Diseño de Puentes de Acero

S T R U C T U R A L E N G I N E E R I N G

D E S I G N

STEEL Ihi DGESTHEORY AND PRACTICE

FOR THE USE OF

CIVIL ENGINEERS AND STUDENTS

BY

F. C. KIJNZ, C. E.COMBER AMERICAN SOCIETY OQ Crvm ENGINEERShfmmtm CANADIAN SOCIETY OF Cmrr. ENGINEBRS‘

FORMERLY DESIGNER FOR THI BRIDGE AND CONSTRUCTION DEPARTMENTor TEE PENCOYD IRON Womcs ,IND TBI A.MERICAN BRIDGB Co.

AND Cams ENGINEER OF THE BRIDGE AND CONSTRUCTION DEPARTMENTOF THE PENNsYLVANIA STEEL co.

FIRST EDITION

MCGRAW-HILL BOOK COMPANY, INC.

238 WEST 3gT$3 STREET, NEW YORK6 BOUVERIE STREET, LONDON, E. C.

1915

STACK

COPYRIGHT, 1915, BY THE

MCGRAW -H ILL BO O K COMPANY , INC.

PREFACE

American literature on the different phases of structural engineeringhas been very extensive in the last twenty years. In addition to the

numerous, valuable monographs published in periodicals, excellent bookshave been written. Most of these are intended as text-books for thestudent, and as such properly lay the greatest stress on a broad treatmentof statics, graphic and algebraic. Concerning the more practical part,however, which is of greatest interest to the practising engineer and theadvanced student, such as the selection and application of the method ofcalculation best suited to a given case, and practical rules, examples,tables, drawings, etc., to be used in designing, very little is available incompact form for immediate use.

As an attempt to fill this want, the author presents herewith a workon “Structural Engineering,” giving data of value to the designer,which the author has collected during thirty years of his professionalpractice.

The present volume on the “Design of Steel Bridges” is the firstof four, each complete in itself, which the author contemplates. It aimsto give the bridge engineer numerical examples and results of the bestmodern practice in designing and estimating of steel bridges, and to servehim as a guide and aid in the calculation of stresses, sections, weights, etc.

The text and the methods of calculation, both graphic and algebraic,have been condensed and fully cross-referenced to the tables and platesto make every part readily accessible, which is important especially in thedesigning room where the quick working out of preliminary calculationsis of the greatest importance.

For graphic calculations preference has been given to the methodsusing influence lines, as they can hardly be surpassed for clearness, andtherefore involve the smallest risk of mistakes. For algebraic methodsthe simplest have been selected.

The only statically indeterminate bridge trusses treated in this volumeare those for swing bridges and two-hinged arches. On account of theirlimited use, arches without hinges and suspension bridges are omitted.The treatment of the various two-hinged arches is full, and is intended tobe of immediate use in design. With the given permissible approxima-tions and numerical examples, it is the author’s hope that this type of

v

vi PREFACE

bridge, with its pleasing appearance, will become more popular withAmerican engineers than it is at present.

Special attention has been given to the usually forgotten influence oflateral forces on the stresses in the main trusses of arch- and cantilever-bridges. This subject will become, in the near future, of great im-portance with the gradual increase of economical span lengths due tohigh-grade, nickel, nickel-chrome, Mayari and other alloy steels.

The descriptions of existing and also of a few completely designedlong span bridges (Chapters XV a,nd XVI), give fully the analyzedweights and also the specifications of the material, loads and permissibleunit stresses which may be found useful for similar designs.

One of the features of the book is the inclusion of 52 plates, many ofwhich are complete in themselves. They are expected to guide the de-signer from dead- and live-load assumptions to the last rivet spacing with-out reference to the text.

The “Additional Information” at the end of Chapters XII, XIII, XIV,XV and XVII, has been carefully compiled from the author’s library andnotes. The “Appendiz” gives original data on Foundation Pressures ofBridges and Structural Work and other useful tables. The Specificationsof the American Railway Engineering Association, First Part, Design,are appended.

The thanks of the author are due to Mr. J. C. Bland and Mr. H. R.Leonard for valuable drawings showing the bridge practice of the Pennsyl-vania Lines West of Pittsburgh, and of the Pennsylvania Railroadrespectively; to Mr. C. H. Mercer for drawings of the Pennsylvania SteelCompany, and to Mr. C. C. Schneider for his permission to reprint partof his paper on “Movable Bridges ” (Trans. Am. Sot. C. E., Vol. 60, 1908),also for many tables and drawings of the American Bridge Company.

The author wishes also to mention Messrs. 0. H. Ammann, S. W.Bradshaw and R. M. Kreutz for their able assistance in the preparationof this volume; Mr. C. W. Reinhardt for the excellent rendering of theauthor’s drawings, and the Publishers for their co-operation.

F. C. KUNZ.PHILADELPHIA, December, 1914.

CONTENTS__-

PAQE

PREFACE ‘ . . . . . . . . . . . . . . . . . . . . . . . . . vLIST OF PLATES . . . . . . . . . . . . . . . . . . . . . xvLIST OF TABLES . . . . . . . . . . . . . . . . . . . . . . . . . xix . . .NOTATION . . . . . . . . . . . . . . . . . . . . . . . XWI

CHAPTER I

EXTERNAL FORCER

ART. 1. General . . . . . . . . . . . . . . . . . . . . . . . . . . . 12. DeadLoad . . . . . . . . . . . . . . . . . . . . . . . . . . . 13. Live Loads for Railroad Bridges . . . . . . . . . . . . . . . . . 104. Live Loads for Highway Bridges . . . . . . . . . . . . . . . . 175. Live Loads for Electric Railway Bridges . . . . . . . . . . . . . 2 06. Distribution of Live Load . . . . . . . . . . . . . . . . . . . 217. Impact and Vertical Vibrations . . . . . . . . . . . . . . . . . 2 28. Wind Pressure . . . . . . . . . . . . . . . . . . . . . . . . 259. Lateral Vibrations . . . . . . . . . . . . . . . . . . . . . . . 2 7

10. Centrifugal Force . . . . . . . . . . . . . . . . . . . . . . . 2811. Braking and Traction Forces . . . . . . . . . . . . . . . . . . 2 912. Snow Load . . . . . . . . . . . . . . . . . . . . . . . . . . 3 013. Temperature Changes . . . . . . . . . . . . . . . . . . . . . 31

CHAPTER II

GENERAL ABOUT REACTIONS AND INFLUENCE LINES

ART. 1. Character of Reactions . . . . . . . . . . . . . . . . . . . . . . 322. Determination of Reactions . . . . . . . . . . . . . . . . . . . 33

3. Influence Lines . . . . . . . . . . . . . . . . . . . . . . . . 35

4. Influence Coefficient . . . . . . . . . . . . . . . . . . . . . . 39

CHAPTER III

M OMENTS AND SHEARS IN SIMPLE SPANS

1. General . . . . . . . . . . . . . . . . . . . . . . . . 402: Influence Lines for Spans without Floorbeams . . . . . . . . . . . 403: Moments and Shears from Uniform Dead Load in Spans without

Floorbeams . . . . . . . . . . . . . . . . . . . . . . . . . 414. Moments and Shears from Uniform Live Load in Spans without

Floorbeams . . . . . . . . . . . . . . . . . . . . . . . . . 43vii

. . .Vlll CONTENTS

ART. 5. Moments and Shears from Moving Concentrated Loads in Spanswithout Floorbeams . . . . . . . . . . . . . .

6 . Maximum Moment in Short Spans without Floorbeams . . .7. Influence Lines for Spans without Floorbeams . . . . .8. Moments and Shears from Dead Load in Spans with Floorbeams . . .9. Moments and Reactions from Live load in Spans with Floorbeams

10. Shears from Uniform Live Load in Spans with Floorbeams . . . . .11. Shears from Moving Concentrated Loads in Spans with Floorbeams12. Floorbeam- and Intermediate Pier-Reactions . . .1 3 . E n g i n e D i a g r a m s a n d W h e e l - L o a d T a b l e s .14. Ecprivalent Uniform Loads . . . . . . ‘. .15. Tables of Moments and Shears . . . . . . . . . . .

CHAPTER IV

STRESSES IN SIMPLE TRUSSES

ART. 1. General . . . ‘. .2. Methods of Calculation .3 . T r u s s e s w i t h P a r a l l e l C h o r d s4. Warren Truss without Verticals . .5. Influence Lines for Trusses with Parallel Chords ,6. Examples for Trusses with Parallel Chords7. Stress Coefficients for Trusses with Parallel Chords .8. Trusses with Polygonal Top Chord . . . . .9. Influence Lines for Trusses with Polygonal Chord

10. Example for Trusses with Polygonal ChordIl. T r u s s e s w i t h S u b d i v i d e d P a n e l s . . .12. Influence Lines for Trusses with Subdivided Panels13. Example for Truss with Subdivided Panels14. Stress Coefficients for Trusses with Polygonal Chord .1 5 . T r u s s e s o f S k e w S p a n s .

CHAPTER V

STRESSES IN BRACING OF SIMPLE SPANS

ART. 1. General . . . . . . . . . . . .2. Lateral System between Straight Chords3. Lateral System between Polygonal Chords4. K-System of Laterals . . . . . .5. Indirect Wind Stresses in Bottom Chords .6 . O v e r t u r n i n g E f f e c t o f W i n d . .7. Stability against Lateral Forces . . . . . .8 . S t r e s s e s f r o m C e n t r i f u g a l F o r c e .9 . S t r e s s e s d u e t o E c c e n t r i c L o a d i n g .

10. S t r e s s e s f r o m B r a k i n g F o r c e11. Stresses in Portals . . . . . . .12. Stresses in End Sway Frames of Deck Spans

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CONTENTS i x

CHAPTER VI

TYPES OF BRIDGES AND PRINCIPAL DIMENSIONSPAW

ART. 1. Classification of Bridges . . . . . . . . . . . . . . . . . . . . 1172. Principal Parts . . . . . . . . . . . . . . . . . . . . . . . . 1173. Types of Main Girders and Trusses . . . . . . . . . . . . . . . 1174. Location of Bridge . . . . . . . . . . . . . . . . . . . . . . . 1205. Data for the Design and Erection . . . . . . . . . . . . . . 1206. Clear Height Below Crossing . . . . . . . . . . . . . . . . . 1227. Grades . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1228. Requirements for Span Length . . . . . . . . . . . . . . . . . 1239. Economical Span Length . . . . . . . . . . . . . . . . . . . . 125

10. Determination of Effective Span Length . . . . . . . . . . . . . 12711. Height Top of Floor to Under Clearance . . . . . . . . . . . . . 12812. Clear Width and Height on Bridge . . . . . . . . . . . . . . . 12913. Arrangement of Cross-section for Bridges Carrying Combined Traffic 130

CHAPTER VII

DESIGN OF FLOOR

A. Railroad Bridges

ART. 1. Open Tie Flooring . . . . . . . . . . . . . . . . . . . . . . . 1322. Ballast Flooring . . . . . . . . . . . . . . . . . . . . . . . . 1323. Solid Steel Floors without Ballast . . . . . . . . . . . . . . . . 1354. Arrangement of Floor System . . . . . . . . . . . . . . . . . . 1355. Principal Dimensions of Stringers and Floorbeams . . . . . . . . . 136

B. Highway Bridges

6. Plank Flooring . . . . . . . . . . . . . . . . . . . . . . . . 1387. Pavement Flooring . . . . . . . . . . . . . . . . . . . . . . 1408. Stringers and Floorbeams . . . . . . . . . . . . . . . . . . . 141

CHAPTER VIII

BEAM AND PLATE GIRDER BRIDGES

ART. 1. I-Beam Bridges . . . . . . . . . . . . . . . . . . . . . . . . 1432. Calculation of I-Beams . . . . . . . . . . . . . . . . . . . . . 1433. Deflection of I-Beams . . . . . . . . . . . . . . . . . . r . . 1444. Shearing Stresses in I-Beams . . . . . . . . . . . . . . . . . . 1445. Plate Girder Bridges . . . . . . . . . . . . . . . . . . . . . . 1456. Number and Spacing of Girders . . . . . . . . . . . . . . . . . 1457. Depth of Girders . . . . . . . . . . . . . . . . . . . . . . . 1478. Calculation of Net Flange Area . . . . . . . . . . . . . . . . . 148

X CONTENTS

ART. 9. Area of Compression Flange1 0 . M a k e - u p o f F l a n g e s11. Length of Cover Plates .12. Web Plate . . . . .13. Stiffeners . .14. Horizontal Flange Rivets15. Vertical Flange Rivets .16. Web Splice . . . . . . .17. Flange Splices . . .18. Bracing . ” . . . .19. Bearings . . . . . . .

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CHAPTER IX

SIMPLE TRUSS BRTDG&

ART. 1. Location of Floor . . . . . . . : . . . . . . . . . . . . . . . 1672. Number of Trusses . . . . . . . . . . . . . . . . . . . . . . 1693. Distance between Trusses . . . . . . . . . . . . ; . . . . . . 1694. General Principles of Truss Design . . . . . . . . . . . . . . . . 1695. Riveted and Pin-Connected Trusses . . . . . . . . . . . . . . . 1706. Parallel and Polygonal Chords . . . . . . . . . . . . . . . . . 1707. Heights of Trusses . . . . . . . . . . . . . . . . . . . . . . . 1708. Web System . . . . . . . . . . . . . . . . . . . . . . . . . . 1719. Shape of Polygonal Chord . . . . . . . . . . . . . . . . . . . 172

10. Panel Length . . . . . . . . . . . . . . . . . . . . . . . . . 17311. Required Area of Section of Truss Members . . . . . . . . . . . 17312. General Principles for the Design of Truss Members . . . . . . . . 17513. Sections for Top Chord . . . . . . . . . . . . . . . . . . . . 17714. Sections for Bottom Chord . . . . . . . . . . . . . . . . . . . 18015. Sections for Diagonals . . . . . . . . . . . . . . . . . . . . . 18116. Sections for Verticals, . . . . . . . . . . . . . . . . . . . . . 18217. Eyebars . . . . . . . . . . . . . . . . . . . . . . . . . , . . 18418. Bracing of Truss Spans . . . . . . . . . . . . . . . . . . . . 18519. Lateral System between the Loaded Chords . . . . . . . . . . . . 18720. Lateral System between the Unloaded Chords . . . . . . . . . . 18821. End Sway Bracing and Portals . . . . . . . . . . . . . . . . . 18822. Intermediate Sway Bracing . . . . . . . . . . . . . . . . . . 18923. Bearings . . . . . . . . . . . . . . . . . . . . . . . . . . . 19024. Gusset Plates of Riveted Trusses . . . . . . . . . . . . . . . . 19025. Chord Splices . . . . . . . . . . . . . . . . . . . . . . . . . 19326. Pin Connections . . . . . . . . . . . . . . . . . . . . . . . . 19327. Size of Pins . . . . . . . . . . . . . . . . . . . . . . . . . . 19428. Bending and Shearing Stresses in Pins . . . . . . . . . . . . . . 19429. Pin Packing . . . . . . . . . . . . . . . . . . . . . . . . . 19730. Calculation of Moments and Shears on Pins . . . . . . . . . . . 20031. Pin Plates . . . . . . . . . . . . . . . . . . . . . . . . . . 20232. Latticing of Compression Numbers . . . . . . . . . . . . . . . 20433. Camber . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

CONTENTS xi

CHAPTER X

SKEW BRIDGES AND BRIDGES O N CURVES

A R T . 1 . S k e w S p a n s - G e n e r a l . .2 . S k e w P l a t e G i r d e r S p a n s . .3. Skew Through Truss Spans with Parallel Chords.4. Skew Through Spans with Polygonal Top Chord .5. Skew Deck Truss Spans . . . j6 . A r r a n g e m e n t o f F l o o r o v e r S k e w P i e r s .7 . S k e w B e n t s .8 . S u p e r - E l e v a t i o n o f O u t e r R a i l o n C u r v e s9 .

10.11.12.13.14.

P r o v i s i o n f o r S u p e r - E l e v a t i o nMiddle Ordinate of Curve . .Width of Deck Spans on Curves .W i d t h o f T h r o u g h T r u s s S p a n s o n C u r v e sDistance between Tracks on Curves .........Arrangement of Stringers on Curves .........

CHAPTER X I

W EIGHTS OB S IMPLE SPAN BRIDGES

A RT . 1 . Tables and Formulas for Steel Weight of Railroad Bridges .2 . Steel Weights of Some Existing Simple Span Railroad Bridges .3 . Weight of Simple Span Highway Bridges . .4 . Weight of Electric Railway Bridges . . . ~ . . . .

CHAPTER XII

VIADUCTS

A. Calculation of Stresses in Towers

ART. 1. External Forces . . .2 . D e a d L o a d S t r e s s e s3 . L i v e L o a d S t r e s s e s4. Wind Stresses . . .,5 . Stresses Due to Centri fugal Force .6 . Stresses.Due t o B r a k i n g F o r c e7 . C o m b i n e d S t r e s s e s8 . U p l i f t9 . E x a m p l e .

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xii CONTENTS

B. Design of Viaducts

A R T . 1 0 . T y p e s o f V i a d u c t s11. E c o n o m i c a l S p a n L e n g t h s1 2 . P l a t e G i r d e r S p a n s1 3 . B a t t e r o f C o l u m n s1 4 . N u m b e r o f C o l u m n s1 5 . C o l u m n S e c t i o n s1 6 . B r a c i n g1 7 . E x p a n s i o n1 8 . A n c h o r a g e s1 9 . W e i g h t s o f V i a d u c t s .

Further Information on Viaducts

CHAPTER XIII

ELEVATED RAILROADS

ART. 1. General . . . . . . . . . . . . . . . . . . . . . . .2. Economical Span Length . . . . . . . . . . . . . . . .3. Arrangement of Cross-Section . . . . . . . . . . . . . .4. Flooring . . . . . . . . . . . . . . . . . . . . . . .5. Longitudinal Girders and Bracing . . . . . . . . . . . .6. Cross Girders . . . . . . . . . . . . . . . . . . . . .7. Expansion Joints . . . . . . . . . . . . . . . . . . .8. Column . . . . . . . . . . . . . . . . . . . . . . .9. Calculation of Stresses in Columns . . . . . . . . . . .

10. Dead and Live Load Stresses . . . . . . . . . . . . . .11. Stresses Due to Wind and Centrifugal Force . . . . . . .12. Stresses Due’to Braking Force . . . . . . . . . . . . .13. Temperature Stresses . . . . . . . . . . . . . . . .14. Combined Stresses . . . . . . . . . . . . . . . . . .15. Anchorages . . . . . . . . . . . . . . . . . . . . . .1G. Weights of Elevated Railroads . . . . . . . . . . . . .

Additional Information on Elevated Railroads, also Subways

CHAPTER XIV

MOVULE BRIDGES AND TURNTABLES

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ART. 1. Types of Movable Bridges . . . . . . . . . . . . . . . . . . . 2732. Center-Bearing Swing Bridge . . . . . . . . . . . . . . . . . . 2763. Rim-Bearing Swing Bridge . . . . . . . . . . . . . . . . . . . 2 7 84. Turning Device . . . . . . . . . . . . . . . . . . . . . . . . 2795.EndLift . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 8 06. Types and Principal Dimensions of Girders and Trusses of Swing

Bridges . . . . . . . . . . . . . . . . . . . . . . . . . . . 2817. Arrangement of Floor and Center Cross Girder . . . . . . . . . 2828. Calculation of Stresses in Swing Bridges . . . . . . . . . . . . . . 282

.*.CONTENTS x111

ART. 9. Dead Load Stresses . . .10. Live Load Stresses .11. Load on Center Pivot and Center Wedges1 2 . W i n d S t r e s s e s13. Example for the Calculations of Stresses1 4 . C a l c u l a t i o n o f D e f l e c t i o n s .15. Power Required to Operate Movable Bridges1G. M o t o r s1 7 . W e i g h t s o f M o v a b l e Bridges1 8 . L o c o m o t i v e T u r n t a b l e s

ART . 1. General . . . . . . .2 . Dead and Live Load Stresses in the Three-Hinged Arch3 . Dead and Live Load Stresses in the Two-Hinged Arch4 . T e m p e r a t u r e S t r e s s e s i n t h e T w o - H i n g e d A r c h5 . A p p r o x i m a t e C a l c u l a t i o n s o f t h e T w o - H i n g e d A r c h6 . Example for the Calculation of Stresses in a Two-Hinged Arch7 . W i n d S t r e s s e s i n t h e T w o - a n d T h r e e - H i n g e d A r c h8 . S t r e s s e s D u e t o B r a k i n g F o r c e9 . S t r e s s e s D u e t o Y i e l d i n g o f F o u n d a t i o n s

PAGE. 283

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294. 294. 296. 298. 300. 301. 305

Additional Information on Swing Bridges . 309

CHAPTER XV

ARCH BRIDGES

A. Calculation of Stresses

B. Design of Arch Bridges

ART. 10. General .11. T y p e s o f A r c h B r i d g e s12. Location of Floor .13. Number and Spacing of Arch Trusses14. Principal Dimensions of Arch Trusses15. Form of Arch Ribs and Web System16. Bents and Hangers . . . .i 7 . F l o o r S y s t e m a n d B r a c i n g . .18. Steel Weights of Arch Bridges1 9 . E x a m p l e s o f A r c h B r i d g e s

Additional Information on Arch Bridges

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CHAPTER XVI

LONG SPAN BRIDGES IN GENERAL AND EXAMPLES

ART. 1. Selection of Design . . 3672. Live Loads and Permissible Unit Stresses . 3 6 93 . C a m b e r . 3 7 14. Examples of Cantilever Bridges . 3725. Examples of Long Span Highway Bridges in New York 3 8 36. Designs for a North River Suspension Bridge (New York) 392

xiv CONTENTS

CHAPTER XVII

CANTILEVER BRIDGES

A. Calculation of Stresses

ART. 1. General . . . . . . . . . .2. Influence Lines for Anchor and Cantilever Arms3. Dead Load Stresses .4. Live Load Stresses .5. Example for the Calculation of a Cantilever6 . W i n d S t r e s s e s .7 . E r e c t i o n S t r e s s e s8 . D e a d L o a d A s s u m p t i o n s

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B. Design of Cantilever Bridges

9. Principal Dimensions . . . . . . . . . . . . . . . . . . . . . 4 1 810. Design of Trusses . . . . . . . . . . . . . . . . . . . . . . . 4 2 011. Bracing . . . . . . . . . . . . . . . . . . . . . . . . . . . 42112. Details and Anchorages . . . . . . . . . . . . . . . . . . . . . 421

ADDITIONAL INFORMATION ON LONG SPAN BRIDGES

(A) Suspension Bridges . . . . . . . . . . . . . . . . . . . . 423(B) Suspension and Cantilever Bridges. . . . . . . . . . . . . 423(C) Cantilever Bridges . . . . . . . . . . . . . . . . . . . . 423(D) Long Simple Span Bridges. . . . . . . . . . . . . . . . 424GENERAL INFORMATION . . . . . . . . . . . . . . . . . . . . 4 2 5

(A) Actual Pressure of Foundations in Bridge Work ‘427

(B) Actual Pressure of Foundations in Structural Work 430((3 Physical Properties of Materials, 433CD) T a b l e s . 4 3 6

(El Properties of Sections 441@‘I R i v e t i n g . 4 4 8

(‘3 Tables 450(H) Specifications, 454

APPENDIX

INDEX . ~ . ‘. 463

LIST OF PLATESPLATE PAC+E

I. DIAGRAM OF E50 FOR SCALING INFLUENCE ORDINATES (See Art. 3,Chap. 2) . . . . . . , . . . . . . . . Following Index

II. INFLUENCE LINER FOR TRUSS WITH POLYGONAL TOP CHORD (SeeArt. 10, Chap. 4) . . . . . . . . . . . . Facing Page 94

III. INFLUENCE LINES FOR TRUSS WITH POLYGONAL TOP CHORD ANDSWDIVIDED PANELS (See Art. 13, Chap. 4) . . . . . Facing Page 100

IV. ECONOMICAL SPAN LENGTHS FOR RAILROAD BRIDGES (SIMPLESPANS) (See Art. 9, Chap. 6) . . . . . . . . . Factng Page 126

V. TYPICAL RAILROAD BRIDGE FLOORINGS (See Arts. l-3, Chap. 7). . . . . . . . . . . . . . . . . . . . See Plates back of Book

VI. TYPICAL FLOOR SYSTEM IN RAILROAD BRIDGES (See Art. 4, Chap. 7). . . . . . . . . . . . . . . . . . . . See Plates back of Boolc

VII. TYPICAL HIGHWAY BRIDGE FLOORS (See Art. 6, Chap. 7) . . .., . . . . . . . . . . . . . . . See Plates back of Book

VIII. TYPICALBIZARINGSOFPLATE GIRDERBRIDGES (See Art.19,Chap.S). . . . . . . . . . . . . . . . . . . . . . . . Facing Page 166

IX. DETAIL DRAWING OF 18-~T. SINGLE TRACK I-BEAM SPAN (See Art.19, Chap. 8) . . . . . . . . . . . . Pacing Page 155

X. STRESS SHEETOF 36-FT.SINGLE TRACK DECK PLATE GIRDER SPAN(See Art. 19, Chap. 8). . . . . . . See Plates buck of Book

XI. STRESS SHEET OF ~&FT. SINGLE TRACK THROUGH PLATE GIRDERSPAN (See Art. 19, Chap. 8) . . . _ . . . See Plates back of Book

XII. GENERAL DETAIL DRAWING OF 116-~T. DOWLE TRACK THROUGH

PLATE GIRDER SPAN (See Art. 19, Chap. 8) Between Pages 166-167

XIII. ELEVATIONS OF NOTABLE SIMPLE SPAN BRIDGES (See Art. 1,Chap. 9). . . . . . . . . . . . . See Plates back of Book

XIV. TYPICAL,TRUSSES FOR SINIPL~ SPANS (See Art. 8, Chap. 9) .. . . . . . . . . . . . . . . . . See Plates back of Book

XV. STREWS SHEET OF 150-FT. SINGLE TRACK RIVETED THROUGHTRUSS’ SPAN (MO. PAC. RY.) (See Art. 28, Chap. 9) . See Plates back of Book

XVI. STRESS SHEET OF 200-FT. SINGLE TRACK RIVETED THROUGHTRUSSSPAN (MO. PAC. RY.), (See Art. 28, Chap. 9). See Plates back of Book

XVII. STRESS SHEET OF 255-FT. DOUBLE TRACK PIN CONNECTED DECKTRUSS SPAN (SUSQUEHANNA, P. R. R.) (See Art. 28, Chap. 9) . .. . . . . . . . . . . . . . . . . . . See Plates buck of Book

x v

xvi LIST OF PLATES

P L A T E P.WeXVIII. STRESS SHEET OF 230-FT. DOUBLE TRACK RIVETED DECK TRUSS

SPAN (STEKJBENVILLE, PIZNNA. L. W.) (See Art. 28, Chap. 9) . ,. . . . . . . . . . . . . . . . . . See Plates back of Book

XIX. STRESS SHEET OF 373-FT. DOWLE TRACK THROUGH PIN CONNECTED

TRUSS SPAN (P. R. R.) (See Art. 28, Chap. 9) . . . . . . . .. . . . . . . . . . . . . . . . . . . See Plates back of Book

XX. STRESS SHEET OF 333-FT. FOUR TRACK TEIROUGH PIN SPAN (PENNA.L. W.) (See Art. 28, Chap. 9) . . See Plates back of Book

XXI. GENERAL DETAIL DRAWING, 150-FT. SINGLE TRACK THROUGH TRUSS

SPAN (MO. PAC. RY.) (See Art. 28, Chap. 9). See Plates back of Book

XXII. GENERAL DETAIL DRAWING, 230-FT. DOUBLE TRACK RIVETED DECK

TRUSS SPAN (STEUBENVILLE, PENNA. L. W.) (See Art. 28, Chap. 9).............a....... See Plates back of Book

XXIII. GENERAL DE T A I L DRAWING , 240-FT. DOWLE T RACK R IVETED

THROUGH Tlzuss SPAN (W. M D. RY.) (See Art. 28, Chap. 9) . . .. . . ...‘... . . . . . . . . See Plates back of Book

XXIV. GENERAL DETAIL DRAWING, 373-FT. DOWLE TRACK THROUGH PIN

CONNECTED SPAN (P. R. R.) (See Art. 28, Chap. 9) .., . . . . . . . . . . . . . . . . . . See Plates back of Book

XXV. GENERAL DETAIL DRAWING, 450-FT. SINGLE TRACK THROUGH PIN

CONNECTED SPAN (M ILES GLACIER, ALASKA) (See Art. 28, Chap. 9). . . . . . . . . . . . . . . . . . . . See Plates back of Book

XXVI. TYPICAL CROSS SECTION OF SPANS ON CURVES (See Art. 9, Chap. 10). . . . . . . . . . . . . . . . . . . . . Facing Page 214

XXVII. ELEVATIONS OF VARIOUS BUILT VIADUCTS (See Art. 10, Chap. 12), . . . . . . . . . . . . . . . See Plates back of Book

XXVIII. STRESS SHEET AND GENERAL DETAIL DRAWING, SINGLE TRACK

VIADUCT (N. & W. RY.) (See Art. 15, Chap. 12). . . .Facing Page 254

XXIX. STRESS SHEET AND GENERAL DETAIL DRAWING, DOWLE TRACK

VIADUCT (N. & W. RY.) (See Art. 15, Chap. 12). . . . Facing Page 255

XXX. STRESS SHEET, SINGLE TRACK VIADUCT (C. C. & 0. RY.) (See Art. 15,Chap. 12) . . . . . . . . . Facing Page 256

XxX1. GENERAL DETAIL DRAWING, DOUBLE TRACK VIADUCT (P. R. R.)(See Art. 15, Chap. 12) . . . . . . . . . See Plates back of Book

XxX11. CROSS SECTIONS AND ELEVATIONS OF ELEVATED RAILROADS (SeeArt. 3, Chap. 13) . . . . . . . . . . . . See Plates back of Boolc

XxX111. TYPICAL TRUSSES OF SWING BRIDGES (See Art. 6, Chap. 14) . .. . . . . . . . . . . . . . . . . . . . . . . . Facing Page 282

XXXIV. TYPICAL CROSS SECTIONS OF CENTER BEARING SWING <BRIDGES(See Art. 7, Chap. 14) . . . . . . . . See Plates back of Book

XXXV. INFLUENCE LINES FOR SWING SPAN (See Art. 13, Chap. 14) . . .. . . . . . . . . . . . . . . . . . . . . . . . . Facing Page 294

LIST OF PLATES xvii

PLATE PaomXXXVI. STRESS SHEET, 94-m. SINGLE TRACK DECK PLATE GIRDER SWING

SPAN (NORF. & SOUTH. RY.) (See Art. 1.4, Chap. 14). . . . . .. . . . . . . * . . . . . . . . . . . . See Plates back of Book

XXXVII. STRESS SHEET, 250~FT. SINGLE TRACK SWING SPAN (G. N. RY.)(See Art. 14, Chap. 14) . . . . . . . . . - See Plates back of Book

XXXVIII. STRESS SHEET, 327-FT. DOUBLE TRACK SWING SPAN (P. R. R.,RARITAN) (See Art. 14, Chap. 14) . . . See Plates back of Book

XxX1X. DETAILS OF MACHINERY, 94-m. SINGLY TRACK DECK PLATE GIRDER

SWING SPAN (See Art. 14, Chap. 14) . See Plates back of Book

XL. GENERAL DETAIL DRAWING, 250-FT. SINGLE TRACK SWING SPAN

(G. N. RY.) (See Art. 14, Chap. 14). . . See Plates back of Book

XLI. DETAILS OF MACHINERY, 250-1~~. SINGLD TRACK SWING SPAN (G. N.RY.) (See Art. 14, Chap. 14). . . . . See PZati?s back of Book

XLII. GENERAL DETAIL DRAWING, 327-FT. DOWLE TRACK SWING SPAN

(P. R. R., RARITAN) (See Art. 14, Chap. 14), See Plates back of Book

XLIII. DETAILS OF MACHINERY, 327-FT. DOUBLE TRACK SWING SPAN

(See Art. 14, Chap. 14) . . , See Plates back of Book

XLIV. STRESS SHEET, 90-FT. DECK TURNTABLE (See Art. 14, Chap. 14). . . . . . . . . . . . . . . . . . . See Plates back of Bo.ok

XLV. STRESS SHEET FOR A 250-FT. ARCH TRUSS (See Art. 4, Chap. 15). . . . . . . . . . . . . . . . . . . . See Plates back of Book

XLVIa. ELWATIONS OF NOTABLE ARCH BRIDGES (See Art. 4, Chap. 15). . . . . . . . . . . . . . . . . . . . See Plates back of Book

XLVIb. ELEVATIONS OF NOTABLE ARCH BRIDGES (See Art. 4, Chap. 15). . . . . . . . . . . . . . . . . . See Plates back of Book

XLVII. GENERAL DETAIL DRAWING, ST. JOHN RIYER HIGHWAY BRIDGE

(See Art. 4 , Chap. 15) . . . . . . See Plates back of Book

XLVIII. ELEVATIONS OF NOTABLE CANTILEVER BRIDGES (See Art. 4,Chap. 16) . . . . . . . . . . Pacing Page 372

XLVIIIa. ELEVATIONS OF THE THREE EAST RIVER SUSPENSION BRIDGES,N. Y. (See Art. 4, Chap. 16) . . , . . . . . . . , Facing Page 388

XLIX. CROSS SECTIONS OF VARIOUS LONG SPAN BRIDGES (See Art. 4,Chap. 16) . . . . . . . . . See Plates back of Book

L. DEAD PANEL CONCENTRATIONS FOR CANTILEVER TRUSS (See Art. 3,Chap. 17) . . . . . . . . . . . . . . . Facing Page 402

LI. STRESS DIAGRAhm FOR AN 1800-FT. CANTILEVER TRUSS (See art. 5,Chap. 17) . . . . . . . . . . . See Plates back of Book

LII. COMPRESSION MEMBERS OF LONG SPAN BRIDGES (See Art. 12,Chap. 17) . . . . . . , . . . . . . . . See Plates back of Book

LIST OF TABLES

CHAPTER ITABLEI P A G E

1 .

2 .3 .4 .5 .6 .7 .8 .

9 .10.11.12.13.

14.

15.

16.17.18.

19.2 0 .

21.

2 2 .2 3 .2 4 .

2 5 .2 6 .

ESTIMATE OF STEEL W EIGHT FOR A SINGLE-TRACK THROUGH-PIN SPAN

150 FT. LONG . . . . . . . . . . . . . . . . . . . . . . . . . 5LNE LOAD FOR RAILROAD BRIDGET IN RECENT SPECIFICATIONS. . . . . . 1 1LNE LOAD FOR RAILROAD BRIDGES IN OLDER SPECIFICATIONS . . . . . . 1 2SHOWING INCREASE IN W EIGHT OF LOCOMOTIVES . . . . . . . . . . . . 1 4HEAVIEST LOCOMOTIVES OF DIFFERENT TYPES . . . . . . . . . . . . . 1 4HEAVIEST LOCOMOTIVES IN ACTUAL SERVICE ON 36 AMERICAN RAILWAYS . 1 5IMPACT IN PER CENT. OF LIVE LOAD . . . . . . . . . . . . . . . . . 2 5CENTRIFUGAL FORCE IN PER CENT. OF VERTICAL FORCE . . . . . . . . . 2 9

CHAPTER III

M OMENTS FROM UNIFORM LOAD . . . . . . . . . . . . . . . . . . . 4 2SHEARS FROM UNIFORM DEAD LOAD IN SPANS WITHOUT FLOORBEAMS . . . 4 2SHEARS FROM UNIFORM LIVE LOAD IN SPANS WITHOUT FLOORBEAMS . . . 4 4SHEARS FROM UNIFORM DEAD LOAD IN SIMPLE SPANS WITH FLOORREAMS . 5 1EXACT SHEARS FRO&I U NIFORM LIVE L OAD I N S IMPLE SPANS W I T H

FLOORBEAMS . . . . . . . . . . . . . . . . . . . . . . . . . 5 4APPROXIMATE SHEARS FROM UNIFORM LIVE LOAD IN SIMPLE SPANS

WITH FLOORBEAMS. . . . . . . . . . . . . . . . . . . . . . . 5 5M OMENTS FOR COOPER’S E-50 LOADING IN UNITS OF 1000 FT.-LB.

FOR ONE RAIL. . . . . . . . . . . . . . . . . . . . . . . . . 5 8M OMENTS FOR COOPER’S E-50 LOADING. . . . . . . . . . . . . . . . 6 0MIAXIMUM M OMENTS, SHEARS AND EQUIVALENT UNIFORM LOADS PER RAIL. 6 4INTERMEDIATE REACTION AND EQUIVALENT UNIFORM LOAD PER RAIL FOR

Two ADJOINING (NOT CONTINUOUS) SPANS . . . . . . . . . . . . 6 5SHEARS IN SPANS WITHFLOORBEAMS AND EQUAL PANELS . . . . . . . . 6 5TABLE GIVING NUMBER OF W HEEL TO BE PLACED AT PANEL POINT OF ANY

PANEL TO PRODUCE MA~IMU~\~ SHEAR IN THAT PANEL . . . . . . . 6 7SHEARS IN GIRDERS WITHOUT FLOORBEAMS FOR COOPER’S E-50 LOADING IN

UNITS OF 1000 LB. FOR ONE RAIL . . . . . . . . . . . . . . . . 6 7BENDING M OMENTS AT EQUIDISTANT PANEL POINTS . . . . . . . . . . 6 8MAXIMUM M OMENTS AND SHEARS FOR ELECTRIC CARS . . . . . . . . . 7 0M AXIMUM M OMENTS AND SHEARS FOR ELECTRIC CARS . . . . . . . . . 7 1

CHAPTER IV

STRESS COEFFICIENTS FOR TRUSSES WITH PARALLEL CHORDS . . . . . . 8 5STRESS COEFPICIENTS FOR TRUSSES WITH POLYGONAL TOP CHORD . . . . 1 0 3

xix

xx LIST OF TABLES

CHAPTER VITABLE

27. DIMENSIONS OF EXISTING PIERS . . . . . . . ~ . . . , .28. BUCKLE PLATES . . . . . . . . . . . . . . . . . . .

CHAPTER VII

29. NOTABLE PLATE GIRDER BRIDGES . . . . . . . . . . . 1 4 6

CHAPTER VIII

. . 1 5 030. CENTER OF GRAVITY OF GIRDER FLANGES . . . . . . . . . . .31. TYPICAL FLANGES OF PLATE GIRDER RAILROAD BRIDGES . . . . . .32.

. . 154STANDARD SINGLE-TRACK DECK PLATE GIRDER SPANS. . . . Facing Page 166

33. NOTABLE SIMPLE SPANS . . . . . . . . : . . . 16834. STRESS SHEET FOR RIVETED TRUSS OF A 160-FT. SINGLE-TRACK THROUGH

35.35a.

36.37.38.39.40.41.42.

43. SUPER-ELEYATION OF OUTER RAIL ON CURVES . . . . . . . . . . . , 215

44. AMERICAN BRIDGE CO.-SINGLE-TRACK DECK PLATE GIRDER SPANS. 22145. AMERICAN BRIDGE CO.-SINGLE-TRACK THROUGH PLATE GIRDER BRIDGES 22246. AMERICAN BRIDGE CO.-SINGLE-TRACK THROUGH PLATE GIRDER BRIDGES 22347. AMERICAN BRIDGE CO.-TRUSS BRIDGES . . . . . . . . . . 22448. AMERICAN BRIDGE CO.-TRUSS BRIDGES, . . . . . . . . . Facing Page 22449. PLATE G IRDER F O R B RIDGES FOR S INGLE -TRACK E-60 L OADING 22650. P E N N S Y L V A N I A RAILROAD-DOUBLE-TRACK B R I D G E S . . . 2 2 7- 5 1 . ILLINOIS CENTRAL R. R. SINGLE TRACK WITH BALLAST FLOOR. . . . 22952. BRIDGES FOR IMPERIAL GOVERNMENT RAILWAYS OF JAPAN-COOPER’S

53.

P A G E

. . 124

. . 139

CHAPTER IX

SPAN. . . . . . . . . . . . . . . . . . . . 176TYPICAL SECTIONS OF TOP CHORDS AND END POST OF RIVETED TRUSSES 178TYPICAL SIGCTIONS OF TOP CHORDS AND END POST OF PIN-CONNECTED

TRUSSES . . . . . . . . . . . . 179T YPICAL SECTIONS O F B OTTOM CHORDS OF R IVETED T RUSSES . 181EYEBARS . . . . . . . . . 183SLEEVE NUTS AND TURNBUCKLES . . . . . . . . . . 184SINGLE-TRACK BRIDGES (SIZE OF BASE PLATES, ETC.). . 191PINS WITH NIJTS . . . . . . 195BEARING VALUES OF PINS. . . . . . . , . . . . . . 196BENDING M OMENTS ON PINS. . . . . . . . . . 197

CHAPTER X

CHAPTER XI

SPECIFICATIONS 1896-LOADING E-30 WITH END FLOORBEAMS . . 230W EIGHTS OF SINGLE-TRACK RAILROAD BRIDGES OF THE PRUSSIAN STATE

RAILROADS . , , , . . . . . . . . . . . . . . . . . . . . 231

LIST OF TABLES xxi

CHAPTER XIITABLE PAGE54. NOTABLERAILROADVIBDUCTS. . . . . . . . . . . . . . . . 24955. TYPICAL COLUMNSECTIONS OF VIADUCTS AND ELEVATED RAILROADS . . 255

CHAPTER XIV

56. PRINCIPAL DIMENSIONS OF SOME EXISTING RAILROAD SWING BRIDGETWITH CENTER-BEARING. . . . . . , . . . . . . . . . . . 277

57. LOADING,~HEAR AND MO~XENTDIAGRAMS-REACTIONS,SHEARS,MOMENTS,DEFLECTIONS OF CONTINUOUS BEAMS . . . . . . . . . . . . . 285

58. INFLUENCE ORDINATES FOR END REACTION z =F FOR SWING SPANSWITH THREE SUPPORTS ANDEQUALARMS . . . . . . . . . . . . . 289

59. DEFLECTIONS 0~324-FT.SWINGSPAN . . . . . . . . . . . . . 297

CHAPTER XV

60. VALUES z =i(k -2/c3+kl) . . . . . . . . . . . . . . . . . . . . 332

61~. NOTABLE SPANDRELBRACED ARCHBRIDGES ............. 36161b. NOTABLE CRESCENTARCHBRIDGES . . . . . . . . . . . . . . . . . 36261~. NOTABLEARCHGIRDERBRIDGES . . . . . . . . . . . . . . . . . . 363’,61d. NOTABLE ARCHTRUSS BRIDGES . . . . . . . . . . . . . . . . . . . 364

CHAPTER XVI

62. CALCULATED DEFLIZCTIONS OF CANTILEVER TRUSS. . . Facing Page 378

CHAPTER XVII

63. PRINCIPALDIMENSIONSOFCANTILEVERBRIDGES. . . . . . . . . . 418

NOTATION

Stresses, Etc.+ tension- compressiond dead load stress1 live load stressi impact stressW wind stresst temperature stresss unit stressP unit shearV, Q, shears stressM bending moment, static moment

Loads, ForcesP load per lin. ft.

IFweight or wind pressure per lin. ft.concentrated loads or forces

w concentrated weights or wind pressures,work of deformation

R reactionsM horizontal reaction of arch spansB braking forcec centrifugal force

Dimensions, Areas, Etc.a, b, c, etc. distancesb width, breadthcl diameter, lever arm, depthY7 e distance of extreme fiber from neutral axis,

eccentricity5 rise of archh height, lever arm, depth1 length of span, length of memberm middle ordinat,e of curver radius, radius of gyration, lever arm2, Y, 2 coordinatesA area . . .

xx111

xxiv

RILEv% P, Y,- &J6P, ?”x

Al

NOTATION

section modulusmoment of inertialoaded lengthsuper-elevation of outer railvolumeanglesdeflectionradius of curvaturepanel lengthchange df length due to deformation

Constants, Coefficients, Etc.E modulus of elasticityG modulus of shearC, K constants, ratiosE coefficient of expansion

;Inumber, ratiodegree of curvature

T 3,14159I: summation

DESIGN OF STEEL BRIDGES

C H A P T E R I

E X T E R N A L F O R C E S

ART. 1. GENERAL

The principal external forces acting on a bridge are: The deadload, live load, and impact.

Forces of secondary importance are: The wind pressure and lateralvibrations, the centrifugal force (where the tracks are on a curve), theforces due to the longitudinal friction of the wheels on the rails (tractionand braking forces) and, in certain cases, the forces due to the changes oftemperature. In exceptional cases asnow load has to be considered.

ART. 2. DEAD LOAD

The dead load of a bridge is made up of the following items:(a) Weight of flooring (track, ballast, planking, etc.), railings, etc.(b) Weight of floor system (floorbeams, stringers, floor bracing)(c) Weight of main trusses, bracing and bearings.The dead load acts vertically and is for ordinary bridges assumed

as uniformly distributed over the whole length of the span. For largebridges every panel concentration has to be determined.

The weight of the steel work (b) and (c) is usually not known and hasto be determined by preliminary designs or by comparison with similarbridges of known weight. For ordinary bridges the dead load can beapproximated by calculating the weight (a) of the flooring and the floorsystem (6) and assuming the weight (c) of the main trusses or girdersand of the bracing.

As an easily remembered fiqst approximation the ratio of average dead load toaverage live load, for single- or double-track railroad truss bridges, open-tie flooring,may be assumed:

for riveted bridgesdead load length of span in feetliveload= 4 0 0

for pin-connected bridgesdead load length of span in feet.-iiveload= 5 0 0

1

2 DESIGN OF STEEL BRIDGES [CH A P . I

Assuming height of trusses as one-sixth of span, unit stress without impact in ten-sion 12,000, in compression 9000, then the largest cross-section will be in the tension

chord Plf-1 and in the compression chord16000 &&j when 1 is the span in feet and p

the sum of the specified live load plus the dead load (see above rule) in pounds perlin. ft. of truss.

An assumed dead load within 4 to 5 ‘% of the actual is satisfactory forthe calculation of stresses in ordinary bridges.

(a) Weight of Flooring and Railings

Railroad Bridges.-The American Railway Engineering Associationspecifies the following weights: Rails and fastenings, 150 lb. per lin. ft..of track; timber, 44 lb. per ft. B. M. (54 lb. per cu. ft.); ballast, 100 lb.per cu. ft.; reinforced concrete, 150 lb. per cu. ft.

Based on the above figures, the weight per foot of various tracks isapproximately as follows:

Tie flooring: (ties 8 in.X8 in. X10 ft.), 400 lb. per lin. ft. of track;‘for every inch increase in depth of ties add 25 lb.; for every foot increasesin length of ties add 2.5 h lb. (h = depth of ties in inches).

Ballast flooring (ballast 13 ft. wide, 14 in. average depth) : 1600 lb. perft. of track; for every inch increase in depth of ballast, add 110 lb. (thisdoes not include steel floor plate or concrete slab).

Highway Bridges.-Oak and yellow” pine timber, 4+ lb. per ft.B. M.; spruce and white pine, 33 lb. per ft. B. M.; ballast 100, stoneconcrete 150, cinder concrete 110, wood block paving 60 lb. per ‘cu. ft.

Trolley rails, fastenings, etc., on ties, 100 lb. per ft. of track.Railings.-For gas pipe railing use 15 lb., for ordinary lattice railing

25 lb., for ornamental railing up to 50 lb. per lin. ft. of railing.

(b) and (c) Weight of Steel Work

The weight of the steel work varies with the live load to be carried,the permissible unit stresses, the length and width of the bridge, thedepth of the floor and main girders or trusses, etc.

Formulas and tables of weights of the various types of bridges aregiven in the respective chapters XI to XVII. For simple spans seepage 221.

For the purpose of estimating the weight of the steel work of simplespans by comparison with similar bridges, the following approximaterules may be used, providing that the span lengths, loadings and unitstresses do not differ considerably and the trusses or girders are of similardesign.

ART.’ 21 EXTERNAL FORCES 3

For the same live load, unit stresses, width and construction of floor,theweight of the floor system per foot of bridge does not change withthe span length, while the weight of the girders or trusses and of thebracing per foot of bridge changes in linear proportion with the spanlength.

For two bridges of the same span length, width of floor and unitstresses but different total loads (p and pl per lin. ft.) the steel weights(w and wl per lin. ft.) have the approximate relation:

Since pl contains the weight wl this has first to be assumed, and if WIobtained by the above formula differs considerably from that assumed,pl has to be corrected and the formula applied again.

In two bridges of the same span, width and load (including impact)designed with different unit stresses s and s1 (tension in chords, otherstresses are in the same proportion) the corresponding steel weights w

and w1 have the relation:

Wl = ; 1+ss,( )

Double-track bridges have a steel weight of about twice as much assingle-track bridges.

The steel weight of deck truss spans with floorbeams and stringers isfrom 5 to 10% greater than that of through spans (the trusses 10 to 15%heavier and the floor lighter).

A decrease of n $& in the economical height of a plate girder increasesits weight by about 0.7 n%.

A decrease of n% in the economical height of a truss increases itsweight by about 0.6 n $&.

A decrease in the economical height of the floor system may increasethe weight of the floor system by up to 25 %.

The weight of a plate girder with stiffeners is obtained approximatelyby calculating the required section at the center of the span, assuming80 to 100% of the flange section constant for the whole length, de-pending upon the variation of the flange section over the length, and add-ing 20 to 25% to the weight of the web and flanges for stiffeners, etc.

The weight of a truss is obtained approximately by calculating theaverage chord section and adding a percentage for web members anddetails. (See pages 233 to 235.) The gross area of riveted tension mem-bers may be assumed 20% greater than the required net area.

4 DESIGN OF STEEL BRIDGES [CHAP. I

If the stress sheet of a bridge is made, the, weight can be estimatedby calculating the weight of the floor system and of the main material ofthe trusses and bracing using the lengths between intersection pointsand assuming a certain percentage for the details. 35oj, of the mainmaterial is an average value for the weight of details of trusses (seealso table 48). Closer results are obtained by calculating the averagepercentages of the weight of the details separately for each kind ofmembers. The following are average values:

Riveted main members: 30 to 35 %.Secondary members latticed on two sides: 40 to 50%.Secondary members latticed on four sides: 100 to 150 %.For eyebars add to the length between centers of pins (2++ w) ft.,

where w = width of bar in inches. For ordinary bridges the pins weighabout 3 y0 of the weight of the trusses; for large bridges 2+ y0 is sufficient.

The weight of the bracing for railroad bridges is about 10 to 157,of the weight of the main trusses; in highway bridges it depends uponthe width of the bridge and usually ranges from 15 y0 in comparativelynarrow and heavy bridges to 30 Y0 in wide and light bridges. The detailsof the bracing usually weigh from 40 to 60% of the main material of thebracing.

In calculating the weight of details, the following allowances in percent. of the whole weight of the main material and details should bemade for rivet heads:

13 to 2% for framed I-beams,2; $& for heavy to 33 ‘% for light longitudinal plate girders,4 y0 for cross girders or floorbeams,3; to 4 y0 for riveted truss members and bracing.The following example illustrates the method of estimating the

weight of steelwork.

ART. 21 EXTERNAL FORCES 5

TABLE 1

Estimate of Steel Weight for a Single-track Through-pin Span 150 ft. Long

Designed for Cooper’s E-50 Loading

4 End posts1 Cov. 24x&.. . ‘ . . . . . . . . .2 15 in. Channels @ 50 lb .2 Flats 4X$. . . . . . . . . . . . . .2 Pinpls. 12x8.. . . . . . . . .2 Pin pls. 14+Xg . . . . . . . . .2 Pin pls. 143x4. . . . . . . . .2 Pinpls. 143X$. . . . . . . . .2 Pinpls. 12in.Xiin.. ..2 Pinpls. 12in.Xiin.. ..2 Tie pls. 24X&. . . . . . . . . .

34 Latt. bars 2+X&. . . . . . .1 Pl. 7x+. . . . . . . . . . . . . . . .

500 Riv. @+lb. . . . . . . . . . . .

35.7 1,360 lb.50.0 3.810

3 7 ft. 1 in. 11.9 ‘8801 ft. 1 in. 15.3 3 01 ft. 6 in. 18.51 ft. 10 in. 18.51 ft. 10 in. 18.51 ft. 5 in.2 ft. 9 in.2 ft. 0 in.2 ft. 9 in.2 ft. 0 in.

15.3 4515.335.7 1::3.7 3408.9 2 0

250

4 Top chord sects., UI-U::Cov. pl. 24X&. . . . . . . . . . . .15 in. Channels @ 35 lb . . . .Flats 4 X 28 . . . . . . . . . . . . . . . . .Pin pls. 12X$. . . . . . . . . . . . .Pin pls. 143 X 3Pin pls. 12x5 . . . . . . . . . . . . .Tie pl. 24X&. . . . . . . . . . . . .Tie pl. 15X& . . . . . . . . . . . . .Spl. pls. 12x; . . . . . . . . . . . . .Spl. pls. 12XQ . . . . . . . . . . . . .Fills 6X& . . . . . . . . . . . . . . . .Spl. pl. 9X$. . . . . . . . . . . . . . .Latt. bars 23 X A. . . . . . . . . .Bent pl. 8X$ . . . . . . . . . . . . . .Riv. @*lb.. . . . . . . . . . . . . .

(2 Top chord sects., Ut-UZ1 Cov. pl. 24X&. . . . . . . .2 15 in. Channels @ 45 lb2 Flats 4X% . . . . . . . . . . . . .6 Pls. 12X+. . . . . . . . . . . . .4 Tiepls. 18X&. . . . . . . . .3 Lat. pls. 12X$. . . . . . . . .3 Lat. pls. 12x;. . . . . . . . .

48Latt.bars24XA . . . . . .600 Riv. @.+lb.. . . . . . . . . .

7,160X4 28,640

23 ft. 8 in. 35.723 ft. 8 in.23 ft. 1 in.

1 ft. 7 in.1 ft. 9 in.2 ft. 0 in.2 ft. 0 in.2 ft. 0 in.1 ft. 0 in.1 ft. 3 in.1 ft. 3 in.2 ft. ‘0 in.2 ft. 9 in.2 ft. 0 in.

53 ft. 5 in.53 ft. 5 in.53 ft. 5 in.

1 ft. 8 in.2 ft. 0 in.2 ft. 6 in.1 ft. 0 in.2 ft. 9 in.

35.011.915.318.525.535.722.315.315.3

6.411.53.7

10.2

8451,660

550

:i!100

7 04 0

:Ft

a:240

1;:

3,915 x4 15,660

35.7 1,91045.0 4,80511.9 1,27015.3 15026.8 21515.3 11515.3 4 53.7 490

300

9,300 x2 18,600

62,900

DESIGN OF STEEL BRIDGES [CRAP. 1

4 Hangers Brought forward 62,900

Pin pls. 11 XQ . . . . . . . . .Fills 11 Xt . . . . . . . . . . . .P i n uls. 15X:. . . .FillllX+.:::. . . . . . . . .P1.32Xa . . . . . . . . . . . . . .Angles3$X3+X+ . . . . . .Riv. @ 3 lb . . . . . . . . . . .

. .

. .

. .

.

. .

. . .

. . .

. .. . .

4 posts Us-L212 in. Channels @ 25 lb . . . .Pin pls. lOXi . . . . . . . . . . . . .Pin pls. 12X$. . . . . . . . . . . . . .

4 Tie pls. 11X$. . . . . . . . . . . . . .4 Angles 32 X3$ XQ . . . . . . . . . .1 Web pl. 10X$. . . . . . . . . . . . .

56 Latt. bars 2$X&. . . . . . . . . .2 Angles 3+X3$X$. . . . . . . . . .2 Angles 3+X3$ Xt . . . . . . . . . .1 P1.36X) . . . . . . . . . . . . . . . . . .

240 Riv. @ +lb.. . . . . . . . . . . . . .

2 posts us-L82 12 in. Channels @ 25 lb2 Pin pls. 1OXt . . . . . . . . .2 Pinpls. 12X+. . . . . . . . .2 Angles 3X3X%. . . . . . . .4 Tie pls. 11X%. . . . . . . . . .4 Angles3+X3+X# . . . . . .1 Pl. 1OXQ.. . . . . . . . . . . . .

56 Latt. 22X&. . . . . . . . . . .2 Angles3+X3$X+. . . . . .2 Angles34X3$X+ . . . . . .1 PI. 36X$. . . . . . . . . . . . . .

240 Riv. @fib... . . . . . . . . . .

. 2 6 ft. 1 1 in.2 6 ft. 1 1 in..3 ft. 2 in.

. 3 ft. 2 in.

. 1 ft. 5 in.

. 3 ft. 0 in.

. 3 ft. 6 in.

. 2 ft. 1 1 in.1 ft. 6 in.

2 9 ft. 4 in.. 1 ft. 6 in.. 1 ft. 7 in.. 1 ft. 0 in.

. 4 ft. 10 in.

. 4 ft. 10 in.

. 1 ft. 4 in.. 1 ft. 0 in.

. 0 ft. 1 1 in.. 3 ft. 0 in.

. . . . . . 29 ft. 4 in.. . . . 1 ft. 6 in.. . . . . 1 ft. 7 in.. . . . 1 ft. 0 in.. . . . 1 ft. 0 in.,..... 4 ft. 10 in.. . . 4 ft. 10 in.. . . . 1 ft. 4 in.. . . . . 1 ft. 0 in.. . . . . 0 ft. 11 in.. . . 3 ft. 3 in.. . . . .

49 Set diagonals2 Bars6Xl&$in.(37ft.6$in.c.c.). 41 ft. 9 in.2 Bars 6 Xl in. (37 ft. 6+ in. c. c.). 41 ft. 9 in.

Counters4 Bars 5 Xl (37 ft. 6; in. c.c.).4 Turnbuckles @ 50 lb.. . .

. 43 ft. 0 in.

46 Bott chords4 Bars6in.Xl$in.(25ftOin. c.c.), 2 9 ft. 3 in.4 Bars6in.Xl$in.(25ft.Oin,c.c.), 29 ft. 3 in.

10.4 1,12010.2 27530.6 19523.4 15014.0 4025.5 15018.754.4 12

8.5 E

2,285 X4 9,140

25.0 1,46512.7 4 025.5 8014.0 . 5 58.5 165

12.72.4 167:8.5 2 08.5

61.2 1;:8 0

2,335 x4 9,340

25.0 1,465

12.715.3 tt7.2

14.0 ::8.5 165

12.7

82.21;:2 0

8.561.2 22

8 0- -2,335 X2

37.0 3,09020.4 1,700

4,790 x4

1 7 . 0 2 , 9 2 02 0 0

3,120X1 3,120

35.7 4,18528.1 3,285

7,470 x4

4,670

19,160

29 ,880

138,210


Pins Brought .forward 138,2104 Pins 63in. 4 (L,). . . . . . . . . . 2ft. 2 in. 1,0204 Pins5iin. (Ll) . . . . . . . . . . . . . . lft lO+in.

lil2::700

4 Pins 5% (L,). . . . . . . . . . . . . . 2 ft: 7 in. 88.3 9102 Pins 5% in. (L3).. . . . . . . . . 2 ft. 10 in. 88.3 4954 Pins 6$ in. (VI) . . . . . . . . . . . . 2 ft. 5 in. 1~U:~ 1,0904 Pins 5: in. (UZ). . . . . . . . . . . . . 1 ft. 10 in. 7002 Pins5$in.(U3) . . . . . . . . . . . . . . lft. 2 in. 88.3 320

32 Nuts for 51 in. pin @J 11.0 lb. 35016 Nuts for 6$ in. pin @ 12.0 lb. 195

5,780X1 5,780

2 Portals2 Angles 5X3+xi.. . . . . . . . . . . . 18ft. 4 in.4 Angles5x3$x$. . . . . . . . . . . . . 9ft. ll$in.2 Angles4x3xi... . . . . . . . . . . . 5 ft. 8 in.4 Angles 4X3x$. . . . . . . . . . . . . . 4ft. 0 in.4 Angles 4X3XQ.. . . . . . . . . . . . . 4 ft. 6$ in.1 Pl. 13x&. . . . . . . . . . . . . . . . . . . lft. 8 in.2 PI. 11X+. . . . . . . . . . . . . . . . . . . . 1 ft. 2 in.2 Pl. 10x;. . . . . . . . . . . . . . . . . . . . 1 ft. 1l)in.2 Pl. 12x$. . . . . . . . . . . . . . . . . . . . 1 ft. 0 in.4 Angles 3+x34x+. . . . . . . . . . . . 1 ft. 0 in.2 Pls. 14x&. . . . . . . . . . . . . . . . . . 1 ft. 8tin.4 Angles 3+X3+xe.. . . . . . . . . . . 1 ft. 3 in.2 Bent pls. 10x3 . . . . . . . . . . . . . . 2 ft. 0 in.2 Bent pls. 25X$. . . . . . . . . . . . . . 2 ft. 11 in.

200 Riv.@+lb.. . . . . . . . . . . . . . . .

3 Top struts4 Angles 3X2+X#. . . . . . . .2 Pls. 15x+. . . . . . . . . . . . . .2 Pls. 15xi.. . . . . . . . . . . . .4 Angles 4X3X+. . . . . . . . .

10 I&t. Bars 22 x &. . . . . . .90 Riv. @$lb.. . . . . . . . . . .

Top laterals8 Angles 3X2+X$ . . . . . . .4 Pls. 7X$. . . . . . . . . . . . . .

70 Rev. @ Q lb . . . . . . . . . . . .

2 4

ii9 0

24

Bottom lateralsAngles 5 x33X$. . . . . . .Angles 5X3$X$. . . . . . .Angles 3$X3+X3 . . . :. .Pls.7X&. . . . . . . . . . . . .Pls.7X&. . . . . . . . . . . . .Riv. @ 3 Ib. . . . . . . . . . .Angles 5x33X+. . . . . . .Angles 5 x3+X+. . . . . . .

. .

. .

. .

. .

. .

14 ft. 3 in.2 ft. 0 in.1 ft. 32 in.9 ft. 0 in.2 ft. 2 in.

10.48.58.58.58.5

26.814.012.815.3

8.526.8

8.512.831.9

380415100135150

50185100

1,835 x2 3,670

2; ;;, 1; f;. . 8 . 9 3 6 . 6 1 , 4 2 0 2 525

2,715

1,470x1 1,470

9; ;;. ; ii. 10.3 10.3 1,1601,120

1 ft: 9 in: 8.5 355

2 ft. 83 in. 13.42 ft. 1 in. 13.4 1:;

2 8 ft. 2 in. 13.6 7::13 ft. 7 in. 13.6 740

--4,365X1 4,365

156,210

8 DESIGN OF STEEL BRIDGES [CHAP. 1

2 End struts2 Angles 4X3)Xg... . . . . . . . .2 Tiepls. 15X;. . . . . . . . . . . . .2 Tie pls. 12X%. . . . . . . . . . . . .2 Pls. 12x2.. . . . . . . . . . . . . . . . .4 Latt. bars 2$x&. . . . . . . . . .4 Pls. 10x2.. . . . . . . . . . . . . . . .

70 Riv. @ +Ib . . . . . . . . . . . . . . .

4 Stringer pedestals4 Angles 5X3$X$. . . . . . . . . . .4 Angles 33 . . . . . . . . . . . . . . . . .4 Fills. 3X+. . . . . . . . . . . . . . . . .1 PI. 13x4.. . . . . . . . . . . . . . . . .1 PI. 13X$ . . . . . . . . . . . . . . . . . .1 PI.19Xj . . . . . . . . . . . . . . . . . .2 Stone bolts 1 in

14 Riv. @ 2 lb:......

::: : : : : : : :

5 Int. floorbeamsAngles 6X4x&...Angles6X4X&....Web 52X$... . . .Ph. 28X& . . . . . . . . .Pls. 23X& . . . . . .Angles 5x3+X&. .Angles5~3+~$....Angles 5X33x&.Fills. 8X$. . . . . . .Pls. 22x;. . .Angles 5X3$X&.Angles 5 x3$ xi..Fills. 31 X’+. . . . . .Riv. @ 4 lb.. . . . “. .

Brought forward 156,21013 ft. 8 in.

. 1 ft. 8 in.

1E 230

. 1 ft. 8 in. 15.3 E3 ft. 03 in. 15.3 90

. 1 ft. 10 in.1 ft. 111 in. 1;:;: ii:3 5

- -580x2 1,160

1 ft. 1; in. 10.31 ft. 3 9 in. 10.3 ii0 ft. Sj in. 10

1 ft 3” i n

223::

1 ft: 34 in: 33.1 it1 ft. 9 in. 48.5 8 50 ft. 9 in. 2.0 5

5- - 1,100

275X4

15 ft. 5 in. 18.1 5 6 013 ft. 5 in. 18.1 48511 ft. 9 in. 88.43 ft. 8 in.

1,07053.5 785

2 ft. 8 in. 34.2 3653 ft. 4 in. 12.0 160

1 ft. 2 in. 10.31 ft. 10 in. 12.0 i::lft . 6”in * . 1.0.2 6 06 ft. 2f in. 37.4 4 7 02 ft. 5 in. 12.0 1152 ftOft:

3’6’

i nin:

10.3 956.0

14:

8 ht. stringers4 Angles 6X6X&. . . . . . . . . . . .1 Web41x+ . . . . . . . . . . . . . . . . .4 Angles 5X3+X& . . . . . . . . . . .4 Fills. 7X&. . . . . . . . . . . . . . . . .

230 Riv. @ 3 lb . . . . . . . . . . . . . . . .

4,325 X8 34,600

4

2 5 0

4 End stringersAngles 6X6X=&,. . . . . .Web 41X$. . . . . . . . . . .Angles 4X3x$. . . . . . .Fills 3X&. . . . . . . . . . .Angles5X3fXA .....Fills. 7X&. . . . . . . . . . .Riv. @ + lb . . . . . . . . . . .

.......

.......

.......

......

......

......

4,450 x 5 22,250

2 5 ft. 0 in. 22.0 2,2002 5 ft. 0 .in. 69.0 1,725

3 ft. 4 in. 12.0 1602 ft. 5 in. 13.4 125

115

2; in.2’ in4’ in:5 in.4 in.5 in.

22.069.08.55.7

12.013.4

2,3101,810

1 1 5

::

1%

4,560 x4 18,240

233,550

ART. 21 EXTERNAL FORCES

2:

6 Sets strg. laterals Brought forwardAngles4X3Xh . . . . . . . . 8ft. 2 in. 290Riv.@$lb . . . . . . . . . . . . . . . . . 10

34

4:

2 Cross framesAngles 4X3X$ . . . . . . . . .Angles 4X3X$. . . . . . . . .Pls. 10X6.. . . . . . . . . . . . .Pl, 7X$ . . . . . . . . . . . . . . . .Riv. @ 3 lb. . . . . . . . . . . .

. . .. . .. . .. . .. . .

2

i

f21

:

100:

2 Roller shoesPls. 12x;. . . . . . . . . . . . .Jaw pls. 19x5.. . . . . . . .Angles 6 X6 X4. . . . . . . .Pl. 19x+. . . . . . . . . . . . . .Angles 3+X3+XQ . . . . . .Fills 3$X$. . . . . . . . . . . .Angle 4X3XQ . . . . . . . .Fill 2$X%. . . . . . . . . . . . .Pl 30x=.F&.2&4................::::Riv. @ +lb.. . . . . . . . . .

..

i6

;3

:

2 Roller nestsRollers 4 in. 4. . . . . . . . . .Bars 23x2 . . . . . . . . . . . . .Flats 5 Xf. . . . . . . . . . . . . .Flat 2% X1$. . . . . . . . . . . . .Rods 12 in. 4 . . . . . . . . . . .Stone bolts la in. + ......Angle 4X3X% . . . . . . . . . .Angle 6X4Xt . . . . . . . . . .

1 Plate 30X;. . . . . . . . . . . . . .70 Riv. 0, 3 lb .. : . . . . . . . . . .

2 Fixed shoes2 Lap pk. 12 Xi ..........4 Jawpls. 23X+. . . . . . . . . .4 Angles 6X6X2.. . . . . . . .2 Fills. 3X&. . . . . . . . . . . . .: ;;g;; cl53 xe. . . . . . . . . .2 Ar;glks3~X3~‘~~;:::;;;2 Fills. 3$ X 34 . . . . . . . . . . . . .1 Pl. 24X:. . . . . . . . . . . . . . .1 P1.30XH . . . . . . . . . . . . . . . . . .1 P!.3oxg . . . . . . . . . . . . . . . . . .3 Stone bolts l$ in. 4 . . . . . . . . .

120.Rivets @+lb.. . . . . . . . . . . .

6 ft. 2 in.5 ft. 9 in.1 ft. 0; in.1 ft. 0 in.

. I ft. 4 in.2 ft. 6 in.

. 2 ft. 6 in.

: 3 1 ft. ft.1 in.7 in.

1 ft. 12 in.2 ft. 1 in.

1 ft. 3ft:9’ ini”in:

. . 2 ft. 6 in.

3 ft. 1 in.2 ft. 6 in.2 ft. 6 in.2 ft. 6 in.3 ft. 2% in.1 ft. 0 in.2 ft. 6 in.2 ft. 6 in.3 ft. 10 in.

. 1 f t . 84 in.2 ft. 6 in.

. 2 ft. 6 in.1 ft. 9 in.

. 2 ft. 0 in.3 ft.. 0 in.

. 1i n

1 ft ft: 111 6’ in:. 2 It. 8 in.

. 2 ft. 8 in.3 ft. 4 in.

. 1 ft. 0 in.

300X6

105100

5 0

t:

285 X2

15.340.4 4;:28.7 28524.28.5 Ii

158.5 2 0

82.9 2::

5;---

1,195x2

42.7 795

2;:

2:: ;t:

2

3%3 5

- - -1,570 x 2

15.348.9 4;:

258:772 8 5

2 08 .5

16.6 28.5 3 58.9

30.6 :I:89.2 2 4 089.2 3 0 0

4.76;

1,670 X2

233,550

1,800

5 7 0

2,390

3 ,140

3,340

244 ,780

1 0 DESIGN OF STEEL BRIDGES [CHAP. I

Packing rings Brought forward 244 ,7804 Rings (63 + in. inside). . . . . 0 ft. 9 in. 143.7 4 6 04 Rings (52 in. inside). . . . . . . . . 0 ft. 1 in. 112.4 45

505x1 505- - -

Weight pf 1 span with end struts, 245,285 lb.

ART. 3. LIVE LOAD FOR RAILROAD BRIDGES

The specified loading consists, as a rule, of two locomotives with theirtenders, followed by a uniform train load. Some railroads still havetheir own standard loadings; the loading proposed by Theodore Cooper(“Cooper’s Loading”) is, however, most frequently used. It consists ofdifferent classes designated by E-40, E-50, E-60, etc. All classes havethe same axle spacing, and the axle loads and the uniform train load ofdifferent classes are proportional to the numbers designating the class,and cause, therefore, stresses in the same proportion. The axle loadsand stresses for E-40, for instance, are 4 of those for E-50. Table 2shows the axle loads and spacing for Cooper’s E-50, E-55, and E-60and for various other loadings of recent specifications. Table 3 gives thelive load of various older specifications and may be of use in investigatingexisting bridges.

In order to provide for the effect of excessive axle loadsp- ,c..q

.Q ()

on short girders, etc., Cooper specifies an alternativeload of 50,000 lb. each on two axles 6 ft. c. to c. forall classes up to E-40 and of 60,000 lb. for all classes

FIG. 1. above E-40. The American Railway Engineering Asso-ciation has adopted an alternative load of 50,000 lb.’

each on two axles 7 ft . c. to c. (Fig. 1) to be used with Cooper’sClass E-40; for heavier clkses the axle load to be changed in proportionto the class number.

In the earlier railroad bridges no provision was made for a futureincrease of the live load. This increase has been so rapid (more than100% in the last two decades) that many bridges had to be replacedafter a short life by heavier ones. Table 4 illustrates the developmentof the engine on the Baltimore & Ohio Railroad. Tables 5 and 6*contain the weights of a number of locomotives in use on differentrailroads.

* Taken from a paper by J. E. Greiner, Bull. 139, Am. Ry. Eng. Assoc.


0

0

00

0

0

0

0

0

0

C

C

c

C

Cc

C

C

-

006%

006s

006fooa

000s

‘000s

~0005

~OOOC

looot

1006I

1006S

1006%

1006t

IOOO!

IOOO!IOOO!

IOOO!

IOOO!

-

: c

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: oc

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: O(; oc

: OS

: O(

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: O(2 O(

; 01

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188;

188:

188:

188:

ISI:

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)91.191.

j91:

188:

1881

YES8

188:

191

191191

191

191

-

z

dE

4eid

2z

da$

.m

z“a

%B

5 01

5 O(

5 O(

5 O(

P O(

P 01

P 01P 01

z 01

z 01

2 01

z 01

z 01

P 01

P 01P 01

P 0’

z o

-

-

091;

091;

OQIi

091:

OZII

OZI!

OZl!OZI!

oz9

0911

0911

091:

091

OZT

OZl'OZI

IOZI

IOZQ

Railroad

1. Deep Water Ry . . . . . . . . . . . . . . .2.B.R.&P.R.R. . . . . . . . . . . . . . . .3. ‘2. R. R. of N. J. . . . . . . . . . . . . . .

4. B. & 0. R. R. . . . . . . . . . . . . . . . . .5.C.H.&D... . . . . . . . . . . . . . . . . . .6. L. V. R . R . . . . . . . . . . . . . . . . . . . .

7. Long Isld. R. R. . . . . . . . . . . . . . . .8 . S . A . L .9. Western Md . . . . . . . . . . . . . . . . . . .

10. L . S .&M. S.. . . . . . . . . . . . . . . . . .11. . . . . . . . . . . . . . . . . . . . . . . . . . . . .12. C. 0. & Gulf . . . . . . . . . . . . . . . . . .

13. S. A. i.. . . . . . . . . . . . . . . . . . . . . . .14. N . Y . C . & H . R . R . R .15. N. Maine Seaport . . . . . . . . . . . . .

16. Chicago & Alton . . . . . . . . . . . . . . .17. Kansas City So . . . . . . . . . . . . . . . .18. Oregon Short Line . . . . . . . . . . . . .

19. Oregon Ry. & Nav. Co . . . . . . . . .20. Southern Pao . . . . . . . . . . . . . . . . . .21. Union Pa0 . . . . . . . . . . . . . . . . . . . .

22. A. T. & S. Fe Ry . . . . . . . . . . . . . .23. A. T. & S. Fe Ry . . . . . . . . . . . . . .24. B. & M. R. R . . . . . . . . . . . . . . . . .

25. Belt Ry. of Chicago . . . . . . . . . . .26. Bess &L. Erie.. . . . . . . . . . . . . . .27. C. P. R . . . . . . . . . . . . . . . . . . . . . . .

28. C.&O . . . . . . . . . . . . . . . . . . . . . . .29. C. R. I. & Pat. . . . . . . . . . . . . .30. Chicago & W. Ind . . . . . . . . . . . . .

31. Chicago & E. Ill.. . . . . . . . . . . . . .32.D.L.&W.R.R. . . . . . . . . . . . . . .33. Erie R. R . . . . . . . . . . . . . . . . . . . . .

-

. .

i1

TABLE 3

Live Load for Railroad Bridges in Older Sloecifications G

Date ofspecif.

19041902

. . . . .

190219041902

190519021903

1904

1900

1902190019041905

190419041904

190419041904

189519021896

189919001901

190219031899

189419031900

TYPO

E-60E-55

T-55+5oc

-.

e

60,00055,000

240,000 15.0 16,000 213220,000 15.0 14,700 195*

.

-l1

E-50 50,000 200,000 15.0 13,300 1773 109.0 6,500

E-45 45,000 180,000 15.0 12,000 1592 109.0 51900

E-40 40,000 160,000 15.0 10,700 1 4 2 109.0 5,200

F-40+450E-38

. .38,000

.152,000 15.0 10,200 135 109.0

I C o m -m o n

stand’d

. ’

55,000 220,000 15.0 14,700 192) 109.0

44,000 176,000 13.5 13,000 139 107.0 5,20066,000 264,000 13.5 19,500 208+ 107.0 7,80035,000 140,000 15.0 9,300 123 112.0 4,400

38,000 152,000 15.0 10,100 133 109.0 4,90060,000 240,000 15.0 16,000 192f 106.0 7,30042,000 168,000 14.25 11,800 1441 111.0 5,400

44,000 176,00044,000 176,00038,000 152,000

158154133

6,1005,6004,900

35,0005o,lmo35,000

140,000200,000140,000

15.015.015.0

13.515.013.5

11,70011,70010,100

10,40013,30010,400

104.0111.0109.0

99.0104.099.0

5,0006,4005,000

%:daxllb.’

Driverbase,

ft.

.oad per 1on drive

blr.s.e,

wt . o fngine am

tender,tom.

-

;ength ofeng. and

tend.,ft.

,oad per ftx eng. am

teg.2

Uniformtrain load,lb. per ft.

3,2004,800

Engines

5,0004,8005,000

Railroad

3 4 . G r a n d T r u n k .3 5 . G r e a t N o r t h e r n . _. _.3 6 . G r e a t N o r t h e r n . .

37. Great Northern. . .3 8 . 111. Cenk-al..3 9 . L . & N . R . R .

4 0 . L . & N . R . R . .4 1 . L . E . & S t . L . C .4 2 . M i o h . C e n t . R . R .

. .4 3 . Missouri P a c . ,4 4 . M e x i c a n Cm..45. N. Y., L. E. & W. _, _.

4 6 . N . Y . , C . & S t . L o u i s4 7 . N . Y . , 0 . & W e s t . . .:48. N. Y., N. H. & H. R. R.. (

49. N.&W.Ry .__.......__..._____50. P. R. R . . . . _.51. P. R. R . . . . _. _.

5 2 . P . & R . . . .53. P.&R _...__......._....._.___5 4 . Rutland R . R . . _. . ,

5 5 . S t . L . & S a n F . R . R . . .5 6 . S o u t h e r n R y .5 7 . S o u t h e r n R y . _. .

58. Texas & Pac. . ,5 9 . W a b a s h R . R . .6 0 . W e s t V a . S h o r t L i n e . _. _.

TABLE 3

Live Load for Railroad Bridges in Older Specifications (Continued)

Date ofspecif. TYPO

190018981898

3898 _.1906-1901

190418961900

190218981891

1904 .~_190019011904

190319011904

190119041904

1 9 0 1 , _.....__18991902

1899 :18961899

IJ

-

45,00034,coo40,000

44,00055,00040,000

45,00035,00050,000

%%35:ooo

“4g:45,000

-2E52:ooo

“5”om;,

40:ooo

45,00042,00045,000

58%40:ooo

--

-

--

wt. on D r i v e rdrivers, base,

lb. ft.

180,000

%:%Pl

15.016.016.0

176,000 16.0220,000 15.0150,000 16.1

180,000140,000200,000

15.015.015.0

15.013.513.5

160,000 15.0180,000 15.25180,000 15.0

20.016.516.5

176,000200,000160,000

13.715.015.0

15.015.015.0

13.513.516.0

-

-

I

;

-

mad per fton driver

bg%

.~

wt. oflgine am

tender,tons

_ _ _

12,000 1658,400 124

10,000 136

11,000 15614,700 lSS$9,300 141

12,000 1729,400 120

13,300 172

:z?1234

10,700 13611,800 15112,000 156

12,50010,70012,600

2%:10:700

:z%12:ooo

13,00014,80010,000

156%

El?

3160

150174142

--T1

-

ength of oad per feng. and 01‘ eng. an

tend., tend.,ft. lb.

109.0 6,100121 .o 4,100121.0 4,500

5,1006,9005,000

112.5 6,100103.0 4,700107.0 6,400

112.0 6,200107.0 5,20099.0 5,000

104.0120.0107.5

.124.0 5,100124.0 6,000

111.8114.0109.0

108.0 5,900106.5 5,200106.5 6,000

104.0105.5117.0

.iT-

-

Uniformtrain load.lb. per ft.

4,5004,0004,000

4,4006,0004,500

5,000 z

4,0004,000 z

5,000 L?5,0004,000 r

4,000 $4,0005,000

5,0005,0005,000

5,0005,0004,500

4,0004,0005,000

4.5004,0004,000

1 4 DESIGN OF STEEL BRIDGES

TABLE 4

Showing Increase in Weight of Locomotives

[CHAP. I

TYPO Weight, tons

Consolidation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Consolidation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Mogul. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Consolidation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Consolidation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Baldwin, lo-wheel . . . . . . . . . . . . . . . . . . . . . . . . . .Consolidation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Consolidation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Electric motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Consolidation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Pacific . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Articulated. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1873188118861887

if

5 8

1888189018921894

iit6 78 0

1895190519061911

1::115231

-

-

TABLE 6

Heaviest Locomotives of Different Types-I

Type

Atlanta. . . . . . . . . .Prairie. . . . . . . . . .

214,800

Consolidation . . .244,700

12 Wheel . . . . . . . .260,100262,000

Decapod. . . . . . .Pacific . . . .

267,000

Mikado . . . . . . .270,000

12 Wheel articulated305,000334,500

10 Coupled. . . . .20 Wheel articulated

361,000

16 Wheel articulated478,000

24 Wheel articulated493,000616,000

12 Wheel electric. . .16 Wheel electric.

300,400320,000

C o o p e r ’ s E-50? .C o o p e r ’ s E-6Ot

225,000270,000

Engine alone

Weight,lb.

30.7934.2526.5027.08

728 ,400807,500

~~~%

29.83 802,00035.20 865,40035.00 960,00030.66 473,800

43.50 1,074,00059.80 703,60040.17 588,00065.92 841,600

38.50 600,80044.22 640,000

23.00 710,00023.00 852,000

W,,i$d

Double-header :

127.76 5,700132.92 6,070131.81 6,520130.15 6,280

127.00 6,320142.48 6,070150.00

64 .‘566,4007,340

161.0099.7082.58

105.82

86.50 6,950102.84 6,220

104.00 6,830104.00 8,190

Weightper ft.

6,6707,0607,1307,950

* Weight and wheel base for articulated engines are given for one engine and tender.t Cooper’s E-50 and E-60 typical consolidation engines me given for comparison.

A R T . 31 EXTERNAL FORCES 15

TABLE 6

Heaviest Locomotives in Actual Service on 36 American Railways

Rai lway

N.Y.,N.H.&H... . . . .B . & M . . . . . . . . . . . . . . .N. Y. C. Lines . . . . . . . . .Erie . . . . . . . . . . . . . . . . . .

P. R. R .. . . . . . . . . . . . . .L. v. . . . . . . . . . . . . . . . . .P . & R . . . . . . . . . . . . . . . .B.&O . . . . . . . . . . . . . . .

N . & W . . . . . . . . . . . . . . .c . &O.. . . . . . . . . . . . . . .Virginian. . . . . . . . . . . . . .S . A . L . . . . . . . . . . . . . . . .

Southern . . . . . . . . . . . . . .A.C.L. . . . . . . . . . . . . . . .L . & N . . . . . . . . . . . . . . . .Wabash . . . . . . . . . . . . . . .

B . & L . E . . . . . . . . . . . . .I . c.. . . . . . . . . . . . . . . . .Pere Marquette . . . .M., St. Paul & S. S. M. .

C . & A . . . . . . . . . . . . . . . .C. & N. W. . . . . . . . . . . .Great Northern. . . . . . . .‘2, M. & St. P. . . . . . . . .

C., B. & Q. . . . . . . . . . . . .A.,T.&S.F.. . . . . . . . . .C.,R.I.&P.. . . . . . . . . .N . P . . . . . . . . . . . . . . . . . .

M. P . . . . . . . . . . . . . . . . .s. P . . . . . . . . . . . . . . . . . .St. L. & S:.F. . . . . . . . . . .M.,K.&T.. . . . . . . . . . .

G r a n d T r u n k . Consolidation .C a n a d i a n P a c i f i c .

211,200M a l l e t . . .

C . N . . . . . . . . . . . . . . . . . .261,900

Consolidation 181,400N . R y s . o f M e x i c o . Nlallet. . . . . 438 ,000

-I- Locomot ive .3 in serv ice I Under consideration

TYPO

Pacific. . . . . . .Pacific. . . . . . .Pacific. . . . . . .Consolidation

Pacific. . 269,800P a c i f i c . 241,400Consolidation 222 ,000Mal le t . . . . 463,000

M a l l e t . 400,000Mallet. . . 392,000M a l l e t . . . 455,000Consolidation 212,000

. . . . . . . . . . . . . . . . . . . . .Mallet 4 0 0 , 0 0 0

. . . . . . . . . . . . . . . . . . . . .1 . . . . . . ..I . . . . . . . . . . .

..................... . . . . . . . . . . . . . . . . . . . .

Mallet. . . 366,000 ..................... . . . . . . . . . . . . . . . . . . . .Consolidation 171,000 ..................... . . . . . . . . . . . . . . . . . . . .Consolidation 224,000 ..................... . . . . . . . . . . . . . . . . . . . .Consolidation 223 ,800 ..................... . . . . . . . . . . . . . . . . . . . .

Consolidation 254,000Consolidation 223 ,000Consolidation 217 ,000Pacific. . . . . . . 253,800

..................... . . . . . . . . . . . . . . . . . . . .MikadoMikado ,280,OOO,280,OOO

. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .

Mallet. . . . . . 323,400Pacific. . . . 238 ,000Consolidation 216,600Mikado. . . . . 260,500

. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .

MalletMallet 463 ,000463 ,000. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .

MalletMallet . . . . . . . . . .. . . . . . . . . . .

M a l l e t . . . 354,500Double Santa Fe, 616,000Consolidation 238 ,900Mallet.. . . . 435!200

Pacific.Mallet.Mallet.Pacific.

. .

. .

.

Weight, lb.I

TYPO Weight, lb.

229,500

I I

Pacific 235 ,000equ2aa$6t;of-43 . . . . . . . . . . . . .

260: 100.....................

Mikado 305 ,000

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

251 ,000437 ,000416,000228 ,000

.....................

.....................

.....................

Mikado 275,000 abt...i=‘ddsbi . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .

It is doubtful whether the weight of these heavy engines will be in-creased in the future by more than 10 to 15%. If so, it will be done byincreasing the length of wheel base rather than the weight on the driveraxles, not only on account of the limited capacity of the tank and roadbed,but mainly on account of the difficulty in concentrating more weight onthe driver axles with the limited height and width of the engines.


The heaviest electric locomotive for freight and passenger traffic on theN. Y., N. H. & H. R. R. weighs 130 tons, or about 5200 lb. per lin. ft.,with maximum axleloads of 45,000 lb. The electric passenger locomo-tive on the P. R. R. weighs 156 tons or 5100 lb. per lin. ft. of its totallength of 65 ft.; it has 8 axles on a total base of 56 ft. and 4 driveraxles each carrying 50,000 lb. on a base of 25 ft;

Typical coal cars now generally in use have a capacity of approxi-mately 100,000 lb. and a total weight on 4 axles of 150,000 lb., which,with a total car length of 35 ft., gives 4300 lb. per lin. ft. of track.Fig. 2 shows the axle spacing of a car of 50 tons nominal capacity.

Weight of empty car. . . . 36,400 lb.Nominal capacity plus LO per cent.. . . 110,000, “

Total weight.. . 146,400 1:Weight per axle. . . . 36,600

A loo-ton, 6-axle coal car of the following weights and dimensionswas put into service on the Norfolk & Western Ry. in 1912 (Fig. 3):

FIG. 3. 1L e n g t h . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46f;liti;.Width............................................ 1 0Height above rail., . . . . . 10 “ 49 ‘(

Weight of empiy car. . . . . . . . . . 65,200 lk.C a p a c i t y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 0 0 , 0 0 0

_ - - -Total weight. . . . . . . . . . 265,200 ”

Weight per axle.. . .Weight per linear foot of track. . .

“$4,:: :i

Ore cars weigh still more and approach, when fully loaded, 6000 lb.per lin. ft. of track.

Passenger curs fully loaded weigh usually about 1500 lb. per ft. oftrack.

AR T . 41 EXTERNAL FORCES 17

Cooper’s loading E-60, therefore, represents about the heaviest nowin use; bridges accordingly designed will be strong enough for anyprobable loading of the near future; Cooper’s E-50 or an equivalent,loading is, however, used by most first-class railroads at the present.

ART. 4. LIVE LOAD FOR HIGHWAY BRIDGES

The live load for highway bridges consists of people, wagons, auto-mobiles, road-rollers, trolley cars, etc. The loads vary considerablyaccording to different localities. They should be selected with due regardto the character of traffic which the bridge is to carry. The probableheaviest concentrated loads or a dense crowd of people should be usedfor proportioning the roadway floor and its supports, while for the maintrusses or girders it will usually suffice to assume a uniform load, repre-senting either a crowd of people or a continuous line of electric cars orboth. Due consideration should be given to the probability of maxi-mum load by assuming a decreasing density of traffic with increasinglength of span (compare abstracts of specifications, Vol. I).

Weight of People.-The usual maximum weight of a crowd of peopleassumed for bridges is 100 lb. per sq. ft., which represents a crowd soclosely packed as to be unable to move, a condition which is impossibleover the whole width of a bridge, except on very short spans in largecities; 80 lb. per sq. ft. may be assumed for small areas for a very densecrowd in the country; 50 lb. per sq. ft. (usually the minimum assumed)represents a crowd hardly able to move.

Weights of Road Vehicles.-Wagons drawn by horses have the fol-lowing average dimensions and weights:

Kind

Weight in lb. parD i s t a n c e

Length, Widt,h, T o t a l sg. ft. of rectanglec. to c. W e i g h t

ft. ft. K&S,G;gC w e i g h t ,

t o n sin lb. p$z

ft. Eyn&$;g Occupiyl* lin. ft.by vehlole

L i g h t w a g o n 15 ?3 1; I 5 1 6 114H e a v y w a g o n 2 0 5 4: 150 i

it 5408 7 0

Very heavy wagon 25 9 1 5 1 5 212 103~

1130

* Obtained by adding for necessary clearance a strip of 1 ft. all around rectangleenclosing vehicle, including horses.

A pair of strong horses weighs about 2500 lb. and occupies a space10 ft. long, which gives, for 7 ft. width of wagon, a weight of only 35 lb.per sq. ft. For very heavy wagons at least two horse lengths shouldbe assumed.

2

18 DESIGN OF STEEL BRIDGES [CHAP . I

Automobiles and motor trucks at present in use have the followingweights and dimensions:

Type

Automobiles:

Runabout . . . . . ......Light touring. . . . ......Heavy touring. . . . . ......Sight-seeing auto. . ......

Motor trucks:

Fire engine . . . . . . . ......3-ton dump wagon. ......5-ton dump wagon. ......

5-ton electric truck7-ton coal wagon . .lO-ton coal wagon .

......

......

......

Length, Width,ft. ft.

-

1,5002,5005,500

20,000

20,00024,00032,000

-I

-

Weight in lb. per sq.ft. of rectangle

E n c l o s i n gY chicle

O c c u p i e d ’)y vehicle1

27 17

iFa 2110 81

76114142

159185285

!i100

111

i-

1,000130 1,170198 1,880

i1 Weight in

lb. perlin. ft. *

1 3 01803 2 08 5 0

4 2 07 0 09 0 0

*Obtained by adding for necessary clearance a strip of 1 ft. all around rectangleenclosing vehicle.

Trucks exceeding 80 lb. per sq. ft. of occupied space are rare.Road-rollers usually weigh from 10 to 20 tons; their dimensions are

approximately as shown in Fig. 4. The smaller dimensions refer tothe lighter rollers. The front wheel usually carries 40% and the backwheels 60% of the total load. The weight per sq. ft. of rectangleenclosing the roller is approximately 130 lb. for lo-ton rollers and250 lb. for 20-ton rollers. If 1 ft. is added all around the rectangle the

FIG. 4.

corresponding figures are 100 and 180 lb. per sq. ft. The road-rolleris, therefore, one of the heaviest road vehicles and is usually specifiedfor the calculation of the floor of bridges carrying no trolley cars.

Combination of Road Vehicles.-From the above figures it followsthat a congestion of vehicles with the necessary clearances will hardly


produce a greater average load per sq. ft. than a dense crowd of peopleor 100 lb. per sq. ft., since a combination of heaviest and fully loadedvehicles is practically impossible. As a rule the average load per sq. ft.from a congestion of vehicles will not exceed 50 lb. per sq. ft. Forthe calculation of the floor system it is therefore sufficient to assumea single heaviest vehicle and the space outside of it occupied by a densecrowd of people; for the trusses it is sufficient to assume a uniform load(see abstract of specifications, Vol. I).

Trolley cars vary considerably and the weights are continuously in-creasing. Those in cities usually weigh from 25 to 35 tons, while thoseon suburban or interurban lines are heavier, weighing from 30 to 40tons, on some lines even 45 to 50 tons or as much as the cars of electricrailways. The trolley cars are about 8 ft. wide but should be assumedto occupy a width of at least 11 (preferably 12) ft. The average weightper lin. ft. of track of a continuous line of fully loaded trolley cars (animprobable condition of traffic) is 1000 to 1500 lb.

For the calculation of the floor system it should always be consideredthat two cars coupled may produce greater stresses than a single car,especially in case of 4-axle cars. The heaviest trolley cars used in NewYork City in 1911 have the following weights and dimensions:

4-wheeled car, 34 ft. long, 143 tonson one truck with 4-ft. &in. wheelbase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 850 lb. per lin. ft.

&wheeled car, 47 ft. long, 30 tons equallydistributed on two trucks, 20 ft. c.to c.; wheel base of truck 4 ft.. . . . . . . . . . 1280 ” ” ” ‘I

Ash car, 38 ft. long, 50 tons, equally dis-tributed on two trucks, 23 ft. 6 in.from c. to c.; wheel base of truck 6 ft. 6 in. . 2630 “ I‘ “ ‘I

While the 8-wheeled car is rather light,the ash car is unusually heavy.

The 8-wheel trolley cars used in Philadelphia in 1910 are 38 ft.8 in. long (4-ft. 6-in., 12-ft. and 4-ft. 6-in. axle spacing), weigh empty21 tons, have a seating capacity of 40 and a total capacity of 125 pas-sengers. Total weight, therefore, 30 tons or 1550 lb. per ft. of track.

Some interurban cars put in service in 1911 weigh empty 46 tons andhave seats for 56 passengers, a baggage compartment, a toilet room, a


motorman’s cab and a closet for a heater. They are 55 ft. 6 in. longand the trucks are 32 ft. 3 in. apart on centers. With a total capacityof probably 100 passengers the total weight is 53 tons and the weightper lin. ft. of track 1900 lb. (Eng. Record, July 22, 1911.)

For specified loadings see abstract of specifications, Vol. I.For the trusses it will usually suffice to assume the weight of the

trolley cars uniformly distributed over the whole length, whereby theload per foot of track obtained by assuming no clear space between carsmay be decreased with increasing span length. For long spans carryingmany lines of traffic the assumption of a clear space of one car length be-tween cars is justified, as this is about the minimum space allowedbetween cars when running at slow speed. For the calculation of theQueensboro Bridge in New York City, the following loads per lin. ft.of each trolley track were assumed: 1000 lb. as a “congested” load and500 lb. as a “regular” load (see page 383).

ART. 6. LIVE LOADS FOR ELECTRIC RAILWAY BRIDGES

The weight of electric railway trains varies from that of a lightelectric passenger train weighing 1500 lb. per ft. of track to that of aheavy freight train hauled by an electric locomotive and weighing 4500lb. and more per lin. ft. of track. The loading to be used for propor-tioning electric railway bridges must, therefore, be selected in each casewith regard to the actual live load to be carried, whereby provision shouldbe made for a possible future increase of that live load.

For typical loadings see extracts of specifications in Vol. I.In 1913 the subway trains in New Yorlc were composed as follows:

Fori(

7 car trains, 5,motor cars and 2 trailers;8 (i l‘

75 (‘ LL (( 3 l‘ ;L < 1o ‘C ‘l ,6 LL il cL 4 (‘ .

One motor car weighs empty 39 tons and one trailer 26 tons. Each car is’52 ft. long and has a seating capacity of 50 and a total capacity of 170passengers; making a total weight of 49 tons for the motor car and 36 tonsfor the trailer. The average weight per lin. ft. of train is about 1800 lb.

The elevated railway trains in Philadelphia have 6 cars. Each caris 50 ft. long and weighs empty 35 tons; seating capacity 46, maximumcapacity 200 passengers; making a total weight of 49 tons per car ,or1960 lb. per lin. ft. of train. The two trucks of a car are spaced 34 ft. 6 in.c. to c. and the axles of each truck 6 ft. 7 in. c. to c.

.

For the calculation of the Queensboro Bridge in New York City the

ART. 61 EXTERNAL FORCEd 21

following loads per lin. ft. of each elevated railway track were assumed:1700 lb. as a “congested” load and 850 lb. as a L’reguZur” load (see page 383).

ART. 6. DISTRIBUTION OF LIVE LOAD

The specifications usually state that the live load should be so placedas to cause the greatest stress in each part of the bridge. For ordinarytrusses, the live load causing the greatest stress in any truss member coversone continuous, that is, uninterrupted stretch. However, in arches,suspension bridges, cantilevers and swing bridges, the live load, causingthe absolute greatest stress in certain members, has to cover certaindiscontinuous, that is, two or more isolated stretches with no live load be-tween these stretches or on any other part of the bridge. Fig. 5 shows the

FIG. 5.

influence lines for several such members, the first is for a diagonal of aswing bridge on three supports, the second for a top chord member nearthe center of an arch with two hinges, and the third for a top chord mem-ber of the island span of Queensboro Bridge. A live load in the stretchab or cd will cause tension in the diagonal or the top-chord membersrespectively, and a live load in the stretch bc will cause compression

in these members. The absolute greatest tension, therefore, wouldbe caused by ab and cd loaded and’ bc unloaded.


Such a condition is, however, highly improbable if not impossible.For railway bridges a more rational way would be to assume a continuoustrain u-d whose part b-c consists of empty cars. As a rule, however, itwill suffice to assume either of the following three conditions whicheverproduces the greater stress:

(1) Full live load extending from a to d, whereby the compressionproduced by load b-c is deducted from the tension produced by loadsu-b and c-d, (2) full live load extending from a to b only, (3) full liveload extending from c to’d only.

In designing bridges for combined railway and highway traffic it is amatter of judgment to assume the proper live loads. The more lines oftraffic, the smaller the possibility of maximum loading. It would, there-fore, not be consistent to assume the same live loads and unit stressesfor combined railway and highway bridges as for railway or highwaybridges separately. If the highway traffic is unimportant, it may beentirely neglected in proportioning the trusses, except that a possiblesnow load of 20 lb. per sq. ft. on the highway floor should be assumed.The safest way is to assume: first, a regular load, that is one whichmay occur at any time under ordinary circumstances and to use theordinary (working) unit stresses in proportioning the truss members,and second, an extreme load composed of the possible greatest loads onall lines of traffic and to use higher unit stresses as this condition ispractically impossible unless special pains were taken to produce it.The latter unit stresses should be below the elastic limit, even if thewind forces or other accidental loads are considered.

ART. 7. IMPACT AND VERTICAL VIBRATIONS

Impact is the difference in the effect on a stress, deflection, etc., duet.o a suddenly applied or dynamic load and that due to a static load ofthe same amount. In bridges the impact is mainly the effect of verticalvibrations caused by the roughness of track or surface, unbalanceddrivers, etc., and is now mostly treated .as an equivalent static loadadded to the live load.

A great number of experiments to determine this impact have beenmade, but they are not sufficient to formulate final conclusions. Themost extensive experiments on railroad bridges have been conducted bythe American Railway Engineering Association. The following con-clusions were derived from these by Prof. F. E. Turneaure (1910) :

‘I With track in good condition the chief cause of impact was found to be theunbalanced drivers of the locomotive.


“When the rate of rotation of the locomotive drivers corresponds to therate of vibration of the loaded structure, cumulative vibration is caused, whichis the principal factor in producing impact in long spans. The speed of thetrain which produces this cumulat ive vibrat ion is cal led the cri t ical speed.

‘(The longer the span length, the slower is the critical speed and, therefore,the maximum impact on long spans will occur at slower speeds than on shortspans . For short spans, such that the critical speed is not reached by themoving train, the impact percentage tends to be constant so far as the effectof the counterbalance is concerned, but the effect of rough track and wheelsbecomes of greater importance for such spans.

“The maximum impact on web members (excepting hip verticals) occursunder the same conditions which cause maximum impact on chord members, andthe percentages of impact for the two classes of members are practically the same.

“The impact caused by balanced compound and electric locomotives wasvery small and the vibrations caused under the loads were not cumulative.”

The impact exerted directly on the rail is assumed by some engineersat 50% of the wheel load, while others based on careful experimentsassume up to 2005!&.

For highway bridges carrying no trolley tracks the impact, beingsmall, is usually neglected. An impact of at least 25% should beadded to the stresses produced in the floor system by trolley cars.

For steam railroads the impact is now mostly provided for by addingan amount

i=(g!&)z

to the live load stress 1, L being the length of the live load in feet of singletrack causing the maximum live load stress 1 in the member. For double-track bridges L is therefore to be taken at twice as much as for single-track bridges. As the impact varies inversely with the loaded length L,it is greatest for the stringers, decreasing for the floorbeams and theirhangers, the counters, the main diagonals andverticals, until it reachesthe smallest amount corresponding to the full length of the span for thechords and end posts. This will make the floor system (or the top chordof a deck truss supporting the ties) and the counters comparativelyheavier, which is desirable as these parts receive the shock directly andsuddenly; and often counters will become necessary in panels where, ifcalculating without impact, none would be needed, which provides tosome extent for an unforseen increase of the live load.

For bridges carrying electric railways only, the impact is assumed as

i=(&)z

24 .DESIGN OF STEEL BRIDGES [CHAP. I

since the rotating motion of a driving wheel of an electric motor causes amuch smaller impact than the reciprocating motion of the connectingrod and the unbalanced driving wheels of a steam engine.

Mr. Theodore Cooper in his specifications for railway and highwaybridges-allows for the different parts of the floor system and its immediatesupports the same unit stresses for the dead and the live load, while forthe truss members he allows for the live load only one-half of the unitstress specified for the dead load, which is practically the same as using auniform permissible stress with an impact of 100 y0 of the live load.

Some specifications make the impact dependent on the dead load,which has the disadvantage that whenever the dead load stresses arechanged in the calculations, the impact also changes. Mr. H. S. Prichardrecommended the formula:

in which cl and 1 denote the stresses from dead and live load respectively.In the specifications of 1904 for steel railroad bridges of the P. R. R.

the permissible unit stress was 16,000 lb. per sq. in. and the impact

This is equivalent to a unit stress of 8000 (1 +Ek. i!trzsi) if no impact

is considered. The P. R. R. specifications of 1906 use the above im-pact formula of Mr. Prichard with a unit stress of 16,000 lb. (comparealso Vol. I).

Table 7 gives the impact in per cent. of the live load for theformula

Assuming the impact, if dead load is zero to be 200% from the enginesand tenders and 25% from the train load, Mr. Gustav Lindenthalderived the following formula for single track. (See Eng. News, Aug.1, 1912.)

12 1200+a“=I+d600+4a

ART. 81 EXTERNAL FORCES

TABLE 7Impact in Per Cent. of Live Load

L

-

97.497.1.96.896.596.195.995.695.495.094.794.494.193.893.593.292.992.692.392.091.891.591.291.090.790.490.189.889.689.389.088.888.588.288.087.7

L

-

87.587.287.086.786.586.286.085.785.585.285.084.884.584.384.183.883.683.483.182.982.782.582.282.081.781.581.381.180.880.680.480.280.079.879.6

L

787980

8"g‘83

it

ii!88ii91iE

w i9697981%105110115120125130135140145150155160

79.479.279.078.878.578.378.177.977.777.677.477.276.976.876.6E:i76.075.875.676.475.275.074.173.272.371.470.669.869.068.267.466.765.965.2

L

165170175180185190195200210220230240250260270280290300325350375400450500550600650700750800850900950

1000. . . . . .

-

.-

25

z

64.563.863.262.561.861.260.660.058.857.756.655.554.5

E:!51.750.850.048.046.244.442.840.037.535.333.431.630.028.627.326.125.024.023.1. .

3001mpilct i = 2 L+300 where 1 = live load, L = loaded length of track in feet.

where a = length of train behind tender for position of maximum stress,in feet. This impact value to be reduced 10% for each additional trackup to four tracks.

ART. 8. WIND PRESSURE

Although numerous observations have been made to determine thewind pressure, no definite conclusions have yet been reached. In gen-eral the wind pressure w per sq. ft. on a small surface has been found tovary in direct proportion with the square of the velocity v

w=kv2where the factor k depends upon the.shape of the exposed surface and de-

26 DESIGN OF STEEL BRIDGES [CEAP. 1

creases with increasing area of the surface. Sir Benjamin Baker foundduring the erection of the Firth of Forth Bridge that under the sameconditions the pressure on a surface of 14 sq. ft. was in the averageabout 50% greater than on a surface of 300 sq. ft., the maximum observedpressures during seven years being 41 and 27 lb. per sq. ft. respect,ively.

Prof. C. F. Marwin found on top of Mt. Washington on surfaces of 4and 9 sq. ft. area the value Ic = 0.004, if w = pound per sq. ft. andv = velocity in miles per hour.

From recent observations made on top of the Eiffel Tower, M. E. Bretconcludes that k is only 0.0029 (Genie Civil, March 12, 1910). The windpressure was generally below 20 lb. per sq. ft. and riever reached 30 lb.per sq. ft. even during severe storms.

It has also been found that the wind pressure is not uniform over theexposed surface and that it is considerably greater in higher altitudesthan near the surface of the ground. Former experiments made on theEiffel Tower showed at the same time a wind velocity at the top three.times as great as that near the bottom of the tower. The storm of 1896in St. Louis showed that the extreme force of wind was generally con-fined to upper stories and roofs and that the intensity was extremelyvariable even on a single building.

Wind pressures of 20 lb. per sq. ft. have been found to extend overconsiderable distances, but higher pressures, such as occur in the pathsof tornadoes, are limited in breadth to a few hundred feet. GeneralGreeley, of the United States Signal Service, in 1890, after the tornadoat Louisville, Kentucky, stated:

“As bearing upon the strength of structures necessary to withstand tornadicwinds, it is important to note that there have been very few cases recorded ofwind velocities in the United States where the pressure of the wind, according tothe latest investigations and accepted formulas, exceeded 16 lb. to the squarefoot.”

Julius Baier in his paper on the St. Louis tornado (Trans. Am. Sot.C. E., Vol. 37, p. 221) found evidences of pressures as high as 60 lb. persq. ft. over a width of 180 ft.

At the St. Charles Bridge, C. Shaler Smith reported a tornado whichexerted a pressure of 52 lb. on small objects on the bridge, and 84 lb. inthe vicinity, but which did not injure the bridge, although its bracingwas only proportioned for 30 lb. per sq. ft. of exposed surface. He alsofound, after following up the paths of several tornadoes, but one casewhere 60 ft. of width was not enough to cover the path in which thecomputed pressures exceeded 30 lb.


Many investigators of the effects of high winds and tornadoes haveconcluded that it is improbable that ,they ever exert 30 lb. per sq. ft. overa space of 150 to 200 ft. at a time.

It should also be considered in designing bridges for wind pressurethat a wind of 30 lb. per sq. ft. is able to overturn an empty railroadcar.

In view of these facts, it seems that ample provision against windpressure is made in the usual practice of specifying for ordinary spans ahorizontal wind pressure of 50 lb. per sq. ft. on the exposed surface of bothtrusses and floor as seen in the elevation when the bridge is unloaded,or 30 lb. per sq. ft. of exposed surface (including a strip of about 10 ft.high for the train) when the bridge is loaded whichever gives greaterstresses. But it should always be remembered that this is only a roughapproximation.

For long spans, the above pressures are excessive since they could notextend over the entire length of the span. The pressure should be se-lected from, case to case; for spans over 1000 ft. a pressure of 20 lb. persq. ft. may be sufficient (see examples, page 372).

In certain cases, especially high structures such as viaducts, a longi-tudinal wind force parallel to the axis of the bridge may have to be con-sidered. Since the exposed area for wind blowing at 45” to the axis israther more than 0.7 of the area seen in the elevation the longitudinalwind pressure should be taken at not less than 0.7X0.7 = 0.5 of thetransverse wind pressure.

ART. 9. LATERAL VIBRATIONS

The lateral vibrations of the locomotives are considerable, while thoseof the cars are unimportant. They affect, therefore, especiallyshort spansand it is important to provide these with substantial bracing and toassume the whole lateral pressure along the floor as a moving force.These vibrations are usually assumed to be included in the wind force,but some specifications make extra allowance by specifying a concen-trated lateral force. The Pennsylvania Steel Co. assumed a concen-trated lateral force at 20% of the load on the drivers. The AmericanRailway Engineering Association’s specifications assumed the lateralforce in the plan of the track at 200 lb. per lin. ft. plus 10% of thespecified uniform train load on one track (see Appendix).

The lateral impact exerted on the rail by a driving wheel is even lessknown than the vertical impact. Some engineers are satisfied with

2 8 DESIGN OF STEEL BRIDGES [CHAP. I

15 ‘%, while others make it two-thirds or even more of the vertical staticload.

Stresses caused by wind and vibrations are generally neglected incombination with dead and live load stresses unless they exceed 25 to30% of the sum of the others (d+Z+i), in which case the permissibleunit stresses are increased by 25 to 30%.

ART. 10. CENTRIFUGAL FORCE

The centrifugal force affects railroad bridges in which the track ison a curve. It acts horizontally at a height of about 6 ft. above base ofrail and is in a certain proportion to the live load, depending upon thedegree of curvature and the speed of the train. It should be remembered,however, that the greatest centrifugal force is as a rule caused by thelighter passenger trains moving at a high speed and not by the heaviestfreight trains.

The centrifugal force C due to a load P is

c=?$

where v = velocity in feet per second, g = acceleration of gravity =32.2 ft. per second, and r = radius of curvature in feet, or

C = 0.0000117 v2D P

5 7 3 0where v = velocity in miles per hour, D = degree of curvature = F*rThe speed is usually specified by the railroad. The American RailwayEngineering Association specifies a speed of

v = 60-2; D

in miles per hour.Table 8 gives C in per cent. of P for various speeds. Owing to the

super-elevation of the outer rail (see table, page 215), the resultant of theload P and the centrifugal force C passe’s through the center line of trackfor a certain speed usually specified by the railroad. For any other speed,the resultant does not pass through the center of track but, since the ele-vation nearly corresponds to the maximum speed allowed on a certaincurve, it is permissible for the calculation of the stresses due to centrifu-gal force to assume this force applied to the structure horizontally atthe height of the rails, if the vertical load is assumed applied alongthe center line of the track.


TABLE 8

Centrifugal Farce in Per Cent. of Vertical Force

VelocityI

Degree of ourvitture

,;‘;$1 sEd 1 lo 1 2” / 3O 1 4’ / 5’ 1 6’ 1 7’ 1 8’ / 9’ / IO0 / ll- 1 12’ 1 13’

E 22.00 14.67 0.12 0.26 0.23 0.53 0.35 0.79 0.47 1.05 0.58 1.31’ 0.70 1.57 0.82 1.84 0.94 2.10 2.36 1.05 2.62 1.17 2.88 1.29 3.14 1.40 3.40 1.51

% 29.33 36.67 0.47 0.73 0.93 1.46 2.19 1.40 2.92 1.86 2.33 3.65 2.79 4.37 3.26 5.11 3.72 5.82 4.18 G.56 4.65 7.29 5.11 8.00 5.57 8.75 9.45 6.05

30 44.00 1.05 2.10 3.15 4.20 5.25 6.30 7.34 8.39 9.4310.4811.5512.5713.61

&7 51.33 58.67 1.43 1.87 2.86 3.73 4.28 5.60 5.70 7.47 9.3411.19 7.13 8.57 13.0614.9016.79 9.9811.4012.82 14.25 18.6520.5022.3724.20 15.70 17.09 18.53

2 73.33 66.00 2.36 2.91 4.72 5.83 7.08 8.7411.6514.5617.5020.3723.3026.1829.0932.00..... 9.44 11.80 14.20 16.5218.9021.2323.5825.9028.28

55 80.67 3.53 7.0510.6014.1217.6521.2024.6928.2031.7335.25_.... .._.60 88.00 4.20 8.39 12.6016.7920.9825.2029.3633.6037.73 ,.._ _._.. ._...

Centrifugal force =&, whore P = vertical force, 2) = velocity in feet per second, T = radius of

5730 -curve - -.

degree of curvature

ART. 11. BRAKING AND TRACTION FORCES

The traction force is equal to the friction of the drivers on the rails,while the braking force is equal to the friction on the rails of all wheelsto which brakes are applied. Both forces may be assumed acting upon thestructure longitudinally at the top of rail. The braking force is the moreimportant of the two, but its effect on bridges is usually overestimated.The braking force is usually assumed at 20% of the maximum verti-cal live load on the span. Now, as a rule, not all wheels in a heavyfreight train are provided with brakes and, further, as pointed out by Mr.S. Blumenthal in a paper before the Canadian Society of Civil Engineers(Oct., 1910), the efficiency of the brakes is variable for the different unitsof a train and considerably less than would be required to produce thefriction corresponding to the maximum vertical load. In order toprevent skidding of the wheels and their consequent flattening, the Amer-ican Air Brake Association has adopted the following maximum pres-sures of the brakes on the wheels as best suited for actual service:

Passenger cars, 90% of the weight of the empty car.Freight cars, 70% of the weight of the empty car.Tenders, 100% of the weight of the empty car.Engines, 75% of the total weight.

Assuming the coefficient of friction at 0.25, which was found by testsfor slow speed, the actual braking force in per cent. of the maximum ver-tical load is for the different units as follows:

Passenger cars (weight empty =0.9 Xtotal weight). . . = 90 X0.9 X0.25 =20.2%Heavy freight cars (weight empty = 0.3 X total weight) = 70 X0.3 X0.25 = 5.2%T e n d e r ( w e i g h t e m p t y =0.4Xtotal w e i g h t ) =100X0.4X0.25 =lO.O%Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...= 75X0.25 =18.8%


From this follows that the braking force is relatively greatest in shortspans (18.8 YO) decreasing with increasing length of span to nearly 5 y0 ifheavy freight trains are considered only.

Mr. Blumenthal derived the following formula for the ratio B ofbraking force to total vertical live load

82000+ 265 L’ = 165000+5000 L

where L = length covered by the live load; for instance for L = 100,B = 16.3%.

The braking force is important mainly for the design of high piers,viaduct towers and similar structures. In viaducts the length of liveload producing greatest braking stresses in the tower is usually small(mostly not over 120 ft.) and the ratio of braking force to verticalload, therefore, nearly a maximum (18.8%). It should be considered,however, that owing to the longitudinal deflection of the tower, part ofthe force is transmitted to the next span and tower, either throughthe continuous flooring or directly as soon as the expansion joint is closedby the deflecting tower. It seems, therefore, justified not to take themaximum braking force for each tower; a method frequently used is tomake the braking force per lin. ft. of track equal to 20% of the specifiedtrain load (0.2X5000 = 1000 lb. for Cooper’s E-50) instead of 20% ofthe average total load per lin. ft.

The braking force is an accidental force, being only rarely applied,except on some bridges near terminals, and on elevated railroads. Whencombined with the dead and live load (as, for instance, in the columnsof viaduct towers) it is justified to use about 25% higher unit stressesthan allowed for dead and live load only. A combination of the brakingforce with the maximum dead load, live load and wind or centrifugalforce is still less probable and may, as a rule, be neglected.

ART. 12. SNOW LOAD

Snow weighs about 10 lb. per cu. ft. when dry and fresh, 30 lb.per cu. ft. when compact and 50 lb. per cu. ft. when saturated withwater. Whether a snow load needs to be considered depends upon localconditions. In southern countries where snow fall is rare, it may beneglected, but even in northern countries it does not seem justified to con-sider it in addition to the live load. On highway bridges it is evident thatthe live load decreases more rapidly than the snow load increases. I f a


bridge carries a railway and a highway, or for a highway swing bridge inthe open position, a snow load of about 20 lb. per sq. ft. on the highwayonly may have to be considered.

ART. 13. TEMPERATURE CHANGES

When a structure is subject to temperature changes it will expandor contract and unless every member can change its length stresses willresult. Statically determinate structures are, practically, not stressed bytemperature changes.

The variation of temperature is different for different countries andshould be assumed accordingly. In the United States a variation ofk60” or -1-75” F. is usually specified. The Canadian Governmentspecifies a variation of Ifr90” F.

C H A P T E R I I

GENERAL ABOUT REACTIONS AND INFLUENCE LINES

ART. 1. CHARACTER OF REACTIONS

A girder in general is a plane structure acted upon by forces in its ownplane only. The external forces cause reactions at the supports of thegirder which must be in equilibrium with the external forces in order toprevent motion of the girder as a whole. If it is possible to determine allthe reactions by means of the laws of equilibrium the girder is calledstatically determinate with respect to the reactions, otherwise it is static-ally indeterminate.

FIG. 1. FIG. 2. FIG. 3. FIG. 4.

If the character of the supports is such that vertical forces cause onlyvertical reactions the girder is called a beam. If a beam has two supportsonly, one at each end, it is called a simple beam or simple span (Fig. 1) ; ifit extends without interruption over more than two supports it is a con-tinuous beam (Fig. 2), and if one or both ends overhang (Fig. 3)) or if, in

FIG. 5. FIG. 6. FIG. 7.

case of several supports, the girder is interrupted by one or more hinges(Fig. 4), it is.called a cantilever beam.

If vertical forces cause inclined reactions the girder is statically eitheran arch (Figs. 5 to 7), which may, according to thenumber of hinges, be ahingeless, one-, two- or three-hinged arch, or it is a suspension truss which

32

ART. 21 GENERAL ABOUT REACTIONS AND INFLUENCE LINES 33

is simply an inverted arch, its reactions acting in opposite sense to thoseof the arch.

The above classification is made entirely with respect to the char-acter of the reactions, it is independent of the shape of the girder(whether straight or polygonal), or of the web system (whether a plategirder, truss, or a combination of these two).

ART. 2. DETERMINATION OF REACTIONS

Each reaction requires for its determination three values, namely: itsamount, direction (angular deviation from a fixed line), and location(distance from a fixed point). One or more of these values of all reac-tions may be given by the conditions of the supports. For instance, ahinge fixes the location of the reaction in that the latter has to passthrough the center of the hinge as otherwise there would be a turningmotion around the hinge. A set of rollers fixes the direction of the reac-tion which must be at right angles to the plane of bearing as otherwise alongitudinal movement would take place.

The laws of equilibrium furnish’ only three equations, namely:(1) the sum of the vertical components of all forces is zero;(2) the sum of the horizontal components of all forces is zero;(3) the static moment of all forces about any fixed point is zero.

These three equations suffice for the calculation of three unknownvalues. A girder is therefore statically de-terminate with respect to the reactions if

,/AY/’ /

the latter have only three unknown values.If there are 4, 5, etc., unknown values the

Ra4

A ,,/” I5

girder is statically indeterminate in the /first, second, etc., degree; the reactions of Rb d$

such girders can only be determined by FIG. 8.

means of the theory of elastic deformation.A simple beam is supposed to have a hinge at each end and to be held

longitudinally at one end A and movable at the other end B. The loca-tion of each reaction is therefore given and, further, the direction of thereaction at B. There remain only three unknown values, viz., the direc-tion of the reaction at A and the amount of each reaction. Simple spansare therefore statically determinate. (Fig. 8.)

In a two-hinged arch (Fig. 7) only the location of each reaction isgiven, the amount and direction or four values being unknown; this archis therefore statically indeterminate in the first degree.

3

34 DESIGN OF STEEL BRIDGES [CRAP. I I

By intermediate hinges a statically indeterminate girder can be di-vided into several statically determinate parts and thereby be madestatically determinate as a whole. If a hinge is supposed to be removedand an external force equal to the stress transmitted by the pin is appliedto each adjoining part of the girder the three equations of equilibrium canbe applied separately to each part. These equations, together with thecondition that the forces transmitted by the pin on the two parts are

FIG. 9.

equal but of opposite sense, suffice for the determination of the reactionsand the force transmitted by the pin. For instance, a three-hinged archABC (Fig. 9) may be divided into the two parts AC and BC if to eachpart the stress R,, transmitted by the center pin, is applied as an externalforce at C. There are now six unknown values (amount and directionof the reactions R,, Rb, and of the force R,) for whose determinationthe six equations of equilibrium furnished by the two parts suffice.The three-hinged arch is therefore statically determinate.

4

R ‘/,b..

R

FIG. 10.

P

If an end of a girder is rigidly fixed to the sup-port (Fig. 10) none of the three values of the reactionR at that support is given; the reaction is thenusually replaced by an equal parallel force R' passingthrough the center of gravity of the end section andby a static moment M = Rr, where r is the distancebetween R and R'. The unknown values are then theamount and direction of R' and the amount of the

moment M. Thus a hingeless arch (Fig. 5) has six unknown valuesand is therefore statically indeterminate in the third degree.

In a continuous beam one bearing is always fixed longitudinally; allothers are movable so that the beam can expand and contract, and eachbearing is hinged to the beam. At the fixed bearing there are two un-known values (amount and direction) * and at each movable bearing only

* The fact that if the loads are vertical the reaction at the fixed bearing is also


the amount is unknown. If there are n bearings in all, there are (n+ 1)unknown values of which (n-t 1- 3) are statically indeterminate and thebeam is therefore statically indeterminate in the (n-2) degree. Theintroduction of (n-2) intermediate hinges (cantilever beam) makes thecontinuous beam over n supports statically determinate; there must,however, not be more than one hinge in any end span .nor more thantwo hinges in an intermediate span (page 396).

ART. 3. INFLUENCE LINES

Influence lines are valuable for finding bending moments, shears,stresses, etc., especially for statically indeterminate trusses, but evenin the simplest cases they greatly assist in forming a clear idea of staticproblems. A mere sketch of an influence line will often be sufficient todetermine the most unfavorable position of the load, or the points of zerostress. Vertical loads only will be considered.

Definition.-If the load unity moves over a span AB and we calculatea certain value 2 (reaction,bending moment, shear, stress,deflection, etc.) for each posi-tion of the load and plot thisvalue as an ordinate x (in$uenceordinate) from a base line A’B’below the corresponding positionof the load, we obtain the in-fluence line A’CB’ of the staticvalue 2 (Fig. 11). The influ-

FIG. 11.

ence line shows therefore the variation of a bending moment, or shear,etc., at a certain section due to a load unity moving over the span,while a moment curve, shear curve, etc., shows the variation of thebending moment, shear, etc., over the span due to a fixed load.

The area between base and influence line is called InJluence Area ofthe value 2. For instance, the influence line for the bending momentat a section D of a simple span AB consists of the two straight linesA’D’ and D’B’ and the influence ordinate at any section X representsthe bending moment M at D caused by a load unity at X.

vertical is not known beforehand but is only proven by one of the conditions ofequilibrium (sum of horizontal forces equal zero); for the other r~ unknown valuesremain therefore only two equations of equilibrium and there arestill n-2 staticallyindeterminate values.

3 6 DESIGN OF STEEL BRIDGES [CHAP. 11

The static value Z caused by aload P is equal to Px, if x is theinfluence ordinate below P. The value Z caused by a series of loadsPI, Pz, P,, etc., is

z = Plxl+Pzzz+ . . . =BPx,

where zl, xg, etc., are the influence ordinates below the correspondingloads.

If all loads have the same value P, we have

z = P(xl+xz+xs+ . . . )= PZX

If the load is uniform and has the value p per lin. ft. (Fig. 12), theload in a short length dx is p dx and the influence of this small load onthe static value Z is xp dx. Integrating over the whole length of the

load we get Z = Psa*z dx = PA,a1

where A, represents the influence area below the load.

i 1 II I

iI

FIG. 12. FIG. 13.

If the load is carried to the main girders or trusses through floor-beams, the influence line is always straight between any two adjoiningpanel points and the ordinates below the panel points are the same as ifthere were no floorbeams (Fig. 13).

Certain influence lines, like those of shears, stresses in web members,etc., consist of a positive part above and a negative part below the baseline (Fig. 14). The point where the influence line intersects the base lineis the point of zero-shear or zero-stress; that is, the shear or stress due to aload at that point is zero. In order to obtain a maximum positive or amaximum negative value Z, the load has to extend respectively only overthe positive or the negative part of the influence area; the point of zerostress marks the limit of the load. The stress due to a uniform load


covering the whole span is equal to the difference of these areas; thismethod of calculating dead load stresses may, however, be inaccurate ifthe difference in the areas is small compared with the areas themselves.

Certain influence lines intersect the base line several times (Fig. 15).In order to obtain the absolute positive value 2, the load should extendfrom a to b, and from c to d. Such a separation of the live load, however,is not customary, and the maximum iis usually obtained either by loadingfrom a to b only, or c to d only, orcontinuously from a to cl, whichevergives the greatest value (see page 21,also examples, page 392).

The method of calculating withinfluence lines is especially simple in

FIG, 14.

spans with constant panel length X and if the load is uniform. As a ruleit is sufficient to assume full panel loads, and the value 2 is then obtainedby adding the influence ordinates x below the panel points and multiply-ing their sum by the product Xp, where p is the load per ft., that is,

If the panel lengths are not alike each influence ordinate x below apanel point is multiplied by the corresponding average length X of the ad-joining panels and the sum of these products multiplied by p, gives

a e

FIG. 15.

For approximate results the influence lines in combination withequivalent uniform loads (see page 62) are often convenient. For thepurpose of determining the equivalent uniform load to be used with aparticular shape of the influence area the rules given below may be fol-lowed, as usually only the equivalent uniform load for the maximumend reaction and the maximum moment at the center ‘of simple spansare known (page 64).

If an influence area of the length b is a triangle whose greatest ordi-

38 DESIGN OF STEEL BRIDGES [CHAP. II

nate is at the middle of the base (Fig. 16 a), the correct equivalent uni-form load is that which corresponds to the maximum bending momentin a simple span b, -whatever the height of the triangle may be. If aninfluence area is a right-angled triangle of the length b (Fig. 16 b) thecorrect equivalent uniform load is that which corresponds to themaximum end-reaction of a simple span b.

It is evident that for influence lines forming triangles as shown in Fig.16 c, intermediate values must be used for the equivalent uniform load,

approaching the more that formoment or that for end-reactionthe nearer the apex of the tri-angle approaches the middle orthe end, respectively, of thebase b. The same applies to in-fluence lines which form one ortwo curves convex toward thebase as shown in Fig. 16 d.

(b)

FIG. 16.

If the influence line is a curveconcave toward the base (Fig. 16e), it is usually on the safe sideto use the uniform load corre-sponding to the maximum mo-ment.

From the rules for these sim-ple cases it will not be difficultto determine the proper uniformload for more complicated casesas shown in Fig. 16 f.

Table 17 contains the equi-valent uniform loads for maxi-

mum moment and end-reaction of simple spans for Cooper ‘s E-50loading.

In case of a series of concentrated loads, a loading diagram is drawnon transparent paper with avertical line through the center of each wheel(Plate I). This diagram is then placed over the influence diagramdrawn to the same scale in such a position that ZPx becomes a maxi-mum. This position is in general most conveniently found by trial.

If a particular loading has to be frequently used it is advisable to drawa scale to each vertical; this scale being obtained from a certain unitscale by multiplying the figures on the latter with the wheel concentra-


tion. For instance, in Plate I for Cooper’s E-50 loading the unit scale5 in. = 1 lb. has been used; therefore, corresponding to the three differentwheel concentrations per rail, we have three different scales; namely,g in. = 12,500, 16,250 and 25,000 lb. respectively, for the truck wheel,tender wheels and drivers; the uniform trainload of 2500 lb. per lin. ft. isreplaced by concentrations of 25,000 lb. 10 ft. apart.

With the zero line of this diagram placed on the base line of theinfluence diagram, the influence ordinates can be directly read off andadded, providing the influence ordinates are drawn to the unit scaleQ in. = 1 lb.

ART. 4. INFLUENCE COEFFICIENT

It will often be convenient if the influence lines of the members of atruss are so drawn that the ordinates of different lines or at least ofdifferent groups of lines have to be multiplied by different values in orderto get the respective stresses. The value with which the influence or-dinates of a certain member have to be multiplied to give the stress inthat member will be called the in$uence coeficient of that member.Often a number of stresses can thus be obtained with a single setting ofthe slide rule. This method is especially convenient for statically in-determinate trusses.

CHAPTER III

MOMENTS AND SHEARS IN SIMPLE SPANS

ART. 1. GENERAL

The external forces (including reactions) are independent of the shapeof the beam and since moments and shears are functions of the externalforces only, the following theory applies to any kind of simple spanswhether they are rolled beams, plate girders or trusses. Vertical forcesonly will be considered here.

The two ends of a simple beam are supported vertically, one endbeing held, the other allowed to move longitudinally under temperaturechanges and deformations. There can, therefore, be only vertical re-actions from vertical forces.

In the following the Bending Moment or simply Moment at a point Xof the span shall denote the static moment of all the external forces tothe left of X about X; it shall be positive when turning &o&wise, andnegative when turning anti-clockwise. Vertical downward forces alwaysproduce positive moments at any section of a simple span and the resultingstresses are always tension below and compression above the neutralaxis of the beam. The shear at a point X shall denote the resultant ofall external forces to the left of X, and shall be positive when actingupward and negative when acting downward.

ART. 2. INFLUENCE LINES FOR SPANS WITHOUT FLOORBEAMS

All influence lines for moments, shears and stresses in simple spansconsist of straight lines. The most important is the influence line forthe end reaction, since from this all others can be derived.

The end reaction R, (Fig. 1 a) due to a load P = 1 (Unity) is

R =I”-“.a 1This equation represents the influence line as a straight line A2B1, whereA 1A2 is equal to the load unity (Fig. 1 6). Correspondingly the lineAIBz (Fig. 1 b) is the influence line for the reaction Rb.

The shear V at any point C is equal to the reaction R, as longas P remains to the right of C and equal but of opposite sign to Rb forloads to the left of C. The influence line for shear is, therefore,

40

Am. 31 MOMENTS AND SHEARS IN SIMPLE SPANS 4 1

identical with that for R, from B to C, and symmetrical about thebase line to that of Rb from C to A (Fig. 1 c).

The, bending moment M at any point C is equal to R, a for loads tothe right and equal to Rb b for loads to the left of C. Multiplying,therefore, the influence ordinates for R, from B to C by a and those for

F IG . 1.

Rb from C to A by b we obtain as influence line for the moment at Cthe line AICIBl (Fig. 1 d).

From these influence lines follows that for maximum end reaction andmoments the load has to cover the whole span, for maximum positiveshear the distance CB and for maximum negative shear the distance AC.

ART. 3. MOMENTS AND SHEARS FROM UNIFORM DEAD LOAD IN SPANSW I T H O U T F L O O R B E A M S

The end reaction from a uniform dead load w per lin. ft. of girder is

R, = Rb zz +l

which is also the maximum shear in the beam (Fig. 2).The shear at any point C is

Vc=R,-wx=w;-x( >

The shear curve represented by this formula is a straight line AzBz withWl Wl

ordinates +T and -2 for x = o and x = 1, respectively. The

4 2 DESIGN OF STEEL BRIDGES [CHAP. III

shear at the quarter-point is one-half of the shear at the end andthe shear at the center is zero.The bending moment at any point G’ is

This equation represents a parabola whose vertex and maximumordinate is at the center and equals

M “rk”maz: = -

FIG. 2.

If the span is divided into a number of equal panels the moments andshears at the panel points can be conveniently determined by means ofTables 9 and 10.

TABLE 9Moments From Uni form Load

For p = 1 and A = 1

TABLE 10Shears From Uni form Dead Load In

Spans wi thout F loorbeamsFor p = 1 and X = 1

Number Panel point

of panelsin span i 1 I! 2 3

:.5 22 32.5 4 4.5

33.5 : K5

2.5 i li.5

2.5 1: ii.5

66u5 :i ::.5

415 /q-F

,... . . .. . ,.... . ,....

8 . . .10

:4”1 2 . 5

16 if.5 ‘is’ :::I!2 z.5 2 ii:5

of panels Iin span

reaction

:::

1:

1.5

g.5

i.544.55

.

Panel point

1/21314150.5 ..,. . . . .

:.5 i.5 : : : : :::: : : : :

x.5 :.5 i.5 : : : : : : : :

i.5 i.5 :.5 z.5 : : : :4 3 2 1 0

..,.’, /. .

Note.-The above tables are calculated for a uniform load unity per lin. ft. and panel pants 1ft. apart. For a load p in lb. per lin. ft. and panel length X in ft. multiply the above momentsby pX? and the shears by pX.

1 2 3 4 5 6 7. ” ” ” ” 4A B

ART. 41 MOMENTS AND SHEARS IN SIMPLE SPANS 43

Example.-A lOO-ft. plate girder carries a uniform load of p = 800 lb. per lin.ft. What are the mcments and shears at the points dividing the span into eight

equal parts? X = loo = 12.5; X2p = 125,000;8 multiplying by 3.5, 6, 7.5, 9

(Table 9) we get moments = 437.5, 750, 937.5 and 1125 in units of 1000 ft.-lb.Xp = 10,000 this multiplied by 4, 3, 2, 1 (Table 10) gives the shears as40, 30, 20, 10, in units of 1000 lb.

ART. 4. MOMENTS AND SHEARS FROM UNIFORM LIVE LOAD,IN SPANSWITHOUT FLOORBEAMS

The maximum reaction and bending moments occur with the spanfully loaded. They can, therefore, be determined by the same formulasas given above for the dead load or by simple proportion from the corre-sponding values from dead load.

M, = g x(1- x), and .M,,, = $

where p is the load per lin. ft. of girder.

FIG. 3.

The maximum positive shear at any point C (Fig. 3) occurs when theuniform load p covers the distance BC and is equal to

v, = & (I - x)” = gf (z-42

This equation represents a parabola with the apex at B’ and the ordi-1

n a t e pg a t A ’ .

The maximum negative shear is obtained by loading from A to C. I tis equal to the maximum positive shear V’, at the point C’ which liessymmetrically to C about the center of the span.

If the span is divided into a number of equal panels the shears at thepanel points are conveniently found by Table 11.

44 DESIGN OF STEEL BRIDGES [CrrnP. III

TABLE 11.

Shears from Uniform Live Load in Spans without FloorbeamsFor p = 1 and h = 1

Number of Endp a n e l s i n reaction

apall A

9

10

-

-

I-

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

Panel points

1

0.670.171.120.121.600.102.080.082.570.073.060.063.560.064.050.05

-

2

. . . .0.500.500.900.401.330.331.780.292.250.252.720.223.200.20

3I 4

I -. . . . . . . . . . . . .. . . . . . . . . ..,.._.,._ . . . . . . . . . .

. . . ~:.,.,-. . . . . .

“i%,_. . . . . . . . . . .

. . . . . . . . . .0.75 . .1.14 . . . . . . . .0.64 . .1.56 1 . 0 00.56 1.002.00 1.390.50 0.892.45 1.800.45 0.80

i1.i .ij

/

-

5

. . . . . . . . .

. . . . . . . . .

. . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . .

. . . . . . . . .

. . . . . . . . .

1.251.25

Note.-Upper figure is maximum positive shear. Lower figure is maximumnegative shear. For a load p in pounds per lin. ft., and panel length h in ft.multiply the above shears by Xp.

I z 3 4 5 6 7. .A 0

Example.-A’lOO-ft. deck-plate girder carries a live load of 800 lb. per ft. Di-viding the span into eight panels of X = 12.5 ft. and multiplying the figures inthe sixth row of Table 11 with Xp = 10,000, we get for the shears at the panelpoints from A to the right: 40, 30.6, 22.5, 15.6, 10.0, 5.6, 2.5 and 0.6 in units of1000 lb.

ART. 6. MOMENTS AND SHEARS FROM MOVING CONCENTRATED LOADSIN SPANS WITHOUT FLOORBEAMS

Shears.-The maximum reaction or end shear R, is obtained bycovering the whole span with loads and placing the heaviest concentra-

F I G. 4 .

tions at and near the end support A. For the usual engine the firstdriver has to be placed over that support (Fig. 4). Moments of all

ART.~] MOMENTS AND SHEARS IN SIMPLE SPANS 45

concentrations on the span are then taken about the support B and thesum is divided by the span length 1:

R, =“y x>

The calculation of Z P(Z - x) for ‘any given series of concentrationscan be greatly reduced by the use of tables as explained on page 57.

The maximum positive shear at any p,oint C (Fig. 5) occurs whenthe load extends from B to the left so far that either the first, second,

_________-_______ -_ 1 f__________---------------

FIG. 5.

ZP(Z- x)third, etc., wheel is over C. If R, = ~~1 is the corresponding

left reaction the shear at C is

V, = R,. . .if first wheel at C.V, = R, - PI. . . .if second wheel at 6,V, = R, - PI - Pz. . . .if third wheel at C, etc.

In order to find out which wheel at C produces the greater shear thefollowing conditions have to be examined:

(1) If PI& > Z+‘. . . _. . .the first wheel has to be at C,

(2) If PIi < iTIP, but P2& > 2&P, the second wheel has to be at C,

(3) If P.& < Z&P, but Pai > i&P, the third wheel has to be at C,

where I&P, &P, 2&P denotes the sum of all loads on the span whilethe first, second, third wheel, respectively, is at C. Should the firstwheel move off the span when the second is at C the latter may pro-duce the greater shear although condition (1) may be fulfilled.

For the usual engine loading with one truck wheel in front of thedrivers the first driver (second wheel) produces as a rule the maximumshear at any point of an ordinary deck plate girder.

Example.-A lOO-ft. single-track deck plate girder span carries Cooper’s E-50loading, each girder therefore one-half the axle loads given on page 60.

46 DESIGN OF STEEL BRIDGES [CHAP. III

Maximum reaction or end shear:

With wheel @ at A wheels @ to @ are on the span,Moments of wheel loads @ to @ about B = 18,750,000,

Max. end shear18,750,OOOyloo = 187,500 lb.

Maximum shear at quarter-point:

Wheel @ at 2 point, 21 P(l to 13) = 265, PI i1 =12.5 y ,= 156 < 290.

Wheel @ at t point, 22 P(l to 14) = 290, P, k =251005 = 500 > 290.

Wheel @ has to be at $ point,

Moment of loads @ to @ about B = 12,070,000,12 070

Shear at $ point = 100 - 12.5 = 108,200 lb.

Maximum shear at center:

Z,P(lto9) = 177.5,P1 ; = 1561

ZS(ltol0) = 1901 P,& = 500

Wheel @ has to be at center.Moment of loads @ to @ about B = 6,170,OOO.

Shear at center = g - 12.5 = 49,200 lb.

Moments.-To find the maximum bending moment at any point Cthe load has to cbver the whole span, one wheel has to be at C and the

2P/ n

=,J \f > D

A*

Ii-------

LIiIIUIC B

--.-- a ___i_ ____ ---.* .___ -__- __________------------ ------- _ _-----..-.---

FIG. 6.

position of the load has to be so that the load per lin. ft. is as near aspossible the same on both sides of point C and equal to the average loadper lin. ft. on the whole span, or the following conditions must befulfilled (see Fig. 6):

; ZPI < ZP . . . . . . . . . . . . . . . . ..(I)

f (PC + ZPJ > z;P. . . . . . . . . . . . . . (2)

As these conditions can sometimes be fulfilled for different wheels P,at C, it is a matter of trial to find the most unfavorable position with thetrain heading in either direction. The most unfavorable position occursusually, but not always, when one of the drivers is at the point C. An ex-perienced calculator can usually find this position by judgment.


In long simple spans (trusses) it may happen that to produce maxi-mum moment at the center of the span the loading has to advance sofar to the left that the head D of the uniform train load p per lin. ft.passes beyond the center of the span C (Fig. 7). The average load on the

left half of the span zp+cp~31 must then be equal to p and the distance

c between C and D must therefore be

1 ZPc=---2 P

This same position of the load produces maximum moments at all pointsbetween C and D.

When the most unfavorable position of the load has been found themoment is obtained as follows:

(1) Find end reaction R, by taking moments of all loads aboutsupport B and dividing by I, that is:

(2) Take moments 2P~(a--2) of all loads to the left of C about C.The bending moment at C is then

M, = R, a -2P&---2)

The point where the absolute maximum moment pccurs is not knownbeforehand, but for all except very short spans it is accurate enough toassume it at the center of the span and, although the maximum momentsfrom dead and live load do not occur at the same point, it is for practicalpurposes permissible to add the two maxima in order to get the total. Fortables of moments and shears from Cooper’s E-50 loading see page 64.

For short spans it is sufficient to calculate only the maximum shearand bending moment; longer spans are divided into a number of, prefer-ably equal, panels 5 to 10 ft. long and the shears and moments are calcu-lated for the panel points.

Example.-A lOO-ft. single-track deck-plate girder span carries Cooper’s E-50loading one-half of which goes to each girder.

Maximum moment at center:There are several positions of the load which satisfy the conditions on page 46;

by trial it is found that wheel @ placed at the center produces the greatest moment;


to the left of the center are then wheels @ to @ or 22’1 = 127.5 and on the wholespan are wheels @ to @ and 10 lin. ft. of uniform load or ZP = 292.5. Since2 X 127.5 < 292.5 and 2(127.5 + 25) > 292.5 the conditions are fulfilled.

Moment of-loads @ to @ about B = 14,567

Moment of uniform load about B = 2.5 X y = 125

- - -

Total moment about B = 14,692

Moment of left reaction about center =14,692X4 = 7 , 3 4 6Moment of wheels @ to @ about center = 3,323

- - - - -

Bending moment at center zz 4,023,OOO ft.-lb.

Maximum moment at 2 point:Assuming wheel @ at t point, we have iXP1 = 62.5, PC = 25, ZP = 322.5 andsince 4 X 62.5 < 322.5 and 4(62.5 + 25) > 322.5 the conditions are fulfilled.Moment of wheels @ to @ about B = 15 ,051Moment of left reaction about + point = 15,051 X $ = 3,763Moment of wheels @ to @ about 2 point = 600

3,163.OOO ft.-lb,

ART. 6. MAXIMUM MOMENT IN SHORT SPANS WITHOUT FLOORBEAMS

For short spans which carry only a few, say up to four, wheel concen-trations, the following simple rule applies for finding the maximummoment:

A

*tR%’ /

,* ______________+ __-___________ + _________ -__-___ +--- _____----

FIG. 8.

Place the center of the span in the middle between the resultant R ofall the loads on the span and one of the loads next to this resultant. Themaximum moment then occurs under the concentration nearest to thecenter (see Fig. 8). It is a matter of trial to find out how many wheels

FIG. 9.

have to be on the span and which of them has to be nearest the center inorder to produce maximum moment.

For equal concentrations, spaced as shown in Fig. 9, the followingformulas cover most practical cases:


N o . of w h e e l s M a x i m u m e n d s h e a r M a x i m u m m o m e n t U p p e r l i m i t s o f s p a non span f o r m a x i m u m m o m e n t

PZ1 P I g 1.7 a1---____ ~-_____

2 2 5 0.4 a~ + 1.82 0.2

____ __-- _--- -__~

3

Example.-A 20-ft. span carries Cooper’s E-50 loading, one-half of which goesto each of the two girders. For the drivers we have al =a2 =5 ft. and (2.16+1.575)5 =18.7 ft. As this is smaller than 20 ft., four wheels have to be on the span for maximummoment, which is

M = 25,000 [g-5] = 258,000 ft.-lb.c

and occurs at t = 1.25 ft. from the center of the span.

ART. 7. INFLUENCE LINES FOR SPANS WITH FLOORBEAMS

The influence ordinates at the panel points are the same as if therewere no floorbeams.

A load applied between two panel points is transmitted to the latterin inverse proportion to their distance from the load. From this followsthat any influence line must be a straight line in any panel.

4

50 DESIGN OF STEEL BRIDGES CHAP. 1111

The influence line for end reaction is the same as for girders withoutfloorbeams, provided there is a floorbeam over the support.

The influence lines for shear and moment at any point C in panelm-n (Fig. 10 b and c) follow directly from Fig. 1, c and d, page 41.

As a rule it is necessary to calculate the moments at the panel pointsonly and for these the influence lines are the same as for girders withoutfloorbeams, that is, the area is a triangle with a corner below the panelpoint.

ART. 8. MOMENTS AND SHEARS FROM DEAD LOAD IN SPANS WITHFLOORBEAMS

The dead load consists of the weight of the floor (flooring and floorsystem) and of the weight of the main girders. The latter is assumed

FIG. 11.

uniformly distributed over the whole span and can, therefore, be treatedin the same way as shown for girders without floorbeams. The weight ofthe floor is applied to the main girders through the floorbeams and repre-sents therefore a series of concentrated loads called panel loads. Forpractical purposes it is permissible to assume also the weight of the maingirder concentrated at the panel points (the shear in the end panel onlymay have to be slightly corrected).

If w is the total dead load per lin. ft. per main girder and h theaverage length of two adjoining panels, the corresponding panel load isw = WA.

-ART. 8] MOMENTS AND SHEARS IN SIMPLE SPANS 61

The reaction R, is obtained by taking moments of all loads W aboutB and dividing by the span length 1. The reaction RI, is then equal toZW-R,. In case the panel loads are symmetrical about the center ofthe span the reactions are

R, = Rc, = $ZW

The shear line is not a straight line as in the case of a girder withoutfloorbeams (see Fig. 2, page 42), but a stepped-off line (Fig. 11 b) whichintersects the straight line (shown dotted) at the centers of the panels.If ZWL is the sum of all panel loads to the left of a certain panel, theshear in that panel is R a - Z Wl.

If there is an even number of equal panels the shear in the mth Ganelfrom the center of span is V, = W(m--3); if there is an odd number ofequal panels’the shear in the mth panel from the center of span, not count-ing the center panel, is V, = W,. Table 12 gives these shears for apanel load unity.

TABLE 12Shears from Uniform Dead Load in Simple Spans with Floorbeams

Forp=landX=l

Number of panelsin span

-i-1

-

End reactionA

1.52.02.5

E4.04.5

E6.0

-i

o-1

:.52.02.53.03.54.04.55.05 .5

Panel

l - 2 2-3 3-4 4-5 5-6

i.5 :::::::: :::::::: :::::::: ::::::::

1.01.5 :.5 : : : : : : : : : : : : : : : : : : : : : : : :

2.02.5 ::: i.5 : : : : : : : : : : : : : : : :3.0 2 .0 1.0 03.5 1.5 0.5 : : : : : : : :

2::

2

3.5 2 . 0 2 . 5 1.0 1.5 i.5

Note.-For a dead load p in lb. per lin. ft. and pane! length X in ft. multiplythe above shears by p X.

The moment curve is a polygon with its corners below the panel

points and circumscribed by a parabola with the maximum ordinate 2

(Fig. 11 c).As a rule the moments are determined at panel points only and can

therefore be found by the formulas given above. If the panels arealike and the dead load uniform the moments at the panel points canbe found by means of Table S.(page 42). If the panels are different or


the loading not uniform the moments and shears are conveniently deter-mined as follows after the reactions have been found:

SheaI M o m e n t s

VW = R, - W, M l = vo-1. X0-I

Vl-2 = V&l - WI Mz = Ml + 8,s,. x1-z

v2--3 = Vl-2 - wz MS = Mz + ‘V2--3. X2--3, etc.

It is advisable to start this calculation from each end of the span andwork toward the center. As a check the moment and shear at or near thecenter are thus determined twice. The values Vo- 1.X0- I, VI- 2. XI- 2,etc., are called the moment increments.

Example.-A girder of lOO-ft. span with ten equal panels of 10 ft. (Fig. 12) carriesa total dead load of 700 lb. per lin. ft. The panel load is 700 X 10 ft. = 7000 lb.

FIG. 12.

Shears MomentsEnd reaction RO = 7x5 ft. =,+ 35.0

PO = - 3.5

Shear VO- 1 = + 31 .5 x10 ft. = + 315 = MlPI = - 7.0

Shear F’1-2 = + 24 .5 XlOft. = + 2 4 5P, = - 7.0 + 5GO =Mz

Shear I’--3 = + 17 .5 Xloft. =+ 1 7 5-Pa = - 7.0 + 735= Mg

Shear Vs-4 = + 10 .5 x 10 ft. =+ 105P4 = - 7.0 + 840 = Mq

Shear V4- B = + 3.5 xloft.-=+ 3 5-+ 875 = Mg

Shears and moments are given in units of 1000 lb. and 1000 ft.-lb.

ART. 9. MOMENTS AND REACTIONS FROM LIVE LOAD IN SPANS WITHFLOORBEAMS

The whole’live load is carried to the main girders or trusses throughthe floorbeams. The maximum end reaction in case there are floorbeamsand the maximum moments at panel points are determined in exactly the

A R T . 101 MOMENTS AND SHEARS IN SIMPLE SPANS 63

same way as for girders without floorbeams (see page 44) except that thepoint of absolute maximum moment is at the floorbeam at or nearestthe center. In case there are no end floorbeams, the end reaction is equalto the shear in the end panel.

If the panels are alike, Table 9, page 42, can be used for uniform loads.For reactions and moments for Cooper’s E-50 loading see page 64.

ART. 10. SHEARS FROM UNIFORM LIVE LOAD IN SPANS WITHFLOORBEAMS

Exact Method.-To determine the maximum shear V,, in a panel CDwe will examine the influence line (Fig. 13).

-_--- ..___ -_ 1 _-- ____.--_------.- ----

b---- .____..__ L ______.---..---...---.--- ,!

FIG. 13.

The maximum positive shear is equa1 to area of triangle I DI BImultiplied by p,, therefore equal to

where X is the length of the panel CD and b the distanceof the right panelpoint D from the support B. The length L of the load is

X & being the distance of the head of the load beyond panel point C.

If the panels-are equal the shears are best determined by Table 13 orby the following formula. If there are N equal panels in all and n denotesthe number of panels to the right of panel point C we get, by substituting

h f o r X a n d ( n - l ) i f o r b,

PlVcd =. a~(~-+- 1)”


Plor if P = mfl = panel load

vcd = 2(N-1)L(“- 1)”

The length L of the load is

The maximum negative shear is equal to the maximum positive shearin the panel located symmetrically about the center of the span.

TABLE 13Exact Shears from Uniform Live Load In Simple Spans with Floorbeams

.Forp=land~=l

N u m b e r o f Endpanels in

3paTl reaction

2 2.0 1.5

i 2.5 3.0i 4.0 3.5

9 4 . 5

:7 5.0 5.512 6 . 0

O-1 / l-2 1 2-3 1 3-4 / 4-5 1 6-6 j 6-7 / 7-8 / 8-9 / 9-10~10-11~11-12

“.I” u.u,,0 . 3 3 0 . 0 8 o.oo.....0.64 0.29 0.07 0.001.00 0.56 0.25 0.061.39 0.89 0.50 0.221.80 1.25 0.80 0.452.23 1.64 1.14 0.73

N&c-For E L live load p in lb. per lin. ft. and panel length A in ft. multiply the above shearsb y p A. (See Table 14.)

Example.-A truss of GO-ft. span and six panels 25 ft. long carries a uniform liveload of 1200 lb. per lin. ft. The shears in the various panels are obtained bymultiplying the figures in the fourth row of Table 13 by 1200 X25 =30,000.

They are in units of 1000 lb.: End reaction 90. Shears: 75, 48, 27, 12, 3 and 0.

Conventional Method.-Although the exact method is very simplethe following approximate method is sometimes more convenientespecially for the calculation of stresses in trusses with polygonal chords.It is sufficiently accurate for practical purposes and the results are on thesafe side.

The load for maximum shear in panel C-D (Fig. 13) is assumed toextend to the middle of that panel but the floorbeam reaction at C isneglected, in other words there are only full panel loads at and to theright of D.

The general formula for shear then becomes

PbV,d = i@+N

ART. 111 MOMENTS AND SHEARS IN SIMPLE SPANS 5 5

If there are N equal panels and n denotes the number of panels to theright of panel point C and P the panel load, we have,

PV,d = B(n-1)n.

Table 14 gives these shears for a uniform live load of 1 lb. per lin.ft. and a panel length of 1 ft.

TABLE 14

Approximates Shears from Uniform Live Load In Simple Spans with FloorbeamsFor p=l and 1=1

Number ofPanel

panels in Endspan

reaction 1 O-1 j l-2 j 2-3 ( 3-4 1 4-5 ( 5-6 ( 6-7 / 7-8 ) I 8-9 ) Q-10 110-11

3 1.50 1 I0.33 ,,,,,. ,,,,,_ ,..... _..,.. . . . . . . . . . . . . . . . . . . . .4 2.00

::;;0.75 0.25 . . _.,,.. . . . . . . . . . .

5 2.50 2.00 1.20 0.60 0 2 . . . . . . . . . . . ..,... ,..... . . . . . . . . . .

:3.00 2.50 1.67 1.00 0. 0 0.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3.50 3.00 2.14 1.43 y3; 0.43 0.14 . . . . . . . . . . .

8 4.00 3.50 2.63 1.88 . . . . . . . . . . .

1:4.50 4.00 3.11 2.33 1.67

0.75 0.38 0.13 .o.ii. :.....1 . 1 1 0 . 6 7 0 . 3 3

5.00 4.50 3.60 2.80 2.10 1.50 1.00 0.60 0:30 6:ib’ :::::: . .

:;5.50 5.00 4.09 3.27 2.55 1.91 1 . 3 6 0 . 9 1 0.55 0.27 0.09 . . . .6.00 5.50 4.58 3.75 3.00 2.33 1.75 1.25 0.83 0.50 0.25 0.08

No&-For a live load p in Ib. per lin. ft. and panel length h in ft. multiply the above shearbypi.

Example.-For the truss and loading of the previous example we get for the shearsby the approximate method: End reaction 90, Shears: 75, 50, 30, 15, 5 and 0.

ART. 11. SHEARS FROM MOVING CONCENTRATED LOADS IN SPANSWITH FLOORBEAMS

From the influence line (Fig. 13, page 53) follows that for maximumpositive shear in any panel CD the load has to advance from the right end

ZP

r \P, Pz p3 p4 Ps etc.

I i ? I

1 I IA% ic iD aB

%------ n_____:

?f---- _________ ..____.-______.-... 1 J--__ _ --...____..---.__._.-----..----- --------

FIG. 14.

at least so far that the first wheel is at D. If this position is the most un-favorable the maximum shear is equal to the left reaction, and is deter-mined by taking moments of all loads about B and dividing by the spanlength 1.

It may happen, however, that a greater shear is obtained by advanc-ing the load so that the second, third, etc., wheel is at D (Fig. 14). In this


case the shear in panel CD is equal to the left reaction minus the panelconcentration at C from the loads in panel CD. The panel load at C isdetermined by taking moments of all wheel loads ahead of D about Dand dividing by the panel length A.

In order to find out which wheel at D produces the maximum shearthe following conditions have to be examined:

(1) If P1 f . . . . . . . . . . . . > Z,P the jirst wheel

has to be at D

(2) If PI ; < ZIP, b u t (Pl+Pz) ; > ZzP, the second wheel

has to be at D,

(3) If (P,+P,) k < T&P, but (P1+P2+PJ f > &P, the third wheel

has to be a t D, etc., where 2&P, I&P, 23P denotes the sum of al lloads on the span while the first, second, third wheel respectively is at D.

The number of the wheels in the panel CD necessary to produce maxi-mum shear decreases for panels from the left to the right. If, therefore,in a panel CD wheel PI at D gives the maximum shear, the same is truefor all panels to the right of panel CD (see example, page 61). Table 19contains the shears from Cooper’s E-50 loading for spans with uniformpanel length. Table 20 gives the number of the wheel to be placed at theright panel point to produce maximum shear in a panel; the moments ofthe wheel loads in the panel about the right panel point are also given,so that the left panel concentration can be easily found. This table isespecially valuable for the calculation of trusses with polygonal chords.

ART. 12. FLOORBEAM- AND INTERMEDIATE PIER-REACTIONS

Influence Line.-The influence area for the center reaction R, of twoadjoining simple (not continuous) spans AC and CB (Fig. 15) is a triangle

FIG. 15.

AIC2B, with a height CICz equal to load unity. It is composed of the twoinfluence areas AICICz and BICICz for reaction R, from span AC and BC


respectively (see page 41). The greatest reaction occurs therefore whenthe load covers both spans.

Uniform Load.-If the load per lin. ft. on span AC is wa and thaton BC is wb per lin. ft. the reaction R, is

If wa = wb = w, R, = wq’

Wheel Loads.-The criterion for maximum reaction R, is the same asfor moment at C of a simple span AB (see Fig. 6 and conditions onpage 46), since the influence area of that moment is also a triangle withits apex below C (see page 41). After the most unfavorable positionis determined the reaction is calculated for each span separately asshown on page 44.

For tables of floorbeam-and pier-reactions for Cooper’s E-50 loadingsee page 64.

ART. 13. ENGINE DIAGRAMS AND WHEEL-LOAD TABLES

Much work can be spared in the calculations of moments and shearsfrom wheel concentrations by the preparation of tables or engine diagrams.It is advisable to make these the more elaborate the oftener the same kindof loading is used.

Many diagrams and tables are inuse; themost common will be givenhere.

In all calculations of shea.rs and bending moments the main workconsists of finding the static moment M of a series of wheel concentra-tions PI to P, about a fixed point X (Fig. 16).

This moment can be found by the formula

M = M, + x&P,

where M, designates the moment of the wheel loads PI to P, about P,,Z,P the sum of the loads PI to P, and x the distance of P, from X. M,and Z,P are independent of the span length or the location of point X

58 DESIGN OF STEEL BRIDGES [ C HAP. III

and can be determined in advance for any series of wheel loads. Havingthe values M, and Z,P for the different wheels tabulated the work offinding the moment M is greatly reduced. On this basis Tables 15 and 16are made.

Table 15 is very simple and should be prepared if only a few calcula-tions with any particular loading have to be made. The table given hereis prepared for Cooper’s E-50 loading.

TABLE. 15Moments fur Cooper’s E-so Loading in Units of rooo”ft.-lb. for One Rail

Distances in ft.

Number ofwheel

Distanceram preced-og wheel, dn

08

5

5

5

956588

5

08

1 3

18

2 3

it4 34 85 66 4

6 9

7 4

7 98 8

El104

109

Wheel load,Pn

12.52 5

2 5

2 5

2 5

16.2516.2516.2516.2512.52 5

2 5

2 5

2 516.2516.2516.2516.25

. . . . .

E lum of wheeloads, ZJ’

12.537.562.5

87.5

112.5

128.75145.0161.25177.5190.0215.0

240.0

265.0

290.0306.25322.5338.75355.0

. . . .

Moment in-croment

Jn = d,tPr1

0100

188

312

438

1,012644870806

1,4201,520

1,075

1,200

1,3252,6101,5311,9351,694

1,775

nbment,ZIIJ

0100

288

600

1,038

2,0502,6943,5644,3705,7907,310

8,385

9,585

10,91013,52015,05116,98618,680

20,455

The last column is most conveniently calculated by the followingformula, starting with wheel 0:

&Pa = lx,-* Pd + d,Z,-1PFor instance, for wheel @ we have

Z,Pd = I&&l + d&P= 287,500 + 5 x 62,500 = 600,000Example to Illustrate the Use of Table X-What is the maximum end shear

in a span of 20 ft.? We place the first driver over A (Fig. 17), wheel @ is then thelast on the span and its distance from support B is r = 20 + 8 -23 = 5 ft.

Moment of wheel loads @ to @ about @ = Z 5 Pd = 1,037,500 ft.-lb.add ~26 P = 112,500 X 5 ft. = 562,500 ft.-lb.


Moment of wheel loads @ to @ about B = 1,600,OOO ft.-lb.deduct moment of wheel load @ about B=12,500 X28 ft. = 350,000 ft.-lb.

Moment of wheels @ to @ about B = 1,250,OOO ft.-lb.

End shear =1,250,OOO =2. 62,500 lb.

Table 16 is very convenient where many calculations have to be madewith the same loading. It differs from Table 15 in that it contains notonly the moments of wheel loads PI to P, about P,, but the moments ofany series of wheel loads P, to P, about P,.

The first horizontal line gives the summation of the wheel loads fromright to left and the second line the same from left to right. The thirdline contains the numbers of the wheels, the wheel concentrations andthe distances between them. The fourth and fifth lines, respectively,

FIG. 17.

give the summation of the distances from left to right and from right toleft. In the other lines any figure to the left of the heavy stepped-offline gives the moment of all wheel loads between and including the loadto the left of the figure and that over the heavy line, about the latterand any figure to the right of the heavy stepped-off line gives the momentof all wheel loads between and including the load to the right of thefigure and that over the heavy line, about the latter.

Using the same example as on page 58 we find from Table 16 to the right of wheel@ the moment of wheels @ to @ about @ or

@ to @ about @ = 750,000add z 2 P = 100,000 X 5 = 500,000

Moment of all wheels about B = 1,250,OOO lb.

An elaborate table has been worked out for Cooper’s E-50 loadingby Mr. Josiah Gibson (see Eng. News, Vol. 55, No. 25, 1906).

TABLE 16

MQlENTS POR COOPER’S E-50 LOADING

,%?$, 8' _, 5' , 5',, 5'.> 9' _ 5'<- 6' ' 5' , 81

8' 13' 18' 23' _ 32' 3 7 ' 43' 43' 56'__ 8' ._ 5'.* 5'., 5'>/ 9'

64'104' 96' 91' 86' 81'

_ 69'. 74'. 79' 88' _ 98' 99'72' _ 67' _ 61' 56' 4 8 ' 40' 35' 30' , 25' 16' 11' 5' $'$

13630 17380 1498012705 10555 8530 7360 6271 5280 . 4370 3770 2770 1896 1145 520 260 811698 98 8 16 115749/134741118244/ 9299 I7399 1 6310 ) 5303 1 4893*'+j 8564- 3026 ~2151~1401 / 776 / 276

3lomnts giJ;iven for one rail in units of 1000 ft.-lbs.


Exam&e to Illustrate the Use of Table 16.-We will assume a span of 100 ft.length with five panels of 20 ft.

Maximum Reaction-We place wheel @ over A,

A I3______-__----------------

FIG. 18.

Distance between wheels @ and @Distance from wheel @ to B; z = 100 I- 96

==

96 ft.4 ft.

Moment of wheel loads @ to @ about @ = 17,380,000zx:p+-18 = 342,500 X 4 ft. = 1,370,000

-Moment of @ to @ about B = 18,750,OOO ft.-lb.

Reaction = 18,750,OOO : 100 ft. = 187,500 lb.

Maximum Positive Shear in Panel l-2.-We try with wheel @ at 2 and examinethe condition (3) on page 56. (P, -I- P,)5 = 187.5; &P=215; (PI -i- P&Pa)5 =312.5; &P = 240.

Condition (3) being fulfilled the position of the load is correct.Distance between wheels @ and @ = 69 ft.Distance from wheel @ to B, z = 60 + 13 - 69 = 4 ft.Moment of wheels @ to @ about @ = 8,385,OOOZZPMZ = 240 X 4 ft. = 960,000

_ _ _ _Moment of wheels @ to @ about B = 9,345,OOO

Left reaction = ‘9 = 93,450

Moment of wheels @ and @ about wheel @ = 288,000Panel load 1 = ‘T zz 14,400 lb.

Shear in panel, l-2 = 93,450 - 14,400 = 79,050 lb.

Maximum Moment at 2.-Advancing the load from the right until it fully coversthe span, we find that the wheels of the tender are near 2 while the engine driversare near 2’. The moment is therefore evidently greater at 2’ than at 2. We willtherefore calculate the moment at 2’ which is the same as if we would reverse theloading and calculate the moment at 2.

By trial we find that wheel @ over 2’ fulfills the conditions on page 46. On thespan are wheels @ to a,.

Distance of wheel @ from B, x = 40 + 69-104 = 5 ft.Moments of wheels @ to @ about @ = 14,980,OOOX2:pa--18 = 317,500 x 5 = 1,588,OOOMoment of wheels @ to @ about B = 16,568,OOOMoment of R, about panel point 2’ =

6 016,568,OOO X m. = 9,941,OOO

Moment of wheels @ to @ about @ = 5,998,OOOMax. bending moment at panel points 2’ and 2 = 3,943,OOO ft.-lb.


ART. 14. EQUIVALENT UNIFORM LOADS

An accurate calculation of stresses is oft.en not warranted, especiallywhere the work can be simplified by an approximate calculation and inthe case of a preliminary design. This can often be accomplishedby the use of equivalent uniform loads; that is, loads which have the sameeffect on certain static values (shears, moments, etc.) as the specifiedwheel loads. The equivalent uniform loads, however, have to beknown and tabulated, as otherwise there would be no advantage overthe accurate methods with wheel concentrations.

Equivalent Uniform Load for Moments.-The maximum moment

from a uniform load p per bin. ft. in a span is PG. If this has to be

equal to the maximum moment &I, produced by the engine load, theequivalent uniform load is I

The moment curve from this load is a parabola and the moment atany point of the span can be determined as shown on page 42. It differsonly slightly from that produced by the engine loads.

Closer results can be obtained by taking the equivalent uniform loadfor the moment at the quarter point of the span. If the wheel loads pro-duce there the moment Ill, the equivalent uniform load is

32 M,P=tj-F

Equivalent Uniform Load for Reaction.-The equivalent uniformload for the reaction of a span 1 is

where R is the maximum reaction produced by the concentrated loads.Equivalent Uniform Loads for Shears.-For calculating shears with

equivalent uniform loads, the following method gives quite satisfactoryresults. For the shear at any point of a span use that equivalent uniformload which in a span equal to the loaded length produces the same endshear as the wheel loads; in a span without floorbeams use as loadedlength the distance between the support and the point where the shear iswanted; in a span with floorbeams use the distance between the supportand the near end of the panel in which the shear is wanted, but assume afull panel concentration at the latter point in calculating the shear.


Equivalent Uniform Loads for Floorbeam Reaction.-The influencearea for a floorbeam reaction from two equal panels A is a triangle of thelength 2 X and with its apex over the floorbeams. The influence area formoment at the center of a span 2 X is a similar triangle. From thisfollows that the equivalent uniform load is the same in both cases. Invery short spans the maximum moment due to concentrated loads doesnot always occur at the center; the equivalent uniform load for thefloorbeam reaction is then slightly different from that for maximummoment. Table 17 contains the equivalent uniform loads for maximummoment and reaction for Cooper’s E-50 loading.

ART. 16. TABLES OF MOMENTS AND SHEARS

The following tables reduce the calculation of moments and shearsto a minimum, it being only necessary to interpolate for intermediatelengths of span or panel. They are self-explanatory.

TABLE 17Maximum Moments, Shears and Equivalent Uniform Loads Per Rail

For Cooper's E-50 loading and 125,000 lb. on two axles 7 ft. apart-

Endshear

83.785.186.5

88.289.891.492.994.3

96.097.699.2

100.7102.1

103.5104.9106.3107.7109.0

111.8114.5117.2119.8122.5

125.2128.2131.2134.8138.1

141.7145.3148.8152.1155.3

1158.61161.8

T-

I

-

Endshear

Cquival. unif.load

597.4 116.7625.5 119.4653.7 122.0

6S5.8 124.4717.9 126.9750.0 129.7783.3 132.3819.5 135.0

855.8 137.6892.0 140.2928.3 142.9964.5 145.6

1,OOl.O 148.3

1,037.o 150.91,073.o 153.41,110.o 156.01,149.0 158.51,189.0 161.0

1,269.0 166.61,351.0 172.51,440.o 178.51,529.0 185.11,624.0 191.5

1,720.o 197.71,819.0 203.61,924.0 209.72,029.O 215.62,134.0 221.3

2,240.O 226.92,350.O 232.42,465.0 238.02,581.0 243.32,700.O 248.6

2,820.O 253.62,946.0 258.7

3,075.O 263.23,205.O 268.33,337.0 273.2

3,471.0 278.03,606.O 282.73,743.0 287.53,882.0 292.04,024.4 296.5

4.185.0 300.84,342.0 305.34.502.0 309.84,672.0 314.24,857.0 318.5

5,035.O 322.85.215.0 327.05,400.O 331.25,580.O 335.75,770.o 340.1

6.245.0 351.2

/for endshear

5,0755,005

4,3904,330

4,945 4,2704,890 4,235 924,855 4,195 944,810 4,150 964,765 4,1204,715 4,100 1%

4,6854,6504,6054,5754,540

4,5004,4654,4404,3904,360

4,0754,045 :;9

4,0153,985 :::3,955 110

3,9253,885 ::4"

3,8553,830 ::i3,805 120

4,3004,2404,1854,1304,085

3,7553,705 :;i?

3,6753,640 23,610 145

4,0404,0063,9753,9653,945

3,9353,9253,9153,9003,885

3,5803,555 ::i

3,5353,515 :z3,485 190

3,455 2003,435 2253,415 2503,395 2753,375 300

3,8703,850

3,355 350

-L3,340 400

-

For end f o rnoment

3,840 3,3303,825 3,3103,810 3,295

3,800 3,2803,785 3,2653,770 3,2503,760 3,2353,750 3,220

$9;;:3:7103,6953,680

2%3:2053,2003,2103,210

3,6703,6553,6453,6303,620

3,2103,2103,2053,205

3,585 3,2003,560 3,1903,530 3,1803,505 3,1653,480 3,155

3,4553,4103,3703,3303,295

I3,140

%:3103;3,005

3,265 2,9703,190 2,8903,130 2,8153,080 2,7603,040 2,715

2,965 2,6602,895 2,620

-

-

I 2 Axles 165.1168.4171.5

174.7178.0181.0184.3187.5

190.6193.6196.6199.5202.5

205.5208.4211.3214.2217.1

224.2231.4238.4245.4252.3

259.2272.8286.3299.7313.0

326.3359.0391.5423.7455.8

519.6583.1

35.138.240.642.645.245.6

E-50loadir

...............

...............

...............

. ,

34.4 50.0 43.7 8,62536.1 58.7 47.2 8,00037.5 70.3 50.0 7,500

40.9 82.0 54.543.8 100.0 58.446.2 118.8 61.648.2 137.5 65.250.0 156.3 68.3

7%7:1106,8906,670

53.155.958.360.562.5

175.0 71.1193.8 73.5

X %:E257.8 81.9

6,640 5,4706,580 5,3656,480 5,2456,370 5,1706,250 5,155

64.365.9

%:271.0

282.4 84.9307.1 87.6331.8 90.4356.5 92.4381.3 94.6

6,1255,990

E%5:680

ES456.91484.91513.1

97.1100.1

E:,"107.9

::;:t

5,585 4,8055,490 4,7305,395 4,6605,305 4,6155,255 4,560

80.5 541.282.1 569.3i

5,1955,130

6,2505,8005,625

5,4205,5655,6205,6105,555

2%5:0204,9504,880

4,5054,450

-!- -

ART. 1’51 MOMENTS AND 8HEARS TN SIMPLE SPANS 65

TABLE 18

Intermediate,Reaction and Equivalent Uniform Load Per Rail for Two Adjoining(not continuous) SpansFor Cooper’s E-50 loading

-Spans, ft. Interm. R~K&I in units of Eq. unif. load in lb. per lin. ft.

3 0 azd 6 0

;; “ T::4 0 I‘ 8 04 0 “ 9 04 0 (‘ 10050 “ 8 05 0 ‘I5 0 Ii 1:::

149.3167.0181.9197.1210.4221.1210.1223.1233.8

3,3163,3403,3073,2853.237311593,2323,1863,117

TABLE 19

Shears in Spans with Floorbeams and Equal PanelsFor Cooper’s E-50 loading in units of 1000 lb. for one rail

P a n e / P o i n t sT 1 I 1 I I

0 I 2 3 4 5 6 e+‘. ‘

N o . o f 2 ~Length of panel in feet

panels 2 8 ( 9 / 10 / 11 ( 12 13( ( 14 / 15 / 16 / 17 1 18 j 19 20 21/ 1

O-l 54.0 59.0 63.0 67.5 71.5 75.5 79.5 84.0 87.5 90.5 94.5 98.OlOZ.OlOG.54 ;; 2;:: 273.;

.3j.i 3%; 34.5 37.0 39.0 41.0 43.0 44.5 46.5 4 0 . 5 5 1 . 5 5 3 . 0

7 . 5 8 . 5 10.0 10.5 11.5 12.5 13.5 1 4 . 5 1 5 . 0 1 5 . 5__-__- --____-__~--

66 DESIGN OF STEEL BRIDGES [CH’AP. 111

TABLE 19

Shears in Spans with Floorbeams and Equal Panels (Continued)-

No. ofpanels I’&

I 22 / 23 / 24 1 25 1 2 G 1 2,“;: i’27’ 3 0 1 31 / 32 / 33 / 34 1 3 5

4

6

6

9


TABLE 20Table giving Number of Wheel to be placed at Panel Point of any Panel 112-n

to Produce Maximum Shear in that Panel-Cooper’s E-50 Loading

Wheel at panel point n

No. of 2 _ _p a n e l s $

I I22

I I

Moment of wheel loads in panelm-n about n in units of lOOOft.-lb.

m n per rail

Length of panel in feet

2423 25 1 26 1 27 / 28 29 30

o - 1 4l-2 3

6 Z-3 33-4 24-5 2

--~o - 1 4l - 2 3

7 2 - 3 33 - 4 34 - 5 25 - 6 2

O-l 3l - 2 32 - 3 3

8 3 - 4 34 - 5 25 - 6 26-7 2

O&l 3l-2 3

o - 1 31-2 32 - 3 33 - 4 3

10 4 - 5 3

;I: ;7 - 8 28-Q 1

TABLE 21Shears in Girders without Floorbeams for Cooper’s E-go Loading in Units of

IOOO lb. for One Rail -

-

-

Span in ft.Shear at 4

pointVn

37.550.0

V.t-e

Shear atcenterv.

v,V,

0.600.630.620.610.610.600.580.570.580.580.580.59

0.280.280.290.290.280.28

2 0 62.5

ii 78.8 94.5i:: 122.5 109.0

i:: 138.1 155.3

l”o; 1 171.5 187.5

22.127.531.134.838.442.145.949.252.255.659.4

58.5

ET82.489.698.4

108.2117.4126.3135.3

0 . 2 70.270.260.260.260.26

110 202.5120 217.1130 231.4

TABLE 22Bending Moments at Equidistant Panel Pdints

For Cooper's E-50 loading in units of 1000 ft.-lb. for one rail.

Panel Poinfs~O I 2 3 4 5 etc

Length of panel in feetNo. of Panel

panels point 8 / 9 ) 10 ~ 11 ; 12 1 13 1 14 1 15 ) 16 / 17 ( 18 j 19 / 20 / 21I 1

982.5 1116.0 1256.0 1401.0 1,552.0 1,709.O 1,871.0 2,061.O 2,273.09';::: 1%: 1263 0 1'440.0 1'617.0 1'826.0 2036.0 2246.0 2463.0/ . j , . 1 , j + j 9 1 1 ~ . 1 8 1 2s6j 2,941.O

0 2,010.O 2,242.0 2,477.0 2,733.0 3,000.00 2.890.0 3.212.0 3.562.0 3.943.0 4,347.05 f I 662:ol 794:ol 930.0 I 1074.0 ’ I 1226.0 ’ I 1396.0’ I 1581.0’ 1788. ’964 0 1,148 0 1,361 0 1,575 0 1,795 0 2,048 0 2,310 0 I 2,599

kcen

TABLE22

Bending Moments at Equidistant Panel Points (Continued)

Length of panel in feetNo. of Panel

panels point 22 1 23 / 24 / 25 1 26 1 /27 28 / 29 j 30 / 31 / 32 / 33 1 34 1 35 g

II 1 I I ii

6 15092.0 I’ 5422.0 I’ 5760.0 /’ ’ ’ ’ 7.228.07,492 0 7,984 0 8,482 Cl

6112.0 18,986 0

6476.0 /9,496 0

6848.0 /1'&010 0 10,590

5,100.o 5,512.08.172.0 8,842.09326.0 10,610.O

5,829.O9,640.O

11,980.O12,790.OI

6,300.O10,430.o12,980.O13,800.O

8,887.014.780.018,430.O19,210.O

10

-__-~~~X7248.0 7,836.0 8,447.0 9,078.O 9.730.0 10,400.O 11,090.O 11,810.O 12,540.O 13,300.O 14,060.O 14,870.O11'520.0 12'500.0 14120.0 15180.0 16280.0 17410.0 17,990.O 19,200.O 20,450.O 21,740.O 23,072.O 24,440.O 25,840.O14'930.0 16'230.0 17'600.0 l9'010.0 20'460.0 21'950.0 23.490.0 25,070.O 26,690.O 20,4GO.O 30,180.O 31,930.O 33,720.O17'090.0 18'530.0 20'020.0 21'540.0 23'100.0 24;730.0 26,390.O 28.090.0 29.880.0 31,720.O 33,GlO.O 35,560.O 37.560.017:520.0 18:920.0 20:410.0 21;950.0 23;540.0 25,190.O 26,900.O 28,660.O 30,500.O 32,410.O 34,380.O 36,410.O 38,500.0

15,700.o27,300.O35,460.O39,620.O40660.0


TABLE 23Maximum Moments and Shears for Electric Cars

In units of 1000 ft.-lb. and 1000 lb. respectively, for one rail. Each wheel load is 10,000 lb.

Loading

-~

Span in feel

-

-

-

o-s” tT’ .‘o.,3-k,’End-shear

15.015.515.816.116.416.716.917.117.217.417.5

:;::

:::9”18.0

. .

. . .

. .

. .

. . .

Momen

28.132.837.642.447.252.157.061.866.771.6

E:186.;91.496.3

101.3106.2111.2116.1121.1126.0131 .o136.0140.9145.9150.9155.9160.8165.8

Znd-shear

14.014.515.015.415.716.016.216.516.716.8

:;::

2217.517.617.717.817.817.918.0

i-

MomenIt 1-

-

End-shear

13.013.614.214.615.015.315.6

2::16.316.516.7

:E17.1

:::;17.417.517.617.7

25.027.530.134.739.444.148.853.658.463.268.172.977.882.787.6

Z102.3107.2112.1117.0

IEnd-shear

10.010.911.712.312.813.313.714.114.414.715.015.215.415.615.816.016.216.316.416.616.7

25.027.530.032.535.037.540.042.547.051.656.261.0

Ki75.280.084.889.794.499.3

104.2

ART. 151 lMOMENTS AND SHEARS IN SIMPLE SPANS 7 1

TABLE 24

Maximum Moments and Shears for Electric CarsIn units of 1000 ft.-lb. and 1000 lb. respectively, for one rail. Each wheel load is 10,000 lb.

I-, I1MU. Beam-

mom.I 1

reactionE n d - Max.

/ I

Beam-Ihear m o m . reaction

- -

SpaIlin feet

End-shear:eam-action

-

-

End- Max.s h e a r m o m . re

-L LI

. .

. . . .

” 16.5,,,.... $ 1 7 . 5

18.5B19.5

4 ;y::

p:;

‘8

$*

3......

; . . . . .

3 . . . . . .

: 8 3 . 591.098.5

:tE121:o

Et!149.0158.5168.0178.0187.5197.5

“2E227.0236.5246.5256.5266.0276.0286.0295.5

?E::325.5

El::355.0365.0375.0384.5394.5404.5414.5424.5434.5

%Ei

24.025.025.526.5

E28.028.5

;::t29.530.0

,.....,.....

,.....

. . . . . .. . . . . .. . . . . .. . . . . .17.518.018.5

:E20.021.021.522.0

%24.024.525.025.525.526.026.527.0

3’7::28.028.028.528.529.029.029.529.530.030.0

)17.518.0

ES!

“2:::22.022.523.524.024.525.026.026.026.527.027.528.028.028.529.029.029.. 530.030.0

,,....

. . . . .128.0135.0142.5

:E::166.0

::Ei195.0204.5

/ 214.0; 223.5: 233.5

243.0‘253.0

262.5272.0282.0291.5301.5311.5

:::::i:i20.021.021.622.0

;i::24.024.5

E::26.026.026.527.027.027.5

I ..............

18.0 . . . . . . . .

! ::::. . . . . . . .. . . . . . . .

:19.5 . . . . . . . .j19.5 . . . . . . . .: ;;:g . . . . . . . .

. . . . . . . .21.0 . . . . . . . .22.0 . . . . . . . .

222.5 . . . . . . . .:23.0 . . . . . . . .123.5 . . . . . . . .L24.0 2.. . . . .

“~2I . . . . . .

y;“5 ff pi

25.5 + 194.526.0 ‘t 203.026.5 m212.526.5 $222.02 7 . 0 2231.52 7 . 5 B 2 4 1 . 527.528.0 * %::20.8 270.0

72 DESIGN OF STEEL BRIDGES [ C H A P . 111

TABLE 24Maximum Moments and Shears for Electric Cars (Continued)

In units of 1000 ft.-lb. and 1000 lb. respectively, for one rail. Each wheel load is 10,000 lb.-

End shexr

2 8 . 52 9 . 0

E3 1 . 03 1 . 53 2 . 03 2 . 53 3 . 53 4 . 03 4 . 53 5 . 03 5 . 53G.O3 6 . 53 7 . 03 7 . 53 8 . 0

2:;3 9 . 039.5 ,4 0 . 04 0 . 04 0 . 54 1 . 04 1 . 04 1 . 54 2 . 04 2 . 54 3 . 04 3 . 54 4 . 04 4 . 54 5 . 04 5 . 54 6 . 04 6 . 54 7 . 04 7 . 54 8 . 04 8 . 54 9 . 04 9 . 04 9 . 55 0 . 0

CD.B.2 .

. . . . . . ..___.

. . . . . . . .

. . . . . . ..___.

. . . . . . . . . . .

. . . . . . . . . . . .

“7 2”;:;5 ZEI$ 2:” 3 2 . 0

k E:i

: i:::“? 3 7 . 0

B %:iz

1 2:::.A 4 2 . 06 4 3 . 0

4 4 . 0m 4 5 . 0. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .

-

End-shear

.!i8f 2:::

4 1 . 54 2 . 04 2 . 04 2 . 54 2 . 54 3 . 04 3 . 04 3 . 54 3 . 5

t 2::” 4 4 . 5g 4 5 . 0m

i-

i

M a x . m o m .

4 9 0 . 05 0 5 . 05 2 0 . 0

-‘i 5 3 5 . 0I 5 5 0 . 0e 5 6 5 . 0+ 5 8 0 . 0

+ 6 4 2 . 52 6 6 0 . 08 6 7 7 . 5

I-

;!I

Be.WIl-reaction

. . . . . . . . .. . . . . . . . .

10. . . . . .

4 ““”

$ iiLO”’2 9 . 5

+ 3 9 . 5t 4 0 . 0

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .

- -

SPZUIin feet

CHAPTER IV

STRESSES IN SIMPLE TRUSSES

ART. 1. GENERAL

In analyzing the stresses in trusses the following general assumptionsare made:

1. The stress is acting in the axis of the member.2. The axes of all members meeting at any joint intersect in a single

point.3. The members are free to turn around this point.4. The external forces are applied at the joints only.The stresses obtained under these assumptions are called the primary

or main stresses. They are either tension (positive) or compression(negative). The stresses resulting from the non-fulfillment of any of the.above assumptions are called secondary stresses. They are bendingstresses and are generally of less importance. The main stresses only willbe considered here. For a detailed discussion of secondary stresses seeVol. I.

Assuming a section through a truss (Fig. 1) acted upon by externalforces and removing the right part from the left, the latter is held inequilibrium if we apply to each cut member an external force equal to thestress in that member.

If it is possible to determine’ these forces by means of the laws ofequilibrium, the truss is called statically determinate with respect to theinternal forces. This will, but not exclusively, be the case when it ispossible to cut each member by a section which cuts only two other mem-bers. We will for the present consider only trusses of this kind.

If the internal forces cannot be determined by the equations ofequilibrium the truss is statically indeterminate, although it may bestatically determinate with respect to the reactions (page 32). Inorder that a truss which is statically determinate with respect to the re-actions be so with respect to the internal forces, the following conditionmust be fulfilled:

2j = mf3

where m = number of truss members and j = number of joints. This73

74 DESIGN OF STEEL BRIDGES [CHAP. 1V

can be proved as follows: Assume successively each joint separated fromthe remaining truss and the stresses acting at that joint applied asexternal forces, these must then be in equilibrium with the load acting atthat joint, that is, the sum of the vertical components and also the sum of

‘ the horizontal components of all forces at that joint must be zero. Eachjoint furnishes two such equations and there are 2j equations for thewhole truss. There are 3 unknown values for the reactions (page 33)and m unknown stresses in the members of the truss or (m+3) un-known values in all, which number must be equal to the number 2j ofequations. If 2j > m+3 the truss is unstable. If 2i < m+3 thetruss is statically indeterminate with respect to the internal stressesand these can only be determined by means of the theory of elasticdeformation. There are, however, statically indeterminate trusses (forinstance, those with multiple web systems) which are usually treated asstatically determinate systems by making simplifying assumptions.

ART. 2. METHODS OF CALCULATION

The usual methods of calculation comprise two principal operations:(1) the calculation of bending moments and shears,(2) the calculation of the stresses in the truss members from the

moments and shears.The first operation is fully discussed in Chapter III, the truss

being treated as a simple girder with floorbeams at the panel points.For this operation the analytical solution is generally preferable.

For the second operation there are principally three methods, eachbeing particularly adapted to certain truss members.

(1) Method of Moments.-If we assume a section through the truss(Fig. 1) cutting three members U, L and D (generally one member ofeach chord and a web member) and remove the right part of the trussfrom the left, the latter is held in equilibrium by the external forcesacting to the left of the section and the stresses in the three cut membersapplied as external forces S,, & and Sd; the static moment of all theseforces about any point must therefore be zero. If we take for this pointthe intersection C of two members D and L, for instance, the lever armsand therefore the moments of Sd and & are zero and the stress S,in the third member is equal and of opposite sign to the static momentof the external forces (bending moment at C) divided by the lever armr,, of U. The intersection point C of the two members D and L iscalled the center of moments for the third member U,

The bending moment at any point between the supports is positive,

A R T . 21 STRESSES IN SIMPLE TRUSSES 75

therefore the moment of the stress in any chord member about its centerof moments must be negative, or anti-clockwise. By inspection of Fig. 1it will be seen that, therefore, in simple spans and under vertical forcesthe top chord members act always in compression and the bottom chordmembers always in tension.

This method of calculating the stresses is convenient for chord mem-bers because the center of moments of a chord member is an oppositepanel point. It can also be used to advantage for web members whosecenter of moments is located near the truss. In most cases, however,center of moments of a web member is located too far outside oftruss, and it is then more convenient to use the following method:

thethe

(2) Method of Shears.-Since we assume only vertical external forcesthe resultant of the three stresses U, D and L (Fig. 1) must also be ver-tical and equal but of opposite sign to the shear (which is the sum ofthe external forces, including reaction, acting on the left part of thetruss). In order to determine the stress in the web member D it isnecessary to find first the stress in one of the chord members for theloading causing greatest stress D and then resolve the shear into thethree stresses U, L and D. This may be performed analytically butmore conveniently graphically.

This method is correct for the dead load stresses and also for the liveload stresses in trusses with parallel chords. In trusses with polygonalchords the position of the live load which causes the maximum shear in acertain panel may be slightly different from the position which causesmaximum stress in the web member of that panel. For practical pur-poses, however, it is sufficient to assume the position for maximumshear.

76 DESIGN OF STEEL BRIDGES [CHAP. Iv

The sign of the stress in a web member can easily be found by inspec-tion. It is, however, convenient to bear the following rules in mind.Assuming that the center of moments of each diagonal is located outsideof the span length (which is the case in the common types of bridgetrusses) any diagonal falling to the right is stressed in tension by loadsto the right and in compression by loads to the left of the panel in whichthat diagonal is located; correspondingly, diagonals falling to the leftare stressed in tension by loads to the left and in compression by loadsto the right of the panel in question. In trusses symmetrical andsymmetrically loaded about the center of the span the diagonals fallingtoward the center are in tension and those rising toward the center arein compression.

(3) Method of Equilibrium at Joints.-Any joint separated by a see-tion from the remaining truss is held in equilibrium by the external forces

53acting at that joint and the stresses in the

5 cut members applied as external forces (Fig.1

Ag

2). If all but two of the stresses are knownthe latter can be determined by a simple

8

4-

2 4

OF

force polygon or analytically by considering

3 - 5that the sum of the vertical components

E3-4 and that of the horizontal components of

\ -3 all forces must be zero. The Maxwell dia-

F I G . 2. grams are based on this method (see Vol. I)and for the dead load stresses in complicated

bridge trusses such a diagram may be most convenient; but for ordinarytrusses methods (1) and (2) are preferable, also for the dead load stresses.For certain truss members, however, method (3) is most convenient.Such members are, for instance, hanger l-2 and post 3-4 (Fig. 2). Thestress in the inclined end post O-l is best obtained from the stress inO-2, their horizontal components being alike. Further, method (3) isconvenient to check certain stresses.

ART. 3. TRUSSES WITH PARALLEL CHORDS

Figs. 3 a and b represent the most common types of trusses with par-allel chords. In the Pratt truss (Fig. 3 a) the verticals or hangers UlLland in the Warren truss (Fig. 3 b) all verticals are stressed only by thepanel load which they carry and form, therefore, no essential part of thetruss.

Chord Members.-Applying the method of moments we find thatthe stress in any chord member of a truss with parallel chords is equal to the


bending moment at its center of moments divided by the depth of the truss.The top chord members are in compression; the bottom chord membersin tension.

For instance, the stresses in members UzU3 and L3Lb of truss Fig. 3 a

are UzU3 = ---%j and LJi4 = +F, where Ms is the bending moment

at panel point 3.In truss (Fig. 3 b) chord members UIUz and UzUa have the same

Mzcenter of moments Lz, the stress in each is -p.h

For maximum chord stress from live load the bending moment at thecorresponding center of moments must be a maximum and therefore the

U, u2 U3 U4 u3 Ul u:

Pratt Truss.

Warren Truss.FIG. 3.

criterions for maximum chord stress are the same as given on pages 41,46, 49, 52, etc., for moments.

Diagonals.-Applying the method of shears (page 75) we find thatthe stress in a diagonal of a truss with parallel chords is equal to the shearin the same panel multiplied by the secant of the angle OL between thediagonal and the vertical, since. the vertical component of the stress ineach chord cut by any section is zero. From this follows that the maxi-mum and minimum stress from live load in a diagonal is obtained byloading the span in exactly the same way as for determining the maxi-mum and minimum shear respectively for girders with floorbeams (seepage 55). This is correct, whether the load is applied along the topchord or bottom chord or both. The diagonals falling toward the re-sultant of the loads (for symmetrical loading the center of the span) arein tension; those rising are in compression For instance, the maximum


tension in diagonal U&S (Fig. 3 a) is caused by a live load extendingfrom the right end B to or somewhat beyond panel point L3 and themaximum compression is caused by a live load extending from supportA to’ or somewhat beyond panel point Lz.

Verticals.-In order to find the stress in a vertical of a Pratt truss, forinstance U&z (Fig. 4), we have to assume a section XX cutting thatvertical and the two chord members UlUz and L,L,. Let us assume forthe present that the whole load is applied only at the panel points of one

u,x\

chord (loaded chord) and none at the otheror unloaded chord. Since the vertical com-

+$------r- ponents of the chord stresses are zero, the$l,\ j

‘\ 1stress in the vertical U2Lz is equal to the

\ shear in panel 2-3 if the bottom chord is

A&

\r

k’ iL2 I-- --

loaded (through bridge), and equal to theshear in panel l-2 if the top chord is loaded

FIG. 4. (deck bridge). Considering equilibrium atthe joints Uz and L, we find also that the

stress in vertical UZLZ is equal to the vertical component of the diago-nal stress U,Ls if the bottom chord is loaded and equal to the verticalcomponent of diagonal stress UIL, if the top chord is loaded.

Expressed in a general way the stress in a vertical of a Pratt trusswith parallel chords is equal to the shear in that panel in which the sec-tion cutting the vertical cuts the loaded chord and is a maximum whenthat shear is a maximum; or, the stress in the vertical is equal to thevertical component of the stress in the diagonal which connects to thevertical at the unloaded chord, and is always of opposite sign to thestress in that diagonal.

For instance, in a through bridge the maximum compression stress inUZLZ (Fig. 4) from live load is caused if the load extends from B intopanel 2-3 and the maximum tension (providing there are no counter-diagonals) is caused when the load extends from A into panel 2-3. I fthe live load were on top it would have to extend from B and A,respectively, into panel l-2.

The stresses thus determined are correct for the live load but must becorrected for the dead load since part of this is applied along the unloadedchord. If the whole dead load has first been assumed at the bottom thecompression stress in vertical U&z must be corrected by adding acompression stress equal to the dead-load concentration at Uz; if theload had first been assumed at the top the correction would be a tensionstress equal to the dead-load concentration at Lz.

ART. 41 STRESSES IN SIMPLE TRUSSES 79

Counter-diagonals.-If the diagonals can take only tension (eyebardiagonals), counter-diagonals are provided in all those panels in which themain diagonals would be in compression when the live load counteractsthe tension from the dead load. When the counter-diagonal acts it hasa compression stress from the dead load equal to the dead-load tensionstress in the main diagonal. The maximum live-load tension in thecounter-diagonal is equal to the maximum live-load compression whichwould occur in the main diagonal if there were no counter-diagonal(see example, page 80).

ART. 4. WARREN TRUSS WITHOUT VERTICALS

In the foregoing analysis of stresses it has been assumed that thejoints of the unloaded chord are vertically opposite the panel points ofthe loaded chord. This is not the case in a Warren truss without verticals(Fig. 5). After the dead loadhas been properly distributed to

,3 5

the joints of both chords, thedead-load stresses are determinedin the same manner as if there were !4, , aa vertical at each joint, the part i -T---b---q

between two such verticals form-

P<O

I I I2 M3

ing a panel. For the live load, I M4I

however, the part between twojoints of the loaded chord must beconsidered as a panel, since the

,

FIG. 5.

load is applied only at these points. The center of moments of anymember of the loaded chord is located between two panel points and theinfluence area for the moment at that point is a quadrangle as shownin Fig. 10 c (page 49) with the corners below the panel points of theloaded chord. The criterion for maximum moment given on page 46does not apply in this case. For practical purposes it is sufficient toassume the line of maximum moments straight between two panel pointsof the loaded chord. If, for instance, Fig. 5 represents the truss of athrough bridge and Mz and M4 are the maximum’bending moments atpanel points 2 and 4 respectively, the maximum bending moment atjoint 3 would be with sufficient accuracy (or even correctly for uniformlive load)

M,=M,+(MrMz)$j

80 DESIGN OF STEEL BRIDGES [CHAP. IV

where a and b are the horizontal distances of joint 3 from panel points 2and 4, respectively.

The vertical component of the maximum live ‘load stress in anydiagonal is equal to the maximum shear in the panel of the loaded chordin which the diagonal is located. For instance, if the bottom chord is,loaded, the maximum stresses in diagonals 2-3 and 3-4 (Fig. 5) are ob-tained from the maximum shear in panel 2-4; if the top chord were loaded,the stress in 2-3 would be found from the shear in panel l-3 and that indiagonal 3-4 from the shear in panel 375 (see example, page 83).

ART. 6. INFLUENCE LINES FOR TRUSSES WITH PARALLEL CHORDS

The influence line for the stress in a chord member is similar to thatof the bending moment at the respective center of moments and. isobtained by dividing the ordinates of the latter by the height h of thetruss.

The influence line for the stress in a diagonal is similar to that of theshear in the same panel and is obtained by multiplying the ordinates ofthe latter by the ratio of the length of the diagonal to the height h ofthe truss (Fig. 10, page 49).

The influence line for a vertical in a Pratt truss is identical withthe influence line for shear in that panel of the loaded chord which is cutby a section cutting the vertical.

ART. 6. EXAMPLES FOR TRUSSES WITH PARALLEL CHORDS

First Example.-Fig. 6 represents a single track through truss span144 ft. long. The bridge shall carry Cooper’s E-50 loading. . All di-agonals will for the present be assumed stiff (no counters).

Dead Load Stresses.[ T r a c k . 400

Assumed dead load 1F l o o r s y s t e m . 480Truss and bracing. 1120

Total 2000 lb. per lin. ft. of bridge

ART 41 STRESSES IN SIMPLE TRUSSES 81

Panel loads: Upper chord11207 X 24 ft. = 6,700 lb. per truss

Lower chord 720 X 24 ft. 17,300 ‘( ” ”=

Total 24,000 lb. per trussEnd reaction 24,000 X 3 = 72,000 lb.

Diagonals (sea 01 = 1.320) Vert ica ls

M e m b e r stress M e m b e r stress

LOU1 60X1320=-79.2 UlLl (Lowrr Ijmel cont.) +17.3UlLE 3 6 X 1 3 2 0 = + 4 7 . 5 UZLZ -12--F.7= -18 .7U2L3 12X1320=+15.8 u&L3 (Upper panel eonc.) - 6.7

Live Load Stresses.-The maximum shears and moments are takenfrom Table 19, page 66 and Table 22, page 68.

Members located symmetrically have same * stress, f. i., UzL3 =+ 100 and - 47.

Stress in U3L3 is zero; stress in UILl is equal to floorbeam reaction= + 92,500 lb.

Max. end reaction = 251,000 lb. if there are end floorbeams.Max. end reaction = Vo-1 = 196,000 lb. if there are no end floor-

beams.Impact Stresses.--These shall be determined according to the

formula

i = (L*o)l

6

82 DESIGN OF STEEL BRIDGES [ C H A P . IV

(see specifications in Appendix). It is sufficient to consider only full ’panel lengths for the loaded length L.

Chord members Web members

212 --I74 . . . . . . .

++‘E.,._..,... .

lJZL3 tJZL2 --Gl;!$?I - 4 0 ‘2

- 1 2 .Y?. r!“‘.

The impact stress in hanger UILl (loaded length 48 ft.) is + 80,000 lb.Total Stresses.-In all chord members, inclined end post and hanger

the maximum stress is equal to the sum of dead, live and impact stressesand the minimum stress is equal to the dead load stress.

The specifications require that wherever the live and dead loadstresses are of opposite character only two-thirds of the dead loadstress shall be considered. Further, in members subject to alternatestresses in immediate succession during the passage of one train eachstress shall be increased by 50 per cent. of the smaller.

We have therefore

Diagonal UIL~ Max. stress + 47+172+130 = +349Min. stress +$ 47- 13 - 12 = -+ 7

Diagonal UZLZ Max. stress + 16+100+ 81 = $197, + + 76 = f235Min.stress +Q 16- 47- 40=- 76,--3 76= - 1 1 4

Vertical U&P Max. stress - 19- 76- 61=-156,++ 54= - 1 8 3Min. stress -$ 19+ 36f 31 =+ 54,+3 54= + 81

All stresses above are given in units of 1000 lb. per sq. in.It will now be assumed that the diagonals consist of eyebars unable

to resist compression. There must therefore be counter-diagonals in thetwo center panels. The counter is supposed to take stress as soon as thedead load tension in the main diagonal is fully counteracted by thecompression due to live load and impact. The total minimum stress inthe main diagonal is therefore zero. The maximum tension in thecounter L2U3 is $47,000 from live load and +40,000 from impact andmust be combined with the dead load compression of - 15,800 X, 8 =- 11,000; so that the total maximum stress is +76,000; the minimumstress is zero.

The minimum compression in verticals ULz and U3L3 is equal tothe dead panel load at Uz and Ua respectively, therefore equal -6700,since the minimum stress in’any diagonal connecting at the top of these


verticals is zero. The maximum compression stress in the verOica1U3Ls occurs at the same time with the maximum tension in the counterand since the dead load stress in the counter is compression that in thevertical is tension equal to + 12( = shear in panel 2-3) - 6.7 = + 5,300 lb.

live load stress (= max. shear in panel 2-3) -36,000 ”impact stress (= max. shear in panel 2-3) -31,000 L(

Total stress considering g of dead load stress - 63,500 lb.Second Example.-Fig. 7 represents a Warren truss without vert-

icals of a highway through bridge 80 ft. long carrying a dead load of350 lb. and a live load of 700 lb. both per lin. ft. per truss.

F I G. 7 .

Dead Load Stresses.-Panel loads: Upper chord 50 lb. X 16 ft. =800 lb. per truss. Lower chord 300 lb. X 16 ft. = 4800 lb. per truss.

End reaetion 23 X 800 + 23 X 4800 = 14,000 lb.

Panel Shear V 1 B o n d i n g I PanelChords

Momentpoint

Member/ s t r e s s =M:8

o - 1 14.0--2.4=11.6 I1-2 11.6-0.8=10.8 ;g:; g:2"

5

G; 5;:::

2-3 10.8-4.8= 6.0 48.0 227.2 Z-43-4 6.0-0.8= 5.2 41.6 268.8 3-5 5;;:;4-5 5.2-4.8= 0.4 3.2 272.0 ‘i 4-4' +34.0

Diagonals (stress = 1.4148)

Live Load Stresses.-The stresses in the top dhord members aretwice as much as the corresponding stresses from dead load (proportionof live to dead load = 2 : 1). The moments at I,3 and 5 are averagesof those at 0 and 2, 2 and 4, 4 and 4’ respectively, therefore the stress inO-2 is one-half of that in l-3, the stress in 2-4 is the average of the

84 DESIGN OF STEEL BRIDGES [CHAP IV

stresses in l-3 and 3-5 and the stress in 4-4’ is equal to that in 3-5.The shears in panels O-2, 2-4, etc., are taken from Table 14, page 55,and multiplied by 700 X 16 = 11,200.

Chord members Diagonals

M e m b e r S t r e s s P a n e l ~ S h e a r 1 M e m b e r ) S t r e s s ) P a n e l 1 S h e a r 1 M e m b e r /

1-3 - 4 4 . 8 JO-1 - 3 1 73 - 5 - 6 7 . 4 O-2 22 .4 1 l - 2 +31.7 4 - 4 ’

o -22 - 4 I%‘: 12-3 - 1 7 . 84 - 4 ’ +67.4 Z-4 12 .6 13-4 f17.8 4’-2’ /

ART. 7. STRESS COEFFICIENTS FOR TRUSSES WITH PARALLEL CHORDS

Table 25 contains the coefficients from which the stresses in the trussmembers due to uniform dead and live loads can be obtained by a simplemuhtiplication. The coefficients of the web members are equal to the shearsand those of the chord members are equal to the chord stresses due to auniform load of 1 lb. per lin. ft. per truss, a panel length of 1 ft. and aheight of truss of 1 ft. For any other load p in lb. per lin. ft. panellength X, height of truss h, length of diagonal s in feet and angle a! be-tween vertical and diagonal the stresses are obtained by multiplyingthe corresponding coefficients by

x2pXtg a o r pkfor the chord members,

pX set a or p X z for the diagonals, and

ph.. . . . for the verticals and the reactions.

d denotes the coefficient for the dead load stress, 1 that for the live loadstress; where only one coefficient is given it refers to both stresses.

The dead load has been assumed concentrated at the bottom chordonly. The dead load stresses in the verticals of Pratt trusses must there-fore be corrected for the dead load concentration at the top chord.

If the diagonals in the center panel of a Pratt truss with unevennumber of panels are stiff they are usually assumed to act simultaneously,each taking one-half of the shear; in this case the stress coefficient of thediagonal and the adjoining vertical has to be divided by two. Thestresses in counters can be obtained from those in the main diagonals.

It is of advantage to prepare similar diagrams for other truss systemswhenever they are frequently used.


ART. 8. TRUSSES WITH POLYGONAL TOP CHORD

Fig. 8 represents a type mostly used for pin-connected trusses. Thediagonals are composed of eyebars, so that counters must be provided in


those panels in which the dead load tension in the main diagonal is re-versed by the live load compression.

a; _______.. -.- .____ _.._______ ___

FIG. 8.

Chord Members.-The stress in a chord member is found by dividingthe bending moment at its center of moments by the perpendiculardistance of the latter from the member. The maximum live loadmoments and, therefore, chord stresses occur when the span is fully loadedunder which condition the counters are not acting. For instance, thestress in top chord UzU3 is equal to the moment MS at L, divided by thelever arm r. For the calculation of the bending moments at panel points,see pages 41, 46, 49, 53, etc.

Main Diagonals.-In order to find the stress i!n the main diagonal ofany panel, we assume a section X

2 through that panel. The stresses inthe three cut members (the counter-diagonal being assumed as not acting)

3 vzm3 ,, must be in equilibrium with the shearin that panel. Knowing the shear andone or both chord stresses determined,the stress in the diagonal can easily be

4 L2L3 I found graphically by a force diagram 1FIG. 9. 2 3 4 (Fig. 9). It can also be found

analytically considering that the alge-braic sum of the vertical components of the stresses in the chords andin the diagonal must be equal to the shear X.

For dead load the shear acting on the part of the truss to the leftof section X is positive (acting upward) for all panels to the left of theresultant of all dead loads on the span (for symmetrical loading thecenter of span). The main diagonals falling toward that resultant are


in tension and the vertical component of the stress is equal to the shearminus the vertical component of the top chord stress.

The maximum tension from live load in the diagonal U&, for in-stance, is obtained from the maximum positive shear in panel 2-3which occurs when the load extends from the right end into panel 2-3.(As stated on page 75, this is not quite correct but sufficiently accuratefor practical purposes.) For this position of the load the stress in thetop chord lJ2U3 or in the bottom chord L2L3 is determined and thestress in the diagonal is then found as explained above.

The stress in UzU3 is found as follows: In calculating the maximumshear in panel 2-3 from live load it has been necessary to determine theleft reaction R. and the moment M’ of the loads in panel 2-3 about LS(this moment divided by the panel length X gives the panel concentrationat Lz which must be deducted from Ro to obtain the shear). The bend-ing moment at La is therefore

Ma=Roaa-M’;

the stress in UzU3 is - $ and its vertical component is

Ma b-h op-~r L

MB ;r set @

where L is the length of &US, p the angle between U,UB and the ver-tical, hz and ha the‘height of truss at panel points 2 and 3.

In Tables 19 and 22 the shears V and moments M’ for Cooper’s E-50

loading are given; the left reaction is then R. = V f F, which is

introduced into the above formula for MR. (See pages 65, 66,68, 69.)For a uniform live load the shears may be calculated by the approxi-

mate method (page 54). There is then no panel load to the left of thepanel in question and the reaction Ro is equal to the shear.

To obtain the maximum compression from live load in diagonal ‘CJ&let us consider the diagonal Lts U’Z symmetrical to UzL3 about the centerof span. The maximum compression in L’3U’z occurs with the maxi-mum positive shear in panel 3’-2’ when the load extends from L’, intothat panel. (This is the same as if we would determine the maximumnegative shear in panel 2-3 for a load extending from Lo into that panel.)

For this position we calculate the stress in U’3U’z

M’, R. ar3=~=~r r

and determine the compression stress in diagonal L’aUt2 by the force


polygon 1 2 3 4 (Fig. lo), or analytically by adding to the shear V,~-,Ithe vertical component

RO a’3= __ cospr

of the stress in Uf3U’~ and multiplying by the secant of the angle a! be-tween the vertical and the diagonal.

Counter-diagonals.-If the compression in the main diagonal fromlive load (including impact) counteracts the dead load tension (or $ ofit according to specifications, see appendix) and the main diagonal cannotresist compression, a counter-diagonal has to be used. The maximumlive load tension in the counter Uf3L's is found as the length 3’ 4’ in theforce polygon 1 2 3’ 4’ (Fig. 10). As can be seen by comparing thispolygon with the truss figure, the stresses in the two diagonals are inproportion to the lengths of the diagonals but of opposite sign. Withthe live load tension in the counter must be combined a dead load com-

5pression equal to the dead load

I.,I

,I-I_

tension in the main diagonal times--. -.$‘i (J’ the ratio of length of counter-diag-

3 li;----- --J.1, --A.3

\-I___ onal to that of the main diagonal.

‘\I \

1 u; U$ --t 2 Verticals.-The maximum live

I \ \

i.:i‘J

load compression in vertical U&z

14% \*

b ‘“,q v3L2, occurs at the same time with maxi-‘\

‘\\ Imum tension in diagonal U&3 and

4can be found as shown by dotted

4’ I!, L!z 1

FIG. 10.lines in Fig. 9. The force polygon2 3 4 5 follows directly from the

examination of the forces at Us which must be in equilibrium; but alsopolygon 1 2 5 4 follows in assuming a section through the truss cuttingthe vertical, the top chord U1lJ2 and the bottom chord L&S. Thestresses in the three members cut (counter-diagonal not acting) mustbe in equilibrium with the shear in that panel in which the section cutsthe loaded chord (here bottom). Analytically the stress in the verticalU&z is found by deducting from the shear V3f-zt the vertical compo-nent of the stress in top chord UIUz.

For the dead load stress in this vertical the method is the same, butto the stress thus obtained by diagram (Fig. 9) from the shear V+.-3 thedead panel load at Uz has to be added.

If there were no counter-diagonal LzlJ3 in panel 2-3 the greatestlive load tension in vertical lT,L, would occur when the diagonal UJ&,gets its greatest live load compression and is therefore represented by the

ART. 81 STRESSES 1N SIMPLE TRUSSES 89

length 4 5 in the force polygon 2 3 4 5 (Fig. 9) or directly from the shearVv3,+ by the polygon 1 2 5 4 (Fig. 10).

If there is a counter-diagonal in panel 2-3 the maximum live loadtension in post UzLz occurs when, of the members connecting at Uz, onlythe two chord members lJ1 UZ and li, Us and the post itself are stressedbut not the diagonal U2L3. The live load tension in the post is thenequal to the difference between the vertical components of the two chordstresses and is a maximum when they are maxima, always providing thereis no stress in any diagonal connecting at the top of the post. Let usconsider the post U'2Lfz on the right half of the truss (Fig. 11).

When the load advances from the right, the counter Uf3L'~ comes intoaction as soon as the dead load tension in the main diagonal is neutral-ized by the live load compression. As long as the counter U'sL'z is actingthe post Uf2L', is stressed in tension and this tension evidently increases,the farther the load advances to the left; when the head of the loadpasses a certain point in panel 3'-2', the stress in the counter begins todecrease until it reaches zero. At this moment, the main diagonal would

FIG. 11. FIG. 12.

come again into action and decrease the tension in the post which there-fore then has reached its maximum. This position of the load could befound by trial but the following approximate method is shorter.

We will assume that the maximum live load tension in counter U'~L'Zhas been determined by means of a force polygon, Fig. 12 (compare Fig.10) for a load extending from L'o into panel 3’-2’. From this stress wededuct graphically the dead load compression in U'$Cfz k distance 4’-5(Fig. 12) which leaves the tension 3’-5 in the counter. In order to re-move this remaining tension, we add to the left of panel 3’-2’ a certainload which causes compression equal to this tension. The shear from theadded load is the end reaction Rro due to this added load and must be inequilibrium with the stresses in the three members cut by section X’.We obtain, therefore, the stress in top chord U'3U'z from this added load


by resolving the diagonal stress 5-3’ (Fig. 12) into a force 5-6 parallelto line U’s L’o and the stress 6-3’ parallel to U’s U’,, L’, being the inter-section of the stress in bottom chord L’,L’z with the end reaction Rro.By adding stresses 2-3’ (Fig. 12) and 3’-6 we obtain the total stress intop chord U’37J’z and finally we resolve this stress 2-6 into stresses6-7 and 7-2 parallel to U’2i7’1 and lJ’&‘s. Stress 7-2 is the desiredmaximum live load tension in U’z L’, or Us Lz. ’

It must be combined with the dead load stress (which may be com-pression) obtained from the top chord stress UfzUrl by a force polygon1 2 3 4 (Fig. 13) in which P’a is the dead panel load at U’s, or fromthe dead load moment at L’, by ‘the formula

Yd - P’, (Fig. 11)

FIG. 13.

This method is not quite correct since when the head of the load.advances from L’, to the left, the load between L’, and L’, may changeand thereby change the tension stress in counter U’&‘z; but the methodis sufficiently accurate for practical purposes, the more so, as the tensionin the vertical is generally small compared with the compression.

ART. 9. INFLUENCE LINES FOR TRUSSES WITH POLYGONALCHORD

The influence area for the end reaction R, is a triangle A’B’A”A’(Fig. 14) with A’A” = load unity. Loads P = 1 to the right of L, of anypanel mn cause stresses X0 in any member cut by a section X throughthat panel equal to

So = R, X 81

where X1 is the stress caused in that member by an upward force 1 at A.Instead of drawing the actual influence ordinates So we show them divided


by S1 and have then Sos = R,, that is, the reduced influence lines to the1

right of L, are identical wit.h the influence line for the reaction R,. Withone part B’n’ thus given the reduced influence line for each membercan be drawn. The stresses determined by these reduced influence lineshave finally to be multiplied by the corresponding influence coefficientS1 (see page 41).

To find the influence line for top chord U,U, draw line A”B’ mak-

-\

FIG. 14.1‘

ing A’,4” = 1 and from the intersection n’ of this line with the verticalthrough the center of moment L, draw a straight line to A’. A’n’B’A’

is the reduced influence area. The influence coefficient is Sl= 1: (Fig.

14 b). A’n’B’ is also the influence line for Bottom chord L,Lo (center

of moments U,), but the influence coefficient is S1 = IF.

To find the influence line for diagonal U,L, draw lini A”B’ (A’A”

= 1) and line A’B” = 2 (for h, an d hb see Fig. 14 a) and connect points


m’ and n’. The intersection point I’ of A’B” and A”B’ must be ver-tically below the intersection point I of chord members U,,U,. andL,L, (Fig. 14 c). Intersection point K’ of the line m’n’ may be foundvertically below point K or K1. K is found by producing top chordU,U, until it intersects the verticals through A and B and drawinglines Aim and Bin. K1 is the intersecting point of line AIB withthe diagonal U,L,. A’m’n’B’ is the reduced influence line for this

diagonal; influence coefficient 231 = 2 where x1 is the distance of I

from A and d the iever arm of the diagonal. If the center of mo-ments I is not conveniently located S1 can also be found as follows:

Cd)

FIG. 15.

An upward force unity at A causes the shear V,, = 1 in panel mn

and the stress in top chord U,U,, the stress X1 in diagonal UnJln

is therefore (l- *D) set a.

Similarly the influence line A’n”o’B’ (Fig. 14 c) for vertical UJ/,is found by considering a section X’. Influence coefficient

X1 = -%-- o rx1+x (1 -FTzq)

It has to be noted that the influence line for a vertical has to intersectthe base line A’B’ in that panel in which section X’ cuts the loadedchord.

The influence coefficients X1 for all members are best found by aMaxwell diagram for an upward force unity at A. If the truss is unsym-


metrical the left half is treated as explained above; for the right halfthe influence lines are derived by using as basis influence line A'B"B'for reaction R,, (where B'B" = 1) and calculating the influence coeffi-cients X1 from a load unity at B.

The maximum tension in a vertical adjoining a panel with counter-diagonals can be determined by means of influence lines as follows:.The considerations as to the most unfavorable position of the load arethe same as for the method on page 88. Let us consider again verticalU& of Fig. 8, page 86. Under the assumption that counter-diagonalLzU3 is acting the influence area for UZLZ is a triangle o 2’0’ (Fig. 15 c)the ordinate at 2 is equal to the difference between the vertical com-ponents of chord stresses UIUz and UzU3 from a load unity at Lz.The chord stress UIUz from load unity at Lp is UIUz = F and the

ordinate 2 2’ can either be found graphically by a force polygon (Fig.15 b) or by the formula

Instead of the actual wheel loads we assume an equivalent uniformload p per lin. ft. which would produce the same end shear in a spanequal to the loaded length x. Since we do not know this length wehave to make a preliminary assumption. We will further assume thatthe influence line for the counter-diagonal LzU3 has been drawn(Fig. 15 d).

When the load has advanced from o to i (Fig. 15 d) it produces themaximum live load tension in the sounter which is equal to area 02imultiplied by p. From this we deduct the dead load compression in thecounter. The remaining tension in the counter is removed when theload has advanced to a point, n for which area i3nn’ times p is equalto the remaining tension. n then marks the head of the load for maxi-,mum tension in the vertical and this stress is found by multiplyingarea onn’2’ (Fig. 15 c) by p.

If the assumed length x is found to differ from the final length so asto cause a considerable difference in the equivalent uniform load p thecalculation has to be repeated.

Plate II shows the influence lines for the truss in Art. 10.

ART. 10. EXAMPLE FOR TRUSSES WITH POLYGONAL CHORD

Fig. 16 represents a truss of a single track through span 200 ft. longcarrying a dead load of 2560 lb. per lin. ft. of bridge and a live load

94 DESIGN OF STEEL BRIDGES [CH A P . IV

of Cooper’s E-50; all diagonals consist of eyebars. Fig. 16 contains thelengths of the members in feet, the secants of the angles which theinclined members form with the vertical and the lever arms of the topchord members. Hate II shows the influence lines for this truss.

Fm. 16.

Dead Load Stresses.Assumed dead load:

Upper chord 440 lb. per lin. ft. per truss = 11,000 lb. per panel.Lower chord 840 lb. “ “ “ “ “ = 21,000 “ “ “

Total, 1280 lb. ” ‘I “ ” “ = 32,000 lb. per panel.End reaction 32,000 X 4 = 128,000 lb.

O-l 128-lG=112 2 8 0 0 2 8 0 0 1

l - 2 112-32= 8 0 2 0 0 0 4 8 0 0 2

2-3 SO-32= 4 8 1 2 0 0 GO003-4 48-32= 16 4 0 0 6 4 0 0 s

Moment BendingShear V i n c r e m e n t moment Panel

AM = 2517 M;=*M$-, PointM e m b e r

Chord members

Vert. compon.of stress=S:sec LI

. . . . . . . . . . . . . .148:5.10=29

168:8.39=20

Diagonals I Verticills____

M e m b e r Stress= (V-~~;~~c;rnp. top o h . )

I I

Member Stress= (-V + vert. camp. top ch.- u pmel- cont.)

112X1341 =@O-29)1341=~~8y-2~~1255 =

=

(Lower penel cont.)-4s+29-ll=-16+20-ll=(Upper panel cont.)

+.a- 3 0

-1:

When the counter-diagonals are acting their dead load stresses are:

LzU3 : -35 gg= -37; L3U4 : -19

PLATE II

__-----I----__--3-

_____-- -j--->-I

-Tc-‘.---_

‘. ----_‘.‘l,

‘.

-< ---_‘, ----___

‘A‘1

‘1‘\

r-----m I /0”“0’3’“‘1.0 1.5 2.0 2.5 3.0

Scale for InfluenceCoefficIents.

I.89

2.74

Influence Lines for a 2004. Truss with Polygonal Top Chord. (Facing paue 94)

ART. lo] STRESSES IN SIMPLE TRUSSES 95

The greatest live load tension in vertical UzLz occucs when thediagonal U&3 is not acting; for this condition we have for the dead loadstress in the vertical:

Vertical component of stress in UIUz = + 2 9Vertical component of stress in U2U3 = 4800 : (32.76 X8.39) = - 17Upper panel concentration at Uz = - 1 1

Stress in vertical UzLz =+ 1

Correspondingly we get for the dead load stress in vertical U3L3 whenneither Lzi73 nor U& are acting:

Vertical component of stress in UzU3 ' = + 2 0Upper panel concentration at Us = - 1 1

Stress in vertical U& =+ 9

Live Load Stresses.YThe maximum moments and shears are takenfrom Table 22, p. 69 and Table 19, p. ‘, 6.

Chord members

Panel point Moment M Member Lever-arm T Stress

6,787

11,240

14,01014,820

LG 28.0:p 3 2 . 3 4 3 3 . 0

I!$;;+341

E213j3 3 5 . 7 4 3 6 . 0 -392Lb?, 36.0

/-412 +390

End Post LoUI.-Shear Vbl= 271.5; stress = 271.5 X 1.341= - 364

Diagonal UILz.-Max. positive shear in panel l-2 = 205.0Panel concentration at L, = 11’.6Moment at Lx = (205.0+11.6)50-11.6X25 = 10,540Vertical component of stress in UIUz = 10,540 : (32.34 X5.1) = 64Max. tension in U,LZ = (+205-64)1.341 = + 190

Max. positive shear in panel 2’-1’ = 6.9Panel concentration at L',- = 4.0Moment at L'2 = (6.9+4.0) 150 = 1635Vertical component of stress in Ur2U'1 = 1635 : (32.34 X 5.1) = 9.9

7

96 DESIGN OF STEEL BtiIDGES [CHAP. IV

Max. compression in Lr2Uf1 or UILz = (- 6.9 - 9.9) 1.341 = - 23

Diagonal UzL3, Vertical UzLa and Counter L,U3.Max. positive shear in panel 2-3Panel concentration at Lz

’

= 147.5= 1 1 . 6

Moment at La = (147.5+11.6) 75-11.6X25 = 11,640Vert. component of stress in iJzU3 = 11,640 : (35.74X8.39) = 39.0Max. tension in U2Ls = (+ 147.5 - 39) 1.255 = + 136

Moment at L, = (147.5+11.6) 50Vert. component of stress in U1 Us = 7955: (32.34 X5.1)Max. compression in lJ2Lz = - 147.5+48.3

= 7955= 4 8 . 3=- 9 9

Max. positive shear in panel 3’ - 2’ = 2 6 . 9Panel concentration at L', = 4.0Moment at L', = (26.9+4.0) 125 = 3870Vert. component of stress in Ufa iYz = 3870 : (35.74x 8.39) = ,12.0Max. compression in LfQ L7'z or UzL3 = (- 26.9 - 12.0) 1.255 = - 4 9

As this is greater than the dead load tension (+35) a counter-diagonal'L&J3 is required and the live load tension in the latter is then

43.8349xq1.4o = j-52

which has to ‘be combined with the dead load compression of -37.The max. tension in vertical U2L, will be determined by the graphical

method shown on page $9. For polygon 1 2 3’ 4’ (Fig. 17) we have

FIG. 17.

already l-2 = V3,-2, = 26.9, besides we have l-4’ = stress in L'ZL', =M’3__ = 3870 : 36 = 107.5 and can now draw this polygon by making

r4’-3’ parallel to U'&', and 2-3’ parallel to U'aUfz. We find as acheck 3’-4’ = live load tension in U'&', = +52; further, we make4’-5 = dead load stress in iT3L', = -37 and then draw 5-6 parallelto U'BL'o and 6-7 parallel to U':!U' 1 and get 2- 7 = live load tensionin U& = + 14, which has to be combined with the dead load stress + 1.

ART. 111 STRESSES IN SIMPLE TRUSSES

Diagonal U,L,, Vertical U3L3 and Counter L3U4.Max. pos. shear in panel 3-4Max. tension in U3L4 = 98 X 1.218

Panel cont. at L,Moment at Lt = (98.0+ 11.6) 75Vert. component of stress in UzU3 = 8210 : (35.74X8.39)Max. compression in U3L3 = -98.0+27.3

Max. pos. shear in panel 4-3’Max. tension in counter La 7Jd = 56.4 X 1.218

97

= 98.0=+119.0- -= 11.6= 8210= 27.3=- 71.0

= 56.4=+ 69.0

which has to be combined with - 19 from dead load.The live load tension in vertical U3L, is obtained in a similar way

as shown for U&g it is 1-33 and has to be combined with the dead loadstress of +9.

Vertical U&Q does not get any tension; its greatest live load compres-sion is equal to the positive shear in panel 4-3’ = -56.

The combination. of stresses is similar as shown for the example onpage 82.

ART. 11. TRUSSES WITH SUBDIVIDED PANELS

The secondary members with part of the main members form trussedstringers (shown dotted) which carry the secondary panel loads to the

FIG. 18. FIG. 19.

main panel points either of the bottom chord (Fig. 18) or of the top chord(Fig. 19).

In panel L~-LG, Fig. 18, for example, the stringer consists of thesubvertical MA, subdiagonal L4M5, part of bottom chord L4Le andpart of main diagonal ilISLe. In removing the stringer nothing ischanged as far as the other members are concerned, the stresses in the


members U&+ U&5 and i!JdUs are determined as if the secondarymembers were omitted and the loads applied to the main panel pointsonly. The maximum live load stress in U~US occurs with the maxi-

’ mum moment at Le, the maximum live load stresses in UdMB and inU& occur with the maximum shear in main panel 4-6.

Submembers.-The tension in hanger LsMb is equal to the panelconcentration at Lg; the compression in subpost USMS equal to panelconcentration at US. The stress in LdMb is obtained by resolving thetotal panel load Pg into stresses parallel to MsLc and L4M5 (Fig. 20).

Diagonal l&L,+-A section X through panel LSLS cuts only threemembers whose stresses must be in equilibrium with the shear in thispanel. The stress in M& is, therefore, determined by resolving theshear into the stresses in the three cut members as shown on page

(486; the maximum and minimum live load stresses occur

c

D

with the maximum and minimum shear in panel 5-6 re-

*(, spectively.p5 Bottom Chord L&.-The center of moments for L&

is at U.+*’

If the subdivision in panel 4-6 were omitted the

9 tension in L&C6 would be equal to the bending moment at(b) Ud divided by the depth of the truss at Ud. To this ten-

FIG. 20. sion must be added the tension in the chord L& as a mem-ber of the trussed ‘stringer L&S due to the total panel

load Pg. This is equivalent to taking moments of all external forces tothe left of section X about Ud (noting that panel load Pg causes a posi-tive moment thus increasing the bending moment at 77,). For themaximum live load stress the same position of the load may be used asfor the maximum bending moment at panel point 5.

Counter-diagonals.-On page 88, it has been proven that for anyloading the stress in the counter of a truss without subdivision is equalbut of opposite sign to that in the main diagonal multiplied by the ratioof length of counter to length of main diagonal. In a similar way itcan be proven that in any panel, for instance panel 2-4 (Fig. al), dividedinto two subpanels of equal length, the stress in counter-diagonal M3U4due to any loading (assuming main diagonal M3L4 not acting) is equaland of opposite sign to the stress in the main diagonal UzM3 (due tothe same loading and assuming LzM3 not acting) multiplied by the ratioof length of M3U4 to length of UzM3.

Correspondingly, the stress in LzM3 (MaL4 not acting) is equal to thestress in MA4 (LzM3 not acting) multiplied by the ratio of length ofLzM3 to length M3L4 (the two stresses are alike in amount if the bot-


tom chord is horizontal and U2M3L4 a straight line, which is usuallythe case).

For maximum live load tension in MS?74 it is therefore only necessaryto determine first the maximum compression which would occur in ’UzM3 from the maximum negative shear in panel 2-3 if LzM3 wereomitted. Correspondingly, the maximum live-loadtension in LzM3 is obtained from the maximum u4

compression which would occur in M3L4 if LzM3 J-4were not acting; in the latter case UzM3U~ formsa trussed stringer which carries the panel load Pato the top chord at Uz and U4. M& can there-

r!xl

/ M3

fore be treated as if there were no subdivision; that ,/”P3

is, its greatest compression stress is obtained from &iLF L4

the maximum negative shear in main panel 24. FIG. 21.

The live load tension stresses in LzM3 and MsUqhave to be combined with the corresponding compression from deadload.

ART 12. INFLUENCE LINES FOR TRUSSES WITH SUBDIVIDED PANELS

The method of influence lines is very convenient for the analysisof subdivided trusses and is less liable to errors than the other methods.Fig. 22 b shows the influence lines for diagonals ‘CJ4Ms and MsLs. Theinfluence line 0 4 6 0’ for U4M5 is found as explained on page 90, fortrusses with main panels only, assuming submembers in panel L&Somitted. The influence line, 0 5 6 0’ for M5Ls (Fig. 22 b) is foundby extending line 0 4 to 5 vertically below MS and connecting points 5and 6.

Assuming that there is no counter-diagonal in panel 4-6, the influenceline for U&4 (Fig. 22 c) is found in the same manner as explained on page90, considering the submembers in panel L& omitted.

The triangle 0 4 0’ (Fig. 22 d) is the influence area for bottom chordL&c assuming the submembers in panel L&s omitted. Triangle 4 5 6is the influence area for chord L& as a member of the trussed stringerL&S. The influence line for total stress in L4L6 is, therefore, 0 4 5 6 0’where, as can easily be proven, point S is obtained by producingline 0 4.

If there is a counter in panel 4-6 and a substrut LzM3 in panel 2-4the tension in vertical U4L4 is found as shown on page 88 and 92.If, however, there is a subtie MsUd instead of a substrut LZMSin panel 2-4 the influence line Fig. 15 c has to be changed as shown in


Fig. 22 e by adding triangle 2’ 3 4’ with the ordinate 3 3’=0.5, that is,equal to the reaction of the trussed stringer U&l3U~ at Uq due to theload unity at L3. Since this reaction causes compression in the vertica.1U4L4 it must be deducted from the tension, that is, ordinate 3’ 3 plotteddownward. The influence area is then as shown hatched. Since theload must extend from o to n the negative area must be deducted from

- - - - - ,-

b---_ - - - - /’/ I

///

!

I/I!

// !I

-4

(a)

(Cl

(d)

(e)

-3

FIG. 22.

the positive area thus leaving a very small tension which, in order toavoid the complicated calculation, can be neglected, the more so as itis only of secondary value.

Plate III shows the influence lines for the truss in Art. 13.

ART. 13. EXAMPLE FOR TRUSS WITH SUBDIVIDED PANELS

For comparison let us assume a truss identical with that shown’ inFig. 16 except that it has subdivided panels and no counter-diagonals.

PLATE III

m 9 IO Feet

scak for kngths

Influence Lines for a 200-ft. Truss with Subdivided Panels.,-


The loading shall be the same, which makes the dead panel load in thesubdivided truss 16,000 lb. or one-half of that in the plain truss (Fig 23).

The moments at the main panel points, the shears in the main panels(assuming the loads carried to the main panel points only) and thereforethe stresses in all top chord members, in diagonals MolU1, U1M12,UzMz3 and U3M34 and in the main verticals are exactly the same asin the truss without subpanels and counters.

.

r, - - - - - - -. _ _ _ _ _ -- _ _ _ _ _ _ _ _ _ _ _ _ _ ,6 fane/s@/2.~~~~~‘C.~~~.~f~~~ --/ ----- _ ---------------____ 4

FIG. 23.

Dead Load Stresses.Subdiagonals .-MolLl and L1M,2 : 316X1.341 = - 10.8; LzMz3 :

316 X 1.255 = - 10.0; LsM34 : 416 X 1.218 = -9.7.Lower Main Diagonals.-Stress in lower diagonal equal to stress in

upper diagonal plus compression in subdiagonal. LoMol : - 150 - 11 =- 1 6 1 ; MI& : +68-11 = +57; M&3 : +35-10 = +25; MsJL4 :+19-10 = +9.

Bottom Chord.-Stress same as in truss without subpanels plustension in chord as member of the trussed stringer due to subpanel load.

16X25.I,& and LILz : +lOO+m

16X25= +107; L&B : +146+- =

16X25+152; LsL4 : +166+ 4x18 = +172.

Live Load Stresses.Subdiagonals.-Maximum panel concentration = 60, stress = 3 60 X

set a; MolLl and L,M,, : -40; LzM23 : -38; L3Ms4 : -36.Lower Main Diagonals.-LoMoI : max. pos. shear in

panel O-01 (wheel @ at LoI) = 294.6Stress in L ,Mol = 294.6 X 1.341 = -395.0

MEL : max. pos. shear in panel 12-2 (wheel @ at Lz) = 196.1Panel concentration at Lls = 8.0Moment at L, = (196.1+8.0)50%.0X12.5 = 10,110Vertical camp. of stress in UlUs = 10,110 : (32.34X5.1) = 61.2


Max. tension in M12L, = (+196-61)1.341 =+x31 .o__--

Max. pos. shear in panel 2’-21’ (wheel @ at L’,,) = 13.8Moment at L’, = 13.8X150 = 2070Vert. camp. of stress in U’lU’, = 2070 : (32.34X5.1) = 12.5Max. compression in L’zWzl or MlzLz = ( - 13.8 - 12.5) 1.341

=- 35

The stresses in .M,&, and M3J4 are found in a similarway; they are:

M&3 : +130 and -68M&4 : +113 a n d - 8 7

Bottom Chord.-LoL1 : this stress occurs simultaneously with thatin LoM,, and is therefore

25395 ____ -37.54 - +263

L,Lz : Position for max. momem at LIZ : wheel @ at LIZ,Moment from this loading at UI = 6675Panel concentration at LIZ = 49.5Total moment at U1 = 6675+49.5X12.5 = 7295Stress in LILz = 7295 : 28 = + 2 6 1

The stresses in the other bottom chord members are found in asimilar way; they are: +353 in LzL3 and +345 in L3L4.

ART. 14. STRESS COEFFICIENTS FOR TRUSSES WITH POLYGONAL CHORD

Table 26 contains for various trusses with a polygonal top chord thecoefficients from which the stresses in the members of similar trusses dueto a uniform dead and live load can be obtained by a simple multiplica-tion. The coefficients are the stresses due to a uniform load of 1 lb.per lin. ft. per truss for a panel length of 1 ft. For any other load pin lb. per lin. ft. and a panel length x the stresses in a similar truss areobtained by multiplying the coefficient by p X.

The dead load has been assumed concentrated at the bottom chordonly. The dead load stresses in the verticals and subdiagonals musttherefore be corrected for the dead load concentration at the top chord.

ART. 151 STRESSES IN SIMPLE TRUSSES

TABLE 26Stress Coefficients for Trusses with Polygonal Top

i

- A -

1 0 3

Chord

ART. i5. TRUSSES OF SKEW SPANS

As the dead load panel concentrations are as a rule unequal they haveto be determined separately for each panel point, at least for those at theends. The weight of the trusses and bracing may thereby be assumeduniform. The reactions and stresses are then determined as shown onpage 80. (See also page 208).

The live load stresses are conveniently determined by the method ofinfluence lines as follows: The influence lines for the truss membersare first drawn as if the span were square and then corrected at the ends


(see shaded areas). Wherever the floorbeam reaction due to a loadP = 2 applied at the center line of the bridge is lc times the load unity, thecorresponding influence ordinate x has to be changed (generally reduced)to x k. For instance, if the floorbeam reaction at f (Fig. 24) is k = 0.4,all influence ordinates below f have to be reduced to 0.4 of their length.From the end floorbeam the influence ordinates decrease to zero below theend of the stringer. The reaction at a is affected by loads between a

I III

End Reaction at 9.

I

II I

(b!

Cd)

(e)

FIG. 24.

and the left end floorbeam h, consequently the influence line continuesfrom a’ to h’ where the ordinate is zero.

The reactions and stresses may be obtained directly from these in-fluence lines or they may be determined as if the span were square, andcorrected by means of the areas shown hatched and an equivalent uni-form load. The latter method is approximate but generally sufficientlyaccurate.

CHAPTER V

STRESSES IN BRACING OF SIMPLE SPANS

ART. 1. GENERAL

The wind pressure (see page 25) is generally specified as a horizontaltransverse load, either in lbs. per sq. ft. of exposed surface or in Ibs.per lin. ft. of bridge. It is made up of the pressure on the movingload, the floor and the trusses. The pressure on the trusses can beassumed concentrated at the panel points along the center lines of thetwo chords. The bending caused in the truss members by the directaction of the wind is ordinarily negligible. In long slender struts, how-ever, this bending effect should be considered. The wind pressure onthe moving load and floor has to be transferred to the lateral systembelow the floor, thereby causing an overturning moment which will beconsidered separately.

Where the pressure per sq. ft. of exposed surface is specified, thearea of that surface has to be determined first. It is the usual practiceto take the exposed surface of the floor plus twice the exposed surface ofone truss, as seen in elevation, deducting that part of the trusses coveredby the floor. As the floor and chords have usually a constant depth, itis accurate enough to calculate the area of an average panel and base onthat area the uniform pressure per ft. of bridge.

The whole wind pressure may be assumed applied at the panel pointsof the windward truss.

If there are two lateral systems, that is, an upper and a lower system,and also sway frames the usual practice is to assume that no lateral forcesare transferred from one lateral system to the other by the sway frames(except by the end frames or portals). This assumption is justified asboth lateral systems are loaded and therefore deflect simultaneouslyand approximately the same amount. If the sway frames have totransfer wind forces, owing to the omission of one lateral system, orif part of the wind forces of one system is assumed transferred to theother, then the resulting overturning moment has to be considered aswill be explained later.

Both lateral systems may be assumed as simple spans. This assump-tion is not correct as one end of the span is fixed longitudinally and,

105

106 DESIGN OF STEEL BRIDGES ’ [CHAP. v

therefore, unable to turn horizontally, but it is on the safe side andthere are always some clearances which permit a slight turning move-ment.

Usually the following loading conditions have to be considered:(1) Wind on trusses and floor, bridge unloaded.(2) Wind on trusses and floor combined with wind on the live load, a

lower wind pressure being used in this case than in the first.The first condition usually affects only the top laterals of through

spans and bottom laterals of deck spans; the second condition thelaterals below the floor and all members which are simultaneouslystressed by the live load and wind.

The wind on the trusses and floor may be specified either as a staticload over the whole length of span or as a moving load; the wind on thelive load should always be treated as a moving load. The static andmoving wind pressures are preferably treated separately.

Since the stresses are reversed by wind forces of opposite direction, itis necessary to calculate only the absolute maximum stresses except inskew spans with unsymmetrical lateral systems or in bridges on curveswhere the centrifugal force has to be considered simultaneously with thewind.

ART. 2. LATERAL SYSTEM BETWEEN STRAIGHT CHORDS

The laterals form with the chords a truss with parallel chords and thestresses are therefore determined as shown in Chapter IV, Art. 3. As

a rule the panels are alike and theshears in the different panels of thelateral truss are then most conve-niently determined by Table 12 (page

v 51) for a uniform static wind pressure

FIG. 1. and by Table 13 or 14, page 54 or 55for a uniform moving wind pressure.

From the shear B in any panel the stress in the corresponding diagonalis found by the formula

s=+v$ if the diagonals act in tension only, and

V d,X=Qa if the diagonals act in tension and compression,

where b = distance between trusses and d = length of diagonals(Fig. 1).

ART.~] STRESSES IN BRACING OF SIMPLE SPANS 107

If the panels are alike and W is the wind pressure per panel allstresses can, therefore be determined by a single setting of the slide rule

in multiplying the value We or y% by the corresponding figures

in Tables 12 and 13 or 14.If the diagonals are acting in tension only, the compression in the

cross strut to the right of any panel of the left half of the span is equalto the shear in that panel since it may be assumed that the wind loadsact along the windward chord only.

The compression in the end strut is equal to the end reaction

If the diagonals act in tension and compression, the stress in any inter-

mediate strut is zero, that in the end strut isR

- -.2The top laterals of through spans and the bottom laterals of deck

spans are generally too long to be designed economically to act in com-pression. Laterals below the floor are usually connected to the stringersand can be designed to take both stresses.

Stresses in Chords.-The maximum stresses in the chord members’ occur when the span is fully loaded; for their calculation the panel loads

from static and moving uniform wind pressures can therefore be combined.The stresses in the chord members are best obtained by multiplying the

bending moments of Table 9 (page 42) by IV k*

If the diagonals act in tension only the stress in a chord member isdetermined from the moment at the intersection of the’ opposite chordwith the acting diagonal of the same panel. If the diagonals act in tensionand compression then the stress in a chord member is the average be-tween the stresses obtained by taking first the one and then the otheropposite panel point as the center of moments.

Example.-Bottom lateral system of 200 ft. single track through span with eightpanels at 25 ft.

Width c. to c. of trusses b = 16.25 ft., length of diagonals d = 29.8 ft.

Static wind pressure 250 lb. per lin. ft.; IV, = 6,250 lb., IV, i = 11,500.

Moving wind pressure 300 lb. per lin. ft.; IV, = 7,500 lb., W, f = 13,750.

Total wind pressure 550 lb. per lin. ft.; Wt = 13,7501b., Wt% = 21,150.

Diagonals act in tension and compression.

1 0 8 DESIGN OF STEEL BRIDGES [CHAp. v

Stresses iidiagonals.

Panel / hii1 1 1-2 1 2-3 1 3-4

StaticCoefficient from Table 12; c = 3.5 2.5 1.5 0.5

wind. stress =$W8t = . . . . . . . . . . k20.1 *14.4 * 8.6 * 2.9

MovingCoefficient ,?rom Table 13; c = 3.5 2.57 1.79 1.14

wind.d

stress = c IV, - =2 b

. . . . . . . f 2 4 . 1 k17.7 1 1 2 . 3 f 7 . 8

Total . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . k44.2 f32.1 *20.9 +10.7

Average coefficient for adjoining panelStresses in points from Table 9; c = . . . 1.75 4.75 6.75 7.75chords.

stress = cW2b = . . . . . . . . . . . . . . *39.1 k10.0 *14.3 k16.4

Stress in end strut or end floorbeam = 7 = - 13,750 X2 = -27.5.

All stresses given in units of 1000 lb.

ART. 3. LATERAL SYSTEM BETWEEN POLYGONAL CHORDS

The stresses in the chords and web members of a lateral systembetween polygonal chords are determined by assuming the system

“$7FIQ. 2.

projected on a horizontal plane. The stresses in this projection areobtained in the same manner as shown above for plane systems, d there-by denoting the projected length of a diagonal. The actual stress inany chord member or lateral is then obtained by multiplying the stressin the projection by the proportion of actual to projected length ofthat member. The actual length of chord members can be scaled from

ART. 41 STRESSES IN BRACING OF SIMPLE SPANS 1 0 9

the elevations. The length D of a diagonal is conveniently obtainedgraphically as shown in Fig. 2.

Slight additional stresses are caused in the chord members and in theother members of the main truss, since at any joint UZ where the chordchanges its direction a certain component Vz of the wind stresses of’ thelateral system goes into the main truss. As can easily be proven byconsidering equilibrium of forces at UZ, this component is vertical andequal to the resultant of the stresses in the chord members meeting atthat point (Fig. 2), assuming these stresses obtained from the moment atthe opposite panel point (U’,), therefore equal to

442 Xl82 = Kl y1 - G y2

( >

where M2 is the bending moment at Uz in the projected lateral truss.When the components V are determined for all joints of the polygonalchord they are assumed to act as external forces and the resulting stressesin the truss members are obtained as for vertical loads.

These stresses are, as a rule, negligible; they should, however, be con-sidered and added to the other wind stresses where the curvature of thechord is considerable.

ART. 4. K-SYSTEM OF LATERALS

This system (Fig. 3) is now frequently used for the lateral bracing.It has the advantage over the double intersection system in that it isstatically determinate and requiresless material if the panels are very Ishort compared with the width be- f-o’

4,’

i, 1 $ 5.

tween trusses.Let us assume a section X through

panel 2-3 cutting the two-chord mem- jbers 2 3 and 2’ 3’ and the two diago-

$fgjlJq

jnals 2” 3 and 2” 3’. As the vertical 0 I

v 3 4components of the chord stresses are FIG. 3.

zero the sum of the vertical compo-nents of the two diagonal stresses must be equal and of opposite signto the shear V in panel 2-3; further, as there must be equilibrium atjoint.2” the horizontal components of the two diagonal stresses must bealike and of opposite sign; if therefore 2” bisects the distance 2-2’ thestresses in 2” 3 and 2” 3’ must be alike but of opposite sign andeach equal to oneihalf the shear times the secant of the angle ar betweenit and the vertical. Each stress may be positive or negative.

8

110 DESIGN OF STEEL BRIDGES [CHAP. V

As the resultant of the two diagonal stresses in 2” 3 and 2” 3’falls into line 2 2’ and intersects the chords in 2 and 2’ these pointsare the centers of moments of the chord members 2’ 3’ and 2-3respectively. The stresses in these members are therefore equal to themoment at panel point 2 divided by the width b.

As can easily be found by considering equilibrium at joints 3 and 3’the stress in vertical 3’ 3” is + 3 V2+ - Wa if the wind acts asshown in Fig. 3 and - 4 V2-8 if the wind acts in the opposited i r e c t i o n .

ART. 6. INDIRECT WIND STRESSES IN BOTTOM CHORDS

The inclined end posts at the movable end of the span can trans-mit only the vertical components of their wind stresses to the bearings ;the horizontal components go into the bottom chords by which theyare transmitted to the fixed bearings. These stresses (tension in theleeward, compression in the windward bottom chord) are constantthrough the whole length. They are seldom considered although theymay be considerable, especially in deck spans where the top lateralforces are carried down by inclined end frames. It should be in-

Wvestigated whether this stress combined

<-! 6. m with the direct wind stress in the wind-

v&L!!: wgj h ward chord does not reverse the dead-loadg I i wl

I f ---__- L< ____ 5 tens1on*

Jx

w wART. 6. OVERTURNING EFFECT OF WIND

Since the wind force W per panel actingon the train (generally assumed 7 ft. aboverail) does not lie in the plane of the lateral

jq . . . . . . b . . . .._.. 4 systemby which it has to be transferred to

FIG. 4.the bearings there results an overturningmoment W.h besides the force W in the

plane of the lateral system. Through the floorbeams this momenth

causes two equal but opposite vertical panel loads W 5 acting on

the trusses or girders (Fig. 4).A similar overturning moment WI hl is caused by the wind WI on the

floor if it does not act in the plane of the lateral system. A furthermoment results if the wind force or part thereof acting along a chordhas to be transferred to the lateral system between the other chords.

The stresses resulting from the forces W i are usually negligible

ART. S] STRESSES IN BRACING OF SIMPLE SPANS 1 1 1

when considered by themselves. They should, however, be added tothe other wind stresses in the bottom chord, as they may cause ex-cessive tension in the leeward chord or reversal of stress in the wind-ward chord. Usually the most unfavorable condition for reversaloccurs when the bridge is loaded with empty cars.

ART. 7. STABILITY AGAINST LATERAL FORCES

The wind pressure tends to move the span laterally and to overturn itaround the leeward pedestals. The former tendency is counteracted bythe friction of the bearings. The overturning moment (moment of windforces about the line connecting the leeward pedestals) is counteractedby the moment of the dead and live load about the same axis. Tohave a margin of safety, the latter moment should be about 50 to100 y0 greater, or, if this is not possible, anchorages should beprovided and so proportioned as to secure at least the same safety.

If the bridge is on a curve, the overturning moment of the centrifugalforce should be added to that of the wind. For the greatest uplift thefollowing conditions of loading should be considered separately wherebyno impact is included in the live load:

(1) Wind (unloaded structure) and dead load.(2) Wind (loaded structure), dead load, live load (empty cars) and

possibly centrifugal force (from empty cars) if any.(3) Wind (loaded structure), dead load, live load (full) and possible

centrifugal force (full) if any.The last condition will not affect the uplift, except possibly in deck

spans on sharp curves. Long through spans may be affected by condi-tion (l), short through spans are usually sufficiently stable.

It is not advisable to make the trusses of simple spans inclined toincrease the stability, as the saving of material does not offset the greatercost of shop work.

ART. 8. STRESSES FROM CENTRIFUGAL FORCE

It has been explained on pages 28 and 214 that the centrifugal forcemay be assumed to act at the height of the rails if the live load isapplied vertically along the center line of track, and, also, that thecentrifugal force is always proportional and distributed similar to thelive load. For proportions of the centrifugal force to live load seepage 29. For stresses in main trusses or girders see Eccentric Loading,Art. 9.


The stresses in the lateral system due to the centrifugal force areobtained by considering that the shear in any panel or the momentat any panel point of the lateral system is equal to 2 lc times the cor-responding shear or moment due to the vertical load in the main truss(assuming two main trusses and symmetrical loading).

The centrifugal force causes an overturning moment similar to thatof the wind, but it can, as a rule, be neglected except for the stresses inthe end frame and the stability of deck bridges.

ART. 9. STRESSES DUE TO ECCENTRIC LOADING

If a load P is applied with an eccentricity e from the axis of thebridge, the truss nearer to P has to carry (Fig. 5)

and the other truss

p (; +f)P (; - %>

This issometimes the case in bridges with unsymmetrical cross-section

Pe/

da

and is always the case in bridges carrying curved tracks.Where the eccentricity is constant through the whole lengththe stresses can, of course, be obtained in the same wayas for symmet,rical loading. In the case of curved tracks,however, the eccentricity is different at different panelpoints and the exact stresses are then obtained most con-veniently by influence lines as follows: The influence lines

b .-.. b-j-e .._. ->are first drawn as for straight track and central loading(

s e ep

ages 40 and 49). Each influence ordinate x below aF IG . 5 . panel point is then corrected by adding or subtracting the

amount x e depending on whether the eccentricity e is on the near orb

far side of the center line with reference to the truss in question. Ifthe method of influence lines with influence coefficients (see page 39) isused, the work of correcting the ordinates is reduced to a minimum.If the stresses for central loading are determined by another method,

then the influence area representing the corrections x 5 may be

multiplied by an equivalent uniform load and the stresses thusobtained added or deducted.

This accurate method is, however, warranted only if the radius ofcurvature of track is small. For girders or the main members of trusses,it will ordinarily suffice to assume a constant eccentricity for the whole

ART. 111 STRESSES IN BRACING OF SIMPLE SPANS 113

length of span by assuming the load applied along a straight line locatedat a distance of about + of the middle ordinate of the curve from thecenter line of track at the center of span. For plate girder spans thatdistance may even be 6 of the middle ordinate, as this gives the actualmoment at the center of span (see page 216 etc.).

For secondary members carrying single panel loads only, the actualeccentricity at the respective panel points must, of course, be used.

For stringers, the average eccentricity of the center line of track inthe panel in question may be used.

ART. 10. STRESSES FROM BRAKING FORCE

As explained on page 29, the braking force B may be assumed to actat the top of rail. It is transmitted, as a rule, by the floor system tothe main trusses at every panel point. It causes a longitudinal reaction

B at the fixed bearing of the span and vertical reactions + B i at

both ends, where 1 = length of span and h = distance of top of railfrom the center of hinge of the bearing. These vertical reactions may,as a rule, be neglected. The stresses in the bottom chord of throughspans should be considered as they may reverse the tension in the endpanels and thus require stiff members. The stress in the bottom chordof any panel is equal to the braking force per truss between the panel inquestion and the fixed bearing, it is therefore a maximum in the panelnearest that bearing where the tension stress is smallest.

There may be cases where the stresses from braking force are con-siderable and may affect the sections if added to the dead and live loadstresses, even if higher unit stresses are used, such as are specified forcombined vertical loads and wind. For instance, if several consecutivespans rest on intermediate bents which cannot transmit the brakingforces to the bearings, the whole braking force from all spans has togo through the end span to the fixed bearing and may cause high stressesin that span.

Stresses from braking force and wind should not be consideredsimultaneously.

ART. 11. STRESSES IN PORTALS

The reaction of the top lateral system has to be transmitted throughthe end sway frame or portal to the bearings. The end posts receivethereby direct stresses and in through spans also bending stresses whichmust be considered as these wind stresses are often high and may affect

114 DESIGN OF STEEL BRIDGES [CHAP. v

the section of the posts. It will be sufficient to analyze the stresses for afew typical portals. To simplify the calculation, the following assump-tions are made:

(1) The end posts are hinged at the bottom (the assumption of fixedbottom ends is not justified).

(2) The lateral force W which is applied at the center line of the topW

strut causes equal horizontal reactions -2 at each post.

(3) The members of the portal are connected to each other and tothe posts by hinges; the posts, however, are continuous from end to end.

As the stresses in the members are reversed for an opposite force W,we need not consider their signs. W produces, besides the horizontal

W hreactions 2, also vertical reactions W ~ of opposite directions.b The

horizontal reaction g causes in the post at any point x a bending

moment z x which is a maximum at point 4 and equal ghl (Fig. 6).

If the column fails laterally, it will ei-ther fail at point 4 by excessive fiber stressor by buckling at a point N, which maybe assumed at a height 0.5 hl above thebottom end. It has, therefore, to be pro-portioned so that the combined direct andbending stress at 4 and at N does not ex-teed the permissible bending stress orbuckling stress respectively (bucklinglength 2 = hl). The bending moments

FIG. 6. M=zhl a t 4 a n d Fhl a t N

can be replaced by an equivalent direct stress

s = M!!l-2

(see page 174), y and r being taken in the lateral direction.To find the stress in a member of the portal, for instance 1-2, we

assume a section cutting that member and 3-3’ and 2-3. The inter-section point 3 is then the center of moments of l-2 and the stress inl-2 is found by taking moments of the external forces on the left of thesection about point 3 and dividing by the lever arm d

STRESSES IN BRACING OF SIMPLE SPANS 115

The vertical component of the stress in 2-3, whose center of ,momenth

is in the infinite, is equal to the vertical reaction WT. As can easily be

proven, the stresses in l-3 and 33’ are zero, and therefore stress in3-4 equal to that in 2-3.

The stresses in the posts of portal shown in Fig. 7 are the same asabove.

If the diagonals are designed to take tension only, then 2-3 is nothacting and the vertical component of the stress in l-4 is W - orb

the stress i&elfhe

x1-4 = +Wbdf-d/ t

The stresses in 1-2 and 3-4 are found by tak- 1,’ing moments about their respective centers of

moments 4 and 1 and dividing by cl

x1-2 = -$ (W+b- ;h,) = -;(h-;)

andW h

s3-4 = --z--J-g--+

If the diagonals can resist tension and com-

I-.

,,I

iT W

i--- - . b . . _ .

h

.i‘$

+-

Fra. 7.

pression, the stress in each is one-half of the above or

SW = s- = *J-c1 4 2 3 2 b d

The stress in l-2 is obtained by considering first point 3 and then 4 asthe center of mcments and taking the mean between the two valuesobtained for the stress. Similarly, stress Z--4 is obtained. It will befound that

Sl-2 = -F a n d X3-4 = 0

ART. 12. STRESSES IN END SWAY FRAMES OF DECK SPANS

In deck span, the end sway frame is made up of a top and a bottomstrut and intersecting diagonals as shown in Fig. 8 and 9. If thediagonals would be very steep, two panels are used. In order to getsimpler details, the frame is sometimes placed in the plane of the verticalend posts. However, as these are generally light, proportioned to carryonly one-half panel load, the bracing should be placed between theinclined end posts.


Fig. 8 shows the stresses if the diagonals resist compression and Fig. 9if they resist tension only. If the top laterals resist compression and

FIG. 8. FIG. 9.

,tension,W Wy is applied at 1 and T at 2 and the stress in member 1-2

is zero in Fig. 8 and -g in Fig. 9.

CHAPTER VI

TYPES OF BRIDGES AND PRINCIPAL DIMENiIONS

ART. 1. CLASSIFICATION OF BRIDGES

Depending on their purpose, we distinguish highway, railway, aque-duct, canal and combined bridges.

We further distinguish in a general way between bridges proper, whichcross a waterway or other obstruction, and viaducts, which cross a valleyor carry an elevated railway or highway through a city.

Depending on whether they are permanently fixed or can be tempo-rarily moved, we call bridges fixed or movable. ,

A bridge consists of the substructure, that is, the foundations, piersand abutments transmitting the loads to the soil, and tbe superstructure,which carries the highway, railway or waterway from support to support.According to the material of the superstructure there are wooden, masonryand steel bridges. In any of these the substructure may be of wood,masonry or steel, but usually a steel superstructure rests on a masonrysubstructure. Steel bridges only will be considered in this Chapter.

ART. 2. PRINCIPAL PARTS

The principal parts of the superstructure are:1. Flooring, which forms the immediate support for the liqe load

(ties, ballast, planking, paving).2. Stringers or joists, which support the flooring and transmit the

load to the floorbeams.3. Floorbeams, which transmit the load to the main girders or trusses.(The stringers and floorbeams constitute the floor system.)4. The main girders or trusses, which transmit the loads to t,he sub-

structure through5. The bearings, which distribute the end reaction over the masonry

and make free expansion and contraction at one end of the span possible.6. The bracing, which stiffens the main girders or trusses and trans-

mits the lateral and longitudinal forces to the substructure.

ART. 3. TYPES OF MAIN GIRDERS OR TRUSSES

Simple Spans.-These are carried by main girders or trusses whichrest on two vertical supports. They form by far the larger part o f

117

118 DESIGN OF STEEL BRIDGES [CHAP. VI

bridges being simpler, stiffer and, as a rule, cheaper than other kinds forspans up to about 700 ft. They are not affected by possible settlementsof the foundations or changes of temperature, and their reactions arestatically determinate.

Depending on the kind of main girders or trusses simple spans areclassified as I-beam-, plate-girder- and truss-spans. The followingkinds are now generally used for the different span lengths.

Railroad bridges Highway bridges

Rolled I-beams . . . . . . . . . . . . .IPlate girders. . . . . . . . . . . . . . . .Trusses. . . . . . . . . . . . . . . . . . .

up to 20 ft. up to 40 ft.15 ft. to 120 ft. 20 ft. to 100 ft.100 ft. up 80 ft. up

Pony trusses, that is, low trusses without upper lateral bracing, may’be used for light highway bridges of 50 to 100 ft. length, especially ifthe cost of freight (railroad or ocean freight, hauling) is considerable, asthey are much lighter than plate girders. They are not advisable forrailroad bridges.

As far as economy in weight is concerned, the above limits could belower. However, as I-beams require less shop work than plate girders,and the latter less than trusses, the saving of metal may be offset by agreater cost of shop work.

Continuous Bridges.-These are carried by main girders or trusseswhich rest on three or more vertical supports. They are rarely used(except for swing bridges) because they are stat.ically indeterminate andany settlement of the masonry changes the reactions, making the stressesuncertain.

Cantilever Bridges.-These arc carried by main girders or trussesresting on more than two vertical supports. They have intermediatehinges which make them statically determinate. They are inferior inrigidity as compared with simple spans and are advisable only forspans over 700 ft. or where the character of the stream, the requirementsof navigation or other conditions make it impossible to erect the struc-ture on temporary falsework, and where conditions are unfavorable foran arch.

Arch Bridges.-These exert inclined pressures on the foundations andare only suitable in cases where the foundations are unyielding. In athree-hinged arch, which is statically determinate, a movement of thefoundations practically does not affect the stresses in the main trusses,while in a two-hinged or hingeless arch, it causes uncertain stresses in

ART. 31 TYPES OF BRIDGES AND PRINCIPAL DIMENSIONS 119

the trusses. In point of rigidity the arch without hinges is the stiffest,and the three-hinged arch the most flexible. The latter has the samedisadvantage as the cantilever in that the deflection line forms a kinkat the hinge and it is, therefore, not well adapted for railroad bridges.The deflection in the center of the three-, two-hinged and hingelessarch from a load in the center is about as 6 to 2 -to 1 respectively.The two-hinged is preferable to the hingeless arch because it has fewerfactors of uncertainty of stress distribution.

Arches are especially adapted for locations which offer natural rockabutments. In such cases the arch weighs less than a simple span or acantilever. In case the conditions do not permit the use of temporaryfalsework, the erection can be accomplished with temporary anchorageson the cantilever principle at comparatively small cost. The advantagesof an arch apply more particularly to the design with the floor on top.Where the floor has to be suspended from the arch trusses, the lateralforces acting on the floor cannot be transmitted satisfactorily to thetrusses through the long and light floor hangers. In order to make thefloor stiff there has, therefore, to be a rigid lateral system in its planewith separate wind chords. In other structures these are formed by thechords of trusses without any or with little additional material andoffer, on account of their greater section, a greater resistance to lateraldeformation.

Arches in which the horizontal component of the reaction is taken upby a tie are not economical.

The arch has a pleasing appearance especially for a deck structure.Where the floor intersects the arch trusses the top chords, if feasible,should be entirely above the floor and there should be end portals, asarches rising out of the floor do not look well as seen from the floor ofthe bridge.

Suspension bridges are carried by main trusses which consist of cablesor chains resting on two vertical towers and are anchored down at theends where they exert an inclined pull. The cable or chain is as a rulestiffened by a stiffening truss which, unless provided with a hinge, makesthe structure statically indeterminate.

Suspension bridges are more expensive than other kinds except forvery long spans or for highway traffic. Owing to their lack of rigiditythey are .not adapted for railroads except for spans over 2000 ft. andmultitrack railroads or for combined traffic where the dead weight isgreat compared with the live load.

Esthetically they afford opportunity for very pleasing structures.


ART. 4. LOCATION OF BRIDGE

The location of smaller bridges is determined by the line of communi-cation, but for large bridges the line itself may be influenced by the mostfavorable crossing of the stream, etc. Economy will, as a rule, be themain factor, besides the following conditions:

Places where the wate): has a swift current or which are subject tosudden highwater, freshets or ice-jams should be avoided. A placeshould be selected where good foundations can be obtained.

Economy in the first cost will be decisive, the cost of maintenancehaving little influence for bridges properly designed and built. Theshortest bridge is not always the cheapest, the economic location beinginfluenced by the cost of the approaches to the bridge.

ART. 6. DATA FOR THE DESIGN AND ERECTION

Skew bridges should be avoided whenever possible.Before any designing is done complete data should be obtained, such

as character of the bottom of the stream, velocity of current, profiles,map, time and levels of low and high water, floods, etc., as these influencethe design, the method of foundation, the erection, etc.

In case a new superstructure has to rest on existing masonry the lattershould be carefully measured and examined with respect to its safetyand bearing capacity.

The following information is suggested for the collection of data.

GENERAL DATA FOR HIGHWAY BRIDGES

1. Crossing over. . . . . . . . Town of. . . . . . . . . . . . . State of . . . . . . . . . . . . .2. Specifications by . . . . . . . . . . . . Classofliveload.. . . . . . . . . . . . . . . . . .:.3. Total length and general description . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4. Number and lengths of spans c. to c. of piers. . . . . . . . . . . . . . . . . . . . . . . . .5. Skew or angle of current with center line of bridge . . . . . . . . . . . . . . . . . . .6. Width of roadway., . . . . . . . . . . Number and width of f ootwalks . . . . . .7. Kind of floor (solid, planking, etc.). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8. Kind of stringers (steel or timber), . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9. Number and location of street car tracks. . . . . . . . . . . . . . . . . . . . . . . . . . .

10. Dimensions of bridge seat and piers, if built. . . . . . . . . . . . . . . . . . . . . . . . .11. Distance floor to high water and to low water . . . . . . . . . . . . . . . . . . . . . . .12. Depth of low water and of ordinary stage . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13. Distance top of floor to lowest point of steelwork, if fixed by local

conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14. Give profile of crossing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15. Character of river bottom. . . . . Velocity of current. . . . . . . . . . . . . . . . . .16. Usualseasonsoffloods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17. Are piles necessary for falsework?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


18. Name of and distance to be hauled from nearest freight station. . . . . . . .19. Is old structure to be removed?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20. Is traffic to be maintained?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21. Lump sum or pound price, f.o.b. what point, or erected ready for

f loor ing, o r ready for traffic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22. Bids are received by whom and until what time? . . . . . . . . . . .23. When shall bridge be completed? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .24. Shall general reaming be done?. . . . . . . . . . . . . . . . . :. . . . . . . . . . . . . . . . .25. Shall edge planing be done and on what members?. . . . . . . . . . .26. Shall floor connection angles be milled after riveting? . . . . . . . . . . .27. Shall field connections be assembled at the shop? . . . . . . . . . . . . . .28. Kind of oil or paint to be used a n d how many coats. . . . . . . . . . . . . . . .29. Otherrequirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

GENERAL DATA FOR RAILROAD BRIDGES

1. Name of railroad . . . . . . . . . . . . . . Location of bridge. . . . . . . . . . . . . . . . .2. Specifications by . . . . . . . . . . . . . . . Class of live load. . . . . . . . . . . . . . . . . .3. Total length and general description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4. Number and lengths of spans c. to c. of piers . . . . . . . . . . . . . . . . . . . . . . . .5. Skew or angle of current with center line of bridge. . . . . . . . . . . . . . . . . . . .6. Number and spacing of tracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7. If bridge is on curve, specify curvature, speed and superelevation. . . . . . .8. Dimensions of bridge seat and piers, if built. . . . . . . . . . . . . . . . . . . . . . . . .9. Distance floor to high water and to low water . . . . . . . . . . . . . . . . . . . . . . .

10. Depth of low water and of ordinary sta.ge . . . . . . . . . . . . . . . . . . . . . . . . . . . .11. Distance top of floor to lowest point of steelwork, if fixed by

localconditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12. Give prof i le of crossing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13. Character of river bottom. . . . . Velocity of current. . . . . . . . . . . . . . . . . . .14. Usual seasons of f loods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15. Are piles necessary for falsework?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .16. Name of and distance to be hauled from nearest freight station. . . . . . . .17. Is old structure to be removed?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18. Is traffic to be maintained? How many trains daily? . . . . . . . . . . . . . . . . .1 9 . Lumpsum o r pound price, f.o.b. what point, or erected ready for

flooring, or ready for traffic?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20. Description of flooring, quality of lumber. . . . . . . . . . . . . . . . . . . . . . . . . . .21. What free transportation is allowed?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22. Floorbolts furnished by whom?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23. Bids received by whom and until what time?. . . . . . . . . . . . . . . . . . .24. When shall bridge be completed? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .25. Shall general reaming be done?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26. Shall edge planing be done and on what members?. . . . . . . . . . . . . .27. Shall floor connection angles be milled after riveting? . . . . . . . . . . . . .28. Shall field connections be assembled at the shop?. . . . . . . . . . . . . . . . .29. Kind of oil or paint to be used a n d how many coats . . . . . . . . . . . . . . . . .30. Other requirements., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 2 2 DESIGN OF’ STEEL BRIDGES [CHAP. VI

ART. 6. CLEAR HEIGHT BELOW CROSSING

For railroads the clear height, from top of rail to the lowest part ofthe overhead crossing, if not specified, should preferably not be less than22 ft., although in extreme cases vertical clearances of as low as 16+ft. have been allowed. (Public Service Commission, New York, re-quires 21 ft., other States allow 18 ft.)

The clear height above highways and electric railways should not beless than 16 ft. except for ordinary country roads, for which 123 ft.suffice (see also Specifications, Vol. I).

For navigable streams the clear height depends upon the height ofthe vessels and the highest water level at which navigation is possible.It is usually fixed by law. ,For the East River in New York it is 135 ft.at the center of the spans. 170 ft. clear height wbs specified for theproposed Sydney Harbor Bridge (Australia), and the proposed NorthRiver Bridge (New York City), on the ground that modern large steam-ers have steel needle masts without detachable top. For the NorthRiver Bridge the Board of Engineers recommended later 150 ft. only(same as for the Firth of Forth Bridge) as it is inexpensive to make thetop of higher masts removable. If the water is not navigable the clearheight above the highest water level should be at least 2 ft. andmore in case the water carries lumber, drift ice, etc.; but in any casethe lowest part of the bearings should be at least 1 ft. above the highwater level.

ART. 7. GRADES

The grade to be adopted on a bridge and its approaches may incertain cases affect considerably the economy of the design. A steepergrade decreases the length and therefore the cost of the approaches(important in cities where the approaches may be located on valuableproperty or require expensive retaining walls). On the other hand, thegrade should be limited to the importance of the traffic over the bridge,the length of the crossing, etc. Steeper grades can be allowed on countrybridges than on city bridges carrying a large amount of traffic; on shortbridges the grade may be steeper than on long crossings.

Railroads usually specify the grade which in general does not exceed0.5 to 1% depending upon the importance of the road.

For roadways the following grades in per cent. should not be exceeded:Asphalt 1%; wood-block 34; planks 4; macadam, stone-block or

brick 5. The grade for footwalks should not exceed 5% for asphaltor 8% for rougher surfaces.

ART.~] TYPES OF BRIDGES AND ,PRINCIPAL DIMENSIONS 123

For long crossings a grade toward the middle improves the appear-ance of the structure.

ART. 8. REQUIREMENTS FOR SPAN LENGTH

For structures crossing railroads the length and arrangement of spanswill depend upon the requirements of the railroad which is to be crossed.For economy any available space between tracks should be utilized forintermediate supports provided these can be protected against derailedtrains and are not otherwise obstructive to the traffic.

Roadways should be cleared entirely if possible. In some cases(El. R. R., New York) columns have been placed on the center line ofwide roadways. If there are sidewalks the columns may be placed onthem near the curb.

For navigable streams the span depends upon the number and thesize of the passing vessels. There should he at least one main opening,not less than 150 ft., where the river makes a curve.

The clear span or the sum of the clear spans of a bridge crossing astream will often depend upon the following two conditions:

(1) The rise of the water level above the bridge due to the piersshould not cause damage to the river banks.

(2) The velocity between the bridge piers should not endanger thefoundations by scour.

If & = volume of discharge in cu. ft. per second,A = area of unobstructed waterway above the bridge in sq. ft.,A, = area of clear waterway between the piers in sq. ft.,B = width of stream at water level above the bridge in ft.,Z = head or rise of water caused by the piers in ft.,V = reduced average velocity cf water immediately above the

bridge in ft. per second,V1 = velocity of water between piers in ft. per second,g = acceleration of gravity = 32.2 ft. per second,c = coefficient of contraction,

we have, according to Prof. Bresse, the following relations:

(1) & = (A+BZ) V or Q( l a ) v = A+%2

(2) Q = cAIVl or (24 v, = $1

(3) 2 = 7 or

c4) z = $ [& - r;Bz)- 21


Pn these formulas &, B, A, c and g must be regarded as knownvalues. In a given design A1 is known and VI and Z can befound by formulas (2a) and (4) respectively and must not exceedthe corresponding permissible values.

TABLE 2-l

Piers for Through Spans

Railroad

Memphis Bridge. . .T h e b e s B r i d g e .P . R . R . . . . . ~

New South Wales.N o r t h e r n P a c i f i c ,C. B. & Q.. . .P . R . R . . . . . . . . . . . . . . .Northern Pacific. ,U n i o n P a c i f i c . . .C.B.&Q . . . .._.......

U n i o n P a c i f i c . . .Peoria & Pekin. . . . .Tenn. Cen. .

I

-

t221

4121

2

:

II

0

-

Xst. 0. cf trumcr

f t . in .

Adjoining Widthspaqs c. c. under

pers coping

ft. ft. f t . in .

30 0 620 62032 0 071 52131 0 540 54028 0 415 41522 0 403 40522 0 402 40230 8 411 15022 9 413 11528 6 250 25016 0 204 202

28 6 248 12016 0 156 15616 0 200 65

Note.;The first two bridges are cantilevers.

P. R. R. . . . . . . . . . . .P. R. R. . . . . . . . . . . .B. & 0. . . . . . . . . . . . .B. & 0. (on curve).B. & 0. .(on curve).Highway . . . . . . . . . . .Railroad . . . . . . . . . . . .Tidewater iR.R. . . . . .

Piers for Deck Spans-

-

1 9 61 9 6

E

3: “013 011 6

-

-

260 260200 200

iii iii

75 75300 300120 120136 136

-

-

13 012 017 0

14 0

i ii10 47 4

i ii

f ::8 0

10 09 0

::

1: :6 68 0

I

--

-

Lengthundercoping

f t . in .

47 639 042 0

42 034 033 0. . .34 043 033 0

43 023 024 0

1Ia

.,

-

Height ofmasonryhove first

stepft.

1529 65 8

1z6 5

. . . .7 0

10933

7 2

;ii

27 Ol 3 227 0 2 816 0 1 018 0 12

18 04 7 6 :;:2 0 0 7 02 7 0 9 0

Piers Carrying One End of Fixed and One End of Swing Span

Fixed ISwing /

span, ft.

P.R.R . . . . . . . . . . . . . 2 31 0 540 330 17 0 42 0L. N. 0. & T. R. R.. i 17 0 300 275

i; i24 6 isi

P. R. R. . . . . . 29 8 260 280 36 0P.R.R .,..............; 29 8 200 280 9 0 36 0 I::

Note.-Length under copying is exclusive of pier-heads

If the minimum permissible A, is desired to fulfill above conditions,Z and V are also given values; A, is then first found from formula (4)


and its value introduced in formula (2~s). If ‘CT1 thus found exceeds itspermissible limit, the latter is introduced in (2) and another A1 is thusfound. Finally 2, if desired, is obtained by formula (4) by trial.

For the determination of & see any book on Hydraulics. The coeffi-cient c is according to Navier:

c = 0 . 9 5 for round and acute-angle pier-heads,c = 0.90 for obtuse-angle pier-heads,c = 0.80 for square pier-heads.

In case the river carries drift ice the spans should not be made lessthan 70 to 100 ft.

Often the location of the piers is determined by places affording goodfoundations.

In case of monumental bridges esthetic consideration may influencethe arrangement of the spans. An uneven number of spans produces abetter effect if the number is small. Two spans with one middle supportgenerally do not look well.

In order to determine the clear waterway the size of the piers mustfirst be assumed. For railroad bridges the following formulas give goodresults:

z+ 1’Single track . . . . . . . . .b = 4 + 100

1 + 1’Double track with 2 trusses . .b = 6 -I- 100

where b width under coping, in feet and 1 and 1’ span length carried bypier, in feet. The batter is usually 1 :24 all around.

Table 27 gives dimensions of existing piers.

ART. 9. ECONOMICAL SPAN LENGTH

If none of the foregoing conditions need to be considered, the mosteconomical span lengths should be used. These depend upon the relativecost of substructure and superstructure. The cost of a pier includingthe foundation, although increasing somewhat with the span length,may be assumed constant for spans of approximately the same length,especially for short spans for which the piers require certain minimumdimensions.

If L = length of crossing (or any portion of the crossing),1 = length of span,

n = : = number of spans in length L,9

126 DESIGh OF STEEL BRIDG’ES [CHAP. VI

G = cost of steelwork per lb.,A = cost of one pier (incl. its foundation),K = total cost of crossing (excluding cost of abutments and floor-

ing, which, being constant, do not affect the span length),w = x; 1 + y = ,weight of steel work in lb. per lin. ft. of crossing

(r and y being constants for the same kind of bridge) we have

K=na(zZ+y)Z+(n-l)A = LazZ+Lay+;A-A

Differentiating for I we get for minimum K

ClK= Lax - i A = 0 and Z =

IA--CZZ TX

when A and a are given in the same units.Giving A in units of 1000 dollars and a in cents per lb., we get

Economical span length I = 316d

Aax’ - - - * * (1)

Coefficient y has no influence on I, which is evident, since it representsa constant *weight per lin. ft. independent of the span length.

Plotting the values c = fi vertically and Z horizontally, we can

represent A-J-

va

; by straight diagonal lines from the origin of the system

(1 = 0 and c = 0) owing to the linear proportion

iI= ;d zi-; x 316

Instead of denoting the ordinates withu’

ia we use the corresponding

values of a and also mark the diagonal lines with the values A instead

of A--\i

-.X

Plate IV shows this applied to simple span railroad bridges.

The above formula for Z applies to any portion of a crossing in whichA is constant for all piers, but since A generally changes, the length Iwould have to vary also in the same crossing. It should be remembered,however, that within a certain limit the cost of steelwork is’the smallerthe more spans are alike and, further, that even a ‘considerable changein the span length affects the total cost only slightly, so that the spanlength should not be varied unless there is a great difference in the costof the piers. For any portion having equal spans, the average cost of apier in that portion should be used for A. Very often the cost of thepiers is an uncertain item and may be considerably greater than the.

3.50

3.75

4.004.254.504.755.00

DECK PLATE-G IRDERSPANS.

Span Lerigth. Span Length

Span Length.0 50 100 150 200 250 390 350

PLATE I'V

THROUGH PLATE-G I R D E R

SPANS.

400

3.003.25

3 . 5 03.75 THROUGH LATTICE

ANDP IN S PAN S.

0 50 100 150 200 250 300 350 4 0 0

S porn Length.Economical Span Lengths (in feet) for Single Track Railroad Bridges (simple spans) for E40 Loading.

N&.-Diagonal Lines Represent Cost of 1 pier in Units of $1000. (Facing page 126)

ART. lo] TYPES OF BRIDGES AND PRINCIPAL DIMENSIONS 127

estimated cost and it is then advisable to make the span Iength greaterthan would seem economical.

If the formula w = % I + y for weight of steelwork per lin. ft. isnot known, the economical span length can be derived approximatelyfrom the known weight of a similar span 11 by making in formula (1)x equal to the weight of trusses plus bracing per ft. of that span dividedby II (y represents then the weight of the floor system per lin. ft.).

ART. 10. DETERMINATION OF EFFECTIVE SPAN LENGTH

As a rule, if there is only one span the clear length I, betweenfaces of abutments, and if there are several spans the length lo betweencenters of piers is given and the effective span I, that is, the distancebetween centers of bearings of the main girders or trusses, has to bedetermined. If a is the length of bearing, b the distance from the edgeof the bearing to the face of the masonry and c the distance betweenthe edges of the two bearings on a pier, we have

FIG. 1. FIG. 2.

1 = &+a+2 b (Fig. 1)or

I = lo-a-c (Fig. 2)

For ordinary bridges b = c = 3 to 6 in. The length of bearing acan be determined by a preliminary calculation of the reaction andthe required area of bearing as explained on page 190. Table 39 onpages 191 and 192 give the size of bearings for various truss bridges.The following formulas give also good results.For R. R. bridges (E-50 loading and open flooring) :

S. T. plate girder spans a = 0.02 Z+O.75 ft.D. T. plate girder spans a = 0.02 z+1.5 ft.5. T. truss spans a = 0.01 Z+l.O ft.D. T. truss spans a = 0.01 z+2.0 ft.

For highway bridges up to 20 ft., wide and spans up to 100 ft., a = 1 ft.;for spans above 100 ft., a . = 0.005 Z + 0.5 ft.


ART. il. HEIGHT TOP OF FLOOR TO UNDER CLEARANCE

This height is, as a rule, measured for railroad bridges from base ofrail (on’ curves from base of low rail); for highway bridges either fromtop of floor at the crown or from top of curb to the lowest part of thesuperstructure, which should not come withi~l the specified clear heightbelow. The arrangement of the floor will in most cases depend upon thisavailable height; the greater this height up to a certain limit, the moreeconomical and stiffer will be the structure. It is of importance for theengineer locating the line to know approximately the economical height

as well as the minimum height for various

+, kinds of bridges:; +- -

Ii j

J=YIT

The total height h (Fig. 3) is made up of

i he the height’ h1 of the flooring (ties, ballast,planking, paving) and the height ha of the

L+--.-- steelwork below the flooring. Figures for hl;3

can be obtained from the data given onu pages 132, etc.

FIG. 3. The minimum values for h are approxi-mately

73 to 11; in. for railroad bridges with open tie floor;20 to 24 in. for railroad bridges with ballast floor;3 to 6 in. for highway bridges with plank floor;

10 to 15 in. for highway bridges with paved floor;53 to 93 in. for electric railway bridges with open tie floor;15 to 20 in. for electric railway bridges with ballast floor.

For hs the formulas on page 129 give good approximate results.Minimum ht is not the smallest possible height, but the smallestheight for a stiff structure, and should preferably not be reduced.

In order to obtain the distance from the top of the floor (or base of rail)to the top of the masonry, the following minimumvalues should beadded to the height h:R. R. Bridges:

1% in. for plate girder spans with sliding bearings (generally spansless than 80 ft.);

1 ft. 6 in.; for plate girder spans with shoes and roller bearings;0.005 I (min. 1 ft.); for single track truss spans;0.007 I (min. 1.2 ft.); for double track truss spans.

Highway Bridges:0.003 Z (min. 0.7 ft.) for spans 100 ft. and more and roadwayup to 20 ft. wide.


Kind of Bridge

Single Traclc R. R., Cooper’s E-50 loading.

I-beam spans’. . . . . . . . .D e c k p l a t e g i r d e r s p a n s . . .Through plate girder spans. .Deck truss spans. . . . .T h r o u g h t r u s s s p a n s . . .

_---_____-

Double Track R. R., Cooper’s E-50 loading:Through pl. girder spans with 2 girdertDeck truss spans with 2 trusses .Through t,russ spans. . . . . . .

~---- - --__---

Highway:

I-beam spans . . . . . . . .Through spans.. . . . . .

- - ~~--

Single Track Electric Railway:

I-beam spans. . . . . . . . . . . . . . . .Deck plate girder spans . . . .Through spans. . . . . . . . . . . . . . . . . . .

*Limited to 10 ft. over all.**b = width of roadway.

Economical Minimumhz hz

0.1 1&Z f 2ft.* 2:

3 ft. 2 ft.%I isz,l

5 ft. 3 ft.-----P-w

4ft. 3 ft.41 81

6 ft. 4:ft.~--

& 1 5’0 1O.lb** + 1 ft. . . . . . . . . . . . .

- - -

ii 1is 13 ft.

ART. 12. CLEAR WIDTH AND HEIGHT ON BRIDGE

Concerning the clear height in through bridges the same applies as in

Am Ry. Eng Assoc., 1909

k-- /f--q

I? RR., 1906

$.. ,&‘!-.;

De$t. R)s & Cana ls ,Canada, 1905.

F I G . 4,

Art. 6. Fig. 4 shows various specified clearance diagrams for railroadbridges on tangent. For bridges on curve see page 217.

The distance c. to c. of tracks is usually 13 ft.


The clear width for electric rcsilroads should not be less than 13 ft.and the distance c. to c. of tracks 12 ft.

The clear width for highway bridges depends upon the density oftraffic. It is desirable to have the same width on the bridge as on theopen road, considering also the probability of a future widening of theroad. Where economy is desired the width may be reduced on the bridge,especially if the traffic is light or the road leading to the bridge is of exces-sive width. The clear width should preferably be not less than 20 ft.,except for unimportant country bridges for which it may be 14 ft. Whenthe distance between curb lines or wheel guards is fixed, at least 1 ft.,preferably 2 ft., should be added to it to get the clear width.

Sidewalks are preferably placed outside of the girders and should havea clear width of at least 5 ft. each. This may be reduced if necessary atcertain points of the bridge, for instance, at extra wide end posts of trussspans, towers, etc. Widths over 8 ft are used on bridges in large citiesonly.

ART. 13. ARRANGEMENT OF CROSS-SECTION FOR BRIDGESCARRYING COMBINED TRAFFIC

Bridges carrying different kinds of traffic can be built at less cost thanseparate bridges. For instance, the additional cost of the superstructureof a bridge carrying a railroad and an unimportant highway over that of aseparate railroad bridge is practically only the cost of the highway flooras the live load on the highway can be neglected for the main trusses.The principal saving, however, is usually in the substructure, especiallywhere the foundations are expensive, as the piers can be the same or onlyslightly longer than for a railroad bridge. Short spans are not built forcombined traffic, the saving being offset by the expense of bringingtogether the different lines of traffic on the same bridge.

The arrangement of the cross-section depends upon the number of thedifferent kinds of traffic (steam and electric railways, trolleys, vehicles,pedestrians) to be accommodated, the width required for each, the availableheight, the approaches, the length of span, etc., so that the conditions varyin almost every case and require special studies and comparative esti-mates. The possibility of future increase of traffic should be consideredand provision made for it at the least expense.

There may be either one or two decks. A single deck will be usedwhere the available height from top of floor to under clearance is limitedand in all cases where the traffic can be accommodated without materiallyincreasing the distance between the trusses. The railroad tracks arealways placed inside of the trusses symmetrically and preferablly in the


center of the bridge. The roadways, footwalks and trolleys are placedeither inside or outside on brackets; trolleys preferably adjacent to thetrusses. The space occupied by the trusses can often be utilized as partof the footwalks. Screens should be provided between the railway andthe roadway in order to prevent the frightening of horses.

Sometimes the same space is used for railway and roadway; thissaves in cost but is not satisfactory in operation. It may be justifiedwhere both kinds of traffic are very light.

If a considerable width between trusses would be required for a singledeck, a double-deck structure may be more economical, especially onaccount of the smaller cost of the substructure. But this saving maybe offset by a greater cost of the approaches. Whether the railroadshould be on the upper or lower deck will depend upon local conditions.

For examples see Chapter XVI, Long-span Bridges, page 333.

CHAPTER VII

DESIGN OF FLOOR(Plates v, VI, VII)

A. RAILROAD BRIDGES

ART. 1. OPEN TIE FLOORING

Most railroads have adopted standards for bridge flooring. Fig. A,Plate V shows a typical tie flooring. The timber is usually yellow pineor white oak. The ties are generally 10 ft. long, 8 in. wide, at least 6 in.(usually 8 in.) deep and spaced with openings of 4 to 6 in. The depthdepends upon the bending moment produced by the greatest axle con-centrationwhich, as sumed distributed over three ties and with 100%impact, should not cause a greater bending stress than 2000 lb. per sq. in.If M is the moment in inch pounds on one tie, the depth in inches is

Footwalks for trackmen are formed by extending every second or thirdtie about 5 ft. to one side covered with 2-in. planking. The ties arenotched at least 4 in. over the stringer flanges and every fourth tie issecured by a j-in. bolt. A guard timber at least8 in. by 6 in. is laidflat at a clear distance of 10 to 15 in. outside of each rail, notched 1 in.over the ties and fastened generally every third tie and at each splice by a$-in. bolt. Track bolts usually weigh from 4 to 5 lb. per ft. of track.

Inside guard rails (of timber, angles or steel rails) with a clear distanceof 8 to 10 in. from the track rails are now generally used, extending about60 ft. beyond the end of the bridge where the trains enter and their endsare draqin together and connected by a frog point.

The weight of the open flooring (rails, ties and fastenings) usuallyranges between 400 and 500 lb. per ft. of track.

ART. 2. -BALLAST FLOORING (See Plate V)

Bridges with ballast flooring are more expensive and require a greaterheight from base of rail to under clearance; but they have the advantageof a smaller impact effect and less noise from the trains and can be made

1 3 2

ART. 21 DESIGN OF FLOOR 133

water- and fireproof. Moreover, the uniformity of the track constructionis not interrupted on the bridge, securing smooth running of the trainsand the possibility of rearranging tracks, switches, etc. For these reasonsseveral railroads have adopted a solid floor for short bridges, especially onstreet crossings in cities. i i

The ballast in which the ties are embedded is supported by the steelfloor construction either directly or by means of a base of concrete, timberor some other material..

Ties and Ballast.-The ties may be of the same size and spacing as forthe open line, generally not less than 8 in. by 6 in. by 8 ft. with a maximumspacing of 24 ft. c. to c. The ballast should be at least 6 in., preferably8 to 12 in. deep under the ties. It is very important to provide for drain-age of the ballast in order to lengthen the life of the ties.

Concrete Base.-The ballast usually rests on a concrete base withsloping surface and holes through which the water can drain off. Thisconcrete base may bc a self-supporting slab resting on the stringers (Fig.B), or on transverse I-beams (Figs. D and 1) and should be properly re-inforced by steel bars and proportioned to carry safely the heaviest loadconcentrations; it should be at least 6 in. thick.

The concrete base may also serve merely as a filling between trans-verse I-beams (Fig. C and 6) or rest on top of a solid steel floor whichmay consist of transverse troughs (Fig. H) or a flat steel plate rivetedto transverse I-beams or buckle plates riveted to the top flanges of thestringers and floorbeams. The concrete filling should extend at least l&,preferably 2+, in. above the steel (including rivet heads).

The mixture of the concrete should be about in the proportion of1:2 :4.

Instead of a concrete base for the ballast some railroads use preservedtimber ties laid close together or planking on top of transverse I-beams.This is, however, not as durable and fireproof as concrete.

Ballast Resting Directly on Steel Floor Construction.-To save theconcrete base or reduce the height of the flooring the ballast is sometimesplaced directly into the troughs or on the steel plates (Fig. E and J),but this is not advisable since the steel surface is not open to inspectionand the water cannot be effectively drained except in the case of invertedb.uckle plates provided with a hole in the center of each buckle. In anysuch case the top surface of the steel should get a heavy coat of a goodpreserving material. The arrangement with ties laid into the troughs(Fig. M), to reduce the height is not satisfactory as it is difficult to tampthe ballast under the ties.

134 DESIGN OF STEEL BRIDGES [CHAP. VII

Steel Floor Construction for Ballast Flooring.-For deck spans a self-supporting reinforced concrete slab (Fig. B) or transverse I-beamsspaced from 15 to 18 in. c. to c. with concrete filling between (Fig. C)are best suited. For through spans with sufficient height from rail tounder clearance to allow the use of floorbeams and stringers, buckleplates are generally cheapest. Buckle plates should be not less than -;I$ in.thick (preferably # in.) and be, as far as possible, of standard sizes,since every new size requires a special die and thus increases the cost.(For sizes of buckle plates see Table 28.) The buckle plates are prefer-ably riveted to the first flange plate if there is any, that plate beingmade about 3 in. wider than the other flange plates.

As a rule the height is not sufficient for floorbeams and sOringers withballast flooring on top; a transverse I-beam or trough floor (Fig.F, G, etc.) is then riveted directly to the main girders. The I-beamfloor is generally preferable as it requires little shop work and is easier forerection. For typical trough sections see Figs. H and 0.

Calculation of Transverse Beams or Troughs.-Corresponding to thecalculations of cross ties for the open flooring the transverse I-beams ortroughs may be proportioned by assuming the axle load uniformly dis-

L+ .__.._...._......_ 3.5 fhb +p&. ..__.-..._.-....... q

max. S!

F I G, 1 .

tributed over a length of track of about 3.5 ft. at the bottom of theties (Fig. 1). This length may be increased by the depth hb of the ballastbelow the ties plus twice the depth of the concrete above the supportingsteel to a maximum of 5 ft. or the distance between the heaviest axle loads.The calculation of a self-supporting concrete slab should be based on thedistribution of the load at the top of the slab. The distribution of theload at right angles to the rails usually does not reduce considerably thebending moment in the transverse beams and it is therefore sufficient,

ART. 41 DESIGN OF FLOOR 135

and on the safe side to calculate as if the rails would rest directly on thebeams or troughs.

The beams and troughs should be proportioned by their section modu-lus (see page 143), proper allowance being made for the rivet holes in thetension flange. For a first approximation the section of a trough floormay be determined by calculating the required effective flange area asfor a plate girder, 8 of the net area of the web being counted in theflange area.

ART. 3. SOLID STEEL FLOORS WITHOUT BALLAST (See Plate V)

Occasionally the available height between rail and under clearance isso small that the arrangement of floorbeams and stringers is not practi-cable even if no ballast is used on top. A transverse I-beam or troughfloor is then riveted to the main girders as mentioned in Art. 2; theI-beams are preferable as the troughs allow water and dirt to collect.

A $oor plate covering the whole floor is connected to the tops of theI-beams and on top of this are riveted two continuous r&l plates about18 in. by 3 in. on which are placed the steel guard angles and the rails,the latter being usually bolted down by means of clip plates (Fig. L).The floor plate should not be less than 28 in., preferably i in. thick.In case of a trough floor which is stiff by itself the floor plate maybe omitted (Fig. K). To reduce the impact, an elastic material(leather, etc.) is often placed between the rail and the rail plate, ortransverse wooden ties are placed into the troughs projecting about aninch above the latter. In some cases the ties have been placed on short ’wooden stringers between the transverse I-beams (Figs. M and N).

ART. 4. ARRANGEMENT OF FLOOR SYSTEM (See Plate VI)

As mentioned before stringers and floorbeams should be used in con-nection with open as well as ballast flooring whenever the available heightpermits it, since shallow floors are in general not economical and not stiff.

In through spans the stringers are as a rule framed into the floor-beams, as the available height does not permit them to be placed on top ofthe floorbeams (Fig. A, etc.).

Fig. G, etc., show various arrangements of floor in railroad decks p a n s . Arrangement, Fig. G (ties resting directly on top chord) isused only for short spans (up to 120 ft.). If floorbeams and stringersare used, the constructions shown in Figs. H and J are the simplestas far as the lateral system is concerned, moreover, for equal depth


of trusses, the piers are lower than for other arrangements, but the distancefrom rail to under clearance is greater. Arrangement shown in Fig. Iis seldom used, as it is more expensive to frame the stringers betweenthan to place them on top of the floorbeams. It has, however, the sameadvantages as the arrangement shown in Fig. L, as the chords formguard rails. In the arrangements, Figs. K and J, the chords form derail-ment stringers or may be used as main stringers in double-track spans.

Spacing of Stringers.-The general practice is to use two stringersfor each track, spaced 6 ft. 6 in. c. to G. Sometimes a wider spac-ing up to .8 ft. or more is used on single-track bridges; this gives amore elastic track and reduces the bending moments on the floorbeams,but requires heavier ties. Some railroads use main stringers 5 ft. apartwith outside safety stringers of about one-half the strength of the mainstringers; this insures greater safety in case of derailment but increasesconsiderably the cost of the floor system.

Panel Lengths.-For through plate girder spans, the economical panellength will approximately be that requiring the least total weight of thefloorbeams and stringers, and is best found by trial calculations. Forsingle-track plate girder spans and economical depth of the floor (seepage 129) 15 ft. is about the best panel length. The smaller the availabledepth, the shorter must be the panels. If the panels are short, it may bedesirable to arrange the laterals over two panels, in which case an evennumber of panels must be used. For truss spans of ordinary length, thepanels are made up to 30 ft. long (see page 173). The panels in anybridge should be of the same length.

ART. 8. PRINCIPAL DIMENSIONS OF STRINGERS AND FLOORBEAMS

The distance between centers of floorbeams should be taken as theeffective span of the stringers, the distance between centers of main girdersor trusses as the effective span of the floorbeams.

In short panels, rolled beams may be used as stringers. For thecalculation of I-beams see page 143. Usually, however, the stringers andfloorbeams are plate girders and are, therefore, governed in their designby the rules given on page 148. The depth of the stringers should, pref-erably, be not less than 6 to 3 of their span to insure rigidity. Cover-plates increase the cost on account of the extra riveting and the extrawork in framing the ties and should, therefore, be avoided if possible;stiffeners may also be avoided to advantage by using a thicker web.The stringers should be braced by a lateral system in the plane of their

ART.~] DESIGN OF FLOOR 137

top chords. One angle 33 X 33 X 3 in. for each lateral will generallybe sufficient. The following stringer sections are economical for Cooper’sE-50 loading and specifications of the American Railway EngineeringAssociation (see Appendix) :

Panel length Depth of stringer Web plate Flange angles

ft. ft. in. in. in.

:: N o stiffeners20 rcguired.

25

i:stipe.5 re-

21 x $628 x &35 x j$43 x 444 x g53 x #62 x 8

6X4Xg*6X4x+sxsx+6X6X&6X6X&6X6XQ6X6x=16 I

*4-in. leg horizontal.

The floorbeams should preferably have a depth of not less than 6 toQ of their span for single-track and 6 to + for double-track bridges,especially where the floorbeams are framed in between the trusses, as thestiffness of the floorbeams adds materially to the lateral stiffness of thebridge.

The depth of the’floorbeam is generally determined by the depth ofthe stringers, if the latter are framed in. In this case, the bottom of thestringer should clear the bottom flange angle of the floorbeam, so as toleave room for a seat angle under the stringer to allow the laterals to passbelow the stringers without being spliced, and to avoid fillers between thefloorbeam flanges. The stringer should not be seated on the bottomflange of the floorbeam except where necessary in shallow floors. Thetop of the floorbeam should be at least 14 in. below the base of rail.

In deck spans where the stringers rest on top of the floorbeams, themost economical depth for the latter can be selected, but for short floor-beams subject to a comparatively small moment and great shear the depthis often governed by the number of rivets required in the end connection.The top flange of floorbeams should preferably be not wider than 14 in.,as otherwise the space between the ties becomes too great.

Trussed floorbeams are not advisable, as the possible saving of ma-terial is offset by the higher cost of manufacture.

Stringers should be rigidly connected to the floorbeams and the lat-ter to the girders or trusses. Where the floorbeams extend considerablybelow the bottom of the stringers, they should have stiffeners at or nearthe stringer connection to transmit the bending caused by the deflectionof the stringer to the flanges of the floorbeam.


B. HIGHWAY BRIDGES (See Plate VII)

ART. 6. PLANK FLOORING

The plank flooring is the cheapest and is satisfactory for light traffic.It is not suitable where a watertight floor is required and where the plankswould require frequent renewal on account of heavy traffic. For lighttraffic, as on country bridges, one layer of planks is sufficient; the planksshould be of hard timber (yellow pine or white oak) about 9 in. wide andat least 3 in. thick or & of the distance between joists; laid with $-in.openings and spiked either to wooden joists or to spiking pieces 3 in. X8 in. or 4 in. X 6 in. which are bolted to the tops of the steel joists. Fortraffic of medium density, as in smaller cities, it is more economical to usea wearing floor of l&in. planks with t-in. openings on top of the 3-in.planks which in this oase are laid diagonally with g-in. openings.

On each side of the roadway, a guard rail 6 X 4 in. is laid flat on theplank floor and bolted thereto to prevent the hubs of wheels from strik-ing any part of the bridge (see Fig. A, etc.).

The footwalk planks should be about 6 in. wide and at least 2 in.thick and laid with i-in. openings.

Steel joists are preferable to wooden joists owing to their greaterdurability and should be used in any case below trolley tracks and forsolid floors. Where wooden joists are used, their width b should notbe less than 3 in. or $ of their depth d. The required depth is

where M = bending moment fcom the specified wheel concentrations ininch-pounds and s = permissible fiber stress.

The treatment of timber by some preservative process materially in-creases its durability.

ART. 61 139DESIGN OF FLOOR

TABLE 28Buckle Plates

(American Bridge Co.)

-I

-

-

S i z e o f b u c k l e ID i eN o .

R iseP

Numbero f buck l e s

in onep l a t e

R a d i i o f b u c k l e s

S i d e L 1-S i d e W F o r L For w-

f’;. i n .3 114 63 1 13 6

ft. in.

f 1;

i 1:

ft . in.6 85

; ii 2

$ ;14 lOi7 1;

10 2

1: x3 7;

10 26 103 log4 10;4 7;3 10;5 ll?T6 3

7”2 # 7,

4 .7i4 lo*3 10;2 62,

1 3 121

ft. in.8 9;6 8%

; ;I7 1;7 1;4 105

10 210 2

l”o “z3 746 103 1024 7%3 lO$3 10%

g 1;-9-7 18”

: 1%Fj l$ 1L

4 7 ;2 6&3 10;5 4 4

1 3 1fQ8 la

Ito 81to 7lto 8lto 9l t o 81 to 101to 81to 8 t1to111to 81 to 1 41to 81 to 101to 1 11 to 121to 1 11to 12lto 9lto 9lto 9l t o 8l,to 91 to 101 to 101 to 101 to 121 to 1 51to 5lto 91to 7

3 9 3 93 cl

i ii3 82 8

2 63 6

i :2 02 G

; ii4 0

Plates are L 25- -7 or & in. thick.4, 10, 8Buckles of different sizes should not be used in the same plate.End flanges to be made alike, minimum 2 in., maximum 1 ft. 6 in. If made wider

than 1 ft. 6 in. use angles riveted across plates for stiffeners.Side flanges Wl and W2 to be made alike, minimum .2 in. maximum 6; in.For fillets L2, minimum 2, maximum 6 in.Connection holes: are generally for 8, Q, or s-in. rivets or bolts.Spacing for holes lengthwise of plate should be in multiples of 3 in. and should

not exceed 12 in. Minimum spacing crosswise 4$, usually 6 in.10


ART. 7. PAVEkENT FLOORING

To allow’the water to drain off the surface all paved roadways shouldbe crowned from curb to center to an amount depending on the longitudi-nal grade and the roughness of the surface, although very frequently noattention is paid to any of these conditions. The steeper the grade andthe smoother the surface the lower should be the crown.

For the parabolic cross-section, which is most frequently used, thefollowing values c of the crown at center in per cent. of the total width ofroadway between curbs give good results:

For level roads withsmooth surface (asphalt, woodblocks) c = 13% ,rough surface (macadam, stone blocks, brick) c = 27;.

For roads on grade the above values of c should be decreased by3 y0 for each per cent. of longitudinal grade to a minimum of 2%. Forinstance, a macadam road with a grade of 4% and a width of 50 ft.should have a crown of c = 0.01 X 50 = 0.5 ft., while on a level itwould have twice as much.

A cross-section consisting of two straight lines connected by a shortcurve at the center is preferable, as the parabolic form is frequently tooflat at the center and too steep near the curbs. For this roof-shapedsurface $ of the values of the crown as given above for the parabolicsurface are sufficient.

If there are street car tracks the two rails of each track should havethe same elevation and their tops be level with the surface. The tracksshould therefore be placed at the center of the road and the crown is thenmade the same as for a width of roadway equal to twice the distancebetween curb and outer rail. The footwalks should slope 1: 50 to 1: 100toward the roadway.

If the grade is not too steep (up to about 13% asphalt makes asatisfactory wearing surface, as it is smooth, light and cheap. Thecost of maintenance, however, is high. The asphalt layer should be atleast lg in. thick.

Woodblocks are generally preferable to asphalt and may be used forgrades up to 3$%. They are usually made 4 to 6 in. thick and rest ona concrete foundation with an intermediate $-in. layer of cement mortar.

Macadam and stone or brick pavements allow grades up to 5%.They are durable and the cost of their maintenance is small; their dis-advantages are the great weight and the rough surface. Stone blocks(usually granite) are made 4 to 6 in. deep and are embedded in a layerof sand, which should be at least 1 in. thick when resting on a concrete

A R T . 81 DESIGN OF FLOOR 141

foundation, and 4 in. when supported directly by a flat steel plate orbuckle plates. Macadam should have a depth of at least 4 in., preferably6 to 8 in.

The wearing floors, as described above, usually rest o.n a layer of con-crete which transmits the load to the steel floor. This concrete may beeither a self-supporting slab at least 6 in. thick between cross-beams orstringers, if necessary reinforced by steel bars, or it may be a layerextending everywhere at least 2 in. above the supporting steel floorwhich may consist of buckle plates or some other kind of continuous steelfloor as described for railroad bridges. Cinder concrete is sometimesused for the filllng to lessen the weight, but as it corrodes the steel it isnot to be recommended. To secure water tightness, especially for wood-block pavement, a bituminous concrete should be used or a layer of water-tight material (cement or asphalt).

The crowning of the roadway is usually obtained by increasing thethickness of the concrete foundations from curb to center. For very wideroadways it is more economical to provide for it in the steel floor. Buckleplates or other steel floor plates should be at least & in. thick.

Concrete for sidewalks should have a least thickness of 4 in. whenself-supporting or 2 in. when resting on buckle plates. The wearingsurface usually consists of 1 to 13 in. cement mortar or asphalt.

The curbs of paved roadways should be of steel, concrete or stone 4 to6 in. deep above the roadway surface.

Distribution of Wheel Loads.-A wheel load may be assumed dis-tributed by the ballast and concrete over a rectangle with the sides b + 2h

FIG. 2.

parallel to the wheel axle and 2 h at right angles to the wheel axle whereb = width of wheel and h = height of ballast and concrete (Fig. 2).

, ART. 8. STRINGERS AND FLOORBEAMS

If the available height is very small, stringers are omitted and ashallowfloor system used similar to those described for railroad bridges.

Whenever possible, however, stringers and floorbeams should be used.

142 DESIGN OF Sl’EEL BRIDGES [CHAP. VII

The stringers are usually placed on top of the floorbeams if the avail-able height permits this. In deck spans the floorbeams are either framedbetween the trusses or placed on top of them. In the latter case theyshould be securely stiffened against overturning if the stringers do notframe into them. Stringers and floorbeams are made of I-beams when-ever possible. The outside stringers carrying a lighter load are usuallymade of channels of the same depth as the I-beams. Steel stringersare spaced about 3 ft. c. to c. for plank flooring and 4 to 5 ft. forpavement flooring. Where I&in. buckle plates are used, the spacingmay be 4 ft. for roadways and 5 ft. for footwalks. If there are trolleytracks, stringers are preferably placed directly under or near the rails.

The economical panel length for ordinary highway bridges is usually10 to 15 ft. for plate girder spans and 15 to 20 ft. for truss spans up to200 ft. The depth of floorbeams is preferably made not less than g ofthe distance between trusses or girders.

Maximum Spans of I-Beam-Stringers Placed 3 ft. Aspart

Size of I-beamFor 5 live load of

___G-ton witgon 15-ton road roller

9 in. - 21 lb. 12 to 16 ft.:: -25 - 315 “ ‘I ‘( “ &‘ I ‘

) 18 to 2 1 ft.;; 2 2 1 7 2 6 2 1 ;; 1; t22 “ <‘

- 42’ ‘1 27 ” 32 ” 3 0 (( 34 I‘- 55 “ 33 “ 4 0 I‘ 35 If 4 0 “

Brackets.-Brackets outside of the main girders or trusses are formedeither by extending the floorbeams or by using separate plate or latticegirders. Latticed brackets are generally used only for footwalks. Theconnection of separate brackets should be such that the bending mo-ment is transmitted to the floorbeams by rivets in shear.

It is sometimes desirable to support the sidewalk stringers by addi-tional brackets between the panel points. In this case, cross frameshave to be placed between the main girders or trusses to resist the mo-ment from the brackets.

CHAPTER VIII

BEAM AND PLATE GIRDER BRIDGES

(Plates VIII to XII)

ART. 1. I-BEAM BRIDGES

For railroad bridges I-beams are used for spans up to 25 ft. If twobeams are sufficient they are spaced 6 ft. 6 in. c. to c. ; if two orthree beams have to be combined to form one girder, the center line of thisgirder should be directly under the rail, so that the load is uniformlydistributed over the beams. The depth of the beams should preferablynot be less than & of the span.

The beams composing one girder should be‘rigidly connected by dia-phragms at each end and about every 5 ft., and by a bed plate at least $in. thick at each end. The two girders should have a rigid lateral systembetween them near the top flange, composed of angle bracing and a strutat each end. SingIe angIes 3s X 34 X $ in. are usually sufficient for thisbracing.

In highway bridges I-beams are used for spans up to 40 ft. and arespaced about 3 ft. between centers for plank flooring and 4 to 5 ft. forpavement flooring. The beams should be connected by end struts and byintermediate bottom struts 10 to 15 ft. apart; their depth should prefer-ably not be less than 7;‘a of the span for the roadway or &, of the span fortrolley tracks.

ART. 2. CALCULATION OF I-BEAMS

The I-beams should be proportioned by the moment of inertia of theirnet section. If M is the bending moment in inch-pounds, s the per-missible stress in lb. per sq. in. and y the distance of the outer fiberfrom the neutral axis (= 3 the depth of the beam if its section is sym-metrical about the neutral axis) in inches, the required moment ofinertia is

I=“y

and the required section modulus is

144 DESIGN OF STEEL BRIDGES [CHAP. VIII

As a rule rivet holes can be avoided in the tension flange, at leastat the section of maximum moment. If this is not the case provisioncan be made for any rivet hole in the tension flange by the following

approximate method: If R, = : is the required section modulus of the

net section, d the diameter of the rivet hole, and t the thickness ofthe flange at the rivet hole, the required section modulus of the grosssection is approximately

If there are two rivet holes in the tension flange at the same section,

Example.--M = 2,800,OOO in.-lb., s=16,000 lb. per sq. in. Required sec-2,800,000

tion modulus 16,000 = 175; if no rivet holes are deducted one 24-in. 1-80 lbwith R = 174 will suffice. If a l-in. rivet hole is taken out in the tension flange therequired section modulus is R = 175 f 1 X 0.84 X 12 = 185, which requires one24%. I-90 lb. with R = 186.6.

ART. 3. DEFLECTION OF I-BEAMS

For deflections of I-beams and their relation to the fiber stress s andthe depth d see Vol I. For the calculation of the deflection due to aseries of concentrated loads the equivalent uniform load producing thesame maximum moment may be used.

It is usually specified that if the ratio 5 of the depth of the beam to

1‘its span length is smaller than a certain value ;, the section shall be

1so determined that the deflection will not be greater than if the ratio c,

had been used. This is equivalent to specifying that for depths smaller

than a,.=; the permissible stress shall be decreased in proportion

to the depth, that is, if so is the permissible stress for the depth d,= fp

the permissible stress for any smaller depth d is

ART. 4. SHEARING STRESSES IN I-BEAMS

As a rule the section of the beam determined by the bending momentis also sufficient for the shear. However, where the vertical shear is

ART. 61 BEAM AND PLATE GIRDER BRIDGES 145

relatively great and the bending moment small the shear may require aheavier beam. Assume first that the vertical shear V is resisted only bythe rectangular section td where t is the web thickness and d the depthof the beam; the unit shear is then

38 *q=2a

If this is smaller than the permissible unit shear (12,000 lb. per sq. in.according to Specifications, see Appendix) the section is sufficient; ifit is greater, calculate more accurately by the formula

where m = $ a, A = area of section, n = distance of center of gravity

of the area above the neutral axis from this axis and I = momentof inertia of the section. If this q is greater than the permissiblevalue the section has to be changed.

For combined bending and shearing stresses see Vol. I.

ART. 6. PLATE GIRDER BRIDGES

Plate girder bridges are in general built in span lengths of 15 to 120 ft.for railroads and 20 to 100 ft. for highways. For notable plate girderbridges see Table 29. Whenever the available height from top of floor tounder-clearance permits it the flooring is placed on top of the girders and’the steel floor construction is omitted. We have then a de& plate girderbridge in distinction to the through plate girder bridge in which the flooris placed between the girders.

ART. 6. NUMBER AND SPACING OF GIRDERS

Railroad Bridges.-For deck plate girder spans two girders for eachtrack are most economical. Adjoining girders of different tracks shouldbe either independent or connected by horizontal struts only, so thateach pair of girders can deflect independently. For single track spanson tangent the following distances c. to c. of girders are recommended:6 ft. 6 in. for spans up to about 65 ft., 7 ft. 0 in. for 70- to 80-ft. spansand 7 ft. 6 in. for spans over 80 ft., but not less than & of the span. Inspans for two or more tracks 13 ft. apart between centers, all girdersmay be spaced 6 ft. 6 in. provided struts are used between girders ofdifferent tracks.

Built

18501848

19051907

1904

1912

1900

1900

191119051909

I

-

Name or railroad

Britannia Bridge..C o n w a y B r i d g e . .Mill Street Bridge.Eastern Bengal Ry..Erie R. R.. . . . . .Lehigh Valley R. R.Riverside Drive BrE r i e R . R . . . .Croton B r i d g e . . .

N.Y.C.&H.R.R.RSixth Str. Bridge.. . .Boston 8-z Albany R. R

Del., L. & W. R. R. .N . Y . C . & H . R . R . R . .C., M. & St. P. Ry.N. Y. C. & H. R. R. R.N.Y..N.H.&H.R.RIndiana Harbor R. R..Tonnelle Ave. Bridge

TABLE 99

Notable Plate Girder Bridges

Location

E n g l a n d . . .England. . . .C i n c i n n a t iIndia. . . . .Hubbard, 0.. . . .Towanda, P a . . .N e w Y o r k . . . .B r a d f o r d D i v . .N. Y. Aqueduct . .

Jersey Shore, Pa.Phi lade lph ia . . .Worcester, Mass..Terre Haute, Ind.Scranton, Pa. . .J o r d a n , N . Y .Janesvil le , Wis.Albany, N. Y.. . . .Waterbury, Corm.. . . . . . .._......_... . . . . . . . . . . . . . . . . .

Over

Menai Straits. . . . . .. . . . . . . . . . . . . . . . . .‘Pilaiee iii;er : : : : : :

Yankee Run. . .S u s q u e h a n n a . ...................................

PineCreek........‘til .$..$ H .pt.H..~

Wabash River. .Lackawanna R.E r i e C a n a l . .Mill Race.. . . .H u d s o n R i v e r .Naugatuck River..Mich. Central R. RD., L. Q W. R. R..

Kind of Greatesttraffic span

D. T. Ry.

l$&wa;: :D. T. Ry.S. T. Ry.D. T. RyHighway.S. T .Ry.Highway.

D. T. Ry 128Highway. 1234 Tr. Ry.. 122 6Highway. 121 66 Tr. Ry.. 115 6D. T. Ry.. 116 0S. T. Ry.. 114 6D. T. Ry.. 116 64Tr. Ry. . 114S. T. Ry.. 110Highway. 109 6

ft . in.

450400213170131 4129 6129 3128 3124 6

Height ofgirders

ft. in.

“iS”0”. . . .

1: :10 39 6

9 at center18% at endi

9” !10 610 0

“iO”0”9 6

10 09 6

ii ii

Type

TubularTubular

ZCY”D e c k .

Ti&&Through

D e c k . .

‘ge&; : :D e c k .D e c k . .

EY”ThroughDeck.. .ThroughThrough

-

-

AI~T. 71 BEAM AND PLATE GIRDER BRIDGES 147

For through plate girder spans two girders per track are generally mosteconomical. If there are two or more tracks, however, there is generallyno room for two girders between two tracks without spreading the latter,which is objectionable. Either two outside girders and one between eachpair of tracks or, in the case of double-track bridges, only two outsidegirders have then to be used. The latter arrangement requires a greaterdepth of floor but only one kind of girders. The minimum distancebetween centers of girders is equal to the specified clear width at thelevel of the top flange plus the width b of the 1”atter in feet, which is(if I = sp%n in feet)

b = 0.9 + 0.0075 1 for S. T. spans,b =,1.2 + 0.0075 I for D. T. spans.

Highway Bridges.-If the floor can be placed on top of the girders, thelatter may be arranged in any economical number. Generally the spacingwill be from 3 to 5 ft. The sidewalks are placed either on brackets con-necting to the outer girders or are supported by separate girders. Deckspans have the advantage in that there is no obstruction between road-way and sidewalks. The several girders should be braced by crossframes 10 to 15 ft. apart, the extensions of which form the sidewalkbrackets.

Through spans should have only two girders, which are placed outsideof the roadway. The sidewalks are preferably supported by bracketswhich form the extension of the floorbeams.

ART. 7. DEPTH OF GIRDERS

The economical depth of plate girders for railroad bridges has beenestablished by practice (see Table 32). For the heavier loadings the

following formula gives close results: h = & + 1.5 ft., where 1 = span

length in feet. The facilities for transportation, however, limit the depthto about loft.; on some roads even to 9ft.

For highway bridges the economical depth varies considerably owingto the greatly varying loads; as a rule it is less than for railroad bridges.The total weight of the girder is approximately a minimum if the weight ofthe flanges equals the weight of the web with the stiffeners and spliceplates; this is approximately the case for

h = 1.1 for plate girders with coverplates,

and h = 1.2 $- for plate girders without coverplates,

148 DESIGN OF STEEL BRIDGES

where k is the depth in inches, M the greatest bending moment inin. lb. (including impact if any), s the permissible bending stress inlb: per sq. in. and t the thickness of the web in inches. A variationof 10% or even more affects the weight less than other conditions,such as number and size of stiffeners, size of coverplates, web splices, etc.

The depth of plate girders should preferably not be less than 2~ ofthe span.for railroad bridges and &J for highway bridges.

The depth of the web should be assumed in full inches and the depthback to back of flange angles 2 in. more for girders without coverplatesand 4 in. more for girders with coverplates so that chipping of the webplate is avoided.

ART. 8. CALCULATION OF NET FLANGE AREA

Shallow plate girders, especially those with heavy flanges, should beproportioned by their moment of inertia. As a rule, however, plategirders are deep enough so that the following method is sufficientlyaccurate:

If ho denotes the e$ective depth of the girder, that is, the verticaldistance between the centers of gravity of the flanges, W, and A’the net areas of the web and of oneflange, respectively, we haveapproximately for the moment of inertia of the section

I = w+ + &+!)2Further, if M = bending moment and s = permissible bending stress,the required moment of inertia is

I=!!+

Equating these two values we get for the required net area of oneflange

or replacing $ W, by $ of the gross area W of the web and as-suming $ IV concentrated in the flange we have for the total requirednet flange area

A =A’+QW = Fho

If the flange area A = A’ + Q W is given, the maximum stress isM


*IV can of course only then be assumed as an equivalent flangearea if the web splices are designed to develop fully the bending re-sistance of the web (see below).

The effective depth h, has in general to be found by trial, that is, acertain flange section has to be assumed and its center of gravity deter-mined and if the effective depth thus found varies considerably from thatof the computed section the calculation has to be repeated. Table 30is convenient for determining the center of gravity of the flanges.

If the distance between centers of gravity exceeds the height of thegirder’ back to back of flange angles the latter should be taken as theeffective depth.

ART. 9. AREA OF COMPRESSION FLANGE

There is no established method of determining the exact distributionof the stresses in the compression flange and therefore the design of thisflange is only a rough approximation. At right angles to the girder thecompression flange acts as a column of the length I between lateral con-nections, but it is partly restrained from buckling by the stiffness of theweb and tension flange and further, the flange stress is not a maximumthroughout the length I as in the case of ordinary columns. The formula

16,000-70 b for the permissible stress in ordinary columns may there-

fore safely be replaced by the formula 16,000-50 f for the stress in

the compression flange of girders. As for flanges composed of two anglesand coverplates (Fig. 1) r is approximately 0.25 of the width b of theflange, we get for the permissible stress

s, = 16,000 - 200 ;

which formula is now largely used (see Specifications, Appendix).

FIG. 1. FIG. 2.

If the compression flange has the shape of a channel (Fig.’ 2) as is

usually the case in crane girders, r = 0.33 b,,and s, = 16,000 - 150 f.

TABLE 30Center of Gravity of Girder Flanges

z=distance center of mavi~ ,f flame from back of flanrt 3 andes-

11III

_

;

;1:

-

:

:22

:.1'

::I2-

-

11

:22

:

:

;

10X$=5.0 10X$=7.5

Area

10x1=10.( 0X12=12.5

9.72 0.312.00 0.504.22 0.666.38 0.808.48 0.910.50 1.02

.OXlf=l5.0Angles onlyTh. ~~

Ares z

2Ls6X4

10X$=3.75

Area z

10.97 1.2113.25 1.3715.47 1.4917.63 1.6019.73 1.6811.75 1.76

0X2$=25.0

E1.723.885.988.00

L

2.224.506.728.880.983.00

z

!2.22!4.50!6.72!8.88LO.9813.00

z

27.0029.2231.3833.4835.50

1.041.221.351.461.551.64

0.130.29 31.720.43 36.880.55 35.980.67 38.00

0.760.951.091.221.321.42

7.22 1.949.50 1.99

5/8 11.72 2.033/4 13.88 2.087/8 15.98 2.12

1 18.00 2.17

0.120.260.380.50

16.3818.48LO.50

0.98:3.00

0.600.18

i-I L

9X8=3.38 9x*=4.5 9X$=6.75 9x1=9.0 X1?=11.25 9x1*=13.5 >Xlf=15.75 9X2=18.0Angles only 9X2$=20.2. QX2f=22.5

0.590.780.931.061.181.28

11.9714.1316.23i8.25 Ii

0.030.180.310.42

0.020.150.27

1 2Ls6X4

1.101.271.401.511.601.68

3.97 0.826.25 1.018.47 1.1510.63 1.2812.73 1.3814.75 1.47

6.228.500.722.884.987.00

8.47 0.380.75 0.572.97 0.735.13 0.877.23 0.999.25 1.09

!0.72!3.00

vi;19:48il.50

0.190.380.540.680.810.92

22.9725.2527.4729.6331.7333.75

3/8 7.22 1.94l/2 9.50 1.995/8 11.72 2.032/4 13.88 2.08

Y8 18.00 15.g8 2.12 2.17

10.60 1.2612.8815.10 :::;17.26 1.6319.36 1.7231.38 1.80

0.010.20 27.500.36 29.720.51 31.880.63 33.980.75 36.00

-

0.030.190.340.470.58

:6.388.480.50

-I ’ I

Angles only 12X*=4.5 12X+=6.0 lZX#=7.5 12x4=9.0 12X%=10.5

6.60 0.048.50 0.150.34 0.242.12 0.323.84 0.39

12x1=12.0 2X1$=16.5 2X13=18.0

2LS5X34

LO.60 0.4212.50 0.51!4.34 0.596.12 0.67L7.84 0.73

0.310.410.500.580.64

0 . 0 80.16

!O.OO 0.06:I.84 0.16!3.62 0.2415.34 0.31

!8.12 0.01!9.84 0.08

0.860.910.951.001.04

3.60 0.215.50 0.327.34 0.409.12 0.480.84 0.55

6.108.00

5/8 9.843/4 11.627/a 13.34

5.10 0.127.00 0.238.84 0.320.62 0.402.34 0.47

1'121

;-

::1.1'1'

1

3

3-

pxa$ 0.08 0.16 26.6226.84 0.23 28.34

-

TABLE 30Center of Gravity of Girder Flanges (Cmtinued)

Angles only 16X$=6.0 ) 1 6 X , -‘ - 8 . 0 ( 16X;=l2.0[ 16X1=16.01 16Xl~=ZO.O~ 16X1+=24.‘\ 16Xlt=28.(Th. I ____I

I 2’ Ai-l?a z AreP. z AreP. z~__

2I

AFeZ%____

T- 16X2=32.0116X2&=36.0 6X2+=40.0

Area

9.852.966 . 0 0

0 . 8 6 3 3 . 5 0 0 . 6 2 3 7 . 5 01 . 0 4 3 6 . 7 2 0.81 40.721 . 2 0 3 9 . 8 8 0 . 9 6 4 3 . 8 81 . 3 3 4 2 . 9 6 1.10 46.961 . 4 5 4 6 . 0 0 1 . 2 3 5 0 . 0 0

Area. z

0 . 0 30.190 . 3 30 . 4 6

0 . 0 1

0 . 0 30 . 2 1 5 2 . 7 20 . 3 6 5 5 . 8 80 . 5 1 5 8 . 9 60 . 6 4 6 2 . 0 0

_ _ _ _

0.210.390 . 5 50 . 6 90 . 8 2

5 . 5 08 . 7 21 . 8 84 . 9 68 . 0 0

44

25

55

_-5

0 . 4 1 4 1 . 5 00 . 5 9 4 4 . 7 20 . 7 5 4 7 . 8 80 . 8 9 5 0 . 9 61 . 0 2 5 4 . 0 0__~

0 . 2 5 2 8 . 7 2 0 . 0 60 . 4 1 3 1 . 5 0 0 . 2 2 3 5 . 5 00 . 5 5 3 4 . 2 2 0 . 3 5 3 8 . 2 20 . 6 7 3 6 . 8 8 0 . 4 7 4 0 . 8 80 . 7 7 3 9 . 4 8 0 . 5 8 4 3 . 4 80 . 8 7 4 2 . 0 0 0 . 6 8 4 6 . 0 0

1 . 6 5 2 1 . 5 0 1 . 4 6 2 5 . 5 0 1 . 1 3 2 9 . 5 01 . 8 0 2 4 . 7 2 1 . 6 2 2 8 . 7 2 1 . 3 1 3 2 . 7 21 . 9 3 2 7 . 8 8 1 . 7 6 3 1 . 8 8 1 . 4 6 3 5 . 8 82 . 0 3 3 0 . 9 6 1 . 8 7 3 4 . 9 6 1 . 5 8 3 8 . 9 62 . 1 2 3 4 . 0 0 1.97 38.00 1.69 42.00

I

0 . 4 7 2 4 . 7 20 . 6 3 2 7 . 5 00 . 7 7 3 0 . 2 20 . 8 8 3 2 . 8 80 . 9 8 3 5 . 4 81 . 0 7 3 8 . 0 0

3/8 8 . 7 2 1 . 6 4 1 4 . 7 2 0 . 9 0 1 6 . 7 2l/2 1 1 . 5 0 1 . 6 8 1 7 . 5 0 1 . 0 4 1 9 . 5 05/8 1 4 . 2 2 1 . 7 3 2 0 . 2 2 l.lG 2 2 . 2 23/4 1 6 . 8 8 1 . 7 8 2 2 . 8 8 1 . 2 6 2 4 . 8 87/S 19.48 1.82 25.48 1 . 3 5 2 7 . 4 8

1 22.00 1.86 28.00 1 . 4 2 3 0 . 0 0

0 . 0 40 . 1 7 4 2 . 2 20 . 2 9 4 4 . 8 80 . 4 0 4 7 . 4 80 . 5 0 5 0 . 0 0

- -

0.000 . 1 20 . 2 30 . 3 3

0 . 0 70 . 1 6 5 8 . 0 0

- -

81.4884.00

---I-- -=7.0 1 14Xpx10.5 1 1 4 X 1 = 1 4 . 0 I14>(1:=17.51 14X1+=2.1 /14X1%=24.,

-1 4 X 2 = 2 8 !14X2$=31.5

0 . 0 40 . 1 6 5 0 . 9 80 . 2 6 5 3 . 5 0

- 4 . 8 8:7.48io.00

0.000.10

!0 . 8 0 1 9 . 2 2 / 0 . 5 4 2 2 . 7 2 0 . 3 2 2 6 . 2 2 0 . 1 30 . 9 5 2 2 . 0 0 0 . 7 0 2 5 . 5 0 0 . 4 8 2 9 . 0 0 0 . 2 9 3 2 . 5 01 . 0 8 2 4 . 7 2 0 . 8 4 2 8 . 2 2 0 . 6 2 3 1 . 7 2 0 . 4 3 3 5 . 2 21 . 1 8 2 7 . 3 8 0 . 9 5 3 0 . 5 8 0 . 7 4 3 4 . 3 8 0 . 5 5 3 7 . 8 81 . 2 7 2 9 . 9 5 1 . 0 5 3 3 . 4 8 0 . 8 5 3 6 . 9 8 0 . 6 6 4 0 . 4 81 . 3 5 3 2 . 5 0 1 . 1 4 3 6 . 0 0 0.94 39.50 0 . 7 6 4 3 . 0 0

3/8 8 . 7 2 1 . 6 4 1 3 . 9 7 0 . 9 5 1 5 . 7 2l/2 1 1 . 5 0 1 . 6 8 1 6 . 7 5 1.09 18.505/8 1 4 . 2 2 1 . 7 3 1 9 . 4 7 1 . 2 1 2 1 . 2 23/4 1 6 . 8 8 1 . 7 8 2 2 . 1 3 1 . 3 1 2 3 . 8 8718 1 9 . 4 8 1 . 8 2 2 4 . 7 3 1.39 26.48

2 2 . 0 0 1 . 8 6 2 7 . 2 5 1 . 4 6 2 9 . 0 0

0.110 . 2 5 3 8 . 7 20 . 3 8 4 1 . 3 80 . 4 9 4 3 . 9 80 . 5 9 4 6 . 5 0

- -

0 . 0 80 . 2 10 . 3 20 . 4 2

-I I ’ I -2Ls 6X4 I I I

0 . 0 60 . 1 4 -!

TABLE 30Center of Gravity of Girder Flanges ~Coontinued~-

~Anglesonly~ 20X+=10 1 20X$=15 1 20X1=20 1 20X1+=25 20X1:=30 120X1$=35I_20X2=40

-

F77,

- -1

20x 2$=45 120X2+=50

Th. IYiG&JEJ I

x Area z AreZt I

0.050.190.330.440.55

0.00

5 Area

1.23 30.50 0.93 35.501.38 34.22 1.08 39.221.51 37.88 1.23 42.881.62 41.46 1.34 46.461.72 45.00 1.46 50.001.80 48.46 1.55 53.46- - - _-

0.06

i

-

I7.88 0.021.46 0.155.00 0.27 80.00 0.118.46 0.38 83.46 0.22 38.46

5 Area z AreaL

T

-

I L

2LsSk8

1/2 15.50 2.195/S

25.5019.22 2.23 29.22

3/4 22.88 2.28718

32.8826.46 30.00 2.32 2.37 36.46

11/840.00

33.46 2.41 43.46

0.45 45.50 0.25 50.50 0.070.62 49.22 0.41 54.22 0.220.76 52.88 0.56 57.88 0.370.89 56.46 0.69 61.46 0.501.01 60.00 0.81 65.00 0.621.11 63.46 0.91 68.46 0.73

0.110.23 46.720.34 49.880.44 52.960.54 56.00- -

T

/-0.060.17 54.880.27 57.960.36 61.00

0.010.110.20

0.74 28.500.86 31.720.96 34.861.04 37.961.12 41.00- -

9.222.886.460.003.46

66.00

--bs8x6I-1/2 13.50 '1.475/S 16.72 1.52

23.60

19.88 1.5626.72

3/4 29.887/8 22.96 1.611 26.00 1.65

32.9636.00

0.50 33.500.62 36.720.73 39.880.82 42.960.91 46.00

I

-0.29 38.500.42 41-.720.53 44.880.63 47.960.72 51.00

- - -I- -!.]Angles only 1 18X+=9.0 1 18Xf=l3.51 18X1=18.0/ 18Xlf=22.5/18X16=27.0~ 18X1;=31.i i 1 8X2=36.0 8X2+=40.5118X2+=45.( 18X2$=49.5--

6:

tt7:

.-

l/25/8

15.50 2.19 24.50 1.29 29.00

3/419.22 2.23 28.22 1.44 32.72

7j/8-I-- 22.88 2.28 31.88 1.57 36.3826.46 2.32 35.46 1.67 39.96

11/S30.0033.46 ,2.37

39.00 1.77 43.5012.41 42.46 1.85 46.96

0.140.300.450.590.710.82

0.070.170.27

1.00 33.50 0.74 38.00 0.52 42.50 0.32 47.001.15 37.22 0.91 11.72 0.69 46.22 0.49 50.721.30 40.88 1.06 45.38 0.84 49.88 0.64 54.381.41 44.46 1.18 48.96 0.97 53.46 0.77 57.961.62 48.00 1.30 52.501.61 51.46 1.39 55.96

5 .5 16 :

iI-

5f6:-

1.38 0.115.96 0.24 71.46I.50 0.36 75.003.96 0.48 78.46

i7 i I i I i I-0.67 34.72 0.47 39.22 0.29 43.72 0.120.78 37.88 0.58 42.38 0.40 46.88 0.23 51.38

0.553l.50~0.34~.00~0.16( ( 1

0.87 40.96 0.68 45.46 0.50 49.96 0.33 54.460.96 44.00 0.77 48.50 0.60 53.00 0.43 57.50


The required gross area of the compression flange, including $ ofthe web, is then

A=%0

but it should not be made less than the gross area of the tension flange.

The latter area is usually sufficient if the ratio k does not exceed

about 12, which is often specified as the limiting ratio in bridge work.

ART. 10. MAKE-UP OF FLAKGES

Table 31 shows typical flanges of plate ‘girders of various areas ofset tion. Flanges consisting of two angles with or without coverplatesare most common in bridge work (Figs. 1 to 7). For very heavy flangestwo vertical side plates between the flange angles and the web are usedin order to reduce the number of coverplates and avoid excessively longvertical rivets (Figs. 8 to 12). In order to simplify the framing of t,heties for railroad bridges some engineers prefer for the top flange thesections 13, 14, 15, composed of four angles with or without side plates.The saving in the ties may, however, be offset by the greater cost of thesteel work owing to the smaller effective depth of the girder and sinceall four angles and the vertical plates between the angles and the webshould extend over the full length of the girder.

Girders of through spans should have at least one coverplate on topextending over the whole length.

The different coverplates should have preferably the same thicknessor decrease in thickness outward from the angles.

Buckle plates, troughs, etc., connected to the flange are preferablyriveted to the first coverplate which is made about 3 in. wider than tkeother flange plates, but this additional width should not be counted inthe effective section.

The vertical leg of the flange angle should be sufficiently wide toaccommodate the necessary rivets which should not be spaced closerthan 3 in. for &in. rivets and 2+ in. for q-in. rivets in any direction (seeAppendix). For a double row of rivets the vertical leg should be 6 in.,for 3 rows 8 in. wide. It is sometimes of advantage to increase thethickness of the web in order to avoid two or three rows of rivets; atthe same time it may thus be possible to omit stiffeners. The thicknessof the flange angle should not be less than X% of the width of theoutstanding flange.


Many specifications require that at least one-half of the flange sectionbe made up by the flange angles or else the heaviest angles be used.

In determining the net area of the flange the holes of all rivets cut byany vertical section should be deducted, but care should be taken that asection connecting the centers of staggered rivets does not cause a smallernet area. See Appendix.

ART. 11. LENGTH OF COVERPLATES

For girders with constant depth back to back of flange angles it issufficiently accurate to assume the effective depth h constant. The re-quired flange area (including Q web) varies then in direct proportionwith the bending moment. In order to obtain the length of the cover-plates the moment curve is drawn to any convenient scale and the maxi-mum ordinate is divided proportionately to the various parts making up

FIG. 3.

the flange at that section as shown in Fig. 3. This division may be con-veniently done on an inclined line mn whose length measured to anyconvenient scale is equal to the area A of the adopted flange section(including + web). To the theoretical length of a coverplate thusfound should be added at each end at least 6 in. for light to 12 in. forheavier coverplates since the full value of the plate is only graduallydeveloped by the rivets.

For a uniform load the moment curve is a parabola and the effectivelength of the nth coverplate counted from the outside can be,found. bythe formula

1, = 1 qyd-

where Z;A, is the sum of the areas of the coverplates outside of andincluding the coverplate whose length is desired. For the first outsidecoverplate we have

11


for the second

-> etc.

For main girders of railroad bridges the moment curve may beassumed to consist of two halves of a parabola AC and DB and astraight l ine CD = 0.11 (Fig. 4). The theoretical length of the nthoutside cover-plate is then

I, = 1 (0.1+0.!3$$)

For a floorbeam of a railroad bridge with two lines of stringersthe moment curve is practically a trapezoid ACDB (Fig. 5) and thetheoretical length of the nth outside coverplate is

I, =.c + (l-c)-ZA,

A

In the top flange the coverplate next to the flange angles is made thefull length of the girder; this is also done in the bottom flange of longgirders.

FIG. 4. FIG. 5.

ART. 12. WEB PLATE

To simplify the shop work the web should be made of the same thick-ness for the whole length of the girder. In girders which have to resist acomparatively great shear for a short length at the end, as, for instance,girders supporting a heavy concentrated load near a support, it may bemore economical to reinforce the web for that length by side plates be-tween the upper and lower flange angles. It should, however, be con-9sidered that a slight saving in weight may be offset by the additional costof shop work,due to the extra handling, riveting of the side plates, etc.

The thickness of the web plate is in the first place determined by thegreatest vertical shear V (page 144). Usually the permissible shearingstress s for the gross area of the web is specified. For ordinary not very


shallow girders the required thickness is therefore with sufficient accuracy

t=$

where h = depth of web.For exceptionally shallow girders the more accurate formula

Vmt=x (see page 145)

should be used, where m = static moment of area of section above theneutral axis about the latter and I the moment of inertia. As alreadymentioned it may be advisable to increase this required thickness slightlyin order to allow a greater rivet pitch in the flange angles and frequentlyalso to avoid stiffeners especially in stringers or shallow main girders(see below),

The thickness should not be less than & in. for highway bridgesand 8 in. for railroad bridges. Further, if the depth of the girder isover 7 ft. the thickness should be at least 5 in. on account of the easierhandling in the shop. The specifications of the Lehigh Valley R. R. of1911 call for a minimum thickness of & of the unsupported distancebetween flange angles.

ART. 13. STIFFENERS

The web plate should be stiffened against buckling by vertical stiffen-ing angles, preferably in pairs, whenever its depth between the flangeangles exceeds 60 times its thickness. The stiffeners should be spaceduniformly with a clear distance not exceeding 5 to 6 ft., or the clear heightof the web, unless a closer spacing is required toward the ends. TheAmerican Railway Engineering Association Specifications (see Appendix)require that the clear distance betwee stiffeners shall nowhere exceedthe following value in inches:

cl = & (12 ,000 - s )

where t = thickness of web in inches,s = actual shear per sq. in. in the web.

In any case there should be stiffeners at points of concentrated loadsand over the bearings to act as vertical posts. The practice of proportion-ing these stiffeners varies considerably; a rational method would be,first, to assume the full concentration carried by the angles only usingthe unreduced unit stress in compression and, second, to consider thefillers and the strip of the web plate covered by these as parts of the stif-


fener and dimension the latter as a column against buckling, the largerrequired section of the angles to be used.

Intermediate stiffeners may be crimped over the flange angles and theoutstanding leg should not be less than & of the depth of the girder plus2 in. End stiffeners and those at local concentrations should have fillersand the outstanding leg should extend as near as possible to the edge of theflange angles.

All stiffeners should have a close fit against the flange angles. Therivet spacing should not exceed about 6 times the rivet diameter and beas nearly uniform as possible and alike in all stiffeners.

If a sufficient number of rivets cannot be placed in stiffeners totransmit the concentrated loads or the reactions to the web, the fillersmay be widened so as to allow additional rows of rivets and make thosethrough the angles act in double shear.

ART. 14. HORIZONTAL FLANGE RIVETS

These connect the flanges to the web and must therefore resist thelongitudinal shear acting between flanges and web. This shear is at any

point H = VT per lin. in. (see Vol. I), where V = vertical shear at

that point,A’h,

m = -2- = static moment of flange area A’ about the

neutral axis (A’ = area of flange not including + of web and ho = cffect-Ahoive depth of girder) and I = 2 = moment of inertia of girder (A = area

of flange including + of web). If r = rivet value (bearing on web or doubleshear whichever is the smaller), the required rivet spacing is therefore

Ahorp=g=A’I/

or the number of rivets required in a distance equal to the effectivedepth of t.he girder is

hn A’V

If the web has been neglected in proportioning the girder flanges therequired rivet pitch is

p=$

or the number of rivets required in a distance equal to the depth ofthe girder is

ho -Vn=-=,P


For girders flange plates between the flange angles andthe web we get for the number of rivets required in a distance equalto the effective depth ho of the girder:

(1) If b web considered in proportioning flange:

ho A’Vn=p - Ar

between flange plate and web,

ho A”Vnl=-=-Pl AT1

between flange angles and flange plate

(2) if web neglected in proportioning flange:ha V

n=-=-P r

between flange plate and web,

ho A”Vn1 zx - =I I-PI A rl

between flange angles and flange plate;

where A = area of flange including Q web,A’ = area of flange excluding Q web, jA” = area of flange excluding flange plates and Q web,r = rivet value either double shear or bearing on the web,

whichever is the smaller,71 = rivet value in double shear.

The rivets through the vertical flanges of the angles can be countedboth for the connection of the angles to the flange plates and for the con-nection of the lat ter to the web. The part of the flange plates notcovered by the angles should, however, get a number of rivets propor-tionately to its area of section.

If the load carried by the girder is applied directly to the top flangean assumption has to be made about the distribution of this load over acertain distance. Jn railroad bridges with ties resting on the flanges thewheel load is usually assumed to be distributed over three ties.L) If P is the concentrated load distributed over the distance d, the re-

quired rivet spacing in the top flange or the number of rivets required inthe distance h,, is found by substituting in the foregoing formulas thevalue

If p becomes very small due to a great local concentration P theweb should be planed so as to bear on a bearing plate transmitting Pand relieving the rivets of the direct stress. The rivet spacing in the

bottom flange is not affected by 5, but for simplicity of shop work it is

made the same as in the top flange.


The distance between centers of rivets shall preferably not be less than3 in. for $-in. rivets and 25 in. for $-in. rivets, nor for angles with asingle row more than 6 and 5 in. respectively; for angles with doublerows of rivets the horizontal distance between the centers of staggeredrivets should not exceed 6 and 5 in. respectively.

ART. 16. VERTICAL FLANGE RIVETS

The longitudinal shear between flange angles and coverplates is muchsmaller than between the angles and the web, and the pitch of the verticalrivets can therefore be considerably greater than that of the horizontalrivets. However, the same limiting values should be observed as for thehorizontal rivets and, further, simplicity in detailing and shop work oftenrequire the same rivet pitch. Figs. 6 to 9 show ‘typical rivet spacings,the two legs of a flange angle being shown developed into a plane.

Her. Leg

Ver!: f eg

At the end of a coverplate the rivets should be closely spaced so thatthe strength of this plate is fully developed at the section where it is re-quired and in as short a distance as practicable. In any case the strengthof a coverplate should be fully developed before the end of the next cover-plate is reached. Coverplates 18 in. wide or more should have four rowsof rivets.

ART. 16. WEB SPLICE

If the web plates exceed th.e length given in Appendix, theyhave to be spliced. The splices should be located symmetrically and pref-erably at places where the bending moment is comparatively small. Thejoint should be covered by splice plates on each side of the web. Whetherproportioned for it or not, the web transmits not only the shear but also apart of the bending moment for which the splice should be designed.

1. Splicing for Shear Only.-As a rule if the web is assumed to resistthe shear only the splices are also designed for the shear only. Two spliceplates are used between the vertical legs of the flange angles (Fig. 10).Their cross-sections should have a combined net area at least equalto that of the web or a thickness not less than &r in. or $ in. each


and the number of rivets connecting the plates to the web on one sideof the joint should be at least

Vn zz --r

where V = greatest shear at the splice and r = rivet value (bearingon web or double shear, whichever is the smaller). The rivets shouldbe uniformly distributed since the shearing stresses are nearly uniformbetween the flange angles. At least two vertical rows of rivets shouldbe used on each side of the joint.

This method of splicing is faulty because the web and, therefore, thesplice actually resists part of the bending moment and the splice rivetsnear the flanges may be overstressed and weakened to resist the shear.

+++ +++++++I++I++I++I

E

++I++I++I++I++I++I++I++

FIG. 10. FIG. 11.

I 1I-- ++++ ff- --I- --*I_- + 2+++I+ ++!I+++I

lYllrl,

tttj+++I,++I++t1+++’it+17-t+--FIG. 12.

2. Common Splicing for Shear and Moment.-Fig. 11 shows the ordi-nary method of splicing the web if it is assumed to resist the shear andpart of the bending moment. The two horizontal plates at each flange(usually 8 in. to 12 in. wide) are’ used to replace the + of the gross areaht of the web assumed as flange area. Since their moment of inertiaabout the neutral axis of the girder must be equal to the moment of inertiaof the web area in the flange, their required total net section on the ten-sion side is

ho2a=+ht-

h2

The plates on the compression side are made the same.

The average stress in these plates is thenhl

s1= s-ho’

where

s = fiber stress at the center of gravity of the flange, since the fiberstress decreases in linear proportion with the distance from theneutral axis. A proportionate decrease has to be assumed for the rivet

i62 DESIGN OF STEEL BRIDGES [CHAP. VI11

value. If, therefore, T = rivet value at the center of gravity of theflange, the rivet value at the center of the splice plate is

and the required number of rivets on one side of the joint connect-ing the horizontal splice plates is

a 81n = - = ErSr1

that is, equal to the full tension value of the two plates divided by thefull rivet value.* If the flange section is not fully utilized the actualflange stress may be used for s instead of the permissible stress.

The vertical splice plates are assumed to replace the web plate inshear. Their thickness and number of rivets are therefore determined asfor splice (Fig. IO).

This method of splicing is more efficient than that of Fig. 10.Nevertheless, it is not rational owing to the arbitrary assumptions re-garding the stress distribution. Further, no provision is made for thelongitudinal shear at sections I and II; in other words, the stressincrement near the splice cannot be transmitted from the vertical tothe horizontal splice plates and from these to ‘the flange angles exceptpartly through the web itself. There may, therefore, be weak points inthis splice and yet waste of material.

Example (See Fig. ll).-Web plate 60 X i = 22.5 sq. in. gr. = 17.2 sq. in. net,flange angles 6 X 6 X 3, flanges are fully utilized, + of web = 2.8 sq. in. assumedin flange area, shear at splice S = 180,000 lb., permissible tension stress in flange =16,000 lb. per sq. in., rivet value T = 7880 lb. for &in. rivets bearing on $-in.web and r = 14,550 lb. for double shear. Effective depth of girder ho = 58 in.

Assume hl = 40 in. between centers of horizontal splice plates.

Required area of two horizontal splice plates a = 2.8 $1 = 5.9 sq. in. net,

use two plates 8 X 3 = 6.0 sq. in. net.Number of rivets on one side of splice

&y = 5.9 x 16,000 = 127 8 8 0

h o*Often the area a is found incorrectly by multiplying Q h t by G and further,

it is neglected that the rivet value decreases with the distance from the neutralaxis of the girder. This would be correct if the web and the splice would resisttheir part of the bending moment independently of the flanges. This is, however,not the case and in order to have the same fiber stresses the splice must replace themoment of inertia of the web.


Required net area of two vertical splice plates = 17.2 sq. in., use two $-in.plates 32 in. long = 24 sq. in. gross = 18 sq. in. net.Number of rivets in vertical splice plates on each side of splice

S 180 000n = - = -F&6 = 23T

use 3 vertical rows at 8 rivets = 24 riv.ets.Total number of rivets in splice = 4 X 12 f 2 X 24 = 96.

3. Rational Method of Splicing Web.-A more correct way of splicingthe web is to splice each part of the web by that part of the splice plateswhich covers it. This requires, as will be seen, two splice plates of thesame depth as the web and uniform rivet spacing (Fig. 12). Wherethese splice plates interfere with the flange angles they are replaced bythe longitudinal splice’ plates covering the vertical legs of the flangeangles.

The uniform rivet spacing is evident for splicing for the shear,but it is also correct for splicing for the bending moment, since themoment of inertia of any longitudinal strip of the splice plate must beat least equal to the moment of inertia of the part of the web which itcovers and the number of rivets required to connect that strip to the webis, as shown above,

therefore dependent only on the required net area a and not ‘on thelocation of that strip; in other words, each strip of equal width re-quires the same number of rivets, that is, the rivets have to be uni-formly distributed over the whole splice. The number of rivets requiredon one side of the joint in the vertical splice plates is, therefore,

N=a f

where a = net area of web between flange angles and r and s as ex-plained before. Correspondingly, the number of rivets in the longitu-dinal flange piate required for splicing for bending moment is

where a is the net area of the part of the web covered by the flangeangles. The total number of rivets should be greater than that requiredfor splicing for shear only (see page 160). In order to splice for thelongitudinal shear at section I the longitudinal splice plate is widened

Tso as to connect to the vertical splice plate by 12’ = 7 rivets on one

164 DESIGN OF STEEL BRIDGES [CEAP. VIII

side of the joint, where approximately T= 5 b = longitudinal shear inhthe length b and r = rivet value in double shear. The same numberof rivets is required for this purpose above the edge of the flangeangle. The total number of rivets required in the longitudinal platesabove the edge of the angle, on one side of the joint, is therefore n + n’,whereby n has to be obtained with r for double shear, but the totalnumber should, of course, not be less than n obtained with r forbearing on web. If it is not possible to place these rivets in the lengthb then the longitudinal splice plates may be extended as shown dotted.

It will rarely be necessary to increase the number of rivets for the com-bined effect of the shear and bending moment. As a rule the shear occur-ring simultaneously with the maximum bending moment at a splice issmall and it should also be remembered that actually the outer rivets takea smaller share of the vertical shear than those nearer the neutral axis,while the latter are less stressed from the bending moment than theformer. In a case where the ccmbined effect of the shear and bendingmoment is considerable as, for instance, when the maximum shear occurssimultaneously with the maximum moment (see Vol. I), the numberof rivets may easily be increased as follows:

Assuming that the bending moment stresses the outer rivets up

to their full value r and the shear J’ to the amount rs = 4 where

n = total number of rivets on one side of the joint, the resultant actionon the rivet is dm. If this exceeds r by my0 then the numberof r ivets should be increased by my0 (an excess up to about 5%may be neglected).

Example (Fig. 12).-We will assume the same conditions as in the previous ex-ample. $-in. splice plates are ample for shear and moment. Area of web coveredby flange = 1.9 sq. in. net, number of rivets required to splice this for bending

n = 1.9% = 4 (rivets bearing on web)

Area of web between flanges = 15.0 sq. in. net, number of rivets required to splicethis for bending

16 000n = 15.0 gjgti- = 30;

use three vertical rows @ 10 rivets = 30 rivets.For shear alone 23 rivets are required.Assume splice plate 20 in. wide (b = 10 in.), longitudinal shear 2’ at I:

180 00010 = 30,000 lb., number of rivets in horizontal splice plate below flange__

n, 30,000- 14,440 - 2

A R T . 171 BEAM AND PLATE GIRDER BRIDGES

Number of rivets required to splice web covered by flange

165

n = 1.9 i$$$ = 2 (rivets in double shear)t

total number of rivets in horizontal splice plate through flange = 12 + n’ = 2 + 2 = 4,or the same as abbve required for rivets bearing on web.Total number of rivets in splice between flanges = 2 X 30 = 60 or 37% less thanfor splice (Fig. 11).

Assuming that the shear of 18D,OOO lb. acts simultaneously with the maximum

bending moment at the splice, causing a bearing stress of %$p = 2370 lb. on

each rivet and that the moment stresses the outer rivets up to the bearingcapacity 7880 lb., these rivets get a resultant stress of

2/78002 + 23702 = 8200 lb.

or only 4% excess, which can be neglected.

ART. 17. FLANGE SPLICES

Splices of the flange angles and coverplates should be avoided when-ever possible. For the maximum lengths of plates obtainable see Appendix.One angle of each flange should be spliced near one end and the otherangle near the other end of the girder, each joint to be covered by two

FIG. 13.

splice angles of at least the same combined net section as that of the splicedangle and a sufficient number of rivets should be used on each side of thejoint to develop the full value of the spliced angle.

I r

---

FIG. 14.

The coverplates should also be spliced preferably near the ends of thegirder but not at the same places as the flange angles. If possible thejoint should be placed so that an extension of the next outside coverplatecan be used as a splice plate. If the splice plate is not next to the splicedplate (indirect splice) a few extra rivets should be used depending on thenumber and thickness of the intermediate plates. The rivets at thesplices should be spaced as closely as possible. Sometimes several cover-plates can be spliced by a single splice plate if the splice is arranged asshown in Fig. 13.


For field splices of long plate girders (f. i. drawbridges) or for exportthe ~~countersunk” splice is used as shown in Fig. 14.

ART. 18. BRACING

Railroad deck plate girder spans up to 50 ft. length usually have only atop lateral system when on a tangent. All spans on a curve and spansabove 50 ft. length should have top and bottom laterals. There should bea frame at each end and intermediate cross frames about 15 ft. apart,each frame to consist of a top and bottom strut and two diagonals.

The lateral system is usually of the simple Warren type, each lat-eral consisting of one or two angles whose inclination toward the girdershould be between 45” and 60”.

In through plate girder spans, the lateral system is composed of thefloorbeams and a double intersectionsy stem of diagonals. The diago-nals should be made up of angles, and should be rigidly connected to thebottom of the stringers at the points of intersection. One angle will, as arule, be sufficient. End floorbeams should be used, or at least rigid endframes made as deep as possible.

Where solid floorplates, troughs, buckle plates or concrete slabs areused, the lateral bracing may be omitted, provided that the lateral forcescan be properly transmitted into the flanges of the girders. The topflange of each girder should be effectively stiffened by knee braces at everyfloorbeam or at distances not over 12 ft. in case of solid floors.

For the calculation of shears in the lateral system see page 106.

ART. 19. BEARINGS

Plate girders should have riveted to each end a base plate of a thick-ness not less than $ in. for railroad and g in. for highway bridges.There should be a bedplate between the base plate and masonry, whichmay be a rolled steel plate of a thickness not less than that of the baseplate for railroad spans up to 75 ft. and ordinary highway spans up to 100ft.; it should not project more than 3+ times its thickness beyond thebottom flange angles. Instead of bedplates, cast steel pedestals arenow more frequently used.

Longer spans should rest on hinged bearings with rollers at one end.The bearings are usually made of cast steel. Plate VIII shows sometypical plate girder bearings.

PLATE Ix

.-_- -_-

Pennsylvania Railroad. 18-k. I-Beam Bridge, Built 1912.

PLATE XII

i-+------- _________.__________! . . . . 6’5 ______ --.- . . ..__..... e ._... -.- ___.__._...._............. -...---------6~ane,~ & ,+‘3”-------r.-

N. Y. C. & H. R. R., General Detail Drawing of 116 ft., Through Plate Girder Span, Built 1904. (Facing plate IX)

C H A P T E R I X

SIMPLE TRUSS BRIDGES(Plates XIII to XXIV)

Simple truss bridges are in general built in span lengths from 100ft. up for railroads and from 80 ft. up for highways. For notablesimple span truss bridges see Table 33 and Plate XIII

The design should proceed in the following order:(1) Flooring (railway ties, solid floor, planking).(2) Stringers and floorbeams or other floor supports.(3) Main girders or trusses.(4) Wind and intermediate sway-bracing.(5) End sway-bracing or portal (check endpost of truss for wind).(6) Pins and rollers.(7) Packing (cross-section) at the panel points.(8) Final layout of eyebars, if any (can then be ordered from the mill).(9) Final sketches of elevation of panel points (main material of the

trusses can then be ordered from the mill).It is important that all main material is ordered from the mill before

the shop drawings are started, to insure prompt delivery. The manu-facture of eyebars requires more time as the rolling mill and the eyeba.rshop may be filled with other orders. For the same reason the commonsizes and shapes should be used, as far as possible, and all special sectionsavoided.

ART. 1. LOCATION OF FLOOR

In a deck truss span (Plate VI), the floor is either placed on top ofthe trusses or between the trusses near the top chords, and in a throughtruss span it is placed between the trusses near the b,ottom chords.If the distance from top of floor to the permissible under-clearance issufficient a deck truss span is, as a rule, more economical than a throughspan, as the supporting piers or towers will be lower and also shortersince in most cases the trusses can be placed nearer together, which more-over reduces the weight of the floor system.Small through spans with low trusses without an -overhead lateral

bracing are called pony truss spans. As they are deficient in lateral stiff-ness, they are seldom used for railroad bridges.

16712

Built

1 19122 19123 19124 19055 18896 18947 18898 18969 1890

10 188511 188512 188813 189114 190915 189016 191017 190918 190419 1910

TABLE 33Notable Sin 1P-

Name or Railroad Location River

I* Nickel steel used.

Municipal Bridge. . . . . St. Louis . . . . . . . . Mississippi . . . .Ken. & Ind. Term R. R Louisville. . . . . . . Ohio. . . . . . . . . .Prussian State Ry’s .. Homberg. . . . . . . Rhine . . . . . . . . .

Elizabethtown‘Kentucky’Cent: k’.‘k: Cincinnati.

. . . Great Miami... . . . . . Ohio . . . . . . . . . .

Louisville & Jeffville.. . . . Ohio. . . . . . . . . .Cincinnati & Cov’gton

Louisville, Ky. . . . . . . . . . . . . . . . . Ohio . . . . . . . . . .

Pennsylvania R. R. . . Philadelphia. . . . . Delaware. . . . .Ohio Con. R. R. . . . . . . Pittsburg . . . . . . . . Ohio . . . . . . . . . .Wheeling & L..E. R. R Wheeling . . . . . . . . Ohio. . . . . . . . .Ohio Bridge . . . . . . . . . Henderson. . . . . . Ohio . . . . . . . . . .Illinois Central R. R . Cairo . . . . . . . . . . . Ohio . . . . . . . . . .Norfolk & Western . . . Ceredo . . . . . . . . . . Ohio . . . . . . . . . .Baltimore & Ohio . . . . Havre de Grace.. Susquehanna.Merchants Bridge . . . . St. Louis. . . . . . . . Mississippi . . . .McKinley Bridge . . . . . St. Louis. . . . . . . . Mississippi. . . .Webster-Donora. . . . . . . . . . . . . . . . . . . . . Monongahela .Tenth Street Bridge .. Pittsburg . . . . . . . . Allegheny . . . . .CopperRiv.&N.W.Ry Miles Glacier, Al. Copper River

I

sic Spans

I -

Kind of traffic 1, Span inft.

D. T. Ry. & Hy . 668D. T. Ry. & Hy . 6 2 0D. T. Ry . . . . . . . SlO+I!$. . . . . . . 586

T.Ry.&Hy./ 5 5 0S:T. Ry.. . . . . . 5463D. T. Ry. & Hy . 542+D. T. Ry . . 5 4 0ET. Ry.. . . . 523

;.’

s:D.

T. Ry . . . . . . . 522T. Ry . . . . . . . . 522T. Ry . . . . . . . . 518T. Ry. . . . . . . . 518T. Ry. . . . . . . . 518

D. T. Ry . . . . . . . . 517*$;‘f. El. Ry & Hy 517s

. . . . . . . . . . . . 515H y . . . . . . . . . . . . . . 4 5 4S. T. Ry . . . . . . . . 4 5 0

-

-

Height / Width of Weight.ofof $my 1 trus;? in ~ ;teee,‘pt.

110 %.5 13,700*11088.6 29.5 .ii,soo..

8 0 32.5 2,190............................................................_...... . . . . . . . .

8 4 32 7,750, , . . .

......... ,.......... ,..........

.............................

.............................

. . . . . . . . .86 31.5

. . . . . . . . . . . . . . . . . .

E29.7 3,23026.7 2,720

fi35.72 4

ART. 41 SIMPLE TRUSS BRIDGES 169

ART. 2. NUMBER OF TRUSSES

Two trusses are preferable, even if some saving of material could beeffected by the use of a greater number, as a uniform distribution of thefloor load over more than two trusses is questionable and different trussesincrease the cost of the shop work. However, certain conditions maywarrant the use of more than two trusses for wide highway bridges,especially if of deck construction. Multiple track railroad bridges may bebuilt as independent spans for each track; this requires, however, gener-ally more metal, longer piers and spreading of the tracks, but may bedesirable if a few tracks are built first and the addition of the others isintended for a future time. In this case, sufficient clearance should beprovided to make the driving of rivets or pins in the new adjoining trusspossible.

ART. 3. DISTANCE BETWEEN TRUSSES

To insure sufficient lateral rigidity and stability against overturn-ing, railroad deck spans with parallel chords on a tangent should have awidth c. to c. of trusses not less than =& of their span nor less than +of their depth c. to c. of chords, with a minimum of 10 ft. for single-and 19 ft. 6 in. for double-track spans. The width c. to c. of trussesin through spans is equal to the required clear width plus the widthb of the end post, but should not be less than & of the span length.The same formulas as given for plate girders (page 147) may be used for apreliminary assumption of b for truss spans.

For spacing of trusses for bridges on curves see page 217.

ART. 4. GENERAL PRINCIPLES OF TRUSS DESIGN

In computing the stresses in a truss, the simplifying assumptions aremade that the truss members are connected to each other by frictionlesshinges and that the straight lines between two panel points are the neutralaxes of hhe members, so that only axial stresses are produced. This con-dition is never realized. The rigidity and in many cases the eccentricityof the connections cause bending or secondary stresses, which may be con-siderable (see Vol; I).

In order to minimize the uncertainties and the secondary stresses, thefollowing principles should be observed :

(1) The truss system should be as simple as possible and all memberswhich make the stress distribution uncertain should be avoided.

(2) The load should be transmitted through the truss system to the

170 DESIGN OF STEEL BRIDGES [CHAP. Ix

supports by the shortest way; a truss so designed will generally be themost economical.

(3) Where possible the loads should be applied at panel points only.(4) The neutral axes of the members should be straight and intersect

at the panel points, and the connection of any member should be sym-metrical about its neutral axis. Where this is not practicable, ampleprovision should be made for the resulting bending stresses.

ART. 6. RIVETED AND PIN-CONNECTED TRUSSES

The principal advantage of pin-connected over riveted trusses ischeaper erection. The disadvantage of pin spans is smaller rigidity, moreparticularly where the dead load is small compared with the live load.Pin trusses are, therefore, better adapted for long spans. With increasingfacilities for field riveting, the upper limit of riveted trusses is steadilyincreasing. For railroad bridges, riveted trusses are now mostly used forspans up to 170 ft. for single track and to 140 ft. for double track;on some roads even over 200 ft. (N. Y. C. & H. R. R. R., Penna. LinesWest, Can. Pac. and others). Trusses of light highway bridges arepreferably riveted up to 200 ft. length. ,

ART. 6. PARALLEL AND POLYGONAL CHORDS

The advantage of polygonal chords compared with parallel chords isthe saving of steel and greater uniformity of the chord and web sectionsthroughout the whole length; their disadvantage is greater cost of manu-facture per lb. of steel. Trusses for single-track railroad and lighthighway through bridges for spans up to about 180 ft., and for double-track railroad and heavy highway through bridges up to about 160 ft.are preferably made with parallel chords; a polygonal top chord is gen-erally more economical for spans above these limits. For deck spans, itmay be more economical to use parallel chords for lengths up to 30.0 ft.and more, since the piers h’ave to be higher if a polygonal bottom chordis used.

ART. 7. HEIGHTS OF TRUSSES

The economical heights of trusses have been established by practice.In general, the greater the live load the greater should be the height fortrusses of equal length. Formulas based on the least theoret,ical weightare of doubtful value on account of practical considerations. It should,however, be noted that that height gives the least total weight for which


the’weight of the chords is about equal to the weight of the web members.This can easily be proven by the fact that the weight of the chords changesapproximately in inverse proportion and that of the web members indirect proportion with the height.

The minimum height for railroad through spans is about 27 ft.c. to c. of chords, depending on the required headroom and the heightof the floor system.

In single-track through spans, the usual ratio of height to span lengthis about W for trusses with parallel chords (short spans) and from &to + at the center for trusses with polygonal top ch,ords, the smallerratio referring to long spans. Long single-track spans should not betoo high on account of the overturning effect of the wind, which maycause considerable stresses in the lower chord.

For double-track through spans, the heights are about 15% greaterthan for single-track.

For trusses of deck spans, the usual ratio of height to length is + tob, but as they should not be higher than twice the width c. to c. oftrusses (see page 169), it may often be better to decrease the heightinstead of increasing the width (cost of substructure).

The minimum height for highway through spans is 16 ft. For lighthighway bridges, through or deck, up to about 20 ft. in width, the heightshould not be less than $$ of the span for trusses with parallel chords,and + at the center for trusses with a polygonal chord. For heavierbridges with greater width, the height should be increased.

ART. 8. WEB SYSTEM

Plate XIV shows typical trusses for simple spans.The Warren truss is mostly used for riveted trusses and sometimes also

for long pin spans; it carries the loads by the shortest way through thetruss system to the supports thus saving in material and in the number ofconnections. The number of panels must, of course, be even.

The Pratt truss is mostly used for pin spans; it permits the use bf eye-bars for all diagonals and counters, which saves in weight and makesthe pin-connections simpler than in the Warren truss. Multiple websystems are now generally discarded, owing to the ambiguity of stressdistribution.

In spans longer than 350 ft., it may be necessary to subdivide thepanels in order to avoid steep and long diagonals or excessively long panels.In shorter spans the subdivision with diagonal substruts is preferable tothat with subties as it makes the truss stiffer.


The subdivision of the panels may be necessary even m short spans ifthe floor is shallow, requiring short panels, or for export work, to limitthe length of the members for easier handling and t&nsportation. I thas to be remembered, however, that any saving of metal may be offsetby the greater cost of shop and field work on account of the greater numberof short members and connections; this is especially the case in rivetedsubdivided trusses.

ART. 9. SHAPE OF POLYGONAL CHORD

The minimum height at the hips and the economical height at thecenter are first determined. Between these three points, the most satis-factory outline of the polygonal chord is that inscribed to a parabola.If for longer spans the inclination of the top chord becomes so steepthat the dead load stresses in the diagonal of the second panel may bereversed by the live load, the height at the hips should be increased.The end post or any diagonal should have an angle of not less than 45”

FIG. 1. FIG. 2.

with the bott,om chord. As in long spans with subdivided panels thiswould require an excessive height at the end; the first one ok two mainpanels may be single and the subdivision begin with the second or thirdpanel (Figs. 1 and 2).

With the height at the end and center determined, the heights at theother panel points are found as follows:

The difference in height between the center point and any other panelpoint of the polygonal chord increases from the center toward the end inproportion of the square of the distance from the center or for equalpanels in proportion of the figures in the sketches given below. Forinstance, if for an uneven number of panels (Fig. 3 6) y5 is the differ-ence of height between hip (this being the fifth panel point from thecenter) and the horizontal center panel, the diff.erence of height betweenthe center chord and the third panel point is

2 5 - l 24Y3 = Y5 sl-1 = y5 8 0 ’ etc.

Fractions of inches should be neglected.


In order to avoid splicing of the chord at the center, the two centerpanels of Fig. 3 a are, as a rule, made straight as indicated by a dottedline.

(a.1FIG. 3.

ART. 10. PANEL LENGTH

The panel length must be the shorter the shallower the floor (seepage 136) ; on the other hand, longer panels require less truss members andconnections, which reduces the weight and shop work of the trusses.According to present practice, 22 to 28 ‘ft. is the usual panel lengthfor single-track railroad spans up to about 200 ft. and for economicalfloor depth; for longer spans and more tracks, the panels are preferablymade longer; the longest existing panels are in the 668-ft. spans of theSt. Louis Municipal Bridge where they increase from 30 ft. at the end to48 ft. at the center of the span. 15 to 20 ft. are usual panel lengthsfor ordinary highway bridges up to 200 ft.; the panel length increaseswith increasing width and span length. To simplify the shop workthe panels should be of equal length as far as possible.

In order to get the truss connections (eyebar packing, gusset plates orsplices) symmetrical, an uneven number of panels is preferable for pin-connected trusses and an even number for riveted trusses.

ART. 11. REQUIRED AREA OF SECTION OF TRUSS MEMBERS

The outline of the truss having been determined (possibly after pre-liminary calculations of weight), the floor is designed first (see page 132)and its weight calculated. The weight of the truss is then assumed (seepages 225, 236, etc.) and the stresses in the truss members and the reac-tions determined (see page 73).

From these the required areas of the members, bearing area onmasonry and length of rollers are then determined.

The axial stresses produced by the dead load, live load and the impacthave to be calculated for all members; wind stresses usually only for thebottom chord of through trusses or the top chord of deck trusses and for

174 DESIGN OF STEEL BRIDGES [CHAP. IX

the end posts. In long spans considerable wind stresses may occur also inthe unloaded chord and, if the latter is polygonal, also in the web members.Stresses from centrifugal force affect only the loaded chord and the endposts of deck spans. Stresses from braking force may affect the sectionof the bottom chord in the end panels. Where the braking force has tobe transmitted through two or more successive spans the resulting stressesmay have to be determined for all members.

Beside the axial stresses the following bending moments may have tobe determined:

(1) In the top chord of deck spans from vertical loads acting on thechord between panel points.

(2) In the end posts of through spans from wind.(3) In long horizontal or inclined members from their own weight.If S is the sum of all axial stresses which according to the specifica-

tions have to be combined (including the addition for alternate stresses)and s the corresponding permissible unit stress, the required area of thesection is

As for compression members s depends on the ratio of the radius of

gyration to the unsupported length of the member,1

( >7 , a preliminary

assumption has to be made which may have to be changed after the sectionis made up.

In deck spans where the flooring rests directly on the top chord, themaximum bending stress ab produced in the latter by its own weight andthat of the floor, and by the live load and impact, has to be added to thedirect unit compression stress sC. It is, however, more convenient tocalculate from the bending moment an equivalent direct stress Sb, whichwould cause a uniform unit stress equal to ab and to add ~5’6 to the otherdirect stresses. If M = bending moment in inch-pounds, y = distanceof extreme top fiber from the neutral axis and r = radius of gyration,both in inches,

ST, = 5. .in pounds

As the chord members are partly fixed at their ends, it is for practicalpurposes sufficiently accurate to determine the bending moment MO atthe center of the panel assuming the member as a simple span and tomake M = 3 MO. Splices should be sufficiently strong to transmit this


moment and the maximum shear. Sb may have to be changed afterthe section is made up if r and y. are different from those assumed.

For wind stresses in end posts see page 113.The bending stresses in a member resulting from its own weight are

seldom considered. According to Cooper’s Specifications, they have tobe considered when they exceed 10 per cent. of the allowed unit stresses,which is the case in very long members only. If W is the weight of themember, L the length between its connections, and a! the angle it formswith the horizontal, the beading moment caused by W is

M = LW cos a~__8

assuming the ends free to turn

andM = LW cos Q

TO- if the ends are fixed.

The equivalent direct stress Sh is found as above, where y = distancefrom the neutral axis to the extreme top fiber, if the member is stressedin compression, and to extreme bottom fiber if stressed in tension. I fthe member is very slender, the more accurate formulas given in Vol. Ihave to be used to determine the bending stress in the outer fiber.

Concerning the permissible unit stress to be used for combined directcompression and bending see ‘Vol. I.

The areas having been determined, the sections of the members aremade up according to the rules given below. The stresses and sectionsare conveniently tabulated as shown in Table 34, page 176, for a 160-ft.single-track span.

ART. 12. GENERAL PRINCIPLES FOR THE DESIGN OFTRUSS MEMBERS

The following general principles should be observed in designing thetruss members :

All surfaces not in contact should be easily accessible for inspectionand painting after erection. Closed sections should therefore be avoided.

Attention should be paid to the facility of riveting up of the severalparts of a member with power riveters, so as to avoid hand-riveting andto allow the driving of \,a11 rivets with parts assembled (see Appendix).

Sections which allow direct connection of the main parts are prefer-able to those which require lug angles or connection plates.

All riveted members should have two or more webs and must besymmetrical about the plane of the truss. Symmetry about the planeperpendicular to the truss plane through the axis of the member isdesirable.

TABLE 34Stress Sheet for Riveted Truss af a 160 ft. Single-track Through Span

Member

Reacbions . . . . .

Stresses per truss

Dead

t 69+124

T

Live

$203

+313

-293+196+ 52-115

272

Im-pact Total Lengl

1

Comp. memb.Unit Area

1 stress req’d, Make up of section @sq. in.

r

f132 +404

+.204 f641

-232 -697

-191+145

r r$

+ 85

177

-584+401

‘2;;’

320 7.8

464 7.8. . . . . .

‘464. ‘5:i“

t205- 9

536

iii’16 12.8 n 4 k 6X4X+8.2 1.1 g 4 IE 6X4X8

.40i 3.6

1:;:

I required.

. Note.-All stresses given in units of 1000 lb. rz denotes net section, g gross section.

‘ihi’

.

:h

-L

rad.gw

r

-

1 6

16

41 I 13.1

5 9 11.9 49 0 g Same as UlU3 except 2 pl. 20 X &. . 16 25.0 n 2-15 in. [,50 lb.‘8i.. ..9:i.

;$qj;zi-i5ih: r,45ib.. . . . . .

‘i:O’

. ..I..

.-2 5 . 2 n 2pl. 1sxt, 4h 4X4X&

40.0 nc

2pl. 18X& 4b4X4X& ‘i2 pl. 9%x&., 2 pl. 16$X$ I

J lcov.24X~, 2b33X3aX6 top\53.3gg 12 pl. 20X%, 2 b 5x3+x% be. i

Area;I:, sq

in.

31.3

52.2

54.0

49.029.4

26.5

14.414:4

-

L’E

i

I

Usedne.4

;q. m.

Lwp/p 1 Only 70% of dead load stress considered.2 Obtained after adding 50% of minimum stress to maximum stress.


The width of the truss members in the plane of the truss should cor-respond to the function and importance of the member, compression mem-bers should be wider than tension members, top chord and end postswider than web members, main members wider than sub-members,etc. The width should, however, not be made unnecessarily great as itmight cause excessive secondary stresses (see Vol. I).

For riveted tension members the net area after deducting the rivetholes is the effective section; the rivet holes should be so arranged thatthe deduction does not exceed about 20% of the gross section. InAmerican specifications no rivet holes are deducted for compression mem-bers (in foreign specifications from 50 to 100% of rivet holes arededucted).

The webs (2, 3, or 4) of heavy and deep compression members shouldbe connected by a continuous longitudinal diaphragm in the axis of themember, or a coverplate in place of the latticing on one side. This appliesespecially to members subject to transverse bending, such as end postswith portals, intermediate posts with floorbeam connections, etc. Thisdiaphragm or coverplate may be counted in the effective area if it isadequately connected.

For economy, compression members should have a small ratio oflength to radius of gyration (chords preferably 30 to 40, web members

60 to 70; for upper limits of i see Specifications, in Appendix).

For practical reasons, the number of different rolled sections in themain material should be as small as possible, preferably plates and anglesor channels only. The latter are especially economical, but as in smallersizes the flanges are too narrow for good connections, no channels lessthan 10 in. should be used for trusses of railroad and not less than 8 in.for those of highway bridges.

ART. 13. SECTIONS FOR TOP CHORD

Table 35 shows typical top chord sections of different areas. Sec-tions (1) to (4) are used for light .highway bridges up to 200 ft. length;(5) to (10) and (13) to (35) f or railroad bridges of up to 200 ft. spanand (ll), (12), (16), to (20) for longer spans.

In top chords (also inclined end posts) which have usually a cover-plate, the neutral axis, especially in pin-connected trusses, should comeas near as possible to the center line of the web, which is accomplishedby the use of horizontal or vertical balance strips riveted to the lowerflange, or when flange angles are used by making the bottom angles


heavier than those on top. If the coverplate cannot be balanced fullyand the neutral axis varies considerably for the different members ofthe chord, the average eccentricity is used for the whole chord.

. The transverse distance between backs of channels or angles shouldbe constant throughout the whole length of the chord so as to keep therivet gage in the horizontal flanges and in the coverplate constant.This necessitates a uniform thickness of the web plate in riveted trussesif fillers at the splices and connections are to be avoided.; The sectioncan be increased by making the angles thicker, or, if this is not practi-cable, by adding outer side plates; but only to such a thickness as willnot interfere with the driving of the rivets in the horizontal flanges.A great number of side plates may reduce the radius of gyrationconsiderably.

TABLE 36Typical Sections of Top Chords and End Post of Riveted Trusses

0.) r7; 12zz,j:$ g} 8 . G sq. in.-

r51.14

(2.) m~=I.81

2-15,t lsa33 Ib, > 27.3 =I. in.1 COY. 20x3

?“=5.88

(5a) Same except: 2%15”[s@40lb. } 31 .Osq. in.

r=5.76

1 cov. 22x*2 pl. 20x1

42.9 sq. in.

(7a) Same plus: 2 pl. 13X 6 } 54.3 sq. in,

r=6.70

1 COY. 26X*2 pl. 22x g2 Ls 4X4X $

1

2Ls 6X4X &$ J 62’7 sq’ in’r=8.65

. . . . . . . . . . . . . . . . . . . . . . . . . .._

(9a) Ssme plus: 2 pl. 14X & } 74.9 sq. in.

~=8.09

152.9 sq. in.

(ila) Same plus: 2 PI. 18X %

(6.)1cov.2ox 82-15” [[email protected] bars 4x5

32.3 sq. in.

r=6.5

(&I) Same except: 2-15”[s@50lb. } 41.9 sq. in.

r=6.07

(8a) Same plus: 2 pl. 14X & } 65.8 sq. in.

. . . . . . .._._........_................................... . . . . . . ..__.....__...._____. I..(loa) Snme Plus: ; $1 ;;k:” } 132;Ez$

Exceptionally heavychord supporting fioor.

231.5e.q. in.

(12a) Same except: 2 PI. 20X%plus: 2 Pl. 40x t } 30).$&i;


The distance between the webs of the top chord in riveted trusses isusually determined by the greatest width of the web members connect-ing inside of the gusset plates with provision for a g-in. clearancenecessary for erection. In pin trusses, it is usually fixed by the pinpacking at the hip of the truss. In no case should the radius of gyra-tion in the transverse direction be less than that in the plane oftruss. In sections with two webs, this condition is approximatelyfulfilled when the width inside of the webs is equal or greater than 2of the depth of the webs. The minimum depth of the webs in pin

TABLE 36A .

Typical Sections of Top Chords and End Post of Pin Connected Trusses

(13.1 1 r1 1 COY. 28X *4Ls4X4X%2 pl. 22x4 54.4 in.sq.2 bars 5x;

1 L r=9.27

(13a) Same except:

(15aY Same plus:2 pl. 18X & I2 pl. 28x0 , 160 1 sq. in.

r=10.23

1 COY. 42X &I 4 L s 4x4x+: 4LsGXBX: lG2.8 sq. in.’ 3pl.34X&

v-=12.73

(17a) Snme p l u s :‘ 2 outs. pl. 24Xt ~ 218.8 sq in,2 ins. pl. 24 X $ J

r=11.5G

r=13.08

(ISa\ Same plus:

2 outs. pl. 24 X %2 outs. pl. 34 xt2 ins. pl. 30 X B I

363.9 sq. in.

T-=12.12

rT-l- 1 cov. 30x R

(‘4.)4Ls4X4Xf2 pl. 26X&

74.6 sq. in.2 pl. 5x1

1 -.L r=10.94

(14a) Same except: 2 pl. 26 X f } 90.8 sq. in.

r=10.42

1 cov. 3GXt81,s 6X4X&; p&y&s 124.1 sq. in.

2 bars 52% Jr=11.85

(1611) Same Plus:2 pl. 30x + out. webs 12 pl. 18X & ins. web I ‘~~“,$~“’

196.0 sq. in.

(18a) Same plus:2 outs. pl. 32Xp } 244.0 sq. in.

r=11.08

Notes: Dotted lines show the increitse in the section of the sane chord.Sections 1 to 4 are for light highway bridges onlyAll radii of gyration given are those with neutral axis through center of gravity parallel to

horizontal axis.


spans is usually fixed by the size of the eyebar heads at the hip of thetruss. Top chords of deck spans which support ties or cross beamsbetween .panel points should have a height of 2~ to Q of the panellength for railroad bridges and & to $2 for highway bridges. Inthis case, ,the radius of gyration laterally can, of course, be smallerif the chord is as strong against buckling laterally as against com-bined buckling and bending in the plane of the truss.

ART. 14. SECTION’S FOR BOTTOM CHORD

Bottom chords of pin spans are generally made of eyebars (see Art. 17)except in the two panels at each end of railroad bridges where they are

FIG. 4.

liable to be stressed in compression from a possible combination of thevertical, lateral and longitudinal forces and are therefore made of rivetedsections.

Table 36 shows typical sections of riveted bottom chords. Sections1 to 4 are used for light highway spans only.

For all riveted tension members it is more economical to turn theflanges inside, as this requires lighter latticing. In bottom chords ofriveted through spans, this affords a simpler connection of the floorbeamsll- than if the flanges were turned out (Fig. 4); for the bottom chord-JL of deck spans, however, it is better to turn the flanges out, as

FIG. 6. this permits a simpler connection of the lower lateral systemand the verticals can extend into the bottom chord (Fig. 4).

A clearance of at least 5 in. should be allowed between the flangeswhen they are turned inside. The section composed of 4 angles (Fig.5) is very economical for riveted tension members, as it requires lattic-ing or tie plates in one plane only, but the thickness of the outstandingleg should not be less than & of its width.


TABLE 36

Typical Sections of Bottqm Chords of Riveted Trusses

(I.) --t-j-. - 4 .&- -A .4.2 Ls 4X3X A=::; sq,;in. n. r=1.27

gr.2Ls5X3;Xi=i:; 1; n . r=1.6

a-.2Ls 6 X 4 X A=;:: 1: n. r=1.92

gr.

(3.) !L Jir 14 Ls 6X4X+ =;::I sq;,in. n . r=2.76

gr.Same plus:

2 pl. 3 x &

~=6.57

Same plus:2 pl. lOXi2~1. 16+X&

r=5.6B

2 pl. 26Xf4 Ls 6X4x+ } ii:“4 sg”n’ p”;.r=9.04

Same plus:2 pl. 14X%2 pl. 24X4

7PI(9.) 2:

4 pl. 32Xt 1 9 7 . 84Ls01 6.6X% 1113.85q;‘in.n. u.

i$ T=ll.oJSame plus:

2 pl. 19*x$4 pl. 30xt

) 8:;:; sq;,in. n .Iv‘.

(2.).ja +-

2 Ls 5X3fX A= 6 . 2 sq. in.n. ~=I.597.1 “ gr.

Same plus: 1 1 2 . 4 “: n. r=2.282Ls5X3+X& 114.1 gr.

(4,) [ 1 2-15” [s @ 40’“.=;;:“5 y.f;;4$,

Same plus:2 pl. 12x+

) ii.2 sq;bL n.gr.

r=4.86

2 pl. 22x * 1 32.3 sq;in. n.4LS5X3hXi J 4 0 . 8 u.

r=7.72

01‘:2 pl. n.4 Ls

22x+ 56.2 sq. in.5x3+x+ 7 4 . 2 “ gr.

2 pl. 12x5 r=6.96

-!! $(8.) g$ I

ii2 pl. 30x* 1 7 3 . 3 sq;kI. n.4Ls6X6Xf I 8 6 . 3 gr.

Lr=10.74

Same plus: 2 pl. lSfX$126.0 sq. in. n.

2 pl. 28x3154.5 “ gr.

r=9.3

- 4~1. 3 6 X ”4 Ls6X6&}:;;:$ “;in’ ;;.

r=12.08Same plus:

Notes: Dotted lines show the increase in the section of the s~lrne chord.Sections 1 and 2 are for light highway bridges only.All radii of gyration given are those for grors section with neutral axis thrdugh center of gravity and

parallel to horizontal axis.

‘ART. 16. SECTIONS FOR DIAGONALS

The inclined end posts of through spans and the diagonal posts inthe end panels of deck spans are made of a section similar to that of thetop chord. All other diagonals of pin spans are usually made of eyebars,counter-diagonals being used in those panels where the greatest live loadreverses the dead load tension (or part of it, see Specifications, Appendix)in the main diagonal. For riveted diagonals the sections are similar tothose of riveted bottom chords; side plates shown dotted in Sections 3to 10 should be avoided as they require extra shopwork, and in com-

182 DESIGN OF STEEL BRIDGES [CIPAP. Ix

pression members reduce the radius of gyration. Section 3 is eco-nomical for tension diagonals; compression diagonals must havesufficient stiffness against buckling in both directions. The, flanges ofchannel sections should be turned inside whenever this gives a sufficientradius of gyration transversely. Heavy compression diagonals shouldhave a continuous diaphragm connecting the two webs.

ART. 16. SECTIONS FOR VERTICALS

The sections of verticals are similar to those of riveted bottom chords.Section 3 without side plates is generally used for light hangers and postscarrying panel loads only. Instead of latticing, a continuous web plate

I

FIG. 6.

or diaphragm is frequently used as shown in Fig. 6 (a), (b) and (c).This arrangement is preferable to latticing, especially in double-trackthrough spans, as it resists the bending of the vertical from the deflectionof the floorbeam and the distortion of the sway bracing from unequaldeflection of the trusses is better resisted. For sections (b) and (c) tieplates should be used to hold the flanges in position and if distance dbetween rivet lines exceeds 6 in., outside latticing should be used in ad-r-----71 dition to the diaphragm.d- In any case there should be a webf i plate or diaphragm for the full height of the floorbeam and at theL--z!lFIG. 7.

sway frame connections; if extending.the full length of the mem-ber it may be counted in the effective section. The verticals

should be so designed that all floor-beams get the same length and fieldconnections and can be easily erected; the transverse width of all verti-cals should be constant and the flanges turned inside. The radius ofgyration in the transverse direction can be much smaller than that inthe longitudinal direction, since the unsupported length is greatly re-duced transversely by the floorbeam and the sway bracing. In trusses


TABLE 37

Eyebars(American Bridge Co.)

Adjustable Eye Bar.

Head Bar Screw endBar

-Thick-

.iIm88

g .3 .Ei

2 1

- -

ag 1

-

-

Max. pin

-

--

--

--

--

47.844.0 $142.8 4;

52.6 4+48.4 546.4 5

44.441.7 i:36.5 51

36.544.052.160.6 ;L- - - a48.039.1 t44.5 6450.556.9 ::

41.1 7

~-42.8 75

46.149.8 f$53.8 8:~-

42.044.1 ;I46.8 8;49.853.1

-

-

_-

_-

40.0

41.7l- Gl -112- 4

37.5l -112- 32- 8

35.02- 12- 83- 3

37.52- 42- G3- 2

35.72- 72-113- 4

37.52- 83- 03- 4

3 8 . 9 2-113- 7

75.0

37.5

3- 53- 94- 1

3- 84- 24- 8

35.7

37.534.4

-

4- 34-105- 5

4-115- 5

87a7t

O-7, “3o-11 , Bl- 4 f

O-10 * al-226 fl- 7 1

l-6 *P

1 - 82- 2 -7

2- 9 g

2- 3 *la2- 62-11 7

2-63- 1 -*I’19

2-103- 3 3

3- 7

3- 3 -

I-

*12 6$-__

I-

-.

*18t 9__-

1819

*20 9

20 7t22 9t~__

22424

*25____

262 1028 11+

*29+ 13

3 14

- -

4 1%

- -

6 2

- -

6 2

- -

7 2

- -

2 2

--

-_

St;rn91I-l.

--

ii12

*34 :“5

36 14*371 16

I-

II-

9 2

_... -

10 2

- -

12 2

_-

14 2

I-

3- 84- 1

Bars marked * should only be used whenabsolutely unavoidable.

3- 94- 44- 8

4- 54-10

_--_

16 2

I- L

184 DESIGN OF STEEL BRIDGES [C H A P. IX

TABLE 38Sleeve Nuts and Turnbuckles

( A l l D i m e n s i o n s i n I n c h e s )

3iam.o fxewu

7

hI

do

Diarro f

.screwu

reighiin lb.

23 03 4

3 8. . . .5 0

. . . .

6 5. . . .9 5

. . . .

1

1.

1. .

1.

1.- -

1..-

1 0 8. .. . . .

- -

S t a n d a r d d i m e n s i o n sLong Insidediam. diam.

B c

l’hiclc-ness Weigh

in lb.t t IHI+ 12’

. . . . . . . . . . .

&5 8: 4%118 6. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .

:1.1X

5 5

t?i

with subdivided panels, long secondary verticals which carry only deadpanel loads are frequently made of four light angles (Fig. 7) with lattic-ing on all four sides.

ART. 17. EYEBARS

The eyebars in a bridge should have the same width as far as pos-sible and be of standard sizes as per Table 37. The thickness should,


preferably, not exceed 22 in., as thicker bars may not be sufficientlyworked in rolling and forging. The width should not be less than thediameter of the pin nor should it exceed it by more than 3.

Eyebars composing a member should be arranged in two groupssymmetrically about the plane of the truss and should be parallel asnearly as possible. Counter-diagonals are made adjustable with screwends and turnbuckles.

The ‘following widths of eyebars and diameters of pins are typicalfor main diagonals and bottom chords of railroad bridges:

Width of bar Diam. of pin For S. T. spans of For D. T. spans of

8 to 1 0 in. 7 to gin. 175 to 300 ft. . . . . . . .10 to 12 I‘ 9 to 11 “ 300 to 500 “ 175 to 300 ft.12 to 14 “ 10 to 12 I‘ above 500 “ 300 to 500 “14 to 16 “ 12 to 14 “ “. . . . . . . .. . . . above 500

Smaller sizes are usual for counter-diagonals in the middle panels,their smallest width being limited by the largest diamkter of pin to whichthey connect.

ART. 18. BRACING OF TRUSS SPANS

The function of the bracing is to secure lateral stability of the trussesas they are unstable by themselves, to stiffen the individual members,especially compression members, by reducing their unsupported lengthand to transmit, in connection with the trusses, lateral and longitudinalforces to the bearings.

The bracing should be so designed as to secure- not only lateralstrength and stability, but also rigidity, especially for railroad bridges.The general rules given for the trusses (page 169) apply also to thebracing. However, since the lateral forces are much smaller and lesscertain than the vertical loads, it is permissible to use statically inde-terminate systems and to make simplifying assumptions for their cal-culation. The following additional rules should be considered:

The neutral axes of the members of the bracing should intersect onthe neutral axes of the chords. The panel points of the bracing shouldcoincide with the panel points of the main trusses. All truss spansshould be provided with a lateral system in the plane of each chord, alsosway bracing or portals at the ends and intermediate panel points andbetween long compression diagonals.


Bracing in a Deck Bridge of Warren Type.

Bracing in a Through Bridgeof WarrenType

Porial and Sway Bracing in a Through Bridge,Pratt Type, with Polygonal Top Chord.

Fig. 8;

Fig. 9.

Fig. 10.


Figs. 8, 9 and 10 show typical arrangements of the bracing in trussspans.

ART. 19. LATERAL SYSTEM BETWEEN THE LOADED CHORDS

This has usually two intersecting diagonals in each panel, the floor-beams forming the posts. The calculation of the stresses is explainedon page 105. Wherever possible, the laterals should be riveted to thestringers. This shortens their unsupported length, enabling them toresist compression as well as tension. In this case, the shear in anypanel may be divided equally between the two diagonals. The stress inany diagonal is then reversed for forces acting in opposite direction; thisreversal need not be considered in designing the diagonal and its connec-tion, since it occurs at long intervals.

The connection of the laterals to the stringers serves also the purposeto transmit the longitudinal forces from the stringers to the truss chords

FIG. 11.

through the laterals; “to avoid bending the latecals at their connectionwith the stringers, a transverse strut should connect opposite intersectionpoints (see Fig. 11).

The laterals in ordinary bridges are made up of one or preferably twoangles riveted back to back. In through bridges, they are connected bymeans of gusset plates to the b6ttom of the floorbeams and the bottomchord (see Plates VI and VII) if the lat’ter is riveted. If the bottomchord consists of eyebars, the gusset is connected to the vertical post,which extends below the pin, and to the bottom of the floorbeam.Plate VI shows various floor arrangements for deck spans. In Fig. G,where the ties rest directly on the top chords, the top laterals are madeas deep as the chord of two or four angles with latticing. Wherefloorbeams and stringers are used, Figs. H and J show the most con-venient arrangement. In Figs. I, K, L the laterals either clear thestringers or they are spliced across the top flanges of the stringers.In the former case, longitudinal timbers are placed between stringers andties so that the lateral angles clear the ties.

188 j DESIGN OF STEEL BRIDGES [CHAP. IX

ART. 20. ‘LATERAL SYSTEM BETWEEN UNLOADED CHORDS

The unsupported length is generally too great to permit the lateralsbetween the unloaded chords to be economically designed as compressionmembers. Each lateral is, therefore, assumed to take the whole shear intension, but should, nevertheless, be built of a rigid shape. Cross strutsare required at the panel points; they generally form a part of the swaybracing and must be designed to resist compression. In case the unloadedchords are polygonal, the stresses in the laterals are found in the same

L4r.7

FIG. 12. ,

irFIG. 13.

li 7rJi Or AL

FIG. 14.

way as for straight cnords, assuming the polygonal chords developedinto (not projected on) a plane.

In light highway bridges the laterals may\ be made of single angles;also in deck spans where the bottom chord is made’ up of four angles(Fig. 12). In this case, the strut is made of two angles so as to be ableto resist compression (Fig. 13). Where the chords are of the box shapemore than 12 in. deep, the laterals and struts should be as deep as the

qhords and made up ‘of two or’four latticed angles

j---

l-ry7-l

(Fig. 14). The struts require four angles or two,/-- channels.

//r

j ‘\\\\ART. 21. END SWAY BRACING AND PORTALS

The end sway bracing transmits to the bear-

/’,-_-‘--_

‘\

lql

ings the lateral forces acting along the toI; chord.In through spans it consists of a stiff overhead

/’ j ‘\,frame, the Portal, connecting the two inclined end

c posts, and is made as deep as the room above theclearance line will allow. The depth of trussed

FIG. 15. portals is,preferably made not less than 5 and thatof full web portals 3 of its length.

Full web portals consisting of a web and four angles are used in shortthrough spans where a shallow portal is desirable in order to reduce thedepth of truss. At the ends it is usually reinforced by full web brackets(Fig. 15).

Trussed portals have usually parallel chords and a ,double or multipleintersection web system; simpler systems, however (Fig. 19), are the mostdesirable. Knee braces are advisable where the depth of the portal is


small. Types 1 and 2 are used for smaller spans; the members are madeI

up of two angles (Fig. 16) so that 7 in any direction does not exceed

the permissible limit. Type 3 is used for longer and heavier spans; the

ir-FIG. 16. FIG. 17. FIG. 18.

members are generally built up of four angles with latticing between(Fig. 17). The top and bottom struts often have a box section (Fig. 18).

For the calculation of stresses in portals, see page 113.

Type 1 Type 2. Type 3.

FIG. 19.

ART. 22. INTERMEDIATE SWAY BRACING

Intermediate sway frames are not needed for statical reasons but theyadd to the rigidity of the bridge. An unequal deflection of the two trussesfrom unsymmetrical loading, especially in double-track bridges, causesa distortion of the rectangular cross-section, and, therefore, bending

FIG. 20.

stresses in all its parts, which are particularly objectjonable in the floor-beam connections; the bracing reduces these bending stresses by equaliz-ing the deflections of the two trusses. The sway frames should be asdeep as possible and be placed preferably in the planes of ‘the main corn-


pression members (not in the plane of slender members) in order to makethe bridge rigid and to reduce the unsupported length of these members.In short truss spans, where the compression members have a sufficient

i laterally, the sway frames are usually placed in vertical planes for

simplicity.The proportioning of the sway bracing is a matter of judgment, as

no assumption will give satisfactory results. As a rule a sway bracingis placed at each panel, its design being similar to that of the end swaybracing or portal. In short spans with small head room the intermediateframes usually consist of the top strut and two knee braces (Fig. 20).

ART. 23. BEARINGS

If R is the total end reaction and p the permissible unit pressure on

the masonry (see Appendix), the required bearing area of the base is a = $*

If the substructure is located, the length of the base has to be approxi-mately determined in order to obtain the length c. to c. bearings. Forthis purpose the values given on pages 127, 191 and 192 are useful.

All truss bridges should have hinged bearings with a nest of rollersat one end. For short and light truss bridges the base plate may be asingle plate not less than lt in. thick. For heavier bridges it should beeither built up or of cast steel with a height preferably not less than 6 in.

RThe required bearing length of the rollers is ns where n = number

of rollers and s = permissible pressure per lin. in. of roller. n is deter-mined by the length of the base plate and the diameter (or width ifrollers are segmental) of the rollers. Rollers of smaller diameter than 6in. are now being discarded as they are ineffective.

The shoes are generally riveted. They should be so designed as todistribute the load equally over the rollers. The depth below the pinshould preferably be not less than one-half of the length of the shoes.Generally the shoe is hinged to the truss by a pin at the intersection ofthe axes of the end post and bottom chord.

Table 39 gives dimensions of bearings of various railroad bridges.

ART. 24. GUSSET PLATES OF RIVETED TRUSSES

The thickness of the gusset plates should be given on the stress sheet.The calculation of the stresses in a gusset plate is a complicated matter,so that it is often neglected. In order to reduce the number of rivets

Roajlgrcd

Am. Br . Co . .

I

C . & M. R . R . . -

P. R . R

M;oK+espt. &P

M&Kept. & ,

M . K . & T . .

A. V. Ry.. .-

Spanc. to c.

end pins

s

ft. in.

100 0

125 0

150 0

175 0

200 0

4 0 0 0

388 0

3 3 8 71

2 9 6 32

250 0

2 2 6 2%

SpaIlundercoping

so

ft. in.

97 5

122 2

147 1

171 9

196 5

396 0

. . . . . .

334 1

292 2

246 9

2 2 2 10

TABLE 39

Single Track Bridges (Size of Base Plates, etc.)

Differ-t2LlCB

s-s,

ft. in.

‘ 2 7

2 10

2 11

3 3

3 7

Lading

a I b

E-50

E-50

E - 5 0

E-50

E-50

E-40

,.........

E - 3 5

ft. in. Ft. in.

1 11 1 3

2 2 1 4

2 3 1 8

2 7 1 8s

2 11 1 log~~

3 41 2 2f

_........

3 10 . . . . . . . .

E-35 3 6

E-30 2 7: 1 106

E-35 2 8% 1 8%

Fixed end

/

I.

I-

c h

ft. in. ft. in.

as+ 1 3

2 6 1 6

2 9 1 10%

2 9 1 lot

1 102_-

2 7f 2 2+

. . 4 9

. .._...... 2 5s

. . . . . . . . . . 2 5%

2 7 1 lot

. . . . . . 1 3

Expansion end

ft. in. f t . in . ft,. in.

1 11 1 9* 2 11

2 2 1 1 0 2 6

2 3 2 0 2 lo*

2 7 2 2 2 105

2 111 2 1 . . . . . .._

3 4-:- i z+ 2 7s

3 6 2 19 . . . .._..

2 74 1 log 2 7

2 8+ 1 1 1 I’... .._..

-

. -

:.L.bott.chord tonasonry.

t2

h’ z

gft. in.

1 3 3c

1 6 %

1 1021 1oq

E

8

1 10;2 2:

zm

4 9

2 5&

2 5i

1 10%

1 3

TABLE 39 (Contkmd)Double Track Bridges (Size of Base Plates, etc.)

-

-

--

-

-

-

--

-

l- Fixed end Expansion end-____

.

Spanundercoping

so

Spanc. to c.

end pins

s

:.L.b&chord t<masry.

Differ-ence

s - s .

Loading

h

ft. in.

1 7+

1 1oq

2 4*

2 7%

2 10%

6 8

4 0

3 11+

1 8

2 gt

2 92

2 1

ft. id. ft. in.

1 9

1 lo* 3 2%

2 1 .

2 2g . . . . . . . . . .

2 3 3 1

a’

ft. in.

3 0

3 4

4 0

3 9

4 2

6 6

5 11

4 0

. .

4 4

4 0

a

ft. in.

3 0

3 4

4 0

3 9

4 2

5 4

5 11

3 10

. . . . . .

4 4

4 0

LI

-II--

ft. in.

96 4

121 0

145 4

170 7

195 2

f t . in .

3 8

4 0

4 8

4 5

4 10

E-50

E-50

E-50

E-50

E-50

f t . in .

100 0

125 0

150 0

175 0

200 0

533 0

407 0

373 1

273 0

255 0

195 5

230 0

ft. in. ft. in. ft. in.

2 1 . . . . . . . . . . 1 7t

2 3 3 3 1 102

2 2a . 2 4;

2 6t . 2 79

2 71 13 3 2 102- - -4102 5 4-:- 6 8

. . . . . . . . . . . . . . . . . . . I5 7*

3 5 35 5 7%

1 2 . ..~ 2 ll$

. . . . . . . . . 2 9+

2 5 . . . . . . . . . . 2 9*

Am. Br. Co.. .

Penna. R. R. .

526 5

. . . . . . . . . . .

3 66 0

267 11

250 0

191 0

. . . . . . . . . . . .

6 7

. . . . . . . . . . .

7 1

5 1

5 0

4 5

. . . . . . . . . . .

E -38

E-47

E-49

E-35

. . . . . . . . . .

E -49

E-50

3 10% 1 4 4

. . . . . . . . ..I . . . . . . . . . .

3 5 3 5

2 1 . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .

2 0 . . . . . . . . . .

2 6 . . . . . . . . . . 2 6 / . . . . . . . . . . 12 3 E

F‘d

E

i

ART. 261 SIMPLE TRUSS BRIDGES 1 9 3

to a minimum the thickness of the gusset should be such that the bearingvalue is not less than the,shearing value of the rivets. If the permissibleunit shear is one-half of the unit bearing the minimum thickness of gussetsfor $ in., Q in. and 1 in. diameter rivets is ;F in., $ in., & in., re-spectively, for single shear connections and Q in., 2 in., 43 in. for thosein double shear. Further, the gusset at the short-est section z-:--x through the last rivet row (Fig. 21),should have an effective area at least as great as

ye

that of the member. If it is not symmetricalV, @ .-.a’

about the center line of the member, additional,+A

area is required so that the moment S.e does not

d

6’. &b”

!?Qyo/:

cause excessive fiber stresses. The thickness of a :IS

gusset may also be determined by the shear alongthe chord due to the component of the stresses in FIG. 21.

the connecting web members parallel to that chord.Thicknesses of & in. and 8 in. are generally used in light highway

bridges, $ in. to 6 in. in short single-track railroad bridges, Q in. to3 in. in heavier and longer bridges. In some very heavy bridges ‘gussetsup to 1 in. and more have been used.

For shearing and bearing values of rivets see Appendix.

ART. 26. CHORD SPLICES

These should be indicated on the stress sheet. Except as explainedbelow for pin spans, chord splices are placed outside of the panel points.If the span is erected on falsework the splice is always placed on that sideof the panel point nearer to the end of the span, that is, on the side of thesmaller stress. In spans erected as cantilevers it is more convenientfor erection to have the splice on the side toward the cantilever end.Bottom chord splices should develop at least the full tension capacity ofthe chord; top chord joints need not be fully spliced if they are facedto bearing, but should have splice plates on all sides to prevent dis-placements; in no case, however, should the splice develop less than50% of the chord stress.

ART. 26. PIN CONNECTIONS

Straight compression chords of pin spans are made continuous overthe panel points. The joints are placed as near to the pins as practicableand spliced as described above, the ends being faced to full bearing.

1 9 4 DESIGN OF STEEL BRIDGES [CHAP. Ix

These are called butt joints, and are also frequently used for curvedchords but are then placed at the center of the pin. The stress trans-mitted from the chord to the pin is only the resultant of the stresses inthe adjoining chord members, and the bearing on the-pin must be suffi-cient for this ,resultant. In pin joints a space of about + in. is left be-tween adjoining chord members which, however, bear fully on the pin andare secured thereto by hinge or wing plates. Pin joints are more favoredfor curved chords, as they do not requ,ire such careful facing and aremore reliable. They require more pin plates as the who!e chord stresshas to be transmitted to the pin.

In designing the pin connections, the diameter of the pins is firstassumed; then the thickness of bearing on the pin is determined for eachmember by dividing the maximum stress in that member by the bearingvalue of the pin per lin. ‘in. (except as noted above for compressionchords with butt joints). After this the packing of the members on thepin is made and the bending moments and shears are determined. If theassumed pin diameter differs considerably from that required for themaximum moment it has to be changed and the whole procedure repeated.For slight changes it will not be necessary to rearrange the packing andrecalculate the moments.

ART. 27. SIZE OF PINS

In the preliminary calculation it is, for ordinary spans,, sufficientto consider only the panel points Lo, U1 and Lz which require the largestpins. The, other pins are now generally made of the same size. Morethan two kinds are rarely used.

A pin provides sufficient bearing for an eyebar if its diameter is atleast $ of the width of the bar ($ to g are mostly used); Q being the ratioof permissible tension to permissible bearing. The bearing value of ariveted tension member is approximately equal to the tension value ofits section through the pin’if the diameter of the pin is Q of the widthof the member.

ART. 28. BENDING AND SHEARING STRESSES IN PINS

(See Tables 41 and 42)

The shear rarely affects the size of the pin. If S = shear in pounds,s = permissible shear in lb. per sq. in., d = diameter of pin in


Sinches, we have the required area of pin section a = ~s, and required

diameter for shear

d=d

4a- = 1.16,\/ra?i-

If M = bending moment in inch-lb., 1 = i 50

4

= moment of

inertia of pin section, we have the maximum fiber stress in bending

Md = s = 10.2 2

and the required diameter of pin for bending

d = 2 .17

TABLE 40

Pins with Nuts

(American Bridge Co.)

Pitch equals six threads per inch.Grip = distance between nuts. For dis-

tance between shoulders add t.o grip theamount given in table. For pins up to lo-in.diam. pin hole X5 in. larger than diam. of pin.

For pins above 10 in. a hole should be boredthrough pin and pin caps used, these oaps tobe held by rods going through pin.

All dimensions in inches.Pins marked * are special.

I Pin

Dimnet? of pinThread

in inches

a bI

. 2 2t 1323 21 2

3 *3+ 34 ag*3a 4 3

,.... *4+ 44 *4f 3i5 *5t 4

5: *5ai 6 4$%a’ *ct 593% 7 51*7*! *7* 5p

*7t 8 *8f 6*s: *sa 9 0

*ok *gJz *et 10 6

11+121%1t1%lflf22222)2t

L.-

Addto,

<rip

-IT

-

-

‘hick.ness

t

-

B113li1:1*1%lfI f1;2Q2a2;

)iamn

2 F#

3%4&4 %5%6a77tsgsg9t

10:ll$

x8,mm

-3)4655 %Gt7x%8%st9l

10IOf11:13

2%3 %3I4%5a5%6+77q88%9 %

10%

1 . 1 211.7 222.5 233.7 244.6 256.2 267.8 279.9 28

11.8 291 4 . 3 3018.6 3123.8 3231.1 33

196 DESIGN OF STEEL BRIDGE [CHAP. IX

P i n

Diam..inches

Area,sq. in.

-

-22,000 24,000

0.785 12,000 15,000 20,0001.227

22,00015,000

1.76718,800 25,000 27,500

18,000 22,500 30,000 33,0002.405 21,000 26,300 35,000 38,5003.142 24,0003.9764.909

27,00030,000

5.940 33,000

30,00033,80037,50041,300

40,00045,00050,poo55,000

44,00049,50055,00060,500'

7.069 36,0008.296 39,0009.621 42,000

11.045 45,000

45,000 jO,OOO48,800

66,00065,000

52,500 70,OOG71,50077,000

56,300 75,000 82,50012.566 48,000 60,000 80,00034.186 51,000 63,800 85,00015.90417.721

54,000 67,50057,000 71,300

90,00095,000

88,000 96;OO093,500 102,00099,000 108,000

104,500 1 114,00019.635 60,00021.64823.758

63,00066,000

25.967 69,000

75,00078,800

100,000105,000

82,500 110,00086,300 115,000

110,000 120,000115,500 126,000121,000 132,000126,500 138,000

28.274 72,00075,000

90,000 120,000,30.68033.183 78,000

93,800 125,00035.785

97,500 130,00081,000 101,300 135,000

132,000137,500

144,000150,000

143,000 156,000148,500 162,000

38.485 84,00041.282 87,00044.17947.173

90,00093,000

105,000 140,000108,800 145,000112,500 150,000116,300 155,000

154,000 168,000159,500165,000

174,000180,000

170,500 186,00050.265 96,00053.456 99,00056.745 102,00060.132 105,000

120,000 160,000 176,000123,800127,500

165,000192,000

170,000181,500187,000

198,000131,300 175,000

204,000192,500 210,000

63.617 108,00067.20170.882

111,00074.662

114,000117,000

135,000 180,000138,800 185,000

198,000 216,000203,500

142,500 190,000222,000

209,000146,300 195,000

228,000214,500 234,000

78.540 120,00082.516

150,000 200,000 220,000123,000 153,800 205,000

86.590225,500

126,00090.763

157,500 210,000 231,000129,000 161,300 215,000 236,500

95.033 132,00099.402 135,000

103.869 138,000108.434 141,000113.097 144,000

165,000 220,000 242,000168,800 225,000 247,500172,500 230,000176,300

253,000235,000 258,500

180,000 240,000 264,000

T A B L E 4 1Bearing Values of Pins

Bearing value of 1 in. thick plate for unit stresses per sg. in. of

24,00030,00036,00042,00048,000 '54,00060,00066,00072,00078,00084,00090,000

240,000246,000252,000258,000264.000270;OO0276,000282,000288,000


TABLE 42Bending Moments on Pins

Momentsin inch-pounds for fiber stresses per aq. in. ofDiam.

-

i-

-

Pin

15,000 / 18,000 1 20,000 1 22,000 1 22,500 / 24,000 1 25,000 1 "

1 I I I I

Area,8q. in.

0.7851.2271.7672.405

3.1423.9764.9095.940

7.0698.2969.621

11.045

12.56614.18615.90417.721

19.63521.64823.75825.967

28.27430.68033.18335.785

38.485

2: TE47.173

50.26553.45656.74560.132

63.61767.201

2: i%

78.54082.51686.59090.763

95.03399.402103.869108.434113.097

1,4702,8804,9707,890

1,770 1,960 2,160 2,210 2,360 2,4503,450 3,830 4,220 4,310 4,6005,960 6,630 7,290 7,460

4,7907,950

9,470 10,500 11,5808,280

11,800 12,630 13,200

11,800 14,100 15,700 17,280 17,70016,800 20,100 22,400 24,600 25,20023,000 27,600 30,700 33,700 34,soo30,600 36,800 40,800 44,900 45,900

39,80050,600

47,700 53,00060,700 67,40075,800 84,20093,2oc 103,500

42%:92:600

113,900

18,800 19,60026,800 28,00036,800 38,30049,000 51,000

63,600 66,30030,900 84,300

105,200129,400

94,200 ll",lOO 138,200113,000 165,800134,200

gygg,;g

157,800 189:400178:900 196,800210,400 231,500

184,100 220,900 245,400 270,000213,100 255,700 284,100 312,500245,000 294,000 326,700 359,300280,000 336,000 373,300 410,600

318,100 381,7OC 424,100 466,500359,500 431,400 479,400 527,300404,400 485,300 539,200 593,100452,900 543,500 603,900 664,300

505,100 506,100 673,500 740,800561,200 673,400 748,200 823,100621,300 745,500 828,400 911,200685,500 522,600 914,000 1,005,400

141,400169,600201,300236,700

276,100319,600367,500419,900

477,100539,300606,600679,400

:zE%157,100

214:700188,400223,700

252,500 263,000

294,500 306,800340,900 355,200392,000 408,300447,900 466,600

508,900 530,100575,200 599,200647,100 674,000724,600 754,800

i%%i,"841,800

994:i300 l,E%i1,096,800 1,142:500

i%~:904:400986,500

904,800 1,005,300 1,105,800992,300 1,102,500 1,312,800

1,085,300 1,205,800 1,336,4001,183,900 1.315,400 1,446,0,00

y;m;

1:356:6001.,479,800

1,206,400 1,256,6001,323,OOO 1,378,2001,447,ooo 1,507,3001,578,500 1,644,200

1,717,700 1,789,2001,864,800 1,942,5002,020,100 2,104,3002,183,900 2,274,900

1,073,500 1,288,200 1,431,400 1,574,5001,165,500 1,398.600 1,554,OOO 1,709,400

;~p7~~~1,262.600 1,515.lOO 1683,500 1,851,300 1:89:'9001,364,900 1.637,900 1,819,900 2,001,900 2,047:400

1,472,600 1,767,100 1.963,500 2,159,800 2,208,9001,585,900 1,903,OOO 2,114.500 2,325,900 2,378,X001,704,700 2,045,700

2,195,30012,273,OOO 2,500,300 2,557,100

1,829,400 2,439,200 2,683,2OG 2,744,100

1,960,100 2,352,1001 2,613,400 2,874,800 2,940,100 3,136,1003,096,800 2,516,100 2,795,700 3,075,200 3,145,100 3,354,8001,239,700 2.687,600 2,986,200 3,284,900 3,359,500 3,683,5001,388,900 2,866,700 3,185,200 3,503,300 3,583,400 3,822,3002,544,700 3,053,600 3,392,900 3,732,200 3,817,OOO 4,071,500

2,356,200 2,454,4002,537,400 2,643,1002,727,600 2,841,2002,927,100 3,049,100

3,266,8003,494,600;.;n;m;

4:241:200

29. PIN PACKING

The main principle in designing pin connections is so to arrange thepacking that the bending moment on the pin is reduced to ‘a minimum.However, expensive shop work, such as cutting of flanges, flattening orcountersinking and chipping of rivet heads, etc., should be avoided asmuch as possible. The following clearances should be allowed forinaccuracies in manufacture: & in. between eyebars, + in. between


eyebar and riveted member (inch rivet heads) and t in. betweenriveted members. Further the thickness of a riveted member shouldbe assumed increased by & in. for each plate if there are more thantwo plates. For the height of rivet heads should be allowed: $ in. fors-in. rivets, 9 in. for g-in. rivets, +$ in. for l-in. rivets; a in. to $j in.if flattened, + in. if countersunk, and nothing if countersunk andchipped.

Eyebars of the same member should be placed parallel as nearly aspossible (see Specifications, Appendix), and as far apart asto allow inspec-tion and painting of all sides after erection (not less than 1 in.). Therequirement cannot always be fulfilled in order to avoid excessive bendingmoments as explained later.

In riveted members pin plates should be added on both sides of themain web if possible. If in an outer pin plate rivets have to be counter-sunk this plate should be at least 25 in. (preferably 4 in.) thick.No rivet should be countersunkin the main material, to avoid extra hand-ling of long pieces in the shop.

The usual requirement for tension members is an excess of 25%of the net area through the pin over that through the body of the memberand a net area back of the pin equal to 75% of that through thepin. The area through the pin can be increased to advantage by makingthe pin plates wider than the body of the member if the width of the bodyis considerably less than three times the diameter of the pin. It isdesirable that each plate has a sufficient net area back of and throughthe pin to develop its full bearing value; if it has not, it should connect tothe other plates with a sufficient number of rivets back of the pin totransfer to them its excess of bearing value (see example, page 202).

In top chords and end posts, attention should be paid to the possi-bility of driving the rivets in the horizontal legs of the angles; if necessarythe outside pin plates may be made only as wide as the distance betweenthe vertical legs of the flange angles.

The thickness of a riveted member should preferably not exceedfour times the diameter of the rivets (see Specifications, Appendix).

Figs. 22 (a and b) show typical packings of the bottom chord panelpoints. The diagonal eyebars (shown dotted) are placed as near to theverticals as practicable. If they consist of two bars (a) both are placedoutside, if of four bars (b) two are placed outside and two inside of thepost. The bottom chord bars (shown full) are placed outside of thediagonals if there are only two bars in the first two panels and four inthe next panel (a). If there are more bars some are placed inside of the


post or of the inner diagonal bars (b). At La (a and b) two bars L&are placed side by side, with a filler ring between, as by this arrangementthe horizontal bending moment becomes smaller than if the bars wereplaced alternately in the one and the other direction. In a case likethis the horizontal moment may also be reduced by making the outsidebar lighter than the others, although it is desirable that all bars of a

LZ L3

FIG. 22.

member have the same thickness. Counter diagonals are placed insideof the post and, if consisting of one bar, at the center of the truss.

Fig. 23 shows the packing at the hip UX if the top chord has only twowebs as in short spans.

In this case there are usuallyo nly two diagonal eyebars UlLs; four barsmake the packing unsatisfactory since, if all are placed inside of the chord,the moment becomes very great and, if two are placed outside (as shown

UI UI

FIG. 23.

dotted), these have to converge considerably toward the bottom, whichis objectionable. Besides it does not look well if bars are packed outsideof the chord. Where four diagonal bars are necessary, it is usuallyadvisable to make the top chord and end post with three or four websand arrange the packing as shown in Figs. 23 b and 24. Instead of acenter web of full length a center diaphragm of a sufficient length totransmit its share of pin bearing to the outer webs has often been used to

1 4

2 0 0 DESIGN OF STEEL BRIDGES [CHAP. IX

advantage. In the case of four webs (Fig. 24) each may be assumed totake one-fourth of the stress in the member; in the case of three webs,however, it is evident that the center web gets much more than the out-side webs and should be designed correspondingly. The pin should beassumed continuous over three supports but the center reaction thusfound may be somewhat decreased.

The packing at the other top chord panel points is similar to thatat the hip; the main diagonal bars are always placed between the webs ofthe top chord and those of the vertical and the counter diagonals areplaced inside of the vertical.

FIG. 24.

Plate XVII shows the complete pin packing for a 255-ft. double-track deck span, and Plate XIX for a 373-ft. double-track throughspan.

30. CALCULATION OF MOMENTS AND SHEARS ON PINS

In calculating the bending moments and shears on a pin, the forcestransmitted by the members meeting at that pin may be assumedapplied at their centers of bearing.

All forces acting on the pin are resolved into two components, usuallyvertical and horizontal. If then M, and n/r, are the moments producedby the ,vertical and horizontal components at any section of the pin, theresulting moment is

It is evident that all forces acting on the pin must be caused by the sameposition of load, hence must be in equilibrium.

Since maximum M, and Mh do not necessarily occur simultaneously,nor at the same section of the pin, it is in general necessary to considerdifferent conditions of loading and to determine the moments M, and .Mhat different sections of the pin in order to find the absolute greatestmoment M.


Assuming a condition of loading causing maximum stress in a certainmember, it is usually necessary to calculate only the stress in one othermember and the remaining forces can be determined by considering thatthe sum of the.horizontal components and that of the vertical componentsmust be zero. The two component forces of each member are thendistributed to the several webs or bars of the member in proportion oftheir bearing area. The individual forces thus found are then assumedacting independently on the pin, which may, therefore, be regarded as asimple span, the two outer forces representing the reactions. Theshears and moments can now be easily determined.

Example.--The greatest moment at lower chord point Lz will usually occur whenthe stress in diagonal UILz is a maximum. Let this stress be 360,000 lb., with a

FIG. 25.

S c a l e : $ “= 5% IOO ,OOO/bs .= 5OqOOO idbs.

Fig. 7.6.

vertical component of 268,000. lb. and a horizontal component of 240,000 lb.For the same condition of loading we determine the stress in bottom chord LIL, at420,000 lb. The stress in UzLz at the pin must then be 268,000 lb. (vert. camp. ofU,L,) and the stress in L2L3 = 420,000 + 240,000 = 660,000 lb. The stressesinthevarious bars are shown in Fig. 28.

The moments can be determined either graphically by a force and equilibriumpolygon (Fig. 28) or analytically by the following table:


Vert ical I Resultantm o m e n t

Momenti n c r e m e n t

MomentM, ~MI,~+MG

1 .l’i~~ .i:ss’ --3d5. 0

2 ‘i’45’ .i:ss’ .‘It:.E;3. -305

3 ‘-‘iid .i:Bi. . -i94’ -222

4 . . . . . . . . . . . . . . . . . . . . -4160 2.0 0 . f134 2.0

5 . . . . . . . . . . . . . . . . . . -416 . . . . . . . . . . . .$268

$268 495,000

The assumption of individual action of the several bars or webs isnot correct, since as soon as the pin deflects the stress distribution ischanged. This tends to decrease the moments on the pin, but some barsare then stressed higher than others of the same member. The foregoingmethod is, however, sufficiently accurate for practical purposes as itshould be remembered that the theory of bending as applied to pins isonly a rough approximation.

The maximum moment at the hip pin U1 usually occurs when theload covers the whole span in such a way as to cause the maximum stressin the hanger iIJ&. The stress in diagonal UILS is then also nearly amaximum.

At any other top chord pin the greatest moment will usually occurwhen the main diagonal and post connecting at that pin get their maxi-mum stresses.

The bottom chord pins get the greatest moments either when thestresses in the chord members or those in the web members are maxima.

31. PIN PLATES

The several pin and web plates of a riveted member transmit stressesto the pin in proportion to their thickness. The pin plates must connectwith a sufficient number of rivets to the main material to transmit theiramount of stress. An excess of rivets is always desirable owing to thebending produced in the rivets especially where the stress has to gothrough several intermediate pin plates.

The determination of the length of pin plates and the number ofrivets is best illustrated by an

Example.-Fig. 29 represents a top chord panel point, the chords are pin bearing.Permissible bearing stress on pin = 24,000 lb. per sq. in.

ART. 311 SIMPLE TRUSS BRIDGES

Pin assumed 6 in. dia. at 7Jz.

2 0 3

Bearing per Iin. in. of pin = 6 X 24 = 144,000 Ib.Bearing required in chord Us - US at Us:

832,000144,000 = 5.8 lin. in.

= 2.9 on each side of chord.

In the chord Uz - lJ3 web is Q in.; 2.9 - .6 = 2.3 in. in bearing plates madeup as shown in sketch.

144 000 lb.To determine the number of rivets required in each plate we have ___‘_i_6_

= 9000 lb. bearing for & in. on pin.

Value’ of one &in. rivet in single shear = 6600 lb.9 0 0 0

Then &-in. bearing requires 6~~ = 1.4 riv. in single shear.

The number of rivets necessary for each plate is then readily determined fromits thickness, noting that rivets are in double shear when connecting pin plates onboth sides of the web, if the unit stress in bearing on the web is not exceeded.

FIG. 29.

In the sketch the 17- and Win. plates, which are f f & + & + 8 in. = 16 in.thick or 3% in., required 30 X 1.4 = 42 single-shear values, whereas we have30 rivets in double shear, and 5 in single shear, giving 65 single-shear vaIues.

In the sketch let us take a section AA. The only reinforcing plate here is thefiller plate 11 X A, and + of its area = 2.4 sq. in. Considering one side only,the area of 3 of the cover plate plus the top flange angle is 6.75 f 2.85 = 9.6 sq. in.

As the web bears directly on the pin, it gets its stress direct, and does not needto be considered. The value of the area 9.6 sq. in. must be made up by the area2.4 sq. in. plus the values of the rivets that pass through the flange angle betweenA and the pin. There are 7 rivets in double shear and one in single shear.Then we have for the value of the 15 rivets in single shear,

(7 x 2) + 1 = 15 x 6600 = 99,000 lb.

This divided by the unit stress used in the chord section is equivalent to

99 ,000m. = 6.8 sq. in.; 6.8 +2.4 = 9.2 sq. in.,>

which is nearly equal to 9.6 sq. in., the required area.

204 DESIGN OF STEEL BRIDGES [CEIAP, Ix

It is necessary that the web should be reinforced by a plate until sufficientrivrts are in the flange to give the latter its full portion of the stress in the member.Thus the flange area 9.6 sq. in. already found must take care of a stress 9.6 X 14,500

or 139,000 lb. This requires139,000~ ~ or 21 single shears to develop it, and6 6 0 0

the last filler or reinforcing plate must be extended until this number of rivet shearhas been introduced into the flange.

Another requirement must be fulfilled, which in many cases will determinethe length of the pin plates. At any section taken through the chord near the pin,there must be a sufficient number of pg. in. of metal on either side of’ thecetter of gravity of the chord section to develop the full strength of that partof t,he section. The angles and coverplate must be assumed to have a value equalonly to that given them by the rivets which hold them in the distance between thesection taken and the pin.

When the stress in a pin plate has to pass through another or several pinplates in order to reach the main section, another condition must be fulfilled.Not only must there be sufficient rivets in the plate in question to transmit itsstress, but there must be enough rivets in this and the next adjoining platetogether, to transfer the stress of the two, and if there are three plates, enoughrivets in the first, second and third plates together to transmit the stress in thethree to the main section.

In &he example given (to the right of U,), for the first plate on the inside(lap plate) which is 3 or & thick, we need 6 X 1.4 = 8.4 rivets in single shear.For the first two plates (Q + & = s), 15 X 1.4 = 21 single shears. For three plates$ + & + & = p$ in.), 24 X 1.4 = 33.6 single shears. Referring to thesketch, we see that this condition is fulfilled.

ART. 32. LATTICING OF COMPRESSION MEMBERS

The minimum sizes of lattice bars are given in the Specifications,see Appendix.

The design of the latticing of columns is largely a matter of judg-ment and any theory advanced can be, only an approximation. Thefollowing method is given merely as a guide in designing the latticing ofheavy compression members. For a more detailed discussion of thissubject see Vol. I.

Any compression formula can be brought into the form

where sb = permissible buckling stress, s = permissible direct stress,

a n d f (i) d1

e n o t es a certain function of 7 and may be regarded as the

bending stress produced by the load P in the deflected column (Fig.30), therefore,


where R = section modulus of the column section. The shear V pro-duced at the end of the deflected column is obtained by resolving Pinto components parallel to the tangent to the elastic lineand normally to P. Assuming the elastic line as a para- P

bola we have V--L%-

v=p+~pG

?

-/2

and introducing the value of P6 from the above formula -s. Ji

for f(i)

;

we find. . . . . . . . . . . . . . . . . , .V=;Rf(;)

~~

I

\(o) Straight-line formula :A

8b = 16,000 - 70 f FIQ. 30.

Since f (4> = 70 s? . . . . . v = 280 ;

2ar2or substituting. . . . . . . . . . . . . .R = ~d(where a = area of column section, d its width in the plane of buckling)

V = 560 f

(b) Rankine formula :s

&, = ___ or

& (1 f f> = 8, therefore, sb = a - sb -$

f(i) = Sb; and v=+&2ar2

-1o r forR=T

V-galc d sb

where for 8b has to be used its permissible value

SbZ s

1 +gi

It should be noted that r is the radius of gyration in the plane of thelacing. a is not the actual area used but the area required for the

206 DESIGN OF STEEL BRIDGES [CHAP . IX

above r and the corresponding 1. In ordinary cases, however, the actualarea may be used.

From the shear V the stress in the lattice bar at the end of the columnis found as follows:

(l)Columns with 2 webs (Fig. 31).If n lattice bars are cut by a section X through the column the stress

vin each bar is ; set CY or

V e- -*n b

For single latticing on two sides, for example, n =2 (Fig. 31), fordouble latticing on two sides n= 4.

a a' a ”

x.

FIG. 31. FIG. 32.

(2) Columns with 3 webs (Fig. 32).

Area of section = a = a’ + 2a”.

&-

FI&. 33.

Longitudinal shear per lin. in. between two ribs is

iVM=-a r2

where M = static moment of outer rib about the column axis = a”b,b being the distance of center of gravity of the rib from the column axis.The longitudinal shear in the panel length L is then

S’ S’ e e V Mand the stress in the diagonal = -2- cosec a or - - - - -2 L - 2 ar2(assuming two sides of latticing).

(3) Columns with 4 webs (Fig. 33).Area of section = a = 2 (a’ + a”)


Longitudinal shear in the panel length L between outer and innerrib

S” = L VM”a?

where M” = arrbr’, and stress in outer diagonal, assuming two sides of

latticing . . . . . . . . . . . .Sft efr err VM”-==2L

-___2 aP

Correspondingly the stress in an inner diagonal is )

S’ e’ e’ VM'-~2 L- 2 a?

where S’ = L VMIa?

and M’ = a”b” f a’b’.

The lattice bar may be stressed in tension or. compression and mustbe designed to resist either stress. Recent tests have shown that the

resistance of flat bars to buckling is very small; it is advisable to use a

permissible compression stress of not more than lO,OOO-40: or to use

lattice angles or channels. The lattice bars must of course be connected

by a sufficient number of rivets to develop the calculated stress.

ART. 33. CAMBER

aA camber is usually provided by making each panel length of the topchord, or i ts horizontal projection 3 in. in 10 ft. , longer than thecorresponding panel of the bottom chord (see specifications, Appendix).If e is the total amount by which the top chord is thus lengthened, h theheight of truss and. 1 the length of span, the camber at the center of spanis approximately

el(1) c = a (all d imensions in same units). This is approximately

equal to the deflection produced by the total load for which thetruss is proportioned.

For a given camber c the top chord has to be lengthened by

for an increase in chord length of ft in. in 10 ft. we have approximatelyc 1 1

e = 0.001 I, and therefore - - __ -1

1 - 8000 h' so that for h between 5 and

8 the camber ranges between & and k. of the span length.


If the truss has single diagonals only, it is not necessary to changethe length of the diagonals. If there are double diagonals, these mustbe lengthened to

d = Jh’f (p+%) 2p being the panel length and a the amount by which the top chord panelis lengthened (Fig. 34).

Sometimes it is specified that the camber should correspond to thedeflection caused by a certain load P, for instance,dead load only or dead plus 3 live load. If Pis the total load on the span, A the area of thechord section, the deflection or camber at thecenter is approximately

FIG. 34.

5 Pl3__‘=ii% EAh2 (E = modulus of elasticity)

and this, introduced in formula (l), gives the re-quired lengthening e of the top chord.

A more correct method, which should be used for large bridges,consists in shortening or lengthening all truss. members by theamount by which they are lengthened or shortened, respectively,under the specified load P. If A is the gross area of the section of amember, L its length, and S the stress produced by P, the change inthe length of the member is

SLAL=m

The resulting deflection or camber and also that due to the pin play,if any, is determined as shown on page 296.

Plate girders are usually built without camber. If a camber isspecified it is obtained by giving the web plates the shape of a trapezoidwith longer sides on top.

CHAPTER X

SKEW BRIDGES AND BRIDGES ON CURVES

ART. 1. SKEW SPANS-GENERAL

Skew spans should be avoided whenever possible, as they are moreexpensive and not as satisfactory in operation and maintenance as squarespans (see Specifications, Appendix). For a slight skew crossing it is pref-

F IG . 1 . F IG . 2 .

erable to build the abutments square and increase the span (Fig. l), orto build the abutments skew, and the superstructure square.

For more than one track it may be possible to build an independentsquare span for each track as shown in Fig. 3 or 3 a.

F IG . 3 .

The above arrangements can, as a rule, be made for deck plate girderspans; for wider bridges, however, ,the masonry becomes too expensive.If a skew superstructure has to be used, complicated details such asskew connections, bent plates, etc., should be avoided. Bearings andends of girders should be square.

2 0 9

210 DESIGN OF STEEL BRIDGES [CHAP. X

ART. 2. SKEW PLATE GIRDER SPANS

In skew de& plate girder spans a square-end cross-frame CF is pref-erably placed at each end B (Fig. 4), the ends A being held by a top

FIG. 4.

and bottom strut AB connected to the flanges by horizontal gussets.Bent connections are thus avoided. The stresses in the laterals arecalculate’d by assuming the lateral truss supported at points B.

All skew through plate girder spans with stringers should be so arranged

FIG. 5.

that each pair of stringers ends square. In through girder spans square-end floor-beams EF (Fig. 5) are placed as near the ends B as possible andthe length BB is divided into an economical number of equal panels.The ends of stringers, which are seated on the masonry, are connected

FIQ. 6.

to each other and to the ends of girders by a strut S which should be asdeep as the stringers allow. If the skew is considerable additional shortfloorbeams are used with one end seated on the masonry and connectedby a strut S to a panel point of the lateral system (Fig. 6).

A R T . 31 SKEW BRIDGES AND BRIDGES ON CURVES 211

The end of a skew shallow floor which consists of cross-beams or troughscan be seated on a continuous masonry plate, the ends of the beams beingconnected to each other by square diaphragms (Fig. 7). A skew faciagirder can generally not be avoided where ballast is used. To avoid theskew floor the main girder may be extended as a cantilever from B to Cor a separate shallow girder BC may be added (Fig. 8).

FIG. 7. FIG. 8.

ART. 3. SKEW THROUGH TRUSS SPANS WITH PARALLEL CHORDS

Skew through truss spans offer particular difficulties on account ofthe skew portals. To avoid skew-end floorbeams the ends may be ar-ranged as shown in Figs. 5 and 6. Skew intermediate floorbeams andframes are avoided by arranging the trusses as shown in Figs. 9 to 12;approximately economical panel lengths should be selected.

FIG. 9. FIG. 10.

Fig. 9 shows the simplest arrangement; it is obtained by dividing thelength AAl into panels equal to the skew s. The truss figure is sym-metrical, all truss panels are equal, and opposite end posts are parallel.It should be used even if the skew is slightly different from that of thesubstructure; if this is not practicable it may be possible to make thepanels BC and BICl equal to the skew s and divide the distance CC1into equal panels slightly different from S.

The arrangement, Fig. 10, has the same advantages as Fig. 9 except


that the first panel of the truss at each end is shorter than the others.It is obtained by dividing the truss span AB or A,B, into a number ofpanels equal to the skew s and two end panels X smaller than s. Theend floorbeams are seated on the masonry with one end or these shortfloorbeams and the supporting hanger may be omitted in certain cases.

If the skew s is much shorter or longer than an economical panellength then Fig. 11 or 12 may be preferable. They are obtained by

FIG. 11. FIG. 12.

dividing length CD (Fig. 11) or BB1 (Fig. 12) into equal panels. Thetruss figure is then unsymmetrical and opposite end posts have differentinclinations. In this case the portal is placed in a plane which intersectsthe plane of the truss in a line whose inclination is a mean between theinclinations of the two end posts, Such a portal, however, can only be‘connected satisfactorily if built single-webbed and if the inclinations ofthe end posts differ only slightly. The connection should be carefullylaid out before the bridge is designed. In such cases the end posts have

FIG. 13.

sometimes been made vertical and the portal placed between them; thefirst diagonal can then be a tension member, that’is, falling toward themiddle of the span.

The same arrangements can be used if the skew extends over two orthree panel lengths, the above end panels, as seen in the plan, simplyrepresenting two or three panels, respectively. Fig. 9, for instance, wouldbecome Fig. 13 for a skew of two panel lengths.

A R T . 61 SKEW BRIDGES AND BRIDGES ON CURVES 213

ART. 4. SKEW THROUGH SPANS WITH POLYGONAL TOP CHORD

The top chords of the two trusses if seen in elevation should covereach other so that the laterals in any panel lie in a plane normal to theplane of the trusses. Trusses with curved chords for skew spans must,therefore, be formed as shown in Fig. 14 (a b c is a straight line), the topstrut of the portal rising from one truss to the other. Such portals are

FIG. 14.

complicated and do not look well and, especially if the skew is consid-erable, it may be preferable to use trusses with parallel chords.

ART. 6. SKEW DECK TRUSS SPANS

Deck truss spans are as a rule provided with skew-end sway bracingand skew-end floorbeams between the vertical end posts. ArrangementFig. 11 is then the simplest; that of Fig. 9 or 12 may also be usedbut the connection of a skew and a square floorbeam at the samepost B is difficult.

ART. 6. ARRANGEMENT OF FLOOR OVER SKEW PIERS

If the adjoining ends are fixed the stringers over the pier are framed‘into the two end floorbeams of the adjoining spans. If one end AIB

FIG. 15.

is movable (Fig. 15) the stringers are seated on expansion pockets con-nected to the end floorbeam of the fixed end. The end strut ABI mustthen not be connected to the stringers.

214 DESIGN OF STEEL BRIDGES CHAP. X

ART. 7. SKEW BENTS

If the superstructure is supported by an intermediate single skewbent BB a change of temperature will change the straight line ABC(Fig. IS) of the girders to a broken line ABICl since the bent movesnormally to its own plane. If the skew is considerable this kinkin the floor is objectionable and to avoid it the main girders or trussesshould be seated at B either on a solid pier or on two single columns

without bracing between them. If columns are used the lateral systemhas to be designed as one single span of the length AC.

ART. 8. SUPER-ELEVATION OF OUTER RAIL ON CURVES

The super-elevation E of the outer rail on curves is, as a rule, specifiedor the speed of trains is given for which the super-elevation has to bedetermined. Table 43 gives the super-elevation for various degrees ofcurvature and speeds on standard gage roads. It is calculated with theassumption that the resultant from the train load and centrifugal forcegoes through the center of the track, using the formula:

E = 0.000055 DV2 in feet

where D = degree of curvature and V = speed in miles per hour. 7 in.or 8 in. is usually the limit for the super-elevation as the speed in sharpcurves is reduced.

ART. 91

-

DeogfPeecur-

vature

SKEW BRIDGES AND BRIDGES ON CURVES 215

TABLE 43Super-elevation of Outer Rail on Curves

Radius

f%in

ieters

730.0 1747865.0 873910.0 582432.5 437146.0 349

955.0 291819.0 2 5 0716.0 2 1 8637.0 194573.0 175

521.0 159477.0 145441.0 134409.0 124382.0 116

109103

il87

PS beginning of Transition Curve.PC beginning of Circ. Curve.PS’ beginning of Trans. Curve.PT beginning of Tangent.

Velocity in miles and kilometers per hour

2 0 25 3 0 35 4 0 45 5 0 55 60 65 7 0 Miles~~

32 4 0 48 56 6 4 72 8 0 88 9 6 104 112 Kilometers

Super-elevation in inches

- . --

:1

-Su~~-$ele$on given in table is between

Length PS to PC or PS’ to PT = super-elevation X 12 X 60 in feet.

(Example: for 2 in. super-elevation, length ofTrans. curve = 2 ft. X 60 = 120 ft.).

Super-elevation in Transition Curve in-creases in a straight line from zero at PSand PT to the full value at PC and P S’.

ART. 9. PROVISION FOR SUPER-ELEVATION

@late XXVI shows different methods of’obtaining the super-elevationof the outer rail. In the open tie floor small super-elevations are gener-ally obtained by tapered ties (A) and greater ones by bolting either alongitudinal shim timber about 12 in. wide to the outer stringer (B),. or a

15 ’


beveled one about 3 ft. long to each tie usually on its bottom side (C).These shim timbers should not be less than 4 in. deep over the stringer.In solid floors the elevation is made up in the ballast (F).

Sometimes the super-elevation is obtained by tilting the whole span(D), or by placing the outer girder or stringer higher than the inner one.(E). The first method may be used for small super-elevations and shortspans by placing shim plates under the bearings; the second methodrequires more expensive construction.

ART. 10. MIDDLE ORDINATE OF CURVE

5 7 3 0If D = degree of curvature, 1 = length of chord in feet and r = 7

FIG. 17.

= radius of curve in feet, the middle ordinate m of the curve is, withsufficient accuracy, in feet (Fig 17)

12 012m = 8r = 45,840

ART. 11. WIDTH OF DECK SPANS ON CURVES

The spacing of the main girders or trusses of deck spans on flat curvesmay be the same as that required for the corresponding spans on tangent(see pages 145 and 169). If the middle ordinate for the span exceeds

~~.__________ -*+---- ----------

FIG. 18.

about 3 in. for small, to 6 in. for larger spans the spacing should beincreased by the middle ordinate of the curve for the span length aschord. This will secure sufficient stability against overturning from thecentrifugal and other lateral forces (page 111).

The trusses are usually located with reference to the curve sothat the center line of bridge bisects the middle ordinate (Fig. 18). I f

PLATE XXVI

--- --__ _ ___-__ (j’6”-- ---__

Fig. A.

Figa F..

----*---_ .--.- ._... _ __..-.

Fig. 0.

k-----_______ 7’0! __________ ..--->i

!

Fig. D.

Fig. C.

Methods for ffaking O&r Rai/as Shown on figs. A, B and D

qenerally Used on Curves- nof exceeding 4 Degrees. -

-_.-._-____ _ .______--_ ___. J7’6”C.toC.Girders u-----------4

Fig. E.

Typical Cross-sections of Spans on Curves. (See Chapter X.) (Facing page 216)

ART. 121 SKEW BRIDGES AND BRIDGES ON CURVES 217

b, = width required on tangent we get for the width on curve b = b, + m.By this method the outer girder gets heavier than the inner one (sincethe live load can be assumed acting along the center line of track asexplained on page 112).

A more rational method would be to locate the center line of bridgeas shown in Fig. 19; assuming a uniform live load along the center line

FIG. 19.

of track equal maximum bending moments are caused in both girders.The distance c. to c. of girders is in this case

b=b,+im

which is slightly greater than required by the first method.

ART. 12. WIDTH OF THROUGH TRUSS SPANS ON CURVES

The spacing of the trusses in through spans depends upon the re-quired clearance which is usually based on the extreme dimensions of acar. If

A = extreme length of car.B = distance c. to c. of trucks.w = clear width required on tangent.H = extreme height of car above rail.h = height of lower corner of clearance diagram above rail.C = extreme length of truss at the height H above rail (not the

span length).E = super-elevation of outer rail.d = distance between centers of rails (= 5 ft.).

m, = middle ordinate of curve for chord A.mb= (( ‘I <‘ ‘I C L “ B.

LC LC ‘I “ “ ‘Im, = c.We find from Fig. 20 that if the car were vertical the clearance w at

the center of the span would have to be increased by ma-mb on the out-side and by mb+mc on the i-nside of the curve. Further, on account of

218 DESIGN OF STEEL BRIDGES [CHAP-X

the tilting of the car (Fig. 21) there must be an additional clearance on

the inside ofH2 E, while that on the outside is decreased by dhE .

We have therefore ‘for the distance from the center line of track atthe center of span to the outside clearance line:

F =;+m.-m,-;h

and for the corresponding distance on the inside:

G=;+ma+m,+fH

FIG. 21.

and the total clear width will be F + G.For the Specifications of the Am. Ry. Assoc. (Appendix) we have

A = 80, B = 60, w = 14, H = 14, h = 4 and further, if D denotes thedegree of curvature, m, = 0.140 D and mb = 0.078 D, therefore

F = 7+0.062D - 0.8E,G=7+0.078D+2.8E+ma,

all dimensions in feet.

ART. 121 SKEW BRIDGES AND BRIDGES ON CURVES 219

To the distances F and G has to be added one-half of the w?dth of thetruss (see page 147) in order to get the corresponding distances from thecenter line of track at the center of the span to the center lines of thetrusses.

FIG. 22.

Example.-125 ft. single-track span (Fig. 22); degree of curvature D = 5, super-elevation of outer rail

E = 4in. = 0.33 ft.

mc===0.982ft.45,840

F=7$0.31-0.27=7.04ft. = 7ft.Oiin.G=7+0.39+0.93+0.98=9.30=9ft.3+in.

Adding on each side 10; in. for the truss we get for the distance c. to c. of trusses18 ft. 1 in.

FIG. 23.

The distance c. to c. of girders of half tlrough spans (pony trusses) is foundin a similar way as shown above for through truss spans except as follows.Denoting with k the height of the top flange above the low rail and with WI( = w for k > h) the width of the clearance diagram at the height k above the low rail,we have:

220 DESIGN OF STEEL BRIDGES [CHAP.X

a n d F = T+na,-ma-$, i f k<h

01‘ F = ~+m,-nz~7;h, i f k>h

WC is here the middle ordinate for a chord equal to the total length of the girder.

ART. 13. DISTANCE BETWEEN TRACKS ON CURVES

Unless otherwise specified the standard distance c. to c. of trackson tangent (usually 13 ft.) should be increased on curves byma (= 0.14D infeet for A= soft.).

ART. 14. ARRANGEMENT OF STRINGERS ON CURVES

On flat curves the stringers may be arranged in the same way as ontangent and proportioned for the resulting eccentric loading as shown on

FIG. 241

page 112. If the middle ordinate for the whole span exceeds about 3 in.the stringer spacing of 6 ft. 6 in. should be increased by the middle ordi-nate. In order to get equal eccentricities for the center and end stringers

F IG . 25 .

the center line between the stringers should bisect the middle ordinateof the curve (see Fig. 24).

On sharp curves it is more economical to keep the distance betweencenters of stringers 6 ft. 6 in. or 7 ft. and offset the stringers as shownin Fig. 25. The offsets should be alike if possible and of such amountthat the connections of two adjoining stringers do not interfere.

CHAPTER XI

WEIGHTS OF SIMPLE SPAN BRIDGES

For weight of flooring (tracks, pavements, etc.), and general rulesgoverning the weight of the steel work, see page 2.

ART. 1. TABLES AND FORMULAS FOR STEEL WEIGHT OF RAILROADBRIDGES

American Bridge Co. of 1902.-Single- and double-track spans up to200 ft. designed according to the American Bridge Co.‘s Specificationsof 1900 and medium steel. The total steel weights and principaldimensions are given in Tables 44 to 47 and detailed weights of thetruss spans in Table 48.

Assumed live loads, Cooper’s E-40 and E-50.

Impact300

i = 1 L + 300(I = live load stress, L = loaded length of single track in feet produc-ing maximum stress).

TABLE 44American Bridge Company’s Single Track Deck Plate Girder Spans

Medium Steel A. B. Co. Spec., 1900

Length,out to

out

Tc2f” c.bearing

f t . in .

13 918 923 628 6

:i :43 348 ,3

xi :63 068 0

;; i

if 9”

i; i102 3107 3

,

Note.-The wzby length “out to out.’

It

-

-

1. TY 1

2

-;hts of steel per lin. ft. were c.”

E-40

1 . to B.of

anglesft. in.

2 2f2 ‘523 213 9+

4 ot

ii ig5 91

3. of R0 msr,ft. in- -

: ‘:6 117 2

t 1:8 0

i :10 010 0

10 910 911 611 6

Totalweight

lb.-320 4,800350 7,000420 10,500430 12,900

477 16,700638 21,500576’ 25,900604 30,200

EE725767

34,90040,90047,10053,700

811864988

1056

XEs4:ooo95,000

1092113112991391

103,700113,100.29,900139,100

-kbt ained 1

E-50

3. of R0 msq

ft . ir

: :10 010 0

10 910 911 611 6

1g t o

347390476483

537583607648

%789850

895954

11081170

1201128813901436

r

-

I

;

3-

Totalweight

lb.

5,2007,800

11,90014,500

18,80023,30027,30032.400

38,80045,500

%:%

67,100npg

.05:300

.14,100

.28,800

.46,000

.58,000

1 steel weight

221

TABLE 45American Bridge Co.- Single Track Through Plate Gtd;,rcWiritE.z 1900

Medium Steel E-40 loading

307 337329 409323 443364 493358 5 1 0

375 569363 592399 649390 683401 727395 786

405399406401407 :401 1

-

861913969029136186

ft. in. ft. in,-

13 9 3 6s13 9 4 3s13 9 4 9+15 0 5 6f15 0 5 93

15 0 6 0;15 0 6 O+15 6 6 6%15 6 6 6;15 6 6 6+15 6 7 o+

15 6 7 6;15 6 7 6+15 6 8 3915 6 8 3%15 6 9 0;15 6 9 0%

- -by d

2 2;2 2;2 2;2 2t2 2;

2 222 2;2 2t2 2;2 2;2 2;

2 2q2 242 2t2 2;2 2;2 aq

-

-

I

.t-

-

f-

1111

-r

L

2

3

4

5

6

7

8

9

-

ft.

2 93 4

2l49

5 4

i:

E79

t. in.

27 932 937 942 947 9

52 957 962 967 972 977 9

82 987 992 997 902 907 9-

#. i n

1 4:1 411 411 4;1 4;

1 4;1 4:1 4:1 4;1 4;1 4:

1 4j1 4:1 4,1 4:

: 2;

No. of panels

3 x10 ft.IXlOft.+2X7ft.6in

4x10 ft.;XlOft.+2X7ft.6 in

5x10 ft.

21 ,70028 ,70033 ,70042,400

47,400

56,40061,90073,50080 ,50090,500

100,500

114,000124,500137,500149,900169,400118,900

24,5OC31,5oc37,ooc46,5OC51,5oc

60,5OC66,5OC78,5OC86,OOC96,00C

105,5oc

123,5OC134,ooc147,5oc160,OOC179,5oc192,5oc

22,5OC29,5oc34,5oc43,5oc48,5OC

57,5oc

2~~~8 2 : O O C92,ooc

102,ooc

L19,OOC.29,5oc!42,5OC156,5OC.76,OOC!88,5OC

:XlOft.+2X7ft.6 in6 X10 ft.

i X10 ft..-&2 X7 ft. 6 in7 x10 ft.

iXlOft.+2X7ft.Bin8 X10 ft.

“XlOft.+2X7ft.6in9x10 ft..

n10x10 ft.

IXlOft.+2X7ft.6in11 x10 ft.

Note.-The weights of steel per lin. ft. were obtain, iding total steel weight by leng "out to oult” of end stringers.

Am. 11 WEIGHTS OF SIMPLE SPAN BRIDGES

-.

.-

--

I

!:

:--;l

TABLE 47

American Bridge Co.-Truss BridgesA. B. Co. Spec.. 1900Medium Steel

Weight per ft. of bridge, lb.Widthc. to 0. Height

truss a tcenter

Height B. of B. B.a t rail to

endangles

masry. (of stgrs.

B. B.angles

,f flbms.

Totalweight

withoutend

floorbm.

ft. in. ft. in. f t . i n . ft. in.I ft. in. / ft in. 1

Single-track Spans--E-40 loading- E27 3 27 3 5 8 3 2: 4 1% 402 614 121,000 116,00027 3 27 3 5 1 1 3 22 4 1; 403 740

ii ::167,000 162,000

5z

28 0 28 0 5 3 3 2; 4 1-1 460 840 9 0 224,000 218,00033 0 28 0 5 3 3 22 4 1: 465 945 105 ::: 282,000 276,00036 0 28 0 5 3 3 2; 4 4t 473 1083 111 63 352,000 346,000

g

Kind!Spanc. to e.

bearingNo. ofpanels

-15 815 916 216 216 2

Thro. lattice 100E

F8

us-Is-501oadingSingle-track S] Pal

27 3 27 3 5 11 3 52 4 4t27 3 27 3 6 2 3 5a 4 4t28 0 28 0 5 6 3 5t 4 4233 0 28 0 5 6 3 52 4 4236 0 28 0 5 6 3 52 4 4$

Double-track Spans--E-50 loading

I468 682 103 138,000 132,0004 7 0 8 6 0 102 :: 193,000 187,000 E520 965 125 258,000 252,000525 1110 140

FiR328,000 322 ,000

E

534 1275 154 67 413,000 406 ,000

T$. lattice 100lattice 125 5”1: pill. . 150 6

175‘l g:: 2 0 0 8’

1 5 1015 1 1

:i ;16 2

Thro. lattice 100 4 29 1 31 0 31 0 8 9 3 5; 6 63 1050 1161 225 164 274,000 260 ,000“ lattice 125 i 29 4 31 0 31 0 9 0 3 5 4 6 6+ 1070 1555 2 3 0 155 390,000 376 ,000:: p i n . . 150 29 8 33 0 33 0 8 2 3 5$ 6 6+ 1180 1765 233 167 518,000 502 ,000

pm. . 175s’

29 8 3 8 0 33 0 8 7 3 5; 6 63 1180 2 0 2 0 282 178 659 ,000 . 641,000“ p i n . . 2 0 0 29 8 41 0 33 0 8 7 3 5t 6 62 1190 2363 270 202 823,000 805 ,000 g

I ,Note.--For detailid weights see Table 48.

/ / / I , I I / 6”

E

TABLE 48American Bridge Co.-Truss Bridges

M e d i u m S t e e l D e t a i l e d W e i g h t s o f S t e e l w o r k f o r E - S O L o a d i n g A. B. Co. Spec., 1900

I Single-track spans I Double-track spans

Length c. c. endpins, ft.

Kind of truss. . . . . . . .

Height c. c. chords.. . . . .Width c. c. trusses. . . . .

(1) Trusses.. . . . . . . . . . . . .(2) Pins.. . . . . . . . . . . . . . .

g(3) Bracing.. . . . . . . . . . . .(4) Floor.. . . . . . . . . . . . . .

2 (5) Shoes.. . . . . . . . . . . . . . .

i Total with end struts . . .

,” (6) Trusses per Ii<:. ::.9 (7) Bracing

(8) Floor “ ‘I “8 (9) Shoes “ “ “

$Total per lin. ft. . . . . . . . . .

Total with end floorbeams.

Timber, ft. B. M.. . . . . . . .

100

Lattice

27”!15’ “lov

Weight y0

67,800 78400 1

10,300 1546,800 69

6,700 10

1,320 100~~138,000

5,sooi

T

-

-Weight

139,7005,000

18,60078,00010,700

- - -252,000

965125520

70

1,680

258,000

8 ,400

58

3 :4-00

-

T200

Pin

100

Lattice

.150

Pin

28’, 33’, 36’I

31’16’ 2 29’ 1 d 2;3;,,

IWeight %

246,000 799 , 0 0 0 4

30,800 13106,800 43

13,406 5

Weight

115,500600

21 ,500105,000

16,400

%75

1

ii14

Weight %

253,800 781 0 , 8 0 0 432,500 13

177,300 702 3 , 6 0 0 9

4 0 6 , 0 0 0

1,27563154 8534 26

67 3

-ij&G-i413,000

11,100

-.

1

259,000

1,161215

1,050164

2 ,590

273,000

11,600

-4 5

4;6

00-

-

-

-

498,000

1,76553216 7

1,180 35159 5

-GaGi->~_514,000

-16,900

200

Pin

33’, 39’, 41,29’ 8”

Weight %

455,000 791 7 , 5 0 0 449,800 11

2;;,;;; 5;7

797,000

2,350 59249 6

1,180 30183 5

- -3,975 100~-815,000

22,200/

A m Note.-% (1) = +; % (2) (3) (4) (5) = f; % (6) (7) (8) (9) = $ Ties 8 in. X 8 in. X 10 ft. 0 in. 14 in. c. to c,SF42%

T = Theoretical weight of trusses. I = Actual weight of items in question. _ c .Guard rails 6 in. X 8 in.A = Actual weight of trusses. S = Actual weight of span. (Facing page 224)

ART. l] WEIGHTS OF SIMPLE SPAN BRIDGES 225

Wind, 50 lb. per sq. ft. on unloaded, 30 lb. per sq. ft. plus 300 lb.per lin. ft. moving on loaded structure.

Material, medium 0. H. steel, ultimate strength, 60,000 to 70,000 lb.per sq. in.

Permissible unit 17,000 lb. per sq. in. for tension, reduced forcompression.

The. total steel weights are approximately’ covered by the followingformulas :

Live loac

E-40E-50

Steel Weight in Pounds

Single-trackdeck plate

S. T. through plate S. T. throughlattice D.T. throughlatticegirder spans

girder spanswith- and pin spans, with- and pin spans, with-

out end floorbeams out end floorbeams out end floorbeamsexcl. bearings excl. bearings

(91+140)1 (111+400)1 1 (6Zf550)Z 1

(10 2 + 160) I (12 Z + 460) Z (71+640)1 (14 z + 1200) z

For end floorbeams For end floorbeams For end floorbeamsadd 3500 lb. add 5000 lb. add 16,000 lb.

For cast-iron bases add1000 lb. average for pl. gird. spans I = length out to out of plate girders.

up to 60 ft.

2000 lb. average for pl. gird. spans’65 to 80 ft.

1 = length c. to c. of endpinsfor trusses.

I

For cast steel shoes, etc., add 6000lb. average for spans 85 tc 110 ft. I

If permissible unit stress is 16,000 lb. per sq. in. instead of 17,000,add 5 y0 to the above weights.

American Bridge Co.‘s Plate Girder Spans of 1910:--These weredesigned according to Specifications of the American Railway En-gineering Association (see Appendix).

300Live load, Cooper’s E-60, impact

i=z Lf300-^----Y permissible unit stress 16,000 lb. per sq. in. reduced

for compression; material, structural 0. H. steel, ult. strength, 55,000to 65,000 lb. per sq. in. The steel weights of plate girder spans up to120 ft. are given in Table 49; they are approximately covered bythe following formulas, not including bearings (1 = length out to out).

Single-track deck plate girder spans. . . . . . W= (13 I + 150) ESingle-track through plate girder spans. . . IV= (15 I + 600) Z

For bearings (pedestals or shoes) add weights given in Table 49.

226 DESIGN OF STEEL BRIDGES [CH A P . XI

TABLE 49

Plate Girder Bridges for Single-track E-60 Loading@p&f. Am. Ry. Eng. Assoc., 19061

T Through spans, weight of steel work in pounds

Lengthout to oul

in ft. Total Girders StringersFloor-

‘earns andbrackets

Bracing Pedestal:rota1 per 1.t. of span

excl.bearings

34 ,200 14,800 9 ,900 7 ,000 9 6 048 ,700 24 ,400 11,000 10,30069 ,100 17,800

1,34035 ,300 12,460 1,460

92 ,100 51 ,300 19,800 16,250113,900 67,200 20 ,400

2 ,49020 ,500 3 ,020

139,000 83 .600 27 ,000 22 ,100 3 ,360

1,0901,1801,3401,5001,5901,700

1::110120

186,900 113,400 29 ,700224,900 138,200257 ,000

35 ,500166,700 35 ,000

313,800 213,900 42 ,300

26 ,200 4 ,62028 ,400 5 ,67032 ,300 6 ,47033,200 7 ,870

1,5401,6602 ,1402 ,2602 ,7802 ,940

Shoes12,98016,53016,53016,530

1,9352,0852 ,1902 ,480

Deck ssans, weight of steel work in pounds--9 ,700 7,000

18,300 14,80029 ,200 24 ,90042 ,900 37 ,30056 ,500 50 ,00075 ,200 66 ,60094 ,200 83 ,300

..........

..........

..........

..........

1,1601,9602 ,7603,4604 ,3605 ,8208,120

1,5401,5401,5402 ,1402 ,1402 ,7802 ,780

Shoes12,98016,53016,53016,530

4105 6 06 9 0815915

1,0351,145

lFl1:110120

126,300 104,700159,200 131,600188,000 159,600235,600 205,400

.......... 8 ,620

.......... 11,070..........

......... I.11,87013,670

1,2601,4251,5601,825

- -

Pennsylvania Steel Co.‘s Standards of 1908.-These cover single-track bridges up to 200 ft. span.

Penna. St. Co.‘s Spec., 1908. Live load, Cooper’s B-50.12

Impact i = mt- permissible unit in tension 16,000 lb. per sq. in.

Material, structural 0. H. steel, ult. strength 55,000 to 65,000 lb.per sq. in.

Single-track deck plate girder spans:

Weight of girders and bracing, w = (10.5 1 + 120) 1;

for spans up to 80 ft. add 13 I for bearing plates;for spans 85 to 95 ft. add 7700 lb. for cast steel shoes;for spans 100 to 110 ft. add 8700 lb. for cast steel shoes.Single-track through lattice and pin spans without end floorbeams:

Total steel weight, IV= (9Z + 410) Z

TABLE 50

Pennsvlvania Railroad--Double Track BridgesSoft E :el ti

I Weight per lin. ft. of bridge i;A p p r o x - m

“it!&? 22t imateShoes weight loading, 2

lb. lb.f t . in .

C o o p e r ’ s o

q

B. of R.to

masonry

ft. in.

11 0

1: ?I8 72

10 3$5 6$6 93

35 7 3

35 76

, of wei;moving

Kind

ft. I I ft. in. / ft. in. 1 ft. in.

533 1 1 6 1 31 0 / 84 0 1 57 0Thro. pin‘I ‘I

I‘ ‘I

‘I ‘I

/Deck pin“ “

353 4,187,800 E-38 i

378 3,;1-99,000 E-47 8M

373 2,789,OOO E-49 m

172 1,400,OOO E-35 ?z

5593 461

6022 4 6 0

4863 490

4 33 5 3; 1450

4 73 5 6$ 1015

4 74 5 64 1750

3 6$ 4 .6+ 1291

3 13 5 0; 734

3 14 5 0;

t of floor.

3206 459

3179 235 11 194 / W600~ z;

* Including end floorbeams. VI

407 1 1 30 8 67 0 37 0

373 1 1 30 8 56 0 36 0

273 9 29 10 49 0 34 0

Note.-Weight of floorbeams about $, of stringers aboutNote.-The greater distance B. of R. to masonry refers t d.

228 DESIGN OF STEEL BRIDGES [CHAP. XI

The trusses and bracing weigh about 8.1 1 lb. per ft. of span and thefloor 480 lb. per ft. of span, bearings 55 X 1 lb. total. For end floorbeamsadd 5000 lb.

Pennsylvania R. R.-Table 50 contains the weights and principaldimensions of various double-track bridges built by the P. R. R. Thelatest double-track bridges of large spans have an approximate

Total steel weight, W = (16Zf 1500) 1

for a loading of about Cooper’s E-50 and P. R. R. Spec., 1906.Pennsylvania Lines West.-Truss spans (Specifications, 1906, live

load per track 5000 lb. per lin. ft. plus a concentrated load of 60,000 lb.for chords and 90,000 lb. for web members) have an approximate steelweight of

w = (10 I + 900) 1 for single trackW = (18 I + 1800) I for double track

N.Y. C. and H. R. R. R.-The Specifications of 1961 (live load approxi-mately E-40) contain the following formulas for the weights of steelworkfor spans up to 200 ft.

Single-track deck plate girders. . . . . . . . . . . . . . . . . . . (10.0 If 350) ISingle-track through! plate girders (floorbeams and stringers). . . . (11.5 If 620) ISingle-track through plate girders (solid floor). . . . . . . . . (14.5 If 850) I

DoubleAtrack through plate girders (floorbeams and stringers). . . (22.5 I +1350) ZDouble-track through plate girders (solid f loor) . . . . . . . (19 .O Z+2350) Z

Four-track through plate girders (floorbeams and stringers). . . . . . . (48.5 Z+2150) ZFour-track through plate girders (solid floor). . . . . . . . . . . . (40 .O Z+4400) Z

Single-track through lattice trusses (floorbeams a.nd stringers). . . . . 15.5 12

Double-track through lattice trusses (floorbeams and stringers). . . . 27.0 Z2Double-track through lattice trusses (solid floor). . . . . . . . . . . . . . . . (21.5 Z+2300) Z

Four-track through lattice trusses (floorbeams and stringers). . . . . . . (19.5 Zf5200) ZFour-track through lattice trusses (solid floor). . . . . . . . . . . . . . . . . (41.5 Z+4200) Z

Illinois Central R. R.-Table 51 gives the steel .weights of plategirder spans for ballast floor (2000 lb. per ft. of track). Live load

Z2approximately E-55, Specification 111. Central, 1905, impact i =I+$

tension 16,000 lb. per sq. in. ; through spans have four lines of stringers.Japanese Railways,-Table 52 gives the weights and principal dimen-

sions of single and double track through and deck truss spans designedby C. C. Schneider and Th. Cooper for the Imperial Government Railwaysof Japan. Loading Cooper’s E-30, Spec. Cooper’s 1896. Narrow gage.

ART. 1 WEIGHTS OF SIMPLE SPAN BRIDGES 229

000 ‘owi

ooo’ozz

000 ‘ooa

000‘081

000 ‘091

+000’0.bl :

L?.r

0 0 0 ‘021 2.F2

000’001 gu-l

000 ‘08

000 ‘09

000 ‘OP

000’02

0

Prussian State Railways.-Table 53 gives the weights of flooringand steelwork for various types of bridges designed for *he Specifi-cations of 1903 (see Vol. I). Live load, two engines, total 137 tonseach, 60 ft. long = 4550 lb. per ft. of track, folloxved by 28 ton cars,20 ft. long = 2800 lb. per ft. of track.

The weights refer to square bridges on tangents and not limited in depth.They are to be increased as follows:(1) If depth of main plate girders Tlq- instead of & of span, their weight increases

20%.(2) If depth of trusses with parallel chords & instead of & of span their weight in-

creases 15 %.

TABLE 53Bridges for Imperial Government Railways of Japan-Cooper’s Specifications 1896-Loading E-30 with End Floorbeams

(Design Cooper-Schneider)

) ft. I ft. in. 1 ft. 1 ft. inI I /

Y$.4 Latt. 100 102 1 1 5 1 5 6

5 g Pin.3x2

150 154 0 7 16 0

s Pin. 200 204 9 9 1 6 0

100 103 9 5 26 3 26 6 26 6

200 205 14 9 27 0 37 0 26 0

Lds Latt. 100 102 1 1 5 12 0

ij Pin. 150 ) 154 0 7 12 0G

Gi Pin. 200 204 9 7 16 0

20 9 20 9

22 0 22 0

28 0 28 0

Heightat

center

ft. in

23 6

25 0

34 0

-

L

-

Heightat

end

ft. in.

23 6

25 0

24 6

B. of railto

masonry

ft. in.

, 5 0;

5 04

5 9

6 9

6 9

25 44

27 0

31 9

B. b. Lof

stgrs.

in.

404

402

42$

404

40t

402

402

24 I

-

8

-

--

-

ofFlbms.

in.

Weight per lin. ft. of bridgeB . b . L s ~ Total

5 2 408 638

5 2 429 840

54 429 988

662 933 1005

65; 9 9 0 1905

52 368 755

52 388 1006

42$ 350 1217

I-

I-

100

112

154 232,000w

271 120 674,000 ;

lil83 63 130,500 %

Note.-The table shows that for deck spans compared with the through spans of same length the truss weight increases about 15 %,while the floor weight decreases about 10 ‘%; this gives a t.otal increase of steel weight of about 6 %.

-6$

Type of bridge

(J) Deck pl. gir&er bridge from33’- 8.2’ BPan.

( 2 ) Ha l f - through Plate girderbridge with me sidewalk.span from S3’-88’

(3) Through riveted truss bridgewithout sidewalks.

-I-0

1

-I

I

B

TABLE 53

Weights of Single Track Railroad Bridges of the Prussian State RailroadsCalculated on the Basis of the Specifications of May 1, 1903, governing bride calculations

Weight of steel in lb. per lin. ft. of bridge

Main girdersr trusses withbrasc;l~sand

= span ;n ft.

I

160+10.82

180+8.82

Floor system

II

9.8 ft. wide, 2551 0 . 8 ft. “ 29012.lft. “ 350

15.7 ft. wide, 4001 6 . 1 f t . “ 4201 6 . 4 f t . “ 450

Total weight of steel,

Col. I and Col. II

III

160+10.82

9.8 ft. wide, 435+8.821 0 . 8 ft. ” 47o+s.s 212.lft. ‘I 530+8.82

Span from 66”131’15.7 ft. wide, 760+5.421 6 . 1 f t . “ 7so+5.4 z1 6 . 4 f t . ” 8101-5.41

Span from 131’-262’15.7 ft. wide, 850+5.411 6 . 1 f t . “ s70+5.4116.4 ft. “ 900+5.4 z

-I1Weight of track in lb. per

lin. ft. of bridge

CYeight of rails, ties, plank:and ballast.

:uard rails excluded, whiclweigh 100 lb. per lin. ft.)

I V

j. 9 ft. c. e. of girders, 430j,6ft. I‘ I‘ --.s 520

9 . 8 ft. wide, 40010.8 ft. “ 42012.1 ft. “ 440

450

-rota1 weight in lb. per lin. ft.

(for calculating stresses)Col. III and Col. IV

V

5.9 ft. e. c. of girders,590+1o.sz

6.6 ft. o. c. of girders,680+10.81

-

9.8ft. w ide , 835+8.811 0 . 8 ft.. “ 890+8.8z1 2 . 1 ft. “ 97o+s.s 1

Span from 66”131’15.7 ft. wide, 1X0+5.4116.lft. “ 1230+5.4116.4ft. “ 1260+5.41

Span from 131’-262’15.7ft. w ide , 1300+5.411 6 . 1 f t . “ 1320+5.4 11 6 . 4 f t . “ 1350+5.41

TABLE 53Weights of Single Track Railroad Bridges of the Prussian State Railroads (Continued)

Calculated on the Basis of the Specifications of May 1, 1903, governing bridge calculations-I I

--7

(

--

-

I<

3 b1

1:

-

Weight of steel in lb. per lin. ft. of bridge Weight of track in lb. perlin. ft. of bridge

rota1 weight in lb. per lin. ft.(for calculating stresses)

Col. III and Col. IV

M a i n g i r d e r sor trusses with

b r a c i n g a n ds h o e s

I = span in ft.

aeight of rails, ties, planksand ballast.Total weight of steel,

Cal. I and Col. IIT y p e o f b r i d g e F l o o r s y s t e m

:uard rails excluded,, whichweigh 100 lb. per hn. ft.

-.-

-

II

I

-

I I I I I V V

8.2 ft. c. c. of trusses, z1080+5.4 I

11.5 ft. c. c. of trusses, g1120+5.42

z

-11-

,(4 Deck riveted truss bridqtwith two sidewalks. 360+5.42 8.2 ft. c. e. of trusses,

690+5.428.2 ft. c. c. of trusses, 33011.5ft. “ “ “ 390

370

With 1 lineo f s t r i n g e r s

andbuckle plates10.8 ft. wide,2520+9.81

12.1 ft. wide,2910+9.81

C5) Half-through plate &de,bridge with ballast floorSpan from 33’-82:

With 3 lineso f s t r i n g e r s

and 3mckle plates

2480+9.81

W i t h 1 l i n e W i t h 3 l i n e so f s t r i n g e r s o f s t r i n g e r s

and andbuckle plates buckle plates.O. 8 ft. wide,

4 5 0 5 1 0.2.1 ft. wide,

5 6 0 630

W i t h 1 l i n eo f s t r i n g e r s

andmckle plates0.8ft. w i d e ,630+9.81

2.1 ft. wide,740f9.81

W i t h 1 l i n e W i t h 3 l i n e s)f s t r i n g e r s

ando f s t r i n g e r s

andmuckle plates buckle platesO.Sft.wide,

1890 17902.1 ft. wide,

2170 1880

W i t h 3 l i n e so f s t r i n g e r s

andbuckle plate:

690+9.82

810+9.82

t180+9.82

2690+9.8 1

/----(6) Plate girders as stringers with

ballast.Weight for 1 ft. in width

110+4.81W e i g h t f o r 1 f t . i n w i d t h

6 1 0 720+4.81 75

$‘d

,x

ART. 21 WEIGHTS OF SIMPLE SPAN BRIDGES 233

Notes.(3) If depth of floor very limited, its weight increases up to 25 %.(3) For skew bridges increase weight of floor up to 15 %.(4) For bridges on a curve with a curvature of less than 6” and spans below 130 ft.,

the increase of total weight up to 12 %.Assumptions for detailing:For (1) Web pIates Q in. thick. Gussem in distances equal to depth of web plate.

Roller shoes used for spans over 50 ft.For (3) Top bracing used for spans over 130 ft. (accounts for the jump in the

weight).For (6) Depth of web plate & of the span.As the width of the bridges is variable, the weight per ft. of width is given.2-in. Planking. Ties 8 in. X 10 in. X 2 ft. 0 in. centers. If main girders 6.6 ft.

apart, ties 9 in. X 12 in.Length of ties 12 ft. 6 in. for deck bridges with 14 ft. 5 in, between railings, for

through;bridges length varies from 9 ft. 2 in. to 13 ft. 6 in. according to width ofbridge.

Ties running alternately from one main girder or truss to a little beyond the foot ofthe gusset of the other.

Sidewalk of plate girder bridges has planking on 2 joists of 6 in. X 6 in.Thickness of balIast above top of floor girders about 1 ft. 2 in. Weight of ballast =

125 lb. per cu. ft.

ART. 2. STEEL WEIGHTS OF SOME EXISTING SIMPLE SPAN RAILROADBRIDGES

633-ft. D. T. Through Pin Span, P. R. R., Built 1896(See Plate XIII)

Specifications P. R. R., 1895. Loading approximately E-38.16 subpanels at about 33 ft. 4 in., height at center 84 ft., at hip

57 ft. 31 ft. c. to c. of trusses. Topchord: 4 webs 31 in., coverplate46 in., heaviest eyebars 12 in., and pins 8: in.

Per cent.of trusses

Per cent.of total

Riveted truss members . . . . . . .Eyebars . . . . . . . . . . . . . . . . . . . .Pins . . . . . . . . . . . . . . . . . . . . . . .

Total trusses . . . . . . . . . . . .Bracing . . . . . . . . . . . . . . . . . . . .Floorbeams . . . . . . . . . . . . . . . . .Stringers and stringer bracing.

Total floor system. . . . . . . . ,.....,.Shoes, rollers, etc. . . . . . . . . . . . . 6.3Field rivets . . . . . . . . . . . . . . . . . . . . . . . . . . .Floor bolts, . . . . . . . . . . . . . . . . . . . . . . . . .

Total steel weight (7850 lb. per . . . . . . .lin. ft.)

1 727 5001:189:800

(Details 23.1%) !

63,700

2,981,OOO 71.1240,800 5.7

274,500485,400

759,900 18.2188,200 4.5

14 ,800 0 .43,100 0.1

. . . . . . . . . . . 4,187,800 lb. 100.0

234 DESIGN OF STEEL BRIDGES [CHAP. XI

420-ft. S. T. Through Pin Span, C., M. & St. P. Ry., Built 1910(See Plate XIII)

Specifications C., M. & St. P. Ry., 1906. E-55 loading.14 subpanels at 30 ft., height at center 65 ft., at hip 48 ft. 23 ft.

c. to c. of trusses. Topchord: 3 webs 34 in., coverplate 42 in., heaviesteyebars 16 in. and pins 12 in. -

IRiveted truss members . . . .Eyebars . . . .Pins . . . . . . . . . . . . . . . . . . . . .

Total trusses . . . . . . . . . .Bracing . . . . . . . . . . . . . . . . . .Floorbeams . . . . . . . . . . . . . . .Stringers and stringer bracing

Total floor system . . . . .Shoes, rollers, etc. . . . . . . . . .Field rivets . . . . . . . . . . . . . . .

Total steel weight (6320 lb. perlin. ft.) . . . . . . . . . . . .

6 9 . 82 7 . 62 . 6

100.07.7

........

........

. . . .5.3

. . . . . .

l,;Z3;,iWt (Details 30.2%)

52;400

1,996,900 74.9149,500 5.6

135,900262,900

. . . . . . . . . . 398,800 15.0103,300 3.9

. . . . . . . . 17,000 0.6

. . . . . . . 2,665,500 lb. 100.0

373-ft. D. T. Through Pin Span, P. R. R., Built 1906(See Plates XIX and XXIV)

Specifications P. R. R., 1904. Loading approximately E-50.11 panels at about 34 ft., height at center 56 ft., hip 36 ft. 30 ft.

8 in. c. to c. of trusses. Topchord: 4 webs 32 in., coverplate 44in., heaviest eyebars 12 in., pins 10 in.

Riveted truss members . . . . . . . . . . .E&bars. . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . .

Total trusses . . . . . . . . . . . . . . . .Bracing . . . . . . . . . . . . . . . . . . . . . . . .Floorbeams . . . . . . . . . . . . . . . . . . . . .Stringers and stringer bracing. . . . . .

Total floor system. . . . . . . . . .IShoes, rollers, etc. . . . . . . . . . . . . . . .

Total steel weight (7480 lb. pelin. ft).

rl

-IPer cent.

of trusses

55.941.5

l,~Ol~,~H~ (Details 26.3 %)

2.6 4s:ooo

100.0 1,814,OOO 65.19.4 170,000 6.1

. . . . . . . 220,000,...a... 446,000

“i:i.. . . . . . . . . . . . 666,000 23.8139,000 5.0

. . . . . . . . . . . . . . . . . . . 2,789,OOO lb. i 100

Anr. 21 WEIGHTS OF SIMPLE SPAN BRIDGES 235

40’7-ft. D. T. Through Pin Span, P. R. R., Built 1903

(See-Plate XIV)

Specifications P. R. R., 1901. Loading approximately E-50.11 panels at 37 ft., height at center 67 ft., at hip 37 ft. 30 ft.

8 in. c. to c. of trusses. Topchord, eyebars and pins as for 373-ft.span given above.

Per cent.of trusses

Riveted truss members . . . . . . . . . . . 58.7Eyebars . . . . . . . . . . . . . . . . . . . . . . . . 39.3Pins . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 . 0

Total trusses . . . . . . .-. . . . . . . . . 100.0Bracing.. . . . . . . . . . . . . . . . . . . . . . . 7.1Floorbeams . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Stringers and stringer bracing. . . . . . . . . . . . . .

Total floor system. . . . . . . . . . . . . . . . . . . .Shoes, rollers, etc. . . . . . . . . . . . . . . . . 6.3

Total s tee l weight (8500 lb . per .lin. ft.)

I,~&&,$)~ (Details 19.2%)

49;600

2,450,500174,000

220 ,000450 ,500

.,......... 3,449,OOO lb.

71.05.1

19.44.5

100.0

196-ft. S-in. D. T. Deck Pin Span, P. R. R., Built 1906

(Compare Plate XVII)

Specifications P. R. R., 1904, loading approximately E-50.7 panels at about 28 ft., height 30 ft. 6 in., 19 ft. 6 in. c. to c. of

trusses. Topchord: 3 webs 36 in., coverplate 36 in., heaviest eyebars10 in.

Per cent.! /of trusses

R i v e t e d t r u s s m e m b e r s . 67.9Eyebars . . . . . . . . . . __........ 29.5

;‘2&Y$ (Details 35.0%)

P i n s ’ . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 16:400

T o t a l t r u s s e s . 100.0 624 ,200 73.1Bracing . . . . . 6.8 42 ,000 5 .0Floorbeams . . . . . . . .._........... . . . . . . . 46 ,400S t r i n g e r s a n d s t r i n g e r b r a c i n g . 101,000

Total floor system. . . . . . . . . 147,400 17.4Shoes, rollers, etc.. . . . 6.1 38 ,000 4.5

- -T&;l$el weight (4340 lb. per . . . .

. .848,600 lb. 100.0

236 DESIGN OF STEEL BRIDGES [CHAP,. XI

ART. 3. WEIGHT OF SIMPLE SPAN HIGHWAY BRIDGES

American Bridge Co. Standard Highway Bridges with TimberFlooring (1909).-These are designed for ordinary country bridges,which do not carry electric trolley lines. Floor planks 3 in. on roadway,2 in. on sidewalks.

Live Load.-On floor and its supports 100 lb. per sq. ft. of floorsurface, or 6 tons on two axles 10 ft. centers and 5 ft. gage, or a 15-tonroad roller. For trusses the following loads per lin. ft. of span:

Length of spa.n, ft.

upto . . . . . . . . . . . . . .76to85 . . . . . . . . . . . . . . .86to 112 . . . . . . . . . . . . . .113 to 156 . . . . . . . . . . . . . .157 to 168 . . . . . . . . . . . . .Over 168 . . . . . . . . . . . . . .

Roadway 12 ft. or 14 ft.

1200 1600. 1100 1400. 1000 1300. . 900 1200. . . 900 1100

800 1000

‘Roadway 16 ft.,20 ft.

1 8 ft. or

The trusses are also proportioned to carry a 15-ton road roller.

Permissible units 16,000 lb. per sq. in. tension, 16,000-70 4 compres-

sion. Material 0. H. structural steel, U. S. 55,000 to 65,000 lb. persq. in. as per Specifications American Railway Engineering Association(see Appendix).

The joists of the plate girder and truss spans are g-in. I-beams for12- to 17-ft. panels, the floorbeams are I-beams 12 in. to 18 in. high,depending upon the width of the roadway.

If 1 = length of span, b = width of roadway (without sidewalks),the total steel weights W are approximately as follows:(1) If no sidewalks are used:

Through plate girder spans 36 ft. to 70 ft. long, roadway 20 ft. wideincluding two lines of railing of about 5 lb. per lin. ft. each:

w = (3.8 1 + 300) I

Through riveted truss spans 36 ft. to 200 ft. long (pony spans up to100 ft.), width of roadway 6 = 12 ft. to 20 ft., including two lines ofrailing of about $4: lb. per lin. ft. each:

w = (0.121+ 12)(1.6 - 0.03b) bl:

For a 20-ft. roadway, the steel weight is therefore

W = (2.4 Z + 240) Z

ART. 41 WEIGHTS OF SIMPLE SPAN BRIDGES 2 3 7

(2) If sidewallcs with steel joists are used, add to steel weight about12 lb. per sq. ft. of sidewalks,

American Bridge Co. Standard Highway Bridges with ConcreteSlab Floors (1910).-These standards comprise beam spans up to 40 ft.and pony truss spans 36 to 102 ft.

The flooring consists of macadam or gravel 3 in. deep at curb to6 in. at the center and rests on a concrete slab 5 in. thick reinforcednear both surfaces by a-in. square rods spaced 6 in. c. to c. (Fig. HPlate VII).

The steel weights are from 40 to 50% greater than the correspondingsteel weights of bridges with timber flooring.

ART. 4. WEIGHT OF ELECTRIC RAILWAY BRIDGES

Bridges with open-tie flooring, carrying exclusively electric carsweighing 30 tons each, have an approximate total steel weight forsingle track :

deck plate girder spans w = (5Zf50)Z

through truss spans w = (2Zf200)Z

CHAPTER XII

VIADUCTS

A. CALCULATION OF STRESSES IN TOWERS

ART. 1. EXTERNAL FORCES

The towers of railroad viaducts have to resist vertical forces (deadand live load including impact), horizontal transverse forces (wind andcentrifugal force) and the longitudinal braking and tractive force.

The columns are stressed by all these forces, the transverse bracingby the transverse and vertical forces or by the former only, if the bentsare symmetrical about a vertical line and the loads are applied symmet-

rically to this line. The longitudinal bracingis stressed by the braking or tractive forceonly.

ART. 2. DEAD LOAD STRESSES

For the steel weight of viaducts see page256. The dead load acting on a bent consistsof the end reactions of the spans which itsupports and the weight of the bent includ-ing one-half of the longitudinal bracing ofthe tower. The weight of the bent may beassumed constant per vertical foot of bent andapplied at the panel points of the latter, ex-cept in very high bents where the actual panel

FIG. 1. loads should be determined. These loadsproduce vertical reactions R, and RT,. R, is

found by taking moments of all loads about B and dividing by 1 (seeFig. 1).

Ra=qb Rb=ZP-Ra

To get the stress in a diagonal l-4 we consider a section X cuttingthat diagonal, take moments of all forces applied above X (P, and Pz)about intersection point I of the columns and divide by the lever arm~1-4. If the moment of these forces P is positive (clockwise) or their

2 3 8

ART. 21 VIADUCTS 239

resultant Rl-z is to the right of I (as in this case) the moment of thediagonal stress must be negative (anti-clockwise), from which followsthat, if the diagonal falls to the right or left, it is stressed in tensionor compression respectively. If there are two diagonals in a panel un-able to take compression only the one stressed in tension is acting. Ifboth can take compression as well as tension the stress in each is de-termined as shown above and finally each stress is divided by two.Thus

if 2-3 is not acting Sl-4 = + 5,

if 2-3 is also acting x1-4 = + ST& andM

~5’2-3 = - 3 __7.2-3

If diagonal 14 only is acting the stress in the intermediate strutis obtained by taking moments of all forces above section X’ (PI, Pz,and P4) about I and dividing by r3--4. If diagonal 2-3 only is actinga section X” has to be considered correspondingly. 3-4 is always incompression if only one diagonal is acting in tension; if both diagonalsl-4 and 2-3 are acting it is not stressed.

The compression stress in top strut l-2 is equal to the horizontalcomponent of stress in column 2-4 if the diagonal l-4 only is acting,and to the horizontal component of the stress in the column l-3 if the

diagonal 2-3 *only is acting, or equal to P2 k and PI $ if1

nz and

1-nl are the batters of the columns 24 and l-3 respectively. It is

zero if both diagonals are acting.Similarly the tension in bottom strut AB is obtained by resolving

the stress in column A-3 or B-4 into a vertical component and oneparallel to AB depending on whether diagonal 3-B or 4-A is acting,and is zero if both diagonals act.

To get the compression in column 2-4 we consider again section X.If diagonal l-4 only is acting the center of moments for 24 is at 1.52-4 is therefore equal to the moment of all forces above X about 1divided by the lever arm rl. Correspondingly if 2-3 only is acting SL4is equal to the moment of the same forces about 3 divided by ~3. Ifboth diagonals act simultaneously, S2-4 is the mean between the twovalues obtained by considering each diagonal acting separately.

If the bent and the loads are symmetrical about a vertical line thereactions are alike and equal to one-half of the sum of all loads. Thepoint I must lie on the center line from which follows that the stresses inall diagonals are zero. The compression in an intermediate strut 3-4 is

240 DESIGN OF STEEL BRIDGES [CHAP. XII

obtained by resolving load P, into components parallel to 3-4 and to thecolumn. The compression at any point of a column is obtained by r8-solving the sum of the loads applied on the column above that point intoa horizontal component and one parallel to the column.

If both columns are vertical the stresses in all diagonals and strutsare zero.

ART. 3. LIVE LOAD STRESSES

The live load reactions of the girders on the columns are determinedas shown on page 56. Table 18 page 65 gives the r’eactions for vari-ous span lengths due to Cooper’s E-50 loading.

FIG. 2.

In order to get the greatest live load stressin a certain member of the bent the most un-favorable position of the load is first determined.In a single-track viaduct there is of course onlyone possibility of a maxirn~m loading, but indouble- or multiple-track viaducts either one ormore tracks may be loaded and in highway via-ducts the whole or any part of the width may becovered with live load.

Assuming the load of one track concentratedat the center of track the maximum stress inany member is obtained by placing loads only onone side of the vertical through the center ofmoments of that member. For instance, in bent(Fig. 2) the maximum compression in column 2-4is obtained by loading part b d, or b da if thecenter of moments is at 1 or 3, that is, if diagonal

l-4 or 2-3 respectively is acting, and the maximum tension results byloading a dr or a da. The maximum stresses in all diagonals and inter-mediate struts occur by loading either from a to c or b to c, where cis vertically below the intersection point I of the columns.

The position of load being determined the stress in the member isfound in the same way as for dead load.

ART. 4. WIND STRESSES

The wind forces acting on a bent (Fig. 3) comprise:(a) the force WI = 4 WI (Z1+Z2) where wl is the specified wind force

per lin. ft. of viaduct on the live load, and II and 12 the lengths of theadjoining spans; for railroads it is usually specified acting 7 ft. abovethe rail.

ART . 41 V I A D U C T S 241

(b) the force W, = + w, (L+Z2) where w, is the wind force per lin.ft. of viaduct on girder and floor acting at the middle height of these;

(c) the forces WI, Wa, etc., which are the panel concentrations ofthe wind acting on the tower.

For railroad viaducts usually three conditions have to be considered(see Specifications, Appendix) :

(1) no live load acting, high wind force on structure,(2) full live load with a lower wind pressure on train and structure,(3) same as under (2) except live load consisting of empty cars.As a rule conditions (1) and (3) affect the possible uplift, (1) and (2)

the bracing and (2) the columns.

F IG . 3 .Rb

F IG . 4 .

F IG . 5 .

The above forces cause a horizontal reaction at the fixed column baseequal to the sum of the horizontal forces, and two equal but oppositevertical reactions which are found by taking moments ‘of all forcesabout the base and dividing by 1 (Fig. 3).

R, = -Ra = + 2 Wh.

The horizontal reaction may also be assumed distributed equally toboth bases owing to the friction between the base and the masonry.

The stresses in the columns and bracing can now be determinedeasily by a Maxwell diagram (Fig. 5) or analytically.


Assuming the diagonals acting in tension only, the stresses in thecolumns are best found analytically as follows:

Wz(hz-hl)+W,(h,-hl) =M,, . . . . . . . M li

: r1=

: rz=

h-3=Ml+(WL+Wg+W1)(~1-~3) =M3, . . . . .M3 1. r4=

M3+(W1+Ws+W1 +TTT3)(h3-w =Ms, M5 J/ :;:II:

1

Wind from

Loft Right

. . . . . .

If both diagonals in a panel are stressed simultaneously, the stressin the column of that panel is equal to the average of the stresses inthat column found by assuming each diagonal acting independently.

Similarly, the stresses in the web members, if diagonals act in tension,are found as follows:

Wz(h-hz) + W,(hi-lb) +Wl(hi-hl)

=M _2 3,. . . . . . .M,-3 : 7.2-3 = +s2-3

MZ-3 + m73(hrh3) =M4--5,. . . . . . . M4-5 ’1 ’ “-’ = +s4-5: (h7h3) = -83-4

M4--5 f. W&i - hd = Me-*, . . . . . . Me+ : “-’ = i-&-A: (hi-h5) = --A%..-6

The forces WZ and W, may be assumed transmitted to the bent asshown in Fig. 4 (for value Ml see above). The upward vertical reaction

causes a tension in top strut l-2 equal to i 71

where ; = b a t t e r

of column. The total compression in l-2 is therefore

The stress in bottom strut AB is - ZW $ ‘, Rb or + k Rb

depending on whether the leeward or the windward column baseis fixed. In ordinary viaducts all four columns are allowed toslide slightly on the masonry and we may assume that the horizontalreaction ZW is equally transmitted by the two columns of the

ART. 61 VIADUCTS 2 4 3

bent through friction, and the stress in the bottom strut is then

- 4 zw+ nIR b.

If diagonals act in tension and compression the stress of eithersense in them is only one-half of the above found tension. The stressesin the struts are zero except in the bottom strut which has to transmit-4 zw.

ART. 6. STRESSES DUE TO CENTRIFUGAL FORCE

The centrifugal force acting on a bent (Fig. 3) is equal to

where c = specified force per lin. ft of track (see Specifications, Appendix).If the centrifugal force is given as a certain percentage p of the live loadper track, then C is equal to the maximum live load reaction on thebent per t,rack multiplied by p times the number of tracks. It is oftenassumed acting 5 ft. above the rail but, as explained on page 28, it isjustified to assume it applied at the,rail if the live load is assumed toact at the center of the track. The stresses are calculated similarly tothose from wind.

ART. 6. STRESSES DUE TO BRAKING FORCE

The braking force is applied to the structure along the rails andassumed at 20% of the live load (see Specifications, Appendix). If p isthe live load per lin. ft. of viaduct and L the length between expansionjoints the braking force B acting on one of the two longitudinal bents ofa tower is B = 0.2 pL +. B causes a horizontal reaction B at the fixedcolumn base and two equal opposite vertical reactions

R,= -Rb=?f ( F i g . 6 )

If the diagonals act in tension only the stresses are as follows:

x1--3 = Szm-4 = - B $

s3--A = ,~‘a-~ = -BP T Ra o r +B+

f&m4 = f&m.-3 = + B +; s3wB = S4+ = +B y

Xs-4 = XAB = - B

If the diagonals act in tension and compression the same applies17


as explained above for transverse bents. It shoulcl be noted thatwhere the columns are battered all values hr, ha, etc., and all lengths ofdiagonals must be measured in the plane of the longitudinal bracingand not in the elevation.

ART. ‘7. COMBINED STRESSES

Care has to be taken to combine only stresses which can occur si-multaneously and to find that condition which causes the greatest com-bined stress in a member. In a double-track viaduct, for instance, thegreatest combined stresses in the bracing are caused by the wind pres-

sure and the live load on one track (usually thewindward track). If the double-track viaduct ison a curve, the greatest stresses in the bracing arecaused by wind combined either with live loadand centrifugal force from one track only, orwith live load and centrifugal force from bothtracks. In a viaduct on a curve with unsym-metrical bents it will, as a rule, happen thatthe total live load causes tension in certain di-agonals and the centrifugal force causes tensionin the counters of those diagonals. If they can

RoFIG. 6.

Rb take only tension it has to be determined firstwhich diagonals are stressed in tension from the

combined forces and for this condition the stresses in the columnshave to be calculated.

Where the centrifugal force causes stresses of opposite sign to thosefrom the live load as, for instance, in the inner columns of viaductson a curve, the tension from the centrifugal force should not be deduc-ted from the dead and live load compression even if impact is added tothe live load as it is conceivable that a train moves slowly and yet withgreat impact; further, the slower the train moves the more the loaddeviates from the center of track toward the inside, thus throwing moreload on the inner columns than assumed.

Stresses from wind and braking force in the columns should not beconsidered simultaneously but only the larger of the two. Some engi-neers consider the braking force only for the longitudinal bracing andnot for the columns.

ART. 8. UPLIFT

The uplift at the base of a column is obtained by deducting fromthe negative reaction produced by either the braking or transverse forces

ART.~] VIADUCTS 245

the positive reaction produced by the dead and live load. Here, too,it is not necessary to combine the longitudinal and transverse forces toget the greatest uphft. The following combinations of reactions shouldbe separately considered; no impact to be included in the live load:

(1) Wind on unloaded structure minus dead load.(2) Wind on loaded structure plus centrifugal force from empty cars

minus dead load minus live load of empty cars.(3) Wind on loaded structure plus full centrifugal force minus dead

load minus full live load.(4) Braking force minus dead load minus full live load (if the braking

force has been derived from the uniform train load then the live loadreaction should also be determined for this load only).

Condition (3) affects only viaducts on sharp curves. See example below.

ART. 9. EXAMPLE

Single-track viaduct on 4” curve carrying Cooper’s E-60 loading,tower spans 30 ft., intermediate spans 60 ft.; total height base of rail tomasonry 82 ft. 6 in. (see Fig. 7); diagonals to resist tension only. Allforces and stresses given in units of 1000 lb.

Dead Load Stresses.-Assumed dead load: 1150 lb. per lin. ft.of 30-ft. span, 1520 lb. per lin. ft. of 60-ft. span, 1000 lb. per verticalft. of tower.

Panel Panel Stress

point load increment ColumnP I=Pseca

gTq-ey

.: 3 2 . 7 3 . 0 3 3 . 2 3 . 0 .::g i -;g:; 3-4 1-2 ~ -5.5 -0.5; 4 . 8 6 . 0 4.9 6.1 5-7 7-9 -41.1 -47.2 5-6 7 -8 -1.0 -0.8

9 3 .0 ,..........._ .,.,..., ,.,,,..,,,,. 9-10 +7.8*

ColumnReaction 1

49.5 = R *= Q(R -P,)

Live Load and Impact Stresses.-Live load E-60; i = I &

Reaction per column from 30-ft. and 60-ft. spans = 179.4.Live-load stress in all column sections I = 179.4 X 1.013 = - 181.7;

i = 181.7 ‘x8 = -139.6

Stresses i n strut 1-2, l179.4

= =6 - 2 9 . 9 , i = 139.6G = -23.3

Stresses in strut 9-10, I = + 29.9, i = + 23.3

2 4 6 DESIGN OF STEEL BRIDGES

Wind Stresses.-Assumed wind forces :

[CH A P . XII

(a) BridgeItJain 400 lb. per lin. ft. of bridge at 7 ft. above B. of R.

loaded girders, 360 lb. per lin. ft. of bridge at 4 ft. below B. of R.(towers, 300 lb. per vertical ft. of tower.

(b) Bridge

i

girders, 600 lb. per lin. ft. of bridge at 4 ft. below B. of R.unloaded towers, 500 lb. per vertical ft. of tower.This gives the panel loads shown in Fig. 7. For the members receiv-

ing live load stress the condition (a) must be considered only.

Column

2- 44- 66- 88-10

FIG. 7.

Wind Stresses In Columns, Top and Bottom Struts-Bridge Loaded

Centerof

momentsMoment M

Lever Columnarm stress

r = M:r

; 718 18X26’+16.2X15’+0.7X10’=718 + (34.9 + 1.8) 14’ = 123111.5 16.1 - - 62.3 76.57 1231 + (36.7 +2.9)24’ = 2181 24.0 - 91 .09 2 1 8 1 +(39.6+3.6)24’ = 3 2 1 8 31.9 -100 .9

ART. 91 VIADUCTS

Reaction R = & 3218 : 32.33 = + 99.7

2 4 7

lS+16.2Stress in top strut l-2 = - 2

18X16+16.2X5-0.7+- -6 xs--- = - 1 0 . 1

Stressinbottomstrut 9-lO=-jZw+iR= -22.@+16.6= -5.9

As can easily be seen by inspection the condition (b) (bridge un-loaded) causes the greater stresses in the diagonals and intermediatestruts. The center of moments for all these members is the intersec-tion point 1 of the two columns (Fig. 7).

Wind Stresses in Diagonals and Intermediate Struts-Bridge Unloaded

Moment at‘ I = M

27X20t+1.2X25’ 1 W;;570+3.0x35675 +4.8 X 49 = 910910$-6.0X73 = 1347

Diagonals

2-3 20.6 $ 2 7 . 74-5 29.0 $ 2 3 . 36-7 38.0 +23.9s-9 63.7 +21.2

I

Member

Struts

.‘32”5-67-s

Lever Stressarm = M:r

T

Stresses from Centrifugal Force.---Speed assumed 50 miles per hour.Centrifugal force = 11.65% of live load (see page 29) = 179.4 X 2 X0.1165 = 41.8 lb. per bent, assumed acting at base of rail.

Member

2-44-66-Ss-10

2-34-56-7s-9

3-4

;I;

Center ofmoments

3

i9

’ Ii

Dist,. ctr. of mom. Lever arm ofS t r e s s

from centrifugal member,force, d r

=41.s;

iw 11.5 16.1 -L - 69.0 85.757 24.0 - 99 .481 31.9 -106 .1

__-

-1 20.6I

16

29.0 : li:!

I 38.0 63.7 ’ + + 17.6 10.5I - -

1 35.0 - 19 .116 49.0 - 13 .7

73.0 - 9.2

41.8 x 8l-2 stress = - 4 41.8 + -8- = . - 13.9

41.8 X 82.59-10 stress = - f 41.8 + ~~32.8 = . . - 3.4


Stresses from Braking Force.-The braking force will be assumedat 20% of the uniform train load or at 600 lb. per lin. ft. per rail =54,000 lb. per longitudinal bent applied at the base of rail.

Compression stress in column 2-6 = 54; = - 59.3

Compression stress in column 6-8 = 54 i; = - 102.5

Compression stress in column 8-10 = 54 3”:, = - 146.0

Stress in struts = - 54.0Stress in diagonals = 54 X 1.28 = + 69.1.

(To be exact the stresses 59.3, 102.5, 146.0, 69.1 have to bemultiplied by secant Q: = 1.013, as the columns have a batter of 1:6)

Uplift.-For the combinations given on page 245 we have to determinethe following reactions:Deadload ( s e e a b o v e ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . + 4 9 a(a)Full live load (see above). . , . . . . Cl79 (b)

Live load of 1200 lb. per ft. of track; L22 X45ft. = . . . . . . . . . . . -l- 27 (c)

Train load of 6000 lb. per ft. of track; y x 45 ft. =. . f135 (d)

Wind on loaded structure (see above) . . . . . . -100 (e)

Wind on unloaded structure;8 9 5 5

(100 - 18 32.8>

3 = . . . - 34 (f)

Centrifugal force from full live load; 41.8 gi =.. . . . -105 (g)

Centrifugal force from 1200 lb. per ft.; 1.2 X 45 X 0.1165 X ig8 = - 16 (h)

Braking force from 6000 lb. per ft.; 54 %65 = . . - 1 4 8 (i)

I INo. Combinations Uplift

(a) + (f) = $49-84 = - 3 5(a) + (cl + (e) + @I = +49+27-100-16 = - 4 0

= +49+179-100-105 == +49+135-148 =

B. DESIGN OF VIADUCTS

ART. 10. TYPES OF VIADUCTS

A viaduct consisting of braced steel towers and plate girder spansis generally the most suitable structure to span a deep valley with onlya small creek.*

*In the competition for designs of the viaduct at Muengsten, Germany, with agreatest height of 320 ft. (see Arch Bridges, Plate XLVI) a viaduct with steel towyersof 66 ft. span and intermediate spans of 100 ft. had been considered, as mosteconomical, but for esthetic reasons the present structure was adopted.

TABLE 54Notable Railroad Viaducts

B u i l

1888

19QC

1892

1900

1909

1890

1905

1907

1909

1901

1907

1908

1904

-

t

I

8’

,

-

Railroad Name or location

A n t o f a g a s t a .

Rgngnqon Mandala:

Southern Pacific R. I

E r i e R . R..

Canadian Pacific Ry

Lima & Oroya Ry

E r i e R . R . .

G u a t e m a l a R y .

Nevada Co. NarrowG. Ry.

Chicago & N. W. Ry.

Erie R. R.. . . . . . . .

Trans Continental Ry

U g a n d a R y . .

I’

t.

. ’

.

-

Loa Viad., Bolivia

Gokteik Viad.,Burma, India.

Pecos Viad., Texal

K i n z u a V i a d . .

Lethbridge Viad.,Alberta, Can.

Verrugas Viad.,Peru.

P o r t a g e V i a d .

Las Vacas Viad.

Bear River, ColfaxCal.

Boone Viad.,Des Moines.

Moodna CreekViad.

Cap Rouge Viad.,Quebec, Can.

2,685

3,200

3,345

Mombasa, Africa 3,000

N. G. denotes Narrow gage.* This weight includes additional girders for future double track (at present only the towers are for double track).

3

Length Heightft. it.

800 3363

2 ,260 3 2 0

2 ,180 320

2,055 301

5,327 293

575 252

819 235743 229

810 190

185

182

173

112i

T Bt highest part

-

3 2

(iii

3 5

389

672

6 0

5 0

4 0

4 0

45

40

4 0

2 0L

Interm.spans

ft.

80

120

6 5

61

100

-140-235

100-118

7 5

60

75

80

6 0

4 0___-

-

No. oftracks

Steelweight

tons

1 N. G

2 N. G

1

1

1

1 N. G

1

1 N. G

1 N. G

2

2

1

1,115

6 ,200

1 ,820

3 ,350

12,200

. . . . . .

1,000

4 8 0

6 ,200

5,790

4 ,500

1 N. G 8,000-

-

*. . .

incl. o n e 80-ft.double tower

incl. o n e 185-ft.cantilever span

.c

Through pl. ?girder spans

z

2u1

incl. o n e 150-ft.cantilever span

incl. one 300-ft.truss span

. . . . .

incl. 3 truss spans125 to 160 ft.

250 DESIGN OF STEEL BRIDGES [CIIAP. XII

Solid concrete piers instead of steel towers make a more substantialand permanent, but generally more expensive, structure and will not beconsidered here. Single bents are substituted for braced towers in lowviaducts in cities where the tower bracing would interfere with thestreet traffic. Such structures are discussed in Chapter XIII on ElevatedRailroads (page 260). Table 54 gives dimensions and steel weights ofsome notable viaducts. (Compare Plate XXVII.)

ART. 11. ECONOMICAL SPAN LENGTHS

The economical length of span for a certain heigbt of a viaductdepends upon the relative cost of the girder spans, towers and masonrypedestals and varies considerably with these changing conditions.However, it Will be found that for given conditions the total cost variesonly slightly for even a considerable change in span length. From thisfollows that in order to reduce the cost of shop work it is advisable tomake the intermediate spans either of the same or of considerablygreater length than the tower spans (30-ft. towers and 40-ft. spans,for instance, are not advisable) and not to vary the intermediate spansin the same viaduct unless the height changes considerably. As arule it is sufficient to use the span lengths established by practice andmake special investigations or comparative designs for extraordinaryviaducts only.

We will assume a single-track viaduct on a tangent with 30-ft.towers designed for a live load of Cooper’s E-50 according to the Speci-fications of the American Railway Engineering Association (see Ap-pendix) .

The steel weight in lb. per lin. ft. of deck plate girder spans of 30ft. to 80 ft. varies according to the formula w = 180 + 9 I, where 1 =length of span in feet.

The steel weight of the towers in lb. per lin. ft . of viaduct isapproximately t = h (17.5 - 0.07 L), where L = length of tower plusintermediate span and h = height of tower, both in feet.

Substituting (L - 30) for I of the intermediate span we get for theweight of spans per lin. ft. of viaduct

13,500w= L + ‘9 [180 + 9 (L - 30)] =

16,200=- f9L - 360

L

Assuming a price of 3$ cents for the spans and 32 cents for the


towers, per pound erected, and a cost of $300 for the pedestals of onetower, regardless of span length, we get the total cost per lin. ft. of viaduct

C = (+ y + 0.29 L - 11.6) + h (0.66 - 0.0026 L) + p

In differentiating we getCZCdL = ( - g + 0.29) - 0.0026 h - g,

and this has to be zero for smallest cost C. From this.______.L= 825 317 000

0.29 - 0.0026 h o r h =112--AL2For intermediate spans 1 from 30 to 60 ft., we find the following

heights h for which these spans are most economical:

Length of inter-mediate span I

Height ;f tower.

in ft. in ft..~_____To;rf;pan

3 0 244 05 0 5;6 0 73

For a height of 73 ft. (60-ft. spans most economical) the total cost C perlin. foot is:

for Z=30 ft., C = 14.5+36.9+5.0 = $56.40for Z=60 ft., C = 20.3+31.1+3.3 = $54.70

This small difference (3%) shows that it would not be warrantedto use span lengths between 30 and 60 ft. For a height of 53 ft. thecost for 30-ft. and 60-ft. intermediate spans is the same.

If the pedestals cost $600 per tower, 30-ft. spans are more econom-ical for a smaller and 60-ft. spans for a greater height than 32 ft.

Assuming 40-ft. towers which have a steel weight in lb. per. lin. ft.of viaduct of t = h (15.8 - 0.045 L) we find

&ength of

intermediate span

Tower span I

r :

Height of tower--

40 ft. in ft.

40

Kl

for cost of pedestals for cost of pedestals

per tower of $300 per tower of $600-___

h h

in ft. in ft.

9": ii120 108

252 DESIGN OF STEEL BRIDGES [[CHAP. XII

If the foundations are difficult and their cost uncertain it is advis-able to use longer intermediate spans than would otherwise be economical.It should be remembered also that the towers are more expensive tomaintain than the plate girders and for this reason longer spans arepreferable even if slightly more expensive in first cost.

In viaducts of small and medium height common lengths of towerspan are 20 and 25 ft. for highways and electric railroads, 30 and 40 ft.for steam railroads. For very high viaducts it is advisable on accountof stability to use long,er tower spans, preferably not less than $ to 6of the height of t,he tower. In low viaducts the intermediate spa,nsshould be made equal to the tower spans; for higher viaducts intermediatespans of about twice the length of the tower spans have proved econom-ical. Common lengths of intermediate spans in higher railroad viaductsare 60 ft. with 30-ft. towers, 70 and 80 ft. with 40-ft. towers. (TheLethbridge Viaduct of the Can. Pac. Ry., 293 ft. high, has 67-ft. 3-in.tower spans and IOO-ft. intermediate spans.)

If the intermediate spans are not longer than about twice the lengthof the tower spans, the same depth of girders can be used for both, whichwill simplify the details, reduce the cost per pound of the steel work andimprove the appearance of the structure.

It is not advisable to make intermediate spans over 80 ft. on accountof the increased cost of erection. The towers of a viaduct should havethe same length and the number of different intermediate spans shouldbe as small as possible in order to reduce the, cost of manufacture.

Skew towers should be avoided.

ART. 12. PLATE GIRDER SPANS

As a rule deck plate girder spans are used in viaducts. In theLethbridge Viaduct, 293 ft. high, for the CanPac. Ry. (see Plate XXVII)through plate girder spans were used to insure greater security of thestructure in case of derailment. In the Gokteik Viaduct, 320 ft. high,in India (see Plate XXVII), which was manufactured in this country,riveted trusses were used for the 120-ft. intermediate spans on accountof their smaller cost of transportation.

If there are only two girders as in single-track viaducts, they areseated directly on top of the columns (Fig. 7). If there are more thantwo girders and only two columns, a cross girder is either framedbetween the tops of the columns to support the inner girders, the out-side ones resting on the columns, or, if the bridge is very wide, the cross

A~~.141 VIADUCTS 253

girder may rest on top of the columns partly cantilevering outside andthe longitudinal girders framed into it. In order to prevent horizontalbending of the cross girder from the braking force a horizontal brakingtruss should be arranged so as to transmit this force from the longitu-dinal girders to.the tops of the columns.

The depth of the girders should be constant throughout, conse-quently the economical depth is somewhat less than that of a singlespan of the same length as the intermediate span. For the girders of

the tower spans coverplates can as a rule be avoided; their depth backto back of angles should in this case be made equal to the depth of theintermediate spans out to out of the coverplates which extend the fulllength of the girder.

The spacing of the girders should be the same for all spans and con-form to the longest span. 7 ft. is usually the minimum for single-trackand 6 ft. 6 in. for double-track railroad deck spans.

Where it is not possible to make adjoining girders of the same depth,the shallow girder should have a pedestal so as to be seated independ-ently, which is important in erection. The tops of battered columns inthe same tower must have the same elevation; if the viaduct is on agrade fillers will be used under the higher end of the tower span.

ART. 13. BATTER OF COLUMNS

The columns of the towers should be made vertical whenever there issufficient lateral stability. This will, as a rule, be the case with lowdouble-track railway or highway viaducts. Single-track railroad via-ducts and other high viaducts require battered columns.

Stability against overturning from the iateral forces without the aidof anchorages is desirable, but, except in low viaducts, this cannot beaccomplished without using a considerable batter, which would makean uneconomical structure. For single-track viaducts on tangent abatter of 1 : 6 has proved satisfactory; a greater batter, up to 1 : 4, isdesirable on curves; for double-track viaducts a batter of 1 : 9 may besufficient. For viaducts on a curve it may be of advantage to use agreater batter for the outer columns than for the inner ones; this, how-ever, increases the cost of shop work and may cause an unsightlyappearance.

ART. 14. NUMBER OF COLUMNS

Two columns per bent are generally most economical and satisfactoryeven for double- or multiple-track viaducts, as the distribution of the load


among more than two columns is uncertain. For low viaducts a verticalcolumn may be used to advantage under each longitudinal girder.

Height of Panels.-The towers should be divided into vertical panelsof a height about equal to the length of the tower span and this heightshould be used throughout,, the variations of height of towers being madeup in the bottom panel as far as possible. Longitudinal and transversebracing should have common panel points except that the top panels ofthe latter may be subdivided so as to avoid steep diagonals.

ART. 16. COLUMN SECTIONS (See Table 66)

The same general principles apply for the design of the column sectionsas for that of truss members (see page 175). The possibility of simple andeffective connections of the bracing and easy access for inspection andpainting is important.

The section formed of 4 angles or Z bars with one system of lacingor a single web is satisfactory for light structures (Figs. 1, 2 and 3).

For railroad viaducts the column sections should be preferably of abox shape. Figs. 4 to 13 show some typical sections for vertical columns.

If the columns are battered they should have a coverplate on theoutside and the usual section is then similar to that for top chords oftruss spans (page 177) ; however, flange plates to balance the coverplateare generally not used (Figs. 14 to 20). In order to keep the rivet lines inthe angles and the distance inside of the webs constant, the thickness ofthe webs should preferably be the same for the whole length of thecolumn; where a variation of section becomes necessary, it can be madeby adding side plates or by changing the thickness of angles at fieldsplices. Shop splices should be avoided. Field splices are locatedabove the panel points so that each story of the tower can be com-pletely erected. Column sections are made in lengths up to 60 ft.

ART. 16. BRACING

Rigidity of the towers is of prime importance, particularly in railroadviaducts which are subject to great vibrations. As the rigidity dependsprincipally on the bracing, the members of the latter should be stiff andhave rigid riveted connections. Rods and adjustable members shouldbe avoided.

The double diagonals in any panel of the longitudinal and transversebracing should be assumed acting simultaneously, one in tension andthe other in compression and should be designed accordingly. As re-

r

ff!idn: ;QIJQ@&. ‘4looo99=%

41 OOOLS- P5u~,,4los +u pU!M:m

,yudfl = 41 om00016 :‘MQQ@OOOBI =7

‘41 ooa19 =p

PLATE XXX

7O'Girder:

Max.5.M.

E a c h F i g .

Max. End Shear :

D. /3.5bfax. End Shear : 113 0

;. 9 9 . 52 2 6 , 0 0 0 l b

Web 84 A &,” = 3 6 . 7 “3~Rivet Pitch Z~"Siag.

Specificafions S o u t h & W e s t e r n R . R.Co’S.1906Maferial:Sfrucfural 0 . H . S+eel

Ties on Viuducf 8% I2 ”Carolina Clinchfield & Ohio Ry.,

Stress-sheet of Single TrackViaduct, Built 1908.

(Fo l l ow ing plate X X I X )

256 DESIGN OF STEEL BRIDGES [CHAP. XII.

versal of stress will not occur in immediate succession, no provisionneeds to be made for it. Where the diagonals are excessively long ascompression members they may be designed to take tension only.

In light structures single angles or two angles riveted back to backmay be used, in al1 heavier work the members of the bracing should bedouble webbed and as deep as the columns and should connect to bothwebs of the latter. 4 angles forming an 1-section (Fig. 8) are eco-nomical for diagonals and short struts. Longer struts are built of twochannels or, if they are very light, of 4 angles with lacing on four sides.

Rigid top and bottom struts should be used in all cases, except that ifthe bottom flange of the tower girder is rigid enough the longitudinal topstrut may be omitted. Intermediate struts have sometimes been

omitted and the diagonals designed for tension and compression; thestruts are, however, valuable as spacers for the columns during erection.

ART. 17. EXPANSION

Expansion of the girder spans should be provided for at every towerby allowing one end of the intermediate span to slide on top of the col-umns. In low viaducts the clearance around the anchar bolts is sufficientto provide for the expansion of the tower base; in high viaducts one col-umn base should be fixed and the other three allowed to slide in thedirection of the lines connecting them with the fixed base.

ART. 18. ANCHORAGES

The anchorages should be able to sustain the greatest possible uplift(see pages 244 and 248) and should be deep enough to engage masonry ofa weight equal to one and a half to two times the uplifting forte.Where the pedestal is submerged the weight of the displaced watershould be deducted.

ART. 19. WEIGHTS OF VIADUCTS

Deck plate girder spans in viaducts weigh somewhat more than singlede& plate girder spans of the same length (see page 221), since the efectivelength of the viaduct span is greater and the uniform depth of the girdersnot economical for all the different spans. About 5% should be added

ART. 191 VIADUCTS 2 5 7

to the weight of those spans which have approximately economicaldepths and about 10 to 15% to those which are considerably deeperas is usually the case with the tower spans. The bearings shouldnot be included, since they are generally omitted in viaducts.

The weight of the towers (including their bracing) varies considerablyfor different specifications, batters, etc. For ordinary heights of via-ducts the weight may be assumed constant per vert. ft. of tower,therefore also constant per eq. ft. of area between the line connectingthe tops and the line connecting the bases of the tower posts as seen inthe elevation.

For towers designed according to the specifications of the AmeritanRailway Engineering Association (see Appendix) the steel weight w inlb. per vert. ft. of tower is approximately as follows:

If L = length of tower span plus intermediate span in feet, we havefor single track on tangent

E-30 loading w = L (14.3 - 0.06L)w = L (17.5 - 0.07 L)

w = L (13.0 - 0.04 L)w = L (15.8 - 0.045 L)

Adding to these weights those of the girders and bracing, the totalsteel weight W of the single-track viaduct per sq. ft. of area includedbetween the lines connecting the tops and the bottoms, respectively, ofthe columns of an average height h in feet, can be expressed by the fol-lowing formulas:

E-50 loading

Weight W inlb. per sq. ft.

f 30-ft. interm. spans30-ft. towers

i 6@ft. ” “.m

L 40-ft. towers40-ft. IL “

80-ft. ” ‘: 10.4+y

Considering that for low viaducts the shorter spans and for high via-ducts the longer spans are used, the above formulas for W can be com-bined into the following formula which gives close results for all heightsand approximately economical span lengths:

LoadingE-50, W= 13+? . . . . . . . . . . . . . . ’ . (2)

where h = average height in feet.

258 DESIGN OF STEEL BRIDGES [CHAP. xII

W ranges from 23 lb. per sq. ft. for h = 50 ft. to 15.5 lb. per sq. ft. forh = 200 ft.; for h below 50 ft. IV increases rapidly.

The corresponding weights for E-40 and E-30 loading are

Loading E-40, w = 12+4~................. . (24

Cl E-30, w = ll+? . . . . . . . . . . . . . . . . .

It would seem that for very high viaducts formula (2) gives some-what too great weights compared with formula (1). It must be con-sidered, however, that the above given weights of towers are for ordi-nary heights and that with increasing height the weight per vert. ft.of tower increases on account of the influente of wind on the lowercolumn sections and the increase in length of the lower transversebracing.

The total weight of steel in the viaduct is obtained by multiplyingthe weight W by the average height h of the columns and the totallength of the viaduct.

Double-track viaducts are approximately twice as heavy as single-track v i a d u c t s .

For viaducts on 4’ curve, the weight increases about 10%; for sharpercurves it may increase as much as 20 ‘%.


FURTHER INFORMATION ON VIADUCTS

(Compare also Elevated Railroads, page 271)

J. E. Greiner, Evolution of Ameritan Railroad Viaducts, Trans.A. S. C. E., Vol. 25, 1891, also Eng. News, June, 1891.

Antofngasta R. R. Viaduct, Bolivia, s. tr., Eng. News, May, 1889,and July, 1890.

Verrugas R. R. Viaduct, Peru, s. tr., Eng. News, May, 1891.Pecos River R. R. Viaduct, Texas, s. tr., Eng. News, Feb., 1892,

and Jan., 1893.C. R. Grimm, Kinzua Viaduct, s. tr., Trans. A. S. C. E., Vol. 46,

1901, with interesting discussion; also R. R Gqxette, Eng. News and Eng.Record, 1900.

Gokteik Viaduct, Burmah, India, s. tr., Eng. News, Feb., 1901, Eng.Record, Jan., 1901.

C. & N. W. Ry., Des Moines (Boone) Viaduct, d. tr., Eng. News,June, 1901. Eng. Record, July, 1901.

Uganda Railway (Mombassa, Africa) Viaducts, s. tr., Eng. Record,Sept., 1904.

Dcepwater Ry., Black Lick Creek Viaduct, s. tr., Railroad Gazette,Aug., 1907.

Stony Brook Glen Viaduct, s. tr., Eng. Record, July, 1908.C. C. & 0. Ry. Viaducts, s. tr., Eng. Eews, Jan., 1909. ’Can. Pac. Ry., Lethbridge Viaduct, s. tr., Eng. News, Sept., 1909.Frank W. Skinner, Methods of Erecting Viaducts, Eng. Record,

April, 1910.P. R. R. Viaducts crossing Terminal Yard in New York, Trans. A.

S. C. E., Vols. 58 and 59, 1910.

CHAPTER XIII

ELEVATED RAILROADS

ART. 1. GENERAL

An elevated railroad is a viaduct located in the streets of a City,intended for city or standard railroads. It generally provides 14 to 16 ft.of vertical clearance for the Street traffic below.

For City railroads (rapid transit) economy of design is of course animportant factor, but consideration must be given to the public require-ments, such as least obstruction to the traffic in the Street, light, leastnoise, esthetic appearance of the structure, etc. Rigidity is of impor-tance on account of the great number of trains and the frequency ofstopping.

As a rule bracing between columns has to be omitted as it interfereswith the Street traffic. To insure lateral and longitudinal stability thecolumns are fixed at the top or bottom or preferably at both ends.

Where part of the elevated railroad is located on private property,braced towers have been used every fourth or fifth span (NorthwesternElevated Railroad of Chicago) ; they improve the rigidity and areeconomical but, to maintain uniformity of the structure, they should beused only if they can be arranged at regular intervals over a great lengthof the viaduct. In such cases the longitudinal forces may be assumedas being resisted by the towers, as shown for viaducts (page 243).

ART. 2. ECONOMICAL SPAN LENGTH

.The length of spans is economical if the cost of the longitudinalgirders and bracing is approximately equal to the cost of the bents in-clusive of the pedestals and their foundation. 45 to 50 ft. have provedmost economical on average soil. The spans should be alike as far aspossible, as duplications of parts are an important factor for reducingthe cost.

ART. 3. ARRANGEMENT OF CROSS-SECTION

The arrangement depends upon the number of tracks and the localconditions. If the Street is wide it may be permissible to locate the

260

ART.~~ ELEVATED RAILROADS 261

columns in the roadway; in narrow streets the columns must be locatedon the sidewalks inside of the curb line. Three or four columns arestiffer and more economical but two columns are preferable since theyoccupy less space; bents with two columns are less affected by unequalsettlement of the foundations. A single line of columns (lower 3d Ave.,New York) should be avoided as it is difficult to make such a structurerigid.

There may be either a column below the center of each track sup-porting the longitudinal girders by means of brackets, girders of ad-joining tracks being connected by cross frames, or the columns may beconnected by a cross girder into which the longitudinal girders areframed. The construction with cross girders requires simpler shopwork and insures greater lateral rigidity than that with cross frames;a longitudinal girder may be placed at the center of the column andconnected to it by longitudinal brackets, making the column fixed atthe top and securing longitudinal rigidity (see Plate XxX11).

One or two girders may be framed into brackets extending outsideof the columns, which reduces the weight of the‘cross girder. On theN. Y. C. Bi H. R. R. R. viaduct on Park Ave., New York, the girderswere placed directly on the columns.

ART. 4. FLOORING

Although ballast would be the ideal flooring for elevated railroadson account of the reduced noise and other advantages mentioned onpage 132, it is rarely used for the whole line because of its greater cost.It is usually restricted to portions on Sharp curves or to importantStreet crossings which it protects from dirt and water.

For the description of various solid floors see page 132. A woodentie flooring similar to that described on page 132 is generally used. Asthe stringers are generally spaced 5 ft. c. to c., the ties have a minimumsize of 8 in. X 6 in. laid flat. The usual practice is to provide guardtimbers on both sides of each rail, substituting steel guard rails oncurves. The inside guard timbers are 6 in. X 6 in. and the outside ones6 in. X 8 in. on edge.

ART. 6. LONGITUDINAL GIRDERS AND BRACING

Longitudinal lattice girders have often been used (Boston, Philadel-~phia, European cities) as they admit more light; plate girders, however,are preferable as they are cheaper in first cost and maintenance. The

262 DESIGN OF STEEL BRIDGES [CHAP. XIII

flooring is, as a rule, placed on top of the girders. In a few cases throughspans were used in order to reduce the height from rail to Street level.Deck girders are usually spaced 5 ft. c. to c. Thegirders should be ofuniform depth throughout. For their design see pages 147 and 252.

The bracing should conform to the requirements of plate girder spansas explained on page 166, except that end cross frames are necessary onlyat expansion ends. There should be a lower lateral system in al1 spanson curves.

Provision should be made to transmit the braking forte from thelongitudinal girders to the tops of the columns to prevent bendingstresses in the cross girders.

ART. 6. CROSS GIRDERS

The cross girder is usually built as a plate girder. For its design thesame rules apply as for floorbeams (see page 136). The deeper it is thegreater its rigidity and also that of the columns, as explained below.

If a cross frame is used instead of a cross girder it should be of heavyconstruction and as deep as possible.

Corner brackets increase the stiffness of the cross girder and columns,if they are of sub,stantial construction. Single angles, especially whenbent, are of doubtful value.

ART. 7. EXPANSION JOINTS

Expansion joints in the longitudinal girders should be provided 150to 200 ft. apart by seating the ends of the girders into expansion ‘pocketsconnected to the cross girder.

ART. 8. COLUMN

The columns are as a rule vertical. They should be designed toresist the direct stresses as well as the bending an.d shearing stresses pro-duced by the horizontal forces.

The sections are similar to those for viaduct columns (see Table 55).Sections with ful1 webs in both directions are preferable. If there is noful1 web in a direction in which a column is bent the latticing should beproportioned to resist this bending as shown on page 204. The flangesof columns 1 to 3 are liable to be injured by vehicles; sections 10 to 13are the most satisfactory.

The column footings should be protected so that no dirt can collect

ART. 101 E L E V A T E D R A I L R O A D S 263

around them, and if located in the roadway should be provided withfender castings.

Whenever practícable the columns should extend to the top of the crossgirder and the latter framed into them.

ART. 9. CALCULATION OF STRESSES IN COLUMNS

If rigidly connected to the ,cross and longitudinal girders and firmlyanchored to an unyielding masonry pedestal the column may be assumedfixed at top and bottom.

The bent formed by the columns and the cross girder or frame isstatically indeterminate as the stresses produced by any load dependupon the relative rigidity of these members. If the rigidity of the crossgirder or frame, expressed by its moment of ‘inertia, is much greater-than that of the columns the simplifying assumption may be madethat the cross girder is absolutely rigid.

ART. 10. DEAD AND LIVE LOAD STRESSES

Under the above assumptions a vertical load P (Fig. 1) appliedbetween C and D causes the same reactions as in a simple span CD :

\

“.- _......“%

1 . ..___ 2

PQ RB

FIG. 1.

d

) 4-c

h

‘PT

FIG. 2.

RA = Pi a n d RB=PF=P-RA

The dead and live load stresses in girder CD are found as in a simplespan and the stress in a column is equal to the reaction of span CDplus the weight of the column. In determining the permissible buckling

2 6 4 DESIGN OP STEEL BRIDGES [CHAP. XIII

stress the total height AC of the column should be taken as the freebuckling length, since points C and D are not held laterally and thecolumn might deflect as shown dotted in Fig. 2.

ART. ll. STRESSES DUE TO WIND AND CENTRIFTJGAL FORCE

Concerning wind and centrifuga1 forces the same applies as for via-ducts (see page 240). On account of the small height the wind actingon the columns may be neglected.

To calculate the stresses from these forces we will first consider a bentwith two columns (Fig. 2). A horizontal forte H causes the columns todeflect as shown by dotted Enes. The point of contraflexure is locatedmidway between the base of the column and its connection to the crossgirder.

The condition of no part is changed if we separate part ECDFfrom AE and BF and apply at points E and F externa1 forces

d-tiH,, Hb a n d V=H-

1

which are equal to the interna1 stresses caused by H at these points(Fig. 3). Since the deflections of the two columns are alike, H, andHb are in the same proportion as the moments of inertia I, and Ib ofthe respective columns

1, > Hb=H

Ib__~Ia+ Ib

and H,fHb =H

or

H, = H, = ; if 1, = Ib.

These forces at E and F cause reactions at A and B equal but of

opposite sign to them and moments M, = H, k and Mb = Hi, k (Fig.

3 b and c). Having these externa1 forces the stresses at any sectionare easily found. The direct stress is + V and - V at any section ofcolumn AC and BD, respectively. The bending moment at a sectionX of column AC, for instance, with the distance z from E is

M, = Haz.

The maximum moment occurs at A and C and is equal to M,. Forthe bending moment we can substitute an equivalent direct stress

A R T . 111 ELEVATED RAILROADS 2 6 5

Sb = 5 as explained on page 174 and, by adding this to the other direct

stresses and dividing by the permissible unit stress, the required area ofthe column is oblained. It has to be noted that there isverse shear at any section equal to H, for column ACcolumn BD for which the column has to be proportioned.

also a trans-and Ht, for

(C.1

FIG. 3.

If a bent has more than two, for instance four, columns (Fig. 4),the horizontal forte H is divided among the four columns in propor-tion to their moments of inertia, since the deflections of al1 columnsare alike and the forte producing a certain deflection is in direct pro-

Hm f

FIG. 4. FIG. 5.

portion with the moment of inertia of the column section. The mo-

ment H (d + i) produces four vertical forces $: Via and + l;b which

are obtained from the conditions

266 DESIGN OF STEEL BRIDGES [CHAP. X111

qv, a +vb b) = H (d + i) and e = %, from which

H (d + ;)va = a(a”pb2) a and

H(d+;)

vb = 2(aZ + b2)b

If a single-track line is supported by a single line of columns(Fig. 5) the bending moment caused by a horizontal forte H at anysection X is M = Hy. It is a maximum at the bottom for y = d + h.

The shear at any section is equal to H.


The braking forte is applied longitudinally at the rails and usuallyassumed at 20% of live load. Considering a portion of the viaductbetween expansion joints, having single bents only, the forte B appliedto it is transmitted by the columns to the foundations in the same wayas shown above for the transverse forces. Since, however, the length ofsuch a portion is usually considerable, the forces V may be neglected.B may for simplicity be divided equally among al1 columns. as they usu-

ally do not differ to any extent.B

If, therefore, ; is the forte applied to

one column (n being the number of columns) the greatest moment occursB h

at the tob and bottom of the column and is M = ; 2, provided the

columns are rigidly fixed longitudinally at top and bottom. The shearB

at any section is ñ. M can again be replaced by an equivalent direct

M Cstress 7, where c and r are taken in the longitudinal direction.

ART. 13. TEMPERATURE STRESSES

Owing to the change of length of the spans due to temperaturechanges the columns are forced to deflect longitudinally, thereby sus-taining bending stresses. If L is the distance between the first and thelast column of a section limited by two expansion joints and + t thetemperature change from the normal, the first and the last column hasto sustain a deflection at its top of

where E = coeflicient of expansion. For t = &- 75” and E = 0.0000065

we get 6 = 0.00049 “2 (6 and L in same units).

ART. 141 ELEVATED RAILROADS 2 6 7

If fixed top and bottom the column deflects as shown in Fig. 6.Each half of the column can be regarded as a beam fixed at one endand loaded at the other, or free, end by a load

deflection, f

T producing the

6 3EITzZ h3 . . . . . . . . . .~

02

T produces a bending moment at A and C ofh .2

M,=TZ = ahZ 6 3EI=

0 5

6sE$

which may be substituted by an equivale&, directstress of

where c = distance of outer fiber from neutralaxis measured longitudinally and a = area ofsection. The shear at any section is equal to T.

. . . (see Val. 1)

p--- 6 -->I

FIG. 6.

ART. 14. COMBINED STRESSES

In combining the stresses from the different forces the same consid-erations apply as for viaducts (see page 244, etc.) except that the tempera-ture stresses have to be added to the dead and live load stresses for theregular condition. If temperature stresses are added, with those fromwind or centrifuga1 forte, a higher unit may be allowed than if the tem-perature were disregarded, since the maximum combined stress occursonly in one comer of the column section.

Engineers differ in their opinion as to whether the braking forteshould be considered in proportioning the columns or not. It is true thatthe assumed friction (20%) may be too high and part of the forte isdistributed over a great length of viaduct by the continuous track; onthe other hand, it seems inconsistent to consider the braking forte forviaduct towers and to neglect it entirely in proportioning the columnsof elevated railroads, where that forte is’ constantly ‘applied. Besides,the assumed fixity of the columns is never realized, which results inhigher stresses than those calculated. It seems therefore proper toconsider the braking: forte, but not simultaneously with the wind, anduse higher units than when it is not considered.

2 6 8 DESIGN OF STEEL BRIDGES [CHAP. 13

ART. 16. ANCHORAGES

There should be four anchar bolts for each column to resist thelongitudinal and lateral bending of the column. The bolts shouldpreferably extend to the bottom of the masonry pedestal and be effect-ively connected to the latter.

For any of the conditions of loading on page 244, etc., there results atthe base of the column an axial downward forte P from the vertical loads,an axial upward or downward forte 8, a horizontal force H, and abending moment M from the horizontal forces. P and V result in anaxial forte R = P rt V (Fig. 7). The friction of the column on the

Amasonry takes care of H. The base of the col-umn has to be of such a size that the maximum

L

bearing pressure sC per sq. in. on the masonry doesR not exceed the permissible value. For a rectan-

gular base with the sides a normal to and b inn

A B the plane of M we have8. _ _ _ __ ._ __ b ’___._. Y R 61M

FIG. 7. & = &+ -& . . . . . . . . . . . . *.~ . . . . . . . (1)

For maximum sC both R and M have to be maxima.Tf for any condition

gb < ‘g, or R; CM

1+le

_________ 8 r -____ ---$ ___________

zFIG. 8.

there occur tension stresses which are taken up by the anchar bolts.The maximum tension will occur for a condition giving maximum Mand minimum R;

To determine the stress 2 in the anchar bolts the following simpleapproximate method may be used.

ART. 161 ELEVATED RAILROADS 269

For a homogeneous section the distrib’ution of the unit stresses wouldbe as shown in Fig. 8, the sum of the tension stresses being

T = ka,,,.

b 6MFrom . . . . . . . . . . . . . . . . . . . x: 2 = at:o~

a b3follows.. . . . . . . . . . . . . x = S1 ___12 M

a2 b3and therefore. . . . . . . . . . . T = rM st2

w h e r e . . . . . . . . . . . . . . . . . . . . . .R 6 M

.st = a-a

Taking moments of T and Z about C we find

2b2s T---x+2b-3e

If there are two anchar bolts at the distance e from A the stress in

each is 2. To be effective the bolts should be as near the edge of the

base as practicable and should engage a sufficient weight of foundationmasonry.

The pressure of the masonry pedestal on thefoundation is calculatedby formula (1) where a and b are the dimensions of the base of thepedestal. R has to include the weight of the pedestal. For no condition

of loading should

in order to avoid tension, that is, opening at the base.

ART. 16. WEIGHTS OF ELEVATED RAILROADS

The steel work of elevated railroads carrying light (passenger) traffic-weighs approximately per lin. ft. 1000 lb. for 2 tracks; 1500 lb. for 3tracks and 2000 lb. for 4 tracks. For standard railroads, the weightsare about 2000 lb. for 2 and 3000 lb. for 4 tracks. T h e abovefigures are based on a wooden tie flooring; if steel flooring and ballastare used about 50% should be added.


The following are the steel weights in lb. per lin. ft. of variouselevated railroads :

Union Elevated R. R. (Myrtle Ave. Line), Brooklyn.Doubre-track 24 ft. c. to c. of tracks, 41; ft. c. to c. of columns.. . 1,150Double-track 24 ft. c. to c. of tracks, 37 ft. c. to c. of columns.. . . 1,050

Lake St. Elevated R. R.; Chicago.Double-track, with provisions for two additional tracks, 12 ft.

c. to c. of tracks, 46 ft. 10 in. c. to c. of columns . . . . . 1,300Four-track 12 ft. c. to c. of tracks, 46 ft. 10 in. c. to c. of columns. 2,000Double-track, with provisions for third track, c. to c. of tracks,

22 ft. 8 in. c. to c. of columns . . . . . . . . . . . . . 1,050Three-track 12 ft. c. to c. of tracks, 22 ft. 8 in. c. to c. of columns 1,420

Boston Elevated k. R. (lattice con&.) double-track. . . . . . . . . . 1,200

Northeastern Elevated Ry.,.Philadelphia (design).Double-track 13 ft. c. to c. of tracks, 36 ft. c. to c. of columns 1,480

Jersey City Elevated R. R. (P. R. R.)Four tracks 12 ft. 2 in. c. to G. of tracks, three lines of columns 18

ft. 3 in. c. to G. . . . . . . . . . . . . . . . . . . . . . . 3,000

Philadelphia & Reading Terminal R. R., Philadelphia.Double-track 12 ft. c. to c. of tracks, 18 ft. c. to c. of columns with

trough floor and ties resting in ballast . . . . 3,300

ART. 161 ELEVATED RAILROADS a71

ADDITIONAL INFORMATION ON ELEVATED RAILROADS,

ALSO SUBWAYS

J. A. L. Waddell, Design of Elevated Railroads, Trans. A. S. C. E.,Vol. 37, 1897.

St. Louis Elevated R. R., Railroad Gaxette, March, 1895.Jersey City Elevated R. R., P. R. R., 4 tracks, Railroad Gaxefte,

Dec. 1890.Brooklyn Elevated R. R., 0. F. Nichols, Trans. A. S. C. E., Vol.

32, 1894.New York, Park Ave. Viaduct, N. Y. C., 4 tracks, Eng. News, May,

1893.New York, Connecting Railroads, Railroad Gaxette, May, 1900.New York Rapid Transit, Eng. News, and Eng. Record, 1902, 1903.New York Rapid Transit, Report of Board of Commissioners, N. Y.,

1903.New York Rapid Transit, Ceo. S. Rice, Proc. Municipal Engineers

of N. Y., Sept., 1903.New York Rapid Transit, Report of Municipal Art Soc., N. Y., 1904.New York Freight Subway Belt Line, Eng. News, Oct. 1908.New York, Samuel Rea, P. R. R. New York Tunnel Extension, Dec.,

1909.New York West Side Traffic Problem, Eng. News, April, 1911.New York Tunnel Extension, Trans. 8. S. C. E., Vols. 68 and 69,

1910.New York Rapid Transit, Eng. News, May, 1912.Boston, Report of Transit Commission, beginning 1895, severa1

volumes (containing also referentes to subways in London, Paris,Budapest, etc.).

Boston Elevated R. R., Railroad Gaxette, July, Oct., Nov., 1899, alsoEng. News, May, 1899.

Chicago Elevated R. R., Railroad Gaxette, March, 1895, Aug. andNov., 1896; Eng. News, Aug., 1896.

Chicago, Northwestern Elevated R. R., 4 tracks, Railroad Gaxette,May, 1896..

Chicago, Elevated R. R., Eng. News, Sept., 1903, and Oct., 1912.Philadelphia, Delaware Ave. Elevated R. R., 3 tracks, Eng. Rec.,

o c t . , 1 9 0 9 .Samuel Tobias Wagner, Elevation of Tracks of P. & R. R. R. in

Philadelphia 4 tracks, Trans. A. 5. C. E., 1913.Philadelphia Report of Transit Commissioner, 1913.San Francisco Rapid Transit, Eng. News, Dec., 1912.London Rapid Transit, Eng. Magazine, Oct., 1901.

19


Paris Elevated Railroad nnd Subway, L. Troske, Papers of GermanSoc. C. E., Berlin, 1903, to 1905.

Paris Elevated Railroad and Subway, Eng. News, Dec. 1900, Sept.,1903, Sept., 1911, Sept., 1913.

Paris Elevated Railroad, Bridge over the Seine at Passy, 2 tracks onupper de&, highway 80 ft. wide on lower deck, steel arches (beautifulstructure), Eng. News, Feb., 1904, Nov., 1906.

Berlin Elevated Railroad, Eng. Record, 1891, and Eng. New-s, Oct.,1898.

Berlin Elevated Railroad, Official Report, Zeitschrift für Bauwesen,Berlin, 1886 (a very valuable publication on Bridges, Train Sheds, etc.):

Berlin Elevated Railroad, Stahi’ u. Eisen, Feb., 1902 (descriptivearticle with many illustrations).

Vienna Elevated Railroad, Papers of Austrian Soc. C. E., Vienna,1897, 1898, 1899.

Hamburg Elevated Railroad, W. Stein, Pnpers of Austrian Soc.C. E., Vienna, June, 1912; also Deutsche Bauzeitung, Aug. 1912.

P. Wittig, Elevated Railroads and Subways in Europe and theU. S. A.: Berlin, 1909.

German Elevated Railroads, F. Steiner, Papers of Austrian SOC.C. E., Vienna, May, 1912.

Henry B. Seaman, Specifications for Bridges and Subways, Trans.A. S. C. E., Vol. 75, 1912.

CHAPTER XIV

MOVABLE BRIDGES AND TURNTABLES *

ART. 1. TYPES OF MOVABLE BRIDGES

A movable bridge is one which may be turned or drawn to one side,lifted up, or let down, so as to permit vessels to pass.

Movable bridges may be classified as follows:1. Swing bridges, turning about a vertical axis;2. Bascule bridges, turning about a horizontal axis, or rolling back

on a circular segment;3. Traversing or retractile bridges, moving horizontally;4.. Lift bridges, lifting vertically;5. Transporter or ferry bridges, consisting of a fixed span with sus-

pended traveler. 7Pontoon bridges (see Eng. News, April 30, 3.908 and Nov.$O, 1913,

Golden Horn Bridge at Constantinople) will not be considered here.It is impossible to give any general rule as to which kind of bridge is

best adapted in a certain case, as there are many factors to be considered.The following general principles should be observed:

(1) When the bridge is closed it should be as nearly as possible afixed span.

(2) The machinery should be designed so that the bridge can beeasily operated while moving. The most simple design which gives theleast firet cost and cost of operation is the best.

(3) The structural and machinery parts of the bridge should beseparate; that is, when the bridgeis closed, acting asa fixed span, themachinery parts should not receive any stress.

Swing Bridges.-A swing bridge turns in a horizontal plane around avertical pivot, and when closed its ends are supported at the proper level.The pivot may be in the center of the span, forming a truss of two equalarms which balance each other and give two openings for navigation;or it may be a”t one side of the center, forming a truss with unequal arms,

* Articles 1, 2, 3, 4, 5, 15 and 16 are condensed reprints from the paper on Mm-able Bridges by Mr. C. C. Schneider, Past-President A. S. C. E. (Trans. A. S. C. E.,Vol. 60, 1908), with his kind permission.

273

274 DESIGN OF STEEL BRIDGES [CHAP. XIV

with the short arm counterweighted to balance the bridge about its pivqt;or there may be two pivots, in which case a locking arrangement has to beprovided at the center where the arms meet, and the shore ends anchoredto the masonry, when the bridge is closed.

Swing bridges are best adapted for long and heavy spans; they havebeen built for single- and double-track railroad structures up to 520-ft.span, and for four-track structures up to 390-ft. span (see page 390). Forspanning two equal openings, a swing bridge is the most economical. Ifthe conditions are such that the pivot pier has to be located near the shore,a bridge of unequal arms may be of advantage, provided space is availablefor swinging the bridge.

Less important types of swing bridges are the shear-pole draw and thejack-lcnife draw. These should be used for temporary structures only.

The shear-pole draw has only one leaf turning around a pivot at oneend; the other end, while swinging, is suspended from the top of a two-legged shear-pole by rods which are attached to a pivot which is ver ticallyin the same line as the pivot below. The shear-pole is stayed by guy rodsfrom the shore end. When the bridge is closed it forms a simple spansupported at both ends.

The jack-knife, or folding draw, differs from the shear-pole draw inthat it consists of two or more separate deck trusses or girders, each girderswinging horizontally around its own pivot at one end. The other end,while swinging, is suspended from a frame or tower by rods attached topivots on top of the frame. The rails are fastened to the top of each gir-der; there are no ties. The girders are connected with hinged links, andfold up close to each other when the bridge is fully opened. When thebridge is closed one end of each girder rests on the shoe at the abutment,while the other end rests on the pivot. This type of bridge can only beused for railroad traffic.

Bascule Bridges.-Bascule bridges turn in vertical planes. Theyrotat,e about trunnions or rol1 back on circular segments, or have a com-bined motion of turning and rolling, and are counterweighted to reducethe power required for operation. There are various patented designs.Bascule bridges are made in one leaf, or in two leaves which meet in thecenter. The two-leaf bridges have a locking device at the ends, andarearranged to act as cantilevers when closed, and sometimes as three-hingedarches.

If the conditions are not favorable for a swing span, a bascule bridge isthe next choice. The conditions which preclude the use of a swing bridgeare:

Am. l] MOVABLE BRIDGES AND TURNTABLES 275

If there is no space available to swing the bridge horizontally about itsvertical axis;

If the number of tracks is such as to increase the width of the bridgeand the size of the pivot pier with its fender to such an extent as to makeit undesirable; or,

If there is a probability that other tracks will be added in the future.The turning space required for swing bridges is a disadvantage, and

increases with the size of the bridge; the bascule bridge has the advan-tage that it can be used in places where there is no space available along-side of the bridge. The bascule bridge can be enlarged or widened byputting up additional spans alongside of it, without interfering with theoperations of the existing span or with navigation; the length of span,however, is limited, as the influente of wind pressure becomes very impor-tant in long spans and is probably 200 ft. for single-leaf, and 350 ft. fordouble-leaf bridges.

The design depends upon many conditions, such as location, dis-tance from the floor to water level, under-clearance required, length ofspan, frequency of opening, speed required, kind of power available foroperating, etc.

(Compare J. E. Greiner, Specifications for Movable Bridges, 1911;B. R. Leffler, Specifications for Bridges Movable in Vertical Plane,Trans. A. S. C. E., Val. 76, 1913.)

Traversing or Retractile Bridges.-They consist of a simple spanacross an opening, extended some distance on one side over the abut-ment. This portion is supported on wheels, and is counterweighted, theoverhanging portion acting as a cantilever when the bridge is moving,and resting on the opposite abutment when closed. Traversing bridgesare constructed so as to be capable of being rolled horizontally backwardin the direction of the center line, or transversely to the center line of thebridge. In the latter case, the bridge generally crosses the opening on askew. In some cases the bridge rolls transversely upon its carriage untilit has passed and cleared the track, and then rolls backward to clear thewaterway. A traversing bridge is not desirable as it requires more powerthan any other kind and is slow of motion. It has been used in only afew cases for railroad bridges, but has proved satisfactory for smallhighway bridges.

Lift Bridges.-A lift bridge consists of a simple span resting on abut-ments when closed. To the ends of the girders are attached ropes orchains which pass over sheaves on top of a frame at each end of thebridge. It is counterbalanced to reduce the pówer required for lifting.


It has the advantage over the bascule bridge in that it can be made of anylength feasible for a simple span, while the span of a bascule bridge islimited. When closed it forms a’simple fixed span. Its disadvantage isthe high first cost and the expensive operation. The vertical-lift bridgeis most suitable for spans which require only a small lift, such as bridgesover canals.

Transporter or Ferry Bridges.-This type (introduced by Mr. Arnodin,France) consists of a fixed span elevated sufficiently to clear naviga-tion. From this fixed span a platform is suspended which travels acrossthe opening from shore to shore, taking the place of a ferry boat. Thefirst bridges of this kind were of the suspension type (see En~ineering,London, February and March, 1900; Eng. News, Oct., 1906; Proceedingsof Inst. C. E., London, Vol. 165, 1907, very valuable paper). Later alsoarches were used, and for the Duluth Bridge built in 1905 the fixed spanis a simple truss bridge 390 ft. long, with riveted const.ruction through-out (C. A. P. Turner, Trans. A. S. C. E., Vol. 55, 1905).

They have the advantage over the ferry boat in that their operationís not obstructed by ice; their capacity, however, is very limited, andthey are not suitable for railroads.

ART. 2. CENTER-BEARING SWING BRIDGE

While swinging, this type of swing bridge is carried entirely on thecenter pivot; the trailing or balance wheels, which rol1 on a track near theouter edge of the pier, are adjusted with a little clearance so that theironly function is to insure lateral stability. When the bridge is closed,the ends are lifted and on the pivot pier independent supports areinserted without lifting so as to carry the live load only. The centersupports and end lifts should work simultaneously and be so adjustedthat, when the center supports just come to a bearing (with no reactionfrom the dead load), the ends are lifted to t.heir proper level. Forthe center supports wedges (usual bevel 1 to 10) have proven mostsatisfactory.

In light highway spans the center wedges are omitted.The balance wheels are from 15 to 20 in. diameter and from 4 to

6 in. tread. Generally eight wheels are used, two on each side of thecenter line of the floor system. The diameter, width of tread, size of axles,and journals of the wheels on each side of the track are proportioned foran over-turning wind pressure of 20 lb. per sq. ft. while swinging, andthe other wheels made the same.

ART. 21 MOVABLE BRIDGES AND TURNTABLES 277

Severa1 kinds of pivots have been used consisting of discs, conicalrollers and balls.. Phosphor bronze discs between two discs of hardenedsteel have given the best results; the wearing is confined to the phosphorbronze disc which can easily be examined and replaced, if necessary,without stopping the traffic over the bridge.

The center-bearing type offers the most advantages and shouldreceive the first ,consideration. It requires less power to turn, has asmaller number of moving parts, is less expensive to construct and main-tain, requires less accurate construction than the rim-bearing type, anddoes not as easily get out of order. The structural ancl the machineryparts are entirely separate, and when the bridge is closed it forms eithertwo independent fixed spans, or a fixed span continuous over twoopenings resting on firm, substantial supports. There are less ambigui-ties in the calculations, and the distance required from base 0-f rail tomasonry is generally less than for a rim-bearing type. Any irregularsettlement of the pivot pier does not affect its operation.

TABLE 66

Principal Dimensions of Some Existing Railroad Swing Bridges with Center-bearing

327323277386

‘hropgh trEss

“ ““ C‘

400195356360

198277176257

250

“ ‘I“ ‘I“ ““ ‘<

‘< II,‘ 6‘‘< ““ ‘I

“ ‘<178 De& pl. gir.220 Through truss1 7 2 Throughpl.gir192 Through truss

Zonstruction Fraclrs

2221

n11

2

21

:1

:

(C

-

(See slso psge 302, et,c.)

2. to c.)f trussin ft.

3 03 03018

2 040;1818

2 82 1

16&

:o’

:B16:

Height oftrusses in

ft.

32 .to 5044 to 6030 to 4227 to 55

26 to 6624 to 3029 to 6526 to 51

z: ti 4:22

28 to 40

30 to 424to 9

28 to 355 to 9

25

rc

-

Base ofrail to

:enter piel

ft . in.-__

14 015 912 816 0

13 0

14 “09 0

10 814 0

13 3

1 : 96 6

,” 0

YL

c !

!

Dia. oflisc ininches

27”

2 :

24

212 0

:8

:8

:7”121216

-

1

1013930775560

537450440440

350340330315

313227205265200

For a very wide bridge the center-bearing type may become imprac-ticable. As the entire weight while swinging is carried on the cross-girders to the center pivot, these girders may become so heavy as tomake their construction impracticable. Center-bearing bridges are


adapted for single-track structures of any span, and have proven satis-factory for double-track through bridges up to 400 ft. and for a four-trackthrough bridge of 235 ft. total span. For principal dimensions, etc., ofvarious built swing bridges of this type see Table 56.

ART. 3. RIM-BEARING SWING BRIDGE

While swinging, this bridge rotates around a center pivot but iscarried entirely by a concentric ring of conical rollers which rollbetween an upper and a lower track. The lower track is secured to thepivot pier while the upper track is usually at tached to a circulargirder (drum) which supports the superstructure. When the bridgeis closed the rollers carry also the live load at the pivot pier. The centerpivot receives no dead or live load.

The most important requirement for a rim-bearing swing bridge is tohave the load equally distributed over all the rollers, as otherwise somerollers will receive more load than others, the drum with the upper trackwill deflect between the points of support, and the bridge will turn hard,with a resulting wearing out of the rollers and track. To obtain an equaldistribution, it is necessary to have practically no deflection in the uppertrack between the points of support. To effect this thepoints of sup-port should be close together and the depth of the drum relatively great.The depth of the drum should not be smaller ,than x%, preferably 3, ofthe distance between the centers of support. Many rim-bearing swingbridges for single track have been built with only four points of supporton the drum. This is inadequate, they should have at least eight points,and double-track bridges pr.eferably more. The distribution should beprecise, and should be arranged so that it will not be affected by the deflec-tion of the distributing girders.

The usual size of rollers is from 18 t,o 30 in. diameter. Experienceshows that as good results are obtained with 1%in. as with 30-in. rollers;as the frictional resistance of the rollers alone is only a small fraction ofthe total resistance, the diameter of the rollers does not appreciably affectthe power required to turn the bridge. Smaller rollers distribute theload better and are lighter. Good practiee requires the use of as manyrollers as can be placed on a circular track. The width of the rollers isdetermined by the permissible unit stress per lin. in. depending upqn thekind of material used for the track and rollers. Generally the axis of therollers is horizontal, so that the upper as well as the lower track are coned.In this case the least distortion of the upper track makes it impossible.

ART. 41 MOVABLE BRIDGES AND TURNTABLES 2 7 9

for the coned wheels to run true. The result is a distortion and rackingstrain upon the radial arms and the guiding center. This has beenavoided in severa1 cases by inclining the axis of the rollers so that thetop line is horizontal and al1 the coning is put in the lower track seg-ments; then a distortion of the bridge cannot affect the perfect actionof the wheels.

The live-ring arrangement for separating and hording the rollers, con-sisting of adjustable radial rods, one end of each carrying a roller, theother end being connected to a ring revolving around the center pivot, isan unmechanical contrivance. The rollers should run between two con-centric circular girders firmly connected to each other; this ring shouldbe connected to the revolving part of the pivot by rigid struts.

The lower circular track should be connected to the pivot with radialstruts so that the track and pivot form one piece which can be fitted to-gether and centered accurately in the shop. In order to prevent the centerpivot from working loose, some engineers have embedded its base in con-crete. This method may keep the pivot in position, but it does-not neces-sarily prevent the pushing, pulling and wearing out around the collarof the pivot.

It is sometimes of advantage to carry a part ($- or less) of the weighton the pivot while the bridge is swinging (combined rim- and center-bear-ing), and this increases the stability of the pivot. The arrangement ofdistribution should be such that the portion of the load which is to becarried by the pivot can be determined accurately.

The rim-bearing bridge generally requires a circular girder or drum(for very heavy spans a double drum) of expensive construction, a ring ofaccurately turned rollers and circular tracks, which require great care intheir construction and delicate adjustment in their erection, as otherwisethe center pivot will work loose. Repairs are expensive, and any irregu-lar settlement of the masonry will throw therturning apparatus out oforder.

The longest rim-bearing swing bridge is the 521-ft. double-track spanover the Willamette River at St. Johns, Ore.

ART. 4. TURNING DEVICE

The usual arrangement for turning swing bridges consists of a rackattached to the lower circular track or roller path and a vertical shaft witha pinion geared into the rack. This pinion shaft is carried by a bracketattached to the drum or rotating part of the bridge. By turning the shaft*


and pinion, the bridge is swung around its center. These brackets carry-ing the pinion shaft should be amply proportioned.

Wire ropes parbuckled around the live ring and operated by hy-draulic machinery have been used successfully for a swing bridge overthe Harlem River, New York City, at Second Avenue.

ART. 6. END LIFT

Al1 swing bridges require an arrangement to lift the ends when closed,so as to make the span continuous over three or four supports, or to maketwo separate simple spans. Various contrivances are used for this pur-pose, such as rollers, screws, cams, eccentrics, toggle joints, wedges,hydraulic rams, etc. The lifting apparatus should fulfill the followingrequirements :

1. The end lift should have power enough to lift the ends to thedesired leve1 with nearly uniform resistance.

2. After the ends are lifted to their final position, they should formsubstantial supports, similar to the end shoes of a fixed span.

That kind of end lift which supports the ends of the bridge, whenciosed, on rollers, toggle joints, or links’is not to be recommended forrailroad bridges. Wedges with a bevel of 1 to 5 or 1 to 6 are themost satisfactory. The mechanism for moving the wedges can be ar-ranged so as to offer nearly uniform resistance in al1 stages of the lifting,and also to lock the wedges to prevent them from sliding backward.Toggle joints and hydraulic rams have also proven satisfactory, if sup-plemented by wedges or other substantial supports after the lifting isaccomplished.

In light center-bearing bridges of moderate span and with unequalarms, it is permissible to have the lifting apparatus under one end only.This end should have an excess of weight, so that, when the end supportsare removed, the bridge will tilt and throw some weight on the balancewheels. The opposite ends of the girders simply rest on bed plates. Thisdispenses with one-half of the lifting machinery.

Such tilting arrangements may also be used in case the lifting isdone in the center by a hydraulic ram under the pivot.

Some swing bridges are designed with sufficient lift at the ends tomake the trusses discontinuous, so that they will form two separate simplespans when the bridge is closed. The operation of this requires approxi-mately ten times as much work as if the ends were lifted only enough toprevent the lifting of the ends from their supports under traffic and their

PLATE XxX111

Center - Bearing Swing Bridge&- -

Single Track

ti- 8 Pan& cá, 22’6” = 180 ’ -+!

Fig. A. Fig. B.

k . . . . ..__.._....._.._.--.-......... **o ’ __- ______.____.__......-.----- ).l

Rim - Bearing B r i d q e s .S w i n q

x-k 6 Pan& @ 25’= 150’ 3117 iík __.__..________________ ---..--- _____ 3/7’ __-._-.- _____ - ____-_... i ---- +/

Fig.J.

Typical Trusses of Railroad Swing Bridges. (Faciny page 280)


consequent hammering. This arrangement, therefore, is a disadvantagein a bridge which has to be opened and closed frequently and quicltly;but may be recommended in cases where the bridge is used almost con-stantly as a fixed span, and where the time required for the opening isof no importance. Lifting the bridge from the center is not as econom-ical as lifting at the ends, on account of the greater power required.A few bridges have been built with a lifting apparatus applied at thetop chord of the center panel, which has the additional disadvantage ofusing a part of the truss which carries stress while the bridge is swinging,also as a part of the lifting mechanism. If the machinery parts getout of order, the structure should not be affected thereby.

ART. 6. TYPES AND PRINCIPAL DIMENSIONS OF GIRDERS AND TRUSSESOF SWING BRIDGES

The following sypes of main girders or trusses are preferably usedfor different span lengths:

For railroad For highwaybridges bridges

Plate girders. . . . . . . . . . . . . . . . . . . . . Up to 160 ft. up to 100 ft.Riveted trusses with parallel chords. . 160 to 200 ft. 100 to 200 ft.Riveted trusses with polygonal chords. 200 ft. upward 200 ft. upwardPin-connected trusses with polygonal 300 ft. upward 300 ft. upward

chords

Since the weight of highway bridges varies considerably, their limitsare given merely as a general guide for ordinary cases.

The distance between girders or trusses and the height of thegirders or trusses in a swing span are approximately the same as in asimple span of the same capacity and of a length equal to twicethe length of the longer nrm of the swing span (see page 277).Plate girders are limited to 10 ft. depth over al1 on account oftransportation.

Plate XxX111 shows some typical trusses of swing bridges.The top chord of the center pane is generahy made horizonta1 to

avoid a long and heavy center post; sometimes it is inclined, whichimproves the appearance and relieves the inclined middle posts of stressat the expense of the center post.

Por long and heavy riveted trusses, the top chord of the center panel,being a tension member, is generally made of eyebars.

Por trusses, about 45 ft. high or more, it is better to arrange themiddle inclined post and also the adjoining diagonal over two panels


and to use stiff subdiagonals to the bottom chord. The latter stiffenthe truss better than tension subdiagonals running to the top chord.

In heavier bridges it is preferable to use a double center post corre-sponding to the double cross girder, the two parts of the post beingrigidly connected by angle latticing.

ART. 7. ARRANGEMENT OF FLOOR AND CENTER CROSS GIRDER

Plate XXXIV shows different arrangements of the center cross girderand tloor system in center-bearing swing spans. In short or lightspans, especially where, the distance from base of rail to top of masonryis limited, the center cross girder is framed into the trusses (Figs. A, B).In long or heavy spans there is as a rule not enough room for the con-nection of the center cross girder to the center post and the truss has,thereforg to rest on top of the cross girder (Pig. C, etc.). The bestarrangement is shown in Figs. D and F, which, however, requires agreat height from base of rail to top of masonry. For the distancefrom base of rail to masonry in other bridges see Table 56.

In short center-bearing swing spans the center cross girder usu-ally rests with its lower flange directly on the center pivot; it iseither a single web or a box girder. In longer spans the centercross girder consists of two separate plate girders connected nearthe center by two longitudinal or “bolt girders” framed into them.The bolt girders which are of the box shape are supported by anumber of heavy bolts which pass through them and are suspendedfrom crossbeams (usually 1-beams) which rest on the center pivot.This “suspension type” of center bearing has the advantage in thatit makes the bridge more stable while swinging, as the point ofsupport is placed as high as practicable; further, it allows a betteradjustment and the center pivot is easier accessible and exchange-able. Unless the two center cross girders are framed into thetrusses they are connected to each other also below each truss bya longitudinal ‘(wedge girder” framed into them. The wedge girder,usually of the box shape, rests directly on the center wedge bear-ings when the bridge is closed and has therefore to transmit theful1 center wedge reaction to the cross girders.

ART. 8. CALCTJLATION OF STRESSES IN SWING BRIDGES

Main girders and trusses of swing bridges are generally calcu-lated for the following loading conditions:


Dead load:(a) Bridge open.(b) Bridge closed, both ends raised by a certain uplifting forte.

Live load:(c) Whole span acting as a continuous beam over three supports.(d) One arm acting as a simple span.

Conditions (u) and (c), or (CA) and (d), or (b) and (c), respectively,may occur at the same time and, therefore, the corresponding stressesin any member have to be combined. so as to give the maximum orminimum stresses in that member.

Conditions (a) and (c) can occur simultaneously when the bridge isclosed and the end wedges driven only so far as to touch the end sup-ports. In this case, however, loads covering one arm only should notbe considered, since the far end would be lifted, thus eliminating thecontinuity of the span; the assumption which is often made that thatend will be held down by a concentrated load is not justified.

Some specifications require the members to be proportioned for rever-sals of stress in immediate succession; others for any possible reversals;and still others for the greatest stress only.

It is practically impossible to maintain in a swing span the exactconditions for which the stresses were calculated; further, the practice ofcombining alternate stresses and of proportioning the members, etc.,varies between such wide limits that any superfine calculation of stressesin swing spans is not justified. In all the following calculations, there-fore, we neglect two factors which would, to a slight degree, affect thetheoretical stresses, that is, the variable moment of inertia of the trussesor girders, and the influente of change of temperature.


Dead load stresses for condition (a), bridge open, are best found ana-lytically, or by means of a Maxwell diagram.

The uplifting forte for condition (b) has to be assumed so that whenthe live fload covers one arm the end of the other arm will not be liftedfrom its support. It is safe to assume the uplifting forte at the end ofthe short arm equal to the maximum negative reaction at that end dueto the live load covering the long arm, plus the corresponding impact.Instead of wheel loads, it is sufficiently accurate to assume an equivalentuniform load which would cause in the long arm as a simple spanthe same maximum bending moment as the wheel loads. If W is the

284 DESIGN OF STEEL BRIDGES [CHAP. XlV

total uniform live load on the long arm of the continuous span ACB(Fig. l), the negative reaction at the end of the short arm is

1Rb=- l l

8 i (; + 1)W

, Adding to this the impact corresponding to the assumed length of loadedtrack we get the uplifting forte B. The uplifting forte at A is then

For equal arms II = 1, Rb = -$ W and the uplifting forte at each

end is A = B = Rb, plus impact.

Tf -....-________ ‘L ____ -------+ ________-

FIG. 1.

Sometimes the uplifting forte is assumed equal to the end reactionfrom dead load, the span being considered continuous over three supports.This reaction can be obtained by the formulas in Table 57, which ísself-explanatory.

If now for any member:D, = dead load stress for condition (a) (span swinging),D, = “ “ i‘ “ “ (b) (ends raised by uplifting

forte),S1 = stress caused by the sole application of a downward load unity

at A, and since the stress from the sole application of the uplifting forteA at A is -ASI, we have the relation

D, = Do-AXI, by means of which the stress D, can be conven-iently determined.

The stresses X1 are easily found analytically, or by means of a Max-well diagram. They will be used again for the determination of the liveload stresses and end deflections.


When the live load consists of wheel concentrations, the simplestmethod to determine the stresses for condition (c) (continuous span overthree supports) is that of the influente lines. Even for a uniform live


TABLE 67

Loading,Shear-and Moment Diagrams Reactions, Shears, Moments, Deflections

Continuous beams over severa1 supports

2 0

Special Case: Z =11

A = P (l-2+$ ; C=P; ( 3 - s )

pos Mma, = A, at P;neg NI,,, = M, = A Z - P ( Z - a )

c=$ z+zl+qg( 1 > ;M,=AZ-pg=negM

If Z < II pos Mm,, = -2P

for 21 = p

neg M,,, = M, = Al - p f

SDecial Case: Z =Z1

A=B=+; C=&Z,pos Mmez=

.@f, = -“s pz2

PA=2 (

11-~~ >

12; B= -pszl(z+zl)

; neg M,,, = M, = BEI

pos Mm,, = - ‘A2P for x= ?,

Soecial Case: Z =ZI

7 P 5A=í6P; B=-i6; C=sP

PI 49Mc=-,,; po~Mmaz=~ PZ for X=:6 z

286286 DESIGN OF STEEL BRIDGESDESIGN OF STEEL BRIDGES [CHAP. XIV[CHAP. XIV

TABLE 67 (Continued)TABLE 67 (Continued)

11

--

Loading,Shear-and Moment Diagrams Reactions, Shears, Moments, Deflections

A m,I‘

A B

.._._..... j< ._.........___ 1 , ----- ..-.11

ss

B = -p 4= -ca8111(Z+L)

pecial Case: 1 =ZI

;

pecial Case: 1 =II

A=PgJ

B= -pgt

A=D=0,4 pl; B=C=l.l pl

neg Mv,,, = Mb= iW,=-0.1 pP

pos M,,, = 0.08 pP between A and B‘I “ = 0.025 ~12 between B and C

i IIv - w

--

If 1x7 = 4 + 8; t-3;

00

A=p [ +(;-2;“) $1

tt EI=; [q+(“-ls) ((5iY;+2

,,C =; [(T-;) (3+2+2$]

P a a3D=K I lsC---J

--

1II11


load the influente Enes are of advantage as they show clearly the loadinglimits for extreme stresses; in this case a sketch of the influente lineswill often suffice.

The following simple method shows the construction of the influenteEnes for a continuous span over three supports, giving at the same timethe influente lines for one arm as a simple span, and therefore, with littleadditional work, the stresses for condition (d).

Assume the continuous span ACB (Fig. 2) with arms 1 and 11 loadedanywhere with the single load P.

FIG. 2. FIG. 3.

This load will cause an interna1 bending moment .iVc at C. By a sec-tion just to the right of C, arm AC is made a simple span, but in orderto preserve equilibrium an externa1 moment M, equal to the interna1bending moment has to be added.

Any stress in arm AC of the continuous span ACB can, therefore, beregarded as composed of the stresses caused separately by load P andmoment M, at C in the simple span AG.

The sole application of M, causes a downward reaction at A equal toMC- -

1 ’ which is the only outside forte acting on the part of the span

between A and any section X (see Fig. 3). The stress in any member

cut by X is therefore equal to - 7 SI, where SI = stress caused

in that member by a downward load un& at A; therefore ifS, = stress in any member of the continuous span due to load P,Sa = l‘ .‘ ‘C i‘ l‘ “ simple span AC “ “ “ ‘i

we have the relation

s, = s o -

or, dividing by SI Sc So Mox, = z l . . . . . . . . . . . . . . . . . . . . . . .

2 8 8 DESIGN OF STEEL BRIDGES [CHAP. XIV

For load P anywhere on arm CB, the stresses X0 in arm AC arezero and equation (1) becomes for members in arm AC:

Sc Mc-=--x1 1

P- l

FIG. 4.

For P = 1 (load unity) the following notations shall be used:

SC-=Sl

yc = ordinate of influente line for continuous span.

ISO-=Sl yo

= ordinate of influente line for simple span.

MC MC~-l--

x = ordinate of influente line for -.1

Equation (1) becomes nowye = yo - 2

Fra. 5.

that is, the injluence area of a stress in the continuous span is epual tothe influente urea of the corresponding -stress in the simple span minus

the in$uence area for F.

Example.-In Fig. 4 the horizontal line A’C’B’ represents the base line and ’ lineA’I C’II B’ the influente line for 7 with ordinates z. If triangle A’XC’A repre-

sents the positive influente area for a stress in simple span AC, then the areashown hatched is the influente area for the corresponding stress in the continuousspan ACB.

ART. lo] MOVABLE BRIDGES AND TURNTABLES 289

The center moment MC can easily be found by the theorem of three momentsand dividing by Z the following formulas are found for the ordinates z:

M, lc-k3“=-1=-2(nfr) for unepual arms 1 and l1 = nl

Mc k--k3x=--=--1 4

for equal arms 1

where k = F (see Fig. 5).

For spans with equal arms and equal panels the ordinates x are tabu-lated in Table 58.

TABLE 68

Influente Ordinates for End Reaction z = 7 for Swing Spans With Three Supports

and Equal Arms

N u m b e r ofmnels inene ann

Influente o r d i n a t e s z at p a n e l p o i n t s

7 0.03500.06560.087 50.09620.08750.0568...... . . . . . .

9. . . .

;;;l;;

0.555610 5188

I I I I 1 I \ , I

For truss spans and for girder spans with floorbeams the influenteline x is a polygon with its corners under the panel points; thelines between panel points being straight.

For trusses with secondary verticals (Fig. S), the polygon has cor-ners under the main verticals only.

The influente lines yo for the simple span can easily be found asexplained on page 49. The influente area yo of the reaction RA at Aof a simple span AC due to a load P= 1 traveling over the span isa triangle A’C’A”A’ (Fig. 6) with A’A” = 1 (load unity).

A load P = 1 to the right of m of any panel nm causes a stress Soin any member cut by section X through panel nm equal to

So = RA XS~. Dividing by SI, we get

SO-- = yo = RASI


which means that the portion of the influente area yo for the stress SOto the right of m is identical with the iníluence area for RA. Withone part C’m’ thus g iven each influente line for simple span AC isdetermined.

A load P placed on the span will produce in any member the stresses:

S, = Py, X1 for continuous span,SO = Py, SI for simple span,

and a series of loads P will cause the stresses:

FIG. 6.

S, = S1 =iPy, for continuous span,S, = ISI LPy, for simple span.

S1 is the in$uence coeficient for that member.Figs. 7 b to 7 g show the influente lines for various moments

and stresses in swing span ACB (Fig. 7 u). For conveniente thebroken line A’C’ B’ is assumed as base line. The influente areas forcontinuous span are shown hatched.

Fig. 6. End Reaction RAInfluente area for simple span = A A’C’A”A’

Intluence coefficient SI = 1

Fig. c. Moment Mz and Top Chord Stress U1-3Influente are8 for simple span = A A’C’B’A’Influente coefhcient for Mz, SI = +z

ART. lo] MOVABLE BRIDGES RND TURNTABLES 2 9 1

Shear 2-3

(f.1

IB’

W- -3

3 ’( b.1

FIG. 7.


Fig. d. Center Moment M, and Top Chord Stress Uh-5Influente area for simple span = 0Influente coefficient for M,, X1 = 1

LL Ii IC u5-5, x1 = -;

Fig. e. Shear in Panel 2-3Influente area for simple span = A’ 2’ 3’ C’A’Influente coefficient X1 = 1

Fig. f. Stress in Diagonal LZUBInfluente area for simple span = A’ 2’ 3’ C’A’

Influente coeflicient SI = $i

where x1 = distance of intersection I of chords from A anddl = distance of I from line L,Us.

If it is not convenient to find the intersection point,s I and I’

line A’2’ can be found by figuring the ordinate C’C” = $ where

h, and h, are given iti Fig. 7 u, or the intersection point ‘Il andvertically under it point 1’1 may be constructed as shown in Fig. 7 asnd f and line 3’11’2’ drawn first and then 2’A’. L i n e C”A’,however, gives point 2’ more accurately.

Fig. 7 g. Center Reaction R,Influente area for simple spans = a A’C’B’A’

1+11Influente c o e f f i c i e n t X1 = I, ( = 2 f o r 1’1 = Z)

Position of Load for Maximum Stress.-In order to get the maximumfor a stress, the live load is placed in such a position on the span asto make ZP a maximum. This position is in general best found bytrial.

A divided train load is generally not considered, for instance, in Fig.7 e the maximum tension in diagonal LzU3 is found either by cover-ing with the load A’Iz only, or arm CB only, or the whole span,deducting .in the latter case LPy, for part I, C.

Stresses for Uniform Live Load.-If the live load consists of a uniformload w per foot of truss, or if for the wheel loads an equivalent uni-form load w is substituted, then the stresses can be found by measur-ing the influente areas and multiplying them by wXS~. For instance,for top chord UlU3 (Fig. 7 a and c) we have

ART. l l ] MOVABLE BRIDGES AND TURNTABLES 293

(1) for simple span: influente area = +(Z -x); influente coeffi(cient

Sr= -a, therefore s t ress S, = +(1-x )z W.

(2) for continuous span arm AC loaded : influente area =+ (E-x) - Z x X when the ordinates x are those below the panel

points of arm AC and X the corresponding average length of the twoadjoining panels. (For equal panels Zx X = X Zx.) The s t ress istherefore

s, = - [+ (z-x)-m] Ew = s o -gw LZX.

(3) for continuous span arm CB loaded: influente area = 2 z X

(z and X for arm CB ), therefore s t ress S, = + 2 w L: xx.

For equal arms and panels, Lx for one arm is given in the las tcolumn of Table 58.

ART. ll. LOAD ON CENTER PIVOT AND CENTER WEDGES

The center pivot has to carry the whole swinging dead load when theb r i d g e is o p e n or closed. The center wedges under the trusses aredriven only so far as to provide a support for the trusses, but not tolift the latter. Consequently, they carry the whole live load center re-

p __...._ a .______ 1’ P

a ------IA ‘3

F.-------------- b

4

A- - - - - - - - - - - - - ~_. . . . . . .._..___ b

i._____ . . ..__~

RVI RC LRWFIG. 8.

action of the trusses. That part of the live load, however, which istransmitted from the stringers to the center cross girder is partlycarried by the center pivot and partly by the wedges.

If in Fig. 8 P denotes the load transmitted by the stringers to thecross girder, the reactions on the center pivot and the center wedge arerespectively

R, = P f (3 -$) and R, = P - i R,

The influente area for the center wedge reaction (see Fig. 7 g, page 291)has, therefore, to be decreased by triangle 5’C”5” by making

294 DESIGN OF STEEL BRIDGES [ CHAP. XIV


For the calculation of the stresses in the lateral bracing and chordsthe following conditions of loading should be considered:

(1) Wind on unloaded bridge while swinging.(2) Wind on unloaded bridge while closed.(3) Wind on loaded bridge while closed.

Condition (1) will usually cause maximum stresses in the laterals andportals near the center pier and may affect the chord sections near thecenter pier of long single-track spans when combined with the deadload for bridge open.

For conditions (2) and (3) the lateral trusses may be assumed con-tinuous over three supports and the reactions determined by theformulas given in Table 57. Condition (2) will affect only the endportals and the laterals near the ends in the system between the unloadedchords, while condition (3) affects most laterals between the loaded chordsand in long single-track spans possibly the chord sections near thecenter pier when combined with the dead and live load for bridge closed.In condition (3) part of the wind load is moving and for each memberthat position of load should be taken which causes maximum stressin that member. This position and the stress itself can easily befound by the influente line as explained for the live load or the posi-tion may be determined approximately by a rough sketch of the in-fluente line, and the stress obtained by calculating the reactions bymeans of the formulas in Table 57, using only full panel concentrations,that is, assuming the load limit always at the center of a panel.

ART. 13. EXAMPLE FOR THE CALCULATIONS OF STRESSES

The dead load stresses and influente Enes for the 324-ft. double-trackswing span shown on Plate XXXV shall be determined. (For thestress sheet see Plate XXXVIII, for General Detail Drawing PlateXLII, for Details of Machinery Plate XLIII,)

Dead Load Stresses.-The trusses carry an assumed dead load of5800 lb. per lin. ft. of bridge or 2900 X 27 = 78,300 lb. per panel pertruss and to provide for the end lift machinery the end panel load isassumed at 46,000 lb. The dead load stresses Do for condition (a),bridge open, are determined by a Maxwell diagram (b) by starting theresolution of the forces at Lo (A in Fig. a). As the truss is symmet-rica1 only one-half of the diagram is needed. In order to avoid inac-

PLATE XXXV

Influente Lines for Center-Bearing Swing Span.

( scok: II’= 10õ’j

( S t r e s s e s gh? h Units o f JOOOlbs.)

Influente Lines for Center Bearing Swing Span(Facing p a g e 2 0 4 )


curacies the stress in U~US has been calculated analytically by takingmoments of all loads on one arm about LS (C in Fig. CL) and dividingby the lever arm as follows:

(46,O X 6 + 5 X 78,3 x 3) 27 : 50 = + 783,000

For a Iive Ioad of Cooper’s E-50 the equivalent uniform Ioad for themoment in a span of 162 ft. is 6200 lb. per ft. per track. The negativereaction due to this live load covering one arm is therefore

6,2 X 16216 = 62,000 lb.

If we add for impact

62. 300,

3oo + 2 x 162 = 30,000 lb.

we get for the uplifting forte at each end of each truss 92,000 lb. Thestresses SI due to a downward forte unity at each end are obtainedby a Maxwell diagram (c); the stress X1 in UsUs must be

1621 5. = 3.24

The dead load stresses D, for condition (b), ends raised by the up-lifting forte of 92,000, are now conveniently found as follows:

Member Stress 8,

-0.84 - 77-1.98 -182-2.70 -248

+1.50+2.41+3.24

tr;;;f274

+1.31-0.98+O.QO

+120- 90+ 83

-0.72+o.ô9-1.14

92.S1

- 66

"103

-

-

-

-

Dead load stress

- 39-246-548

+ 38

-30;:

;tE4+7s3

- 10

:508

+ 61 - 59-136 - 46+218 +135

-258+335-498

-1925;;"5

Ends raisedD, = Do - 92 SI

Influente Enes.-From Table 58 we get the ordinates x for thepanel points of the left arm from left to right and for the right armfrom right to left as follows: .0405, .0741, .0938, .0926 and .0638 and


plot these downward from the base line A’C’B’ (d and e), the ordi-nate A”A’ being made equal to unity. The influente lines for all chordmembers and the end post A U1 can now be drawn at once as shownby Fig. 7 c; all these lines are combined in Fig. d.

The influente Enes of the web members are combined in Fig. e;since the chord members U~UG and L& are parallel the influenteline for U& is identical with that for shear in panel L&. For the in-fluente lines of the web members in panels 1 to 5 we produce top chordU1U5 to its intersections with the verticals through A and 6; theseintersection points are 27.5 and 54.5 ft. respectively above the bottom

chord. The ratio54.527 = 1.98 is the ordinate C’C” and the influente

lines for the web members can now be drawn as shown by Fig. 7 f.Center Wedge Reaction.-Let us assume that a live load unity per

track on the center cross girder causes a reaction of 0.3 at the centerwedge. This reduces the influente ordinate for the center wedge reac-

tion (Fig. cl) at the center to C’C”= f 0.3 = 0,.15,

ART. 14. CALCULATION OF DEFLECTIONS

The vertical downward deflection 6 of the end A of the swingingspan from dead load is, according to the principie of virtual velocities:

where SI = Stress in that member from a downward load unity at Atherefore equal to the influente coeficient (see page 39and 297, etc.),

LDoAL = z = distortion of any member,

L = Length of member in in.,a = Gross area of member in sq. in.,E = Modulus of elasticity (30,000,OOO lb. per sq. in. for steel),

Do = Stress in member from dead load for bridge swinging,in Ib., therefore,

Correspondingly, the upward deflection from the end lifting forte Ais equal to

since ASI = stress due to forte A.

h3T. 141 MOVABLE BRIDGES AND TURNTABLES ,297

The part 6 - 61 of the deflection 6 is eliminated by a correspondingshortening in the geometric length of the center top chords, so that theend supports are brought to their proper geometric elevation when thebridge is closed.

Table 59 shows this applied to the 324-ft. riveted swing span on.Plate XXXV.

In pin-connected trusses, the pin play has to be considered in calculat-ing the deflection.

If E denotes the elongation or reduction in length of the memberon account of pin play (= difference of diameter of hole and diameter ofpin) the downward deflection at end is

6 = Z E & or if the pin play is the same for all pins 6 = E Z X1.6 has to be given the same sign as the dead load stress.

The upward deflection from forte A on account of pin play is6 = 2 Z E X1 but here only those members are considered in which thedead load stress is reversed under the action of the uplifting forte A.Z: E S1 is doubled because in reversing the stress the movement of thepin is 2 6. E is generally assumed to be between $3 and 20 in.

TABLE 59

Deflections of 324-ft. Swing Span

Melll. Lin ft.

ansq. in.

ll-3 54.75 82.3u3-5 54.75 96.3U5-6 27.0 100.0

LO-224:;

68.8 - 39 -0.84484.0 -246 - 1 . 9 882.2 -548 -2.70

LO-U1Ul-L2L2-u3U3-L4L4-u5Uá-Lô

54.054.054.0

37.137.944.344.350.051.3

91.252.957.979.798.B

105.2

Idl l

Ll

!-

D =oad i r300 lh

$32+784

1

-

llb: 4at Lo

+3.24

1-1.31-0.985+O.QOl-0.7235;: 14'

-

ILLa

128540686

3::971

66150104117277

-

Deflections duo todead load

S,LD 12000TX-E-

in.

jo.560 = 9,‘16

10.542 = 17/32

i(0.322 = 5/16

/1

1.424 = 1 13/32

1.503.302.83

0.56

4%

0.700.690.62

0340.63

.L

Deflections due to100,000 lb. at LaSlZL 1200000ãXE

in.

10.316

:0.326

!, 0.131

10.773in.

To develop uplift of 85,000 lb. at Lo, Lo must be raised 0.85 X 0.773 in. = 0.657 in. = # in.Bars U5-6 must be shortened

on acct. of defleotion (1.424 - 0.657),$ = 0.237 in. = $ in;

on aoot. o f pin play, & in., to ta l = * in.Note.-For lengths of web members distances o. to c. of rivet groups are used.Chord mem are increased ta allow for effect of gusset and splioe platea.


ART. 16. POWER REQTJIRED TO OPERATE MOVABLE BRIDGES

The determination of this power is a matter of practica1 experienceand judgment. Experiments have been made to ascertain the fric-tional resistance; but, as the conditions are not the same, the frictionalresistance varying in different bridges even of the same type, and as themachinery sometimes rusts up, a wide margin for the required powershould be allowed. This adds only a small percentage to the totalcost, does no t increase the operat ing expenses, and increases the re-liability of the machinery.

The resistances to be overcome in tuyning a swing bridge are:1. Resistance due to friction;2. Resistance due to the inertia of the bridge;3. Resistance due to the action of the wind.

Resistance due to Friction.-For rim-bearing swing bridges, the ratioof the total frictional resistance in the live ring to the load on the rollerswas found by C. Shaler Smith to vary from 0.004 to 0.008, by Messrs.Boller and Schumacher as 0.0035 for the Thames River Bridge, and byTheodore Cooper as 0.0038 for the Second Avenue Bridge. For center-bearing bridges C. Shaler Smith found the frictional resistance at thecircumference of the pivot to be 0.09 of the weight turned. C. C.Schneider found the coefficient of frictional resistance at the circumfer-ence of the pivot, on hardened-steel and phosphor-bronze discs, with theusual working pressure of about 3000 lb. per sq. in., to be 0.067 at thestart, and 0.045 to keep the bridge moving at a uniform speed. Forthe total frictional resistance, including that of the shafts and gearingrequired for hand-operation, the highest coefficients observed on newbridges were 0.115 for the starting and 0.08 for keeping the bridge inmotion. For bridges which have been in operation for some time, thesecoefficients would be smaller.

Resistance due to Inertia of Bridge.-The power required to over-come the inertia of the swinging mass and develop the desired velocitydepends upon the time allowed for opening or closing. It is the usualpractice to assume that the acceleration is developed in half the time ittakes to open the bridge. With this assumption, the following formulaswill give sufficiently correct results:

WP’ = 20.5 t2’


P = Forte applied to the center of gyration necessary to pro-duce the required acceleration;

HP = Number of horse-powers;W = Total moving weight in pounds;

1 = Length uf bridge in feet;b = W i d t h o f b r i d g e ‘( “

p = Radius of gyration = b2+lZzi-

__12 ’ approximately 0.29 1;

t = Time in seconds in which maximum velocity must be ob-tained, or one-half the number of seconds required to openthe bridge;

v = Maximum linear velocity at the center of gyration, in feet

per second = g$

Resistance due to Action of Wind.-Some engineers assume anunbalanced wind pressure of 4 or 5 lb. per sq. ft. to act entirely onone arm of the bridge at right angles, and follow the bridge in its variouspositions while swinging, which is an arbitrary assumption. It has beenobserved, however, that with a strong wind it may require nearly twice ’the power to turn the bridge than without wind. For swing bridges withtwo equal arms the resistance due to the action of the wind may, there-fore, be included in the coefficient of friction. If the bridge has arms ofunequal length the resistance due to wind pressure should be calculatedand included in the total resistance.

For rim-bearing bridges the total resistance to be overcome in turn-ing, including frictional resistance of the gearing and wind pressure, maybe assumed about twice the value given by C. Shaler Smith for frictionalresistance, therefore 0.015 of the load on the rollers acting in the centerline of the track; and for center-bearing bridges ,0.15pivot when swinging, acting on the circumferenceestimating the capacity of the motor, he assumes athe center line of the track, of twice that required forin the specified time, or expressed by formulas:

P = 0.015 W for rim-bearing;

P = 0.15 W $ for center-bearing;

of the load on theof the pivot. Inlinear velocity, inturning the bridge

P vHP = sso;

RTv = 2t;

21


where P = Forte required to turn the bridge, acting in center line oftrack;

W = Total moving weight in pounds;HP = Number of horse-powers;

T = Radius of pivot in feet;R = Radius of center line of track in feet;v = Maximum linear velocity in center line of track, in feet

per second;t = Time in seconds in which maximum velocity must be

obtained, or one-half the number of seconds required toopen the bridge.

The foregoing formulas for frictional resistance generally give resultsin excess of the power actually needed under ordinary conditions. Inestimating approximately the power required to operate swing bridges,1 h. p. may be assumed for each 15 tons of swinging weight.

If the bridge is to be operated by hand-power, the gearing has tobe arranged either for the time allowed to operate the bridge or for thegreatest possible number of men. The work which can be performedby an average man on the hand-lever for any considerable time is about40 lb. with a speed of 200 ft. per minute. In calculating the strengthof the gearing, etc., however, the power of one man should be taken at125 lb., as this is about the forte a strong man with a foothold canexert for a short time.

If a swing bridge is run by motor, pulling out the wedges and pushingthem home can be done in 10 seconds. Turning the bridge 90” can beaccomplished in one minute if necessary. For a hand-turning draw,say of 200-ft. span, single track, the same operations would take twominutes and three minutes respectively.

In bascule and lift bridges the wind pressure is an important factorin increasing the frictional resistance of the machinery, but the pressuremay be assumed smaller than that specified for the structural parts ofthe bridge, as navigation would be impossible in such a case. 20 lb.per sq. ft,, corresponding to a velocity of 70 miles per hour, is sufficient.

ART. 16. MOTORS

Steam and hydraulic power were in former years the only motivepowers used for operating movable bridges. At present electric andinterna1 combustion motors have replaced the steam engine.

Electric motors are the most convenient where electric power can be

ART. 171 MOVABLE BRIDGES AND TURNTABLFS 301

obtained satisfactorily, more particularly if the company owning thebridge also controls the power (electric railway bridges, steam railroadbridges situated near a power-house, or City bridges). Electric powerl iseconomical, the motors occupy little space and can be put in the mostconvenient places very close to the point where the power must beused, so as to reduce the transmission by long shafts.

Where electric power cannot be obtained, a gasoline motor is thebest. Gasoline motors up to 80 h. p. have been ked successfully foroperating movable bridges.

Hydraulic motors can be used to advantage in exceptional casesonly, where the bridge is operated from the shore and where waterwith sufficient pressure can be obtained. Hydraulic pawer is generallyused with hydraulic rams in connection with accumulators. They havethe advantage that a great amount’of power can be accumulated witha small motor during the time the bridge is not in operation.

ART. 17. WEIGHTS OF MOVABLE BRIDGES

(See also Table 56)

Swing Spans.-The steel weights of swing spans vary considerably,depending upon the type (rim-bearing, center-bearing or combinationof both) and upon the machinery used for turning.

For the weight of the floor system, trusses or girders and bracing,the same rules apply as for simple spans (see page 221). The weight ofthe trusses of a swing span is from 60 to 70% of the weight ofsimple trusses of the same total length, while the weight of bracing andfloor system is approximately the same as in a simple span. Assumingthe weight of trusses of a simple span about $ of the total steel weight,the weight of the trusses, bracing and floor system of a swing span wouldbe about 70 to 80 y0 of the total weight of a simple span of the samelength. This weight may be assumed uniformly distributed: over thewhole length.

The center cross girders, wedge and bolt girders, bolts, machinerysupports, etc., of center-bearing swing spans weigh approximately 10 y0of the weight of the trusses, bracing and floor system. This weightis mainly at the center and is to be considered only for the calculationof the center girders, pivot, etc.

The machinery of center-bearing swing spans may be assumed at

1 Amperes X Volte = Watts; 1 English h. p. (550 ft.-lb. = 76 mkg. per sec.) =746 Watts; 1 metric h. p. (75 mkg. per sec.) = 736 Watts; t.hereforr, approximately,

number of Wattsnumber of horse-power = $j PP*1000


15% of the weight of the structural part. About 50% of the machinery(rack, track, pivot, bearings, etc.) rests directly on the masomy anddoes not affect the calculation of the stresses and deflection, about 20%should be assumed concentrated at the two ends of the span and 30% atthe machinery supports near the center.

The total steel weight (including machinery) of a center-bearing swingspan is, therefore, approximately 90 to 100% of the weight of a simplespan of the same length.

The weight of distributing girders, drum, radial girders, machinerysupports, etc., of rim-bearing swing spans is about 15% of the weightof the trusses, bracing and floor system and the weight of the machineryis 10 to 15% of the weight of the structural part.

The engine house floor and its supports weigh about 25 to 50 lb. persq. ft.

STEEL WEIGHTS OF SOME EXISTING MOVABLE BRIDGES(The item Machinery includes steel and other metal)

134-ft. S. T. Deck Plate Girder Swing Span (Center-bearing)6 ft. 6 in. c. to c. girders, E-40 loading, A. C. L. R. R. Spec., 1902.Main girders and bracing.. . . 142,900 lb.Center girders and machinery supports. . . . 11,600Machinery (hand-turning) . . . . . . . . . . . . . . , . . 15 ,600

Total . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170,lOOlb.

172-ft. S. T. Through Plate Girder Swing Span (Center-bearing)Main girders, floor system and bracing. . . . . . . . . . . . . . ‘319,900 lb.Center girders and machinery supports . . . . . . . . . . . . . . . 25 ,700

Total structural part. . . . . . . . . . . . . . . . . . . . . . . . . . 345 ,600Machinery (hand-turning). . . . . . . . . . . . . . . . . . . . . . . . 63 ,600

T o t a l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409,200Ib.

200-ft. S. T. Through Swing Span (Center-bearing, Riveted Trusses) Built 1907Specifications G. N. Ry. Similar ín design to 250-ft. span given below

S t r u c t u r a l s t e e l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353,300Ib.Machinery (hand-turning) . . . . . . . . . . 63 ,500

__-T o t a l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416,SOOlb.

260-ft. S. T. Through Swing Span (Center-bearing, Riveted Trusses) Built 1907lSpecifications G. N. Ry. about E-50 loading

T r u s s e s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308,500Ib.B r a c i n g :53 ,300Floorbeams and stringers. 129,400C e n t e r g i r d e r s a n d m a c h i n e r y s u p p o r t s . 45 ,700Engine-house floor . 17,200

Total structural steel. 554,100Machinery (hand- and power-turning) . 72 ,200

Total. 626,300 lb.

1 See Plates XXXVII, XL and XLI.

ART. 171 MOVABLE BRIDGES AND TURNTABLES

327-ft. D. T. Through Swing Span (Center-bearing, Riveted Trussesf)

Specifications P. R. R. 1904. Loading E-55 followed by 5000 lb.

T r u s s e s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 952,000 lb.B r a c i n g . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ’ 1 0 3 , 2 0 0F l o o r s y s t e m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Center girders and machinery supports. .

518,700

E n g i n e h o u s e f l o o r a n d s u p p o r t s .166,000

54,200

303

Total structural. . . . . . . . . . . . . . . . . . . . . . . . . . . . .Machinery

1,794,100. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232,800

Total.. . . . . . . . . . . . . . 2,026,900 lb.

267-ft. S. T. Through Swing Span (Rim-bearing, Riveted Trusses)

Specifications C. R 1. & P. Ry. 1903

T r u s s e s . . . . . . . . . . . . . . . . . 279,000 lb.B r a c i n g . . . . . . . . . . . . . . . . . . 46 ,500Floor s y s t e m . . . . . . . . . . . . . . . . . . 122,500Center distributing girders . . . . . . . . . . . . . . . . . 28 ,000Drum and machinery supports. . . . . . . . . . . . . . . . . . 46 ,500E n g i n e f l o o r . . . . . . . . . . . . . . . . . . 6 ,000

T o t a l s t r u c t u r a l .Machinery .J,

Total. . ,

................ 528,500

. . . . . . . . . . . . . . . . 53,100__-

. . . . . . . . . . . . . . . . 581,600 lb.

220-ft. D. T. Through Swing Span (Rim-bearing, Riveted Trusses)Specifications W. 111. R. R. 1903. Loading E-50

Trusses .B r a c i n g . . . . . . . . . . . . . . . . . . . . . . .F l o o r s y s t e m .Center distributing girders.Drum and machinery supports.

Total structural. . . . . . . . . . . .Machinery . . . . . . . . . . . . . . . .

. . . . . . . . 482,600 lb.

..,.........,.. 82 ,700

..,,....<<..... 279 ,200

. . . . . . . . . . . . . . . 90;OO0. . . . . . . . . . . . . . . . 72 ,400

................ 1,006,900

. . . . . . . . . . . . . . . . 86 ,000

Total.. : 1,092,900 Ib.

260-ft. D. T. Through Swing Span (Rim- and Center-bearing, Riveted Trusses)Specifications N. Y., N. H. & H. R. R. 1900

Trusses, floor and bracing.. . . 974,500 lb.Center distributing girders. . . . . . 98,000Drum and machinery supports. . 148,000Operator’s house.. . . . , . . 22 ,500

- -Total structural. . . 1,243,OOO

M a c h i n e r y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189,000--__

Total .’ 1,432,OOO lb.

1See Plates XXXVIII, XLII and XLIII.

3 0 4 DESIGN OF STEEL BRIDGES [CHAP. XIV

lOO-ft. D. T. Scherzer Rolling Lift Bridge (Riveted Trusses)

Spccifications N. P., N. H. & H. R. R. 1901 revised. Operated by two 50-h. p.Westinghouse motors (Fig. 9)

L i f t s p a n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Track span.

776,700 lb.

Machin~ry.....................................228,200

O p e r a t i n g h o u s e .21 ,100

R a c k s a n d rack s u p p o r t s .27 ,800

G r i l l a g e s a n d anchar b o l t s . .54,00027,100

_ _ _ _ _Total. 1,134,900 lb.

Areas given in sq. in. per trussFIG. 9.

146-ft. D. T. Scherzer Rolling Lift Bridge (Riveted Trusses)

Specifications N. & W. 1903. Loading about E-50. Operated by two 50-h. p.Westinghouse motors (Fig. 10)

L i f t s p a n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,067,6001b.T r a c k s p a n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .M a c h i n e r y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

326,200

Operating house. .30,900

Racks and rack supports.. . . . .8,400

Grillages and anchar bolts. .I. .68,40080 ,300

T o t a l . 1,581,800 lb.

Areas giwn in sq. in. per trussFIG. 10.

ART. 181 MOVABLE BRIDGES AND TURNTABLES 3 0 5

64-ft. D. T. Scherzer Rolling Lift Bridge (Four Plate Girders)

Specifications N. Y., N. H.‘& H. R. R. 1901 revised. Operated by two 30-h. p.gasoline engines (Fig. ll)

Lift span (including 564,000 lb. cast iron counter-weight blocks). . . . . . . . 1,395,200 lb.

Trackspan.....................................M a c h i n e r y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

210,300

Operating house.. . .81,700

Grillages and anchar bolts. .84,200

Chain supports. . .537,100

Bolsters, etc.. . . . . . .55,60019,200

T o t a l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,883,3001b.

Areas given in sq. in. per truss

FIG. ll.

ART. 18. LOCOMOTIVE TURNTABLES

Calculation of Stresses.-Figs. 12 and 13 explain the action of a turn-table for balanced and unbalanced loads respectively.

The center pivot C supports longitudinal girders DE, these supportcross girders D and E which in turn support .the main girders. Themain girders are supported at one end only for an unbalanced load.

.ri

(Balanced Load.) ( Unbalanced load.)

FIG. 12. Fm. 13.

For the calculation of the stresses in the main girders, the followingconditions.of loading have to be considered:

(1) Dead load (balahced).(2) Live load (balanced).(3) Live load (unbalanced).

306 DESIGN OF STEEL BRIDGES [ C H A P . XIV

For maximum stresses conditions (1) and (2), or (1) and (3), have tobe combined.

Maximum Moments in Main Girders.-In arder to obtain the re-quired flange areas, vertical sections about 5 ft. apart, or at floorbeamconnections for through turntables, are assumed; the bending momentsfor all sections are calculated and plotted from a horizontal line AIB(Fig. 14) to represent the curves of moments. Curve 1 represents thatfrom the dead load and is obtained by assuming AD and EB as canti-levers. Curve 2 represents the maximum negative moments from liveload (condition 2). To obtain ‘it the engine is placed on the span sothat the resultant of all wheel loads goes through C, the moments arethen calculated at the wheels by assuming AD and EB as cantilevers andplotted (Fig. 14, left side). The moment curve thus obtained gives

F IG . 14 .

the moments at points 1, 2, 3, etc., where it intersects the verticalsthrough these points.

The maximum positive moments from live load (condition 3) areobtained by considering AC or BC (not AD or EB) a simple span andplacing the load on this span only. Curve 3 is the moment curve for thiscondition.

Finally the curve for maximum moments (shown full) is found byadding the ordinates of curve 1 to those of curve 2 or deducting themfrom those of curve 3, whichever gives the greater result.

Shear at End of Main Girder.-The dead load shear at A or B ispractically zero. The greatest live load shear at A occurs when AC isconsidered a simple span with heaviest loads at and near A.

Shear at Center of Main Girder.-Fig. 15 shows the condition ofloading for maximum live load shear at a section immediately to the


right of E. The heavy loads are placed at and near E, but so that re-action B is positive and reaction A zero, that is, the resultant of allloads must be to the right of C. The shear at E is then found asfoilows: take moments of al1 wheel loads about B and divide by 1(not I’), t,his gives the reaction 6; deducting therefrom the loads tothe left of C gives the shear at E. The dead load shear is equal tothe weight of span between E and B.

Tender Engtiz _j

AAAiBk ________ ______ , __._._.__.. ._____: ________ 1 _-.-----.- -

FIC. 15.

Maximum Reaction R at Center Pivot C.-This occurs with condi-tions (1) and (2), is therefore equal to the total swinging weight of theturntable plus weight of pivot plus the total weight of the live load.

Longitudinal Girders DE (Fig. 16)

Maximum Shear = 2 .

Maximum Moment = 2 “2” per girder, where RI = R

minus weight of pivot, assuming two longitudinal girders.Cross Girders D and E (Figs. 16 and 17)

R2Maximum Shear = 4

Rz b - aMaximum Moment = x 2 per girder, where R2 = RI

minus weight of longitudinal girders, assuming two cross girders.

FIG. 16.

Weights of Turntables.-The following table gives the total steelweights of standard turntables for various lengths and loadings. Thethrough turntables have no floor system, the ties resting on shelfangles riveted to the main girders.

Permissible units: tension 10,000 lb. per sq. in,, shear on rivetsand web gross section 7500 lb. per sq. in. No impact considered.

308

Lengthc. tro c.

LoE%. . .E-50....E-60....

DESIGN OF STEEL BRIDGES [CHAP. XIV

65 ft.

50 ,10053 ,700. . . . . .

1

Deck turntables

70 ft. 80 ft.

54,70056,300 60,200 66,20061,600 66,600 72,000

l

Through turntables

70 ft.‘I I

75 ft. 80 ft.

73,400 76,400ii;&i ’ 80,300 81,100-

These weights include the machinery, which amounts to about10,500 lb. for deck and 9000 lb. for through turntables.

For an interesting review of turntable design see papers of the Ameritan Rail-way Bridge and Building Association, 1912; also Eng. News, Dec. 5, 1912.


ADDITIONAL INFORMATION ON SWING BRIDGES

Interstate Bridge, Omaha, Neb., rim-bearing, 1600 tons turningweight, cl. tr. and highway, 520 ft. long, Eng. News, Dec., 1893.

Dutchkills Bridge, L. 1. R. R., center-bearing, 550 tons turningweight on bronze disc 20 in. dia., 3 tracks, 195 ft. long, trusses 41 ft.c. to c. (widest center-bearing span with two trusses) weight of metal450 tons, built 1894.

Alton Bridge, Ill., rim-bearing, 1100 tons turning weight, d. tr.,453 ft. long, Eng. News, June 1894.

Third Ave. Bridge over Harlem River, N. Y., rim-bearing withdouble drum, 2500 tons turning weight, 300 ft. long, 88 ft. wide, 4trusses, Railroad Gaxette, Sept., 1893, and May, 1895, Eng. News, Nov.,1895. (First double drum ever used.)

N. Y. C. & H. R. R. R., Harlem River, N. Y., rim-bearing withdouble drum, 2500 tons turning weight, 4 tracks, 389 ft. long, 3 trusses,26 ft. clear, ballasted floor, Railroad Gaxette, Aug., 1893, Eng. News,June, 1893, and April, 1896, Eng. Record, Jan., 1896.

The Design of Movable Bridges, Eng. News, Nov., 1896. .iC. R. 1. & P. Ry., Mississippi, rim-bearing, 2 decks (highway and

d. tr.), 366 ft. long, Eng. News, Dec., 1896.C. & N. R. R., Drainage Canal, rim-bearing, d. tr., 474 ft. long,

Eng. Record, Oct., 1897, Eng. News, Dec., 1897, Railroad Gaxette, Dec.,1897.

Duluth Bridge, rim-bearing, 2000 tons turning weight, highwayand d: tr., 491 ft. long, Railroad Gaxette, March, 1897, Eng. Record,Aug., 1897, and June, 1900.

N. Y. C. & H. R. R. R., Hudson River, Albany, combined rim-and center-bearing (bronze disc 15 in. dia.), d . t r . , r iveted trusses,270 ft. long, with three 25-h. p. electric motors, Railroad Gaxette, NOV.,1899.

P. R. R., Delaware River, center-bearing, 1000 tons turning weighton bronze disc 27 in. dia., d. tr., 323 ft. long, weight of metal 930 tons,steam engine 50 h. p., Eng. Record, Nov., 1899. Paul L. Woelfel inProceedings of Engineers’ Club of Phila., 1897. F. C. Kunz in” Allgemeine Bauzeitung,” Vienna, 1900 (with many plates).

Rock Island Bridge, Mississippi, rim-bearing, s. tr. 438 ft. long,Eng. News, Jan., 1900.

Willis, Ave. Bridge over Harlem River: N. Y., rim-bearing, 66 ft.wide, 304 ft. long, Eng. Record, Nov., 1900.


Charlestown Bridge, Boston, rim-bearing, 1200 tons turning weight,2 decks (highway 100 ft. wide on lower de& and 2 El. R. R. trackson upper deck), 240 ft. long (widest swing span), Eng. News, Jan ,,1901.

Marquette Bridge, combined rim- and center-bearing, s. tr., 360ft. long, Eng. Record, May, 1901.

C. & W. Indiana R. R., Drainage Canal, rim-bearing, 1660 tonsturning weight, 4 tracks, 334 ft. long, weight of metal 1350 tons, Eng.News, Sept., 1901.

Interstate Bridge, new span, Omaha, Neb., rim-bearing, withdouble drum, 2200 tons turning weight, d. tr. and highway, 520 ,ft.long, Eng. News, Jan,, 1903, Eng. Record, Jan., 1903.

Lincoln Bush, Passaic River Bridge, D. L. & W. R. R., center-bearing, 1000 tons turning weight, 2 decks, 4 tracks, 220 ft. long,floated into place, floating weight 2000 tons, Eng. News, Dec., 1903. li

Fraser River Bridge, Canada, rim-bearing, 2 decks (highway ands. tr.), 380 ft. iong, Eng. Record, April and May, 1904, l&g. News,June, 1905.

C. H. Cartlidge, Notes on the Design of Swing Bridges, Eng.Recod, May, 1906.

Kingsbridge, N. Y., Harlem Canal, center-bearing, 1500 tons turn-ing weight on bronze disc 30 in. dia., 2 decks (highway 54 ft. wideon lower and 3 Rapid Transit tracks on upper de&), 270 ft. long,trusses 39 ft. c. to c., weight of metal 1200 tons, two 35-h. p. electricmotors, floated into place, floating weight 2000 tons, Eng. Record,Aug. and Sept., 1906.

New York & Long Island R. R., Raritan River, center-bearing,1400 tons turning weight on bronze disc 33 in. dia., d. tr., 327 ft. long,weight of metal 1013 tons, Railroad Gaxette, May, 1908.

Portland & Seattle Ry., Willamette River, rim-bearing, 2700 tonsturning weight, weight of metal 2400 tons, d. tr., 521 ft. long (longestswing span), built 1908.

J. E. Greiner, Specifications for Movable Bridges, 1911.L. Schaller, Comparison of Movable Bridges, Papers German Soc.

C. E., Berlin, 1911.Louis H. Shoemaker, Conneaut Bridge, Ohio, B. & L. E. R. R.,

center-bearing, 1400 tons turning weight on bronze disc 34 in. dia., 4R. R. tracks, 235 ft. long, 4 riveted trusses (widest center-bearingspan), weight of metal 1100 tons, built 1910, Trans. A. S. C. E., Vol.75, 1912 (novel method of center support).


E. I. Schneider, Dumbarton Bridge, Central California Ry.‘, cen-ter-bearing, 1300 tons turning weight on bronze disc 36 in. dia., d. tr.,310 ft. long, Trans. A. S. C. E., 1913.

Northern Pacific Ry., St. Louis Bay, rim-bearing, 2200 tons turn-ing weight, d. tr., 479 ft. long, weight of metal 2000 tons, built 1912.

Kentucky & Indiana Terminal R. R., Ohio River, center-bear-ing, 2200 tons turning weight on bronze disc 42 in. dia., d. tr., andhighway, 400 ft. long, weight of metal 1600 tons, built 1912.

Sacramento River Bridge, Cal., Southern Pacific Ry., center-bear-ing, 3000 tons weight on bronze disc 52 in. dia., 2 decks (highway 28ft. wide with concrete floor on upper deck, d. tr. R. R. on lower deck),390 ft. long, heaviest swing span, weight of metal 2400 tons, built 1912.

CHAPTER XV

A R C H B R I D G E S

A. CALCULATION OF STRESSES

ART. 1. GENERAL

The arch differs from the beam in that it exerts inclined pres-sures on the bearings while those of the beam are vertical. The inclinedreaction is most conveniently represented by a vertical and a hori-zontal component. The horizontal component may be resisted by a tieconnecting the two bearings, the arch with the tie acting as a beam asfar as the reactions are concerned, the stresses, however, are determineda s in an arch. This system has been frequently used in Europe forbridges, but in this country only for large roofs, as railroad terminals,assembly halls, etc.

Combining the inclined reactions with the externa1 forces we obtainthe equilibrium polygon. If there are hinges in the arch this polygonmust pass through the hinges as the bending moment at the center of ahinge is zero. According to the number of.hinges, we have the followingcases :

The arch with three hinges (at center and at bearings), the two-hinged arch (hinges at bearings) and the hingeless arch or arch withfixed ends. The last type is rarely used and will not be treated here.

ART. 2. DEAD AND LIVE LOAD STRESSES IN THE THREE-HINGED ARCH

The arch with -three hinges is statically determinate since, for anygiven loading, the equilibrium polygon can be drawn so that it passesthrough the three hinges, thus determining the reactions. The stressescaused by the changes of temperature and siight settlements of thefoundations are secondary stresses and can in general be neglected.

Influente Line for Horizontal Reactions.-For a single verticalload P to the left of C (Fig. l), the equilibrium polygon consists of twostraight lines BCD and AD. The reactions and their components areobtained by resolving P as shown in Fig. 1 b. From this follows thatthe two horizontal reactions H at A and B are equal but of oppositedirection, as they are the only horizontal forces and must be in equilib-

312

Am. 21 ARCH BRIDGES 313

rium. From the condition that the vertical forces must be in equilib-rium follows that the vertical reactions V, and Vb are the same as fora simple beam AB:

The condition that the bending moment at C must be zero gives

Hf= Vbb = P;b, and

xbH=PW~......................

FIG.. 1.

For P= 1, we getx b

H=z=~

(b.)

which represents a straight influente line A1C1 for H between A andC (Fig. 1 c) with the ordinate

ab2, = íf at C.

By reversing, it follows that the influente line to the right of C is astraight line CIB1.

I f n=b=;, H=P$1- -or 2. - tj and xc = Q

A~B,CI is the influente area for H and is equal to $* The horizon-

314 DESIGN OF STEE¿ BRIDGES [CHAP. XV

tal reaction for a uniform unit load p covering the whole span is,therefore,

abH = pf ; for a = b = %,

12H = pgf . . . . . . . . . . .

As the influente area for the bending moment of a simple beam isalso a triangle, it-follows that the condition for maximum H due to aseries of concentrated loads is the same as given on page 46 and 52 formaximum moment at C, and also that an equivalent uniform load cor-responding to that moment can be used for determining H.

The influente lines for the vertical reactions are the same as for asimple span AB (see page 40). That for V,, for instance, is shownin Fig. 1 cl, its ordinate at A is equal to load unity.

p ------ c( ._____ -.+----- b _.__ >f If the two bearings A and B (Fig. 2)

m

are not on a horizontal line, all deduc-tions and formulas apply, if for H the

$c-:- -component of the reaction in the line AB

__--.____ t __-____ ._.__ y is taken and for f the normal distance of

FIG. 2.C from this line.

The forte transmitted through thehinge C is determined by the direction of the equilibrium polygon atthat point and its vertical component is equal to the shear at C con-sidering AB a simple span, and its component parallel to AB is equalto H.

Influente Lines for Truss Members.-The stress in any member ofthe arch truss due to a certain loading is the algebraic sum of the stressin that member due to that loading, the arch being considered as asimple span, and the stress caused by the sole action of the horizontalreaction H.~ The influente area for that member is, therefore, the geo-metric sum of the influente area.for simple span and the influente areadue to H. The influente area due to H is a triangle with apex at C,since the stress caused in the member by H varies in linear proportionwith H.

In Fig. 3 b, for instance, triangle Al 3’B1 represents the positive in-fluente area for chord member 2-4, the truss being considered a simplespan AB and triangle AlClB1 is the negative influente area dueto H. The resulting influente area is shown shaded; it is positivewhere Al 3’Bl overlaps AlClB1 and negative where AlClB, overlapsA 13’B1. A load P= 1 at C causes a horizontal reaction of

abH= <f

1(= q if the arch is symmetrical)


j I /’i 1 /’

1’3- ,/’

FIG. 3.

22

316 DESIGN OF STEEL BRIDGES [CHAP. XV

and the ordinate at C of triangle A1CJ?1 (Fig. 3 b) is equal toab Ythe stress caused in member 2-4 by this reaction or equal to - - -.Ifd

The influente area A13’B1 is found as shown on page 90 for simplespans except that the ordinat,e A A1 2 is not made equal to the loadunity but equal to the stress caused in the member by an upwardload unity at A, the t&s being assumed Gxed at B, since the ordinatesof triangle AICIB1 also represent the correct or unreduced stress dueto H, in other words, the influente ordinates are shown to the correctsize and need not be multiplied by an influente coefficient.

As a check we may find point K1 of zero stress as follows:Connect A (Fig. 3 a) with center of moments 3 and produce this line

to its intersection K wit,h line BC. Kl (Fig. 3 b) then lies vertically below.K because a single load at K produces reactions of the directionsAK and BK and since AK passes through point 3 the moment atthis point and, therefore, the resulting stress in 2-4 is zero. The lineB,3’ can, therefore, also be drawn by using point K1 without calculat-ing the ordinate A1A2. It is, however, always advisable to check theinfluente line in some way. Point 3’ may also be found by calculating

the ordinate B1B2 = 2 (d being the lever arm of 2-4 to point 3) which

follows by reversing the two sides of the truss; this ordinate BIB, beingthe stress in 2-4 caused by a vertical upward load unity at B, the trussbeing assumed fixed at A.

,Fig. 3 c shows the construction of the influente area for diagonal 3-4.The center of moments is I,-, (intersection of chords 3-5 and 2-4which are cut simultaneously with 3-4 by a section through the truss),

A horizontal forte H = @ produces in 3-4 a stressw

$ i (e being the

lever arm of 3-4 to point 13-4) which is the height of triangle AICIBl,the influente area due to H. The influente area for simple span is

Al 3’4’Bl; it is obtained by calculating the ordinates AlA = : and

B1B2 = 27 which are the stresses in 34 caused by an Upward load

unity at A and B respectively. Point Az locates line B14’ and point Bzline Al 3’ finally 3’ is connected with 4’. The point Ll of zero stress isvertically below L which is the intersection of lines AIS-4 and BC sincea load at L produces no stress in 3-4. The intersection point 11 of linesBIAz and A1B2 lies below the center of moments 13-4 similarly as formember 2-4. The difference between these two influente areas is thedesired influente area (shown shaded).

ART. 21 ARCH BRIDGES 317

In a similar way the influente area for vertical 4-5 is found as shownin Fig. 3 d. It has to be noted that intersection point M1 of A1C1 andB1A2 is found in a similar way as Ll for diagonal 3-4. MI, however,is not the point of zero stress, since M is located to the right of C; thereaction af d due to a load at M therefore passes through C and notthrough 14-5 (Fig. 3 a). It is assumed that the load is applied at the topchord; if it were applied at the bottom chord, we would have to drawline 5’C’ instead of 3’-4’, that is, the intermediate line has to be drawn

’ in that panel in which the loaded chord is cut by a section cuttingsimultaneously the vertical.

It is evident that for any member the intersection point of BICIand A1Ba could be found instead of the intersection point A1C1 andB1A2; it must be vertically below the intersection of line AC with theline connecting B with the center of moments of that member.

From the above follows that any influente line can be found andchecked in various ways.

As a rule the following method is the best:1. Determine the stresses Sh .in al1 members due to a horizontal

forte

abH=ls

This is most conveniently done by a Maxwell diagram (if the truss issymmetrical only one-half needs to be drawn). These stresses give theheights of the triangles AICIBI.

2. Determine the stresses S, in al1 members due to a vertical up-ward forte V, = 1 at A, the truss being assumed fixed at B. This isbest done by another Maxwell diagram starting the resolution of stressesat A and ending at B. These stresses are the ordinates A1A3 locat-ing lines B1A2.

3. Determine in a similar way the stresses Sb in all members dueto a vertical upward forte Vb = 1 at B. These stresses are the ordi-nates B1B2 locating Enes AIBz. A separate diagram for the stresses Sbis not necessary if the arch is symmetrical, since the stress Sb in anymember to the left of C is equal to the stress S, in the correspondingmember to the right of C.

4. As a check see that the intersection of AIB and B1A2 lies ver-tically below the center of moments of the member.

5. If any of the ordinates becoine very great or the center of momentis inconveniently located (which is generally the case for the web mem-

318 DESIGN OF STEEL BRIDGES [CHAP.XV

bers if the top chord is also curved) find as explained above either ofthe intersection points of A1C1 with BlA2 or of BICI with AIB,.

It has to be remembered that if there are many members the Maxwelldiagrams often become inaccurate. To avoid this it is advisable to checkanalytically the last stress obtained by the diagram.

In order to determine the sign of the influente area, it is necessary togive the proper signs to the stresses Sh, S, and. Sb. If S, has the samesign as Sh, then these ordinates must be laid off on different sides of thebase line A1B1 since the areas must be added; if they have oppositesigns they must be plotted on the same side. Correspondingly Sb isplotted on the opposite or on the same side as Sh if its sign is the sameor opposite to that of Sh (see page 315).

If there are many members the method can be somewhat simplifiedby making the height of triangle AICIB, the same for al1 members and

abequal to - and reducing the other ordinates in the same proportion.

uSevera1 influente lines can then be shown in one figure. The sum ofordinates has finally to be multiplied by a certain influente coefficientin order to obtain the stress. This method is especially convenient andis explained more in detail for the two-hinged arch (see page 323).

Dead Load Stresses.-The dead load stresses could be found by meansof the influente lines. The following method, however, is usually more

convenient. As a rule, the dead load can be assumed uniform and thehorizontal reaction is then calculated according to formula (2), page 314.The vertical reactions are found as for a simple span. With thesereactions the stresses in the truss members are determined by means ofa Maxwell diagram as shown in Fig. 4 for one-half of the arch. If thetruss is not symmetrical, the diagram must be drawn for the wholetruss. The dead load is here assumed applied at the top chord; thestresses in the verticals have, therefore, to be corrected by the bottomchord panel load, which is added algebraically as tension.


If the dead load is not uniform, formula (l), page 313 is applied foreach panel concentration and the results are added to obtain the totalhorizontal reaction H.

If the dead load is uniform and the chord containing the center hinge(usually the bottom chord) is parabolic, the horizontal component of thestress in any member of this chord is equal to the horizontal reaction.The stresses in the other members are zero, except that the verticalscarry the t,op chord panel loads in compression.

Live Load Stresses.-The live load stresses are most convenientlydetermined by means of the influente lines as these show in what mannerthe load has to be placed to produce maximum strésses. For the de-termination of the stresses from the influente Enes see page 314.

ART. 3. DEAD AND LIVE LOAD STRESSES IN THE TWO-HINGED ARCH

Reactions.-The two-hinged arch is externally statically indeter-minate in the first degree, that is! the calculation of the reactions

requires one more condition besides the equations of equilibrium (seepage 32).

Any vertical load P (Fig. 5 a) causes inclined reactions R, and Rbwhich may be resolved into vertical components V, and Vb andcomponents H in the line connecting the two hinges. As this line isusually horizontal, H will simply be called the horizontal reaction.

320 DESIGN OF STEEL BRIDGES [CHAP. xv

From the condition of equilibrium follows that the vertical reac-tions are found as for a simple span (Fig. 5 a).

The two horizontal reactions H must be equal and opposite. Theyare determined by the condition that when the arch deflects its lengthbetween the hinges does not change (or in case of an arch with a tie,the change of length is equal to the elongation of the tie).

If one end would be free to move longitudinally, the arch truss wouldbe a simple span (Fig. 5 b). Any vertical loading would produce a longi-tudinal deflection of the free end of

61= 2 TA- = PZ g$ . . . . . . . . . . . . . . . .(P&> SlL (1)where for any member

XO = stress caused by the vertical load unity (P = 1) if the arch isconsidered as a simple span (Fig. 5 b).

SI= stress caused by the sole application of a horizontal forteunity acting in the line of the horizontal reaction but inopposite direction to the latter (Fig. 5 c).

L = length of member, A = the area of its gross section andE = modulus of elasticity.

Correspondingly, a horizontal forte H applied at the free end of thetruss causes a horizontal deflection at this point of

csz= -pyp~,=H~~~ . . . . . . . . . . . . . . . . . . (2)

Equating these values for 61 and & and assuming E constant for allmembers we find the horizontal reaction produced by the vertical load P

H = P . . . . . . . . . . . . . . .

To be correct the summations. should extend over al1 truss members;-_-.- .____- _- ..-... “, __.. .-it is, however, permissible to consider only the chord members, sincei. . -.~-.- _ -.. ._ -_ - .the mfluence of the web members on H is very small, especially forarches with a curved top chord. This does not apply to the calculationof the deflections on which the web members may have a considerableinfluente.

A R T . 31 ARCH BRIDGES 321

Influente Line for Horizontal Reaction H.-The influente line f%r Hcould be found by placing successively a vertical load unity at the dif-ferent panel pojnts and applying for each position of this load (P = l),formula (3). This method is simple but the work can be considerablyshortened as follows:

MThe stress SO in any member is equal to 7 where M = bending

moment at the center of moments of that member due to the verticalload unity, the ar’ch truss being considered as a simple span, and r =lever arm of that’ member. The numerator of formula (3) can there-fore be written as

Now suppose the arch to be a simple span Ioaded with a vertical load

FIG. 6.

unity at m (Fig. 6 u). For any chord member U the value Mw would be

(1 p X) w, where x: is the distance of the center of moments I from the

bleft support. According to Maxwell’s theorem 7 xw is also equal

to the bending moment produced at m by a vertical load w at I. Thesum Z Mw is therefore equal to the bending moment M, at m pro-duced by al1 loads w placed at the respective centers of moments ofthe chord members; from this follows that the moment polygon due tothe loads w is the influente line for the value M, = IZ Mw. The hori-zontal reaction due to a load unity at a point m is then

H = !&. . . . . . . . . . . . . . . . . . . . . . . . (4)

3 2 2 DESIGN OF STEEL BRIDGES [CHAP. XV

where the summation being extended over all chord

members. The influente line of H is obtained by dividing the ordi-

nates of the moment polygon due to the loads w by zSlz g = Zv.

The moment polygon due to the loads w may be found either analytic-ally, as shown in example on page 325, or graphically by a forteand equilibrium polygon. If the pole distance of the forte polygonis D (Fig. 6 c) and the ordinates of the equilibrium polygon are x

(Fig. 6 b), we get M, = Dx and H=k!E=DzLv 2%

If we make D = Zv we get H = x, tha t is, the equilibrium polygonis then the influente line for H.

The values v and w are conveniently found by means of a table asshown on page 337 and can be calculated sufficiently accurate with theslide rule. The dimensions r (lever arm) and L (length of member) can

be scaled in an elevation of the arch truss. The stresses S1 = “, from

a horizontal forte unity may be found analytically or by a Maxwelldiagram or preferably by both, since they will be used again for findingthe stresses in the truss members from vertical loads and from tem-perature changes. Al1 dimensions must be given in the same units orthe end result must be corrected if different units (for instance sq. in.for A and, feet for r, L and y) are used; except that for finding H fromvertical forces, A may be given in any unit since it appears in both thenumerator and denominator of the formula for H. In order not to gettoo large figures for w and v, it is even advisable to divide all areas bya convenient figure or, if many chord sections are alike, by t,he area oftha t sec t ion so tha t A = 1 for those members.

No attention needs to be given to the signs, since the values w andv have the same sign for al1 chord mekbers.

If the truss is symmetrical, the calculation extends over one-half ofthe truss only, as the influente line for H is then also symmetrical. I tshould not be forgotten, however, to multiply Zv by 2 if the summa-tion has been extended over one-half truss only. Concerning the Max-well diagram for S1 see page 325.

This general method of finding the influente line f?r H can in certaincases be considerably simplified, as shown on page 328.

Influente Lines for Truss Members.-With the influente line forthe horizontal reaction H determine& the influente line for the stress

ART. 31 ARCH BRIDGES 3 2 3

in any member of the arch truss is easily found in a similar way as shownfor the three-hinged arch (see page 315)‘. The stress in any member isthe algebraic sum of the stress if the arch is considered as a simple span‘and the stress caused by the horizontal reaction H. The influente arreafor that member is, therefore, the geometric sum of the influente area forsimple span and the influente area due to H. Since the stress due toH in any member changes in direct proportion with H the influente ordi-nates for that stress are obtained by multiplying the influente ordinatesfor H by a coefficient X1 which is equal to the stress in that membercaused by a horizontal forte H = 1. The influente po lygons thusobtained replace the triangles AICIB1 for the three-hinged arch (Fig. 3,page 315) but the influente lines for simple span are exactly the same.

This method has the advantage that the influente ordinates for allmembers are shown to the same scale and give directly the stress due toa load unity. It can, however, be considerably shortened in the follow-ing manner : :

Instead of multiplying the ordinates of the influente Ene, for H byS1, we use this influente polygon for al1 members but divide the influ-ence ordinates for simple span by X1. Any stress obtained in this man-ner has to be multiplied finally by the corresponding influente co-

_ efficient !S; (page 325). :If for any member (for instance chord 7J Fig. 7 a):

y = distance of center of moments from line AB,x = LL Ii ‘l (( ic ” vertical through A,x,= ll LL <c <l l‘ Ll ‘l “

B,r = lever arm (distance of member from its center of moments),

w e have 81 = 1 F = Influente coefficient; the ordinate AlA is

ob ta ined as the s t ress caused by an upward forte uni ty at A ( thetruss being fixed at B) divided by SI, or

AIA2 = f : +- = +

and correspondinglyx’ . Y 2’BJi?z = 7 . y- = y

Fig. 7 shows the influente Enes for various members obtained bythis method. The polygon AICB1 is the influente line for H. LinesAIB and B1A2 determine the influente line for simple span, in allcases the intersection point of lines A1B2 and B1A2 must lie verticallybelow the center of moments of the member. The area shown shaded

3 2 4 DESIGN OF STEEL BRIDGES [CHAP. XV

is the reduced influente area of the stress for the member, its ordinatesbeing the influente ordinates for a load unity. (In Fig. 7 d the dotted

lineA13’4’B1 would be the influente line if the load were applthe bottom chord instead of at the top chord.)

( b.1

( d.1

i j Influente Coefficient:

Vertical V.Influente coefficient:

S,=k

,h - -Diagonal E.

FIQ. ‘7.

.ied a t


The inlluence coefficients X1 = F can be determined with the slide rule

or by a Maxwell diagram. The values i for the chords and those web

members for which the centers of moments I are conveniently locatedare best determined with the slide rule, x and y being scaled off inthe elevation of the truss; if I lies outside of the drawing, which is the

case for web members if the chords are nearly parallel, f may be found

as follows:If U and L (Fig. 8) are the chord members, draw any two verticals

ab and cd, not too near each other (truss verticals may be used to

41,“/

4 j

c I/‘,/

U I

P

//a

1’ /’

,! LJd1

,‘,/ b 1/

‘/ Y:i

//i//

I I

A#------’ - _ _ _ _ _ _ _ _ _ -ii+------- . .._...... XI ti. - . . . . .ii------ ---.-------.-.------ --.............._......~..-. x ______________________ _ ___.__________.._. J

F IG . 8 .

advantage), connect hinge A with a and b, then draw parallels to theseconnecting lines through c and d respectively and scale off the ordinates

x1 and yl of the intersection point I’. i is equal toX1E as can easily be

proved geometrically.If many of the points 1 are outside of the drawing, it is more conven-

ient to find, by means of a Maxwell diagram, the stresses due to anupward forte LLunity” at A (the truss being assumed fixed at B). These

stresses are the values 5 and, dividing them by the influente coefficient

SI = r we obtain t. Correspondingly we. find the values $ by a

forte diagram from an upward forte ‘<unity” a t B. I f the t russ isX ’

symmetrical, only one diagram is necessary, since the stress T in aX

member is equal to.the stress ; in the symmetrically located member

and vice versa.

326 DESIGN OF STEEL BRIDGEi [CHAP. xv

Concerning the Maxwell diagrams see page 325.The signs of the stresses caused by the forces V, or Vb and H are

best determined by inspection. If the forte V, or Vh (Fig. 7) causesa stress in any member of opposite sign to that from H, then the

ordinat,e must be laid off on the same side as the ordinates H. If it

has the same sign, it is laid off on the opposite side.Assuming the ordinates H laid off downward from line AlB1, the

ordinates AlA, and BlB2 for a certain member have to be laid off asshown in Fig. 9, whereby the directions given in that one of the sixspaces apply, in which the center of moments of that member is located.

For all chord members, for instance, t must be laid off on the same

side as H, since their centers of moments are in the space between andabove AB.

By this method severa1 influente Enes can be shown in one figure; forinstance, those for the chord members (Fig. 7 b) in one and those for the

A, A,: dOWfl do wnB, B,: dZr7 down

n _,<------------ , ---------,

I “JJ

“ ‘A,,, H

B- ’

A,A,: down UP8, B2: up “’ up “ , ufJdown

FIG. 9.

web members in another (Fig. 7 c and d), so that the intluence polygon forH needs to be drawn only twice. It may happen, however, that forcertain members, for instance, the web members near the center of archeswith curved top chord, the influente ordinates for simple span would getexcessively great and the influente coefficient very small, the productbeing possibly inaccurate. For such a member the method explainedon page 316 for the three-hinged arch (Fig. 3) should be used, that is, the

ordinates A1A2 and B1B2 are made equal to F andI

F which are the

stresses caused in that member by an upward vertical forte unity atA and B respectively (the other end being fixed); and the ordinates Hare shown multiplied by the stress S1 caused in the member by ahorizontal forte H = 1. The influente coefficient then becomes “unity,. ”

Table XLV shows the method applied to a 250-ft. arch (see page 335).Often the live load is applied to the arch truss through a few vertical

bents. It is evident that all influente lines are straight between the bents.


The influente line for the chord member ahown heavy, for instance,would be as shown in Fig. 10 b. The influente polygon for H has evi-dently corners only below the bents.

In such cases the following simple method may be used. A verticalload unity is applied successively at each bent and the horizontal reactionis calculated as shown above or by means of formula (3) page 320 and thestresses and their signs.in al1 members determined by means of a Max-well diagram, separately for each load. These stresses are the influenteordinates. For the arch of Fig. 10, if symmetrical, only three diagramswould be necessary. This method is easily comprehensible and may beused for every arch, but it involves too much work if the load is appliedat many panel points.

FIG. 10.

Dead Load Stresses.-To determine the dead load stresses it is mostconvenient to calculate first the vertical reactions as for a simple span andthe horizontal reaction, by multiplying the influente ordinates for H withthe respective panel concentrations and adding the products and thendraw a Maxwell diagram as shown for the three-hinged arch (page 318).

Live Load Stresses.-These are best determined by means of influ-ence lines. For the determination of the stresses from fhe influentelines see page 322.

ART. 4. TEMPERATURE STRESSES IN THE TWO-HINGED ARCH

A change of temperature causes reactions in Bthe direction of the lineAB (Fig. 6). We will designate these reactions, which must be equal andopposite with H,. If one end were free to move longitudinally it wouldmove under a change of temperature of +t” by

& = * atz

328 DESIGN OF STEEL BRIDGES [CHAP. X V

where CY = coefficient of expansion. The forte H, causes a deflectionas per formula (2) on page 320.

Equating these values for ~5~ and 62 and assuming E constant foral1 members we find

H, = +aEi!l orEt1

zii

= +-zv . . . . . . . . . . . . . . . . (5)

1For steel (CX = 160000-7 E = 29,000,000), Hz = & 180-

ZV

(al1 dimensions in in., t in degrees Fahrenheit , Ht in lb.).

The value of Zv (v = S12$) is the same a s used for the calcula-

tion of H from vertical forces, that is, the summation extends only overthe chord members, neglecting the web members, which is permissible.

With the reaction H, given, the temperature stresses 8, in the archmembers can be determined by a Maxwell diagram, but if the abovemethod of finding the influente l ines for vertical forces has been

followed, the stresses S1 = +7 due to a forte H = 1, have already been

determined and the stresses Xt can be found by multiplying the values

$ with H, with one setting of the slide rule.

ART. 6. APPROXIMATE CALCULATIONS OF THE TWO-HINGED ARCH

If the sections of the members are unknown, as is the case in a newdesign, a preliminary calculation of the horizontal reaction has to be madebased on simplifying assumptions. It is necessary only to determine thesections, of a few chord members by this preliminary method; those of theother chord members can be found with sufficient accuracy by interpola-tion, and those of the web members are not required since their influenteon the horizontal reaction is negligible.

There are various approximate methods of calculating the horizontalreact ion depending upon the shape of the arch and the degree ofaccuracy desir$d.

(a) The assumption of a constant area of section A for all chordmembers gives good preliminary results and may even be sufficientfor the final calculation if the found chord areas do not vary con-


siderably; this is usually the case in flat arches with parallel chordsand in flat crescent arches. The formula (3) for the horizontal reactiondue to a vertical load unity at any point m becomes then

The notations are the same as given on page 320, and the summations areextended over -all chord members. The influente line for H is obtainedas shown on page 321 except that the values w and v now become

SIw=rL and 2, = S12L.

The area of the chord section A does not appear in the above formulasand, therefore, does not need to be known for the calculation of thestresses due to the vertical forces. It must, however, be assumed forthe calculation of the horizontal reaction Ht from a temperature changeaccording to formula (5) which can be simplified (since A is constant) to

aEtlA.Ht=fmx=f “~tvzA ( w h e r e v = S12L)......(7)

or for steel (c$ I 180),’ H, =‘rF: 180+$

As the chord members near the crown have the greatest influenteon the reaction Ht, A may be made equal to the chord area at thecrown (page 328).

(b) As a further simplification the length L of al1 chord members maybe assumed constant. In the case of a spandrel-braced arch with astraight top chord, L may be made equal to the panel length, providedthis is constant; since the chord members near the crown have thegreatest influente on the horizontal reaction. For other arches Lshould be made equal to the average chord length. The horizontalreaction due to a vertical load unity at any point m becomes thenusing formula (6) :

H = ~SoS1.. . . . . . . . . . . . . . .z SI2

(for So and S1 see page 320), and the values w and v for determining theinfluente line for H become

SIwzz- and u = sI2r

330 DESIGN OP STEEL BRIDGES [CH A P. XV

The horizontal reaction due to a temperature change is

altEAHt = Lz;x12 = aEtlA.. . . . . . . . . . . .(g)

LLV

or for steel (olE = 180),tíA

Ht = + 180 m

(c) For arches with parallel chords the lever arm r of the chordM

members is constant; if A and L are assumed constdnt, since SO =y

(see page 321) and SI = F: the above formulas become:

Horizontal reaction due to vertical load unity:

f&?MYZY

2.......................

and values of w and v for determining the influente line for H:w = y and v = y2.Horizontal reaction due to temperature change:

Ht

= olEtlA+LL:y2 . . . . . . . . . . . . . . . . . . . . .W)

(cl) For crescent arches with a comparatively yreat rise (Fig. ll)the general method can be simplified as follows:

For every panel determine the values

w - &L

rAand

FIG. 11.

takjng for L the length of the axis of the arch in that panel, for ythe ordinate of the axis at the middle of the panel, for r the averagenormal distance of the two chord members and for A the sum of the areasof top and bottom chord sections; and assume the value w applied at thecenter of the panel. Thus far the method can be used even for the finalcalculations by using formulas (4) and (5). For a preliminary designthe calculation can be further simplified as explained under (a) and (b).


4.fab(e) For paraboíic arches (equation y = 7) with approximately

parallel chords (Fig. 12) the following formulas give good results.(They are also useful for a first approximation for any two-hinged

arch.)Horizontal reaction due to a vertical load P = 1:

H = 9% ab (Z2 + ab) 1 . . . . . . . . . . . . . . . . . . (12)

FIG. ’ 12.

Horizontal reaction due to temperature change:H

t= 15aEth2A116f2 ;. . . . . . . . . . . . . . . (13)

where15 h2

c = lfzp

For steel (aE = ISO), Ht = 170 ylc

(f = rise of axis of arch, h = depth of arch truss at Crown).Horizontal reaction due to a uniform load p per lin. ft. covering the

whole span (panel length assumed infinitely small)

H=P!?I8f . . : . . . .c (14)

Forh

1= O.l

0.2 0.3 0 .4 0.5

c = 0 . 9 9 5 0.981 0.960 0.930 0.895

This shows that in all ordinary cases 1 d’ff1 ers

I .

but slightly from unity so that

the inaccuracy is small if the coefficient ; IS omitted in the above formulas.

If we make a = Icl, then b = (1- k) Z, and we can write formula(12) in the form

23

H =; (k-21c3+k4) ;J where K = a . . . . . . (12a)1

332 DESIGN OF STEEL BRIDGES [ C H A P . XV

Table 60 gives the values of x = ; (le - 2k3 + 164) for various

numbers.of equal panels; the injkence ordinates for H are obtained byIl

multiplying the values x by - - *f c

TABLE 60

Values of x = i (76 - 21c3 + k4), wherelc - F

NllIIlbWPanel point

of

_ pan& 1 1 2 m[m;-ml[- 4 i 5 / 6 1 7 1 8 1 9 F

20 0 . 0 3 1 0 . 0 6 2 0 . 0 9 019-

1817-

16 .0390.0760.1100.1390.1630.18115 0 . 0 4 1 0 . 0 8 1 0 . 1 1 6 0 . 1 4 6 0.179 0 . 1 8 6 0.194

- 1 4 0.045- 0.086 0.1230.1540.17ô0.19113-0.0480.092 0 . 1 3 1 0 . 1 6 1 0 . 1 8 3 0 . 1 9 4

12 0.0820.0990.139-11-

--~~. 0.056 0.107 0.148 0.178 0.193

10 0.062- 0.116 Tis- 0.186 0.195

(f) For parabolic arches (trusses or plate girders) with approximatelyparallel chords (including crescent arches) the influente area for H(for P = 1) may be replaced. with sufficient accuracy, by that of a para-bola of the equation

H = ” -; = ; f (lc - l?) ;- . . . . . . . . . . .

where values k and c are the same as under (e).The middle ordinate is

intermediate ordinates can be found by one setting of the slide ruleusing Tablein Appendix which gives the ordinates of a parabola withthe middle ordinate 1.


For a uniform purtial load p per lin: ft. over length a from A (Fig.12 a), we get for H the following equation:

Hz- i8.f \

;. . . . . , . . . . . . . . . . (16

(equal to the shaded influente area of H multiplied by p).

k:::-:.4.x:-,... a’. ___. i -...,I

py

,A ’ -----k---+--------- ’ Bl+ . . . . . . + ___i . .._....... + . . . .._... + . . . ..-... “ Il l I,I I /

l

FIG. 12~.

For a = ‘zjpl2 1

H = KjC

For a = 1, [email protected] c

If the uniform load extends between points C and C’ only, the result-ing H is found as the difference of the two values from formula (16)with abscisske a’ and a, respectively.

FIG. 13. \\

Formulas (15) and (16) are extremely simple, especially if we assumec = 1, and gíve useful results also for Rat circular arches with parallelchords.1

1 During construction of the Niagara-Clifton Arch Bridge, span 840 ft. (see PlateXLVI), the author had occasion to check the stresses, using formula (15), and found itvery accurate for that case.


(Formula (15) was first proposed by Prof. Mueller-Breslau, in his“Graphic Statics.“)

A relation Worth while to remember is that, for a parabolic influenteline for M, the LLrea~tion Zocus” (locus of the intersection point C of thecorresponding reactions R, and Rb due to a vertical single load P) is a

straight line MN parallel to AB in a vertical distance i f c, therefore,

approximately i f from AB (Fig. 13).

(For an arch with a semicircular instead of a parabolic axis this

vertical distance would be approximately i f, where f = ib)

Th% gives the direction of R, and Rb and if we resolve P into R,parallel to AC and Rb parallel to BC, the horizontal component repre-sents H caused by P. j

íl > , 3, Mox. ,>77 II

F I G . 13a.

The shaded area (Fig. 13) between triangIe ACB and the axis ofthe arch represents the area of moments caused by P; the moment atD, for instance, is equal to the ordinate DD1 multiplied by the Hcaused by load P.

By means of the reaction locus we can quickly find what parts of thearch have to be loaded with the live load in order to cause at a certainpoint X or X’ of the arch (axis, chord, panel point of trusses, kernelpoint of plate girders), the maximum or the minimum (max. of oppositekind) moment. Produce lines AX and BX (or AX’ and BX’) to inter-section with MN; these points (C or C’) are the loading points for zeromoment at X (or X’) where the moment changes its sign (fig. 13a).

For a uniform load over the whole length 1 and a parabolic axis of thearch the moment for all points of the axis is zero.


For a partinl uniform load the maximum moment occurs approxi-mately at the quarter points of the span if the uniform load extendsover one-half of the span from the hinge and is equal to

+ (2” p for the loaded half and - i 2 p0 for the unloaded

half.ART. 6. EXAMPLE FOR THE CALCULATION OF STRESSES

IN’ A TWO-HINGED ARCH(See Plate XLV)

Fig. (a), Plate XLV, shows the truss of a 250-ft. single-track two-hinged spandrel braced arch designed for Cooper’s E-50 loading. Thedead load i@NN,.,Jb. per lin. ft. of span. The figures given on theright half of Fig. (a) are the areas of the gross section of the chordmembers.

SI LInfluente Line for Horizontal Reaction H.-The values w = 7 2

Land v = Sl A are first calculated for the chord members as shown in

Table A. As a check the influente coefficients SI have been deter-mined by a Maxwell diagram, Fig. (b), as the stresses due to a hori-zontal forte H= 1. The lengths L and r have been scaled from theelevation; the areas A are given in units of 100 sq. in. in order to getconvenient figures for w and v. By adding the value v for one-half of thetruss and multiplying by,2 the value ZIJ is obtained.

The values w are now assumed as vertical loads applied at the cor-responding centers of moments and the bending moments M, due tothese loads are calculated as shown in Table B, whereby the upper andlower load w, and wI at any vertical have been combined to a singlepanel load.

The influente ordinates for H are obtained by dividing the momentsM, by the value ZV. The sum of the influente ordinates at the panelpoints is 4.282 (= twice the sum of ordinates 3 to 9 plus ordinate ll);the panel length being 25 ft., the influente area for H is 25 X 4.28 = 107.For comparison the influente line for H has been determined also by

51the formula (12a) H = uf (k - 21c3 + lc4) ‘, and by the parabolic

formula (15) H = $ (k- hz) i (see pages 331 and 332).

Dead Load Stresses.-The assumed dead load is 1400 lb. per lin.ft. per truss or 1400 X 25 = 35,OOOlb. per panel (23,000 lb. at the topchord and 12,000 lb. at the bottom chord). The horizontal dead load

3 3 6 DESIGN OF STEEL BRIDGES [ CH A P . XV

TABLE ACalculation of Values w and

-

LWWann

kngth Area ofo f gross

mmbel section,

-

1

n

-

Infl.c o e f f i c . ,

SI=;

v

LA

W=

SI Lr A

2) = sI2 4

Ordi.nate,ctr. of

mo-ments,

Y

ft. ft.

72.0 54.572.0 41.572.0 30.072.0 20.572.0 14.0

21.7 50.238.5 33.550.5 21.557.7 14.260.0 12.0

-

*

- -

L A

ft. 1OOsq.i~.33.1 0.71030.1 0.65927.7 0.65926.0 0.50525.1 0.471

25.025.025.025.025.0

0.3620.3620.498

1.32 46.5 lY21.74 45.6 1.912.40 42.0 3.363.52 51.5 8.855.15 53.2 19.60

0.43 69.11.15 69.. 12.35 50.14.05 33.35.00 33.3

Wl0.592.375.499 .5013.85

Ccntero f

nomenta

468

10

Chordmomber

o- 22- 44- 66- 8S-10

l- 33- 55- 77- 99-11

lY8242638

1405

13

2;:5 4 7832

4263 x 2= 8526 = Zv

TABLE B

Calculation of Influente Ordi na tes for Horizontal Reaction

Panelp o i n t

S h e a r

V W

IIlflUtXKXordinate IIlflUl3INX

ordinstes HII= Mw:Zv f r o m T a b l e 601

1.12

2.50

5.73

14.34

29.10

13.85

66.64

66.64

i;i:ii

fii:ili

iii: ib’

ii:%’

ii:&

. . . . . . .16.38

. . . 16 .3815 .75

. . . . 32.1314.32

. . . . . . 46.4510.74

. 57.19. . . . .3.46

. . . 60.65

. . . . EH= . . . . . . .

0

0.192

0.377

0.545

0.671

0.712

4 . 2 8 2

1)

0 0

0.228 0.252.

0.433 0.447

0.592 0.588

0.692 0.670

0.728 0.700

, 4.618 4.614

51 1 5x250IH = sf(7c-2k3+7c4)c.= 8x66 (k-Z7$+k4)+ / 1zH=4f(k-k2)C 31 1 wnere c= 1 +

15.12232.662 = 1015.

Ordimtes 11Eor parabolic

infl. linez

reaction is Hd = 1400 x 107 = 150,ÓOO lb. and the vertical reaction Vd =

1400 X 125 = 175,000 lb. The resultant reaction is Rd = d1502 + 1752= 231,000 lb. With these reactions a Maxwell diagram is drawn givingthe dead load stresses (Fig. g).

As a check for the accuracy of the diagram the stress in member9-11 is calculated analytically as follows: the bending moment at point

Calculation of influente ordinates Al.42 and BIBZ Calculation of Iive load stresses 22

Sum o f influenteordinates

ZB

-

-

Approx. equivalentuniform load

lb. per ft.Live load stress inUnits of 1000 lb.

P

+ - POS.stress

hfl.ooeff.

SI

=25 p s1zz

z-

-t-

'O:ii'0.260.340.30

4.282.881.801.000.43

'4iOO‘350032503200

2800 1.322900 1.743050 2.403100 3.523200 5.15

'ii'

E123

0.780.550.330.08

1.641.471.201.000.92

3150315032003300

. . . . . . . . . . . .

3200 0.433200 1.153200 2.353100 4.052800 5.00

26

i;27

. . . . . .

1.651.311.071.172.86

0.79 3350 3150 0 . 9 5 1310.43 3350 3200 1.20 1320.12 3350 3200 1.58 1420.26 3350 3900 1.97 1931.93 3450 3750 1.05 258

0 . 7 9 2.22 31500.43 2.37 32000.12 2.03 32000.15 2.07 43501.20 4.33 3900

3750 0.873350 0 . 9 73350 1.023300 0 . 9 83400 0.46

2:10

54

4.28 . . . . . . 2800-

-

-1.00 l 300

Influente ordinates

Ordinate

rY

ft.

Abscissa

2

fe.

AIAZ

5Y

BIBZ

x’=-

Y

Memberì-

-

I-

--LL-

0 .“iiii“’ 0.347200 0 . 6 9 4175 1.04150 1.39

. . . .3.13

2.782.432.08

225 1.15 10.38200 1.30 5.20175 1.48 3.47150 1.74 2.60125 2.08 2.08

167150130101-8

/same as l- 2same as 3- 4same as 5- 6same as 7- 8same as 9-10

1.151.391.672.073 . 5 9

2.322.081.811.41

i.-

. . .. . . .. . ... . .

-0.11

. . . . . .

. . . . .

. . .

. . . .

. . . . .

1.-

-

72.072.072<072.072.0

25:o50.075.0

100.0

21.7 25.038.5 50.050.5 75.057.7 100.060.0 125.0

83.0100.0120‘. 0149.0258.0

395363328 k273177 g

1’

l- 23- 45- 67- 89-10

56135 E

225 ü314322

8m

72.072.072.072.072.0

4:5;

190

181192174167170

.........

.........i

-Horizontal reaction

co-J


10 is ~ X1400 2502

-8 150,000 X 60 = 1,950,OOO ft.-lb. and the stress in

9-11 is, therefore, 1,950,OOO : 12 = -162,000 lb.Live Load Stresses.-These will be determined by means of influente

lines and, to simplify the calculation, approximate equivalent uniformloads will be used, according to, the rules explained on page 62.

In order to obtain the influente lines, we calculate for al1 memberst

the ordmates A1A2 = f and B1B2 = ti x and x’ being the hori-

zontal and y the vertical distance of the center of moments of a memberfrom the hinges A and B. The centers of moments of the web mem-bers are obtained by producing the bottom chord members to theirintersection with ‘the top chord. The ordinates A,A, and BlB2 arethen plotted from the base line AIB,. The center of moments of al1members except 8-9 and 9-10 lies in the asgace between and above Aand B, the ordinates A1A2 and BIBZ have therefore to be plotteddownward, that is, on the same side of the base line as the influentepolygon for H (see page 326). For members 8-9 and 9-10 whose cen-ter of moments is to the right of B ordinate A1A2 has to be plotteddownward and B1B2 upward. The influente line for any member isnow easily obtained by drawing line AIB~ and BlA2 and as ‘a checkthe intersection of these two Enes must lie vertically below the center ofmoments.

The influente lines for the bottom chord members are shown in Fig.(d), those for the top chord members in Fig. (e), and those for the webmembers in Fig. (f).

The signs of the influente areas are easily found by inspection; forinstance, in member 5-7, a horizontal forte H at A causes tensionand a vertical forte Ti, at A compression, therefore, where the ordi-nates of the H polygon exceed those for simple span, the area is positive(tension) and vice versa.

To obtain the greatest live load stress of a certain sign in a memberthe influente ordinates x below the panel points are added (assumingonly full’ panel loads except at the end panel point for member O-l).ZX multiplied by the panel length X = 25 ft. gives the influente areaand this multiplied by the equivalent uniform load and by the in-fluente coefficient SI taken from diagram (Fig. b) gives the stress(Table C).

The greatest horizontal reaction is obtained by multiplying theinfluente area for H by the equivalent uniform load:

Ha = 107 x 2800 = 300,000 lb.


The greatest vertical reaction is found as for a simple span:

VS = 125 x 3130 = 391,000 lb.

The reactions ¡?L and VI are caused by different positions of thelive load but may be combined for the calculation of the bearings. TO

determine the stability of piers, however, both positions of load formaximum horizontal and maximum vertical reaction should be consid-ered separately.

Temperature Stresses.-The horizontal reacti’on due to a change of

temperature of t = 575” is obtained by the formula Ht = + 180 gU

(page 328). Since in calculating the values v= SI2 i we have given Lin feet and A in units of 100 sq. in. and since al1 dimensions should

be given in inches we have to multiply the result by $:

H = +18o75X25OX12 100t - 8406 - X 12 = +40,QOO lb.

The temperature s t resses in the members are now obtained bymultiplying the values SI by 40,000. Por instance, the stress in 5-7is St = 40,000 X 2.35 = f94,OOO lb.

ART. 7. WIND STRESSES IN THE TWO- AND THREE-HINGED ARCH

The members of the arch trusses are stressed by the wind owing tothe polygonal chords. These stresses may be quite considerable, espe-cially in high arches, and cannot be neglected as in simple truss spans withcm-ved chords. The correct analysis of wind stresses in arches is compli-cated; approximate calculations are sufficient since the wind forces andthe stress distribution are uncertain. ‘As a rule there is a lateral systemalong each chord and one along the floor; for simplicity we may assumeall systems to work independently of each other. Further, to determinethe maximum stresses in the laterals, we may assume the lateral systemas a simple truss span developed into a plane; in other words, calculate thestresses in the projection of the lateral truss as a simple span and mul-tiply the stresses by the proportion of the actual to the projected lengthof the members as explained for simple spans with curved chords (page108). The stresses thus obtained are correct for a symmetrical wind loadbut not for a moving wind load, since the lateral systems are actuallyJixed at the ends. The error is greatest, but on the safe side, for thelaterals of the center panels where the etresses are smallest and where

3 4 0 DESIGN OF STEEL BRIDGES [CEAP. xv

the laterals have usually an excess in section; the error decreases to zerotoward the ends where the stresses are greatest.

To determine the stresses in the truss members it is sufficient to assumea uniform wind load over the whole span. As a rule the trusses are sym-metrical and the reactions can then be easily calculated. Their direc-tions are as shown in Fig. 14.

FIG. 14.

The horizontal transverse reactions are T = f 2 IV. The horizontal

longitudinal reactions H and the vertical reactions R, are found asfollows :

(a) Arch Trusses Are in Vertical Planes.-We will assume a longitu-dinal vertical section separating the two trusses and cutting the diagonalsand struts of the lateral system. The equilibrium is not changed if we ap-ply the stresses in the laterals and struts as externa1 forces. We assume

for the present a single system of lateralswith struts normal to the plane of thetruss and consider the leeward truss. IfV = shear in any panel m - n (Fig. 15)and 11 the length of the diagonal, the stressin the latter is

Its component normal to the plane of the truss (equal to V) mustbe in equilibrium with the stress in the etrut at n and therefose doesnot affect the trusses; its horizontal and vertical components h andv in the plane of the truss affect the latter only. If x = panel lengthand yn - ym = difference of ordinates of points n and m, we have

h=V% and v= vYn-Ym-b

We calculate these components at all panel points and apply them tothe arch truss as externa1 forces. This loading is symmetrical; the forcesh are acting outward and the forces v downward at the leeward truss:


If the laterals act in tension only, the stresses in the windward truss areslightly different from those in the leeward truss, but they can be assumedalike but of opposite sign.

The shear V in any panel is found as for a simple span; it is equal tothe transverse reaction T minus the wind forces to the left of the panel,or in this case of symmetrical loading equal to the sum of wind forcesbetween that panel and the center of the arch (including one-half thepanel load at the center panel point, if any).

It is simpler to consider the horizontal forces h separate from thevertical forces v for the calculation of the reaction H.

- - - - --.-- .____ . . . .._........

FIG. 16.

The vertical reaction R, and the horizontal reaction H, due to theforces v are obtained in the same manner as shown in Arts. 2 and 3 forvertical loads. For the two-hinged arch the influente line for H ismost convenient. For the three-hinged arch the horizontal reaction due

to any two symmetrical loads v is found by the simple formula H = VT

(Fig. 16). The horizontal reaction Hh due to the forces h may be obtainedfor the two-hinged arch by formula (3), page 320, where S = stress causedby the forces h if the arch is considered a simple span, the’ summationbeing extended over the chord members only. It is, however, sufficientlyaccurate to assume a hinge at the center of the arch truss in the middlebetween the two chords and calculate the reaction Hh as for the three-hinged arch, that is: The reaction due to any pair of symmetrical loads

h (Fig. 16) is H = h “f” and the sum of all these reactions gives Hh.

Finally, H, and Hh added give the total reaction H. With al1 externa1forces R,, H, v and h thus given, the stresses in the truss membersare best found by a Maxwell diagram. If there are two lateral systemsbetween polygonal chords, the externa1 forces are determined separatelyfor each, but are preferably combined for the stress diagram.

3 4 2 DESIGN OF STEEL BRIDGES [CHAP. xv

If the wind acting on the train and floor is carried to the arch rihthrough bents or sway frames (Fig. 17) each bent or frame transmits to

the truss besides the transverse forte W, a vertical forte II+ which

must be added to the forte o at that panel point.If there are two diagonals in a panel act#ing simultaneously, the

average shear V in the two adjoining panels should ‘be used in the for-mulas for h and v at,any panel point.

(b) Arch Trusses Are Inclined.-We will con-+ ’ sider a single lateral system between polygonal

i chords of the trusses whose planes intersect in a-4’

,’

R

./_. /,,,.b.” horizontal line 1, which lies in the vertical planethrough the center line of the bridge as shown by

+W-t; a cross-section through the bridge (Fig. 18b). Let

W6 us assume transverse sections X and X through

FIG. 17.’ any two symmetrically located panels mn and m’n’(Fig. Ba), each cutting the two trusses and one

lateral. The part between these two sections is held in equilibriumby the wind forces acting on it and the stresses in the cut membersapplied as externa1 forces. Since all forces are symmetrical, we con-sider only one-half of part XX and denote the resultant of the windforces acting on this half with V (it is equal in amount to the shearin panel m-n), and its vertical distance from the line I with i,. Wenow assume the stress D in the lateral resolved at its intersection Kwith the vertical plane through line I into a component in this planeand one normal to it. .The latter or horizontal transverse component ist = D COSCY cos /3 where CY = angle between the lateral and its projection ona horizontal plane and p f angle between this projection and a trans-verse strut. The vertical distance of point K from line 1 shall bedenoted with i; it is approximately equal to the average distance of thestruts at m and n from line I. Since there must be equilibrium againstturning about line I and since all forces, except V and t, pass throughthis line, we find

Vi, = t i = D i COSCY co@

and D=+“;G~&

or if b = average transverse distance c. to c. of chords in the panel inquestion, and L = length of lateral, stress

ART. 81 ARCH BRIDGES

The stress D, in the transverse strut at n is correspondingly

D, = -+

343

where i, = vertical distance of thestrut at n from line 1.In order to find the stresses in the trusses, we assume a longitudinal

vertical section separating the two arch trusses and consider the far(leeward) truss. It is in equilibrium if we apply the stresses in the lat-erals and struts as externa1 forces. To be accurate we would have toresolve these forces at all panel points into components in the plane ofthe truss and normally to it, the former only affecting the truss. Since

I.-.-.-.-.-. -.-.-_- .___________________.-.-~-. -.-.1

ta.1

FIG. 18.C b.)

however the trusses are generally only slightly inclined, it is permissible‘to assume them vertical and to resolve the above lateral forces into thetransverse components t and vertical and horizontal componenta v and hrespectively, in the plane of the truss. The transverse component t ofthe stress D in the lateral m - n approximately counterbalances thestress in the strut at n. The components h and ZJ of the stress D arerespectively

h= T’;g and i, yn-- ymv = vi -9b

where p denotes the panel length and y% - ym the difference of ordi-nates of panel points n and m. Having determined these forces at al1panel points, we proceed in the same manner as shown for the case ofvertical trusses.


A horizontal longitudinal forte B causes inclined reactions R, and RT,(Fig. 19) which intersect in the line of the forte B. The horizontal and

3 4 4 DESIGN OF STEEL BRIDGES : [CHAP. X V

vertical components of these reactions are found by the laws of equi-librium as follows:

v, = -Vb = ll; . . . . . . . . . . . . . . . . . . . . (1)

H,+H1> = B. . . . . . . . . . . . . . . . . . . . . . . (2)

If B is applied at the center C of a symmetrical arch then H, = Hb = ie

For the two-hinged arch the theory of elastic deformation furnishes,similarly as shown on page 320 for vertical forces, the formula

KlF IG . 19.

where for any member So = stress caused by a forte B = 1, the arch be-ing considered a simple span with end A free to move longitudinally; theother values are the same as given on page 320. Again we consider only

the chord members and may express SO by F where for any member

r = lever arm and’ M = bending moment at its center of momentsdue to the forte B = 1, the arch being considered a simple span. W ehave then

&f!!! 4r A

H, =ZMw=-

ZV

where w and v are exactly the same values as given Eor vertical forces(page 321). To apply this formula it is, therefore, only necessary to cal-culate the moments M for the chord inembers which is best doneanalytically. The moment M at any point x to the left of m (Fig. 19)is

M = V,x = B;,

.‘iRT. 8] ARCH BRIDGES 345

and at any point x to the right of m

M = v, x - B(h - y).

Atfention should here be given to the signs of the values Mw sincethey may be different for different members. This is best done byinspecting the eigns of the stresses SO and SI since their product deter-mines the sign of Mw. SI is always positive (tension) for bottom-chord and negative for top-chord members; therefore, Mw has thesame sign as So for bottom-chord members and the opposite sign fortop-chord members.

Useful approximate values of H, and Hb are obtained by assuminga hinge at the center in the middle between the two chords and proceed-ing as for the three-hinged arch (Fig. 20) as follows:

If forte B is applied on part CB of the arch, the reaction R, mustpass through hinges A and 6. Taking moments about C we get

H,f= Va; . . . . . . . . . . . . . . . . . . . . . . . (4)

This combined with formulas (1) and (2) givesh

H, = B %f a n d H,=B-H,=B(l-4)

Having determined al1 reactions the stresses in the arch truss arebest found by a Maxwell diagram.

The braking forte should correspond to that length of live loadwhich produces the maximum live load stress. Since this length differsfor different members, it is best to calculate first al1 stresses for abraking forte B = 1 and multiply these by the corresponding loadedlength in feet and the braking forte per lin. ft. per truss.

In the case of an,arch with horixontal top chord where the brakingforte is applied at the panel points along the top chord, it is per-missible, in order to simplify the calculation, to assume the whole forte

Bapplied at the center, causing horizontal reactions H, = Hb = --2 and cor-


recting t,he stresses in top chord thus found as followsi Assuming thatthe train advances from the right and extends to the left a certaindistance beyond the top chord member in question, a tens ion equalto the braking forte cgrresponding to that distance to the left of themember has to be added algebraically to the stress (provided that theforte at the center was also assumed acting from right to left).

For opposite direction of the braking forte, the stresses are of opposites ign .

The effect of the braking forte has to be considered, especially inarches with great rise; in flat arches it is not important .

ART. 9. STRESSES DUE TO YIELDING OF FOUNDATIONS

It is sometimes desirable to know the stresses produced by an increasei of the span length due to yielding of the foundations. The stressesare exactly the same as those produced by ,a decrease of temperaturewhich would reduce the length of span by i if one end were free tomove. For the three-hinged arch these stresses are of secondary im-portance and may be neglected, provided the settlement of the foun-dations is only slight. For the two-hinged arch the horizontal reactionis found by the formula for H, (page 327) if g is substituted for OL Zt.The stresses are found similarly as those from temperature changes.

B. DESIGN OF ARCH BRIDGES

(See Plate XLVI and Table 61)

ART. 10. GENERAL

The arch is adapted to almost any’length óf span, provided the localconditions are favorable.

AS far as stiffness is concerned the arch is inferior to the simple trussspan. Since an arch rib proportioned for a symmetrical load (deadload and total live load), is not stiff for an unsymmetrical load (partiallive load), it requires & stiffening system which will be the heavier thegreater the live load cokpared with the dead load. For ordinary spans,therefore, the arch type is more adapted for highway than for railroadbridges. I t is especially economical for aqueduct or canal bridgeswhich require practically no stiffening system.

In cases where other types would require high piers the arch may bethe most economical as the height of the piers can be reduced to a mini-mum and the loads are transferred on the shortest way to the foun-

ART. ll] ARCH BRIDGES 347

dations. The foundations must be unyielding since a movement ofthe bearings may cause high stresses in the two-hinged or the hingelessarch and objectionable deformation of the floor at the center of thethree-hinged arch. The arch type is, therefore, best suited for crossingsover deep ravines which offer natural abutments.

An important advantage of the arch is the possibility of its erection asa cantilever by means of temporary anchorages.

The horizontal component of thereaction can be eliminated by usinga tie connecting the two bearings or two other points of the arch truss.This makes the arch a simple span as far as the externa1 forces areconcerned (vertical reactions). (Rhine Bridges at Bonn, Duesseldorf,Worms, etc.)

ART. ll. TYPES OF ARCH BRIDGES

Depending upon the number of hinges there are three-hinged, two-hinged and hingeless arches.

The three-hinged arch has the advantage of being statically deter-minate and practically unaffected by changes of temperature or slightsettlements of the foundations (see page 32); this is of great impor-tance f o r fla,t arches.

The three-hinged arch truss requires the least depth at the center andmay be used in preferente to the other types where the available heightat the center is small. (Alexander III. Bridge and Mirabeau Bridgein Paris, Stephanie Bridge in Vienna.)

Its principal disadvantages are the greater deflection and smallerstiffness, and as its elastic line forms a kink at the center hinge it is not‘adaptecL&r..railroad traffic.

The three-hmged arch is lighter than the two-hinged arch, but thesaving of material may be offset -by the additional cost of the centerhinge.

The two-hinged arch is statically indeterminate in the first degree.It is the type mostly used at present, being stiffer than the three-hinged arch and less uncertain in its stress distribution than thehingeless arch. The erection of the two-hinged arch truss can be sim-plified by providing a temporary hinge at the center and leaving outthe opposite .chord member, which can afterward be inserted withoutinitial stress so that the arch truss acts as three-hinged for the greatestpart of the dead load and as two-hinged for the rest of the dead loadand the live load. The costly erection of the center chord memberwith initial stress (by means of toggles, hydraulic jacks), which would

24


be necessary to make the arch two-hinged for dead load, is t husavoided and any effect of settlements of the foundations during erec-tion eliminated.

The arch without hinges is statically indeterminate in the third degree.It is very stiff but sensitive to settlements of the foundations andchanges of temperature. It is rarely used. (Eads Bridge in St. Louis,Miingsten Viaduct in Germany.)

The arch bridges are further classified according to the main girderswhich may be solid plate girders or riveted trusses. Pin connections arenot suitable for arch trusses owing to the reversa1 of stress in the webmembers.

(a) Plate girder arches are suitable for short spans, especially forwide highway bridges with more than two girders in cases where theavailable height below the floor in the center of the span is too smallfor simple deck spans.

The depth of the arch girders ,is usually 4% to 60 of the span.(b) Arch trusses with parallel chords are generally used for long spans.

They are not stiff since the depth at the quarter point,, where the greatestdeflection occurs, is insufficient; they are therefore’ not well adaptedfor railroad traffic. (Niagara Clifton Bridge, Eads Bridge, St. Louis.)

(c) Arch trusses with chords diverging from the’ center toward the endsare preferably used for bridges’ with fixed ends since the grea tes tmoments occur at the ends. (K orn h aus Bridge, Berne,j Switzerland; andMüngsten Viaduct in Germany.)

(d) Crescent (or sickle) arches in which the chords converge for the tvo-hinged arch from c’enter toward the bearings and for the three-hingedarch from quarter points toward the hinges, are very pleasing in ap-pearance and have been built for great spans and heights. The depthof the truss at the center of the two-hinged arch and at the .quarterpoints of the three-hinged arch can be made greater than in the archwith parallel chords. They are, however, not as st iff as spandrelbraced arches and require a greater depth at the center . (DouroBridge, Portugal; Garabit Viaduct, France; Grünenthal Bridge, BalticCanal, Germany.)

(e) Spandrel braced arches are the stiffest kind and therefore bestsuited for railroads. Their appearance is not very favorable, if there areadjoining simple spans. The bottom chord which contains the hingescarries the main portion of the load, the top chord with the web membersacting mainly as a stiffening truss. If the floor rests on top, which isusually the case, the upper chord is made horizontal; this type is best


suited for medium span lengths, as for long spans or a great rise theweb members near the ends become too long, causing waste of material.The erection by the cantilever method is very simple as the truss canbe held at the end of the top chord. Many arches of this type havebeen built.

(f) Cantilever arches usually consist of three spans. The center span isa three- or two-hinged spandrel braced arch, each side span is a canti-lever arm forming a continuation of the arch truss and supporting insome cases a suspended span, one end of which rests on the abutment.If the side span has no hinge the system is statically indeterminate ina higher degree than if it has a hinge; on the other hand, the systemwith the hinge has the disadvantages of a cantilever, that is, a great-deflection and the kink of the elastic line at the hinge. The suspendedspan should have a length of about 2 to 8 of the side span to reducethese deflections. The side spans are usually about one-half of themain opening.

Cantilever arches are stiffer and more economical than cantilevers.They can be erected in the same manner as cantilevers, with falseworkunder the side spans. They have the advantage over arches in reducingthe horizontal thrust from dead load and, therefore, making narrowerpiers possible or permitting a smaller rise of the arch; in the MirabeauBridge in Paris, for example (see Eng. News, Nov. 12, 1896), the 326-ft.center span has a rise of only 20 ft. = $5 1; other. examples of canti-lever arches are the Viaur Viaduct, France; Rio Grande Bridge, CostaRica;,Stephanie Bridge in Vienna, etc.

Plate XLVI and Table 61 contain the elevations and the principaldimensions of notable arch .bridges.

ART. 12. LOCATION OF FLOOR

Whenever possible the floor should be located àbove the arch trussesor in case of spandrel braced arches be framed in between the topchords.

Where the íloor intersects the arch trusses their top chord should beentirely above the floor and have end portals, as arches, rising out of thefloor cause complicated details.

ART. 13. NUMBER AND SPACING OF ARCH TRUSSES

Concerning the number of trusses the same principles apply as forsimple spans (page 169).


If the floor rests on top the distance between centers of vertical archtrusses should not be less than & of the span length, nor less than 3,preferably 3, of the total height from bearings to top of floor. If thisis not possible without increasing the spacing of the trusses beyond thatnecessary for the floor, the trusses may be given a batter. It must beremembered, however, that this causes more expensive shopwork whichmay offset the saving in weight. + to 2% are common values for thebatter.

If the floor is suspended from the arch, the spacing of the trusseswill be determined by the width of the floor. It should, however, not beless than 2~ of the span length nor less than + of the height of the archtrusses from bearings to crown.

In any case the overturning moment due to the wind on the loadedas well as on the unloaded bridge should be calculated in order todetermine the stability. If the anchorages are not relied upon (as isusually the case) the moment of the vertical loads should be equal tonot less than the double moment of the wind forces, otherwise theanchorages must be designed for the resulting tension.

ART. 14. PRINCIPAL DIMENSIONS OF ARCH TRUSSES

The span length and the rise of the arch are usually determined bylocal conditions. The proportion between rise and span length rangesordinarily between a and 112; extremevalues are + and &. The mosteconomic proportion will differ in each case. The flatter the arch thegreater the horizontal thrust and the temperature stresses; + is anaverage economical value.

The height of the arch truss at the center ranges from $5 to 6’ó ofthe span length. The economical height depends upon the spán length,rise, loading, kind of arch truss, etc. ’ For two-hinged arches for railroadsthe height at the center is preferably assumed as follows’:

For spandrel arches and arches with parallel chords, $5 to 2’0;For crescent arches, l’a to &Y of the span length.If the available height at the center is limited the height of spandrel

braced arches may be made as small as z$ of the span in which case it isoften preferable to use a solid web plate in the center panels.

For highway arches these depths may be reduced by 25 to 40 %.

ART. 16. FORM OF ARCH RIBS AND WEB SYSTEM

The bottom chord of spandrel arches and both chords of the othertypes have usually the form of a parabola which not only 1001~s well


but is also economical as it conforms to the equilibrium polygon for fulluniform load. For high arches an average curve between parabolaand circle may look better (Kornhaus Bridge in Berne).

The chords should be straight between panel points.Arch bridges whose chords extend straight over severa1 panels are

cheaper and may be commendable where appearance is of no importance.[Arch Bridge of the White Pass & Yukon Ry., Alaska (Enq. News,Vol. 45) ; the Assopos Viaduct, Greece (Enq. News, Nov. 4, 1909) ; theBlaauw Krantz Viaduct, Cape Colony (Am. Engr., 1885); the Nam-TiArch, French China (Enq. News, Apri121, 1910).1

The web system should be simple; multiple systems are objection-able. The panel length should be constant as far as possible and soselected that at the quarter points of the span the diagonals form anangle of approximately 45” with the verticals. In crescent arches itlooks generally better if the panel length decreases toward the hinges.Otherwise the same general principies concerning the design of thetrusses apply as for simple spans (page 169).

ART. 16. BENTS AND HANGERS

If the floor is on top of arches with a curved top chord it is preferablysupported by only a few bents instead of at every panel point of the archtruss’. If the floor is suspended hangers e ..__......... d . . . . .._....__._..___ tiare arranged at every panel point.

Owing to the expansion of the floor !\, 4 ,“.,b:

due to temperaturé changes .the postsor hangers, if fixed at their ends, sus-tain bending stresses for which provi-

Ty

?, ”:

I+

sion must be made in their section. FIG. 21.

If h = length of post or hanger in inches,d = its distance from the point of connection of the floor to the

arch trusses in inches,b = its width in the plane of the truss in inches,t = change of temperature in degrees Fahrenheit,

a! = coefficient of expansion,the longitudinal deflection in inches of the post or hanger is e = t a!d.

Assuming the post fixed top and bottom, each half deflects as a beamfixed at one end and loaded at the other by a single load causing a

deflection a. The maximum bending stress cnused thereby is in lb.

bEper sq. in., s = 3 e h2 .

352 DESIGN OF STEEL BRIDGES [ C H A P . x v

If the post is fixed at one end only and hinged at the other, s isonly one-half the above.

ART. 17. FLOOR SYSTEM AND BRACING

Concerning the design of the floor and bracing of arch spans, thesame principles apply as for simple sians (page 167). Owing to the usu-ally considerable wind stresses in arches, it is important to have thecenter lines of lateral members intersect with the center lines of the archribs. The lateral members should be stiff to insure lateral rigidity,although the details are more complicated than if rods were used.

The arrangement of the floor system is in many cases complicated,as provision must be made for the expansion of the floor independentof the trusses so that the stringers are not stressed by the thrust of thearches due to temperature changes. Further complications are causedwhen provision has to be made for the transmission of braking ortractive forces to the arch trusses.

A lateral system in the jloor plane should be used to resist ‘the lateralforces acting on the floor.

In the case of spandrel braced arches with horizontal top chords, th’efloor braeing forms the lateral system between the topchords and trans-mits. the lateral forcés to the end frames and the abutments. I t isarranged in the same way as for simple deck truss spans (page 185). Thebraking forte can be transmitted to the trusses at each top chord panelpoint. In three-hinged spandrel braced arches with floor on top, thisbracing has to be interrupted at the center so as not to make the centerhinge ineff ective. The lateral system forms then two simpIe trusses sup-ported by the sway frames at the center and the ends of the arch. Thereaction at the center frame is transmitted, to the lower latera1 system.which mus-t be continuous over the hinge, or the top chord .wind loads ,,may be assumed transmitted through the sway frames to the lower lateralsystem at every panel point. In no case of loading should the compres:sion from the vertical forces at the pin be less than the tension caused bythe wind in order to prevent an opening of the hinge. If there is a possi-b’ility ‘of a resulting tension by increasing the wind load 50 to lOO%,the width between trusses should be increased or provision made to pre-vent an opening.

In short spandrel braced arches with floor on top, no special provisionneed be ,made for the expansion of the floor, but in long ,spans it isadvisable to provide expansion pockets for the stringers at intervals ofabout 150 ft. fo prevent undue horizontal bending of the floorbeams.


In arches with curved top chord and floor on top, the lateral forcesacting on the floor should be carried to the arch ribs through the bentssupporting the floor. The floor lateral system is the same as for plategirder spans (page 166). The floor is fixed to the arch trusses at the crownand allowed to expand toward the ends, thereby deflecting the bents..The stringers or longitudinal girders must be designed for the trans-mission of the braking forces, if any, to the crown of the arch trusses.

Where the floor is suspended from the arch the simplest way would beto transmit the .floor lateral forces to the arch rib through sway frames orportals between the floor hangers. This is, however, not satisfactorywhere the hangers are long. In that case the forces must be transmittedby a lateral truss system with separate wind chords to the intersec-tions C (Fig. 22) of the flóor with the arch truss.

The floor including wind chord may be interrupted for expansioneither at points C (the lateral truss can then be assumed as a simple

FIG. 22.

span CC causing at C only transverse reactions which are transmittedto A through the bottom lateral system AC), or the floor may be inter-rupted at any two other panel points D between points C. In this casethe floor lateral system forms a cantilever, with suspended span DD,cantilever arms CD and anchar arms EC, whose reactions at C and Eare transmitted to.A by the lateral system AC and the end sway frameEA. This arrangement can be used only if there is an anchar armEC of at least one panel length and CD should not be longer than E?.

For railroad bridges, there should be provision for the transmission ofthe braking forces acting on‘ part DD of the jloor to points C of thetruss. This can be done by connecting lateral truss DD to one armDC by a vertical pin on the center line of the bridge and providing forexpansion at the other end D only; it has the disadvantage that thehangers near the expansion point have to deflect considerably.

354 DESIGN Oi? STEEL BRIDGES [CHAP. XV

At any place where the braking forte is transmitted from the stringersto the main trusses, or to a separate chord, there should be a horizontalbraking truss so that the floorbenms are not bent longitudinally.

Arch Bracing.-There should be a lateral system along the top chordand one along the bottom chord and rigid end frames or portals. Al-though not necessary for the strength of the bridge, there should be swayframes at least at those panel points at which the load is applied.

In smaller bridges one of the lateral systems may be omitted, especiallyif the chords are close together and if sway frames are used at all panelpoints.

In railroad bridges with suspended floor, for two or more tracks, it isadvisable to use sway frames or portals even between the hangers, asan additional means to stiffen the floor and prevent high secondarystresses at the floorbeam and lateral connections due to the unequaldeflections of the arch trusses under one-sided loading.

ART. 18. STEEL WEIGHTS OF ARCH BRIDGES

The steel weights of arch bridges up to 300 ft. span and economicalrise (about + to +) may be assumed the same as for simple spans (seepage 221). The weight of the arch trusses increases with decreasing rise sothat for a rise of only $C the weight increases by 30 to 40% Archesof greater span and economical rise weigh from 10 to 15y0 less thansimple spans.

For preliminary calculations and for ordinary spans the dead load maybe assumed uniformly distributed over the whole span for calculating thestresses. For long spans and high arches the final calculation of thestresses should be based upon an accurate estimate of the dead load panelconcentrations as the weight of the arch trusses and bracing may in-crease considerably from the center toward the ends

ART. 19. EXAMPLES OF ARCH BRIDGES

(See Table 61 and Plate XLVI)

(1) Arch Bridge across Mississippi at St. Louis, built by Eads, 1868-1874, total length with approaches 4885 ft.; center span 520 ft., two sidespans 502 ft., four lines (16$ ft. apart) of circular arch trusses withparallel chords 12 ft. apart, rise Ti of span, no hinges. The bridgecarries two decks: upper for highway 52 ft. wide, lower for two tracks ofrailroad 33 ft. wide.

The first bridge in which steel was extensively used.


Designed originally for a live load of 4000 Ib. on upper and 4000 lb. onlower deck, both per lin. ft. of bridge, each arch truss therefore 2000 lb.per lin. ft. No impact or reversa1 of stresses considered. The speci-fications required for chords hard steel (chrome steel) of 100,000 ultimateand 40,000 elastic limit; the obtained material was not uniform. U n i t

stresses in lb. per sq. in., 30,000 for compression, 20,000 for tension.The chords are circular tubes 18 in. diameter. Each arch truss weighsabout 1000 lb., and the steel and iron of the bridge about 7000 Ib., bothper lin. ft.

Erected wilhout falsework.The bridge has been strengthened in 1888 and again in 1902.

For further information, see:Woodward, A History of the St. Louis Bridge, St. Louis, 1881. (An excellent review

of the cslculations, the construction, erection, tests of material, etc.)N. W. Eayrs, Railroad Gazette, Aug. 31, 1888, Reconstruction of railroad floor.J. C. Bland, Reinforcing óf Eads Bridge 1902 and 1903 (with many plates),

contnined in Report of Chief Engineer to the Terminal R. R. Assoc. of St. Louis, 1905.Carl Gayler, Eng. News, June 10, 1909 (notes on the design).

(2) Washingtoñ Arch Bridge across Harlem River, New York, built1886-1889, two spans at 510 ft., 6 lines (14 ft. apart) of circular plategirder arches 13 ft. deep and 90 ft. rise. Each girder has two hinges.The bridge carries a highway on asphalt paving 80 ft. wide.

Dead load: 225 lb. per sq. ft. for paving plus 15,000 lb. per lin.ft. of bridge for the metal weight of the structure.

Live load: A 20-ton road roller or 100 lb. per sq. ft.; the 8000 lb.per Iin. ft. for the arches were assumed over the whole.length or onlyfrom abutment to center, whichever gave a greater stress.

Wind load: 1200 lb. per lin. ft. of bridge.Unit stresses: For arch flanges (steel) 18,000 lb. in tension and 15,000

lb. in compression. Ultimate of steel: (62,000 to 70,000 lb. per sq. in.Erected on falsework.Weight of metal in the entire strudure:

Steel in arches and bracing.. . . . . . . . . . 7,550,OOO lb.W r o u g h t iron in posts, b r a c i n g a n d f l o o r . 5,928,OOO “Cast and wrought iron in cornice and balustrade.. $234,000 “

- - - - - -14,712,000 lb.

(equals 180 lb. per sq. ft. of floor.)

For further information, see:Eng. Netus, Feb. 20 and Apr. 10, 1886, June ll, 1857.Eng. Record, July, 1888.

(3) Railway Arch Bridge across Niagara, built by L. L. Buck, 1895-1897, span 550 ft., the two trusses are apart c. to c. at crown 30 ft.,SS--

356 ‘DESIGN OF STEEL BRIDGES [CHAP. XV

and at springing 57’ ft. (batter 1 : 10). Bottom chord parabolic withriae of 114 ft., top chord horizontal. Depth of trusses at center 20 ft.,at springing 134 ft. Each t russ has two hinges. The bridge carries

two decks: upper for two tracks of railroad 32 ft. wide, lower for high-way 48 ft. wide.. Bottom chord 48 in. deep, greatest section 633 sq.in. Top chord 30 to 36 in. deep, greatest section 352 sq. in.

Live load for arch trusses: two Engines Cooper’s loading E-40, fol-lowed by a uniform load of 3500 Ib. per lin. ft. on each track for therailroad, plus 50 lb. per sq. ft. on the highway surface, making a totalof 10,000 lb. per lin. ft. of bridge.

Erected without falsework as three-hinged arch, center top chordinserted with initial stress.

The steel weiglzts of the 550 ft. arch ipan are as follows:

A r c h t r u s s e s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3,391,OOO lb.L a t e r a l b r a c i n g ( = l l % o f t r u s s e s ) : 376,000 “Raihoad floor system ( = 1600 lb. per lin. ft.) , 879,000 <‘Highway floor system (.= 1030 Ib. per lin. ft.) 567,000 “

-__-Total.. . . . . . . . . .f . . . . 5,213,OOO lb.

áverage steel weight 9480 lb. per Iin. ft. of arch span.

For further information, seé :R. S. Buck, Trans. Am. Soc. C. E., Val. 40, 1898 (with extensive discussion).Eng. Record, April 24, 1897.

(4) Highway Arch Bridge across Niagara at Clifton, built by L. L.Buck, 1896-1898, span 840 ft . , two trusses apart c. to c. at crown30 ft. and at springing, SSi ft. (batter 1 : 8). Trusses are parabolic withparallel chords 26 ft. apart, rise 150 ft.! and have two hinges. Thebridge carries a highway on plank flooring with two tracks of electricrailway, total width 46 ft.

Live load for trusses: 1500 lb. per lin. ft. of bridge of the railway(max. axle load 14,600 lb.) plus 50 lb. per sq. ft. of remaining surface,making a total of 2700 lb. per lin. ft. of bridge. Wind pressure a tfloor leve1 200 lb. per lin. ft. carried down to the abutments throughthe end bents; and 30 lb. per sq. ft. of 12 times the exposed, surface ofone truss (including the bents) carried through the truss laterals to theabutments:

Unit stresses for trusses: for live load stresses in chords 12,000 lb.;in web members 10,000 Ib.; for dead load plus temperature stresses 24,000lb. in chords and 20,000 Ib. in web members; al1 reduced for compression

according to “* Ultimate of steel: 60,000 to 68,000 lb. per sq. in.

ART. 191 .ARC%I BRIDGES 357

Erected without falsework as three-hinged arch, center top chordinserted with initial stress corresponding to two-hinged arch.

The steel weights are as follows:

Arch trusses. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,673,400 lb.Bents between floor and trusses . . . . . . . . . . . . . . . . . . . 450,500 “Longitudinal struts . . . . . . . . . . . : . . . . . . . . . . . . . . . . . . 150,700 “

- -Total trusses. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2,274,600 lb.

Lat.eral bracing ( = 17 % of trusses) . . . . . . . . . . . . . . . . 383,500 (IShoes and bearings ( = 10 % of trusses) . . . . . . . . . . . . . 226,600 ”Floor system ( = 900 lb. per lin. ft. of bridge). . . . . . . . 766,300 “

d--Total . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3,651,OOO lb.

Average steel weight 4350 lb. per lin. ft. of arch span (equal to 95 lb.per sq. ft. of floor).

For further information, see:L. L. Buck, Papers of Ins<itution of C. E., London, 1900.F. C. Kunz, Papers of Austrian Soc. of C. E., Vienna, July 28, 1899.Engineering, London, May, 1899.

(5) He11 Gate Bridge across East River, New York, for 4 R.R.tracks, being, built by Gustav Litidenthal for the N. Y. Connecting R. R.Actual construction begun in 1911, to be completed 1915. Span 1000 ft.,between hinges 977+ ft. The two-arch trusses are 60 ft. apart, 40 ft.deep at center, 140 ft. deep at en+, with a rise of bottom chord of 220 ft.,and have two hinges.

Dead load: 15,000 lb. per lin. ft. of bridge for ballasted floor, trackconstruction and electric conduits, plus steel weight of structure.

Li& load: On each track Cooper’s loading E-60. Impact considered(see page 24), but reduced 30% for stresses due to live load on al14 tracks. Wind pressum: 500 lb. per lin. ft, at the plane of thetracks plus 30 Ib,. per sq. ft, ‘on exposed surface of bridge.

Unit stresses: For tension in floor system (structural steel) 20,000 lb.;in trusses (hard steel) 24,000 lb.; for compression reduced according to

4.r

Hard steel for trusses 66,000 to 76,000 lb. Ultimate, min. elonga-

%ion 20% in 8 in.Estructural steel for floor system 62,000 to 70,000 Ib. Ultimate, min.

elongation 22% in 8 in.TO be erected without falsework as & three-hinged arch, center top

chord to be inserted without initial stress,

358 DESIGN OF STEEL BRIDGES [C H A P. XV

The estimated steel weights are as follows:Bottom chords. 11,400,OOO lb.T o p c h o r d sW e b m e m b e r s .

3,600,OOO “

Hangers..5,200,OOO “

G u s s e t p l a t e s .1,400,OOO “1,400,OOO “

Total trusses . 23,000,OOO lb.L a t e r a l b r a c i n g ( = 16% o f t r u s s e s ) . .Floor system ( = 9300 lb. per lin. ft. of bridge)

3,700,OOO ::

Shoes and bearings ( = 8.7% of trusses).9,100,OOO2,000,OOO “-___

Total.. . . . . . . .37,800,000 lb.

equal to an average of 38,500 lb. per lin. ft. of bridgeand a proportion of steel weight to working live loads(24,000 lb. per ft.) of 1.6.

From the calculated vertical deflections of the bottom chord of thearch truss and of the floor line, for a live load of 10,000 Ib. per lin. ft. oftruss, diagram Fig. 23 has been prepared, which shows the stiffness ofthe design. The maximum range of deflection of the floor line (down-ward plus upward) for the imaginary condition of the live load on theleft half and then on the right half of the span is 4.78 in. which isonly &o of the span.

Shows Botion Chord of Truss and Floor-Line underno Load.- - - - - ------_ PT C. Lof Botfom Cbordand FJomLine under Load cm onehalf ofSpan.--:----- ,, 9, 7, I> 1: 7, 7, 11 7, ,> 9, wbole span.

Scale for Deflect/ons, $“- 1:”

FIG. 23.

For further information see:Connecting Raihoads in New York City, Railroad Gazette, May 11, 1900 (gives

location of bridges).Projected Steel Arch Bridge over East River at Hell Gate, New York City, Eng.

Neux, May 30, 1907 (with photographs of design); a!so Jan. 8, 1914.Design of Proposed Hell Gate Bridge, Eng. Rec., June 8, 1907 (criticism of arch

design) .


Proposed Steel Arch Bridge, N. Y. Conn. R. R., Railroad Gazetle, May 31, 1907(with photographs of design).

Steel Arch Bridge over East River, N. Y. Conn. R. R., Railroad Gazette, Nov. 10,1911.

(6) Arch Bridge across St. John River at St. John, N. B., Canada,designed in 1912 by C. C. Schneider and F. C. Kunz, being built by theProvincial Government of New Brunswick, Canada, actual constructionbegun in 1913, to be completed 1914. Span 565 ft. (the longest spandrelbraced arch); the two-arch trusses are vertical and 41 ft. apart, 8$ ft.deep at center, 60% ft. deep at springing, with a rise of bottom chordof 61$ ft., and have two hinges. The bridge carries a roadway 36 ft.wide and two sidewalks at 7 ft., total 50 ft. Roadway pavement creo-soted wood blocks 4 in. deep on reinforced concrete slabs 6 in. deep.

To be erected without falsework. Each bottom chord is provided forerection with an additional center hinge to act as a three-hinged arch whenthe two halves are connected, with a traveler and most of the dead load oneach half; the top chord at center to be inserted without initial stress, sothat the rest of the dead load and the entire live load will be carried by atwo-hinged arch.

Web plates of bottom chord are 42 in., of top chord 30 in. deep.Dead load for trusses after completion (in lb. per lin. ft. of bridge):

Flooring, rails, water pipes, etc.. . . . . . . . . . 5,100Floor system. . . . . . . . . . . . . . . . . . . . . . . . 1,300Trusses and bracing (average) . . . . . . . . . . . . . . . . . . . . . . 5,600

- -Total (average), . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12,000

Maximum load during erection:Stringers, floor beams and rails.. . . . . . . . . . . . . . . . . 700Trusses and bracing (average) . . . . . . . . . . . . . . . . 5,600

- -Total (average). . . . . . . . . . . . . . . . . 6,300

Traveler 75 tons total.For the erection stresses and the dead load stresses the actual panel

concentrations during erection and after completion were used.Live load for Floor System:

For roadway stringers: One 20-ton road roller.For trolley stringers and floorbeams: On each track 2 coupled

8-wheel 4 0 - t o n cars p l u s 25y0 impact.For sidewalk: 100 lb. per sq. ft.

Live load for Main Trusses:3000 lb. per lin. ft. of bridge (this represents 60 lb. per sq. ft. between

railings or on each trolley track 40-ton cars, 40 ft. long and 40 ft. apartand 36 lb. per sq. ft. of remaining surface).

360 DESIGN OF ‘STEEL BRIDGES [CHAP. XV

Wind Pressure:On finished structure and during erection, 50 lb. per sq. ft. on both

trusses and floor in elevation, and on finished structure 30 lb. per sq.ft. plus a moving lateral load of 150 lb. per lin. ft. along the topchord.

Temperature change of f 75” Fahrenheit to be considered.Rsversal of stresses not to be considered except in connections.Unit stresses‘ (in lb. per sq. in.) :For dead and live load stresses plus impact (if any), tension 16,000,

compression 16,000 - 70 i, bearing on masonry 600.

For (d + E + t) stresses as kell as for erection stresses, the abovepermissible stresses may be increased +

Ultimaté for structural steel, 56,000 to 64,000, for steel castings min.65,000.

The estimated steel weights of the arch span are as follows:

Bottom chords. . . . . . . . . . . . . . . . . . . 1,641,OOO lb.T o p c h o r d s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Web members. . . . . . . . . . .

792,000 “.333,000 “

Total for trusses. . . . . . . 2,766,OOO lb.Bracing ( k 14% of trusses) . . .Floor system ( = 1280 Ib: per lin. ft. of bridge). . . . .

379,000 ‘¡

Fences.........................................721,000 “

Shoes and bearings ( = 5% of trusses). . . . .59,000 “

140,000 “___-

Total.. . :. . .: . 4,065,OOO lb.

equal to an average of 7200 per lin. ft. iof bridge or145 lb. per sq. ft. of floor.

Yearo f

COmtìuC-tion

1913-191

1 8 9 5 - 1 8 9

1 9 0 4 - 1 9 0

1888

1 8 9 0 - 1 8 9

1909

1895-189

-

1

L5

k

7

5

1

1

-

16

-

Highway LOCPJame and location

tionor

railway o f5oor

-

1

St. John River Highway TOPmidge at St. John,N. B..* Canada.Niagara Railway wtraok T O Pbridge,* U. S. A. Highway Top

~_~Vi$coo~èl$alls, 2 traok R. R. TOP*

Rhodesia, Alrica. ~___2ake Street bridge, Highway Top

Minneapolis,Minn., U. S. A. ~_~

Driving Park Ave. Highway TOPbridge over theGenessee river atRoche$eST, AN. Y.,

Fort Snelling Highway TOPbti,dge out?~

Mims”‘~~Af’““’~_~

Panther Hollowbridge, Pittsburgh, Highway Top

Pa., U. S. A.

TABLE 61-s

Notable Spandrel Braced Arch Bridges

50.0 560 60

3 2 . 247.5

30.0 -‘90500 /

30.0

35.0

48.0 - 7 4 . 5364

60.0

RiseWidth

SPl””f

be+sm

'"'it~""~ft.

-

Depth oftruss

ft.

9.3 2 8.5 60.0 165.8

4 . 8 2 -134.027.5-20.0

- - - - -5.5 2 15.0 105.0 53.3

5.1 3 6.0 96.0-76.0

~-- ~__ ~ ~6.3 3 12.0 79.0 35.0

4.9 3 10.0 84.5 36.4

-~~~~

8.0 3 4.75 49.75 76.0

-

tt

-

-

-

Distance b e -tween trusses

pum- I-

2

/ i

41.0 41.0

-30.056.82

2 - 5 3 . 7 527.5

-29.02 i / 29.0

1 /2 -46.020.0

-36.036.02

-13.5-.l li 13.54 13.5 13.5

13.5 13.5

-

- 1

I!

i

-_1

-.

-.

-

[nclina-tion oftruss

hne toverticalplane

: 10

1:8

0

‘1 : 6.1

Literature

lhans. Am. Soc.C. E., Vol. 40.

1898; Eng,x

Reco;~$pnl, bJ

-Eng_Recor;i,H

Sept., 1905.

Ens.Record,Dec., lS95.

Eng. Record,July, 1891; R.R . Gazette,July, 1891.

7zzpz&Record,June, 1909

Eng. Record,June, 1898.

Note.-Bridges marked * were erected without falsework.

TABLE 61-b

Notable Crescent Arch Bridges-

17

2.5

3.z

7.:

7.c

1.:

-

-

NUEPber of

trussesin

span

6.S 2

-

zh

1

í

-

1

1:

-

t

-

-

/

-

. -

-

Span Risez l f

LiteratureMH

%

Distance be-tween trusses

Inclina-tion of

at at truss

orown spring- plane t oing line vertical

plane

ft.

Depth oftruss

YB,¡-of

construc-tion

LOCPtionof

floor

Top

Widthbetweenrailings,

ft.

16.4

ft.

541 187

15.0 525 139

4 3 . 2 513 77

Highway\Jame and location or

railway

/--

-

ft.-

20.3 I 65.6 ( 1 :9 E n g . News,Aug., 1884.

Eng. R e c o r d ,aMarch, 1891. _

r

3 2 . 2 I 0Garabit Viaduct S i n g l e t r a c kat St. Flour, * R. R.

FEWHX.

1880-1884

1876-1877

1891-1892

1904-1905

1897-1899

.l-

TOP

Mid-d l e

2 3 2 . 8 0Douro Bridge, Pia Single trackMaria, * Portugal. R. R.

Grunenthe! bridge Sple, traa:over the Baltio

canal, Germany. highway

13.0 4 9 . 2 1 : 8.6

~___1:s Eng. News,

Aug., 1891.ü

Fe: ““9> .


2 13.5 1 0

l-Austerl+z b r i d g e Elevatedover S&~&e;~?ans, R.R. 2 tracks.

Mid-d l e

2 7 . 3 4 6 0 ô5.1 6 : 2 5 . 6

3 4 . 5 347 36

3 . 7 0

Bridge over the HighwayRhine at Worms,

Germany.

TOP 6.9/ O

LNote.-Bridges marked * were erected without falsework.

TABLE 61-c

1886-1889 Washington brideover Harlem river

at New York ,U. S. A.

1899 Bridge AlexanderIII oveì Seine at

Paris, France.

-

L

-

_-

-

HighwayOì

r$lway

Highway

Highway

-

-

-.

-

Loca-tion

offloor

TOP

TOP

Widthbetmm

ra1ftgs9

80.0

131.2

-

-

-.

-

Notable Arch Girder Bridges-

SPlan Risef

ft.

509 90be-

tweenhinges____

3 5 2 2 0 . 6_-

1

l-

Depth oftruss

,661 2 1 1 3 . 0 (........ / 3 8 . .

NUN?ber of

trussesin

E3p*ll

f3

15

--:-.l1


Inylf-

at at trusscrown spring- plane to

ing line verticalplane~__

ft./

ia14.0 5a14.0 0

- - - -14f.úJ9.4 [email protected] 0

l i

-

-

-

Litersture

E

TABLE 61-cNotable Arch Truss Bridges

-

2LOCa-tion

o ffloor

-

t

i

Width,etweenrailings,

ft._ _ -

Mid-dle

93.0 977 220to 4.4

cho"

TOP 46.2

TOP 14.75 176 14.1

Bot-tom

46.0 614

- -Bot-tom

46.5 595.-

.,-

-j-91 to t6.5‘

ch%?

TOP 27.9 588 219 i2.5‘

TOPBot-tom

26.3 566

-

- -

/

-

0

1

Depth oft russ


“py

at at trusscrown spring- plane t e

ing he verticaplane

ft.

NUIIPb e rof

trusseIin

SPZUl

YSXTco,,&,- Name and location

Highwayor

tion railway

Nllm-ber of at athinges

’ !

“‘h”” BP”ng-mg hne

ft.

Literature

l 1-L

-

1 9 1 1 - 1 9 1 5 East River bridge 24.4 2

-60.0 60.0 0

32.:

64.5

30.0

17.6

68.5 1:s

39.2

37.7

29.5

31.8

109.5 1:4

29.5 0

-lo31.8

42.5 2 16.4 84.4 1:7

2 1 . 5 19.7 52.5 1 1:Q.ô

L- -

Eng. News,May , 1907 .R. R. Gazette

May, 1907 andNov., 1911.

2 40.0 140.0

1 8 9 6 - 1 8 9 8 Niagara,andClift,op,~bpe, *

Highway

1896-1902 Viaduct over the SinglftrackViaur* Carmeux-

Roda Ry.. France.

2 126.11 P.I. C.E.,1900.Engineering,May, 1899.

3 / 11.0 1 88.0 Eng. New.S e p t , 1 9 0 0 .

2 1 15.7 1 34.6 Eng. News,April, 1899.Engineering,March, 1900.

1 8 9 7 - 1 8 9 8 Rhine bridgeat Bonn, Germany.

Highwsy

_____2 15.7 34.6

--I-

1 8 9 7 - 1 8 9 8 Rhine bridgeat Düsseldorf

Highway

Germany. ’

Eng. News,April, 1899.Engineering,Maroh, 1900.

Eng. New,S e p t , 1 8 9 8 .


0 / 13.1 1 40.01893-1897 Wupper Viaduct 2 Trackat Müngsten, * R. R.

G e r m a n y .

1881-1885 Douro bridge,Luia 1, * Portugal.

HighwayHighway

2 55.0

Bellows Fallsbridge over Con-

necticut river,u. s: A.

Bot-tom

26.0 27.8 ’ 0

I /

Eng. Record,Nov., 1904.

Trans. A m .Soc. C. E., Vol.

61, 1908.

Highway 2 14.0 0 38.5 2 27.8

51.0 2

31.9 2

_ _ _ _

43.4 4

~__2

35.6 2

44.0 2

~__2

_ _ _71.8 2

_ _ _ _ _16.S 2

_ _ _ _.lû.O 7

41.3

27.9

‘16 5’ lö:5(16.5

16.0

20.3

26.3

23.0

_-

_-

_-

_-

_-

i,

L

29.5

1904-191

1893-18!

1 5

-24

-

7 4 ;

.-

-18

-.

-

Mid.d l e

33.5

. .

52.033.0

69 .0 17.7 1 0 . 5 1 8 . 4

16.4 32.S

Levensau bridgeover the Balticcanal, Germany.

41.3 0

27.0’ 0 Eng. News,Deo., 1912.

Single trackR. R.; high-

way.

Singlf Back Bot-t o m

5231911-19:

1868-18:

aga river,Kamerun, Africa.

52016.5) ( R. R. Gasettc,

:e:ij OAug> 1888.

Eng. News,June, 1909.

-__-5 6 . 4 _......_ Eng. Record,

oct., 1888.R. R. Gasette,

Sept., 1888.

Mississippi bridgeat St. Louis,* Mo.,

U. S. A.

Highway2 track R. R.

68.5

__-123 4.0

-.

-.

-.

-.

-.

-

1888%18Z Adda Viaduct atPaderno, Italy.

HighwaySingle track

R. R.

TOPTOP

23.0 492

TOP 16.0 4 4 9 56to’S.Obottomc h o r d .

38.8 Eng. News,oct., 1902.

1902 Rio Grande bridgePacificRK;;* Costa

1

Single trackR. R.

31.4 Eng. News,May, 1907.

TOP 34.0

--Bot-tom

44.0

41.3

383

TOP 377

TOP 1 7 . 7 336

TOP 65.0 326

-

-

-

Oakland bridge,Pitts$,uighi.Pa.,

Highway 1 4 . 0

38.0Bridge over theRhine at Worms,

G e r m a n y .

2 trackR. R.

29.5 0 Eng. News,Oct., 1901.

Eng. RecordSept., 1901._~

44.3 1 : 12 2 Eng. Record,Junc, 1899.

_ _ _ _ _43 1 : 10 Eng. News,

Aug, 1894.Eng. Record.

/_ _ _ _ _ _

___- Feb., 1904.

0l ! Eng. News,Nov., 1896.Eng. Reoord,April, 1903.

.1895-l% Highway 103 13.6i 0 1 3 . 8Kornhausbridgeo~eì the Aar at

Bern, Switserland.

Stony Creekbridge,. Canadian

Paafic Ry.

sing$ pk1893

1896-

J.-

20.3 114.4 3

-

2.8Mirabeau bridge,Paris, France.

Highway

Note.-Bridges marked * were erected without falsework.

366 DESIGN OF STEEL BRIDGES [CHAP. X V

ADDITIONAL INFORMATION ON ARCH BRIDGES

Max am Ende, Arch Design for Firth of Forth Bridge, Engineering,London, 1880, 1, p. 168.

Max am Ende, Bridge across North River in New York, crescentarch, 2800-ft. span, The Engineer, London, 1889, 1, p. 411.

Berlin Elevated Railroad, Official Report, Zeitschrift für Bauwesen,Berlin, 1886 (contains valuable information with many plates on archbridges and train sheds).

Leibbrand, Charles Bridge at Stuttgart, Highway 60 ft. wide, plategirder arches, 165 ft. max. span, rise Tl-i, Zeitschrift für Bauwesen, 1895(complete description with many plates concerning the design, calcula-tion, masonry, foundations, etc., of this unusually beautiful structure).

Report on Connecticut Ave. Bridge across Rock Creek (competitivedesigns), Washington, D. C., 1898, also Eng. News, Jan., 1898.

Frank H. Cilley, Statically Indeterminate Frameworks, Trans. A. S.C. E., Vo’. 43, 1900 (with extensive discussion, also by EuropeanEngineers).

C. W. Hudson, Comparison of Weights of a Three-hinged and aTwo-hinge,d Spandrel Braced Arch, Trans. A. S. Cr E., Vol. 43, 1900.

Lewis D. Rights, Erection of the Bellows Falls Arch Bridge, Trans.A. S. C. E., Vol. 61, 1908.

Crooked River Bridge, Oregon, s. tr., spandrel branch arch with twohinges. span 340 ft., depth of truss 12 ft. center to crown, 72 ft. atends, 12 panels. Trusses spaced 18 ft. at top chord, 30ft. at hinges(batter 1 to 12). Eng. News, March, 1913.

C. R. Grimm, .Arch Principie for Long Spans, Trans. 8. S. C. E.,Vol. 71, 1911 (with interesting discussion).

CHAPTER XVI

LONG SPAN BRIDGES IN GENERAL AND EXAMPLES

ART. 1. SELECTION OF DESIGN

Long span bridges are of the arch, cantilever or suspension type.Simple truss bridges cease to be economical for spans over 700 ft.,

if the length of span is selected for smallest total co&, as the conditionswhich make an intermediate pier too expensive increase also the cost oferection falsework and the chances of accident during erection. TheIongest built simple spans up to 1910 were 550 ft. for a railway (KentuckyCentral R. R., Cincinnati, built 1889) and 586 ft. for a highway (overMiami River, built 1905); in 1911 spans of 668 ft. were completed (seePlate XIII).-

Probably the longest designed simple span is 850 ft. for a double-track railway;steel weight about 14,000,OOO lb., which conforms approximately to th‘e fo?muIa ofsteel weight w in lb. ‘per lin. ft. dcduced from steel weights of well propor-tioned double-track through bridges for E-50 loading up to 500-ft. spans, viz.:w = 17 1 + 1500, where 1 = span in feet, see Chap& XI.

Arch bridges may be economical for any span, whether 200 or 2000 ft.providing that the conditions of the ground are favorable. If the groundcan resist the horizontal thrust of the arch truss without much additionalmasonry, considerable cost compared with a simple truss can be saved.Their great advantage is the possibility of erection without falsework,especially for deck arches, and, if properly designed, their small deflec-tions under dead Zocd and, therefore, small secondary stresses. Thelongest existing arch span of 840 ft. (see page 356) carries a light highwaytraffic across the Niagara River. An arch span of 977; ft. to carryfour heavy railway tracks on a solid floor is being built for the N. Y.Connecting R. R. Longer arch spans have been designed: 2000-ft.spans for Detroit (tunnels built), 3100-ft. span for New York across theNorth River (tunnels built), 3300-ft. spans between Italy and Sicily, etc.

Cantilever bridges are generally economical for spans between 600 and1600 ft.; the longest existing span (Firth of Forth Bridge) has 1710 ft.A double-track cantilever with a center span of 1800 ft. is being builtnear Quebec, Canada (see Plate XLVIII). Their principal disadvantage(especially for shorter spans and railroad traffic) is the great deflection

367

368 DESIGN OF STEEL BRIDGES [CHAP. ‘XVI

of the cantilever arms and the expensive mechanical devices for adjust-ment during erection. Many cantilever bridges have been built wheresimple spans would have been cheaper and more rigid. (Market StreetBridge in Philadelphia, Eng. News, March 3 and April 7, 1888; Pough-keepsie Bridge across the Hudson River, Trans. Am. Soc. C. E., Vol. 18,1888; also Eng. Record, June 16, 1888, RaiZroadGaxette, July 1, 1887, Eng.Record, Aug. 18, 1906.)

For shorter spans where falsework cannot be used, it may be prefer-able to design simple truss spans which can be erected on the cantileverprincipie (see Eng. Record, June 3, 1905) : Bridge across Columbia Riverin California (Eng. News, 1895, p. 266), Atbara Bridge in the Soudan(Railroad Gaxette, May 26, 1899, and Engineering, London, June 30,1899), Ohio Bridge at Benwood for the B. & ‘0. R. R. (Eng. Record,June 10, 1905, and Trans. Am. Soc. C. E., Vol. 55, 1905), Bridges forWestern Maryland Railroad and B. & 0. R. R. (Eng. Record, July 15,1905), Miles Glacier Bridge for the Copper River Railway, Alaska (Eng.Record, August 6, 1910), Celilo Bridge, Oregon Trunk Line (Eng. Record,Jan. 20, 1912, and Eng, News, Feb. 22, 1912)

or, by jloating them into place: The 523-ft. span of the Ohio Con-necting Railway at Pittsburgh, floating weight 1000 tons, height of baseof rail to water 80 ft. (Eng. News, Sept., 1890), the 415-ft. spans of theHawkesbury Bridge in Australia (Railroad Gaxette, Aug. 10, ISSS), CoteauBridge in Canada (Eng. News, April 12, 1890, Eng. Record, Dec. 3, 1910),Miramichi Bridge in Canada (Eng. News, Mar. 26, 1903, and Eng. Record,March 14, 1903), Passaic Draw, D. L. & W. R. R., floating weight2000 tons (Eng. News, Dec. 31, 1903), Fraser River Bridge in Canada(Eng. News, June 22, 1905), French River Bridge, C. P. R. R. in Canada,“end launching” erection (Eng. News, July, 1908), Sanaga Arch Bridge,South Africa (Eng. News, Dec. 26, 1912), etc.

For spans from 1600 to 2000 ft., the stiffened suspension bridge may becheaper than the cantilever, depending on the live load, the permissibleworking stresses, etc. (for light highway traffic even for much shorterspans) ; the cantilever, however, is preferable to suspension bridges onaccount. of its greater rigidity.

For spans over 2000 ft. and especially for railroad traffic, a cantileveris at present hardly ‘practicable, and if extraordinary conditions do notfavor an arch, the stifened suspension bridge is the only possible type.The longest exist,ing suspension span of 1600 ft. has the WilliamsburgBridge in New York (Brooklyn Bridge center span 1595 ft., ManhattanBridge 1470 ft.) (see Plate XLVIII a).

ART. 21 LONG SPAN BRIDGES IN GENERAL AND EXAMPLES 369

The longest suspension span carrying its own weight only was used fora telephone cable in Switzerland built in 1895 across the Walen Lake witha span of 7900 ft., which was broken by a storm for the third time in 1898and then removed. The three wires composing the cable had an ultimatestrength of 250,000 lb. per sq. in. Considering the dip of the cables of590 ft. and the difference of 750 ft. in the elevation of the cable supports,the stress from the steel weight was about 54,000 lb. per sq. in. only.This shows the influente of snow, ice and wind at the time of itsdestruction.

If we call 7900 = 2, 590 = j, 750 = d, we have length of cable

from c we determine the total weight of the cable and this ‘divided by 1

gives weight w per lin. ft. of span. Then, the horizontal thrust H = $ and the

vertical component I’ of stress at the upper and V’ at the lower support is

V = g + f H and V’ = $ - f H and the maximum stress on cable at the

upper support S = 2/H? + Vz.

If d = 0 and 4 = 10, a wire from its own weight would have a stress

of 60,000 lb. per sq. in. for a free span of 13,000 ft., of 120,000 Ib. for 26,000 ft.,

of 180,000 lb. for 39,000 ft., etc. If “f = 8, thespans are about 12% greater.

Complete designs for suspension bridges with a center span of morethan 3000 ft. have been made.

It is impossible to compare the so-called factor of sufety of long spanbridges of different spans, even of the same type, as that depends not onlyon the kind of steel used, the assumed live loads and the permissible unitstresses, but also dn the truss system, the lengths of the live load causingthe greatest stresses in the different members, the consideration of alter-nate stresses, the assumed impacts, the relation of the assumed to theactual working live loads, the possibility of a future increase in the latter,the arrangement and sections of the details in the heaviest members, etc.

ART. 2. LIVE LOADS AND PERMISSIBLE UNIT STRESSES

For the Jloor system, if only one kind of traffic is to be carried, the liveloads and permissible unit stresses as given in Standard Specificationsfor Railway and Highway Bridges should be used (see Vol. 1).

For the trusses of long span bridges, however, if they carry more thanone line of traffic, the live loads are often reduced or the permissible unitstresses raised as the probability of a concurrence of the greatest stresses

370 DESIGN OF STEEL BRIDGES [CHAP. XVI

caused by the different lines of live load decreases with the number of lines.For Railway bridges the snow load is neglected if the floor is open (ties) ;if wind load is considered for the trusses the usual increase of 20 to 30%of their otherwise permissible unit stress is sufficient. For Highwaybridges, to assume wind or snow load together with the maximum liveload seems unnecessary; assuming a strong wind or a thickness of com-pact snow of 12 in., or of ice and sleet of 4 in., representing about 15 Ib,per sq. ft., a corresponding amount may safely be deducted from themaximum live load on footwalks and roadway. (A crowd of peopleweighing 50 lb. per sq. ft. is hardly able to move.) For long span bridgeswith combined trajk it will often be sufficient to assume a heavy workingVive load (plus impact) on the railway tracks and 20 lb. per sq. ft. onthe roadway (representing a moving snow load or light highway traffic)with 16,000 to 20,000 lb. unit stress in tension for structural steel, depend-ing on span length and traffic; but to analyze also the case of a heavierextreme live load on the tracks (providing for a probable future increase),plus a live load on the highway of 50 lb. per sq. ft. and limit the per-missible unit stress in tension to from 20,000 to 24,000 lb. depending onthe live load assumptions, the span, the truss system, etc., in orderto ascertain whether the stresses in the truss members, especially thosein tension which could be made up of eyebars, would be reversed bya considerable,future increase of live load and other unforeseen circum-stances (see also page 383).

Unit stresses for Niclcel steel in tension members can be increasedSO’%, for compression members only about 15 to 40 Yo, depending onwhether the member is light and slender or heavy and short.

It should be considered that every additional Pound of dead load overthat absolutely necessary is merely additional live load and will, forvery long spans, add severa1 pounds of steel weight (see Examples andChapter XVII, page 410), making a certain design not only expensive butsometimes even impracticable.

It will often be possible to modify the outline of the trusses so as toreduce the section of the heaviest members in order to reduce thegreatest single weights to be handled in the shop, during transporta-tion and erection.

A design of a long span may on the stress sheet appear stronger butbe comparatively inferior in its details to a lighter bridge, since the detailsare mostly based on experience with similar smaller bridges and may notbe able to develop the ful1 strength of the member which its cross-sectionindicates. For greater area of cross-section than 1200 sq. in., for example,

Am. 31 LONG SPAN BRIDGES IN GENERAL AND EXAMPLES 371

our present compression truss members of a box section will have to beradically changed, which will increase the cost of ehop work, difficultiesof transportation, inaccuracies of stress distribution, etc., and maymake a reduction in the intended live load and the dead load the bestsolution to produce a satisfactory bridge of that particular design (QuebecBridge). On the other hand, the so-called “skinning” of a design andespecially of the details based on experience with existing bridges ofmoderate size, is exceedingly dangerous in long span bridges as thereare very few precedents to follow.

In designing long span bridges, especially for densely built-up cities,ít should be remembered that the cost of the bridge proper may be thesmaller part of the total cost of the entire crossing (in the BrooklynBridge crossing it is only one-third of the total cost), the cost of ap-proaches, terminals and right of way being the greater part. A changein the originally assumed “local elements” (clear height for navigationand permissible maximum grades, difficult foundations; expensive land,etc.), may easily double the estimated total cost of a crossing. Thetotal cost of the Brooklyn Bridge crossing, for example, is $17,000,000,while the first estimate of the total cost was only one-third; the Firth ofForth crossing costs $16,000,000, while the first cost estimate was onlyone-half.

The clear height for navigation (see page 122) is an important factorin the cost of the approaches to the bridge. Some recent specificationsfor long span bridges seem extravagant in this respect. The East Riverbridges in New York have only 135 ft.; the Firth of Forth Bridge, theQuebec Bridge, the proposed North River Bridge, and the first proposedSydney Bridge (span 1350 ft.) 150 ft.; the recently modified specifica-tions for the Sydney Bridge (span 1600 ft.) cal1 for 170 ft.; for the pro-posed Channel Bridge between England and France 180 ft. are specifiedand a proposed bridge (span 2700 ft.) at Liverpool shall have 200 ft.

For literature on Long Span Bridges, see page 423.

ART. 3. CAMBER

The corre& method to provide a camber in the trusses of a Iong spanbridge is to shorten or lengthen each member by the amount by which itis lengthened. or shortened, respectively, due to a certain loading. Arational way would be to use the dead load and one-half of the live loadcovering the whole bridge. In very large bridges, it is even sufficient todetermine the camber for dead load only, as was done in the Blackwell’s

3 7 2 DESPGN OP STEEL BRIDGES [CHAP. XVI

Island and Quebec Bridges. Concerning the calculation of deflectionssee page 296.

It is the usual practice to manufacture and erect continuous compres-sion chords of long span bridges so that the joints are open on one sidewhen the members are put in place and will gradually close until the deadload reaches its ful1 value, the turning being made possible by the use ofdrift-pins and bolts of a smaller diameter than that of the holes for thefield rivets. If the turning should actually occur, part of the secondarystresses from dead load are thus eliminated after the bridge is erected,but bending stresses occur during erection. These may be quite seriousin large bridges, especially cantilever bridges, in which the erectionstresses are considerable, and instead of running the risk of over-stressingduring erection, while the connections are not yet riveted and the jointsnot closed, it seems advisable to give the members such bevels that thejoints are closed at the beginning and to make provision, if necessary, forthe secondary stresses due to the ful1 dead load.

ART. 4. EXAMPLES OF CANTILEVER BRIDGES

(See Table 63 and Plates XLVIII, XLIX, LII)

(1) The Firth of Forth Bridge, built by J. Fowler and B. Baker,1882-1889, 5350 ft. long; greatest span 1710 ft., with a suspended span of350 ft. (height 41 and 51 ft.), the anchar arms 690 ft. each (height oftruss at pier 330 ft.) and a greatest width of 120 ft. c. to c. of trusses(inclined toward the top), was designed for two tracks, each for a liveload of approximately Cooper’s loading E-22, allowing a compres-sion of 17,000 lb. per sq. in., with no reduction for buckling, and a ten-sion of 16,350 lb. per sq. in.; no impact was considered. The steel inspecimens had a tensile ultimate for compression members of 76,000 to83,000 and for tension members 67,000 to 74,000. The specified elonga-tion in 8 in. is rather low; so that the “Coefficient of quality” (productof elongation in 8 in. and ultimate, see Val. 1) is about 1,500,000, thatis the same as specified by Standard Specifications of today for steelwith a lower ultimate.

The steel weight comparable to that of other Cantilever bridges withonly one center span is as follows:

S u s p e n d e d span ( 3 5 0 f t . l o n g ) . . . 1,840,OOO 16.2 Cantilevers (each 680 ft. Iong) . . . 24,080,OOO2 Anchar arms (each 690 ft. long) . 24,376,OOO ::2 Towers (each 145 ft. long) . . . . . . . . . 21,575,OOO

-Total . . . . . . . . . . . . . . . . . . . . . . . . . 71,871,OOO lb.

or per lin. ft. (3380 ft.) ‘21,260 lb.

PLATE XLVIII A

-Thebes Bridge over Mississippi -(Double Track. - BuiIt~l~OZ-05.)

- Cantilever Bridge for 4 Tracks -( Design Mude in 1904.)

- Beaver Bridge over Ohio River-( Doub\e Track I Built 1910.)

- Highway Bridge overRhine River between Ruhrort and Homberg, Germany-( Built 1907.)

-Memphis Bridge cwer Mississippi River-( Built 1915.)

Elevations of Notable Cantilever Bridges. (Facing page 37:

PLATE XLVIII A (Cont’d.)-I

- Fi r th o f Forth Br idge -( DoubleTrack; Built 1882-89. )

-Qvebec Br idge over St. Lawrence Kiver-(Double T r a c k > In Con5truction.)

- Quebec B r i d g e over S t . Lawrence River-( Double Track, 2 Trolleysi Design Madein 1 9 0 9 . )

Elevations of Notable Cantilever Bridges. page 373)


Therefore a proportion of steel weight to working live load (4480 lb. perft. of bridge) of 4.7.

Total steel weight (length 5350 ft.) 114,000,000 lb.

For further information, see:Westhofenin Engineering (London), Feb. 28, 1890, and Prof. Barkhausen, Papers

of German Society of C. E., Berlin, 1888, as the most elaborate.

(2) The Memphis Bridge, built by G. S. Morison, 188661892, 2258 ft.4 in. long; greatest span 790 ft., was designed for one track for a liveload of approximately Cooper’s E-40, allowing a compression of 14,000

lb. persq. in., with no reduction for i smaller than 45, and no considera-

tion of impact; and a tension of 20,000 lb. per eq. in., adding animpact of 100% to the live load. The steel in specimens has a tensileultimate for compression members of 69,000 to 78,000 and for tensionmembers of 66,000 to 75,000. “Coefficient of quality” 1,250,OOO to1,400,000.

The steel weighfs are:lAnchorspan621ft.+in . . . . . . . . . 5,122,OOOlb.2 Suspended s p a n s a t 451 f t . 8 in.. 4,658,OOO “3 C a n t i l e v e r arms a t 1 6 9 f t . 41, in.. . . 3,798,OOO ”1 Anchar arm 225 ft. 10 in.. . . . . . . . 1,606,OOO “

--~-Total. . . . . . . . . . . 15,184,OOO lb.

or an average of 6700 lb. per lin, ft. of bridge and a proportion of steelweight to live load of 1.7.

Weight of track and wooden Aoor 650 lb. per lin. ft. of bridge.

Cost: Steelwork per lb. delivered 4.53 cents, erection 1.30 cents per lb., paint-ing 0.05 cents per Ib., track and wooden floor $7.46 per lin. ft.

For further information, see:G. S. Morison’s Report on Memphis Bridge, 1894, also Trans. A. S. C. E., Val. 29,

1893, and Eng. News, Sept. 15 and Dec. 1, 1892.

(3) The Monongahela Bridge, buiIt by Boller and Hodge, 1901-1904,1504 ft. long, center span 812 ft. with a suspended span of 360 ft.(height 60 ft.), was designedfor two tracks, each for a live load of approxi-,mately Cooper’s E-45, with 100% impact, allowing a compression of

21,000 lb. per sq. in. with no reduction for $ smaller than 40, and a ten-

sion of 22,000 lb. per sq. in. The steel in specimens had a tensile ulti-mate for riveted members of 60,000 to 70,000 and for eyebars 63,000t10 73,000. “Coefficient of quality” about 1,500,OOO.

Estimated steel weigkt given in the specifications is 13,520,OOO lb.divided as follows:

374 DESIGN OF STEEL BRIDGES [ C H A P . XVI

Rers.Compression members.Pins, rings, e t c . . . . . . . .Lower latcrals.. .U p p e r laterals a n d

transverse b r a c i n gFloorsystem..........C a s t s t e e l shoes.

Totals .

SuspendedQXl,Il

250,0001,104,000

35,00030,000

109,000463,000

<...<..,<._-_1,991,ooo

TWOcantilever

arms

1,308,0001,630,OOO

60,00072,000

153,000562,000

.,,..<..<<

3,785.OOO

-

-

TWOtowerpanels

77,0001,033,000

40,000. . . .

120,00018,000

114,000-__1,432,OOO

1,912,ooo 104,000 3,651,OOO2,846,OOO 104,000 6,717,OOO

92,000 12,000 239,000100,000 202,000

220,000 620,000886,000 1,929,ooo

. . . < . 36,000 180,000

6,056,OOO 256,000 13,520,OOO

-

FOUFanchor-

ages

-

The actual total steel weight is i4,350,000 lb.,or an average of 9500lb. per lin. ft. of bridge and proportion of steel weight to live load 1.1.

For further information, see:Eng. News, Nov., 1902; Eng. Record, Jan., 1903, and March, 1904.

(4) The Thebes Bridge, built by R. Modjeski, 1902-1905, 2750 ft. 4in. long, greatest span 671 ft., was designed for two tracks, each for aloading of approximately Cooper’s E-50, for the floor and E-45 for the

trusses and, adding 100% impact to the live load stresses, allowing a1

compression of 21,000 lb. per sq. in., with no reduction for ; smal-ler

than 45, and a tension of 20,000 lb. per sq. in. Wind pressure 1000lb. per lin. ft. of bridge. The steel in specimens had an ultimate of62,000 to 70,000. “Coefficient of quality ” about 1,500,OOO.

The steel weights are:2 Anchar spans @ 521 ft. 2 in.. . . 10,960,000 lb.3 Suspended spans @ 366 ft . . . . .4 Cantilever arms @ 152 ft. 6 in.. . . .

6,340,OOO “

Bearings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4,830,OOO “

900,000 “

A total of 23,030,OOO lb. or an average of 8350 lb. per lin. ft. ofbridge and a proportion of steel weight to live load of 0.9.

For further information, see:Eng. News, Nov., 1902, and May, 1900; Eng. Record, July and Sept., 1905.

(5) The first Quebec Bridge, built 1900-1907, 2800 ft. long, centerspan 1800 ft., with a suspended span of 675 ft. (height 97 and 130 ft.),the two anchar arms 500 ft. each (height of truss at piers 315 ft.) and awidth of 67 ft. c. to c. of trusses, was designed for two roadways,each 17 ft. wide, to carry a total snow load of 1600 lb. per lin. ft. ofbridge and two railway tracks to carry :

1. A working live loa,d of about Cooper’s E-30, with a permissibleunit tension and compression reaching 21,200 Ib., which includes impactby means of a “minimum over maximum” formula, but no reduction for

buckling made for + smaller than 50; or,


2. An extreme live load of 50% more (about E-45), with a per-missible unit stress of 24,000 lb. in tension and also in compression for

the chords and main diagonals, and of 24,000 - 100 f in compression

for the posts.Wind pressure for the bridge unloaded to be 25 lb. per sq. ft. along

its total length, and an added moving wind pressure of 25 lb. per sq. ft.,500 ft. long. The wind pressure for the bridge loaded to be 25 lb. persq. ft. In estimating the wind surface, the area of both trusses wasused, 50 $% added to the exposed surface of the floor, and 50 $?$ added toone screen. The wind to act either horizontally or at an angle of 30degrees above or below the horizontal, but the horizontal componentonly considered.

The steel in specimens 60,000 to 70,000 ultimate and a “Coefficient ofquality” of 1,500,OOO.

The actual steel weights were as follows:

R i v e t e d t r u s s e s a n d b r a c i n g . 6,759,OOO 17,449,OOO 16,285,OOO 40,493,OOOEye-bars.Pins.............................

6;:‘;CHl; 6,;57,;;; 6,418,OOO 14,042,OOO

Floor system. 2,430:OOO 4,134:OOO4 5 9 , 0 0 0 1,189,OOO

3,557,OOO 10,121,OOO

9,947,OOO 29,179,OOO 26,719,OOO ô5,845,000Two center post systems. . 5,417,oooB e a r i n g s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $ 6 2 9 , 0 0 0Anchorage-eye-bars and pins . . . .Anchar s h e l l s and b r a c i n g . . . .

446,000744,000

T o t a l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74,081,OOO

or an average of 26,500 lb. per lin. ft. of bridge and a proportion ofsteel weight to live load (6000 lb. per foot) of 4.4.

Price of steel average 5.6 cents per lb. erected.For further information, see:Report of Canadian Roya1 Commission and of C.C. Schneider, 1908 (also Eng.

News, Aug., 1908); Eng. News, March, 1908; and Engineering Record, 1908.

(6) A cantilever design for the Quebec Bridge made by the Pen-coyd Iron Works in 1899 under similar specifications as given abovehad a total length of 2600 ft.; greatest span 1600 ft. with a sus-pended span of 600 ft. (height 100 and 125 ft.), the two anchar arms500 ft. each (height of truss at piers 260 ft.) and a width of 70 ft. c. to C.trusses.


The estimated steel weights were as follows:

Trusscs. . 4,192,OOO 12,987,OOO 14,959,OOQ 32,138,OOOB r a c i n g b e t w e e n t r u s s e s . 5 5 3 , 0 0 0 2,535,OOO 2,535,OOO 5,623,OOOF l o o r s y s t e m . . . 2,275,OOO 3,820,OOO 3,825,OOO 9,920,OOO

--___7,020,OOO 19,342,OOO 21,319,OOO 47,681,OOO

T w o c e n t e r p o s t s y s t e m s .Bcarings..................................................

~,~W&,~Kl~

A n c h o r a g e s . . 1:209:000- - - -

Total................................................... 56,755,OOO

If the suspended span were to be erected as a cantilever instead of by“$oating in,” add to the weight of trusses of suspended span 1,200,OOOlb. This added, gives an average of the total steel weight of 22,400 lb.per lin. ft. of bridge, and a proportion of steel weight to live load of 3.7.

(7) A cantilever design for the Quebec Bridge, made by F. C. Kunzin 1909, total length 2800 ft., center span 1800 ft. with a suspended spanof 600 ft. (height 100 and 125 ft.), anchar arms 500 ft. each (height oftruss at piers 275 ft.) and a width of 75 ft. c, to c. trusses (the existingpiers limited this dimension, 85 ft. would have been better for greaterrigidity and to be able to make the trusses at the piers 300 ft. high).The bridge carries between the trusses a double-track railroad in thecenter and two highways; outside of the trusses a trolley track on eachside, making the distance between the end stringers 107 ft. Panel lengthof truss is 50 ft. throughout.

Dead load (in lb. per lin. ft. of bridge) :Tracks and timber floor 2800, steel in floor system 4400.The steel in trusses and bracing for dead load averages 27,600 lb.( suspended span 9900, cantilever arms 29,700, anchar arms35,840).

Live Load for Floor Xystem:On each R. R. track, Cooper’s E-50 loading plus impact.On each trolley track, two coupled 40-ton trolley ears plus impact.On two roadways (eich 17 ft.) and two sidewalks (each 8 ft.),100 lb. per sq. ft.

Live Load for ikfain Trusses:One Cooper’s E-40 loading per truss plus impact.

ART. 41 LONG RPAN BRIDGES IN GENERAL AND EXAMPLES 3 7 7

Impact :3 0 0

For railroad loading i = 1 L+300 1 i = impact, Z = live load stress,

150___For trolley loading i = Z L+300 1

L = loaded lengthof track in feet.

Wind pressure:Suspended span 30 lb. per sq. ft. on both trusses and floor plus300 lb. per lin. ft., 7 ft. above rail.Anchar or cantilever arms 20 lb. per sq. ft. on both trusses andAoor, plus 200 lb. per lin. ft., 7 ft. above rail.During erection, 30 lb. per sq. ft.

Snow: 1600 lb. per lin. ft. of bridge.Unit stresses for (d + 1 + i) (in lb. per sq. in.):

Tension in structural steel (56-64,000 ultimate). . . .16,000

Compression in structural steel (56-64,000 mtimate) 16,000 - 70 f

Tension in nickel steel (85-95,000 ultimate). . . . . . . .24,000

Unit stresses for (d + 1 + i + w) or, during erection plus wind:increase above stresses 30%.

Unit stresses for (d + l+ 1 + 1; i + w + s) : increase stresses assumedfor (d + Z + i) by 50%.(d denotes stresses from dead load, ì from live load, i fromimpact, w from wind, s from snow.)Reversa1 of stresses not considered.Heaviest truss members: Pier diagonal 1520 sq. in., 54 in. deep;

bottom chord 990 sq. in., 48 in. deep; top chord 28 nickelsteel eyebars 16 X 18 = 840 sq. in., pin 16 in. diam.

The estimated steel weights are as follows:

Eyebars of struct. steel. . . . . . . . . . . . .Eyebars of nickel steel.. . . . . . . . . . . .Riveted truss members. . . . . . . . . . . . .Pins (mostly nickel steel). . . . . . . . . . .Bracing . . . . . . . . . . . . . . . . . . . . . . . . . .Floor system . . . . . . . . . . . . . . . . . . . . . .

One T w o Twosuspended cantilever anchar Totals in

span arms arms pounds2

6 4 , 0 9 0 . . 64,000’ %%i,%& 7,852,OOO 6,720,OOO 15,161,OOO“,;%;,iWl, 22,75;,0”;; 23,800,OOO 50,759,OOO

7 5 0 , 0 0 0 1,603,OOO693;OOO 4,550:OOO 4,550,OOO 9,793,OOO

2,640,000 5,280,OOO 4,400,OOO 12,320,OOO

- 8,580,OOO 140,900,OOO j40,220,000 189,700,OOOBearings and anchorages.. . , . . , . . , . . , . . . . . . . . . 4,500,000

T o t a l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94,200,OOO

26


The total steel weight represents an average of 33,600 lb. per lin.ft. of bridge and a proportion of steel weight to working live load (8000lb. per ft.) of 4.2. If no nickel steel had been used this proport ionwould be about 4.6.

The above steel weight includes l,lOO,OOO lb. additional material(nickel and structural steel) required for cantilever erection of the sus-pended span.

Another design with a suspended span of 800 ft. (height 120 and150 ft.) gave approximately the same total steel weight.

,(8) A cantilever design for a four-track R. R. Bridge, made byF. C. Kunz for the Pennsylvania Steel Co., in 1904, total length 1386ft., center span 855 ft. 6 in., with a suspended span of 367 ft. 6 in.(height 50 and 65 ft.), the two-anchar arms 265 ft. 3 in. each (heightof truss at piers 135 ft.), and a width of 59 ft. c. to c. trusses, withtwo outside sidewalks 6 ft. wide.

Live load: on each track Cooper’s loading E-50. Impact: L$3%o 1

Dead load: (in lb. per lin. ft. of bridge):

T r a c k c o n s t r u c t i o n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2000Ib.Ballasted floor.. . . . . . . . . 4600 “Steel in flooi- system (average). . . . 4800 “

The steel in trusses and bracing for dead load averages 12,400 lb.( suspended span 8100, cantilever 13,200, and anchar arms 14,300).Plate L gives the distribution of the total dead load between the panelpoints.

Wind pressure: 5001b. on top chord, 500 lb. on bottom chord, and500 lb. on floor moving, al1 figures per lin. ft.

Unit stresses: those of the Ameritan Railway Engineering Associa-tion (tension 16,000 lb.) except for trusses of suspended span (tensionof 18,400 allowed) and for trusses ofd anchar and cantilever arms (ten-sion of 19,200 allowed). Al1 eyebars are 16 in. wide (max. thickness2$ in.) and of nickel steel for which 50$& higher unit’ stresses wereassumed. All pins are 14 in. diameter and of nickel steel. The com-pression chords have 42-in. webs (radius of gyration 14 to 16 in.).

Table 62 shows the calculated deflection of the trusses for two ex-treme positions of the maximum live load of 20,000 lb. per ft. ofbridge. The greatest deflection of 8 in. a t the center of the sus-pended span is only i$Tò of the length of the center span.

The estimated steel weights are as follows:

TABLE 62Calculated Deflections of Cantilever Truss of Example (8)

I.Due+oa Live Load of 20000lbs.per lin.ft.of Bridge between lowers.

Depresslon of end of canf(lever at-rn, caused by fbe fower.posf shorfeninDepression of end of canfllever arm, caused by fhe cotnbtqed u

of fhe anchar arm, and downward deflecflon o Pward de, lecflon -~~~~~~~~~~-~~~

c$ g..-..-

fhe canf/lever arm . .._._........Deflecfion of cenfer of suspended span, coused by load on suspended span . . . .._____._. -_. . ..- --_ __ z?&Tofaf downward deflec+ion af fhe cenfer of suspended span ____..______..._________________________. -.---_ 8”

2. Dueto a Live Load of 20,000 Ibs.per lin.-Ft.of Bridge on one Anchar At-m.

ress&I

DeEI& f’

ofend of canfilever arm caused by fhe towerJ-

Pos+ shorfening &; . .._..... . ..___ ---. 0%.

v a ,on of end of canfilever arrn caused by downward def ecflon of anchar arm _ _-_ _. -- .-_ . _. __ 05/s”tiful elevafion of end of can fijever armz+al elevafion

__ -. - . ._. _. - _.______ ._...._ _ -. _...__. --- _.._ ____________._. o %;of cenfer ofsuspended span -_ __ _. .--_-..--. ..- .._____ -- ..______ -- ._.__ -.-- -._____ - _.._..___ 0 7;

(Facing page 378

ART.~] LONG SPAN BRIDGES IN GENERAL AND EXAMPLES 379

Eyebars (nickel steel) . . . . . . . . . . . . . .Riv. truss members . . . . . . . . . . . . . . . .Pins (nickel steel) . . . . . . . . . . . . . . . . .Bracing . . . . . . . . . . . . . . . . . . . . . . . . . .Floor system. . . . . . . . . . . . . . . . . . . . . .

Two center post systems.. ..,,.<..<.,..<,.....<....<.,..B e a r i n g s

1,977,ooo..<.......,<.............<....

A n c h o r a g e s .808 ,000

. . . . .< . . . . . . . . . . .< . . . . . . .< . . . 1,128,OOO

-

-

Onewspended

span

546 ,0002,154,OOO

88,000188,000

1,653;OOO

2,016,OOO 2,185,OOO 4,747,OOO3,931,OOO 4,868,OOO 10,953,OOO

140,000 130,000 358 ,000405,000 410,000 1,003,000

2,408,OOO 2,558,OOO 6,619,OOO

4,629,OOO 1 8,900,OOO /10,151,000 123,680,OOO

Total..................................,................ 27,593,OOO

equal to an average of 19,900 per lin. ft. of bridge and a proportion ofsteel weight to working live load (20,000 lb. per lin. ft.) of P.O. Ifno nickel steel had been used the proportion would have beenabout 1.1.

(9) The Ohio Bridge at Beaver, Pa., designed by Albert Lucius forthe P. & L. E. R. R., built 1908-1910, 1409 ft. long, center span 769 ft.,suspended span 285 ft. (height 57 ft.), two anchar arms 320 ft. each(height 145 ft. at main piers, 64 ft. at abutments), width 34 ft. 6 in.c. to c. of trusses, designed for double-track railway.

Dead load: averages about 20,500 lb. per lin. ft. of bridge (open tiefloor) .

Live load: on each track, E-60 for floor and 10% less for trusses.trusses. Impact for hangers 100% of live load stress; for other mem-bers according to formula

(Z = live load stress, d = dead load stress). In members subject toreversa1 of stress 2 of the smaller stress was added to both stresses.

Wind pressure: 300 lb. per lin. ft, on trains, 30 lb. per sq. ft. onaboth trusses and floor (20 lb. of latter assumed as a moving load).

Unit Xtresses:

Tension Max. comprcssion

For dead load, Iive load and impact. . . . . . . . . . . 16,000 14 booFor wind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Por dead load, live load, impact and wind. . . . . . .For erection., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

18’00020 ,00022 ,000 20:ooo18,000 16,000

3 8 0 DESIGN OF STEEL BRIDGES [CHAP. XVI

The compression stresses not to exceed the value

for (d + 1 + i), and correspondingly reduced values for the otherconditions.

Ultimate strength of steel specimens for riveted members 56,000to 64,000 lb. per sq. in., elongation 26%.; ultimate strength for full-size eyebars 56,000 to 66,000 lb. per sq. in., elongation 12 Yo in 10 ft.

Steel weights in pounds:

Anchar and cantilever arms (1124 ft.):( Trusses and bracing incl. bearings nnd anchorages 21,027,OOO) Floor system.. . . . . . 3,455,ooo

-_I__T o t a l anchar a n d c a n t i l e v e r arms..

Suspended span (285 ft.) :24,482,OOO

1 Trusses and bracing. , .~Floorsystem................................

2,217,OOO875,000

_____Total suspended span. . 3,092,ooo

___-T o t a l s t e e l w o r k . 27,574,OOO

This gives an average of 19,500 lb. per lin. ft. of bridge of which3070 lb. for floor system.

For further information, see: Eng. News, 1910, 1911; Eng. Record, 1910, andAlbert R. Raymer, Trans. Am. Soc. C. E., Vol. 73, 1911.

(10) The Danube Bridge at Budapest, Hungary, built 1894-1896,1094 ft. long; center span 574 ft. with a suspended span of 154 ft.(height 14 and 10 ft., top chord, therefore, concave to’ produce the effectof a suspension span) ; the two anchar arms are each 260 ft. long,while the cantilever arms are only 210 ft. each; 42.3 ft. c. to c. oftrusses, height of truss at river piers 71.5 ft.; was designed for a road-way 37.7 ft. wide in the clear and two sidewalks each 9.5 ft. wideon outside brackets, total width between railings 68 ft.

Live load for fioor system: two wagons each 8 ft. wide and weighing26 tons on two axles 13 ft. apart, on any part of the roadway.

Live load for main trusses: 90 lb. per sq. ft, on roadway and side-walks (this gives 5100 lb. per lin. ft. of bridge).

Wind pressure: 50,lb. per sq. ft,. for bridge unloaded.Unit stresses (in lb. per sq. in.): for floor system 11,400; for main

tarusses 17,000.No reversa1 of stress is considered except for rivets for which the per-

missible shear is 9200 lb. per sq. in. with and 10,700 without reversal.


Ordinary structural steel is used, except for counterweights which areof cast iron.

Steel weights in pounds:

15,000 1 3,708,OOO / 4,854,OOO 1 9,457,OOO 127B e a r i n g s . . 302 ,000 4Connections for counterweights., . 94 ,000 1

Total structural steel.. . . . . . . . . . . . . . . . . . . . . . 9,853,OOO 132Counterweights (cast iron). . . . . . . . . . . . . . . . . . . . 2,680,OOO 3 6

Total . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12,533,OOO 168

equal to an average of 9100 lb. per lin. ft. of bridge (without counter-weights) and a proportion of steel weight to live load (5100 lb. per lin.ft.) of 1.8. This high figure is caused by the heavy pavement andheavy water pipes, the riveted tension members and the extremely shallowtrusses (especially of the suspended span whose top chord is curved down-ward) which were designed more for appearance than for economy.

The above figures do not include railings and ornamental work.To obtain a more favorable distribution of the dead load, the weight

of the flooring (wood blocks on roadway, asphalt on sidewalks, both onsteel troughs) is in the anchar arms 170 lb. and in the center span 140lb. both per sq. ft. The track construction of the two trolley tracks,plus the gas and water pipes, give a weight of 2000 lb. per lin. ft. ofbridge; if we add this to the live load, we get proportion of steel weightto live load of 1.3 instead of 1.8.

The bridge has been erected on fixed falsework (also the suspendedspan), excepting a portion of the cantilever arms which was erected inplace on floating falsework (see also Papers of Austrian Soc. C. E., 1897,p. 124).

(ll) The Rhine Bridge at Ruhrort, Germany, built 1904-1907,1487 ft. long; center span 667 ft. with a suspended span of 443 ft. (height36 and 47 ft.), the two anchar arms 399 and 421 ft., respectively; 37 ft.c. to c. of trusses, height of trusses at river piers 82 ft.; was designedfor a roadway 30 ft. wide in the clear for four lines of vehicles, and twosidewalks each 8 ft. wide on outs ide brackets ,

382 DESIGN OF STEEL BRIDGES [C H A P. XVI

railings 52 ft. Clear height for roadway 16 ft., clear height for navi-gation 30 ft.

Live load for Jloor system: one wagon weighing 24 tons, on two axles10 ft. apart, on any part of the roadway, but no other live load. Onsidewalks 100 lb. per sq. ft.

Live load for main trusses: 100 lb. per sq. ft. on roadway and side-walks (this gives 4600 lb. per lin. ft. of bridge).

Wind pressure: 50 lb. per sq. ft. for bridge unloaded or’30, lb. persq. ft. plus a moving wind pressure of 250 lb. per lin. ft. .of bridge.

In case of reversa1 of stress, one-half of the smaller is to be added tothe greater stress.

Unit stresses (in lb. per sq. in.):For Aoor system 13,000.F?r main trusses 16,400 without and 20,700 with wind.Ordinary structural steel is used.

Estimated steel weights in pounds.

i

Trusses and bearings . . . . . . . . .Bracing . . . . . . . . . . . . . . . . . . . .Floor system . . . . . . . . . . . . . . . .

One Twocenterspan 1 a;;m 1 Totals

Total per sq.ft. of floor be-tween railings

4,670,OOO 4,986,OOO 9,656,OOO 125

equal to an average of 6500 lb. per lin. ft. of bridge and a proportion ofsteel weight to live load (4600 lb. per lin. ft.) of 1.4. This high figureis caused by the heavy pavement, the riveted tension members and theshallow trusses. To obtain the dead load the weight of the flooring wasassumed as 100 Ib. per sq. ft. for the roadway (wood block pavement onconcrete) and 35 lb. per sq. ft. for the side’walks (asphalt).

For the erection of the center span a temporary wooden tower supporthas been built in the middle of the river and the span erected cantileverfashion proceeding from one pier only.

The outline of the trusses in the center span is very pleasing, on ac-count of the unusually long suspended span, which was made possibleby the adopted method of erection.

For further information, see: Eng. News, July 14, 1904, and Papers of German Soc.of C. E., Berlin, 1904 and 1905, with competitive designs.


ART. 6. EXAMPLES OF LONG SPAN HIGHWAY BRIDGES IN NEW YORK

(See Plates, XLVIII, XLVIIIa and XLIX)

(1) The Queensboro Bridge, built by the Department of Bridges,New York City, in 1903-1908, 3724 ft. 6 in. long, a cantilever without asuspended span; greatest span 1182 ft. (for clear width of floors seePlate XLIX), designed to carry ultimately on upper deck 4 elevated rail-road tracks and two footwalks, on lower deck 4 trolley. tracks and aroadway 354 ft. wide, no allowance for impact was made, but the per-missible unit stresses were assumed different for the different memhérsdepending on the traffic and the probability of simultaneous loading.

The live loads to be added to the dead load and the permissible basicunit etresses were as follows:

(a) For stringers for the elevated railroad and the trolley tracks,also the side brackets carrying trolley tracks (one kind of traffic in oneline) a unit stress of 10,000 lb. for car loads of 52 tons on 4 axles forthe elevated railroad and 26 tons on 2 axles for the trolley.

(b) For upper floorbeams carrying 4 elevated railroad tracks (onekind of traffic in 4 lines) a unit stress of 15.000 lb. for car loads on eachtrack as specified under (0).

(c) For roadway stringers (slow traffic) a unit stress of 15,000 lb.for a wagon load of 24 tons on 2 axles.

(d) For lower floorbeams carrying 2 trolley tracks and the roadway(2 kinds of traffic in 6 lines) a unit stress of 15,000 lb. for car loads oneach trolley track, one wagon load of 24 tons on 2 axles on any part ofthe roadway and 100 lb. per sq. ft. on remaining portion of the tloor.

(e) For secondary truss members carrying the lower and the upperfloorbeam and the brackets (4 kinds of traffic in 14 lines) equal to 16,000lb. per lin. ft. of bridge a basic unit stress of 18,000 lb.

(f) For main truss members:(1) For a worling live load of 8000 lb. per lin. ft. of bridge (4 kinds of

traffic in 14 lines) a basic unit stress of 20,000 lb. without wind and24,000 lb. with wind.

(2) For an extreme live load of 16,000 lb. per lin. ft. of bridge (4elevated trains at 1700 lb., 4 lines of trolley cars at 1000 lb., 35.5-ft. roadway at 100 lb. per sq. ft., 22 ft. -of footway at 75 lb. per sq. ft.;total 16,000 lb.) without wind a basic Unit stress of 24,000 lb.

For nickel steel the permissible unit stresses were raised 50%.Permissible unit stress for shear is 2 of the above, for bearing twice

that for shear; unit stress for compression reduced in .a straight line.

PHYSICAL REQUIREMENTSSpecimen Tests for Nickel Steek

1. tUltImate tenslle Elastio limitstrength

Elongetion Reduotion Cold bend without fractureo f area

Charac ter o f _fracture

Pounds per Pounds per Min. p e r cent. in Min. % Material less than Mater ia l 1 in. orsq. in. sq. in. 8 in. in 2 in. 7% 1 in.-in thickness morc in thiokness

E y e b a r s (unsnnealed) 100,000 min. 55,000 min. {1,600,000

Ultimate tensile

E y e b a r s (annealed). _. 1 8 5 , 0 0 0 min. 4 8 , 0 0 0 min. { / ”

P i n s (unannealed) 90,000 min. 50,000 min. 20

18O0 around a pin with diameter = 3t.

Pieces of bar not less than 4 in. w - i d o180’ around pin witb diamcter = 2t.

Plates, shapcs and bars f o rriveted work.

E y e b a r s snd pins..

R i v e t steel.

Rivet steel, nicked andbent around bar of samedismeter as rivet red.

Steel castings (annealed)

60,000 desired

60,000 desired

50,000 desired

. . . . . < . . . . .

65,000 min.

-‘3

-

Specimen Tests for Structural Steel

,O,OOO min.1,500,000

Ultimate tensile

i of ultimate‘L

22

) of ultimate‘<

.._..... ._.....____.......

k o f ultimate18

-l

.l

i

.........

.........

.........

.........

.........

-

i

Silky/

1800 flat

<<

/

II

1‘ l

Fine, silky, grad-ua l break-uni-form fracture.

Silky or fine 90” around pin withgranular. diameter = 3t.

180° a round pinwith diameter = 2t.

‘1

t is the thickness of the specimen.

PHYSICAL REQTJIREMENTSFull Size Tests for Nickel St.4

Pounds per sq. in. Pounds per sq. in., w

E y e b a r s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . < . . < . . . . . _ . . . . . ___....<<<...,_... _.......______.... ..,,._............180’ around pin with

E y e b a r s ( a n n e a l e d ) 85,000 min.48’ooo min’ 1 i n e l u d i n g frac&.

9 % in ISft. I ,To be

diameter = 3t.

recorded Mostly silky and In the neck of the bar,free from toarse 90” around pin withcrystals. diameter = 2tt.

Ful1 Size Tests for Structural Steel

E y e b a r s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .._.___...____.... .,,.,_..__....,,._ .._...__....____ ._,,.___ ._,....______.,.__.. 180’ a r o u n d pin w i t hdiameter = 2t.

E y e b a r s ( a n n e a l e d ) . 56,000 min. f of ultímate 102 FaJody Fine, silky.

A n g l e s $ in. or less in thickness... . . . . . . . . <<....<............. Open flat, under blowsof a hammer.

A n g l e s + in. or less m. thlckness.. . . ..___._ __,....,........... . Bent shut, under blowsof u hammer.

t is the thiokness of the piece.

386 DESIGN OF STEEL BRIDGES [CHAP. X V I

(g) Wind bracing 20,000 lb. per sq. in. for a moving wind load of2000 lb. per lin. ft.

The preceding table gives the physical requirements for the steel.

Heaviest Members are:Riveted shoe on top of tower 9 X 12 X 15 ft., 140,000 Ib., mostly nickel steel.Tower post, 5 X ll X 194 ft,., 1396 sq. in. cross-section, 126,000 lb., equal to

6460 lb. per lin. ft.Diagonal, 3 ft. 6 in. ?( 105 ft. long, with four ribs, 180,000 lb.Bottom chcrd, 4 X 6 X 59 ft., 1120 sq. in. cross-section, 240,000 lb., equal to

4100 lb. per lin. ft.Top chord 48 ft. long, composed of twenty nickel steel eyebars lö X 26 in. equal

to 680 sq. in., cross-section.Pedestal under each tower post 10 ft. 3 in. high, composed of three courses of steel’

castings, maximum weight of single casting, 41 tons. Weight of one pedestal, 140 tons.Nickel steel pin, 16 in. dia. and 10 ft. long, weighing 7000 lb.Length of floorbeams, 55 ft., and maximum weight of one floorbeam, 36,000 lb.

The steel weight in pounds as estimated in 1905 is divided as follows:

Fl00r

M a n h a t t a n anchar mm.. 3,943,oooManhattan cantilever arm.. 4,859,OOOCant. arm west of Island span. 4,883,OOOIsland span.. . . 5,304,oooCant. am e?st of Is!and span 4,096,OOO

ueens cmtdever arrn.. 4,081,OOOueens anchar arrn. 3,840,OOO

T o t a l 31,006,OOO

Trusses

8,280,OOO;,;;;m;

13:779:0005,093,ooo4,992,ooo5,644,OOO

53,557,ooo 5,449,ooo ( 12,527,OOO 1 102,539,000

Brscing TOWCXS

647,000859,000856,000

..<........3,360.0002,711,ooo

..<........2,980,OOO

<..........

16,346,OOO13,715,ooo16,871,OOO22,840,OOO

9,885,OOO12,747,OOO10,135,000

The actual steel weight is 105,150,OOO lb. or an average of 28,230lb. lin. ft. of bridge.

Weight of rails for 4 trolley tracks and fastenings . . . . . .Weight of pipes, etc. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ;;g 1:

Weight of handrails. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 “Weight of ties and rails for 4 El. R. R. tracks . . . . . . . . . . 1,370 “Weight of paving (wooden blocks on concrete). . . . . . . . . 3,400 “

Average total weight.. 33,720 lb.

CONTRACT PRICE IN CENTS PER POUND ERECTED

N i c k e l s t e e l e y e b a r s . . 8.03N i c k e l s t e e l pins.. . 10.03Structural s t e e l e y e b a r s .Structural s t e e l p i n s .

8 59

Structural r i v e t e d work., . 5.64S t e e l c a s t i n g s . , 9 .Ol

This rcpresents an average of 6.06 cents per Pound of total metal work erected.Por further information, see:


Report of Messrs. Boller and Hodge and Prof. Burr, New York, 1908 (partlyreprinted in Eng. News, Nov. 12, 1908, and Eng. Reo., 1908).

Report of F. C. Kunz to the Pennsylvania Steel Co., Steelton, Pa. (partly re-printed in, Eng. News, June 17, 1909, and Eng. Rec., 1909).

Engineering (London), 1909.Prof. Barkhausen, Papers of German Society of C. E. (Berlin), 1911.Connel-Ferry cantilever, Scotland, Eng. Record, July 2, 1904 (similar design).

(2) The Manhattan Suspension Bridge, built by the Department ofBridges, New York City, in 1904-1909, center span 1470 ft., total clearwidth of the two floors 160 ft., was designed for the same traffic capacityand loading as the Queensboro Bridge.

Specified unit stresses in pounds per sq. in:

(a) Floor system of roadway and footways.. . . . . 15,000(b) Floor system of railroad and trolley tracks. . . 10,000(c) Anchorage eyebars in tension.. i . . . . . 16,000

@JTowersand I::~~~~~~~~:~~~stiffeningtrusses

(e) Wire suspenders. . 30,000 for condition 1.(f) Wire cables.. . 60,000 for condition 1, and 73,000 for condition 2.For nickel steel in stiffening trusses. . 40,000 for condition 2.

The sections are proportioned for the greatest stresses (reversa1 ofstresses neglected).

Shear 2 of above; bearing twice shear; compression reduced instraight line.

The structural steel has 60,000 to 68,000 ultimate.The nickel steel specimens for plates and shapes (there are no eyebars

of nickel steel) ultimate 85,000 to 95,000; elastic limit 55,000 minimum;“Coefficient of quality” 1,600,OOO (same as Queensboro Bridge) andreduction of area 40% minimum.

The nickel steel for rivets, ultimate 70,000 to 80,000; elastic limit45,000; “Coefficient of quality” 1,600,OOO.

For the steel wire was specified an ultimate of 215,000 lb. per sq. in.before, and 200,000 after galvanizing and an elongation of 2% in 12inches. Actual tests gave a yield point of 136,000 to 138,000 lb.

3 8 8 DESIGN OF STEEL BRIDGES [CHAP. XVI

Estimated (1905) steel weights:

N i c k e l s t e e l . ‘.Struct. steel ’ 1,336,OOOWire :!.S u s p e n d e r s , e t c . ,E y e b a r s . . ,I 3,732,OOOSteel castings 2 ,000Iron castings 19,000Pins, bolts, nuts, 308,000

/_---Totals.

./5,397,ooo

-1

.21,334,OOO 30,000

12,177,OOO1,154,ooo

3,385,OOO 1,745,ooo189,000 55 ,000119,000 383 ,000

___-__25,027,OOO 15,544,ooo

16,248,OOO21,050,OOO

.

42 ,00032 ,00033 ,000

37,405,ooo /

16,248,OOO43,750,ooo12,177,OOO

1,154,ooo3,732,0005,174,ooo

295 ,000843 ,000

83,373,OOO

Therefore per lin. ft. between anchorages (2920 ft.) 28,600 lb. andproportion of total steel weight to working live load (8000 lb.) 3.6.

SusyTnded :qeel wc\Fht of the floor. . . . . . . . . . . . . . . . . . .5,800 lb. pzr 1;:. ft.of the cables and suspenders.. . . . .5,300 “

“ ‘L “ of the stiffening trusses and their bracing. . .7,200 “ “ “ “

For further information, see:Ralph Modjeski, Report on Manhattan Bridge, Department of Bridges, N. Y.,

1909.Eng. News, July, í904, Aug., 1905, April, 1906, April, 1908.Eng. Record, July, 1905.Eng. News, 1903.Eng. Record, 1903. \

Railroad Gazette, 1903. t

Interesting discussions on Nickel Steel Eyebars versusWire Cables.

Making the Cables of i large suspension bridge, Cassier’s Magazine, May, 1905(excellent article with many illustrations).

Cable Construction, Eng. Record, 1908, p. 637.Manhattan Bridge, Der Eisenbau, 1911 (very complete article).Suggestions for Suspension bridge Design, Der Eisenbau, 1911.

(3) The Williamsburg Suspension Bridge, designed by L. L. Buck,built 1897-1904, total length 2793 ft., center span 1600 ft., clear widthof floors 135 ft. (two tracks El. R. R., four tracks trolleys, two roadwaysat 20 ft., two footwalks at 10 ft., two bicycle paths at 7 ft.). The endspans are not supported from the back stays. Live load assumed fordesign: 5700 lb. per lin. ft. of bridge for the cables and 4500 lb. for thestiffening trusses. Wind pressure 30 lb. per sq. ft.

The unit stresses specified for the trusses without wind 20,000and with wind 30,000 (reduced for compression) and for the towers

(20,000-70;) for compression and 16,000 for net section in tension.

The structural steel has 60,000 to 68,000 ultimate with elongation 20%in 8 in. and’the finished wire 200,000 ultimate with elongation 23 y0in 5 ft. and 5% in 8 in.

PLATE XLVIII B

1 M a n h a t t a n

!3levatlons of the three East River (New York) SuspensionBridges. (Facing page 388)

ART.~] LONG SPAN BRIDGES ,IN GENERAL AND EXAMPLES 389

Dead load assumed for design:

Weight of cables and suspender-s throughout. 3,500 lb. pn lk. $.

Other weight ( in s u s p e n d e d span. . . 13,000 lb.1 in end spans. 12,000 lb. “ “ “

The steeì weights are approximately:2 Anchorages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6,200,OOO lb.2 Towers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12,200,OOO “

Cables and suspenders. . . . . . . . . . . . . . . . . . . . . . . . . . . . 10,000,000 “Centerspan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15,540,OOO “

2 End spans. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12,280,OOO “

56,220,OOO lb.

Therefore per lin. ft. between anchorages (2793 ft.) 20,200 lb. andproportion of total steel weight to working live load (4500) 4.5.

For further information, see:Eng. News, 1898 to 1904, Feb. 17, 1908.Eng. Record, 1898 to 1904, Aug. 21, 1909.Specifications for Cables, Eng. Record, Nov., 18, 1899.Cable making, Eng. Record, 1, p. 418, 1902.L. L. Buck, Design of Bridge, Railroad Gaxette, Dec. 25, 1903.Reinforcing under Traffic, Eng. News May 14, 1914.

(4) The Brooklyn Suspension Bridge, built according to design ofJohn A. Roebling, 1870-1883, center span 1595 ft., designed originally(see Report of Mr. W. A. Roebling, 1882) for a live load of 1790 lb.per lin. ft. of bridge, consisting of two tracks of cable railway, tworoadways each 18 ft. wide and ‘a footwalk 15 ft., a total clear width of75 ft.; in 1892 a trolley track was placed on each roadway. Specifiedultimate for steel wire 160,000, for other steel 75,000 Ib.

In 1898, Mr. C. C. Martin reported (see Eng. News, Dec. 15, 1898,and Eng. Record, Dec. 10, 1908) that the live loads at that time were asfollows:

(cz) Bridge trains: one motor car 44 tons (on 8 wheels) and 3cars loaded 28 tons each, total 128 tons, 193 ft. long; as trains run 45sec. headway and ll.3 miles per hour they are 748 ft. apart c. toc., and since there are two tracks the total is 690 lb. per lin. ft. ofbridge.

(6) Trolley cars: 14 tons each, spaced 102 ft. c. to c., and åincethere are two tracks the total is 550 lb. per lin. ft. of bridge.

(c) Carriageways: one continuous line of trucks 20 ft. long andweighing 2.75 tons, on each of the two roadways, total 550 lb. per lin.ft. of bridge.

(d) Footwalk: 50 lb. per sq, ft. (“although Mr. John A. Roeblingsays in his original report of 1870 that 30 lb. per sq. ft. is a maximumload of a moving mass of people”).

27

390 DESIGN OF STEEL BRIDGES [ C H A P . X V I

Total live load from these four sources is 2540 lb. per lin. ft. ofbridge.

In 1908 (see Eng. News, April 9, 1908) Prof. W. H. Burr reportedthat the heaviest bridge trains of 236 tons (4 motor cars at 46 and 2trailers at 26 tons) and even future trains of 276 tons each (6 motorcars) spaced 700 ft. in the clear are permissible on the bridge. As suchtrains would be about 300 ft. long, the 690 in item (u) would change to1100 lb. per ft.

In 1908 (see Eng. News, Nov. 12, 1908) Prof. Burr in his report on the Queens-boro Bridge states that the loaded heaviest motor car (elevated OP subway)weighs 53 tons and trailer 35 tons. A train composed of 5 motor cars and 3trailers would, therefore, qveigh 370 tons, and be about 400 ft. long. If we assumethese figures and 700 ft. in the clear, item (a) would change from 690 to 1350 lb.per ft.

In 1901 the loaded heaviest trolley cars were 19 tons (37 ft. long); in 1904,22 tons; in 1908’, 31 tons (43 ft. long). The last weight (spaced 102 ft. c. to c.)changes the 550 in item (b) to 1220 lb. per ft.

Item (c) represents about 30 lb. per sq. ft.; if we raise this on account ofheavier modern trucks to only 50 lb., the 550 lb. would change to 900,lb.

Total live load would, therefore, change from 2540 lb. to 3970 lb. per ft. ofbridge (or 4220 if heaviest trains used in 1908 in New York City were assumed).

The dead load is as follows (see report of Mr. E. Duryea and Mr.Jos. Mayer, Eng. News, Oct., 1901, and Eng. Record, Oct., 1901):

St‘eel i‘n s u s p e n d e d superstructure.m a i n c a b l e s .<‘ “ s u s p e n d e r s a n d c o n n e c t i o n s . .‘L <‘

“ “ s t a y s .other wire work (wind, cables, etc.)

T i m b e r , f l o o r i n g , tracks, e t c .

T o t a l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2;;;1. p:&r 15. ft.LI

‘230 Ib: ‘I ‘I L‘250 lb. “ “ “310 lb. “ “ “

- - - -;,“,;“, $. p:; li$. ft.“,---___8,190 lb. per lin. ft.

The report states that with this dead load and 2600 Ib. per lin, ft.rof live load (no wind load) the cables are stressed to 75,000 lb. per sq,in, neglecting action of stays, and to 71,000 lb. making a rough allow-ante for the stays.

With the above-mentioned imaginary live load (3970 or 4220) these unit stresseswould be about 15% greater.

For further information see also:F. Collingwood and other interesting contributions, Railroad Gazette, Dec., 1898.Wilhelm Hildenbrand, Safety of Brooklyn Bridge, Eng. News, Jan. and Feb.,

1902.New Terminals of Brooklyn Bridge, Eng. News, May 9, 1912.

Bridges Over The East River, New York, General Data

TYP~

Length, river spsn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Length, main bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Length, Manhattan spproach . . . . . . . . . . . . . . . . . . . . . . . .Length, Brooklyn approach . . . . . . . . . . . . . . . . . . . . . . . . .

Width, ovw all . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Tracks, elevated railway . . . . . . . . . . . . . . . . . . . . . . . . . . . .Tracks, surface railway . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Roadways . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Footwnlks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Main t r u s s e s , h e i g h t , o . t o o . p i n sM a i n t r u s s e s , p a n e l l e n g t h . .Elevation above M . H . W.-cable a t tower..

Cables, number of wires . . . . . . . . . . . . . . . . . . . . . . . . . . . .Cables, diameter, each wirc . . . . . . . . . . . . . . . . . . . . . . . . .Cables, total diameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Cables, length, center to center anC&rage pins.

Total steel in main bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . .Total steel in Manhattan approach. . . . . . . . . . . . . . . . . . .Total steel in Brooklyn approach . . . . . . . . . . . . . . . . . . . .Total masonry, main piers . . . . . . . . . . . . . . . . . . . . . . . . . .Total masonry, anchorages . . . . . . . . . . . . . . . . . . . . . . . . . .Cost of construction, including terminals . . . . . . . . . . . . . .

~Construotion of masonry, piers started . . . . . . . . . . . . . . . .Construction of steelwork stnrtod. . . . . . . . . . . . . . . . . . . .Rosdways opened . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Trains first operated, elevated. . . . . . . . . . . . . . . . . . . . . . .

Travel, both directions for 24 hours, Dec. 28, 1911:Elevatcd railway cara. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Surfaoe railway ca-s. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Vehicles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Passengers, elevated rzxilway oars . . . . . . . . . . . . . . . . . . .Passengers, surface ra1lway cars. . . . . . . . . . . . . . . . . . . . . .Passengers, vehioles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Pedestrians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Total number of people crossing . . . . . . . . . . . . . . . . .

-

-

-

1I

-

Brooklyn bridge Manhattan bridge Williamsburg bridge Queensboro bridge

Suspension

1,595.5 it. 1,470.o ft. 1,600.O ft. 1,182 and 984 fe.3,455.5 ft. 2,920.o Et. 2,793.0 ft. 3,724.5 ft.1,562 5 ft. 2,067.O ft. 2,650.O ft. 1,052.O ft.

998.0 ft. 1,868.O ft. 1,865.0 ft. 2,672.2 ft.

86.0 ft.,

;Two, 16 ft. 9 in. wideO n e , 15 Et. 7 in. wide

1 2 2 . 5 ft.44

O n e , 35 ft. 0 in. w i d eTwo, 13 ft. 7 in. w i d e

118.0 ft.

z

89.5 Et.

Two, 19 ft. 11 in: wideTwo, 17 ft. 8 in. wide

2 (Dec. 31, 1912)

One, 53 ft? 3 in. wideTwo, 16 ft. 4 in. wide

17 ft. + in. 24 Et. 0 in.7 ft. 6 in. 18 ft. l* in.+272.0 +322.5

5,296 9,4720.165 in. 0.192 in.15.75 in. 21.25in.

3,578.5 ft. 3,224 ft.

/

11,920 tons10,000 tons85,160 cu. yd.

. . < . . . . . . . .$16,091,580

Jan. 3, 1870May 29, 1877May 24, 1883Sep t . 24, 1883

342,277

Suspension l Suspension

Oct . 1 , 1901Apr. 30, 1906Dec. 31, 1909

. . < . . . . . . . . . .

. . . . . . . . ...<<4,399

..<..<<.<............

. . . < < < .7,2441,911

9,155

40 ft. 0 in.19 ft. ll in.

f333.0

7,696 . . . . . . . ..<<....__....0.192 in. . . . . . . . . . . .._........18.625 in. . . . . . . . .._...._._....2,985 Et. . . . . . . ..<..<........_

28,700 tona 54,200 tons10,500 tons 6,000 tons

6,085 tons45,500 cu. yd.

13,600 tons

112,800 cu. yd.53,000 cm yd.

$14,181,56053,000 cu. yd.$13,496,500

Nov. 7, 1896F e b . 21, 1899Dec. 19, 1903Sep t . 16, 1908

230,925 47,694

Cantilever

45 to 185 Et.43 to 80 ft.

+323.0 top chord pin

July 19, 1901Nov. 20, 1903Mar. 30, 1909

. . . . . . ...<....2,2842,342

. . . . . . . . . . .42,3874,;;;


ART. 6. DESIGNS FOR A NORTHRIVER SUSPENSION BRIDGE (NEW YORK)

(1) MR. GUSTAV LINDENTHAL (see Appendix to Report of Board ofEngineers upon New York and New Jersey Bridge, 1894) assumed for a3100-ft. center span and 8 tracks a live load of limited length of3000 lb. per lin. ft. of track (two suburban, two through passengeror express, two freight, two electric railways). 6 rapid-transit tracks(live load 2000 lb. per lin. ft. of track) could be added later on anupper deck; the cost of the heavier anchorages and towers caused bythis addition would be only 9% of the total while the section of thecables could be increased later. Width of bridge 115 ft.

Assumed wind pressure 2400 lb. per lin. ft. of bridge.Maximum stress in towers (including bending stresses) 30,000 lb.

per sq. in. Steel for towers, ultimate 100,000 lb. per sq. in.Maximum stress in the cables which are composed of pin-connected

wire links 60,000 lb. per sq. in.; ultimate for wire 180,000 lb. per sq. in.Maximum stresses in anchorages for 8 tracks 20,800, for 14 tracks

27,600 lb. per sq. in.; steel for anchorages ultimate 60,000 lb.

Tqfal st:el we$ht in anchorages. . . .in towers. . . . .

24,800,OOO )“ ‘6 “ 49,560,OOO

in superstructure. . . 190,120,OOO264,480,OOO lb.

Therefore between anchorages (6800 ft.) 4860 lb. per lin. ft. of track(8 tracks) and proportion of total steel weight to live load 1.6.

(2) MR. THEODORE COOPER (see New York and New Jersey BridgeCo.‘s General Specifications 1895, also Eng. News, March 7, 1895) as-sumed for a 3100-ft. center span and 6 railroad tracks a live loadof 3000 lb. per lin. ft. of track, covering al1 the tracks from tower totower moving slowly, or on each track one train of the same weight but1000 ft. length and moving at high speeds. Wind pressure 25 lb. persq. ft. of surface when bridge loaded; or 100 lb. per sq. ft. of surface ofthe bridge only, for a moving length of 300 ft.

Permissible unit stress in lbs per sq. in. in towers 20,000-70 i.

For chords of stiffening trusses in tension 18,000 without and 22,500with temperature and wind. For reversa1 of stresses in stiffening trussesthe stress is reduced. Stringers 10,000, floorbeams 15,000, eyebars inanchorage 20,000, cables 54,000, suspenders 30,000.

Steel for wire 180,000 ultimate with elongation of 4% in 1 ft., othersteel 60,000 to 68,000 ultimate with 20y0 elongation in 8 in.

(3) MR. G. S. MORISON (see Trans. Am. Soc. C. E., Vol. 36, 1896)assumed for a 3200-ft. center span and a width of 92 ft. clear between


stiffening trusses (100 ft. c. to c.) a live load for the stiffening trussesof 11,000 lb. per lin. ft. of bridge, equivalent to:

(a) a live load of 120 Ib. per sq. ft. of floor, or(5) 8 passenger trains of 1375 lb. per lin. ft. per track, or(c) 8 freight trains 1400 ft. long at 3000 lb. per lin. ft. per track, or(d) 4 maximum freight trains of any length, weighing total 11,000

lb. per lin. ft. of bridge, with no provision for unequal weight on therapid-transit tracks.

Wind pressure for top laterals 500, for bottom laterals 750 lb. perlin. ft., this representing 30 lb. per sq. ft. for loaded bridge.

The cables are made of twisted ropes and are proportioned for a deadweight of the cables and suspended superstructure of 39,000 and a liveload of 11,000 lb. per lin. ft. of bridge, which causes a unit stress of59,000 lb. per sq. in. Ultimate of wire 180,000 lb. per sq. in.

Maximum stress in towers (not considering bending and wind stresses20,000 lb. per sq. in.; the material is ordinary structural steel.

Maximum stress in stiffening trusses is 20,000 lb. per sq. in. forordinary structural steel and &20,000 for members of the chords wherereversals of stresses occur, which are of nickel steel with 60,000 elasticlimit (3+ y0 nickel).

Total steel weight in anchorages . . . . . . . . . . . . . . . . . . . 8,784,OOO lb.t o w e r s . 47,800,OOO “

Wire work and connections. . . . . . . . . . . . . . . . . . . . . . . . . 65,793,OOO “Other .superstructure ( ordinary steel. . . . . . . . . . . . . . . . . 88,320,OOO “

\ nickel steel.. . . . . . . . . . . . . . . . . . 16,000,OOO “226,697,OOO lb.

therefore per lin. ft. between anchorages 5450 lb. per lin. ft. of track(8 tracks assumed) and proportion of total steel weight to liveload of limited length (3000 lb.) 1.8. The weight of the anchorages issmall because it is formed by the continuation of the cables and not ofeyebars.

(4) MR. Jos. MAYER (see Trans. Am. Soc. C. E., Val. 48, 1902) as-sumed in 1902 for 2800-ft. center span and a width for 12 tracks a liveload for the stiffening trusses as follows:

(a) Two tracks each loaded with one freight train 1000 ft. longat 3000 lb. per lin. ft. per track.

(6) Two tracks each loaded with one passenger train 1000 ft.long at 1500 lb. per lin. ft. per track.

(c) Six tracks each loaded with one rapid transit electric train 500ft. long at 1200 lb. per lin. ft. per track.

(d) Two lines of surface cars (20 tons each car) 100 ft. c. to c.,equal to 400 lb. per lin. ft. of track.


Total approximately ‘8500 lb. per lin. ft. of bridge covering the

whole length of the main span and considered for cables, anchorages andtowers. For the stiffening trusses the surface cars were neglected asthey are nearly uniformly distributed over the length of the bridge.

Wind pressure assumed with 1600 lb. per lin. ft. of bridge (30 lb. persq. ft. exposed surface).

The cables are made of straight wire, 200,000 ultimate and 70,000 Ib.per sq. in. unit stress.

Unit stress in stiffening trusses 22,500 tension, 22,500-904 com-

pression, no reversa1 considered. Unit stresses increased 20% whenwind stresses included. Ultimate 80,000 to 90,000.

(5) The following table gives some additional information. It isimpossible to obtain accurate figures from published data, especiallyfor the purpose of comparing the different designs. In design (1) the twoside spans are suspended from the backstays (as in the old Brooklyn andalso the Manhattan Bridge), in the other designs the end spans supportthemselves (as in the Williamsburg Bridge). Designs (II), (III) and(IV) have stiffening trusses with parallel chords suspended from thecables. Design (1) uses the cables as chords of stiffening trusses runningalong the cables. * In designs (1), (II) and (III) the cables in the anchor-

INFORMATION C~N~ERNING DIFFE;;~T~~;~;GN:GNSFOR A NORTH RIVER BRIDGE,

DesignG. Linden-thal, July,

18941

Length o f centor span in f t . . _. _.L e n g t h between anchorsges in f t .

3,1006,800

Number of tracks znd live load per ft., .

Stress per sq. in. of wire cables in lb..

8X3000of limited

length

60,000

59,700132,240

4,860

Tons of wire work between anchorages..T o t a l tons o f steel.,Steel weight in lb. per lin. ft. of me traol~

Proportion o f t o ta l s t ee l pe r lin. f t . tlive load per lin. ft.

Cost of steel.. . . . . . . . . . . . . . . . . . . . . . . .Cost of substructure . . . . . . . . . . . . . . . . . . .

Total oost . . ..:. . . . . . . . . . . . . . . . . . . . . . . .

Cost per ft. between anchorages . . . . . . .

e6 X 3000aoh 1000 f

longt. E

6X3000 8X3000:ach 1500ft. oach 1400 ft.

long long

60,000 60,000 59,000

35,370 28,160 32,900131,200 108,910 113,350

8,500 7,260 5,440

2.8 2.4

816,059,OOO $10,403,000P14,684,000 $11,784,000

ã30,743,000 $22.187,000

$5,900 $4,400-

1< I

0

Board of U.Y. Engineer

OffiCfXS,otober, 1891

III

3,220 3,2005,150 5,000

~.KtMh&30”,1896 ’IV

,

1

1

3,2005,200

(Stiffen..IYmses, 4100)

1.8(Nickel steel)

$13,000,00069,000,000

$22,000,000

$4,230-

* An interesting example of such a type is the Grand Ave. Bridge, St. Louis,Mo., 4W ft. span built by Carl Gayler, see Eng. News, July 18, 1891.


ages consist of eyebars, in design (IV) the wire cables extend into theanchorages. The rigidity of the four designs cannot be compared asthe respective calculations were not published. The table shows thatthe greater the number of tracks the smaller the proportion of steelweight Co live load, therefore the greater the economy. Al1 the designsassumed on,each track only one train at the same time and practicallythe same unit stresses. The costs are for present-day use entirely toosmall, not only on account of the assumed low unit prices, but also ofthe assumed low live loads which have doubled in the last twenty years.They refer to the bridge proper under certain assumed local conditionsand exclusive of cost of approaches, terminal facilities and land.

For literature on Long Span Bridges, ,see page 423.

CHAPTER XVII

CANTILEVER BRIDGES

A. CALCULATION OF STRESSES

ART. 1. GENERAL

A cantilever is a continuous beam (or truss) made statically deter-minate by the introduction of hinges in certain spans. If there are nsupports (n-2) hinges are required to make the beam statically deter-minate; however, not more than two hinges can be in any one span.A hinge at any point has the effect of reducing the bending moment atthat point to zero. In trusses the chord member opposite the hingemust be omitted or provided with a sliding end so that it is not stressed.

A cantilever bridge is, as a rule, made symmetrical for better ap-pearance. This is possible only with an uneven number of spans. Fig.1 shows typical arrangements.

H K47 4L ca.)

A B C D

A~--A”--”( W

A B c D E ‘F

dA

Aia ( CJ

B C D

*A

AL ‘ 2 ”0 c D E

0 Cd.)F

FIG. 1.

The supports divide the bridge into anchar spans or anchar arms(AB and CD, Fig. 1 a) which contain no hinges, and cantilever spans(BC, Fig. 1 a) which are divided by the hinges into the suspended span(HK, Fig. 1 a) and the cantilever arms (BH and KC, Fig. 1 a).

The suspended span is a simple beam affected only by the londswhich it supports. The stresses are therefore determined as shown inChapter IX. The stresses in the cantilever and anchar arms are mostconveniently determined by means of influente lines.

ART. 2. INFLUENCE LINES POR ANCHOR AND CANTILEVER ARMS

It will suffice to consider part AK (Fig. 1 a) as agencia1 case. Loadsto the right of K have no influente on this part. Similarly as shown

3 9 6

Am. 21 CANTILEVER BRIDGES 397

for simple spans (page 40 and 49) the influente lines of shears,moments or stresses can be easily derived from those of the reactions.

Reactions.-For a load unity applied anywhere between A and H(Fig. 2) the reaction at A is

(negative for londs to the right of B or x > Zr). This represents a straightinfluente line A,HI with ordinate equal zero at B. A load unity on the

suspended span HK causes a reaction in the cantilever arm at H of l-13

and this causes a negative reaction

x’ 1,R, = - 1 t3 G

Ca.)

fi

( b . ) L

I

P cFIG. 2.

<t Reaction R,

Reaction Rb

(2

which represents a straight influente line HlK1. Similarly the influenteline for reaction Rb is found to consist of two straight lines AsH2 andHzK2 (Fig. 2 c).

Shears and Moments in Anchar Arrn.-The shear in any panel mnof the anchar span is equal $0 the reaction R, for loads to the right of nand equal but of opposite sign to reaction Rb for loads to the left of m.From this follows the influente line AImlnlHIK1 for the shear in panelmn (Fig. 3 b).

The moment at m is aR, for loads to the right of m; the part ofthe reduced influente line for M, to the right of m is therefore identicalwith that for the reaction R,, the influente coefficient being equal to aj

398 DESIGN OF STEEL BRIDGES [CHAP. XVII

that is, to get the moment the sum of ordinates has to be multipliedby a (see page 41). As can easily be proven the part of the influenteline to the left of m is a straight line A 2m2 (Fig. 3 c). The influente areafor moment at B consists of the single triangle B&12Kz, the influentecoefficient being II (Fig. 3 c).

ShearV,,,,=z

M o m e n t M,=za

FIG. 3.

Shears and Moments in Cantilever Arm.-These are independentof the reactions and influenced only by loads on the cantilever arm andsuspended span. The shear in a panel mn due to a load between nand H is equal but of opposite sign to that load; the influente line is,therefore, a straight line nIH1 (Fig. 4 b) parallel to and at a distancefrom the base line equal to unity. Loads to the left of m have no in-

IP-1

( b.1Shear ‘.&Pz

M o m e n t M,=z.a

HsFIG. 4.

fluente on the shear V,,. The straight lines mlnl and HlK1 of the in-fluente line are easily explainable.

The influente line for the bending moment at m is a triangle mzHzK2with ordinate la below H, or, if we make this ordinate equal to unity(Fig. 4 c), we have to multiply the sum of ordinates x by the influentecoefficient a in order to get the moment.


Stresses in Anchar Arrn.-Fig. 5 shows the reduced influente Enes forthe stresses in various members of the anchar arm AB. Any of theseinfluente lines can easily be derived from the influente line for end reac-tion R, (Fig. 2 b) considering that the part between A and B is identicalwith the influente line for that member if the anchar arm is assumeda simple span (see page 90). The part of the influente line between B

and K is, for al1 members of the anchar arm, a triangle with the height k

below H, that is, identical with the influente line for R,. The influente

,(e:) /7-----y \i I d 1lnfh?ce , coeffxent = 1 1

Diagonal LEUS

Diagonal U,B

FIG. 5..

coefficient for any member is equal to the stress produced in that memberby an upward forte unity applied at A, the truss being assumed fixed atB. These influente coefficients are conveniently found by a Maxwelldiagram, starting the resolution of forces at A and ending at B.

To obtain the influente line for a web member LzU3 it is necessary todetermine either its center of moments I (by producing the two chordmembers UzU3 and L2L3 cut by a section cutting that web member) or


the influente ordinate 2 at B (for hl and hz see Fig. 5a), or preferably

both.Fig. 5 e shows the influente line for a diagonal whose center of moments

1, lies between A and B. (For UJ3, r1 = lever arm of UdB from itscenter of moments 1, and dr = horizontal distance of 1, from A.)

If the center of moments 1 is near to A this method becomes inac-curate or even impracticable, since the ordinates become very greatand the influente coefficient very small. In such a case the influentearea is shown reduced by making the ordinate at’ B equal to unity

11sinstead of k or

1; and the influente coefficient is then equal to the

3

stress produced in the member by a downward forte unity at B, thetruss being assumed fixed at A. For instance, for diagonal L,Ua

ll-Id(Fig. 5 c) the influente coefficient would become 7 and for diagonal

L-d1U4B (Fig. 5 e) it would be ~1.

If 1 lies at A the member is no t s t ressed by loads to the r ightof k.

Stresses in Cantilever Arm.-The influente line for a chord memberL& is identical with the influente line for the bending moment (Fig.4 c) at the center of moments of that member; the influente coefficient,however, has to be divided by the lever arm of the member (Fig. 6 b).

A load unity at H’produces in the diagonal U1L2 a stress G (Fig. 6 a).

If we make the influente ordinate at H equal to the load unity (Fig. 6 c)

the influente coefficient becomes 4 or equal to the stress in U1L2 due

to a load unity at H. To the right of H the influente line must evidentlybe a straight line HzK2. To the left of H the ordinates must increasein proportion to their distance from the center of moments 1; the in-fluente line 2”H, is therefore obtained by producing line I,Hz. This

line may also be determined by calculating the ordinate ; below B

(Fig. 6 c). A load at or to the left of Ll causes no stress in tie diagonal,from which follows influente line 2”l’.

In a similar way the influente line for vertical U& is obtained.Its center of moments is the same as for UILS, and the ordinate at B is

h,also G, the influente line is therefore identical with that for UlL, to the

right of La. Section X’ cutting the vertical, cuts the loaded (lower)

A R T . 21 CANTILEVER BRIDGES 401

chord in panel 2-3, from which follows line 3’2’. The influente coef-d

ficient is za.

Fig. 6 cl shows the influente line for a diagonal ,whose center ofmoments lies between B and H. (For BUI, rl = lever arm of BUIfrom its center of moments and cl, = horizontal distance of the centerof moments from a vertical through H.)

The influente lines for the cantilever arm are, therefore, obtainedsimilarly to those of the anchar arm. The influente ordinate below H

(C.)

(d.1

Diagonal U,Lz

Vertical UZLZ

F I G . 6 .

is made equal to unity and the influente coefficients are obtained (mostconveniently by a Maxwell diagram) as stresses due to a load unity atH, the truss being assumed fixed at B. If the center of moments 1of a web member lies near to H, the ordinates become very greatand should be shown reduced by making the ordinate below B equal to

hlunity instead of to c and the influente coefficient is then the stress

2

produced in the member by a load unity at B, the truss beingassumed fixed at H.

4 0 2 DESIGN OF STEEL BRIDGES [CHAP. XVII

Plate LI shows the influente lines for a cantilever truss with sub-d i v i d e d panels.

S t resses in Intermediate Anchar Spans.-To obtain any influenteline for an intermediate anchar span AB (Fig. 7) we draw the part ofthe influente line to the right of A in exactly the same way as shown inFigs. 2, 3 and 5. Then we extend the straight line immediately to theright of A as far as its intersection with the vertical through hinge HIand from this point draw a straight line to the intersection of the baseline with the vertical through hinge Kl. This is shown in Fig. 7 b and c

( b.)

FIG. 7.

for the shear in panel mn and for the moment at m (compare Fig. 3 band c on page 398).


The dead load of cantilever bridges, except for the suspended spans,cannot be assumed uniform as for simple spans, since the weight ofthe trusses and bracing increases considerably toward the intermediatepiers. This weight is assumed as shown on page 410. After the panelconcentrations have been determined, and conveniently tabulated (seePlate L) the dead load stresses can be found by means of the influentelines. To obtain the stress in any member the panel concentrationsare multiplied by the corresponding ordinates of the influente line andthese products added algebraically, and, finally, the sum is multiplied bythe influente coefficient.

Generally, however, the following method is shorter:The reactions are first determined either by the influente lines

or by a forte and equilibrium polygon or analytically. For instance,taking moments of all loads about suppor t A and dividing by thelength II of the anchar arm, the reaction Rb is obtained. R, is then

PLATE L

Dead Load Panel Concentration for Cantilever Truss. (Facing zq-e 402)


found as the difference between Rb and the sum of all loads. Qne-half of the weight of the suspended span is assumed concentrated atthe hinge. The stresses are then determined by a Maxwell diagramas shown in Fig. 8. It is simpler to assume the panel loads concen-trated at the chord nearest the floor and make afterward the correc-tions in the stresses of the verticals due to the loads at the other chord.

0w 5 - 7 5I d I

s,*- 7 - B l---17

FIG. 8.


Al1 chord and most web members of the anchar arm (Fig. 5) gettheir maximum stress if the cantilever arm and suspended span arefully loaded with the greatest concentrations at and near hinge H.,provided only loads continuous in one s t retch are considered. Sinceal1 these members have the same influente area between B and K it isnecessary to determine only once the sum of the products of wheelconcentrations and influente ordinates. By multiplying this sum bythe respective influente coefficients the stresses can be obtained withone setting of the slide rule.

For a uniform load the influente line needs not even be drawn; the12

influente area between B and K is 4 (Z, + ZS)E,’ This is to be multi-28


plied by the uniform load w per lin. ft. and by the correspondinginfluente coefficient. Instead of the wheel loads an equivalent uniformload may be used which causes the same bending moment in a simplespan of the length lz+&.

Some web members near B whose center of moments I lies outsideof AB get their maximum stress if the load covers the whole length AK.For instance, for diagonal LzU3 (Fig. 5 c), the negative area is smallerthan the positive area to its left; therefore, the resultant area is stillpositive or of the same sign as that to the right of B.

The minimum stresses are caused by loads on the anchar arm andmay be determined.either by means of the influente lines or analyticallyconsidering the anchar arm a simple span.

The stresses in the cantilever arm are best determined by the influentelines.

ART. 5. EXAMPLE FOR THE CALCULATION OF A CANTILEVER(See Plate LI)

Fig. (a) shows the anchar and cantilever arms of a cantilever bridge2800 ft. long with a main opening of 1800 ft. The anchar and cantileverarms are 500 ft. each and the suspended span 800 ft. Al1 loads will beassumed applied at the panel points of the bottom chord.

Influente Lines.-The influente line for the reaction R, at Awhich consists of the two straight lines A2H1 (passing through BI) andHIK1 is first drawn by making the ordinate at Al equal to unity and,since the anchar and cantilever arms have the same lengths, the or-dinate at H1is also equal to unity. Theinfluence lines for the membersof the anchar arm between A and B are now drawn as for a simplespan as explained in Chapter IV (page 90) by keeping the ordinate atA1 and HI constant for al1 members and equal to unity; the influenteline BIHIK1 is then identical for al1 members (Figs. b and c). If therewere no subdivision of panels between LS and B the intluence linefor the bottom chord members LsB would consist of the two lines Al4and 4B1 (Fig. b). Owing to the subdivision by the secondary trussesLsMsL4 and L4M2B the influente area for the bottom chord LsL~ isincreased by the triangle 4 6 8’ and that for LdB by the triangle 4 2 BI;owing to another subdivision by the trussed stringers L~MTL~J, LGM~L~,etc., the influente area for LLS is increased by triangle 8’ 7 6, that forLsL4 by triangle 6 5 4, etc. As can easily be proven points 6 and 7are obtained by producing line B14, points 2 and 1 by producingline A14, point 5 by producing line 8’6 and point 3 by producing

ART. 61 CANTILEVER BRIDGES 405

line B12. Similarly the influente lines for the diagonals LsM7 andM7Me follow from that for MsU4 by the addition of triangles 8 7 4and 8 6 4 respectively (Fig. c). Line A,8 (Fig. c) is obtained byproducing the top chord UgU4 to its intersection I with the bottomchord, drawing a vertical through I until it intersects the line B,A,in í1 and producing line IIAl; as a check this line must intersect

the vertical through B at a distance of s = 6.00 from BI (the values

390 and.65 follow from Fig. a).The signs of the influente areas can easily be found by inspection.

For any member the parts of the influente area above the base lineAIBIK1 are of opposite sign to those below that line. The influentecoefficients are obtained by a Maxwell diagram (Fig. f) as the stressesdue to a load unity at A.

In a similar way the influente Enes for the members of the cantileverarm are found by making the ordinate at H1 equal to unity for al1members (Figs. d and e). The influente coefficients are then equalto the stresses caused by a load unity at H and are found in the sameMaxwell diagram (Fig. f) as those for the anchar arm.

The live load stresses can now be obtained from the influente lines asexplained on page 90.

The dead load stresses have been found by a Maxwell diagram (Fig.g) after the panel concentrations had been determined. As the panelloads are assumed concentrated at the bottom chord the stresses in theverticals must be corrected for the loads applied at the upper and inter-mediate panel points. ,


In a rationally designed cantilever bridge the wind forces are trans-ferred to the supports along the shortest way through lateral systemsbetween the chords. High wind stresses in the web members of themain trusses should be avoided. Wind stresses are caused in the webmembers if the chorde, along which the wind forces are assumed to act,are polygonal since at any point where the chord changes direction acomponent of the wind stress in the chord goes into the main truss. Thewind stresses in the main trusses are sometimes considerable and cannotbe neglected as is generally done in the case of simple spans withpolygonal chords. The overturning effect of the wind may also causehigh stresses in the main trusses and should be considered.

The assumption is generally made that the bridge acts as a canti-

406 DESIGN OF STFE¿ BRIDGES [CmIJ. XVII

levcr horizontally in the same manncr as vertically, the lateral trussesbeing hinged between the cantilever arm and the suspended span andprovision being made at all piers to transmit to them effectively thelateral wind reactions. This assumption is justified if the piera ortowers on which the bridge is fixed longitudinally are high and slenderand therefore able to twist a certain amount owing to the horizontaldeflection of the bridge. If the pier is comparatively low the windstresses should be calculatecl with the assumption that the bridge is fixed(unable to turn horizontally) at that pier. The difference due to theseassumptions affects only the stresses in the .lower lateral system of theanchar arm.

Usually the following wind forces have to be considered.(a) Wind on trusses and jloor; bridge unloaded.(b) Wind on trusses and jloor combined with wind on the moving train,

a lower wind pressure being used in this case than in the jkst.The wind acting on the trusses and floor may be specified either as a

static load over the whole length or as a moving load (in the latter caseonly continuous stretches of great length need be considered).

Either of the above conclitions may cause the maximum stresses in thelaterals and members not subjected to live load stresses. For a mem-ber receiving live load stresses, only the second condition needs to beconsidercd and the position of the wind acting on the train should corre-spond to that of the train causing the maximum live load stress in themember in question.

The wind acting on the trusses and jloor should be treated separatelyfrom that on the train.

Since the stresses due to wind forces of opposite direction haveopposite signs, it is necessary to calculate only the absolute maximumstresses and it will then usually suffice to consider the following positionsof the moving wind forte.

1. Suspended span and cantilever arm fullyloaded, causing maximumstresses ,in all members (lateral and main) of the cantilever arm and inmost members of the anchar arm. This follows from the influente Enesfor shears and moments (Figs. 3 and 4).

2. Anchar arm as simple span, partly or fully loaded, causing maxi-mum stresses in the top lateral system of the anchar ‘arrn (includingportals) if this system is independent of the cantilever arm.

3. Bridge fully loaded, csusing maximum stresses in the bottomlaterals of the anchar arm near the mnin pier and possibly in certainweb members of the main truss.


We will assume a through bridge with polygonal top and bottomchord (Fig. 10) and with lateral systems in the planes shown by henvylines. The wind forces are concentrated at the panel points along thechords and the floor; the suspended span and the cantilever arm arefully loaded. Al1 wind forces acting on one-half of the suspended spanare transferred to the bottom chord of the cantileverarm. Their resultant R can be substituted by anequal lateral forte R at L7 and two equal but opposite

vertical forces R ab’We assume further that the wind forces F acting

along the floor are transferred by sway frames to thebottom chord, thereby causing two equal but opposite

hvertical forces F x at ench pancl point (Fig. 9 for

”

panel point L,).FIG. 9.

The vertical forces R % and F h affect only the main trusses.

The resulting stresses are bes-t found by a Maxwell diagram after thevertical reactions have been determined as explained on page 404.

The lower lateral system forms a cantilever LsLOL,LT (Fig. ll).

FIG. 12.

Fra. ll.

The horizontal forces acting along the bottom chord of the canti-lever arm cause a horizontal reaction HS at LG of.the anchar arm (ob-tained by talring moments of al1 forces abput a vertical through Loand dividing by íJ and HO at Lo (equal to the sum of HC and thehorizontal forces).

408 DESIGN OF STEEL BRIDGES [CISAP.XVII

The shears and moments are now determined by assuming the lateralsystem projected on a horizontal plane. The shear V in any panel ofthe anchar arm is equal to He and that in a panel of the cantilever arm isequal to the sum of the forces to the right of that panel. Assuming

only one lateral diagonal in any panel acting, its stress isd

V 6, where

d is its actual (not projected) length and b = width c. to c. of trusses.If M is the bending moment at any panel point the stress in the

opposite chord member is M fo- -, where fi = chord length and f itsb f

projection or the panel length.Owing to the change of direction in the bottom chord at Lb the re-

sultant Vd (which, as shown on page 404, is in any case vertical) of the twochord stresses L3L4 and LL5 (both being obtained from moment at L.s)(Fig. 12) acts on the main truss as an externa1 forte causing stresses inall members to the left of U2L4. It may be treated either separately

FIG. 13.

hor combined with the vertical forces Rz and Fb noting that it acts

in opposite sense to these. It causes vertical reactions at the bearingsLS and LO. If the chord changes its direction at more than one point,the same method is applied to each of these points. The change ofdirection of the bottom chord at Lo affects only the vertical reactionLo by an amount equal but of opposite sign to the vertical componentof the chord stress L0L1 (this stress being determined from thebending moment at Lo).

The wind forces Uz to U”r are transferred to Lo throuyh the bracingLoi which forms, with the top lateral system U2U7 and the bottomlateral system of the anchar arm, a cantilever (Fig. 13), which is to betreated in exactly the same manner as that shown for the bottom lateralsystem (Fig. ll).

The stresses in the bottom laterals of the anchar arm from the load-


ing of Figs. ll and 13 have to be added, and for every member of themain truss the stresses due to the horizontal and vertical forces haveto be combine& special attention being given to their signs, as thestresses may have different signs from different sources. To avoidmistakes one truss only, for instance, that on the near side, should beconsidered. The stresses in the far truss are of opposite sign for thesame direction of the wind.

For loads on the anchar arm, the latter is treated as a simple spanas shown on page 399. They cause maximum stresses in the top lateralsand portals, including members LcU5 and U&o. Together with thewind on the cantilever arm, the wind covering the whole anchar armcauses maximum stresses in the bottom laterals of the right half of theanchar arm (this follows from the influente Iine for shear, Fig. 3 b).

ART. 7. ERECTION STRESSES

If the suspended span is erected as a cantilever considerable erec-tion stresses are caused in the members at the ends of cantilever arm andsuspended span which require extra material. They are generallygreatest when the traveler has reached its extreme position. However,

I 1 H

%

K--_--*

+

FIG. 14.

certain web members of the cantilever arm may get their greatest stresseswhen only part of the cantilever is erected. The most unfavorableposition of the traveler and the corresponding stresses are easily derivedfrom the influente lines whereby the suspended span has to be treated,not as a simple span but as part of the cantilever arm. The influentelines (Figs. 2 to 6) change only in so far as the line to the right of Hbecomes a straight continuation of the line immediately to the left of H.For instance, Fig. 6 d would become as shown in Fig. 14, and it is evidentthat the greatest erection stress in diagonal BUI might be caused if thetraveler stancls somewhere between B and 1. A Maxwell diagram will beconvenient to determine the stresses for the extreme position of thetraveler.

The following approximate weights of various travelers may beuseful for preliminary designs:


Bridge ,zt . . . . . . . . . . . . . . . . . . . . . . . . .

A crms . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 ,Quebco / Blackwell’s 1s. 1 Pittsburg 1 zz;;pi

St Lawrenco East River N . Y . Monongahela

Length of main span ín ft. .Width c. to c. trusses in ft., .Weight of traveler incl. engine,

1800 1182 812 67167 6 0 3 2 32

etc., in tons . . . . . . . 1125 550 2 9 0Kind of traveler.. . . Outside Inside Inside

100

:l?o%’

Attention should also be paid to the mind stresses during erection,as these may require extra material when combined with the othererection stresses. The wind on the traveler must be considered.

ART. 8. DEAD LOAD ASSUMPTIONS

In simple spans the dead load can, with sufficient accuracy, beassumed uniformly distributed over the whole length; in canti leverbridges, however, the weight of the trusses and bracing increases con-siderably from the ends of the cantilever and anchar arms toward themain piers and a difference in the assumed load from the actual mayresult in considerable errors in the stresses in certain members. Further,in bridges of moderate span the dead load is usually smaller than thelive load, so that a small difference in the assumed dead load will notinfluente the weight of the trusses.

In long span bridges, however, the dead load is more important thanthe live load, as every Pound additional weight may require severa1pounds of additional steel. The proportion of steel weight (not of deadload) to the assumed working live load (without impact) is, for thefollowing cantilever bridges,. approximately (compare pages 372, etc.).

Greatestspan in

ft.

Proportion ofstcel weight toworking live

load

T h e b e s b r i d g e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 671M i n g o b r i d g e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 0 0Beaver bridge. . . . . . . . . . . . . . . . 769Memphis bridge (ene track only and shallow trusses). . 7 9 0Monongahelabridge. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 812A design for four R. R. tracks (nickel steel eyebars used) . 855A design for four R. R. tracks (no nickel steel used). 855Queensboro bridge (nickel steel eyebars used but heavy floor). 1182A design for first Quebec bridge (high unit stresses). 1600Firth of Forth bridge (riveted connections, light live load). 1710First Quebec bridge (high unit stresses) . . . . . . . . . . . . . . . . 1800A design for second Quebec bridge (nickel steel eyebars used).. 1800A design for second Quebec bridge (no nickel steel used). . . . 1800

0.91.01.71.71.11.01.13.53.74.74.44.24.6

Am. SI CANTILEVER BRIDGES 411

The heavier the live load per foot of bridge, the smaller the “Propor-tion”; the heavier the additional dead load, as pavement, etc., the greaterthe “Proportion.”

In order to arrive at a sufficiently accurate assumption of the deadload for the design, it is necessary to make one or more trial calculations,as no formula or comparison with similar bridges can give satisfactoryresults, on account of the different relation of live and dead load, per-missible unit stresses, relation of span lengths, height and form of, trusses,panel length, etc. A superfine mathematical investigation has no prac-tical value, as it is either too complicated or, on account of simplifyingassumptions, not better than an empirical formula.

The weight of the suspended span is found as for ordinary simple spans.If this span is erected as a cantilever about 10% should be added tothe weight of the trusses for extra material required for the erection butremaining permanently in the bridge.

F IG . 15 .

The weight of the floor is, of course, independent of the span ,lengthand can be determined accurately as soon as the panel length and thedistance between centers of trusses have been assumed.

The weight of the trusses of the cantilever and anchar arms of athree-span cantilever bridge may, for a first calculation, be assumedto increase uniformly from the ends H and A respectively toward themain pier B (Fig. 15) and be determined by the following formulas:

If Z1 = length of suspended span in ft.,12 = length of one cantilever arm in ft.,13 = length of one anchar arm in ft.,w1 = average weight of trusses of suspended span in ib. per lin. ft.,pl = total load (dead + live load + impact) on suspended span

in lb. per lin. ft.,p2 = average total load (dead + live load + impact) on cantilever

arm in lb. per lin. ft.,the weight of the trusses of one cantilever arm is, in lb. per lin. ft.,-___

w2“= 3.2 w1 at the main pier B. . . . (1)

412 DESIGN OF STEEL BRIDGES [CHAP . XVII

w2 = 1.6~1 -p,2;+- in the average. . . . (2)1 p1 ll2

and w2l = 2w2 - wz’l at the hinge H.. . (3)

The weight of the trusses of one anchar arm is, in lb. per lin. ft.,w3” = w2” atthemainpier B . . . . . . . . . . . . . . . . . . . (4)

w3 = s = 0.7wa” in the average.. . . . . . . . (5)

w3’ = 0.4 w3” at the anchar pier A.. . . (6)

The weight of the trusses of one cantilever arm at a distance x: fromthe hinge H is per lin. ft.

WZ = w2’+(wz”-w2’) ;. . . . . . . . . . . . . (7)2

and the weight of the trusses of one anchar arm at a distance y fromthe end A is per lin. ft.

W v = wa’+(w3”-w3’) f... . . . . . . . . . . . . . . . (8)3

The panel concentrations are obtained by multiplying the weight perlin. ft. at the corresponding panel points by the average length of theadjoining panels. The weight is thus distributed first to the main panelpoints only and corrections are then made for the concentrations at thesecondary panel points.

A full panel concentration should be assumed at the end H of thecantilever arm owing to extra erection material at that point. Thetotal weight of the trusses of one cantilever arm is therefore

w2 =w2’;w2” 12 ; “a’ x

where X is the panel length at the end H.The total weight of the trusses of one anchar arm is

w3 = .u131;11ig” l3 = 0 . 7 w31q3.

As explained below this formula is not applicable to unusually longanchar arms; for these the weight approaches that of a simple span ofthe same length.

The weight of trusses of an intermediate anchar span in a long canti-lever bridge of more than three openings may be assumed uniform perlin. ft. and equal to

Anr. SJ CANTILEVER BRIDGES 413

assuming that the length of the cantilever arm and suspended apan oneach side of the anchar span are respectively alike. If they are differentthe weight of the trusses of the anchar span per lin. ft. may be assumedto change lineally between the two different values ~2~‘.

In order to get the total average load pz the weight of the cantileverhas first to be guessed, and if the weight found by the formula does notcorrespond to this assumed weight, pz has to be corrected and the formulaapplied again until the assumed and calculated values agree closely, whichthey generally will after the second application.

The weight of the bracing may be assumed similarly distributed asthat of the main trusses and can, therefore, be found by the sameformulas or, if w1 includes the weight of the bracing of the suspendedspan then the above formulas give the weight of trusses plus the bracingof the cantilever and anchar arms.

It is, of course, assumed that the same material is used throughout. Ifcertain members are of nickel steel an allowance should be made for thecorresponding reduction in weight.

The weight of the bracing depends upon the width of the bridge; forrailroad bridges it is about 10 to 15% of the weight of the maintrusses.

The shoes and anchorages of a three-span cantilever weigh approxi-mately 5% of the total steel weight.

Por detailed weights of various cantilever bridges see pages 372 to386.

Another useful method of determining the weight of the trusses and-bracing of the cantilever and anchar arms is given in “De Pontibus,”by Mr. J. A. L. Waddell.

Example.- Let us apply the foregoing formulas to the three-span cantileverdescribed on page 376. Since part of the truss members is of nickel steel for which50% higher unit stresses were allowed, we get an approximate equivalent weightof carbon st,eel by adding 50% to the weight of nickel steel; we have then thefollowing average loads per lin. ft.

Weight. of trusses and bracing . . . . . . . . . . . . . . . . . .Weight of flooring and steel floor. . . . . . . . . . . . . . .Approx. live load and impact. . . . . . . . . . . . . . . . . . .

Total load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


Applying formulas (1) to (6) we get for the weight of the trusses and bracing of thecantilever arm per lin. ft.

W2 ” = 3.2X10,4002/1+1.77 = 54,300

W2 = 1.6X10,4002/:!+1.77 = 31,700 (actual 33,000)

WZ’ = 2X31,700 - 54,300 = 9,100

and for the weight of the trusses and bracing of the anchar arm per lin. ft.

w3” = 54,300W3 = 0.7x54,300 = 38,000 (actual 39,200)W3’ = 0.4x54,300 = 21 ,700

Al1 panels are 50 ft. long; by formulas (7) and (8) me get the panel concentrationsin units of 1000 Ib. as follows, considering a full panel load at the end of the cantileverarm:

Cantilever arm: 455+643+832+1020+1209+1397~1615+1773+1961+2150+2338+2527+$ 2715 = 19,277,OOO (estim. wght. 19,773,OOO).

Anchar arm: f 1085+1248+1411+1574+1737+1900+2063+2226+2389+2552 +$ 2715 = 18,996,OOO (estim. wght. 19,590,OOO).

* * m * * B *

The above formulas are derived as follows:Providing the trusses have an approximately economical height it is

reasonable to assume that the distribution of the steel weight is similar asin a beam with rectangular cross-section, constant width and uniformresistance. In such a beam the height and, therefore, the weight variesin direct proportion with the square root of the bending moment.

For a uniform load pl on the suspended span the weight of the latterchanges according to an ellipse (Fig. 16) and has an average value of

Wl = k. +f1/3PIll’

where k is a constant depending upon the unit stress, the width of thebeam and the unit weight of the material.

For a concentrated load Ei at H and a uniform load p2 on the

cantilever arm the weight of the latter changes according to a parabolawith the equation

W = k 2/w3p2X2

12whicli gives for x= -2

w2 = 1c.a dBp1l1 l2 i- 3p2122

a n d f o r x=1=w21’ = lc d3p1h& i- 3Pdz2

ART. SI CANTILEVER BRIDGES 415

For the parabola we may substitute a straight line (shown dotted)with the ordinates wz” and w2 and get for

x = 0, wz’ = 2wz - w’zr’.

Between the weights w1 and wz we have now the relation

By comparison with the weights of a number of cantilever bridges it isfound that the coefficient 1.27 has to be increased to 1.6 which givesformuia (2). This increase may be explained by the fact that, for practica1

FIG. 16.

reasons, the height of the trusses is generally made less than would be eco-nomical as far as the weight of the trusses is concerned. Formula (1)is derived similarly.

The formula for the weight of the anchar arm is derived with the sim-plifying assumption that only the suspended span and cantilever arm areloaded. This assumption is reasonable for ordinary lengths of anchararm since the reduction of weight in the chords due to the dead loadin the anchar arm is to a great part offset by the additional weight ofcertain web members due to the dead and live load in the anchar arm.For comparatively long anchar arms the formulas give somewhat exces-sive weights. The height and therefore the weight of the beam of uni-form resistance due to the negative reaction

changes according to a parabola tiith the formula


and if we substitute again a straight line as shown dotted we get for y = Z3,____~

w3” = k2/3p,&+-3p212~ = wzl’13and for ~‘-2’

andw3’ = 2w3-w3/’ = 0.4 w3”

* * * * * * *

The above formulas are useful for determining the variation of thesteel weight due to a variation in the span lengths and also for the deter-mination of the most economical span length in a given case. It will befound that the proportions of span lengths thus found agree closely withthose established by practice (see page 418). For the purpose of deter-mining the variation of weight of trusses and bracing the formulas aretransformed as follows:

The weight of the trusses and bracing per lin. ft. of the suspendedspan, as shown on page 414, is

If we substitute wr+q for pl, where p is the total load per lin. ft.excluding the weight of the trusses and bracing, introduce the valuem = 0.231 k2112 and solve for WI, we get

w1 = mf2/m2-+2mq. . . . . . . . (1)

For the weight of the trusses and bracing per lin. ft. of the canti-leuer arm we get by introducing

into formula (2) Art. 8:

w2

If we again substitutethe values n = 0.595 k2L2

= k 1.0942 p,1112+p2122

wl+q for pl and w2+q for p2, introduceand r = 2.38 lc2 1112 and solve for ~2, we get

w2 = n+ dn2+2nq+r(q+wl). . . . . . . . . (II)

Correspondingly we find for the weight of trusses and bracing perlin. ft. of the anchar arm, w3 = k 1.52d/p,lllz+pz

or w3 = do.97 rp1+3.88 np2 . . (III)

ART . SI CANTILEVER BRIDGES 4li

and for the corresponding weight of an intermediate anchar span

w4 = Ic 2. 17~p11~l~fp2122 = 2/2 rp1+7.9npz

In any given case the coefficient k can be found from the weight of thetrusses and bracing of the suspended span or any other simple spandesigned for the same loading, unit stresses, etc. The most economicalproportions of span lengths are found by a few trial calculations since thedifferentiation of the above formulas is too complicated.

The application of the above formulas gives the following approximaterules :

(a) The total weight of the trusses and bracing of the suspended spanor any simple span of the length II changes in direct proportion with 1:where x =2 for ordinary spans up to about 500 ft . and increasesslightly with increasing span length approaching 3 for excessively longspans; x is comparatively greater for lighter loadings.

(b) For given lengths of center span and anchar arms the total weight oftrusses and bracing changes only slightly if the length of the suspendedspan changes from about + to 4 of the center span and is approxi-mately a minimum for II = 0.4 of the center span practically independentof the relation between dead and live load.

(c) For given length of center span and suspended span the totalweight of trusses and bracing of the anchar arm changes in direct propor-tion with the length of the anchar arm; it is therefore the smaller theshorter this length, but with decreasing length the cost of the anchor-ages increases and the most economical length of anchar arm dependstherefore largely upon local conditions.

(d) For given proportions between suspended span, center span andanchar arms, in other words, if the center span and total length betweenanchar piers are changed in the same proportion as the suspended span,the total weight of trusses and bracing changes in the same proportionas the weight of trusses and bracing of the suspended span.

(e) For given total length L between anchar piers and economical pro-portion of suspended span to center span but variable length of centerspan, the total weight of trusses and bracing changes within certain limitsin about the same proportion as the length of the center span. I t istherefore the smaller the shorter the center span and is a minimum if thelength of one anchar arm is about 2 L. With further increasing lengthof anchar arm the total weight increases on account of the weight of theanchar arm approaching that of a simple span of the length + L.

(f) In a cantilever bridge with more than three spans, for a given


total length L of one main span (including the suspended span) plus oneintermediate anchar span, the total weight of trusses and bracing of thesetwo spans changes approximately in the same proportion as the length ofthe main span and is a minimum if the anchar span is about 0.4 ofthe length L or g of the main span. A further increase in the lengthof the anchar span increases the total weight as the weight of theanchar span approaches that of a simple span of the length L.

B. DESIGN OF CANTILEVER BRIDGES

ART. 9. PRINCIPAL DIMENSIONS

For example of long cantilever bridges see Chapter XVI (Long spanbridges). Whenever the height from clearance line to top of floor per-mits it, a deck structure should be used, as it is more economical consider-ing the smaller cost of the substructure.

Table 63 contains the principal dimensions of a number of builtor designed cantilever bridges; see also examples of long span bridges(page 372).

Besides economy in the material there are important practica1 con-siderations which determine the proper arrangement of span lengths.Often the span lengths are influenced by local conditions, such as specifiedclear openings, suitable foundations for the piers, etc.

If the length of the center span of a three-span bridge is given, as isusually the case, it becomes necessary to determine the proper length ofthe suspended span and of the anchar arms.

The economical length of the suspended span varies for different con-ditions, but within certain limits the total steel weight changes onlyslightly. It is well to consider that the longer the suspended span thesmaller the total deflection. For cantdever erection (without fixed orfloating falsework) a suspended span of # to 3 of the c,enter span is thebest; the smaller value to be used for longer center spans; for very longcenter spans even + might be preferable as, on account of the adjustingdevices between the suspended span and the adjoining cantilevor arms,the difficulties of erection increase with the weight of the suspended span.If the bottom chord is straight, a longer suspended span with a polygonaltop chord 1001~s better, especially if the trusses are comparatively highover the main piers. The average proportion of suspended span lengthsto center span in Table 63 is 0.4.

The length of the anchar arm should be governed by the conditionthat the negative dead load reaction at the anchar pier (causing tension


in the anchorages) is not reversed when the live load covers the anchararm, especially in railroad bridges, in order to avoid “hammering” ofthe ends and reversa1 of stresses in the chord members, so that the wholetop chord of the anchar arm can be built of eyebars. The proportionbetween length of anchar arm and center span depends, therefore, upon therelation of dead and live load and the length of the suspended span. As-suming the latter to be about 0.4 of the center span, the anchar arm willin general have to be + to $ of the center span. There should be a cer-tain margin for the negative reaction so that it cannot be reversed b y apossible increase of the live load including its impact. The total steelweight (excluding anchorages) is the smaller the shorter the anchar arms(if center span is given) even if the greater length of the approaches isconsidered. However, with decreasing length of anchar arm the negativereaction and, therefore, the cost of the abutment pier and anchoragesincreases; if there are natural abutments affording cheap anchoragesthe length of the anchar arm can be much shorter than if an expeneiveanchar pier has to be built. Further, very short anchar arms do notgive a favorable appearance.

If the total length between anchorages of a three-span cantileverbridge is given, and the intermediate piers can be located anywhere, themost economical length of anchar arm is about 2 of that length. A sim-ple span bridge of three openings, if not too Iong, may, however, be moreeconomical even if the center span has to be erected as a cantilever, besidesit is superior to the cantilever as far as stiffness is concerned. Thequestion whether a simple span bridge partly erected by the cantilevermethod is cheaper than a cantilever bridge depends largely upon the rela-tion between dead and live load. For a highway bridge (live load smallcompared with dead load) the erection of a simple span as a cantilevermay require a comparatively great amount of extra material remainingpermanently in the bridge, while in a railroad bridge, owing to the greatlive load, the sections are usually strong enough so as to require little or noaddition for the cantilever erection. It should be considered further thatin a simple span bridge there are usually more duplicate parts which reducethe cost of the shopwork.

For a long cantilever bridge with more than three spans navigationoften requires a certain minimum length of anchar span. As far as econ-omy in material is concerned a length of $ of the main span is approxi-mately most favorable for the anchar span, but, if possible, the upperlimit of the length of anchar span should be governed by the condition thatthe live load (including impact) in the anchar span shall not reverse the

29

4 2 0 DESIGN OF STEEL BRIDGES [CHAP. XVII

dead load stresses in the chord members of this span, and the lower limitby the condition that the live load in any adjoining cantilever span shallnot cause a negative reaction at the piers.

Table 63 shows the principal dimensions of cantilever bridges.For the arrangement of the croa-section, the same applies as for simple

spans (page 129) ; see also examples of long span bridges (page 372). Thedistance between centers of trusses should preferably be not less than &-of the main opening nor less than 2 of the height of the trusses over themain piers. It is not advisable to place the trusses into inclined planesas this increases the cost of shop work and also of erection considerably,which offsets the saving of material. In the Queensboro Bridge (NewYork) the main posts over the piers were made inclined since the pierswere built originally for a wider bridge. This necessitated the use ofinside travelers for the erection, which is undesirable since the swayand top bracing in any panel has to be omitted until the traveler haspassed that panel.

ART. 10. DESIGN OF TRUSSES

Concerning the design of the trusses the same general principiesapply as for simple spans (page 169). Polygonal bottom chords of theanchar and cantilever arms should be avoided whenever possible; theweb system should be simple so that the loads are transferred to thesupports by the shortest way, and the diagonals should have an inclina-t ion of not less than 45”. Multiple systems are objectionable. Thegreater part of the main pier reaction should come down the inclinedposts instead of being carried to the top of the vert ical post . Theproper height of the trusses over the main pier is practically independentof the relation of the length of suspended span to the center span andshould be about 6 to + of the center span, but not more than 4 timesthe width between trusses.

A variation from a shorter panel length at the end of the cantilever andanchar arm respectively to a longer one at the main pier is of advantage asfar as economy in material and appearance are concerned. Such a varia-tion should, however, be limited to a small number of different panellengths as otherwise the saving in material would be offset by the greatercost of shop work due to the different sizes of floor pieces.

The suspended span should have a curved top chord, convex upward,not only for economy but also to make it appear as an independent spanand not a continuation of the cantilever arm. Its design is governed bythe rules given for simple spans, its depth at the center being made from


6 to + of its length. Its panels should have preferably the samelength throughout. The suspended span is usually connected to thecantilever arm by means of a hanger free to move longitudinally at thebottom to allow for the expansion and contraction of the main span. Inrailroad bridges the lower end of the hanger should also be heldlongitudinally at one end of the suspended span so that the tractionand braking forces are transmitted from the suspended span to the‘cantilever arm.

ART. ll. BRACING

Lateral rigidity is very important. There should be an efficienttop and bottom lateral system and sway frames between all main com-pression members whether vertical or inclined.

The lateral forte from the suspended span should be transmitted tothe lower lateral system of the cantilever arm, except in deck spanswhere it might be more advantageous to transfer the top lateral forces tothe top lateral system of the cantilever arm. The top lateral forces of thecantilever and anchar arms should be transmitted through swaybracing between the inclined posts to the main piers instead of going tothe top of the vertical post.

ART. í2. DETAILS AND ANCHORAGES

Concerning the make-up of sections of truss members see page 173.Plate LI1 shows typical sections of various large cantilevers.

Pin connections should be used for all main members; in order tosimplify the packing of the pins, submembers may often be riveted to themain members by gusset plates. For the design of pin connections seepage 193. A splice of a riveted chord should always be close to the panelpoint on that side toward which the erection proceeds.

The following table gives the sizes of pins and eyebars used in variouslarge cantilevers:

Location of bridge River Mainspan, lWidesteyebar, W$Gefe;1 >

ft. in. in.

Quebec. . . . . . . . . . . . . . . . . . . St. Lawrence 1800::

12 and 14Queensboro, N. Y.......... East River 1182 14 to lS+Pittsburgh . . . . . . . . . . . . . . . . Monougaheia 812 12 and 14 12 and 14

.Memphis . . . . . . . . . . . . . . . . . Miy;s;ppl 790Minto. . . . . . . . . . . . . . . . . . . . 700 12 a: 14 12 a$ 14Thebes . . . . . . . . . . . . . . . . . . . 671 14 10 to 14Beaver . . . . . . . . . . . . . . . . . . .

M$sis&ppi769 16 12 to 16


The ends of anchar arms are usually anchored down by means ofeyebars extending into the anchar piers and secured thereto by rivetedcross girders. The anchorages should be entirely enclosed in concreteto prevent corrosion.

The lateral system should be anchored to the masonry independentlyof the main trusses. This anchorage is best located on the center line ofthe bridge where it does not interfere with the main anchorages.

The anchorages should engage masonry of a weight of at least 13 to2 times the greatest uplifting forte.

A~~.121 LONG SPAN BRIDGES 423

ADDITIONAL INFORMATION ON LONG SPAN BRIDGES

(a) Suspension Bridges :

Long Span Bridges and North River Bridge, Eng. News, Jan., 1888.Report of Board of Engineers on New York and New Jersey Bridge, Washington,

D. C., 1894.Report of Board of Engineer Officers on Maximum Span of Suspension Bridges,

Washington, D. C., 1894.Prof. Barkhausen, North River Bridge, Papers German Soc. C. E., Berlin, 1894.G. S. Morison, Suspension Bridge across North River, N. Y., Trans. Am. Soc.

C. E., Vol. 36, 1896 (with extensive discussion).Bridges or Tunnels (across East River, N. Y.), Eng. Record, Dec., 1899.C. W. Buchholz, North River Bridge and Tunnels in N. Y. (with discussion and

editorial), Eng. News, May, 1901.Joseph Mayer, Stiffening Trusses for Railway Trains, Trans. Am. Soc,. C. E., Vol.

48, 1902 (with interesting discussion).Gustav Lindenthal, Stiffened euspension Bridge, Trans. Am. Soc. C. E., Val. 55,

1905 (with extensive discussion).Leon S. Moisseiff, Computation of a Three-span Suspension Bridge with Braced

Cable, Trans. Am. Soc. C. E., Val. 55, 1905.Comparative Merits of Bridges and Tunnels, Eng. News, July, 1909.Gustav Lindenthal, North River Bridge, Eng. News, Dec., 1912 (also editorial).Boller and Hodge, Proposed Highway Suspension Bridge across the North River,

Eng. News, April, 1913.

(b) Suspension and Cantilever Bridges :

Gustav Lindenthai, Long Span Bridges, Address before A. A. A. S., 1890.Prof. Burr, Long Span Bridges, Proceedings Engineers’ Club, Phila., 1899.Suspension and Cantilever Bridges, The Engineer, London, January, 1902.Gustav Lindenthal, Quebec Bridge, Eng. News, Nou., 1911.Proposed Sydney Bridge, N. S. W., span 1350 ft., Report of Harbor Bridge

Advisory Board, 1904, and Report of Commission, 1909; also Eng. News, Sept., 1904,July, 1909, and F. Bohny, Engineering, London, Aug., 1904.

Types of Long’Span Bridges, Eng. Record, Sept., 1908.Outlines of Long Span Bridges, Eng. News, Aug., 1912.Proposed Sydney Bridge, N. S. W., span 1600 ft., Eng. News, July, 1913.

(c) Cantilever Bridges :

Report on New York and Long Island Bridge across the East River at Blackwell’sIsland (competitive designs for d. tr. R. R.), N. Y., 1877.

Early Designs for the Firth of Forth Bridge, Engineering, London, 1880, 1, p. 168.Proposed Bridge over St. Lawrence River at Quebec, Engineering, London, April,

1885.C. C. Schneider, Niagara Cantilever Bridge, Trans. A. S. C. E., Val. 14, 1885,

also Eng. Record, Oct., 1900 (strengthening by means of new center truss).John F. O’Rourke, Poughkeepsie Bridge, Trans. A. S. C. E., Vol. 18,1888.Proposed Cantilever Bridge between England and France, total length 24 miles,

120 piers, depth of water 180 ft., height for navigation 180 ft.., main spans 1640 ft.,alternating with anchar spans 984 ft., trusses inclined, greatest height of trusses 213ft., Eng. News, Nov., 1889.

Cantilever Bridges, Engineering, London, 1890, 1, p. 216.

4 2 4 DESIGN OP STEEL BRIDGES [CHAP. XVI

Henry H. Quimby, S. M. Rowe, S. W. Robinson, Red Rock Cantilever Bridge,Trans. A. S. C. E., Val. 25, 1891, also Eng. News, Sept. and Oct., 1890.

Gustave Kaufman and F. C. Osborn, The Cantilever Highway Bridge at Cincin-nati, O., Trans. A. S. C. E., Vol. 27, 1892, also Eng. Rewrd, Sept., 1892.

Designs for Proposed Montreal Bridge, Eng. News, Jan., 1897.Proposed New Orleans Bridge, Eng. News, Jan., 1897, Railroad Gazette, April, 1897.Proposed Cantilever Bridge over East River at Hell Gate, N. Y., d. tr., Railroad

Gazette, May, 1900.Interprovincial Bridge, Ottawa River, Eng. Record, Dec., 1901.Mingo Bridge, Ohio River, Eng. Record, July and Dec., 1904.Report of Canadian Roya1 Commission and C. C. Schneider on Quebec Bridge,

Ottawa, 1908.Charles Worthington, Design for Quebec Bridge, Eng. News, May, 1910.C. A. P. Turner, Design for Quebec Bridge, Eng. News, May, 1913.F. C. Kunz, Report on Blackwell’s Island (Queensboro) Bridge, in New York,

Steelton, Pa., 1909 (also comparison of long span bridges).Prof. Marburg, Compression Members in Long Span Bridges, Eng. Record, Sept.,

1910.A. W. Buel, Sewickley Cantilever Highway Bridge, Ohio, Trans. A. S. C. E.,

Vol. 76, 1913, also Eng. News, Feb., 1913.Tsinanfu Cantilever Bridge, China, Yellow River, 4100 ft. long s. tr., center span

540 ft., Eng. News, Jan. and Sept., 1913.Hanoi Cantilever Bridge, China, Red River, 5700 ft. long, nine anchar spans at

245 ft. and ten cantilever spans at 350 ft., Papers German Soc. C. E., Berlin, July, 1909.C. R. Grimm, Arch Principie for Long Spans, Trans. A. S. C. E., Val. 71, 1911

(with interesting discussion).New Memphis Cantilever R. R. Bridge, Eng. News, Nov., 1913.Ralph Modjeski, Long Span Bridges and the New Quebec Bridge, Journa of

Franklin Instifute, Phila., 1913.

(d) Long Simple Span Bridges :

Report on Connecticut Ave. Bridge across Rock Creek (competitive designs),Washington, D. C., 1898.

Delaware River Bridge, P. R. R., d. tr., 533-ft. span, Paul L. Wölfel, in Proceedingsof Engineers’ Club of Phila., 1897, also Eng. Record, Nov., 1899, and F. C. Kunz, in“Allgemeine Bauzeitung,” Vienna, 1900 (with many plates).

Rankine Bridge, d. tr., 500-ft. span, very heavy, Eng. Record, Nov., 1901.Allegheny River Bridge, Penna. Lines West, Pittsburgh, 4 tracks, 333-ft. span,

Eng. Record, Oct., 1902, Eng. News, Nov., 1902.Clairton Bridge, d. tr., 501-ft. span, Eng. Record, March, 1904.Highbridge, Kentucky River, s. tr., 3 spans at 353 ft., Eng. News, March, 1905,

and- April, 1911 (erected as cantilever).J. E. Greiner, Benwood Bridge, B. & 0. (erected as cantilever), Trans. A. S. C. E.,

Val. 55, 1905.Report on Terminals and Bridges at St. Louis, Eng. News, July, 1906.Steubenville Bridge, Ohio River, Penna. Lines West, d. tr., 240-ft. deck span, Eng.

Record, Nov., 1910.Miles Glacier Bridge, s. tr., 450 ft. longest span (erected as cantilever), Eng.

Record, Aug., 1910.Mobridge, C. M. & St. Paul, s. tr., 420-ft. spans, Eng. Record, June, 1910.McKinley Bridge at St. Louis, d. tr. Electric Ry. and Highway, 518 ft. longest

span, Eng. Record, April, 1910, Eng. News, July, 1910.

ART. 121 LONG SPAN BRIDGES 425

Municipal Bridge at St. Louis, d. tr. Ry. and Highway, 668 ft. longest spans (thelongest built), Eng. News, March, 1911, and Feb., 1912, Eng. Record, Oct., 1909, andDec., 1910.

Kentucky & Indiana Terminal R. R., Ohio River, d. tr. and Highway, ô20-ft. span,Eng. News, Feb., 1912.

Susquehanna Bridge, B. & 0. R. R., d. tr., 51%ft. span, Eng. Record, April, 1912.North Side Point Bridge, Pittsburgh, highway 60 ft. wide, span 531 ft. (heavy

construction), Eng. News, Oct., 1912.Dirschau Vistula Bridge, d. tr., 430-ft. max. span, riveted trusses, both chords

polygonal, Zeitschrift f iir Bauwesen, Berlin, 1895 (very complete description withplates).

Homburg Rhine Bridge, d. tr., 610 ft. longest span, riveted trusses of Warrentype, heaviest top chord 405 sq. in., heaviest bottom chord 450 sq. in., both chords54 in. deep, erected on floating falsework, Schaper, Zeitschrijt für Bauwesen, Berlin,1911 and 1912 (very complete description with plates), also Eng. Record, Feb., 1912.

Lower Ganges Bridge, at Sara Bengal, India, d. tr., 15 spans at 359 ft., rivetedtrusses 52 ft. deep at center, weight per span 1250 tons. Erected by means of a“service span” which can be floated from span to span. Difficult foundation. Totallength 5890 ft., Eng. News, Nov., 1911, Sept., 1913, and Jan., 1914.


GENERAL INFORMATION

B. Baker, Long Span Railway Bridges, London, 1873.Theo. Cooper, Ameritan R. R. Bridges, Trans. A. S. C. E., Vol. 21, 1889.Prof. Barkhausen, Review of Modern Bridges, Papers German Soc. C. E., Berlin,

Sept., Oct., Nov., 1889.J. A. 1,. Waddell, Bridge Design (with extensive discussion), Trans. A. S. C. E.,

Val. 26, 1892. .

Erection of Ameritan Bridges, Papers German Soc. C. E., Berlin, 1899, pp. 801 and834.

Prof. Mehrtens, Hundred Years of German Bridge-building, Edition in English,Berlin, 1900 (a valuable review of German and Ameritan bridges). I

Bridges in Theory and Practice, The Engineer, London, Dec., 1901.A. W. Farnsworth, Constructional Steel Work, Notes on the Practica1 Aspect and

the Principies of Design, London, 1905. (Enlarged reprint of very timely articlesfrom The Engineer, London, 1900 to 1904, on relation of design to manufacture, costsietc.)

William Henry Thorpe, Anatomy of Bridge Work, London, 1906. (Reprintof articles from Engineering, London, relating to design from the standpoint ofmaintenance.)

Henry S. Prichard, Proportioning of Bridge Members, Proceedings Engr. Soc.of W. Penna., July, 1907, also Eng. Netas, Sept., 1907.

Henry B. Seaman, Specifications for Bridges and Subways, Trans. A. S. C. E.,Val. 75, 1912.

The Engineer, London, Annual Reviews (published in the first part of each Val.).Report of the Department of Bridges, City,of New York, 1905-1912, published

1913.

APPENDIX A

ACTUAL PRESSURE OF FOUNDATIONS IN BRIDGE WORK, IN TONSPER SQ. FT.

New York.-Brooklyn Bridge.River Piers: hard pan bottom for Brooklyn pier, 7 tons (45 ft.

below water level); rockbottom for Manhattan pier, 7 tons (78ft. below water level).

New York approach 34 to 46 tons.New York.-Manhattan Bridge.

River piers are founded 92 ft. below water level, of which 60 ft. issand. Brooklyn pier on boulders and cemented grave1 3 ff.thick overlying bedrock (settlement $ in.) ; Manhattan pier oncompact sand and grave1 25 ft. thick overlying bedrock (settle-ment 4 in.). Pressure 8 tons per sq. ft. considering buoyancyand skin friction (500 lb. per sq. ft. of surface).

Neti York.-Williamsburg Bridge.Foundations of Anchar Piers: Brooklyn side 5 to 53 tons (set’tle-

ment 1: in.) ; Manhattan side, rests on piles.River piers on rock bottom 5 tons (Brooklyn pier 88 ft., Manhattanpier 66 ft. below water level).

Poughkeepsie R. R. Bridge.-River piers, on grave1 3 tons.Mississippi Arch Bridge at St. Louis, on rock 19 tons, 103 ft. below water

level.Mississippi Bridge at Memphis 5.3 tons or 2.9 tons, without or with

buoyancy and skin friction, respectively.Cleveland Central Viaduct, Ohio.-Blue clay bottom 2 tons.Cincinnati Suspension Bridge, on toarse grave1 12 ft. below water leve1

4 t o n s .New Orleans, on soft alluvial soil less than 1 ton.

Sydney Bridge design (New South Wales).-Bottom of caisson 8 tons.Berlin Elevated R. R. (Germany) .-Pier foundation of masonry arches, on

dry sand 4+ tons. Tests have shown that the soil could be pressedconsiderably higher when the pressure is uniformly distributed andsmall settlements permissible; 4+ tons, however, were used, since thepressure varies in intensity and direction for different positions ofthe live load.

Rhine, Kuilenburg (Holland), on compact sand, 3 tons.4 2 7

428 A P P E N D I X A

Viaur R. R. Viaduct (France), on gneiss rock, 7 tons.Coarse sand, Nantes Bridge, Seine, 73 tons.Coarse gravel, Seine viaduct, Point du Jour near Paris, 5 tons.

Yellow sandy clay, R. R. from Busigny to Somain, 3 tons.Unstable sand, R. R. from Busigny to Herson, 2 tons.Compact clay, Quaywall in Antwerp (Belgium), 5 tons.

London compact blue clay for foundations:Thames, Tower bridge piers, Highway, 44 tons.Thames, Charing Cross bridge, R. R., 5 tons.Thames, Blackfriars bridge, R . R., 53 tons (settled).Thames, Cannon Street bridge, R. R., 6% tons.Thames, Old Westminster bridge, 53 tons (failed).

Excluding exceptional cases the author recommends as follows:

LIIb%ITS OF FOUNDATION PRESSURE

Tons TonsS o f t a l l u v i a l s o i l . . f to 1 \Compact hard clay ,Confined soft clay 1

I t” 2Dry toarse sand

C o n f i n e d w e t s a n d (

( . . . . . . 4to5

G r a v e 1 w i t h toarse s a n d . 5 to 7Medium dry clay ’

2 to 3C o m p a c t g r a v e l . 7to 10

Dry sand mixed with clay H a r d r o c k . . . . . . . . . . . . . . . . . . . loto 2 0C o m p a c t d r y clay. . 3 to 4

Greater pressures than these have produced no settlement when thefoundation bottom is at great depth in the soil, for instance:

Hard sand, Ganges delta, Gorai R. R. Bridge, 9$ tons, 100 ft. belowwater level.

Yellow clay, Bengal, Hugli R. R. Bridge, 10 tons, 100 ft. below waterleve1 of which 60 ft. silt and 10 ft. clay.

Sand, Ganges at Benares, Dufferin R. R. Bridge, 12% tons, 190 ft.below water leve1 of which 80 to 160 ft. sand.

Sand, New South Wales, Hawkesbury R. R. Bridge, 10 tons, 162 ft.below water leve1 of which 120 ft. mud and sand.

These pressures however have been determined without consideringthe skin friction and buoyancy and are therefore greater than the actualpressures.

Iron Piìes.-Morecambe Bay (England), Leven and Kent Viaductsfounded on hollow iron disc piles in 20 ft. of sand, 42 tons per sq. ft.(actual tests gave 5+ tons as safe).

Other examples of screw or disc piles in very deep sand give 8 tonsper sq. ft.

APPENDIX A 429

Timber Piles.-Bridge at Neuilly (France), 12 in. dia., in grave1 restingon rock, 30 tons per pile. Bridge at Orleans, 30 tons per pile (one piersettled 10 in.). Bridge at Tours, 42 tons per pile (pier failed). Locks inHolland, 7 tons per pile. Quaywall in Rotterdam, 4 tons per pile.

Perronet used for 10 in. dia. 25 tons, for 12 in. dia. 50 tons, in goodsoil.

Bridge over Tyne at New Castle (England). Piles in grave1 70 tonsper pile.

New London Bridge, England (built 1831), 80 tons per pile, total pres-sure 5 tons per sq. ft. on soil (settled).

New Westminster Bridge, 12 tons per pile, total pressure 2 tons persq. ft. on soil.

Piles in moderately compact clay, 12 tons; in hard cIay, 25 tons; ingravel, 70 to 80 tons, approaching their compressive ultimate stress themore, the harder the surrounding soil.

APPENDIX B

ACTUAL PRESSURE OF FOUNDATIONS IN STRUCTURAL WORK, IN TONSPER SQ. FT.*

New York CitySana.-Around the Post-Office most buildings, 4 tons.St. Paul Building (concrete and steel grillage), 3.2 tons.World Building (inverted arches), 4.7 tons.Central Bank Building (concrete pedestals), 4 tons (settlement & in.).Municipal Building, 6 tons, 80 ft,. below Street level.Pine Xand.-Pressures of 3 to 4 tons have been used, with settlements

of t to ?j in.Hard Pan.-Church at 37th St. and 5th Ave., 7 tons.Ro&-Hotel Waldorf-Astoria, rock foundation at smnll depth;

pressure of concrete on rock, 15 tons.Pneumatic Foundations on Bedrock.Singer Building, 15 tons, 90 ft. below st,reet level, 70 ft. below water line.Manhattan Life Building, ll tons.Gillender Building, 12 tons.Ameritan Surety Building, 7 tons.Disc Piles.-Coney Island Pier, 8 tons per sq. ft. of disc on sand.Timber Piles.-Park Row Building, 16 tons per pile.

ChicagoBoard of Trade Building, 23 to 32 tons per sq. ft.Grain elevators on piles up to 6 tons per sq. ft. on soil.Auditorium Building (steel and concrete grillage), 2$ tons.Fair Building, 3 tons.Monadnock Block, 3% tons.

ClevelandPope Building, l& tons on wet sand.

PittsburgUnion Station, on gravel, 4 tons.Farmer’s Ba& Building, on sand and gravel, 3 tons.Frick Building, on rock, 15 tons.

* See Preface.4 3 0

APPENDIX B 431

CincinnatiUnion Central Life Insurance Building, 53 tons on dry sand, 50 ft.

below Street level.

BostonIn the higher parts of the City, on sand and grave& 5 tons per sq. ft.For friction piles with 20 ft. penetration in stiff sand, 10 tons per pile.Boston South Terminal, 8 to 10 tons per pile.Bunker Hill Monument, on hard sand and gravel, 5; tons per sq. ft.

(no settlement).Boston Elevated R. R., on good soil 3 tons; on clay overlying softer

blue clay 2 tons; on piles 10 tons per pile.

WashingtonWashington Monument (weight 90,000 tons), on fine sand 2 ft. thick

resting on gravel, 8 tons or 12 tons, without or with wind pressure,respectively.

Congressional Library, on yellow clay mixed with sand, 2+ tons.

AlbanyCapitol, on clay 3 ft. below surface, 2 tons.

NorfolkCoa1 pier on disc iron piles, 50 tons per pile ( = 5 tons per sq. ft. of

disc).

New OrleansOn soft alluvial soil less than 1 ton.

SeattleSmith Building, Raymond concrete piles, 20 ft. long, through sand

and clay, pressure 30 tons on each ( = 4 tons per sq. ft. on clay bottom).

San FranciscoSpreckles Building, on compact wet sand (grillage), ‘2+ tons, 25 ft.

below Street level.Phelan Building, on sand (grillage), 3 tons, 25 ft. below Street level.Panama Exposition Buildings, on sand and clay, 20 tons per pile.

London, EnglandNelson Column, on stiff clay, 1.3 tons (weight 4665 tons on 60 ft.

square, concrete and brick).

Paris, FranceEiffel Tower, on gravel, 18 ft. thick, 3 tons.Exposition buildings, on grave1 and stiff clay.: when grave1 10 ft.

432 APPENDIX B

thick, 3 tons; when grave1 5 to 10 ft. thick, 2: tons; when grave1 less than5 ft. thick, piles were used.

Hamburg, GermanyTower of Waterworks (290 ft. high, weight 5000 tons), on quicksand

between sheet piling, 2 tons.Bismarck Monument (weight 8000 tons), on sand, 2.3 tons.

Berlin, GermanyStone foundations of the elevated railroad on dry sand, 4+ tons.Church steeples on dry sand, 3 tons.

Cremona, ItalyCampanile (395 ft. high), on pliocene gravel, 12 tons.

Pisa, ItalyLeaning Tower (178 ft. high, weight 12,000 tons), 4 tons.

Venice, ItalyOld Campanile (weight, 15,800 tons), pressure without wind 6: tons,

with wind 83 tons, on the underlying clay and sand which was com-.pressed by driving into it short piles (the usual foundation method inVenice).

New Campanile, pressure 2 tons smaller than in old one.

APPENDIX C

Physical Properties of Metals

MaterialT

1

2Soft, 0.10% c.. . . . . .M e d i u m , 0 . 2 0 % C . . _.

3, Rail, 0.70% C.. . . . .pi

:Rail, 1.00% C.. . _.3;s N i , 0 . 4 0 % C . . _. .

* Soft......................

B iM e d i u mHard.....................

~- - -SofI?......................

% M e d i u m .6 H a r d

N i c k e l . . . . . . . . . . . . . . . . . . . .

%&

i

Wrought . . . . . . . . . . . . . . . . . . . . . .Wire, unannealed . . . . . . . . . . . . . . .Cast , 34% C, compression.Cast, 31% C, tension. . . . . . . . . . .Cast, 33% C, bending . . . . . . . . . . .Cast, malleable, tension. . . . . . . . .

Bronze, compression . . . . . . . . . . . . . . . .Bronze, tension. . . . . . . . . . . . . . . . . . . .Brass, cast, compression. . . . . . . . . . . . .Brass, cast, tension . . . . . . . . . . . . . . . . .Copper, cast. . . . . . . . . . . . . . . . . . . . . . .Copper, forged. . . . . . . . . . . . . . . . . . . . . .Copper wire . . . . . . . . . . . . . . . . . . . . . . .Zinc, cast, tension. . . . . . . . . . . . . . . . . .Lead, compression . . . . . . . . . . . . . . . . . .Lead,.tension. . . . . . . . . . . . . . . . . . . . . .Alummum, cast . . . . . . . . . . . . . . . . . . . .Aluminum bars. . . . . . . . . . . . . . . . . . . .Aluminum wire, . . . . . . . . . . . . . . . . . . .Platinum, wire. . . . . . . . . . . . . . . . . . . . .

Notes.-

I

--

.1:

- -

- -

-

Tltimate/ Yield point ~~~a$$&

:n tension or compression in unitsof 1000 lb. per sq. in.

5 0

110140100

90-140140-200200-280

60-9040-60

3 0

30

50

4.872-4

10-1520-2530-40

5 0

60-100loo-160160-220

5:i, + ultimateI

>+ ultimate

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .I

t+ ultimate

.

29,500

J30 ,500

1 29 ,500J

,i 30 ,000

30 ,500

28 ,000

1

30 ,000

13,000

1 2$00

1 13,000 11,000

1 16,000

i

14,000

700

10,000

24 ,000

490

1 480

450

-

-I

-,

>I1

1-

530

500

5 0 0

440

710

165

1350

Ultimate of steel(acc. to H. H. Campbell) =

Basic: 37,430 + 95C + 8.5Mn + 105PAcid: 38,600 + 121C $ 89P

Ultimate of steel(acc. to A. C. Cunningham) =

Basic : 40 ,000 + 1,OOOC where C inAcid: 45,000 + l,OOOC > 0.01%.

4 3 3

1 where C,Mn, P inunits of0.001 yo.

units of

4 3 4 APPENDIX C

For Nickel steel add to Ultimate and Yield point 5000 lb. for each 1% of nickelup to 4%.

Elongation in 8 in. of soft and medium rolled steel = 1,500,OOO divided by UltimateElongation in 8 in. of rolled nickel steel = 1,7Oq,OOO divided by UltimateReduction for rolled carbon steel equals approximately twme the>Elongation in 8 in.Elongation in 2 in. of cast carbon steel = 1,200,OOO divided by Ultimate.Reduction of cast carbon steel equals approximately 1.3 times the elongation in

2 in.Elongation in 2 in. of cast nickel steel = 2,000,OOO divided by Ultimate.Reduction of cast nickel steel equals approximately 1.6 times the Elongation in

2 in.Ultimate in shear for stecl and wrought iron equals + to & of the Ultimate in

tension.Modulus of Elasticity in shear for steel and wrought iron (also cast iron and bronze)

equals Is of Modulus for tension given above.Linear Expansion by Heat of 100” F.: Steel and iron 0.0007, brass and copper

0.001, lead 0.0016, Mercury 0.0033.For Permissible Unit Xtresses see Vol. 1.

Physical Properties of Stones and Concrete

Material

Ultimate Modulus ofcompression U, elasticity

In pounds per sq. in.

T r a p ( b a s a l t ) .G r a n i t e , porphyry. .L i m e s t o n e , m a r b l e .S a n d s t o n eB r i c k . . . . . . . . . . . . . . . . . . . . . .B e s t brick.L i m e m o r t a r . . . . . . . . . . . . . . . .Portland cement, slow bindingPortland cement, quick bind-

ing.

15,000 to 40,00010,000 to 30,0007,000 to 20,0005,000 to 15,000

800 to 1,5003,000 to 8,000

6 0 02,000 to 4,0001,000 to 2,000

Portland cement mortar

1

3 ,000after 28 days.

;;; 2 ,600‘3 r>nn1:3 Y,Y””C i n d e r c o n c r e t e . 600 to 8OCStone c o n c r e t e .Slate. .

2,000 to 4,000

A s p h a l t ( p a v i n g ) .10,000 to 20,000

G l a s s8 ,000

20 ,000

‘3,iiii,O%‘io’ fi;oiio,ooo2,000,000 to 4,000,0001,000,000 to 2,000,000

500 ,000 to 1,000,0001,000,000 to 2,000,000

2,000,000 to 4,000,000

3,000,OOO to 6,000,OOO 120

600,000 to l,OOO,OOO2,000,000 to 3,000,000

7,000,000. . . . < < . . . . . . . . . . . . . .

11,000,000

Weightin lbs.per

cu. ft.

175170160145120140100

9 0

9 0145175130160

Notes.-Por natural stones Average ultimate in tension Ut = & U,(acc. to Bauschinger) Ultimate in shear

Ultimate in bendingu, = &U, = 2Utu1, = iu, =4Ut

For concrete and mortar(acc. to Bauschinger)

Ut = ¡$j to 0 u,u, = iu,U& =auc =2Ut

Modulus of elasticity usually assumed i2,000,OOO for compression1 ooo ooo for tension

Strength after- 1 year about 50% greater than after 28 days.

For brick Ut = &U,

Ultimate of masonry structures about 0.5 of the ultimate of specimens.For Permissible Unit Stresses see Vol. 1.Linear Expansion by Heat of 100” F. = granite, limestone and sandstone 0.0005,

brick 0.0003, slate 0.0006, cement 0.0008.

APPENDIX C 435

Physical Properties of Seasoned TimberStresses given in pounds per sq. in

Material

Ash (Ameritan) . . . . . .

Cedar (White). . . . . . . . . . . . . . . . . . . . .Cedar (Red). . . . . . . . . . . . . . . . . . . . . . .Chestnut . . . . . . . . . . . . . . . . . . . . . . . . . .

Douglas Spruce (Oregon Pine) . . . . . . .

Fir . . . . . . . . . . . . . . . . . .

GIlID. . . . . . . . .

Hemlock . . . . . . . . . . . . . . . . . . . . . . . . . .Hiokory . . . . . . . . . . . . . . . . . . . . . . . . . . .

Mahogany (Spanizh). . . . . . . . . . . . . . . .Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Oak (White). . . . . . . . . . . . . . . . . . . . . . .Pine (Southern Yellow, long lealed). ..Pine (White). . . . . . . . . . . . . . . . . . . . . . .Poplar. . . . . . . . . . . . . . . . . . . . . . . . . . . .spruce...............: . . . . . . . . . . . .

Walnut (Black). . . . . . . .

Ultim.;ension

17,000

.10,80011,500

13,000

13,000

8,70019,600

14,90011,150

13,60013,00010,0007,000

11,000

Ultim.comp.length

7200

520060005300

82007150

85008000540050006000

7500

-

1

-

1900

700

800

1800

22001260700

1100

400,..<..

500

1300

800

4001100

. . . . .500

1000835400

400

6 2 8 0 1,640,OOO 3 9

1370 9 1 0 . 0 0 0 2 3. . . . . . . . . ...<1533 1,140,000 4 1

1,680,000 32

1,530,000 . .

5 8 9 0 1,700,OOO 3 7

50”

49

For Permissible Unit Xtresses see Val. 1.

APPENDIX D

Table of Intermediate Ordinates y of a Parabola with a Middle Ordinate = 1

(For a middle ordinate f the

intermediate ordinate = fv).

--

-.

-.

-.

-

.-

_-

_-

..-

-

z-m

2m

z-m

0.510.520.530.540.55

Y Y Y Y

0.7399

0: E910.70840.6975

0.760.770.780.790.80

0.42240.4071

Et80:3600

0.010.020.030.04

0.9999 0.93240.260.270.280.290.30

0.310.320.330.340.35

0.99960.99910.9984

0.92710.92160.91590.91000.05 0.9975

0.99640.99510.9936

0.90390.89760.8911

E%

0.56 0.6864 0.81 0.34390.57 0.6751 0.82 0.32760.58 0.6636 0 . 8 3 0.31110.59 0.6519 0.84 0.29440.60 0.6400 0.85 0.2775

0.0;0:080.090.10

0.99190.9900

_-0.360.370.380.390.40

0.8704 0.610.620.630.640.65

0.62790.61560.6031

EE%

0.860.870.880.890.90

0.26040.2431

Ek%Y0.1900

0.110.120.130.140.15

0.98790.98560.9831

0:%8

0.86310.85560.84790.8400

0.410.420.430.440.45

0.8319 0.66 0.5644 0.91 0.17190.67 0.5511 0.92 0.15360.68 0.5376 0.93 0.13510.69 0.5239 0.94 0.11640.70 0.5100 0.95 0.0975

0.160.170.180.190.20

0.97440.97110.96760.9639

0.82360.81510.80640.79750.9600

0.9559

0: if%Z0.9424

__0.460.470.480.490.50

0.78840.7791

0: EiY0.7500

0.710.720.730.740.75

0.96 0.07840.97 0.05910.98 0.03960.99 0.01991.00 0.0000

0.210.220.230.240.25 0.9375

-This table can be used in case of a uniform load for calculating the

Bending Moment (or required Section Modulus) at intermediate points, fromthe Bending Moment (or Section Modulus) at center; for determining the outlineof a Parabolic Arch; for the ordinates of a parabolic Influente Line for theHorizontal Thrust of Arches, etc.

(Approximate length of a Parabola 1’ = Z (1 + iii), where 2 = span, f = rise;

this is more correct for flat curves, for example:

for $ = $j, Z’ = 1.02606 Z, accurate 1.02667 Z

for i = jl+ Z’ = 1.09823 2, accurate 1.10667 Z

for + = $, 1’ = 1.47894 Z, nccurate 1.66667 Z).

436

APPENDIX D 437

Approximate Radius of Gyration r of Compound and Singlefor Columns and other ComDression Members

Sections,‘used

1 1% IIIIP [III]I-S 0.38 h =(j h)

3 IK’II IKr=0.35h

r= 0.37 h

HZl-= 0 . 2 0 w

b-4 Fr=o.z3w

]H[‘iHr=0.31 h

r= 0.30 h

Is1;

4.r= 0.35w

lt-- Zh 1r=0.30h

T-3r= 0.32 h

‘y-d4 r=0.33d

Td&

r=0.25d

+” r=0.40h

r- 0.29 h

F¿¿õr Box Secfions the radiiofgyrat/‘on in the otbermaindirecfion wl’ll he abouf tbesame as given ubove whenfhe sections are arrangedas follows,

a-0.7h a-0.8 h a.0.8h a-0.9 h a.0.6h

Nofe:,v fndicates fb<fXl;hould bifaken from oufside ofA.ites

F o r orm?&ary apprAx.imate Moment of Inertia = 0.13Ah2,where A= area, h = heightapproxlmate Section Modulus = 0.25Ah (al1 in inches)

(For symmetrical rails the two yalues are O.llAh2 and 0.22Ah respectively.)

438 APPENDIX D

Properties of Circular Sectionsd = diameter

- - A =.rea=$

J = Moment of Inertia = $

R = Section Modulus = z

A J R

0.7854 0.0491 0.09823.1416 0.7854 0.78547.0686 3.976 2.651

12.566 12:57 6.28319.635 30.68 12.2728.274 63.62 21.21

38.48 117.950.27 201.163.62 322.1

%Y71.57

78.54

95.03113.1132.7

490.9~~

‘%71402

- 98.17

130.7169.6215.7

153.9 1886 269.41 7 6 .7 2485 331.3

201.1 3217 402.1

227.0 4100 482.3254.5 5153 572.6283.5 6397 673.4

314.2 7854I---.

9547ll49913737

785.4 10

38C415.5

9Ei~1194

452.4490.9530.9

:%22432

135715341726

572.6615.8660.5

x34719

193221552394

706.9 39761 2651

704::855.3

YE:!1018

:1;41195

45333 292551472 321758214 3528

385942094580

91998 4973102354 5387113561 5824

1257

132013851452

125664

138709152745167820

676672747806

1521 183984 83631590 201289 89461662 219787 9556

:7%1886

239531 10193260576 10857282979 11550

1963 306796 12272

d

:3

79

10

:;13

:216

17

:9"

20- - - -

za23

30-__

8;* 33

8;36

8739

40- -

4;43

44

4:

47

49

50

d J RA

:a53

204321242206

EJo:387323

130231380414616

z";56

2290 417393 154592376 449180 163342463 482750 17241

5759

255226422734

%%594810

181811915520163

60- -

61

63

2827_ - - -

292230193117

636172 21206

679651 22284725332 23398773272 24548

i566

321733183421

8%40931420

257362696128225

S869

E23739- - -3848

989166 295271049556 308691112660 32251

33674

395940714185

1178588--__

:E%1393995

351383664438192

4301 1471963 397834418 1553156 414174536 1637662 43096

465747784902

1725571

:9:197:L

448204658948404

5027__-

515352815411

2010619

2113051 521742219347 541302329605 56135

k%5809

2443920 581892562392 602922685120 62445

5945 2812205 646486082 2943748 669036221 3079853 69210

90 6362 3220623

Yi93

6504 33661656648 35165866793 3671992

71569_----

739827644878968

815428417386859

979899

100

75437698

4345671 896014527664 924014715315 95259

7854 4908738 98175

APPENDIX D 439

I

--

h

- - -

R RI h J

~-R RI

-0.0833 0.1667 0.11791.3333 1.3333 0.94288.7500 4.5000 3.1820

3233

76960 4965.2 3510.987381 5461.3 3861.798827 5989.5 4235.2

4 21.333 10.6675 52.083 20.8336 108.00 36.000

111361125052139968

4632.05052.85498.4

79

200.08 57.167 40.423341.33 85.333 60.340546.75 121.50 85.913

9739

156180 8442.2173761 9145.3192787 9886.5

10 833.33 166.67 117.85 40 213333 10667

5969.56466.76990.7-~~

7542.5

:a13

1220.1 221.83 156.861728.0 288.00 203.842380.1 366.17 258.92

4;43

235480 11487 8122.3259308 12348 8731.3284900 13251 9369.9

14

:B

3201.3 457.33 323.38 44 312341 14197 100394218.8 562.50 397.74 45 341719 15188 107395461.3 682.67 482.71 4G 373121 16223 11471

17

:;

6960.18748.010860

818.83972.001143.2-__-

1333.3

579.00687.30808.33

47

:9"

406640 17304442388 18432480400 19608

20 13333 60

a::23

16207 1543.519521 1774.723320 2027.8

942.81~___

1091.41254.91433.9

51

:3

520833 20833__- __-563767 22109609301 23435657540 24813

122361303313865

- -14731__-

156331657117545

$526

27648 2304.0 1629.232552 2804.2 1841.438081 2929.3 2071.3

5556

708588 26244 18557762552 27729 1960781Q541 29269 20696

27

ZY

44287 3280.5 2319.651221 3658.7 2587.058940 4064.8 2874.2

E859

879667 30866 21825943041 32519 22994000780 34230 24204

SO 67500 4500.0 3182.0 60 080000 36000 25456

Properties of Square Sections

J = Moment of Inertia for any center axis = g

R = Section Modulus for center axis parallel to

the sides = z

45RI = Section Modulus for diagonal center axis = 12 h3

-

440 APPENDIX D

Moments of Inertia of Rectangular Sectionsjc-b-.e!

Moment of Inertia = E12(Example: for 8in. x 4in., Moment of Inedia = 4 X 42.67)

Width of reotangle,ininohesDepth,

ininchest H 1- - -

4.50 G.75 9.007.15 10.72 14.29

10.67 lG.00 21.3315.19 22.78 30.38

ll.25 13.5017.8G 21.4426.67 32.0037.97 45.56

10

ll

:z

:B

:I:

:Y

20.83 31.25 41.67__-

55.4672.0091.54

114.33140.63

-

-

15.75 18.0025.01 28.5837.33 42.6753.16 60.75

52.M 62.50 72.92 83.33

27.7336.0045.7757.1770.31

___-85.33

102.35

14;: 90

41.5954.00

85%105.47

69.3290.00

114.43

:7"5: "78

83.18108.00137.31171.50210.94

97.06126.00160.20

248: 09

110.92144.00183.08228.67281.25

128.00 170.67153.53 204.71182.25 243.00214.34 285.79

213.33255.89

35"7 : 74

256.00

361: EO428.68

298.67 341.33358.24 409.42425.25 486.00500.14 571.58

168.67 250.00 333.33 416.67 500.00 583.33 666.67

192.94221.83253.48288.00325.52-__

366.17410.06457.33508.10

289.41

%X2432.00458.28

482.34 578.81665.50

764%976.56

675.28

X8

%9:32

771.75887.33

1013.921152.001302.08

549.25 732.33 915.42615.09 820.13 1025.16686.00 914.67 1143.337G2.16 1016.21 1270.26

:230: 181372.001524.31

1281.58 1464.671435.22 lG40.251600.G7 1829.331778.36 2032.42

30 562.50 843.75 1125.00 1406.25 1687.50 1988.75 2250.00

620.65682.67

7%: 8:893.23

930.97 1241.30 1551.62 1861.94 2172..26 2482.601024.00 1365.33 1706.67 2048.00 2389.33 2730.671123.03 1497.38 1871.72 2246.06 2620.40 2994.761228.25 1637.67 2047.08 2456.50 2865.92 3275.331339.84 1786.46 2233.07 2679.68 3126.30 3572.92

31 :39

972.00

VE: f::1235.81 1

1458.001582.90

%8: 72

1944.00 2430.00 2916.002110.54 2638.17 3185.802286.33 2857.92 3429.502471.82 3089.53 3707.44

3402.003893.44

4%: å4

3888.004221.084572.674943.24

40 1333.33 2000.00. 2666.67 3333.33 4000.00 4686.67 5333.33- -- --

APPENDIX E

EXTREME LENGTH IN FEET QF PLATES AND STANDARD SHAPES(Rolled by Carnegie Steel Co.)

(Subjectto clmnge)

~ ____~----~~~~__--~__~~~_---~~25 45 0 50 0 100 0 100 095 090 0 85 O 75 070 065 0 60 0 50 045 040 033 3 31 2 31 2 30 0 26

~'~~__--~~-__-~-_~~'___I'-~_Q 30 ,_.. 45.8 90.0 90.095.090.085.065.0G0.050.040.040.037.536.234.632.932.9 30.8 30* ___~-__----~-~----_---___---~; 35 . .._ 43.3 90.0 90.09O.CI90.080.0 65.0 50.045.040.0 38.337.536.235.0 33.8 32.9 30.8 36'd ___~__a 40 . .._ 41.7 80.0 80.090.090.080.065.050.041.740.038.337.535.834.632.932.(130.8 40___------_------~-_--___--~__

____________---__------__-__50 29.235.0 41.7 46.746.74G.746.746.845.043.342.540.837.535.033.331.230.021.7 50

54 29.235.0 41.7 46.746.746.745.845.043.841.740.838.3 35.833.3 30.830.027.9 20.8 54-__ -__60 27.9 34.6 42.5 46.74G.745.844.642.941.239.2 37.9 35.8 33.331.229.8 27.9 26.2 20.0 60_______-___---_--________

76 22.929.2 36.2 38.339.G38.737.536.735.0'32.530.829.6127.525.8 24.6 22.9 21.217.5 76

64 21.225.0 35.0 37.738.337.535.434.232.531.329.627.9125.824.622.921.720.016.7 84__________---_-----__~~__---90 18.722.9 31.2 35.435.835.833.331.729.628.727.926.724.623.321.720.019.615.8 90___________------____-_-~96 17.522.1 27.5 30.830.833.331.229.628.3 27.128.725.023.722.120.4 19.2 18.3 15.0 96___~-_____ -_~-_~~__-_---102 16.220.8 25.0 28.729.231.730.027.9 27.526.725.824.222.9 20.8 19.2 18.3 17.515.0 102

108 12.519.2 22.1 24.225.025.025.425.425.025.024.223.321.719.Gl8.3 17.516.714.2 108

114 .., 16.7 20.4 22.122.9 23.324.624.623.723.322.9 22.120.018.7 17.5 16.7 15.8 12.9 114______--__-_--________-_~120 .___ 15.0 18.3 19.2 20.420.821.722.122.121.720.820.4 19.2 17.5 16.7 15.8 15.0 12.5 120__________-_--_-__-_--~126 ..__ 15.8 16.717.5 17.9 18.719.219.218.718.717.9 17.516.7 15.815.014.212.1 126______~_~--_----___-_~132 _.__ ..,< . . . . . 15.416.716.718.318.317.917.917.917.516.215.'815.014.613.7 _... 132

441

4 4 2 APPENDIX E

EXTREME LENGTH IN FEET OF STANDARD SHAPES

Size in inches 1,ength in fee Size in inches

5 to 8

9

10 to 24- -

5

6 to 9

10

12 to 15

8 X 8

6X6

4 x 4

34 x 36

3 x 3

75

70 ‘

7 5

75

9 0

7 0

7 5-

125

8 5

85

85

75

6X4

6 X 3%

f, 5 x 3;

5 x 3

4 x 3

31 x 3

3; x 2s

3 x 23

(Rolled by Carnegie Steel Co.)-

1

-

,ength in feet

8 5

85

85

8 5

8 5

75

75

7 0

APPENDIX E 443

AREAS OF SECTIONS OF STANDARD ANGLES, ROL¿ED BYCARNEGIE STEEL CO., IN SQUARE INCHES

Area of Angles with Equal LegsAnnles marked * are smcial

Inches QI

8 X86 X 6

*5 x 5

4 x 434 x 3f3 x 3

24 x z+2 x 2

*8 X 6*; GJ 4.40!8:UU 7% 6%

3.61 4.18 4.75 5.;1 5.86

9.32 9.94 10.76 ll:62 12.35~ 13.06' *8 X 6'6.75 7.31 7.87 8.42 8.97 9.50 *7 X3f6.41 6.94 7.47 7.99 8.50 9.00 6 X4

6 X 3 2*5 x 45 X3f

5 x 3

$4 X3"

3.42 3.97 4.50 5.03 5.55 6.06 6.56 7.06 7.55 8.03 8.50 6 X3$3.23 3.75 4.25 4.75 5.23 5.72 6.19 6.65 7.11 *5 x 4

2.56 3.05 3.53 4.00 4.47 4.92 5.37 5.81 6.25 6.67 5 x3+

2.40 2.86 3.31 3.75 4.18 4.612.25 2.67 3.09 3.50 3.90 4.302.09 2.48 2.87 3.25 3.62 3.98

5 x3*4 x3+4 x3

34 x 3 1.93 2.30 2.65 3.00 3.34 3.67 4.00 4.31 4.623$ x 23 1 . 4 4 1.78 2.11 2.43 2.75 3.06 3.36 3.65 ;"

3*z ; :

3 X2$ 1.3 1 1.62 1.92 2.22 2.50 2.78 x 2)

*3 x2 1.1 9 1.47 1.731.31/1.55

2.00 2.25 *3 x 22$ x 2 0. 8 1 1 . 0 6 1 . 7 8 2.00 2g x 2

7.75 8.68 9.61 10.53 ll.44 12.34 13.2:4.36 5.06 5.75 6.43 7.11 7.78 8.44 9.09 9.743.61 4.18 4.75 5 . 3 1 5.86 6 . 4 1 6.94 7.47 7.9I

2.40 2.86 3.31 3.75 4.18 4.61 5.03 5.44 5.842.09 2.48 2.87 3.25 3.62 3.98 4.34 4.69 5.03

1 . 4 4 í.78 2.11 2.43 2.75 3.06 3.36

Area of Angles with Unequal Legs

Andes marked * me special

+i l /t

+$ 1

4.12 15.0(0.37 ll.O(8.50 9.0(

-

11

;

-

-l&

5 . 8

-

l7

-

lnches

8 X 86 X 6'5 x5

4 x 4

i" 5 3"

2* x 2t2 x 2

Inches

444 APPENDIX E

WEIGHTS OF ABOVE STANDARD ANGLES IN POUNDS PER LIN. FT.

Weight of Angles with Equal Legs in Pounds Per Lin. Ft.

Inches +l

6 2;*5 x 5

4 x 43* x 3+3 x 3

2* x 2;2 x 2

Anglas marked * are mecial

26.4 29.6 32.7 35.8 38.9 42.0 45.014.9 17.2 19.6 21.9 24.2 26.5 28.7 31.0 33.112.3 14.3 16.2 18.1 20.0 21.8 23.6 25.4 27.2

8.2 9.8 11.3 12.8 14.3 í5.7 17.1 18.5 19.97.2 8.5 9.8 11.1 12.4 13.6 14.8 16.0 17.1

4.9 6.1 7.2 8.3 9.4 10.4 l l . 5

3.1 4.1 5.0 5.9 6.8 7.7z.5 3.2 4.0 4.7 5.3

g 1

48.1 51.035.3 37.428.9 30.6

-lhT54.0

*8 X 6

*i :: 4 ”

6 X3t*5 x 45 x3*

5 x 3

*4 Xi”

Weight of Angles with Unequal Legs in Pounds Per Lin. Pt.

-

&

2.-

A n g l e s m a r k e d * are specialI/ I / I

5.0 5.9 6.8 7.74.5 5.3 6.1 6.8

-Fk

2.00.58.9

7.3

-

4.482.3,0.6

8.9

-

Inches

8 X86 X 6

*5 x 5

4 x 43t x 343 x 3

2+ x 2*2 x2

% X 6

“: x 4”

6 X3+:5 x 45 x3t

5 x 3:4 x 3+4 x 3

;i 2;

3% x2j

:;- $2;

Notes.-For center of gravity of angles see Table 30. page 150.of angles, me page 437.

For approximate radius of gyration

APPENDIX E

Properties of Standard 1-Beams

445

VeightPerfooL

,ounds

S:i7.5

8::9.5

10.5- -9.75

12.2514.75

12.2614.7517.26

15.017.520.0

17.7520.2t22.7t25.2:- -21.025.030.035.0

25.030.035.040.0

31.535.040.0

42.045.050.055.050.0

55.060.065.070.0

65.070.075.0

80.085.090.095.0

100.0

-

ea

_-

_-

_-

-

Sectionmdulusxis l-lnches

-1

Ba

Raofus:yrstimxis l-linohes

1.7

::i:: 9f1.15

3.03.23.43.6

L--4.8

8:";

1.641.591.541.52- -2.051.941.87

7.38.08.7

_-----10.411.212.1

_ - - -14.215.016.017.0

2.462.352.27

2.862.762.68- -3.27

3:::3.03

18.9 3.6720.4 3.5422.6 3.4024.8 3.30

24.426.829.331.7

- - -36.038.041.0-__

58.9

64268.171.8

4.073.903.773.67.__4.834.714.57

88.4

9%102.4

5.955.875.735.625.52__-

2:

6:79

117.0122.0126.9

173.9180.7186.5

E:";

%7.58

- - .

9:819.209.098.99

-

Widtho f

Flange,inches

-

1

-

-.

b

I

ia

i

%Ft.nertiaxis 1-1,nches”

0.170.260.36

2.332.422.52

2.52.72.9

0.190.26

E4

0.210.360.50

2.662.732.812.88-__3.00

Z%

6.06.46.77.1

12.113.615.1

3.61 0.234.34 0.355.07 0.47

3.33

E::

21.824.026.2

0.250.350.46

3.66

;:;1:

0.270.350.440.53~-0.290.410.570.73

4.00

4::;4.26

36.239.242.2

_---.56.960.264.168.0

4.33 84.94.45 91.94.61 101.94.77 111.8

--0.310.450.600.75

0.350.440.56__-

0.410.460.560.660.75-__

0.460.560.640.72

4.664.804.955.10__-5.005.095.21

__-~5.505.555.655.755.84

122.1134.2146.4158.7_--.215.8228.3245.9__-441.8

48;::511.0538.6

6.00 795.66.10 841.86.18 881.56.26 921.2

0.50 6.25 1169.50.58 6.33 1219.80.65 6.40 1268.8

0.500.57

0%0.75

-.

-

7.007.07

7%7:25

-~

-

%:82238.42309.02379.6

-

,

,-

)

l.-

f>-8

;

t

-

-

-

,,

iii,

-~

-

-

- -

-

4rea ofo f

mtion,t3q”Meinches

Ra:fius:yration.xis 2-2,inches

0%0.52- - - -0.590.580.580.57

0%0:60

0.77

0%1.01

1.23

SO

1.852.092.36__-

2.672.943.24

3.754.044.364.71-~

5.165.656.427.31

- -6.89

22"9.50

__--9.50

10.0710.95

14.6215.09

Ei18.17

- -21.1922.3823.4724.62

27.8629.0430.25

42.8644.35

KO48.55

Depthof beam,inches

3

-_-

4

5_--

6

7

8

-__

9

10

12- -

15

13

- -

20

24

:::12.21

--.2.212.502.793.09

2.87

23:

0.650.630.63___-

0:790.68

0.780.760.74__-0.84

ET0.80

4.425.155.88

--~5.33

6%7.43

__-6.31

Eir10.29

7%

?ZR

9.26

%f~---

12.4t

:4:;1

XE

0.900.88

0%

0.970.930.910.90__--1.010.990.96

:::i;1.111.09~-

1.21

:::7

15.9217.6:19.1:20.5E_---19.0120.5<22.0(__-

23.3:25.0(26.4:27.9z29.41

1.361.331.311.301.28

Note.-For special sections aee Manufacturers’ Handbooks.

446 APPENDIX E

Properties of Standard Channels

- -

3ectiol 1 Rrzdiu smodu.

IU8of gyr: %.-

tionmis l- 1 axis l- -1 iinches 3 incheí 1

Mo-ment cinerti:axis 2-inchez

1.11.21.4

- -

1.171.121.08

-

IfI-23

,0.200.250.31

0.210.240.27

1.92.12.3

1.561.511.46

-.

1.951.831.75

- -2.342.212.132.07

-.

2.722.592.502.442.39

--

3.102.982.892.822.76

._

3.493.403.213.10

.-

3.873.66

3%3.3s__~4.614.434.284.174.09

-.

5.625.575.445.325.235.16

0.320.380.44

0.290.320.35

3.03.54.2

0.480.640.82

0.700.881.071.28

0.500.570.650.74

297::9.5

0.981.191.401.621.85

E0.790.870.96

8.19.0

10.011.011.9

1.33 0.791.55 0.871.78 0.952.01 1.022.25 1.11

10.511.313.515.7

1.77 0.971.95 1.032.45 1.192.98 1.36

13.41 .7!1 .2

20.623.1- -21.424.026.929.932.8

2.30 1.172.85 1.343.40 1.503.99 1.674.66 1.87

3.91 1.754.53 1.915.21 2.095.90 2.276.63 2.46

312.6 41.7319.9 42.7347.5 46.3375.1 50.0402.7 53.7430.2 57.4

8.23 3.168.48 3.229.39 3.43

10.29 3.63ll.22 3.8512.19 4.07

- - -

MO-ment <inertiaxis l-inche

1.C1.t2.1

Deptfo f

channeinches

3

9

10

12

15

WeighPer

foot,poundr

-

t

ì L

,/I

Area Thic,g nesstion, w-eh3q. in. inch

4.006.006.00

5.256.257.25

1.19 0.1’1.47 0.211.76 0.31

1.55 0.121.84 0.2:2.13 0.3:

6.509.00

11.50

8.0010.6013.0015.50

1.95 0.152.65 0.3:3.38 0.4t

_ _ _ _2.38'0.2C3.GQ 0.3:3.82 0.444.56 0.5(

9.7512.2514.7517.2519.75

11.2513.7516.2518.7521.25

13.2516.cDO20.0025.00

16.0020.00

%Z35.00

2.85 0.213.60 0.3::4.34 0.425.07 0.535.81 0.63____3.35 0.224.04 0.314.78' 0.405.51 0.496.25 0.58____3.89 0.234.41 0.295.88 0.457.35 0.61_ _ _4.46 0.245.88 0.387.35 0.538.82 0.68-0.29 0.82

20.5025.0030.0035.0040.00

33.0035.0040.0046.0050.0055.00

1

11

-

6.03 0.287.35 0.398.82 0.510.29 0.64-1.76 0.76_ _ _ _9.90 0.4C

10.29 0.43ll.76 0.5213.24 0.6214.71 0.7216.18 0.82

l

-

-

k- Widto f of‘> f l a n ge s inchl

-

733

353

;3

)>Li

1.411.5c1.6C

1.581.651.73- -1.751.892.04

1.922.042.162.28

2.092.202.302.412.51

2.262.352.442.532.62

3.403.433.523.623.723.82

ll

-

2.432.492.652.81

2.602.742.893.043.18

2.943.053.173.303.42

-

-

.h

e,?S

-

>fa-1 Lg4

>3

,

,

Radiu,af gyrz

tionaxis 2-inches

0kl2

Distilncsd of

oentcr ofg;“y

outsideof web,inches

0.41 0.440.41 0.440.42 0.46

0.45 0.460.45 0.460.46 0.46

0.500.490.49

0.490.480.51

0.54 0.520.53 0.500.53 0.520.53 0.55

0.59 0.550.57 0.530.57 0.530.56 0.550.56 0.58

0.63 0.580.62 0.560.61 0.560.60 0.570.60 0.59

0.67 0.610.66 0.590.65 0.580.64 0.62

0.720.700.680.670.67

0.810.780.770.760.75

0.640.610~620.650.69__-0.700.680.680.690.72

0.910.910.890.880.870.87

-

.-

-

0.790.790.780.790.800.82

3.54.:4.f

7.48.9

10.4

13.015.117.319.5

21.124.227.230.233.2

32.336.039.943.847.8-__47.350.960.870.7

66.978.791.0

103.2115.5

128.1144.0

:79:3196.9

Note.-For special sections see Manufscturers’ Handbooks.

APPENDIX E 447

Weights and Dimensions of Ameritan Standard Rail Sections(Rolled by Carnegie Steel Company)

*11010095

8:80

T”o66

:550

4:36

i52016

Ares insquareinchcs

10.89 . 89.3

8.88.37.8

7.46.96.4

6.95.44 . 9

4 . 43 . 93 . 4

3 . 02.52 . 01.6

Width ofbase andheight in

inohes

Web ininches

Width Dhead ininches

Ieight of centef gravity abovbase in inches

2.9 55.22.8 43.82.7 38.6

252:4

34.030.026.2

:1:010.0

2.4 22.92.2

9.319.6 8.2

2 . 2 lG.9 7.4

2.1 14.5 6.72 . 0 11.9 5.81.9 9.8 4 . 9

-

*Note-N& rolled by Cernegia Steel Co.

*Note.-Approximately: Moment of inertia = 0.13

Approximately: Section Modulus = 0.25 Ah%.,

Where A = aea, ih = height (al1 in inches).

Axis z-z

Moment Sectionof inertia modulus

:7:S13.3

8.0 4.26 . 6 3 . 64 . 8 2.8

3.52.41.71.1

2.31.71.30.97

APPENDIX F

Areas to be Deducted for Rivet Holes In Tension MembersArea to be deducted = (Diam. of rivet + t in.) X thickness of metal

Thiclrnessof

metal

%H+z

-

-

-

-

Areas to be deducted in square inches

$ in. 4 rivel

0.0391 0.0469 0.0547 0.0625 0.07030.0781 0.0938 0.1094 0.1250 0.14060.1172 0.1406 0.1641 0.1875 0.21090.1563 0.1875 0.2188 0:2500 0.2813

0.1953 0.2344 0.2734 0.3125 0.35160.2344 0.2813 0.3281 0.3750 0.42190.2734 0.3281 0.3828 0.4375 0.49220.3125 0.3750 0.4375 0.5000 0.5625

0.35160.3906

. . . . . . . . .

. . . < . . .

0.4219 0.4922 0.5625 0.63280.4688 0.5469 0.6250 0.70310.5156 0.6016 0.6875 0.77340.5625 0.6563 0.7500 0.8438

. . . .< . . .<

..<......

. . . ..-

0.7109 0.8125 0.91410.7656 0.8750 0.9844

. . . . . . . . . 0.9375 1.0547

..<........ 1.0000 1.1250

5 in. $I rivet $;n. 4 rivet

\--~

$ in. 4 rivet 1 in. 4 rivet

Conventional Signs for Riveting

Fiattened to M”or Countersunk and

not chippedFlattened to X” FIattened to X1’

APPENDIX F

Sizes of Rivet Heads and Clearances A for Machine Driving

449

Diam. of Area 3;rivet in_____ sq. ;s

Frac. Dec'l in. 2;

a ,500 1963 216(0 .625 ,306s 337(t ,750 .4418 4861

Rivet

Diam.

Shearing and Bearing Value of RivetsBearing Val‘ys above ca to right of upper zigzag liys are grm;x than d;;i;;e shear.“ below “ “ left “ lower “ “ CL 1‘

Bearing value for different thickmss of plate at 15,OOOlb. per sq. in.

Beming value for different thickness of plate at 22,000 lb. per sq. in.

,500 .1963 2 3 6 1.625 ,306s 3 6 8 1750 ,441s 5301

Bearing value for different thickness of plate at 24,000 lb. per sq. in.

3750 4690 5620 6560 7500 8440 9370*500 5620 6750 7870 9000 10120 11250 12370 13500

Kj37X$jy 1.00~ H ( 1 1 ( 1

525065607870 9190 10500 11810 13120 14440 1575017060183706000 75009000 10500 12000 13500 1500016500 18000 1950021000 22500 24000

APPENDIX G

, Cosines, Tangents and CotaneentsVatural Sin

Sine

es- - -Cosine Tangent Cotangent

0.000 1.0000.017 1.0000.035 0.9990.052 0.9990.070 0.9980.087 0.996

0.0000.0170.0350.0520.0700.087

575928.6419.0814.30ll.43

0.105 0.995 0.105 9.5140.122 0.993 0.123 8.1440.130 0.990 0.141 7.1150.156 0.988 0.158 6.3140.174 0.985 0.176 5.6710.1910.2080.2250.2420.259

--

--

--

--

--

-

0.9820.9780.9740.9700.966

0.1940.2130.2310.249

5.1454.7054.331

0.2684.0113.732

0.2760.2920.3090.3260.342--__0.3580.375

" 0:4n30.423

0.9610.9560.9510.9460.940

0.287 3.4870.306 3.2710.325 3.0780.344 2.9040.364 2.747

--

E::0.9210.9140.906

0.4380.4540.4690.4850.500

0.8990.8910.8830.8750.866

--

-,-

--

--

--

--

.--

--

0.384 2.6050.404 2.4750.424 2.3560.445 2.2460.466 2.1450.488 2.0500.510 1.9630.532 1.8810.554 1.8040.577 1.732

0.5150.5300.5450.5590.574

0.8570.8480.8390.8290.819

0.601 1.6640.625 1.6000.649 1.5400.675 1.4830.700 1.428

0.5880.6020.6160.6290.643

0.8090.7990.7880.7770.766

0.7270.7540.7810.8100.839

1.3761.3271.2801.2351.192

0.6560.6690.6820.6950.707

0.7550.7430.7310.7190.707

0.8690.9000.9330.9661.000

Cosine

--

-

1.1501.1111.0721.0361.000

Sine Cotangent Tangent Degrees

Degrees 1‘

0

a3

9089

:1:8686

46

:B:416

:t:92021

83SS26ai8930

328436

535160

48'4745

450

MillimetersI‘

Centimeters“MetersI‘

I LKilometers

lb6‘

Square Millimeters“ L‘Square Centimeters

“ “Square MetersSquare KilometersHektares

Cubic Centimetersi‘ <‘I‘ Meters“ <‘‘L L‘

Grams:: ( w a t e r )

<‘ per cu. cent.Kilograms

<‘

APPENDIX G

Metric Conversion Table

‘L per sq. cent.KilogrammetersKilograms per lineal meterKiloq;ams pl; lin. centim$er

i‘ II squarecubic “

gip&atts

CL

Cheval-vapeur1 Dollar = 4.198 Marks1 Cent‘per lb.

“1 Dollar per cu. yd.

10 Dollars per Mille foot B.M.1 Cent per lin. ft.

X 0.03937 = inches.+ 25.4 = <(x 0.3937 = i*

‘Lj;. 3;:;$ 1 IIX 3 .281 = feet .X 1 .094 = yards .X 0.621 =+ 1.6093 =

mi$s.

X 3280.7 = feet.

X 0.~~~“~ 1 sq;;are iny‘hes.+x 0.155 = ‘< ;:+ 6.451 = ”x 10.764 = “ feet.

x247.1 = acres.2.471 = acres.

+ 16.383 = cubic inches.L 3.69 = fluid drachms.x 35.315 = cubic feet.x 1.308 = cubic yards.X 264.2 = gallons (231 cu. in.).

X 15.432 = grains.t 29.57 = fluid ounces.+ 28.35 = ounces avoirdupois.+ 27.7 = pounds per cubic inches.x 2.2046 = pounds.+ 907.2 = tons (2000 Ib.).+ 1016.0 = gross tons (2240 lbs.).

X 14.223 = pounds per sq. in.X 7.233 = foot-pounds.X 0.672 = pounds per lineal ft.X 5.6 = pounds per lin. in.X 0.205 = pounds per square ft.X 0.062 = pounds per cubic ft.

X 1.34 = horse-power.+ 746 = “ (‘X 0.7373 = foot-pounds per second.X 0.9863 = horse-power.= 5.183 francs.= 93 Shillings per gross ton (2240= 9,236 Pfennig per kilogram.= ll. 403 centimeters per kilogram.= 5.491 Marks per cu. meter.= 17.79 Marks per cu. meter.zz 13.8 Pfennig per lin. meter.

851

lb.)

* Atmosphere is the pressure of a column of 76 centimeters of mercury at thetemperature of melting ice at Paris, where it is equal to 1.0333 kilos on a squarecentimeter.

452 APPENDIX G

Average Weight in Pounds per Cu. Ft. of Various Substances

(See also Tables paw 433, 434, 435)

Anthracite, solid. . . . . . . . . . . . . . . . . . . 93c< broken, loose. . . . . . . . . 5 4

Asphaltum . . . . . . . . . . . . . . . . . . . . . . . . . 8 7Clay, potters’, dry . . . . . . . . . . . . . . . . . . 119

” in lump, loose. . . . . . . . . . . . . . . . 63Coal, bituminous, solid . . . . . . . . . . . . 8 4‘L L< broken, loose. . . . 49Coke, loose . . . . . . . . . . . . . . . . . . . . . . . . 26Earth, common loam, dry, loose . . . . . . 76Earth, common loam, dry, moderately

rammed . . . . . . . . . . . . . . . . . . . . . . . . . 95Earth, as a soft flowing mud . . . . . . . . . 108Glass, common window . . . . . . . . . . . . . 157Grain. at 60 lb. ner bushel.. . . . . . . . . 4 8Gravel, see sand.’Gypsum (plaster of Paris). 142Hornblende, black. 203I c e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 8 . 7Lime, quick, ground, loose, or in small

lumps........................... 5 3Lime, quick, ground, loose, thorough-

l y s h a k e n . . . . . . . . . . . . . . . . . . . . . . . 7 5

Masonry, of granite or limestone,well dressed 165

Masonry, of mortar rubble. 154

sclbbled; . ,dry .’ y .(.we” 138Masonry, of sandstone, well dressed 144Mercury, at 32” Fahrenheit . . 849M i c a . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 8 3Mortar,hardened................ 103Mud, dry, close.. .80 to 110Mud, wet, f luid, maximum. . 120Petroleum, . . . . 5 5Quarta, common, pure.. 165Sa;d, of pure quartz, dry, loose 90 to 106

well shaken.. .99 to 117

Sntw freshly fallen.. .5 to l-2perfectly wet.. ,120 to 140

“ moistened and Compacted byrain.......................15to50

T a r . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2Turf or Peat, dry, unpressed. .20 ta 30WEter, pure r a i n or d i s t i l l e d . 62

s e a . . . . . . . . . . . . . . . . . . . . . . 6 4

APPENDIX G 453

Decimal Parts

Decimal parts of a foot

& 1. 1, 1, !. 1. !. 1. j . 1 .6745! .7578/ .8411/ .9245! &0078 0911 1745' 2578 3411' 4245 5P78 5911

i .0104 .0938 .1771.2604 .3438 .4271.5104 .5938 .6771 .7604 .8438 .9271 4& .0130 .0964 .1797 .2630 .3464 .4297 .5130 .5964 .6797 .7630 .8464 .9297 &i% .0156 .0990 .1823 .2656 .3490 .4323 .5156 .5990 .6823 .7656 .8490 .9323 &

.0182 .1016 .1849 .2682 .3516 .4349 .5182 .6016 .6849 .7682 .8516 .9349 &~~~--~-t ,020s .1042 .1875 .2708 .3542 .4375 .5208 .6042 .6875 .7708 .8542 .9375 22% .0234 .1068 .1901.2734 .3568 .4401 .5234 .6068 .6901 .7734 .8568 .9401 &L:1 .0260 .1094 .1927 .276G .3594 .4427 .5260 .6094 .6927 .7760 .8594 .9427 -156TB .0286 .1120 .1953 .2786 .3620 .4453 .5286 .6120 .6953 .7786 .8620 .9453 +$L-------.~_____~~__~_% .0313 .1146 .1979 .2813 .3646 .4479 .5313 .6146 .6979 .7813 .8646 .9479 $18Ty .0339 .1172 .2005 .2839 .3672 .4505.5339 .G172 .7005 .7839 .8672 .9505 82 .0365 .1198 .2031 .2865 .3698 .4531 .5365 .6198 .7031 .7865 .8698 .9531 1',32 .0391 .1224 .2057 .2891 .3724 .4557 .5391 .6224 .7057 .7891 .8724 .9557 %____-t .0417 .1250 .2083 .2917 .3750 .4583 .5417 .6250 .7083 .7917 .8750 .9583 %3 .0443 .1276 .2109 .2943 .3776 .4609 .5443 .6276 .7109 .7943 .8776 .9609 E

.0469 .1302 .2135 .2969 .3802 .4635 .5469 .6302 .7135 .7969 .8802 .963532 .0495 .132X .2161.2995 .3828 .4661.5495 .6328 .7161 .7995 ,882s .9661

&32p-__-p-____---- __-

% .0521 .1354 .2188 .3021.3854 .4688 .5521 .6354 .718X .8021 .8854 .9688 2.0547 .1380 .2214 .3047 .3880 .4714 .5547 .6380 .7214 .8047 .8880 .9714 +i$.0573 .1406 .2240 .3073 .3906 .4740 .5573 .6406 .7240 .8073 .8906 .9740 +k.0599 .1432 .221X .3099 .3932 .4766 .5599 .6432 .721X .8099 .8932 .9766 $$

f .0625 .1458 .2292 .3125 ,395s .4792 .5625 .6458 .7292 .8125 ,895s .9792 t% .0651.1484 .2318 .3151 .3984 ,481s .5F51 .6484 .731! .8151 .8984 .9818 $$

.0677 .1510 .2344 .3177 .4010 .4844 .5677 .6510 .7344 .8177 .9010 .9844 g

.0313

.0625,093s- -,125.15s3.1875.2188

.25

.2813

.3125

.3438

,375.4063.4375,468s

.5

.5313

.5625,593s

,625.6563.6875,718s

.75

.7813

.8125,843s

,875,906s.9375,968s

APPENDIX H

AMERICAN RAILWAY ENGINEERING ASSOCIATION

General Specifications for Steel Railway Bridges

PART FIRST-DESIGN

1. GENERAL

1. The material in the superstructure shall be structural steel, except Materialsrivets, and as may be otherwise specified. pq

2. When alinement is on tangent, clearanccs -: -------’shall not be less than shown on the diagram; the sheight of rail shall, in al1 cases, be assumed at 6 in.

T- -_-

The width shall be increased so as to provide theij

same minimum clearances on curves for a car 80ft. long, 14 ft. high, and 60 ft. center to center oftrucks, allowance being made for curvature andsuper-elevation of rails. -- __-

ti*3. The width center to center of girders and

5 rl

-- -_-__

T&F $.>

5p3 in.1*

-- ----- -4

8

.---_-_ __

trusses shall in no case be less than one-twentieth Top ‘If Yai’

of the effective span, nor less than is necessary to prevent overturningunder the assumed lateral loading.

4. Ends of deck plate girders and track stringers of skew bridges atabutments shall be square to the track, unless a ballasted floor is used.

5. Wooden tie floors shall be secured to the stringers and shall beproportioned to carry the maximum wheel load, with 100% im-pact, distributed over three ties, with fiber stress not. to exceed 2000lb. per sq. in. Ties shall not be less than 10 ft. in length. They shall bespaced with nor more than 6-in. openings; and shall be secured againstbunching.

II . LOADS6. The dead load shall consist of the estimated weight of the entire

suspended structure. Timber shall be assumed to weigh 44 lb. per ft.B.M.; ballast 100 lb. per cu. ft., reinforced concrete 150 lb. per cu. ft.,and rails and fastenings, 150 lb. per lin. ft. of track.

7. The live load, for each. track, shall consist of two typical enginesfollowed by a uniform load, according to Cooper’s series, or a system ofloading giving practically equivalent strains. The minimum loadingto be Cooper’s E-40, as shown in the following diagrams:

Clearances

SpacingTrusses

SkewBridgesFloors

Dead Load

Live Load

x 0

and the diagram that gives the larger strains to be used.

4 5 4

APPENDIX H 455

HeavierLoadingImpact

8. Heavier loadings shall be proportional to the above diagrams onthe same spacing.

9. The dynamic increment of the live load shall be added to themaximum computed live load strains and shall be determined by the

3 0 0formula Z = X L+y

where Z = impact or dynamic increment to be added to live load strains.S = computed maximum live load strain.L = loaded length of track in feet producing the maximum strain

in the member. For bridges carrying more than onetrack, the aggregate length of al1 tracks producing thestrain shall be used.

Impact shall not be added to strains produced by longitudinal, cen-trifugal and lateral or wind forces.

Lateral 10. Al1 spans shall be designed for a lateral forte on the loaded chord

Forces of 200 lb. per lin. ft. plus 10 per cent. of the specified train load on onetrack, and 200 lb. per lin. ft. on the unloaded chord; these forcesbeing considered as moving.

Vhnd 11. Viaduct towers shall be designed for a forte of 50 lb. per sq. ft. on

Forte one and one-half times the vertical projection of the structure’unloaded;or 30 lb. per sq. ft. on the same surface plus 400 lb. per lin. ft. of thestructure applied 7 ft. above the rail for assumed wind forte on trainwhen the structure is either fully loaded or loaded on either track withempty cars assumed to weigh 1200 lb. per lin. ft., whichever gives thelarger strain.

Longitudinal 12. Viaduct towers and similar structures shall be designed for a

Forte longitudinal forte of 20 per cent. of the live load applied at the top of therail.

UnitStresses

Tension

Compres-sion,

Bending

13. Structures located on curves shall be designed for the centrifuga1forte of the live load applied at the top of the high rail. The centrifuga1forte shall be considered as live load and be derived from the speed inmiles per hr. given by the expression 60 - 2$D, wlxxe “D” = degreeof curve.

III. UNIT STRESSES AND PROPORTION OF PARTS

14. Al1 parts of structures shall be so proportioned that the sum ofthe maximum stresses produced by the foregoing loads shall not exceedthe following amounts in pounds per sq. in., except as modified inparagraphs 22 to 25:

15. Axial tension on net section. . . . . . . 16;OOO

16. Axial compression on gross section of columns. 16,000 - 70:

with a maximum of.. . . . . . . 14,000where “Z” is the length of the member in inches, and “r” isthe least radius of gyration in inches.Direct c o m p r e s s i o n on s t e e l castings.. . . 16,000

17. Bending: on extreme fibers of rolled shapes, builtsections, girders and steel castings; net sect ion. . 16,000on extreme fibers of pins.. . . . . . . . . . . . . . . . . . . 24,000

456 APPENDIX H

18. Shearing: shop driven rivets and pins . 12,000field driven rivets and turned bolts. . . 10,000p l a t e g i r d e r w e b s ; gross s e c t i o n . 10,000

19. Bearing: shop driven rivets and pins. 24,000field driven rivets and turned bolts. . . . 20,000expansion rollers; per lin. in. . . 600~2where ‘(~2” is the diameter of the roller in inches.o n m a s o n r y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 600

20. The lengths of main compression members shall not exceed 100times their least radius of gyration, and those for wind and sway bracing120 times their least radius of gyration.

21. The lengths of riveted tension members in horizontal or inclinedpositions shall not exceed 200 times their radius of gyration about thehorizontal axis. The horizontal projection of the unsupported portionof the member is to be considered as the effective length.

22. Members subject to alternate stresses of tension and compressionshall be proportioned for the stresses giving the largest section. If thealternate stresses occur in succession during the passage of one train,as in stiff counters, each stress shall be increased by 5Oq¿ of thesmaller. The connections shall in al1 cases be proportioned for the sumof the stresses.

23. Wherever the live and dead load stresses are of opposite char-acter, only two-thirds of the dead load stresses shall be considered aseffective in counteracting the live load stress.

24. Members subject to both axial and bending stresses shall beproportioned so that the combined fiber stresses will not exceed theallowed axial stress.

25. For stresses produced by longitudinal and lateral or wind forcescombined with those from live and dead loads and centrifuga1 forte, theunit stress may be increased 25% over those given above; but thesection shall not be less than required for live and dead loads and cen-trifugal forte.

26. In proportioning tension members the diameter of the rivet holesshall be taken + in. larger than the nominal diameter of the rivet.

27. In proportioning rivets the nominal diameter of the rivet shallbe used.

28. Pin-connected riveted tension members shall have a net sectionthrough the .pin-hole at least 25 % in excess of the net section of thebody of the member, and the net section back of the pin-hole, parallelwith the axis of the member, shall be not less than the net section ofthe body of the member.

29. Plate girders shall be proportioned either by the moment ofinertia of their net section; or by assuming that the flanges are con-centrated at their centers of gravity; in which case one-eighth of the grosssection of the web, if properly spliced, may be used as flange section.The thickness of web plates shall be not less than &J of the unsupporteddistance between flange angles (see 36).

30. The gross section of the compression flanges of plate girdersshall not be less than the gross section of the tension flanges; nor shallthe stress per sq. in. in the compression flange of any beam or girder

1exceed 16,000 -200 ~7 when flange consists of angles only or if cover

Shearing

Bearing

LimitingLength ofMembers

AlternateStresses

CombinedStresses

Net Sectionat RivetsRivets

Net Sectionat Pins

PlateGirders

Compres-sionFlange

APPENDIX H 457

FlangeRivets

DepthRatios

OpenSectionsPockets

SymmetricalSections

Counters

Strength ofConnections

MinimumThickness.Pitch ofRivets

EdgeDistance

MaximumDiameter

1consists of flat plates, or 16,000-150 -f if .cover consists of a channelbsection, where Z = unsupported distance and b = width of flange.

31. The flanges of plate girders shall be connected to the web witha sufficient number of rivets to transfer the total shear at any pointin a distance equal to the effective depth of the girder at that pointcombined with any load that is applied directly on the flange. Thewheel loads, where the ties rest on the flanges, shall be assumed to bedistributed over three ties.

32. Trusses shall preferably have a depth of not less than one-tenthof the span. Plate girders and rolled beams, used as girders, shallpreferably have a depth of not less than one-twelfth of the span. Ifshallower trusses, girders or beams are used, the section shall be increasedso that the maximum deflection will not be greater than if the abovelimiting ratios had not been exceeded.

IV. DETAILS OF DESIGN

33. Structures shall be so designed that all parts will be accessiblefor inspection, cleaning and painting.

34. Pockets or depressions which would hold water shall have drainholes, or be filled with waterproof material.

35. Main members shall be so designed that the neutral axis will beas nearly as practicable in the center of section, and the neutral axes ofintersecting main members of trusses shall meet at a common point.

36. Rigid counters are preferred; and where subject to reversa1 ofstress shall preferably have riveted connections to the chords. Adjust-able counters shall have open turnbuckles.

37. The strength of connections shall be sufficient to develop thefull strength of the member, even though the computed stress is less, thekind of stress to which the member is subjected being considered.

38. The minimum thickness of metal shall be Q in., except forfillers.

39. The minimum distance between centers of rivet holes shall bethree diameters of the rivet; but the distance shall preferably be not lessthan 3 in. for g-in. rivets and 2% in. for $-in. rivets. The maximum pitchin the line of stress for members composed of plates and shapes shall be6 in. for %-in. rivets and 5 in. for s-in. rivets. For angles with two gagelines and rivets staggered the maximum shall be twice the above in eachline. Where two or more plates are used in contact, rivets not more than12 in. apart in either direction shall be used to hold the plates well to-gether. In tension members, composed of two angles in contact, a pitchof 12 in. will be allowed for riveting the angles together.

40. The minimum distance from the center of any rivet hole to asheared edge shall be l+ in. for &in. rivets and lt in. for $-in. rivets, andto a rolled edge la in. and 16 in., respectively. The maximum distancefrom any edge shall be eight times the thickness of the plate, butshallnot exceed 6 in.

41. The diameter of the rivets in any angle carrying calculated stressshall not exceed one-quarter the width of the lcg in which they are driven.

458 APPENDIX H

In minor parts H-in. rivets may be used in 3-in. angles, and $-in. rivets in2+in. angles.

42. Rivets carrying calculated stress and whose grip exceeds fourdiameters shall be increased in number at least 1% for each additional$3 in. of grip.

43. The pitch of rivets at the ends of built compression membersshall not exceed four diameters of the rivets, for a length equal to oneand one-half times the maximum width of member.

44. In compression members the metal shall be concentrated asmuch as possible in webs and flanges. The thickness of each web shallbe not less than one-thirtieth of the distance between its connections tothe flanges. Cover plates shall have a thickness not less than one-fortieth of the distance between rivet lines.

45. Flanges of girders and built members without cover plates shallhave a minimum thickness of one-twelfth of the width of the outstandingleg.

46. The open sides of compression members shall be provided withlattice and shall have tic-plates as near each end as practicable. Tic-plates shall be provided at intermediate points where the lattice is inter-rupted. In main members the end tie-plates shall have a length notless than the distance between the lines of rivets connecting them to theflanges, and intermediate ones not less than one-half this distance. Theirthickness shall not be less than one-fiftieth of the same distance.

47. The latticing of compression members shall be proportioned toresist the shearing stresses corresponding to the allowance for flexure foruniform load provided in the column formula in paragraph 16 by the

term 70 + The minimum width of lattice bars shall be 23 in. for f-in.

.rivets, 2$ in. for $-in. rivets, and 2 in. if s-in. rivets are used. Thethickness shall not be less than one-fortieth of the distance between endrivets for single lattice, and one-sixtieth for double lattice. Shapes ofequivalent strength may be used.

48. Three-fourths-inch rivets shall be used for latticing flanges lessthan 2+ in. wide, and $-in. for flanges from 24 to 3; in. wide; s-in. rivetsshall be used in flanges 39 in. and over, and lattice bars with at least tworivets shall be used for flanges over 5 in. wide.

49. The inclination of lattice bars with the axis of the member shallbe not less than 45 degrees, and when the distance between rivet Enesin the flanges is more than 15 in., if single rivet bar is used, the latticeshall be double and riveted at the intersection.

50. Lattice bars shall be so spaced that the portion of the flangeincluded between their connections shall be as strong as the memberas a whole.

51. Abutting joints in compression members when faced for bearingshall be spliced on four sides sufficiently to hold the connecting membersaccurately in place. All other joints in riveted work, whether in tensionor compression, shall be fully spliced.

52. Pin-holes shall be reinforced by platea where necessary, and atleast one plate shall be as wide as the flanges will allow and be on thesame side as the angles. They shall contain suflicient rivets to distributetheir portion of the pin pressure to the ful1 cross-section of the member.

Long Rivets

Pitch atEnds

Compres-sionMembers

MinimumAngles

Tie-Plates

Lattice

FacedJoints

Pin Plates

ForkedEnds

Pins

Bolts

IndirectSplices

Fillers

Expansion

ExpansionBearings

FixedBearings

Rollers

Bolsters

Wall Plates

Anchorage

InclinedBearings

FloorBeams

APPENDIX II 459

53. Forked ends on compression members will be permitted onlywhere unavoidable; where used, a sufficient number of pin plates shallbe provided to meke the jaws of twice the sectiona1 area of the member.At least one of these plates shall extend to the far edge of the farthesttie-plate, and the balance to the far edge of the nearest tie-plate, but notless than 6 in. beyond the near edge of the farthest plate.

54. Pins shall be long enough to insure a ful1 bearing of al1 the partsconnected upon the turned body of the pin. They shall be secured bychambcred nuts orbe provided with washers if solid nuts are used. Thescrew ends shall be Iong enough to admit of burring the threads.

55. Members packed on pins shall be held against lateral movement.56. Where members are connected by bolts, the turned body of these

bolts shall be long enough to extend through the metal. A washer atleast &in. thick shall be used under the nut. Bolts shall not be used inplace of rivets except by special permission. Heads and nuts shall behexagonal.

57. Where splice plates are not in direct contact with the parts whichthey connect, rivets shall be used on each side of the joint in excess of thenumber theoretically required to the extent of one-third of the numberfor each intervening plate.

58. Rivets carrying stress and passing through fillers shall be increased50% in number; and the excess rivets, when possible, shall be outsideof the connected member.

59. Provision for expansion to the extent of +in. for each 10 ft.shall be made for al1 bridge structures. Efficient means shall be pro-vided to prevent excessive motion at any one point.

60. Spans of 80 ft. and over resting on masonry shall have turnedrollers or rockers at one end; and those of less length sha 1 be arrangedto slide on smooth surfaces. These expansion bearings shall be designedto permit motion in one direction only.

61. Fixed bearings shall be firmly anchored to the masonry.

62. Expansion rollers shall be not less than 6 in. in diameter. Theyshall be coupled together with substantial side bars, which shall be soarranged that the rollers can be readily cleaned. Segmenta1 rollers shallbe geared to the upper and lower plates.

63. Bolsters or shoes shall be so constructed that the load will bedistributed over the entire bearing. Spans of 80 ft. or over shall havehinged bolsters at each end.

64. Wall plates may be cast or built up; and shall be so designed asto distribute the load uniformly over the entire bearing. They shall besecured against displacement.

65. Anchar bolts for viaduct towers and similar structuree shall belong enough to engage a mass of masonry the weight of which is at leastone and one-half times the uplift.

66. Bridges on an inclined grade without pin shoes shall have the soleplates beveled so that the masonry and expansion surfaces may be level.

BLOOR SYSTEMS

67. Floor beams shall preferably be square to the trusses or girders.They shall be riveted directly to the girders or trusses or may be placedon top of de& bridges.

460 APPENDIX H

68. Stringers shall preferably be riveted to the webs of all inter- Stringersmediate íloor beams by means of connection angles not less than g-i,.in thickness. Shelf angles or other supports provided to support thestringer during erection shall not be considered as carrying any of thereaction.

69. Where end floor beams cannot be used, stringers resting on Stringermasonry shall have cross frames near their ends. These frames shall be Framesriveted to girders or truss shoes where practicable.

BRACING

70. Lateral, longitudinal and transverse bracing in al1 structures shall Rigidbe composed of rigid members. Bracing

71. Through truss spans shall have riveted portal braces rigidly Portalsconnected to the end posts and top chords. They shall be as deep asthe clearance will allow.

72. Intermediate t,ransverse frames shall be used at each panel of Transversethrough spans having vertical truss members where the clearance will Bracingpermit.

73. Deck spans shall have transverse bracing at each end propor- Endtioned to carry the lateral load to the support. Bracing

74. The minimum sized angle to be used in lateral bracing shall be Laterals36 by 3 by Q-in. Not less than three rivets through the end of the anglesshall be used at the connection.

75. Lateral bracing shall be far enough below the flange to clear theties.

76. The struts at the foot of viaduct towers shall be strong enough Towerto slide the movable shoes when the track is unloaded. Struts

PLATE GIRDERS

77. If desired, plate girder spans over 50 ft. in length shall be built Camberwith camber at a rate of &-in. per 10 ft. of length.

78. Where flange plates are used, one cover plate of top flange shall zitelangextend the whole length of the girder.

79. There shall be web stiffeners, generally in pairs, over bearings Webat points of concentrated loading, and at other points where the thickness Stiffenersof the web is less than 2~ of the unsupported distance between flangeangles. The distance between stiffeners shall not exceed that given bythe following formula, with a maximum limit of 6 ft. (and not greaterthan the clear depth of the web):

tcl = 4o (12,000 - s),

Where d = clear distance, between stiffeners of flange angles.t = thickness of web.s = shear per sq. in.

The stiffeners at ends and at points of concentrated loads shall beproportioned by the formula of paragraph 16, the effective length beingassumed as one-half the depth of girders. End stiffeners and those underconcentrated loads shall be on fillers and have their outstanding legs aswide as the flange angles will allow and shall fit tightly against them.

APPENDIX H 461

Intermediate stiffeners may be offset or on fillers, and their outstandinglegs shall be not less than one-thirtieth of the depth of girder plus 2 in.

Stays for 80. Through plate girders shall have their top flanges stayed at eachTop Flanges end of every floor beam, or in case of solid floors, at distances not exceed-

Camber

R&MembersEyebars

PonyTrusses

ing 12 ft., by knee braces or gusset platea.

TRUSSES

81. Truss spans shall be given a camber by so proportioning thelength of the members that the stringers will be straight when the bridgeis fully loaded.

82. Hip verticals and similar members, and the two end panels ofthe bottom chords of single track pin-connected trusses shall be rigid.

83. The eyebars composing a member shall be so arranged thatadjacent bars shall not have their surfaces in contact; they shall be asnearly parallel to the axis of the truss as possible, the maximum inclina-tion of any bar being limited to 1 in. in 16 ft.

84. Pony trusses shall be riveted structures, with double webbedchords, and shall have al1 web members latticed or otherwise effectivelystiffened.

INDEX

A

Actual pressure of foundations in bridgework, in tons per sq. ft., 427

Actual pressure of foundations in struc-tural work, in tons per sq. ft.,430

Ameritan Railway Engineering Bssocia-tion Specifications for steel rail-way bridges, 454

Ameritan standard rail sections, weightsand dimensions of, 447

Angles, areas of sections of standard, 443weights of standard, in pounds per

lin. ft., 444Approximate radius of gyration T of

compound and single sections,437

Areas to be deducted for rivet holes intension members, 448

Arch bridges, 312design of arch bridges, 346

arch ribs and web system, formof, 350

arch trusses, number and spac-ing of, 349

arch trusses, principal dimen-sions of, 350

bents and hangers, 351bibliography of arch

bridges, 366examples of arch bridges, 354

arch bridge across St. JohnRiver at St. John, N. B.,Canada, 359

He11 Gate bridge acrossEast River, New York,357

highway arch bridge acrossNiagara at Clifton, 356

Mississippi at St. Louis, 354railway arch bridge across

Niagara, 355Washington bridge across

Harlem River, NewYork, 355

4

Arch bridges, design of, floor, locationof, 349

floor system and bracing, 352arch bracing, 354lateral system in floor

plane, 352general, 346steel weights of arch bridges, 354types of arch bridges, 347

arch trusses with parallelchords, 348

arch without hinges, 348arches diverging toward the

ends, 348cantilever arches, 349crescent arches, 348plate girder arches, 348spandrel braced arches, 348three-hinged arch, 347two-hinged arch, 347

stresses, calculation of, 312approximate calculations of the

two-hinged arch, 328arches with parallel chords,

330constant area for al1 chord

members, 328constant Iength of all chord

members, 329crescent arch with great

rise, 330for arch with semicircular

axis, 334for a uniform partial load,

333influente line for H as- a

parabola, 332parabolic arches with

approximately parallelchords, 331

dead and live load stresses inthe three-hinged arch,312-319

dead load stresses, 318influente line for horizontal

reactions, 312

464 I N D E X

Arch bridges, stresses, dead and live load,influente l i n e s f o rtruss members, 314

best method, 317live load stresses, 319

dead and live load stresses inthe two-hinged arch, 319


reaction 321influente lines for truss

members, 322-327live load stresses, 327reactions, 319

general, 312stresses due to braking forte,

3 4 3for arch with horizontal

top chord, 345for the three-hinged arch,

3 4 5for the two-hinged arch,

3 4 4stresses due to yielding of

foundations, 346stresses in a two-hinged arch,

example for the calcula-tion of, 335


reaction H, 335live load stresses, 335temperature stresses, 330

temperature stresses in the two-hinged arch, 327

wind stresses in the two- andthree-hinged arch, 330

arch trusses are inclined,3 4 2

arch trusses are in verticalplanes, 340

Average weight in pounds per eu. ft. ofvarious substances, 452

B

Beam and plate girder bridges, 143bearings, 166bracing, 166

Warren type, 166compression flange, area of, 149coverplates, length of, 155flange area, calculation of net, 148

Beam and plate girder bridges, flangesplices, 165

flanges, make-up of, 153girders, depth of, 147girders, number and spacing of, 145

highway bridges, 147railroad bridges, 145

deck plate girder spans, 145through plate girder span,

1 4 7horizontal flange rivets, 158

for girders with vertical flangeplates, 159

1-beam bridges, 143for highways, 143for railroads, 143

I-bea&, calculation of, 1431-beams, deflection of, 1441-beams, shearing stresses in, 144plate girder bridges, 145stiff eners, 157

specifications of Ameritan Rail-way Engineering Association,157

vertical flange rivets, 160web plate, 156web splice, 160

common splicing for shear andmoment, 161

rational method of splicingweb, 163

splicing for shear only, 160Bearing value of rivets, shearing and, 449Bridges, arch, 312

beam and plate girder, 143cantilever, 423long span, in general and examples, 392simple span, weights of, 221simple truss, 167skew, and bridges on curves, 209types of, and principal dimensions, 117

classification of bridges, 117clear height below crossing, 122

for highways and electric rail-ways, 122

for navigable streams, 122for railroads, 122

clear width and height on bridge,129

for electric railroads, 130for highway bridges, 130for railroad bridges, 129

I N D E X 465

Bridges, types of, cross-section for bridgescarrying combined traffic, arrange-ment of, 130

design and erection, data for, 120highway bridges, 120raih-oad bridges, 121

grades, 122height, top of floor to under clear-

ante, 128highway bridges, 128railroad bridges, 128

location of bridge, 120main girders or trusses, types of,

117arch bridges, 118

advantages and disadvan-tages of, 119

cantilever bridges, 118continuous bridges, 118simple spans, 117 ’suspension bridges, 119

principal parts, 117span ‘length, determination of effect-

ive, 127span length, economical, 125span length, requirements for, 123

for navigable streams, 123Bresse’s equations for, 123

for structures crossing railroads,123

monumental bridges, 125Bridge work, actual pressure of founda-

tions in, in tons per sq. ft., 427limits of foundation pressure, 428

C

Cantilever bridges, 396design of cantilever bridges, 418

bibliography of long spanbridges, 423-426

cantilever bridges, 423long simple span bridges,

4 2 4suspension bridges, 423suspension and cantilever

bridges, 423bracing, 421details and anchorages, 421principal dimensions, 418

arrangement of cross-sec-tion, 420

Decimal parts, of a foot and an ínch,4 5 3

/ Design of floor, 132

Cantilever bridges, design of, principaldimensions, economicallength of suspended span,418

length of anchar arm, 418trusses, design of, 420

stresses, calculation of, 396cantilever, example for the cal-

culation of a, 404influente lines, 404

dead load stresses, 405live load stresses, 405

dead load assumptions, 410approximate rules, 417derivation of formulas, 414example, 413weight of bracing, 413weight of floor, 411weight of suspended span,

411weight of trusses, 411

dead load stresses, 402erection stresses, 409general, 396influente lines for anchar and

cantilever arms, 396reactions, 397shears and moments in

anchar arm, 397shears and moments in

cantilever arm, 398stresses in anchar arm, 399stresses in cantilever arm,

400stresses in intermediate

anchar spans, 402live load stresses, 403wind stresses, 405

on trusses and floor, un-loaded, 406

on trusses and floor com-bined with load, 406

reactions, 407shears and moments, 408

Channels, properties of standard, 446Circular sections, properties of, 438Conventional signs for riveting, 448

D

466 I N D E X

E Externa1 forces, live loads for electric rail-way bridges, 20

New York subway, 20Philadelphia Elevated Railway,

2 0Queensboro bridge, New York,

2 0live load for highway bridges, 17

people, weight of, 17road vehicles, combination of,

18trolley cars, 19

road vehicles, weights of, 17automobiles and m o t o r

trucks, 18road-rollers, 18wagons, 17

live load for railroad bridges, 10coa1 cars, typical, 16Cooper% loading, 10electric locomotive, heaviest, 16ore cars, 16passenger cars, 1 6

snow load, 30temperature changes, 31wind pressure, 25

Extreme length in feet of plates andstandard shapes, 441

Elevated railroads, 260anchorages, 268bibliography of subways and ele-

vated railroads, 271columns, 262combined stresses, 267cross girders, 262cross-section, arrangement of, 260dead and live load stresses, 263economical span length, 260expansion joints, 262flooring, 261general, 260longitudinal girders and bracing,

261stresses in columns, calculation of,

263stresses due to braking forte, 266stresses due to wind and centrifuga1

force, 264266temperature stresses, 266weights of, 269

Externa1 forces, 1braking and traction forces, 29

maximum allowable brake pres-sures, 29

Blumenthal’s formula, 30centrifuga1 forte, 28dead load, 1

defined, 1flooring and railings, weight of,

highway bridges, 2railings, 2railroad bridges, 2

steel work, weight of, 2general, 1

defined, 1impact a n d vertical vibrations,

2 2for electric railway bridges,

2 3for highway bridges, 23for railway and highway bridges,

Cooper’s specifications,2 4

Lindenthal’s formula, 24Prichard’s formula, 24

for steam railroads, 23lateral vibrations, 27live load, distribution of, 21

F

Floor, design of, 132highway bridges, 138

pavement flooring, 140concrete for sidewalks, 141curbs, 141wheel loads, distribution

of, 141plank flooring, 138stringers and floorbeams, 141

brackets, 142railroad bridges, 132

ballast flooring, 132ballast resting directly on

steel floor construction,133

calculation of transversebeams or ‘troughs, 134

concrete base, 133steel floor construction for,

134ties and ballast, 133

I N D E X 467

Floor, design of, railroad bridges, floorsystem, arrangement of,135

panel lengths, 136spacing of stringers, 136

open tie flooring, 132solid steel floors without ballast,

135stringers and floorbeams, prin-

cipal dimensions of, 136stringer sections for

Cooper% E-50 loading,137

Foot, decimal parts of, 453Foundations in bridge work, actual pres-

sure of, in tons per sq. ft., 427limits of foundation pressure, 428

iron piles, 428timber piles, 428

Foundations in structural work, actualpressure of, in tons per sq. ft.,430

1

1-beams, properties of standard, 445Inch, decimal parts of, 453Influente lines, reactions and, 32

L

Long span bridges in general and ex-amples, 367

camber, 371cantilever bridges, examples of, 372

cantilever design for four-trackrailroad bridge, 378

Danube bridge at Budapest ,Hungary, 380

Firth of Forth bridge, 372Memphis bridge, 373Monongahela bridge, 373Ohio bridge at Beaver, Pa., 379Quebec bridge, a cantilever de-

sign by the author, 375Quebec bridge, cantilever de-

sign for, by Pencoyd IronWorks, 375

Quebec bridge, first, 374Rhine bridge at Ruhrort,

Germany, 381Thebes bridge, 374

design for a North River suspensionbridge (New York), 392

Long span bridges, Cooper% specifica-tions, 392

Lindenthals’ specifications, 392Mayer’s specifications, 393Morison’s specifications, 392

live loads and permissible unitstresses, 369

extreme live load, 370floor system, 369nickel steel, 370trusses, 369

long span highway bridges in NewYork, examples of, 383

Brooklyn suspension bridge, 389Manhattan suspension bridge,

3 8 7Queensboro cantilever bridge,

383Williamsburg suspension bridge,

388selection of design, 367

arch bridges, 367cantilever bridges, 367factor of safety, 369simple truss bridges, 367suspension bridges, 368

M

Metals, physical properties of, 433Metric conversion table, 451Moments of inertia of rectangular sec-

tions, 440Moments and shears in simple spans,

4 0engine diagrams and wheel-load

tables, 57floorbeam- and intermediate pier-

reactions, 56influente line, 56uniform load, 57wheel loads, 57

general, 40influente lines for spans with floor-

beams, 49for end reaction, 50

influente lines for spans withoutfloorbeams, 40

bending moment, 41end reaction, 40shear, 40

maximum moment in short spanswithout floorbeams, 48

X E D N I

Moments and reactions from live loadin spans with floorbeams, 52

moments and shears from dead loadin spans with floorbeams, 50

momentsand shears, tables of, 63moments and shears from moving

concentrated loads in spans with-out floorbeams, 44

moments, 46shears, 44

moments and shears from uniformdead load in spans mithout floor-beams, 41

bending moment, 42shear, 41

moments and shears from uniformlive load in spans without floor-beams, 42

shears from moving concentratedloads in spans with floorbeams, 55

shears from uniform live load inspans with floorbeams, 53

conventional method, 54exact method, 53

uniform loads, equivalent, 62for floorbeam reaction, 63for moments, 62for reaction, 62for shears, 62

Movable bridges and turntables, 273bibliography of, 309-311center-bearing swing bridge, 276-278dead load stresses, 283deflections, calculation of, 296end lift, 280floor and center cross girder, arrange-

ment of, 282live load stresses, 284-293

position of load for maximumstress, 292

stresses for uniform live load,292

load on center pivot and centerwedges, 293

locomotive turntables, 305maximum moments in main

girders, 306shear at center of main girder,

306shear at end of main girder, 306stresses, calculation of, 305weights of, 307

Movable bridges and turnt,ables, motors,300

electric, 300gasoline, 301hydraulic, 301

power required to operate movablebridges, 298

resistance due t,o action ofwind, 299

resistance due to friction. 298resistance due to inertia of

bridge, 298rim-bearmg swing bridge, 278stresses, example for the calculations

of, 294center wedge reaction, 296dead load stresses, 294influente lines, 295

swing bridges, calculation of stressesin, 282

dead load, 283live load, 283

swing bridges, types and principaldimensions of girders and trussesof, 281

turning device, 279types of movable bridges, 273

bascule bridges, 274lift bridges, 275siving bridges, 273

jack-knife draw, 274shear-pole draw, 274

transported or ferry bridges, 276traversing or retractile bridges,

275weights of movable bridges, 301

steel weights of some existingmovable bridges, 302

swing spans, 301wind stresses, 294

N

Natural sines, cosines, tangents andcotangents, 450

New York City Bridges, 383

0

Ordinates of a parabola, table of inter-mediate, 436

P

Parabola, table of intermediate ordinatesof, 436

I N D E X

Physical properties of metals, 433Plate girder bridges, beams, 143Plates and standard shapes, extreme

length in feet, 441Properties of circular sections, 438Properties of standard channels, 446Properties of standard 1-beams, 445

R

Radius of gyration T of compound andsingle sections, approximate,

4 3 7Rail sections, weights and dimensions

of Ameritan standard, 447Reactions and influente lines, 32

influente coefficient, 39influente lines, 35

definition, 35reactions, character of, 32

arch, 32cantilever beam, 32continuous beam, 32simple beam or span, 32statically determinate girder, 32statically indeterminate girder,

3 2suspension truss, 32

reactions, determination of, 33equations of equilibrium, 33in simple beam, 33in two-hinged arch, 33

Rectangular sections, moments“ of .-in-ertia of, 440 &,l .#+j

Rivet heads and clearances for machinedriving, sizes of, 449

Rivet holes in tension members, areas tobe deducted for, 448

Riveting, conventional signs for, 448Rivets, shearing and bearing value of, 449

S

Shapes, extreme length in feet of stand-ard, 442

Shearing and bearing value of rivets, 449Simple span bridges, weights of, 221

electric railway bridges, weight of,2 3 7

high>vay bridges, simple span, weightof, 236

:simple span highway bridges, AmeritanBridge Co. standard highwaybridges with concrete slabfloors (1910), 237

Ameritan Bridge Co. standardhighway bridges with timberflooring (1909), 236

live load, 236steel weights of some existing simple

span railroad bridges, 233deck pin span, P.R.R. , built

1905, 235through pin span, C., M. & St.

P. Ry., built 1910, 234through pin span, P.R.R., built

1896, 233through pin span, P.R.R., built

1906, 234through pin span, P.R.R., built

1903, 235tables and formulas for steel weight

of railroad bridges, 221Ameritan B r i d g e Co.‘s stand-

ards of 1902, 221-225Ameritan Bridge Co.‘s plate

girder spans of 1910, 225Illinois Central R.R., 228Japanese railways, 228N.Y.C. and H.R.R.R.,228Pennsylvania lines West, 228Pennsylvania Railroad stand-

ards, 228Pennsylvania Steel Co.‘s stand-

ards of 1908, 226Prussian state railways, 229

Simple spans, moments and shears in, 40stresses and bracing of, 105

Simple truss bridges, 167bearings, 190bottom chord, sections for, 180camber, 207chord splices, 193compression members, latticing of,

204columns with four webs, 206columns with three webs, 206columns with two webs, 206Rankine formula, 205straight-line formula, 205

diagonals, sections for, 181end sway bracing and portals, 188eyebars, 184

470 I N D E X

Simple truss bridges, floor, location of, 167in de& truss span, 167in through truss span, 167

gusset plates of riveted trusses, 190intermediate sway bracing, 189lateral system between loaded

chords, 187lateral system between unloaded

chords, 188panel length, 173parallel and polygonal chords, 170

pin connections, 193butt joints, 194pin joints, 194

pin packing, 197-200pin plates, 202-204pins, bending and shearing stresses

in, 194pins, calculation of moments and

shears on, 200-202pins, size of, 194polygonal chord, shape of, 172riveted and pin-connected trusses,

170top chord, sections for, 177-180truss design, general principies of,

169secondary stresses, 169

truss members, general principiesfor the design of, 175, 177

truss members, required area ofsection of, 173

bending moments, 174determination of stresses, 173

truss spans, bracing of, 185trusses, distance between, 169

in deck spans, 169in through spans, 169

trusses, heights of, 170for highway bridges, 171for railroad deck spans, 171for railroad through spans, 171

trusses, number of, 169verticals, sections for, 182web system, 171

Pratt truss, 171Warren truss, 171

Simple trusses, stresses in, 73Sines, cosines, tangents and cotangents,

table of, 450Sizes of rivet heads and elearances for

machine driving, 449

Skew bridges and bridges on curves, 209de& spans on curves, width of, 216middle ordinate of curve, 216outer rail on curves, ‘super-elevation

of, 214skew bents, 214skew deck truss spans, 213skew piers, arrangement of floor

over, 213skew plate girder spans, 210

deck plate girder spans, 210skew shallow floor, 211through plate girder spans, 210

skew spans, general, 209skew through spans with polygonal

top chord, 213skew through truss spans with

parallel chords, 211stringers on curves, arrangement of,

220super-elevation, provision for, 215through truss spans on curves,

width of, 217-219tracks on curves, distance betiTeen,

220Specifications for steel railway bridges,

Ameritan Railway EngineeringAssociation, 454

Square sections, properties of, 439Standard channels, properties of, 446Standard 1-beams, properties of, 445Standard shapes, plates and, extreme

length in feet, 441Steel railway bridges, Ameritan Railway

Engineering Association speci-ficatiocs for, 454

Stones and concrete, physical propertiesof, 434

Stresses in bracing of simple spans, 105braking forte, stresses from, 113centrifuga1 forte, stresses from, 111eccentric loading, stresses due to,

112end sway frames of deck spans,

stresses in, 115g e n e r a l , i05

wind. pressure, 105indirect wind stresses in bottom

chords, 110R-system of laterals, 109lateral system between polygonal.

chords, 108

I N D E X 471

Stresses in bracing of simple spans, lateralsystem between straightchords, 106

chords stresses in, 107’ overturning effect of wind, 110

portals, stresses in, 113portals, to find stress in member of,

114stability against lateral forces, 111

Stresses in simple trusses, 73example for truss with subdivided

panels, 100general, 73

compression, 73primary or main, 73secondary, 73statically determinate, 73statically indeterminate, 73tension, 73

influente lines for trusses withloaded chords, 80

influente lines for trusses with poly-gonal chord, 90

influente lines for trusses with sub-divided panels, 99

methods of calculation, 74method of equilibrium at joints,

76Maxwell diagrams, 76

method of moments, 74method of shears, 75

stress coefficients for trusses withparallel chords, 84

stress coefficients for trusses withpolygonal chord, 102

truss with subdivided panels,examplefor, 100

dead load stresses, 101bottom chord, 101lower main diagonals, 101subdiagonals, 101

live load stresses, 101bottom chord, 102lower main diagonals, 101subdiagonals, 101

polygonal chord, 102counter-diagonals, 98dead load stresses, 101

trusses with parallel chords, 76chord members, 76counter-diagonals, 79diagonals, 77

Stresses in simple trusses, dead loadstresses, 80, 83

examples for, 80impact stresses, 81live load stresses, 81, 83total stresses, 82verticals, 78

trusses with polygonal chord, ex-ample for, 93

dead load stresses, 94live load stresses, 95

trusses with polygonal top chord, 85chord members, 86counter-diagonals, 88main diagonals, 86verticals, 88

trusses with subdivided panels, 97counter-diagonals, 98submembers, 98

trusses of skew spans, 103Warren truss without verticals, 79

Structural work, actual pressure offoundations in, in tons per sq.ft., 430

T

Tension members, areas to be deductedfor rivet holes in, 448

Timber, physical properties of seasoned,435

Turntables, movable bridges and, 273bibliography of, 309-311center-bearing swing bridges,276-278dead load stresses, 283deflections, calculation of, 296end lift, 280floor and center cross girder, ar-

rangement of, 282live load stresses, 284-293

position of load for maximumstress, 292

stresses for uniform live load,292

l o a d on center p i v o t a n d centerwedges , 293

locomotive turntables, 305maximum moments in main

girders, 306shear at center of main girder,

306shear at end of main girder, 306stresses, calculation of, 305

I N D E X

Turntables, locomotive, weights of, 307motors, 300

electric, 300gasoline, 301hydraulic, 301

power required to operate movablebridges, 298

resistance due to action of wind,299

resistance due to friction, 298resistance due to inertia of

bridge, 298rim-bearing swing bridge, 278stresses, example for the calculations

of, 294center wedge reaction, 296dead load stresses, 294influente lines, 295

swing bridges, calculations of stressesin, 282

dead load, 283live load, 283 j

swing bridges, types and principaldimensions of girders and trussesof, 281

turning devicé, 279types of movable bridges, 273

bascule bridges, 274lift bridges, 275swing bridges, 273

jack-knife draw, 274shear-pole draw, 274

transporter or ferry bridges, 276traversing or retractile bridges,

275weights of movable bridgés, 301

steel weights of some existingmovable bridges, 302

swing spans, 301wind stresses, 294

Types of bridges, and principal dimen-sions, 117

vViaducts, 238

design of viaducts, 248anchorages, 256batter of columns, 253bibliography of viaducts, 259bracing, 254-256column sections, 254

economical span lengths, 250single-track viaduct, 250

expansion, 256number of columns, 253

height of panels, 254plate girder spans, 252types of viaducts, 248weights of viaducts, 256-258

stresses, in towers, calculation of, 238combined stresses, 244dead load stresses, 238example, 245

dead load stresses, 245l i v e l o a d a n d impact

stresses, 245stresses from braking forte,

248stresses from centrifuga1

forte, 247uplift, 248wind stresses, 246

externa1 forces, 238live load stresses, 240stresses dueto braking forte, 243stresses due to centrifuga1 forte,

243uplift, 244wind stresses, 240-243

W

Weights and dimensions of Ameritanstandard rail sections, 447

Weights of simple span bridges, 221

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