f6r BOUGHT WITH THE INCOME FROM THE SAGE ENDOWMENT FUND THE GIFT OF $iettrg W. Sage 1S91 f?.ai.g(>H-l ;Ll|ttl3 Cornell University Library TG 260.S76 The theory of structures. The date shows when this volixme was tKea. To renew this book copy the call No. and give the librarian ^^ * HOME USE RULES. All Bosks subject to Recall All books must be re- turned at end of college year for inspection and repairs. Students must re- turn all books before leaving town. Officers should arrange for the return of books wanted dtuing their absence from town. Books needed by more than one person are held on the reserve list. Volufties ef periodi- cals and of pamphlets are held in the library as much as possible. For special purposes they are given out for a limited time. Borrowers should not use their library privileges for the bene- fit of other persons. Books of special value and gift books, when the giver wishes it, are not allowed to circulate. Readers are asked to report all cases of books marked or mutilated. Do not deface books by marks and writing. 1924 015 400 389 w Cornell University Library The original of tiiis book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924015400389 THE THEORY OF STRUCTURES PubSished hy the McGraw - Hill Book. Company N< ■ Yoirle: ^Succe^sol^s to tKeOook. Departments of tKe McGraw Publishing Company Hill Publishing' Comfwny FViblJahers of &ooks foi^ Electrical World The Engineering and Mining Journal Engineering Record American Machinist Electric Railway Journal Coal Age Metallurgical and Cliemical Engineering Power THE THEORY OF STRUCTURES BY CHARLES M. SPOFFORD, S.B. HAYWARD PROFESSOR OB' CIVIL BNGINEEKINO AT THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY MEMBER OF THE AMERICAN SOCIETY OF CIVIL ENGINEERS FIRST EDITION SECOND IMPRESSION — CORRECTED McGEAAV-HILL BOOK COMPANY 239 WEST 39TH STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E.G. 1911 7^ Copyright, 1911, liY CHARLES M. SPOFFORD THE SCIENTIFIC PRESS ROBERT DROMMOND f BROOKLYN, N. Y. PREFACE The purpose of this book is to present in a thorough and logical manner the fundamental theories upon which the design of engineering structures is based and to illustrate their apphca- tion by numerous examples. No attempt has been made to treat of the design of complete structures, but the design of the more important elements of which all structures are composed is fully considered. The subject-matter is confined almost entirely to the treat- ment of statically determined structures, it being the writer's purpose to deal with indeterminate cases in another volume; the commonly used approximate methods for some of the more ordinary types of indeterminate structures are, however, included. While flie theories presented are for the greater part only such as have been in common use for many years, the method of treatment frequently differs considerably from that found in other books. Special attention maj'^ be called to the early introduction of the influence line and to its use in deriving and illustrating analytical methods, as well as to the chapter upon deflections. The author wishes particularly to acknowledge his indebted- ness to Professor George F. Swain for the logical and inspiring instruction received from him as a student. July 18, 1911. TABLE OF CONTENTS CHAPTER I OUTER AND INKER FORCES ART. PAGE 1 . Definitions 1 2. Live and Dead Load 4 3. Outer Forces 4 4. Weight of Structure 4 5. Weight of Railroad Bridges f) 6. Approximate Truss Weights 7 7. Weight of Highway Bridges 10 8. Weight of Roof Trusses 12 9. Weight of Steel-frame Buildings 14 10. Live Loads for Railroad Bridges 15 11. Live Loads for Highway Bridges 17 12. Live Loads for Buildings 19 1.3. Wind Pressure 20 14. Snow Load 27 15. Centrifugal Force and Friction 28 16. Impact on Railroad Bridges 28 17. Impact on Highway Bridges and Buildings 30 1 8. Inner Forces 30 19. Factor of Safety 32 CHAPTER II LAWS OF STATICS, REACTIONS, SHEARS AND MOMENTS, INFLUENCE LINES 20. Laws of Statics 34 21. Reactions 35 22. Computations of Reactions — Method of Procedure 36 vii viii TABLE OF CONTENTS ART. PASE 23. Reaction Conventions 38 24. Point of Application of Loads and Reactions 38 25. Solution of Reaction Problems 39 26. Shear and Bending Moment Defined 44 27. Method of Computation, Shear and Bending Moment 44 28. Signs of Shear and Bending Moment 44 29. Shear and Moment, Common Cases 45 30. Curves of Shear and Moment, Defined and Illustrated 45 31. Shear and Moment, Distributed Load 47 32. Shear and Moment, Uniformly Varying Load 48 33. Location of Section of Maximum Moment 51 34. Theorem for Computing Moments 51 35. Beams Fixed at Ends 52 36. Effect of Floor Beams 53 37. Typical Curves of Shear and Moment 54 38. Influence Lines and Tables Defined 61 39. Examples of Influence Lines 63 40. Properties of Influence Lines 65 41. Neutral Point 66 42. Position of Loads for Maximum Shear and Moment at a Definite Section 66 43. Maximum Moments and Shears. Structures Supported at Ends. ... 67 44. Approximate Method for Maximum Shear 69 CHAPTER III CONCENTRATED LOAD SYSTEMS 45. Shear at a Fixed Section. Girder without Floor Beams 75 46. Moment at a Fixed Section 79 47. Shear. Girder with Floor Beams 81 48. Formula for Position of Ijoads for Maximum Shear for Intermediate Panels. Girder with Floor Beams 86 49. Maximum Moment. Girder with Floor Beams 87 50. Moment and Shear at the Critical Section 88 51. Moments and Shears. Floor Beams and Transverse Girders 92 52. Moment Diagram 96 CHAPTER IV BEAM DESIGN 53. Formulas 100 54. Method of Design 102 TABLE OF CONTENTS ix ART PAGE 55. Wooden Beams 102 56. Steel Beams 103 57. Examples of Beam Design 104 58. Composite Beams 106 59. Stiffness 106 CHAPTER V PLATE GIRDER DESIGN 60. Plate Girders Defined 108 61. Plate Girder Web Theory 109 62. Plate Girder Flanges. Theory 110 63. Degree of Approximation of Flange Formula 113 64. Degree of Approximation of Shear Formula 118 65. AUcJwance for Rivet Holes 120 66. Example of Girder Design 122 67. Rivets and Riveted Joints 122 68. Flange Rivets. Ordinary Method of Computation of Pitch 126 69. Flange Rivets. Precise Method of Computation of Pitch 128 70. Flange Rivets. Example in Computation of Pitch 128 71. Direct Web Stresses 131 72. Web Stiffeners 134 73. Flange Plates 138 74. Connection Angles and Fillers 141 75. Web SpUces 143 76. Flange-angle SpHce 146 77. Cover-plate Splice 148 CHAPTER VI SIMPLE TRUSSES 78. Trusses Defined 150 79. Classification 150 80. Theory 151 81. Methods 151 82. Analytical Method of Joints Described 151 83. Character of Stress 152 84. Determinate and Indeterminate Trusses 153 85. Mode of Procedure. Analytical Method of Joints 154 86. Application of Analytical Method of Joints 154 87. Graphical Method of Joints Described 157 X TABLE OF CONTENTS ART. PAGE 88. Mode of Procedure. Graphical Method of Joints 157 89. Application of Graphical Method of Joints 158 90. Ambiguous Cases. Graphical Method of Joints 159 91. Analytical Method of Moments Described 161 92. Mode of Procedure. Method of Moments 161 93. Application of Method of Moments 162 94. Method of Shears Described 163 95. Mode of Procedure. Method of Shears 163 96. Method of Shears. Application 164 97. General Rules for Determination of Truss Stresses 165 98. Counters 166 99. Types of Trusses 167 100. Systems of Loading 170 101. Index Stresses 171 102. Computation of Stresses. Pratt Truss 173 103. Computation of Stresses. Warren Truss 178 104. Computation of Stresses. Subdivided Warren Truss 180 105. Computation of Stresses. Bridge Trusses with Non-parallel Chords. Uniform Load 182 106. Computation of Stresses. Bridge Trusses with Xon-parallel Chords. Concentrated Load System 189 107. Computation of Stresses. Bridge Trusses with Parabolic Chord . . . 196 CHAPTER VII BRIDGE TRUSSES WITH SECONDARY WEB SYSTEMS, INCLUDING THE BALTIMORE AND PETTIT TRUSSES 108. Secondaiy Systems Described 203 109. Computation of Maximum Stresses in Pettit Truss 208 CHAPTER VIII TRUSSES WITH MULTIPLE WEB SYSTEMS, LATERAL AND PORTAL BRACING, TRANSVERSE BENTS, VIADUCT TOWERS 110. Trusses with Multiple Web Systems 226 111. Approximate Determination of Maximum Stresses in a Double Sys- tem Warren Truss 227 112. Approximate Determination of Maximum Stresses in aWhipple Truss 235 113. Skew Bridges 239 114. Lateral and' Portal Bracing 240 TABLE OF CONTENTS xi ART. PAGE 115. Lateral-bracing Trusses 242 116. Approximate Determination of Maximum Stresses in Lateral Bracing 242 117. Portals. Approximate Solution 244 118. Portals, Miscellaneous 249 119. Transverse Bents in Mill Buildings 249 120. Viaduct Towers 252 CHAPTER IX CANTILEVER BRIDGES 121. Types of Structures for Long-span Bridges 258 122. Cantilevers Described 258 123. Equations of Condition 259 124. Anchorage 261 126. Reactions, Cantilever Trusses 261 126. Shears and Moments, Cantilever Trusses 265 127. Bar Stresses, Cantilever Trusses 266 CHAPTER X THREE-HINGED ARCHES 128. Characteristics of the Arch 269 129. Types of Arches 269 130. Reactions, Three-hinged Metal Arches 272 131. Maximum Stresses in Elastic-Arch Ribs 273 132. Parabolic Three-hinged Arches 276 CHAPTER XI DESIGN OF COLUMNS AND TENSION MEMBERS 133. Columns. General Considerations 285 134. Condition of Ends 286 135. Formulas for Columns of Ordinary Lengths 287 136. Typical Formulas for Columns of Ordinary Lengths 288 137. Formulas for Long Colunms 290 138. Tests of Steel Columns 291 139. Cast-iron Columns 292 140. Timber and Concrete Columns 294 xil TABLE OF CONTENTS ABT. PAGE 141. Typical Column Sections 294 142. General Dimensions and Limiting Conditions 295 143. Method of Design 296 144. Determination of Cross-section of Typical Steel Colunms 298 145. Lattice Bars and Batten Plates 301 146. Rivet Pitch 306 147. Eccentric Forces 307 148. Effect of Combined Flexure and Thrust 309 149. Building Colunms under Eccentric Loads 310 150. Design of Cast-iron Columns 311 151. Design of Iron and Steel Tension Members 314 CHAPTER XII PIN AND RIVETED TRUSS JOINTS 152. Bridge Pins Described 318 153. Arrangement of Members on Pin 318 154. Minimum Size of Pins 321 155. Stresses Causing Maximum Moment and Shear 322 156. Computation of Maximum Moment and Shear 323 157. Computation of a Top Chord Pin for Truss Shown in Fig. 245. . . . 323 158. Computation of a Bottom Chord Pin for Truss Shown in Fig. 245.. 327 159. ETect upon Pin of Change in Arrangement of Members 331 160. Fin Plate Rivets 332 161. Pin Nuts 332 162. Packing Rings 332 163. Riveted Truss Joints 333 CHAPTER XIII GRAPHICAL STATICS 164. Graphical and Analytical Methods Compared 337 165. Force and Equilibrium Polygon 337 166. Characteristics of the Equilibrium Polygon 341 1 67. Reactions 341 168. Graphical Method of Moments 343 169. Graphical Method of Moments with a Concentrated Load System . . 345 170. Graphical Method of Shear 346 171. Equilibrium Polygon through Several Points 349 TABLE OF CONTENTS xiii CHAPTER XIV DEFLECTION AND CAMBER ABT. PAGE 172. Elastic and Non-elastic Deflection 355 173. Truss Deflection. Trigonometrical Method 355 174. Truss Deflection. Method of Rotation 356 175. Truss Deflection. Method of Work 358 176. Truss Deflection. Illustrated - . . 362 177. Deflection of Beams and Girders 364 178. Graphical Method of Deflection. Williot Diagram 368 179. Correction of Williot Diagram 373 180. Camber Defined 377 181. Rules for Computing Camber 378 CHAPTER XV CONTINUOUS AND PARTIALLY CONTINUOUS GIRDERS AND SWING BRIDGE REACTIONS 182. Definitions 379 183. Reactions on Continuous Girders. Method of Computation 379 184. Derivation of the "Three-moment" Equation 380 185. Application of the " Three-moment" Equation 384 186. Reactions, Shears, and Moments for Common Cases of Continuous Girders 386 187. Partially Continuous Girders. Method of Solution 386 188. Theorem of Least Work 395 189. Determination of Reactions on a Partially Continuous Girder 396 190. Types of Girders for Centre-supported Swing Bridges 397 191. Influence of End Supports upon Swing Bridge Reactions 398 192. Tables of Reactions for Continuous and Partially Continuous Girders Used for Swing Bridges 399 STRUCTURES CHAPTER I OUTER AND INNER FORCES 1. Definitions. A structure as defined in the " Century Dictionary " is, " a production or piece of work artificially built up, or composed of parts joined together in some definite manner." As used in this book, however, its meaning will be restricted to a part or combination of parts constructed to hold in equi- librium definite forces, with special reference to bridges and buildings. Structures may be either statically determined or statically undetermined. Statically determined structures are those in which the reactions and primary stresses can be computed by statics. Structures for which these functions cannot be obtained by statics belong to the second class. A bridge is a structure built to provide transportation across some natural or artificial obstacle, such as a river, ravine, street or railway. The term includes not only the superstructure of wood, metal or masonry, but also the substructure which may consist of masonry piers and abutments, or of steel towers. The superstructure may consist of simple beams supported at the ends directly on the masonry, or in case of long spans sup- ported on cross beams which are themselves supported at the ends by girders, trusses or arches.' In the latter case the longi- tudinal beams are known as stringers and the cross beams as floor beams. As a clear conception of the function of the stringers ' For a clear understanding of girders and trusses see Figs. 1 to 4 and Arts. tiO and 78. OUTER AND INNER FORCES Art. 1 and floor beams is essential to the understanding of the matter which follows, the student is advised to study carefully Figs. 1, 2, 3 and 4, and to examine some of the bridges in his vicinity. Deck bridges are those in which the floor is at the top of the lateral bracing Bottom lateral 1)Eacmg shown bj dotted lines The two trusses are abed and efyh An intermediate floor beam End Floor Beam Fig. 1. — Framework of a Through Railroad Truss Bridge. Masonry Abutment (Sub-structure) CROSS SECTION — Spati-Ceutre to Centre of Bearings- SIDE ELEVATION Fig. 2. — I-Beam Bridge for Single Track Railroad. (A Deck Structure.) main superstructure, as in the simple I-beam bridge shown in Fig. 2. Such bridges if of considerable width require the use of floor beams and stringers, but for narrow bridges these may often be omitted. Art. 1 DEFINITIONS Half-frhrough bridges are those having the greater part of the superstructure above the floor level but with insufficient CROSS SECTION eiiiiuiiiusi^iiisiiiiisii StriDger i Masonry pedestal Floor Beam resting on abutment ■ Vj' "^W//A and supporting end of jaaaonry^^ stringer. Sometimes tbia ^ti\>tw^n)0 is replaced by an end floor beam LONGITUDINAL SECTION Fig. 3.— Half-Through Single Track Plate Girder Railroad Bridge. Note. — Portion of bridge between floor beams measured along axis of bridge is called a panel. Stringer^ .^End Hollei-s SIDE ELEVATION OF-END OF A PIN CONNECTED TRUSS Fig. 4. — A Single Track Through Railroad Bridge. depth to permit the use of overhead bracing. Lateral stability in such bridges may be obtained by the use of brackets or knee braces, as shown in Fig. 3. 4 OUTER AND INNER FORCES Art. 2 Through bridges are those in which the greater p&rt of the superstructure is above the floor level and in which overhead lateral bracing may be used between the trusses to obtain lateral stability. Such a bridge is shown by Fig. 4. Whether a deck or through bridge should be used for a given location depends upon the external conditions. In general, bridges of considerable span are built as through structures unless the approaches on either side are at a considerable eleva- tion above the obstacle to be crossed. The solution of this question for a given case is usually obvious and will not be con- sidered here. 2. Live and Dead Loads. The forces to be considered may be divided into two classes: outer and inner. The outer forces consist of the applied loads and the resultant reactions, and maybe divided into two distinct types : live or moving loads and dead or quies- cent loads. The inner forces are the molecular forces which are brought into action by the outer forces and hold them in equilibrium. The dead load includes the weight of the struc- ture itself, and all of its permanent quiescent load such as the pavement on highway bridges; the rails and other track appur- tenances on railroad bridges; the floors, walls, roofs and par- titions in buildings. The live load consists of all forces which are applied intermittently. For bridges these may be locomotives and cars, vehicles, pedestrians, snow and wind ; for buildings they con- sist of people, snow, wind, ofiice furnishings and partitions; for dams and retaining walls of water pressure and earth pressure. 3. Outer Forces. The determination of the intensity, dis- tribution, and point of application of the outer forces is often difficult and requires mature judgment based upon extensive experience. For structures of great magnitude the question is particularly complicated and of vital importance; the design of such structures should never be attempted without a thorough study of this problem in its relation to the structure in question. In the following articles some of the difficulties in the way of an exact solution of the question will be presented and data given for use in the solution of the more common cases. 4. Weight of Structure. It is impossible to determine accurately the weight of a given structure before the completion of the design. It is equally impossible to design the structure with precision until its weight is known. It is therefore neces- Art. 5 WEIGHT OF RAILRf)AD BRIDGES 5 sary in all cases to make use of approximate methods of solu- tion, first assuming the weight, next designing with the assumed data, then computing the weight and revising the design in the light of the new information thus obtained. For the more common types of structure data accumulated by experience may be used by the designer, and the first assumption made with sufficient accuracy to make revision unnecessary.. For structures out of the ordinary, and particularly those in which the weight of the structure itself is a large percentage of the total load, several revisions are sometimes necessary, and a final compu- tation of the weight after the completion of the detailed draw- ings and before the commencement of shop work, should never be omitted. The failure to do this for the huge Quebec bridge which failed during erection in 1907, resulted in serious errors in the stresses for which the structure was designed. In all cases the designer should first design completely the minor portions of the structure and determine their weight carefully so as to eliminate as much uncertainty as possible. For example, in the design of a railroad bridge, the stringers should first be figured and their weight carefully d-etermined, the floor beams may then be designed, and finally the lateral bracing, thus giving considerable information as to the total weight of the bridge and throwing the uncertainty into the main girders or trusses. 5. Weight of Railroad Bridges. It is possible to make a more accurate preliminary estimate of the weight of such bridges than can be done for other types of important structures, since there is less variation in loads and other conditions. Current practice on first class American railroads differs but little, and it is believed that the diagrams given in Figs. 5 to 10 inclusive give reasonable values for the total weight of the steel in such struc- tures. The total weight of the bridge includes also the weight of the ties, rails and other accessories, which should be added to the values given in the diagrams. For the ordinary railroad bridge-floor with wooden ties this weight may be taken, in the lack of a specific design, as from 400 to 450 lbs. per linear foot. For solid ballasted floors this weight is of course much greater. These diagrams were furnished by the Heath & MiUigan Mfg. Co., Paint and Color Makers, of Chicago, 111., U. S. A., for whom they were prepared by consulting engineers con- OUTER AND INNER FORCES Abt. 5 nected with one of the large railroad systems of the country, and are for carbon-steel bridges designed for the typical loco- motives shown in Fig. 11, and are known as Cooper's Eeo loading. Where other loadings are to be used, these weights DECK PLATE GIRDER SPANS WEIGHTS, DIMENSIONS, ETC. LOADtNG- Total end E thousands rail and includes live and dead loads. 60 . ()5 .. |5 n SO I' 85 " 90 .1 100" 8-0" Erector's Note; W- Total weight of one single track span 25' to 6 65'" no' ■3 is J5 .in 2 § .to g Fig. 5. may be changed approximately in the ratio of the locomotive weights. For weights of cantilever spans and nickel steel bridges the reader is referred to the paper by Dr. J. A. L. Waddell entitled Art. 6 APPROXIMATE TRUSS WEIGHTS " Nickel Steel for Bridges " published in the Transactions of the American Society of Civil Engineers, Vol. LXIII, pages 165 to 172. For bridges designed for either heavier or lighter loads these weights may be altered in a somewhat less propor- FiG. 6. tion than the live loads in accordance with the designer's judg- ment. 6. Approximate Truss Weights. Bridges differing materially from those previously considered and for which other data are not available may be estimated by the following rule devised 8 OUTER AND INNER FORCES Abt. 6 by Clarence New York. W. Hudson, Consulting Engineer, 45 Broadway, Fig. 7. Let L= maximum live stress in bottom chord. / = impact in member in which L occurs. D-i = dead stress in same member due to known weight of floor. Da = dead stress in same member due to weight of truss and bracing (guessed). /( = allowable unit stress in tension. Art. 6 APPROXIMATE TRUSS WEIGHTS Let Ai =area in square inches of member in which L occurs. A2 =area in square inches per linear unit of one truss. Tr= weight per linear foot of one truss and its bracing. WEIGHTS OF SINGLE TRACK THROUGH PIN OR RIVETED SPANS. Spans 300 Ft. to 000 Ft. In Length LOADING-COOPER'S E. 60 FOR DOUBLE TRACK SPANS-INCREASE CURVE WEIGHTS 85 PER CENT Fig. 8. Then ^1 = L+I+R1+D2 ft 50 42=5Ai and W=—Ai. O • (1) The above is based upon an allowance of 1.25^1 for the upper chord, 1.254i for the web members, Ai for details, and .5Ai for bracing. The weight of steel is used in round figures 10 OUTER AND INNER FORCES Art. 7 as 10 lbs. per square inch of cross section for a bar one yard in length. This method is said to give a very close approximation to the actual weight. WEIGHTS OF SIGNAL BRIDGES -2000 -2000 2000 -2000 2000 X'XXX*XXX'X \7 i ^Top of Kail 1 X 1 "ma 1 1 r J± 'i V-ikww^teggy'-'^.^ifaTO^^'^^^Hg^MWVs^^ ^ ^ ^.^A in ' 1 A.\\ spans designed to carry a signal, weighing 2000 lbs. ^ jj It/' • ■• h over each track. 86 00 — -: — -- / 1 1 1 ' t ~l 1 1 s r^ § '- i § •^ - J 1 ■■$■ 1 ~ ^" H 00 :^ /- —1-1- ■4" ZL __]_ -^ 1 / "* ' 1 1 / 1 , A ; , 90 'XJa^ - r -,„ - LJ-/[-]-l- -n' / ■ : / 14 000 / / 1 5- 2 ; [ ■~ 80 ^ f ; ^ - 1 / / / - 10 000 - / / / ,/, V 30 10 00 CO Erector's Note:- W= Total weight of one span and two bents. .63 W=Weight of one span. .38 w= Weight of two bents. Fig. 9. 7. Weight of Highway Bridges. The weight of highway bridges is less easily determined than that of railroad bridges since the width, span, floor covering and character of loading are subject to wide variations. Formulas have been deduced Art. 7 WEIGHT OF HIGHWAY BRIDGES 11 for special cases, but these are of little value and will not be quoted. The designer should proceed step by step as previously stated, and if experienced should obtain good results. The method given WEIGHTS OF SINGLE TRACK DRAW SPANS aioo LOADING 1 - ' /' ~t "'^1 COOPER'S E 60 - ;: — 7 t/H ^ 180U - ._ - -l-j- 1 / 1 - ■- 1 ' / «? "" J ^ 1100 / - - / 1 •^ / ^ 1000 f / iSO / / 600 / / ^^ / 100 s? ^ 100 200 / ^ A / c^ ' - -- 350 1 IBO 800 260 S 00 Note:- Curve weights include all structural steel and machinery. W=Total weight of single track span. 1,85 W= Total weight of double track span. Drum and machinery comprise 15 per cent of total weights. Fig. 10. in Art. 6 may be applied in order to obtain an approximate truss weight. The following figures for weights of the steel in actual bridges in the city of Boston may be useful in making estimates for 12 OUTER AND INNER FORCES Akt. 8 city bridges. These bridges are all modern structures designed for heavy street car traffic. The floor in all cases consists of yellow pine underplanking, 5 or 6 inches in thickness, water- proofed on top, and supporting a 6-inch block pavement on a sand cushion. These values together with those of paving materials were furnished the writer by Mr. Frederic H. Fay, Engineer of Bridges and Ferries of the Public Works Department of the city of Boston. WEIGHT OF STEEL IN TYPICAL HIGHWAY BRIDGES IN THE CITY OF BOSTON Bridge. Type. Span. Weight of Steel per Square Foot. Broadway Plate girder Pin trusses 67 ft. 8i ins. 72 ft. ins. to 85 ft. ins. 82 ft. 11 ins. 80 ft. ins. 146 ft. 9| ins. 52 . lbs. 55 2 " Charlestown 57 6 " Craigie (proposed) Northern Avenue 65.3 " 67.0 " WEIGHT OF PAVING MATERIAL Note. — B.M. =Board Measure. Hard (yellow) pine, 4 lbs. per ft. B.M. (Where protected by waterproofing and always dry. Otherwise use 4J- lbs.) 48 lbs. per cu.ft Creo-resinate yellow pine paving blocks 65 Spruce and white pine, 2^ lbs. per ft. B.M 30 Bricks, pressed and paving 150 Portland cement concrete 160 Tar concrete (base for asphalt walks, etc.) 125 Silician rock (Simpson Bros.) 140 Trinidad asphalt (Barber Asphalt Co.) refined 74 As laid 140 Granolithic or artificial stone 150 Pavements (exclusive of sand cushion) : 6-inch granite block 80 lbs. per sq.ft. 4-inch brick 50 " 4-inch wood block (creo-resinate) 22 " " Roadway waterproofing: 1} ins. thick (felt, roofing pitch, sand, and road pitch). 12 lbs. per sq.ft. Ruckle plates 10 to 20 " " 8. Weight of Roof Trusses. The weight of roof trusses depends upon the span, distance apart of trusses, roof covering and roof pitch. The conditions are somewhat more uniform Art. 8 WEIGHT OF ROOF TRUSSES 13 than for highway bridges, and formulas for approximate weights may be used with some degree of success. The formula 1 which follows was deduced by N. Clifford Ricker from the weight of 50 roof trusses designed for spans from 20 to 200 ft. in length. The distance apart of trusses varied from 10 to 30 ft. and the rise from yV to i the span. The roofs were assumed as covered with | ins. wooden sheathing carrying painted tin. The snow load was taken as 20 lbs. per horizontal square foot and wind as 30 lbs. per square foot on a vertical plane, properly reduced to allow for pitch of roof. Trusses of yellow pine with steel verticals, of white pine with steel ver- ticals, and entirely of steel were included. Let PF= weight of truss in lbs. per square foot of horizontal projection of roof. *S=span in ft. 25^6000- • • ■. ■ • ^"^^ Steel trusses of shorter spans than 100 ft. probably weigh somewhat more than the value given by the formula, and white pine trusses are somewhat lighter than those of yellow pine. For roof trusses under other conditions, the value given by the formula may be modified or the designer may use his judgment. The following table gives the approximate weights per square foot of roof surface for some of the roof coverings in common use (see also latter portion of Art. 14) : Lbs, per sq.ft. Pine shingles 2 Corrugated iron, without sheathing. 1 to 3.7.5 Felt and asphalt, without sheathing 2 Felt and gravel, without sheathing 5 to 10 Slate, i in. thick 9 Tin, without sheathing 0.7.5 Skylight glass, ^ to J in., including frames 4 to 10 White pine sheathing, I in. thick 2 Yellow pine sheathing, J in. thick 4 Tiles, flat 15 to 20 Tiles, corrugated 8 to 10 Tiles, on concrete slabs 30 to 35 Plastered ceiling 10 to 15 ' See "Bulletin No. 16" of the University of Illinois, Engineering Experi- ment Station, entitled "A Study of Roof Trusses," by N. Chfford Ricker, D.Arch., Professor of Architecture. 14 OUTER AND INNER FORCES Art. 9 9. Weight of Steel-frame Buildings. The weight of such buildings is largely dependent upon the weight of walls, floors, partitions and fire-proofing and these can be estimated in detail from the architect's plans. The weight of the steel is, however, so variable that no attempt will be made to give values for it, but no difficulty need arise in designing, since the weight of the steel, in any given member forms but a very small per- centage of the load which it has to carry. The table which follows may be used in determining the weight of hollow tile floors and walls. WEIGHTS OF HOLLOW TILE FLOOR ARCHES AND FIREPROOF MATERIALS Hollow Brick for Flat Arches, Side Construction Width of Span between Beams. Depth of Arch. Weight per Square Foot. 6 inches 7 " 8 " 9 " 10 " 12 " 27 pounds 29 " 32 " 4 " " 4 " 6 " 4 " 6 " 5 " " 5 " 6 " 6 " " ... 30 " 6 " " 6 " 6 " 39 6 " 6 " 7 " " 44 " Partitions Hollow brick (clay) partitions. Porous terra-cotta partitions. Thickness. 2 inches 3 Weight per Square Foot. H pounds 14 " 15 " 19 " 20 " 27 " 16 " 19 " 22 " 23 " 33 " Reproduced from " Cambria Steel " by permission or Cambria Steel Co. Art. 10 LIVE LOADS FOR RAILROAD BRIDGES End Construction, Flat Arch 15 Width of Span between Beams, Depth of Arch. Weight per Square Foot. 5 feet to 6 feet 8 inches 9 " 10 " 12 " 27 pounds 29 " 6 " 7 " 7 " 8 " 33 " 8 " 9 " 38 " FpRRiNG, Roofing, and Ceiling Porous terra-cotta furring, roofing . ceiling . Thickneaa. Weight per Square Foot. 2 inches 8 pounds 2 " 12 " 3 " 15 " 4 " 19 " 2 " 11 " 3 " 15 " 4 " 19 " 6-inch segmental arches, 27 pounds per square foot. 8-inch segmental arches, 33 pounds per square foot. 2-inch porous terra-cotta partitions, 8 pounds per square foot. Reproduced from "Cambria Steel" by permission of Cambria Steel Company. Concrete for building work may be made with cinders, broken stone or gravel, and its weight may be taken as follows: Cinder concrete, 112 lbs. per cubic foot. Trap rock or gravel concrete, 150 to 155 lbs. per cubic foot. For concrete reinforced with steel add 4 lbs. per cubic foot to above weights. In practice the minimum weight of a fireproof floor may be taken as 75 lbs. per square foot except for office buildings where 10 lbs. should be added to provide for movable partitions. Fireproofing for columns or beams may be either of terra- cotta or concrete. The thickness should be not less than two inches. The weight per foot depends upon the size of the member to be protected. 10. Live Loads for Railroad Bridges. It is possible to deter- mine definitely the weights of the locomotives and cars used upon a given railroad. In consequence the actual live loads crossing a given bridge can be ascertained with considerable 16 OUTER AND INNER FORCES Art. 10 exactness, though it is necessary to make due allowance for the effect of high speed, irregularities in track, and other dynamic effects which do not occur when the loads are at rest. These dynamic forces are considered in Arts. 16 and 17 and will be neglected for the present. In the design for a new bridge it is also desirable to make due allowance for possible increase in Weight of locomotives and cars, hence the loads for which bridges are designed may be somewhat heavier than those which are in actual use at the time of construction, though the factor of safety (see Art. 19) provides to some extent for such increase. As to the type and number of locomotives and character of train loading, American practice is fairly uniform. Two combinations are usually considered : (a) Two consolidation locomotives followed by a uniform load per foot. (b) A pair of axles with loads somewhat heavier than those of the consolidation engine and no uniform load. The former loading gives the maximum stresses for most cases, but the latter is sometimes the controlling factor for stringers, short beam spans, and minor truss members. In designing, the effect of rails and ties in distributing the locomotive load is usually neglected, the wheel loads being con- sidered as applied at points. As the actual variation in wheel spacing and loads for loco- motives of different makes is often slight, it has become in recent years the custom of many railroads to specify the typical locomotives, first proposed by Theodore Cooper, Consulting Engineer, of 45 Broadway, New York. In these locomotives the distance between axles are in even feet and the wheel loads in even thousands of pounds. WhEe these loads and spaces may not represent actual cases they agree closely with average locomotives, and are much simpler to deal with than loadings in odd hundreds of pounds and axle spacings in feet and inches. Moreover, the uncertain developments of the future and the unknown effect of impact make the use of such typical loads but little less accurate than the use of actual wheel loads. Fig. 11 shows Cooper's Eeo locomotive, which is suitable for loads carrying heavy traffic. For other conditions types known as ^50, E^o, -Ean, etc., are used, these differing from Art. 1] LIVE LOADS FOR HIGH^\'AY BRIDGES 17 the ^eo type in weight only; for example, in the £'40 type the driving-wheel axle loads are 40,000 lbs. instead of 60,000 lbs., while the other loads are proportionate. This has the advantage of allowing tables of moments and shears made for one type to be readily used for another type by multiplying bj"^ a simple ratio. Such tables are now incorporated in numer- ous handbooks and specifications. Befoi'e leaving the subject a few M'ords as to other methods of loadings are desirable. Some years ago there was consider- able agitation in favor of adopting a uniform load, in order to simplify computation. The advantage is obvious to those who are familiar with such work, but the disadvantage is that in order to obtain properly proportioned trusses this load must necessarily vary, for different spans and for different memlsers TjOad=i are axle loads. s sssg gggs i gggg. iigg o 0000 ojociS" =• 000= 200SS '^ AAXX """" " =??LS, ""«" Unilorm Load Fig. 1L — Diagram of Cooper's EgQ Standard Loading. in the same span. This offsets to a considerable extent the advantage gained. Moreover, the adoption of the above stand- ard loadings has further simplified the labor of computation for actual wheel loads, so that at the present time it is believed that for ordinary structures the advantage in using a uniform load is too small to consider. For complicated trusses a combination of the two methods is perhaps best; viz., the use of wheel con- centrations for web members and of a uniform load for the chords, since the approximation for the chords b}^ using a uniform load is less than for the web members. 11. Live Loads for Highway Bridges. The magnitude and character of such loads depend almost entirely upon the location of the bridge. If it be a large city or in a district where heavy manufactured articles or quarried stones of great weight are to be transported it is quite probable that wagon loads from 20 to 30 tons may at times pass over the bridge. Electric-car loads of from 40 to 60 tons should also be assumed, as such cars are already in use in some portions of the United States, while the construction of interurban electric lines in many sections of the country indicates a future widespread extension of heavy 18 OUTER AND INNER FORCES Akt. 11 street car traiSc. The amount of foot travel and ordinary vehicular traffic on highway bridges also requires careful study. The weight per square foot from a crowd of people may reach the high figure of 150 lbs., which is probably heavier than the weight per foot from horses and wagons on the roadway. To assume, however, that such a load is likely to occur over the entire surface of an ordinary bridge is absurd, and the longer and wider the bridge the less the load that should be taken. In fact, every considerable highway bridge is in itself a problem in loading and should be carefully studied. The following clause from the Massachusetts Railroad Commissioner's " Specifications for Bridges Carrying Electric Railways," prepared by Professor George F. Swain, consulting engineer of the commission, may, however, be used as a broad general guide for the determination of loads for ordinary highway bridges: " (a) For city bridges, subject to heavy loads: " For the floor and its supports, a uniform load of 100 lbs. per square foot of surface of the roadwaj' and sidewalks, or a concentrated load of 20 tons on two axles 12 ft. apart with 6 ft. between wheels. In computing the floor beams and supports, the railway load shall be assumed, together with either (1) this uniform load extending up to within 2 ft. of the rails, or (2) the above-described concentrated load alone. " For the trusses or girders, 100 lbs. per square foot of floor surface for spans of 100 ft. or less, SO lbs. for spans of 200 ft. or over, and proportionally for intermediate spans. This uniform load is to be taken as covering the floor up to within two feet of the rails. " (6) For suburban or town bridges, or heavy country high- way bridges: " For the floor and its supports, a uniform load of 100 lbs. per square foot, or a concentrated load of 12 tons on two axles 8 ft. apart; these loads to be used as described under (a). " For the trusses or girders, 80 lbs. per square foot of floor surface for spans of 100 ft. or less, and 60 lbs. for spans of 200 ft. or more, and proportionally for intermediate spans; to be used as described under (a). (See d). " (c) For light country highway bridges: " For the floor and its supports, a uniform load of 80 lbs. per square foot; this load to be used as described under (a). (See d.) Aht. 12 LIVE LOADS FOR BUILDINGS 19 " For the trusses or girders, 80 lbs. per square foot of floor surface for spans of 75 ft. or less, and 50 lbs. for spans of 200 ft. or more, and proportionally for intermediate spans; to be used as described under (a). " (d) All parts of the floor of a highway bridge should also be proportioned to carry a road roller weighing 15 tons, and having three wheels or rollers, the weight on the front roller being 6 tons, and the weight on each rear roller to be 4.5 tons. The width of the front rollers to be taken as 4 ft. and of each rear roller 20 in.; the distance apart of the two rear rollers to be 5 ft. centre to centre, and the distance between front and rear rollers 11 ft. centre to centre. In using this roller, the fibre stresses allowed shall be 30 per cent above those specified, and, if the stringers are not over 2^ ft. apart on centres, each load shall be considered distributed equally on two stringers.'' Snow and ice load must also be considered in computing the stresses in draw spans when open since such stresses may attain considerable importance. The magnitude of these loads in the vicinity of New York will probably not exceed 10 lbs. per square foot. For fixed span bridges snow need not be taken into account since the maximum wagon and other loads will not occur simultaneously with the snow load. 12. Live Loads for Buildings. The proper loads for buildings depend upon the purpose for which the building is to be used, and in the larger cities is generally prescribed by law. The live loads which follow are the minimum live loads prescribed by the present New York city building laws and represent good practice: FLOORS Dwelling-house, apartment house, or hotel 60 lbs. per sq.ft Office building, first floor 150 Office building, other floors 75 School house 75 Stable or carriage house 75 Place of public assembly 90 Ordinary stores, light manufacturing, or light storage 120 Stores where heavy materials are kept, warehouses, and factories 150 " ROOFS All roofs with pitch less than 20°, 50 lbs. per sq.ft. of surface. All roofs with pitch more than 20°, 30 lbs. per sq.ft. of horizontal pro- jection of surface. 20 OUTER AND INNER FORCES Art. 13 " For columns of dwellings, office buildings, stores, stables, and. public buildings when over five stories in height a reduc- tion of the live loads may be made as follows: " For roof and top floor use full live load; for each succeeding lower floor reduce live load b}' 5 per cent, until 50 per cent of the live loads fixed by this section is reached, when such reduc- tion or such reduced loads shall be used for all remaining floors." For further information upon live loads for buildings the student is referred to the article by C. C. Schneider in the Trans- actions of the American Society of Civil Engineers, Vol. LIV, page 371 et seq., with the ensuing discussion. 13. Wind Pressure. Wind pressure is a subject upon which little exact information exists, although many experiments have been made and much study given to the subject by engineers and scientists. Among the unsettled questions are: a. The relation between pressure and velocity. b. The variation of pressure with size and shape of exposed plane surfaces. r. The direction and intensity of pressure upon non-vertical surfaces. (/. The intensity of pressure upon non-planar surfaces. e. The total pressure upon a number of parallel bars or other members placed side by side. /. The decrease of pressure upon leeward surfaces. g. The lifting powers of the wind. a. In comment upon this subject it may be said that the pressure varies about as the square of the velocity and that the results given by different experimenters vary from p = .00572 to P=. 003272, of which the latter value represents the result of unusually care- ful experiments by Stanton i upon the intensity of pressure on plates varying in size from 25 to 100 sq.ft. and is probably more nearly correct than the higher value. In these formulas P= pressure in pounds per square foot, 7=velocity in miles per hour. ' See Minutes ot Proceedings of the Institute of Civil Engineers, Vols. CLVI and CLXXI. Akt. 13 WIND LOAD 21 In the Stanton formula the values are reduced to correspond to a temperature of 60° F. and an atmospheric pressure of 14.7 lbs. per square inch. b. The vai'iation of pressure with size and shape of exposed surface is important and is not well understood, although it is sure that the resultant pressure on a large surface may be taken as less per square foot than that on a small surface, since the maximum intensity of the wind is due to gusts of comparatively small cross-section. c. The pressure upon vertical plane surfaces may be taken as normal to the surface and equal in intensity to the assumed wind pressure. Upon surfaces which are not vertical, the pres- sure is usually considered to be normal to the surface but lower in intensity than upon vertical surfaces. The variation in pres- sure with respect to the slope is not well understood and a number of empirical formulas are in use, among which may be noted the much used Duchemin formula 2sini ^"-^l+sin^i' ^-^^ and the Hutton formula P„=P(sini)(i-84-=°8.-i)^ (4) in which P„ = intensity of normal pressure upon the given surface, P= intensity of normal pressure upon the vertical sur- face. i= angle made by surface with the horizontal. A theoretical formula may be deduced by making the assumption that the wind always blows in horizontal lines, and that if the pressure be resolved info normal and tangen- tial components, the tangential component may be neglected. Upon this basis the following formula may be derived, using the nomenclature just given P„=P sin2t (5) This formula may be demonstrated in the following manner: Let the wind be assumed as blowing horizontally against the surface ab of height be and making an angle i with the horizontal. 22 OUTER AND INNER FORCES Art. 13 Let the length of the surface be one foot, perpendicular to the plane of the paper. (See Fig. 12.) Let P= intensity per square foot of the horizontal wind force on a vertical surface. Fp= intensity per square foot of the normal force acting on surface ab. Pt =intensity per square foot of the tangential force acting on surface ab. The total horizontal pressure on surface ab then equals Ph. The normal component of this pressure =Ph sin i. The intensity of the normal component = r — • But ab = ~, .-. Pn=Psm^i. smi This formula gives lower values than the empirical formulas (3) and (4) and probably gives too low results since it makes no allowance for the reduction in pres- sure on the leeward side which is known to exist, and which may in part be attributed to the influence of the tangential component. Jt should also be noted that the wind does not blow Fig. 12. uniformly in horizontal lines but may deviate considerably from this direction. The values given by these three formulas are tabulated for comparison on page 23, using an assumed value of 30 lbs. per square foot for P In the absence of further experience upon this phase of wind pressure it would seem wise to use one of the empirical formulas instead of the theoretical one, and the Hutton formiila (4) is used quite generally in England and the United States. The following theorem relating to the wind pressure upon plane surfaces is particularly useful in determining reactions upon roof trusses : The horizontal component of the total normal pressure upon a plane surface equals the intensity of the normal pressure multi- plied by the area of the vertical projection of the surface, and the vertical component of the total normal pressure equals that intensity multiplied by the area of the horizontal projection of the surface. ^^V -] "f ir. Art. 13 WIND LOAD 23 This theorem applies to any surface subjected to a uniformly distributed normal pressure and may be proven as follows: Let Pn = intensity of the normal force acting upon surface ah. Pa= horizontal component of total normal force upon ah. Pr= vertical component of total normal force upon ah. he = vertical projection of ah. ac= horizontal projection of ah. 9 =angle between ah and horizontal. Assume surface ah to be of length unity perpendicular to paper. Then total normal pressure on ab=PnXab. be hence Pa =P„ Xa6 Xsin ^=P„Xa&X^=P„X6c ao Fig. 13. and Pv=PnXahXcose=P„XahX^=P„Xae. ao Since he and ae are the vertical and horizontal projections of ah, the theorem is proven. TABLE FOR WIND PRESSURE. P = 30 LBS. PER SQ.FT. 1 P sin' i p 2 sin i I +sin2 i PCeini) (1.84 cos i-l) Duchemin Formula. Hutton Formula 5° 0.0 5.2 3.9 10° 0.9 10.1 7.3 15° 2.0 14.6 10.5 20° 3.5 18.4 13.7 25° 5.3 21.5 16.9 30° 7.5 24.0 19.9 35° 9.9 25.8 22.6 40° 12.4 27.3 25.1 45° 15.0 28.3 27.0 50° 17.6 29.0 28.6 55° 20.1 29.4 29.7 60° 22.5 • • . . 65° 24.6 Above 60° Above 60° 70° 26.4 use 30 lbs. use 30 lbs. 75° 2S.0 80° 29 . 1 85° 29.7 90° 30.0 24 OUTER AND INNER FORCES Art. 13 It will be observed that if is greater than 45°, Ph will be larger than F„; if less P„ will be the larger. It is obvious that this is correct since the steeper the roof the greater the horizontal component. When ^=90°, Pb=0 and when 0=0, Pa=0. The application of this method to the solution of a problem is given in Art. 25. d. The pressure upon non-planar surfaces is important in the case of chimneys, stand-pipes, and other similar objects. If the assumptions that were made in the deduction of formula (5) be also made for curved surfaces the total pressure upon such surfaces can be easily figured. The following demonstration shows the solution for the case of a vertical cylinder. Let P = intensity of pressure on a vertical plane. P„ = intensity of pressure on a plane making an angle of 90°-/? with the direc- tion of the wind = P sin^ (90° -0). Pi = tangential component of pressure on same plane. The normal pressure on the dif- ferential area ab subtended bv the K a >- h .',ii:!l 1 Ml 'i: ilfrl ll'!" I' angle -d)^4dd. Fig. 14. '=Psin2 (90°- As the tangential component P< is neglected by hypothesis and the component of P„ acting upon surface ab in a direction parallel to ef is balanced by an equal and opposite component upon cd, the force tending to overturn the cylinder is the summation of the components of P„ parallel to gm. and is given by the following expression: / "^ Psin2 {90°-d)h-^dd cos (9=— P Phd f2 COS0 {\-sm^ d)dd- Phd/-i 2 \3 cos3 edd 'rPhd = two-thirds of the total pressure on a plane diametrical section. Aht. 13 WIND LOAD 25 In a similar manner the pressure on a spherical or conical surface may be computed. The pressures obtained by this method lack experimental proof but are probably more nearly correct than the pressure obtained by the same method upon plane surfaces. The value given for the cylinder is quite generally used. e. With respect to the total pressure upon a number of paral- lel bars placed side by side it may be stated that experiments previously referred to indicate that the total pressure on a pair of circular plates placed 1^ diameters apart is less than that on one plate from which the conclusion is drawn, that the pressure on the leeward plate is in a direction opposite to current. When plates were placed 2.15 diameters apart the resultant pressure on the two plates was found to equal that on a single plate and the shielding effect was found to be well maintained with wider spacing, since at a distance of five diameters the total pressure was only 1.78 that on a single plate. /. The pressure upon the windward side of an exposed surface is a function of the density and velocity of the air currents. The pressure on the leeward side is also a function of the shape of the surface, and has been shown by numerous experiments to be less than the static pressure of the air current. The resultant total pressure upon a surface is in consequence not only a function of the direct pressure on the windward side but also of the pressure on the leeward side, which in turn is a function of the form of the surface. It is therefore doubtful if an algebraical formula can be deduced which will give the pressure on surfaces of vary- ing shape with any considerable degree of precision. g. In the case of a very rapid reduction of atmospheric pres- sure, as in a tornado, it ,is often observed that building roofs are lifted and walls blown outward. This phenomenon is due to the air in the building, which is under more or less restraint, changing pressure less rapidly than the outside air and thereby producing a difference in pressure. This lifting action doubtless occurs to a greater or less degree whenever the external pressure is reduced, and should be guarded against by anchoring roofs securely to the walls. The many uncertainties connected with wind pressure make worthless the attempts to specify with precision its magnitude and direction. In the lack of additional information and fur- 26 OUTER AND INNER FORCES Abt. 13 ther theoretical studies there seems to be no reason for devia- ting from the common rules which have been used in bridge design for many years with satisfactory results. These rules as applied to bridge engineering may be stated as follows : The portal, vertical and horizontal bracing shall be pro- portioned for a wind pressure of 30 lbs. per square foot on the surface of the a])plied load, and on the exposed surfaces of the floor system and both trusses. The pressure on the applied load shall be considered as a moving live load, and the other pressure, as a dead load. For structures of ordinary spans the wind stresses shall also be computed upon the unloaded structure for a pressure of 50 lbs. per square foot. In the design the max- imum stress computed by either of the above methods shall be used. The wind stresses in main truss members shall also be com- puted, but if the combined stress in any members due to dead load, vertical live load and wind load does not exceed by more than 20 per cent the allowable unit stress no allowance in the main members need be made for the wind. In computing the area of exposed surface take twice the front surface of members composed of many bars, and 1.5 that of bars in pairs. The pressure upon the ends of ties, and upon the guard timbers should not be neglected and ma}^ be con- sidered as one square foot per linear foot of bridge. For wind pressure on roofs and buildings it is common prac- tice to allow 30 lbs. per square foot acting horizontally upon the sides and ends of buildings, or on the vertical projection of roofs. It is also very important to figure the wind stresses on the steel frame considering it as an independent structure without walls, floors or partitions, since failures often occur in erection. For lateral pressure on steam railroad bridges and trestles, due account should be taken of the sidewise vibration of the train in addition to the wind force. The following paragraphs from the general specifications of the American Railway Engineering and Maintenance of Way Association for steel railroad bridges may be used as a guide. All spans shall be designed for a lateral force on the loaded chord of 200 lbs. per linear foot plus 10 per cent of the specified train load on one track, and 200 lbs. per linear foot Art. 14 SNOW LOAD 27 on the unloaded chord; these forces being considered as moving. Viaduct towers shall be designed for a force of 50 lbs. per square foot on one and one-half times the vertical projection of the structure unloaded; or 30 lbs. per square foot on the sama surface plus 400 lbs. per linear foot of structure applied 7 ft. above the rail for assumed wind force on train when the structure is either fully loaded or loaded on either track with empty cars assumed to weigh 1200 lbs. per linear foot, whichever gives the larger strain. 14. Snow Load. The weight per foot of snow and ice varies greatly with climatic conditions. The following rule suggested by C. C. Schneider in the paper recently referred to gives reason- able results for conditions similar to those existing in Boston and New York: " Use for all slopes up to 20° with the horizontal 25 lbs. per square foot of horizontal projection of roof. Reduce this value by one pound for each additional degree of slope up to 45°, above which no snow need be considered." To determine the maximum stresses in a truss member, wind and snow must be properly combined. The following combina- tions may exist and should be considered : Dead load with snow on both Sides. Dead load with snow on one side and wind on the other. Dead load with ice at 10 lbs. per square foot, properly reduced according to slope, on both sides, and wind on one side. The maximum stress as determined by either of these com- binations should be used. For roof trusses of short span it is becoming the custom to combine the snow, wind, and dead load by using a value suffi- cient to cover them all. The following are suggested by Schneider as minimum loads per square foot of exposed surface to provide for combined dead, wind and snow loads on spans less than 80 ft. These loads are to be taken as acting vertically. Lbs. Gravel or composition, on boards, flat slopes, 1 to 6 or less 50 Gravel or composition, on boards, steep slopes, more than 1 to 6 . 45 Gravel or composition, on 3-inch flat tile or cinder concrete 60 Corrugated sheeting on boards and purlins 40 Slate on boards and purlins 50 Slate on 3-inch flat tile or cinder concrete 65 Tile on steel purlins 55 28 OUTER AND INNER FORCES Art. 15 Where no snow is likely to occur these values are to be reduced by 10 lbs., but no roof is to be designed for less than 40 lbs. For roofs with other coverings than those above use 30 lbs. per square foot of horizontal projection for combined effect of snow and wind on all slopes. 15. Centrifugal Force and Friction. For railroad bridges on curves the effect of centrifugal force must be considered. This may be computed by the following formula: C = 0.03 IFD for a curvature up to 5°, ... (6) where C = centrifugal force in pounds; TF = weight of train in pounds; D = degree of curvature. The coefficient for centrifugal force (0.03) to be reduced 0.001 for every degree of curvature above 5°. On trestle towers and similar structures the longitudinal thrust of the train must be considered. This may be taken as having its maximum value when the brakes are set and the wheels sliding, and may be computed assuming a value of 0.2 for the coefficient of sliding friction. 16. Impact on Railroad Bridges. It is easy to determine the weight per wheel applied to a railroad bridge by the locomotive or cars of a given train when at rest, but when in motion the effect of unbalanced locomotive drivers, roughness of track, flat, irregular or eccentric wheels, rapidity of application, and cen- trifugal force induced by deflection of structure, cannot be deter- mined theoretically and has not yet been precisely determined by experiment. Neither is the distribution of these loads by rails and ties a matter which can be easily ascertained. In consequence the engineer is compelled either to use a low imit stress or to increase the live stresses by an allowance for " impact" sufficient to cover these uncertainties. The latter method is more scientific and is coming into general use. Impact is used in mechanics to mean the dynamic effect of a suddenly applied load, but as used in bridge engineering it stands for the increased stress produced in a member not only by the rapid application of the load, but also by the other causes just mentioned, and the term " coefficient of impact " is given to the factor by which the live stress must be multiplied to obtain the impact. No Art. 16 IMPACT ON RAILROAD BRIDGES 29 rational formula for determining this coefficient of impact has yet been deduced, but several empirical formulas are in more or less common use. It is proven in mechanics that a load when instantaneously applied to a bar produces a stress exactly double that caused by the same load when gradually applied. In the ordinary structure the maximum load is, however, never applied instan- taneously, though in short railroad bridges the length of time required to produce maximum moment or shear is very small. In consequence sudden application alone is never sufficient to double the live stresses as computed for quiescent loads. Many engineers, however, use for short spans a coefficient equal to unity, assuming that the effect of vibration and other uncer- tainties is balanced by the difference between the stress due to instantaneous application and that due to the very rapid but not instantaneous application caused by a railroad train. For longer spans the coefficient is generally reduced. The two following formulas represent two different types of impact formulas: From Specifications of the American Railway Engineering and Maintenance of Way Association, r o 300 From Specifications of the Department of Railroads and Canals, Canada, '-'sTD («) In these formulas /= impact. L = length in feet of distance which must be loaded to produce maximum live stress in member. S = that maximum live stress, and D = th.e dead stress. It will be seen at once that for short spans the coefficient in the first formula is very nearly unity, and that this also applies to the second formula unless the bridge is unusually heavy or the live loads very light. The first formula has been quite generally adopted by American engineers, and while purely empir- ical agrees reasonably well with such experiments as have been 30 OUTER AND INNER FORCES Art. 17 made, and is in a logical form. It will be used hereafter in this book. For full discussion of impact upon railroad bridges with experimental data, see Bulletin No. 125 of the American Railway Engineering and Maintenance of Way Association. 17. Impact on Highway Bridges and Buildings. Allowance for impact upon these structures may usually be less than for railroad bridges. On highway bridges of moderate length the practice of allowing 25 per cent for impact is. not unusual, thi« being intended to cover the jolting effect of wagons on rough pavement, and the impact of electric cars. It should be noted that the loads on highway bridges are probably seldom or never applied with sufficient speed to make rapidity of application an important element, and that the impact of electric cars is not as severe as that of locomotives since the motion of the machinery is rotary rather than reciprocating. For long span highway bridge trusses it is usual to make no allowance for impact. For buildings it is customary to make no allowance for impact, except where moving cranes or other shock-producing machinery are used. Before leaving the subject of impact it should be noted that it is probable that the effect of impact upon wooden beams is less injurious than upon steel beams owing to the greater elasticity of the wood, and that some engineers disregard impact in design- ing wooden structures. 18. Inner Forces. The allowable working unit stresses for a given structure depend upon the material, the character of loading, the precision with which the stresses can be computed, and the uses to which the structure is to be put. If proper allowance for impact be made the character of loading may, how- ever, be neglected. The following unit stresses represent good practice for ordinary structural steel structures, provided proper allowance for impact be made. Aut. 18 WORKING STRESSES 31 WORKING STRESSES. STRUCTURAL STEEL TThimate tensile strength from 56,000 to 64,000 lbs. per sq. inch Live Load to be Increased to Allow for Impact Tension on net section, direct compression, and ex- treme fibre stress in bending 16,000 lbs. per sq.in. I t 16,000 — 70— lbs. per sq.in. Compression in columns .^th a "^maximum of I 14,000 lbs. Shear on net section of plate girder webs and on machine-driven shop rivets 12,000 lbs. per sq.in. Bending on extreme fiber of pins 24,000 ' ' Bearing on pins and shop-driven rivets 24,000 ' ' Bearing on hand-driven rivets 18,000 ' ' Shear on hand-driven rivets 9,000 ' ' Modulus of elasticity 28,000,000 In the expression for the compression in columns — = maximum r value of ratio of the unsupported length of column to radius of gyra- tion, both values being expressed in inches. This ratio should be re- stricted by the form of the column so that it will not exceed 100 for main members and 120 for lateral and other secondary members. The following values, with the exception of that for tension, are recommended for timber for railroad bridges by the American Rail- way Engineering and Maintenance of Way Association, and may be used for green timber and without allowance for impact. For high- way bridges and trestles these figures may be increased by 25 per cent, and for buildings when protected from weather and reasonably free from impact by 50 per cent. For these values and for timber other than yellow pine see "Proceedings of the American Railway Engineering and Maintenance of Way Association," Vol. 10, Part I, p. 564. WORKING STRESSES. LONG-LEAF YELLOW PINE No Allowance for Impact Required Bearing along grain 1,300 lbs. per sq.in. Compression in columns, length over 15 diameters 1,300 ( 1 — ^^TTT ) Tension parallel to grain 1,400 lbs. per sq.in. Bending, extreme fibre stress 1,300 ' ' Shearing along grain in beams 120 ' ' Shearing along grain in chord blocks, etc 180 ' ' Bearing across grain 260 ' ' Modulus of elasticity 1,600,000 — = maximum value of ratio of unsupported length of column to least diam- d eter, both values being expressed in inches. For deflection of yellow-pine beams under long-continued loading, use for modulus of elasticity 800,000 lbs. 32 OUTER AND INNER FORCES Art. 19 The following values for bearing on masonry represent good practice : WORKING STRESSES. BEARING ON MASONRY Live Load to be Increased to Allow for Impact Granite masonry and Portland cement concrete 600 lbs. per sq.in. Sandstone and limestone 400 19. Factor of Safety. The unit stresses given in the previous article are all much less than the breaking strength of the material, the ratio between the breaking strength and the allowable unit stresses being known as the " factor of safety." The necessity for using such a low value is due to the following facts : 1. Material can not be stressed with safety above the elastic limit, which is generally not more than one half the breaking strength. 2. The magnitude, point of application, and distribution of the live loads as well as the allowance for impact is approximate. 3. The material is variable in quality, and may be injured in fabrication. 4. The effect of changing conditions can not be predicted. This applies to character and amount of loading and to deteri- oration of material through rust or rot. 5. The common theories give primary stresses only and neglect 'the secondary stresses due to distortion of the structui-e, these additional stresses being sometimes of considerable im- portance. PROBLEMS 1. Using the Hutton formula, determine the horizontal and vertical components of the total wind force on the side L11U2 of roof truss A, for an assumed wind pressure of 30 lbs. per square foot on a vertical .surface. Direction of wind shown by arrows. 2. Determine the horizontal wind pressure per square ft. required to overturn the car, assuming the direction of the wind to be perpendicular to the side of the car, and the. centre of gravity of the car to be 8 ft. from top of rail. 3. Compute the impact in pounds by formulas (7) and (8) for a bridge member subjected to the following conditions : Maximum live stress 200,000 lbs. Dead stress 100,000 " Loaded length when stress is a maximum. . . . 100 ft. Art. 19 FACTOR OP SAFETY 33 4. Estimate from the diagram of Fig. 8 the total weight of steel in u double-track through pin bridge of 150 ft. span, and determine its total cost, assuming the steel to cost four cents jjer pound, erected and l)ainted. 6. State whether each of the beams shown is staticall}' determined with respect to the outer forces and give reasons. Assume that the magnitude and position of applied loads, and position of points of sup- port are known in all cases. ' Shoe to he faBteued to abutment Peob. !). BUILDING Truss "A" Prob. 1. -10'- Weight of car 20 tons c^r> I r^ K-8M H Punn ■'> Standard i-ROB. _. Gauge. 1 Read Articles 20 and 21 before solving this problem. CHAPTER II LAWS OF STATICS, REACTIONS, SHEARS AND MOMENTS, INFLUENCE LINES 20. Laws of Statics. The theory of structures is based upon the fundamental principles of statics, and these the student must thoroughly understand. For the present only structures in a plane and with the applied loads acting in the same plane will be considered. Such structures will be in equilibrium if the following conditions are satisfied: 1. The algebraic sum of the components of all the forces acting parallel to any axis in the plane of the forces miist equal zero. 2. The algebraic smn of the moments of all the forces about any axis at right angles to the plane of the forces must equal zero. If the forces be resolved into components parallel to two rectangular axes, OX and OY, and the algebraic sum of the forces parallel to OX be designated as HX and of these parallel to F as 2F, the first of above conditions will be fulfilled when SX=0 and I1F = 0, hence the two principles stated above are fully comprehended by the three following equations: SX=0, ^¥-=0, 11M = 0. If the forces acting upon a body do not satisfy all of these three equations, then the body cannot be in equilibrium. For example, if 2X=0 and "LY^O, but SAf does not, the body must be in a condition of rotation about a stationary axis. If SF = and SAf = 0, but SX does not, then the body has a motion of translation in a direction parallel to the X axis but no other motion. In practice it is common to use horizontal and vertical axes for which case the first two equations may be written SH = and SF = 0. 34 Art. 21 REACTIONS 35 21. Reactions, Each of the reactions upon a structure may have three unknown properties, viz., magnitude, direction, and point of application. Usually, however, the point of application of each reaction is fixed in position and the direction of at least one of the reactions is known. If this condition exists when there are two points of support, i.e., two reactions, as is the case in most structures, there remain but three unknown properties of the reactions, all of which may be computed by the three equations of statics, and the structure is statically determined with respect Fig. 15. Fig. 16. to the outer forces, whether it is or is not possible to determine the inner stresses by statics. If there are more than three unknown properties of the reactions, e.g., if only the points of application are fixed; or if the structure is supported on more than two points, then it is statically undetermined with respect to the outer forces, unless some special form of construction is adopted, as in the three-hinged arches and cantilever bridges considered later. If there are fewer than three unknowns, then Fig. 17. Pig. 18. the structure is in general unstable and wiU tend to move bodily under the applied loads unless these fulfil certain special con-. ditions. Illustrations of the above conditions are afforded by the structures shown in Figs. 15, 16, 17 and IS, for all of which the position and magnitude of the applied loads, and all of the dimen- sions of the structure are supposed to be known. 36 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 22 Fig. 15 represents the ordinary masonry arch in which each reaction is unknown in direction, magnitude, and point of appli- cation. In consequence the structure is indeterminate with respect to the outer forces in a three-fold degree. Fig. 16 shows a two-hinged arch which has the point of application of both reactions determined, but the magnitude and direction of neither are known, hence it is indeterminate in the first degree with respect to the outer forces. In Fig. 17 a set of rollers is shown at one end. The function of these rollers is to make the reaction at that end perpendicular to the supporting surface since the rollers, if in good condition, can offer but little resistance to motion along this surface. This structure is, therefore, statically deter- mined with respect to the outer forces since the points of appli- cation of both reactions and the direction of one are known. In Fig. 18 rollers are shown at both ends, hence the direction of both reactions are known. Unless these reactions meet on the line of action of the resultant of the applied loads, equilibrium can not exist and the structure will move, therefore the structure is unstable. 22. Computation of Reactions — ^Method of Procedure. It is evident that if the horizontal and vertical components of a reac- tion which is unknown in direction and magnitude, or if either component of a reaction which is known in direction but not in magnitude be determined, the reaction itself may be at once obtained. In consequence the determination of the reactions in a structure which is statically determined with respect to the outer forces and hence has but three unknowns, may be accomplished by computing the horizontal and vertical com- ponents of one reaction and either component of the other. This method often, though not always, simplifies the solution of reaction problems and will bo used hereafter. Its adoption makes it desirable to use the horizontal and vertical components of the outer forces and these also can frequently be computed more easily than the actual forces. With these components of •the outer forces known the solution of the problem may be accomplished by the application of the three statical equations. The following mode of procedure is suggested for the use of the beginner, who is advised to follow it exactly until he -has mastered the method thoroughly. For structures in which the reactions are not parallel to the forces or in which the character Art. 22 COMPUTATION OF REACTIONS 37 of the unknown reactions can not be easily predicted, even the experienced computor should not omit any of the steps in the process : 1. Draw a careful sketch of the structure and show on it the horizontal and vertical components of the outer forces. This sketch need not be to scale but should not be materially dis- torted. 2. Indicate on the sketch by arrows, and by the letters H and V, the assumed components of the reactions, using letters R and L as suffixes of H and V to indicate right and left reactions. The direction of the components of the reactions which are unknown in direction may be assumed at random, e.g., the hor- izontal component may be assumed as acting either to the right or the left and the vertical component either up or down, but the components of the reaction the direction of which is known must be so assumed as to be consistent with this known direction. 3. Determine the unknown H and V components by the solution of the equations TiH = 0, TiV = 0, and T,M = 0, con- sidering as positive, forces acting upwards or to the right and clockwise moments. A positive result shows that the component in question acts in the direction originally assumed, and not necessarily that it acts up or to the right. With the magnitude of all components known, the magnitude of either reaction may be obtained by computing the square root of the sums of the squares of its two components. Its direction is determined by the direction of the components. The beginner is more likely to make errors bj^ omitting some of the forces than in any other way. Particular attention may well be called to the fact that horizontal forces may produce vertical reactions and vice versa. If the load, or any portion of it, be distributed over a con- siderable distance instead of being applied at a point, the result- ant of this portion of the load may ordinarily be used in the com- putations as a concentrated load. This method, however, should be used only in reaction computations; it would in general be incor- rect for the -determination of shears, moments, and truss stresses. It is also incorrect for the determination of reactions in three- hinged arches. It is always desirable to obtain a check by twice applying the equation, SM = 0, once about each point of support. This 38 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 23 gives an independent check for at least one of the reaction com- ponents, which in the case of a simple beam with vertical loads is sufScient and conclusive. 23. Reaction Conventions. Hereafter, in both text and problems, structures supported at one end upon a set of rollers or by a tie-rod will be considered as having the reaction at that point fixed in direction. The reasons for this in the case of rollers is stated in Art. 21. For the tie-rod, it is sufficient to recall that such a rod is little better than a stiff rope and is incapable of carrying bending or compression, hence the reaction which it carries must act along its axis and produce tension in the rod. Rollers will be indicated by this conventional symbol, /°//°/ and the reaction in this case is always to be considered as per- pendicular to the supporting surface, whether the surface be horizontal, inclined or vertical. When the point of application of a reaction is fixed but not its direction this symbol, j— will be used. This is not intended to represent a knife edge bearing since the reaction may act in any direction, i.e., up, down, horizontal or inclined. If this symbol be combined with rollers, then both point of application and direction of reaction are to be considered as fixed. If the reaction be carried by a tie-rod, the rod will be so marked; in this case the point of application should be taken at the point where the rod is fastened to the structure. 24. Point of Application of Loads and Reactions. In practice it is seldom that the point of application of load or reaction is definitely fixed; it is, however, in many cases fixed within such small limits that no error arises in considering it as located at a definite point. This is the case when the structure is supported on steel pins, as in most bridges of considerable size; the reaction in such a case passes through the pin, which is generally but a few inches in diameter and its resultant will pass through the pin centre, or nearly so, unless the pin be badly turned or the bearing surface upon which it rests imperfect. With wheel loads the load acts at the point of tangency of wheel and bearing surface, which is practically a point, but as the wheel does not rest directly on the structure but has its load distributed by rails and ties, or by the floor if a highway bridge, it is not applied to the struc- Art. 25 SOLUTION OF REACTION PROBLEMS 39 ture itself at a point, though it is generally so considered, as the error thus arising is small and on the safe side. For ordinary beams which rest at the ends upon steel-bearing plates inserted to distribute the load over the masonry supports, the assumption that the reaction is applied at the centre of bearing is by no means an exact one. The actual distribution of the reaction in such a case is a function of the relative elasticity of the beam and support. If both beam and support were to be absolutely rigid — an impossible case — ^the reaction would pass through the centre of bearing; if the support alone were to be rigid the reaction would pass through the edge of the bearing plate; in the actual case where both beam and bearing surface yield to some extent, the reaction is distributed over the entire surface and its intensity varies uniformly or nearly so, as shown in Fig. 19. It will be noticed that the resultant pressure acts at a point between the centre of bearing and inner edge of the masonry. The common assumption for such cases is to assume the reaction as applied at the centre of bearing. This assumption is on the safe side in design- ing the beam as a whole, but on the unsafe side in proportioning the area of bearing. However, the error for short beams which deflect but little, is not serious. For long girders which deflect considerably the end bearing is usually made by a pin which is supported upon a shoe which in turn rests upon rollers, thus ensuring a uniform distribution of the reaction. 25. Solution of Reaction Problems. The application of the methods of Art. 22 is illustrated in the problems of this article. Problem. Compute horizontal and vertical components of reac- tions on beam shown in Fig. 20. Neglect weight of beam itself. Solution. First apply 2^=0. This gives the equation Hj, = 0, since the applied loads are all vertical and in consequence have no horizontal components. Now apply SM=0, taking for origin of moments the point of appli- cation of either reaction, thus eliminating one unknown. The equation which follows is derived by taking moments about the right end. -10,000X26-1-1671,-5000X12-10,000X2=0. .•. 16Fa - 340,000 ft.-lbs. and 7i= -h 21,250 lbs. Fig. 19. 40 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 25 Since the value of Vl is positive the reaction acts in the direction assumed in the figure. The application of S7=0, using the value of Vl just obtained, gives the value of Vr and completes the solution of the problem. The equation follows : -10,000 + 21,250 -5000- 10,000 + yij = 0; .-. 7^ =+ 3750 lbs., and acts upward as shown. To check this value apply SM=0, using the left point of support for the origin of moments. The expression thus obtained is -10,000X10+5000X4 + 10,000X14-167^ = 0; .-. 7ij =+ 3750 lbs., which checks the value obtained by the application of 27=0 and hence checks the value of Vl since this was used in the original determination of 7fl. Problem. Compute horizontal and vertical components of reac- tions on beam, shown in Fig. 21, neglecting weight of beam. /This force will be repla^red ons 14.14 tone „„ Jl_^ in the computations by its components which are also shown , k-HR Fig. 21. Solution. In this problem Vl and Hl are independent of each other in magnitude and direction and each may be assumed as acting in either direction. Vr and Hr are, however, mutually related both in Art. 25 SOLUTION OF REACTION PROBLEMS 41 direction and magnitude since their resultant must act at right angles to the supporting surface, and hence make an angle of 60° with the horizontal. To fulfil this condition if Vr is assumed as upward, Hr must be assumed to the left. The ratio of their magnitude equals the ratio of the sides of a 30° triangle, as indicated by Fig. 22, hence Vr = Hr cot30°=1.73^fi. To solve this problem apply the equation S.l/=0, taking moments about the point of application of the right-hand reaction. The follow- ing equation results : -10X254-2071,-10X16-14.14X10=0. The solution of this gives 7z,= +27.57 tons; .-. 7x, acts upward as assumed. It will be noticed that the lever arms about the origin of moments of all the horizontal forces are zero, hence these terms do not appear in the equation. Had the inclined force been resolved at the top instead of the bottom of the beam, this condition would not have existed, but the value of the reaction would not have been changed since the moment of the horizontal component would have been neutralized by the change in the moment of the vertical component due to its altered lever arm.^ The equations sy=0 may now be used. This gives the following expression, -10+27.57-10-14.14 + 7^ = 0, hence Vr= +6.57 tons and acts as shown. From Fig. 22 it is evident that Hr= 7fltan 30°=0.5777h; :. Hr = 0.577X 6.57=3.79 tons. The application of Si7= completes the solution by giving the value of Hi^. Theequationisi/i — 14.14 — 3.79 =0; hence Hl = 17.93 tons and acts to the right. To check the value of Vr take moments about the left point of support. This gives the following expression: Fig. 22 -10X5 + 10X4 + 14.14X10-207jj = 0, ' The device of resolving a force into its components at a point where the lever arm of one of the components is zero is a very useful one, and frequently saves considerable labor. Its correctness is evident since the eflfect of a force upon a body as a whole always equals that of its components no matter at what point the force is resolved, nor what may be the direction or length of the lever arms of the components, hence if the lever arm of one of the components is zero the moment of the force equals the moment of the other component. 42 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 25 whence Vr^ +6.57 tons, thus checking the value previously obtained, and in consequence the value of Vl- As an independent check of Hr and H^ cannot readily be made a second computation of their value should be carried through, or the original computations carefully reviewed, the former being the safest method. Problem. Compute horizontal and vertical components of the reactions for the truss shown in Fig. 23 for an assumed wind pressure of 30 lbs. per square foot on^ vertical surface. Solution. Since the slope of the roof surface in this problem is about 30°, it will be assumed that the normal intensity of the wind pressure is 20 lbs. per square foot. (See table in Art. 13, Hutton's formula.) The roof trusses are 20 ft. between centres, hence the area of the windward side of the building supported by one truss has a length of 20 ft. Wind pressure 20 lbs. — ■ ^ per eq. ft.nonual to roof surface 12 OOOllis.- =20 1 30 X 20^ Fig. 23. for intermediate trusses, and 10 ft. for end trusses. The reactions upon an intermediate truss will be computed. Using the method of Art. 13 the horizontal and vertical components of the total wind pressure on the windward side are found to be Pft= intensity of normal pressure multiplied by the vertical projection of roof surface = 20X30X20= 12,000 lbs. P^= intensity of normal pressure multiplied by the horizontal projection of surface = 20 X 50 X 20= 20,000 lbs. The truss may now be considered as loaded with the two forces of 20,000 lbs. and 12,000 lbs. acting at centre of windward surface, and the reactions due to these forces computed in the following wav : Applying SM=0 about right end gives 100y£, + 12,000X15-2b,000 X75, whence Vl= +13,200 lbs., acting up as assumed. Applying S/7= gives 12,000 -Hr = Q, whence Hr= + 12,000 lbs., acting to left as assumed. Applying sy=0 gives 13,200 -20,000 + 77e = 0, hence 7b=+6800 lbs., acting up as assumed. Applying 2ilf=0 about left end as a check gives -1007^+20,000 Abt. 25 SOLUTION OF REACTION PROBLEMS 43 X 25 + 12,000X15=0, whence 7ie=+6800 lbs., acting up as assumed and agreeing with value previously obtained. Problem. Compute horizontal and vertical components of the reactions on crane shown in Fig. 24. Neglect weight of structure itself. Solution. The direction of the reaction at the top of the crane is fixed by the tie-rod, hence Vi, and Hi cannot be assumed to act at random but must be so chosen that their resultant will act along the tie-rod. Their magnitude will, of course, be equal since the tie-rod makes an angle of 45° with the horizontal. Applying 2M=0 about the bottom gives -35^i,-l-5000X(20+30) + 20,000X40=0, hence Hl= +30,000 lbs., acting as assumed. Since the two components of the tie-rod stress are equal 7^=30,000 lbs., also acting as assumed. Applying Si7 = 0, using the value previously found for H^, gives 30,000+//k= 0, hence Hr= -30,000 lbs. acting to the riqht and not as assumed. Applying 27 = 0, using the value previously found for Vl, gives -30,000-5000-5000-20,000 + 7;j = O, hence Vr = +60,000 lbs. acting up as assumed. Applying 2M = 0, about the top as a check gives +3577e +5000X20 +5000X30+20,000X40 =0, hence 7/ij=-30,- 000 lbs. checking the value previously obtained. It is always advisable to assume the reactions as acting in their probable directions to avoid complications. The opposite assumption was made for Hr in above problem in order to illustrate the solution with an incorrect assumption. The results will be found to agree in any case, provided the work is correctly done, but it is confusing to have the reaction incorrectly indicated on the sketch. Sometimes, however, it is impossible to foretell the actual direction of a reaction. In this problem the actual value of the reaction at the top should be found, since this gives the tension in the tie-rod. 44 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 26 This value = -r-^ — :: =42,430 lbs. approximately. sin 45° ' ^'^ ^ This should equal V'ViJ' +Hi/, which may be used as a cheek. 26. Shear and Bending Moment Defined. Shearing force or shear at any section of a body is that force which tends to produce slipping along the given section. The bending moment at any section of a body due to a set of co-planar forces is the resultant moment about an axis passing through the centre of gravity of the section, of all the forces on either side of the section, it being understood that the section and the axis are perpendicular to the plane of the forces. Fractures due to shear are due either to transverse fracture of the grains or fibres, or to the slipping of the fibres upon one another. Of the ordinary structural materials wood is the only one of a fibrous character and shearing failures in this material ordinarily occur by longitudinal slipping of the fibres. Fractures due to bending are caused by longitudinal failure of the fibres, either by tension or crushing. 27. Method of Computation, Shear and Bending Moment. The magnitude of the shear upon a given section due to a set of co-planar forces may be readily computed as follows: Resolve each force into two components parallel and perpendicular, respectively, to the given section. The algebraic sum of the components parallel to the section of all the forces upon either side of the section equals the shear. That either side of the section may be considered is evident from the fact that for structures in equilibrium 2 F = 0, hence the algebraic sum of the forces on one side of the section and parallel to the F-axis must be equal in magnitude and opposite in direction to the corresponding term for the other side of the section. The magnitude of the bending moment upon a given sec- tion due to a set of co-planar forces may be computed by resolving the forces into horizontal and vertical components. For this case, however, it is necessary to include the moments of both sets of components, though again it is immaterial which side of the section is considered in computing the moment. 28. Signs of Shear and Bending Moment. The signs for shears and bending moments must be used with care or errors will occur. Any reasonable convention may be adopted, but it Art. 29 SHEAR AND MOMENT, COMMON CASES 45 should be carefully observed that positive shear may represent forces acting in exactly opposite directions and that positive bending moment may stand for either clockwise or counter- clockwise moment, depending in both cases upon the side of the section considered in making the computation. The dis- tinction between the moment of forces in general as used, for example, in determining reactions, and the moment upon a crosa- section of a beam should be carefully observed. In the former case clockwise moments should always be taken as of the same sign, since the effect of such a moment upon the body as a whole is the same no matter upon what part of the body it may act. In the beam, however, clockwise moment upon the left of "a given section produces the same effect upon the fibres as does counter clockwise moment upon the right. In both cases compression is produced in the fibres of the upper portion of the section and tension in those of the lower portion. 29. Shear and Moment, Common Cases. In ordinary prac- tice it is seldom necessary to compute shears or moments, except for vertical sections of horizontal beams and for trusses car- rying vertical loads. For such cases the following conventions may be adopted. Shear. The shear upon a vertical section of a beam or truss equals the algebraic sum of all the outer forces (including reac- tions) upon either side of the section. It is positive when the resultant is upward on the left of the section or downward on the right. Moment. The moment upon a vertical section of a beam or truss equals the algebraic sum of the moments of all the outer forces (including reactions) upon either side of the section, about the neutral axis of that section. It is positive when the moment of the forces on the left of the section is clockwise, or when the moment of the forces on the right of the section is counter-clockwise. 30. Curves of Shear and Moment Defined and Illustrated. A curve of shears or of moments is a curve the ordinate to which at any section represents the shear or moment at that section due to the applied loads. If the load be uniformly distributed the curve may be a continuous smooth curve, a series of smooth curves, or a series of straight lines. If the loading consists of a series of concentrated loads the curve will always be com- 46 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 30 posed of a series of straight lines. If the loading be a combi- nation of concentrated and distributed loads the curves may be composed of a combination of straight and curved lines. It should be thoroughly understood that these curves repre- sent the effect of loads which are fixed in magnitude and position. The shear and moment due to a set of moving loads constantly CURVE OF MOMENTS Weight of beam neglected Fig. 25. Curve a b is a parabola Efiuation is y^ —500x2 Ciu've h C CZ 16 a parabola Equation is y=— 500X^+12 100 (i Fig. 26. -6J (Read Art. 31 before studying this figure). vary and hence cannot be represented by such curves except for a certain definite position of the loads. The effect of moving loads is shown more clearly by influence lines which are explained later. Typical curves of shears and moments are shown in the figures which follow. It should be noticed that in all cases the ordinate to the curve Akt. 31 SHEAR AND MOMENT 47 of shears at any section equals the algebraic sum of the forces acting on either side of the section, and that the curve of moments reaches both its maximum positive and maximum negative values at points where the curve of shear crosses the axis. This latter relation always exists and is demonstrated in Art. 33. The computation of the values of the ordinates to the ciirve of moments at points a, h, c, and d of Fig. 25 are given below for illustration. At a, 7.62X5 = +38.10 ft.-tons. b, 7.62X13-5X8 =+59.06 c, 7.62X19-5X14-10X6= +14.78 d, -5X8 = -40.0 " Note that a point of maximum or minimum moment occurs in all cases where the curve of shears crosses the axis. 31. Shear and Moment. Distributed Load. In determining reactions it has been stated that a distributed load may be replaced by its resultant and the latter used as a concentrated load. This method is incorrect for shear and moment, and should never be used for such cases unless the distributed load lies wholly on one side of the section under consideration. The reason for this may readily be seen. Both shear and moment are functions of the forces on one side only of a section, and all such forces must be included in the determination of either of these quantities. It is evident that if the structure be loaded with a distributed load its resultant may act on either side of a given section, say on the right, while a considerable portion of the actual load may be on the left. If the shear or moment be computed for the forces on the left of the section with the distributed load replaced by its resxiltant, the serious error of neglecting a considerable portion of the loads will be made. For reactions, on the other hand, it is the influence of the load as a whole which is to be considered, hence the result- ant may properly be used. To illustrate the difference between the correct curves of shear and moment for the case of a beam carrying a uniformly distributed load, and the same curves if drawn in accordance with the erroneous assumption that the load may be replaced in magnitude and position by a concen- trated load, see Fig. 27. 48 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 32 32. Shear and Moment. Uniformly Varying Load. It is frequently necessary to determine shears and moments for a beam or girder loaded with a uniformly varying load. Such a condition may occur with a vertical member subjected to Uaifoi-m load 1000 lbs. per lineal ft. Fig. 27. hydrostatic pressure, as in a canal lock, or in a diagonal floor girder in a building. The curves of shear and moment for such a girder are shown in Fig. 28, and the necessary computations follow. Let the load be represented in intensity by the trapezoid abed, the area of which equals the total load on the beam. If the trapezoid be divided into two parts by a line ce parallel to the axis ad the effect of each portion may be treated separately and the problem simplified. Art. 32 SHEAR AND MOMENT 49 s%.S A uuifoi-iuly varying load lege Fig. 2S. Magnitude of force represented by triangle bce = ^s.x20= 8000 lbs. Magnitude of force represented by rectangle adce = 200 X20 = 4,000 " Total load = . 12,000 lbs. = area of trapezoid abed. 50 lAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 32 Reaction. The computation of the reactions should be divided into two operations: the determination of the reactions due to the load represented by the rectangle aecd and the deter- mination of the reactions due to the load represented by the triangle bee. 200 X20 For the first case both reactions = = 2000 lbs. = V. To determine the reactions due to the load represented by triangle hce it is advisable to determine the position of the resultant of this load. This passes through the centre of gravity of the triangle and hence is ^- ft. from the line ah and ^- ft. from ci. The left reaction F^" due to this load may now be deter- mined by applying SM = about right hand end of beam. The following expression results: -8000X Y +^//'X20 = 0, hence 7j;^" = 5333 lbs. The total left reaction, Y ^, therefore equals y^" + 7 = 7333 lbs. To obtain the right reaction apply SF = 0. This gives 7333- 12.000 + Fjj = hence F_„ = 4667 lbs., which may be checked by applying SAf = about the left end of the beam. The curve of shears may now be drawn. Its equation referred to rectangular axes passing through point d with % positive to the left and y positive upwards is 2/= — -1667 +200x+20a;2, in which the term 200a: equals the area of a rectangle of height cd and length x, and the term 20a;2 equals the area of that portion of the triangle, hce, comprehended between its vertex, c, and a vertical line drawn at a distance x from the vertex. This curve cuts the axis at a point 11.1 ft. from right end, as may be seen by placing ?/ = and solving for x. The curve of moments may be obtained in a similar manner. 200a;^ 20a;^ Its equation referred to the same origin is y =46670- -— . This equation may be written directly from the shear equation by multiplying each term in the latter, which represent forces, by the distance of the particular force from the section. Thus, 4667 equals the right reaction and hence should be multiplied by x\ 200a; equals that portion of the load represented by a rectangle extending a distance, x, from the right reaction, and X hence should be multiplied by -; 20a;2 equals that portion of the Art. 33 LOCATION OF ' POINT OF MAXIMUM MOMENT 51 load represented by a triangle of length x, and with its vertex at the right reaction, and hence should be multiplied by 5-. o 33. Location of Section of Maximum Moment. It is a well- established principle of mechanics that the first derivative of the moment equals the shear, hence the moment must have either a minimum or a maximum value at every section where the curve of shears crosses the axis of the beam. The following rule may therefore be stated : The maximum moment always occurs at a section where the curve of shears crosses the axis of the beam; i.e., where the shear equals zero. This rule may also be proven by the use of the theorem of Art. 34, since it is evident that the moment Ma begins to dimin- ish when Sa changes from positive to negative, i.e., passes through zero. The reader will observe that if the equation for the curve of moments in Art. 32 be differentiated with respect to x the equation for the curve of shears will be obtained. In the light of what has just been stated, this is correct, and such a result should always be found. The converse of this is also true, viz. : That the moment curve is the integral of the shear curve with respect to x. It follows that the ordinate to the curve of moments at any section equals the area of the shear curve between the end of the beam and the section. An inspection of the numerous shear and moment diagrams on the following pages will show that this relation occurs in every case. The student in testing this by integration must not forget the constant of integra- tion. 34. Theorem for Computing Moments. In computing moments at a number of consecutive points, as is often necessary in dealing with concentrated loads, the following theorem may be used to great advantage: The moment at any section 6 of a structure loaded with parallel forces, either concentrated or distributed, is equal to the moment at any other section a, at a distance x from 6, plus (algebraically) the shear at a multiplied by x, plus (algebraic- ally) the moment about h of the loads between a and h. This may be expressed as follows: 52 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 35 Let ' Floor beam 200 " " Gilder 300 •• Track (rails, ties etc.) 100 Iba. per ft. per ti'ack 1 J/VHEEL| loads! Positicjn and magnitude o^oads actually applied to gii'der 3000 lbs. 4500 lbs. 4500 lbs. ISOOlba. 3000 Ibg. Uniftirm load4300 Ibs.p'er ft. ;: i — -i i — 15 7^0 lbs. I I ! 15 7^0 lbs. I I DEAD LOADS j ■ Fositiojn and magnitude qf loads actjually applied to gij-der Ordinates ra()=Zm = 12700 1bs. \cd=ik =9750 " demji = 5250 " [fa=yh = 2250 '• CURVE OF MOMENTS Concentrated loads only. Moment at panel points Is same as If the loads were appl led di- rectly to the girder. CURVE OF MOMENTS Dead load only Locus of points a, b, c, d and e Is a parabola but moment curve Is not a parabola. Fig. 38. Note. — Floor beams are ordinarily riveted to sides of girders. Above construction is adopted here for sake of clearness. a load of unity as it moves along the structure. This may be done graphically by plotting a line called an influence line, or analytically by preparing an influence table in which are set 62 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 38 down the values of the function under consideration when the load is at various governing points, such as the panel points of a truss bridge. The following simple illustration shows clearly the character of line and table. Distance of Load from Right Shear at a. Reaction. 1ft. + 1/6 2 ft. + 2/6 3 ft. + 3/6 3.9 ft. + 39/60 4.1 ft. -19/60 5 ft. -1/6 Influence table for shear at a of beam shown in Fig. 39. Fig. 39. — Influence line for simple beam. Shear at a. The influence line is the locus of the values in the second column of the influence table and is merely the graphical repre- sentation of the equation for the shear at a due to a load of unity passing along the beam. If x be the distance of the load from the left reaction and y the ordinate, the equations of the influence line will be as follows: and J/= — ^, X varying between and 2' 6 — X y=—o—, ^ varying between 2' and 6', The difference between an influence line and the curves given in the preceding articles should be carefully observed. A curve of shears, or moments, is a curve, the ordinate to which at any point shows the shear, or moment, at that point caused by a set of loads, flxed in magnitude and position. The ordinate to the influence line shows instead the shear or moment at the section for which the influence line is drawn, due to a load of unity acting at the point where the ordinate is measured. The examples in Art. 39 serve to illustrate influence lines for the more common cases of simple beams and girders. Art. 39 EXAMPLES OP INFLUENCE LINES 63 The actual employment of influence lines and tables in prac- tice seldom occurs except for complicated structures where they are frequently almost indispensable. In this book the influence line will, however, be used with freedom, partly for purposes of illustration and demonstration and partly that the student may better familiarize himself with the behavior of various structures under moving loads. 39. Examples of Influence Lines, a. Simple Beams and Girders. "1" L a ^^^ ^^ ..t3^ 1 1 1^ ■e- X -L =>j Fig. 40. — Influence line for shear Fig. 41. — Influence line for moment at section a. at section a. I^^r^ -«— LrJ !Hrfrq2f t^ h'^-^''--" -Ui^*- ■>f— 1-3.J Fig. 42. — Influence line for shear Fig. 43. — Influence line for moment at section a. at section a. 3; TT Fig. 44.— Influence line for shear at section a. ff Fig. 45. — ^Influence line for moment at section a. b. Girders with Loads Applied through Floor Beam^, as in Fig. 46. Note that the usual form of construction for such bridges 64 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 39 is that in which the floor beams are riveted to the girder webs and the stringers to the floor beam webs. The type shown in iC I T r Panel 1 Ponel 2 Panel 3 Panel 4 i5s — i panels @ p ^ SIDE ELEVATION Fig. 46. pi-awe— i)—i»ta-»t , i t t i, CROSS SECTION Fig. 47. — Influence line for shear in panel 1. 1 / •r-j /______ — i- Fvi Fig. 48. — Influence line for shear in panel 2. Fig. 49. — Influence line for shear in panel 3. Fig. 50. — Influence line for shear in panel 4. Fig. 51. — Influence line for moment Fig. 52.^Influence line for moment at panel point a. at panel point b. Art. 40 PROPERTIES OF THE INFLUENCE LINE 65 the figure is chosen here for clearness in presentation. The influence lines, moments, shears, etc., would be identical in the two cases. Fig. 53. — Influence line for moment at panel point c. Fig. 54. — Influence line for moment at center of panel 2. 40. Properties of the Influence Line. The following theorems may often be used to advantage. 1. The value of a given function due to a single load in a fixed position equals the product of the magnitude of the load and the ordinate to the influence line measured at the point where the load is placed. This needs no proof, but follows directly from the definition of the influence line. 2. The value of a given function due to a uniformly dis- tributed load equals the product of the intensity of the load and the area bounded by the axis of the beam, the influence line and the ordinates drawn through the limits of the load. If this area be partially positive and partially negative the algebraic sum of the two should be used. Fig. 55. This theorem may be proven as follows: Let bee represent an influence line for a portion of a given structure of length L. Let w equal the intensity of a uniformly distributed load. 66 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 41 Then the total load on a section of length dx = wdx and the effect of this portion of the load upon the given function = 'U)2/da;. Integrating between the limits and L gives w I ydx the entire distance L. ] for the effect of a load covering the entire distance L. But ydx is the area of the infinitesimal strip subtended by dx, and ( ydx is the area abecd, hence w I 2/da; = wXarea abecd. Jo 3. The value of a given function due to a set of concentrated loads equals the algebraic sum of the product of each load and its corresponding ordinate to the influence line. This is a corollary of 1. 41. Neutral Point. The influence lines shown in Figs. 47 to 50 inclusive cross the axis of the beam in each case except for shear in the end panels. The point of intersection is called the neutral point since a single load placed at this point pro- duces no shear in the panel where the intersection occurs. The neutral point for the end panels is at the ends of the beams. 42. Position of Loads for Maximum Shear and Moment at a Definite Section. The following important laws may be deduced from the influence lines given in Art. 39. 1. For a simple beam supported at the ends a single concen- trated load causes maximum shear at a section when placed an infinitesimal distance on one or the other side of the section, and maximum moment when placed at the section. A uniformly distributed live load produces maximum shear at a section when applied over the entire distance between the section and one or the other end of the beam, and maximum moment when applied over the entire length of the beam, 2. In a girder or truss loaded by means of floor beams, a single concentrated load produces maximum shear in a panel when placed at one end of the panel, and maximum moment at a panel point when placed at that point. A uniformly distributed live load produces maximum shear in a panel when applied over the entire distance between the neutral point of that panel and one, or the other, end of the structure, Art. 43 MAXIMUM MOMENTS AND SHEARS 67 and maximum moment at any point when applied over the entire length of the structure. 43. Maximum Moments and Shears — Structures Supported at Ends. In the preceding article moments and shears at particular sections have alone been considered, and no attention has been given to the maximum values of these fimctions. These max- imum values must, however, be computed before the structure can be designed. For single concentrated loads and for uniform live load the value of these quantities can be easily determined as follows, for beams supported at ends. Case 1. Maximum shear, single concentrated load, beam without floor beams. The influence line shows that the maxi- mum value of the ordinate occurs either when x = L, or L—x = L, and equals unity, hence the maximum shear due to a load P, occurs with the load at either end of the beam. Its value equals P. Case 2. Maximum moment on beam under same conditions as Case 1. Here the ordinate to the influence line is a maximum at the load and equals -j- {L—x). This can be easily shown to ij be a maximum when x=L—x, hence, the maximum moment due to a load P occurs when the load is at the centre of the beam. Its value is — r-. 4 Case 3. Maximum shear on same beam due to a uniform live load of intensity w. It is evident that the area between the influ- ence line and the axis will be a maximum if section a is at either end, hence the maximum shear equals— jr-. Li Case 4. Same as Case 3, but maximum moment instead of shear. The maximum moment occurs for load over entire beam, and occurs at the section where the ordinate is a maximum, which has already been shown in Case 2 to be at the centre. The moment at the centre equals \w'L?. Case 5. Maximum shear. Single concentrated load. Girder with floor beams and equal panels. The maximum evidently occurs in the end panel; its value depending upon the number of panels. If n equals number of panels and P the load the , P(n-l) maximum shear = . LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 43 -^ (n — 1) where p Case 6. Same as Case 5, but for uniform load w per foot instead of concentrated load. Maximum shear occurs in end panels and with a load over the entire structure. Its value is panel length. Case 7. Same as Case 5, but maximum moment instead of maximum shear. Place load at panel point nearest centre. Maximum moment occurs at this panel point and equals -- J ( ^ I if number of panels is even, and -j^ {n^ — 1) if number of panels is odd. Case 8. Same as Case 6, but maximum moment instead of maximum shear. Maximum moment occurs at panel point nearest centre with load over entire span. Its value is ^wL^, when number of panels is even and ^wL^l 1 — ^ j; when number is odd. In deriving these two quantities the following theorem may be used: " The moment at a panel point of a girder with floor , beams equals that at the cor- responding point of a simple \ beam under the same load." I The proof of the theorem is as follows: Let Fig. 56 StrlDgei's f j5 j 9f- FiG. 56. represent a portion of a girder carrying floor beams. Let Mi, = moment at panel point b. Ma = moment at panel point a. jS = shear in panel to left of given panel. Then in accordance with rule given in Art. 34 Mi = Ma+Sp-p(-]p = Ma+Sp-Px This is also the value of the moment at b with the load P applied directly to the girder at the point c. Of the formulas in this article the student is advised to Art. 44 APPROXIMATE METHOD FOR MAXIMUM SHEAR 69 memorize that for maximum moment at the centre due to a uniform load, viz., Af = iwL2. (10) This formula is applicable not only to simple beams, but also to girders with floor beams provided the number of panels is even. Since the moment at a panel point equals that at the cor- responding point of a simple beam under the same load, the locus of the moments at the panel points for a uniform load over the entire beam is a parabola, with a centre ordinate equal ^wIJ^, hence the ordinate at any panel point of a girder with an odd OHgin of iiarabola iHorizoDtal axis of pacabola Fig. 57. number of panels may be deduced from this value by applying the equation of a parabola. This is illustrated by Fig. 57. The ordinate, y, equals \wL^[l — 'f ^ j = maximum moment on girder. 44. Approximate Method for Maximum Shear. In practice it is common to determine the maximum shear produced by a uniform load on an end supported girder with floor beams by the following approximate but safe method. Compute the maximum positive shear in a panel as if all panel points to right were loaded with full panel loads and panel points at left with no load; for maximum negative shear reverse this process. This method is illustrated by the following example: Let the problem be the determination of the maximum positive shear in panel cd of the girder shown in Fig. 58 due to a uniform live load of 3000 lbs. per foot. 70 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 44 By the approximate method the shear should be computed for full panel loads at d, e, and /, and no loads at b and c, and will therefore equal ( ^"^^^^^ ^45,000 lbs. =45,000 lbs. By the exact method the girder should be loaded from the right end up to the neutral point, m, in panel cd. From the similar triangle of Fig. 58 it is evident that md nk id ik i+i but nk = cd=15'. .-. md = 9', hence the area of the triangle mt5r = 54Xi = 13.5. Since the maximum shear equals the area of the triangle mig multiplied by the intensity of the load per linear foot, its value is 3000X13.5 = 40,500 lbs., or considerably less than the value obtained by the approximate method, a relation which will always occur for the intermediate panels of an end supported girder. For the end panels the neutral point occurs at the end of the panel, hence for such panels the exact and approximate methods give identical results. Art. 44 REACTION PROBLEMS 71 PROBLEMS In Problems 6 to 22 inclusive, compute the horizontal and vertical components of each reaction. 10 tODB 20 tODS J- ^10^— H»-6i<» IS)'- — litZil Prob. 6. Pkob. 7. I (ODD II tonB 10 ODS 10 bma 20 lonfl |«aO4t«0^0^Off-l£^16il Prob. 8. WindjircBBUTB _ SU^sq.ft. "^ -formal to , Burfaoe 30 U- 6'-*K- 5'-»4- 1 0^ — »1*-B^ Prob. 9. Prob. 10. Prob. 11. Thia truss is an intermediate truss of a series. Trusses spaced 20' between centres. SOOOOff 15 00* BOOot^ * '"'' ' 10 ooot A.^-''^^- 1 s 10 ooot^ L^''^^^^^^ ® 60O04^./^^-^^*~^'jl.!^ I ^ _ 4 k, Pbob. 12. 10 OM ID tons ■ ^Olfr, 1 10 one ^^ 1 1 1 1 _ i_ 1 IspacesOU)^-^ Pbob. 13. £.f0?a 10 tons 10 tons 10 to°B 10 tooB Wt. of car =10 tons. Car is held in equilibrium by rope. ^ Neglect friction ^-^ V' between car ^ and truss. > ■'^"' Pbob. 15. ona 1(rtoDB S 10 toss Peob. 14. wid BideofthetoverSy J^ Pbob. 16. 72 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 44 Wlnd=201bB. per gq, ft. normal to inclined roof Burfaoe and bO lbs. per Bq. ft. noTmBl, to Tertloal surface Prob. 17. This truss is one of tfie end trusses of a series. Dis- tance apart of trusses equals 20' centre to centre. 20 tons : K — 3-panelB ® 2 Prob. IS. Wind pressure same as in Prob. 17 Prob. 19. Prob. 20. Prob. 21. This truss is one of the inter- This truss is an end truss of mediate trusses of a series. a series. Distance apart of Distance apart of trusses trusses equals 20' centre to equals 30' centre to centre. centre. lnUnBrtr-=f2000 IbB. per ft. IntenBlty=1000 lbs. per ft. Uoirormlj yurjing Prob. 22. Prob. 23. Kod of etringei ia supported and not fixed ' Diulanue froni a to point of supj)0rt.=12 ' i!-4 1 }^ be d J<- StringOTB Floor QcamB 1 panels @ 1^^ 5 Prob. 24. 23. a. What is the magnitude of the shear at sections a and c witl: a concentrated load of 10,000 lbs. at b? 6. What is the magnitude of the shear at sections a, b and c witl a uniform load of 1000 lbs. per linear foot over the entire beam? ABT. 44 SHEAR AND MOMENT PROBLEMS 73 24. o. Where should a single concentrated load be placed to cause maximum shear in panel del In panel a6? h. What is the magnitude of the shear at section c of the girder with a single concentrated load of 20,000 lbs. applied to the stringer at the centre of panel hdt 25. (In the following problems, relating to curves of moments and shears, and to influence lines, positive values should be plotted above the axis, and numerical values given for ordinates at all points where the curves change direction.) Plot the curve of shears for beam shown in Prob. 23 with a uni- formly varying load extending over the entire beam. Intensity of load at free end of beam 2000 lbs. per foot; at fixed end, 1000 lbs. per foot. 26. (See Prob. 24 for figure for this problem.) a. Plot the curves of shears and moments for a uniform live load of 1000 lbs. per foot extending from the free end to the centre of panel ah and applied to the stringers. h. Compare the moment at each floor beam for the loading stated in a with that which would exist if there were no floor beams and the same load were applied directly to the girder (i.e., a uniform load of 1000 lbs. per foot, extending 42 ft. from the free end of the girder). 27. a. Draw curves of shear and moment for one girder. 6. Draw similar curves for a uniform load of 3000 lbs. per foot applied to the stringers and extending over entire span, and compare the moments at the floor beams with those which would occur at similar points if the load were applied di- rectly to the girder. c. Determine position of a single con- centrated load for maximum shear at section a. For maximum moment at same section. Load to be applied at the stringers. d. Draw the curves of dead moment and shear for following assumed weights : Stringers, 300 lbs. per foot per stringer (this includes weight of bridge floor) . Hoor beams, 100 lbs. per lineal foot of floor beam. Girders, 200 lbs. per lineal foot per girder. 28. (See Prob. 23 for figure for this problem.) a. Plot the influence lines for shear at sections a and h. b. Plot the influence lines for moment at sections a and b. 29. (See Prob. 24 for figure for this problem.) a. Plot the influence lines for shear in panel ab and in panel ef of girder. A.ssume girder to be directly under a stringer and load to be applied at the stringer. i: <3fi< 6- UsS CROSS SECTION 20 000 IbB. n B. h ) 000 IbB. eninB 1^ |<-6'*1 ELE -4a>anels@14 ' =56'- Prob. 27. 74 LAWS OF STATICS, REACTIONS, SHEARS, ETC. Art. 44 6. Plot the influence lines for moment at sections a and d. c. From an inspection of the influence line determine over what portion of the beam a uniform load should extend in order to produce maximum shear in panel ab, and compute the magnitude of this shear, assuming the uniform load to equal 1000 lbs. per linear foot and to be applied at the stringers. d. Same as c, except substitute moment at section a for shear in panel ab. 30. a. Plot the influence lines for shear at sections a, h and c. b. Plot the influence lines for moment at sections a, b and c. c. From an inspection of the influence lines determine where a single load should lie to give maximum shear at section c. To give maximum moment at section a. d. From an inspection of the influence lines determine what portions of the beam should be loaded with a uniform load per foot to give maxi- mum shear at section c. To give maximum moment at section a. 1 1 i J J rT~T J< 7 panels @ 12' >j I Prob. 30. Pkob. 31. Sections a and c are to be assumed as an infini- tesimal distance from the adjoining point of support. e. Compute the maximum shears at sections a and c due to a uniform live load of 2000 lbs. per foot, and state in each case whether the shear is positive or negative. /. Compute the maximum moments at sections a and 6 due to the load given in e and state whether positive or negative. 31. a. Plot influence lines for shear in panels 0-1 and 1-2. Make same assumption as to relative position of stringers and girders as in Prob. 29, and assume loads to be apphed at stringers. 6. Plot influence lines for moment at sections 1 and 2. c. From an inspection of the influence line determine where a single concentrated load should lie to cause maximum positive shear in panel 1-2 and maximum positive moment at section 2. d. Compute by the " influence-hne method" the exact maximum positive shear produced in panel 1-2 by a uniform live load of 2000 lbs. per foot, and check this result by computing the shear analytically. e. Compute the maximum positive live shear in panel 1-2 by the approximate method given in Art. 44. CHAPTER III CONCENTRATED LOAD SYSTEMS 45. Shear at a Fixed Section. Girder without Floor Beams. To determine the position of loads which will produce maximum shear at a given section of a simple end-supported beam or deck-girder a method of trial may be employed. Stated briefly this consists of moving the loads along the beam by intervals corresponding to the distance between wheels and writing expres- sions for the change in shear thus produced. This process is repeated until the shear is found to decrease. This method is based upon the fact that the maximum shear at a given section of a simple beam carrying concentrated loads occurs with one of the loads at an infinitesimal distance from the section. The proof of this proposition and the application of the method to a definite case will now be given. Loads are in thousands of Ibs^per wheel. k— 8^->'<-5W^Vk-5U«— 9^->f<^%-t<5->l<-5->i till III' I I I - I I 111 Fig. 59. Let Fig. 59 represent a typical set of concentrated loads, in this case a single consolidation locomotive, and let it be desired to compute the maximum shear at section a, for the beam shown in Fig. 60. The influence line for the section is shown in Fig. 60 and shows clearly that for maximum positive shear at section a most of the heavy' loads must be to the right of a. 75 76 CONCENTRATED LOAD SYSTEMS Art. 45 To prove that one of the loads should lie an infinitesimal distance to the right of the section, or practically at the section, proceed as follows : Suppose the loads to be on the beam as shown in Fig. 61. Fig. 60. As the shear due to a set of concentrated loads in any position equals the summation of the product of the loads and their ordinates, it is evident that starting with loads in the position Fig. 61. shown in Fig. 61 the shear at a will be increased by moving the loads to the left until load (1) reaches the section. If the loads are moved still further until load (1) passes to the left Fig. 62. of the section there will be a sudden decrease in the shear due to load (1) crossing the section. The new position is shown by Fig. 62 from which it is again evident that if the loads be moved Art. 4.') SHEAR AT A FIXED SECTION 77 still further to the left there will be an increase in shear until load (2) comes to the section, and that the result of load (2) crossing the section will be another sudden decrease in shear, after which the shear will again increase till another load reaches the section, and so on. It is also clear that the effect of a load coming on the span at the right or going off at the left during the process of moving up the loads will not affect the above conclusions. Fig. 63 is a graphical illustration of the changes in shear at a of the beam shown in Fig. 60 as the loads move to the left. The ordinates represent this shear with load (1) at the point |.34Uj<— 8^-4«4i<-5^5^j<-rA|< — ii^-»i<-6i4e-6^»|«^ Fig. 63. where the ordinate is shown. In consequence the line 1-2 shows the increase in shear at a due to moving load (1) on the span until load (2) reaches the right end; 2-3 shows the increase due to moving to the left the first two loads until load (3) reaches the right abutment, and so on up to 9-10, which shows the effect of moving the first 9 loads, i.e., all the loads, until the first load reaches the section a. When the first load crosses the section the shear drops suddenly by 10,000 lbs. and then increases again, as shown by 11-12, until the second load reaches section a. As this load crosses the section the shear is diminished by 20 000 lbs. and then increases, as shown by 13-14, until the first 78 CONCENTRATED LOAD SYSTEMS Art. 45 load passes off the span. This produces no sudden change in shear but changes the slope of the line, as shown by 14-15. From the preceding discussion it is evident that the following method may be used to determine the location of locomotive loads for maximum positive shear at any section of a simple beam: Starting with all the loads to the right of the section and with load (1) at the section, write an expression for the change in shear due to moving load (2) to the section. If this expres- sion shows a decrease it is evident that load (1) at the section gives the maximum shear. If, on the other hand, the expression shows an increase it wUl be necessary to write another expres- sion for the increase due to moving up load (3) and so on until a decrease is finally obtained. In practice the operation is simple, as is shown by the fol- lowing example for the beam and loads of this article. It will be noticed that instead of writing an equation for the change in shear the method used is to write an inequality one side of which shows the increase in the left hand reaction due to moving up those loads which are on the span to begin with and remain on or which come on during the process of moving, and the other side of which shows the effect of a load crossing the section or going off the span to the left. The numerical expressions for the case in question will now be given. With (1) at section move up (2). 146x|>10. As the left hand quantity is greater than the right it is evident that the shear is increased by moving up load (2), With (2) at section move up (3). (146-10)^+5 + (10)^<20. Since in this case the left hand side of the inequality is less than the right hand it is evident that there is no further increase and that the maximum shear will occur with load (2) at section a. As the left hand side of the above expression may not be entirely clear a few words of explanation may be added. The first term shows the increase in the left hand reaction due to Art. 46 MOMENT AT A FIXED SECTION 79 moving up those loads which are on the span to begin witli and which remain on the span. The second term, o, represents the slight increase in the shear at the section due to loads whicli may have come on the span at the right end of the bridge during the process of moving up the loads. This term is always small and may generally be ignored. Its value in the present case is 0. The third term gives the shear caused by load (1) when load (2) is at section a. This shear being negative and disappearing during the movement on account of the load going off the span, an increase in shear is obtained which is, therefore, placed on the left-hand side of the inequality. Having in above fashion determined the position of the loads for maximum shear, it remains simply to compute this shear in the ordinary manner by figuring the left-hand reaction and subtracting therefrom the loads between it and the section. 46. Moment at a Fixed Section. The method of determining the position of loads for maximum moment differs somewhat from that used in determining the position for maximum shear, and is as follows: Let the original position of the loads be as shown in Fig. 64. Fig. 64. Let JM= increase in moment at m due to moving all the loads a short distance d to the left. Then, since the change in the moment at m caused by the movement of the load system equals the summation of the product of each load by the change in length of the influence line ordinate corresponding to that load, the following expression for the increase in moment may be written: JM={P2+P3+Pi+P!i)dtsind-{Po+Pi)dtixnB = {P2+P3+P4+P5)dj^-{Po+Pi)d^ AMPj+Pz+Pj + Ps Po+Pi hd L—a a 80 CONCENTRATED LOAD SYSTEMS Art. 46 This equation shows that the moment at m will be increased by moving the loads to the left if the average load per foot on the right of m be greater than the average load per foot on the left. The converse of this proposition is also true. It should be noted that if the average load per foot on the right equals the average load per foot on the left there will be no change in the moment by moving the loads. The above equations are true, provided the relative position of the loads does not change; that is, if no load comes on from the right, or goes off to the left, or passes the section. It may be readily seen, however, that if the average load per foot on the right exceeds that on the left a movement to the left suffi- cient to bring another load on from the right or to cause a load to go off to the left serves to increase the moment more rapidly, and hence does not vitiate the conclusion that the loads should be moved to the left. It is also evident that the movement to the left should be continued until P2 reaches the section, hence we have the following theorem: For maximum moment at any section one load must lie at the section, and the loads must be so located that with that load just to the right of the section the average load per foot on the right is greater than that on the left, while with that load just to the left of the section the average load per foot on the left is greater than that on the right. A special case of the above is when the average load per foot on one side equals the average load per foot on the other side. In this case a load does not have to lie at the section, but if it does lie at the section the moment will be equal to the maximum, hence the theorem applies for this case also. It should be noticed that the proof of this theorem would be equally applicable to any case where the influence line is composed of two straight lines, and that in consequence the theorem is very useful for many cases other than that of moment on a simple beam. The application of this theorem is simple, but it sometimes happens that several loads of the same system will be found to satisfy the above criterion. This is explained by the fact that a different set of loads may be on the span for each position and consequently there may be several maxima. In such cases it is usually necessary either to compute the value of each Art. 47 SHEAR. GIRDER WITH FLOOR BEAMS 81 maximum, or else to compute the change in moment due to moving the loads from one maximum position to another. A numerical example of the determination of the position for maximum moment will now be given. Let the loads and span be as in Art. 45, and let the problem be to find the position giving maximum moment at a. ,-,. . ^ , , „ Av. load per Av. load per i irst try load 2 : ft. on left. ft. on right. Load (2) to right of section — < Load (2) to left of section — > Load (2) gives a maximum: 10 ^ 50 30 116 10 ^ "50 „ , , ,„, Av. load per Av. load per lryload(3): ft. on left. ft. on right. Load (3) to right of section . ^ < Load (3) to left of section — > ft. on righ 116 10 ^ 50 40 96 10 ^ 50 Load (3) also gives a maximum. It is seen by inspection that in this case it is unnecessary to try load (4) and that loads (2) and (3) are the only ones giving maximum moments. To determine which of these is the larger it is advisable to compute both independently and then check the results by computing the change in moment produced by moving from load (2) at a to load (3). That the maximum moment at a given section due to a set of concentrated loads always occurs with a load at the section is apparent from the fact that the maximum moment for a given position of loads occurs where the shear curve crosses the axis; i.e., where the shear equals or passes through zero, and that this can never happen except at one of the loads. 47. Shear. Girder with Floor Beams. For such girders the maximum shear in every panel must be computed. The method of determining the position of the loads differs in detail from that given in Art. 45, although the same general method may be used. To illustrate this case the bridge shown in Fig. 65 is chosen. Here again, for greater clearness, the stringers and floor beams 82 CONCENTRATED LOAD SYSTEMS Art. 47 Po, Pi and P2are floor beam reactions. These vary in magnitude for different positions of the loads. ' are shown above the girders, though, as already explained, such construction is uncommon. End floor beams are also used, but this makes no difference in the method or its application. Consider first the position of loads for shear in panel 1. In this case it is clear that the maximum shear occurs with the same condition which would produce maximum moment at panel point a, since the proof given in Art. 46 applies equally well here. In consequence, one of the loads must lie at a. To determine which, either the method of moving up the loads explained in Art. 45 may be used, or that of ob- taining the position of the loads for maximum moment at a. If the latter plan be adopted it may happen that more than one position will be found to give a maximum, and hence an extra computa- tion will be needed. This latter is, however, useful as a check and is not a conclu- sive argument against the method since an approximate check computation with an- other load at the section should invariably be made. The fact that the maxi- mum shear in the end panel and the maximum moment at the first panel point occur simultaneously is important. It follows that since none of the live loads can be applied to the girder between panel points, the maximum live moment at the first panel point equals the product of the maximum live shear in the end panel and the length of that panel. For intermediate panels the latter method can not be used since it is incorrect, except for cases where the influence line Tnfluegce line — shear in pajiel cross section Fig. 65. AST. 47 SHEAR. GIRDER WITH FLOOR BEAMS 83 is composed of two straight lines forming a triangle with the axis of the beam, and for intermediate panels this condition does not occur. For such panels therefore the method of Art. 45 wUl be adopted. Examination of the influence line shown in Fig. 65 for the shear in the second panel, which is typical of the influence lines for all intermediate panels, shows that the loads when brought on from the right must at least be moved to the left until the first load reaches b. Further movement to the left will cause a decrease in the shear due to the first load, but an increase due to the loads on the right. If the result is a net increase, the loads should be moved until load (2) reaches b. This conclusion is uninfluenced by the action of other loads which may come on the span from the right, or by the fact that load (1) may pass a. Further movement to the left produces an additional increase in shear due to loads to the right of b, but a decrease due to load (2), and either an increase or a decrease due to load (1). If the expression for the change is positive it will remain so until load (3) reaches the section unless load (1) passes ofl the bridge, which would lower the rate of increase and perhaps cause a decrease. This condition is, however, not likely to occur and may be neglected. In view of the foregoing it may be stated that for maximum positive shear in either end or intermediate panels, one of the loads must lie at the panel point to the right. Before proceeding to a numerical illustration of these prin- ciples, the student should observe that the increase in shear in panel 1 equals the increase in Rj^ minus the increase in P^; that the increase in shear in panel 2 equals the increase in R^ minus the combined increase in Pg and Pi, and similarly for other panels. A load which passes off the span in the process of moving up should always be considered by itself. It should be noted that the change of shear, or of any other function, due to removing a load from a structure, is equal to the shear or other function caused by the load when on the structure. Hence, to find the change in shear due to a load passing off the span compute the shear due to it in its original position before the loads are moved. The application of these principles to the structure shown in Fig. 65 will now be given for the locomotive shown in Fig. 59. 84 CONCENTRATED LOAD SYSTEMS Akt. 47 Shear in End Panel. Method of Moving^p the Loads. Start with load (1) at panel point a. Increase in Rl Increase in Po 8 8 Move load (2) to a, UQX-^+S > lOX^^ .-. shear is increased. 3 = in this case. Move load (3) to a, 136X^+5 > 20X^ + 10Xj2>^5- * .-. shear is increased. 5 = in this case. The fact that the increase in moving up load 3 is very sUght and that the next step of moving up load -1 would materially in- crease the change in Po without increasing that in Pi makes it evident that load (3) at section gives the maximum shear. The value of the maximum shear in the end panel may now be computed. The expression for it is 90 14 5 g(53 +48 +43 +38) +gQ(29+24 +19 + 14) -20X^2 = 73 (thousand lb. units.) To show that above conclusions are correct the shear with load (2) at a will be computed. |5(48 +43+38+33) +^(24 + 19 + 14+9) +10X^x| = 72.07 (thousand lb. units.) The value of this is less than that for load (3) at section in accordance with the conclusions of the previous method. * The last term in the above expression gives the shear due to load (1) when load (2) is at a. Its value is obtained by computing the floor beam reaction Fj and the shear due to it. The reaction Pq may be ignored since it produces no shear in the girder. The same result should be obtained by the usual method of computing TJ^, and subtracting Pq from it; thie gives /56 8 (-; ) 10, which equals the value already found. Art. 47 SHEAR. GIRDER WITH FLOOR BEAMS 85 Shear in End Panel. Average Load Method. Av. load per Av. load per ft. on left. ft. on right. Load (2) to right of panel pt. a, — < ^ 1 t j /on ■ 1 4; I •. Load (2) gives a max. Load (3) gives a max. Load (2) to left of panel pt. o, — > — — Load (3) to right of panel pt. a, — - < — 1 4 40 96 Load (3) to left of panel pt. a, — > — T J rA\ i. ■ ui. c 1 J. 40 ^ 96 .-. Load (4) does Load (4) to right of panel pt. a, -- > --- . ' 1 4 not give a max. From these expressions it is seen that by the application of the average load criterion, loads (2) and (3) are found to give maxima, and that it is necessary to calculate both to determine the greater. It should be noted that in the application of the average load method the average shear per panel instead of the average shear per foot has been used. This is simpler and gives the same result when the panels are of equal length as in the bridge under consideration. If the panels are of unequal length this method would be incorrect. Shear in Second Panel. Method of Moving up the Loads. Start with load (1) at panel point b. Increase in Rl Increase in (Po+^i) Move load (2) toft, 104 X~+^ > 10 X^^, ■'■ ^^^^"^ ^^^"■ 60 12 creased 9 . . . 5=14X;T7rbut it IS evident that this value need not have been 60 computed since it is too small to alter the result. Move load (3) to b, Increase in R^^ Increase in (Pj +Pi) 132x|:+^< 10xl+20X^ .-.Load (2) 60 12 12 gives a maximum 2 5=14X^ (necessary to compute in this case since otherwise results would be doubtful). 86 CONCENTRATED LOAD SYSTEMS Art. 48 The right-hand side of above inequality may require some explanation. With load (2) at b, Po = and ^1 = -^^ 10. With load (3) at b, Po = ^ 10 and Pi = -^ 10 +-| 20. .-. Increase in (Pq +Pi) = '^Ox(M -J^ +^ 20. That is, the increase in (Pq+Pi) when load (1) is moved from the second into the first panel, equals the reaction on the floor beam at b due to load (1) when load (2) is at b, plus the increase in the reaction on the floor beam at a due to moving the second load into the second panel. The value of the maximum shear in the second panel equals 4-4-90 14- 1 V S 10x|^ + ~(36+31+26+21)+iQ(12+7+2)— ^ = 43.56 (thousand-lb. units). As an approximate check the corresponding shear with load (3) at b has been computed and found to be 43.37. If the increase in shear as determined from the expression for the increase due to moving up load (3) be added to this the result should equal the shear with load (2) at b, thus giving an exact check. The student will observe that in all cases where no load goes off or comes on the span, or goes out of the panel, the distance which the loads are moved appears on both sides of the inequalitj' and may be omitted. Moreover, the denominator of the left- hand term equals the span length and that of the right-hand term the panel length. Hence we may say that for the special condi- tions noted the shear will be increased by moving up the loads whenever the average load per foot on the entire span exceeds that on the given panel. 48. Formula for Position of Loads for Maximum Shear for Intermediate Panels. Girder ivith Floor Beams. The method of moving up the loads as used in the preceding article is simple and very general. It is applicable not only to the determina- tion of the position of loads for maximum shear but to the determination of position for many other functions. The student should understand it thoroughly and apply it to many different Aht. 49 MAXIMUM MOMENT 87 cases until he thoroughly comprehends the influence of such load systems upon the various portions of the girder. For the practitioner who may wish a definite formula for determining the position for a general case, the following may be of use for intermediate panels. Let P be any load which may be on the span or which may come on during the process of moving up the loads. Let a be the distance which any load may move on the span. For all the loads which are on the span at the beginning and which remain on, a is the distance between the two loads the effect of which is being compared. For a load which comes on or goes off the span during the moving-up process a is the distance which the load actually moves on the span. Let Pi be any load which may be in the panel under con- sideration or which may come into it or go out of it during the process of moving up the loads. Let Oi be the distance which any load Pi may move in the panel. This equals a if no load moves out of or into the panel during the process of moving up the loads. Let L = length of span. p = length of panel under consideration. The shear will be increased by moving up the loads provided 2^>2^1^ (U) L p 49. Maximum Moment. Girder with Floor Beams. For girders with floor beams it is customary to compute maximum moments at panel points only. If, for any reason, the maxi- mum moment between panel points is desired it may be obtained with sufficient accuracy by interpolation. For uniform live loads and for concentrated loads which are fixed in position interpolation gives exact results since the curve of moments for such loads consists of a series of straight lines. The same is also true for moments due to the weight of the floor system, but is slightly in error for the weight of the girder itself. For a system of moving loads this method is somewhat inaccurate but is on the safe side, and hence may be used with security. This is shown by the following demonstration which refers to Fig. 66. Let the ordinate bA represent the maximum live moment 88 CONCENTRATED LOAD SYSTEMS Art. 50 at any panel point, b, due to a concentrated load system. For the position producing this maximum the moment curve for the portion of the girder between b and c will be the line AB, Be representing the moment at c for the position of the loads giving maximum moment at b. If the loads be now moved so as to give the maximum moment at c we shall have cB' and bA' as the ordinates for moments at c and b respectively for this new position, and B'A' will be the moment curve between b and c. It is evident from the figure that interpolation between the maximum moment at b and that at c wUl give a safe value for the maximum moment at any point in the panel, since the line AB can never rise above AB' nor the line B'A' above B'A; therefore, the ordinate xx' for the moment at x can never be less than the actual maximum moment at x. It wUl readily be seen by drawing an influence line for the moment at x that for maximum moment some load should lie at either panel point b ^^ Fig. 66. or cf that the moment at c with the loads in the position neces- sary for maximum moment at b can never exceed the maximum moment at c and wUl almost invariably be less than that; and that this principle holds good for the condition when the moment at 6 is a maximum. This proof is perfectly general and applicable to any panel. 50. Moment and Shear at the Critical Section. The cases already treated have been for shear and moment at stated sections of simple beams and for panels and panel points of girders with floor beams. For the latter it is necessary and sufficient to compute the maximum shear in each panel and the maximum moment at each panel point, since thereby the maximum values of these functions will be obtained. For beams or girders which do not support floor beams it is always necessary to compute maximum values of shears and moments, and in addition, for long girders, the values at intermediate sections taken with sufficient frequency to insure a good design. Art. 50 MOMENT AND SHEAR AT THE CRITICAL SECTION 89 In order to determine the maximum values it is necessary to locate the sections at which they occur, that is, the critical sections. For shear the critical section is an infinitesimal distance from one of the points of support. This needs no demonstra- tion, as an inspection of influence lines for various sections including one at the end furnishes sufficient proof. For moment with uniform load it has already been shown that the maximum moment occurs at the centre and equals IwL^, when w equals the load per foot and L the span. With a system of concentrated loads the maximum moment may not occur at the centre though the critical section will be very near the centre. To treat this case it is necessary to H Mo+d+.r)- itr ^b Fi-fct;i=KMrtion I H- FlG. 67. make use of the already established principle that for maxi- mum moment at any section of a beam under a system of con- centrated loads one of the loads must lie at the section. If, therefore, it is possible to determine the location of the system of loads as they cross the span such that the moment at any one load is a maximum, the problem can be solved by trying a sufficient number of loads and computing the different maxima. As will appear later the critical section is always near the centre of the span, hence, as a rule, only loads need be tried which are found to give a maximum moment at the centre. Consider the set of loads shown in Fig. 67, and let the prob- lem be the determination of the position of these loads in order that the moment at P3 may be a maximum. Let R be the resultant of the loads Pi to P5 inclusive and n its distance from the last load P5. 90 CONCENTRATED LOAD SYSTEMS Art. 50 Let X be the distance from P5 to the right support when the loads lie in the proper position for maximum moment at P3. Then the moment at P3 is given by the equation M3 = i2^t^[L-(c+d+a;)]-Pi(a+6)-P2&. For maximum value of M^ differentiate with respect to x and put the first derivative equal to 0. This gives dx ^|[-n+L- -c-d-2x\ = <). Therefore, in order to find the maximum moment at P3 as the loads cross the span, P3 must be so located that -n+L-c—d-2x = or L — (c+d+x) =n+a;. That is, the resultant of the loads on the span when the maxi- mum moment at P3 occurs must lie as far from the right support as the load itself lies from the left support, or in other words the centre of the sTpan must lie half way between the resultant and the load. The following examples serve to illustrate the application of this principle : Problem. Compute the absolute maximum moment on a simple beam of 12-ft. span due to two wheel loads of 10,000 lbs. each spaced 6 ft. between centres. Solution. In this case there are two equal loads, hence it is imma- terial which load is considered. For maximum moment at load (1) the loads should lie as shown in Fig. 68, the centre of the span being half way between load (1) and the resultant of the two loads. The moment at the first load will then equal 20,000 ' ■' = 3.3,750 ft.-lbs., Abt. 50 MOMENT AND SHEAR AT THE CRITICAL SECTION 91 The maximum moment at the centre for this beam would be 30,000 ft.-lbs., hence the absolute maximum moment exceeds the maximum centre moment by over 10 per cent. It should be particularly noted that the demonstration which has been given only serves to fix the position for maximum moment at a given load with certain assumed loads on the span, and that if a different set of loads be on the span the position will be different. To illustrate this consider the same loads as in the p-revious examples and a span of 10 ft. There are then two positions of the first load which give maximum moment. First, assume only the first load on the span; in this case it should be placed at the centre and the moment would be 25,000 ft.-lbs. Second, assume two loads on the span; in this case the centre of the span should be half way between the resultant of the two loads and the first load, and the maximum moment at the first load will equal 20,000 ''^~Q*^ - = 24,500 ft.-lbs., which is somewhat less than with one load at the centre. In such a case the length of span can easUy be determined for which one load at the centre gives a moment at the load equal to that with two loads on the span. In general it is necessary to consider both cases when dealing with two loads. The absolute maximum moment on spans above 25 or 30 ft. in length does not materially differ from the maximum centre moment, and in practice the latter is generally used. For the loads previously considered with a 30-ft. span the absolute maximum moment =20,000--^^^— =121,500 ft.-lbs., while maximum centre moment = 20,000 ( — 115=120,000 ft.-lbs. The difference is about one per cent, which is so small as to be negligible. The following example serves to show the application of this principle for a locomotive loading: Problem. Determine the maximum moment on a simple beam of 21-ft. span due to the locomotive given in Art. 45. Solution. First determine which load or loads give maximum moment at the centre, as it is probable that one of these loads will give the 92 CONCENTRATED LOAD SYSTEMS Abt. 51 absolute maximum moment. By applying the criterion for maximum moment, loads (3) and (4) are found to give maxima, but it is clear that the centre moment with load (3) at the centre will equal the centre moment with load (4) at the centre, and that it makes no difference whether we use one or the other load. Let the maximum moment therefore be determined at load (3), assuming loads (2-5) inclusive on the span. The position for maximum moment will then be as shown in Fig. 69 and the moment at load (3) will equal 80,000 (10 ^ -Ij)' 21 -20,000X5=226,9.50 ft.-lbs. In this case it is impossible to get more than four loads on the span at once. If three loads are on the span the resultant coincides with load (3), hence for a maximum for this assumption load (3) should lie at the centre, but this is inconsistent with three loads being on Fig. B'J. the span, hence a maximum at load (3) with only three loads on the span cannot be obtained, and the case considered gives the absolute maximum moment. The maximum centre moment for these loads occurs with either load (3) or load (4) at the centre and equals 80,000X — X 10^-20,000X5= 220,000 ft.-lbs., so that in this case the difference is only 2.7 per cent. 51. Moments and Shears. Floor Beams and Transverse Girders. As a preliminary step in the examination of this case the influence lines shown in Figs. 70 and 71 have been drawn. These are influence lines for stringer reactions on floor beams. Since the stringers are simple beams of a length equal to one panel and are supported at the ends upon the floor beams it is evident that a load moving along the bridge causes no reaction on a floor beam unless it is on the stringers in one of the panels Art. 51 MOMENTS AND SHEARS 93 adjoining the floor beams in question. Fig. 70 represents the stringer reactions on an intermediate floor beam and Fig. 71 on an end floor beam. It will be noticed that the influence line shown in Fig. 70 has the same characteristics as the influence line for moment ^ -P'- FiG. 70. — Influence line for stringer reaction on floor beam at 6. at any section of a simple end-supported beam, hence the demon- stration of Art. 46 is applicable. The conclusion may therefore be at once drawn that for maximum reaction on an intermediate floor beam one load must lie at the beam, and that load must be one which, when placed just to the right of the given floor Fig. 71. — Influence line for stringer reaction on floor beam at a. beam, makes the average load per foot on the stringers in the right hand panel greater than on those in the left panel, and when placed just to the left of the floor beam reverses this condition. For the end floor beam, the maximum reaction occurs for the loading giving maximum stringer reaction and equals that reaction. It remains to consider the actual moments and shears on 94 CONCENTRATED LOAD SYSTEMS Abt. 51 the floor beams. Curves of moments and shears for a floor beam due to stringer reactions are shown in Fig. 72. Both shear and moment are direct functions of the stringer reactions. The maximum moment must occur at one of the stringers, since the floor beam is in the condition of a girder loaded with concentrated loads and the curve of shears can cross the axis only at a load. The case illustrated is not the usual one, since the stringers are unsymmetrically placed with respect to the centre of the floor beam. Were the floor beam symmetrical the maximum moment would occur at both stringers and at all points between. Since, in the actual design the dead 'Curve of Moments I Fig. 72. load of the floor beam would also have to be considered, the maximum combined live and dead moment for the ordinary symmetrical floor beam occurs at the centre. For floor beams where the stringers in one of the adjoining panels are not of equal length, that is, where the panel is a skew panel, special treatment is necessary. It is usually advisable to treat this case by means of influence lines without attempting to apply special rules. The application of the methods just given to the determina- tion of the maximum moment and shear on a floor beam (or a transverse girder, such as a cross girder in an elevated railroad structure) , wUl now be illustrated. Art. 51 MOMENTS AND SHEARS 95 Problem'. Determine maximum moment and shear on floor beam b of Fig. 73 for loads shown in Art. 45. It may be easily seen that a given load when equidistant from the floor beam b produces greater reaction if on the longer stringer, hence it is probable that the maximum reaction in this case will occur with the nr Stringers, t 1 ^^ 1 -I r- -1 r- n r BtrlDger h 1^^ ■ ^ k 4'->i< — - 7' — »V4'-»i CROSS SECTION -16' >K 10'- ELEVATION 15' 10' <^. i '4 ^^—^ PLAN Fig. 73. greater number of loads on the 15-ft. panel. Let the loads, therefore, be brought on from the left. Av. load per Av. load per ft. on left. ft. on right. Load (2) to left . Load (2) to right. Load (3) to left . . Load (3) to right. Load (4) to left . . 60 15 > 10 ■ 10 Load (2) does not 60 15 > 30 To , give a maximum. 60 15 > 20 10 Load (3) gives a 40 15 < 40 10 maximum. 54 < 40 Load (4) does not 15 10 give a maximum. Load (3) at 6 evidently gives the maximum floor-beam reaction. Its value is given by the expression. 20X^1-V204t^=50, 10 15 that is, the reaction on the floor beam at each stringer connection equals 50,000 lbs. The floor beam is then in the condition shown by Fig. 74. The maximum shear= 50,000 lbs. and the maximum moment = 200,000 ft.-lbs. 96 CONCENTRATED LOAD SYSTEMS Art. 52 Before concluding this article the beginner should be cautioned to avoid the mistake that is frequently made of adding the maxi- mum live reactions on two adjoining stringers to determine the floor beam load at a point such as m in Fig. 74. The fact that the maximum reaction on a stringer occurs when one of the heavy loads lies at the end of the stringer is sufficient to show that the same condition cannot exist on the adjoining stringer in the next panel, because such a condition would necessitate two wheel loads occupying practically the same place at the same time. ^ s s Diagram showing loads on floor beam. Fig. 74. 52. Moment Diagram. To save repetition of computations for a given set of concentrated loads when used for varying spans, it is customary to use a moment diagram upon which certain quantities frequently required and unaffected by the length of spans are placed once for all. Upon this diagram the loads are plotted to a convenient scale at top and bottom of sheet for convenience in reading, and the quantities desired are placed between. The diagram is used in connection with another sheet upon which the span is drawn to the scale used in plotting the loads. The diagram shown in Fig. 75 is of a convenient form and is self-explanatory. The use of the diagram can be readily understood from the simple example that follows. Let the problem be the compu- tation of the moment at the second panel point from the left of a span having 5 panels at 20 ft. when load (8) is at the given panel point. Place the plotted span on a separate sheet so that load (8) is over panel point 2; the ends of the span will then be in the positions shown at bottom of Fig. 75. Since the desired moment equals the moment of the left reaction due to loads (2) to (17) inclusive about load (8), minus the moment of loads (2) to (7) inclusive about the same point, it is necessary Art. 52 MOMENT DIAGRAM 97 MOMENT DIAGRAM— COOPER'S E40 LOCOMOTIVE Distance to resultant of loaila Loads In thousands of lbs. per wheel = tons per aKle Fig. 75. Moments are in thousands of foot-pounds per rail and are the summations of the mo- ments of all the loads to the left of and including that under which they appear about the load indicated in right-hand columns; e.g., the value 102.56 underload 10 and opposite load 16 in right-hand column is the moment about load 16 of the loads 1 to 10 inclusive. Moments to right of zig-zag line are moments about load under which they appear of all loads to left thereof. Application of Diagram. To find moment about a given load, say load 14 of a certain number of loads to left, say 6 to 13 inclusive, proceed as follows: Find in diagram, moment about 14 of all loads to left; subtract from it moment about S of all loads to left of S plus the product of the sum of loads 1 to 5 inclusive and distance from load S to load 14. The result will be the desired moment. The expression thus obtained = 8728— [830-1-90X561 = 2858. The moment of the same loads about a point between 14 and 15 and distant X ft. from 14 can be found by adding to quantity just obtained the product of the sum of the loads 6 to 13 and the distance x. See Article 34. 98 CONCENTRATED LOAD SYSTEMS Akt. 52 to compute these two quantities. The moment of the left reaction equals the moment of the loads on the span about the right reaction divided by 5 panel lengths and multiplied by 2 panel lengths. This equals |[13,589-990 + (271-10) X4] = ?- 13,643 = 5457.2. The moment of loads (2) to (7) inclusive about panel point 2 = 2851-10X43 = 2421. Hence, the moment desired = 5457. 2— 2421 = 3036.2 expressed in units of thousands of pounds per rail or tons per track. This method involves the application of the principle stated in Art. 34, which should be thoroughly understood. PROBLEMS 32. a. Compute the maximum positive shear in thousands of pounds at sections a, b and c for the system of moving loads shown. 6. Compute maximum moment in foot-pounds at 6 and c for the system of moving loads. c. Compute uniform live load per foot which would give a maximum moment at section 6 equal to that found for the system of moving loads. [n'Mi'M 10 20 20 SO 20 K-t- 17^— k6M»-12'-+-6'-j Loads in thousaads of lbs. SectluD a ia an inBnltcalinnl diatimce to rigUt of pt. of BuppoEt. -r ;-i 1 I -15' >!< w,m Peob. 32. Loads'in thousands of lbs. Pbob. 33. 33. Draw curves of shear and moment for one girder for the con- centrated loads when placed in the position shown in the figure. (Bridge has two girders located symmetrically with respect to the loads.) 1 a (1)6(2) c m d /. 1 i i A PLAN -•) 54. Method of Design. Frequently the design of beams requires merely the application of formula (14), and the shearing strength need not be considered. In the case of comparatively short 'beams, however, the shearing strength is more important and should be investigated. In wooden beams this is especially important, since the resistance of wood to longitudinal shear is small and such beams may fail by splitting longitudinallj'. The design of reinforced concrete beams also requires the appli- cation of formiila (13). 55. Wooden Beams. In selecting wooden beams care should be taken to use commercial sizes only. The following table gives such sizes: Spruce: 2X 3, 2X 4, 2X 5, 2X 6, 2X 7, 2X 8,2X10,2x12. 3X 4, 3X 6, 3X 8,3X10,3X12. 4X 4, 4X 6, 4X 8,4X10,4X12. 6X 6, 6X 8, 6X10,6X12. 8X 8, 8X10, 8X12. 12 ft. to 22 ft. are ordinary lengths. 23 ft. to 26 ft. are less common. 27 ft. to 32 ft. are obtained with difficulty. Yellow Pine : Sizes about the same as for spruce, also 2X14, 2X16. 3X14, 3X16. 4X14, 4X16. 6X14, 6X16. 8X14, 8X16. 10X14, 10X16. 12X14, 12X16. 14X14, 14X16. 16X16. Art. 56 STEEL BEAMS 103 Lengths of yellow pine sticks are longer than for spruce and run up to 40 ft., and it is usually possible to obtain even 50 ft. lengths except for the largest sizes. The cost of wooden beams depends upon the price of lumber per board foot. This is subject to considerable variation, and if a close estimate is desired, a dealer should be consulted. 56. Steel Beams. Such beams are usually made with a cross-section of the shape of the letter I in order to obtain a large moment of inertia from a comparatively small amount of material. They are rolled from a solid piece of steel in varying heights and thicknesses. In selecting such beams the manu- facturer's handbooks should be consulted, and sections marked " standard " chosen since the selection of a " special " section is lilcely to cause delay in filling the order. These handbooks give all the properties of the beams, such as area, weight, moment of inertia, etc., and may be relied upon as accurate. The cost of steel beams is dependent upon the weight of the beam, and upon the amoimt of punching, riveting and other work which has to be done. The price is usually figured on a " cent per pound " basis, the price of the unfabricated beam being taken as the base price, and the other prices added thereto. Other things being equal, the lightest beam having the requisite strength and stiffness is most economical. The base price is published from time to time in such engineering papers as the Engineering News, Iron Age, etc., e.g., in the Engineering News of Dec. 2, 1909, the f.o.b. Pittsburg price was quoted as 1.55 cents per pound for 3 inch to 15 inch I-beams and channels and 1.60 cents for depth greater than 15 inches. The freight rates for carload lots from Pittsburg as quoted in this same issue were as follows: To New York 16 cents, and to Boston 18 cents per 100 lbs. This price is for beams cut to length with a variation not to exceed f in. more or less than specified. For cutting to more exact length and for other work the following schedule adopted in January 1902 gives the extra cost in cents. 1. I'or cutting to length with. less variation than plus or minus f 0.15 per 100 lbs. 2. Plain punching one size hole in web only 0.15 " 100 ' ' 3. Plain punching one size hole in one or both flanges 0.15 " 100 ' ' 4. Plain punching one size hole in either web and one flange or web and both flanges 0.25 " 100 " 5. Plain punching each additional size hole in either web or flange, web and one flange, or web and both flanges . . 0.15 " 100 ' ' 104 BEAM DESIGN Akt. 57 57. Examples of Beam Design. Problem. Design wooden and steel beams for 12-ft. span. Beams to be supported at ends and to be loaded with a total uniform load (live and dead) of 1000 lbs. per foot. Allowable unit stresses to be those given in Art. 18 for steel and yellow pine. Fifty per cent of total load to be added in the case of the steel beam to allow for impact. Solution. Maximum moment is at centre of beam and equals i lOOOX 12X 12= 18,000 ft.-lbs. 8 Maximum shear is at end of beam and equals 6000 lbs. For wooden beam M=^sbh^, :. 18,000X12 = ^ ISOObh^. 6 6 .-. 6^2 = 997. Either an 8 X 12-inch, 6 X 14-inch, or 4 X 16-inch beam has a value of bh^ greater than that required and may be used. The area of cross-section needed to carry shear may be determined by Eq. (16) and is given by the following expression: , 18,000 ^^ . Evidently the 4 X 16-inch beam is too small, and one of the other beams should be selected. The 6 X 14-inch is the cheapest and should be chosen if conditions permit. The longer side should always be placed parallel to the plane of the loads, i.e., vertical if the loads are vertical. This was the position assumed in solving for bh^, and none of 'the beams selected would be strong enough if not so placed. The bearing area on the abutment should also be determined. If the reactions were uniformly distributed over the bearing surface there would be needed = 23 sq.ins. To allow for unequal distribution .iOU 50 per cent will be added to this, giving 34.5 sq.ins. The 6 X 14-inch 34.5 ins. beam would therefore need to extend or, say, 6 ins. over the D abutment. For the steel beam the moment after allowance for impact is made = 27,000 ft-.lbs. . /_27,000X12_ The term — is known as the section modulus. Values of this for c various beams are given in the handboolis issued by steel makers, and the hghtest beam having a modulus equal to or greater than the above figure should be selected. A 10-inch I-beam, weighing 25 lbs. per foot has sufRcient strength and will be chosen. A 9-inch, 25-lb. beam is also strong enough, but as this is just as expensive as the 10-inch beam Art. 57 EXAMPLES OF BEAM DESIGN 105 and not so strong, the 10-inch beam should be selected, provided condi- tions do not require the use of a shallower beam. The distribution of shear over an I-beam is more complicated than in a rectangular beam. It will, however, be shown later that the shear is practically all carried by the web over which it is distributed almost uniformly. Making this assumption the area required in the web would be ' of a square inch, and as the actual area in the J ^,UUU beam selected is far in excess of this the beam is evidently strong enough to carry the shear and hence may be used with safety. The sizes selected were based upon the assumption that the beam would have no rivet or bolt holes and no other reductions in the cross-section area. If such reductions occur the value of / should be corrected to allow for the reduction in section, and the value of c also changed if the position of the neutral axis be shifted by the change in area. Methods of making such corrections will be given in full in Chapter V. Another important element to be considered in selecting beams is that of vertical and horizontal stiffness. This will be considered in Art. 59, it being assumed for the present that the beams designed in this article are supported laterally where necessary, and that their vertical deflection is not excessive. Problem. Design wooden and steel beams for a single track electric railway bridge of 12-ft. span carrying the electric car shown in Fig. 76, Loads ore Y^eel Loads 5 -f « 15 -5 -I Fig. 76. Assume track (ties, rails, etc.) to weigh 400 lbs. per lineal foot (200 lbs. per foot per rail), and each beam to weigh 40 lbs. per lineal foot. Allow- ance for impact to be 25 per cent. Unit stresses as in Art. 18. Solution: Maximum Moment Maximum Shear Dead =-240X12X12 = 4,300 ft.-lbs. Dead 240X6= l,4401bs. 8 Live =20,000— -=7P^= 37,600 " Live 10,000X- = 15,830 lbs. 12 1-^ Irapact= 9,400 " Impact= 3,960 " Total moment= 51,300 " Total shear= 21,230 " Steel Beam. For steel beam assuming no reduction due to bolt holes / 51.300X12 c 16,000 = 38.5. 21,230 , ^, . Web area needed for shear= T^nr^^ '^■'' s | liiV i l k^ii U'ea^e^oVJ (b (^ mm fh fK (h fh fb ft) fh — ^ Axle loads Loads are wheeUoads Pkob. 38. Prob. 39. 38. Design a steel I-beam stringer for an electric railway bridge. Bridge to be a single track bridge with 15 ft. panels and stringers located symmetrically with respect to the rails. Assume total dead weight of track and stringers to be 600 lbs. per lineal foot of bridge. Use live loads shown and allow 25 per cent for impact. Unit stresses as given in Art. 18. 39. Design steel I-beam stringers for the bridge of Prob. 38 for the electric locomotive shown. All other conditions to be the same as in the previous problem. CHAPTER V PLATE GIRDER DESIGN 60. Plate Girders Defined. For lengths greater than are admissible for steel beams or where beams of sufficient strength cannot be obtained other types of structures must be adopted. Of these the plate girder is the next higher form. A plate girder is essentially an I-beam made, not out of one solid piece of metal, but out of a number of pieces riveted together. Fig. 3 shows a plate girder bridge, and Fig. 77 shows 3^ Cover platea -Flange L^ Fig. 77. — Cross-section of a Plate Girder. .^ n P J ' '=> — 'J cr- — ^ — (-, — » ^ ) ( r J ) ( L " — i_i — ' Fig. 78. — Cross-sections of Two Box Girders. the cross-section of a typical plate girder with the different parts named. Plate girders are rarely made of greater depth than 10 ft. 6 ins. owing to difficulties in transportation by raU, and a length of 100 ft. is seldom exceeded for the same reasons although girders of 125 ft. in length' have been made and shipped in one piece. Occasionally plate girders are made in sections and spliced in the field, but this expedient is not common and should not be adopted except to meet some unusual condition. 108 Art. 61 PLATE GIRDER WEB THEORY 109 Plate girders are sometimes made with more than one web, as indicated in Fig. 78. Such a girder is called a box girder. It is used in situations where great strength with limited depth is required. 61. Plate Girder Web Theory. Plate girder webs may be proportioned on the assumption that all the transverse shear is uniformly distributed over the net area of the web, and that this may be taken as three-fourths the gross area without material error. That this assumption is essentially correct is shown by the following demonstration: Let Fig. 80 represent the square prism, abed, from the web of the plate girder shown in Fig. 79 and let it be assumed that :n I Fig. 79. Fig. 80. there are shearing forces acting on all four surfaces, and direct stresses on the two vertical surfaces. Let t = its thickness = thickness of web. S;^ = intensity of the shearing force on the surface ab. s^' = intensity of the shearing force on surface cd. s^= intensity of shearing force on surface bd. Sb' = intensity of shearing force on surface ac. /= intensity of the direct stress on surface 6d (assumed compression for convenience). /' = intensity of the direct stress on surface ac (also compression) . Application of the equations of equilibrium give the following results: ShUx+fUx = Sh'tJx+f'Ux. .-. sh = sh'+{f'-f). (skUx) Jx - (sJJx) Ax + [ (/-/') tAx]~ = 0. As the distance dx becomes infinitesimal zero, hence at the limit Sft = Sft' = s„. Sfc=S,; + (/'-/) /'-/ approaches no PLATE GIRDER DESIGN Art. 62 It therefore follows that the intensity of the horizontal shear in the web of a plate girder at any point on a vertical plane equals the intensity at the same point on a horizontal plane. Since the intensities of the vertical and horizontal shears are equal, it is evident that the distribution of the vertical shear can be determined by the application of formula (13), from which it at once follows that the vertical shear is distributed over the web with approximate uniformity since Q is the only term in the equation affected by the distance from the neutral axis, and its value changes much more slowly than does the distance from the axis. The numerical examples given in Art. 64 show the degree of approximation of this assumption for certain typical girders. In many girders the thickness of the web is determined by imposing restrictions upon its minimum thickness to prevent undue corrosion. For railroad bridges it is common to specify that the web shall be not less than three-eighths inch thick. 62. Plate Girder Flanges. Theory. Formula (14) applies to girders as well as beams. It is, however, in inconvenient form for use and is replaced in practice for symmetrical plate girders by a less accurate but more easUy applied formula. The formula recommended for plate girder flanges is as follows: M 1 ^=^-12^^^^ (^7) The derivation of this formula is as follows : Let ^=net area in square inches of tension flange (through rivet holes). /i = distance in inches between centres of gravity of the two flanges, s = allowable unit stress in bending. i = thickness of web in inches. hi = depth of web in inches. lf = maximum bending moment on given section in inch- pounds. 7 = total moment of inertia of gross cross-section about neutral axis. ^1 = gross area of each flange. ' This formula should not be used tor very shallow girders with heavy flanges, where the distance between centre of gravity of flanges is much less than the total depth of the girder, nor for other abnormal cases. Art. 62 PLATE GIRDER FLANGES. THEORY 111 ^2 = depth out-to-out of flanges, ^c.g. = moment of inertia of each flange about its own centre of gravity. The following equation for I may now be written: / = 2/,,.+24i(|)'+liAi3. The term 2/^^ is small in comparison with the other terms and may be omitted without serious error, this being on the safe side. In consequence the value of - may be written thus 2^a^ , 1 , 3 7_ 4 +12^^^ _A^h^ 1 th^^ C h2 /l2 6 /l2 ■ 2 ^ ^ I M M Aih^ 1 thi^ But — = — . .•. — = — 1 — • c s s h2 6 h2 ' nence, ^i_ ^ ^^ _______ _______ For ordinary girders the value of h is seldom larger than that of hi while it is usually smaller, i If ~ be therefore assumed as unity the last term in above equation will ordinarily be less than its true value, and since this term is small compared with the term involving M, the slight change in its value by the above approximation will affect the value of A but little, and that on the safe side, since it reduces the value of the negative term. This approximation will therefore be made. The assumption that -^ = unity will also be made. This is on the unsafe side, since h is always less than /12, and to assume ' Most specifications forbid the use in design of a, value for h greater than hi even if it actually exists. It is good practice to proportion girders so that such a condition will not occur. 112 PLATE GIRDER DESIGN Art. 62 it equal gives a smaller value for Ai than is required. The error in making this assumption is largest in shallow girders having large flanges as may be seen in the numerical examples given later. By making the above approximation the formula becomes For material in compression it is customary to make no deduction whatever for rivet holes since it is assumed that the rivet which is driven while hot and ordinarily under high pres- sure fills the hole so completely as to become an integral portion of the material. This is open to some doubt in the case of thick material or hand-driven rivets, and may be vitiated at any section by a loose rivet, but for most cases this assumption is probably a reasonable one. For sections in tension full allow- ance for rivet holes must be made, since under no circumstances can tension be transmitted through a rivet hole. The last term in formula (17) represents the bending resist- ance of the web. As there are usually vertical rows of rivets in the web for floor-beam connections, stiffener angles, etc., and as these may occur at the section carrying maximum moment they must be considered. To allow for such holes, it may be assumed that a vertical row of holes one inch in diameter and 2| inches apart may occur in the tension-half of the web. This would decrease the moment of inertia by j-\ approximately, thus making the. last term in the equation ^\ thi or say \ thi . Allowance for rivet holes in the tension flange must also be made. This may be done by substituting A for Aj, which is in reality equivalent to providing for rivet holes in both flanges. This may seem excessive, but some excess is necessary since the section has been considered as solid and with its neutral axis at mid-height, whereas in reality the influence of the rivet holes in the tension portion is to shift the neutral axis from the centre, thus diminishing the moment of inertia and increasing the distance from neutral axis to extreme fibre. The substitu- tion of A for ill is, however, more than sufficient for this pur- pose and helps to diminish the error made in placing -^ = unity. Art. 63 DEGEEE OF APPROXIMATION 113 This modification gives the following formula, which is adopted by many engineers: The last term in this equation represents the resistance of the web to bending. Owing to the difficulty in satisfactorily splicing the web many engineers disregard its resistance to bending and use the formula -I It is believed, however, that ample provision has been made in formula (17) for insufficient web splices in long girders by putting the term for web resistance as -^thi. The assumptions made in deriving formula (17) are of such a character as to make the formula inaccurate for girders having great depth in proportion to their length. Such girders are not common in bridges but are sometimes used in architectural work, and should be solved by the direct application of formula (14). It should be stated, furthermore, that experimental knowledge of the distribution of stress in plate girders is insufficient to permit a confirmation of the accuracy of formulas of the type of (17). Formulas of this character have, however, been in use for many years with satisfactory results, and may well be con- sidered as safe working formulas. Formula (17) is more con- servative than that usually employed. 63. Degree of Approximation of Flange Formxjla. In order to show the degree of approximation of formula (17), in com- parison with formula (14), the problems which follow have been inserted. Problem. Compute allowable bending moment, M, for the girder shown in Fig. 81. Assume no intermediate web stiffeners, and hence only one rivet hole (flange rivet) in tension half of girder. Allowable unit stress=s. Allowable moment by beam, formula. Area in sq. ins. Top angles, 2-6" X 4" X i", at 4.75, 9 . 5 gross Bottom angles, 2-6"X4"Xi", at (4.75-0.5) = 8.5 net Web, 29"XV'. =14.5 net Total effective area of cross-section =32.5 sq. ins. 114 PLATE GIRDER DESIGN Akt. 63 Distance of centre of gravity of cross-section above axis xy IXUXlSi 32.5 -= 0.6 in. /i= 30.25 ins. -3.98 ins. = 26.27 ins. Let 7x7/= moment of inertia of gross section about axis xy and /e.g. = moment of inertia of any piece about an axis parallel to xy and passing through the centre of gravity of the piece. _L. )-- C::> 6x4x J^ Ls 6 leg: vertical y 6 X 4 jt J^ L3 6 leg: vertical Jl m ^ Fig. 81. Ixy of webs=yV4-30.30-30 FiQ. 82. Aih^ 7:tj,of angles = 7c.g. of angles + ^— = 4x17.4 +4X4.75X (13.13) = 1125 = 3345 Total h Deduct for flange rivet hole Ijxl X(13.13)2 7 of net section about axis xy To obtain 7 for net section about axis passing through centre of gravity deduct, 32.5X0.62 = 12 = 4470 = 259 = 4211 = 4199 267. / of net section about neutral axis 7_ 4199 c~ 151 + 0.6 ins.' .-. M=s — =267s. c Allowable Moment by Formula (17). By transformation of terms in the formula, M is found to be given by the expression M={A +^ih,)sh= (8.5 + 1 .25) (s) (26.27) = 256s. Since the allowable moment as computed by formula (17) is less than that given by the beam formula, the formula for this girder is on the safe side. The approximation is about 4 per cent. Aet. 63 DEGREE OF APPROXIMATION 115 Problem. Compute allowable bending moment for the same girder, assuming a row of rivet holes in the web at the point of maximum bending moment in addition to the flange rivet holes. .Solution. For practical reasons the web rivets nearest the flanges should be located not less than IJ ins. from the edge of the flange angles, or for this girder 7i ins. from the backs of the angles. In order to get even rivet spacing let this distance be made 7f ins., thus giving 15 ins. for the distance between these rivets and permitting the use of five spaces at 3 ins. for the horizontal rivets between flanges. Assume that only the rivet holes in that portion of girder below the neutral axis, i.e., the tension half, need be deducted. Net area of cross-section=32.5— 3Xi=31.0. The position of the centre of gravity of the crosss-section above the axis xy= — — =0.85 m. The deduction from I^y to allow for rivet holes will be lXHX(13.13)2 + iX(7.5^+4.5^ + 1.5^) = 298. .". / of net section about axis a;2/= 4470— 29S = 4172 and about neutral axis =4172 -31.0X0.85' =41.50 / 4150 _ -7= 9fin ■■ c 15.12+0.85 si and Jlf=— =260s. c The approximation in this case is somewhat less than 2 per cent and is also on the safe side. Problem. Compute allowable bending moment for the same girder assuming 2-lOxi 'w- plates to be added to each flange and no web rivets at the critical section, see Fig. 82. (Note that the horizontal and vertical flange rivets are usually staggered and hence a section containing the \'ertical rivets may not contain horizontal rivets ; also that the material in the horizontal legs of the flange is thicker than that in the vertical leg, hence the reduction due to rivet holes is larger.) Allowable Moment by Beam Formula. Area in sq.ins. Top angles, 2-6"X4"Xi"at 4.75 = 9.5 gross Bottom angles, 2-6"X4"Xi" at (4.75 -0.5) = 8.5 net Top plates, 2-10" Xi" at 5 = 10.0 gross Bottom plates, 2-10"Xi" at (5-1) = 8.0 net Web30"Xi" =15.0 gross Total eilective area of cross-section = 51.0 sq.ins. Computation of h. Gross area of one flange = 19.5 sq.ins. From back of angles to centre of gravity of angles = 1.99" 116 PLATE GIRDER DESIGN Art. 63 Hence from centre of gravity of angles to centre of gravity of flange IPX (1.99 +0-5) , „_„ 19:5 -^■^' • :. /(=30.25"-2x(1.99"-1.27") =28.81" Allowance for rivet holes = l XliX2 = 3 sq.ins, hence centre of gravity of cross-section above axis xy 3X(15J + i) 51 = 0.9" Allowable Moment by Beam Formula. Ixy of web Ixy of angles 7:,„of plates = 20x(lo|)2 ■Total I of gross section about axis xy Deduct for rivet holes 3x (151)'' = 1125 .= 3345 =4883 = 9353 = 709 Correction for / about neutral axis = 51X0.9'= 41 I of net section about neutral axis /_ 8603 c~ 16 J + 0.9 750 =8603 - = 505. .-. M=505s. Allowable Moment by Formula (17) Jl/= (A H-i^,- B Fig. 87. 6 -^ E directly with such an area. It is, however, wise to make liberal allowance for rivet holes and if the pitch CB is less than 2^ ins. to ordinarily allow for three holes in the section as shown in Fig. 84. Attention should be called to the fact that while the maximum moment on a girder ordinarily occurs at only one section, and that at this section the rivet pitch may be a maximum, a maxi- mum flange stress is developed wherever a cover plate ends, and the rivet pitch at such a point may be little if any larger than the minimum allowable pitch. 122 PLATE GIRDER DESIGN Art. 66 66. Example of Girder Design. An example of the complete design of the cross-section of a girder will now be given. Problem. Determine cross-section of a girder to carry a maximum bending moment of 1,250,000 ft. lbs. and a maximum shear of 160,000 lbs. Depth of girder back to back of flange angles = 481 ins. Allowable unit stresses, Bending 16,000 lbs. per sq.in. Shear 12,000 Solution. Web. Net area of cross-section required = — 7;^— ■ =13..3.3 sq.ins. 13.33 Depth of web 48 ins. Thickness of web ■— — = 0.37 in. or f in. Flange. Assume h to equal depth of web =48 ins. Trial section : 2 angles 6"X4"X|" at 5.86 = 11.72-2.50= 9.22 sq.ins. 2 plates 14"X|" at 5.25 = 10.50-1. 50 = _9.00 " (Rivets in flange assumed as in Fig. 83) 18.22 Computation of h for this section. Centre of gravity of angles from back of angles = 1.03" (from handbook). Correction to allow for cover plates — '■ ^ — = 0.66". 10.54-11.72 .-. /i = 48.5"-2X(1.03"-0.66")=47.8". Hence, original assumption for value of h, while slightly too large, is sufficiently accurate and the trial section may be used. 67. Rivets and Riveted Joints. The flange rivets form the only connection between the flange and the web, hence the deter- mination of the proper size and distance apart of these rivets is an essential feature in girder design. The diameter of the rivet is ordinarily fixed by practical considerations, the common practice for structural work being to use I in. rivets. The distance apart of the rivets has to be computed. The question of riveted con- nection between two members is also of great importance. Thorough treatment of rivet spacing is found in treatises upon mechanics and will not be given here. The essential points with which the structural designer must be thoroughly conversant are as follows. A riveted connection may fail in one of the following ways: a. By the shearing of the rivets. Art. 67 RIVETS AND RIVETED JOINTS 123 b. By the crushing of the rivets or of one of the pieces upon which they bear. c. By the tearing of the rivets through one of the connected pieces. Under a it should be noted that the allowable shearing value of the rivet may be found by mxiltiplying its cross-section area by the allowable shearing stress per square inch, and that the area of a |- in. rivet is 0.60 sq.in., and of a | in. rivet 0.44 sq.in. In designing rivets to resist shear the plane upon which the maximum shear occurs must always be determined. If the maximum shear be equally distributed over two planes the rivet is said to be in double shear. The permissible bearing, or crushing, strength of a rivet against a given plate is determined by multiplying the allow- able bearing strength per square inch by the diameter of the rivet and the thickness of the plate in question. To satisfy the requirements stated in c, use the following empirical rule: Rivets may not be spaced closer than three times the diameter, and the distance of a rivet from the edge or end of a piece may not be less than 1\ in. for a ^ in. rivet if the edge in question be rolled or planed, or lA in. if it be sheared, though where possible this distance should be at least twice the diameter of the rivet. For other sizes of rivets proportional allowances should be made. The two following examples show the application of these rules to some simple cases: Problem. Determine number of | in. rivets needed in row a to connect plates shown in Fig. 88. Allowable shearing stress = 7,500 lbs. per sq.in. Allowable bearing stress =15,000 " Htofl -rivets a f^ Vplate 100 000 _^ r ""• S^s; I v!3 -60 000lb». - 60 000 Ibi. 5o I'late Fig. 88. Solution. The maximum shear on a plane through the rivets = 50,000 lbs. As the rivets are f in. diameter, one rivet will carry in shear 7500X0.6 = 4500 lbs., hence, if the strength of the joint is limited by the shearing strength of the rivets there are needed -~z — = 11 (+), or 12 rivets. These rivets are limited in crushing strength by the ^-in. 124 PLATE GIEDER DESIGN Art. 67 plate which carries 100,000 lbs. The value of the rivet in bearing against this plate equals |X 2X15,000 = 6560 lbs., and the number required = — — ^•— = 15(+), or 16. As this is larger than the number needed 6560 to prevent shearing 16 rivets must be used. Problem. Determine number of |-in. rivets required to connect plates in joint shown by Fig. 89. Use same rivet values as in previous problem. Solution. The maximum shear =100,000 lbs. and occurs between plates 2 and 3, or 4 and 5. The number of rivets needed to carry this shear = i^° = 22( + ) or, say, 23. 60 000 ibB.-<— c: 100 0001bs.-«— C 100 000 lbs.-< — d C:\ C\ Plains t:7 150 000 lbs 1^ C7 Thickness of Plates*l-?s' o-/ia Fig. 89. -It- iOOOMlba Ipi ' 'Plate'*! The bearing strength is evidently limited by plate No. 2, which carries 150,000 lbs. and has a thickness of i in., this producing a greater stress on the rivets than would be the case for the ^ in. plate earrving 100,000 lbs. The number required for bearings ' ' =22(+), or 23, hence for d5dO this joint the number of rivets is limited by either shear or bearing. The examples just given illustrate methods of computing connection rivets for plates carrying direct stress. Sometimes, however, it is necessary to transmit torsion as well as direct stress by means of rivets. Such a condition often occurs in steel-frame building construction where the connections of girders to columns must be given considerable rigidity to provide proper transmission of the wind stresses. The condition commonly occurring in such a case is represented diagrammatically by Fig. 90, in which the load P is applied at a distance x from the centre of the group of rivets, thus producing a torsion Px which must be carried by the rivets in addition to the direct load P. If the torsion were to be produced by a couple, as in Fig. 91, then the vertical load upon the rivets would be zero and a rivet Art. 67 RIVETS AND RIVETED JOINTS 125 might be legitimately assumed to offer a resistance to torsion varying directly with its distance from the centre of gravity of the group of rivets, and acting at right angles to the line con- necting it with the centre of gravity. Upon this basis the rivets at a, should each be computed as stressed equally and up to the allowable working load, while the other rivets would carry such Fig. 90. Fig. 91. proportion of the working stress as the distance cb is to ca when c is the centre of gravity of the group of rivets. The resistance to torsion of such a group of rivets may there- fore be expressed as follows: Let r = the allowable working value of the most stressed rivet. / = summation of the squares of the distances from the centre of gravity of the group of rivets to each rivet. d = distance from centre of gravity of group of rivets to the most stressed rivet. R = resistance to torsion of the group of rivets. Then T a (18) For the case shown in Fig. 90 the above method must be modi- fied to allow for the effect of the vertical load. To make this correc- tion it is only necessary to determine the allowable resistance to torsion consistent with the rivet carrying its share of the vertical load. This may best be done graphi- cally by the method indicated in Fig. 92. In this case the total vertical load is 20,000 lbs., of which each rivet is assumed to carry Fig. 92. 1500 lbs. With an allow- 126 PLATE GIRDER DESIGN Art. 68 able total working value of say 7500 lbs. per rivet, the com- ponent at a perpendicular to the line ac is found graphically to be 5800 lbs. The corresponding allowable components of the stress in the rivets at d, b, and e are larger, hence the rivets at a furnish the minimum resistance to torsion for the given working value and consequently give the limiting value of r in equation (18). Tables giving the resistance to torsion of various groups of rivets have been prepared by E. A. Rexford and are published by the Engineering News Publishing Co. 68. Flange Rivets. Ordinary Method of Computation of Pitch. Since it is through the rivets that stress is transmitted into the flanges it is evident that the number of rivets needed is a direct function of the variation in the flange stress, which in turn is a direct function of the variation in the moment, hence the distance apart of the rivets or the pitch (the pitch is to be considered the distance apart of the rivets measured along the flange, i.e., in the direction of the stress) varies inversely as the variation in the moment. In a girder supported at the ends and carrying a uni- form, load the curve of moments is a vertical parabola, with its vertex on the vertical line passed through the centre of the span, hence the variation in the moment is a minimum at the centre and a maximum at the ends and the variation in rivet pitch fol- lows the same law. This is also approximately true for such girders when loaded with other than uniform loads. A knowledge of the variation of the pitch is, however, insuffi- cient; it is necessary to determine the pitch itself. The following method of doing this is obvious. Compute the total stress in the flange at any section and compute also the total stress- at a section one inch from the first; the difference between the two stresses gives the increase in flange stress per longitudinal inch at that portion of the girder and this increase must be carried into the flanges through the rivets. If one rivet can carry p lbs. and if the increase in flange stress is x lbs., the proper rivet pitch . V to use m that portion of the girder is — . If the portion of the bending moment carried by the web be neglected in the determination of rivet pitch, this error being small and on the safe side, the increase in flange stress in the ordinary plate girder may be found by dividing the increase in moment by the distance between the centres of gravity of the Art. 68 FLANGE RIVETS 127 flanges. It has already been stated that the first derivative of the moment equals the shear and as it also equals the increase in moment, it follows that the rate of increase in flange stress at a given section equals the shear at that section divided by the distance between centres of gravity of the flanges. The following formula for rivet pitch in the flanges of a girder may therefore be given: V-^, (19) in which p = maximum allowable pitch in inches at section under consideration. /J = allowable load on rivet in pounds (usually the value of rivet in bearing on web). A = distance between centres of gravity of flanges in inches. F = maximum external shear on given section in pounds. Owing to the fact that if the distance between rivets be too great the different pieces in a compression member may wrinkle, it is customary to specify a maximum pitch not greater than 6 in. or 16 times the thickness of the thinnest plate connected. This restriction is frequently the controlling factor in determining the rivet pitch, and is commonly applied to tension members as well as compression pieces. Equation (19) is applicable only to rivets through the vertical leg. These are the rivets which carry the stress into the flange. The rivets through the horizontal leg serve to transmit a part of this flange stress into the cover plates and in consequence may have a larger pitch. It is customary, however, to use the same pitch for the vertical as for the horizontal rivets,^ hence the method given is, in general, all that is necessary. One special case must, however, be mentioned, as it is of frequent occurrence, viz., where the whole load or a part of it is transmitted into the girder directly through the flanges, such being the case in bridge stringers, in girders carrying brick walls, etc. The effect of such a loading is to impose upon the rivets a vertical load as well as a horizontal thrust, the stress per rivet being thereby increased and the allowable rivet pitch decreased. In solving such a case it is necessary to find the resultant of the ' At the ends of the cover plates it is customary to place the vertical rivets at "a small pitch for a, distance of one or two feet to ensure that the stress may be properly carried into the plate. 128 PLATE GIRDER DESIGN Akt. 69 increase per inch in the horizontal flange stress and in the vertical load per longitudinal inch, and to divide the value of the rivet by this resultant, the quotient thus obtained giving the pitch. 69. Flange Rivets. Precise Method of Computation of Pitch. The method represented by Eq. (19) is the approximate method of figuring rivet pitch which is generally used in plate girder design. In order to thoroughly understand the question of rivet pitch and to be able to properly figure the pitch in other cases which may arise, such as columns carrying bending, it is neces- sary to develop a more exact method. Such a method may also be well employed in investigating existing girders the strength of which may be in doubt. To obtain such a method the formula for horizontal shear may be used. Referring to Fig. 93, it is evident that the function of the rivets at a is to prevent the flange angles from sliding along the web; that is, the rivets must resist the longitudinal shearing tendency of the angles. Hence, if this tend- ency can be computed the rivet pitch necessary to withstand it can be determined. This com- (■■ O^ putation can be easily made by multiplying the intensity of the longitudinal shear at the bottom of the angles by the thickness of the web, using y^^ gg for Q ux thc determination of the intensity, the statical moment about the neutral axis of the girder of the angles and cover plates combined, but not of the por- tion of the web included between the flange angles, since the stress in this is carried by the web itself. This gives the shearing force per longitudinal inch in the flange which equals the increase in flange stress, and this can- be used as before in figuring the pitch. By the same method the correct pitch for the rivets at b can be determined by computing Q for the cover plates only. 70. Flange Rivets. Example in Computation of Pitch. To illustrate the application of these methods the girder shoA\'n in Fig. 83 will be considered. Problem. Determine the rivet pitch required at a section where the shear is 300,000 lbs., assuming that at this section all the cover plates are needed. Solution. The rivets through the vertical legs of the angles will first be considered, assuming that the outer force is applied directly to the web and not through the flange. Let the bearing value per Art. 70 RIVET PITCH COMPUTATIONS. EXAMPLES 129 square inch of the rivets be taken as 24,000 lbs. and the shearing value as 12,000 lbs. The strength of the rivet wiU then be limited fflther by bearing on the i^-in. web, whiph equals iX|-X 24,000 = 10,500 lbs., or by double shear, which equals 0.60X12,000X2 = 14,400 lbs. As the bearing value is smaller it should be used. Increase in flange stress per linear unit: Approximate method: Exact method : 300,000 90.4 =3320 I«=300,000 16-88(45.25-1.78)+30.0(45.25+0.94)^^g^^ Since one rivet can carry 10,500 lbs. the required pitch by the approx- imate method is 10,500 „,„„ -3^-=3.16",or,say,3", and by the exact method "^^=3.67", or, say, 3^". It is evident that the approximate method is decidedly on the safe side in this case. Were the required pitch Jess than three diameters of the rivet it would be necessary to locate the rivets in two rows as shown in Fig. 94, where a pitch of 2 ins. is assmned. To determine the pitch of the vertical rivets the exact method should be used. The increase in flange stress per inch is 30 X (45.25 +0.94) 222,295 X300,000 = 1870 lbs. The value of each rivet in this case is evidently its strength in single shear, but as there are two vertical rivets in each cross-section the pitch may be obtained by dividing the value of one rivet in double shear by the increase in flange stress per inch. This gives 14,400 1870 = 7.7", or, say, 7i". Fig. 94. As this exceeds 6" (see Art. 68) the pitch of these rivets should be made 6" or less. It will be noticed that the pitch is the distance between rivets measured along the axis of the angle. The vertical dis- tance between the rows of rivets must be sufficient to make the 130 PLATE GIRDER DESIGN Art. 70 distance d equal to or greater than three times the diameter of the rivet or 2f ins. for a ^ in. rivet. The distance e should not be less than IJ ins., as already noted, and should preferably be 1^ ins. or If ins. The distance c is determined by the amount of room needed for driving the rivet. The standard values for different angles are given in the steel-makers' handbooks. This example shows that the pitch of the vertical rivets may be considerably greater than that of the horizontal rivets. It is, however, usually made the same for practical considera- tions. There remains one other case to be treated — ^that of a girder supporting a load on the upper flange. To illustrate the method required for this case let the girder shown in Fig. 81 be considered, and let it be assumed that this girder is a railroad bridge stringer with ties resting directly upon the top flange. Let it also be assumed that the maximum wheel load crossing the stringer is 24,000 lbs.; that the maximum end shear including impact is 100,000 lbs.; and that the pitch of the rivets at the end of the stringer is to be determined. With the allowable unit stresses previously used the limiting value of the rivet is |Xi X24,000= 10,500 lbs. Using the approximate method the increase in flange stress is found to 100,000 be = 3800 lbs. per linear inch. This value must be com- W' 1 I i-3"-; 26.27 bined with the vertical load, carried by the rivets. Since the rail has considerable strength as a ;< m J beam, it is evident that a wheel load will not be carried entirely by one tie, but will be distributed over several. The common assumption is that one wheel load is distributed over three ties. If this assumption be made one wheel load will be distributed over the rivets in the space m. Fig. 95. If the ties are 8 ins. wide and spaced 6 ins. apart in the clear, three ties will occupy a total distance of 42 ins., hence the vertical load per inch which 24,000 ^^^ „ , . ^2 °^ 5"0 lbs., neglectmg the dead weight, which is so small compared with the live load as to be Fig. 95. the rivets must bear is Art. 71 DIRECT WEB STRESSES 131 negligible. To allow for impact this value should be doubled since these rivets are more directly affected by the shock of the locomotive than any other portion of the structure. To obtain the rivet pitch, it is therefore necessary to divide the value of one rivet by the resultant of 3800 and 1140. This result- ant may be obtained quickly and with sufficient accuracy by the graphical method indicated in Fig. 95. Its value is found to be 3970, hence the proper pitch at the end is 10,500 3970 = 2.65" or, say, 2|". This pitch can be used without difficulty for the girder under consideration since two rows of rivets should always be used in a 6 in. angle leg, and the actual distance apart of the rivets will, there- fore, be considerably greater than the nominal pitch. Were the 4 in. leg vertical instead of the 6 in. leg, a pitch of 2 1 ins. could be used but this is the minimum allowable value, and the adoption of the minimum value except where unavoidable is not recommended, a better plan being either to increase the depth or thickness of the web to permit larger pitch, or else to use a wider legged angle with two rows of rivets. It frequently happens tKat the determination of the flange section of girders is materially influenced by the ques- tion of rivet pitch, and the experienced designer will always look into this before selecting flange angles. 71. Direct Web Stresses. It has previously been shown that the intensity of the horizontal shear at any point in the web of a Fig. 96. Fig. 97. girder equals the intensity of the vertical shear and that these reach their maximum values at the neutral axis. Consider again an infinitesimal prism at the neutral axis. The shearing forces acting on this prism are shown in Fig. 96. These forces will develop internal forces of tension and compression, the value of which may be found as follows: 132 PLATE GIRDER DESIGN Art. 71 Let the thickness of the prism at right angles to the paper be unity, and let v equal the intensity of the shear. Then the total shearing force on each side = vdx. Resolving these forces into vdx components the value of each is found to be — j=, acting as indicated in Fig. 97. The effect of these components is to pro- duce on the diagonal plane bd a total tension = — =^. Since v2 the length of db = \^2dx the intensity of the tension on it is 2vdx _ V2V2dx In the same manner a compressive force may be shown to act on ac the intensity of which is also v. It therefore follows that at the neutral axis there exists a tension and compression acting upon planes at right angles to each other and at 45° to that axis, and that the intensity of these forces is equal to that of the shear. If the prism in question had been taken above or below the neutral axis these conditions would have been modified somewhat, through the introduction of direct fibre stresses. The effect of the shear in producing direct stresses would not be changed, that is, the shearing forces would develop direct stresses as before, but the final value of the tension or compres- sign upon any plane would have to be obtained by combining the direct stresses due to shear and the direct fibre stresses due to bending. The expression for the maximum direct stress for this more general case is developed in books on mechanics, and is as follows : p,^P^^1^4^2 + (p^_p^)2. In this equation p' = intensity of the maximum direct stress occurring at a given point on any plane, Vx and py are the intensities of the direct stresses acting at the same point on two rectangular planes passing through the point, and V is the intensity of the shear on each of these two latter planes. Fig. 98 illustrates this condition. It is evident that if point a is at the neutral axis of a beam subjected to bending but not to direct stress, p ^ and p,, are both zero and p' = v. Art. 71 DIRECT WEB STRESSES 133 The expression for the angle 9 between the x plane and the plane upon which the maximum intensity occurs is also derived in mechanics, and is as follows for a beam subjected to bending only : tan 6 = — . At the neutral axis of such a beam or girder v =p', hence 6 =45°. That tension and compression act as shown in Fig. 97 is also evident from the distortion produced by the shearing forces. It is plain that under the action of horizontal and vertical shear the prism which is rectangular when unstressed will take the shape shown, greatly distorted, in Fig. 99, and hence the line ac will be lengthened, and the line bd shortened. These changes can be produced only by tension and compression at 45° to the axis. l\.<- FiG. 98. Fig. 99. In plate girders the existence of this compression at 45° to the axis is of considerable importance since, if it be not recog- nized and proper means taken to provide for it, failure will occur through sidewise buckling of the web. To prevent such failure the web must be made of such thickness that there will be no danger of excessive compression, or else the buckling tendency must be restrained by other means. The latter is the common method, and is accomplished by the use of stiffeners in the form of vertical angles riveted to the web and extending from top to bottom of the girder. Sometimes, however, it is more economical of material as well as of labor, to increase the web thickness rather than to use stiffeners. In reinforced concrete beams the diagonal tension is an important factor since concrete is very weak in tension and means must generally be taken to provide against failure by rupture at 45° to the axis, either by steel rods placed at approximately 45° to the axis, or by vertical stirrups. 134 PLATE GIRDER DESIGN Art. 72 The combination of the direct stress due to shear with that due to bending gives a resultant compression in the web acting in the direction indicated by the dotted line in Fig. 100. The shape of this line ^ is dependent upon the relative value of the shear and the direct stress. At the end of a girder, where the shear is a maximum and the bend- ing moment a minimum, it would Pjq ^00 ^^^ ^^ ^^° ^^ *^® ^^^^ throughout its entire length. At the section where the moment is a maximum the shear is zero, hence at this section there is no direct stress in the web at the neutral axis, and the direct stress above or below this axis is parallel to it. 72. Web Stiff eners. The subject of stiff ener spacing is complicated and no accurate theory has yet been developed. Experimental results have been inconclusive, but indicate that the ordinary methods of practice are safe if not precise. The only theoretical method that seems rational is to treat a strip of the web as a column and to make an assumption as to the influ- ence of the remainder of the web upon this column. By this method an equation can be deduced for the distance apart of the stiffeners which should be in rational form, and which, if it does not give results exceeding the limits of good practice, may be used with security. Let Fig. 101 represent a portion of the web near the end of a girder where the shear is a maximum. Since the bending moment at the ends of the girder is small the direct web stresses act at approximately 45° throughout the entire depth of the girder, hence the strip of web to be considered is taken at a 45° slope. Its length is restricted by the flange angles and it is partially restrained against sidewise buckling by direct web tension at right angles to its axis as indicated in the figure by arrows : Fig. 101. Art. 72 WEB STIl'TENERS 135 Let unity = width of strip. i = thickness of web. i = length of strip. d = distance apart of stiff eners in clear. r = radius of gyration. 7= moment of inertia of strip. A = area of strip. Then l = dV2 and r=yjl=yj^j 12 t v/i2' P I A column formula of the form -i-=16,000— c— will now be A r applied to the strip. Since c for the ordinary unsupported pin- ended column may be safely taken as 70, it would not seem un- reasonable to reduce this materially here since the column is fixed at the ends by the flange angles and held sidewise through- out its entire length by the direct web tension. The amount which it should be reduced to allow for these restraining influences is unknown, but the value c = 25 will be adopted as a conserva- tive value, giving results which do not exceed the limits of ordinary practice. Substituting this value for the constant and expres- sing the value of — in terms of t and d gives the following equation: ^= 16,000-25^24= 16,000 - 120^ (very nearly), P .... in which -j- equals the allowable mtensity of axial compression in the strip. For the case in question this equals the shearing intensity, v, per square inch, hence the formula may be written thus, V= 16,000 -120y (20) (Formula (20) should be used only when the live shear has been properly increased for impact.) This formula may be used in either of the following ways: 1. To determine distance apart of stiff eners for a given shear and web thickness. 136 PLATE GIRDER DESIGN Art. 72 2. To determine the required web thickness for a given shear and distance apart of stiffeners or for the case where no stiflfeners are used. It should be noted that stiffeners placed further apart than the clear distance between the flange angles would not reduce the length of the strip .shown in Fig. 101, and hence would theo- retically be of no service at the ends of the girder where the shear has its maximum value. It is, however, customary to use stiffeners on all except shallow girders in order to stiffen the girder during fabrication and transportation, and a common requirement is that the maximum clear distance between stiffeners shall be the depth of web plate between flange angles, and shall not be greater than 5 ft. The following example illustrates the method of using this formula. Problem. Determine the required spacing of web stiffeners in the following girder : Depth, 40i" back to back of angles. Web, 40" Xi". Flange angles 6"X4"X V with 4" leg vertical. Maximum shear (live, dead, and impact) = 120,000 lbs. Solution. « = 1=^ = 16,000 -120p .-. d = 33ins. 4 ■ 40 • 5 7 As the clear distance between the flange angles is 40^"— 8" = 32i" it is evident that no stiffeners are needed, although the girder is just on the line. Had the web been thinner than Y', stiffeners would have been required. For example, with a f " web, stiffeners would be required at intervals of 16f" in the clear. For the portion of a girder where the bending moment is large and the shear relatively small the conditions in the web differ materially from those assumed in developing formula (20). For example, at the point of maximum bending moment the shear is zero, and in consequence no web compression exists in the half of the girder between the neutral axis and the tension flange, while the web compression in the other portion of the girder is parallel to the flange and increases in intensity as the distance from the neutral axis increases. Between the section of maximum shear and that of maximum moment the condition varies from that assumed in developing the formula to that just Art. 72 WEB STIFFENERS 137 stated. While it is evident that the formula does not apply very closely to all these conditions, the fact that it gives a grad- ually increasing distance apart of the stiffeners as the shear diminishes, is probably consistent with actual conditions. The size of intermediate stiffeners cannot be determined theoretically. A good rule is to make the outstanding leg equal to or greater than two inches plus one-thirtieth the depth of girder. The other leg should be of sufficient width to permit of proper riveting. There is perhaps no point in plate girder design upon which engineers differ so greatly as that of stiffener spacing, and carefully conducted experiments are greatly needed to estab- lish the necessary constants. The writer claims no special merit /o I on o I o |0\ o o o o o o o o o o o o o o o o — r o I o o o o o o o o o o o o o o o o o o o o O O o o o o o O ryr m i i ^ 1 Fig. 102. for his formula other than that it is derived from the column formula in common use at the present day, and that it gives conservative values. The ratio between the unsupported length and radius of gyra- tion of the' column strip shown in Fig. 101 should be restricted, as otherwise the column formula used would be inapplicable. If this ratio be restricted to 300, the corresponding value of - is 60, a commonly specified limiting value for girders without stiffeners. WhUe the ratio between the allowable unsupported length and radius of gyration may seem unduly high it should be remembered that this column is quite different from the ordinary column in being held throughout its entire length by the web tension. In addition to the angles required to stiffen the web against buckling, stiffener angles should be used at all points where 138 PLATE GIRDER DESIGN Art. 73 concentrated loads of considerable magnitude are applied to the girder, in order to transmit these loads into the web without over- stressing the flange rivets. The design of such stiff eners consists in selecting angles of sufficient area in the outstanding legs to with- stand the load without crushing and with sufficient total area to carry the applied load as a column, using the formula of Art. 18, and considering the unsupported length to be approximately one-half the depth of the girder. The number of rivets necessary to trans- mit the load into the web must also be determined, the value of the rivet being limited either by bearing on the web or by double shear. Both types of stiffeners are indicated in Fig. 102.* 73. Flange Plates. Flange plates are used to increase the flange area and thereby give a variable and more economical flange. It is not considered good design to use many cover plates. In general the total area of cover plates should not exceed one-half the total flange area, unless the largest sized angles are jFlange plates -20- Dead weight of girder = 300 lbs. per ft. ^ Fig. 103. Fig. 104. used. As the length of rivets should be limited in order to ensure good results, the thickness of the metal in the flanges should not exceed 4J ins. In case a larger flange area is required, vertical flange plates may be used, as shown in Fig. 103, or a box girder. To determine the proper location of the ends of a cover plate it is necessary to equate the bending moment which the girder can carr3r without the cover plate to the external bending moment at the end of the plate. The following example serves to illustrate this method: Problem. How far from the ends of the girder shown in Fig. 104 may the cover plates be cut? Girder to consist of a 48" X A" web, 6"X6"Xf" flange angles, and two 16"Xi" cover plates on each flange. ' For the results of experiments on the buckling of plate girder webs see article by Tumeaure in the "Journal of the Western Society of Engineers for 1907," Vol. XII. Akt. 73 FLANGE PLATES 139 Solution. Effective area of the tension flange members : sq. in. Two angles, 6"X6"X|" at 7.11 = 14.22 -2..50=U. 72 Two plates, 16"Xi"at 8.00 = 16.00-2.00=14.00 Web 1.48. A =2.25 To locate end of outside cover plate proceed thus : Effective flange area after plate is cut =11. 72 + 7.0 + 2.25 = 20.97 sq.ins. Distance from back of angles to e.g. of flange /i=48.5" -2.0 ins. = 46.5" Bending moment which girder can carry with one cover plate on flange. 20.97X16,000X46.5 12 = 1,300,000 ft.-lbs. ft.-lbs. Let a: = distance in feet from end of girder to point where plate may be cut. 3QQJ.2 Bending moment at s = 106,000x — „„„ 300a;^ 20.97X16,000X46.5 .-. 106,000x -—= 12 The value of a; as determined from this equation is 12.5 ft. The actual length of the cover plate should be somewhat longer than the theoretical length, in order that its stress may be properly carried into it. A foot is usually allowed at each end for this purpose. If this allowance be made, the cover plate in question would begin 1 1 .5 ft. from the end of the girder and its length would be 17 ft. The value of x for the cover plate nearest the flange is given by the following expression : aOOx' 13.97X16,000X45 106,000x —= ^ . In the case of girders subjected to moving concentrated load systems the following graphical method may be used to advantage. Plot the span and the external bending moments at each panel point, connecting these latter points by a smooth curve, which will be the curve of bending moments. This curve is prac- tically a series of straight lines and may be so used with safety if desired, the influence of the weight of the girder being offset by the fact that a straight-line curve for moving concentrated 140 PLATE GIRDER DESIGN Art. 73 loads gives excess moments throughout except at panel points. (See Art. 49.) Compute the allowable moment, Af ,i by the applica- tion of formula (17) for the controlling conditions, viz., no cover plate; one cover plate on each flange; two cover plates on each flange; and so on up to the maximum number of cover plates used less one. Plot these moments to the same scale as the external bending moments. Since these moments for each case are con- stant throughout the length of the girder, each may be represented graphically by a straight line parallel to the girder axis, the points of intersection of which with the curve of bending moments locate the points where the cover plate may be cut. This method is shown for a girder with two cover plates by Fig. 105. Ma, Ml and M^ are the external bending moments; Fia. 105. Mo is the allowable moment without cover plates ; ilf j with one cover plate. (The moment with two cover plates need not be plotted.) The cover plate nearest the flange should extend from m to n, the outer cover plate from o to p. These are theo- retical lengths and the actual plate should be made somewhat longer, as previously stated. The flange width is an important feature and should be care- fullj' considered in selecting angles and plates. It is common in railroad bridge practice to specify that the compression flange should be supported laterally at intervals not greater than twelve times its width, this being accomplished in half-through bridges by brackets attached to the floor beams and in deck spans by cross frames and horizontal bracing. In case it is necessary to deviate materially from this rule the flange should be figured as a column. For the sake of appearance it is usual to select cover plates of sufficient width to project slightly beyond the flange angles on either side. They should, however, project not more ' The allowable moment which the girder can carry may be called the moment of resistance. Art. 74 CONNECTION ANGLES AND FILLERS 141 than 2 ins. For example, flanges with 6"X6" angles should have plates not less than 13 ins. and not more than 16 ins. in width. Plates with a width in even inches should preferably be chosen. 74. Connection Angles and Fillers. It is necessary either to use fillers under plate girder stiff eners, or else to crimp the stiff ener angles over the flange angles. There is but little difference in the cost of the two methods, but the former is generally preferred. One objection to the use of fillers is that unless the filler is riveted to the web plate by an inde- pendent row of rivets, thus becoming practically a portion of the web (this type of filler is frequently called a tight filler), the rivets connecting the stiffener to the web are reduced in strength since they have to carry stress through the loose filler plate and thus are subjected to some bending. This is of no importance in intermediate stiffeners which serve merely to stiffen the web, but should be considered in the case of stiffeners carrying a con- centrated load into the web. In such cases if loose fillers are employed an excess of rivets, say 50 per cent, should be used. The use of tight fillers is also advisable in some cases in order to increase the bearing value of rivets which otherwise would be limited by bearing on the web instead of by Ww. The following example illustrates this: Problem. Determine whether sufficient rivets are used in the connection of stringers to floor beams shown in Fig. 107. Allowable unit stress per sq. in. upon rivets: Bearing Shear Machine 24,000 12,0.00 Hand 18,000 9,000 For J-in. rivets above units give the following working values: Machine-Bearing on ^-in. plate = 7875 lbs. Shear = 7200 lbs. Hand-Bearing J-in. =787.5 " " =o400 " Assume that the rivets shown in Fig. 107 are all that can be used in the angles. Field rivets, which are hand rivets, are shown thus (•). Solution. To carry the shear of 40,000 lbs. from stringer a, there are required =5.1 or 6 rivets to connect the stiffener angles to the web. 7o75 SECTION SHOWING BOTTOM FLANGE Fig. 106. 142 PLATE GIRDER DESIGN Art. 74. As indicated in the figure, the largest number of rivets that can be used is 6, but it is inadvisable to count upon those in the flanges, which are frequently fully stressed by the flange stress, and additio al rivets should be added if the filler is to be a loose one, hence it is necessary cither to use wider stiffener angles with two rows of rivets, or else a wide filler to increase the bearing value of the connection rivets. The latter o o o o o * nl * * n * * HI * * lH * • ml • Maximum end shear: stringer a = 40,000 lbs.; stringer b=30,000 lbs. Maximum reaction on floor beam from both strinEers=B5,000 lbs. Fig. 107. would be cheaper and consequently advisable. If the filler, therefore, be made wider and connected to the web by two extra rivets an addi- tional stress equal to that which two rivets can carry can be taken from the web into the filler and by that carried into the rivets connect- ing the stiffeners. As these rivets would, however, have to carry a considerable bending moment in addition to the direct shear it is advisable to make a liberal allowance, hence it would be well in this case to use four rivets in the fillers placed directly opposite those in the stiffeners. The stringer would then be as shown in Fig. 108. One other point yet remains to be con- sidered, viz., the shearing value of the rivets. The connection has so far been designed to carry the stress from the web plate into the rivets. Can the rivets carry this stress into the angles? As the thickness of the angles is not restricted sufficient bearing area can be obtained, but can the rivets carry the stress without shearing off? As the rivets are in double shear there , , 40,000 are needed . . .„„ = 2.8 or 3 rivets, hence the O O o o o o "6 o o o 1 o 1 o o o o o o Fig, 108. 14,400 number needed for bearing is also sufficient for shear, otherwise it would be necessary, despite the wide filler, to use wider stiffener angles. The connection of stringers to floor beam may be treated in a similar manner. Ten hand rivets are shown. These have to carry in single Art. 75 WEB SPLICES 143 shear the maximum shear in a single stringer, i.e., 40,000 lbs. They also have to carry 65,000 lbs. in bearing upon the web. 40,000 ;-.Q» <10, hence there are enough in .shear; 65,000 „, , y(iye =8C+) or 9) hence there are also enough in bearing. 75. Web Splices. Owing to the limited length of plates obtainable it is frequently necessary to splice the webs of long and deep girders. Fig. 109 shows several methods of making such splices. Of these the type used in A is best in appearance and is recommended for use. The design of such splices requires two distinct operations, viz., the determination of the size of the splice plates, and the o o o jo O Of lo o!o o I 10 do o f o o|o A\ O 0]0 o o o|o o O Q 0;0 O O I o o olo o o o o I m \o n Q ;o o o I 'eb Splices. In all cases ''plice plates are used on both sides of tlie web Fig. 109. determination of the number and location of the splice rivets. The former question involves the selection of plates that are of sufficient strength to carry not alone the shear at the section where the splice is to be located, but also the bending resistance of the web as given by formula 17, viz., ^ its gross area multi- plied by the product of the allowable unit stress and the distance between centres of gravity of flanges. For a splice of the type shown by A, Fig. 109, both of these considerations are usually satisfied by plates of the minimum allowable thickness, although for thick web plates or shallow girders the thickness of the plates should be carefully computed by the method used in the follow- ing example. The width of the plates if! usually determined by the number of rivets needed, and requires no computation. The rivets must be sufficient to carry the shear and bending moment which the splice plates are required to resist. Their computa- tion involves the application of the method given in Art. 67 for the strength of rivets in torsion. 144 PLATE GIRDER DESIGN Art. 75 One of the well recognized and important rules of good design is to so proportion the member that it will be equally strong at joints and other critical sections as in its main portion. The application of this rule to the design of web splices involves making the splice of sufficient strength to carry all the shear and bending moment which the web plate is capable of carrying. For many girders this would give excessive strength since the web is not called upon to resist maximum moment and shear simultaneously. It is, however, a safe rule to follow in all cases and should not be deviated from unless the location of the splice can be so fixed that it will surely come at a point where it will not be subjected to maximum conditions of both kinds simul- taneously.^ The example which fol- lows illustrates the design of a web splice for a girder, the web plate of which is supposed to be fully stressed in shear and bending. 1^ FiG. 110. Fig. 111. Problem. Design a web splice for a girder with a 6"xA" web, 6"X3|"XA" flange angles with 3J" legs vertical, and two 16"xf" cover plates on each flange, using the unit values specified in Chapter T. Solution. First assume the number and location of rivets which can be used in one ■\'ertical row. As showTi b)- Fig. Ill, eighteen rivets may be used, spaced 3 ins. apart. Had the minimum allowable spacing of three times the rivet diameter been adopted, a few more rivets could have been inserted in a row, but it is inadvisable to use the minimum pitch if it can be avoided. Next assmne that the minimum allowable thickness of material is -h in. and determine if this thickness will prove sufficient for the splice plates. If these plates are assumed to be fitted to the edges of the ver- tical legs of the flange angles their length will be 54^- ins. and the net area of the two splice plates through a row of rivets will equal 2 (54J — 18) ft ' Such a condition could exist at section a, in the girder shown by Fig. ] 10, provided a cover plate stopped a little beyond this point. Art. 75 WEB SPLICES 145 = 22.8 sq.ins. The net area of the girder web equals (61 -20)A = 17.9 sq.ins., hence A in. plates are ample to carry the shear. To determine whether their strength in bending is sufficient, the allowable resistance of the girder web as used in formula 17 should first be determined. This equals (Jj-61 • A)sA = 2.22X61.lXs = 135.62s. If there were no rivet holes in the splice plates their resistance to bending would be given by the formula, M = ^shK' = \s ■ (f ) (54i) ' = 309s. As already stated the allowance for rivet holes is commonly made by using a coefficient of \ in the formula for M instead of J. Making this allowance gives the following value for the resistance ' to bending : M= isbh'= 1 309s = 232s. This value is much larger than necessary, hence the J^ in. plates are of sufficient strength to carry bending and shear. To determine the number of rows of rivets required the allowable shearing and bending resistance of the girder web must be computed. These values are as follows: Shear, 61 X A XfX 12,000 = 240,000 lbs. Bending, 135.62x16,000 = 2,170,000 in.-lbs. If two rows of rivets are assumed on each side of the splice the vertical load per rivet will equal ' =6666 lbs. 2 X lo The method of Art. 67 may now be applied, but is somewhat laborious and no essential error will be made if the resistance of each rivet to tor- sion be assumed to vary with its distance from the neutral axis of the girder instead of its distance from the centre of gravity of the group of rivets, and to act in a direction parallel to this axis. Making this assump- tion the resistance of the outermost rivet to bending will equal: /' 9187^ -6666^ = 6320 lbs. The value of / in formula 18 has already been computed for a half row or rivets. (See foot-note.) The resistance to bending of the two rows of rivets may now be written (ROOf\\ -—-) (2X2180) =2,161,000 inch-lbs. 25.5/ This is practically equal to the value previously found for the allowable, web resistance, viz., 2,170,000 in.-lbs. Were the resistance to bend- ' The actual effect of the rivet holes in the tension half of the splice plates in this case is to reduce the value of / for the gross area by the following amount: |(25.5H22.5H19.52-M6.5^-M3.52-h 10.52 -l-7.52 + 4.52-M.52) = |(2180) = 1362. The value of / for the gross area =tV-|-'(54 5)'=8431; hence the reduction made by using the coefficient J instead of J is ample. 146 PLATE GIRDER DESIGN Art. 76 ing of the rivets less than the allowable bending moment of the web another row of rivets might be used or the splice located at a point where the web is not fully stressed in bending and shear simultaneously. A good location in such a case would be at a point a slight distance toward the centre of the girder from the end of a cover plate. At such a point practically all the cover plate area would be in excess and could be coimted upon to make up the deficiency in the strength of splice. The method of calculation illustrated by the previous example is not strictly accurate, and probably less so than for the cases given in Art. 67, where the number of rivets in a vertical row was much less. It should be noted that for such cases the dis- tribution of the shear over the rivets is probably by no means uniform, the rivets near the neutral axis, where the shear is a maxi- mum, probably carrying more than those nearer the outer fibres. In practice it may be found desirable to use splice plates thicker than those required by computation. If the splice plate be used as a filler it should be as thick as the flange angles. It is, however, possible in such a case to make up the total thick- ness required by the use of a ^-in. splice plate and a filler, an arrangement frequently used. 76. Flange Angle Splice. In very long girders it is frequently necessary to splice the flange angles. When this is to be done only one angle in each flange should be spliced at a section. A common practice is to splice the top angle on one side of the girder and the bottom angle on the other side at a section a little to one side of the centre of the girder, and to reverse this process for a corresponding section on the other side of the centre. The splice should always be made by another angle the cross-section area of which should be equal to that of the angle to be spliced. In order to simplify the construction the splice angles for the tension flange should be exactly like those for the compression flange, hence the net area of the splice angle should equal the net area of the main angle. In order to obtain a splice of neat appearance and which answers the above requirements it is usually necessary to select an angle with the same width of legs as the main angle, but fg in. or I in. thicker, and to plane off the projecting legs so that they may be flush with the main angle. The following example illustrates this: Determine the splice angle required for a 6"X6"X2-" flange angle. The net area of Art. 76 FLANGE ANGLE SPLICE 147 the main angle = 5.75 -1.00 = 4.75 sq.ins. The net area of a 6" X 6"XA" angle planed to fit the 6"X6"Xi" angle = 6.44 -1.12 - 2(AXi)=4.76 sq.in., hence this angle has just the right area and should be used. Fig. 112 shows by cross-hatching the portion of the angle to be cut off. The outer corner must also be rounded off as indicated to fit the fillet of the main angle. The number of rivets required in the splice angle may be determined, if there are no cover plates, by computing the stress which the angle can bear and dividing it by the value of one rivet, the strength of the rivet being generally limited by single shear. If the angles are equal-legged one-half the number of rivets needed should be used in each leg. If the legs are unequal O (: ( — w ) FiQ. 112. Fig. 113. each leg should have its proportional part of the total rivets required; e.g., if a 6" X4" angle is to be spliced and if 20 rivets are needed in all, put ^^o-20 = 12 rivets in the 6-in. leg, and 3^-20 = 8 rivets in the 4-in. leg. Cover plates are generally required in girders which require flange angle splices, and in such girders the number of splice rivets needed may be somewhat in excess of the number required to carry the total stress which the angles are good for, since the increment in stress in a distance equal to the length of the splice angle must be taken care of. Referring to Fig. 113, it is evident that the rivets at a must carry from the flange into the splice angle in the distance mn one-half the increment in flange stress in that dis- tance (the other half going through the same rivet to the flange angle on the left-hand side) plus one-half the stress in the main angle at m (since the angle is equal-legged). The rivets at c should be computed to carry the same amount, since it is proper to assume that all the increment in flange stress is carried by the 148 PLATE GIRDER DESIGN Akt. 77 cover plates, the angle being fully stressed before cover plates are added. The rivets at both a and c are limited usually by single shear, and should be designed accordingly. Since the splice would ordinarily be placed near the centre of the girder where the increment in flange stress is small, it is usually sufficient to determine the number of rivets required to splice the angle, assuming it to be stressed to its full value and to add one or two rivets to carry the flange stress increment. If no cover plates are needed it is unnecessary to consider the increment in stress since if the splice rivets be determined for the full value of the angle they will surely be sufficient to carry the stress in the angle at m plus the increment in mn. 77. Cover-plate Splice. A cover-plate splice may always be made by the addition of a splice plate of the same size as the plate to be cut, and the use of sufficient rivets to transmit the full stress from one plate to another, with the addition of a liberal percentage of extra rivets, say 33 per cent for each plate intervening between the plate to be spliced and the splice plate. Such a splice is shown in Fig. 114. The disadvantage of long rivets, subjected perhaps to bending moment because of the intermediate plates, together with the unsightly appearance of such a splice, makes it desirable if the girder has more than one cover plate to splice one cover plate by means of another. This may be done by properly choosing the section where the splice is to be made. This is illustrated by Fig. 115, in which the lower cover plate , r\ r^ f^~^ r\ ] ^, '-^-^ '^ • \ ^ / ■o- W - \ Fig. 114, If lower comer plate were not to be Ct splk'ed top cover plati; should begip gt ft Fig. 115. IS to be spliced. If this plate be cut at a, where the top cover plate should begin, and if the top plate be extended to b, making the distance ab such that enough rivets can be put between a and b to carry the stress that the plate is good for, the splice will be ^K^- 77 COVER-PLATE SPLICE 149 properly made. If the top plate be thinner than the bottom plate the splice would have to be located nearer the end of the girder at a point where one cover plate of the thickness of the top plate would be just sufficient to carry the stress. It will be observed that this method is based upon the transfer of all the stress from the end section of the plate to be spliced into the plate immediately above it. The intermediate section of the spliced plate instead of the upper plate will then take the additional increment of flange stress. PROBLEMS 40. a. Compute - for this girder with respect to the neutral axis and to the axis ZQ, and compute maximum fibre stress for a total uniformly distributed vertical load of 3200 lbs. per foot over the entire girder. Span = 40 f t. z 6. Compute the maximum fibre stresses in ./^' botli flanges for the loading given in a, using s i-..o". 6-i vl formula (17), Art. 62. Allow for holes for |-in. rivet. y^ c. Compute by both the approximate and w.b3o\M:. exact methods the required pitch of the hori- _ _ r zontal and vertical flange rivets at end of top '■■■»« "y^g flange assuming all the load to be applied ^uto «•,)<■ directly through the flange, and one cover plate on each flange to extend to end of girder. Use Prob. 40. unit values given in Art. 18, and |-in. rivets. d. Determine distance from end at which the cover plates may be cut if desired. e. Determine whether intermediate stiffeners are needed. /. Determine the size of stiffeners needed on the girder to support the top flange under a concentrated load of 200,000 lbs. Allow 20,000 lbs. per square inch bearing on stiffeners and assume that their outstand- ing legs only are effective. CHAPTER VI SIMPLE TRUSSES 78. Trusses Defined. A truss is a structure composed of a number of separate pieces connected at their intersections only. The connection is sometimes made bj' plates to which the mem- bers are riveted, such a truss being called a riveted truss, or some- times less properly a riveted girder; or else by large bolts or pins, the latter type receiving the name of pin truss. The points of intersection of the members are called the joints, and the outer forces should be applied so far as is possible at these joints only. This is accomplished by the use of floor beams or in a roof by purlins. As the depth of plate girders is limited by the avail- able width of plates and by the inability to ship by rail single pieces wider than 10 ft. 6 in., or thereabouts, it is necessary to use trusses where economy or rigidity require greater depths. The common practice in the United States at the present time is to use beams or girders up to lengths of 90 or 100 ft.; riveted trusses above these lengths up to 150 or 175 ft.; and pin trusses above this length. The use of shorter pin-truss spans for rail- road bridges has been given up because of lack of rigidity and consequent early wearing out of the bridge. A typical pin truss is illustrated by Fig. 4. 79. Classification. All trusses may be divided into two general classes based upon the methods necessary for the deter- mination of the stresses in the members; if these stresses can be determined by statics the truss is statically determined; other- wise it is statically undetermined. It should be noted that a truss may be statically undetermined with respect to the outer forces, i.e., the reactions cannot be determined by statics, and yet be statically determined with respect to the inner forces and vice versa. The former is usually the case with draw bridges the latter with the double intersection trusses sometimes used in simple span bridges. 150 Art. 80 THEORY 151 80. Theory. The theory upon which the computation of truss stresses is based assumes that the members are connected at the intersections of the centre of gravity lines by frictionless pins, and that in consequence the stresses are direct stresses. That this deviates considerably from the truth for riveted trusses is evident; the error in pin trusses is less, but not negligible, hence the common theory of trusses is by no means an exact one. The secondary stresses produced by resistance to motion at the joints are, however, small in well-designed trusses, as compared with the primary stresses, as the stresses computed by the above assumption may be called, and experience shows that for simple spans of ordinary length these primary stresses are sufficiently exact to be used in designs where the ordinary factor of safety is applied. 81. Methods. The methods necessary for the computation of the stresses in statically determined trusses are very simple, and consist merely of the application of the three equations of equilibrium to portions of the truss, these portions being chosen in such a way as to enable the stress in a given bar or bars to be immediately found. There are in common use three methods of accomplishing this result; the method of joints, the method of moments, and the method of shears. All of these are applica- tions of the general method and differ only in detail. In the computation of a truss it is often advantageous to employ all three methods, choosing for each bar that which is best adapted to it. The method of joints is the most general of these methods and will be considered first. 82. Analytical Method of Joints Described. Fig. 116 repre- sents a simple truss carrying a load at the apex. Let a section be taken around the joint at a and the remainder of the truss removed. As the f entire truss is in equilibrium the portion /"XT f" enclosed by this section, shown by a circle ^ \/ \^^ h in the figure, must also be in equilibrium ( "^—\ -^-1- andtheproblemresolvesitself into that of \^^y l ^ determining the forces in the bars con- T sistent with equilibrium of the various Fig. 116. portions of the truss cut off by similar sections taken at a sufficient number of joints to permit the determination of all the unknown stresses. Fig. 117 shows the 152 SIMPLE TRUSSES Abt. 83 condition which exists at joint a, assuming that the stresses in the bars are axial stresses, ohis being in accordance with the general theory. It should be carefully observed that this method deals with the stresses in the bars rather than with the bars them- selves. Referring to Fig. 117 it is evident that as there are but two unknown forces, Si and S2, the two equations of equilibrium, S^ = and 2F = 0, will be just sufficient to determine these, and since all the forces meet at a point the equation I;j1/ = will be satis- fied by any value of Si and 1^2 and need not be considered. AA'ith the stresses in bars ah and Fig 117 ^^ thus determined, a section may next be taken at either joint h or joint c, and the stress in bar he computed in a similar manner, thus completing the necessary computations for this truss. 83. Character of Stress. The determination of the character of the stress is often more important than that of its magnitude, as a bar designed for tension may fail if the stress is compression even if its magnitude is small. To determine the character of the stress in any bar it is sufficient to arbitrarily assume the direction of the stresses before applying the equations of equilib- rium. If the solution gives a positive restilt for a stress it shows that this stress acts in the direction originally assumed. In this connection it should be carefully observed that the inter- nal stresses in a bar subjected to tension continually tend to pull the ends together; that is, to shorten the bar, hence tension in a bar always acts away from the joints at hoth ends, and compression toward the joints. Fig. 117 illustrates this. *S2 is shown acting away from the joint, that is, in tension; Si on the other hand is assumed in compression and is shown acting toward the joint. If the joint at the top of the truss should next be investigated it would be necessary to show Si as acting toward that joint also, since the computations give a positive value for Si, thus indicating that it is compression. It is safer for the beginner to assume the stress in each bar as tension, or away from the joint. Positive values will then indicate tension and negative values compression. This is in accordance with the common but not universal convention of representing tension (which increases the length of a bar) by a plus sign. Art. 84 DETERMINATE AND INDETERMINATE TRUSSES 153 84. Determinate and Indeterminate Trusses. For the truss under consideration there are three unknown bars. The stresses in two of them have been determined by considering one joint only; the stress in the other may be found by taking either of the other joints. Since by taking both of the other joints there would be four equations and only one other unknown bar it would seem as if there were too many equations. This is not correct, however, as these equations must suffice to determine the unknown reactions as well as the unknown bars, since equilib- rium of each joint involves equilibrium of the entire structure; that is, for this particular structure and in general for all struc- tures which are statically determined with respect to the reactions there must be three more equations than there are bars. In other words, the six equations of joints for such a truss are not independent but are related in such a manner as to satisfy the three general equations of equilibrium for the truss as a whole, viz., SX=0, 2F = and IlAf = 0, which may for most cases be replaced by the more common equations ^^■ = 0, 27 = 0, SM" = 0. There are therefore for the truss shown in Fig. 116 but three independent equations which can be used in determining the bar stresses, hence these stresses are determinate. In general it may be said for all planar trusses which are statically determined with respect to the outer forces, that if n equals the number of joints, 2n— 3 equals the number of bars which the structure must have to be determinate. If it has more bars the stresses can not be computed by statics; if less it will not be rigid and will collapse except under special conditions. If it be desired to build a structure which because of the number of points of support or other reasons would ordinarily be statically undetermined with respect to the outer forces, it may be possible to make the structure determinate in all respects by properly choosing the number of members. For example, in the case of a cantilever truss, a diagonal over a pier is sometimes omitted for this reason. In swing spans diagonals are often omitted or made very small in order to reduce the numbers of unknowns. If the unknown components of the reac- tions be four it is evident that there can be only 2n— 4 bars if the structure is to be made determinate. 154 SIMPLE TRUSSES Akt. 85 85. Mode of Procedure. Analytical Method of Joints. In the solution of problems by the analytical method of joints the following mode of procedure should always be adopted: 1. Compute reactions. 2. Select a joint at which only two bars meet. 3. Assume the stresses in these bars to be tension, that is, to act away from the joint; and apply the equations of equilib- rium. If the stress in either bar is found to be negative it indicates that the bar is in compression instead of tension. 4. Consider any other joint at which only two unknown bars meet and determine the stresses in these bars in the same manner and proceed thus until all the stresses have been determined. 86. Application of Analytical Method of Joints. The follow- ing numerical example has been worked out to show the applica- tion of this method ; 6000 Fig. 118b. Fig. 118c. Joint B: SV=0: Fi +27,500-5000 = 0, Fi= -22,500 lbs. 40 ffi = —Fi= -30,000 lbs, 30 50 »Si=—7i = -37,500 lbs. 30 Art. 86 APPLICATION OF ANALYTICAL METHOD OF JOINTS 155 SH=0: Hi+S5 = 0, o - £ 00 OO o c-j CO 00 Oi (N T-H en »o CO rr ^ (M i*^! CO o 1-H + + 1 + 1 1 + + + 1 ' + o S ^ v: GO -^ rH o CO t^ o o o CO CO o r^ t^ -f -^ d CO I* CD cj o CO r^ .— M CI -t^ ■* lO ^ CO (N T)< Ol I-H I-H 3 + + 1 + 1 1 + + + 1 1 + 9 11 II II II II II II II II II II II o 'O lO 'O lO Olti O |0 lO lO C-. |0 c; |0 OV |0 CO |0 -; oi \n CI Iro lO Iro to Ico t~ Ico CO Ico CO Ico CO Ico CO Ico o X X X X X X X X X ■^ o o o o o o o o o o o ^ •^ 'Jl ■* Tf -t^ 'I" TP ■* ^ Tt< -*< a X X X X X X X X X X X s lO |!£) .C |tO rt< |a5 -f |cO CO |co >o |co T), |C0 co|co CO [cO CI |co (N |cO i + + 1 + 1 1 + + 1 1 + ifi 0) _> 1—5 1 (M Dl o o T— 1 o> >o lO o o ■ o 00 5 to CO o o ^ CO M t^ lO >o l^ 00 g lO •o lO UO 00 Tt< CO Oi t^ t- CO «^ I— 1 i-H M N IN IN T-H .2 + + 1 + 1 1 + + + 1 1 + "c II II II II II II 11 II II II II II u o o o o ■ O liO ■O liO ira i.fs ira [ic lO i»o lO ILO Oi lO 0> lO C31 |0 1- Ico W lc<3 t^ Ico t^ 1^? t^ Ico t^ Ico CO Ico CO Ico 1 — 1 i-H 1 — 1 CO Ico X X X X X X X X X X X X o 03 Ol t^ t^ M rj-j 'O LO o lO lO lO t^ t^ t^ t~ r^ m CO CO T— I r-H CO CO T— 1 2|co CO |co CD|cO CO |co CO |co 5 T— ( 1 — 1 ^ m a; + + I + 1 1 + + + 1 1 + > •^ i o CJ r- t- CI tXl CO oo o lO o ca C~] IM CD O CO X' 01 lO t^ d ■a m t> t^ y-\ cc I-H --D 01 IN (M I-H L. '^ i-H o 0^ + + 1 + 1 1 + + + 1 I c Q p — a o o o O o o o o o o o CO « CO CO CO CO CO o o OS ^^ \. ^ IC lO lO LO lo c» c: o a "3 (N (N (N M (M CO CO CO '-^ -*J i-i o _!z: m s in lO o o >o »o lO lO o lO o t> S [^ r^ o c= r^ r^ CI t^ lO r^ o Oj 00 oo ^ -^ lO CC lO (M (N I-H X I-H S "S + + 7 + 1 1 + + + 1 1 ^ S3 o ° C o P3 4" b (-i* tT i4' 178 SIMPLE TRUSSES Art. 103 103. Computation of Stresses. Warren Truss. Uniform Load with Locomotive Excess. Problem. Determine the maximum stresses of both kinds for aU the bars of the truss shown in Fig. 143 with the following loads: Dead weight of bridge, 600 lbs. per ft. per truss, top chord = 9,000 lbs. per panel. 200 " " bottom chord = 3,000 " panel. Uniform live load, 2000 " " top chord _= 30,000 " panel. Locomotive excess, " " =25,000 lbs. Fig. 143. — Bead Panel Loads and Index Stresses. Single-Track Deck Warren Truss Railroad Bridge. Index Stresses. These and the panel loads are shown in Fig. 143 in units of 1000 lbs. The net reaction at left end evidently equals 9X2+|3 = 25.5, which checks the index stress in UqLi. To check the index stress in the centre member of the top chord moments should be taken about L5. Since this is not a panel point for both chords this moment does not equal ipL^, but may be found from the moment of the reaction minus the moment of the panel loads. By this method the stress in the centre top chord bar equals (25.5X2^-3X3-9X2) ^ = 73.5. Since the diagonals make an angle of 45° with the horizontal the chord index stress equals the actual stress and is therefore correct. The index stress in the centre member of the bottom chord plus the index stress in ?74L5= 73.5 = the index stress in the centre member of the top chord, and is therefore correct. Art. 103 WARREN TRUSS 179 For this truss the live stresses cannot be computed from the dead index stresses since the bottom chord joints are not directly under the top chord joints; as the chord stresses for the uni- form live load have maximum values for full loading they can, however, be determined by the method of index stresses, and Fig. 144 shows these stresses for a full uniform live load. The moment at 174 for this case equals f^j-Ci-pL^), using Li+120 L3+I8O Lj Fig. 144. — Panel Loads and Index Stresses. Full Live Load. the method of Art. 43, hence the tension in bar L3L5 in thousands of pounds _24 75X75 -25 *^ ^:5^-^^"- This equals the index stress in this member and also in U^Ue as should be the case since the index stress in diagonal [741/5 = 0. Position of Loads for Maximum Ldve Stresses: C/qLi and L1U2, full uniform load with E at C/2. U2L3 and I/3C/4, uniform load from right up to and including U4, E at Ui. C/4L5, uniform load from right up to and including Uq, E at Ue. -^5^76 = maximum compression in UiLs, uniform load from right up to and including Us, E at Uq. C/ei'? = maximum tension in U^s, full panel load and loco- motive excess at Ug. U0U2 and LiLz, full uniform load, E at Ui. U2 U4 and L3L5, full uniform load, E at U4. U4U6, full uniform load, E at f/4 or Ue. Maximum Stresses. These may now be computed and are given in thousand-pound units in the following table in which all the necessary computations are shown: 180 SIMPI-E TRUSSES MAXIMUM STRESSES— WARREN TRUSS. Art. 104 Bar. Dead Stress in Units of 1000 lbs. Live Stress in Units of 1000 lbs. u,u. -25.5 - 60+|25\ = - 80.0 u,u, -61.5 - ^50+^25X3^ -- 195.0 VJJ, -73.5 - 'lS0 + |25X5^ = -230.0 L,L> + 48.0 + 'l20+-|25X2^ = + 160.0 L,L, + 72.0 + 'l80+|25X4^ = = +240.0 U,L, +-«■;?■ = + 36.1 + W|-)^' = + 113.1 L,U, — '-'-^f- = -36.1 - ;"4-)';:r = -113.1 UrL, +»«w= = + 19.1 + S»4-)^' = + 72.1 L,U, -»-'?f= = -14.8 - (1-4-)^' = - 72.1 U,L, . .-5x!M! = = + 2.1 + ;i»+HW = + 39.6 *L,U, Same as f jLj = = + 2.1 - ;i»+H^' = - 39.6 *U,L, Same as f/jLj = = -14.8 + .i,»..«-«' = + 15.6 * In tills truss no counters are used, lience it is necessary to compute the maximum stresses of both Icinds in all diagonals in which the live stress may tend to reverse the dead stress. This is easily done in the manner shown above. 104. Computation of Stresses, Subdivided Warren Truss. Uniform Load with Locomotive Excess. Problem. Determine the maximum stress of both kinds in all the bars of the truss shown in Fig. 145 with the following loads: Dead weight of bridge, 1000 lbs. per ft. per truss, top chord =25,000 lbs. per panel 500 " " ' bottom" =12,500 " " Uniform live load, 2,000 lbs. per ft. per truss, top chord =50,000 lbs. per panel Locomotive excess, = 30,000 " Art. 104 SUBDIVIDED WARREN TRUSS 181 Index Stresses. These are shown in Fig. 145. Their com- putation involves no difficulty. To check the index stress in UiU^ use the method of moments as follows: U- -10-Paiiels-@-25-0-=-2iJO— siUo lUl 4-U2 -iUa 4U4 ■(•Us ■ Wi iUi lUs *Uo Wii - vLa ■IL4 ^Li _. m u? >o U3 lO lA ii^ ^ lA c4 c4 c^ g Fig. 145. — Dead Panel Loads and Index Stresses. Single-Track Deck Subdivided Warren Truss Railroad Bridge. Stress in U^Us^iXl 500 250-250 30 = 390,625 lbs. From the index stress the actual stress in this bar = 468,750 X 25 30 = 390,625, hence index stresses are correct. Position of Loads for Maxhnum Live Stresses. For the chords the uniform live load should extend over the entire span and the stresses due to it may be computed directly from the dead stresses. The locomotive excess should be placed as follows: Bars U„U, and U,U,. E at i7,. Bar.s LyL., and Z/j^S' ^ ^t [/,. Bars TJ-^V^ and U^Ui, E at U,. Bars LjL, and LJj^. E at XJ^. Bar U^U^, E at U^. For the diagonals the uniform load and locomotive excess should be placed to give maximum shear in the different panels, i.e., full uniform live panel loads to the right of the panel contain- ing the bar in question, and the locomotive excess at the nearest panel point to the right. For the verticals it is evident that the maximum stress in all odd numbered bars like UzL^ will occur with full live panel load and locomotive excess at top panel point, while the even numbered verticals will have no live stress. Maximum Stresses. All necessary computations for these, together with the final values, are given in the following table in units of 1000 lbs: 182 SIMPLE TRUSSES Art. 105 MAXIMUM STRESSES— SUBDIVIDED WARREN TRUSS Bars. Index Stress. Multiplier. Dead Stress. 1000 lb. Units. Live Stress Due to Uniform Load. 1000 lb. Units. u„u,x U,UJ -168.75 25 30 -140.6 50 -140.6X37-5= -187.5 U,UJ -393.75 25 30 -328.1 50 -328.1X31^= -437.5 u.u. -468.75 25 30 -390.6 50 -390.6X37-5= -520.8 + 300.00 25 30 + 250.0 50 + 250.0X37-g = + 333.3 + 450 25 30 + 375.0 50 + 375.0X37-5= + 500.0 U,L, + 168.75 39 30 + 219.4 + 219. 4X3^— = +292.5 L,U, -131.25 39 30 -170.6 36 39 -5OXJ0X30 =-234.0 U,L, + 93.75 39 30 + 121.9 28 39 + 50XjqX3(j = + 182.0 L,U, - 56.25 39 30 - 73.1 21 39 -5OXJ0X3Q =-136.5 U,L, + 18.75 39 30 + 24.4 + ™XI5X| =+ 97.5 *L,U, + 18.75 39 30 + 24.4 -50Xjqx|5 =- 65.0 * U,L, - 56.25 39 30 - 73.1 + ^OXj^x| =+ 39.0 *L,Us + 93.75 39 30 + 121.9 3 30 -S°XjoX3o =- 19.5 U,L,^ U,L, - 25.0 1.0 - 25.0 = - 50.0 U,L,i + 12.5 1.0 + 12.5 0.0 * The live stresses in these bars are maximum stresses of the opposite character to those occurring in the corresponding bars in the other half of the truss. (Table continued onnextpage.) 105. Computation of Stresses. Bridge Trusses with Non- parallel Chords. Uniform Load with Locomotive Excess. To compute the stresses in such trusses it is necessary to modify somewhat the procedure adopted in the simple parallel chord trusses hitherto treated. This is due to the fact that the web stresses can no longer be directly determined by the method of shear, owing to the influence of the inclined top chord. Although the modification is in mode of procedure rather than in principle Aht. 105 SUBDIVIDED WARREN TRUSS 183 MAXIMUM STRESSES— SUBDIVIDED WARREN TRUSS Live Stress Due to Locomotive Excess. 1000 lb. Units. Total Live Stress. 1000 lb. Units. 9 25 -IqXSOX 3o=-22.5 -210.0 7 75 -IqXSOX 3o=-62.5 -490.0 u,u. 5 125 -joXSOXgo- 62.5 -583.3 8 50 + j;oX30X 3o=+40.0 + 373.3 6 100 + ^OX30X30-+60.0 + 560.0 U,L, + ^X30X | = +35.1 + 327.6 LJJ, 8 39 -joX30X 3o=-31.2 -265.2 U,L, 7 39 + ^X30X 30= +27.3 + 209.3 L,U, -roX3«xS=-23.4 -169.9 U,L, + jjjX30X 30=4-19.5 + 117.0 *L,U, -n)X^°x|=-i5.6 - 80.6 *U,L, + 5^X30X S=+11.7 + 50.7 ''L.U, -roX^°xi=-^« - 27.3 -30.0 - 80.0 U,LJ 0.0 0.0 it seems desirable to illustrate the necessary computations for such a truss, hence the following example is given. Problem. Compute the maximum stresses for the truss shown in Fig. 146 with the following loads: Dead weight of bridge, 600 lbs. per ft. per truss, top chord= 15,000 lbs. per panel. 1200 " " " " " bottom " =30,000 " " Uniform live load, 3000 lbs. per ft. per truss, bottom chord = 75,000 " " Locomotive excess, =40,000 184 SIMPLE TRUSSES Art. 105 The determination of index stresses for this truss requires some explanation. The inclination of top chord members adds vertical forces at joints Ui, U2 and U^, hence the vertical com- ponents in the inclined chord members must be determined before the index stresses for bars meeting at these joints can be written. In the trusses previously considered all the diagonals had the same slope, and multiplication of the chord index stresses by the ratio of horizontal to vertical projection of the diagonal gave actual stresses. It is obvious that in order to follow this same method in the truss under consideration, some modification must be adopted. The simplest method in this case is to correct the index stresses in bars ^72^3 and C/3L4, before writing chord index stresses. The best method of accomplishing this is to multiply Fig. 146. — Dgad Panel Loads and Index Stresses. Single-Track Through Non-Parallel Chord Pratt Truss. the index stress in each of these bars by the inverse ratio between its vertical projection and that of diagonal U1L2, that is, multiply the index stress in U2L3 by f , and that in UsLa by f .* * The correctness of this method is illustrated hy the following example: Let Vi, VI, and V3 be the vertical components in the diagonals of truss shown in Fig. 147, and S^, II2 and IT3 the stresses in the bottom chord Evidently /f,= rix J3, = (F,-FF,)|-. and H,= {V, + V2)^+V,xf .L.,r.+v.,£+(,,x,*)| FxG.147. =r>+^^^+^^3X/;)F For trusses in which the height is constant, but the panel length variable, the same method may be applied with the panel lengths substituted for the heights. Art. 105 BRIDGE TRUSSES WITH NON-PARALLEL CHORDS 185 The effect of this is to reduce the chord index stresses to the values they would have if all the diagonals had the same slope as UiL'i- The computation of the vertical components due to dead load in the inclined top chord members follows: V.C. Bar C/jC/a-Dead Load (157.5X^-45X^^^ = 38.6 ^25\ '35/50 (157.5X|-. V.C. Bar t/zt/g-Dead Load fl57.5X^ -45x1^')^ = 42.2 \ 40 40/ 50 With these known, the vertical components in the web members may be written at once, beginning at the centre in the usual . manner and obtaining a check at the end where the vertical component in the inclined end-post LqUx is found to equal the net dead reaction. It will be noticed that the effect of the vertical component in U2U3 is to cause tension in bar U^Ls. In a parallel chord truss this member, with the other verticals, would always be in compression under dead load. The corrected values of the diagonal index stresses which are to be used to determine the chord index stresses are as follows : Bar [/2L3 = 25.3X1=21.7, Bar C73L4 = 22.5X1=16.9. These values should be substituted for the actual diagonal index stresses in determining the chord index stresses. For example, the index stress in Bar L3L4 = 231.4 -1-21.7 = 253.1, and that in {73t/4 = 253. 1-1-16.9 = 270.0. To obtain a final check of these index stresses, the top chord dead stress, as computed by the method of moments, should be compared with the value as obtained from the index stresses. Dead stress in U3 C/4 by method of moments = (i- 1800 -200 -200) ^ = 225,000 lbs. Dead stress in U3U4 from index stress 25 = 270X57^ = 225 thousands of lbs. 186 SIMPLE TRUSSES Art. 105 Position of Loads for Maximum Live Web Stresses. The dead stresses and chord and inclined end-post stresses due to uniform live load may be determined directly from the index stresses, and wUl be given later. To determine the position of the live loads for maximum web stresses, the method of shear previously used should be replaced by the method of moments. In deter- mining the actual stress, once the position of loads is known, the method of shear may be used, provided the shear be corrected by the amount of the vertical component in the top chord, or the method of moments may be used directly. The individual bars will now be considered. Bar UiLi : Maximum live stress with full load at Li, E at Li. Bar U1L2: Place load to give maximum counter-clockwise moment about the intersection of the inclined top chord, prolonged, and the bottom chord. Evi- dently a load at Li will cause a clockwise moment about this origin, since the moment of the reaction due to a load at Li will be less than the moment of the load itself as its lever arm and magnitude are both less, hence this point should not be loaded, but all other panel points from the right up to and including L2 should be loaded (note that this con- clusion would not necessarily be correct for a con- centrated load system). The locomotive excess should be placed at L2. Bars U2L2 and U2L3: In this case load from right up to and including L3, with E at L3, since this condition produces the maximum counter-clockwise moment about 0. Bars UsLs and U3L4^: Load from right up to and including L4 with E at L4, Bars UiLi, and C/4L5 (counter) : Load from right up to and including L5 with J5 at L5. Bar f/sLe (counter) : Load from right up to and including Le with E at Lq. The tables which follow show the maximum stresses in all bars with all necessary computations. Art. 105 BRIDGE TRUSSES WITH NON-PARALLEL CHORDS 187 CQ m h-i o o o i-H &^ O CO H hH 1^ t) 6; 02 C3 w =1 m a W Fl rt y H M >* ^ H c«3 M o S? P-, c 1 '^ w i o H W ^ ^ -2 HH J3 ?i .2 ^ .a t— 1 M H Pi ^ -r_ o CM C-. 05 lO CO t- d oi CM <© Tt< ^ •^ in CM Ttl ^ i 1 1 1 1 + + + II II II II II II II X w lO m • lO ■ in lO lO Ol vn c^ > IN CM o X X i O lO o im m lo 8° m |o o |in m lo s CO ICO lO Ico t^ I-* CM Ico in Ico 1^ Ith 3 X X X ^X X X X - o o o o o c o CG t^ 00 •rH CO ^ m |oc> ■ri T-l CM i>.|oc CO |-t< in cc (U > 3 tH 00 to O ;o ^ in 1— 1 l^ 00 in 00 1—1 T— 1 ■^ tH M in r^ CM m CO CO CO CO Oi CO CO a 1 1 1 1 + + + g II II II II II II II ira |io lO IIO in |in in im in |in in lie in |in o t^ iTtI w 1^ t^ M t^ \^ r^ 1'^ 1> lTt< t^ \^ *2 X X X X X X X & 00 i>- ^ CM 00 a> ■<^ to in m ^ c^i d o a> 1— ( Cl CO o IN IN CM J-t i-H CM CO t~ »— 1 o CM CO cn ■^ CO U5 in I— 1 CM O o Ol CM CO cn CM 1 T-l 1 CM 1 CM 1 I— 1 + 1— 1 + CM + S' II II II II II II II 3 lO in ■ in ■ in H >0 04 m (N tn C<1 CM X X Oi lo •a |0 in lo in |o m |0 m lo m |0 CO Ico M Ico c^ Ico oa Ico CM Ico CM Ico CM Ico X X X X X X X >» ^ T— 1 in Tf I— 1 t> T-H CO o t^ T-H CO in CO in t^ in CO in 1-t CM CM c^ CM CM «i lO "*! ^ o m ^ rt ■oi r^ ^ CO d t^ 1— t CO o£ fo CO m t^ in CO in -s M CM CM " CM CM i IS lo l£' ►4" 188 SIMPLE TRUSSES Art. 105 a o o o re P 3 rn m W .^ r/l a; V) a W ^ ttl C3 H CO rn IS o m f„ w ^ -Q al H > A h-1 H ^ P s 1— ( XI CO cq 1—1 CO CO T-l LO iC o GO t^ T-H ^ I-H 3; CO 1—1 T-H • 00 00 T— I (N T-l T-H i-H + + ' + II 1 + II 1 + II + II >. (M CM M ^ O lO CO |iO ■ o o • o CO Ico Tj- Ico t^ rj^ t^ -^ I^ ^ g 3 ^ f . ^ X X X X X a lO 00 CO CO (N CO CO T-i d s CO 2 ;^ 1^ l-^ + + + + + T-H CO lO f— ( CO CO CO 00 c ■ (N CO CO rH CO ,— ( ,— ( o « S CO o t^ 1—1 t- t^ t^ T— 1 1— < T-H T-H II 00 II r~^ !l t^ " CD LO lO CO (M CO to •— 1 00 (M T— 1 rH CO _. 1 00 cc >-< 1-- l^ 'd ■^ t^ IN (M (M t CD ^ CD ^ CO b b tT "^2 tD K- b b lb "£ \£ £ £ , , CD o CO cr '■«£ "* CO o CO CO If o t CM CO CO a> §•5 a fM ^^ CO 1 *^ p^ W! en • b b b o.„ 'O-^ d b b b" b b" o ^ t^ ^< i-t t-( hJ ^ fl CS pq ca & Oi CD CO 00 00 CO CO T-i CD lO VO CO CO 1-H ,—1 00 03 CO CO rH Ir^ t^ CO •A M (M I— 1 tH 1] 1— 1 ■^ 11 1] 11 II O o II 11 o II 11 o II ■^ tH K? 51^ kJ t b t o CO ^ »0 00 b" >0 ■ CO |qo b" ^1^ *o + + TP + to + ■f + d »o lO ^ VO « - lO 03 1* fS S^l^ c 2I« CO |oo m CO |cc ^-, --- , ta e-> u kj >-J kJ I--J sj b K-:] h-jkj >~^Ka b tS' b' k:j b b b'^ bb bb Anr. 106 BRIDGE TRUSSES WITH NON-PARALLEL CHORDS 189 DEAD WEB STRESSES IN UNITS OF 1000 LBS. This table shows all necessary computations. Bar. Index. Stresses. Dead Stresses. Bar. Dead Stress. U,L, + 73.9 3Q 73.9X3JJ =+96.1 U,L, + 30.0 U,L, + 25.3 25.3x|| = + 31.1 U,L, -43.9 U,L, + 22.5 22.5X^ = + 26.6 40 f'.i'a + 4.7 UJ-A -22.5* 22-5X^=-26.6 c^At -15.0 -2S.9t 47 2 28.9X-7-„- = -'^^ 1 40 * ^^■ith live load placed to produce maximum stress in tJ^Lh the main diagonal, C/'sZ/, will be thrown out of action and dead shear in panel 4-5 will be carried by the counter U^Lb. This would tend to produce compression in this bar the vertical component of which is 22.5, but this is balanced by some of the live stress, hence the bar does not actually carry compression, as its sign would seem to indi'eate. A similar condition exists with bar LhL%. t Owing to the counter action, the dead load when truss is loaded to produce maxi- mum live compression in bar U^Lx tends to cause in the bar a tension of 22.5— 15.0 = 7.5. This value should be combined with the live stress to obtain maximum stress. t+.s.«=.....-9o-(-™^ri«x^)i?. For convenience it is common to write the maximum stresses in a diagram called the stress diagram and Fig. 148 is given to illustrate such a diagram. D'- 106.1 Lo Li La L3 L4 Fia. 148. — Stress Diagram, Maximum Live and Dead Stresses. Uniform .Live Load with Locomotive Excess. 106. Computation of Stresses. Bridge Trusses with Non-Par- allel Chords. Concentrated Load Systems. In the parallel chord trusses hitherto studied, the stresses due to concentrated load systems were not determined, since this would have involved merely the substitution of maximum shears and moments caused by such loads for the maximum shears and moments due to 190 SIMPLE TRUSSES Art. 106 uniform loads, and the method of determining such shears and moments was discussed with sufficient thoroughness in Chapter III. For non-parallel chord trusses, however, the conditions differ sufficiently to warrant special consideration; in consequence, the computations for the maximum live stresses due to Cooper's E40 loading will be given for the truss shown by Fig. 146. The computations for all the bars will be given, to make the solution complete, although in a number of cases it involves nothing more than the determination of maximum moment or shear. The moment diagram of Fig. 75 is used in all computations, and units are in thousands of pounds. Position of Loads for Maximum Stresses: Bars LqUi, LqLi and L1L2: Maximum stress occurs when shear in panel — 1 is a maximum. Trv load (3) at Li and move up load (4) . (284 + 2 X 79) j^- + 50 X -. 8p p Try load (4) atLj and move upload (5). (284 +2 X84)— -|-5< 70 X-. Sp p ■. Load (4) at Lj gives maximum. Bar UiLi: Maximum stress occurs with the loads placed in the position giving maximum moment at the centre of a beam 50 ft. long. See Article 51. Try load (3) 66>50 .-. not a maximum. Try load (4) 59<70 .•. a maximum. Try load (5) 59<70 .'. not a maximum. .•. Load (4) at Li gives maximum. Bar U1L2 : Maximum stress occurs with the loading, giving max- imum counter-clockwise moment about (inter- section of U1U2 prolonged and bottom chord). This position may be determined by method of moving up the loads. The distance from Lq to may be found as follows : Chord bar U1U2 drops 5 ft. in one panel, or 30 ft. in six panels, hence its intersection with the bottom chord is six panel lengths from Li, or five panel lengths from L^. Start with load (2) at L2 and move up (3) . Increase in moment about of left reaction = (284 -h 2 X 49) |- X 5p -F5 = 382 ('y ) + 8. Art. 106 BRIDGE TRUSSES WITH NON-PARALLEL CHORDS 191 Increase in moment about of floor beam reaction Li = 30X^X6p= 900^=900. Since 3821-^ j + S>900, load (3) gives greater moment than load (2). Start with load (3) at L2 and move up load (4) . (284+2X54)^X5p + ^<50X-X6p. .'. Load (4) should not be moved up. Load (3) at I/2 gives maximum. Bars L2L3 and Ui U2 ■ Maximum stress occurs with loading, givmg maximum moment at L2. This position is found in the usual manner as follows : Trial Load. Av'g. Load on Left. Greater or Less Than Average Load on Right. Max- imum? Load (6) toleft of ij 103 2 < 284 + 73X2-103 327 6 6 No Load (7) to left of Lj 116 2 > 430+10-116 324 6 6 Yes Load (8) toright of L, 116 2 > 440+12-116 336 6 6 No Load (11) to right of Lj.... 102 2 < 2S4 + 2X105 -1.52 342] 6 6 Yes Load (11) to left of L2 122 2 > 342-20 322 6 6 J Load (12) to right of L,.. . . 102 2 < 322+2x5 332 6 6 Yes Load (12) to left of i^ 122 2 > 332-20 312 6 6 J Load (13) to right of 7.2.... 102 2 < 312+10 322-] 6 6 Yes Load (13) toleft of Lj 122 "2" > 322-20 302 6 6 J Load (14) to right of L^.... 122 > 302+10 312 6 6 No 192 SIMPLE TRUSSES Art. 106 Maximum moment may occur with either load (7), (H), (12) or (13) at L2- Computations show that load (7) at 7.2 gives maximum, as would be expected from inspection of the loading. Bars XJ2L2 and U2L3: Maximum stress occurs with maximum counter-clockwise moment about origin of all forces to left of a vertical section in panel 2-3 (or a diagonal section cutting bars U1U2, U2L2 and 1/2^3)- Determine the position for maximum moment by method of moving up the loads. Start with load (2) at L3 and move up load (3) . (284+24x2)X^X5p + 5>30X-X7p. 8p V 2X5X2.5 8p X5p. In this case the slight increase in moment due to the term 8 is sufficient to cause a larger moment with load (3) than with load (2). .-. Maximum stress occurs with load (3) at L3. Bars L3L4 and U2U3: Maximum stress occurs with loading causing maximum moment at L3. This position is found in the usual manner, as follows: Trial Load. Av'g Load on Left. Greater or Less Than. Average Load on Right. Max- imum ? Load (11) to right of L^. . . 152 3 < 284 + 80X2-152 292 1 5 5 Yes Load (11) to left of L., 172 3 > 272 5 Load (12) to right of L.,... 172 3 > 272+10 282 5 5 No Load (13) to right of L,... 192 3 > 282-20+10 272 5 5 No Load (11) at L3 gives maximum. Art. 106 BRIDGE TRUSSES WITH NON-PARALLEL CHORDS 193 Bar f/aLs: Maximum stress occurs with loading giving maximum moment about 0, of forces to left of a diagonal section cutting bars U2U3, C/3L3, and L3L4. This position may be determined by method of moving up the loads. Start with load (1) at L4 and move up load (2). 27lX^X5p + a>10X8X-^. :. Move up load (3). 284x|-X5p + 5<30X5X — . 8p ^ p .-. Load (2) at L4 gives maximum stress. This loading is evi- dently consistent with main diagonals, U2L3 and C/3L4, being in action. Bar [73L4: Maximum stress occurs for loading giving maximum positive shear in panel 3-4, and is determined as follows : Start with load (2) at L4 and move up load (3) . 284x|- + 5>30X- 8p p Move up load (4) . 292X7^ + 5 < 50 X^. 8;) p .-. Load (3) at I/4 gives maximum. Bars U4LS = UiLs, and C/4L4 : Maximum stress occurs for loading giving maximum positive shear in panel 4-5, and is determined as follows: Start with load (2) at L5 and move up load (3) . 232X8^ + ^<30Xy .-. Load (2) at L5 gives maximum. 194 SIMPLE TRUSSES Akt. 106 Bar UsLe (counter) = C/3L2. Maximum stress occurs with load- ing giving maximum clockwise moment about 0', the point of intersection of U^Uq prolonged and the bottom chord prolonged, of forces to left of vertical section through C/5L6. Determine position by method of moving up the loads. Start with load (1) at Lq and move up load (2). (l42X^)l3p + i?>10X-X8p. Move up load (3). (^152X^)l3p + 5>30x|x8p. Move up load (4). ( 152 X^] 13p + 5< 50 X-X8p. .•. Load (3) at Lq gives maximum moment. Note that one locomotive followed by uniform load will cause a larger stress than two locomotives. Bar UsUi'. Maximum stress occurs for loading giving maximum moment at L4. This position is found in the usual manner as follows: Trial Load. Av'g Load on Lett. Greater or Less Than. Average Load on Right. Maxi- mum. Load 202 4- Yes Load (14) to right of L^ 212 4 = 212 4 Yes Art. 106 BRIDGE TRUSSES WITH NON-PARALLEL CHORDS 195 Maximum moment occurs with either load (13) or load (14) at L4. Computations show that load (13) causes the maximum. The necessary computations for maximum stresses in all bars are shown in the two tables which follow. MAXIMUM LIVE WEB STRESSES, IN UNITS OF 1000 LBS. Cooper's Ejo Loading Bar. Position of Loads. All Necessary Stress Computations. (Z/22=left reaction. V.C. = vertical component.) L„V, 4 at Z/i Shears (16,364 + 284X84+84X84) -=-200 V.C. Stress = 217.2 Stress = 217.2X1.3 = 480 25 = 217.2 -282.4 U,L, 4 at L, s.„.-io(j,)+.o"'+»;-+->+.3 (11 + 6) 25 + 75.6 U,L, 3 at L2 Lr = [16,364+ (284+ 54)54] -h 200 Floor beam load at Li = 230-=-25 Stress = [(173.1X5p-9.2X6p)H-7p]X1.3 = 173.1 9.2 + 150.4 U,L, 3 at L3 Lk = [16,364+ (284+ 29)29] -=- 200 Floor beam load at L^ Stress=(127.2X5p-9.2X73a)^7p ^ 127.2 9.2 -81.7 U,L, 3 at L3 V.C. = (127.2x5p-9.2X7p)^8p Stress= (71.5X43) -f- 35 = 71.5 + 87.8 IhL, 2at L, Z,;j =(16,364- 284) -^ 200 Floor beam load at ^3= (lOXS) H-25 Stress = (80.4 X5p- 3.2 X8p)-^8p ^ 80.4 3.2 -47.1 U,L, 3 at L, Shear = [16,364 + (284 + 4)4] -f- 200 - (230 -f Stress = (78.4X47.2) -^40 -25) = 78.4 + 92.5 1 2 at is Shear= (8,728 + 232X4) -^ 200- (10X8) -f- Stress = (45.1X47.2) ^40 25 = 45.1 + 53.2 V.L, 2at L; Stress = V.C. in bar U^L^ = -45.1 UrL, 1 3 at L* Lk= [3,496+ 142X15+20X5] ^200 Floor beam load at 1,5 = 230-^25 Stress=[(28.6X13p-9.2X8p)-h7p]Xl.l8 = 28.6 9.2 + 50.3 * One locomotive followed by uniform load. 196 SIMPLE TRUSSES Art. 107 MAXIMUM LIVE CHORD STRESSES, IN UNITS OF 1000 LBS. Cooper's 'E,„ Loading Bar. Position of Loads. All Necessary Stress Computations. (See previous table for some of the values used in table.) L„i, 4atL, = (217.2X25) -=-30 = -1-181.0 L,L, Tat Lj Mom. atL2 = [16,364+(284-|-78)78]x|--2155 = 8995 Stress =8995-^35 = -1-257.0 U,U, 7atL, 257 X 25.5 H- 26 = -262.1 L,L, 11 at L3 Mom. at ^3= [16,364+ (284-1- 80)80] x|- -5848= 11208 Stress =11208h-40 = +280.2 *U,U, 11 at L3 Stress = 280.2 X-^ = -285.8 u,u. 13 at Lj Mom. at L4 = [16,364+ (284 + 65)65] X^ -7668 = 11856 Stress =11856^40 = -296.4 * This stress would be incorrect if the loading used were to throw counter U3L2 or LiUi into action. To decide whether this is the case, the shear in panel 2-3 due to this loading may be computed, and the vertical component in top chord U2U3 subtracted from it. If the result is positive, or negative, but less (with due allowance for impact) than the dead shear in panel, the counter will not be in action. The computations follow, making use of previous computations and the moment diagram. Sl.ea.=l|if-n6-ll0--±-,3=.89.0. V.C. C/j (73 = 280.2 X .'. Counter is not in action and stresses are correctly determined. 107. Computation of Stresses. Bridge Trusses with Parabolic Chord. Uniform Load v/ith Locomotive Excess. The methods used for the truss considered in the two previous articles were perfectly general and may be used for any non-parallel chord truss. If the panel points on either or both chords lie upon a parabola passing through the end panel points, the truss has, however, certain characteristics which may be taken advantage of in making the computations. Such trusses are not commonly used in railroad bridges, but the same special features occur in certain trussed arches, hence it seems desirable to give an example of the computations for such a truss. Akt. 107 BRIDGK TRI?SSES WITH PARABOLIC CHORD 197 Problem. Determine the maximum stresses in all the members of the truss shown in Fig. 149 with the following loads: Dead weight of bridge, 1000 lbs. per ft. per truss, top chord =25,000 lbs. per panel. 400 " " " " " bottom " =10,000 " " Uniform live load, 2000 lbs. per ft. per truss, top chord =50,000 lbs. per panel. Locomotive excess, " " =25,000 lbs. In the computations of this truss the following points should be noted: 1. When the truss is loaded uniformly throughout its length the ordinates representing the bending moments at the panel -6 Panels-© 26=150— -6-Paii8l8 =160- 'S-U ol iu Fig. 149. — Truss with Paraboli? Bottom Chord Showing Dead Panel Loads. points are ordinates of a parabola, hence the bending moment at the panel points divided by the depth of truss is constant. 2. Since the horizontal component of the stress in any bottom chord member equals the moment at a panel point divided by the depth of the truss at the same panel point, the horizontal component of the bottom chord stress under uniform load is constant throughout. 3. For the same reason the top chord stress under uniform load is constant throughout. 4. It follows from 2 and 3 that under uniform load the hori- zontal components in the diagonals will be zero, and the verticals will all carry compression equal to the top panel load. 5. Since under the dead load (if a uniform load) the stress in the main diagonals is zero, it follows that the live load can always 198 SIMPLE TRUSSES Art. 107 be so placed as to produce compression in any diagonal, hence if the diagonals are to be tension members counters will be required in every panel. Dead Stresses. For the given truss the dead stresses in imits of 1000 lbs. will be as follows: Top chord, stress = i(l. 4)^^^^-^ = -145.8 Bottom chord, horizontal component = +145.8 Diagonals, stress = Verticals, stress = — 25.0 To confirm the correctness of the conclusions reached for web stresses the diagonal stresses will be computed in the usual manner. Shear in panel 1-2 =87.5-35 =52.5 V.C. in bottom chord L1L2 = ( — ^p 1;^ = 52.5 V.C. in diagonal C/1L2 =52.5-52.5 =0. Shear in panel 2-3 = 87.5 -70 = 17.5 Y— V.C. in diagonal C/gLs =17.5-17.5 =0 V.C. in bottom chord L2L3 = ( "• -5X50^ 3 5X25 \ /^\ ^ ^^^ Counters. Parabolic Trusses. It has been stated that counters are needed in every panel. The truth of this may easily be tested by actual computation. For example, to determine whether counters are required in panel 1-2 assume the section XY, and see if the live load can be so placed as to produce com- pression in bar U1L2. The stress in this bar may be computed by taking moments about the origin 0. If a load be placed to the left of XF it will produce a reaction less than itself, and the moment of this reaction about will be less than the moment of the load itself not only because of its smaller value but because its lever arm is less, hence any load to the left of XY will produce clockwise moment about of the forces to the left of XY and Art. 107 BRIDGE TRUSSES WITH PARABOLIC CHORD 199 thereby cause compression in U1L2, therefore a counter will be needed in that panel. As this method is perfectly general, it follows that counters are needed in every panel since the live load can always be placed so as to produce compression in the main diagonals, and the dead stress in these members is zero. Live Chord Stresses. The maximum live chord stresses occur with the uniform live load extending over the whole truss and can be computed from the dead stresses by multiplying the latter by the ratio of live load to dead load. The chord stresses due to the locomotive excess are as follows: MAXIMUM STRESSES DUE TO LOCOMOTIVE EXCESS IN UNITS OF 1000 LBS. Bar. Position of Load. Computationa. U,L, E at Ui H.C. Stress =1-25 X^= +34.7 D 15 U,U^ Eat Ui Stress = |-25xff=-34.7 15 L,L, ■ E at U^* H.C. Stress = 1-25 xf| = + 27.7 D lO u,u. E at I/j* Stress = 1-25 X |5 =- 34 . 7 L,L, Eat Us H.C. Stress = i-25 X |5 = -f 26 . u,u. Eat Us 1 71: Stress = ^25X^= -34.7 * Note thatif .Bwere to be placed at Ui the counter L1U2 would be brought into action hence the horizontal component in bar L1L2 would equal^25X-^rr = +17.4, and the 5 25 o Z4 stress in U1U2 would equal ^ 25 X 7^=34,7. o lo 200 SIMPLE TRUSSES Art. 107 Live Web Stresses. The maximum live web stresses occur with partial loading. The position of the loads may be deter- mined by the methods previously used. The necessary com- putations for maximum stresses are given in the following table : MAXIMUM LIVE WEB STRESSES, IN UNITS OF 1000 LBS Bar. Panel Points Loaded with Uniform Load. Position of B. Computations, Vertical Components of Maximum Live Web Stresses. U,L, U, or {7, to f/j incl. u. 50 + 25 =-75.0 U,L, U., to U^ inclusive u, 10 4 Shear in panel 1-2 =— 50+-25 =100.0 V.C. in L,L, =100X?|x^= 60.0 V.C.inU^L^ =100-60 =+40.0 U,L, f/^or [A, to t/j incl ^■h - 75.0 U,L, V, to U, inclusive u. Shear in panel 2-3 = |- 50 + 1-25 = 62.5 b D Y.C.in L,L, =62.5x|5xA= is.^ V.C.in U^L, =62.5-15.6 =+46.9 u,h U,ov C/ito (75 incl. Us - 75.0 U,L, V, and V^ u< 3 2 Shear in panel 3-4 = -50 + -25 = 33 .3 D 6 V.C.ini3r.. =33.3xgx^= 11.1 Y.C.mU,L^ =33.3 + 11.1 =+44.4 U,L, u, u. Shear in panel 4-5 = — 75 = 12.5 V.C. in LA -12.5X24"x2'g-18.75 Y.C.mUJ., =12.5+ 18.75= +31.25 For actual locomotive loads the computations for this truss should present no more .difficulty than for the truss of the pre- vious example. Art. 107 TRUSS PROBLEMS 201 PROBLEMS 41a. State which of the trusses shown in the figure are statically determined with respect to the inner forces, and give reasons. 416. Draw influence line for stress in bar a of truss H. 42. State which of the structures shown in the figure are statically undetermined with respect to the inner forces, and give reasons. Trestle Tower All diagonals are tension rods and cannot caiTy com- Prob. 41. Pbob. 42. 43. a. Compute by the analytical method of joints the vertical components in all diagonal members, and the actual stress in all other members of this structure. Tabulate results in order according to bar numbers. Designate tension by ( + ) and compression by (— ). 6. Determine stress in all members by graphical metiiod of joints (Bow's Notation). Tabulate results in same order as in a. 10 10 000* Prob. 43. Prod. 45. 44. Compute by method of moments the stress in bars a, b, c and d. and state whether stress is tension or compression. 202 SIMPLE TRUSSES Art. 107 45. a. Compute by method of moments the stress in bars a, b, c and d, and state whether tension or compression. 6. Same as a, but direction of reaction is not fixed by rollers. (Assume both reactions to act parallel to direction of applied loads.) 46. Compute maximum stress in each member due to following loads applied at top chord: 1. Dead, 30 lbs. per horizontal square foot. 2. Snow, 20 lbs. per horizontal square foot. 3. Wind, 30 lbs. per sq. ft. normal to surface. Tabulate stresses for each kind of loading, and determine maximum stresses, arranging results according to bar numbers as given on diagram. Indicate tension thus (+) and compression thus (— ). Use any method of computation desired. .Horizontal tie rod ~0 tons 100 tons K- 6 panela .@ 15'«:9o'-ij * 3?rueBofl l&'Cto C. ' Pbob. 46. Pbob. 47. 48. 47. Compute stresses in tons and state whether tension or com- pression for the following bars: Bar a, by method of joints. Bar 6, by method of moments. Bar c, by method of joints. Bar d, by method of moments. Uniform live load, 2000 lbs. per foot, on top chord. Locomotive excess, 20,000 lbs. on top chord. Dead load, top chord, 600 lbs. per foot. Dead load, bottom chord, 200 lbs. per foot. Determine panels in which counters are needed and compute maximum stress in each member of this truss. Number bars as shown in figure and arrange results in order according to bar numbers. Stresses to be given in pounds. Tension to be denoted by ( + ) and compression by ( — ) . Peob. 48. Pbob. 49. 49. Uniform live load, 2000 lbs. per foot on top chord. Locomotive excess, 20,000 lbs. top chord. Dead load, 1000 lbs. per foot top chord. Dead load, 500 lbs. per foot bottom chord. No counters are to be used. Compute maximum stresses of both kinds in all members of this truss. (Rules as to arrangement of results, etc., as in previous problems.) CHAPTER VII BRIDGE TRUSSES WITH SECONDARY WEB SYSTEMS, INCLUDING THE BALTIMORE AND PETTIT TRUSSES 108. Secondary Systems Described. The bridge trusses hereto- fore treated have all been of such simple types that the application of the ordinary methods of joints, moments, and shear required no special explanation. For spans of considerable length, how- ever, the frequent subdivision of the main panels and the addition of a secondary set of diagonals and verticals produce complica- tions the effect of which will be explained in this chapter. The Baltimore and Pettit trusses, illustrated by Figs. 132 and 134, are the common forms of such trusses and will alone be considered. An examination of one of these trusses shows that the stresses in the secondary verticals may easily be determined by the method of joints, the real complication occurring in the secondary diagonal stresses. In order to study the stress in one of these diagonals, consider the portion of such a truss shown in Fig. 150, and let the problem be the determination of the stress in diagonal L2M3. In order that the case may be perfectly general, let it be assumed that the dead loads are applied at middle as well as top and bottom panel points, although such an accurate division of the dead panel loads is not generally required. An examination of the forces acting at joint M3 shows that the function of the secondary diagonal M3L2 is to support the main diagonal, C/2^4, under the loads Pj, P2 and P3, which, if the secondary diagonal were not inserted, would cause the member {72^4 to bend and thereby produce the collapse of the truss. If no loads be applied at the secondary panel points U3, M3 or L3, 203 Fig. 150. 204 TRUSSES WITH SECONDARY WEB SYSTEMS Aet. 108 there will be no stress in either the secondary diagonal M3L2, or the secondary verticals U3M3 and M^Lz, since these members take no part in the transmission to the abutment of the loads at other panel points. It is therefore necessary to consider only the loads Pi, P2 and P3, in determining the stresses in the secondary members meeting at the joint M3, and the effect of these members upon the main diagonal stresses. The stresses in the secondary verticals are evidently equal to the panel loads applied a* their ends; that is, the compression in U3M3 = Ps, which ordinarily is merely the dead weight of the top chord acting at this point, and the tension in M3Lz = Pi. This is equivalent, so far as the secondary diagonal is concerned, to the application at Mz of a resultant downward vertical force equal to Px +P2 +^3- For simplicity this resultant will here- after be called R and considered as acting directly at M3. The stress in M3L2 may then be computed by the method of joints by resolving the force R along two axes coincident with U2Li and L2M3. With this stress known, the effect of R upon U2MZ may also be readily determined by the same method. The stress in M3L4 is evidently unaffected by the secondary system and may be determined by the method of shear in the usual manner, since its vertical component equals the shear in panel 3-4. While the method of joints for this case as applied in the ordinary manner presents no special difficulty, the following combination of graphical and analytical methods is somewhat simpler. In Fig. 151, let ah represent R in direction and magnitude. (This Pjq ^c,i requires merely the choice of a suitable scale.) Draw he parallel to af, and draw the horizontal line ce. The sides 6c and ac then equal the components of R parallel to af and ag respect- ively, therefore he represents the compressive stress in af. he ge_hg Now -T= — "^rr- ah ag fg But be equals the vertical component of the stress be. Abt. 108 SECONDARY SYSTEMS 205 bg bg Hence the vertical component in af=j-R= {Pi+ P2 + P3)-r- It is evident that the same value would be obtained if the secondary diagonal were to extend from M3 to C/4, Fig. 150, instead of from M3 to L2, with the difference that its stress would be tension instead of compression. The following proposition may therefore be stated. The vertical component of the compression in bar (1) in the case shown by Fig. 152, or the vertical component of the tension .Pi +P2+P3 in bar (2) in the case shown by Fig. 153 = ^ . It follows from the above rule that the vertical component of the maximum stress in a secondary diagonal in a Baltimore truss with equal panels and horizontal chords equals one-half the maximum panel load. With the vertical component in the secondary diagonal known, the vertical component in the main diagonal in the same panel may be found by subtract- ing this value from the shear in the panel if the bars be as shown in Fig. 152, or by adding it to the shear for the case shown in Fig. 153, provided in both cases that the shear is positive. In case the shear is negative, the question of counters must be investigated in accordance with the methods of the following article. The demonstration just given is simple, but is applicable to trusses with parallel chords only. A similar secondary system is however frequently used in trusses with non-parallel chords (Pettit trusses) ,and in orderto cover this case also a more general demon- stration, based upon the method of moments and applicable to both parallel and non-parallel chord trusses, will now be given. This method consists in first deriving an expression for the sum of the horizontal components in L2M3 and L2L3, Fig. 154, called hereafter for convenience 3 3 t ^2 \ : \5/ 1 \ - 173: . 11 105 Y105 105 6 panels © 30' = 951.5 net Fig. 159. — Index Stresses and Dead Panel Loads for Truss Shown in Fig. 158. computations for bars in which the index stresses are at all com- plicated follow. V.C. inM3t/4=173x|^x|^ = 103.8 = 294.1 ^,„ . ^- ^, /951.5X 120 -173X150X15 V.C. m U,U, = (-- ^g jgQ V.C. in UiLi =259.5 + 103.8+68.0-294.1 (method of joints) =137.2 V.C. in [72^1 = 191.5 +311.4 +294.1 +68.0 (method of joints) =865.0 Corrected index stress in M3 U4,, for use in determining index stress in bar C/4C/5, = 103.8 X3»/45= 69.2 Corrected index stress in U^Ms, for use in determining index stress in bar C/4C/5, =259.5x60/75 = 207.6 To check top chord stresses determine combined horizontal component in UsUe and M^Ue by dividing centre moment by 210 TRUSSES WITH SECONDARY WEB SYSTEMS Am. 109 centre height, and add to the value thus obtained the horizontal component in M^Uq. ^. . „„ 951.5X180-173X5X90 ^ _- 30 ,„, . „ Stress m UaUs = ;— 1-86. 5 X;t7;-^= 1314.8. 75 37.5 This equals the index stress in U^U^, as should be the case, since the latter was determined for diagonals sloping at 45°. Dead Stresses. The actual dead stresses are given in the following table in which the columns headed " ratio " give the length of each web member divided by its vertical projection and of each chord member divided by its horizontal projection. DEAD STRESSES IN UNITS OF 1000 POUNDS Bars. Index Stress. Ratio. stress. Bars. Index Stress. Ratio. Stress. L„M, - 951.5 1.414 -1345.4 ^^f, + 105.0 1.000 + 105.0 M,U, - 865.0 1.414 - 1223 . 1 L,M, + 105.0 1.000 + ro5.o u,u. -1176.4 1.031 -1212.9 L,M, + 105.0 1.000 + 105.0 U,V, -1176.4 1.031 -1212.9 U,M, - 68.0 1.000 - 68.0 VJJ, -1314.8 1.000 -1314.8 U,M, - 68.0 1.000 - 68.0 u.u. -1314.8 1.000 -1314.8 M,L, - 86.5 1.414 -122.3 LM + 951,5 1.000 + 951.5 L,U,, + 191.5 1.000 + 191.5 L,L, + 951.5 l.COO + 951.5 U,M, + 311.4 1.414 + 440.3 L,L, + 865.0 1.000 + 865.0 M,L, + 242.2 1.414 + 342.5 h,L, + 865.0 1.000 + 865.0 M^U, + 103.8 1.202 + 124.8 L,L, + 1107.2 1.000 + 1107.2 U,L, -137.2 1.000 -137.2 ^5^. + 1107.2 1.000 + 1107.2 U,M, + 259.5 1.280 + 332.2 M,U, + 86.5 1.280 + 110.7 M,L, + 173.0 1.280 + 221.4 L,U, -241.0 1.000 -241.0 Counters. Before computing the live stresses, and even before determining the position of live loads for maximum stresses, it is necessary to decide in what panels counters are required. Panels 4-5 and 7-8. Evidently counters will be needed in these panels if the resultant shear due to live, dead and impact in panel 7-8 is ever positive. The application of the method of moving up the loads shows that load (2) at I/g gives maximum positive shear in this panel. Its magnitude per truss for Ejq equals Fig. 160. -X2 16364 + (284 + 19) 19" 360 -^X2X^°- a46.9. Art. 109 MAXIMUM STRESSES IN PETTIT TRUSS 211 If impact be computed by formula (7), its value will be 1^146.9=102.9, hence the live shear plus impact = 102.9+146.9 = 249.8. The dead shear in this panel equals —259.5, but the difference between this and the live shear plus impact is so small that the counter should be used. Panels 2-3 and 9-10. Counters will be needed in these panels if the live compression plus impact in bar Mgllio exceeds its dead tension. The position of loads which will give the maximum compression in MgUio will be that which will give maximum clockwise moment of forces to left of vertical section through panel 9-10 about 0', the intersection of top chord bar UsUqUio prolonged and the bottom chord prolonged. To determine this position start with load (1) at Lio and move up load (2), using for convenience p to represent the panel length, Q O 152XT-^-l'8p + 5>10X-X9p. .-. move up load (2). 12p p Now try moving up load (3). 172Xt|--18p+,?<30X--9p. 12p P .■. load (2) at Lio gives maximum. V.C. live stress in bar MqUio with load (2) at L^ for E50 ^fxyJ-oH-n^xJI-^: = 75.7. This value is so much less than the vertical component of the de stress in the bar that no counter is needed. Position of Loads for Maximum Live Stress in all Membe B.^R U2M3. Load for maximum moment about of loa. to left of a vertical section through panel 2-3. Start with load (2) at L3 and move up load (3). (284 + 169X2)y|-X6p+5>30X-X8p. 212 TRUSSES WITH SECONDARY WEB SYSTEMS Art. 109 Move up load (4). (284 + 174x2)-|-X6p+.?<50X-X8p. .-. Load (3) at L3 gives maximum. Bar M3L4. Let ^4/44 = moment about U4 of forces to left of a vertical section through panel 3-4 divided by height of truss at L4. Let ilf 2,M2 = moment about U2 of forces to left of a verti- cal section through panel 2-3 divided by height of truss at L-z- Since the horizontal component of the stress in MzL^^Mn/hi —M 2/112, the position of loads for maximum stress in the bar is that giving the maximum values of this quantity. Fig. 161 Tan J- xi. = -X 10 Up 600 '^*°"io"2i."5oo Fig. 161. — Influence Line for Vertical Component in MJ^,^. is the influence line for the vertical component in this bar, and shows that one of the loads should lie at Z/4. To determine the position for maximum stress use the method of moving up the loads, multiplying the loads to right of Z;4 by the product of the distance moved and the tangent I/6OO, and those in panels 2-3 and 3-4 by the product of the distance moved and the tangent 7/600. Start with load (3)' at L4 and move up load (4). (234+.144x2)A + ,>50x5XgIg. Move up load (5). (214+149X2)gi^ + .>70X5Xg^. Move up load (6). (194+154x2)^-+.^<90X9Xg^^. Load (5) at L4 gives a maximum. Art. 109 MAXIMUM STRESSES IN PETTIT TRUSS 213 It is possible that this bar may be brought into compression by loads coming on from left, hence the position giving max- imum compression should be determined. Start with load (2) at L2 and move up load (3), bringing loads on from left, Move up load (4) Move up load (5). 142X5XA + ,>30X5X^-^. 142x5x4 + .^>50x5Xg^^. 142 X5 X A + ,< 70X5x4. .•. Load (4) at L2 gives a maximum. Bar U^M^. Load for maximum shear in panel 4-5. Start with load (2) at L5 and move up load (3) : (284 + 109 X2) 4- + 0^ >30 X-. Move up load (4). (284 +114X2)-!^ + o<50X-. 12p p .-. Load (3) at L5 gives maximum. Bar M^Lq. Load to give the maximum value of the resultant of the positive shear in panel 5-6 and the vertical component in bar MsUe. Start with load (3) at Le and move up load (4) . (284+84x2)-^~ + 5>i(50X^). Move up load (5). (284+89X2) A +5 >l(^70x^). Move up load (6). (284+94X2)-|^ + 5<^(90X^).i Load (5) at Lq gives maximum. ' The right-hand side of this inequality equals the increment in the sum of the panel load at L^ and the vertical component of the stress in the secondary 2H TRUSSliS WITH SECONDARY WEB SYSTEMS Akt. 109 Bar M'jLs. If this bar be in action the condition shown in Fig. 163 will exist. Place loads so that the sum of the positive I. Fig. 162. shear in panel 7-8 and the vertical component in bar MjUs will be a maximum. Start with load (2) at Lg and move load (3). (284 + 19x2)-|- + 5>^ 30 X^. 12p 2 p Move up load (4). (284+24x2) 12p 1 5 f>^ 50 X-. 2 p Move up load (5). (28^ Load (4) at Lg gives a maximum. (284+29x2)-|- + (?<^ 70 X-. I2p 2 p diagonal M^Uf due to the movement of the loads. If no load passes Lj it i.s obvious that this change equals one-half the sum of the product of the loads moving in panel 5-6 and the distance which they move. That this is also true, provided no load passes L^, may be readily proven as follows : Let the original position of a load P be as shown by the full circle in Fig. 162, and assume that in moving the loads, P passes L5 to the position shown by the dotted circle. The following equations may then be written: Original position of loads: Second position of loads: 2 p 2p The mcrease m B. + -^ = P — [-P^ — -~ = — ^:r -. 2 p 2p 2p 2p This demonstration applies equally well to two corresponding panels in any other position of the truss. Art. 109 MAXIMUM STRESSES IN PETTIT TRUSS 215 Bars LqMi, LqLi, LiLn. Load for maximum shear in panel 0-1. Start with load (3) at Li and move up load (4). (284+234X2)-|-+a>50X-. Move up load (5). (284+239X2):^ +5< 70 X-. 12p p .'. Load (4) at Li gives maximum. Bars MilJi, ^^2^3 and L3L4. Load for maximum moment at t/2. Try load (7) at L2, 624/10 > 116/2. Not a maximum. Try load (8) at L2, 636/10 > 116/2 and 623/10 < 129/2. A maximum. Try load (9) at L2, 633/10K29/2. Not a maximum. .-. Load (8) at L2 gives a maximum. Bar C/2I/2- This bar is really a part of the secondary system and is affected by loads at Li and L2 only. The influence line for ^.---^^'^'^^^^ this bar is shown by Fig. 164, ^— """^^ ! ^\. and has the same form as the in- '"" ^' '"^ "■• fluence line for moment at a point ^'°- ^'^s;;;^^^^'''^;"'^ ^'"^ ^°' 30 feet from the right end of an ' 2 ■ end-supported 90-ft. span; hence the criterion for maximum moment may be applied to determine the position of loads which should be brought on from the left. Try load (3) at L2, 142/60 > 50/30. Not a maximum. Try load (4) at L2, 142/60 > 70/30. Not a maximum. Try load (5) at L2, 142/60 < 90/30. A maximum. .-. Load (5) at L2 gives a maximum. Bars C/2C^3 and UgUi. Load for maximum moment about Li of forces to left of vertical section through panel 2-3. The influence line for the horizontal component of the stress in these bars is shown in Fig. 165. Evidently for a maximum one of the loads should lie at -L3. AVhile the influence line for the stress in this case is not com- posed of two straight lines and the criterion for maximum moment cannot be applied, it is evident that the loads will lie somewhat 216 TRUSSES WITH SECONDARY WEB SYSTEMS Art. 109 as for the ordinary case of maximum moment at a panel point, and one of the second-engine loads will probably give the max- imum. Fig. 165. — Influence Line for Horizontal Component in UiUz and U^Ui- The following expression for the change in the stress may be written, using the method of moving up the loads. Start with load (11) at L3 and move up load (12) : (112 +225 X2):j^^ + d >56 X5 X^^ + lK'^A+2X22-5)-*-103X5X2|. Move up load (13). (92 +230 X2)2|5 + a< 63 X.3 X 2I5 '225J + 13(4X2^ + lX2i^)+116X5X2i5. .-. Load (12) at -L3 gives maximum. Bar UiJLi. Assuming that counter M^L^ is not in action, the influence line for stress in this bar will be as given in Fig. 166, Fig. 166. — Influence Line for Stress in U^L^. and shows that for maximum compression the load should come on from the right, and for maximum tension from the left. Position for maximum compression, load coming on from right: Art. 109 MAXIMITM STRESSES IN PETTIT TRUSS 217 Start with load (1) at Ls and move up load (2), making iise of the tangents to the influence line, as was done with bar M3L4. (274 + 10lX2)A+,>iox8X^. Move up load (3). (254 + 109X2)--+^<30x5X^^. .". Load (2) at L5 gives maximum compression. Position for maximum tension, loads coming on from left : For this case heavy loads should be placed at both L4 and L2. These panel points are 60 ft. apart, hence if the heavy loads of the first locomotive are placed near L4, the heavy loads of the second locomotive will be located near L2, this giving a favorable position for maximum stress. Start with load (2) at L4 and move up load ( 3) : 86X5xi^+(92 + 2X19)X5xA+20Xlx4 + ^>30X5xi^^+56X5XgJ-^ + 20X4x4 Move up load (4). 79X5Xe-oo+(72 + 2X24)X5Xgo-, + 20Xlx4 + a<50X5X^^ + 63X5Xg^+20X4Xg^. .'. Load (3) at I/4 gives maximum. The above condition for maximum stress will not be correct if in either case counter M5L4 be in action. That this bar is not in action for the position of loads giving maximum compres- sion is, however, evident from inspection. For the position for maximum tension the negative shear in panel 4-5 is given by the following expression: Shear in panel 4-5, load (3) at L4, loads coming on from left ' ^^^5r i6364 + (284+ 24m_:g^,^145.7. [ 2301 30 J This is considerably smaller, even after impact is added, than the positive dead shear in the panel and the counter will not be in 218 TRUSSES WITPI SECONDARY WEB SYSTEMS Abt. 109 action; hence the assumed condition is consistent with the position of the loads as determined. Bars UiUa and UsUq. Load for maximum moment about Le of loads to 'eft of vertical section through panel 4-5. The influence line for the stress in this case consists of three straight Fig. 167. — Influence Line for Stress in U^U^ and U^Ur. lines, as shown in Fig. 167, and shows that the maximum stress occurs with one of the loads at L^. Start with load (14) at L5 and move up load (15), making use of the tangents to the influence line. (52+180X2)-!- + 5 >142Xy^+lo(A+7x ^ Move up load (16). ^150 150 150 +80X9X 150' (39+189X2)^^ + ^<152xA+93x5xA. .-. Load (15) at L5 gives maximum. Bars L4I/5 and L^L^. Load for maximum moment at U4, assuming counter L^M^ to be out of action. 23'7 479 Try load (14) to left of L4, -— < -^. Not a maximum. 24'5 477 Try load (15) to left of L4, -— >— - A maximum 4 5 945 487 Try load ( 1 6) to right of L4, ^ > ^- Not a maximum. .-. Load (16) at L4 gives a maximum. The shear in panel L4L5 for this condition 16364+ (284+219)219 58 360 ~^'^^ ~30^^ -2Xj -3>] = + 196.5. Hence counter LiM^ is not in action for this loading. Bar UeLe. Two cases must be considered for this bar These are shown in Fig. 168. Art. 109 MAXIMUM STRESSES IN PETTIT TRUSS 219 Case I. Maximum stress will occur with the loading giving the maximum value of the algebraic sum of the positive shear on section XY and the vertical component in diagonal M^Uq. Try load (2) at L7 and move up load (3) . (284+49X2)-!- + 5 >30X-. 12p p Move up load (4) . (284 + 54 X 2)-|- + 5 < 50 X-. 12p p .'. Load (3) at L7 gives a maximum. Case I Fig. 16S. Case II. The maximum stress in this case cannot exceed twice the vertical component of the maximum stress in one of the secondary diagonals; i.e., it will not exceed the maximum panel load. Since the stress in Case I is likely to be greater than this limiting value, the position of loads should not be determined until after the stress for Case I has been computed. If it then becomes necessary to determine the position, the method of influence lines will be used. Bars MiLi, M1L2, M^Lz, UiM^, MgLs and UeMs. Maximum stress in these bars is a function of the maximum load at a secon- dary panel point. This has the same value in aU cases, and may be found for any one of these panel points, such as Li , by placing the loads so as to give the maximum moment at the centre of a 60-ft. span. Try load (12) at Li, 86>56 and 66<76, hence a maximum. Try load (13) at Li, 79 >63 and 72< 83, hence a maximum. Try load (14) att Li, 72 >70 and 52<90, hence a maximum. .-. Maximum stress occurs with either load (12), (13), or (14) at a secondary panel point. (Note that load (14) gives same moment as load (5).) 220 TRUSSES WITH SECONDARY WEB SYSTEMS Art. 109 MAXIMUM LIVE STRESSES IN MAIN DIAGONALS IN UNITS OF 1000 POUNDS This table shows all necessary computations. (Note that 16,364 -;- 360 = 45.-15.) Bar. Position of Loads. . Computations. C/.Ms 3 at Li, Shear in panel 2-3 Max. 16,364+ (L>S4+ 174)174 230 Tension. 360 30 = 266.8-7.7 = 2,V.).l Ycrt. Comp. in U2U3 = i^, (266.8X4 -7.7 X2)g = 105.2 Vert. Comp. in U^M, in tons for E„ = 153.9 Tension for Es„ = 153.9 X 1 .414 X^ X 2 = 544.0 M,L, 5 at L, Shear in panel 3^ Max. 16,364+ (284+ 154)154 830 Tension. 360 30 232.8-27.7 = 205.1 Vert. Comp. in CjC/, ->'-'x< = 93.1 Vert. Comp. in M^Ut 45/8.30\ /30\ 30 \ 30 / 1,75/ = 16.6 ^'ert. Comp. in M^Lt in tons for E,,, = 205.1-93.1 + 16.6 = 128.6 TensionforE5„= 128.6X1.414XxX2 = 4, =54 .6 M,h, 4 at Lj • Max.Comp., Shear in panel 3-4 8728-212 Loads 360 = 23.6 coming Vert. Comp. in U^Ut on from left. 60r^-^^^+30;75 Vert. Comp. in M^Ut 480 30 45 = 20.5 \{ \{ — 9.6 30 75 30 Vert. Comp. in Mjl,, in tons for E^o = 23.6 + 20.5-9.6 = 34.5 ('ompre.ssion for E5„ = 34.5Xl.414X-^X2 = 121.9 This is so much smaller than the dead tension that com- pression will never actually occur in this bar. U,M, 3 at L^ Shear in panel 4-5 Max. 16,364+ (284 + 114)114 230 Tension. 360 30 • = 171.5-7.7 = 163.8 Tension for E5,, = 163.8Xl.280X-|-X2 = 524.2 AsT. 109 MAXIMUM STRESSES IN PETTIT TRUSS 221 MAXIMUM LIVE STRESSES IN MAIN DIAGONALS— Continued Bar. Position of Loads. Computations. M,L. 5 at Lj Max. Tension. Shear in panel 5-6 16,364+ (284+94)94 830 360 30 = 144.1-27.7 = 116.4 Vert. Comp. in M-JJ^ Vert. Comp. in MJj, in tons for E,„ = 116.4+13.8 = 130.2 Tension for E5o= 130.2X1.280X^X2 = 416.6 M,L, 4 at Ls Max. Tension. Shear m panel 7-8 16,364+ (284 + 29)29 480 360 30 = 70.7-16.0 = 54.7 Vert. Comp. in M^U^ = S,.Q Vert. Comp. in MjLj in tons for Ej,, = 54.7 + 8.0 = 62.7 Tension for Ejo = 62.7X1.280X^X2 = 200.6 222 TRUSSES WITH SECONDARY WEB SYSTEMS Abt. 109 MAXIMUM LIVE STRESSES IN INCLINED END-POSTS, CHORDS, AND MAIN VERTICALS IN UNITS OF 1000 POUNDS This table shows all necessary computations. Bar. Position of Loads. Computations. LJU 4 at Lj Shear in panel 0-1 16,364+ ("284 + 239)233 480 360 30 = 392.7-16.0 Compression in L„AIi for Ejq = 376.7X1.414X^X2 4 376.7 1331.6 L„L, 4 at Li Tension tor E;, =376.7 X-j-XS 941.8 M,U, 8 at L, Moment at L, 16,364+ (284 + 234)234 2851= 20,078 Hor. Comp. in MJJ^ in tons for E,^ _ 20,078 _ 60 ^^^-^ 5 Compression for £50 = 334.6 X 1.414X^X2 = 1182.8 L3L, 8 at i. Tension for E,, 334.6X-rX2 4 836.5 LJJ, 5 at L2. Loads coming on from left. Panel load at L2 10X^+80X^+52X^ = Panel load at L, 52X — + 80X^+10x|^ = oU oU oU Tension for E., 84.9 + «4iUx^ 39. 1\ 5 2 j 4 84.9 69.1 298.5 U2U, and 12 at L, Moment about L, of left reaction 16,364+ (284 + 230)230 3 Moment about L^ of loads to left of section 2155+116X62+^X60 = 44,861 Hor. Comp. in bar in tons for E. _44,861- 11,029 9S "^ +26X=— X60= 11,029 75 = 451.1 tons. Compression for Ej„ = 451.1 XI.O3IX-27X2 = 1162.7 Abt. 109 MAXIMUM STRESSES IN PETTIT TRUSS 223 MAXIMUM LIVE STRESSES IN INCLINED END-POSTS, CHORDS, AND MAIN VERTICALS— Conimwed Bar. Position of Loads. Computations. U,L, U,L, and and L5L5 U^L^ 2atL5 Max. Com- pression. 3 at L, Max. Tension Loads coming on from left. 15 at Z/. 15 at L^ 3atZ/, Moment of left reaction about 16,364+ (284-1- 109)109 Panel load at L^ = 80 30 12p X6p Compression in bar for. Ej,, = 29,600 2.67 = 240.0 M'oment about XJ^ of all forces to right of section through panel 3-4 = ^[16,364-1- (284+ 24)24] -230 = 15,607 Hor.Comp.(BarLsL4+M3iJ=^|^ = 208.1 Moment about TJ^ of loads to right of section through panel 2-3 =^^[16,364+ (284+24)24] -(7668 -192) = 19,797 -7476 = 12,321 12 321 Stress in L^L^ in tons for E4„= — ~~ = 205.3 Hor. Comp. in tons for £40 in MJ^^ = Vert. Comp. = 2.8 tension. Panel load at L, (Load 3 at L^ 230 / 25+20 \ 13(11 + 6) "30"+ ^^ \^~) 30 = 79.7 Tension in f/jL, for Ej„ = 2x|-X(79.7-2.8) = 192.2 = 50- Moment about L^ of left reaction 16,364+ (284+ 189)189 ■~ 2 Moment about i, of loads to left of section 52,880 Q A N/ 1 ft 5 ^4632+152X62+ \;„ X60 = 16,696 30 Compression in bar for E^t, 52,880-16,696^ 5 75 ^^T^^ = 1206.1 Moment about L^ 16,364+ (284+219)219 3 Tension in bar for E50 31,358 5 -10,816= 31,358 = 1045.3 Stress in panel 6-7 16,364+ (284+ 54)54 230 ~ 360 30 = 96.2-7.7 = 88.5 Stress in M5t/e = Compression in bar= 88.5 X2X^ = t221.2 * Note that sliear for this loading in panel 4-5 is positive, hence coimter Miht is not in action. t Note that this is larger than maximum panel load, hence is max imum stress. 224 TRUSSES WITH SECONDARY WEB SYSTEMS Akt. 109 MAXIMUM LIVE STRESSES IN SECONDARY MEMBERS, IN UNITS OF 1000 POUNDS This table shows all necessary computations. Bar. Position of Loads. Coznp utations. 13 at L„ L. orL. It has been previously determined that a maximum occurs with either load (12), (13), or (14) at a secondar}' panel point, hence panel loading for each case is com- puted below. . . ,,„.13(4+9+ll + 6) , ,.^17 Load (12) 3^ +IOX3Q 25 + 30+25+20 + ''' 30 Load(13)15(!±l|+lL+l) + i0X3^ + 20 m L^ (i4,1521±|±i«±S+i„x3li + 20 (15+20+25+30) 30 Tension in bar forE5|, = 86.3X^X2 = 85.4 = 86.3 84.8 215.7 M,L., 13 at L, Compression in bar for E5, i.3X-r-X1.414 = 152.5 4 M,U, 13 at L, 5 45 Tension in bar for £50= 86.3X-r-X=r^X2Xl.202=155.6 4 75 yhu. 13 at L. Tension in bar for Es,, = 86.3 X-^^X 1.280 = 138.1 PROBLEMS 50. Uniform live load, 2000 lbs. per foot on bottom chord. Locomotive excess, 20,000 lbs. " " Dead load, 800 lbs. per foot on bottom chord. Dead load on top chord and intermediate panel points as shown in figure. 1 -9-Ll0^1lll2 2 7 ra-r»- B H 25 26 P 4t.V N*4t. 2 2 2 Itoiu lil£=„4^^id/ifc^sa4?i 8 asX A ' Thiahalf 38 R T. /of trass same IT-J-lS^^sd^as halt shown Warren Tmssrlfo CoonteiB 20 Paneb @ 20^=400-^ — Prob. 50. Compute maximum stress in each member, following rules given in previous problems as to arrangement of computations, using special care to number and letter the bars exactly as in figure. Art. 109 PROBLEMS 225 51. Uniform live load, 2000 lbs. per foot on bottom chord. Locomotive excess, 20,000 lbs. Dead load, Dead load, 1200 lbs. per foot on bottom chord. 600 lbs. per foot on top chord. l-Panel8-@- ! Pbob. 51. a. Draw influence line for stress in bar a and compute its maximum value for above loading. 6. Draw influence line for stress in bar 6 and compute its maximum value. c. Draw influence line for stress in bar c and compute its maximum value. 52. Draw influence line for stress in bar a of trusses shown in Prob. 41 . a. Truss I. Truss has 12 panels at 25 ft. and height of 60 ft. b. Truss .1. Truss has 8 panels at 20 ft. and height of 30 ft. 12 panels @ 30 '- Peob. 53. 53. Dead load, top chord, 2250 lbs. per ft. per truss = 135,000 lbs. per panel point (approx.). Dead load, bottom chord, 3500 lbs. per ft. per truss = 105,000 lbs. per panel point. Uniform Uve load, bottom chord, 3000 lbs. per ft. per truss. Draw influence lines for stresses in a, b, c and d and compute maxinium values .of hve stresses. CHAPTER VIII TRUSSES WITH MULTIPLE WEB SYSTEMS, LATERAL AND PORTAL BRACING, TRANSVERSE BENTS, VIADUCT TOWERS 110. Trusses with Multiple Web Systems. Trusses of this type are statically undetermined, but are frequently built for spans of moderate length, as many engineers believe that more rigidity is thereby obtained. The trusses shown in Figs. 169 and 170 represent the more common types of such structures. Double System "Warren Tiiuss Fig. 169. The fact that such trusses are indeterminate makes it impossible to .correctly determine the stresses by methods previously given. Methods of accurately computing such stresses will be given later in fuU; but it may be said here that these methods can only be applied to trusses in which the areas of the various members are known or assumed in advance, hence, if used in design they must be applied through a series of approximations, the areas being first determined approximately, the stresses then computed, and the areas revised if necessary, this process being continued until a sufficiently accurate design is finally obtained. The accuracy of the approximate method ordinarily employed for such trusses is, however, sufficiently 226 Art. Ill STRESSES IN A DOUBLE SYSTEM WARREN TRUSS 227 high to make unnecessary the employment of more exact methods for the simple types of trusses shown in this article. The approximate method in common use consists of the division of the web members into separate systems, each of which is considered to be entirely distinct from the others. This amounts in reality to dividing the truss into two or more separate trusses with common top and bottom chords. The maximum web stresses in each of these trusses may then be computed in the ordinary manner. The method of determining the maximum chord stresses is clearly shown in the examples which follow. 111. Approximate Determination of Maximum Stresses in a DoublesSystem Warren Truss. Problem. Let the problem be the determination of the maximum stresses in all the bars of the truss shown in Fig. 171, with the following loads : Dead weight of bridge, 800 lbs. per ft. per truss, top chord =12,800 lbs. per panel. 400 " " " " " bottom " = 6,400 " " " Uniform live load, 3000 lbs. per ft. per truss, top chord =48,000 lbs. per panel. Locomotive excess, =40,000 lbs. Un Ui Ug U^ U4 Us Ug u? U s /Ub Li Ls Lg L4 Lg Lo L7 — '8 -panels @ 16 ==*13S ' I Fig. 171. The two web systems into which this truss is assimied to be divided are shown by the full and dotted lines respectively. The number of redundant bars in the structure may be deter- mined in the usual manner by comparing the total number of bars with twice the number of joints less three. This comparison shows that the number of bars is one in excess of the number needed for statical determination. Inspection shows that any one of the web or bottom chord members might be omitted without endangering the stability of the truss which would then be statically determined. Index Stresses. These may be written for each web system separately in the ordinary manner, considering the full sys- 228 TRUSSES WITH MULTIPLE WEB SYSTEMS Abt. Ill tem to carry only such loads as act at even numbered top chord panel points, and the dotted system to carry all other panel loads. With the web index stresses known, the chord stresses may be written in the ordinary manner, by adding the diagonal stresses at each joint successively, both systems being considered. Fig. 172 shows the index stresses for one half the truss. Were this truss to have an odd number of panels it would be necessary to write the index stresses for the web members in both halves of the truss, since neither system would be symmetrical. The index stresses were written as usual by beginning at the centre of the truss. The left reaction = 4i(6.4) +34(12.8) =73.6, X 6.1 12.8 12.8 12.8 1 12.8 panel loads 12.0 ri-SS.e l'-'^-188.o|^ 3-U7l2ly4 Fig. 172. which checks the web index stresses. The chord index stresses may be checked by the method of moments as in the ordinary truss, provided due allowance is made for the stress in the diagonal cut by the section selected. In this case the stress in the centre panel of the bottom chord may be checked by computing the moments about C/4 of the external forces to left of section XY, and subtracting from it the moment of the stresses in diagonal UgLi, making use of the fact that the moment about U4 of the stress in this diagonal equals the product of its vertical component, i.e., its index stress, and the panel length. The stress in L3L4 as determined from the index stresses = 150,400X^=120,300 lbs. By the method of moments the stress in L^Li^j- 1200-—^ "^ ^^ =120,300 lbs. This value agrees with that obtained from the index stresses, and consequently shows the correctness of these stresses. The Akt. Ill STRESSES IN A DOUBLE SYSTEM WARREN TRUSS 229 actual dead stresses may be computed from the index stresses in the usual manner and will not be given. Maximum Live Web Stresses. To determine the maximum live web stresses consider each system as an independent truss, and determine the stresses in the usual manner by the method of shear. MAXIMUM LIVE WEB STRESSES IN UNITS OF 1000 POUNDS Bar. Truss System. Uniform I.,oad at Panel Points. Loco. Excess at Panel Points. Vert. Comp. in Bars. L h Stress. U„L„ Full u,-u,-u,-u. Uo ^^48 + 24 + 40 = 136.0 1.00 -136.0 U^, Full u,-u,-u. U, 12 6 — 4S+-|40 =102.0 1.28 + 130.6 L,U, Full u,-u,-u. u. 12 6 ^4S+-|40 =102.0 1.28 -130.6 U,L, Full u,-u. u, -|48+-|-40 =56.0 1.28 + 71.7 L,U, Full u,-u. u. -|48 + -|40 = 56.0 1.28 - 71.7 U,L, Full u. u, |-48+-|40 = 22.0 1.28 + 28.2 L.U, Full u, u. |48+|40 =22.0 1.28 - 28.2 L,U, Dotted u,-u,-u,-u, V, ^48+^0 =131.0 1.28 -167.7 U,L, Dotted u,-u,-u, t'3 |-48 + -^40 = 79.0 1.28 + 101.1 L,U, Dotted u,-u,-u, l\ |-48+|-40 =79.0 1.28 -101.1 U,L, Dotted UrU, u. 448+1-40 =39.0 1.28 + 49.9 L^U, Dotted u,-u, u, i4S + -|40 = 39.0 1.28 - 49.9 U,L, Dotted u, u, -i+S + iiO - 11.0 O O 1.28 + 14.1 230 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. Ill As the truss is a Warren truss no counters are needed, but the maximum stress of both kinds should be computed in all bars in which reversal of stress may occur, since the area of such bars is dependent upon the magnitude of both kinds of stresses. Maximum Live Chord Stresses. For the maximum stresses due to the uniform live load, the index stresses should be written and the maximum stresses computed in the ordinary manner. It should be observed that for this truss the live stresses cannot be obtained from the dead stresses by multiplying by the ratio between the two loads since the live stress is not distributed in the same manner between the top and bottom chord. To determine the maximum stresses due to the locomotive excess it is necessary to decide in which system the bar should be considered in order that the stress may have its maximum value. This can usually be settled by inspection, but if doubt exists the maximum stresses for both systems should be written and the larger value used. The following table gives the maximum stresses due to locomotive excess in all bars: Bar. System, Load at stress. U,U, Full u. >^l-'' u,u. Dotted u. f40x|=-40 L\U^ Full u. |40x|=-48 v.u. Dotted V, |-40x|=-48 L,L, Dotted u. -^40xi5=+28 />jij2 Full u. |-40x|=+48 L-Jj^ Dotted u. ^«xS=+«'^ L,L, Full u. i-40x|=+64 Art. Ill STRESSES IN A DOUBLE SYSTEiM WARREN TRUSS 231 Concentrated Load System. The position of loads for the maximum stresses in this truss due to a concentrated load system may be determined by the use of influence lines. A complete solution for all bars will not be given, but the typical example which follows includes all the important points which are likely to arise. Bar UqLi Position of Loads for Maximum Stress for Cooper's E40. The influence line for the vertical component in this case is shown in Fig. 173 and indicates that heavy loads should lie in panels 1-2 and 2-3, with one of the loads at point 2. The method of moving up the loads, making use if necessary of the tangents of the angles 6 1 2 3 J T V between the influence lines and influence Lme tor Wrtical component in Bar U„ L, the horizontal, will enable us to Fig. 173. determine which load should lie at panel point 2. As the loads in panels 1-2 and 2-3 will be of the most importance in deciding this question, it is advisable to first determine the position, considering only the loads in these two panels, and then investigate to see whether a change in position will diminish or increase the stress. Since the influence line for these two panels is composed of two straight lines, the loads in these panels should be placed so as to give maximum moment at the centre of a 32-ft. span. It is evident from inspection that this occurs with load (3), at panel point 2. For this position the total load on the left panel of each of the other two-panel segments is greater than that on the right, and move- ment to the left until load (4) comes to panel point 2 will change this relation in one case only, hence it is evident that load (4), at panel point 2, gives a smaller stress than load (3). Movement to the right until load 2 is at the panel point will decrease mate- rially the stress due to loads in panels 1-2 and 2-3, but will increase the effect of the loads in the other panels. This will probably decrease the stress in the bar, but as the effect of this change cannot be so readily determined by inspection as in the other case, both cases will be computed, as this is simpler than to attempt to determine the exact change by the process of moving up the loads. 232 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. Ill Vertical component of stress in bar UqLi. Load (2) at U2'- Load at panel point 2. 10 X^ +20 ( ^^^^Iq^"^^ ) =47.5; Load at panel point 4. 13 X Load at panel point 6. 20 X 8 + 13 + 13+8 16 8 + 13 + 14+9 16 = 34.1; = 55.0; V.C. in bar from influence line ordinate 47.5X-?-+34.1x4-+55X^ = 66.4; 4 2 4 Load (3) at U2 Load at panel point 2. 10 X^ +20 ^ ^^ +^^1"^^ +^ = 56.9- 16 16 oioi-iiiio P Load at panel point 4. 13 X ttt '~'^^^^Jr =34.0; Load at panel point 6. 20X ^ "^^^^^ "^^^ + 13 X^ =51.6; V.C. in bar 56.9X^+34.0Xy +51.6X^ =72.6; This latter value is the maximum and should be used in the design. The position of load for the other web members may be determined in a similar manner. Bar U3 U4. Position of Loads for Maximum Stress for Coopers E40. The influence line for this bar is shown in Fig. 174. V N>. yf-- ^X^ A' ft|.5 =,|^ n.|^ 0.H H«Sv. y^ % «=|^ "1" MJCJ o,|„ "1- H^V |,I23t66 "^ Influence Line^ Stress in.'Bar UgUi Fig 174. The values of the ordinates are given by the following computa- tions, the bar in question being considered as a part of the dotted system for loads at odd-numbered panel points, and as a part of the full-line system for loads at other panel points. Art. hi STRESSES IN A DOUBLE SYSTEM WARREN TRUSS 233 Load at 7— Bar in dotted system— Ordinate=—X4^ = ^. 8 h 2h Load at 6— Bar in full system— Ordinate = — x3t = tt- 8 h 4th Load at 5 — Bar in dotted system — Ordinate = — -X4^ = -2_ 8 h 2h Load at 4 — Bar in full system — Ordinate = — X3f = ;rf. 8 h 2ih Load at 3 — Bar in dotted system — Ordinate = — -X4^ = '^. Load at 2 — Bar in full system — Ordinate = -^X5r = :^. 8 ft 4/1 Load at 1 — Bar in dotted system — Ordinate = —-x4^ = ^. 8 h 2h Inspection shows that for this case the moment will cer- tainly increase as the loads come on from the right until load (6) reaches panel point 3. As the loads move still further it is more difficult to determine exactly the position for maximum moment. An approximate determination based upon the assumption that the sloping' influence lines coincide with the dotted lines may be used, the error thus introduced being com- paratively small. Assuming this condition, the position for maximum moment will occur with the load on panels 5 to 8 inclusive, equal to that on panels 1 to 3 inclusive. Try load (6) to left of 3 : Load on 1-3=103; load on 5-8=118. .-. move up (7). Try load (7) to left of 3 : Load on 1-3 = 116; load on 5-8=108. .-. load (7) at panel point 3 will probably give the maximum value. It should be noticed that for this position load (12) is at panel point 5. * To determine the stress for this position compute the panel loads at panel points 1, 2, 6 and 7. Compute also the panel loads at 3 and 5 due to loads in panels 2-3 and 5-6. Multiply each of these panel point loads by the corresponding ordinate to the influence line, and multiply the loads in panels 3^ and 234 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. Ill 4-5 by the influence line ordinate in these panels. The sum- mation of these quantities gives the stress in the bar. Stress in t/3 C/4. Load (7) at panel point 3 : Load at panel point 1. 10 X I-20X = V V V Load at panel pomt 2. 20 X hl3X- = V V V Load at panel point 3, (loads in panel 2-3 only). 20x^+^'^ =1^ V V V Load at panel point 5, (loads in panel 5-6 only). 20xii+-^ =M2 V V -r , . , •.« „„^^5 + 10 , ,„13+8+2 599 Load at panel pomt 6. 20 X f-13 = V V V T J . 1 ■ X -7 ,0 ..^3 +8 + 14 ,13X13 , 16X4 558 Load at panel pomt 7. 13 X 1 V = V V V V Strccc; in bar- / 590+5 58\ y 825 5p 599 3p on cob 111 Odil — I I ^, n -rr T^ -rr \ p /2h p 4h p 4:h / 183+340 + 89Xl6 \3p 4975 V P )2h~ 20 - -248.7 As the method used for determining the position in this case was not a rigid one, the stress in the bar with load (13) at panel point 5 will be computed for comparison: Load at panel point 1. lOX— +20X^'^"^^^ "'"^"'"^= — P P p Load at panel point 2. 20 x^-^^— + 1 X 3^*^=^ P P p Load at panel point 3, (loads in panel 2-3 only). 13X^±i^ ^^ P p Art. 112 MAXIMUM STRESSES IN A WHIPPLE TRUSS 235 Load at panel point 5, (loads in panel 5-6 only). _246 ,14 + 13+7+2 568 1 1 9 20X-+13X- V P Load at panel point 6. 20X- + 13X- P Load at panel point 7. 13 X 3+9 + 14 P +2X P 13X6.5 P 507 Stress in bar=^^^+-^07W_^675.5^^ /2ft p in P 568 Sp p ih p I'lh p , /' 221+246+96Xl6 \3p 4938 „ . . „ 20 or considerably less than the value previously obtained. 112. Approximate Determination of Maximum Stresses in a Whipple Truss. The Whipple truss shown in Fig. 170 may be treated in a similar manner to the double-system Warren truss. The two systems into which the truss may be divided are shown in Fig. 175 by dotted and full lines, respectively. La L3 L4 Ls Le \-^ Lg -.10 panels @ 20 = 200 ' — -. — Fig. 175. This truss has one redundant member assuming that only one of centre diagonals of full system can act at once and the removal of any one of the web members except the end diagonals or end verticals would make the truss staticallj'' determined. This truss has, however, one element of uncertainty which does not exist in the double-system Warren truss previously treated, viz., that the end verticals TJiLi and C/9L9 do not distinctly belong to either system. This ambiguity is troublesome in determining how to place the live load for maximum stresses. The usual solution in this case is to use these verticals in such a manner as to give the maximum stiess in the bar under considera- tion. For example, if the problem be the determination of the maximum tension in bar C/2-^4, the bar -L9C/9 should be con- sidered as a part of the full system, and the bar UiLi as a part of the dotted system and the truss loaded accordingly. The fol- lowing example illustrates the method of solution for such a truss : 236 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. 112 Problem. Let the problem be the determination of the maximum stresses in all the bars of the truss shown in Fig. 176. Dead weight of bridge, 1200 lbs. per ft. per truss,' bottom chord = 24, 000 lbs. per panel. 600 " " " " " top " =12,000 " " Uniform live load, 3000 lbs. per ft. per truss, bottom chord = 60,000 lbs. per panel. Locomotive excess, =40,000 lbs. , (Notl Fig. 176. Index Stresses. The index stresses may be written for the dotted system by beginning at the centre, the bar C/3L5 carry- ing one-half of the centre panel loads, the dotted system being symmetrical, and panel point 5 at its centre. For the full system the shear in the centre panel is zero, and the stresses in bars U^Lg and L^Uq will each be considered as zero. It should be noticed that above conditions are based on the assumption that the dead panel load at both Li and Lg is equally divided between the two truss systems. The index stresses present no special difficulty. The only point to which attention should be called is the necessity for correcting the index stresses in the diagonals in the same manner as in the inclined chord trusses previously considered. In this problem the diagonal index stresses are corrected to conform to the slope of the diagonal U1L2] i.e., the stresses in the other diagonals are each doubled before the chord index stresses are written: Check calculations, 1 1800X200X200 Stress in UiU^ by method of moments = Stress in UiU^ in 1000 lb. units from index stresses 8 30 = 300,000. = 450x|^ = 300. Art. 112 MAXIMUM STRESSES IN A WHIPPLE TRUSS 237 The actual dead stresses are given in the following table in which the column headed ratio gives for each web member its length divided by its vertical projection; and for each chord member the fraction f , which equals the horizontal projection of the diagonal U1L2 divided by its vertical projection. DEAD STRESSES IN UNITS OF 1000 POUNDS Bar. Index stress. Ratio. Dead Stress. Bar. Index Stress. Ratio. Dead Stress. L,U, -162 1.201 -194.6 L„L, + 162 3" + 108 U,L, + 72 1.201 + 86.5 L1L2 + 162 2 ■3" + 108 U,L, + 24 1.000 + 24.0 L,L, + 2:34 2 3" + 156 U,L, - 48 1.000 - 48.0 L,L, + 342 2 3 + 228 u,h - 30' 1.000 - 30.0 L,L, + 414 2 3 + 276 U,L, - 12 1.000 - 12.0 U,U, + 342 2 3" -228 U,L, - 12 1.000 - 12.0 u,u. + 414 2 T -276 U,L, + 54 5 3 + 90.0 u,i\ + 450 •7 -300 U,L, + 36 + 60.0 u,u. + 450 '7 3" -300 U,L, + IS 3" + 30.0 U,L, 5 3 Before computing the live stresses the necessity for counters will be investigated. To do this consider each system separately. Maximum live compression in t/sLs— load Li and L3—E 4 3 at L3 V.C. = ~ 60 + — 40. This is considerably larger than the corresponding figure for dead tension, hence a counter LsC/fi is required. 238 TRUSSES WITH MULTIPLE WEB SYSTEMS Abt. 112 Maximum live compression in C/2l'4— load Li and ^2 at L2 V-C- = -^ 60+^40 = 26. This with impact added would be larger than the corresponding figure for dead tension, hence a counter L2f^4 should be used. LIVE WEB STRESSES IN UNITS OF 1000 POUNDS This table shows all necessary computations. Bar. Uniform Load at Panel Points. E at Panel Point. Vertical Component of Maximum Stress. Ratio. Stress. iof/l Li to L, incl. il 60X4^+^0 = 306 1.201 -367.5 U,L, L2, L„ I/j, Xg, L9 ^2 l^+lV*^ 158 1.201 + 189.8 U,L, L, L, 60+40 =100 1.000 + 100.0 . U,L, Li, Lg, Le, ^9 Li g60+fg40=102 5 3 + 170.0 U,L, Lf, io, Lg, Lj Li ;^60+^0 102 1.000 -102.0 u,h Lf, Lg, Lg L, l60+fg40= 58 5 3 + 96.7 U,L, ■^8, -^S) L« L, >+±40= 58 1.000 - 68.0 U,L, Ls,L, is f^60+l40= 26 5 3" + 43.3 u,h L3, is, L^, Lg is > + l,40 = 124 5 3" + 206.7 U,L, -^5; ^7, -^9 is 1>+1> "^ 5 3 + 123.3 U,L, ■^5. i?, -^9 is 1>«+1> ^4 1.000 - 74.0 U,L, i„i. i. i>+^0 36 5 3 + 60.0 V,L, L„L, L, ^60+^40= 36 1.000 - 36.0 Art. 113 SKEW BRIDGES 239 LIVE CHORD STRESSES IN UNITS OF 1000 POUNDS This table shows all necessary computations. Bar. Live Stress Due to Uniform Load = ^ of Dead Stress. Pfwition of E. stress Due to E. Total Maximum Live Stress. L,L, 108x|5= + l80 I-, >X3 +24-0 + 204.0 L,L, 108X^= + 180 L, S)"°X3 +-'■'' + 204.0 L,L, on 156X^=+260 U 8.40x|-=+42.7 + 302.7 L,L, Of) 228Xt^=+380 I-. ^40x|-=+56.0 + 436.0 L,L, 276Xt^=+460 lo L^ ^40xf= + 56.0 + 516.0 UJJ, Of) 228Xt^=-380 lo L^ 1>X3 ''■' -436.0 u,u. on 276X^=-460 lo L, ^40xf=-64.0 -524.0 V,U, OQ 300Xt3=-500 lo L, 1>X3"- ''■' -567.7 u^u. 300X?^=-500 L, >X^=-6«.7 -567 7 The determination of the maximum stresses in a Whipple truss for a concentrated load system should be made in a manner similar to that employed for the Warren truss, making use of influence lines to determine the position of loads. Computa- tions for such loads will be omitted as involving no new methods. 113. Skew Bridges. It is often necessary to construct bridges the abutments or piers of which are not at right angles to the bridge axis. Plans of such bridges are shown in Figs. 177 and 178. In structures of this sort the trusses are frequently unsymmet- rical, as is evidently the case for the trusses shown in Fig. 177. The trusses shown in Fig. 178 are symmetrical, but the panel loads are affected somewhat by the skew of the ends. If it is desired to use inclined end diagonals for such trusses, they should both 240 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. 114 have the same inclination to the horizontal in order that the end portal may lie in a plane. For simplicity in construction the floor beams should be located at right angles to the trusses. In order to satisfy both of these conditions it is frequently desir- able to place the end hangers at an inclination to the vertical, as shown in Fig. 179. The computation of stresses in such trusses may be made in the same manner as in the trusses already considered, and requires no special treatment. If difficulties occur in determin- ing the position of the loads, they may usually be solved by using the influence line. %'"m Fig. 177. " f Tiuss A. Elevation Truss A. Pla 1 showing floor- 'ipams i Truss B. 1 / \ 1 \^ X / \ Fig. 178. / Truss B. Elevation Fig. 179. 114. Lateral and Portal Bracing. It is evident that a bridge in which the floor beams form the only connection between the trusses would be unstable laterally, especially if of long span. This instability would be due partially to its inability to with- stand the force of the wind acting upon the truss itself and upon the train or other live load which may be upon the bridge, and partially to the lateral vibration to which it may be subjected by the live load, this being especially severe for railroad bridges exposed to swift and heavy train service. In addition to the insecui'ity of such a structure as a whole another disadvantage would be the fact that the top chords would have to be made much heavier than would be the case were they to be rigidly braced, since they would be in the condition of very long columns unsupported laterally, and the extra material used to give them Art. 114 LATERAL AND PORTAL BRACING 241 sufficient strength would, in most cases, be more than sufficient to provide for lateral bracing. For these reasons it is considered necessary to use lateral bracing in all bridges. In through bridges this bracing should consist of a horizontal truss in the plane of the bottom chord, another in the plane of the top chord when the depth permits (trusses of insufficient depth to permit the use of overhead brac- ing are called pony trusses and should be avoided) , and vertical bracing between the verticals of as great a depth as the allowable clearance permits.^ In deck bridges a horizontal truss may be used in the plane of the upper chord and vertical sway bracing : ^ixixixixMf : Plan.of Top Lateral Syatena ^ Intermediate Section F^ End View ShowiDg Portal /KTK M/ ■^ Plan of Bottom Xateral System. Fig. 180. between the vertical members, no horizontal bracing being used in the plane of the bottom chord, or all three systems of bracing may be used. In through bridges the end reactions of the top lateral truss cannot be directly transmitted to the abutments owing to the necessity of preserving a suitable opening for the traffic, hence portal bracing is required in the plane of the end posts, the purpose of this bracing being to tie the end posts together and make thereby a rigid frame by which the end reactions can be transferred to the abutments. Figs. 180 and 181 show the lateral bracing in through and deck bridges respectively. ' One of the reasons for using vertical bracing when both top and bottom lateral systems are used is to assist in distributing unequal train loads between the trusses. 242 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. 115 115. Lateral-bracing Trusses. Lateral trusses may be either statically determinate, or statically indeterminate, according to whether the diagonals are tension rods, or riveted members capable of carrying both tension and compression. In the former case the maximum stresses may be easily determined, once the wind panel loads are known, by dividing the truss into two S3'stems, as was done in the multiple system trusses previouslj' considered. In the latter case, the cross struts of the top system and the floor beams in the bottom system (in a through bridge) con- nect the two sets of diagonals so rigidly that it is impossible to divide into separate trusses ; a reasonable assumption for such ^ ?f End View -/-f/'"'^ aud Intermediate Section Showing Sway Bracing Tlan of Bottom Lateral System Fig. 181. a truss is to consider the shear in a panel to be divided equally between the two diagonals, one being brought into tension and the other into compression. It should be said that the present-day practice is to use riveted laterals in both top and bottom systems of railroad bridges in order to secure rigidity, but that tie rods are frequently used for highway bridges. Where the wooden floor bridge is con- tinuous, as in many highway bridges, or where a continuous steel floor is used, the principal use of the lateral rods of the loaded chord system is to assist in erection by holding the trusses in line. 116. Approximate Determination of Maximum Stresses in Lateral Bracing. Problem. Let the problem be the determination of the maximum stresses in the bottom lateral system of a through bridge with eight panels at 25 ft. and with trusses spaced 18 ft. between centres, assum- ing that the laterals are stiff members and able to carry both tension and compression. The horizontal lateral truss is shown in Fig. 1S2. Art. 116 MAXIMUM STRESSES IN LATERAL BRACING 243 I Birec^n. o{ wind lyixixixixMMxi L, L„ 1-3 L, 8i)an6la@2ff' Fig. 182. Solution. The lateral force acting at the bottom chord will be assumed as a moving force of 500 lbs. per lineal foot=12,500 lbs. per panel. It is unnecessary to compute the lateral stresses in the floor beams, since the addition of a slight direct stress in these would be of no importance, hence it is immaterial whether this lateral force be assumed to be dis- tributed between the two chords or be applied entirely to the wind- ward chord. The latter condition will be assumed, however, for ease in computation. For conven- ience, the components of the diagonal stress at right angles to the axis of the truss will be spoken of hereafter as vertical components, and those along the truss axis as horizontal components. Index Stresses. These will be written for the full load, this being the simplest method of getting the chord stresses, and are shown in Fig. 183. The actual chord 25 stresses will be -- of the index lo stresses. It should be noted that the lateral-truss chords are also the chords of the main truss, and that the wind stresses in them are some- times of sufficient size to re- quire additional area in these members, although it is cus- tomary to permit higher unit stresses for the combination of live, dead, and wind loads than would be allowable for live and dead stresses only. Maximum Diagonal Stresses. The vertical components of the max- imum diagonal stresses in 1000 lb. units will be as follows, assuming the shear in each panel to be divided equally between the two diagonals : Fig. 183. Panel 0-1, imX12.5j =+21.9; Panel 1-2, V|^X12.5 ) = +16.4; Panel 2-3, |(^X12.5 ) =+11.7; Panel 3-4, i(^^|xi2.5J = + 7.8. 244 TRUSSES WITH MULTIPLE WEB SYSTEMS Aet. 117 117. Portals. Approximate Solution. The portal bracing and end posts of a tlirough bridge must be designed to carry to the abutment the reaction from the top lateral system, and also to withstand the wind pressure on the end posts themselves, the former being the more important factor. This combination of brac- ing and end posts is called the portal, and is a statically inde- terminate structure. Accurate solutions of such structures may be made by the method of least work, but the approximate solu- tion which follows is sufficiently accurate for all ordinary cases. Fig. 184 shows a common type of end portal for a through bridge. The statical indetermination is due to the condition at the bottom of the end posts and to the rigidity of the portal bracing. Neither of the posts is pin-ended; that is, neither has a pin at Fig. 184. right angles to the plane of the portal, the main truss pins being in the plane of the portal. The ends of the posts are, how- ever, really fixed to a considerable degree, since they bear upon the foundations, although they are not usually rigidly fixed thereto, and the dead weight of the structure is sufficient to offer a very considerable resistance to overturning under the action of the wind forces. If the weight of the bridge is sufficient, it is evident that the posts may be treated as if they were fixed at the bottoms. More- over, if the knee braces Mq/™ and ilf'o/'« be rigidly fixed to the posts, the latter may be considered as fixed at points Mq and M'o also. Assuming that such is the condition, the posts will bend under the action of the wind forces as shown in Fig. 185 and points of inflection will occur at a point in each post between the bottom of the knee brace and the bottom of the post, a and b indicate these points of inflection. If the position of these Art. 117 PORTALS— APPROXIMATE SOLUTION 24.5 points of inflection be known, and if the horizontal reaction at the bottom of the posts be also known, the stresses in the structure become determinate, since the moment at the point of inflection must equal zero-.'- It is commonly assumed that each point of inflection occurs midway between the bottom of the knee brsipe and the bottom of the post. It is also commonly assumed that the portal brac- ing is so rigid that the distance apart of the posts remains unchanged under the action of the wind forces, and that in consequence the horizontal reaction at the bottom of each post equals one-half the sum of the applied loads. Neither of these assumptions is more than approximately correct, but in the ordinary structure the error introduced thereby into the design of the end posts is small, since the wind stresses in these mem- bers are in themselves small compared with the live and dead stresses, and the percentage error in consequence is still smaller. The portal bracing itself is frequently made considerably larger than is necessary, owing to the comparatively small magnitude of the wind forces, and the difficulty in choosing members with small enough areas which are also suitable in other ways. With the points of inflection and the distribution of the horizontal reactions between the two posts known or assumed, the computation of the stresses in the various members may be easily made. The structure, however, differs somewhat from those which have been previously treated, since it consists of a combin- ation of columns, carrying direct stresses and bending, and a truss. The example which follows illustrates the method of com- putation based upon these assumptions. Problem. Let the problem be the determination of the stresses in the portal of the bridge shown in Fig. 186. ' This may be proven as foUo-ws: Let R = radius of curvature at any section of a member exposed to bending; M = bending moment at this section due to external forces; I =the moment of inertia at this section; E =the modulus of elasticity. , . 1 M From mechanics, R~~ffJ' At the point of inflection the beam must be straight, since at this point the curvature changes, hence iS = infinity, and n =0, .". ^ = 0. 246 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. 117 Solution. The wind force on top chord at, say, 200 lbs. per Uneal foot of bridge equals 2500 lbs. per panel point per truss. The force applied by the lateral truss to the portal at m equals the vertical component in diag- onal mo plus the panel load at m. The sum of these two forces equals 5 5000X— + 1250=13,750 lbs. There will also be a force of 1250 lbs. at n. In addition to the wind force acting along the top chord, there will be 1 Top Lateral System ^^ 1. o p* fi panels® 35 ' '^, Fig. 186. a uniformly distributed wind force applied directly to the end posts. This will be assumed as 100 lbs. per lineal foot of the member. The outer forces acting upon the portal will then be as shown in Fig. 187, assum- ^J» isBoibK. ''^S points of inflection and dis- 7P T" tribution of horizontal forces as -|- " previously stated. The vertical forces and bending moments at the bottoms of these posts may be computed as follows: Let the moment at the bottom of each I _; --^ Inflection "T:^ ^- post = M, and the vertical force = I -^ s ^ V. Then, since the moment at the point of inflection = zero, the moment about an axis through the point of inflection in each -i -Vm post of the forces below that point ^ ,„., ^ =10,500X11 -lOOXllX^-M. Fig. 187. 2 .-. i\/ = 109,450 ft.-lbs. The direction of the moment in each case must be counter-clockwise, as shown, to balance the clockwise moment due to the horizontal forces. In order that equilibrium may exist, the moment of the couple due to the vertical forces must equal the moment of the external forces about any axis minus 2M. Taking the origin of moments at the bottom of either post, the following equation may therefore be written : L 0^10 500 Iba. 15,000 X 30+ 6000 X 15 -109,450X2 - .-. 7 = 20,070 lbs. ■167 = 0. The next step is the determination of the stresses in the portal members themselves, and the direct stresses, bending moments, and Art. 117 PORTALS— APPROXIMATE SOLUTION 247 : V2 100 IbB. per ft. ^Mo shears in the end posts. It is evident that each main post is a con- tmuous member without hinges. That is, the joint at Mo can in no sense be considered a pin joint so far as the two sections UoMo and il/o^o of this member are concerned, since the stability of the entire struc- ture depends upon the lateral stabihty of these end posts. Indeed the moment in the post at this point, according to our hypothesis, equals 109,450-10,500X22 + 100X22Xll = -97,350ft.-lbsi The other joints may, however, be pin joints, and will be so considered. Moreover the joint il/o will also be considered a pin joint so far as the stress in Molm is concerned; that is, the stress in M Jm will be assumed to be direct stress. To compute the stress in the portal bars it is necessary to treat the post U^L^ as a beam supported at the point M, by a truss bar, the direction of which determines the direction of the beam reaction at this point, and at the point U^ by a reaction which is unknown in direction, and which equals the resultant of the unknown stresses in UoTd and UJm. This beam is loaded by a uniform load of 100 lbs. per foot o^^er its entire length, and by the horizontal forces of 10,500 lbs. at L„ and of 13,750 lbs. at Uo- It is also subjected at L„ to a bending moment of 109,450 ft.-lbs., and a tension of 20,070 lbs. This condition is shown by Fig. 188, in which the reactions at {/» and Tl/o are represented by their horizontal and vertical components. The ratio of V^ to H2 is determined by the slope of the portal bar MJm. Since this makes an angle of 45° these two components are equal. The ordinary equation of statics may now be applied. Application of the equation 271/ =0, using Uo as the origin of moments, gives the following equation : 10,500X30-3000X15-109,450-8/^2=0; .-. i/j =+20,070 lbs. =yj. — i ^- t 20 0TDJbs. Fig. 18S. ' This value may be verified by considering the portion of the post between M^ and the point of inflection. The shear at the point of inflection in pounds =10,500 — 1100 = 9400, and the moment is zero, therefore, the 11 moment in foot-pounds at M„ = 9400Xll-100XllX-^ = 97,350. 248 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. 117 Application of the equation Si/=0, gives the following equation: -/^i -20,070-3000+10,500-13,750=0; .-. i/i = -26,320 lbs. Application of SF=0 gives the following equation: 7i -20,070 +20,070=0 .-. T^=0. Hence the stress in bar UJm=0, from which it follows that the stress The actual stress in i¥„/m= stress in/mr„=20,070X1.414= +28,380 lbs. Proceeding in a similar manner with the other post, the following results are obtained : Stress in r„/™' and Im'M/ =-28,380 lbs.; Stress in Im'U^'= 0; Stress in T^U^' =+13,820 lbs. The computations may be checked by considering the joint T^ and T„ „„ applying the equations of equilibrium. The forces acting at the joint are as shown in Fig. 189, and evidently satisfy the equations of equilibrium. Since the stresses in UJm, Imlm', and Im'Uti' are zero, it might perhaps be thought that these bars should be omitted, but it should be remembered that the computations are approximate, and that the stresses as determined by more exact methods may not be zero. Moreover the appearance of the portal is improved somewhat by the inclusion of these bars. In addition to the determination of the stresses in the portal bars, the maximum moments and shears in the posts should also be obtained. These are alike for both posts and are shown by the curves of Fig. 190. The maximum direct stress in each post =20,070 lbs. It is tension in L„Mo, com- FiG. 189. in Lo'Mo', and zero in MoUo, and pression M„'f/„'. Before leaving should be called wind forces cause 105001b8. Pig. 190.— Curves of Mo- ment and Shear in column. Full Line shows Moment. this subject, attention to the fact that the stresses in the main truss members. These stresses are rela- tively small compared with the stresses due to the vertical loads, but may attain high absolute values in large trusses. In the windward trusses these stresses tend to cause compression in the bottom chord which in conjunction with the stresses due to longitudinal thrust caused ' This value may be checked by taking moments about Tg of the forces to the left of a vertical section through this point. It will be found that this moment = 0, hence the stress in ImW equals zero. Art. 118 PORTALS— MISCELLANEOUS 249 by the tractive force may even reverse the normal tension in the end bottom chord members, which are frequently made as columns to resist this compression. 118. Portals — Miscellaneous. The portal treated in Art. 117 represents a common type of portal which is statically deter- mined with respect to the inner forces. Portals are frequently built, however, which are statically undetermined with respect to the inner as well as the outer forces. For such cases the methods used in the treatment of double-system trusses maj' sometimes be applied. For more complicated portals special methods may have to be devised, but the construction of such portals should be avoided. Portals which lie in a plane inclined to the vertical, as would be the case for a bridge with inclined end posts, may be treated in the same manner as vertical portals, care being taken to use the correct lengths along the posts and not the vertical projec- tions of these lengths. 119. Transverse Bents in Mill Buildings — Approximate Method. A typical structure of this type is illustrated by Fig. 191. The stresses due to the vertical forces may be figured in the ordinary manner, assum- ing vertical reactions from the truss upon the columns at points b and i, and assuming zero stress in knee braces ac and hk. For the horizontal wind forces an approximate method similar to that used in computing portals is com- monly employed, the horizon- tal reactions at the bottoms of the columns being assumed equal and points of inflexion midway between bottoms of columns and points of connection between columns and knee braces.^ As in the portals all joints Fig. 191. ' This assumption should not be made unless warranted by the conditions existing at the bases of the columns. In many structures of this character the resistance to bending-moments offered by the column footing is very slight; in such cases the, point of inflexion may be assumed as occurring at the base of the column. 250 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. 119 are assumed to be pin-joints except those at a and k, while these latter are also so considered with respect to the stresses in the knee braces themselves, which are assumed to act along the axes of these bars. The stresses in bars ac and kh may be determined as in the portal by applying the equation, i;Af=0, to the two columns hm and in, using for the origin of moments points h and i. The horizontal and vertical forces required at points 6 and i may then be obtained by the appHcation of the equations, 2F = and 'LH = Q, to the two columns. With these values determined, the roof truss may be treated as anj^ simple truss, the outer forces being the applied wind loads, the stresses in the knee braces, and the reactions at the column tops. The complete determination of the stresses in such a struc- ture by the approximate method will not be given, the problem which follows including all the essential points. An accurate determination of these stresses may be made by the theorem of least work, but will not be given here. Problem. Compute the horizontal and vertical components of the truss reactions at points h and i, and of the knee-brace stresses in the transverse bent, shown in Fig. 191, for a horizontal wind force of 600 lbs. per lineal foot on hm, and a normal wind force of 400 lbs. per lineal foot on hf. Solution. The applied loads will be as shown in Fig. 192. As in the portal the horizontal components at TO and n are each assumed to equal one-half the total horizontal force on the structure, thus having a value of 16,50t) lbs. each. For this case the moment at point m will not equal that at point n, since no wind force is assumed to act on the lee- ward column. Each moment may be found, however, by applying the equation, SAf =0, about the point of inflection of the forces below that point. The equations for these moments will be as follows : Fig. 192. 15 From which (16,500-600X 15)15 + 600X 15 Xy-il/m = 0, Mm =-l- 180,000 ft.-lbs. In a similar manner Mn = 16,500X15 = 247,500 ft.-lbs. Art. 119 TRANSVERSE BENTS IN MILL BUILDINGS 251 The vertical reaction Vm may now be determined by the application of the equation, SM=0, using for an origin the point n. The result- ing equation is as follows : -180,000 + eOFm + 600 X 45 X 22.5 + 6000 X 52.5-12,000 X 45-247,500 = 0. From which Vm= +750 lbs. Application of 27=0 gives Fn = 12,000 -750= +11,250 lbs. The horizontal components in bars ac and hk may next be computed by applying the equation, SM = 0, using for origins paints b and i respectively. The equations thus obtained are as follows: 180,000+600X45XY + -'^C'('0"rfflc)X 15 -16,500X45 = and 247,500-//C (bar /i/c)X 15 -16,500X45 =0 From which HC (bar ac) ■■ 3,000 lbs. and HC (bar hk) = - 33,000 lbs. The vertical components of these forces equal the horizontal com- ponents, since the bars have a slope of 45°. Hb 3 000 lbs. - — 3 000 lbs. 33 000 33.000 Iba, ' ' Res ultant Win d force=600 x ib = 27 000 16 500 Iba. 1^ >I180 000 ft. lbs. 16 500 Ibs -y ^847 500 £t. Ibi '760 "lbs. Fig. 193. ' 11 260 lbs. The reactions at points b and i may now be determined by applying the equations "SH — O and I1F=0 to each column as a whole. The forces acting on the columns are shown in Fig. 193, hence, ^6=27,000-19,500= +7500 lbs. Vb=+ 2,250 lbs. /?i= +16,500 lbs. 7i= +21,750 lbs. The forces acting on the truss will therefore be as shown in Fig. 194, and the truss may now be computed in the ordinary manner. The 252 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. 120 moments, shears, and direct stresses in the columns may be deter- mined as in the portal columns previously treated. 1 501 copo sooot- 3 000 — " 1 '600 + 7 500 [ I 3 00o"f j^OOlb,. 2 260 3 000 33 000 lbs. Fig. 194. 120. Viaduct Towers. In the determination of the stresses in such towers it is necessary to consider the vertical forces due to the weight of the structure itself and the superimposed load, and the horizontal forces due to wind, centrifugal force if tie structure be curved, and tractive force. Such towers are usually composed of four columns, braced transversely and longitudinally. To obtain sufficient width at the base to prevent excessive uplift at the windward columns when the structure is either unloaded, or loaded I13' an empty train, the latter are usually built to a batter, transversely, of one horizontal to six vertical. For symmetry a double system of bracing should be used, and the structure will, therefore, be statically undetermined unless the bracing consists of rods, which is not common in railroad viaducts. The stresses due to the horizontal forces may be computed in a manner similar to that used for the wind bracing systems already computed, i.e. by assuming each diagonal to carry one-half the stress it would be called upon to carry if but one system of diagonals were used. If the vertical loads are symmetrical with respect to the central axis of the tower they will cause no primary stress in the bracing, and the vertical components in the columns will therefore equal the vertical loads. If the vertical loads are not symmetrical with respect to the tower, as may be the case with a structure buUt on a curve, stresses due to these loads will be caused in the diagonals as well as the columns. The necessary computations for the loaded structure are clearly illustrated by the problem which follows. Problem. Compute stresses for the tower shown in Fig. 195 due to the following assumed loads: Dead load, 600 lbs. per foot per rail for girder and track. 200 lbs. per foot in height of tower for eaph column. Live load, 3000 lbs. per foot per rail. Art. 120 VIADUCT TOWERS 253 Wind load, 600 lbs. per lineal foot of structure applied at base of rail and 50 lbs. per foot of height of tower for each column. Tractive force, 20% of live load applied at base of rail. Longitudiual View Transverse View Viaduct Tower Fig. 195. The spans on each side of tovi^er are each 56 ft. in length centre to centre of end bearings. Dimensions and load concentrations are shown in Fig. 196. _j-;B".>p^llPjy3*i"5.20O lbB.=:WlDd force on train and elrders = traction Eorre r L'r\ H > , i._i.AT 71 800 v:. Fig. 196. 254 TRUSSES WITH MULTIPLE WEB SYSTEMS Art. 120 And the necessary computations follow: COMPUTATION OF STRESSES IN VIADUCT TOWER Transverse Bracing Bars. Horizontal Component. - Strea3. tt 25,200X14 + 500X21 2X25.04 363,300 50.08 = 7250 1 o oo 7250 X^^ = ± 11,000 8.D7 b 363,300+1150X31 2X36.37 399,000 72.74 = 5500 5500xJ|^=± 7900 c 399,000 + 1800X44 2X53.12 478,200 106.24 = 4500 on CO d 478,200 + 2300X67 2X76.81 632,300 153.62 = 4100 4100X^;|5=± 5500 The column stresses are shown on next page. The maximum uplift on the windward column should also be deter- mined. For the wind load previously considered the uplift at base of column =71,800 lbs. To this should be added the uplift due to the tractive force. Assuming an unloaded train at 1200 lbs. per linear foot the uplift due to this force = --— X84 x 0.20 X^ =27,300 lbs. The total reaction on one column due to live and dead loads =39,000+^^X42=64,200 lbs., hence the net uplift = (71,800+27,300) -64,200=34,900 lbs. It is also common to determine the uplift on the unloaded structure due to an assumed wind force of 50 lbs. per sq.ft. on one and one-half times the vertical projection of the structure. In the design of viaduct towers it is common to assume that the combination of dead, live, wind, and tractive forces will seldom, if ever, occur simultaneously, and that in consequence a higher unit stress may be used for these combined forces than would be employed for live and dead stresses only, a common practice being to increase the unit stress 25%. For example, if the allowable unit stress for dead load, and live load corrected for impact, is 16,000 lbs., a value of 20,000 lbs. would be used for the maximum stresses due to live, dead, wind, and traction. If centrifugal force exists the stresses due to it should be con- sidered as live stresses, but need not be corrected for impact. Art. 120 VIADUCT TOWERS 255 i O H O w 02 W H o o H -^ P O o o o Q o •* O ■* o -* o Tt( o •— t o t-H (N 1— 1 r- 1—1 IN o TtT q oo" q a q ai' 1 1— I CO tH ■* >o 1—1 to X HH X +1 X +1 X -H ^ o 11 o o fi o II m o o o ti o o to_ a> IN cc" w oo" 00 CO Tf lO CD o (-, 1 ^ 1 o 1 1 o o r^ w .o T— ( o £ (X> cc 1— ( 05 00 IN 1 ^ 1 cc lO t^" oi 00~ gj. oo" S •^ CO ■«< m to "o 5 q II ^Y" II ^ |co II .-( |co II S- + + + a 1-1 \m CO r^ CO ?J 1 CO q CO g + d n^ c^ Pi t^ T— 1 1— ( , "^ 1 1 1 1 •1- o •1- o o T o "ri o o o o IX) u N ■o q^ t^_ .9 T-T 1—1 t^ c^ rH ^ .a o CO o T-H o CO o o IJO oi OJ o 1 X o 1« X o co" X o o CO o 1—1 T— K 1—1 oc O o q lO CO co" T— 1 c^ CN c^ -i X + o 4- O + -S o o O a o B °. co"" ltT 00 i> Tj- 1 r r-^ t-" DQ X o 1-1 1 01 1 1—1 1 O o 1 .t o 11 y^ 1—1 ^ o -^ Q '^ Q ^ O 1—4 o ,-H 1—1 1— 1 o m q CD_ q Oi q to q (M^ S ,—1 ecT 1— ( oo" 1—1 (M" tH l>r ^ X N X being sustained during erection by the weight of the anchor arms mn and oq and by anchorage at m and q. A number of large bridges of this type have been con- structed in the United States, among which may be mentioned the Red Rock Bridge, the Poughkeepsie Bridge, and the Beaver Bridge, the outline of which is shown in Fig. 195. The Har- vard Bridge at Boston is a plate-girder cantilever. The Forth Bridge in Scotland, with a clear span of 1710 feet, the longest clear span in the world, is also a cantilever bridge. If the 258 Art. 123 EQUATIONS OF CONDITIOX 259 suspended span be omitted and the cantilever arms connected, the bridge becomes indeterminate. The Queensboro Bridge in New York city is the most important example of a bridge of Ancihorage Ancbonge Fig. 195. this type. Such a bridge can be built along more graceful lines than if a suspended span is used, and troublesome details at the connection of the suspended span and the cantilever arm avoided. 123. Equations of Condition. Let Fig. 196 represent, diagrammatically, a bridge of three spans similar to that shown in Fig. 195. As shown, there are eight unkno\\'n components of the outer forces, viz., two at each point of support. Evidently this structure is statically indeterminate to a high degree. The most obvious method of reducing the degree of indetermination is to fix the direction of some of the reactions. Since one of the H4 IVa v. tv3 tv, Fig. 196. Fig. 197. reactions, at least, must have a horizontal as well as a vertical component to give stability, it is possible in this manner to elim- inate only three of the eight unknowns, this being insufficient to make the structure statically determined. The remaining equations necessary to secure statical determination must be obtained by the method of construction and are called equations of conditions. Such equations may be obtained by introducing hinges in the end spans, as indicated in Fig. 197, these hinges being so constructed as to make it impossible to transmit bend- ing moment through them. This construction therefore gives the two following additional equations: I,Ma=0 and 2il/i,=0. These equations signify that the moment of all the forces on either side of either hinge about an axis passing through the hinge at right angles to the plane of the forces equals zero. These 260 CANTILEVER BRIDGES Art. 123 equations should not be confounded with the general equation SM =0, which is also applicable at the hinges, but which means that the moment of all the forces on both sides of any section about any axis perpendicular to the plane of the forces equals zero. With regard to the moment about the hinges it should be noted that although the moment about a of all the forces to the left or right thereof = 0, it should not be supposed that this gives two independent equations, since if the moment about a of all the forces to the left of a equals zero, the moment of all the forces to the right of a about the same point must also equal zero, such a result following at once by the subtraction of the former equation from the general equation I!M=0. There are then but five entirely independent equations, hence five and only five unknown quantities can be determined, and the structure is, therefore, determinate. Were there more than five independent equations the structure would be unstable; were there less it would be statically undetermined. The simplest method of providing a hinge in a truss is to omit a chord bar in one panel as is done in Fig. 198. Evidently the Fig. 199. moment about an axis through a of all the forces on either side of a must equal zero, since the structure can by its construction offer no resistance to moment at this point. For a plate girder cantilever a hinge may be constructed as shown in Fig. 199. As it is uneconomical in practice to have a long cantilever bridge restrained horizontally at one point only, — such a condition involving the transmission of a horizontal force, if applied at one end of the structure, throughout the entire length of the bridge — it is common to fix the structure horizontally at more than one Art. 124 ANCHORAGE 261 pier and to omit a bottom chord bar at one or more of the hinges, thereby preventing the transmission of horizontal forces across the bridge. For the case illustrated by Fig. 198 the omission of bar ab would accomplish this result. The application of these methods to the structure shown in Fig. 195 would involve the omission of bars cc', d'd, and ab, and of the rollers at o. In practice the bars mentioned would not actually be omitted, as they are necessary in erection and improve the appearance of the structure. They should, however, be made adjustable and incapable of resisting a horizontal thrust. 124. Anchorage. Since the load on the suspended span or on the cantilever arms causes negative reactions at points such as m and q, Fig. 195, which may exceed the dead reactions at these points, it is necessary to anchor the structure to the masonry and to provide sufficient weight in the piers to equal this uplift. The anchorage usually consists of girders embedded in the pier and fastened to the structure by eye-bars. The freedom to move horizontally may be obtained by rollers or other devices. 125. Reactions — Cantilever Trusses. The reactions upon structures of this type inay be determined by the application of the three equations of statics combined with the equations of condition in the same general manner as for simple trusses and girders. The problem is, however, more complicated than for structures supported at two points, and in consequence influence lines for certain of the reactions in typical cantilevers will be given and methods of determining the reaction values stated. Consider first the structure illustrated by Fig. 195. For this cantilever the trusses mbc and db'q evidently act like beams supported at two points only and supporting the suspended span, bb' at their ends. That this follows from the application of the equations of equilibrium and condition may be proven in the following manner : Assume first a concentrated load, P, upon the truss db'q. For this condition the forces acting to the left of b are the same as the forces acting to the left of b', viz., Vi and V2, Fig. 200, and the moment of these forces about each of the hinges b and b' = 0. .-. the following equations may be written: I. Vi{Li+L2) + V2L2 = 0. II. Vi{Li + L2 + L3) + V2(L2 + L3)=0. 262 CANTILEVER BRIDGES Abt. 125 Subtracting I from II gives F1L3 + VzLg = 0. .-. Fi+F2 = 0, lience (Fi + 72)^2 = 0. Subtracting this latter equation from I gives ViLi = 0, therefore, Fi = 0, ^2 = 0, and Vi' + V2' = P, hence the span db'q when loaded, acts like a simple beam, since the moment at each end is zero, and the sum of the reactions equals the load. Influence Liui.- for Vj Fig. 200. Now consider a load, P, on span bb' at a distance x from b' . The following equations may be written for the moments about b and b' respectively. III. Vi{Li + L2) + V2L2 = 0. IV. Vi{Li + L2 + Ls) + V2{L2 + L3)-Px=0. Px Subtracting III from IV gives Fi + F2 = -f^=positive shear at J'. J-'3 The moment at n equals the moment at b plus the product of the /Px\ shear at b and the lever arm L2; it therefore equals— I y— JL2, the moment at b being zero by construction. This moment also equals— ViLi, hence the following equation may be written: m' whence Vi- Aht. 125 REACTIONS— CANTILEVER TRUSSES 263 This is identical with the reaction that would be obtained if the span bb' were assumed to be supported on the ends of the two simple beams mb and qb'. In a similar manner the reaction at q may be shown to equal the reaction that would exist if a similar assumption were to be made ; hence the proof of the state- ment that the reaction in a structure such as that shown is identi- cal with the reactions which would occur if the structure were to be considered as composed of two independent beams mb and qb', supporting the simple span bb' at their ends. The influence lines in Fig. 200 show clearly the reactions due to loads in the different portions of the structure. It should be noted that these influence lines would be unchanged and statical determination accomplished by omitting the rollers at o and bar ab in Fig. 195. The cantilever shown in Fig. 201 differs somewhat from that of Fig. 195, some of the equations of condition for the structure Fio. 201. being established by the omission of diagonals over the central piers. The structure as shown has eight unknown reactions; six vertical and two horizontal. To determine these unknowns there are available in addition to the three general equations of statics, five equations of condition. Two of these condition equations are obtained by the insertion of hinges at a and b, i.e., by the omission of upper chord bars at these points; two by the omission of diagonals over the centre piers,i and one by the omission of the bottom chord bar in the cantilever arm adjoining the hinge at &. Of these condition equations the latter is of use in determining reactions due .to horizontal forces only (wind or ' Diagonals may be used in the towers for purposes of bracing, but they should be of small size and offer sUght resistance to distortion. This same device is frequently employed in partially continuous draw spans. 264 CANTILEVER BRIDGES Abt. 125 longitudinal thrust of train), and in combination with the equa- tion 11H=0 is sufficient for this purpose. The remaining four equations of condition and two of statics may be used in the following manner to determine the six vertical reactions. Let S = shear in panel nn' = by construction. M„ = moment at n. M„' = moment at n'. Ma = moment at a = by construction. Mf, = moment at 6 = by construction. Sa = tension in hanger aa'. Case 1. Load on suspended span ab. Consider the load P at distance x from b, and the portion of the structure between sections XY and ZQ. The following equation may be written: ]\h = Px-Sa{Ls)=0, Oa — ^ — ■ But Sa is the supporting force at the left end of the suspended span and equals tlie reaction at the corresponding end of a simple end-supported span. It is evident, therefore, that the suspended span ab acts like a simple end-supported truss, since the moment at each end equals zero and the reactions are inversely propor- tional to the distance of the load from either end. It should be observed that no stress is caused in the hangers at a and b by a load unless it is applied to the suspended span. Case 2. Load P on cantilever arm n'a at distance x from n'. For this case Ma=Mn'+{S+ Vs)L2-P{L2-x), Mn' = Mn+Sp, Sa==0. By construction Ma=0, and S=0. M„' = M„ and 7i=-F2, and M„' + F3L2 - P (La - x) = 0, Art. 125 REACTIONS— CANTILEVER TRUSSES 265 hence, ViLi + {V3-P)L2+Px=0. But Vi + V2 = S=0, and V3-P=S, since !« bs »l^b2 » V jSO-panelB-®^- ■• 1^ ^ Influeuce Line for V3 Fig. 202. 127. Bar Stresses — Cantilever Trusses. The determination of the bar stresses in cantilever trusses involves no special dif- ficulties and may be accomplished by the use of the methods employed for simple trusses. Influence lines may be employed for determining the position of the loads, if concentrated load systems are to be used. For structures of the magnitude and weight of such bridges, however, the actual use of concentrated load systems for the stresses in the main truss members is gen- erally -unnecessary, an equivalent uniform load giving nearly if not quite as accurate results. The determination of the position of a uniform live load for maximum stress in each bar and the computation of that stress may be accomplished by the use of influence lines if desired. An influence table similar to that prepared for the three-hinged arch given later showing the stress in each bar for a load at every panel point should, however, generally be prepared to facilitate Art. 127 INFLUENCE LINES— CANTILEVER, TRUSSES o 267 vit II _v4 II — r I — f^^^ — ■ 8| paoelB I 10 panels i 8 panels 10 panels --Ml I ViT 1 1 vi 10 l>aDe1s ^ — |-+®-30M Influence Line for Moment at panel point (13) Fig. 203. (Refers to truss on page 259.) TO (3)(1) Ww'(i)Xl°)«.^Hinse-^6 O o' m' Influence Line for Moment at panel point i(0) Fig. 204. (Refers to truss on page 263.) 268 CANTILEVER BRIDGES Aet. 127 the computation of the stress due to the dead load, which in a large structure should not be taken as uniformly distributed, and with this table once prepared no advantage would be gained by using influence lines. The influence line in Fig. 205 is given to illustrate the variation in stress in a particular bar rather than for its aid in computing the stress. This statement also applies to the influence lines of the previous articles. If it be desired to use influence lines to check the tabular results the actual stress may be determined most readily for uniform loads by multiplying the areas between the influence line and the horizontal axis by the proper load. a,_^Hmge^6 m' T \i ip >lp|<-4j)->k ap — >\—ip- -ip- FiG. 205. — Influence Line for Stress in Barn" n'", Truss shown in Fig. 201. Referring to Fig. 205, it is evident that the maximum stress in bar n" n'" will occur with the truss loaded with the uniform live load from n' to h, and that its value equals the product of the area n'a'h and the combined live and dead loads per foot, provided these are uniformly distributed. PROBLEM 69. a. Show that this structure is statically determined with respect to the outer forces. 6. Draw influence line for reaction at ijj. All panel points of top chord lie on a pareibola D J* — 8 panda @ 30^— >t< 8 panela @ 30^ — lk~^ panda Prob. 59. c. Draw influence line for stress in bar a and compute its maximum value for a uniform live load of 3000 lbs. ]ier ft. CHAPTER X. THREE-HINGED ARCHES. 128. Characteristics of the Arch. The essential difference between the ordinary arcli and the girders and trusses that have hitherto been investigated is that the stresses in an arch may be confined to compression and- shear, while in trusses and girders large tensile stresses are also developed. This possible elimination of tensile stress in the ordinary arch rib is due to the fact that both ends of the arch are fixed in position by construction, hence each reaction has a horizontal component even under vertical loads; in consequence the reactions converge, and if the shape and thickness of the arch rib be properly chosen, the resultant force at each section for any given position of the loads may be made to pass through the centre of gravity of the section and therefore cause no bending moment, or so near the centre of gravity that the tensile fibre stress due to the bending moment caused by the eccentricity is insuflficient to overcome the com- pression due to the thrust. The advantage of the arch form was well known to the ancients, as is shown by the many stone arches constructed by the Romans and even by older races, and the arch remains to the present day one of the most useful and graceful of structures, its employment being frequently dictated both by aesthetic and utilitarian con- siderations. 129. Types of Arches. Up to a comparatively recent period the arch was always constructed as a statically undetermined structure, similar .to that shown in Fig. 206, which represents the conventional masonry arch with neither reaction fixed in direction, magnitude, or point of application, the arch being in consequence statically undetermined in a three-fold degree, having six unknowns. With the application of iron and steel to bridge construction came a recognition of the advantage of statical determination, 269 270 THREE-HINGED ARCHES Art. 129 and metal arches began to be constructed in which some of the unknowns were eliminated by the insertion of hinges. Such Fig. 206. — Masonry Arch. arches are shown in Figs. 207 and 208. If in the arch shown in Fig. 208 a hinge be inserted at the centre similar to that of the arch shown in Fig. 207, the arch becomes a three-hinged arch and Fig. 207.— Metal Arch with One Hinge. is statically determined. The ribs of metal arches may be formed 'either of plates and angles as in plate girders; of cast iron or cast steel segments riveted together; or of riveted trusses. Fig. 208.— Two-hinged Metal Arch. In recent years a considerable number of three-hinged masonry arches have been constructed, and since the adoption of plain Art. 129 TYPES OF ARCHES 271 and reinforced concrete to arch design, the custom of applying the loads at fixed points to the arch rib by transverse walls has also been adopted in many long-span bridges, thus doing away with some of the uncertainty which formerly occurred in such cases, and securing many of the advantages of the metal arch. Fig. 209 illustrates such an arch. It should be said, however, that the common type of masonry arch remains that without hinges, shown by Fig. 206. Fig. 209. — Three-hinged Masonry Arch. One other important type of metal arch — the spandrel- braced arch, is shown by Fig. 210. This structure is in reality a combination of a truss and an arch rib. As will be shown later, if the arch rib in the three-hinged spandrel-braced arch be con- structed to a parabolic curve, the diagonals and top chord "nill not be in action under a full uniform load, the arch rib in that case acting like the arches previously described, the loads being applied through the vertical posts. Fig. 210. — ^Three-hinged Spandrel-braced Metal Arch. Like the other arches the spandrel-braced type is frequently constructed as a two-hinged arch. The three-hiaged arch is the only type which will be considered here, the statical indetermina- tion of the other forms requiring the development of other than statical methods as a preliminary to their investigation. 272 THREP^HINGED ARCHES Art. 130 130. Reactions— Three-hinged Metal Arches. These may be computed for any position of the load by the application of the three general equations of statics combined with the equations of condition established by the hinges. If the end supports are at the same elevation, as is generally the case, the horizontal components of the reactions balance and hence have no effect upon the vertical reactions, which would be the same as for a simple truss or girder of the same span. To obtain the horizontal reactions it is necessary to make use of the equation of condition, viz., that the moment about the centre hinge of all the forces on either side of that hinge equals zero. The application of this equation is so simple as to need no expla- nation. The influence lines for the vertical and horizontal com- ponents of the reactions are given in Figs. 212 and 213 for the "T^ ^ ' f , y ,k- 1^ //>'- •<- ' 1 , ■ V 1 ~ =l-~ k^'-r 1 ,-t' 1 , +- 1 '^T^^ Influence Line lor Vl Fig. 212. Influence Line tor H„ and Hi, Fig. 213. arch shown in Fig. 211 in order to show clearly the variations in the reactions as the load crosses the structure. These lines are also correct for the spandrel-braced arch shown in Fig. 210, pro- vided the arch rib has the same dimensions, since the construc- tion above the arch rib has no influence upon the value of the reactions. For a uniform load the maximum value of both the horizon- tal and vertical components, and hence of the actual reaction. Art. 131 MAXIMUM STRESSES IN ELASTIC ARCH RIBS 273. evidently occurs when the entire structure is loaded, while the maximum reaction for a concentrated load occurs when the load is placed at the centre of the span. In designing the piers it is as important to know the direction of the reaction as its magnitude. Both may be determined graphically for any position of the loads by the methods shown in Fig. 212, in which the sloping dotted lines showing the direction are determined by laying off at the foot of each vertical the corresponding horizon- tal ordinate as obtained by scale from Fig. 213. It will be observed that the direction of the left reaction is constant for loads on the right half of the structure. This is not accidental, but is due to the effect of the centre hinge. Since \\ith a load on the right half of the structure the only force to the left of the centre hinge is the left reaction, and since the moment about the centre hinge equals zero, the left reaction must pass through it. This principle may be stated as follows: For a load to the right of the centre hinge the direction of the left reaction coincides with a line drawn through the left and centre hinges, and vice versa. It is evident that whUe the reaction at one end due to the live load on the other half of the arch may pass through the end and centre hinges, the actual reaction will not have this direction, since such a condition would involve the entire absence of dead load in the half of the structure adjoining the reaction in question. With a concentrated-load system the maximum vertical reactions may evidently be determined as for any simple beam, while the shape of the influence line shows that the maximum horizontal component will occur for that position of the live loads which would give a maximum moment at the centre of a span of the length of the arch, and hence may be easily determined. The exact position for the maximum value of the reaction itself is less easily determined, but an equivalent uniform load may be used with safety to determine the actual maximum reaction. 131. Maximum Stresses in Elastic Arch Ribs. The maximum fibre stress at any section of the arch rib of a structure like that shown in Fig. 211 may be determined if the direction, point of application, and magnitude of the resultant force at the section are known. It is a well-known principle of mechanics that a force P applied at one of the principal axes OF of a cross-section of a straight elastic bar in the manner shown in Fig. 214 causes a 274 THREE-HINGED ARCHES Art. 131 direct fibre stress at a distance c from the other principal axis OX, which may be expressed by the equation: ^' , AT- C S = -r±Nv-j-, in which iV= normal component of the force P. (SI= transverse component. A = area of cross-section. r= distance of point of application of force from the axis OX. / = moment of inertia of cross-section about axis OX. The shearing stress due to the transverse component, S, may for such a case be computed in the same manner as in an ordinary beam. Longitudinal Section o£ Bar. Fig. 214. N Nv , ., Re For a curved bar, such as an arch rib, the formula just given for the value of s is not strictly correct, but should be replaced by the following equation: f R+y ■J RyHA' in which R — radius of curvature of the axis of the arch. 2/ = distance of any fibre from axis OX. For arches the radius of curvature, R, is always very large, compared with the dimensions of the cross-section, hence CR^y , , r 1 1 , fie and RtfclA "^"""" '"'-^ """"■' J yHA I """ E^c V is also small compared with R in any well-proportioned arch; hence the second term of above expression for s may be neglected with but little error, giving for a final value Art. 131 MAXIMUM STRESSES IN ELASTIC ARCH RIBS 275 the same expression as for a straight bar.i in this formula A'^f = external bending moment on the section, hence the formula may be written: N , Mc in which M= Nv. In order to determine the maximum compression at the cross- section of any arch rib, it is necessary to determine the position of the loads which will produce the maximum value of the expres- N Mc sion -J + -J-, and to determine the maximum tension (or minimum compression) , the position of loads giving the maximum negative value or minimum positive value of ^ — j- must be determined. These equations are applicable for arches which can carry both tension and compression. If the arch can carry compres- sion only, as in the case of the ordinary stone arch, they are cor- rect only when the value of -^ j- is positive. Masonry arches should, however, be so proportioned that this condition will always exist. For uniform loads the simplest method of determining the position for maximum direct fibre stresses is by an influence table, in which the maximum values of the direct tension and compres- sion at various sections for load unity at each panel point are given, a sufficient number of sections being chosen to ensure economy and safety in the design. For arches carrying concentrated load systems, influence lines may be drawn for maximum stresses of both kinds at as many sections as may be desired, and the position of the loads deter- mined in the manner previously used for trusses, or an influence table may be employed and the maximum stresses determined by trial, the value of the panel loads for probable positions being first tabulated. Examples of the computations for such an arch will not be given, as it involves nothing but the application of '■ The error made by these approximations is extremely small, even for arches with as sharp a radius of curvature as an ordinary sewer arch. The general formula should, however, be employed in determining the stress in a curved bar such as a crane hook. 276 THREE-HINGED ARCHES Art. 132 the principles already thoroughly illustrated for other structures, and the student who is familiar with these principles should have no difficulty in applying them to such a structure. 132. Parabolic Three-hinged Arches. In practice three- hinged arches are frequently constructed either with a parabolic axis or with panel points lying on a parabola. If the end pins of such an arch are at the same elevation, and if the load is vertical, uniformly distributed, and applied to the arch by vertical posts, the moment at any panel point equals zero. ^ Hinge Bottom chord panel points Ue on a parabola Fig. 215. Ve This proposition may be proven as follows : Let H^ and Vj^, Fig. 115, be the horizontal and vertical com- ponents respectively of the left reaction. X and y equal the abscissa and ordinate respectively of any panel point on the arch axis referred to the left hinge. M = moment at this point due to a uniform vertical load over the entire span. M^ = moment at same point due to the vertical loads and vertical reactions only. M";j = moment at the same point due to the horizontal reac- tion, H^. The vertical reactions in such a structure are the same as for an end-supported beam, therefore M„ equals the moment on such a beam, hence it varies as the ordinates to a parabola. (See Art. 43, Case 8.) Since Hj^ is constant for the loading under consideration, and y is the ordinate of a parabola, Mh= {Hj^) (y); therefore it also varies as the ordinates to a parabola. But M=M^—Mh; therefore it also varies as the ordinates to a para- bola, therefore — is constant for every panel point. At the center Akt. 132 PARABOLIC THREE-HINGED ARCHES 277 M hinge ilf =0. .". — =0 at this point and consequently at every panel point of the arch. For sections between panel points M^ varies as a straight line. (See Art. 36.) If the arch itself be straight between panel points, {Hl)(jj) also varies as a straight M line, hence — varies as a straight line between panel points and in consequence equals zero, hence the arch rib carries direct com- pression only. This is the ordinary condition for spandrel- braced arches, hence under a full uniform load the stresses equal zero in top chord and diagonals of such an arch, i.e., an arch conforming to the conditions stated at the beginning of this article; the stress in the verticals equals the panel load, and the stress in the bottom chord is direct compression throughout and has a horizontal component equal to the horizontal component of the reaction. If the arch rib be curved between panel points the bending moment in it will be zero at the panel points only. For partial loads the moments at the panel points will not equal zero and the arch rib will be subjected to bending moments throughout its length. It should be observed, however, that the maximum positive moment at any panel point due to a uni- form live load will equal the maximum negative moment at the same point due to the same load. This is due to the fact that the portion of the structure which should be loaded with a uniform load for maximum positive live moment at any section should be unloaded for a maximum negative live moment at the same section and vice versa, hence the combined loading for maximum positive and maximum negative moment is equivalent to a full uniform load, therefore the maximum positive live moment plus the maximum negative live moment equals zero. For spandrel-braced arches a partial load causes stress in the diagonals and in all the top chord bars except the adjustable one, and the maximum tension in these members under uniform live load equals the maximum compression for the reasons already given. For this type of arch the bottom chord, or arch rib, carries only direct stress if straight between panel points, the structure acting like any other framed structure. With a concentrated load system the maximum positive bending moment will not equal the maximum negative moment, nor will they be equal for a uniform load with locomotive excess. 278 THREE-HINGED ARCHES Art. 132 These conclusions for a spandrel-braced arch are confirmed by the problem which follows : Problem. Compute the maximum stresses in all members of the spandrel-braced three-hinged parabolic arch shown in Fig. 216. Hiui^e' Botlom chord panel points lie on a parabola Fig. 216. Dead weight of bridge, 1000 lbs. per foot per truss, top chord =20,000 lbs. per panel. 400 lbs. per foot per truss, bottom chord = 8000 lbs. per panel. Uniform live load : 2000 lbs. per foot per truss, top chord =40,000 lbs. per panel. Locomotive excess, 25,000 lbs. This problem may be solved either by use of influence lines or an influence table. The latter will be employed here in order to illustrate its use. The following laws concerning the magnitude and direction of the left reaction when the load is to the right of the centre hinge are of mate- rial assistance in preparing such a table. a. The line of action of the left reaction passes through Lq and L4. 6. Its vertical and horizontal components and hence its magni- tude varies directly with the distance of the load from C/g. c. The moment about each of the panel points to the left of the centre is counter-clockwise, hence the stress in the top chord is tension and that in the lower chord compression. It follows from the above rules that the magnitude of the stress in all the lower chord bars in the left half of the arch varies uniformly as the load moves from Us to U^, hence the magnitude of the stress in the web mem- bers of the left half of the arch also varies uniformly, since the stress in each of these members is a function of the combined vertical components of the left reaction and the stress in one of the bottom chord members. With the load on the left half of the arch the stresses in the bars on the left half of the structure will not vary uniformly, and may be either tension or compression, since the left reaction varies in magnitude and direction. The influence table will now be given, and a table giving maximum stresses in all bars follows. Art. 132 INFLUENCE TABLES 279 ?l'^ ^l-'' SI'O §1"= X X X X 1— 1 1— 1 ,— 1 + 1 + 1 i SI-" Sl'^ ;2|.o gl-= SI-" Sl"^ 21'-'' o\ v"' Q en X ,o X o «o Ix o 1 o 1 CI X Q X ,„ ;2i=c.i X Q IC |(M O ^ 1 ^1 7 ^ a + 6 1 d + i; 1 d + -; 1 d + !^ O W o SI"' 1 SI"' §1"' 1 Sh^ Sl-^ 1 Sh §!'» + X II X II X II X II X II X II X II X t^ 1 yj w |co CO |Tt< CO |-* m |o3 lO j:/3 ^ Ic^i rHJOl '^ 03 1 + 1 + 1 + 1 + m H b ►J b hj b ■-C b ►J W b -4' b ■-f b "^ b l-T O o g|« Sh S X X O 0} 1—i 1— 1 o ID + 1 ^A J^jco ?x\'^ £3|»3 §1" ^'1°° Sl=° 2|oo Sl« , CO C « >< 00 >< CO >< .o >< u. ■o^ =^ ^^ ^ ^ o ^ o O d 05 ^|00JH ■"hSS .o|^g '^ 1 CO 2k^g 1 (M 6 (B W + d 1 d + ^ 1 d + d 1 w + ^ 1 vi g °l«> 1 §|00 + Sl<» 1 §1* + §l« 1 §1°° 1 §l« + Sl« 1 g M P X II X II ^|-x X II CO \-« X II col-^ X II X II ■o |oo X II X II 1 + 1 + 1 + 1 + Q o ;z; ^ e-i pj M -< CI b ►-T b l-l b b b S CB 2 O - b 4' b i4' b "-T b ►4' *^ O N t-l" O ICO 1^1" O ICO t^jco o jco r- 1" O ICO (-H » ■^IrH (M IrH '^I'H IM l-l ■^ l-H C-l Irt li-i (M 1— 1 O X X X X X X X X si 1 O ^ loo !» <^\^Trj 1 Oi -I"*! ^1«?5 "5 |(N t-^ .oImE:: 02 + 2 O IfO 1 1 d g|2 + + d O ICO , 1 d O ICO 1 o|co| IM Irt + 1 2 O|co7 + d g|2 + 1 % O ICO 1 d (M I.-H 1 (N l-H 1 CV1 1^ 1 c-l Irt 1 O) Irt 1 tf W X II X II X II X II X II X II X II X II o i^ |oo I- J-Xi co|^ co|^ m jyj o 1 x rt |(N 1 + 1 + 1 + 1 + w J _ 5 kl b kJ" b Kj" b 4' b ^-^ ^ 03 c ffi b ►4" b 4" b b b •--^ H lO Ico "5 \-i< 2I«> lO Ic-i o ' ^ II II II II W o O \fO O ICO OIO O |CD S-. 00 If-H 00 I.-H Vj \^ 00 1—1 2; ■^ X X X X ^ H rf loo -1^ col:o ^ |(M 1— 1 rt '-^ ,-|-^ CO |-t< '- !y: -Hj-I S I b b b b o 280 THREE-HINGED ARCHES Akt. 132 Q O ■n H 2; W Z O 1 *1 o < H Z O N O w Q m O H O < O H -I P3 <1 H O t) o o lO o C^ lO 1^ o I— 1 o (^ o CO o m 1— i •yi o c-i o T-H g + 1 + 1 + 1 M II II II II II II d o o o o o o o o o c M iq o IQ o iq ^ ,-H ^ r-i TtH rA -t X X X X X X "9 r,|- CO ]^ ^1^, -|o, -h --H -r + 1 + 1 + 1 pa o Kj K^ t3 ►J" b ►-f ta' >--: b ►J" b ►-f ;_j g "i i~ Ol ICO lO CI X C2 (M t- CO H CO c/q^ « c/j CO C-. '/ 6 oi O ^ d d li, 1 + 1 + 1 + 1 o ■n II II II cfi d II o II o 11 o H lO lO lO UO «:> •o fc K C-I r^ Ol c^ c^ t--. W ,— i co •—i CO rH CO o X X X X X X fc CO \^ CO |-I^ -|c, r-* |CI ^l-t. -1-^, o + 1 + 1 + o J f. p. fi <: >-.' o ■-f b ^ b >-J H n a a l«^ ►4', ti. 4' b ^ O t4 IS o s CO 00 03 00 -t^ — CO O 00 ^ -+ ^ s '•- Tt^ t^ CO t^ (M t^ »o t^ ^ t~ 1^ t- -J t^ -4 t^ -^ t^ t^ — ; (^ m IQ O O .IM lO o O '-' lO o o o d d + C^ 1 6 + ci 1 6 + CO 1 a / II X II X II X II X II X II CO |~*^ CO I'* ^|o, rt 1^1 ^ |-H ^ I"* + 1 + 1 + 1 - 01 b '^ b 1^ b 4' CQ :::J ►4" t3 ^- b t-J O 2|«) "0 1^ "0 [oo H 1 *A CO |go -H [^ ^|x ^^ ' lO (O ^ 3 3 t) t) tT J 2, (3 O o3 += 3 '- 2 O J3 T3 . a M C3 *" 3 Ol ^ ra 3 =9 -,^ a 0^ ,j3 C3 »T3 & s g g S 2 «> — SJ3 tn ^ QJ 3 S.2 S3 2 J =^£ Art. 132 INFLUENCE TABLES 281 INFLUENCE TABLE FOR VERTICAL COMPONENTS IN DIAGONALS Vi= shear in panel containing diagonal. ^2= vertical component in bottom chord bar in panel as determined from previous table. Fs=F,±y2=vertical component in diagonal. Load at Bar UoLi. Bar UiLi. Bar I/iLs. Bar U3L1. u. yi = + 0.875 r2= -0.21(1 1',= +0.656 -0.125 + 0.096 -0.029 -0.125 + 0.047 -0.078 -0.125 0.000 -0.125 V, F,= +0.750 rj=_0.437 r3=+0.313 + 0.750 -0.192 + 0,55S -0.250 + 0.094 -0.156 -0.250 0.000 -0.250 u. T-,= +0.625 1-2= -0.657 ■ 1^3= -0.032 + 0.625 -0.481 + 0.144 + 0.625 -0.234 + 0.391 -0.375 -0.000 -0.375 u, ri = + 0.500 ^2= -0.875 1-3= -0.375 + 0.500 -0.769 -0.269 + 0.500 -0.563 -0.063 + 0.500 -0.200 + 0.300 u, 1'3= -0.281 -0.202 -0.047 + 0.225 u. T3= -0.187 -0.134 -0.031 + 0.150 U-. 1-3= -0.094 -0.067 -0.016 + 0.075 Under full load the vertical component in each diagonal equals the algebraic sum of the tabular values. This sum should and does equal 0, thus checking all the tabular values. 282 THREE-HINGED ARCHES Art. 132 INFLUENCE TABLE FOR STRESSES IN VERTICALS Fs= vertical component in diagonal running to joint at top of vertical. See previous table. ^4= panel load at top of vertical. F5= stress in bar. Load at Bar UoLo. Bar UiLi. Bar U2L2. Bar UsLi. Bar UiLt. u„ F3= Fi= -1.000 F5= -1.000 0.000 0.000 0.000 0.000 u. F3 = +0.656 F.= 0.000 F5= -0.656* 0.029 -1.000 -0.971 -0.078 0.000 + 0.078 -0.125 0.000 + 0.125 0.000 u. F3=+0.313 V,= 0.000 F.= -0.313 + 0.558 0.000 -0.558 -0.156 -1.000 -0.844 -0.250 0.000 + 0.250 0.000 u. Fs= -0.032 V,= 0.000 1*5= +0.032 + 0.144 0.000 -0.144 + 0.391 0.000 -0.391 -0.375 -1.000 -0.625 0.000 u, F3= -0.375 Vt= 0.000 F5= +0.375 -0.269 0.000 + 0.269 -0.063 0.000 + 0.063 + 0.300 0.000 -0.300 1.000 u. F5= +0.281 + 0.202 + 0.047 -0.225 0.000 u, F5 = + 0.187 + 0.134 + 0.031 -0.150 0.000 u, T'5= +0.094 + 0.067 + 0.016 -0.075 0.000 * Note that ( + ) stress in diagonal gives the atreas in each vertical equals unity. ( — ) stress in vertical and that for full load Art. 132 INFLUENCE TABLES 283 INFLUENCE TABLE FOR STRESS IN EACH BAR FOR LOAD AT Bar. u. Ui U^ Ui U, Ui Us Uv u,u. 0.000 -1.010 -0.481 + 0.048 + 0.577 + 0.433 + 0.289 + 0.144 u,u. 0.000 -0.938 -1.875 -0.312 + 1.250 + 0.937 + 0.625 + 0.312 u,u. 0.000 -0.625 -1.250 -1.875 + 1.500 + 1.125 + 0.750 + 0.375 u,u. 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 L,L, 0.000 -0.662 -1.325 -1.990 -2.650 -1.990 -1.325 -0.662 L,L, 0.000 + 0.397 -0.793 -1.980 -3.170 -2.380 -1.590 -0.793 L,L, 0.000 + 0.317 + 0.632 -1.580 -3.790 -2.840 -1.900 -0.950 L,L, 0.000 0.000 0.000 0.000 -4.000 -3.000 -2.000 -1.000 U,L, 0.000 + 1.204 + 0.575 -0.059 -0.688 -0.515 -0.343 -0.172 U,L, 0.000 -0.078 + 1.500 + 0.388 -0.725 -0.545 -0.362 -0.180 U,L, 0.000 -0.322 -0.644 + 1.614 -0.260 -0.194 -0.128 -0.066 U,L, 0.000 -0.637 -1.275 -1.913 + 1.530 + 1.148 + 0.765 + 0.382 U„L, -1.000 -0.656 -0.313 + 0.032 + 0.375 + 0.281 + 0.187 + 0.094 U,L, 0.000 -0.971 -0.558 -0.144 + 0.269 + 0.202 + 0.134 + 0.067 U,L, 0.000 + 0.078 -0.844 -0.391 + 0.063 + 0.047 + 0.031 + 0.016 U,L, 0.000 + 0.125 + 0.250 -0.625 -0.300 -0.225 -0.150 -0.075 U,L, 0.000 0.000 0.000 0.000 -1.000 0.000 0.000 0.000 The values in this table may be verified by the same methods used for preceding tables. From the influence table for stress in each bar, the maximum live stresses, due to imiform load, may be easily obtained for any given bar by summing up the total positive and negative values for the bar and multiplying each sum by the live panel load. The stress due to the locomotive excess may be computed by multiplying the maximum value for each bar by the excess load. The table which follows shows the stresses thus obtained : 284 THREE-HINGED ARCHES Art. 132 TABLE FOR MAXIMUM LIVE STRESSES IN ALL BARS Tension. Corapres sion. Uniform Live at Eat stress, Lbs. Uniform Live Load at E at stress, Lbs. v„c, I' 3 to [/, inc. u, 74,100 U, and U, u, 84,900 u,u, U, to [/, inc. u. 156,200 U-i to U3 inc. u. 172,900 u,u, C', to (7; inc . u. 187,500 Ui to Ug inc. u. 196,600 XJi to U-j inc. u. 490,400 L,L, l\ u, 25,800 U2 to !7, inc. u. 507,500 L..L, U, and U^ u. 53,800 f/3 to {/, inc. u, 537,200 L,L, XJt to [/, inc. Ui 500,000 f',/-, U, and f'. u, 101,300 f/, to 'U^ inc. u, 88,300 l\L, [/j and U, f': 113,100 r, and U^ to C/, inclusive u. 93,700 U,L, u. u. 104,900 (7, and U^ and i7< to U^ inclusive l\ 80,700 U,L, Ui to [/, inc. u. 191,300 f7, to C/j inc. u. 200,600 U„L„ [/j to U^ inc. u. 48,200 U„ so f72 inc. u„ 83,800 U,L, (7, to (7; inc. u, 33,600 Ui to [/j inc. u. 91,200 U,L, Ui and !7i to U., u. 11,400 t/2 and U^ u. 70,500 U,L, Ui and [/, u. 21,300 1/3 to [7, inc. C7, 70,700 (\L, u, u. 65 000 The dead stresses are as follows: Top chord bars and diagonals End verticals 10.000 lbs, Intermediate verticals 20,000 lbs. Bottom chord bars (horizontal component) 280,000 lbs. PROBLEM 60. a. Draw influence line for horizontal components of reactions. h. Draw influence line for vertical component of stress in bar a. Pkob. 60. CHAPTER XI DESIGN OP COLUMNS AND TENSION MEMBERS 133. Columns — General Considerations. A column is a mem- ber designed primarily to resist compression, although it may also be subjected to transverse loads causing flexure. For the present columns of the first type only will be considered. Com- pressive tests of blocks of plastic material of such proportions that the length does not greatly exceed the minimum lateral dimen- sion show that failure occurs by lateral flowing of the material with no well-defined ultimate strength; a definite elastic limit, however, exists beyond which the material simply expands laterally and contracts longitudinally under increasing loads. On the other hand, compressive pieces in which the ratio of length to least lateral dimension is high, fail by lateral bending, when subjected to compression, even when the load is applied along the longitudinal axis passing through the centre of gravity of the bar, and the bar is originally straight and of homogeneous material without initial I I stress. The ultimate load per square inch for the latter class of columns may be much less than the product of the elastic limit and the cross- section area. A good illustration of such a condition is presented by a straight bar of tempered steel of very small cross-section. A short piece of such a bar would sustain a high load per square inch without showing * signs of failure, while a long piece would | I collapse by bending laterally under a com- ^,^^ ^^y paratively light load, the column bending in one of the ways indicated by Fig. 217. The columns used in engineering structures generally have a slenderness ratio midway between these two extremes, hence failure may be expected either by crushing or bending, or by both together, even if the 285 286 DESIGN OF COLUMNS AND TENSION MEMBERS Akt. 134 column be originally in an ideal condition so far as material, shape, and loading are concerned. The ideal column, however, does not exist in practice. The load is seldom if ever applied exactly at the centre of gravity or along the column axis; the process of fabrication in a metal column is sure to leave the column with some distortion, and with the material in a condition of initial stress, and columns of timber or concrete are equally sure to be imperfect. Moreover, the material is never homogeneous, and in a built-up steel column, such as is generally used in important structures, the behavior of the column as a whole is dependent upon the integ- rity of its cross-section, which may or may not be preserved by the rivets, tie-plates, lattice bars, and other devices required to hold together the main pieces. In view of these many uncertainties, the economical and efficient design of columns is one of the most serious problems which the engineer has to confront, especially when dealing with unusual cases. The difficulties are heightened by the lack of sufficient experimental data. In recent years the study of the subject has, however, received a decided impetus, due in great part to the failure of the compression chords of the Quebec bridge, and many valuable data are being collected. 134. Condition of Ends. If the ends of a column are unre- strained against turning, it is said to have hinged ends; this condition, however, seldom exists. Columns in which the loads are applied by pins at the ends, as in many American bridges, are said to have pin ends. Columns in which the ends are subject to such restraint that the tangents to the elastic curve at the ends remain parallel to the column axis when the column deflects laterally are said to have fixed ends. If the ends of the column are square and bear upon flat surfaces, they are said to be square or flat-ended; this condition closely approximates the condition of fixed-ended columns when the columns are short, and pin-ended columns when long. That the condition of the end may affect the strength of the column is apparent from a study of Fig. 218, which shows the curves which round ^ended and fixed -ended columns would take under vertical loads before failure by bending. These two cases are somewhat analogous to free-ended and fixed-ended beams. A fixed-ended beam is materially strongei Art. 135 FORMULAS FOR COLUMNS OF ORDINARY LEXGTHS 287 than one with ends simply supported; and the same is true with columns. The portion cd of the fixed-ended column corresponds to the entire length of the round-ended column, the points c and d being points of contraflexure. The distance between c and d equals one-half of ab, since the portion ce of the column is in the same condition as ac; that is, the tangent to the elastic curve at e is parallel to the original axis of the column, and so also is the tangent at a. It follows that in comparing fixed - ended with round-ended columns it may be considered that the unsupported length in one case is half that of the other. Columns with ends entirely free to turn do not exist in actual structures. The nearest approach to this condition probaljly occurs in the ordinary pin-ended column, but such pins are by no means friction- less; indeed, in some cases after exposure to weather, with the consequent rusting which takes place, the pins are so restrained that it is with great difficulty that the members can be turned about them. It is also seldom that structural columns are rigidly fixed at the ends, """ pj^ 218 since the piece to which the column is riveted is seldom so rigid that it will not yield somewhat under the influence of the bending tendency. Owing to these facts the former practice of using column formulas based upon the end conditions has been abandoned by most American engineers. 135. Formulas for Columns of Ordinary Lengths. No entirely satisfactory formula for proportioning columns of lengths such as are common in ordinary structures has yet been developed. Two types of formulas are, however, in common use, the Gordon (or Rankine) formula, and the straight-line formula. Of these the first rests upon an imperfect theoretical basis. The latter is purely empirical. These two types of formulas are as follows: Let P= total allowable load on the column. A = area of cross-section in square inches. 7^ = allowable compression per square inch in a short prism. — = maximum ratio of unsupported length to radius of gyration. (Note that L and r should both be ex- pressed in the same units.) c = an experimental constant. 288 DESIGN OF COLUMNS AND TENSION MEMBERS Art. 136 P /c Gordon formula: P 1 / Straight-lme formula : -j=fc ( c \ r P „ l/L r The latter formula is of the first degree, and consequently P gives a straight line for values of -r. It is simpler to use than the A. P Gordon formula, although tables giving the value of -r as de- A termined by the Gordon formula for different ratios of — are published in the various structural handbooks. Of these two formulas the former is the older and has been more generally used by American engineers and is still favored by many. The growing recognition of the fact that the strength of a column is largely dependent upon accidental conditions, which cannot be expressed theoretically, such as initial stress, lack of straightness, and unhomogeneity of material, coupled with the fact that experiments fail to demonstrate that a formula of the Gordon type corresponds more closely to experimental results than the straight-line formula, has led to the adoption of the latter by many engineers, so that an examination of the bridge specifications of twenty-seven typical American railways shows that at the present time thirteen of these specify the straight- line formula. 136. Typical Formulas for Columns of Ordinary Lengths. The two formulas which follow are typical of the Gordon and the straight-line formulas and represent present day American practice for ordinary structures. In both the condition of the ends is ignored. Formula from Massachusetts Railroad Commission's " Specifi- cations for Bridges Carrying Electric Railways." For structural steel having a required ultimate strength of from 55,000 to 65,000 lbs. Live stress to be corrected for impact: P 12,000 ,^.. 20,000 V r Abt. 136 FORMULAS FOR ORDINARY LENGTHS 289 Formula from American Railway Engineering and Mainten- ance of Way Association's " General Specifications for Steel Rail- way Bridges." For structural steel having an ultimate tensile strength of about 60,000 lbs. per square inch. Live stresses to be corrected for impact: ^=16,000-70-. .1 r (22) Both of these formulas are for main members ^ and should be used for columns for which the value of — does not exceed 100. r 16,000* 15,000" 14,000* 13,000* ll-T- 12,000 1 1, 1 ,» ^% 1 1 ' 20,000 straight line tormula -?=l!6,000- 70 V % ^^. ■"% \ ^ \, ^--^ \. \ \ \ \ \ \ N ^ \ ^ ^ ^1^^,000" »11,000* 1^10,000* 9,000* 8,000 « T,000* e,oflo* ^•''™ 10 20 30 40 50 60 70 80 90 100 110 120 130 140 130 Values of ^ Fig. 219. The term — in these formulas should equal the maximum r ratio of unrestrained length to radius of gyration. If the column is restrained against lateral deflection in all directions at the two ends and at no intermediate point, then L would be the ' Secondary members, such as lateral struts, are commonly designed for a somewhat higher unit stress, and a larger value of — is permissible. The following values may be used for such members: -<120. T = 1-20 ( 16,000 -70-). r A V r/ 290 DESIGN OF COLUMNS AND TENSION MEMBERS Art. 137 total length of the column between lateral supports, and r the least radius of gyration of its transverse section, provided the column is of constant cross-section between supports, as is usually the case. If the column is held against lateral deflection in all directions at one or several intermediate points, the value of — r to be used should be the largest value obtainable for any portion of the column between points of support. If the column is held at an intermediate point in one direction only, then the value of — to be used in the formula should be the maximum obtained r by using for L either the length of either section or 'the total length of the column, and for r in each case . the radius of gyration referred to the axis about which the column is free to bend. E II For example the maximum value of — for r the column shown in Fig. 220 may be either L L Fig. 219 illustrates graphically the rela- p tive values of -j as computed by formulas (21) and (22), and shows the slight deviation of the former from a straight line for all values of — above 40. r 137. Formulas for Long Columns. The term " long column," as used in this article, refers to columns of length such that failure tends to »ccur by lateral bending before the material has reached its elastic limit. The collapsing load which such columns will carry without yielding, when centrally loaded, can be closely determined by mathematical in- vestigations,! provided the columns are perfectly straight and homogeneous, and the load axially applied, and is dependent upon the elasticity ' See "Applied Mechanics," Lanza, Edition 9, pp. 330-333. r=»i-adiu3 of gyration about axis X-Y r.'^radius of gyration abou^t axis Z-Q I Fig. 220. Art. 138 TESTS OF STEEL COLUMNS 291 rather than the crushing strength of the material. The formula commonly used for such columns is known as the Euler formula, and is given in treatises on mechanics, as follows : P / r\2 Columns with fixed ends : -j = 4t:'^E{j\ . . . (23) Columns with round ends: -r^ t:'^FAj\ . . . (24) .4 V^, p In these formulas E equals modulus of elasticity and j is the axial stress required to hold the column in equilibrium if slightly deflected laterally. If a load greater than P be applied to such a column, the column will collapse; if a smaller load, the column will spring back to its original condition when the lateral forces are removed. If Euler's formula be applied to columns composed of material with an elastic limit of 33,000 lbs. and a modulus of elasticity of 29,000,000, both of which are reasonable p values for structural steel, the value of -7- for round-ended columns A would exceed the elastic limit whenever — <94, hence Euler's r formula should not be used for such columns when the ratio of length to radius of gyration is less than this limit. For fixed- ended columns, on the other hand, the use of the same constants would give for — the value of 186. r Since columns as used in structures are in an intermediate condition between round -ended and fixed-ended, it would seem that for columns for which this ratio exceeds about 150, Euler's formula should be used. The fact that the value of — is usually r restricted in bridges to a ratio of 100 is apparently a conservative custom and agrees with the column formula commonly employed. 138. Tests of Steel Columns. There are comparatively few carefully conducted tests of well-proportioned full-sized steel columns available for stud}^ Among the most complete and reli- able of these tests may be cited those made upon the 2,000,000 lb. testing machine of the Phcenix Iron Company, at Phoenixville, Pa., and conducted by James E. Howard. The results of these 292 DESIGN OF COLUMNS AND TENSION MEMBERS Art. 139 tests were published in the " Proceedings of the American Society of Civil Engineers" for February, 1911, and show an ultimate strength of 30,000 lbs. per square inch for columns having a ratio of — of 47.1 and composed of plates and angles the elastic r limit of which varied in the different pieces from 29,400 to 37,300 lbs. per square inch, the former value occurring in one of the plates and being the minimum value found. The conclusion reached by Mr. Howard from these tests is that " the minimum value of the elastic limit, as found in the component parts, chiefly modifies the ultimate resistance of the columns, and that varia- tions of its value would overshadow the considerations which find expression in empirical formulas for strength and take nc account of such features." The entire subject of column strength is now being investigated by a committee of the American Society of Civil Engineers, and it may be hoped that before long sufficient reliable data will be available to permit the establishment of better formulas with more reliable constants. 139. Cast-iron Columns. Cast iron is unsuitable for struc- tural members exposed to tension or bending because of its low tensile strength and brittleness. It may, however, be used for compression pieces if these are properly designed, cast-iron columns being frequently used for interior columns in buildings. Such columns cannot be made in long lengths and the different sections cannot be fastened as rigidly together as steel columns, hence they are decidedly inferior in rigidity to the latter; more- over, the fact that it is difficult to secure uniform thickness of shell and that the material is often very variable in composition, may contain flaws, and is frequently in a state of initial stress, is against their use. It is also difficult to obtain good connections of transverse beams and girders. A set of very valuable tests was made upon cast-iron columns at the works of the Phoenix Iron Company in 1896-97, and a formula based upon these tests is probably as reliable as anything that can be obtained, although the results of the tests were so variable that a large factor of safety should be used in applying the formula. A study of the tests shows that a straight-line formula conforms to the results, as well as any other type of formula. Such a formula is derived by Professor Wm. H. Burr Art. 131) CAST-IRON COLUMNS 293 in "The Elasticity and Resistance of the Materials of Engineering," and is as follows: Ultimate strength per square inch for circular flat-ended columns: -r = 30,500 -160^. ... . (25) 111 this formula L = unsupported length and d = diameter. A formula based upon this same set of tests, as given bj- Johnson in " The Materials of Construction," differs somewhat from the above and is as follows: 2 = 34,000-88- (26) In this formula r = least radius of gyration. A factor of safety of five, which is none too much for a material as uncertain as cast iron, reduces Burr's formula to the following form: j = 6100-32j (27) This value is very much less than the corresponding value for steel columns in spite of the high compression strength of cast iron, and in consequence, while cast iron is cheaper per pound than steel, there is little if any economy in the use of properly designed cast-iron columns except for light loads for which it may be difficult to obtain steel columns of sufficiently small cross-sec- tion. It should be noted, however, that the building laws of the large cities permit, in general, the use of much higher unit stresses than those given by either of the above formulas when properly reduced by a factor of safety, and that the use of cast-iron columns in buildings will probably continue until the legal unit stresses are reduced. The employment of cast iron in bridges was abandoned many years ago both because of its treacherous character and the difficulty of making satisfactory connections between members. The limiting lengths to which these formulas are applicable are stated to be as follows: 294 DESIGN OF COLUMNS AND TENSION MEMBERS Art. 140 Johnson: — <120 r = radius of gyration. Burr: J — '^^ d = diameter of circular column or shorter side of rectangular column. 140. Timber and Concrete Columns. Timber columns resemble cast-iron columns in being very variable in strength. This is largely due to the presence of knots and other defects. Wide variations in the results of tests are notice- able, hence a straight-line formula is probably as well adapted to such columns as any other, and that given in Art. 18 may be used. It is important to note that timber columns made by bolting a number of sticks together are no stronger than if each stick were to be separate and loaded by its share of the total load. This has been shown by tests and may be readily understood, since the bolts cannot be counted upon as holding the indi- vidual sticks In place, owing to the small bearing value of wood across the grain and the difficulty of keeping nuts tight. Concrete columns will not be considered here. The student is referred to books upon concrete structures such as " Concrete Plain and Reinforced," by Taylor and Thompson, and " Prin- ciples of Reinforced Concrete," by Turneaure and Maurer, for full treatment of such columns. 141. Typical Column Sections. Fig. 221 represents the cross-sections of a number of types of columns. A and B are columns frequently used in bridge construction, the latter set representing the common type for upper chords of pin bridges, the horizontal plate on the top flange being used to give lateral rigidity. C shows some very heavy column sections, used in the Queensboro Bridge, the Metropolitan Tower, and the Bankers' Trust Building, all in New York city. D shows columns sometimes used in elevated railroad construction in which the central diaphragm is useful both in adding to the cross-section and in preserving the integrity of the column. E is & type of column frequently used for verticals of riveted trusses, i^ is a Z-bar column much used in building construction. G is the well-known Phoenix column, made by the Phoenix Iron Com- pany and once ^\-idely used for bridges and elevated railroads. L is the Larimer column made by Jones & Laughlins, Limited. Akt. 142 GENERAL DIMENSIONS AND LIMITING CONDITIONS 295 H shows H -section columns made by the Bethlehem Steel Com- pany. / is an ordinary I-beam column; and J is an angle column. ^===--^"1^ #--— !^ l^-"--^ l^----^ JL G L " ' ' Tig. 221. — Column Types. Dotted Lines Represent Latticing. , 142. General Dimensions and Limiting Conditions. In designing a column the first thing to be determined is the type and general dimensions of the member; that is, the width and depth, provided these are limited by other considerations than those of strength, as is often the case. For example, the com- 296 DESIGN OF COLUMNS AND TENSION MEMBERS Art. 143 pression chords in bridges must be of sufficient width and depth to permit of proper connections to the web members, and the verticals must be of such a size as to give suitable floor-beam connections; in buildings the columns are frequently limited in size because of the space available or the character of the neces- sary connections. Ease of construction must also be considered, and this is frequently a ruling factor, as in the case of channel columns with flanges turned toward each other, where the space between the flanges should not be less than 4 ins. and lattice bars 1 must be far enough apart to permit insertion of hand. This is illustrated in Fig. 222. \ /(OK "/6) 1^ \/ /^ Space ^v / for hand \ /^ ^^_ X2K^ \i Nothess th^D 1 ^K. Fig. 222. In determining the dimensions of the individual pieces there are also restrictions due to practical considerations. If the web or cover plates are too thin they may wrinlde under com- pression, hence it is common practice to limit the thickness of webs ai)d cover plates to not less than from -^^ to ^j^^ of the distance between connecting rivets. It is also desirable to so proportion the column that the centre of gravity will be near the centre of section. If a cover plate is used on the top flange as in the chords of Fig. 221 B, unequal-legged angles with wide leg horizontal, or narrow flange plates, vertical or horizontal, are often used on the bottom flange to lower the centre of gravity. 143. Method of Design. With these restrictions in mind, an approximate design of the column may be made, either by assuming the value of the minimum radius of gyration or of the allowable \init stress. The actual allowable unit stress for the section thus obtained may then be computed, and the column redesigned if this stress varies too widely from the stress which the column actually carries. It is usually economical to place the ' Lattice bars are diagonal members such as shown in Fig. 222 intended to hold the two column halves in line and make the column act as a solid piece. These bars are of great importance and will be fully treated later. Art. 143 METHOD OF DESIGN 297 webs such a distance apart that the radius of gyration about the principal axis parallel to the webs will equal that about the principal axis perpendicular to the webs, that is, if the unsupported length of the column is the same with respect to both axes. This involves the con- dition, that the moments of inertia h about each of these axes should be equal, since the radii of gyration will ' then be equal, and the value of — 9 Axis e.g. passes through centre of gravity of oae channel and axes hh and vv through centi-e of gravity of CohimiL' Fig. 223. 223 is simple, and (This is unalterable and consequently ot the allowable unit stress will be the same in both directions. The following method of accomplishing this for a column com- posed of two channels as shown in Fig. illustrates the problem sufficiently. Let rft = radius of gyration about axis hh. for any given channel.) r-»= radius of gyration about axis vv. /„ = moment ot inertia of cross-section about axis vv. Ih = moment of inertia of cross-section about axis hh. /c J = moment of inertia of each channel about its axis eg. A = area of one channel. To determine the moment of inertia about axis vv the follow- ing principle of mechanics may be used: The moment of inertia of a section about any axis equals the moment of inertia of that section about a parallel axis passing through its centre of gravity, plus its area multiplied by the square of the distance between the two axes. From the application of this principle the following equation results: Hence Ih should equal 2{Icg + Ax^). But h = 2{An^) hence 2Arh^ = 2I,g+2Ax'^. 2ArH^-2h, Hence "^ " 2A "" A- 298 DESIGN OF COLUMNS AND TENSION MEMBERS Art. 144 In the case of channel columns the value of leg is usually small compared with A, hence the error involved in omitting the last term of the equation is small and is on the safe side, there- fore it may be neglected and the value of x made equal to ru- The proper distance between channels to secure equal rigidity about either axis is given in some of the steel manufacturers' handbooks and need not be computed; but for more complicated sections, such as plate and angle columns, it must usua,lly be determined in the manner indicated, although the approximation mentioned is not always allowable, and-, in the case of top chords with cover plates would be considerably in error and should not be made. 144. Determination of Cross-section of Typical Steel Columns. Problein. Design a channel column for the following assumed con- ditions : Total applied load (live, dead, and impact) = 250,000 lbs. Unsupported length = 2.5 ft. P 16,000 / 1 ilA Allowable stress A "^ 20,000 Vr P Solution. Determine trial section by assuming — = 14,000 lbs. This gives a trial area of ' = 17.9 sq.ins., which could be obtained by the use of two 15-inrh channels at 33 lbs., having a total area of 19.S sq.ins. The radius of gyration for this section about an axis perpendicular to the web=5.62 ins., hence the allowable value of '- 20,000X5.62' The actual unit stress if the proposed section should be used equals - — ^— = 12,600 lbs. It follows that the column can stand considerably 19.8 ^ more than the applied load, hence it is safe and may possibly be decreased in size. The next smaller channel is 12 in., 30 lb., hence try this. The ' P allowable value of — for a column composed of these channels equals '''''' 12,800 lbs. 25'X12' 20,000X4.28' Art. 144 CROSS-SECTION OF TYPICAL STEEL COLUMNS 299 The actual unit stress, if these channels be used equals 250,000 17.64 14,200 lbs.; hence these channels are too small, and the 15-inch channels should be chosen. These should be placed so that the distance x in Fig. 223=5.62 ins. "Cambria Steel " gives the proper distance between the webs as 9.51 ins., which agrees closely with the corresponding value when x = 5.62 ins. Problem. Design a top-chord member of a bridge using a top cover plate : Minimum clear distance between web plates= 10 ins. Minimum clear depth =17 ins. Total applied load (live, dead, and impact) = 430,000 lbs. Unsupported length = 25 ft. Column formula, -=16,000-70-. A r Thickness of web to be not less than to distance between horizontal flange rivets. Thickness of cover plates to be not less than 4V distance between vertical flange rivets. Solution. In a compression piece of this type, r, with respect to the horizontal axis, is usually about to the depth of the member, therefore this value for the radius of gyration will be tried instead of assuming the allowable unit stress, as was done in the previous example. Making this assumption, and assuming also the minimum depth of 17 ins., and a distance apart of webs of 10 ins., gives 6.8 ins. as the trial value of r. Substituting » this value in the cofumn formula gives I -=16,000-70 25X12 6.8 13,000 lbs. \lt Using this value . 430,000 „„ gives -T-^—;:x=o6 J a pre- V 1 CoTor Pl>ly Fig. 226. The fact that the strength of latticed columns is largely dependent upon the proper design of the latticing requires that the proportioning of the latticing should be as carefully studied as the design of the main cross-section. Unfortunately the theo- retical treatment of such details is more obscure than that of the columns them- selves. It is evident, however, that if the column were to remain absolutelj' Fig. 225. Fig. 226. straight under loading, no latticing would be needed, and the stress in such lattice bars as might be used would be merely the secondary stress due to the shortening of the column as a whole and the consequent distortion of the lattice bars. On the other hand, if the column bends somewhat under loading, as would probably be the case unless the column were of very short length, bending moment, and consequently transverse shear would Art. 145 LATTICE BARS AND BATTEN PLATES 303 occur, which would cause stresses In the lattice bars, and if the value of this shear can be determined the lattice bars maj^ be easily proportioned. The magnitude of the bending moment in ordinary columns of the limited lengths consistent with good practice is largely- dependent upon the unintentional eccentricity of the load due to the initial condition of the column with respect to its straightness, initial stress, or homogeneity of material, and cannot be determined. It may be estimated, however, to con- form to the conditions assumed in proportioning the column, thereby securing a consistent design, by deternuning the external shear equivalent to the bending moment which causes the fibre stress in such a column to exceed the fibre stress in a short prism, and assuming the lattice bars to act as web members of a truss subjected to this bending moment. Such a method, while approximate, is perhaps as accurate as any yet derived, and will be developed to correspond to the bending moment obtained b}- the straight-line formula. For other formulas a similar method may be adopted. Let /= fibre stress due to bending; 1(1 = an assumed load uniformly distributed and applied at right angles to the column axis; c= distance from neutral axis to extreme fibre of column; M = external bending moment due to load w; ;S = maximum external shear; A = area of cross section ; / = moment of inertia of cross section about proper axis; r= radius of gyration corresponding to /; L = unsupported length of column. Assume that the bending moment in the column equals that which would occur if the column were loaded uniform]}- at right angles to its axis throughout its length by the load w per foot, this giving a larger shear than would occur with any other reasonable assumption, such, for example, as a concentrated load applied at the centre. The assumed distribution of stress over the cross-section of the column corresponding to the value given by the column formula is shown by Fig. 227, from which it is evident that /= 16,000-^. 304 DESIGN OF COLUMNS AND TENSION MEMBERS Art. 145 But Also Hence ■ 16,000 A~ Mc 70L 70L J y ' 1 ^Ac 1 wL'^c w = 560' L A, 1 wL^c 8 Ar^ and 70L ,S= wL = 2S0 Ar The total stress in a lattice bar, if single latticing is used, may now be taken as equal to one-half the product of S and the cosecant of the angle which it makes with the longitudinal axis of the column. This method gives the stress in the end lattice bars, but it is common to use the same size bars throughout the column. The following problem illustrates this method : ji j^ ^_ lejooo* Fig. 227. Problem. Determige stress in lattice bars for the lo-in. 33 lb. channel column designed in Art. 144. Solution. For this column A=19.i the beams based upon the assumption just p^^ 23,3. made. The distance d may be determined by dividing the maximum beam reaction liy one-half the allowable crushing strength per lineal inch. For the case under consideration, this gives 7V' approximately. The value of x is H" + l" + (§)(7i") = 7.5". The eccentricity under the partial loading is then (7.5) (0.14) = 1.05 ins. The eccentricity due to bending of the column will be neglected here, as being an unnecessary refinement for a material as variable as cast iron, hence the fibre stress due to the eccentricity will be 4X70,000X1.05 .049(8* -64) -^^"^"-- 314 DESIGN OF COLUMNS AND TENSION MEMBERS Aut. 151 To determine \Ahether the column is safe this eccentric stress should be added to the maximum stress due to the direct load as determined by the column formula, and the sum should not exceed the allowable unit stress in a short column. If formula (27) be applied, the maximum fibre stress due to direct load is given by the expression Ad 22 8 The eccentric stress added to this gives a total of 5990 lbs., which is less than 6100 lbs., the allowable unit stress by the formula for short columns, hence the column has an area that is only slightly in excess of the required amount and may be used. 151. Design of Iron and Steel Tension Members. The design of tension members involves little more than the selection of bars with sufficient net area to carry the total stress without exceedin'g the allowable unit stress. Steel or iron tension mem- bers may be divided into two general types: viz., solid bars rectangular or circular in cross-section, and built-up members composed of structural shapes riveted together. Solid bars are used generally in pin trusses for diagonals and bottom chord members, and in Howe trusses for verticals. Built-up members are generally employed for tension members in riveted trusses and for the end hangers in pin trusses. Of the first t>pe of member the eye bar shown in Fig. 234 is used most commonly. Such bars are made by most of the large Fig. 234. steel manufacturers and are fully described in their handbooks. The heads of these bars are designed so that the bar if tested to destruction, will fail in the body rather than in the head, and the engineer should specify that full-sized tests should give this result and not attempt to proportion the heads. In deter- mining clearance, the dimensions of the heads given by the makers may be used, noting that the diameter of the head de- Art. 151 DESIGN OF IRON AND STEEL TENSION MEMBERS 315 pends upon the size of pin hole. Eye-bars may be manufactured to any thickness above the minimum size quoted by the makers, but a thickness above 2 ins. should not generally be employed, since such thick bars are not likely to be of the best material. A good rule to observe in selecting bars is to keep the thick- ness between one-sixth and one-third the width. Eye-bars are generally used in pairs, since an odd number of bars would give a TurDbuckle o D o Fig. 235. poor arrangement on the pin. For counters, adjustable eye-bars such as those shown in Fig. 235 may be used, the two bars being connected by a turnbuckle or sleeve nut; iron rods with loops formed by welding such as those shown in Fig. 236 may be Fig. 236. used if the counter stresses are small. For the verticals of Howe trusses iron rods, with screw ends fastened by nuts bearing on washers supported by the top chord, are generally employed. In proportioning adjustable members allowance must be made J Plan 3 Plan -e- Side Elevation Fig. 237. Side Elevation Fig. 238. for the decrease in section due to the screw threads. It is usually advisable to upset the screw end, that is, to make it of larger diameter than the body of the bar, so as to give sufficient area at 316 DESIGN OF COLUMNS AND TENSION MEMBERS Art. 151 1^- -d- -^ ^ A ,> >r ~Tie plates 1 „ a ^ o o\ o o o o o _o 3 lY ^ ' " ^ " ^Tie plate. ^ ^ ^ ^ ' Side Elevation Distance d may be made several feet Fig. 239. O <: 1^ J End View h d ^ r^ &, r. .i: -Tie plateB . . „ I .- Side Elevation Distance d may be made several feet Fig. 240. End "View Plan ,_£! Q Q C^ ^ a £^ Ci_, ^ Q CL ^ r\ _ r^ ■^ O O O O o o o o o o 4 .■,^^^! ,v^:jj>;: ^ 111 ll 1, ! O 1 O 1 [' O 1 O '1 i o 1 o 1 \ o 1 o 1 irr4 |\ N \ i /^ i^@>l\ - — ^ < i3M^^ H ( L +7^ pin fe^ Size of eyebars Fig. 242. If rivets are countersunk, this distance may be reduced by 1^ in. If rivet heads are flattened or are countersunk but not chipped the distance between channels may be varied accordingly. (Xote that rivet heads if countersunk but not chipped usually project ^ in. above the surface and that rivet heads are frequently flattened without being countersunk, so that they project but |- in. above the surface.) 320 PIN AND RIVETED TRUSS JOINTS Art. 153 2d. Arrange eye bars so that their centre lines will be paral- lel or nearly so with the centre line of the truss. It is seldom possible if compact joints and small pins are to be obtained to follow this rule very closely, but it is common to specify that no bar shall deviate from the centre line of the truss by more than fg in. per foot in length of the bar. In cases where a greater allowance than this is necessary the bar should be bent to the proper slope before being annealed. As it is sometimes difficult to arrange the different members so that all the above conditions will be observed, the student is advised to lay out to a large scale, say 1^ in. = 1 ft., the different joints of each chord. The distance apart of the various joints should also be plotted to scale, but this scale may be much smaller than the scale of details. The different members can then be drawn from joint to joint and the deviation from the centre line can be determined by scale. This method is indicated in Fig. 243 for two joints of a bottom chord. To carry out the method , Chord Bar ^ Chord Bar^ Chord Bar ^ P''>g°-'L Chord Bar Chord Bar Center Line of Chord Fig. 243. completely the chord should be drawn on a sheet of suffi- cient length to show all the joints, or if the truss is symmetrical, all the joints up to and including the centre joint, and the top chord should be plotted above the bottom chord in a similar manner. It will be noticed that, in order to secure the above arrange- ment, the channel flanges may have to be cut away. This is undesirable, but is frequently necessary in order to avoid the necessity of using a pin of large diameter. If this is done, the channels should be reinforced by web plates extending throughout the distance over which the flanges are cut, unless the channel webs alone without the flanges are of sufficient strength to carry the compression. In investigating this case in a compression member, the column formula should be applied, using for the unsupported length the distance from the centre of pin hole to the first row of rivets beyond the point where the flanges are cut. Art. 154 MINIMUM SIZE OF PINS 321 154. Minimum Size of Pins. Before it is possible to com- plete the arrangement of the various members, it is necessary to make some assumption as to the size of the pins, since it is usually necessary to add pin plates to the riveted members either for the purpose of strengthening the member against crushing on the pin, or to make up for the area taken out by the pin hole, since the pin, unlike a rivet, does not completely fill the hole and hence cannot be counted upon to carry compression. To deter- mine the size approximately requires some experience; the lowest limit is, however, usually fixed by the width of the widest eye-bar connected to the pin as shown below. Let /), = allowable bearing stress per Square inch on pin. ft = allowable tension stress per square inch in bar. w=width of widest bar coming on the pin. i=thickness of same bar. rf= diameter of the pin. Then /fcrfi=bearing value of the bar on the pin, and /(Wi= tensile strength of the bar. Putting these equal gives fbdt- =flWt. Therefore, d > vf~. For example if /«= 16,000, 76=24,000, and the width of the widest bar coming on the pin is 6 in., the diameter of the pin should not be less than 777X6 =4.0 in., hence the pin in this case should be assumed as not less than 4.0 in. in diameter. Whether it should be assumed as larger is a matter which can only be estimated by experience, but it should be noted that it is wiser to assume the pin too small rather than too large, since, in the former case, pin plates, which are somewhat thicker than are needed, will be selected at first and these may be easily reduced in thickness if it be found that the diameter of the pin should be larger than that assumed. The only exception to the statement " that it is usually on the safe side to assume the pin too small " 322 PIN AND RIVETED TRUSS JOINTS Art. 155 is when a reinforcing plate is not needed to increase the bearing resistance on the pin, but is required to make up for reduction in section by the pin hole. This sometimes happens near the centre of the top chord of a simple truss, but in arranging the members on a pin it is wise to always allow for at least one pin plate upon the chord at every joint; a f-in. plate if rivets do not require countersinking, and a ^-in. plate if the clearance is so small as to make countersinking necessary. 155. Stresses Causing Maximum Moment and Shear. After the arrangement of the members is satisfactorily accomplished it is necessary to compute the maximum stresses which act simultaneously on each pin, and which seem likely to produce critical bending moments and shears. In order to obtain these simultaneous stresses it is some- times necessary to calculate anew the stresses in a number of bars under the loading which produces the maximum in one of them. For example, for the truss shown in Fig. 244, the maximum Fig. 244. moment on pin L^ may occur under the loading which produces maximum stress in chord L1L2, diagonal U1L2, or chord L2L3, hence it becomes necessary to compute the stress in the bars connected by pin L2 under all of these conditions of loading. In all cases it is the horizontal and vertical components of the stresses which are desired, and the results should be checked by noting whether the pin is in equilibrium under the action of these components, that is, whether HH =0 and 11V=0. There is one point here which may cause trouble. The floor beam in an ordinary bridge is frequently connected to the post above the pin, hence the post stress which reaches the pin is not the stress in the post as a whole, but is the stress in the post below the floor beam. To avoid confusion, no attention should be paid to the actual stress in the post, whether the floor beam is Art. 1.-)6 MAXIMUM MOMENT AND SHEAR 323 al)Ove or below the pin, but the stress coming to the pin from the post should be placed equal to the vertical component of the diagonal stress. After the stresses are found, it is desirable to make sketches for each joint showing the stresses in the bars meeting at the joint. 156. Computation of Maximum Moment and Shear. The next step is to determine the maximum bending moment and shear on the pin for each loading. This can best be accomplished for the moment by plotting the curves of vertical and horizon- tal moments, and determining by inspei'tion or trial the section where the maximum resultant moment occurs. This resultant can be determined with sufficient accuracy Ijy graphical methods, since its value equals that of the hypothenuse of a right-angled triangle, the sides of which equal the vertical and horizontal moments respectively. It is seldom that the maximum shear needs to be carefully figured, since ordinarily the bending moment determines the size of the pin. The shearing stress should always be investigated, however, and in doubtful cases its maximum value determined by the method given above for bending moment. If the size of the pin as computed differs materially from that assumed, the thickness of the pin plates should be investi- gated and revised if necessary. This should not be done too hastily, however, since it is customary to use but few different sized pins, in a truss, and it may happen that the pin as com- puted may not be the one which it is finally decided to use. Examples of pin computation will now be given. 157. Computation of a Top Chord Pin for Truss Shown in Fig. 245. Problem. Determine the size of pin and thickness of bearing area of chord and vertical at joint U2, using following allowable unit stresses : Bearing on pin, 22,000 lbs. per square inch, Bending on pin, 22,500 lbs. per square inch, Shear on cross-section, 10,000 lbs. per square inch. Solution. For this pin the only loading which needs to be considered is that which produces the maximum stress in diagonal U2L3. The reason for this is that the top chord is continuous at the joint, and spliced elsewhere as shown. This is inconsistent with the theory upon which the computation of truss stresses is based, but is the common practice and probably does not affect the stresses materially, while it simplifies 324 PIN AND RIVETED TRUSS JOINTS Art. 157 greatly the construction. Under this condition the duty of this pin is to connect the diagonal to the top chord and vertical, the horizontal Fig. 245. (For composition of members, see opposite page.J component of the diagonal stress being transmitted by the pin to the chord, and the vertical component to the post. The actual stress in the chord, therefore, is not an element in the pin design, and needs to be considered only in investigating the strength of the chord at the cross-section through the pin hole. Fig. 246 shows the maximum stress in the diagonal with its vertical and horizontal components. The allowable unit stresses of 16,000 lbs. per square inch tension, and 22,000 lbs. per square inch bearing give for the minimum size pin required for bearing on the 6-in. diagonal bar, — X6"=4.35". Since the stress in the diagonal is large the size of pin which will be assumed in determining bearing areas will be taken as somewhat larger than the minimum size, or say 5 J in. The bearing area required by this assumption may then be computed, by assuming the stress to be distributed uniformly over a surface equal to the plane diametrical section of the pin. Total thickness of bearing required on chord = 196,000 22,000X61" Total thickness of bearing required on vertical 243,000 Fig. 24G. 22,000 X5i' 7=2.10" Art. 157 SIZE OF TRUSS MEMBERS 325 o I— ( '^ o W CO CO o m K O O o CO ^ X X J3 XI ^ CO CO o ^< x? !£i iC iC CD r^ r^ t^ t^ I— ( I— ( 1—1 1-H

^x co^ lb cq -jH xr XXXX 00^ lb cq J xxxxx 01 - V 5 ^ :; ,-H r^ o i>- 1- i>- X XXXX » M c t- o3 O tn I til ® S g 2 rt & & & OHHH be ® « g •Eg'a ^-1 ^ O -^, r/j *^ ^ i^ — — > ^ o c-* o o -^ — ^ ^ (V, o o o o yy ,r, 01 m K CS O ,j3 S S & & 3-^ c o ^ ^ ^ r^ o) Q) a; f,^ iy: 1% bepween riv 15 -33 "channel-^ (Web .40") ^S ; heads / i 6"x l^a diagonal \^ ', I — Rivet but so W'<'\-k.^M leads coiujtei-sunk 1 chipped' ^" high T pin plate 28 X y,a Web 9H" Center Line of Fig. 247. The forces acting on the pin were assumed, in determining the required bearing area, as distributed uniformly over a plane surface equal to the diametrical section of the pin. In computing moments, however, it is the usual custom to consider these forces as concentrated at the centre of the bearing areas. The distance between the points of application of these stresses should be computed upon this basis, and the bending moment and shear on the pin determined. The results of these computa- tions are shown in Fig. 248. It is evident that the maximum moment in this case is the resultant of the maximum horizontal and the maximum vertical moments, since these both occur at the same section. This is found graphically, as shown by Fig. 249, and equals 262,000 in. lbs. Art. 158 COMPUTATION OF A BOTTOM CHORD PIN 327 With an allowable fibre stress in bending in the pin equal to 22,500 lbs., a 5-in. cin is required to carry this moment. (See table for Curve of Horizontal g Bending Moments ol»-a= X(13«)=lf" i i ■as Curve of Vertical S Benfliiig Moments t Cgn ter line o f Pin i ' ^ ..,- ' S s^ ■ - « o -a I S § a 3 a s ^ ■S S 3 Center line of Pin K(l«)+«+O.M =I.60"say 1^ Fig. 248. Fig. 249. "Maximum Bending Moments on Pins" in "Cambria Steel," 1907 edition, page 312). This pin may be used, provided it is strong enough to carry the maximum shear. (The -pin plates used on post and chord are both somewhat in excess of the size needed for a 5j-in. pin, and hence need not be recomputed.) The area of a 5 in. pin is 19.6 sq.ins. which at a unit stress of 10,000 lbs. per square inch assumed to be uni- formly distributed over the cross-section, is good for 196,000 lbs. shear. The resultant shear may be determined in the same manner as the resultant moment, and is found to be 150,000 lbs. approxi- mately, which is less than the allowable shear, hence the 6-in. pin satisfies all the necessary requirements and should be used. 158. Computation of a Bottom Chord Pin for Truss Shown by Fig. 245. Problem. Determine the size of pin and thickness of bearing area on vertical at joint L^, using same unit stresses as for pin U2. Solution. For this pin two loadings must be considered. 1st. That which produces maximum stress in chords L1L2 and L2L3. 2fl. That producing maximum stress in diagonal UiL,. For the first case, the chord stresses are identical with the maximum stresses since the chords of this truss were computed for a uniform load 328 PIN AND RIVETED TRUSS JOINTS Aet. 158 per foot, and hence these stresses may be written down at once. The difference between the chord stresses equals the horizontal component of the diagonal stress, from which the vertical component is readily obtained. The stress transferred to the vertical from the pin equals the vertical component of the diagonal stress. The stresses for this case are shown by Fig. 250. For the second case it is necessary to compute the stresses in the chords produced by the loading which causes maximum stress in the diagonal. This computation requires but little additional wori, even if a concentrated load system is used, since the position of the loads is known and the left reaction would have been determined in making the shear computations. The stresses for this case are shown by Fig. 251. Had the maximum chord stresses been computed for a concentrated load system it might have been necessary to compute the pin for three instead of two cases, since the position of the loads for maximum stress in chord LiL^ might have been different from the position for maximum stress in the chord LJ^,. It should be noted, however, that in pin com- I)eaa=178 0001b«. Dead =178 000 lbs. I.v= = «0 0001b.. Liv,.=160 000ta. Dead =178 000 lbs. \T Liye = UO 000 lbs. , \| Dead=178 000 lbs. Eead-M4 000lV. Ilead=144 OOOJbs. Live =100 000 lbs. Live=113 0001b& ^ riead=226 000 Iba. Deiid=389 000 lbs. Dead = 225 000 lbs. Dead = S6'9 000 lbs Live =170 000 lbs. Live = 289 000 lbs. Live = 137 000 lbs. live = 264 000 lbs. Fig. 250. Fig. 251. putations a uniform load giving the same maximum stress as ttiat occur- ring in any one of the bars connected by the pin may be used if desired in determining the simultaneous stress in the other bars, the error being slight. It will be noticed that impact is not included with above stresses. The reason for this is that the allowance for impact if figured by the formula / =S I I , which will be used for this case, would give different percentages for the different bars, with the result that the forces on the pin would not balance. It is necessary, therefore, in such a case to compute the dead and live moments separately, and deter- mine the impact as a factor of the moments and not of the bar stresses. If the impact is computed on the basis of the percentages of loaded length, it might if desired be included in the stresses, since the loaded length is the same for each case for all the bars concerned. 1 fi The minimum size of pin for this joint is ^ X7"= 5.1". The stresses are large, hence it would seem reasonable to assume a pin somewhat larger than this, and a 6-in . pin will be taken. For the loading of the second case, the post stress, with impact added, equals 414,000 lbs. ; hence the Art. 158 COMPUTATION OF A BOTTOM CJIORD PIN 329 total thickness required for bearing on the 6-in. pin is '■ = 3 14 in. ^ f 22,000X6 ''•^^"'- The thickness of the channel web is 0.40 in., hence to each web must be J 7 I 2'C 17 I IJi"DiiiBi)nal | \ ^7x1 Diagonal stj^ ^^ be count^jreank aid chipped Chord Cento ■ Line of 7 I lis' Chord s:^^ .^pin plate X6'-33*chamiel (web=.40") A' pin plate Pig. 252. added 1.17-in. pin plates or say one A-iii- plate and one f-in. plate. proposed arrangement of the different members coming on the The pin is D=92,0OOlbs. // L—TZ.OOOlbs. / D =112,000 IbB, L= 88,000 lbs. D= 10,000 lbs. L=31,000 lbs. S \ \ D =32,000 in. lbs\\ L =26,000 in. lbs. _121i000m._ll»_ IM.OOO in. IbB. PZSl^; 71,000 in. IbB.1 D = 40,0001b8. L=39,000 lbs. 1^ ||. L ^3j (^ n^ yi i'.^bsN.S 18.000 in. lbs. 15,000 in. Iba. Centerlineof Fin ■f D= 89,000 lbs. jj L =70,000 lbs. Fig. 253. — Curves of Moments for Case 1. Full Lines are Dead Moments. shown in Fig. 252. This arrangement is one which gives a satisfactory location of the bars as regards the other joints of the truss. 330 PIN AND RIVETED TRUSS JOINTS Akt. 158 Figs. 253 and 254 show curves of moments for both loadings, and should be understood without difficulty. It will be noted that in deter- mining distances between loads each eye-bar is assumed to be -h-in. thicker than its nominal size. "MOO j. lbs. 121 200 in. lb«, -1.32^ ■M'sOMOtba. Fig. 254. — Curve of Live Moments for Case 2. It is evident that the maximum moment on the pin occurs for Case 1 and equals : D = 175,000 in.lbs., L = 137,000 in.lbs., / = 60,000 in-lbs. (by formula (8) ). 372,000 in.lbs. The size of pin required to carry this moment with unit stress of 22,500 lbs. is 5|-ui (see Cambria Handbook) . This is somewhat smaller than the size assumed in computing the thickness of the bearing plates on the post. As the thiclaiess of these bearing plates has no influence upon the maximum moment on this pin which occurs at the next to the outermost chord bar, it is evident that the 5f-in. pin may be used without recomputation by making one of the pin plates on th'e vertical somewhat thicker than the size required for a 6-in. pin, and that no other change is necessary. Shear. The allowable shear on the 5f-in. pin at 10,000 lbs. per square inch equals 248,500 lbs. Art. 159 ARRANGEMENT OF MEMBERS 331 A slight study of the pin and its applied loads shows that the maxi- mum shear for Case 1 is : Z)= 92,000 lbs. L= 72,000 " 7=32,000 " 196,000 For Case 2, the shear is still less, hence the 5|-in. pin is strong enough to carry the shear. 159. Effect Upon Pin of Change in Arrangement of Members. The student should consider carefully the comparatively great effect upon the moment of a slight change in the arrangement of the members on pin L2. If the 2-in chord bar were to be interchanged in position with the l^-in. diagonal, the maximum dead moment due to horizontal forces would be increased by 54,000 in. lbs. and 1 1 1 Fig. 255. Fig. 256. the live moment proportionally. The effect of an interchange of the 2-in. chord bar with the adjoining l|J^-in. bar would be to increase the horizontal dead moment by 363,000 in. lbs. and the live moment proportionally. It would also make the maximum horizontal moment occur at a section where vertical moment would exist, hence the maximum would be a resultant of hori- zontal and vertical moments instead of a single moment, as is now the case, and this would still further increase the moment. It is desirable to use as small pins as possible, so that the size of the eye-bar heads may be kept within reasonable limits, hence the arrangement of the bars should be carefully studied, and the designer should bear in mind that an arrangement which will produce both positive and negative moments will usually give a satisfactory result. For example, if the arrangement of bars shown in Fig. 255 be changed to correspond to that shown 332 PIN AND RIVETED TRUSS JOINTS Art. 160 in Fig. 256, the moment will be reduced, since in the first case the moment continually increases while in the second case the moment varies from positive to negative and then back to posi- tive, its maximum value being far below that reached in the first arrangement. The thickness of the bars also has an important effect upon the size of the pin, and a reduction can often be made by reducing in thickness one bar of a member and increasing another by the same amount. This, of course, cannot be done if the member is composed of only two bars, since in such a case both bars must be equal in size to preserve the symmetry of the truss. 160. Pin Plate Rivets. The determination of the number of rivets required in the pin plates sometimes requires careful study. The student should, however, have no difficulty in solv- ing this problem if he is careful to use enough rivets to carry from each plate the stress which it receives from the pin, assum- ing that it receives that proportion of the total stress which its thickness bears to the total thickness of bearing. Due allowance should be made in case several pin plates are needed for the effect of intermediate plates upon the strength of the rivets, and it is often found desirable to make plates of different lengths so that something of the effect of a tight filler may be obtained . 161. Pin Nuts. The nut commonly used on bridge pins is a special nut which is much thinner than the ordinary hexagonal or square nut, since its function is not to carry tension into the pin, but merely to hold the bars in place. It should be held in position by a cotter pin, since nuts not held have been known to be loosened by the impact of trains, and to fall off. On very large trusses, nuts are sometimes replaced by washers which are held in place by a rod passing through a hole boreid along the longitudinal axis of the pin. 162. Packing Rings. In order that the bending moment on the pin may not differ from the computed value, it is necessary that the eye-bars be held in the position assumed in the computa- tions. To do this, it is necessary to use washers or collars between some of the bars. These are sometimes made of thin plates bent around the pin, and sometimes of short pieces of iron pipe. When the bar is restrained by the other members so that the clearance is not more than J in. to ^ in. the use of such washers is unnecessary. Art. 163 RIVETED TRUSS JOINTS 333 163. Riveted Truss Joints. The design of the joints of riveted trusses is of equal importance with the design of the main members and should receive most careful study. The observance of the following rules is necessary in order to prevent eccentricity in the application of the forces to the members meeting at a joint and consequent increase in fibre stress in the main members. 1. Centre of gravity lines of members meeting at a joint should intersect at a point. 2. Connection rivets in a member should be arranged sym- metrically about the axis passing through its centre of gravity, with as few rivets as practicable in a line parallel to its longitu- dinal axis. 3. Members composed of a single angle, or of two angles back to back, should be connected to plates by means of lug angles in the manner illustrated by Fig. 257, or else the allowable unit stress in the member should be reduced to provide for the eccentric application of the load. The use of the lug angle is often desirable, not only to prevent the eccentricity of applica- tion of the load, but also to decrease, the size of the connection plate which would otherwise be necessary. Fig. 257. 4. If stress at any joint is to be transferred from one member into a gusset plate and thence transferred to another member, the group of connection rivets in the second member should have its centre of gra^^ty coincident as nearly as possible with the point of intersection of the two members. 334 PIN AND RIVETED TRUSS JOINTS Akt. 163 5. The arrangement of the connection rivets in a tension member should be such as to reduce the cross-section area of the member as little as possible consistent with economy in the connection plate. In order that this result may be obtained Max. QompreB siont.__ij- rj _ i 160 000 lbs. i Ls Si^'x Sj/i %" I Gross area 1- plate O'i Jf,"J =12,77 Eq.ii 2 it is usually desirable to have not more than two rivets at right angles to the line of action of stress in the row farthest from the point of intersection of the members meeting at the joint, and to make the distaiice between this row and the row next to it as much as 5 ins. in order that rivets in outstanding legs of angles or in channel flanges may stagger completely with the connection rivets. It should be observed that the connection rivets gradually trans- Art. 163 RIVETED TRUSS JOINTS 335 fer the stress from a member to a plate, and that in conse- quence the required net area of the member decre.ises in passing from the edge of the plate toward the end of the member; hence as the latter point is approached, the reduction in area of the member due to rivet holes may be very large without reducing the strength of the member. 6. The size and thickness of connection plates should be determined by the following considerations : (a) The size of connection plates must be sufficient to enable the rivets necessary for connecting the different members to be properly located. In general it is desirable to use a small rivet pitch; usually for |^-in. rivets a 3-in. pitch, except where a larger pitch is required by the application of rule 5. (6) The net section across the plate at right angles to the line of action of a member must be sufficient to carrj- that pro- portionate part of the stress in the member which is transferred to the plate by the rivets between the given section and the end of the plate. (g) If the resultant stress upon any section of a connection plate is eccentrically applied, as determined by assuming each rivet on one side of the section to carry to the plate its proportion- ate part of the total stress in the connecting member, the plate must be made of sufficient tliickness to withstand the effect both of this eccentricity and the direct stress upon the section. Fig. 258 shows a typical joint in a riveted truss and illustrates the application of some of these principles. It also shows a splice in a tension member in which the connection plate is used as a splice plate. This is a common practice, and whether the splice be of a tension member or a compression member, suf- ficient rivets should be used in the splice plates to carry the entire stress, no dependence being placed upon the abutting of the ends of the members. The following example illustrates the character of the compu- tations necessary to determine the thickness of such a plate: Problem. Determine the necessary, thickness of the connection plate shown in Fig. 258, using an allowable unit stress in bending of 16,000 lbs. per square inch. Solution. Inspection of the plate indicates that section xy is probably the critical section, since it contains many rivet holes, and the resultant stress on either side of it is large in magnitude and applied at some dis- 336 PIN AND RIVETED TRUSS JOINTS Art. 163 tance below the centre of gravity of the cross-section. The strength of the plate at this section will therefore be investigated. The forces acting to the right of xy are the proportionate part of the chord stress carried into the plate by the chord rivets, and the total stress in the diagonal. Evidently the stress due to the chord is the more important factor, since its line of action is further from the centre of the cross-section, hence the condition of loading corresponding to maximum stress in this chord bar wiU be assumed. Computations show that for this condition the stress in the diagonal is 100,000 lbs. The stress passing into the plate from the chord rivets will be taken as the prod- uct of the number of rivets to the right of xy and- the allowable stress per rivet, it being assumed that the total number of rivets is little if any in excess of the number actually needed. The assumption that the thick- ness of the plate will be such as to cause the rivets to be limited by shear rather than bearing and that the allowable unit stress in shear is 12,000 lbs. per square inch, gives a total force of 14X7200=100,800 lbs., the rivets cut by the section being included, since they bear upon the portion of the plate to the right of xy rather than to the left. The forces acting upon section xy of one of the two connection plates will, therefore, -6-H.c.=3uoo ^6 f^^ shown in Fig. 259. It has already been shown in considering |3--(B^M) , 100800 ib^ plate girder web splices that the effect of a row of rivet holes such as exist at section xy in reducing the strength in bending will Fig. 2,59. be amply allowed for if the moment of inertia be considered as three-quarters of the value for the gross cross-section. If this allowance be made the maximum stress in the plate, assuming its thickness as t will be gi\-en by the following expression : 100800-31400 , 4 6(100800X14-31400X11) f= " " + {40-10)t 3 i(40)2 69400 106580 7642 sot 20t ~ t ' 7642 Since the allowable value of /is 16000 lbs. t= =y= required thick- ness. This thickness would develop more than the shearing value of the rivets, and is consequently sufficient, at least for the section investigated. A more accurate determination could be made if thought desirable by actually determining the net moment of inertia, and other sections may be tested in a similar manner if doubt exists as to the critical section. CHAPTER Xin GRAPHICAL STATICS 164. Graphical and Analytical Methods Compared. It is gen- erally possible to solve by graphical methods all statical problems which can be solved analytically, while for certain classes of problems such methods are somewhat simpler and more rapid than analytical methods, such, for example, being the case in the problems of Arts. 89 and 90. As a general rule, however, analytical methods are more satisfactory both in accuracy and speed. The engineer should nevertheless be thoroughly familiar with the principles of graphical statics so that he may be prepared to apply them, particularly in checking analytical computations. A loiowledge of them is also necessary in order that engineering literature may be read intelligently. For a comprehensive treatment of the subject the student is referred to " Graphische Statik," Ijv Muller- Breslau. 165. Force and Equilibrium Polygons. The most obvious method of determining graphically the mag- nitude, direction, and point of application of the resultant of a set of coplanar forces may be briefly stated as follows : Plot the correct position and direction of the forces as indi- cated in Fig. 260 by Fi, F2, and i^3. Prolong any two forces, such as Fi and F2, until they meet, thus obtaining the point of appli- cation of the resultant of these two forces. Determine the magnitude and direction of this resultant force by the 337 > Fig. 260. 338 GRAPHICAL STATICS Art. 165 parallelogram of forces. In a similar mamier combine this resultant with one of the other forces, and continue the process until the resultant of all the forces has thus been determined in direction, point of application and magnitude. This process may be continued indefinitely if the forces are not parallel, but fails for parallel forces, since for such forces it gives only the magnitude and direction of the resultant, the point of applica- tion being indeterminate. This method is simple in its applica- tion, but the fact that it fails in the case of parallel forces and that it does not give compact diagrams makes the following general method more desirable: Let the force Fi be resolved into am- two components, such as OP and PI of Fig. 261, and let the force F2 be resolved into the two components O'P' and P'l'. Since the effect of any force is equal to that of its components it is evident that OP and PI may be substituted for Fi and O'P' and P'l' for F2 with- YiQ, 261. out changing the conditions, hence the resultant of Fi and F2 equals the resultant of the four components OP, PI, P'l', and O'P'. Since Fi and F2 may be resolved into components at any point, and since Pi and P'l' may be made parallel, it is evident that Pi and P'l' may be made to coincide in direction. If they can also be made equal, then the resultant of Fi and F2 equals the resultant of OP and O'P' and acts in the same direction. The components corresponding to Pi and P'l' wUl be equal, parallel, and opposite in direction if the forces Fi and F2 be resolved, as shown in Fig. 262, in which Fi and F2 are given in direction and magnitude but not in position, P being taken at any con- venient point. In Fig. 263 the forces are shown in their correct positions and the components OP, PI, IP, and PS are plotted so that Pi and IP coincide in position and are equal in magnitude and opposite in direction, therefore the resultant of Fi and F2 acts at the intersection of OP and PS, its magnitude and direction being given by the side 02 of the triangle of forces, OlS, in Fig. 262. Had the forces Fi and F2 been parallel the same method could have been used, as is illustrated by Fig. 264. Akt. 165 FORCE AND EQUILIBRIUM POLYGONS 339 In this method the point P is called the pole, 012 the force polygon, the figure ahcd the equilibrium or funicular polygon, the lines PO, Pi and P2 connecting the pole and the apices of the force polygon the rays, and the corresponding lines in the equilibrium polygon the strings. The force polygon serves to determine the magnitude and direction of the resultant while the Fig. 262. Fig. 263. equilibrium polygon fixes its position by determining a point on its line of application. It is evident that the method is simple, compact, and perfectly general. It may be briefly stated as follows: To find the resultant of a series of co-planar forces, lay off the forces Fx, F2, . . . , Fn to any desirable scale, thus forming Fig. 264. the force polygon, locate the pole P at any desirable point, draw the rays PO, PI, ... , Pn, and the strings PO, PI, . . . , Pn parallel to these rays. In constructing these strings draw PO till it meets F^, PI till it meets F2, P2 till it meets F3, etc., each string being drawn from the point of intersection of the previous string and the appropriate force. The resultant wUl act through the point of intersection of the first string PO and the last string Pn and will be given in magnitude and direction by On of the force polygon. If the forces are in equilibrium. 340 GRAPHICAL STATICS Art. 165 the force polygon must b$ a closed figure, i.e., point and point n must coincide, since only imder this condition can 21ff=0 and 21F=0. The equilibrium polygon must also close, that is, the string PO, and the string Pn must coincide, since otherwise the resultant force which equals the resultant of the two com- ponents represented by these strings would be a couple. For concurrent forces, i.e., forces all of which meet at a point, closure of the force polygon is sufficient to show that equilib- rium exists, since such forces are in equilibrium if HZ? =0 and SF=0, a condition which obviously exists if the force polygon closes. Fig. 265. It is clear that unless the pole be located on the line On the coincidence of the first and last strings of the equilibrium polygon will involve the closure of the force polygon. This is illustrated by Fig. 265, in which the equilibrium polj^gon, shown by full lines, closes, since PO and P4 coincide, a result which obviously would not occur if and ^ in the force polygon were not to coincide. Were the forces in this case to consist of Fi, F2, and F-i only, the first and last strings of the equilibrium polygon could not coincide unless the pole were to be located on the line OS, or in the more general case, on the line On. If F^ were to act in the direction indicated by the dotted line marked F\, the force polygon would close as before, but the last string of the equilibrium polygon would not coincide with the first string, but would instead have the dotted position P'4, and the resultant of the forces OP and P'Jj. would be a couple with a value of (PO)a = {Fi)b. Art. 166 CHARACTERISTICS OF THE EQUILIBRIUM POLYGON 341 166. Characteristics of the Equilibrium Polygon. The strings of the equilibrium polygon represent the bars of a framework which would be held in equilibrium by the applied forces, and in all of which the stresses would be either direct tension or direct compression. An infinite number of such frameworks can be selected, their position and shape being determined by the location of the pole. Since each intermediate string represents two forces which are equal in magnitude and opposite in direction, the resultant of all the forces will be held in equilibrium by the forces repre- sented by the extreme strings, hence this resultant acts at the point of intersection of the extreme strings, as has already been stated. The resultant of any set of consecutive forces is also held in equilibrium by the extreme strings, corresponding to that particular set of forces, hence it acts at the point of intersection of these extreme strings. This may be illustrated by Fig. 265, in which the resultant of Fi and F-2 acts at the intersection of PO and P2; of Fi, F2, and Fs at the intersection of PO and P3; of Fi, F2, F3, and F4, at the intersection of PO and P4, that is, at any point along PO or P4, an obviously correct conclusion, since the resultant of Fi, F2, F3, and F4 equals zero, these forces being in equilibrium. The following general rule may be applied to the determina- tion of the point of application of the resultant: — The resultant of any set of consecutively numbered forces acts through the point of intersection of the two strings, one of which is designated by a number equal to that of the highest numbered force, and the other by a number one lower than the lowest numbered force. For example, the resultant of a series of forces, F^ to Fj inclu- sive, acts through the point of intersection of P3 and P7. In applying this rule the forces and strings should be numbered in the exact manner used in the illustrations. 167. Reactions. Since in order that a set of forces may be in equilibrium the force and equilibriiun polygons must close, it is evident that the reactions of a given structure may be determined graphically if their values are such to make these two polygons close. The method of doing this is clearly shown by the following examples. Problem. Determine by the equilibrium polygon the reactions for the beam shown by Fig. 266. 342 GRAPHICAL STATICS Art. 167 Solution. In order that the equilibrium polygon may close, the first and last strings must lie on the same line. To insure that such will be the case draw the string Po through the point of application of the left reaction, since this is the only known point on the line of action of this reaction. •i r. r. Fig. 266. Draw the remainder of the equilibrium polygon in the usual manner, and draw also the line marked " closuig line," which should connect the point of intersection of the string P4 and the reaction Rr, with the point of intersection of PO and Rl- The first and last strings PO and Pn of the equilibrium polygon may now be made to coincide by drawing the line PK in the force polygon parallel to the closing line. Fig. 267. and fbdng the position of K by drawing from 4 a line parallel to the reaction the direction of which is known, that is, to R r. 4K will equal the right-hand reaction and KO the left hand one, since by using these as applied forces and drawing the equilibrium polygon for the six forces Fi, F2, F3, Fi, R R, and R l, the first and last strings will coincide. Problem. Determine by the equilibrium polygon the reactions for the truss shown by Fig. 267. Solution. For this case the left reaction is the one that is known in direction. The equilibrium polygon has therefore been constructed by drawing Ps first, thus re^'ersing the usual method of construction. The closing line is drawn from the right point of support to the inter- section of Po with the left reaction. The value of the right reaction is given by 3K and of the left reaction by KO. Art. 168 GRAPHICAL METHOD OF MOMENTS 343 Further examples need not be given, as no new methods are required. The essential thing for the student to grasp is that the closing line should connect the points of intersection of the reactions and the extreme strings (that is, the strings holding the applied loads in equi- librium), and that the point K in the force polygon should be so located as to enable the reactions and forces in the force polygon to be read consecutively beginning with the left-hand force (or reaction). 168. Graphical Method of Moments. It is evident that the moment of any set of forces about a given axis may be obtained by scaling the distance from the axis to the line of action of the resultant of the given forces and multiplying this value by the resultant. To illustrate: Let point a (Fig. 268) represent the trace of the axis and let the problem be to find the moment about a of the forces Fi, F^, and F^. The resultant of these Fig. 268. forces = 03, and it acts through i, the point of intersection of PS, and PO produced, hence its moment about a=OS to the scale of force multiplied by ab to the scale of distance. The above method is very simple, but the following modi- fication of it is more useful. Draw through a a line, ma, parallel to 03. Then the moment about a of the given forces equals 03 multiplied by ns, the distance from ma to the line of action of the resultant. Draw from P in the force polygon a line PA' perpendicular to 03. Then the A ntm is similar to A P SO; .-. ^—1. (altitudes of two As are to each as their bases). ns mn .-. (PX) (mn) = (03) (ns) = moment desired. The theorem thus deduced may be stated as follows: To find the moment about any point of any number of co- planar forces, draw through the point a line parallel to the resultant of the forces, and find its intercept between the strings 344 GRAPHICAL STATICS Art. 168 holding the resultant in equilibrium. This intercept measured to the scale of distance multiplied by the perpendicular distance, hereafter called H, from the pole of the force polygon to the resultant of the given forces measured to the scale of force equals the desired moment. For a horizontal beam carrying vertical Equilibrium Polygon Scale 15^=1" Force Polygon Scale 40.000*=!" Fig. 269. loads this equals the product of the intercept of the vertical ordinate through a and the horizontal pole distance. The character of the moment can usually be determined by inspection. If doubt exists the point of application of the resultant of the forces on one side of the section should be located, and with this known and the direction of the resultant given in Abt. 169 MOMENTS WITH A CONCENTRATED LOAD SYSTEM 345 the force polygon, the character of the moment can be easily seen. For a horizontal beam with vertical loads, the fact that the moment is zero wherever the closing line intersects the equilibrium polygon and hence changes sign at such a point, may often be used to advantage in determining the character of the moment, as is illustrated in the problem which follows. Problem. Determine by the graphical method of moments the bending moment at section a of the beam shown in Fig. 269. Solution. This problem involves the determination of the moment of the forces Rl, Pi, F2, and F3 about a horizontal axis passing through a. Since the forces are all vertical, draw through a a vertical line and measure to the scale of distance its intercept between the strings holding the given forces in equihbrium. These are the closing line and Ps, hence the moment equals the product of mn to scale of distance and H to scale of force. The result thus obtained equals 0.8.5X100,000= -85,000 ft.-lbs. This is negative, since it is of the same character as the moment in the overhanging end, the point of zero moment occurring to the left of point a. 169. Graphical Method of Moments with a Concentrated Load System. The application of the graphical method of moments for a system of moving wheel loads may be easily made as follows : Lay off the forces to any convenient scale and locate the pole of the force polygon so that its normal distance from the force line measured to scale of force equals some even number, say 100,000 lbs. Plot the loads to any convenient scale, and draw the equilibrium polygon considering the uniform load as equiv- alent to a series of equal concentrated loads equally spaced. If the given load system is likely to be used for a number of spans the equilibrium polygon should be made comprehensive enough to permit its use for any span likely to be investigated. Such an equilibrium polygon is shown by Fig. 270, the force polygon being omitted. With the equilibrium polygon constructed, the operation of finding the moment with any load at a given point of an end- supported span is very simple. Suppose it be desired to deter- mine the moment at the centre of a 60 ft. span with load F13 at the centre. Lay off on the equilibrium polygon the distance 30 ft. to the left of i^i3, and an equal distance to the right of the same load, and draw verticals to intersect the equilibrium polygon at s and t. 346 GRAPHICAL STATICS Art. 170 The ordinate, mn, to the scale of distance multiplied by the distance, i?, in the force polygon equals the desired moment. The moment thus obtained = 19.2 X 100,000 = 1,920,000 ft.-lbs. In this manner several loads may be tried, and that giving the largest ordinate will give the maximum moment at the centre of this span. This method may be very conveniently used to verify the results of analytical computations, and a diagram once pre- pared for a standard loading, like Cooper's Ego, and a long span, should be of material value to the designing engineer. Fig. 270. — Equilibrium Polygon for Cooper's Ej, Loading. Force Polygon Omitted. iT of Force Polygon = 100,000 lbs. Loads are Wheel Loads. 170. Graphical Method of Shear. Since the shear at any section of a beam equals the resultant parallel to the given section of the forces on either side of the section, its value may be determined graphically by the force polygon, the reaction having previously been determined by the method of Art. 167. The following method, however, is somewhat better adapted to the treatment of concentrated load systems and should be thoroughl}' understood. Consider the beam shown in Fig. 271, and let the problem be to determine the shear at a distance x from the left end with Art. 170 GRAPHICAL METPIOD OF SHEAR. 347 the first load of a concentrated load system at x. Draw the force and equilibrium polygons in the usual manner making PO horizontal, prolong the string PO, and draw the vertical, be. Then the A abc of the equilibrium polygon is similar to the A POK of the force polygon; hence be KO ab PO' , iab){Kl .-. be = iPO) but ab L PO~H' hence if H be made equal to L, be will equal KO. But KO =Rj^ equals the shear at x with Fi at x, hence the following theorem may be stated : Fig. 271. For a simple beam supported at the ends and loaded with vertical loads the shear at a distance x from the left end with the first load at x equals the vertical ordinate measured to the scale of force between the equilibrium polygon and the first string produced at a distance L—x from the first load provided the pole be so chosen as to make H and L equal, the latter condition being most readily secured by constructing the force polygon with the loads at one point of support and the pole at the other. The vertical ordinate between the first string, PO, and the equilibrium polygon at a distance, L — x, from the first load has the same value whether the loads, force, and equilibrium polygons are laid off, as in Fig. 271, or as in Fig. 272, since one of these equilibrium poh'gons may be superposed upon the other 348 GEAPHICAL STATICS Akt. 107 Fig. 272. if drawn to same scale, by inverting it and turning it end for end. If, therefore, the given loads are laid off, as in Fig. 272, the shear at a distance x from the left end of a simple end-sup- ported beam may be found when the first load is at x by laying off the distance L—x to the left of Fi and scaling to the scale of force the ordinate between PO and the equilibrium polygon. In order to determine the maximum shear at a given section due to a concentrated load system it may be necessary to trj^ several loads at the section. If load (2) be moved up to the section it is evi- dent that the first load will be at a distance x — d from the end, assuming that d is less than x, hence the shear at the first load may be found by scaling the ordinate at a distance L — {x — d) from the first load or L—x from the second load. If Fi goes off the span during the process of moving up, the ordinate should be measured at the distance L — x from Fi, but the intercept should be the vertical distance be- tween the string PI and the equilibrium polygon. It should be noted that this method is, in fact, merely a method of deter- mining the left-hand reaction; this is equal to the shear at x when the first load is at x, but if one of the other loads is at x, then the shear at that section will be the reaction minus the load or loads that may lie between the reaction and the given section, provided there are no floor beams. If floor beams are used the shear must, of course, be determined by subtracting from the reaction the proper percentage of the loads between the support and the panel point at which the load under considera- tion is located. If the load system is to be used for a number of spans the diagram should be drawn for the longest span, and the scale for any given span determined by proportion. The application of the graphical method of shear is clearlj' shown by Fig. 273. It should be noticed that the equilibrium polygons will be exactly alike, whatever the span chosen, 'provided the ratio of the scales of forces be inversely proportional to the spans, e.g., the equilibrium polygon of above figure is constructed for a 200 ft. Art. 171 POLYGON THROUGH SEVERAL POINTS 349 span to a scale of forces of 60,000 lbs=l in., but if constructed for a 100 ft. span with a scale of forces of 120,000 lbs. = 1 in. as also indicated, the same equilibrium polygon would be obtained. It follows that this polygon may be used for any span by multi- plying the scaled ordinate by the ratio between the span for which the polygon is constructed and the given span. Scale of Distance 100 ft.= 1 in. Shew, Fl at 10 £t. from Iffe end of 20O'span=a6 =52200lba. *' Fi "10" " .. .. ..100' " =c(( J' 2 = 300OOlb9. " F, " 10 • 50' " =e/x 1 = 16000 lbs. Fig. 273. 171. Equilibrium Polygon through Several Points. It is clear that if a hinged framework be constructed identical in shape to a given equilibrium polygon and with its bars designed to carry the string stresses of the polygon, it will be in equilib- rium under the loads for which the equilibrium polygon is drawn. This fact may be used in determining the proper outline for an arch intended to carry a set of fixed loads, but cannot be used for moving loads. On the other hand, if an equilibrium polygon cannot be drawn within the limits of an arch the shape of which has already been decided upon, it is reasonable to suppose that such an arch will not be stable, and it is upon this hypothesis that the commonly accepted theory of stone arches is based. The theory of such arches will not be taken up at this point, but since in studying them it is often useful to be able to draw an equilibrium polygon through certain fixed points, methods of doing this will be derived. Equilibrium Polygon through One Point. Since the pole may be chosen anywhere, and any string of the equilibrium polygon drawn through a given point, it is evident that an infinite 350 GRAPHICAL STATICS Abt. 171 number of equilibrium polygons may be drawn through one point. Equilibrium Polygon through Two Points. Since the first and last strings of an equilibrium polygon must always meet on the line of action of the resultant of the set of forces for which the polygon is drawn, it is evident that an equilibrium polygon may be drawn through two points by first drawing any equilibrium polygon through the first point and plotting the resultant of the set of forces indirection and position; the last string of the desired polygon may then be constructed through the second point in question and the intersection of the resultant and the first string. This method is illustrated by Fig. 274, in which Fig. 274. the original equilibrium polygon is constructed with a pole located at random, the string PO being drawn through point a. The final equilibrium polygon may then be constructed so that P'4 will go through point b, PO remaining unchanged in position, by drawing P'4 through point b and the intersection of PO and P4. P'3 may next be constructed by connecting the point of intersection of P'4 and F4, with the point of intersection of P an J PS, and in a similar manner P'2 and P'l may be located. This construction is evidently consistent with corre- sponding strings of the two equilibrium polygons intersecting on the resultant of the forces held in equilibrium by these strings. If the intersections as obtained by the method just given are not on the sheet the second equilibrium polygon may be con- structed by locating a new pole P', and constructing an entirely distinct polygon. The method of doing this is illustrated by Fig. 275, in which the line of action of the resultant of all the forces is plotted and the new strings PO and P'4 drawn at Art. 171 POLYGON THROUGH SEVERAL POINTS 351 random through points a and h respectively to meet upon the line of action of this resultant. The new pole may then be located by drawing from and 4 in the force polygon rays parallel to P'O and P'4 respectively, their intersection locating the new pole P' . If it be desired to have other than the first Fig. 275. and last strings pass through the two points, the resultant of the forces held in equilibrium by the desired strings should be used. The following important theorem is also of use at tinaes, viz.: That, for any set of loads, the intersection of correspond- FiG. 276. ing strings of two equilibrium polygons drawn with different poles will lie on a line parallel to the line joining the poles. To prove this consider the two equilibrium polygons, shown in Fig. 276, with the poles P and P' . The force Fi may be resolved into the two components OP and Pi, consequently that force may be replaced by these components without changing existing conditions. The force Fi will be held in equilibrium by the two forces IP' and P'O, consequently the resultant of these two forces is equal and 352 GRAPHICAL STATICS Art. 171 opposite to the resultant of the two forces OP and PI, hence the forces OP, PI, IP' , and P'O are in equilibrium, therefore the resultant of OP and P'O is equal and opposite to the resultant of Pi and IP'. The resultant of OP and P'0=P'P and acts at the intersection, a, of the strings PO and P'O; the resiiltant of PI and IP' =PP' and acts at the intersection b, of the strings PI and IP'. Since these resultants are in equilibrium they must not only be equal but act along the same line, that is, both must act along the line ab, hence ab must be parallel to the actual direction of these forces; that is, to PP'. In the same way it may be shown that the resultant of PI and IP' acts in the same line but in the opposite direction to the resultant of P'3 and SP, this direction being parallel to Fig. 277. PP', hence the point of intersection of P'2 and 2P also lies on the prolongation of the line ab. In the same manner the inter- section of all the corresponding strings may be proven to be on the line, ab, hence the theorem is proven. Equilibrium Polygon through Three Points. Application of the theorem just stated enables an equilibrium polygon to be passed through three points. The following is the mode of procedure: Construct a polygon through either two of the given points, say, the points a and b. Fig. 277, and connect these points by a straight line. If this line be made the line of intersection of the corresponding strings it is evident that if a new polygon be drawn it will also pass through the points a and b, hence it is merely necessary to draw a new string through the third point, c, and the intersection of the corresponding string of the first poly- gon with the line ab, and finish the polygon by the method Art. 171 POLYGON THROUGH SEVERAL POINTS 353 given for a polygon through two points. The figure shows the application of this method, assuming that the polygon PO Pi, etc., has already been drawn through two points. It will be noted that in the construction it was necessary to locate the pole P' in order that P'2 might be drawn, since the intersection of P2 and the line ah does not come on the sheet. The remainder of the polygon was drawn by using the inter- sections of the strings of the original polygon and the line ab. Alternative Method for Equilibrium Polygon through Two and Three Points. Another simple and useful method of drawing an Fig. 27S. equilibrium polygon through two points, a and b, is as follows: Assume the forces which are held in equilibrium by the strings which are to pass through the given points to act upon a beam supported at a and &, Fig. 278. Assume the reaction at b to be fixed in direction, vertical in this case, and determine graphically both reactions for these loads, drawing one of the strings through a. The equilibrium polygon thus drawn will pass through point a by construction and the value of the reactions will be independent of the position of the equilibrium polygon. If a new equilib- rium polygon be now drawn with its pole at any point on a line drawn from the closing point K of the force polygon parallel 354 GRAPHICAL STATICS Art. 171 to the line ab and with the same string passing through a, as in the original polygon, this polygon will pass through both points, since its closing line passes through a, and is parallel to ab. This construction is illustrated by the figure in which the pole P' of an equilibrium polygon with the string P'O passing through a and P'4 through b might lie at any point on a line drawn from K parallel to ab. To extend this method to a third point, c, proceed as follows: Draw a vertical through c till it intersects the original equilib- rium polygon at d. From the pole P draw a line parallel to ad to its intersection, K', with a vertical through 2. Draw from K' a line parallel to ac. The intersection of this line with the line KP' locates P" which is the pole of an equilibrium poly- gon passing through a, b, and c. This method consists essentially of the determination of the reactions upon a beam, ac, due to the forces Fi and F2, and the location of the closing point K' of the force polygon corresponding to these reactions. An equilibrium polygon with its pole at any point on a line from K' parallel to ac will pass through points a and c. Since, as has already been shown, an equilibrium polygon with its pole at any point on KP' will pass through a and b, it is evident that a polygon with its pole at the intersection of these two lines will pass through the three points a, b, and c. CHAPTER XIV DEFLECTION AND CAMBER 172. Elastic and Non-elastic Deflection. The deflection of a truss is due to changes in length of the members and may be divided into two parts, elastic and non-elastic. The former may be caused by stresses, or by differences in temperature of the various members, and disappears upon the removal of the loads or the return to uniform temperature; the latter is due to play at the joints and occurs when the falsework used in erec- tion is removed, being particularly important for pin connected trusses in which considerable play usually exists in the pin holes. A knowledge of the deflection is often desirable, particularly in proportioning the lifting devices at the ends of a swing bridge and in planning the erection of cantilever structures. A method based upon deflections also furnishes a convenient mode of deter- mining stresses in a statically indeterminate structure. 173. Truss Deflection. Trigonometrical Method. A rigid truss is composed of triangles all the properties of which may be determined if the lengths of the three sides are known. The vertices of the triangles coincide with the joints of the truss, hence the various positions of a joint with respect to a pair of rectangular axes may be determined for any length of the sides of the triangles, that is, of the members of the truss. It is evident, therefore, that to flnd the vertical elastic deflection of a joint, sa}- the centre joint of the bottom chord, it is only necessary to compute by trigonometry its position with respect to a fixed horizontal axis both before and after the application of the load causing the deflection. If the axis passes through the original position of the joint, the vertical movement under the load will be found without the former computation. The horizontal movement of an}- joint ma}' be determined in a similar manner, using a vertical instead of a horizontal axis for reference. 355 356 DEFLECTION AND CAMBER Akt. 174 The length of each member after the application of the load may be foimd by adding to its original length the change of length due to tension, or by subtracting the change of length due to compression. If the non-elastic deflection be desired the same method may be used, but for this case the change in length of a member will equal the average play in the pin holes at the two ends instead of the change due to stress. While this method is simple in theory and application, it is very laborious and is not used in practice. 174. Truss Deflection. Method of Rotation. The deflection of any joint of a simple truss due to a change in length of one bar only may be readily determined by investigating the result- ing rotation of one portion of the truss with respect to the other. Fig. 279. the latter being assumed as fixed. By considering the effect of each bar separately and summing up the results, the final deflection may be determined. To illustrate this, consider the truss shown by full lines in Fig. 279, and let it be desirgd to deter- mine the vertical deflection, d^, of panel point e, due to a decrease in length, 8, of bar he. Evidently hce is the only triangle which will be changed in shape by the change in length of this bar. If the portion, ahe, of the truss be assumed for the time being to remain fixed in position, the figure abc'f'h'ea will represent the new position of the truss. The value of S for any ordinary change of leng-th is very small compared with the original length of the bar, e.g., for a steel bar subjected to a stress of 15,000 lbs. per square inch, d is approximately -^-^ of the length of the bar. It follows that the angular rotation of bar ec is extremely small, Art. 174 TRUSS DEFLECTION. METHOD OP ROTATION 357 and in consequence the distance cc" in Fig. 280 may be assumed as equal to 5 as shown, the error being infinitesimal. The angular movement, d, of bar ce evidently equals the Pi ri angular movement of eh. The sine of this angle equals — =— , c e ce But eh' = eh, eh hence h'k= — u ce the vertical hence h'k = eh' —. ce deflection of point h with respect to the axis ae, which is assumed to remain unchanged in position. Actually, however, point h remains on the abut- ment and ae changes its position. The correct position of the dis- torted truss may be found by rotating it about a horizontal axis passing through a until ah' becomes horizontal. This rota- tion will cause e to drop below its original position by the amount which it is now below ah', that is, by one-half of h'k (neglecting the effect of the slight difference in length between ek and eh'), hence the vertical deflection, de, of point e due to the change, d, in bar be, will be given by the following equation: Fig. 280. . = h ..eh In a similar manner the deflection of other points due to the change in length of this or other bars may be obtained. In order to illustrate this method more completely its applica- tion to the problem of determining the vertical deflection of point e resulting from an increase, d, in the length of bar ab, will also be given. For this case the portion, dbfhed, of the truss will be con- sidered stationary. The condition of the truss after the change in length of the bar will be as shown, greatly distorted, in Fig. 281, and e'e^^a'm will be the actual vertical upward deflection of point e. The value of a'm may be determined as follows: Let the distortion of the triangle, abd, be as shown, greatly exaggerated. 358 DEFLECTION AND CAMBER Art. 174 in Fig. 282. Then bn equals the new length of bar ab, and the new position of a will be at the intersection of arcs swung from b and d as centres, with radii bn and da respectively. The fact that d is very small compared with the sides of the triangle makes Fig. 281. the error in assuming the tangents na' and aa' to coincide with the corresponding arcs negligible, and gives the condition shown in Fig. 2S2, from which it at once follows that aa' equals sin a But bd , a = — hence ab = (),— and bd therefore ee' <■ ™= method is evidently mufh simpler in application than the trig- onometrical method previously mentioned, but is nevertheless awkward for general use, since it involves an entirely separate solution for each bar. It is given here to illustrate graphi- cally the distortion due to change in length of a single bar. 175. Truss Deflection. Meth- od of Work. The method which follows is a modification of one usually attributed to Prof. Otto Mohr, and provides a very simple and ingenious solution of the prob- lem of determining the deflection of a truss. The method is based upon the fact that if a truss is loaded by a single load of unity at any point and is then deflected by the application of other forces, the external work done by the load Fig. 2,S'2. Art. 175 TRUSS DEFLECTION. METHOD OF WORK 359 of unity equals the internal work done by the bar stresses caused by this load of unity. This is in accordance with Bernoulli's law of virtual work, which states that a system of concurrent forces in equilibrium may be moved a small distance by an external force without the performance of work iDy the sj'stem. Such a condition occurs at each joint in a truss, the forces being the bar stresses due to the load of unity, and the movement of the joint being that due to the external forces producing the deflection of the truss. A slight approximation actually occurs in the application of the method, since it is assumed that the bar stresses due to the load of unity remain constant during the distortion of the structure; actually these change slightly owing to the change in the angles between the various members meeting at the joint, but the error is extremely small for trusses formed of material with a high modulus of elasticity, since the change in these angles is inap- preciable for a safe working load. The method is inapplicable for a truss which would distort greatly under load, such, for ex- ample, as a truss formed of spiral steel springs or rubber bands. It will be noticed that the load of unity is merely a measuring device and has no influence upon the deflection of the structure by theapplied loads, since it and the stresses caused by it are in equilib- rium before the forces causing deflection are applied, and remain in equilibrium during the application of these forces. More- over the load of unity may be expressed in the smallest possible units, say milligrams, and could under no circumstances have an appreciable effect upon the deflections. In order to apply this method consider the truss shown in Fig. 283 and let it be desired to determine the deflection, ^2, of panel point L2, due to the shortening of bar Ui U2 by the amount d, this shortening being due to any cause whatever, such for example as a stress in the bar, a difference of temperature in the bar as compared with other bars in the truss, or an adjustment of its length by a turnbuckle. Fig. 283 shows by full lines the truss before the length of bar U1U2 is changed, and by dotted lines the distorted position of the truss. Evidently the external work which would be performed by a vertical load of imity hanging at 1/2 during the change of length of the bar would be unity X ^2- This load would cause a compression, s, in bar U1U2, and this internal force would have to move the distance, d, during the change in length of the bar, hence would perform an internal 360 DEFLECTION AND CAMBER Akt. 175 work of sd. Equating the internal and external work gives 82 = sd; therefore the vertical deflection, 82, of point L2, due to a change in length, §, of bar U1U2, equals the product of the stress, s, in Ui U2, and the change in length of the bar. Were the load of unity inclined instead of vertical, s would have a different value, and ^2 would be the deflection along the inclined line of action of the load of unitJ^ A comparison of the results obtained by this method and the method of rotation shows them to be equal. The signs must be carefully considered. If tension and increase in length are both denoted by positive signs, the deflection will be in the direction joi action of the load of unity if the resulting value of sd is positive, and in the opposite direction if this product is negative. For the case considered s is compression, and d is a shortening, hence each has a negative sign and the product iS 1*^ Uj 1 U2 I 3 \,^_ y X y jV ~ r — A^\ ^ N \. j^y^ I -^ 1 /y ' // ' ^\ I // I '^ \ // 1 N \ . // ' '^ 1 // // ^X \ \ 1 C\ // \ \ 1 A' X ^x // \ L3 '^ |.L ■ /^^-^r--- w V////A ' Unit Load //M. Fig. 283. will be positive; therefore the deflection will be in the direction of action of the load of unity, that is, downward. The correctness of this method of dealing with the signs may be readily seen by examining the case of a single vertical bar with a force of unity acting downward at its lower end. The stress, s, in this case is tension and hence has a positive sign. If the length of the bar be increased by the amount, 5, the pro- duct, sb, will also have a positive sign, showing, according to our convention, that the lower end of the bar deflects in the direc- tion of action of the unit force; that is, downward. If the bar be shortened, s8 will have a negative sign indicating a deflection of the lower end of the bar in a direction contrary to the direc- tion of action of the unit force; that is, upward. Both of these conclusions are obviously correct, and the direction of the deflec- tion with respect to the direction of action of the unit force Art. 175 TRUSS DEFLECTION. METHOD OF WORK 361 would evidently be unchanged if the unit force were to be ap- plied to the bar through a series of other bars instead of directl)', and if it were to be inclined or horizontal instead of vertical. To apply the method to all bars of a truss it is only necessary to obtain the summation of the various products. The final formula for deflection may then be written, in which 5„=the component of the deflection of any joint, n, of the truss in any desired direction; s = stress in any bar of the truss due to a load of unity acting at joint under consideration and in the direction of the desired deflection; ^ = change in length of the bar of the truss in which the stress s occurs; Ss5 = summation of the products, s§, for all bars of the truss. If Hsd is found to have a positive value the deflection will be in the direction of action of the force of unity; if a negative valu§, it will be in the opposite direction. If the actual deflection of a given joint is desired the deflection in both a horizontal and vertical direction must be obtained and the resultant found. The usual problem is to determine the deflection in a given direction of a given joint due to applied loads such as the weight of the structure itself, or a given position of the live loads. For this case d will be the change in length due to the stress caused by the applied loads, hence the formula may be written in which (^„=deflectioninfeet of any joint in any desired direction; L = length of any bar in feet; /I = area of same bar in square inches; P = stress in pounds in same bar due to applied loads; £■= modulus of elasticity; s = stress in same bar due to force of unity, acting at joint under consideration and in direction of desired deflection. 362 DEFLECTION AND CAMBER Art. 176 If E is constant for all bars of the truss, as is usually the case, it is simpler to express the change in length of each bar in terms of E and substitute the final numerical value of E after the summation is complete. 176. Truss Deflection Illustrated. The following example illustrates clearly the application of this method. Problem. Let the problem be the determination of the vertical deflection of point Li of truss shown in Fig. 284 for a uniform live load of 2000 lbs. per foot over the entire truss. Solution. The simplest method of solution of such a problem is to prepare a table in which separate columns are assigned for the terms s, P, L, and A; for the change in length of the bar; and for the product of the change in length of the bar and the stress, s. The table on page 363 is prepared in this manner and is self-explanatory. 2-15"Ls36* 2-15"Cs35* S-lS'CsSS* UlM8'x X''Cov. Fl."2l-18'i[ X'Cov. Pl.Ual-lS'x ^'Cov. FlMi Fig. 284. The summation of the numbers in the last column of the table 2,170,000 gives + E and equals the vertical deflection downward of Li. Were this summation to have a negative sign it would equal the upward deflection of this point. The numerical value of the deflection may be obtained by substituting the value of E. If this be taken as 29,000,000, ^ = 2lmm =°-°^^^ ^*- =°^°" =^" 'iPP'-°''i«iately. An inspection of the table shows that the stresses due to load unity should be computed before the change in length of the bars, since if the stress in any bar caused by this load is zero the deflection due to this bar is zero and its change in length need not be figured. Had the problem been that of computing the horizontal movement of the roller-end the load of unity should have been applied horizontally at that end. The only truss bars which Art. 176 TRUSS DEFLECTION ILLUSTRATED TABULAR VALUES FOR DEFLECTION OF POINT Lj SHOWN IN FIG. 284 (Slide rule used throup;hout.) 363 OF TRUSS Bar. Stress due to Load of Unity at 1/2 = «. LoUi 0.848(-) Stress due to Applied Load = P. 169600(-) Length of Bar in Ft. = L Area of Cross- section inSq.In. --A. 42.4 30.27 Change in Length of Bar PL nFt. = . = £|. 23800 E (-) Deflection due to each Bar in Ft. = si. 20200 E (+) u.m 1.200(-) 180000(-) 30.0 27.33 198000 (-) 23800 E (+) U2U3 1.200(-) 180000(-) 30.0 27.33 19800 E (-) 23800 E (+) U,Ui 0.800(-) 180000(-) 30.0 27.33 19800 E (-) 15800 E (+) UiLi 0.565(-) 169600(-) 42.4 30.27 23800 E (-) 13400 E (+) LoL\ 0.600(+) 120000(-|-) 30.0 13.50 26700 E (+) 16000 E (+) L1L2 0.600 (+) 120000( + ) 30.0 13.50 26700 E (+) 16000 E (+) LizJua 0.800(+) 180000(+) 30.0 20.25 26700 E (+) 21400 E (+) LzLi 0.400(+) 120000(+) 30.0 13.50 26700 E (+) 10700 E (+) Ldiij^ 0,400(+) 120000 (+) 30.0 13.50 267000 E (+) 10700 E (+) ViLi 0.000 60000(+) 30.0 necessary UtL^ 0.000 30.0 UzLi 0.400(-) 30.0 I/4Z/4 0.000 60000(+) 30.0 C/iLj 0.848(+) 84800(+) 42.4 11.25 32000 E (+) 27100 E (+) C/2i3 0.000 42.4 necessary f/siz 0.565(+) 42.4 UiLz 0.565(+) 84800(+) 42.4 11.25 32000 E (+) 181000 E (+) 364 DEFLECTION AND CAMBER Art. 177 would be stressed by this load would be those in the bottom chord in which the stress would be 1.00 (+) were the load taken as acting to the right. The deflection would then be found by the summation of the changes in length of the bottom chord bars, which equals (+) 0.00920X5= (+)0.046, hence the horizontal movement of the roller-end of the truss under the load of 2000 lbs. per foot = 0.046 ft. or 0.55 in. to the right. Had it been desired to find the elastic deflection due to the dead load the dead stresses should have been used instead of the stresses due to the load of 2000 lbs. per foot. For the non-elastic deflection due to play in the pin holes the change in length of the bar could be written directly, and the third, fourth and fifth columns omitted. For example, if the allowable play in the pin holes at L\ and L2 is ^ in., the change in length of bar L1L2, that is, the change in distance centre to centre of pins, would be g^ in. (-|-), and this value should be written in the sixth column. Deflection at centre= ' f''- 48 E I Deflection at free end Deflection at tree end= 1 wL* s E I Fig. 285. 177. Deflection of Beams and Girders. Simple formulas for the deflection of beams and girders of constant cross-section supported at both ends or fixed at one end are derived in all standard works on mechanics. The more common cases are represented by Fig. 285, in which the deflections given are the maximum deflections. For more complicated cases of loading, or for girders with variable cross-section the method of work Art. 177 DEFLECTION OF BEAMS AND GIRDERS 365 may be applied in the same manner as for trusses, the fi!bres of the beam being substituted for the bars of the truss. A general formula for the deflection may be developed by this method in the following manner : Let M = moment at section, de, of the given beam (sec Fig. 286) due to the external forces causing the deflection of point a; fd T Width of beam = 6 2! =Ff'*!/ r^a dx Unity _i„_ Fig. 286. m = moment at same section due to load unity acting vertically at a; 5 = longitudinal distortion, due to the external forces, of the prism, fdeg at a distance y from the neutral axis; ^a = deflection of point a due to the external forces; /i =- longitudinal fibre stress due to moment, /», at a dis- tance y from the neutral axis ; fz = fibre stress at same point due to moment .1/ ; w = internal work done in prism of length dx, depth dy, and width b, with its centre at a distance y from the neutral axis of the beam, Ijy the load unity, during the distortion of the beam by the appli- cation of the external forces ; lF = total internal work done in the beam by the stresses due to the load of unity during the beam's distor- tion by the external forces. Then ' E ~ El ' 366 DEFLECTION AND CAMBER Art. 177 and w={fibdy)(-^dx but hence /- y'^bdy _ I Jo E] w= < '«=.,. The external work due to the load of unity = 1 X ^a, hence The application of this equation to a beam of constant cross- section is illustrated by the following problem: Fig. 287. Problem. Let it be required to find the deflection at the load for the case shown in Fig. 287. Solution. Consider a section at distance x from left end. Then Px X ilf = — J- and rn='—. 2 2 Art. 177 DEFLECTION OF BEAMS AND GIRDERS 367 Since the beam is symmetrical the integral for the entire length of beam may be taken as double that for the left-hand half, therefore, the value of da is given by the following equation : ^X L 2 ' 2'lfl~4S~E7' Were the beam to be loaded for its entire length with a load of p per foot, il/ in the above equation would be —x —^r, hence the equation for deflection would be i "~{Lx—x''){x)dx 5 pL^ -IX'" EI 384 EI ' For beams of variable section formula (29) may be applied by inte- grating for different portions of the beam and then adding the results Suppose, for example, that in the first case the middle half of the beam had the value /i for the moment of inertia and the two end quarters each had the ^'alue /j. The total deflection would then be expressed by the following equation : L_ LpI^+xY -4 Pi' dx „ r4 \4 / dx -X*^l^-X' Eh' The case just given illustrates the method necessary for an end-supported girder with a single cover plate on each flange extending over the central half of the girder. If more cover plates are used it is necessary to write more terms, but the same general method is applicable. If the girder varies in depth, as well as in flange ai-ea, it may be divided . into as many sections as seems desirable and the average moment of inertia of each section used, the equation for the deflection including as many terms as there are moments of inertia. Before leaving this method it should be observed that both by it and the elastic curve method, by which the results shown in Fig. 285 are usually Fig. 288. obtained, the deflection due to shear is neglected. The effect of positive shear at the section under consideration would be to distort the prism, fdeg, in the manner shown by Fig. 288, and hence cause some deflection. The value of the deflection due to shear is, however, relatively very small and may be neg- 368 DEFLECTION AND CAMBER Art. 178 lected. For a full discussion of this method the reader is referred to a paper by Clarence W. Hudson in the " Transactions of the American Society of Civil Engineers," Vol. LII. 178. Graphical Method of Truss Deflection. Williot Diagram. The method of work given in Art. 176 furnishes a simple and accu- rate method of determining the deflection in a given direction of a particular joint in a truss. It is, however, occasionally desirable to determine the actual distortion of several or all Joints, a problem which can be solved somewhat more readily by graphical than by anal3-tical methods. It is obvious that the actual movement of any truss joint, due to changes in the bar lengths, may be determined graphically in the following simple manner : Fig. 289. Let abed, Fig. 289, be the truss before the bar lengths are changed, point a being fixed in position. Let the new lengths of the bars be ab', b'c', c'd', d'a, and b'd'. Assume for the present that bar ab remains unchanged in direction. If its length be now changed to ab', b will move to b', and if from a and b' as centres arcs with radii ad' and b'd' be swung these arcs will intersect at d', which will be the new position of d on the assumption that ab remains imchanged in direction. In a similar manner arcs swung from b' and d' with the new lengths of the other bars as radii will give the new position of c at c', hence ad'c'b' will be the actual shape of the truss after dis- tortion takes place. Its position is of course incorrect, since point c should remain on the abutment, hence the line ac' should actually be horizontal, and by so considering it the deflections of all points may be obtained; e.g., the vertical deflection of Art. 178 GRAPHICAL METHOD OF DEFLECTION 369 point b equals the normal distance from b' to ac', the horizontal deflection of point c = ac' — ac, etc. This method, which may be extended to cover anj' form of truss, is impracticable in practice, since accurate results cannot be obtained without the use of a very large scale, owing to the very small changes in bar lengths for materials having the high moduli of elasticity of structural materials. To overcome this difficulty a modification of this method by which changes of length only are dealt with was developed by the French engineer, Williot, and will now be given. Let the truss shown in Fig. 290 be identical with that given in Fig. 289, and assume the same changes in bar lengths to occur. Assume also as before that bar ab is fixed in direction, and that b moves to b' when distortion occurs. Fig. 290. Let b'e be parallel to bd and equal to it in length ; " ee' be the change in length of bd (an increase) ; " dfhe the change in length of ad (a decrease). If normals to ad and b'e' be erected at points / and e' they will meet at d' , which will be the new position of d for any rigid truss of ordinary structural materials, since the distortions are so small compared with the bar lengths that the normal e'd' may be considered as coinciding with the arc swung from b' as a centre with radius b'e', and normal fd' may also be con- sidered as coinciding with the arc swung from a as a centre of with radius af. Since the figure dee'd'f may be drawn to any scale without reference to the truss itself, as shown in Fig. 291, it is evident that the actual displacement of point d with reference to axis ab may be found with great accuracy. In a similar manner each triangle of which the truss is composed may be dealt with and the displacement of each vertex found with reference to anv one of the sides of the triangle as' an axis. 370 DEFLECTION AND CAMBER Art. 178 cV If this process be carried out for the entire truss each displace- ment may be determined with respect to bar ab as an axis, by using for each new axis the new position of that bar which is common to any triangle pre- viously considered and that imder consideration; e.g., to find the new position of point c, with reference to axis ab, use the new position of the bar bd, that is, b'd', as an axis. This is illustrated in Fig. 292, where ab'd' is the new position of the triangle, abd, resulting from the changes in length of bars ab, bd, and ad. To find the position of c' due to the combined distortion of the two triangles, abd and bed, it is evidently neces- sary to determine the intersection of two arcs, one swung from b' as a centre with the new length of bar be, and the other from Fig. 291. d' with the new length of bar dc. Here, as before, it is essen- sially correct to replace the arcs by their tangents. The process may be accomplished by laying off en equal to the combined increase in length of ab and be, eo equal and parallel to dd', op equal to the decrease in length of ed, and nc' and pc' per- pendicular respectively to be and de. The point of intersection, c', of these normals is the new position of point c. It should be observed that the new position of point c is a function of the change in shape of triangle abd as well as of triangle bed, even if bar ab actually remains horizontal, since bar bd cannot be changed in length without influencing the shape of triangle bed. Art. 178 GRAPHICAL METHOD OF DEFLECTION 371 Since dd' of Fig. 291 equals co of Fig. 292, and de of Fig. 291 equals the change in length of bar ab, it is evident that Fig. 291 may be used as a basis for determining the movement of point c. The resulting diagram is called a Williot diagram and is shown in Fig. 293. Fig. 293. In this diagram d/= decrease in length of bar ad; de= increase in length of bar ab; en = increase in length of bar be; ee' = increase in length of bar bd; (i'p = decrease in length of bar dc. If all of above changes are plotted parallel to the original directions of the bars, then dd' will equal the displacement of the point d, and dc' the displacement of point c. In order to avoid confusion it is desirable to letter the initial point the same as the panel point used for an origin, in this 372 DEFLECTION AND CAMBER Art. 178 case point a, hence if in the diagram letter d is replaced by a^ the displacement will in every case be the distance from a to 7iW Diagram is drawn assuming a to be fixed in position and ah in direction. Distances from a to 6, h, c, g, d, f, and e equal actual movements of these points referred to axis ab. Figures in brackets indicate bar numbers and are placed on the lines which show tlie changes in length of the corresponding bars. Fig. 294. the intersections of the proper normals. A Williot diagram for a more complicated truss is shown by the full lines of Fig. 294. The following rules for the construction of the Williot diagram may now be given : Art. 179 CORRECTION OF THE WILLIOT DIAGRAM 373 1. Draw a diagram of the truss, and select one member as an axis, or fixed reference bar, assuming this bar as fixed in direction with one end fixed in position. Choose, if possible, a bar which does not change its direction under the given loading. 2. From some convenient point plot the axial deformation of the bar chosen as axis parallel to and in the direction of the motion of its free end. Letter the origin and the point just plotted, hereafter called the second point, to correspond with the lettering of the fixed and free ends respectively of the chosen axis of the truss diagram. 3. Select two bars which form a triangle with the assumed axis. From the origin, on a line parallel to that oiie of the two bars which is hinged to the fixed end of the axis, lay off the axial deformation of this bar in the direction of the motion of its far end. From the second point, similarlj- lay off the deformation of the bar hinged to the free end of the axis. At the extremities of these plotted deformations erect perpendiculars. The distance from this point of intersection to the origin equals, the distortion of the apex of the triangle under consideration with reference to the fixed axis and should be lettered to correspond to this apex in the truss diagram. 4. Consider the bars forming a triangle, of which the dis- placed positions of the ends of one leg are given by two of the thi-ee points now plotted. From each of these two points, on a line parallel to the bar of the triangle which is connected thereto, lay off the axial deformation of this bar in the direction of the motion of its far end. At the extremities of these plotted deformations erect perpendiculars. The distance from this point of intersection to the origin equals the distortion of the apex of the triangle under consideration with reference to the fixed axis, and should be lettered to correspond to this apex in the truss diagram. 5. Continue thus till all points are located. 179. Correction of the Williot Diagram. It is evident that the WilHot diagram shows the actual movement of the joints only when the bar which is assumed to be fixed in direction actually remains fixed dm'ing the change in shape and when the origin also remains fixed. The latter condition is readily obtained by selecting a point which is not on rollers, and the former may sometimes but not always be secured, e.g., in the 374 DEFLECTION AND CAMBER Art. 179 truss shown by Fig. 294 if loaded with a uniform load per foot, bar eg will remain vertical while the truss deflects. As in many cases no bar remains fixed in direction this method would be incomplete unless some means can be found for correcting the displacements thus obtained to provide for the rotation about the assumed axis. If the displacements found by the Williot dia- grams for truss of Fig.294be plotted, the truss will appear as shown by dotted lines in Fig. 295 with the distortion exaggerated owing to the plotting of the displacements on a different scale from the truss diagram. - Since point e should remain on the abutment, its true movement being horizontal, the correction necessary to apply to the diagram must be such as would be produced by rotating the whole truss about a until e' drops to the horizontal line through a, that is, until e drops through the distance e'k (since here again the arc swung with a as a centre and ae' as a radius differs in position from the tangent by only an infinitesimal amount, the correct distance e'k being very small) . The movement of the other points of the truss due to this rotation will bear the same relation to the movement of e as the distance from a to these points bears to the distance ae. In Fig. 296 all the full lines are perpendicular to the corre- sponding bars of the actual truss shown in Fig. 297. In con- sequence any triangle such as a'd'e' is similar to the corresponding triangle ade, hence a'd' ad ae '^a'e'X- ad Art. 179 CORRECTION OF THE WILLIOT DIAGRAM 375 In a similar manner it may be shown that a'd = a'e' X — , a'V = a'e' X -, etc . ae ae Hence if e'a' equals the movement of point e due to rotation, d'a' will equal the movement of point d, c'a' the movement of point c, etc. To obtain, therefore, the true displacement of the various points the displacements determined in Fig. 294 must be corrected by these amounts. A simple method of accomplish- ing this has been devised by Prof. Mohr and is illustrated by Fig. 294. It consists of the insertion in the Williot diagram of a figure corresponding to Fig. 296 with a' at a and e' on a horizontal through e. The correct displacement of a point will then be given in direction and magnitude by the dis- tance from the corresponding point of the in- serted truss, shown dotted, to the same point as located by the Williot diagram, e.g., the Fig. 296. Fig. 297. correct displacement of point c = c'c. The truth of this is easily seen; oc= movement shown by the diagram, c'a = movement due to rotation, hence c'c equals actual displacement of point c. In other words rotation causes point c to move from c' to a, and distortion from a to c, the resultant movement equalhng c'c. This method of correction for rotation is simple and no confusion need arise if the following rules be observed: 1. Draw a line through the displaced position, as given on the Williot diagram, of the truss joint at the expansion point of support, parallel to the known direction of movement of this point, i.e., in general, parallel to the surface upon which the expansion rollers move. 376 DEFLECTION AND CAMBER Art. 179 2. Draw a line through the point on the Williot diagram corresponding to the joint at the fixed point of support of the structure, perpendicular to the line in the truss connecting the two points of support of the truss, and determine its point of intersection with the line previously drawn. 3. Insert in the Williot diagram a truss diagram drawn with all its bars perpendicular to corresponding bars in the actual truss, i.e., drawn in a position peipendicular to the original position of the truss. The location and scale of this new truss Fig. 298. — Displacement Diagrams for Truss Shown in Fig. 294. Bar eg assumed to be fixed in direction. Fig. 299. — Displacement Diagram for Truss Shown in Fig. 294. Point h assumed fixed in position and bar kg in direction. Correction diagram i", b", d'', e" is drawn on assumption that point a is free to move hori- zontally instead of point e. diagram is fixed by locating the joint corresponding to the expansion point of support at the point of intersection previously determined (see 2) and the joint corresponding to the fixed point of support at the corresponding point on the Williot diagram. 4. The correct displacement of each joint of the truss may now be determined in magnitude and direction by scaling the distance from the joint as given on the correction diagram drawn Art. 180 CAMBER DEFINED 377 as described under (3) and the position of the joint as given on the Williot diagram. Figs. 298, 299, and 300 illustrate more fully the graphical method as applied- to the truss shown in Fig. 294, In Fig. 298 bar eg has been assumed as fixed in direction. This agrees with the actual condition if the truss is loaded uniformly and the displacement diagram needs no correction for rotation. Fig. 299 is drawn with hg fixed in direction and point h in position. This needs to be corrected for rotation and the correction diagram is given both for the truss shown, and for the same truss with point a free to move horizontally. Fig. 300 is drawn to show Scale = Fig. 300. — Displacement Diagram for Truss Shown in Fig. 294. Point a and bar ah assumed fixed in position and direction respectively. Change in length assumed to occur in bar 2 only. (Decrease of yOts ft.) the effect of a change in length of one bar only. The correction diagram must also be drawn for this case and is shown in the figure. 180. Camber Defined. A structure is said to be cambered when so constructed that it will not assume its theoretical form until fully loaded. For beams and short-span girders no provision for camber need be made. Girders of long span are often cambered 1)\' being slightly arched. Trusses, except in the case of com- paratively short riveted spans, should always be cambered. This is particularly important in the case of pin bridges as ordinarily constructed with the splices of the top chord dependent for their strength upon the intimate contact of the planed ends of the chord sections. 378 DEFLECTION AND CAMBER Art. 181 181. Rules for Computing Cambers. Short span parallel chord trusses are ordinarily cambered by the following more or less empirical rule: Make the top chord panel lengths longer than those of the bottom chord by i in. for every 10 ft. in length. This process necessitates a corresponding change in the diagonals, but the verticals and bottom chords are unaffected. For long spans or for trusses with curved top chords the cambering should be accomplished by decreasing the length of the tension members and increasing that of the compression members by an amount equal to their change in length under the dead stresses and the live stresses due to full live load with due allowance for pin hole play, using for a basis the geometrical lengths of the bars, as given in the truss diagram. The lengths of the bars thus obtained correspond to the outline of the structure when assembled on false work and not carrying its own weight. When the false work is removed the structure will deflect by an amount equal to the non-elastic deflection plus the deflection due to its own weight. If cambered in this manner the application of the live load, under which the changes in length of the bars were computed, should cause the truss to take the shape of the theoretical diagram. CHAPTER XV CONTINUOUS AND PARTIALLY CONTINUOUS GIRDERS AND SWING BRIDGE REACTIONS 182. Definitions. Continuous girders i are structures sup- ported at more tJian two points, and capable of carrying bending moments and shear at all sections throughout their entire length. Such structures are commonly used for swing bridges and in reinforced concrete buildings, but their employment for ordinary bridges is inadvisable owing to the difficulty in obtaining rigid supports, a slight settlement of one pier changing materially the magnitude of all the reactions. Continuous structures are also subject to both positive and negative live moments over portions of their length and in consequence may require additional material to provide for the resulting reversal of stress; they are also stressed by changes of temperature. Partially continuous girders are structures supported at more than two points, but so built that the continuity is interrupted at one or more sections. Such gird-^ ers are generally trusses in which the continuity is broken by the omission of diagonals in one panel, as was noted in connec- tion with cantilever trusses. 183. Reactions on Continuous Girders. Method of Computa- tion. The reactions on continuous girdere can be accurately determined by the " Theorem of Three Moments," if the moment of inertia and modulus of elasticity of the material are constant throughout, a condition which sometimes exists for beams. If the moment of inertia and the modulus of elasticity of the material are not constant throughout, the reactions cannot be accurately computed until the cross-section areas are known, hence an accurate determination of the stresses for such structures can only be accomplished by a series of approximations, the reactions ' As used in this chapter the word girder is intended to cover all cases of continuous structures and is used indiscriminately for beams, plate girders, and trusses. 379 380 CONTINUOUS GIRDERS Art. 184 first being approximately determined, the stresses and areas com- puted, and the computations revised to correspond to the new areas, the process being repeated as often as is necessary to obtain sufBciently precise results. A common custom, however, for continuous girders is to design the structure on the assumption that the moment of inertia is constant throughout, the " Three- moment Equation," derived from the differential equation of the elastic line being used to determine the reactions.^ 184. Derivation of the " Three- moment Equation." Let the girder shown in Fig. 301 have n spans. There will then be Span „ Span 1 Span 2 Span 3 n-l Span n Ri Ra Ra R4 Rn R(rH-l) Fig. 301. (w + 2) unknown reactions, all the supports but one being on rollers, and hence (n — 1) equations must be obtained from other conditions than those of statics. These equations may be deduced (Py M from the differential equation of the elastic curve, -^ = -p^, by the method which follows, each of the n — l resulting equations connecting the moments at three adjoining supports. ■ Let Fig. 302 represent a portion of a continuous girder with M,i Pi Ml, Ps -fci-ti — H 'i^ ^-^-^"- H -tir- FiG. 302. a constant moment of inertia and modulus of elasticity, the entire structure having n spans, and the section imder con- sideration including any portion of the girder supported upon three adjoining supports. The axis of the unloaded beam is assumed to be straight and the supports level. The assumption that each load acts at a distance, kL, from the adjoining support, is adopted from Merriman's " Mechanics of Materials," and simplifies greatly the resulting equations. ' See also Art. 190. Art. 184 " THREE-MOMENT EQUATION " 381 Let Ma, Mft, and Mc be the moments upon the beam at the three adjoining supports, and let these be assumed as positive, when the moment of the forces on the left of the section is clockwise. Let Sa, Sb_i and Si+i be the shear at infinitesimal distances from the supports, a and 6, and let these be assumed as positive when acting as shown. Let M equal the moment at any section of the girder. Let t^, ife+i, and tb_i equal the tangents at c and at infinitesimal distances on either side of b of the angle between the neutral axis of the deflected girder and its original position. Let /i = normal distances of points a, b and c above some assumed axis parallel to abc and below it. For the portion of the girder between b and c, the moment at a distance, x, from b is given by the following equations : M=EI—=Mb + Sb+iX (for portion of girder between b and d) . (30) M=EI^=Mb+Sb+iX- P2(x-k2L2) (for portion of girder between d and c). (31) From (30) by integration we obtain EI^=M.x+^^ and El'^=MbX+^^^^ + C,EI, (32) dx i Ely=^^^+^^^.+CiEIx + C2EL . . (33) When x = 0, dy d ■ hence ^ = 4,1 and y = h; .'. Ci = fe+i and C2 = h, dx EIy = ^ + ^^^ + th+,EIx + EIh. . . . (34) From (31) by integration we obtain El^=MbX + ^^^-^+P2k2L2X + C,, . . (35) dx I z ^Z,=^+^-^^^^^ + C3.-^C.. . (36) 382 CONTINUOUS GIRDERS Art. 184 When x==k2L2, EI-^ in (35) = corresponding value in (32), .-. Cs = h+iEI-^{k2L2)^ (37) When .r = L2, y = h, +~L2^-^k2L2^-t,,+lEIL2+^k^L2^. Hence EIy = 'Y/i-o--~L2^) +^^-'(x3- L^S) + --^{k2L2x){x~k2L2) +tb+iEI{x—L2) + EIh + ^k2L2^{k2-l) (38) The value of y given by (34) equals its value as given by (38) when x^k-iLi] equating these values gives = -^12^-^12^ - ^L2^ {k2^ - 1) - fe+i£;/L2 Ar^k2L2^ (^2 ^ 1) , hence But h+^EI = ^^ (1 - fca^ + 3/C22 -3^2) - "2-^2 - '^^L2^. M, + St+iL2~P2L2{l-k2) = M, .... (39) .-. by substituting for ^ Spa|n 3 S Span 4 ^ t t-fi^ t^5^^ t t I I ,' I k 4 panels ® 10 sH I I Fig. 303. Solving the three equations thus derived for the three unknowns, ^[2, M3, and Mi, gi^'es the following values: .1/2=- 5,250 ft.-lbs.; .1/,= -12,600 ft.-lbs.; M,= - 1,537 ft.-lbs. ButM2 = i?iX]0, .'. 72i=— 525 1bs. (arting down) , and .1/4=7^5X10, .■. 7^5= —154 lbs. (acting down). Moreover, .l/3 = 20iSi+107?2-40,000 = 207?5 + 1072.-25,000, hence 107?2 =40,000 -12,600 + 10,500 = 37,900; .-. 7;2 = ( + )3790 (acting up), and 107?4 = 25,000 -12,600+ 3,080 = 15,480; .". 7?, = (+)1550 (acting up). Application of I,V = gives 7?3 = (+) 10340 (acting up). In a similar manner the reactions for any number of spans, whether cjual or unequal in length, and for any loading may be readily computed. 386 CONTINUOUS GIRDERS Art. 18(1 186. Reactions, Shears, and Moments for Common Cases of Continuous Girders. In order to simplify the determination of reactions, shears and moments for certain common cases of continuous girders, the diagrams of Figs. 304 to 312 inclusive have been prepared. Inspection of these diagrams shows that for a continuous girder of either two or three equal spans loaded with a uniform live load, w, per foot the maximum live moment PCl-fc)- ■k''} Curve of momoiits for position of load giving maximum positive moment Fig. 304. — Continuous Girder. Reactions and Moments for a Single Concentrated Load. occurs at a support and is negative, its value equalling, for the two-span girder, that of the positive live moment on an end- supported span, and being slightly less than this for the three- span girder. The maximum positive moment equals ^wL^ for both cases, or about three-quarters of the value it would have for an end-supported span. 187. Partially Continuous Girders. Method of Solution. The only case of a partially continuous girder which will be treated Art. 187 CONTINUOUS GIRDERS 387 Curve of moments for uniform load=i« per ft.over entire structure. This loading gives maximum negative moment. Tr.itD^ Curve of moments for uniform load=t(' per ft. on one span only, .This loading gives maximum positive moment. Fig. S05.-Cvnti>iuous Glrdc: Curves of Moment for Uniform Load. 388 CONTINUOUS GIRDERS Abt. 187 ivL Curve of shears for uniform load =w per ft. on one span only- Curve of shears for uniform load = i« per ft. over both spans. This loaillnir gives maximum shear. Fig. 306. — Continuous Girder. Curves of Shears for Uniform Load. Art. 187 CONTINUOUS GIRDERS 389 y^ggj g»g^?g=s=^^^^s^r^:;^^Si^j^^gv % 15 ' I 15 15 15 Curve of moments for position of load] giving maximum positive moment. Curve of momenjts for position of load] negative moment. p(i-feii- iP(k-r ) 15 givmg maximum u""^'-^ p(k-ry 15 Curve of shears for load on side span. Maximum shear would eijual P with load at either end. Fig. 307. — Continuous Girder. Curve of Shears and Moments. Single Concentrated Load. 3S0 CONTINUOUS GIRDERS P J< kL Art. 187 ? Span 1 L Span 2 L t Curve of moments for position of load giving maximum negative moment in middle span. Curve of momentS-for position of load giving maximum positive moment In middle span. Curve of shears for concentrated load in middle span. Maximum shear would equal P with load at either end of middle span. FiQ. 308. — Continuous Girder. Curves of Moments and Shears. Single Concentrated Load. Art. 187 CONTINUOUS GIRDERS 391 Curve of moments for live load =i(; pei-_ft. on spans 1 and 3 only. iThls loading gives maximum positive moment on spans 1 and 3^ Curve of moments for live load=T(J per ft. over entire girder. Fig. 309. — Continuous Girder. Curves of Moments for Uniform Load. 392 CONTINUOUS GIRDERS Art. 187 Curve of moments for live load =iy per ft. on spans 1 and 2, TliiB Joading gives maximum negative moment on structure. rowL Curve of moments for live load=iy per ft. on span 2 only. Tills loading:ffIves maximum. positive moment on span S. Fig. 310. — Continouus Girder. Curves of Moments for Uniform Loads. Aht. 187 CONTINUOUS GIRDERS 393 Curve of BhcHiia for uulfoi:ui load=u) per ft on opans 1 and 3 only. Curve of shears ior unifoian load =w per ft. over entire Etructuue- Fig. Sll.—Continuovs Girder, Curves of Shears for Uniform Load. 394 CONTINUOUS GIRDERS Art. 187 Curvo oCBboara for unifDrniJoad=M) per ft. on Epan 2 only. Curvo of shears for uniform Ioad=l[) per ft. on spans 1 and 2 only. Ibis loading gives maximum sliear in middle and end spans. Fig. 312. — Continuous Girder. Curves of Shears for Uniform Load. Art. 188 THEOREM OF LEAST WORK 395 here is that of a truss supported at four points and without diagonals in the centre panel, a type of structure frequently used in swing bridges. Such a truss is illustrated by Fig. 313. Evidently the three-moment equation is inapplicable to this truss, since its development is based upon the transmission of shear across the centre span, an assumption which is inconsistent with the conditions for a partially continuous girder. Equations for the reactions upon a partially continuous girder, assuming E and I to be constant, as was done for the continuous girder, may Fig. 313. however, be developed bj- the theorem of least work, which will now be explained. 188. Theorem of Least Work. This useful theorem was first brought to the general attention of engineers by Castigliano in his book entitled " Theorie de I'Equilibre des Systemes Elastique," published in 1879. It may be briefly stated as follows : The internal work done in a structure by the application of ex- ternal forces will be the least possible consistent with equilibrium. This theorem may be used to determine reactions or bar stresses in any indeterminate structure, though its application is frequently more laborious than other methods. To apply it, it is necessary to express the internal work in terms of the unlinowns remain- ing after the application of the equations of statics; the first derivative of the work with respect to each unknown may then be put equal to zero and the resulting equation solved. The following example illustrates clearly the application of the theorem to a simple case : Rih — Uniform load = W per ft. 4^ Problem. Determine by the theorem of least work the reac- tions on the beam shown in Fig. 314. Solution. For this case there are four unknown reactions and three Moment of inertia of beain = /. Modulus of elasticity = £?. Fig. 314. 396 CONTINUOUS GIRDERS Aet. 189 statical equations, hence only one unknown need be obtained by the theorem of least work. Let this unknown be taken as Ri. By statics, Rs^Ri, R2=2wl-2Ri. The internal work must now be expressed in terms of Ri. The equation for internal work in a beam is /j\Pdx "p. ■ (See treatises on Mechanics.) ! case mider consideration the total be double that in the left span, hence TF = j For the case mider consideration the total work in the beam will M'dx EI But M=R,x-'-^, 2 hence I 2 + 4 20 T^ . . , dw b or a mmimum value — must equal zero, hence 2R^L' wL' ^ ^ ^, 3wL _ ._.o and R,^^. 189. Determination of Reactions on a Partially Continuous Girder. For a truss similar to that shown in Fig. 3 13 the method p —h Li >\ T Rl R2=P-Ri R8 =RrLi4-PLl(fc-l) Ri= RiLi+PLi(fc-l) 1-3 La Fig. 315. of the previous article may also be applied. For this case there are five unknown reactions. All of these, however, can be expressed in terms of Ri by applying the equation of statics accompanied by the equation of condition, viz., that the shear in the centre panel equals zero. The resulting values of the reactions will be found to be as shown in Fig. 315. The internal work in this girder is given by the following Art. 190 CENTRE-SUPPORTED SWING BRIDGES 397 equations, it being assumed that E and / are constant throughout: '2 Determining the value of ^ from above equation and equating to zero gives the following value for Ri : Ri=P{l-k)^ 1 For the special case usually found in swing bridges Li = Lg = tiL^ where n equals the number of panels in each arm of the truss^ For this case, therefore, we may substitute L for Li and L3 and - for L2 giving the following expressions for the reactions, the signs indicating the actual direction of the reactions as compared with the directions assumed in Fig. 315. ^ ' 4n + 6 ' R^^Rk^P^^^l 4n + 6 ' „ Pn{k-k3) P _ Pnjk-k^) 190. Types of Girders for Centre-Supported Swing Bridges. Swing bridges are usually constructed with continuous or partially continuous girders, such structures being more economical to operate than the statically determinate types occasionally built. The girders used may be divided into three general classes : a. Continuous girders supported at three points when closed ; b. Continuous girders supported at four points when closed; c. Partially continuous girders supported at four points when closed. 398 CONTINUOUS GIRDERS Art. 191 The points of support of the main girders are usually upon cross-girders. In the class a type one such girder is required, the cross girder itself being supported on a pivot resting on the central pier. In classes 6 and c two cross-girders a panel distance apart are needed, these girders being supported either directly upon a circular girder or drum, or upon other girders so arranged as to distribute the reactions more imiformly over the drum than could be done by the two main girders. The circular girder or drum is itself supported upon a ring of conical rollers running upon a track supported on the central pier. All such structures are statically determinate when opened, but are statically indeterminate when closed. If the girders correspond to the conditions assumed in develop- ing the three-moment equation, that is, if their moments of inertia and moduli of elasticity are constant throughout their length, except as noted in the case of partially continuous girders, then the reactions may be accurately computed by the methods of this chapter. The moment of inertia is, however, seldom constant, since cover plates are generally used on plate girders, while trusses are made deeper over the centre support than at the ends. The variation from the assumed conditions is seldom large and the reactions as computed by these methods are usually sufficiently accurate to permit of safe designs, and in all cases to give closely approximate designs to which more accurate methods, based upon the deflection of the structure, may be applied if desired, and the design corrected i. 191. Influence of End Supports upon Swing Bridge Reactions. The continuous and partially continuous girders hitherto con- sidered have been assumed to be level and supported on level supports. It is evident that if this condition exists for a swing bridge when closed, when the bridge is opened the trusses will deflect at the ends below the level of the centre supports and hence when the bridge is again closed will have to be raised to reach their original level, the force required to accomplish this equalling the dead reactions which would exist when the truss is closed. If the ends be not raised there will ' See Journal Franklin Institute, 1883, article by George F. Swain, entitled "On the Application of the Principle of Virtual Velocities to the Determination of the Deflection and Stresses of Frames." Art. 192 TABLES OF REACTIONS 399 be no dead reactions at the ends when closed, the dead stresses being the same as when the bridge is. open. If this latter con- dition exists, however, a partial live loading which would pro- duce negative reactions at the ends would cause the ends to rise unless latched down, a serious objection, especially for a double- track railroad bridge. It is common, therefore, for railroad swing bridges to raise the ends when the bridge is closed, this being accomplished by means of levers, toggle joints, or hy- draulic jacks. In case the ends are raised sufficiently, the reac- tions for the closed bridge for both live and dead loads will be given by the formulas already deduced. If the ends are latched down and not raised, the end dead reactions will be zero and the live reactions will be given by the formulas of this chapter. If the ends are neither latched down nor raised, the end dead reactions will again be zero, and the live reactions will be those given by the formulas provided none of the end reactions be negative, and that the negative live reaction at the centre pier does not exceed the positive dead reaction at that point. If the latter conditions are not fulfilled the structure becomes a girder supported at two points if of the class a type, and at three points if of types h and c, and the formiilas are inapplicable. It should be added that the partially continuous truss has an advantage over the continuous girder upon four points of support, in having the negative reaction due to live loads occur at one end where it may be properly taken care of instead of at the centre support, where it might cause the drum to lift from the rollers. 192. Tables of Reactions for Continuous and Partially Con- tinuous Girders Used for Swing Bridges. The actual stresses in girders used for swing bridges may be computed by the methods already given for simple girders and trusses, once the reactions are determined, and hence no example of the computation of such stresses will be given. It may be noted, however, that influence tables or influence lines may be employed to advantage. In order to facilitate the computation of the reactions the following tables have been prepared for girders with equal panels. These will be found sufficient for many structures. For bridges not covered by these tables the formulas previously developed should be employed. 400 CONTINUOUS GIRDERS Art. 192 REACTIONS FOR UNIT LOAD— GIRDER CONTINUOUS OVER TWO EQUAL SPANS Moment of Inertia and Modulus of Elasticity Assumed to be Constant Positive signs indicate upward reactions. Formulas used in deriving these values are determined by the three- moment equation and are as follows: K.-(l-«-(^). B , = 1-(B, + R,). k + + R, k R, + R, + Rs 1/10 2/10 3/10 4/10 5/10 6/10 n= 0.8752 0.7520 0.6318 . 5160 0.4062 0.3040 0.2108 . 1280 0.0572 10 0.1495 . 2960 0.4365 0.5680 0.6875 0.7920 0.8785 0.9440 0.9855 0.0247 0.0480 0.0683 0.0840 0.0937 0.0960 0.0893 0.0720 0.0427 1/7 2/7 3/7 4/7 5/7 6/7 Total n= 0.8222 0.6487 0.4840 0.3324 . 1982 0.0860 = 7 0.2128 0.4169 0.6035 0.7638 0.8893 . 9709 0.0350 0.0656 0.0875 0.0962 0.0875 0.0569 7/10 8/10 2.5715 3.8572 0.4287 9/10 1/6 2/6 3/6 4/6 5/6 Total n— 6 Total 3.8812 5.7375 0.6187 0.7928 . 5926 0.4062 0.2407 0.1030 0.2477 0.4815 0.6875 0.8519 0.9606 0.0405 0.0741 0.0937 0.0926 0.0636 1/9 2/9 n= 0.8615 0.7250 . 5926 0.4664 0.3484 0.2407 . 1454 0.0645 -9 . 1659 0.3278 0.4815 0.6228 0.7476 0.8519 0.9314 . 9821 0.0274 0.0528 0.0741 0.0892 0.0960 0.0926 0.0768 0.0466 3/9 4/9 2.1353 3 . 2292 0.3645 5/9 6/9 7/9 8/9 1/5 2/5 3/5 4/5 Total n-- 0.7520 0.5160 0.3040 . 1280 = 5 0.2960 0.5680 0.7920 0.9440 0.0480 0.0840 0.0960 Total 3.4445 5.1110 0.5555 0.0720 1.7000 2.6000 0.0305 0.0586 0.0806 0.0937 0.0952 0.0S20 0.0513 0.3000 1/8 2/8 3/8 4/8 5/8 6/8 7/8 0.8442 0.6914 0.5444 0.4062 0.2798 0.1680 0.0737 0.1866 0.3672 0.5362 0.6875 0.8154 0.9140 0.9776 1/4 2/4 3/4 Total n 0.6914 0.4062 0.1680 = 4 0.3672 0.6875 0.9140 0.0586 0.0937 0.0820 1.2656 1.9687 0.2343 1/3 2/3 Total ■n 5 Total 3.0077 4.4845 0.4919 0.5926 0.2407 0.4815 0.8519 0.0741 0.0926 0.8333 1.3334 0.1667 Art. 192 TABLES OF REACTIONS 401 REACTIONS FOR UNIT LOADS— CONTINUOUS GIRDER WITH FOUR SUPPORTS AND EQUAL SIDE SPANS Moment of Inertia and Modulus of Elasticity Assumed to be Constant. Centre span = - side span. Positive signs indicate upward reactions. Formulas used in deriving these values are determined by the three- moment equation and are as follows: (fc-fc=')(2n)(n + l) in^ + Sn + S {k-k'){n)(2n' + 5n + 2) in^ + 8n + 3 «,= (.k-h')n R,= — {k-k^)n(2n' + 3n + l) 4re2 + 8n + 3 «i( + ) R,{ + ) B,{-) «.( + ) «i( + ) R2{ + ) «3(-) R,{ + ) 1/10 2/10 3/10 4/10 5/10 6/10 7/10 8/10 9/10 Total 0.8549 0.7125 0.5757 0.4470 0.3292 0.2251 . 1374 0.0688 0.0221 n=10 0.6166 1.2018 1.7243 2.1531 2.4567 2.6036 2.5628 2.3027 1.7922 0.4735 0.9183 1.3057 1.6070 1.7936 1.8367 1.7075 1.3774 0.8179 3 . 3727 17 . 4138 1 1 . 8376 . 0512 0021 0040 0056 0069 0078 0079 0074 0060 0035 1/7 2/7 3/7 4/7 5/7 6/7 Total 0.7956 0.5991 0.4177 0.2596 0.1320 0.0430 2.2470 0.6615 1.2581 1.7250 1.9975 2.0107 1.7001 0.4609 0.8643 1 . 1524 1.2678 1.1524 0.0038 0.0072 0.0096 0.0106 0.0096 0.7492 0.0063 i 9.3529 5.64700.0471 1/9 2/9 3/9 4/9 5/9 6/9 7/9 8/9 Total 0.8394 0.6825 0.5330 0.3946 0.2712 . 1662 0.0836 0.0269 2.9974 n=9 0.6285 1.2181 1.7301 2.1257 2.3660 2.4126 2.2261 1.7683 14.4754 4703 9054 2697 5284 6459 5871 3166 7995 9.5229 .0025 .0048 .0067 .0081 .0087 ,0084 .0069 .0042 0.0503 1/8 2/8 3/8 4/8 5/8 6/8 7/8 Total 0.8202 . 6455 0.4813 0.3328 0.2052 . 1037 0.0336 n = 8 0.6433 1.2368 1.7318 2.0790 2.2287 2.1316 1.7387 2.6223 11.7899 0.4667 0.8882 1.2212 1.4210 1.4433 1.2435 0.7773 0.0031 0.0058 0.0080 0.0093 0.0094 0.0081 0.0051 7.4612 0.0488 1/6 2/6 3/6 4/6 5/6 Total 0.7635 0.5391 0.3385 0.1738 0.0570 1.8719 n = 6 0.6852 1.2814 1.7000 1.8518 1.6480 0.4537 0.8296 1.0499 1.0369 0.7128 0.0050 0.0091 0.0115 0.0114 0.0078 1664 4.0829 0.0448 1/5 2/5 3/5 4/5 0.7194 0.4590 0.2389 0.0792 n=5 0.7169 1.3046 1.6338 1.5754 0.4430 0.7754 0.8861 0.6646 Total 1.4965 5.2307 2.76910.0419 0.0067 0.0117 0.0134 0.0101 1/4 2/4 3/4 0.6553 0.3485 0.1174 n = 4 0.7612 1.3182 1.4660 0.4262 0.6819 0.5966 Total 1.1212 3.5454 1.7047 0.0379 0.0095 0.0151 0.0133 1/3 2/3 0.5538 0.1922 n = 3 0.8272 1.2840 0.3951 0.4938 Total 0.7460 2.1112 0.8889 0.0317 0.0141 . 1076 402 CONTINUOUS GIRDERS Art. 192 REACTIONS FOR UNIT LOADS — PARTIALLY CONTINUOUS GIRDER WITH FOUR SUPPORTS AND EQUAL SIDE SPANS Shear in centre panel =0 Moment of Inertia and Modulus of Elasticity Assumed to be Constant Centre span = — side span. Positive signs indicate upward reactions. Formulas used in deriving these values are derived by the method of least work and are as follows: fli = P(l-fc) R Pn{ki-k^') 4n + 6 R^ = Pki Pn{ki-ki') 4™ + 6 4n + 6 R, + Ri + R, R, R, Ri + R3 + R, 1/10 2/10 3/10 4/10 5/10 6/10 7/10 8/10 9/10 Total 0.8785 0.7583 0.6406 . 5269 0.4185 0.3165 . 2224 0.1374 0.0628 i. = 10 1215 2417 3593 4730 5815 6835 7776 8626 9372 0215 0417 0593 0730 0815 0835 0776 0626 0372 3 . 9619 5 . 0379 . 5379 . 5379 .0215 .0417 .0593 .0730 ,0815 .0835 .0776 .0626 .0372 1/9 2/9 3/9 4/9 5/9 6/9 7/9 S/9 Total 0.8654 0.7325 . 6032 0.4791 0.3621 0.2540 0.1564 0.0711 3 . 5238 ,1346 2675 3968 5209 6379 7460 8436 ,9289 4.4762 0235 0453 0635 0764 0823 0794 0658 0400 0.4762 .0235 .0453 .0635 .0764 .0823 .0794 .0658 .0400 0.4762 1/S 2/8 3/8 4/8 5/8 6/8 7/8 . 8491 0.7007 0.5571 0.4210 0.2948 0.1809 0.0818 0.1509 . 2993 0.4428 . 5789 0.7052 0.8191 0.9182 0259 0493 0678 0789 0802 0691 0432 .0259 .0493 .0678 .0789 .0802 .0691 .0432 1/7 2/7 3/7 4/7 5/7 6/7 Total 1/6 2/6 3/6 4/6 5/6 Total 1/5 2/5 3/5 4/5 Total 1/4 2/4 3/4 Total 1/3 2/3 0.8283 0.6603 0.4994 0.3493 0.2137 0.0960 n = 7 0.1717 0.3397 0.5006 0.6507 0.7863 0.9040 2.6470 3.3530 0288 0540 0720 0792 0720 0468 0.3528 .0288 .0540 .0720 .0792 .0720 .0463 0.3528 0.8009 0.6074 0.4250 0.2592 0.1157 2 . 2082 n = 6 0.1991 0.3926 0.5750 0.7407 0.8842 0.0324 0.0593 0.0750 0.0741 0.0509 0.0324 0.0593 0.0750 0.0741 0.0509 2.79160.29170.2917 0.7631 0.5354 . 3262 . 1446 n = 5 0.2369 0.4646 0.6738 0.8554 1.7693 0.0369 0.0646 0.0738 0.0554 2.2307 0.2307 0.2307 0.0369 0.0646 0.073S 0.0554 0.7074 0.4318 . 1903 n = 4 0.2926 0.5682 0.8097 0.0426 0.0682 0.0597 0.0426 0.0682 0.0597 1.3295 1.6705 0.1705 0.1705 Total 3.0854 3.91440.41440.4144 Totil 0.8889 1.11110.11110.1111 n = 3 . 6173 . 3827 . 0494 . 0494 0.2716 0.7284 0.0617 0.0617 INDEX PAGB Adjustable bars, reasons for use in cantilever bridges 261 Anchorage, for cantilever bridges 261 Anchor arms, for cantilever bridges 259 Arch, characteristics of 269 relation between shape of arch and equilibrium polygon 349 Arches, hinges in 270, 271 influence table for a three-hinged 279-283 masonry, described and illustrated 269, 270, 271 metal, described and illustrated 270, 271 one-hinged 270 parabolic, characteristics of 276, 277 reactions for three-hinged 272 spandrel braced, described and illustrated 271 three-hinged, described and illustrated 270, 271 maximum stresses in 278-284 parabolic 276, 277 two-hinged, described and illustrated 270 Arch rib, distribution of stress over cross-section 273, 274, 275 Baltimore, truss, described and illustrated 169 influence of secondary diagonals 203-208 Batten plates 301-307 Beams, composite 106 commercial sizes of timber 102 cost of steel 103 examples in design 104, 105, 106 fixed-ended 53 formulas for 100, 101, 102 lateral rigidity of 107 stiffness of 107 Beaver Bridge 258 Box girder, described and illustrated 108, 109 Bridges, defined 1 span of, illustrated 2 403 404 INDEX PAGE Bridges, systems of loading compared 170 See also Deck, Half-through, I-beam, Through and Skew bridges. Bridge pin, computation for a bottom chord pin 326-330 curve of bending moment for 326, 328, 329 effect of change in arrangement of members upon 330 nuts 331 packing rings for 331 Bridge pins, arrangement of members on 31S computation of a bottom-chord pin 327-331 computation of a top-chord pin 323-327 computation of maximum moment and shear 323 described 318 minimum size 321 stresses causing maximum moments and shears 322 Camber, defined 377 rules for computation 378 Cantilever arm, illustrated 259 Cantilever bridges, adjustable bars in 261 anchorages for 261 anchor and cantilever arms, illustrated 259 described 258 suspended span, illustrated ; 259 use of falsework in erection 258 Cantilever trusses, equations of condition for 259 influence lines for moments upon 267 influence lines for shears upon 267 reactions upon 261 Cast-iron columns, design of 311-314 formulas for 292, 293 tests upon 292 Centrifugal force 28 Channels, distance apart when used for columns 297, 298 Columns, cast-iron 292, 293 ' concrete 294 eccentrically loaded 310, 311 ■ effect of end conditions 286 287 effect upon of combined flexure and thrust 309, 310 example in design of steel 298-301 fixed-ended, described 286, 287 flat-ended, described 286 formulas for 287-291 general considerations 285 286 general dimensions and limiting conditions 295, 296 integrity of cross-section 286 lattice bars and batten plates 301-307 methods of design 296, 297 298 INDEX 405 PAGE Columns, methods of failure 285 pin-ended, described 286, 287 Columns, rivet-pitch in 306, 307 round-ended, described 286 steel, tests of 291, 292 typical sections 294, 295 timber 294 variations from ideal conditions 286 Condition, equations of, defined 259 Connection angles 141, 142 Continuous girders, defined 379 method of computation of reactions 379 reactions, moments and shears for common cases. 386-392 tables of reactions 400-402 Cooper's standard locomotive loading, described 16, 17 equililjrium polygon for Ego 346 moment diagram for 97 used in computations, 189-196, 210-224 Counters, described 166 in parabolic trusses 198 in trusses with secondary web systems 210, 211 Critical section, determination of maximum moment and shear at ... . 88-92 Dead load, defined 4 Deck bridge, described and illustrated 2, 4 Deflection, computation of for trusses, illustrated 362-364 elastic defined 355 graphical method, illustrated 368-377 method of rotation for trusses 356-358 method of work for trusses 358-362 non-elastic defined 355 of beams and girders 364-368 reasons for computing 355 trigonometrical method for trusses 355, 356 Eccentric forces, effect upon building columns 310 effect upon fibre stresses 274, 307, 308 Electric cars, allowance for impact 30 weight of 17 End-hanger, illustrated 2 Equilibrium polygon, characteristics of 341 described 337, 338, 339 for Cooper's Ego loading 346 relation between equilibriiun polygon and arches. . 349 through one point 349 several points 349 three points 352, 353, 354 406 INDEX PAGE Equilibrium polygon, through two points 350-354 Euler formula 290, 291 (^FactOT_ofsafetJ^ _32, Falsework, use of in erecting cantilevers 258 Fillers for plate-girders 141, 142 Fixed-ended columns described 286, 287 Flat-ended columns described 286, 287 Flexure and thrust on columns, effect of when combined 309, 310 Floor-beams, computations of moments and shears upon 92-96 defined 1 effect of, on shears and moments in girders and trusses 53 illustrated 3 Force polygon, described 337, 338, 339 Forth Bridge 258 Friction, coeiBcient of, for use in bridge computations 28 Funicular polygon, defined 339 Girders, see Box, Continuous, Partially Continuous and Plate Girders. Gordon formula for columns _. 287-290 Graphical and analytical methods for solving problems in statics com- pared 337 Graphical Method of Moments 345, 346 Graphical Method of Shear 346, 347, 348, 349 Half-through bridge, defined and illustrated 3 Harvard Bridge 258 Hinges, use of in arches 270, 271 cantilever trusses 259, 260 Howe truss, described 167 I-beam bridge, illustrated 2 Impact on bridges and buildmgs 28, 29, 30 Inclined end-post, illustrated 2 Index stresses, defined 171, 172, 173 corrected for non-parallel diagonals 184 examples in computation of, 174, 178, 179, 181, 184, 209, 228, 236, 243 Influence lines, defined 61, 62 for cantilever trusses 262, 266, 267, 268 illustrated 63, 64 properties of 65 Influence table, defined 61, 62 for a three-hinged parabolic arch 279-283 illustrated 62, 279-281 Inner forces, defined 4 INDEX 407 PAQE Joints, analytical method of, applied 154, 155, 156 described 151 graphical metlioj of, applied 158, 159 described 157 Joints for riveted trusses, example of design 335, 336 for riveted trusses, rules for designing 333 Kneebraces for plate-girder bridges 3 Lattice bars 301-307 Lateral bracing 240, 241, 242, 243 Least work, theorem of 395, 396 Live loads, defined 4 values for bridges and buildings 15-19 Loads, point of application of 38, 39 Locomotive excess, defined 170 example in computation of 174 Locomotives, wheel loads «nd spacing, 16, 17 Masonry arches, described and illustrated 269, 270, 271 Metal arches, described and illustrated 270, 271 Metropolitan tower, column section 294 Moment, absolute maximum, computation of 90-91 common cases 45 concentrated load system, girder with floor beams 87, 88 without floor beams ... 79, 80, 81 defined 44 diagram, defined 96 diagram for Cooper's E40 loading 97 distributed load 47 effect of hinges upon 259, 260 formulas for maximum on end-supported beams antl girders, 67, 68, 69 location of section of maximum 51 method of computation of , 44 for bridge pins 323 theorem for computations of 51, 52 uniformly varying load 48, 49, 50 Moments, curve of, defined 45 curves of, illustrated 46 curves of, typical 54-61 method of 161, 162 graphical method, general 343, 344 graphical method with a concentrated load system 345, 346 Neutral point, defined . . '. 66 New York Building Laws, allowable live-loads 19 Non-parallel chord truss, computation of maximum stresses 182-186 408 INDEX PAGE Non -parallel chord truss, modification of methods for parallel cliurd trusses 182 Nuts, for bridge pins 332 One-hinged arch, described and illustrated 270 Outer forces, defined 4 Packing rings, for bridge pins 332 Panel, bridge panel defined 3 Parabolic arch, characteristics of 276, 277 chord trusses, characteristics of 196, 197 computations of maximum stresses in 196-200 counters in 198 Partially continuous girders, defined 379 determination of reactions 396, 397 method of solution 386, 395 table of reactions 402 Pettit truss, computations of maximum stresses in 208-224 described and illustrated 169 Pin-ended columns, described 286, 287 Pin plate, rivets in 332 Pins, arrangement of members upon 318, 320 Pins, computation of maximum moment and shear upon 323 described 318 minimum size of 321 stresses causing maximum moments and shears 322 Plate girder, allowable length 108 bridge, illustrated 3 cantilever, hinge for, illustrated 260 defined 108 deflection of, see Deflection. depth of 108 example of determination of cross-section 122 illustrated 108 flange plates, illustrated 138 location of ends 139, 140 flanges, allowance for rivet holes 120, 121 approximation in formula for 113-118 formula 110 theory 110-113 width 140 rivets. See Rivets. stiffeners, computation of spacing 136 derivation of formulas 134, 135 illustrated 137 to carry concentrated loads 138 web, approximation in theory 118 INDEX 409 PAGE Plate girder, web, direct stress in 131-134 minimum thickness 110 theory 109, 110 Pole, of force polygon, defined 339 Portals, approximate solution for stresses 244-249 illustrated 3 miscellaneous 249 Poughkeepsie Bridge 258 Pratt truss, computation of maximum stresses in 173-177 described 168 Queensboro Bridge 259, 294 Rankine column formula 287-290 Rays, of force polygon, defined 339 Reactions, cantilever trusses 261 , 262, 265, 266 computations 36, 37 conventional symbols 38 graphically 341, 342 point of application of 38, 39 solution of problems 39-43 three-hinged arches 272 Red Rock Bridge 258 Riveted joints, investigation of rivets m 122-126 Rivet pitch, defined 126, 129 example in computation for plate-girder flanges 128-131 ordinary method of computation for plate-girder flanges 126, 127 precise method of computation for plate-girder flanges 128 Riveted trusses, design of joints 333-336 Rivets, allowable unit stresses 31 conventional rules for location and spacing in girders 123, 127 in columns 307 determination of allowable stress 123 in pin plate 332 methods of failure 122, 123 resistance to torsion 125 single and double shear defined 123 Round-ended columns 286, 287 Shear, approximate method for maximum 69, 70 common cases 45 concentrated load system, girder with floor beams 81-87 without floor beams 75-79 curves of, defined 45 illustrated 46 typical 54-61 defined 44 410 INDEX PAGE Shear, distributed load 47 formulas for maximum on end-supported beam 67, 68, 69 graphical method of 346, 347, 348 method of computation 44 for bridge pins 323 position of concentrated loads for maximum at a section 66 imiformly varying load 48, 49, 50 Shears, method of, applied 164, 165 described 163 Skew bridges, defined and illustrated 239, 240 Spandrel-braced arches, computations for 278-284 described and illustrated 271 Splices, of flange angles 146, 147, 148 plates 148 web plates 143-146 Statical determination by inserting hinges in arches 270, 271 by omitting chord bars in cantilever trusses. 261, 263 diagonals in cantilever trusses 263 Statically determined structures defined 1 Statically undetermined structures " 1 Statics, laws and equations 34 Stiffeners. See Plate-girders. Stresses, allowable for masonry 32 Stresses, allowable for steel 31 timber 31 Stringers, defined 1 illustrated 2-3 flange rivets 127 Strings, of equilibrium polygon, defined 339 Structures, defined 1 types of, for long-span bridges 258 unstable, defined 35 Substructure, defined 1 Superstructure, defined 1 Suspended span, cantilever bridge.s 259 Suspension bridge 258 Sway bracing 241, 242 Swing bridges, influence of end supports :V.)S, 399 tables of reactions 400-402 types of girders for centre supported bridges 397, 398 Tension members, design of, iron and steel 314-317 Tests, cast-iron columns ' 292 steel columns 291 Theorem of least work 395, 396 three-moments 380-385 Three-hinged arches, computations for maximum stresses 278-284 INDEX 411 PAGE Three-hinged arches, described and illustrated 270, 271 parabolic 276, 277 Three-moment equation 380-385 Through bridges, defined and illustrated 2, 3 when used 4 Timber columns 294 Transverse bents for buildings 249-252 Truss, pin-connected, illustrated 3 Trusses, classified and defined 150 deflection of. See Deflection. theory 151 types of bridge 167-170 roof 170 with secondary web systems, described 203 Truss stresses, ambiguous cases 159, 160 by analytical method of joints 154, 155, 156 graphical method of joints 157-161 method of moments 161, 162 shear 163-165 general rules for determination of 165 influence of secondary web system upon 203-208 methods of computation 151 theory 151 Two-hinged arch, described and illustrated 270 Viaduct towers ^ 252-255 Warren truss, computations of maximum stresses in single system, 178, 179, 180 Warren truss, computation of maximum stresses in subdivided system 180, 181, 182 computation of maximum stresses in double-system. . .227-235 double system described 226 single system described 168 subdivided type described 169 Whipple truss, computation of maximum stresses in 235-239 described 226 Williot diagram, correction for rotation 373-377 defined and illustrated 368-372 rules for construction of 373 Wind bracing. See Lateral and Sway bracing; also Portals. Wooden beams, impact allowance 30 sizes of 102