 X 15 = 3075000; and the product of the latter (see Article 29) is 12 600 X 36.7 X 12 = 5 550 000. Since the latter is greater than the former, the post is hinged, and the bending moment at the foot of the portal strut, which joins the end-post 28.2 feet from the end, is 6300 X 28.2 X 12 - 2130000 pound-inches. The stress in the extreme fibre due to this bending moment is: 2 130000 x 11.5 3256.3 - 410500 X (36.7 X 12) a 10 X 28 000 000 241 232 BRIDGE ENGINEERING = 8 250 pounds per square inch tension or compression. In computing this stress due to the wind moment, care must be taken to take y t equal to one-half the width of the cover-plate, and to take 'the moment of inertia as that about the axis perpendicular to the cover-plate. In computing the total stress on the extreme fibre, it must be noted that the stresses due to weight and eccentric loading do not stress the same extreme fibres as the stress due to wind, the former stressing the extreme fibres on the top and bottom of the post, while the latter stresses those on the inner and outer sides. The total direct unit-stress is 8310 pounds per square inch; and this, added to the 8 250 pounds per square inch due to the wind, gives a total of 16 560 pounds per square inch on the extreme fibre only. oc 7 v 1 9 The allowable unit-stress is 16 000 - 70 X- - = 12 200 8. 11 pounds per square inch when wind is not taken into account, and (23) is \\ X 12 200 = 15 250 pounds per square inch when the wind is taken into account. The difference between this and the actual stress is 16 560 15 250 = 1 310 pounds per square inch, which shows that the section is not strong enough. The section can be in- creased by widening the cover-plate or by making the plates thicker ; but as this excess is due to wind only, the section being amply suffi- cient under the other stresses, and is fixed to some extent by the floor-beam connection, no change will be made. The pin at each end of the end-post will be the same namely, 6] inches in diameter and therefore the pin-plates will be the same at each end. The total stress in the post is 410 500 pounds, which 410 500 makes a required bearing area of 1777.^ = 17.2 square inches for ^.4 UUU both sides, or 8.6 square inches for one side, and the total required Q * thickness of ' = 1.375 square inches for one side. Since the thickness of the web plates is \ inch, this leaves a remainder of 1.375 0.5 = 0.875 inch for the thickness of the pin-plates. One plate f inch thick and one plate \ inch thick will be used. The proportion of the total stress which is taken by the f-inch 242 BRIDGE ENGINEERING 233 0.375 410500 56 000 pounds; and that taken by the |-inch plate is -^ - X 205250 = '74 600 pounds. The number of rivets required to transfer the stress from the f-inch plate to the |-inch plate is = 8 shop rivets in single shear; and the number of rivets required to transfer the stress from both pin-plates to the web is 56 000 + 74 600 * i =18 shop rivets in single shear. As in the case of the top chord, one pin-plate should extend over the angle, and the number of rivets required in that pin-plate should go through the pin- plate and the angles (see Fig. 188). The |-in. hinge plate is used for erection purposes, and is not considered as a pin-plate. It is omitted at L . Since this section is the same as that of the top chord, the tie- plates and the lattice bars must be the same size. 88. The Pins. The design of the pins requires a simple but quite lengthy computation. Sim- ple Pratt railroad trusses for single-track bridges usually have the same arrangement of tension and compression members; that is, the same tension members occupy relatively the same positions with respect to the compression mem- bers. Also,while theoretically a different sized pin will be required at every joint, it is not customary to make them so. In practice the pins at the joints U^ and JL are made of the same diameter, and those at the remainder of the joints are also made in diameter equal to each other but different from those at U l and Z/ , the pins at U\ and L usually being larger in diameter. On account of the above conditions and facts, it is unnecessary to design the pins in spans under 200 feet, since usually they are the same for any given span and loading. Table XXVI gives the diameters of pins for spans of 100 up to 200 feet for loading E 50. Fig. 188. Plates and Riveting at Upper End of End-Post. 234 BRIDGE ENGINEERING TABLE XXVI Pins for Sing!e=Track Bridges Loading E 50 DIAMETER OF PIN 17, and L All Others 100 feet 4i inches 4 inches 125 " 54 " 5 150 " 6J " 5* " 175 " 6| " 51 200 " 7 6 For E 40 loading, decrease the above values by \ inch; for E 30 loading, decrease them by inch. The diameter of pins for spans not given in the table can be interpolated from the given values. No pin should be less than 34 inches in diameter. The span of this bridge is 147 (say 150) feet, and the diameter of the pins at U^ and L is 6- | = 6j inches; and the diameter of the pins at the other panel points is 51 | = 5 inches. It should be noted that no pin is required at point L v as the two mem- bers which join here are built-up members and are riveted together. The above table is for single-track bridges only. The diameters of pins for double-track bridges are given in Table XXVII. These values are for E 50 loading; and for E 40 and E 30 loading, deduc- tions must be made as required in the case of Table XXVI. TABLE XXVII Pins for Double-Track Bridges Loading E 50 DIAMETER OP PIN E/, and L All Others 100 feet 6 inches 51 inches 125 " 8 64 " 150 " 9 74 " 175 " 91- " - 81 " 200 " 94 " 84 No pin in a double-track bridge should be less than 4J inches in diameter. Pins for highway bridges are usually much less in diameter than those for railway bridges, except in the case of first-class trusses for heavy interurban traffic or for city bridges carrying paved streets^ 244 BRIDGE ENGINEERING 235 where they should be taken equal to those given for E 30 loading. Table XXVIII gives the diameters of pins for different length spans of simple highway bridges designed for 16-ton road-rollers or farm wagons and 100 pounds per square foot of roadway. TABLE XXVIII Pins for Country Highway Bridges DIAMETER OF PIN 17, and Lower Chord Upper Chord 50 feet 100 " 150 " 200 " 2i inches 3" " f :: 2 inches 2i " 2| " 3 89. The Portal. In order to have a clearance of 21 feet (2) above the top of rail, it is necessary that the portal be as shown in Fig. 189. The stresses are found by methods of Article 54, Part I, the wind load being computed according to (10). It must be remembered that the column is hinged. In case the members of the portal braces bend about one axis, their length will be equal to the distance from one end to the other. In case they bend about the other axis as indicated by the broken line in Fig. 189, their length will be one-half of what it was in the first case. The portal struts or diagonals will be designed first. Their length is 8.5 X 1.414 - 12 feet, or 144 inches. This is the total length. Although the Specifications do not men- tion it, the ratio of the length to the radius of gyration should not exceed 120. This means that the radius of gyration in this Fig. Portal Dimension and Stress Diagram. 245 236 BRIDGE ENGINEERING 144 case should be greater than - = 1.2. The section of the strut will be composed of two angles placed back to back. Two angles 3^ by 3 by f- inch, with an area of 4.6 square inches and r 2 equal to 1 . 72 see Carnegie Handbook, p. 146, and (72) will be assumed to be sufficient to take the stress, and they must now be examined to see if the assumption is correct. The allowable unit-stress (23) is 25 per cent greater than in the case of live or dead loads. This makes the unit-stress as computed from the formula: P = AG 000 - 70 X ^7j) It = 12 68 pounds per square inch. 38 500 The required area is - = 3 . 05 square inches ; and since this is \-Zt DoU less than the given area, the angle will be amply sufficient. The re- quired area is over one square inch less than the given area, but this angle must be used, since it is the smallest angle allowed by the Specifi- cations. Note that unequal legged angles should be used, as this will make the radius of gyration about one axis larger than about the other; and this will prove economical, since, when one axis is considered, the length of the member is greater than when the other is considered. The above angle should also be examined for tension, it being considered that one rivet-hole is taken out of the section of each angle. The net section of the two angles will now be 4.60 2 ( + ) X 38 500 = 3 . 85 square inches ; and the area required for tension is T^y- r = 1 .93 square inches, which shows that the angle is amply sufficient. It should be noted that these Specifications do not require that only one leg of the angle shall be efficient unless both legs are connected. In case this strut had been designed according to Cooper's Specifica- tions, two angles 5 by 3 by |-inch would have been required, and the 5-inch leg would have been placed vertically and the angle connected by this leg alone. While it is not within the province of this work to discuss the question of connecting angles by one or by both legs, yet it might be said that tests made on angles connected with one leg only, seem to indicate that the ultimate strength in tension is about 60 per cent of that obtained from the same angle when tested with both legs connected. 246 BRIDGE ENGINEERING 237 While according to (20) the alternate strains in the wind bracing do not have to be considered, since they do not occur very closely together, yet in framing the connections it is required that the sum of both positive and negative stresses shall be added. In this case the stress for which the connections must be designed is 2 X 38500 = 77 000. It must be remembered that in this case also, the unit-stresses are increased 25 per cent over those allowed for live and dead loads. The number of rivets required in the end connections will be governed by bearing in the connection plates, and these plates are usually made f-inch thick. The number of rivets required is 77 000 77 000 7-880^0* - 8 Sh P metS ' r 6560-YU = The portal bracing is riveted up in the shop and brought to the bridge site, where it is connected to the trusses by field-riveted con- nections at its end. Therefore the end of the portal struts which connect with the top piece will have 8 shop rivets, and the other end which connects with the end-post will have 10 field rivets. Since the angles are small, all the above rivets must go in one line, and this will cause the connection plate to be quite large. It will probably be better to connect both legs of the angle by means of clip angles and thus reduce the size of the connection plates. The top part of the portal bracing will consist of two. angles. Two angles 3-j by 3 by f-inch will be assumed and examined to determine if the area is sufficient. The length of this strut is the distance center to center of trusses, and is equal to 17 X 12 = 204 ^04 inches. The least radius of gyration is therefore ~ - = 1 . 70. The radius of gyration of the two angles assumed is 1 . 72 when referred to an axis parallel to the shorter leg when the two angles are placed back to back and One-half inch apart. The unit-stress is now computed: (9Q4 \ 16 000 - 70 X Y~^) H = 9 625 pounds per square inch. 27 200 The required area is^-^- = 2.825 square inches. This is con- siderably less than the area given by the two angles; but as these are the minimum angles allowable, they must be used. Since the stress in this case is less than in the previous -case, and since the angles 247 238 BRIDGE ENGINEERING used are the same, it is evident that these angles are safe in tension. The number of rivets is determined by the bearing in the f-inch connection plates, and is : 2 X 27 200 7 880 X 1 . 25 2 X 27 200 6560 X 1.25 = 6 shop rivets, and = 10 field rivets. As in the case of the lateral strut, this member should be connected by both legs of the angle in order to reduce the size of the connection plates. Fig. 190 gives the details of the portal bra- cing and its method of connection to the end- post. The full circles represent shop rivets, and the blackened circles rep- resent field rivets. Some engineers connect the portal bracing to the top cover-plate of the end- post. This produces an excessive eccentricity in the end-post and is bad practice. Those members of the portal bracing which do not take any stress will be made of single angles, and the size of these angles will be taken 3$ by 3 by f-inch. 90. The Transverse Bracing. This bracing will be the same general style as the portal bracing, except that the top member will consist of two angles placed at a distance apart equal to the depth of the top chord, and these angles will be joined together by lacing. As in the case of portal bracing, those members which do not take stress will be made of one angle 3| by 3 by f-inch. The general outline is shown in Fig. 191, and the stresses are com- puted from (10) and by the methods of Article 54, Part I. In design- ing this top member, the top angle only is supposed to take the stress. The length in this case is 204 inches. Two 34 by 3 by f-inch angles will be assumed as sufficient and will be examined. These angles Pig. 190. Details of Portal Bracing and Connection to End-Post. 248 BRIDGE ENGINEERING 239 give a total area of 4.60 square inches. In examining these it will be found that they are amply sufficient, in fact so much so that it will be better to see if one single angle at the top will not be better. According to the length, the smallest radius of gyration which can be used is 1.7. In looking over the tables of angles, it is seen that the T.|.9*" -ii Oil T-126" Hi Section Fig. 191. General Outline of Transverse Bracing. first angle to fulfil this condition is a 6 by 3^ by f-inch, and it has a radius of gyration of 1.94. The allowable unit-stress is computed as follows: P = (l6 000 - 70 X p^)H = 10 780 pounds per square .inch; 9 100 and the required area is = 0.85 square inch. This is con- 249 240 BRIDGE ENGINEERING siderably smaller than the area of the angle, which is 3.97 square inches; but since this is the smallest possible angle which will fulfil the conditions of the Specifications, and since it is much smaller than the two angles as first assumed, it will be used. Fig. 192 gives a cross- section of this member. Since this angle is joined to the cover-plate by one leg, the joints will be weak in single shear, and the number of rivets required will be: X 9 100 = 2 shop rivets, or 6013 X 1J = 3 field rivets. According to (45), the width of the latticing must be 24 inches ; and according to Table XXV, the thickness must be -j 7 s inch, the distance c be- ing 1 foot 11 inches. The length of the knee-bracing is 144 inches; but on account of the small stress, one angle will , , i i - . T be used. One 4 by 3 by f-inch angle, with an . , area of 2 . 48 square inches and a radius of gyra- tion 1.26, will be assumed as sufficient. The radius of gyration is greater than the minimum allowable, which is 1.2. The allowable unit-stress is: Fig. 192. Cross- Sec- tion of Top Member of Transverse Bracing. 1C 000 - 70 X The required area is 12300 10 000 pounds per square inch." = 1.23 square inches. The required area is much less than the given area; but this angle must be used, since it is the only one allowed on account of its radius of gyration. Two of the minimum sized angles might have been used; but their total area, 4.60 square inches, is much in excess of that of the angle used. This angle must be examined for tension. The net area is 2.48 ~(i + i)Xf = 2.1 square inches. The required net area in 12 300 tension is = 0.615 square inch, which shows this angle lo 000 X lj to be amply sufficient. 250 BRIDGE ENGINEERING 241 The number of rivets required will be governed by the shear, since the angle is connected by one leg only; and it is: 2 X 12 300 7 220 X 2 X 12 300 91. The Lateral Systems. The stresses in these systems must be computed according to (10) and Article 54, Part I. They are given on the stress sheet, Plate III (p. 251). Since according to (68) these members must be constructed of rigid shapes, it is customary, in com- puting the stresses, to assume that one-half the shear is taken by each of the diagonals in any given panel; that is, one diagonal is in tension, and the other diagonal is in compression. The stresses given on the stress sheet are computed by making this assumption. Also, since both diagonals in each panel are considered as acting at the same time, the stresses in all the verticals are zero. The section of the upper lateral members will be made up of two angles placed apart a distance equal to the depth of the top chord. Fig. 193 shows the section. The radius of gyration about the axis parallel to the long leg will be consider- ably larger than that about an axis parallel to the shorter leg. In fact, it is so much greater that the strut will not need to be examined with respect to this axis. The diagram of the first panel is given in Fig. 194. The radius of gyration is to be taken about the horizontal axis if the entire length is to be taken; and the radius of gyration is to be taken about the vertical axis if one-half the length is taken, in which case it will bend as shown by the broken line in Fig. 194. The members are designed for the latter conditions only, since they are amply safe in regard to the first condition if they satisfy the latter. The length in this latter con- dition is 13.5 feet, which requires a radius of gyration not less than 13. 5 X 12 Fig, Upper Lateral Men 120 1.35. Two angles 5 by 3 by f-inch, with a total area of 5 . 72 square 251 242 BRIDGE ENGINEERING inches and a radius of gyration equal to 1.61, will be assumed and investigated to determine if they are sufficient. 1 o c y -to The unit-stress is computed to be P = (16 000 - 70 X ' p.) X n nr\c\ 1 J = 11 000 pounds per square inch, and the required area is = 1 1 UUU 0.61 square inch. The required area is very much less than the given area; but the angle chosen must be used, since this is the smallest one which conforms to the requirements of the Specifications. Fig. 194. Outline Diagram of First Panel in Upper Lateral System. Fig. 195. Outline Diagram of First Panel in Lower Lateral System. The width of the lattices (46) must be 2| inches; and according to Table XXV, the thickness must be & inch, the distance c being 23 inches. Single shear governs the number of rivets required. In accord- ance with (20) and (23), their number is = 2 rivets. Field 6 013 X 1 5 rivets 3 in number are used in all places, since the lateral system is riveted up after the trusses are swung into place. Since this is the minimum sized angle which will give a radius of gyration greater than 1 . 35, it must be used in the remainder of the panels of the top chord. Four angles of the minimum size might have been used, and would have been satisfactory, except that the area would have been excessive. The stresses in the lower lateral system are computed according to (10), a similar assumption to that for the upper lateral system being made namely, that both diagonals in each panel are stressed at the same time, one taking tension and the other taking compression. Fig. 195 shows the first panel of the lower lateral system. These 252 BRIDGE ENGINEERING 243 diagonals are connected to the stringers wherever they cross them, and also to each other where they cross in the center. This reduces the length which must be used in computing the cross-section of the member. In this case it is the distance C-A, and is equal to 90 inches. Since the angle is free to move about either axis, angles with even legs should preferably be employed, since this will give greater 90 economy. The radius of gyration must be greater than = . 75. One angle 3 1 by 3^ by f-inch, with an area of 2 .48 square inches and a radius of gyration of 1.07, will be assumed and investigated. 00 The allowable unit-stress is P = (16 000 - 70 X ) H = 12650 on C:AQ pounds per square inch, and the required area is = 2 . 38 square \2i boO inches. This is nearly equal to the given area, and therefore the angle chosen will be taken for the section. This angle must now be investigated for tension, one rivet-hole being taken out of the section. The net area is 2.48 (f + ) f = SO 500 2.10 square inches. The required net area is -y =1.53 i u (JUU X J- X square inches, which shows the angle to be sufficiently strong. Single shear determines the number of rivets to be required. These are: All rivets in the lower lateral system are field rivets, since this system also must be riveted up in the field after the trusses are swung into place. The total stress in the second panel is 21 500 pounds, and a 3^ by 3 by f-inch angle, with an area of 2.30 square inches and a least radius of gyration of . 90, will be assumed and examined. The 90 allowable unit-stress in compression is \\ (16 000 70 X ) = 21 500 1 1 250 pounds per square inch, and the required area is = 1 . 91 1 1 ^-)U square inches. Since this is less than the given area, and since the size of the angle (72) is the smallest allowable, this angle must be used. 253 244 BRIDGE ENGINEERING It is required that this member shall have a net area of 21 500 1.08 square inches in tension. The net area of the 16 000 X angle, one rivet-hole being taken out, is 1.92 square inches, which shows the angle to be safe in tension. The number of rivets required is determined by single shear, Fig. 196. Two Types of Hearings. since they tend to shear off between the member itself and the con- necting plates. The number required is 500 0013 XH Since the above angle is the smallest that can be used, and since the remaining angles of the panel of the lateral bracing have smaller stresses than the one just designed, it is evident that this size angle must be used in all panels of the lower lateral system other than the first. 92. The Shoes and Roller Nests. For bridges of short spans and for plate-girders whose spans require rocker bearings to be pro- vided (80), several different classes of bearings are in use. Two such bearings are shown in Fig. 196 (a and 6). The type illustrated by a is seldom used on any spans except plate-girders. That shown in b 254 BRIDGE ENGINEERING 245 may be used on either plate-girders or small truss spans; it is the invention of Mr. F. E. Schall, Bridge Engineer of the Lehigh Valley Railroad, who uses it on plate-girders. It has given very great satis- faction; and for simplicity of design and also for economy it is to t)e recommended. Some railroads have used a bearing which consisted of a lens-shaped disc of phosphor-bronze, the faces of which fitted into corresponding indentations in both the masonry and the bearing plates. One advantage of this bearing is that it allows movements due to the deflection of the girder, and also lateral deflection of the floor-beam. It is claimed to have given satisfaction. A bearing which is used on both short-span and long-span bridges Fig. 197. Bearing Adapted to Bridges of Both Short and Long Span. is shown in Fig. 197. This class of bearing will be used. The end reaction of the bridge proper is equal to the vertical component of the 30 1 stress in the end-post, and is ^ X 410 500 = 336 500 pounds, 336 500 which requires a bearing area (19) on the masonry, ot ' = 561 square inches. According to the table on page 193, the masonry plate will be 28 inches long. The total bearing area for one of the vertical plates is : 255 246 BRIDGE ENGINEERING and the total required thickness is: 7.0 - , = 1 . 12 inches, 6.2o a 6 j-inch pin being used at L . Since the vertical plates will be made f inch thick, this leaves a remainder of f inch to be made up of pin- plates. The amount of stress which is carried by the f-inch pin-plate is n *?7^ ^*^fi ^oo -^ X " = 56 100 pounds. These plates will tend to shear 56 100 off the rivets at a plane between the plates, and therefore = 8 shop rivets will be required to fasten them to the vertical plate. Since the length of the masonry plate is 28 inches, and the total 561 area required is 561 square inches, the required width is = 20 inches. The actual width will be greater than this, since it must be sufficient to allow for the connecting angles and also for the bearings of the end floor-beam. The connecting angles should be f inch thick, and should not be less than 6 by 6 inches; and the plates to which they are connected should not be less than f inch in thickness, and likewise they should not be greater, on account of the punching. The bottom plate should extend outward about 3 inches, in order to allow suf- ficient room for the anchor bolts, which should be f inch in diameter and should extend into the masonry at least 8 inches. In addition to the reaction of the bridge proper, the masonry plate must be of sufficient area to give bearing for the end reaction of the end floor-beam. The maximum end reaction (see Article 83, p. 197) is 104 740 pounds. The bearing area required on the masonry 104740 is = 17o square inches; and assuming that the base of the 600 bearing will be 12 inches long (see Fig. 197), the required length will be 14.6 inches. Usually, however, the bearing is extended the entire length of the masonry plate, which is 28 inches in this case. The distance from the center of the pin to the top of the masonry will be the same for both the fixed and the roller end. This distance should be such that the angles of the shoe will clear the bottom chord member and allow the floor-beam to rest upon the plate as shown. Since the first section of the bottom chord is 18$ inches deep, the top of the angles of the two must be at least 9| inches from the center line 256 BRIDGE ENGINEERING 247 of pins. This requires that the distance from the center line of the pin to the base of the angle shall be at least (9j + 6) = 15J inches, or more. The tops of all floor-beams are at the same height, and the bottoms of the intermediate floor-beams must be on a level with the bottom of the first section of the lower chord (see Fig. 174). This requires that the bottom of the intermediate floor-beams shall be 9^ inches below the center line of pins, and this brings the top of the floor-beams (4S-J- 9|) = 39 inches above the center line of the pins. Since the end floor-beam is 52 \ inches deep, back to back of angles, the hijj 6uld 5-. p i^ Seqmzntal Kollew eV^a" llflh Guide Bars ej-ifj If i-L s"*3 i"* f|Cho,BoU * ,, II U 1 Lsa: Fig. 198. Type of Bearing Construction where End Floor-Beam Does Not Rest Directly on Bearing or Masonry Plate. Grillage of Iron Bars Used instead of Cast-Steel Pedestal. lower flange will be (52f 39) = 13|- inches below the center line of pins. In case the end floor-beam does not rest directly upon the bearing plate or the masonry plate, the intervening space is filled out with a grillage of iron bars or a cast-steel pedestal, as shown in Figs. 197 and 198. The small plates upon the side of the shoe, going entirely around the pin, are called the shoe hinge-plates. These do not take any stress, and require only sufficient rivets to hold them in position. They are used during erection to keep the end-post in line; and after erection their function is to keep the end-post on the shoe, and to prevent it from having any upward motion due to the vibration of the structure. 257 248 BRIDGE ENGINEERING The rivets through the vertical legs of the shoe angles are in double bearing in the f-inch angles, in single bearing in the vertical plate, and in double shear. A rivet in double shear has a less value than in bearing in the plates. This value is 14440 pounds, and therefore the number of shop rivets required through the vertical legs of the angles is : 336 500 2 X 14 440 = 12 rivets. The rivets which go through the horizontal leg of the angle and through the cap plate and cap angles, do not take stress. The num- ber of rivets put in is that demanded by the detailing, the rivets in the horizontal legs of the angles usually staggering with those in the verti- cal legs. The cap plate tends to keep the vertical plates in line, and to keep out the dust and dirt and other deteriorating influences of the elements. Wherever the rivet-heads tend to interfere with other members or project beyond surfaces which are required to be flat as, for example, the bottom of the masonry or bearing plates they must be countersunk (see Carnegie Handbook, p. 191, and "Steel Construc- tion," Part III, p. 192). The space for the anchor bolts, that for the connection angles, and that for the bearing of the end floor-beam, require that the total width of the masonry plate for the fixed end shall be 2 X f + 14^ + 2 X 6 + \ + 3 + 12 = 3 feet 1\ inches. The design of the roller end requires that the length of the masonry bearing, the size of the vertical plates and angles, and also the number of rivets shall be the same as that for the fixed end. The width of the masonry plate is determined by the length of the rollers and their connections at the end. The rollers (60) are required to be 6 inches in diameter, and the unit-stress (19) per linear inch is 6 X 600 = 3 600 pounds, which requires : 336 500 gf .- = 93 . 5 linear inches. 6 600 This is for the reaction of the bridge alone; and in addition to this, there are required for the floor-beam reaction : r- = 29.0 linear inches. 858 BRIDGE ENGINEERING 249 The total number of linear inches is 93.5 + 29.0 = 122. 5; and if 5 rollers are used, they must be at least -^- =24.5 inches long. The 5 masonry plate is only 28 inches long, and therefore cylindrical rollers cannot be used, since they would occupy a space 30 inches or over. Segmental rollers (see Fig. 199) must be used. The determination of the sizes of the angles which go at the end of the rollers, and also of the guide-plates, is a matter of judg- ment and experience. Those sizes indicated in Fig. 198, repre- sent good engineering practice, and will be used. The distance from the center line of pins to the top of the ma- sonry can now be determined, and is 16J + | + 6 + | = 23| inches. On account of putting in Fig. 199. Segmental Rollers Used for Deal- ings in Space under 30 Inches. sufficient connections and angles as shown in Fig. 198, the masonry plate must be considerably wider than that theoretically determined. According to Fig. 198, the total width must be as follows, and the width should be computed in two parts, as the plate is not symmetrical about the center line of the truss: From center line to outer edge: ^ + \ + 6 + * + (3 - f = 2f ) + (3* - f = 3|) + 3 = 1 ft. Hi in., (say, 1 ft. 11 in.). From center line to inner edge: 2 + i + 6 + 1 + 12 + * + 2 + 3i + 3 = 3 ft. Oi in., say, 3 ft. in. Total width 4ft. 11 in. Allowing guide-plates and guide-bars of dimensions as shown in Fig. 198, and assuming ^ inch as clearance at the ends, the total length of the rollers is: (4 ft. 11 in. ) - 2 (3 + 3* + J + J + $) = 44. 5 inches. This shows them to be amply long enough, as only 22 inches is 259 250 BRIDGE ENGINEERING theoretically required. Here, as in most cases for single-track spans up to 200 feet in length, the width of the masonry plate is determined by the detail and not by the unit bearing stress. The guide-plates are small bars riveted to the top of the bottom plates, and serve to keep the rollers in line. The guide-bars are con- nected to rollers at their ends, and serve to keep the rollers equi- distant, therefore causing them to roll easier and keeping them from becoming worn by contact with each other. The expansion (57) must be allowed for at the rate of | inch for every 10 feet in length of span. This makes a total allowed for tem- perature of expansion of -.--- X I = If (say 2) inches. No slotted holes are to be provided for the anchor bolts, since they do not go I ' Fig. 200 Binding of Insufficiently Spaced Fig. 201. Computation of Spacing for Seg- Segmental Rollers. mental Rollers. through that part of the bridge which slides. The shoe slides over the rollers, and is kept in place by the angles at the end, which are riveted to the masonry plate (see Fig. 198). Unless sufficient room is allowed between the segmental rollers, they will tend to bind when the bridge has reached the extreme posi- tion for expansion or contraction (see Fig. 200). This distance can be computed from proportions as indicated in Fig. 201, and from the following formuhe*: in which e is the amount allowed for the change of temperature, and D is the diameter of the rollers, both being taken in inches. The angle < is in degrees. In the present case, eis2 inches; D is 6 inches; * Derived by Prof. Frank B. McKibben of Lehigh University, and published in Engineering News, December, 1896. 252 BRIDGE ENGINEERING and (f>, computed from the above formula, is 9 30'. Substituting in the equation giving the value for y, there is obtained 1 .02 inches (say \\ inches) for the distance between rollers. Rollers must not be less in thickness than the total expansion allowed for temperature. Since there are 5 rollers, there are 4 spaces between them. Also, since the rollers must occupy a space of 28 inches, the length of the masonry plate, each roller must be: - = 4.6 inches (say 4 inches) wide. The width of the guide-bars must be such as to allow freedom Fig. 202. Details of End Floor-Beam Connections. of motion for the rollers. The maximum width allowable is given by the formula :* in which and D are indicated above. This requires the bar to be . W = -|- X . 985 = 2 . 96 (say 24) inches wide. 93. The Stress Sheet. Plate III shows the stress sheet of the bridge which has been designed in the preceding articles. This sheet represents the best current practice among the larger bridge * Derived by Prof . Frank B. McKibben of Lehigh University, and published in Engineering News, December, 1896. 262 BRIDGE ENGINEERING 253 corporations. It will be noted that very few details are given upon the sheet; also that few rivets are noted, and that sketches showing the manner in which the parts go together are entirely wanting. The shears and moments for the stringers and floor-beams, as well as the reactions and the number of rollers re- quired, are given. This is to save the drafts- man the trouble of recomputing values which have necessarily been determined by the designer. The details of the various members, and also - the manner in which the different members are con- nected, are left to the draftsman, who is un- der the direct super- vision of the engineer in charge of the draft- ing room, upon whom rests the responsibility for good details. The figures given in the text indicate the best current practice. Figs. i ' " 1 l . 5s . . 1 I _ H ..J 263 254 BRIDGE ENGINEERING 202 to 204 show details of the end floor-beam connections, and also the packing of the members of the upper and the lower chord. The arrangement here given may be said to be standard for single-track Pratt truss spans up to 200 feet in length. BIBLIOGRAPHY The following books are recommended to the student in case it is desired to pursue further the study of the subjects of Bridge Analysis and Bridge Design : The Theory and Practice of Modern Frame Structures. JOHNSON; BRYAN, and TURNEAURE. John Wiley & Sons, New York, N. Y. Roofs and Bridges. MERRIMAN and JACOB Y. John Wiley & Sons, New York, N. Y. Design and Construction of Metallic Bridges. BURR and FALK. John Wiley & Sons, New York, N. Y. Influence Lines for Bridges and Roofs. BURR and FALK. John Wiley & Sons, New York, N. Y. Details of Bridge Construction. FRANK W. SKINNER. McGraw Pub- lishing Company, New York, N. Y. Steel Mill Buildings. Milo S. Ketchum. Engineering News Publish- ing Company, New York, N. Y. Statically Indeterminate Stresses. HIROI. Engineering News Pub- lishing Company, New York, N. Y. Stresses in Frame Structures. A. JAY DuBois. John Wiley & Sons, New York, N. Y. Die Zusatzkrdfte und Nebenspannungen eiscrner Fachwerkbriicken. FR. ENGESSER. Julius Springer, Berlin, Germany. Bridge Drafting. WRIGHT and WING. Engineering News Publishing Company, New York, N. Y. It must not be presumed that the above is a complete list of the books which have been published relating to the theory and practice of Bridge Engineering; neither must it be presumed that the obtaining of information relative to bridges is limited to textbooks on the subject. One of the best sources of information is found in the cur- rent engineering periodicals and the "Proceedings" of the various technical societies. The great advantage of these sources is that they give the most up-to-date information, and usually they are very pro- fusely illustrated. 264 ROAD CONSTRUCTION IN THE PHILIPPINES Section of 35-mile road built by American engineers, connecting the seaport of Dagupan with the mountain village of Baguio. province of Benguet, island of Luz6n, and affording a cool and healthful retreat from the heat and malaria of the lowland re*ions. Dagupan lies 120 miles north of Manila, with which it is connected by rail. This view reveals some of the engineering difficulties to be overcome, masonry and concrete work of the type shown being necessary at many points. HIGHWAY CONSTRUCTION PART I. COUNTRY ROADS. GENERAL CONSIDERATIONS. Object of Roads. The object of a road is to provide a way for the transportation of persons and goods from one place to another with the least expenditure of power and expense. The facility with which this traffic or transportation may be conducted over any given road depends upon the resistance offered to the movement of vehicles. This resistance is composed of: (1) The resistance offered by the roadway, which consists of (a) "friction" between the surface of the road and the wheel tires; (6) resistance offered to ths rolling of the wheels, occasioned by the want of uniformity in the road surface, or lack of strength to resist the penetrating efforts of loaded wheels, thus requiring the load to be lifted over projecting points and out of hollows and ruts, thereby diminishing the effective load the horse may draw to such as it can lift. This resistance is called "resistance to rolling" or "penetration;" (c) resistance due to gravity called "grade resistance;" (2) The resistance offered by vehicles, termed "axle friction;" (3) Resistance of the air. The road which offers the least resistance to traffic should com- bine a surface on which the friction of the wheels is reduced to the least possible amount, while offering a good foothold for horses, to enable them to exert their utmost tractive power, and should be so located as to give the most direct route with the least gradients. Friction. The resistance of friction arises from the rubbing of the wheel tires against the surface of the road. This resistance to traction is variable, and can be determined only by experiment. From many experiments the following deductions are drawn : (1) The resistance to traction is directly proportional to the pressure, Copyright, 1908, by American School of Corespondent. 267 HIGHWAY CONSTRUCTION (2) On solid, unyielding surfaces it is independent of the width of the tire, but on compressible surfaces the resistance decreases as the width of the tire increases (but there is no material advantage gained in making a tire more than 4 inches wide). (3) It is independent of the speed. (4) On rough, irregular surfaces, which give rise to constant concussion, it increases with the speed. The following table shows the relative resistance to traction of various surfaces: TABLE 1. Resistance to Traction on Different Road Surfaces. Traction Resistance. Pounds per ton. In terms of load. Earth road ordinary condition Gravel 50 to 200 50 to 100 100 to 200 30 to 100 30 to 50 15 to 40 A to T V rir to A A to T V & to A A to^ T*T to J a Sand Macadam Steel Wheelway These coefficients refer to the power required to keep the load in motion. It requires from two to six or eight times as much force to start a load as it does to keep it in motion, at two or three miles per hour. The extra force required to start a load is due in part to the fact that during the stop the wheel may settle into the road surface, in part to the fact that the axle friction at starting is greater than after motion has begun, and further in part to the fact that energy is consumed in accelerating the load. Resistance to Rolling. This resistance is caused (1) by the wheel penetrating or sinking below the surface of the road, leaving a track or rut behind it. It is equal to the product of the load mul- tiplied by one-third of the semi-chord of the submerged arc of the wheel; and (2) by the wheel striking or colliding with loose or pro- jecting stones, which give a sudden check to the horses, depending upon the height of the obstacle, the momentum destroyed being oftentimes considerable. The rolling resistance varies inversely as some function of the 868 HIGHWAY CONSTRUCTION diameter of the wheel, as the larger the wheel the less force required to lift it over the obstruction or to roll it up the inclination due to the indentation of the surface. The pow r er required to draw a wheel over a stone or any ob- stacle, such as S in Fig. 1, may be thus calculated. Let P represent the pow r er sought, or that which would just balance the weight on the point of the stone, and the slightest increase of which would draw it over. This power acts in the direction C P with the leverage of B C or D E. Gravity, represented by W, resists in the direction C B with the leverage B D. The equation of equilibrium s Fig. 1. will be P X C B = W X B D, whence = \\ 7 BD CB v CD 2 -BC 2 CD -AD Let the radius of the wheel = C D - 26 inches, and the height of the obstacle = A B = 4 inches. Let the weight W = 500 pounds, of which 200 pounds may be the weight of the wheel and 300 pounds the load on the axle. The formula then becomes P-500 676 - 484 26-4 500 13.85 - = 314.7 pounds. The pressure at the point D is compounded of the weight and the power, and equals fD 9fi W^ = 500 Xg = 591 pounds, and therefore acts with this great effect to destroy the road in its collision with the stone, in addition there is to be considered the effect of the blow given by the wheel in descending from it. For minute accuracy the non-horizontal direction of the draught and the thickness of the axle should be taken into account. The power required is lessened by proper springs to vehicles, by enlarged wheels, and by making the line of draught ascending. HIGHWAY CONSTRUCTION The mechanical advantage of the wheel in surmounting an obstacle may be computed from the principle of the lever. Let the wheel, Fig. 2, touch the horizontal line of traction in the point A and meet a protuberance B D. Suppose the line of draught C P to be parallel to A B. Join C D and draw the perpen- diculars DE and D F. We may suppose the power to be applied at E and the weight at F, and the action is then the same as the bent lever E D F turning round the fulcrum at D. HencePrW ::FD :DE. But FD :DE ::tanFCD:l, A~ ^B and tan F C D = tan 2 (DAB); therefore P = W tan 2 (DAB). Now it is obvious that the angle DAB increases as the radius of the circle diminishes; and therefore, the weight W being constant, the power required to overcome an obstacle of given height is diminished when the diameter is increased. Large wheels are therefore the best adapted for surmounting inequalities of the road. There are, however, circumstances which provide limits to the height of the wheels of vehicles. If the radius AC exceeds the height of that part of the horse to which the traces are attached, the line of traction C P will be inclined to the horse, and part of the power will be exerted in pressing the wheel against the ground. The best average size of wheels is considered to be about 6 feet in diameter. Wheels of large diameter do less damage to a road than small ones, and cause less draught for the horses. With the same load, a two-wheeled cart does far more damage than one with four wheels, and this because of their sudden and irregular twisting motion in the trackway. Grade Resistance is due to the action of gravity, and is the same on good and bad roads. On level roads its effect is immaterial, as it acts in a direction perpendicular to the plane of the horizon, and neither accelerates nor retards motion. On inclined roads it offers considerable resistance, proportional to the steepness of the incline. 270 HIGHWAY CONSTRUCTION The resistance due to gravity on any incline in pounds per ton 2000 is equal to T rate ot grade The following table shows the resistance due to gravity on dif- ferent grades. TABLE 2. Resistance Due to Gravity on Different Inclinations. Grade 1 in 20 30 40 50 60 70 80 90 100 200 300 400 Rise in feet per mile ... 264 176 132 105 88 75 66 58 52 26 17 13 Resistance in Ib. per ton .11274* 56 4538322825 22 11 J 7 5* The additional resistance caused by inclines may be investigated in the following manner: Suppose the whole weight to be borne on one pair of wheels, and that the tractive force is applied in a direction parallel to the surface of the road. Let A B in Fig. 3 represent a portion of the inclined road, C being a vehicle just sustained in its position by a force acting in the direction CD. It is evident that the vehicle is kept in its position by three forces; namely, by its own weight W acting in the vertical direction C F, by the force F applied in the direction C D parallel to the surface of the road, and by the pressure P which the vehicle exerts against the surface of the road acting in the direction C E perpendicular to same. To determine the relative magni- tude of these three forces, draw a horizontal line A G and the vertical one B G; then, since the two lines C F and B G are parallel and are both cut by the line A B, Pig. 3. they must make the two angles C F E and A B G equal ; also the two angles C E F and A G B are equal ; therefore, the remaining angles F C E and BAG are equal, and the two triangles C F E and A B G are similar. And as the three sides of the former are proportional to the three forces by which the vehicle is sustained, so also are the three sides of the latter; namely, AB or the length of the road is proportional to W, or the weight of the vehicle; B G, 271 6 HIGHWAY COXSTKUCTION or the vertical rise in the same, to F, or the force required to sustain the vehicle on the incline; and A G, or the horizontal distance in which the rise occurs, to P, or the force w r ith which the vehicle presses upon the surface of the road. Therefore, W : A B : : F : G B, and W :AB : :P : A G. If to A G such a value be assigned that the vertical rise of the road is exactly one foot, then p= W _W_ = W -sin4 AB i/AG 2 +l and _ W-AG W-AG P = = . = W cos A, AB l/AG-'+l in which A is the angle BAG. To find the force requisite to sustain a vehicle upon an inclined road (the effects of friction being neglected), divide the weight of the vehicle and its load by the inclined length of the road, the vertical rise of which is one foot, and the quotient is the force required. To find the pressure of a vehicle against the surface of an inclined road, multiply the weight of the loaded vehicle by the horizontal length of the road', and divide the product by the inclined length of the same; the quotient is the pressure required. The force with which a vehicle presses upon an inclined road is always less than its actual weight; the difference is so small that, unless the inclination is very steep, it may be taken equal to the weight of the loaded vehicle. To find the resistance to traction in passing up or down an incline, ascertain the resistance on a -level road having the same surface as the incline, to which add, if the vehicle ascends, or subtract, if it descends, the force requisite to sustain it on the incline; the sum or difference, as the case may be, will express the resistance. Tractive Power and Gradients. The necessity for easy grades is dependent upon the power of the horse to overcome the resistance to motion composed of the four forces, friction, collision, gravity, and the resistance of the air. All estimates on the tractive power of horses must to a certain 272 HIGHWAY CONSTRUCTION extent be vague, owing to the different strengths and speeds of animals of the same kind, as well as to the extent of their training to any particular kind of work. The draught or pull which a good average horse, weighing 1,200 pounds, can exert on a level, smooth road at a speed of 2^ miles per hour is 100 pounds, equivalent to 22,000 foot-pounds per minute, or 13,200,000 foot-pounds per day of 10 hours. The tractive power diminishes as the speed increases and, per- haps, within certain limits, say from f to 4 miles per hour, nearly in inverse proportion to it. Thus the average tractive force of a horse, on a level, and actually pulling for 10 hours, may be assumed approximately as follows: TABLE 3. Tractive Power of Horses at Different Velocities. Miles per hour. Tractive Force. Lb. Miles per hour. Tractive Force. Lb. 3 333 33 21 Ill 11 1 250 2 i 100 li 200 90 91 1$ 166 66 3 83 33 142.86 3J 71.43 2 125 4 62 50 The work done by a horse is greatest when the velocity with which he moves is of the greatest velocity with which he can move when unloaded; and the force thus exerted is 0.45 of the utmost force that he can exert at a dead pull. The- traction power of a horse may be increased in about the same proportion as the time is diminished, so that when working from 5 to 10 hours, on a level, it will be about as shown in the following table: TABLE 4. Hours per day Traction (pounds) Hours per day Traction (pounds) 10 ..... ..... 100 7 .......... 9 .......... 111^ 6 .......... 8 .......... 125 5 .......... The tractive power of teams is about as follows 1 horse .............. =1 2 horses .............. . 95 X 2 = 1 . 90 3 " ........ ...... 0.85 X 3 - 2.55 4 " 0.80X 4-3.20 146| 166J 200 273 HIGHWAY CONSTRUCTION Loss of Tractive Power on Inclines. In ascending in- clines a horse's power diminishes rapidly; a large portion of his strength is expended in overcoming the resistance of gravity due to his own weight and that of the load. Table 5 shows that as the steepness of the grade increases the efficiency of both the horse and the road surface diminishes; that the more of the horse's energy is expended in overcoming gravity the less remains to overcome the surface resistance. TABLE 5. Effects of Grades Upon the Load a Horse can Draw on Different Pavements. Grade. Earth. Broken Stone. Stone Blocks. Asphalt. Level 1.00 1.00 1.00 1.00 1 : 100 .80 .66 .72 .41 2 : 100 .66 .50 .55 .25 3 : 100 .55 .40 .44 .18 4 : 100 .47 .33 .36 .13 5 : 100 .41 .29 .30 .10 10 : 100 .26 .16 .14 .04 15 : 100 .10 .05 .07 20 : 100 .04 .03 Table 6 shows the gross load which an average horse, weighing 1,200 pounds, can draw on different kinds of road surfaces, on a level and on grades rising five and ten feet per one hundred feet. TABLE 6. Description of Surface. Level. 5 per cent grade. 10 per cent grade. \sphalt 13 216 Broken stone (best condition) 6 700 1 840 1 060 " (slightly muddy) 4,700 1,500 1,000 ' " (ruts and mud) 3 000 1 390 890 " (very bad condition) . . 1,840 1,040 740 Earth (best condition) 3,600 1,500 930 (average condition) .... 1,400 900 660 (moist but not muddy). 1,100 780 600 Stone-block pavement (dry and clean) 8,300 1,920 1,090 " (muddy) 6,250 1,800 1,040 Sand (wet) 1,500 675 390 " (dry) 1,087 445 217 The decrease in the load which a horse can draw upon an incline is not due alone to gravity; it varies with the amount of foothold 274 HIGHWAY CONSTRUCTION afforded by the road surface. The tangent of the angle of inclination should not be greater than the coefficient of tractional resistance; therefore it is evident that the smoother the road surface, the easier should be the grade. The smoother the surface the less the foothold, and consequently the load. The loss of tractive power on inclines is greater than any inves- tigation will show; for, besides the increase of draught caused by gravity, the power of the horse is much diminished by fatigue upon a long ascent, and even in greater ratio than man, owing to its anatom- ical formation and great weight. Though a horse on a level is as strong as five men, on a grade of 15 per cent, it is less strong than three; for three men carrying each 100 pounds will ascend such a grade faster and with less fatigue than a horse with 300 pounds. A horse can exert for a short time twice the average tractive pull which he can exert continuously throughout the day's work; hence, so long as the resistance on the incline is not more than double the resistance on the level, the horse will be able to take up the full load which he is capable of drawing. Steep grades are thus seen to be objectionable, and particularly so when a single one occurs on an otherwise comparatively level road, in which case the load carried over the less inclined portions must be reduced to what can be hauled up the steeper portion. The bad effects of steep grades are especially felt in winter, when ice covers the roads, for the slippery condition of the surface causes danger in descending, as well as increased labor in ascending; the water of rains also runs down the road and gulleys it out, destroy- ing its surface, thus causing a constant expense for repairs. The inclined portions are subject to greater wear from horses ascending, thus requiring thicker covering than the more level portions, and hence increasing the cost of construction. It will rarely be possible, except in a flat or comparatively level country, to combine" easy grades with the best and most direct route. These two requirements will often conflict. In such a case, increase the length. The proportion of this increase will depend upon the friction of the covering adopted. But no general rule can be given to meet all cases as respects the length which may thus be added, for the comparative time occupied in making the journey forms an 275 10 HIGHWAY CONSTRUCTION important element in any case which arises for settlement. Disre- garding time, the horizontal length of a road may be increased to avoid a 5 per cent grade, seventy times the height. Table 7 shows, for most practical purposes, the force required to draw loaded vehicles over inclined roads. The first column ex- presses the rate of inclination; the second, the pressure on the plane in pounds per ton; the third, the tendency down the plane (or force required to overcome the effect of gravity) in pounds per ton; the fourth, the force required to haul one ton up the incline; the fifth, the length of level road which would be equivalent to a mile in length of the inclined road that is, the length which would require the same motive power to be expended in drawing the load over it, as would be necessary to draw over a mile of the inclined road', the sixth, the maximum load which an average horse weighing 1,200 pounds can draw over such inclines, the friction of the surface being taken at -fo of the load drawn. TABLE 7. Rate of grade. Feet per 100 feet. Pressure on the plane in Ib. per ton. Tendency down the plane in Ib. per ton. Power in Ib. required to haul one ton up the plane. Equivalent length of level road. Miles. Maximum load in Ib. which a horse can haul. 0.0 2240 .00 45.00 1.000 6270 0.25 2240 5.60 50.60 1.121 5376 0.50 2240 11.20 56.20 1.242 4973 0.75 2240 16.80 61.80 1.373 4490 1. 2240 22.40 67.40 1.500 4145 1.25 *2240 28.00 73.00 1.622 3830 1 . 50 2240 33.60 78.60 1.746 3584 1.75 2240 39.20 84.20 1.871 3290 2 2240 45.00 90.00 2.000 3114 2 25 2240 50.40 95.4'0 2.120 2935 2^50 2240 56.00 101.00 2.244 2725 2.75 2240 61.33 106.33 2.363 2620 3 2239 67.20 112.20 2.484 2486 4 2238 89.20 134.20 2.982 2083 5 2237 112.00 157.00 3.444 1800 6 2233 134.40 179.40 3.986 1568 7 2232 156.80 201.80 4.844 1367 8 2232 179.20 224 . 20 4.982 1235 9 2231 201.60 246.60 5.840 1125 10 2229 224.00 269.00 5.977 1030 * Near enough for practice; actually 2239.888. Pressure on the plane = weight x nat cos of angle of plane. Axle Friction. The resistance of the hub to turning on the axle is the same as that of a journal revolving in its bearing, and has 276 HIGHWAY CONSTRUCTION 11 nothing to do with the condition of the road surface. The coefficient of journal friction varies with the material of the journal and its bearing, and with the lubricant. It is nearly independent of the velocity, and seems to vary about inversely as the square root of the pressure. For light carriages when loaded, the coefficient of friction is about 0.020 of the weight on the axle; for the ordinary thimble- skein wagon when loaded, it is about 0.012. These coefficients are for good lubrication; if the lubrication is deficient, the axle friction is two to six times as much as above. The traction power required to overcome the above axle friction for carriages of the usual proportions is about 3 to 3^ Ib. per ton of the weight on the axle; and for truck wagons, which have medium sized wheels and axles, is about 3 j. to 4^ Ib. per ton. Resistance of the Air. The resistance arising from the force of the wind will vary with the velocity of the wind, with the velocity of the vehicle, with the area of the surface acted upon, and also with the angle of incidence of direction of the wind with the plane of the surface. The following table gives the force per square foot for various velocities: TABLE 8. Velocity of wind in miles per hour. Force in Ibs. per sq. ft. Description. 15 1.107 Pleasant Breeze 20 25 1.968\ 3.075/ Brisk Gale 30 35 4.428\ 6.027J High Wind 40 45 7.872) 9.963\ Very High Wind 50 12.300 Storm Effect of Springs on Vehicles. Experiments have shown that vehicles mounted on springs materially decrease the resistance to traction, and diminish the wear of the road, especially at speeds beyond a walking pace. Going at a trot, they were found not to cause more wear than vehicles without springs at a. walk, all other conditions being similar. Vehicles with springs improperly fixed cause considerable concussion, which in turn destroys the road covering. 277 12 HIGHWAY CONSTRUCTION LOCATION OF COUNTRY ROADS. The considerations governing the location of country roads are dependent upon the commercial condition of the country to be traversed. In old and long-inhabited sections the controlling ele- ments will be the character of the traffic to be accommodated. In such a section, the route is generally predetermined, and therefore there is less liberty of a choice and selection than in a new and sparsely settled district, where the object is to establish the easiest, shortest, and most economical line of intercommunication according to the physical character of the ground. Whichever of these two cases may have to be'dealt with, the same principle governs the engineer, namely, to so lay out the road as to effect the conveyance of the traffic with the least expenditure of motive power consistent with economy of construction and main- tenance. Economy of motive power is promoted by easy grades, by the avoidance of all unnecessary ascents and descents, and by a direct line; but directness must be sacrificed to secure easy grades and to avoid expensive construction. Reconnoissance. The selection of the best route demands much care and consideration on the part of the engineer. To obtain the requisite data upon which to form his judgment, he must make a personal reconnoissance of the district. This requires that the proposed route be either ridden or walked over and a careful examina- tion made of the principal physical contours and natural features of the district. The amount of care demanded and the difficulties attending the operations will altogether depend upon the character of the country. The immediate object of the reconnoissance is to select one or more trial lines, from which the final route may be ultimately deter- mined. When there are no maps of the section traversed, or when those which can be procured are indefinite or inaccurate, the work of reconnoitering will be much increased. In making a reconnoissance there are several points which, if carefully attended to, will very considerably lessen the labor and time otherwise required. Lines which would run along the imme- 278 HIGHWAY CONSTRUCTION 13 diate bank of a large stream must of necessity intersect all the tribu- taries confluent on that bank, thereby demanding a corresponding number of bridges. Those, again, which are situated along the slopes of hills are more liable in rainy weather to suffer from washing away of the earthwork and sliding of the embankments; the others which are placed in valleys or elevated plateaux, when the line crosses the ridges dividing the principal water courses will have steep ascents and descents. In making an examination of a tract of country, the first point to attract notice is the unevenness or undulations of its surface, which appears to be entirely without system, order, or arrangement; but upon closer examination it will be perceived that one general prin- ciple of configuration obtains even in the most irregular countries. The country' is intersected in various directions by main water courses or rivers, which increase in size as they approach the point of their discharge. Towards these main rivers lesser rivers approach on both sides, running right and left through the country, and into these, again, enter still smaller streams and brooks. The streams thus divide the hills into branches or spurs having approximately the same direction as themselves, and the ground falls in every direction from the main chain of hills towards the water courses, forming ridges more or less elevated. . The main ridge is cut down at the heads of the streams into depressions called gaps or passes; the more elevated points are called peaks. The water which has fallen upon these peaks is the origin of the streams which have hollowed out the valleys. Furthermore, the ground falls in every direction towards the natural water courses, forming ridges more or less elevated running between them and separating from each other the districts drained by the streams. The natural water courses mark not only the lowest lines, but the lines of the greatest longitudinal slope in the valleys through which they flow. The direction and position of the principal streams give also the direction and approximate position of the high ground or ridges which lie between them. The positions of the tributaries to the larger stream generally indicate the points of greatest depression in the summits of the ridges, 279 U HIGHWAY CONSTRUCTION Fig. 4. Contour Lines. HIGHWAY CONSTRUCTION 15 and therefore the points at which lateral communication across the high ground separating contiguous valleys can be most readily made. The instruments employed in reconnoitering, are : The compass, for ascertaining the direction; the aneroid barometer, to fix the ap- proximate elevation of summits, etc. ; and the hand level, to ascertain the elevation of neighboring points. If a vehicle can be used, an odometer may be added, but distances can usually be guessed or ascertained by time estimates or otherwise, closely enough for pre- liminary purposes. The best maps obtainable and traveling com- panions who possess a local knowledge of the country, together with the above outfit is all that will be necessary for the first inspection. The reconnoissance being completed, instrumental surveys of the routes deemed most advantageous should be made. When the several lines are plotted to the same scale, a good map can be pre- pared from which the exact location of the road can be determined. In making the preliminary surveys the topographical features should be noted for a convenient distance to the right and the left of the line, and all prominent points located by compass bearings. The following data should also be obtained: the importance, magnitude, and direction of all streams and roads crossed; the character of the material to be excavated or available for embankments, the position of quarries and gravel pits, and the modes of access thereto; and all other information that may effect a selection. Topography. There are various methods of delineating upon paper the irregularities of the surface of the ground. The method of most utility to the engineer is that by means of "contour lines." These are fine lines traced through the points of equal level over the surface surveyed, and denote that the level of the ground throughout the whole of their course is identical; that is to say, that every part of the ground over which the line passes is at a certain height above a known fixed point termed the datum, this height being indicated by the figures written against the line. The intervals between the lines vertically are equal and may be 1, 3, 5, 10 or more feet apart; where the surface is very steep they lie close together. These lines by their greater or less distance apart have the effect of shading, and make apparent to the eye, the undulations and irregularities in the surface of the country. HIGHWAY CONSTRUCTION 60.00 1-55.00 - 54.20 - 54.4 - 54.62 Fig. 4 shows an imaginary tract of country, the physical features of which are shown by contour lines. Map. The map should show the lengths and direction of the different por- tions of the line, the topography, rivers, water courses, roads, railroads, and other matters of interest, such as town and county lines, dividing lines between property, timbered and cultivated lands, etc. Any convenient scale may be adopted; 400 feet to an inch will be found the most useful. Memoir. The descriptive memoir should give with minuteness all information, such as the nature of the soil, character of the several excavations whether earth or rock, and such particular features as can- not be clearly shown upon the map or profile. Special information should be given re- garding the rivers crossed, as to their width, depth at highest known flood, velocity of current, character of banks and bottom, and the angle of skew which the course makes with the line of the road. Levels. Levels should be taken along the course of each line, usually at every 100 feet, or at closer intervals, depending upon the nature of the country. In taking the levels, the heights of all existing roads, railroads, rivers, or canals should be noted. "Bench marks" should be established at least every half mile, that is, marks made on any fixed object, such as a gate post, side of a house, or, in the absence of these, a cut made on a large tree. The height and exact 55.10- 55.00- Fig. 5. Preliminary Profile HIGHWAY CONSTRUCTION 17 description of each bench mark should be recorded in the level book. Cross Levels. Wherever considered necessary levels at right angle to the center line should be taken. These will be found useful in showing what effect a deviation to the right or left of the surveyed line would have. Cross levels should be taken at the intersection of all roads and railroads to show to what extent, if any, these levels will have to be altered to suit the levels of the proposed road. Profile. A profile is a longitudinal section of the route, made from the levels. Its horizontal scale should be the same as that of the map; the vertical scale should be such as will show with distinct- ness the inequalities of .the ground. " Fig. 5 shows the manner in which a profile is drawn and the nature of the information to be given upon it. Bridge Sites, The question of choosing the site of bridges is an important one. If the selection is not restricted to a particular point, the river should be examined for a considerable distance above and below what would be the most convenient point for crossing; and if a better site is found, the line of the road must be made subordinate to it. If several practicable crossings exist, they must.be carefully compared in order to select the one most advantageous. The follow- ing are controlling conditions: (1) Good character of the river bed, affording a firm foundation. If rock is present near the surface of the river bed, the foundation will be easy of execution and stability and economy will be insured. (2) Stability of river banks, thus securing a permanent concentration of the waters in the same bed. (3) The axis of the bridge should be at right angles to the direction of the current. (4) Bends in rivers are not suitable localities and should be avoided if possible. A straight reach above the bridge should be 'secured if possible. Final Selection. In making the final selection the following principles should be observed as far as practicable. (a) To follow that route which affords the easiest grades. The easiest grade for a given road will depend on the kind of covering adopted for its surface. (6) To connect the places by the shortest and most direct route commensurate with easy grades. (c) To avoid all unnecessary ascents and descents. When a 18 HIGHWAY CONSTRUCTION road is encumbered with useless ascents, the wasteful expenditure of power is considerable. (d) To give the center line such a position, with reference to the natural surface of the ground, that the cost of construction shall be reduced to the smallest possible amount. (e) To cross all obstacles (where structures are necessary) as nearly as possible at right angles. The cost of skew structures increases nearly as the square of the secant of the obliquity. (/) To cross ridges through the lowest pass which occurs. ( inches, requir- ing 61 to the square yard, and weighing 6i pounds each; "Metropoli- tan", 3X4X9 inches, requiring 45 to the square yard, and weighing 9\ pounds each. Advantages of Brick Pavements. These may be stated as follows : (1) Ease of traction. (2) Good foothold for horses. (3) Not disagreeably noisy. (4) Yields but little dust and mud. (5) Adapted to all grades. (6) Easily repaired. (7) Easily cleaned. (8) But slightly absorbent. (9) Pleasing, to the eye. (10) Expeditiously laid. (11) Durable under moderate traffic. Defects of Brick Pavements. The principal defects of brick pavements arise from lack of uniformity in the quality of the bricks, and from the liability of incorporating in the pavement bricks of too soft or porous a structure, which crumbles under the action of traffic or frost. Foundation. A brick pavement should have a firm foundation. As the surface is made up of small, independent blocks, each one must 96 HIGHWAY CONSTRUCTION be adequately supported, or the load coming upon it may force it downwards and cause unevenness, a condition which conduces to the rapid destruction of the pavement. Several forms of foundation have been used such as gravel, plank, sand, broken stone, and concrete. The last mentioned is doubtless the best. Sand Cushion. The sand cushion is a layer of sand placed on top of the concrete to form a bed for the brick. Practice regarding the depth of this layer of sand varies considerably. In some cases it is only half an inch deep, varying from this up to three inches. The sand cushion is very desirable, as it not only forms a perfectly true and even surface upon which to place brick, but also makes the pavement less hard and rigid than would be the case were the brick laid directly on the concrete. The sand is spread evenly, sprinkled with water, smoothed, and brought to the proper contour by screeds or wooden templets, properly trussed, mounted on wheels or shoes which bear upon the upper sur- face of the curb. Moving the templet forward levels and forms the sand to a uniform surface and proper shape. The sand used for the cushion-coat should be clean and free from loam, moderately coarse, and free from pebbles exceeding one-quarter inch in size. Manner of Laying. The bricks should be laid oh edge, as closely and compactly as possible, in straight courses across the street, with the length of the bricks at right angles to the axis of the street. Joints should be broken by at least 3 inches. None but whole bricks should be used, except in starting a course or making a closure. To provide for the expansion of the pavement, both longitudinal and transverse expansion-joints are used, the former being made by placing a board templet seven-eighths of an inch thick against the curb and abutting the brick thereto. The transverse joints are formed at intervals varying between 25 and 50 feet, by placing a templet or building-lath three-eighths of an inch thick between two or three rows of brick. After the bricks are rammed and ready for grouting, these templets are removed, and the spaces so left are rilled with coal-tar pitch or asphal- tic paving cement. The amount of pitch or cement required will vary between one and one and a-half pounds per square yard of pavement, depending upon the width of the joints. After 25 or 30 feet of the pavement is laid, every part of it should be rammed with a rammer 364 HIGHWAY CONSTRUCTION weighing not less than 50 pounds; and the bricks which sink below the general level should be removed, sufficient sand being added to raise the brick to the required level. After all objectionable brick have been removed, the surface should be swept clean, then rolled with a steam roller weighing from 3 to 6 tons. The object of rolling is to bring the bricks to an unyielding bearing with a plane surface; if this is not done, the pavement will be rough and noisy and will lack dura- bility. The rolling should be first executed longitudinally, beginning at the crown and working toward the gutter, taking care that each return trip of the roller covers exactly the same area as the preceding trip, so that the second passage may neutralize any careening of the brick due to the first passage. The manner of laying brick at street intersections is shown in Fig. 59. 365 98 HIGHWAY CONSTRUCTION Joint Filling. The character of the material used in filling the joints between the brick has considerable influence on the success and durability of the pavement. Various materials have been used such as sand, coal-tar pitch, asphalt, mixtures of coal-tar and asphalt, and Portland cement, besides various patented fillers ; as "Murphy's grout", which is made from ground slag and cement. Each material has its advocates, and there is much difference of opinion as to which gives the best results. The best results seem to be obtained by using a high grade of Portland cement containing the smallest amount of lime in its composi- tion, the presence of the lime increasing the tendency of the filler to swell through absorption of moisture, causing the pavement to rise or to be lifted away from its foundation, and thus producing the roaring or rumbling noise so frequently complained of. The Portland cement grout, when uniformly mixed and carefully placed, resists the impact of traffic and wears well with brick. When a failure occurs/repairs can be made quickly; and, if made early, the pavement will be restored to a good condition. If, however, repairs are neglected, the brick soon loosens and the pavement fails. The office of a filler is to prevent water from reaching the founda- tion, and to protect the edges of the brick from spalling under traffic. In order to meet both of these requirements, every joint must be filled to the top, and must remain so, wearing down with the brick. Sand does not meet these requirements. Although at first making a good filler, being inexpensive and reducing the liability of the pavement to be noisy, it soon washes out, leaving the edges of the brick unprotected and consequently liable to be chipped. Coal-tar and the mixtures of coal-tar and asphalt have an advantage in rendering a pavement less noisy and in cementing together any breaks that may occur through upheavals from frost or other causes; but, unless made very hard, they have the disadvantage of becoming soft in hot weather and flowing to the gutters and low places in the pavement, there forming a black and unsightly scale and leaving the high parts unprotected. The joints, thus deprived of their filling, become receptacles for water, mud, and ice in turn; and the edges of the brick are quickly broken down. Some of these mixtures become so brittle in winter that they crack and fly out of the joints under the action of traffic. The Portland cement filler is prepared by mixing two parts of 366 HIGHWAY CONSTRUCTION cement and one part of fine sand with sufficient water to make. a thin grout. The most convenient arrangement for preparing and dis- tributing the grout is a water-tight wooden box carried on four wooden wheels about 12 inches in diameter. The box may be about 4 feet wide, 7 feet long, and 12 inches deep, furnished with a gate about 8 inches wide, in the rear end. The box should be mounted on the wheels with an inclination, so that the rear end is about. 4 inches lower than the front end. The operation of placing the filler is as follows : The cement and sand are placed in the box, and sufficient water is added to make a thin grout. The box is located about 12 feet from the gutter, the end gate opened, and about 2 -cubic feet of the grout allowed to flow out and run over the top of the brick (care being taken to stir the grout while it is being discharged). If the brick are very dry, the entire surface of the pavement should be thoroughly wet with a hose before applying the grout; if not, absorption of the water from the grout by the bricks will prevent adhesion between the bricks and the cement grcut. The grout is swept into the joints by ordinary bass brooms. After about 100 feet in length of the pavement has been covered the box is returned to the starting-point, and the operation is repeated with a grout somewhat thicker than the first. If this second applica- tion is not sufficient to fill the joints, the operation is repeated as often as may be necessary to fill them. If the grout has been made too thin, or the grade of the street is so great that the grout will not remain long enough in place to set, dry cement may be sprinkled over the joints and swept in. After the joints are completely filled and inspected, allowing three or four hours to intervene, the completed pavement should be covered with sand to a depth of about half an inch, and the roadway barricaded, and no traffic allowed on it for at least ten days. The object of covering the pavement with sand is to prevent the grout from drying or settling too rapidly; hence, in dry and windy weather, it should be sprinkled from time to time. If coarse sand is employed in the grout, it will separate from the cement during the operation of filling the joints, with the result that many joints will be filled with sand and very little cement, while others will be filled with cement and little or no sand ; thus there will be many spots in the pave- ment in which no bond is foimed between the bricks, and under the action of traffic these portions will quickly become defective. 367 100 HIGHWAY CONSTRUCTION The coal-tar filler is best applied by pouring the material from buckets, And brooming it into the joints with wire brooms. In order to fill the joints effectually, it must be used only when very hot. To secure this condition, a heating tank on wheels is necessary. It should have a capacity of at least five barrels, and be kept at a uniform tem- perature all day. One man is necessary to feed the fire and draw the material into the buckets; another, to carry the buckets from the heat- ing-tank to a third, who pours the material over the street. The latter starts to pour in the center of the street, working backward toward the curb, and pouring a strip about two feet in width. A fourth man, with a wire broom, follows immediately after him, sweeping the sur- plus material toward the pourer and in the direction of the curb. This method leaves the entire surface of the pavement covered with a thin coating of pitch, which should immediately be covered with a light coating of sand ; the sand becomes imbedded in the pitch. Under the action of traffic, this thin coating is quickly worn away, leaving the surface of the bricks clean and smooth. Tools Employed in Construction of Block Pavements. The principal tools required in constructing block pavements comprise hammers and rammers of varying sizes and shapes, depending on the material and size of the blocks to be laid; also crowbars, sand screens, an:! rattan and wire brooms. Cobblestones, square blocks, and brick require different types of both hammers and rammers for adjusting them to place and forcing them to their seat. A cobblestone rammer, for example, is usually made of w r ood (generally locust) in the shape of a long truncated cone, banded with iron at top and bottom, weighing about 40 pounds, and having two handles, o;ie at the top and another 'on one side. A Belgian block rammer is slightly heavier, consisting of an upper part of wood set in a steel base; while a rammer for granite blocks is still heavier, comprising an iron base with cast-steel face, into which is set a locust plug with hickory handles. For laying brick, a wooden rammer shod with cast iron or steel and weighing about 27 pounds is used. A light rammer of about 20 pounds' weight, consist' ing of a metallic base attached to a long, slim wooden handle, is used for miscellaneous work, such as tamping in trenches, next to curbs, etc. Concrete=Mixing Machine. Where large quantities of concrete are required, as in the foundations of improved pavements, concrete can be prepared more expeditiously and economically by the use of 868 HIGHWAY CONSTRUCTION 101 mechanical mixers, and the ingredients will be more thoroughly mixed, than by hand. Thorough incorporation of the ingredients is an essen- tial element in the quality of a concrete. When mixed by hand, how- ever, the incorporation is rarely complete, because it depends upon the proper manipulation of the hoe and shovel. The manipulation, although extremely simple, is rarely performed by the ordinary laborer unless he is constantly watched by the overseer. Several varieties of concrete-mixing machines are in the market. A convenient portable type is illustrated in Fig. 60. The capacity of Fig. 60. Concrete Mixing Mat-blue. the mixers ranges from five to twenty cubic yards per hour, de-lending upon size, regularity with which the materials are supplied, speed, etc. Gravel Heaters. Fig. 61 illustrates a device commonly employed for heating the gravel used for joint filling in stone-block pavements. These heaters are made in various sizes, a common size being 9 feet long, 5 feet wide, and 3 feet 9 inches high. Melting Furnaces, for heating the pitch or tar for joint filling, are illustrated by Fig. 62. Various sizes are on the market. WOOD PAVEMENTS Wood pavements are formed of either rectangular or cylindrical blocks of wood. The rectangular blocks are generally 3 inches wide, 102 HIGHWAY CONSTRUCTION 9 inches long, and 6 inches deep; the round blocks are commonly G inches in diameter and 6 inches long. The kinds of wood most commonly used are cedar, cypress, juni- per, yellow pine, and mesquite; and recently jarrah from Australia, and pyinyado from India, have been used. The wood is used in its natural condition, or impregnated with creosote or other chemical preservative. The blocks of wood are laid either -on the natural soil, on a bed of sand and gravel, on a layer of broken stone, on a layer of concrete, ' n ig. 62. .Melting Furnace. or, .sometimes, on a double layer of plank. The joints are filled either with sand, paving-pitch, or Portland-cement grout. Advantages. The advantages of wood pavement may be stated as follows : (1) It affords good foothold for horses. (2) It offers less resistance to traction than stone, and slightly more than asphalt. (3) It suits all classes of traffic. (4) It may be used on grades up to five per cent. (5) It is moderately durable. (0) It yields no mud when laid upon an impervious foundation. (7) It yields but little dust. 370 HIGHWAY CONSTRUCTION 103 (8) It is moderate in first cost. (9) It is not disagreeably noisy. Defects. The principal objections to wood pavement are: (1) It is difficult to cleanse. (2) Under certain conditions of the atmosphere it becomes greasy and very unsafe for horses. (3) It is not easy to open for the purpose of gaining access to underground pipes, it being necessary to remove rather a large surface for this purpose, which has to be left a little time after being repaired before traffic is again allowed upon it. (4) It is absorbent of moisture. (5) It is claimed by many that wood pavements are unhealthy. Quality of Wood. The question as to which of the various kinds of wood available is the most durable and economical, has not been satisfactorily determined. Many varieties have been tried. In England, preference is given to Baltic fir, yellow pine, and Swedish yellow deal. In the United States the variety most used (on account of its abundance and cheapness) is cedar; but yellow pine, tamarack, and mesquite have also been used to a limited extent, and cypress and juniper are being largely used in some of the Southern States. Hardwoods, such, as oak, etc., do not make the best pavements, as such woods become slippery. The softer, close-grained woods, such as cedar and pine, wear better and give good foothold. The wood employed should be sound and seasoned, free from sap, shakes, and knots. Defective blocks laid in the pavement will quickly cause holes in the surface, and the adjoining blocks will suffer under wear, the whole surface becoming bumpy. Chemical Treatment of Wood. The great enemy of all wood pavements is decay, induced by the action of the air and water. Wood is porous, absorbs moisture, and thus hastens its own destruction. Many processes have been invented to overcome this defect. The most popular processes at present are crcosoting and modifications of the same, known as the "creo-resinate" and "kreodine" processes. These consist of creosote mixed with various chemicals which are supposed to add to the preserving qualities of the creosote. Creosoting. This process consists in impregnating the wood with the oil of tar, called creosote, from which the ammonia has been ex- pelled, the effect being to coagulate the albumen and thereby prevent 371 104 HIGHWAY CONSTRUCTION its decomposition, also to fill the pores of the wood with a bituminous substance which excludes both air and moisture, and which is noxious to the lower forms of animal and vegetable life. In adopting this pro- cess, all moisture should be dried out of the pores of the timber. The softer woods, while warm from the drying-house, may be immersed at once in an open tank containing hot creosote oil, when they will absorb about 8 or 9 pounds per cubic foot. For hardwoods, and woods which are required to absorb more than 8 or 9 pounds of creosote per cubic foot, the timber should be placed in an iron cylinder with closed ends, and the creosote, which should be heated to a temperature of about 120 F., forced in with a pressure of 170 pounds to the square inch. The heat must be kept up until the process is complete, to prevent the creosote from crystallizing in the pores of the wood. By this means the softer woods will easily absorb from 10 to 12 pounds of oil per cubic foot. The most effective method, however, is to exhaust the air from tht cylinder after the timber is inserted; then to allow the oil to flow in; and when the cylinder is full, to use a force pump with a pressure of 150 to 200 pounds per square inch, until the wood has absorbed the requisite quantity of oil, as indicated by a gauge, which should be fitted to the reservoir tank. The oil is usually heated by coils of pipe placed in the reservoir, through which a current of steam is passed. The quantity of creosote oil recommended to be forced into the wood is from 8 to 12 pounds per cubic foot. Into oak and other hard woods it is difficult to force, even with the greatest pressure, more than 2 or 3 pounds of oil. The advantages of this process are : The chemical constituents of the oil preserve the fibers of the wood by coagulating the albumen of the sap ; the fatty matters act mechanically by filling the pores and thus exclude water; while the carbolic acid contained in the oil is a powerful disinfectant. , The life of the wood is extended by any of the above processes, by preserving it from decay; but such processes have little or no effect on the wear of the blocks under traffic. The process of dipping the blocks in coal tar or creosote oil is injurious. Besides affording a cover for the use of defective or sappy wood, it hastens decay, especially of green wood ; it closes up the ex- 372 HIGHWAY CONSTRUCTION 105 terior of the cells of the wood so that moisture cannot escape, thus causing fermentation to take place in the interior of the block, which quickly destroys the strength of the fibers and reduces them to punk. Expansion of Blocks. Wood blocks expand on exposure to moisture; and, when they are laid end to end across the street, the curbstones are liable to be displaced, or the courses of the blocks will be bent into reserve curves. To avoid this, the joints of the courses near the curb may be left open until expansion has ceased, the space being temporarily filled with sand. The rate of expansion is about 1 inch in 8 feet, but Caries for different woods. The time required for the wood to become fully expanded varies from 12 to 18 months. By employing blocks impregnated with the oil of creosote, this trouble will be avoided. 'Blocks so treated do not contract or expand to any appre- ciable extent. The comparative expansion of creosoted and plain wood blocks after immersion in water for forty-eight hours, in percentage on orig- inal dimensions, was: Expansion of Wood Paving Blocks On length of block.. On width " ' . On depth " " . Manner of Laying. The blocks are set with the fiber vertical, and the long dimension crosswise of the street, the longitudinal joints being broken by a lap of at least one-third the length of the block; the Fig. 63. blocks should be laid so as to have the least possible width of joint. Wide joints hasten the destruction of the wood by permitting- the fibers to wear under traffic, which also causes the surface of the pavement 373 106 HIGHWAY CONSTRUCTION to wear in small ridges. The most recent practice for laying blocks on 3 per cent grades, has been to remove from the top of one side of each block a strip inch thick and 1 inches deep, extending the length of the block. When the blocks are laid and driven closely together, there is a quarter-inch opening or joint extending clear across the street in each course. These joints are filled with Portland cement grout. Fig. 63 shows a section of pavement having this form of joint. Filling for Joints. The best materials for filling the joints are bitumen for the lower two or three inches, and hydraulic cement grout for the remainder of the depth. The cement grout protects the pitch from the action of the sun, and does not wear down very much below the surface of the wood. ASPHALT PAVEMENTS. Asphaltic Paving Materials. All asphaltic or bituminous pave- ments are composed of two essential parts namely, the cementing material (matrix) and the resisting material (aggregate). Each has a distinct function to perform; the first furnishes and preserves the co- herency of the mass ; the second resists the wear of traffic. Two classes of asphaltic paving compounds are in use, namely, natural and artificial. The "natural" variety is composed of either limestone or sandstone naturally cemented with bitumen. To this class belong the bituminous limestones of Europe, Texas, Utah, etc., and the bituminous sandstones of California, Kentucky, Texas, Indian Territory, etc. The "artificial" consists of mixtures of asphaltic cement with sand and stone dust. To this class belong the pavements made from Trinidad, Bermudez, Cuban, and similar asphaltums. For the artificial variety, most hard bitumens are, when properly prepared, equally suitable. For the aggregate, the most suitable mate- rials are stone-dust from the harder rocks, such as granite, trap, etc., and sharp angular sand. These materials should be entirely free from loam and vegetable impurities. The strength and enduring qualities of the mixture will depend upon the quality, strength, and proportion of each ingredient, as well as upon the cohesion of the matrix and its adhesion to the aggregate. Bituminous limestone consists of carbonate of lime naturally cemented with bitumen in proportions varying from 80 to 93 per cent of carbonate of lime and from 7 to 20 per cent of bitumen. Its color, when freshly broken, is a dark (almost black) chocolate brown, the 374 HIGHWAY CONSTRUCTION 107 darker color being due to a large percentage of bitumen. At a tem- perature of from 55 to 70 F., the material is hard and sonorous, and breaks easily with an irregular fracture; at temperatures between 70 and 140 F. it softens, passing with the rise in temperature through various degrees of plasticity, until, at between 140 and 160 F., it begins to crumble; at 212 it commences to melt; and at 280 F. it is completely disintegrated. Its specific gravity is about 2.235. Bituminous limestone is the material employed for paving pur- poses throughout Europe. It is obtained principally from deposits at Val-de-Travers, canton of Neufchatel, Switzerland; at Seysell, in the Department of Ain, France; at Ragusa, Sicily; at Limmer, near Hanover; and at Vorwohle, Germany. Bituminous limestone is found in several parts of the United States. Two of these deposits are at present being worked one in Texas, the material from which is called "lithocarbon"; and one on the Wasatch Indian Reservation. These deposits contain from 10 to 30 per cent of bitumen. The bituminous limestones which contain about 10 per cent of bitumen are used for paving in their natural condition, being simply reduced to powder, heated until thoroughly softened, then spread while hot upon the foundation, and tamped and rammed until compacted. Bituminous sandstones are composed of sandstone rock impreg- nated with bitumen in amounts varying from a trace to 70 per cent. They are found in both Europe and America. In Europe, they are chiefly used for the production of pure bitumen, which is extracted by boiling or macerating them with water. In the United States, exten- sive deposits are found in the Western States; and since 1880 they have been gradually coming into use as a paving material, so that now up- wards of 150 miles of streets in Western cities are paved with them. They are prepared for use as paving material by crushing to powder, which is heated to about 250 F. or until it becomes plastic, then spread upon the street and compressed by rolling; sometimes sand or gravel is added, and it is stated that a mixture of about 80 per cent of gravel makes a durable pavement. Trinidad Asphaltum. The deposits of asphaltum in the island of Trinidad, W. I., have been the main source of supply for the asphal- tum used in street paving in the United States. Three kinds are found there, which have been named, according to the source, lake-pitch, 375 108 HIGHWAY CONSTRUCTION land or overflow pitch, and iron pitch. The first and most valuable kind is obtained from the so-called Pitch Lake. The term land or overflow pitch is applied to the deposits of asphaltum found outside the lake. These deposits form extensive beds of variable thickness, and are covered with from a few to several feet of earth ; they are considered by some authorities to be formed from pitch which has overflowed from the lake; by others to be of entirely different origin. The name cheese pitch is given to such portions of the land pitch as more nearly resemble that obtained from the lake. The term iron pitch is used to designate large and isolated masses of extremely hard asphaltum found both within and without the bor- ders of the lake. It is supposed to have been formed by the action of heat caused by forest fires, which, sweeping over the softer pitch, re- moved its more volatile constituents. The name epurce is given to asphaltum refined on the island of Trinidad. The process is conducted in a very crude manner, in large, open, cast-iron sugar boilers. The characteristics of crude Trinidad asphaltum, both lake and land, are as follows : It is composed of bitumen mixed with fine sand, clay, and vegetable matter. Its specific gravity varies according to the impurities present, but is usually about 1.28. Its color, when fieshly excavated, is a brown, which changes to black on exposure to the at- mosphere. When freshly broken, it emits the usual bituminous odor- It is porous, containing gas cavities, and in consistency resembles cheese. If left long enough in the sun, the surface will soften and melt, and will finally flow into a more or less compact mass. Refined Trinidad Asphaltum. The crude asphaltum is refined or purified by melting it in iron kettles or stills by the application of indirect heat. The operation of refining proceeds as follows : During the heat- ing, the water and lighter oils are evaporated; the asphaltum is lique- fied; the vegetable matter rises to the surface, and is skimmed off; the earthy and siliceous matters settle to the bottom; and the liquid asphal- tum is drawn off into old cement or flour barrels. When the asphaltum is refined without agitation, the residue remaining in the still forms a considerable percentage of the crude material, frequently amounting to 12 per cent; and it was at one time considered that the greater the amount of this residue the better the 376 HIGHWAY CONSTRUCTION 109 quality of the refined asphaltum. Since agitation has been adopted, however, the greater part of the earthy and siliceous matters is retained in suspension; and it has come to be considered just as desirable for a part of the surface mixture as the sand which is subsequently added. The refined asphaltum, if for local use, is generally converted into cement in the same still in which it was refined. The average composition of both the land and lake varieties is shown by the following analyses : Average Composition of Trinidad Asphaltum CONSTITUENTS LAKE LAND Hard Soft Per Cent 07 ^5 Per Cent 34.10 25.05 6.85 34.50 Per Cent 26.62 27.57 8.05 37.76 Inorganic matter Organic non-bituminous matter 26. H8 7.63 38.14 Bitumen When the analyses are calculated to a basis of dry substances, the composition is: Inorganic matter Organic matter not bitumen 100.00 100.00 38.00 9.64 52.36 100.00 36.56 10.57 52.87 37.74 10.68 51.58 Bitumen The substances volatilized in 10 hours at 400 F . .. The substances soften at 100.00 100.00 100.00 3.66 190 F. 200 F. 12.24 170 F. 185 F. 0.86 to 1.37 200 to 250 3 F. 2 10 to 328 F. flow at The characteristics of refined Trinidad asphaltum are as follows: The color is black, with a homogeneous appearance. At a tempera- ture of about 70 F., it is very brittle, and breaks with a conchoidal fracture. It burns with a yellowish-white flame, and in burning emits an empyreumatic odor, and possesses little cementitious quality. To give it the required plasticity and tenacity, it is mixed while liquid with from Ifi to 21 pounds of residuum oil to 100 pounds of asphaltum. The product resulting from the combination is called asphalt paving-cement. Its consistency should be such that, at a temperature of from 70 to 80 F., it can be easily indented with the fingers, and on slight warming be drawn out in strings or threads. Artificial Asphalt Pavements. The pavements made from Trini- dad, Bermudez, California, and similar asphaltums, are composed of mechanical mixtures of asphaltic cement, sand, and stone-dust. The sand should be equal in quality to that used for hydraulic cement mortar; it must be entirely free from clay, loam, and vegetable 110 HIGHWAY CONSTRUCTION impurities; its grains should be angular and range from coarse to fine. The stone-dust is used to aid in filling the voids in the sand and thus reduce the amount of cement. The amount used varies with the coarseness of the sand and the quality of the cement, and ranges from 5 to 15 per cent. (The voids in sand vary from .3 to .5 per cent.) As to the quality of the stone-dust, that from any durable stone is equally suitable. Limestone-dust was originally used, and has never been entirely discarded. The paving composition is prepared by heating the mixed sand and stone-dust and the asphalt cement separately to a temperature of about 300 F. The heated ingredients are measured into a pu^-:nill and thoroughly incorporated. When this is accomplished, the mix- ture is ready for use. It is hauled to the street and spread with iron rakes to such depth as will give the required thickness when compacted (the finished thickness varies between 1^ and 2j inches). The re- duction of thickness by compression is generally about 40 per cent. The mixture is sometimes laid in two layers. The first is called the binder or cushion=coat ; it contains from 2 to 5 per cent more cement than the surface-coat; its thickness is usually -| inch. The object of the binder course is to unite the surface mixture with the foundation, which it does through the larger percentage of cement that it contains, which, if put in the surface mixture, would render it too soft. The paving composition is compressed by means of rollers and tamping irons, the latter being heated in a fire contained in an iron basket mounted on wheels. These irons are used for tamping such portions as are inaccessible to the roller namely, gutters, around man- hole heads, etc. Two rollers are sometimes employed; one, weighing 5 to 6 tons and of narrow tread, is used to give the first compression; and the other, weighing about 10 tons and of broad tread, is used for finishing. The amount of rolling varies; the average is about 1 hour per 1,000 square yards of surface. After the primary compression, natural hydraulic or any impalpable mineral matter is sprinkled over the sur- face, to prevent the adhesion of the material to the roller and to give the surface a more pleasing appearance. When the asphalt is laid up to the curb, the surface of the portion forming the gutter is painted with a coat of hot cement. Although asphaltum is a bad conductor of heat, and the cement 378 S i MS '8 |! 1 If I iil 2! * P 1 I, il! si >< o SSl II F II il 111 i|| 11 HIGHWAY CONSTRUCTION HI retains its plasticity for several hours, occasions may and do arise through which the composition before it is spread has cooled; its con- dition when this happens is analogous to hydraulic cement which has taken a "set," and the same rules which apply to hydraulic cement in this condition should be respected in regard to asphaltic cement. The proportions of the ingredients in the paving mixture are not constant, but vary with the climate of the place where the pavement is to be used, the character of the sand, and the amount and character of the traffic that will use the pavement. The range in the proportion is as follows: Formula for Asphaltic Paving Mixture Asphalt cement 12 to 15 per cent. Sand 70 to 83 " Stone-dust .5 to 15 " A cubic yard of the prepared material weighs about 4,500 pounds, and will lay the following amount of wearing-surface: 1\ inches thick 12 square yards. 2 " " IS " 1J " " 27 " One ton of refined asphaltum makes about 2,300 pounds of asphalt cement, equal to about 3.4 cubic yards of surface material. Foundation. A solid, unyielding foundation is indispensable with all asphaltic pavements, because asphalt of itself has no power of offering resistance to the action of traffic, consequently it is nearh always placed upon a bed of hydraulic cement concrete. The concrete must be thoroughly set and its surface dry before the asphalt is laid upon it; if not, the water will be sucked up and converted into steam, with the result that coherence of the asphaltic mixture is prevented, and, although its surface may be smooth, the mass is really honey- combed, so that as soon as the pavement is subjected to the action of traffic, the voids or fissures formed by the steam appear on the surface, and the whole pavement is quickly broken up. Advantages of Asphalt Pavement. These may be summed up as follows: (1) Ease of traction. (2) It is comparatively noiseless under traffic. (3) It is impervious. (4) It is easily cleansed. (5) It produces neither mud nor dust. (6) It is pleasing to the eye. 379 112 HIGHWAY CONSTRUCTION (7) It suits all classes of traffic. (8) There is neither vibration nor concussion in traveling over it. (9) It is expeditiously laid, thereby causing little inconvenience to traffic. (10) Openings to gain access to underground pipes are easily made. (11) It is durable. (12) It is easily repaired. Defects of Asphalt Pavement. These are as follows: (1) It is slippery under certain conditions of the atmosphere. The American asphalts are much less so than the European, on account of their granular texture derived from the sand. The difference is very noticeable; the European are as smooth as glass, while the Ameri- can resemble fine sandpaper. (2) It will not stand constant moisture, and will disintegrate if excessively sprinkled. (3) Under extreme heat it is liable to become so soft that it will roll 01* creep under traffic and present a wavy surface; and under ex- treme cold there is danger that the surface will crack and become friable. (4) It is not adapted to grades steeper than 2-V per cent, although it is in use on grades up to 7.30 per cent. (5) Repairs must be quickly made, for the material has little coherence, and if, from irregular settlement of foundation or local vio- lence, a break occurs, the passing wheels rapidly shear off the sides of the hole, and it soon assumes formidable dimensions. The strewing of sand upon asphalt renders it less slippery ; but in addition to the interference of the traffic while this is being done, there are further objections namely, the possible injury by the sand cutting into the asphalt, the expense of labor and materials, and the mud formed, which has afterwards to be removed. Although pure asphaltum is absolutely impervious and insoluble in either fresh or salt wafer, yet asphalt pavements in the continued presence of water are quickly disintegrated. Ordinary rain or daily sprinkling does not injure them when they are allowed to become per- fectly dry again. The damage is most apparent in gutters and adja- cent to overflowing drinking fountains. This defect has long been recognized; and various measures have been taken to overcome it, or 380 HIGHWAY CONSTRUCTION 113 at least to reduce it to a minimum. In some cities, ordinances have been passed, seeking to regulate the sprinkling of the streets; and in many places the gutters are laid with stone or vitrified brick (see Figs. Fig. 64. 64 and 65), while in others the asphalt is laid to the curb, a space of 12 to 15 inches along the curb being covered with a thin coating of asphalt cement. Asphalt laid adjoining center-bearing street-car rails is quickly broken down and destroyed. This defect is not peculiar to asphalt. All other materials when placed in similar positions are quickly worn. Granite blocks laid along such tracks have been cut into at a rate of more than half an inch a year. The frequent entering and turning off of vehicles from car tracks is one of the severest tests that can be Asphalt 3 rich Gutter Curb Fig. 65. applied to any paving material; moreover, the gauge of trucks and vehicles is frequently greater than that of the rails, so one wheel runs on the rail and the other outside. The number of wheels thus travel- ing in one line must quickly wear a rut in any material adjoining the center-bearing rail. To obviate the destruction of asphalt in such situations, it is usual to lay a strip of granite block or brick paving along the rail. This pavement should be of sufficient width to support the wheels of the widest gauge using the street. 381 114 HIGHWAY CONSTRUCTION The burning of leaves or making of fires on asphalt pavements should not be permitted, as it injures the asphalt, and the paving com- panies cannot be compelled to repair the damaged places without compensation. Asphalt Blocks. Asphalt paving blocks are formed from a mix- ture of asphaltic cement and crushed stone in the proportion of 8 to 12 per cent of cement to 88 and 92 per cent of stone. The materials are heated to a temperature of about 300 F., and mixed while hot in a suitable vessel. When the mixing is complete, the material is placed in moulds and subjected to heavy pressure, after which the blocks are cooled suddenly by plunging into cold water. The usual dimensions of the blocks are 4 inches wide, 3 inches deep, and 12 inches long. Foundation. The blocks are usually laid upon a concrete founda- tion with a cushion-coat of sand about | inch thick. They are laid with their length at right angles to the axis of the street, and the longitu- dinal joints should be broken by a lap of at least 4 inches. The blocks are then either rammed with hand rammers 01 rolled with a light steam roller, the surface being covered with clean, fine sand ; no joint filling is used, as, under the action of the sun and traffic, the blocks soon become cemented. The advantages claimed for a pavement of asphalt blocks over a continuous sheet of asphalt are: (1) That they can be made at a factory located near the materials, whence they can be transported to the place where they are to be used and can be laid by ordinary paviors, whereas sheet pavements require special machinery and skilled labor; (2) that they are less slippery, owing to the joints and the rougher surface due to the use of crushed stone. Asphalt Macadam Bituminous Macadam. Recently it has been proposed to use asphalt as a binding material for broken stone. There are two patented processes the Whinery and the Warren which differ slightly in details. The advantages claimed for these methods are: (1) The first coat will be materially less; (2) it will offer a better foothold for horses; (3) it will be at least as durable as the ordinary sheet asphalt; (4) it will not shift under traffic and roll into waves; (5) it will not crack; (6) it 982 HIGHWAY CONSTRUCTION 115 can be repaired more cheaply and with less skilled labor than can the ordinary sheet asphalt. Tools Employed in Construction of Asphalt Pavements. The Fig. 66. Steam Roller. Fig. 67. Asphalt Tools. tools used in laying sheet asphalt pavements comprise iron rakes; hand rammers; smoothing irons (Fig. 67); pouring pots (Fig. 69); 116 HIGHWAY CONSTRUCTION hand rollers, either with or without a fire-pot (Fig. 68); and steam rollers, with or without provision for heating the front roll (Fig. 66). These rollers are different in construction, appearance, and weight Fig. 68. Hand Hollers. Fig. 69. Pouring Pots. from those employed for compacting broken stone. The difference is due to the different character of the work required. The principal dimensions of a five-ton roller are as follows : Front roll or steering-wheel 30 to 32 inches diameter. Rear roll or driving-wheel 48 " Width of front roll 40 " " " rear " 40 " Extreme length 14 feet. " height 7 to 8 feet. Water capacity 80 to 100 gallons. Coal " 200 pounds. 384 HIGHWAY CONSTRUCTION 11? FOOTPATHS CURBS GUTTERS. A footpath or walk is simply a road under another name a road for pedestrians instead of one for horses and vehicles. The only difference that exists is in the degree of service required ; but the con- ditions of consruction that render a road well adapted to its object are very much the same as those required for a walk. The effects of heavy loads such as use carriageways are not felt upon footpaths; but the destructive action of water and frost is the same in either case, and the treatment to counteract or resist these elements as far as practicable, and to produce permanency, must be the controlling idea in each case, and should be carried out upon a common principle. It is not less essential that a walk should be well adapted to its object than that a road should be; and it is annoying to find it impassable or insecure and in want of repair when it is needed for convenience or pleasure. In point of economy, there is the same advantage in constructing a footway skilfully and durably as there is in the case of a road. Width. The width of footwalks (exclusive of the space occupied by projections and shade trees) should be ample to accomodate com- fortably the number of people using them. In streets devoted entirely to commercial purposes, the clear width should be at least one-third the width of the carriageway; in residential and suburban streets, a very pleasing result can be obtained by making the walk one-half the width of the roadway, and devoting the greater part to grass and shade trees. Cross Slope. The surface of footpaths must be sloped so that the surface water will readily flow to the gutters. This slope need not be very great; | inch per foot will be sufficient. A greater slope with a thin coating of ice upon it, becomes dangerous to pedestrians. Foundation. As in the case of roadways, so with footpaths, the foundation is of primary importance. Whatever material may be used for the surface, if the foundation is weak and yielding, the surface will settle irregularly and become extremely objectionable, if not danger- ous, to pedestrians. Surface. The requirements of a good covering for sidewalks are : (1) It must be smooth but not slippery. ' (2) It must absorb the minimum amount of water, so that it may dry rapidly after rain. 118 HIGHWAY CONSTRUCTION (3) It must not be easily abraded. (4) It must be of uniform quality throughout, so that it may wear evenly. (5) It must neither scale nor flake. (6) Its texture must be such that dust will not adhere to it. (7) It must be durable. Materials. The materials used for footpaths are as follows: Stone, natural and artificial; wood; asphalt; brick; tar concrete; and gravel. Of the natural stones, sandstone (Milestone) and granite are ex- tensively employed. The bluestone, when well laid, forms an excellent paving material. It is of compact texture, absorbs water to a very limited extent, and hence soon dries after rain; it has sufficient hardness to resist abrasion, and wears well without becoming excessively slippery. Granite, although exceedingly durable, wears very slippery, and its surface has to be frequently roughened. Slabs, of whatever stone, must be of equal thickness throughout their entire area; the edges must be dressed true to the square for the whole thickness (edges must not be left feather- ed as shown in Fig. 70); and the slabs must be solidly bedded on the foundation and the joints filled with cement-mortar. Badly set or faultily dressed flagstones are very unpleasant to walk over, especially in rainy weather; the unevenness causes pedes- trians to stumble, and rocking stones squirt dirty water over their clothes. Wood has been largely used in the form of planks; it is cheap in first cost, but proves very expensive from the fact that it lasts but a comparatively short time and requires constant repair to keep it from becoming dangerous. Asphalt forms an excellent footway pavement; it is durable and does not wear slippery. Brick. Brick of suitable quality, well and carefully laid on a concrete foundation, makes an excellent footway pavement for resi- 380 HIGHWAY CONSTRUCTION 119 dential and suburban streets of large cities, and also for the main streets of smaller towns. The bricks should be a good quality of paving brick (ordinary building brick are unsuitable, as they soon wear out and are easily broken) . The bricks should be laid in parallel rows on their edges, with their length at right angles to the axis of the path. Curbstones. Curbstones are employed for the outer side of foot- ways, to sustain the coverings and form the gutter. Their upper edges are set flush with the footwalk pavement, so that the water' can flov, over them into the gutters. The disturbing forces which the curb has to resist, are : (1) The pressure of the earth behind it, which is frequently augmented by piles of merchandise, building materials, etc. This pressure tends to overturn it, break it transversely, or move it bodily on its base. (2) The pressure due to the expansion of freezing earth behind and be- neath it. This force is most frequent where the sidewalk is partly sodded and the ground is accordingly moist. Successive freezing and thawing of the earth behind the curb will occasion a succession of thrusts forward, which, if the curb be of faulty design, will cause it. to incline several degrees from the vertical. (3) The concussions and abrasions caused by traffic To withstand the destructive effect of wheels, curbs are faced with iron; and a concrete curb with a rounded edge of steel has been patented and used to some extent. Fires built in the gutters deface and seriously injure the curb. Posts and trees set too near the curb, tend to break, displace, and destroy it. The use of drain tiles under the curb is a subject of much differ- ence of opinion among engineers. Where the subsoil contains water naturally, or is likely to receive it from outside the curb-lines, the use of drains is of decided benefit; but great care must be exercised in jointing the drain-tiles, lest the soil shall be loosened and removed, causing the curb to drop out of alignment. The materials employed for curbing are the natural stones, as granite, sandstone (bluestone), etc., artificial stone, fire-clay, and cast iron. The dimensions of curbstones vary considerably in different localities and according to the width of the footpaths; the wider the path, the wider should be the curb. It should, however, never be lass than 8 inches deep, nor narrower than 4 inches. Depth is necessary 387 120 HIGHWAY CONSTRUCTION to prevent the curb turning over toward the gutter. It should never be in smaller lengths than 3 feet. The top surface should be beveled off to conform to the slope of the footpath. The front face should be hammer-dressed for a depth of about 6 inches, in order that there may be a smooth surface visible against the gutter. The back for 3 inches Slo. Fig. 71. from the top should also be dressed, so that the flagging or other paving may butt fair against it. The end joints should be cut truly square, the full thickness of the stone at the top, and so much below the top as will be exposed , the remaining portion of the depth and bottom should be roughly squared, and the bottom should be fairly parallel to the top. (See Figs. 71 and 72). Artificial Stone. Artificial stone is being extensively used as a footway paving material. Its manufacture is the subject of several patents, and numerous kinds are to be had in the market. When manufactured of first-class materials and laid in a substantial manner, with proper provision against the action of frost, artificial stone forms a durable, agreeable, and inexpensive pavement. 888 HIGHWAY CONSTRUCTION Pig. 73. Tamper. The varieties most extensively used in the United States*are known by the names of granolithic, monolithic, fcrrolithic, kosmocrete, metalithic, etc. The process of manufacture is practically the same for all kinds, the difference being in the materials em- ployed. The usual ingredients are Port- land cement, sand, gravel, and crushed stone. Artificial stone for footway pave- ments is formed in two ways namely, in blocks manufactured at a factory, brought on the ground, and laid in the same manner as natural stone; or the raw materials are brought upon the work, pre- pared, and laid in place, blocks being formed by the use of board moulds. The manner of laying is practically the same for all kinds. The area to be paved is excavated to a mini- mum depth of 8 inches, and to such great- er depths as the nature of the ground may require to secure a solid foundation. The surface of the ground so exposed is well compacted by ramming; and a layer of gravel, ashes, clinker, or other suitable material is spread and consoli- dated ; on this is placed the concrete wearing surface, usually 4 inches thick. As a protection against the lifting effects of frost, the concrete is laid in squares, rectangles, or other forms hav- ing areas ranging from 6 to 30 square feet, strips of wood being employed to form moulds in which the concrete is placed. After the concrete is set, these strips are removed, leaving joints about half an inch wide between the blocks. Under some patents these joints are filled with cement; under others, with tarred paper; and in some cases they are left open. Tools Employed in Construction of Artificial Stone Pavements. Tampers (Fig. 73). Cast iron, with hickory handle; range from 6 by 8 inches to 8 by 10 inches. Fig. 74. Quarter-Round. Fig. 75. Jointer. 122 HIGHWAY CONSTRUCTION Quarter-Round, (Fig. 74). for forming corners and edges. Fig. 76. Cutter. Made of any desired radius. Used Jointer (Fig. 75). Used for tiimming and finishing the joints. Cutter (Fig. 76). Used for cut- ting the concrete into blocks. Gutter Tool (Fig. 77). Used for forming and finishing gutters. Imprint Rollers (Figs. 78 and 79). Here are shown two designs of rollers for imprinting the surface of artificial stone pavements with grooves, etc. SELECTING THE PAVEMENT The problem of selecting the best pavement for any particular case is a local one, not only for each city, but also for each of the various parts into which the city is imperceptibly divided; and it involves so many elements that the nicest balancing of the relative values for each kind of pavement is required, to arrive at a correct conclusion. In some localities, the proximity of one or more paving materials determines the character of the pavement; while in other cases a careful investigation may be required in order to select the most suitable material. Local conditions should always be considered; hence it is not possible to lay down any fixed rule as to what material makes the best pavement. The qualities essential to a good pavement may be stated as follows : (1) It should be impervious. (2) It should afford good foothold for horses. (3) It should be hard and durable, so as to resist wear and dis- integration. (4) It should be adapted to every grade. (5) It should suit every class of traffic. (6) It should offer the minimum resistance to traction. (7) It should be noiseless. (8) It should yield neither dust nor mud. 390 HIGHWAY CONSTRUCTION 123 (9) It should be easily cleaned. (10) It should be cheap. Interests Affected in Selection. Of the above requirements, numbers 2, 4, 5, and affect the traffic and determine the cost of haul- age by the limitations of loads, speed, and wear and tear of horses and ve- hicles. If the surface is rough or the foothold bad, the weight of the load a horse can draw is decreased, thus ne- cessitating the making of more trips or the employment of more horses and vehicles to move a given weight. A defective surface necessitates a reduc- tion in the speed of movement and consequent loss of time; it increases the wear of horses, thus decreasing their Fig. ?8. imprint Roller. jjf e serv i ce an d lessening the value of their current services; it also increases the cost of maintaining vehicles and harness. Numbers 7, 8, and 9 affect the occupiers of adjacent premises, who suffer from the effect of dust and noise; they also affect the owners of said premises, whose income from rents is diminished where these disad- vantages exist. Numbers 3 and 10 af- fect the taxpayers alone first, as to the length of time during which the covering remains serviceable; and sec- ond, as to the amount of the annual repairs. Number 1 affects the adjacent occupiers principally on hygienic grounds. Numbers 7 and 8 affect both traffic and occupiers. Problem Involved in Selection. The problem involved in the selection of the most suitable pavement consists of the following factors: (1) adaptability; (2) desirability; (3) serviceability; (4) dura- bility; (5) cost. Adaptability. The best pavement for any given roadway will 801 124 HIGHWAY CONSTRUCTION depend altogether on local circumstances. Pavements must be adapt- ed to the class of traffic that will use them. The pavement suitable for a road through an agricultural district will not be suitable for the streets of a manufacturing center; nor will the covering suitable for heavy traffic be suitable for a pleasure drive or for a residential district. General experience indicates the relative fitness of the several materials as follows: For country roads, suburban streets, and pleasure drives broken stone. For streets having heavy and constant traffic rectangular blocks of stone, laid on a concrete foundation, with the joints filled with bituminous or Portland cement grout. For streets devoted to retail trade, and where comparative noiselessness is essential asphalt, wood, or brick. Desirability. The desirability of a pavement is its possession of qualities which make it satisfactory to the people using and seeing it. Between two pavements alike in cost and durability, people will have preferences arising from the condition of their health, presonal pre- judices, and various other intangible influences, causing them to select one rather than the other in their respective streets. Such selections are often made against the demonstrated economies of the case, and usually in ignorance of them. Whenever one kind of pavement is more economical and satisfactory to use than is any other, there should not be any difference of opinion about securing it, either as a new pavement or in the replacement of an old one. The economic desirability of pavements is governed by the ease of movement over them, and is measured by the number of horses or pounds of tractive force required to move a given weight usually one ton over them. The resistance offered to traction by different pave- ments is shown in the following table: Resistance to Traction on Different Pavements TRACTIVE RESISTANCE KIND OP PAVEMENT Pounds per ton In terms of the loud Asphalt (sheet).... 30 to 70 1 t o a Brick 15 " 40 6 T to ^ 7 1 1 Cobblestones Stone-block 50 " 100 30 " 80 rsnr BTF i " i Wood-block rectangular 30 " 50 r ?w i " i Wood-block round 6 7 fW 1 " 1 TO ?T 392 HIGHWAY OBSTRUCTION 125 Serviceability. The serviceability of a pavement is its quality of fitness for use. This quality is measured by the expense caused to the traffic using it namely, the wear and tear of horses and vehicles, loss of time, etc. No statistics are available from which to deduce the actual cost of wear and tear. The serviceability of any pavement depends in great measure upon the amount of foothold afforded by it to the horses provided, however, that its surface be not so rough as to absorb too large a per- centage of the tractive energy required to move a given load over it. Cobblestones afford excellent foothold, and for this reason are largely employed by horse-car companies for paving between the rails; but the resistance of their surface to motion requires the expenditure ot about 40 pounds of tractive energy to move a load of 1 ton. Asphalt affords the least foothold; but the tractive force required to overcome the resistance it offers to motion is only about 30 pounds per ton. Comparative Safety. The comparison of pavements in respect of safety, is the average distance traveled before a horse falls. The materials affording the best foothold for horses are as follows, stated in the order of their merit : (1) Earth, dry and compact. (2) Gravel. (3) Broken stone (macadam). (4) Wood. (5) Sandstone and brick. (0) Asphalt. (7) Granite blocks. Durability. The durability of pavement is that quality which determines the length of time during which it is serviceable, and does not relate to the length of time it has been down. The only measure of durability of a pavement if the amount of traffic tonnage it will bear before it becomes so worn that the cost of replacing it is less than the expense incurred by its use. As a pavement is a construction, it necessarily follows that there is a vast difference between the durability of the pavement and the durability of the materials of which it is made. Iron is eminently durable; but, as a paving material, it is a failure. Durability and Dirt. The durability of a paving material will vary considerably with the condition of cleanliness observed. One 126 HIGHWAY CONSTRUCTION inch of overlying dirt will most effectually protect the pavement from abrasion, and indefinitely prolong its life. But the dirt is expensive, it injures apparel and merchandise, and is the cause of sickness and discomfort. In the comparison of different pavements, no traffic should be credited to the dirty one. Life of Pavements. The life or durability of different pavements under like conditions of traffic and maintenance, may be taken as follows : Life Terms of Various Pavements MATERIAL LIFE TERM Granite block 12 to 30 years Sandstone Asphalt Wood . .. (> ' 12 10 ' H 3 7 Limestone Hrii-k 1 ' 3 5 ' ? Macadam 5 ' ? Cost. The question of cost is the one which usually interests taxpayers, and is probably the greatest stumbling-block in the attain- ment of good roadways. The first cost is usually charged against the property abutting on the highway to be improved. The result is that the average property owner is always anxious for a pavement that costs little, because he must pay for it, not caring for the fact that cheap pavements soon wear out and become a source of endless annoyance and expense. Thus false ideas of economy always have stood, and undoubtedly to some extent always will stand, in the way of realizing that the best is the cheapest. The pavement which has cost the most is not always the best; nor is that which cost the least the cheapest; the one which is truly the cheap- est is the one which makes the most profitable returns in proportion to the amount expended upon it. No doubt there is a limit of cost to go be- yond which would produce no practical benefit ; but it will always be found more economical to spend enough to secure the best results, and this will always cost less in the long run. One dollar well spent is many times more effective than one-half the amount injudiciously expended in the hopeless effort to reach sufficiently good results. The cheaper work may look as well as the more expensive for the time, but may very soon have to be done over again. Economical Benefit. The economic benefit of a good roadway is 394 HIGHWAY CONSTRUCTION 127 comprised in its cheaper maintenance; the greater facility it offers for traveling, thus reducing the cost of transportation; the lower cost of repairs to vehicles, and less wear of horses, thus increasing their term of serviceability and enhancing the value of their present service; the saving of time; and the ease and comfort afforded to those using the roadway. First Cost. The cost of construction is largely controlled by the locality of the place, its proximity to the particular material used, and the character of the f6undation. The Relative Economies of Pavements whether of the same kind in different condition, or of different kinds in like good condition are sufficiently determined by summing their cost under the following headings of account : (1) Annual interest upon first cost. (2) Annual expense for maintenance. (3) Annual cost for cleaning and sprinkling. (4) Annual cost for service and use. (5) Annual cost for consequential damages. Interest on First Cost. The first cost of a pavement, like any other permanent investment, is measurable for purposes of comparison by the amount of annual interest on the sum expended. Thus, assum- ing the worth of money to be 4 ff /<>, a pavement costing $4 per square yard entails an annual interest loss or tax of $0.16 per square yard. Cost of Maintenance. Under this head must be included all out- lays for repairs and renewals which are made from the time when the pavement is new and at its best to a time subsequent, when, by any treatment, it is again put in equally good condition. The gross sum so derived, divided by the number of years which elapse between the two dates, gives an average annual cost for maintenance. Maintenance means the keeping of the pavement in a condition practically as good as when first laid. The cost will vary considerably depending not only upon the material and the manner in which it is constructed, but upon the condition of cleanliness observed, and the " quantity and quality of the traffic using the pavement. The prevailing opinion that no pavement is a good one unless, when once laid, it will take care of itself, is erroneous; there is no such pavement All pavements are being constantly worn by traffic and by the action of the atmosphere; and if any defects which appear are 395 128 HIGHWAY CONSTRUCTION not quickly repaired, the pavements soon become unsatisfactory and are destroyed. To keep them in good repair, incessant attention is necessary, and is consistent with economy. Yet claims are made that particular pavements cost little or nothing for repairs, simply because repairs in these cases are not made, while any one can see the need of them. Cost of Cleaning and Sprinkling. Any pavement, to be con- sidered as properly cared for, must be kept dustless and clean. While circumstances legitimately determine in many cases that streets must be cleaned at daily, weekly, or semi-weekly intervals, the only admis- sible condition for the purpose of analysis of street expenses must be that of like requirements in both or all cases subjected to comparison. The cleaning of pavements, as regards both efficiency and cost, depends (1) upon the character of the surface; (2) upon the nature of the materials of which the pavements are composed. Block pave- ments present the greatest difficulty; the joints can never be perfectly cleaned. The order of merit as regards facility of cleansing, is: (1) asphalt, (2) brick, (3) stone, (4) wood, (5) macadam. Cost of Service and Use. The annual cost for service is made up by combining several items of cost incidental to the use of the pave- ment for traffic for instance, the limitation of the speed of movement, as in cases where a bad pavement causes slow driving and consequent loss of time; or cases where the condition of a pavement limits the weight of the load which a horse can haul, and so compels the making of more trips or the employment of more horses and vehicles; or cases where conditions are such as to cause greater wear and tear of vehicles, of equipage, and of horses. If a vehicle is run 1,500 miles in a year, and its maintenance costs $30 a year, then the cost of its maintenance per mile traveled is two cents. If the value of a team's time is, say, $1 for the legitimate time taken in going one mile with a load, and in consequence of bad roads it takes double that time, then the cost to traffic from having to use that one mile of bad roadway is $1 for each load. The same reasoning applies to circumstances where the weight of the load has to be reduced so as to necessitate the making of more than one trip. Again, bad pavements lessen not only the life-service of horses, but also the value of their current service. Cost for Consequential Damages. The determination of conse- quential damages arising from the use of defective or unsuitable pave- 396 HIGHWAY CONSTRUCTION 129 ments, involves the consideration of a wide array of diverse circum- stances. Rough-surfaced pavements, when in their best condition, afford a lodgment for organic matter composed largely of the urine and excrement of the animals employed upon the roadway. In warm and damp weather, these matters undergo putrefactive fer- mentation, and become the most efficient agency for generating and disseminating noxious vapors and disease germs, now recognized as the cause of a large part of the ills afflicting mankind. Pavements formed of porous materials are objectionable on the same, if not even stronger, grounds. Pavements productive of dust and mud are objectionable, and especially so on streets devoted to retail trade. If this particular disadvantage be appraised at so small a sum per lineal foot of frontage as $1 .50 per month, or six cents per day, it exceeds the cost of the best quality of pavement free from these disadvantages. Rough-surfaced pavements are noisy under traffic and insufferable to nervous invalids, and much nervous sickness is attributable to them. To all persons interested in nervous invalids, this damage from noisy pavements is rated as being far greater than would be the cost of sub- stituting the best quality of noiseless pavement; but there are, under many circumstances, specific financial losses, measurable in dollars and cents, dependent upon the use of rough, noisy pavements. They reduce the rental value of buildings and offices situated upon streets so paved offices devoted to pursuits wherein exhausting brain work is required. In such locations, quietness is almost indispensable, and no question about the cost of a noiseless pavement weighs against its possession, When an investigator has done the best he can to determine such a summary of costs of a pavement, he may divide the amount of annual tonnage of the street traffic by the amount of annual costs, and know what number of tons of traffic are borne for each cent of the average annual cost, which is the crucial test for any comparison, ;is follows: (1) Annual interest upon first cost $ (2) Average annual expense for maintenance and renewal . . . (3) Annual cost for custody (sprinkling and cleaning) (4) Annual cost for service and use (.5) Annual cost for consequential damages Amount of average annual cost Annual tonnage of traffic Tons of traffic for each cent of cost Gross Cost of Pavements. Since the cost of a pavement depends 397 130 HIGHWAY CONSTRUCTION upon the material of which it is formed, the width of the roadway, the extent and nature of the traffic, and the condition of repair and clean- liness in which it is maintained, it follows that in no two streets is the endurance or the cost the same, and the difference between the highest and lowest periods of endurance and- amount of cost is very con- siderable. The comparative cost of the various street pavements, including interest on first cost, sinking fund, maintenance, and cleaning, when reduced to a uniform standard traffic of 100,000 tons per annum for each yard in width of the carriageway, is about as follows : Comparative Cost of Various Pavements MATERIAL ANNUAL COST PEK SQ. YD. Granite blocks Asphalt street Brick... Wood $0.25 0.40 0.35 0.60 398 REVIEW QUESTIONS. PRACTICAL TEST QUESTIONS. In the foregoing sections of this Cyclopedia nu- merous illustrative examples are worked out in detail in order to show the application of the various methods and principles. Accompanying these are examples for practice which will aid the reader in fixing the principles in mind. In the following pages are given a larg. num- ber of test questions and problems which afford a valuable means of testing the reader's knowledge of the subjects treated. They will be found excel- lent practice for those preparing for Civil Service Examinations. In some cases numerical answers are given as a further aid in this work. 399 REVIEW QUESTIONS ON THK SUBJKOT OK BRIDGE ENGINEERING 1. Write a short history of early bridges. 2. Define: Truss, bridge truss, truss bridge, girders, and girder bridges. 3. Draw an outline of a through bridge, and also an outline of a deck bridge. 4. Make an outline diagram of n truss, and write the names of the various parts on the respective members. 5. Make an outline diagram of a Warren, Howe, Pratt, bowstring, and Baltimore truss. 6. Compute the weight of steel in a 130-foot highway bridge whose trusses are 16 feet center to center, given W = 34 + 226 + 0.1G6Z + 0.71. 7. Compute the weight of steel in a deck plate-girder span of 100 feet. Loading, E 50. Given W = 124.0 + 12. 01. 8. WTiat are equivalent uniform loads? 9. WTiat is Cooper's E 40 loading? 10. Prove that the stress in a diagonal of a horizontal chord truss with a simple web system is V sec <. 1 1 . Prove that the chord stress is M -r- h, where M is the moment at the point, and h is the height of the truss. 12. Prove that the load must be on the segment of the span to the right of the section to produce the maximum positive shear. 13. Compute the maximum positive and negative live-load shears in a 13-panel Howe truss, the live panel load being 40 000 pounds. 401 REVIEW QUESTIONS O IV THE S IT B J K <3 T O K BRIDGE ENGINEERING 'ART II 1. Write an essay of 200 words on the economic considerations governing the decision to build and the decision as to what kind of bridge to employ. 2. What determines the height and width of railroad truss bridges? '3. Draw a clearance diagram for a bridge on a straight track, and state what allowance should be made if the bridge is on a curve. 4. Describe a stress sheet, and tell what should be on it. 5. Make a sketch of a cross-section of a deck plate-girder, showing the cross-ties, guard-rails, and rails in place. G. Make a sketch showing how tracks on curves are con- structed. 7. What is the span under coping, the span center to center of bearings, and the span over all? 8. Design a tie for Cooper's E 50 loading. 9. If the end shear of a plate-girder is 394 500 pounds, design the web section, it being 108 inches deep. 10. If the dead-load moment is 8 489 000 pound-inches and the live-load moment is 30 010 000 pound-inches, design the flange, if the distance back to back of flange angle, is 7 feet 6i inches, it being assumed that the web does not take any bending moment. 11. If, in the girder of Question 10, above, the web were 90 by ^-irich* design the flange, considering | of the gross area of the web as effective flange area. REVIEW QUESTIONS <* THE HUB .1 K C T < > IT 11 1 G 11 WAY CONSTRUC r r ION 1 . Fjxm what does the ease with which a vehicle can he moved on a road depend? 2. \\ hat kind of a road surface offers the greatest resistance to traction? 3. How may the power required to draw a vehicle over a pro- jecting stone be calculated? 4. What effect has gravity on the load a horse can pull? 5. Under what condition is the tractive power of a horse de- creased? G. What are the best methods for improving sand roads? 7 State briefly how earth is loosened and transported and the conditions under which each method is most advantageous? 8. What are the essential requisites for securing a good gravel road? 9. How should gravel roads be repaired? 10. State the considerations that control the maximum grade. 11. How should different grades be joined? 12. What considerations control the width of a road? 13. What is the essential quality of a stone used for road covering? 14. What should be the shape and size of broken stone? 15. For a light traffic road what thickness should the layer of broken stone have? 16. How should the foundation for the broken stone be pre- pared? 403 REVIEW QUESTIONS ON THE SUBJECT OF HIG-HWAY CONSTRUCTION 1, How should the natural soil be prepared to receive a pavement ? 2. In ramming blocks in the pavement, what point requires to be watched ? 8. How is a sand cushion prepared for use ? 4. What influences the durability of a granite? 5. How are rectangular stones laid on steep grades ? 6. How is the surface and sub-surface drainage of streets provided for? 7. What are the principal objections to wood pavements? 8. What determines the best width for a street? 9. In filling the joints with gravel and bituminous cement, what should be the condition of the material? 10. What controls the maximum grade for a given street? 11. AVhat varieties of wood give the most satisfactory results? 12. To what tests are stones intended for paving subjected? 13. Do cobblestones form a satisfactory pavement? 14. What properties should a stone possess to produce a sat- isfactory paving block? 15. How are expansion joints formed in a pavement? 16. What is the most suitable material for the foundation of a pavement? 17. Under what class of traffic may wood be used? 18. Upon what does the durability of a pavement depend? 19. What materials are employed for filling the joints be- tween the paving blocks? 404 INDEX The page numbers of this volume ivill be found at the bottom of the pages; the numbers at the top refer only to the section. Page Page A Bridge Engineering Asphalt pavement 374 weights 18 advantages of 379 bridge design 141 defects of 380 clearance diagram 147 foundation 379 economic considerations 141 materials for 374 economic proportions 143 tools used in construction of 383 floor system 150 Asphaltic paving mixture, formula for 379 practical considerations 155 Axle friction 276 specifications 150 B stress sheet 150 Baltimore truss 74 weights and loadings 148 Belgian block pavement Bituminous limestone 352 374 problems Bridge trusses, definition of 135-139 13 Bowstring truss 64 Bridges, loads for 22 live loads 22 Brick pavements 360 wind loads 24 absorption test advantages of 361 363 Bridges, weights of formula 1 for 18 19 cross-breaking test crushing test defects of 362 363 highway spans railroad spans Broken-stone roads 22 21 332 foundation for 363 rattler test 360 C Bridge analysis 11 Catch-basins 347 early bridges 11 City streets 341 truss bridge development 12 arrangement of 341 Bridge engineering 11-264 asphalt pavement 374 bibliography 264 Belgian block pavement 352 bridge analysis 11 brick pavements 360 definitions 13 catch-basins 347 descriptions 13 cobblestone pavement 352 girder spans 117 curbstones 387 history 11 drainage 346 loads 22 footpath 385 st resses 26 foundations 348 theory 26 grades 342 trusses 16, 53 granite block pavement 352 Note. For page numbers see foot of pages. 405 II INDEX Page Page City streets D gutters 347 Deck bridges, definition of 14 stone-block pavements 350 Drainage of roads 298 transverse contour 346 Drains, fall of 301 width of 341 E wood pavements 369 Cobblestone pavements 352 Earth 313 Concrete 349 Earth roads 316 Concrete-mixing machine 3 68 Earthenware pipe culverts 308 Country roads 207 Earthwork 310 drainage () balancing cuts and fills 310 culverts 304 classification of 313 fall of drains 301 embankments on hillsides 314 side ditches 301 prosecution of 314 of surface 302 shrinkage of 312 water breaks 303 slopes 311 earthwork 310 Embankments on hillsides 314 general considerations Engine loads in computing stresses 83 axle friction. 276 maximum moments, position of wheel effect of springs on ve hides 277 loads for 91 frict ion 267 maximum shear, position of wheel loss of tractive power on inclines 274 lo:lds for 89 object of roads 267 F resistance of air Final stresses 1 10 resistance to rolling Floor-l>eams, moments and shears in 117 tractive power and gr Clients 272 FloQr systems 18 location of Footpaths 385 bridge sites Forces, resolution of 27 cross levels 283 examples 284 G final selection 283 Girder bridges, definition of 13 intermediate towns 286 Girder spans 117 levels moments in plate-girders 119 map moments and shears in floor-l>eam 117 memoir shears in plate-girders 128 profile 283 stresses in plate-girders 134 reconuoissance 278 Girders, definition of 13 topography 28 ! Grades road coverings OOQ establishing 294 maximum 291 transverse contour 296 minimum 293 width undulating 293 Creosoting 371 Gradient 291 Culverts 304 Grading tools 31!) jointing 308 c:lrts 323 materials for 307 draining-tools 32!) Curbstones 387 dump cars 324 Note. For page numbers see fo ot of pages. 406 INDEX III Page Pago Grading tools Mountain roads 287 dump wagons 325 alignment 289 horse rollers 329 construction profile 290 mechanical graders 325 establishing grade 294 picks 319 final location 290 ploughs 319 gradient 291 road machine 327 halting places 288 scrapers 321 level stretches 294 shovels 319 loss of height 288 sprinkling carts 330 maximum grade 291 surface graders 327 minimum grade 293 wheelbarrows 322 undulating grades 293 Granite block pavement, 352 water on 28 K Gravel heaters 309 /.ig/ags 290 Gravel roads 330 P Glitters 347 Parabolic truss H4 11 Pavements 348 Hard pan 313 asphalt 374 Highway bridges, live loads for 22 Helgian block 352 Highway construction, 2G7 -398 brick 30 city streets 341 cobblestone 352 country roads 207 concrete foundation for 349 drainage 298 cost 394 earthwork 310 granite block 352 grading tools 319 selecting 390 mountain roads 287 stone block 350 pavements 348 wood 309 road coverings 330 Plate-girder railway-span design 1 5fi Highway spans, actual weights of 22 bearings 199 Howe truss 59 cross-frames 181 I determination of class 166 Inclines, loss of tractive power on 274 determination of span 157 Iron pii>e culverts 309 flanges lt>3 lateral systems 181 K masonry plan 168 Knee-braces, definition of 15 stiffeners 192 L stress sheet 202 ties* and guard-rails 159 Lateral bracing, definition of 15 web, economic depth of 12 Live load, position of, for maximum web splice 195 positive and negative shears 38 Plate-girders LOOSJ rock 313 moments in 1 19 M shears in 128 Maximum and minimum stresses 50 stresses in 134 Mechanical graders 325 Pony-truss bridge, definition of 14 Melting furnaces 369 Portals, definition of 15 Moments in plate-girders 119 Pratt truss 48, 53, 93 Note. For page numbers see foot of pages 407 IV INDEX R Railroad spans, actual weights of Railway bridges, live loads for Road coverings broken-stone gravel Road machines Roads country drainage of earth mountain sand Roadways on rock slopes Sand roads Shears, maximum live-loud Shears in plate-girders Shoes and roller nests Side slopes Slopes covering of form of inclination of Snow-load stresses Solid rock Steam-rollers Stone block pavements Street grades snow-load wind-load Stresses in bridge trusses in chord members counters engine loads live-load moments live-load shears maximum and minimum moments method notation used resolution of forces method Warren truss under dead load Warren truss under live load in web members Stresses in plate-girders Note. For page numbers see foot of Page 21 23 332 330 327 267 298 316 287 318 '316 318 38 128 254 311 311 312 311 311 103 313 338 350 342 103 104 26 33 47 83 44 38 50 30 35 27 35 45 31 134 Surface graders Sway bracing, definition of Page 327 15 Tables bridges, formula for weights of 19 bridges, types of for various spans 142 dead-load chord stresses 53. 61 dead-load stresses in diagonals 54 deck plate-girders, weights of 21 floor-beam reactions 119 force requked to draw loaded vehic!es over inclined roads 276 force of wind per sq. ft. for various velocities 277 grade data 292 grades, effects of, upon load horse can draw on different pavements 274 gross load horse can draw on different kinds of road surfaces 274 impact coefficient, values of 102 impact stresses in a Pratt truss 103 lacing bars, thickness of 229 loads, equivalent uniform 24 masonry bearings, length of 158 maximum moments in a deck plate- girder 126 maximum moments, determination of position of wheel loads for 93 maximum shears in a deck plate- girder 133 maximum shear, determination of position of wheel loads for 90 pavements, comparative cost of various 398 pavements, life term of various 394 pins for country highway bridges 245 pins for double-track bridges 244 pins for single-track bridges 244 plate-girder bridges,, width of, for various spans 145 reactions for a deck plate-girder 126 reactions for a through plate-girder 123 resistance due to gravity on different inclinations 271 resistance to traction on different pavements 392 408 INDEX Page Tables resistance to traction on different road surfaces 268 rise, suitable proportions for different paving materials 297 rivet spacing in bottom flange 177 rivet spacing in top flange 179 safe spans for I-beams 151 specific gravity, weight, resistance to crushing, and absorption power of stones 352 standard gauges for angles 1 75 stresses in a Pratt truss 100 tension members 217 tractive power of horses at different velocities 273 Trinidad asphaltum, average com- position of 377 trusses, chronological list of 16 wheel position, moments in deck plate-girder 125 wheel position, moments in a through plate-girder 1 22 wheel position, shears in a through plate-girder 1 29 wind stresses in Pratt truss 1 1 3 Through bridges, definition of 14 Through Pratt railway-span design 203 determination of span 204 end-post 239 floor-beams 207 intermediate posts 225 lateral systems 251 masonry plan 203 pins 243 portal 245 shoes and roller nests 254 Note. For page numbers see foot o/ pages. Page Through Pratt railway-span design stress sheet 262 stringers 205 tension members 215 top chord 231 transverse bracing 248 Transverse grade 345 Trinidad asphaltum 375 Truss bridge development 1 2 Truss, members of 14 Truss bridge, definition of 13 Trusses, classes of 16 chord characteristics 16 web characteristics 18 Trusses, definition of 13 Trusses under dead and live loads 53 Baltimore 74 bowstring 64 Howe 59 parabolic 64 Pratt 53 Trusses under engine loads 93 Pratt 93 W Warren truss under dead loads 35 under live load 45 Water breaks 303 W T eb splice 195 Wind-load stresses 104 bottom lateral bracing 107 overturning effect of wind on train 110 overturning effect of wind on truss 1 09 portals and sway bracing 112 top lateral system 104 W T ind loads 24 Wood pavemenos 369 409 UNIVERSITY OF CALIFORNIA LIBRARY Los Angeles This book is DUE on the last date stamped below. Form L9-100m-9,'52(A3105)444 JIH.7I 1 ' V