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 arV18356 
 
 A handy book tor the cs!S"''fi°,!;,,9L^^^^^^ 
 
 3 1924 031 263 902 
 olin.anx 
 
PLATE I. 
 
HANDY BOOK 
 
 FOK THE CALCULATION OF 
 
 STEAINS IN GIKDEES 
 
 AND SIIILAR STRUCTURES, AND THEIR STRENGTH; 
 
 CONSISTIKO OF 
 
 FORMULAE AND CORRESPONDING DIAGRAMS, 
 
 WITH 
 
 NUMEROUS DETAILS FOR PRACTICAL APPLICATION, 
 
 ETC., ETC. 
 
 By WILLIAM HUMBER, ^ssoc. Inst. C.E. 
 
 AUTHOR OP "a FXtACTlCAL TREATISE ON CAST AKD WBOnOHT IBOK 
 
 BRIDGE COHSTEUCTION," "a RECORD OP THE PROGRESS 
 
 OP MODERN EKOIKBEBIKO," 
 
 SIC., ETC 
 
 NEW YORK: 
 
 D. VAN N08TRAND, PUBLISHER, 
 
 No. 23 MuBKAT AND 27 Wabrbn ST8. 
 
 1869. 
 
Pholo-lithographic Beprodnction 
 
 BY THE 
 
 New York Lithograrhing, Engraving and rriuting Co., 
 16 and 18 Fatk Flaee. 
 
PEEFACE. 
 
 Notwithstanding that the subject of Strains has' been ably 
 treated of again and again, it is dif&cult in submitting this 
 little work to the public to avoid the almost stereotyped ex- 
 pression that " the design has been to supply a want long felt 
 in the profession ; " for the numerous volumes which have 
 appeared on the subject have not, principally on account of 
 their elaborate investigations, been calculated to afford that 
 ready assistance which in the ordinary run of office and other 
 work is being continually needed, while on the other hand, most 
 of the general Engineering Pocket-books, not having been able 
 to afford svifficient space to do justice to the subject, have been 
 compelled to leave its treatment incomplete. It is hoped, 
 therefore, that by devoting a small work, in a handy form, 
 entirely to Bridge and Girder Calculations, without giving more 
 than is absolutely necessary for the complete solution of prac- 
 tical problems, both the above obstacles to quick and satisfac- 
 tory manipulation may be overcome. 
 
 One of the chief features of the present work is the exten- 
 sive application of simply constructed diagrams to the. calcu- 
 lation of the strains on bridges and girders, the advantages of 
 which as a system, most ■ undeniably far outweigh its disad- 
 vantages. The parabola (anything but a troublesome figure 
 to draw) and a few right lines are all that are required. 
 
Tliere is, again, a more general application of the Moments 
 of Buptore, and Shearing Forces, to open-webbed girders of all 
 kinds, than has hitherto been attempted. 
 
 It was originally intended to divide the whole work into 
 three sections or chapters, similar to those actually adopted 
 only for the middle portion (pp. 24 to 60), which chapters 
 should correspond with the yanous processes in the design of 
 a bridge, thus making the yery arrangement of the work a 
 general gmde. But it was afterwards deemed advisable,- as 
 win be seen, to place the Moments of Bupture, and Shearing 
 Forces, by themselves at the commencement, as a basis upon 
 which the remainder is principally founded. 
 
 In the following pages will be found, almost necessarily, 
 many omissions, but care has been taken as far as possible 
 to. avoid inaccuracies. It wiU be observed that attention has 
 been paid to the arrangement of the matter in different types, 
 so as to facilitate as far as possible the manipulation of the 
 contents. The work is not advanced with the pretensions of 
 a treatise, as no investigations whatever are given, but merely 
 their results; and these, it is hoped, in an intelligible and 
 practical form, suited to the wants of the Engineer, Architect, 
 Draughtsman, or Builder. 
 
CONTENTS. 
 
 STRAINS IN BEAMS. 
 
 PARAGRAPH PAOK 
 
 1, Inibosuotort.— Stabilit; of a loaded beam . > . ' . 1 
 
 MOMENTS OF RUPTURE, 
 
 2, 3. AbbreTiatioDS 1 
 
 i. Positive and negative moments 2 
 
 5, 6. Diagrams of tbe moments 2 
 
 7. Semi-beam loaded at extremity . . ... . .2 
 
 8. Semi-beam loaded at several points . . . . -2 
 
 9. Semi-beam uniformly loaded .I 
 
 10. Semi-beam uniformly loaded for part of its length onljr . . 3 
 
 11. Semi-beam uniformly loaded, and also with a weight at its 
 
 extremity 3 
 
 12. Beam loaded at the centre 4 
 
 13. Beam loaded at any other point i 
 
 14. Beam loaded with several weights ..... 4 
 
 15. Beam loaded with two equal weights equidistant about the 
 
 centre 5 
 
 16. Beam loaded with four equal weights symmetrical from the 
 
 centre . . . 5 
 
 17. Beam loaded with a concentrated rolling-weight . , . 8 
 
 18. Beam loaded with two concentrated rolUng-weigbts > . 6 
 
 19. Beam uniformly loaded 6 
 
 20, 21. Beams with a load uniformly distributed oyer part of their 
 
 lengths . . ' . . . . ■ . .7 
 
 22. Beam with a distributed rolling load 8 
 
 23. Points of contrary flexure 8 
 
 24. Beam of uniform section, or uniform strength, fixed at both 
 
 ends and loaded at centre . . . . . .9 
 
 26. Beam of uniform section, fixed at both ends, and uniformly 
 
 loaded 9 
 
 26. Beam of uniform strength, fixed at both ends, and uniformly 
 
 loaded • 1" 
 
 27. Beam of uniform section, uniformly loaded, the ends being 
 
 subject to known moments of rupture . - . . .10 
 
 28. Beam of uniforiu section, fixed at one end only, and uni- 
 
 formly loaded . , 11 
 
VI OOSTKHTS. 
 
 PAKAOBIPH PAQB 
 
 29. Beam of uniform strength, fixed at one end only, and uni- 
 
 formly loaded .11 
 
 30. Continuous beam, with uniformly distributed btationary 
 
 load 12 
 
 31. Continnons beams with varying loads 12 
 
 32. Beam continuous over one pier, subject to a stationary, and « 
 
 also a moving load . . . . . . .12 
 
 33. Beam continuous over two or more pierB, subject to a sta- 
 
 tionary, and also a moving load 14 
 
 SHEARING F0RGE3. 
 
 • 
 34,35. Abbreviations and. diagrams 16 
 
 36. General rule for the shearing force 16 
 
 37. Semi-beam loaded in any manner 16 
 
 38. Semi-beam uniformly loaded 17 
 
 39. Semi-beam loaded uniformly, and with a concentrated 
 
 weight 17 
 
 40. Beam loaded at centre 17 
 
 41. Beam loaded at any point , 17 
 
 42. Beam loaded with a concentrated rolling weight . . .17 
 
 43. Diagram of the shearing forces produced by a concentrated 
 
 weight on a beam . . ; 18 
 
 44. Beam uniformly loaded . -■. 18 
 
 45. Beam with distributed moving load 18 
 
 46. General formulae for the shearing forces in continuous beams . 19 
 
 47. Beam fixed at both ends, and uniformly loaded . . .19 
 
 48. Beam of uniform .strength, fixed at one end only, uniformly 
 
 loaded 19 
 
 49. Beam of uniform section, fixed at one end only, uniformly 
 
 loaded . .20 
 
 50. Continuous beams with uniformly distributed stationary 
 
 loads 20 
 
 51. Beam continuous over one pier, subject to a stationary, and 
 
 also a moving load. ....... 20 
 
 52. Beam continuous over two or more piers, subject to a sta- 
 
 tionary, and also a moving load 21 
 
 FLAITSED GIRDERS, ARCHES, AND SUSPENSION BRIDGES. 
 
 53. Processes in the design of abridge 24 
 
 SGCTIOK I. — ^DKIERHIKAIIOir OF THK NAItTSB, FslHOIPAL DlHBK- 
 8I0HB, EIO., OF THB SiKVOIDBB. 
 
 54, 55. Kind of bridge and general cross-seotion . . . .24 
 66 — 59. General proportions 25 
 
 Section U.—OaloviiAiioii of ihb Stbaiih on thb taxioits Pabts. 
 60, 61. Approximate estimation of the weight of the structure , 25 
 
.FLANGED GIRDERS WITH THIN CONTINUOUS WEBS. 
 PABAQRAPa PAOE 
 
 62, Distinct fanctions of the flanges and the web . . .26 
 
 6S— 6{i. Strains in the flanges generaUy 26 
 
 66. Strains in the web generally 27 
 
 Oirden ivith Parallel Straight Flangta. 
 
 67—71. Flanges 27 
 
 72—74. Web 28 
 
 OirderB imth Curved or OUique Flanget. 
 
 75—78. Flanges 28 
 
 79—83. Web 28 
 
 GIRDERS WITH WEBS 01" OPEN BRACING. 
 aXNEBAIi BCLES. 
 
 84—87. Booms 30 
 
 88—95. Web 80 
 
 CaLODLAIIOK by means of the MoUXNTS of BnPTDRB AND Shxabiho 
 FOBOES. 
 
 96. Loads concentrated at apices 81 
 
 Girderi viiih, PwrallA Straight Boona, 
 
 97, 98. Notes 31 
 
 99. Warren girder, loaded on one boom . . . .31 
 
 100, 101. Uounter strains in the web 32 
 
 102. Warren girder, both booms loaded . . . .32 
 
 103. Girder with vertical struts and inclined ties . . .32 
 
 104. Warren girder with scalene bracing, loaded on one boom . S3 
 105,106. Method of applying the diagrams 33 
 
 107. Lattice girder loaded on one boom . . . .34 
 
 108. Lattice semi-girder loaded on one boom . . . .35 
 109, 109a. Lattice, whole or semi, girder loaded equally on both booms 36 
 
 110. Concentrated loads on lattice girders . . . .36 
 
 Oirdera mth Curved or Oblique Boona. 
 
 111. Curred or oblique whole or semi-girder, single-triangular 
 
 web, loaded on one boom 37 
 
 112. Curved or oblique whole or semi-girder, single-triangular 
 
 web, loaded equally on both booms . . . .37 
 113, 113a. Continuous girders 38 
 
 114. Fixing the points of inflection in continuous gurders . 38 
 
 115. Continuons girders with varying depths. , . .38 
 
 CaLOCLATION BT the CoMPOBIItOir AMD UlSOIiUTIOH QF FoBOES. 
 
 116. Method # . 38 
 
 117. Keaction of supports , , 39 
 
 118, 119. Composition and resolution of forces — parallelogram — 
 
 triangle 39 
 
 120. More than one concentrated load on a girder . . .39 
 
Till C0KTEKT8, 
 
 PARAOBAFB _ _ PAS" 
 
 Example — Btrains in a bent girder (roof principal) 
 
 121. Calculation b; the iDomentB of mptare and shearing forces 39 
 
 122. Calculation by the composition and resolntion of forces . 41- 
 
 KeTBODS or CALOTTLAIIOir FOUSDID OH THE FAEALIiELOGnAH OF. FoKOXS. 
 
 123. General law of the strains in the booms . . .41 
 124.~ Note on General Rules for booms and bracing . . 42 
 
 125. Concentration of loads'at the apices . . .42 
 
 126. Trigonogietrical functions 42 
 
 127. Straight semi-girder, loaded in any manner . .42 
 
 128. Example 42 
 
 129. 'Straight Warren semi-girder, loaded at extremity . . 43 
 
 130. 'Straight Warren semi-girder, loaded uniformly on one boom 43 
 
 131. Straight Warren girder, loaded at any point . . .43 
 
 132. Same, with weight at centre 44 
 
 133. Straight Warren girder, with a concentrated roiling load . 44 
 
 134. 'Any Uraight Warren or lattice girder, with any load 
 
 symmetrically disposed about the centre . . ^ . 44 
 
 135. Any straight Warren or lattice girder, with a uniformly 
 
 distributed moving load .45 
 
 136. Example 45 
 
 187. Dead and live loads on girders , , ... .45 
 
 138. Any straight Warren, lattice, or other girder, with an 
 
 nnsymmetrical load . ...... 46 
 
 139. Lattice girders with the bars fixed at their intersection . 46 
 
 140. Simple truss ; central load 46 
 
 141. Same; load not at centre 47 
 
 142. Note on the above two cases 47 
 
 143. Simple truss; load distributed . . ■ . . .47 
 
 144. Compound' truss 47 
 
 ARCH BRIDQES. 
 Akches with Sfandril Bbaoino. 
 
 145. Uniform horizontal load 48 
 
 146. Moving load 48 
 
 Unbbaosd Arches. 
 
 147. Neutral surface or ourv^ 49 
 
 148. ][jine gf pressures 49 
 
 U8^154. Stability of arches . , 49 
 
 1S4a. Arched bp^ge pf several spans 50 
 
 SUSPBNSJON BBIDetlSB, 
 
 155. Suspension bridge of one span, nniform horizontal load , 51 
 
 156. Suspension bridge of more thftn oq? spa^, uniform hori- 
 
 zontal load ,...,... 51 
 157._S9Spen8ion In-idge with sloping rods, uniform horizontal 
 ^ load . ..... . . .52 
 
 158—1610. Suspension bridges with moving loi^ds . . .52 
 
 162. Abutments and Fiers ....... 53 
 
CONTENTa. ix 
 
 Section hi. — ^Distribution of Material to £ksi3i the 
 Caloulates Straihb. 
 
 PARAGRAPH PACE 
 
 163, 164. Strength of structures 6:{ 
 
 165. Uniform strength 54 
 
 166. Units 54 
 
 FRiiraiPAL Strains to be met, eto. 
 
 167. Tension . . .* ' 64 
 
 168. Compression 54 
 
 169. Breaking weights of columns 55 
 
 170. Long struts 66 
 
 171. Shearing .66 
 
 172. Coefficients of safety 66 
 
 173. Modulus of elasticity 66 
 
 JOINTS. 
 
 174, 175. General rules , . . 67 
 
 Ikon Joints and Fastevinqs. 
 
 176. Rivets 67 
 
 177, 178. Bolts 57 
 
 179. Pin joints in tension bars 57 
 
 Riveted Joinit in Tenaion. 
 
 180, 181. Effective section, &o., of a plate [,S 
 
 182.' Lap-joints 68 
 
 183. Fish-joints 69 
 
 Riveted Jtmitt in Compreiiion. 
 
 184. Lap-joints 
 183. Butt-joints 
 
 186. Gibs and cotters 
 
 Joints in Timber Strtiotures, 
 
 187. General note .... 
 
 188. Joints in tension — fished and scarfed 
 
 189. Joints in compression 
 
 69 
 59 
 59 
 
 . 59 
 
 . 59 
 
 . 60 
 
 190. Shouldered tenon for attaching cross to main beams . 60 
 
 BEAMS OP VARIOUS SECTIONS. 
 
 191. General remarks ^ . 60 
 
 192 — 195. Designing a beam 60 
 
 196. Stability of a loaded beam 61 
 
 197. Abbreviations 61 
 
 198, 199. Neutral axis 61 
 
 200, 201. Notes «1 
 
 202. Curved beams ffl 
 
 203. Modulus of rupture 62 
 
MouENis OF Ihekila. ahd ResistasoS' of Beams or 
 
 VARIOUS SECTIONS. 
 I-ABAORAFH PACK 
 
 204. Beam of solid rectaBgnlar section 62 
 
 20.'). Beam of hollov rectangular section . . ,62 
 
 206. Beam of solid circnlar section 62 
 
 207. Beam of hollow circular section 63 
 
 208. Beam of solid elliptical section 63 
 
 209. Beam of hdllow elliplical section 63 
 
 210. Beam with one flange 63 
 
 211. Beam with two equal flahjies . . .64 
 
 212. Beam with two unequal flanges 64 
 
 213. Beam of any section 64 
 
 214. Similar beams 64 
 
 215 — 218. Beams of rectangnlar section and of uniform strength . 65 
 
 219. How to cut the strongest and stiffest heam from a cylin- 
 
 drical log 66 
 
 220. To find the centre of gravity of any cross section . . 66 
 
 DEFLECTION. 
 
 221. Definition 66 
 
 222. Camber 66 
 
 223. Girders of uniform section ..,',, 66 
 
 224. Flanged girders of uniform strength , . , .66 
 
 225. Flanged semi-girders of nniform strength . . .67 
 
 226. Continuous girders and whole girders fixed at one or both 
 
 ends . 67 
 
 BBEAEING AND SAFE LOADS FOR BRIDGES AND 
 OIRDERS. 
 
 227, 228. Breaking or safe load found from amount of material, 
 
 span, &c. 
 229, Example I.— Flanged girder 
 280. Btample II. — Semi-beam of rectangular section 
 
 231. Table of the strength and elasticity of materials 
 
 232, Methods of constructing parabolas . . 
 
 67 
 68 
 68 
 68 
 70 
 
 Directions to Binder. 
 
 Plate I. to form .... Frontispiece. 
 
 ,, II. to face Page 58 
 
 „ III. to face 70 
 
FORMULAE AND DIAGRAMS 
 
 FOK THE 
 
 CALCULATION OF GIRDERS, &o. 
 
 STRAINS IN BEAMS. 
 
 I. The stability of a loaded beau or girder is.fonnded on the 
 equality that must always exist between the resultants of all the various 
 external forces tending to cause its rupture, and the sum of the molecular 
 reactions which resist the same. The former may be resolved — (1) hori- 
 zemtally into strains, depending for their value upon what are known as 
 Moments of Supture, or Bending Moments, tending to cause the failure of 
 the beam, by tearing asunder its fibres in one part' and crushing them 
 together in another (4) : and (2) vertically into what are known as Shear- 
 ing Forces, due to the transmission of the vertical pressure of the load to 
 the points of support, and tending to cause contignous vertical sections in 
 the beam to slide over each other (171). The values of the molecular 
 reactions are theMoments of Sesistance, for which see (804 — 91*}. 
 
 MOMENTS OF KUPTURE. 
 
 9. Abbrevlatloug adopted .tn tlie Formnla;. 
 
 Mx = moment of rupture at any point {x). 
 Mai Ms = „ at points of support (A, B). 
 
 Mc = „ at centre of span. 
 
 I = length of clear span = distance between supports in a whole 
 beam =. distance between W and support in a semi-beam loaded 
 with W. Where need in any other way explanation will be 
 made. 
 X = horizontal distance between the left abutment (except where 
 
 otherwise stated) and the point at which HI is to he found. 
 W = concentrated load at any point (61). 
 w = distributed stationary or dead load per unit of length, 
 «!' = ,, moving or live „ „ (61). . 
 
 Bef. = maximum deflection (2%1) for a beam of uniform section (ggs). 
 I = momenli of inertia of the section of the beam (for value see 
 
 304— «14). 
 E = modulus of elasticity (ITS) ; for value see 831. 
 
 Other abbreviations will be explained as they occur, 
 
 3. Note, Z, z, and other horizontal distances when occurring in the same 
 case, must be all of the same denomination ; and so also must w and w'. 
 
stuains in beaus. 
 
 4. Note.— The value of a fornrala being (^neg|^^e(!!!J) show* that 
 
 the aotion of the load makes, oi tends to make the upper surface of the heam 
 
 ( ^^'^''^' \ and therefore compresses together the fibres in the ( j™.. ) 
 
 part, and stretches them in the ( lower \ pj^yj._ 
 Vupper/ 
 
 5. In the Diaskaus, the . ordinates (the vertical distances from the 
 horizontal, or other lines) to the curves, &c., as shown, thereon, correspoml 
 to the values of th# formulae accompanying them. If the diagram be 
 Jiawn to scale in the manner directed, the. Moments of Supture may be 
 obtained by direct measurement. 
 
 Fig. A. Fig. B. 
 
 The vertical lines, as in fig. A, correspond to positive, and those as in 
 fig. B to negative values in the formulae (4). 
 
 6. ))7hen a diagram is used in the calculation of the moments, it should 
 be drawn immediately under, or over, and to the same horizontal scale as 
 the outline sketch (5S, HI., and &9) of the girder itself, so that the ordi- 
 nates in the diagram Will correspond with the several points in the length 
 of the girder to which they apply. 
 
 7. Semi-lSeam Fixed at One Knd, and l.oaded ivith » Concen- 
 
 trated Weigbt at tlie Othen 
 
 Kg. 1. 
 
 M.= 
 
 -WZ . . . 
 
 (*.) 
 
 M« = 
 
 
 (4.) 
 
 Def. 
 
 "3EI 
 
 DiASKAU. — Let A B be the beam (5, 6). Draw 
 A C = W J. Join C to B. Then the vertical dis- 
 tances between A B and B will give the moments of rapture. 
 
 S* Beml-Beam Fixed at One Knd, and loaded witli several Con- 
 centrated WelBlits, W, W,, W„ (87). 
 
 Fig. 2. 
 
 M^=-(Wt-l-W,?, + W,Z,) . . . (4.) 
 
 M, -- {W(Z-a!)-hWi (ii-a:) + W,(Z,-a:)} (4.) 
 
 When (I— x) or (ij— a:), or so on, is negative, 
 it is to be omitted. 
 
 DiASKAM.— Let A B be the beam (5, 6). Draw 
 AD = W,i„ DC = Wi«„ and AE = Wi. Join 
 D to B, E to W, and C to F, in the manner shown. Then the vertical dis- 
 tances between C F B and E W B wiU give the moments of rupture. 
 
MOMENTS OF Rt;t>!rtTlt£> Z 
 
 9. Beml-Beaiu Fixed at One Knd, and loaded ITnlformljr Ita 
 Entire tensth (Bg). 
 
 M i= — — g- ; ome-W/ t/to< iIik to the same 
 load (w 2) concentrated at the end. 
 
 M.= --sil-!^ 
 
 2* 
 Def. = 
 
 (4.) 
 
 8£X' 
 
 DiASKAH. — Let A B be the beam (5, 6). Draw A C = 
 
 wH' 
 
 Draitr the 
 
 parabola C B, whose vertex is at B (asa). Then the vertioal dUtanceB between 
 A. B.and C B will give the moments of rapture, 
 
 10. 8eml<Beam Fixed at One End, ivitb a load Vnlfonnly Dis- 
 tributed over Part of its Kentrtli* 
 
 Let z = the length of the load ;- 
 
 Fig.*. 
 
 Ma = — W S U ■ 
 
 ■» 
 
 When X ia less than or equal to (2 — z),''&iea 
 
 M« = - w s A - "2 - a: Y ' 
 When X is greater than ({-— z), then 
 
 (4.) 
 
 SrAOBAU. — Let A B be the beam (5, o). At A erect A C = wz 12 — n*)- 
 
 Join to a point in A B at the mid length of the load z (as in the Fig.). 
 Draw the semi-parabola D B (238) the same as for a beam of the length z, 
 fully loaded (9). The vertical distances between A B and D B will (?ive the 
 moments of rupture. 
 
 11. 8eml>Beam Fixed at One End, and Loaded wltli a IJni. 
 formly Dlstrlbnted load, and also a Cokieentrated IiOad at Its 
 Extremity (39). 
 
 M.= -^W+ gjt .... (4.) 
 M. = -(w(2-a!)+|(2-!F)») (4.) 
 
 Fig. 5. 
 
 wP 
 
 BlAOluir. — Let A B be the beam (5, 6). Draw A = -7 , and A D= 
 
 W I. Draw the parabola OB (as in 9) with its vertex at B (832 J. 
 Then the vertical distances between D B and O. B wiU''=give the moments of 
 riiptore. This is but a combination of ( T) and (9j, 
 
 B 3 
 
1%. 
 
 STBAIKS IK BEAMS. 
 
 Supported at Botb Enda, and Loaded at the Centre (40). 
 
 Fig. 8. 
 
 M^ = M„ = 0. 
 
 Mc = 
 
 Wl 
 
 ' 2 
 the nearer pier) 
 
 Ma: = ~s~ ^"^ l*™? me'osnred from 
 
 Def. 
 
 -W l^ 
 
 ~48BI 
 DiAGBAu. — ^Let A B be the heam 
 
 W I 
 <S. 6), At mid span erect C W = -/-. Join C to A and B. Then the 
 
 vertical distances between A B and A C, C B Will give the moments of rupture. 
 
 IS. Beam Supported at Botii Knds, and loaded nitli a Con. 
 centrated Wcisbt at any Point (4l). 
 
 The greatest moment will always be obtained at the point oT application 
 
 : ^'^: ''■ of the load, and = — -, — 
 
 Mi = M. = 0.' 
 
 Wxb 
 
 Between A and W, Mj, = — . — ■ 
 
 W xa 
 
 Between B and W, Mx = 
 
 I 
 
 'Mote particularly that x must, be measured from the pier which is on 
 the same side of W as x is, 
 
 W Or b 
 
 BiAGBAu. — Let AB be the beam (5, 6). AfWerectWC= — j— . Join 
 
 to A and B. Then the vertical distances between A B and A 0, C B will give 
 the moments of rapture. 
 
 Fig. 8. 
 
 14. Beam Supported at Both Ends, and loaded with any Number 
 
 of Concentrated Weiebts at any 
 Points. 
 
 The moment of rupture at any 
 point produced by oUl the weigJUs ii 
 ilie SUM of the viomehfa produced at 
 thai point by each of .the weights 
 sepa/rately (13 find 33a). 
 
 Let there, be three weights, W, 
 W], Wj, and let the segments into 
 which they each divide the beam be 
 
 respectively o, 6; a,, b,; a^, b^; then supposing that x is talten between 
 
 Wl and Wj. 
 
MOMKKTS OF KBPTOm. 
 
 M, = -J (W o I + W, a, a + Wj 6j x), having regard to the note in-. (ia)i 
 
 M^ = M. = 0. 
 
 DuoRAM. — Let A B be the beam (5, 6). Draw A B,. A,M> B, 
 and A K B as for three separate cases, by (IS). Produce W GniJl 
 WL=WF + WG + WC. Produce W, DtillW, N=W, H + W, I + W; D; 
 and so on for tlie \yeighf .W„ making W, P = W^ K + Wj J + WjE. • Join Jt 
 to t, h to N, N to P, and F to B. Then the ordinates from A B to the poly^ 
 gonal figure A L N P B will Bive the moments of rupture. 
 
 Note. If the weights be all equal, the vertieals at the weights representing 
 the mnments produced there by those 'weights (as WOj WjD, W^B) will 
 all be ordinates to a parabola (S32) drawn as for (II). 
 
 The following cases (15, 16) are adaptations of the above scheme to con- 
 ditions frequently met with in practice, 
 
 15. Benin Supported at Both ISiids, nnd landed with Two EqunI 
 Woiplits placed Kquldlstant n-oiii • 
 
 the Centre.* ^'^-^ 
 
 The moment fur any point between 
 the weights is a constant quautity 
 
 W (« - S) 
 = -^2 — -= Wa = WJ. 
 
 Between the weights and the sup- 
 ports Uj: = Wa: . M^ = M, = 0. 
 
 DiAQRAM. — Let AB be the beam (S, 6). At the weights erect WC and 
 W D each equal to (W a). Join A to C, C to D, and I) to B. Then the ver- 
 tical distances between A B and A G X> B will give the moments of rupture. 
 
 16, Beam Supported nt Both Ends, nnd l.ondcd with Four Eaunl 
 Weiffhts ss'minctrlcally disposed 
 nbont the Centre, f 
 
 M at Wi = M at Wj =i W (2 a -^ a') ; 
 constant from W, to Wj. 
 
 M at W = M at W, = 2 W*-. 
 
 DiAflSAU. — Let A B be the beam 
 (6, «). At W and W, erect W C and 
 WjF, each equal to (2Wa); and at Wi and Wj erect W, D and 
 W, E, each equal to W (2 o + a'). Join A to C, C to D, D to B, K to P, 
 and F to B. Then the vertical distances between A II and A C D E F B will 
 g;ive the moments of rupture. 
 
 * As in the caf^e of a cross girder cnrrylng a single line of railway, 
 t Case of a crobs girder carrying a double line of railway. 
 
aiBiiKS iH beahs. 
 
 II. Beam Supported at Both Ends, loaded with, a Coneentrated 
 
 RoIIlUK WelEbt (49). 
 
 — — — '- The mBzimom moment at any 
 
 point— 
 
 M^ = M. =0. 
 
 DiAOKAH. — ^Let A B te the lieam 
 (5, 6). DraT the iiarabola A D B 
 
 (238), -whose ordinate at centre (C D) = -7— Then the vertical distanceB 
 
 betvreen A3 and the parabola ADB will give the mazimam moments of 
 ruptnie. 
 
 18. Beam Supported at Both Knds, and loaded with Two 
 
 Welehts moTlne slmnltaneoiuly 
 VifS-'ii- In either direction over the 
 
 Let V and v, be the two weights. 
 The valne of the maximum moinent 
 produced at any point is 
 
 M, = 7{(w+wJ (Z-a!)-Wi«} 
 
 X jbeing measured from the nettrest 
 
 pier. The positioii of w causing ihe ' greatest moment is when 
 
 I 'w S IS 
 
 X -! 5"— g/w4-'w \ ' ^^ '^ *''* *^° weights be equal ; when * = o — 4* 
 
 Mi = M, = 0. 
 
 DiASBAK. — Let A B be the beam (5, 6) . Draw the parabola A C B (S3S), 
 
 ^ (w + w,) I. At A and B erect AD and BB = 
 
 whose ordinate at centre = - 
 
 Kg. IS. 
 
 W, S. Join A to B and D to B. .Then the vertical distances between AFB 
 and the paiabola A C U will give the masimnm moments of rupture. 
 
 19. Beam Bnpported at Both Knds, and loaded nnlfonnly Its 
 
 entire lencth (41). 
 
 % a: V) fl 
 
 M. = -J- {l-x). Mc = -g-. 
 
 Mx = M. = 0, 
 
 one half the moment at centre, and | 
 the deflection produced by the tame 
 load cono^rated at.the centre. 
 DiAOBAK. — Let A B be the beam (S, 6). pn A B draw the parabola 
 
 * As tn the conpled driving wheels of 4 locomotive. 
 
MOMSHTS 07 KUPTURB. 
 Wl* 
 
 A B (23%) wbose ordinate at centre = 
 
 8 
 
 Then the vertical dis- 
 
 tances between A B and the parabola A C B will give the moments of rupture. 
 
 Flg.U, 
 
 SO. Beam supported at Both Ends, subject to a I«ad milformly 
 Iklstrlbated over a certain leugtb 
 trotu one Support (15). 
 
 . iet 3 tqual the length of the load, 
 <md let X be mecaured from the abut- 
 ment from which the Uiad advcmcee. 
 
 The greatest moment produced liy a 
 given length of load trill be at its extre- 
 mity! or when x—z, that is, provided the 
 
 load does not extend beyond the centre of the span; for should it depassthat 
 point, the ^eatest stram will remain constant in position at the midspan, in- 
 creaajng izi intensity until the load completely covers the sp9in, 
 
 When X = z, 01 more than z, ^^ : 
 
 When X is less tl(an z, 
 
 M^ = 
 
 Stl 
 
 wa: (z ( 
 
 II-?) 
 
 Mi = M. = 0. 
 
 SiAosiH. — Iiet 4 B be the beam (5, 6). At the extremity (0) of the load 
 
 draw C D = ^' ! - • '^°'° ^ *° ^' "^^ ^ *" ^- ^^*^ ^^ pM»boUi 
 
 A EC (938), irhose ordinate at its centre = !^ ; the same aa if for a 
 
 8 
 nniform load on a beam A 0, supported at A and 0. (19). Then the vertical 
 dJBtonoes between A D B and A H C B will give the moments of rujptfir?. 
 
 SI. Beun Supported at Botb Knds, subject to a Load unlfonail)r 
 .Distributed over a certain lengtb noti.-lExtendlns; to. eitjhi^r Suppffftt 
 
 Let z =^ the length of the toad. Pi- ^g, 
 
 Zet V =:tht dittance from the load * ' ' 
 
 to the left ifwpp&rt; and let x be mea- 
 
 im'ed frqm the tame tnpport. 
 When X ■= v, or less than v. 
 
 When a; is equal to or or greater than (^ + z), 
 V I z\ 
 
 *^ = T ("+2] ('-==)• 
 
 When X is greater than v, and less than (c -I- z), 
 
 w X { z (z+ 2 «) ) w »• 
 M, = — ^2z + 2v - X j- — j ^' 
 
 Mi = M. = 0. 
 
STRAINS IK BEAUS. 
 
 DiAQRAH.— Let A B be the beam (6, 6). And let the load extend over 
 
 wzab 
 
 the length z. At.K, the centre of the load, erect K H = iilfp^. Join 
 H to A and H to sB. At C and D, the extremities of the load, draw per- 
 pendicnlwB to a:B, intersecting HA and HB in P and G respectively. 
 Join P lK) 9, Op C 3), 4ra» tbe parabola G^ p (83?), whose ordinate at 
 
 centre iiJ eqo»l to J™, tJiS same "» fW % dwWbnted load on a beam of the 
 
 length f, and »lipport|d.»t Cand D (19). Then the vertical distances be- 
 tween A 6 S D ^ and Aff G B ■wia give the moments of rupture. 
 
 as. Beam Supported at Both Ends, snbjeet to a RoUlnK Blstri- 
 
 buted I,aad of a Eencth less tkan 
 Tlg.W. that of tbe Beam. 
 
 Zet i = the length of the load. 
 Maximnm moment at any point, 
 mie(l — a:) wz' 
 
 llils vin cease to gi^g the correct value 
 when'x approaches nearer either support 
 
 than i , beyond which limits 2 in the for- 
 
 mola must be taken equal to 2 x. 
 
 DiAOBAii. — Let A B be the beam (5, S). On A B draw the parabola A D B 
 
 (»sa), -irboae ordinate at centre = -r— ^ Make CE = — . Through S 
 
 draw E F parallel to A'B. Make the horizontal distance of F and E from 
 
 z 
 A, (md B respectively (as C B) = -. The diagram will'be acea)i«te from F. 
 
 to E, But for the construction beyond those limits proceed as follows. Draw 
 the semi-panabqla E B (833) whose vertex is at B. Divide C B into a number 
 of parts 11 (say fiv^), and at tlje divisions draw YWtiiSfis, At the first division 
 
 from B tal^ ~ (= i) of the vertical distance between the pantbolic arcs IS B, and 
 
 D B messoring fiiom E B, At the second division take — and so on, counting the 
 
 divisions from the abntmept, and measuring ftom (he smsller (lower) parabola, 
 E B. ^e points thus obtained will enable the ctirve, as shown in the diagram, to 
 be trace4 through. Repeat the same operation on the opposite end of the beam. 
 And then the vertical distances between AHDKB and AFEB will give the 
 moments of fupture. 
 
 9S. Points of Contrary Flexnre, or of inflexion, or of "no-: 
 oorvatnre," as they are sometimes called, are points at which the npper 
 and lower surfaces change from convexity to concavity (4), and vice vend 
 (see fig. 23). At these points, as there is no curvature there is no moment of 
 rupture, for the moments of rupture are the intensities of the curving or 
 bending forces (I). 
 
MOMENTS OF KUPTUKE. 
 
 Fisr. 17. 
 
 94. Beam of Eqaal nnd llnlforra Section, or Beam or ITnlfonii 
 StrenictU (1G5), Fixed iiorizoutally at tiie Ends, and loaded at 
 tbo Centre. (See 30.) 
 
 The length ff it identical with 
 (18), ami the parlt A/ and B/'' 
 viih (T). 
 
 M. = -g-. 
 
 . . (4.) 
 
 Where M. = there are points of emdrmry fiexmre, distant from their 
 respective piers by \ I. 
 
 PiAQQAU.— 'Let A B be the beam (5, 6). At mid-span erect C W =7 . 
 
 WZ 
 
 At A and B erect A J> and B E, each = Join D to E, ^nd C to A 
 
 
 and B. Then the vertical distances between DE and A.CB will give the mo- 
 ments of rapture. 
 
 The j>aint> of contrary jUxare (33) are at the intersection of A Q #n4 
 OB, withDE. 
 
 35. Beam of Eanal and Vniform Section, Fixed liorlzontaliy at 
 Botii ^iM|SuA>ul Iioaded nnlformi]' Its entire I«nEtli (4T). (^^e 3a.) 
 
 The length f i> identicai wilk (19), and the parts A/ and B/' mth 
 (11), the Gontsentrated load at the ex- 
 tremity of each semi-beam A/, Bf being Pig. 18. 
 equal to half the distributed load on// . 
 
 M. 
 
 
 wx wl^ 
 
 ■wl' 
 Mi =M, = — -jg- . . 
 
 (4.) 
 
 Where M, = 0, there are the points of contramj flexure (l^S), distant 
 from A and B respectively by '2112. 
 
 5v>l* ., , , 
 
 Wk ~ 1 S36 B I ~ ""' V^i^^ "^' "/ '^ same ieaifii. if not fixed at die 
 
 endsT 
 
 DiASBAU. — liet AB be the beam (5, 6). On AB draw the parabola 
 ACB (832), whose ordinate at centre CD = -5-. At A and B erect 
 
10 
 
 STRAIH3 IN BEAMS. 
 
 A K and B F respectively, each equal to 
 
 12 ■ 
 
 Join E to F. Then the 
 
 vertical distaneeB lietween E F and A.G B will give the momenta of rupture. 
 
 The points of contrary flesaire (S3), are at the intersection of A C B 
 .•Brith E.P. 
 
 Fig. 19. 
 
 !!6. Beam of Tniform Strensth (I65), Tlxed hoTlzantally at Both 
 Ends, and loaded nntfonnly Its entire lenstta (4T). (See 30.) 
 
 The length ff {= ^ij w identical vnth {19), amd the parts A/ amd 
 
 Bf wt<A (11) ; the Gonoentrated Io»d 
 at the extremity of each being equal 
 
 wi 
 to — = half the load on // . 
 
 ^^ °° 32"' 
 
 M. = -2-(«-a!)--3a- 
 
 ZwV 
 
 (4.) 
 (4.) 
 
 M^ = Mb = - 
 
 Where M, = 0, there are the 'ptnvtt of contrary fiexwe (83). 
 
 DiA9BAX.^I<et AB be the beam {p, 0), On A B dntw the pjabola 
 
 wl^ 
 A B (999), whose ordinate a.t centre (0 D) = ^. At A and B erect 
 
 AE and B F respectively, each = -™' . Join E to F. ThV^e vertical 
 
 distances betvreen E F and A C B wiU ^vs th# moments of rupture. The poi/ntt 
 of contrary flexure (83) ar» at tibM latei^ctioit of A C B with E F. 
 
 ST. Beam of Vnlfbnn and Eanal Seotloa, n«tln(c on Two gnpnorts 
 
 und imttCplrinly loaded, the Ex- 
 ^'?- ^- tremltteg Itflns Subjected to two 
 
 knawn Moments of Rnpture (M^ , 
 Mb) actlns in a Contrary Blree- 
 tion to those due to the load 
 (46), 
 
 Mx = ^ (« - *) - Ma + 
 
 J 
 
 (M^-Mb). 
 
 When Mi = 0, there are the fointt of eonlrary jkxwe, 
 SiAORAX.— Let A B be the beam (9, 6). Qh A B draw the parabola 
 A B (itSS), whose ordinate at centre C D ~ At 4. and B erect A E 
 
UOUERIS OF BCPTUBE. 
 
 11 
 
 and B F re8i>ectiTely, makin(r A E = 11 ;,, and B F = M. . Join E to F. 
 Then Uie vertical digtanoes between B F and A O B will give the moments of 
 rapture. When S F interaeets A C B, there will be the pointt of contrary 
 fiaatre (S3). 
 
 >8. Beam ut VnironB and Knnal Beetton, anpported at One Bnd 
 (A), and Fixed horizontally at the Other (B)*, nnirormlr leaded 
 over Ita entire Eencth (49). 
 
 Thi» east it identical v/ith the 
 Ungthf^df(M). 
 
 w X mix 
 
 M, = -g- (J — a) - — g— (a being 
 
 measared from the nnfixed 
 end). 
 
 M» ?= 0, M. = - -g- . (4.) 
 Tbepoia( of eontrary flexure (83) ia where Mi = 0. 
 
 82 
 
 M at midway between A and / = . ' loa ' ' 
 
 DuasAK.— Let A B be the beam (S, 6). On A B draw the panboU 
 A C B (831), whose ordinate at centre CD = ^ . At B, the fixed end of 
 
 a 
 
 the beam, erect B E = ^ . = C 0. Join A to E. Then the vertical diatances 
 
 8 
 between A. E and A C B will e^ve the moments of rnptnre. Where A E inter- 
 sects C B, there will be the point of contrary fltxm-e (83). 
 
 29. Beam of Vnlform Btrencth (16S), inpported at One End (A) 
 and Fixed horlzontallr at the other .(B)*> nnlformly I«aded over 
 tta entire Kensth (48). 
 
 2%i( cate w identical wif^ tJie length fB of (26). 
 
 wa mix . 
 
 Mj, = -J" (' — as) — -g~i [x being 
 
 measured from the nnfixed end). 
 M^ = 0. M, = - -g- . (4.) 
 
 M at midway between A and / 
 
 = "iF" 
 
 The paint of amtrary flexure (23) ia where Mz => 0. 
 
 Tig. 22, 
 
 A/=|'- 
 
 • See 39. 
 
 f The valaea in this case are slightly erroneous. The oonect valae for A/ ia 
 •783*. 
 
12 STKAINS IN BEAMS. 
 
 DjASKAM., — LetAB be the team (5, «). .On AB draw the parabola 
 
 ACS (»3!S), whose ordinate at centre (C D) = ^. At B, the fixed enll 
 
 of the girder, erect B E = M, = — p-. Join A to E. Then the vertical 
 
 6 
 distances between A E andA OB will give the moments ofnipture. Where AE 
 intersects A C B, there is the point of contrary flexure (83). 
 
 30. Benin continuous Tor two of more Rltrlitly Pritportioned * 
 Spans, sultject to a Stationary Load, Vulforinly Dlslributcil over 
 its entire JLenetli (50). 
 
 All such cases may he regarded as comhinations of some of the cases 
 previously given (24 — 29). For if, in any of the latter, -the beam, instead 
 of being fixed atone or both ends, be .continued over a support where 
 originally fixed, and subjected to the action of a load, which shall produce 
 at the point of support a moment equal to that produced there by the 
 first load when the beam was fixed at that support, then the moments in 
 the original length of the beam will remain as they were^ and will not be 
 affected by the substitution of the continuation for the fixing. 
 
 Fijj. 23. 
 
 Let AF.be a beam continuous over, a number of equidistant piers, 
 B, C, &c. 
 
 If the beam be of uniform strength (l«S), the outer spans (A B and 
 E F) should be J the length of the otUeis. If of uniform section, the outer 
 spans .should be 789 the length of the others.. (See USA.) 
 
 The end spans may be regarded as identical with (28) or (20), and the 
 remaining spans with (24, 29, or 2G), so that the moments of rupture may 
 be obtained from the formulae or diagrams there given. 
 
 .tl. Continuous Bcains, not of Vniforin Section, subject to Taryins 
 Loads. 
 
 Tt would, perhaps, be impossible to give mathematically accurate formulse for 
 the moments of nipture in continnous beams, witih moving loads, that would be 
 worth anything for practicaV application. A well-known author t has even pro- 
 nounced the case ** too complicated for investigation." 
 
 The following approximations, however, may. be relied on for tafeiy vnihnvt 
 enetravagance. 
 
 32. Beam of Vnlform Strenetb .(I6S) for the Blaxiuium Strainn, 
 
 • By riphtly proportioned is meant, proportioned so that if the beam were 
 fixed on any one of the piers instead of continuous over it, the moments iwodnoed 
 there by the two adjacent loaded spans would be eqnal. Then Tf thin condition 
 bfobnerveU the cabre will include beams not uniformly loaded throughout. 
 
 t J. H. Latham, Esq., M. A.— "iron Bridges." 
 
MOMKNTS OF EUriDRE. 13 
 
 rontlnnoiis over one Pier, formlQe: two Evoal Spans, subject to a 
 Fixed Load Viiiformly Oistrlbuted, and also to a Itlovlns load (Si). 
 (See 113 A.) 
 
 The greatest moment over the pier will be pvoduoed when both spans ave 
 fallyloaded. Each spaa will then nearljr correspond to ■'.99). 
 
 The greatest posiiive (4) moment will obtain in the span fully loaded, 
 when the other spau bears only the fixed loud. 
 
 Let w = fixed load per unite of length. 
 
 a/= varying ,, ,, 
 
 and let X be always measured from an abutment, and not from the 
 pier. Then, 
 
 X {rb + w') (I — x) (2 TO + M)*)*- 1 
 Maxm. positive (4) moment, Mj, = 5 — j^ . 
 
 (to + w') t* 
 Masm. negative (4) moment over pier (C) = „ 
 
 Max. neg. mom. [M. = g (M + «/) (2 2 - 3 x), or 1 ^^^ ^^^^^^ ^,_ 
 
 at any other/ ^^ ^^ ) be taken. 
 
 P^"*. j^M« = Y (^ - =") - la '^ *" + "'>' J 
 
 Any positive value of the last two, and any negative^ value of the first of 
 these four equation^ must of course be disregarded. . 
 m:» = Mb = 0. 
 
 By making Mx = in the first and last equations, and then finding the 
 Value of a,, the limits of deviation of the points of contrary flexure (aS) 
 may be obtained.* * 
 
 Fig 24. 
 
 DiAQKAM. — Let A C B be the beam (5. «). On A C draw the parab4|('ADC 
 SS9), whose ordinate at centre D E = [w ^ w') -;. and on C B draw the 
 
 o 
 
 V)P 
 
 parabola C F B (asa), whose ordinate at centre F G = — . At the centre 
 
 . Xt/tl.\ be not greater than (' "I"" — ), the beam will require holding 
 down to the abutments. 
 
14 STRAIHS IK BEIUS. 
 
 pier erect C H = e'^'~t *"''> nie»s»ring from C, make C K = 
 
 (2 10 + W) i» ^ 
 
 — jg . Join H to A, and E to A and B. Then the vertical distances 
 
 between the parabolic arc A D L and A L will give the maximum poBitive (4) 
 moments. The vertical distances between N H and the arc K L G, or those be- 
 tween K M and F M, whichever be the greaUr^ will give the maximum negative (4) 
 moments. The points of corUrarg jtexttre \1SS) will approach as near the 
 centre pier as L, and recede from it as far as M. (See foot note, p. 13.) 
 
 NoTS. — The various values, &c., given above, apply equally to both the spans. 
 
 The diagram ab8ve is drawn to scale, on the supposition that the rolling load 
 is i the intensity of the flxed.load. 
 
 33.' Beam of Vnironn Strenetli (165) for the Maximiuii Strains,* 
 eontlnuons over two op more Piers, subject to a Fixed load Vnl- 
 rorinly Sistrlltated, and also to a.Hoviiiic load (SIS). (See 113 A.) 
 
 The maxinmm momeni over any pier will obtain, wlien only the two 
 a^jaceht spans, and every alternate span from them, are simultaneansly 
 loaded with the greatest load, the remaining spans sustaining only the 
 -fixed load. 
 
 The maximum moment at the centre of any span will ohtain when it and 
 the alternate spans from it are iuUy loaded, the remaining spans sustaining 
 only the fixed load. 
 
 Let w = the fixed or dead load per unit of length. 
 . «/= themoving or live ,, „ 
 
 I' = either outer span. 
 2 = any other span.* 
 
 Then, the meximnm negative (4) moment over any pier, B or C, 
 
 Maximum negative (4) moment between any two piers (t.e., in any 
 inner span, Q, 
 
 (positive valaes of which must he disregarded). 
 
 Kaximom potitive (4) moments between any two piers (i.e., in any inner 
 span, 1), 
 
 (w + vf ) , . ZwP 
 
 M. 2~" <^ - *) "* - 'W 
 
 (negative values of '#fiich must be disregarded). 
 For^her outer span, the maximum negative moments, 
 a(ro + i»7 . x.P /2w i^v 
 
 (positive values of which must he disregarded). 
 
 * If there be but thfee spans, a modification of the values hereafter given will 
 be necessary, wiiich see. 
 
MOMKHTS OP KDPTURE. 15 
 
 The maximum positive moments, 
 
 x{w + w') SwxP 
 
 M* = 2 (l - X) —327-. 
 
 X being measured from the abutment. 
 If any of the foregoing expressions for M, be made equal to 0, the value 
 of X obtained from them will givS the positions bi the points of isontrary 
 flexure («S).* 
 
 JSIo'CE.^Jf the beam be eantinuout for three epani only, Z, as a coefficient 
 . P /2 w vf\ xP /2w v/\ 
 
 in the expression -3-1~+y1, or in sTT ( "»-+-2"l. must have a value 
 
 l + l' 
 given to it = —5 — 
 
 Fig. 26. 
 
 DiAGKAH. — Let ABO (6, 6) be part of the beam. On BC draw the 
 
 parabola B D (2S8), whose ordinate at centre (DE) = (" +.^ . 
 
 8 
 On A B draw the parabola A F B ( S3S )j whose ordinate at centre 
 
 (w +■ v/) I'i ' 
 
 (P G) = — — g-!— • At B and C erect B H and C Jt respectively^ 
 
 P /2 w v/\ 
 each.= - [_-+ — ). Join A to H and H to J. Measuring from B and 
 
 ^ S w P 
 
 0, make fi'E and C L, each = -^^— . Join A to K and K to L. 
 
 The vertical distaiices between AFO P and AF give the maximum positive 
 moments for either outer span ; and those between M N D N' Mi , and M Mi give 
 those for any inaw span. The vertical distances between O H and O P B ^ive 
 the maximum negative moments for the outer spans^ and those between H N 
 and BM N, or between N' J and N'Mi C, give the maximum negative* mdments 
 for any inner span. 
 
 The_ points and P, and M and N or M, and N', show the limits of 
 deviation of the points of comirary flexure (83). 
 
 The diagram above is drawn to scale, on the supposition that the intensity of 
 the ToUing load is one-half that of the Sxed load. 
 
 * If ^__j be not greater than f-^ — ), the beam will require holding 
 
 down to the- abutments. ^ 
 
 t N.B. — The note given above (for the value of 2 in th^mbrmula) must be 
 nbserved here. 
 
16 
 
 STUAIKS IN BEAMS. 
 
 S3 A. Note. — The moment of rupture at any point, produced by several 
 loads acting simultaneously on a beam, is equal to the sum of the moments 
 produced by the several loads acting separately. 
 
 SHBAEING FORCES (I). 
 34. Abbr^yiations : — 
 
 Let F and P — the reactions on the supports due to the total load 
 on the beam between those supports. 
 S Hi = the shearing force at any point x. 
 
 w = distributed load per unit of length (61). 
 W = total load concentrated a,t any point (61). 
 X = distance from left-band support to the section at 
 
 which S H is required. 
 { = length of span. 
 
 Other abbreviations Will be explaified when they 
 
 35. In the Diasbaus, the ordinates (the vertical distataces from the 
 horizontal or other lines) to the curves, &c., as shown thereon, correspond 
 to the values of the formulse accompanying them. If the diagram be drawn 
 to scale in the manner as directed, the shearing -forces may be obtained by 
 dwed measwrement. See also (6). 
 
 36. Geneial Itnle ibr deterininlne the Sliearins Force at any 
 part or a Beam and nnder any Disti'Ilintlon of load. 
 
 Fig. 26. 
 
 Let it be required to find the shearing force at any point (C) of a beam. 
 
 Let W = the load between A and C. 
 W"= „ BandC. 
 
 Then, SHatC = 
 
 S Hx = P - W, or = F - W" ;— 
 
 the greater of the two values to be 
 talcen. 
 
 At the supports, W or W" = ; so that the shearing forces there are 
 always equal to P or P'. The above values hold good for semi-beams. 
 
 37. Semi-beam Fixed at One End, ft-co at tbe Otber, and loaded 
 
 In any Manner (g). 
 Fig. 27. ^ ' 
 
 The shearing force at any point P is equal to 
 all the load between that point and the unsup- 
 ported extremity. 
 
SHEARIKO FOHCES. 
 
 17 
 
 »8. Semi-beam Fixed at One End only, and uniformly loaded ItK 
 entire Leneth (9). ' 
 
 S Hx = w (« - x). 
 
 5 Hi = Ml!. 
 
 Diagram.— Let AB be the beam (35). At 
 A erect A C = w I. Join C to B. Then the 
 vertical distances between AB and B will give the 
 shearing forces. 
 
 Fig. 29. 
 
 39. Beam Fixed nt One End only, loaded uniformly its entire 
 lenetli, and also with a Concentrated 
 W||Bl>t nt Its tree Extremity (11). 
 
 S H, = W + w (l-x). S Hi = W + M Z. 
 
 DiAORAH. - Let A B be the beam (35). Make 
 A C = w i. Join C to B (as in 38). Make 
 A D and B E = W. Join D to E. Then the 
 vertical distances between B and D E will give the 
 shearing forc^. 
 
 40. Beam Supported at Both Ends, and loaded at the Centre (1%). 
 At any point in the beam, 
 
 W 
 
 ' 2" 
 
 DlAQBAU. — In Pig. 30, the shearing forces (35) for this case are given by the 
 
 W 
 vertical distances between C D and A B, A C and B D being each = -—•■. (43.) 
 
 SH = 
 
 41. Beam Supported at Both Ends, and loaded with a Con- 
 centrated WelEht at any Point (13). 
 
 Let a and b be the distances of W from the supports A and B respec- 
 tively. 
 
 6 
 SHi = P=W2==SH,, oonsUint between A and W. 
 
 a 
 S H, = F = W 7 = S H, , constant between B and W. 
 
 DlASKAH. — In Fig. 30, the shearing forces (35) for this case are given by 
 the vertical distances between A B and B P Q iS, W being supposed at Q. 
 
 AB^Wy. BS=W-2... (43.) 
 
 4%. Beam Supported at Both Ends, with 
 movlnc: in either Blrectlou (17). 
 
 Concentrated load* 
 
 Or, 
 
 S H. = W y I 
 S Hi = S H, = W. 
 
 The greater of these two -ralnea to be taken. 
 
 DlAORAX. — In Fig. 30, the verticals A E and E F at either extremity of 
 
 
 
IS 
 
 STBAIMS IS BXAMS. 
 
 the beam are made equal to W j B joined to B, and A toP. The vertical dis- 
 tances between B K F and A. B give the shearing forces. 
 
 43> IHaKmiii of the Stacarlne Forces In u whole Beam with s 
 
 Concentrated Iioad. 
 
 Fig. 30. 
 
 A B is the beam. A E and B F are 
 equal to W. 
 
 When W is in the centre of the beam 
 
 (K), the shearing force for the vhole 
 
 W 
 length of the beam equals — = the 
 
 ^d 
 
 vertical distance between C D 
 AB. 
 
 When W is at any other point (Q), 
 the: shearing forces in the two segments, R F, Q S, are inversely as the 
 lengths of the segments. The vertical distances between A B and H P Q S, 
 give those shearing farces. 
 
 When W rolls from end to end of the beam, the shearing forces are as 
 the vertical distances between A B and £ K -P. The points F and Q will 
 always be points in either K B or A F. * 
 
 44. Beam 'Hui»pfl»rte4l, at Uoth Ends, and l^Mided nnlTormly Itit 
 entire Leneth (19). 
 
 Pis.31. SHi = SH. = P = F' = "' 
 
 •2 
 
 SH« 
 
 '(.--)• 
 
 Pig. 32. 
 
 The sign of the result to he disregarded. 
 At mid-span, S H = 0. 
 DiAQBAii. — Let A B be the beam (35). At A and B erect A C and B D, 
 
 ' 10 1 -3 n 
 
 each equal to „-. Join G and D to the mid-span, £. Then the vertical 
 diatnnees between A B and C E D will give the shearing forces, 
 
 4.1. Beam Sniiported at Both Kiids, subject to » DIstrlbntcd 
 I.ond atlvancine trom clher Support * 
 
 (ao). 
 
 The greatest shearing force will be 
 developed at the point of junction of the 
 loaded and unloaded segments ; in which 
 case let x also represent the length of the 
 load. Then, 
 
 ■ .»)" ) 
 
 The greater of the two values to be 
 taken. 
 
 ml 
 At mid-span, S H = -g- 
 
 « As iu the case of a railway train crossing a bridge. 
 
SIIKARING FORCES. 19 
 
 DiAORAM.— iet A B be the beam (35). At A and B erect A C and B D 
 . w I 
 respectively, eacU equal to ^. Draw the semi-parabola, C E B («.■*«), whose 
 
 V) I 
 vertex is at B, and ordinate B D =< y. Draw the semi-parabola A E D 
 
 corresponding exactly to C E.B. Then, as the load advances from either 
 pier (say A), tlie sheijj-ing force developed at the junction of the loadft 
 with the nnloaded apex will be represented by the vertical distances be- 
 tween A B and A E D. When the load covers the whole span, the shearing 
 forces will be ^s in (44). 
 
 As the loacj is liable to advance from the pier B as well as from A, the 
 maximum shearinj; forces for all positions of the load will bo given by the vertical 
 distances from A B to C {! 0. 
 
 46. Cencntl Foniiulie for Itetcrmlnlng the Keartlon or the 
 Supports, and the ShenrlUB Forces In tli« r^tse at Continuous 
 ■Icams uuil Beamii whose Extrcinilics arc Fixed, »r subject ti> 
 the Action or known Moments of Kuptiire (ST), 
 
 Let A B be a beam, subject to the action of the moments M^ , Mj , and 
 let tlie beam be uniformly loaded. 
 
 The notations as before. 
 
 ^ = T+ I • 
 
 wl M» — Mi 
 
 ^ = T+ ~ r ■ 
 
 The shearing force at any point distant x from either pier, is found by 
 subtracting (w x) from the re-action of that pier produoed by tb^ load between 
 A and B. 
 
 The values of F, F' are the pressures on the piers produced by the load 
 between A and B only. Should there be a load on the beam continued 
 beyuud these points, the pressures similarly found must of course be added 
 to those above for the total pressures on the piers (IWt). 
 
 47. Beam Fixed at noth Extremities, ami Eoaded uuiroruil.v 
 
 (*.=>, SO). 
 
 Exactly the same as for (44). 
 DiAQBAU. — The same as in (44), which see. 
 
 48. Beam of Uniform Strensth {IHH), supported nt One End, 
 and Fixed horizontally at the other, uniformly (.omled IM eutire 
 lenslh (99). 
 
 «> I 
 At the unfixed end, S H^ = -^' 
 
 2 wl 
 At the fixed end, S Hj, = -^s — 
 
 a 
 
2(| STRAINS W BKAHS. 
 
 S Hi = w f r — x), z being measured from the unfixed end. The sign 
 of the result to be disregarded. 
 
 DlAGBAH. — Let A B (fig. 34) be the beam (35). At the unfixed end (A) 
 
 w I 
 erfft A C = „-, aud at the fixed end (B) erect B D = 2 (A G). Join 
 o 
 
 and D to E, distant I I from A, or half-way between A and the point of 
 contrary flexure, / (23). The vertical distances between A B and G G D will 
 give the shearing &rceg. 
 
 The length A/is identical leilh A/ (fig. 22), and tlie length f B with 
 /B (fig. 22). 
 
 49. Beam of irnifonn and Eonal Section, supported at One Knd 
 and Fixed Uorizontally at. the Other, 
 "'^' unlformlr Iioaded Its entire I«nsth 
 
 («8). 
 
 3 w 2 
 At the unfixed end, S H^ = — g— • 
 
 5 w I 
 At the fixed end, S Hj = -v— 
 9 
 fSl \ 
 8 Hi = JO ( -g- — a: 1, a; being measured from the unfixed end ; and the 
 
 sign of the result being disregarded. 
 
 DiAGBAH. — Let A B be the beam (3S). At the unfixed end (A) 
 3 w Z 5 w I 
 
 erect A C = —5—, and at the fixed end (B) erect B D' = —3 — . Join C 
 o o 
 
 and jy to E', distant 1 1 from A, or mid-way between A and the point of 
 contrary flexure, /' (83). Then the vertical distances between A B and C K'D' 
 will give the shearing forces. 
 
 The length A/' w identical viith A/ (fig. 21), and /' B with f B 
 (fig. 21). '■ 
 
 80. Contlnnons Beams with Fixed nntlormly Blgtrltonted loads 
 
 (SO). 
 
 If the spans be rightly proportioned (see foot-note, (•) p. 12), case (48) or 
 (49) will apply to the outer spans, and case (4T) to the remaining spans. 
 
 51. Beam of Uniform Strenicth (165) for the maximum strains, 
 continuous for Two eunal Spans, 
 ^'g- •''^- suhjcet to a Fixed lAad nnlt'ormir 
 
 Distributed, and also to a Moving 
 load (3«). (See 1I3A.) 
 
 The maximum shearing fijrce ai either 
 abutment will obtain when its span only 
 sustains the moving load. 
 
 The maximum shearing force at the 
 centre pier will obtain when both spans are fully loaded. 
 
SBEAniirO TOKOKS. 21, 
 
 The total maximum presanre on the centre pier, when hoth apans are 
 folly loaded, will be twice the above maximum shearing forceat the pier. 
 
 Letw' = the moving load per unit of length, the other notations as before. 
 The following ue the maximum values* : — 
 
 At oither abntment, S H^ = — - (4 u; + 6 «/), or 
 
 2 I 
 At the centre pier, S Ho = — (is + «/), or 
 
 3 
 
 At an; other point (a;), 
 
 S H, = Tj (4 w + 5wf) — x(a + jo'), The greatest value to be 
 • I taken ;t x to be mea- 
 
 Or, =(g-a)(w + w'), 
 
 1 _ \ /._ . . j> / Bured from the abut- 
 
 ment, and the sign of 
 I I the result to be disre- 
 
 Or, = j2(4w — w') — wat-t I garded. 
 
 As the load may be supposed to advance from either abutment right 
 across the beam, an addition muit be made to the above values for S Hi { 
 
 bemg equal to -g- when x is about equal to } I, and gradually diminishing 
 
 as X gets more or less than -. (See the dotted curve in the diagram.) 
 3 
 
 DiAORAK. — Let A C (6g. 35) be one span of the beam (3S). At the 
 
 abutment A erect AD = ^(4jo+ 5 »')• Make A E = r- (tB + w'). 
 
 12 o 
 
 At C erect C F = twice A E. Join E and F to M, distant 1 1 from A. 
 Through D draw D N parallel to E M. Sketch in a curve similar to that 
 (dotted) in the figure, making an additional depth to the ordinates, at the 
 
 point of minimum shearing force, of W -. t Then the vertical distances be- 
 tween A and D S T F may he considertd to Bive the maximum shearing forces 
 for either span. 
 
 S9, Beam of Tnirorm atrenarth (165) for the ^axlmiiin Strains, 
 Continuous over two or more Piers, snbJe«|'to'a Fli^ea Koad 
 uniformly Distributed, nad also to n MovIuk' Koad (33). (See 
 USA.) 
 
 * .^)pivoximate values corresponding with the moments of rupture 32. (See 31.) 
 t ^e last of the three valiieS fs the shearini; force in one span, when the other 
 only is flilly loaded, the addition'however made for the load being a moving one 
 will entirely cover it. ' 
 
 t An exact expression for the value of the shearing forces dev|Blopod by a load 
 gnidually advancing across a continuous beam of (about) uniforiA ^tf ength, would, 
 even if it could be obtained, be most domplicatfld. The piroos^S here auggested, 
 however, though necessarily only approximate, may be regarded as practically 
 safe. 
 
22 
 
 STRAINS IN BRAHS. 
 
 The maximum shearing force at any pier (B or C) will obtati simulta 
 neously vibh the maximum moment of rupture over that pier (33). 
 
 For any inner span, 
 
 SH. = SHc-i(g + -J-)- 
 
 /'to 2 w)'\ 
 SHx = Z(- + -— I -a;(w + ie'), 
 
 X being measuredl^pm the nearer pif}'; 
 
 For either outer span' — 
 Ajb either abutment, 
 
 Or, 
 
 SH, = («. + «/) 2 - 
 
 32 2' 
 
 At the pier, 
 
 
 S Hb = 1 («) + W). 
 
 
 At any other point, 
 
 
 SH, = (2 -a:)(«'+OT') - 
 
 Swl? 
 S2l' 
 
 = («' + «;') (3 -4 
 
 
 The greater of these va- 
 lues tg be taken, and 
 X tn ']be measured from 
 the abutment. 
 
 At t}ie middle of the inn.er spans, and at J I' from the abutments in the 
 outer 8p;uis, the values obtained by the above formulse must be increased 
 
 Wl ' w' I' 
 by — g- and —7- respectively ; and this addition, gradually diminished, 
 
 must be ma4.e for some distance on either aide of those points (see foot 
 note (j:), p. Zi) as shown in the diagram below. 
 
 If the bear)t be conlimious for three spang only, the above formuise 
 must be modified as directed on the next page. 
 
 Fig. 36. 
 
 DiAQBAU. — Let A B C be part of the beam (3S). 
 
 /' w 
 
 \T 
 
 For any inner span 
 (as l).—ht B and C erect B G and C H, each= ( "j" + ""3" /• *'*''® ^ ' 
 and C E each = — (w + «/). Join D and E to the mid-span, -and draw 6 K 
 
 and K H parallel to D F and F E respectively. For either outer span (as V) : — 
 
 2 /,' ' . 3 / 
 
 At B erect a perpendicular = — — (la + w'), -wljich, if I' = —, will 
 
SUKARINS FOKCUS. 23 
 
 coincide with B S. At A erect A L = 4 S B. Join D and L to M, 
 
 I' Z V) I' 
 distant J V from A. Make A = - (to + w') 55-t; . Draw N paral- 
 
 lei to L M, At T and K sketch in curves,* similar to those in the figure, 
 
 giving additional depths to the ordinatea there of — r- and -3- respec- 
 . I 8 8 
 
 tively. Then the vertical distances between O T D and A B may be considered 
 to Kive the maximum shearing forces for either outer span, und those between 
 G y H and B C for the remaining spans. 
 
 If the beam be conlitiitous for three spans only, the value given in the 
 formal^ for S H , and S He , and in the diagram for B Q and C H, must be 
 ,, ^ {IS w + 16 w')l V(iii)v)'\ ■ ^ ^, ^^ , . 
 altered to ^ ^^r + S"i V "7" "*■ 'o' ) > *'"1 further, the value given 
 
 to S H» for the inner span must be altered to = S Hb — a! (w + to'), 
 in which latter expression S Hb must have the value just assigned to 
 
 .1 + 1' 
 .t. I- =-3-- 
 
 See foot note (t), p. 21. 
 
24 DESIRN OF BBinCES AKU CIRUKKS. 
 
 FLANGED GIRDERS, AECHES, AND SUSPENSION BRIDGES. 
 
 53. In the nesien or » tiridee, girder, or otber similar structnre, 
 certain parts of which are supposed to resist certain strains, * the varions 
 processes are followed out in a progressive order. When the data do not 
 extend beyond the amount and nature of the load, and the width, &c., of 
 the obstacle to be crossed, the processes will be as follows : — 
 
 I. Determination of the Mnd of bridge or girder (54). 
 II. Determination of the general cross section, and major proportions 
 of the structure (65 — 68). These enable — 
 
 III. An outline sketch to he drawn (59). 
 
 IV. Appraeijitate estimation of the weight of the strudure (60, 61). 
 
 V. Calculation of the strains on the various parts (62 — 168) (which 
 strains must be figured on to the outline sketch), at as many points 
 as will be found necessary for the accurate — 
 VI. Determination of the amount of material to retist the strains on 
 the various parts (163 — 173). 
 VII. Diatrilmtion of Uie matei^l in the vairious parts into a form of 
 cross section best suited to resist the kinds of strain brought upon 
 them (174 — 190), having regard to — 
 
 a. facility of construction. 
 
 b. adaptation of one })art to another in contact with it. 
 
 SECTION I. 
 
 DKTERMINATtOS 07 THE NATURE, PbINCIFAL DiHXNSIORS, STC, 07 THE 
 BBISOE OB GlBDEB. 
 
 (Embodying processes, I, II, III, above.) 
 
 54. The first operation (I, 53), that of determining the kind of bridge 
 or girder, must be left entirely to the discretion of the designer ; at least 
 no rules can be laid down for bis guidance : the several conditions which 
 would influence the decision, such as nature of site, obstacles to be crossed, 
 facilities of construotion, &o., being infinitely diversified. 
 
 55. Tbe areneTnl cross section of a bridge must also be left to the 
 experience and discretion of the designer, as no definite rules can be giveu 
 
 * The system of reftnrdin!^ particular members or parts of a girder as resistiug 
 the particular strains for which they are adapted, and these alone, is perhaps the 
 most generally received. (See 6J3 and 191.) 
 
OALCHLATION OF STRAINS. 
 
 upon the questions of tlie relative advantage and economy of the many 
 systems which have been adopted and suggested. 
 
 GENKRAL PROPORTIONS. 
 
 56. Ttie ce<itral depth of straight independent girders may be made 
 from ^ to J, of the span. The greatest economy , of material is perhaps 
 obtained at J^. 
 
 For continuous girders, or girders fixed at the ends, the depth may vary 
 from fj to ^ of the span. 
 
 57. \Vhere convehieni it is in most cases advisable that girders with 
 fixed loads and with oblique or curved flanges or booms, should have their 
 depths varying as the moments of rupture (1), that is, as the ordinates in 
 the dlagriiiAs (5) given for several cases (T — 33) of loading, &c. (T6), 
 
 It follows from the above, that bow-string girders, arches, suspension 
 chains with stationary uniformly distributed loads, should (the latter will 
 nearly) have the form ot a parabola {9Si). 
 
 58. Diagonal bracing will, as a rule, be most economically disposed at' 
 an angle of 45° with a vertical, 
 
 59. The design having, as supposed, advanced thus far, it is advisable 
 that a skeleton elevation of the bridge or girder be drawn to a moderately 
 large scale, that the strains on the various parts about to be calculated 
 may be figured thereon. 
 
 SECTION II. 
 
 Calodi.vtion of the Stkaiks oif THE Various Parts. 
 
 (Embodying Processes IV, V, p. 24.) 
 
 60. Apitroxlniat-c Estimation of tlic 'IVcferlit of the Sirilctnre. 
 
 1st. If no other source of information be at hand* (COA), assume a pro- 
 bable weight from the data of experience. 
 
 2nd. Calculate a few sections by the formulae, fee, given hereafter, on 
 the supposition that the bridge or girder is loaded with the just-assumed 
 weight uniformly distributed, and the maximum extraneous load that is to 
 be brought upon it. 
 
 •3rd. Make a second approximation of the weight from the few sections 
 just calculated ; allow a percentage for contingencies (which may vary 
 from 5 to 25), and if the total be at all near the first estimate, it will 
 generally be sufficiently correct to stand for the weight of the structure in 
 the final calculation of the strains, &c. 
 
 • See B. Baker's diagrams and tables giving weiglus of girders up to 200 feet 
 span. 
 
26 
 
 OALCOLATIOK OP STRAINS. 
 
 or WrougUt-lrou Girder Railway 
 
 GOA. Approxliuntc iVclelitH 
 ftrltiges. 
 
 Single Line of Way. 
 For ■ 30 feet spans .5 owt. per foot run. 
 ■I 60 ,, 6 ,, „ 
 
 „ 100 „ 9 „ 
 
 „ 150 „ 12 „ „ 
 „ 200 „ 15 „ 
 
 61. The weight of the beam, girder, ibc, must alwayt he introduced 
 into the calculation, for the atraim. It may generally be considered as a 
 uniformly distributed (of course stationary) load. Where the load is 
 stationary, and also uniformly distributed, (viz., cases 9, 19, 25 — 30, and 
 38, 44, 41 — SO), the unit of weight of the girder may be combined with 
 that of the load proper. But in iiearly all other cases it will be necessary 
 to regard them separately. 
 
 FLANOBi) Girders, with Thih Continuous Webs. 
 Fig. 36a. . 
 
 OS. Distlnci tUttetlolu of the Flanses and the Weh. In pages 
 25 to 29 inclusive, the web is considered to take no part in resisting 
 the horizontal strains (I), the whole of which will be provided for in the 
 danges. 
 
 Though this is not theoretically correct, the error is practically so small as to 
 be disregarded with safety. 
 
 Neither are horizontal BaAgei considered to take part in resisting the 
 the shearing forces (1), the whole of which will be provided for in weh 
 (04, 80, 191). 
 
 Strains in, Planget Generally. 
 
 63. Flanges or parts of flanges, perpendicular to the action of the load 
 on 2k girder, have to resist only the bending effect of the load, which 
 depends on the moment of rupture (I). 
 
 64. When, however, a flange or part of a flange is not as just supposed, 
 it stiSiirg ab additional strain, which is part or all (as the case may be) of 
 the shearing force (1)) the amount of additional strain depending on the 
 inclination of the flange (75). 
 
 fhds in prdttaary )>racti(!al cases, where the action of the load is vertical, the 
 straia on the itange increases the more it wanders from th6 horizontal position. 
 The extra strain it soffers, however, iUocUaes proportionally the strain on the 
 web (80, 81). 
 
FOLL-WBBBED StfcAIOHT GIRDEBSi 27 
 
 65. Nature of the strains. At any vertical section of a girder, the 
 strains in the two flanges are of different kinds (4) : — 
 
 1. When the action of the load tends* to malce the beam convex on 
 its lower surface, then — 
 
 the upper flange is in compression (168) ; and 
 the lower flange is in tension (161). 
 
 2. When the action of the load tends* to make the beam convex on 
 its upper surface, then — 
 
 the upper flange is in tension, and 
 the lower flange is in compression. 
 
 Strains in the Web Generally. 
 
 66. Thestrains borne' by the web are the shearing forces (I, .34), due to 
 the transmission of the Tertieal pressure of the load to the points of sup- 
 port. Their amount (and sometimes manner of action) are greatly modified 
 by the longitudinal form of the adjoining flanges (64, SO). 
 
 GIRDERS WITd EARALLKI. STRAIOHI FLANGES. 
 
 Flanget, 
 
 6T. To And the Amount of Strain on either Flanare at any 
 
 vertical section (6S), — nivld« the Moment of Uuptnre, as found from 
 the formulie or diagrams, pp^ 2 to 15, by the depth of the srirdeF, i. e. 
 
 by the distance between th6 centres of gravity of the sections of the two 
 flanges., 
 
 68. At any vertical section, the strains on the two flanges are equal in 
 amount, but opposite in their nature (65). 
 
 69. The strains in the flanges will vary throughout as the moments of 
 rupture (1, jt), and therefore as the of dinatcs in the diagrains (S). 
 
 TO. Bt Diagram. — If the ordinate, in the diagram for the case (s), *at 
 any point of the girder, be made to represent on a scale of parts the 
 strain on the flanges at that point, the strain at any other point may be 
 MttuUVed off from the diagram. 
 
 71. The strains are either direct tension (16T), or direct compres- 
 sion (168). 
 
 * The action of the load ma girder supported at both ends, and having its 
 lower surface concave (see Plate I.) will lessen the concavity, and so ttnd to cause 
 convexity. 
 
28 OAlODLlTIOIt OF SIBA.III8. 
 
 Web. 
 
 19. The Strain upon any Vertical Section of the Web 1* eanal 
 to the " Shearlnir I'orce " (1, 34) developed at that section, for tbe 
 Tfttes eee (84 — 62). 
 
 T3. If the girder have more irebs than one,* the strains as fbnnd l>y (72) 
 mnst be divided by the number of webs for the strain in each, 
 
 74. The nature of the strain in tiie web is a shearing (proper) (17l). 
 
 OlkCXttS WITH OUBTED OB OBLIQVE FLAD&^S. 
 
 Flanges. 
 
 76. To find the Amount of Strain on either Fiance at any point, 
 — Divide the moment of rupture (formulse and diagrams, pp. 2 to l6) by 
 the depth of the girder, the vertical distance between the centres of gravity 
 of the sections of the two flanges, and multiply the quotient by the secant 
 (186) of the angle inhich ihejlomge, or a tangent to U at the point, makes 
 mith a horizontal, f 
 
 76. Tf the depth of the girder vary throughout as the moments of 
 rupture (I, 57), i. e., as the ordinates shown in the diagrams for those 
 moments (S), then 
 
 (a), the strains in the flanges mil vary as the secants (see 77, 186) of 
 the angles ef inclvaatiim to the horiion. So that 
 
 (h), if the strain at a horizontal parti be hmum, the strain at any 
 other part may be found by multiplying the fanner by the secant (see 77, 
 1S6) o/ the angle of inclination of the latter. 
 
 77. IToTB. I'or the operation of muldplying by the secant of an angle there 
 
 may be Bubstituted a geometrical process. Let 
 Yig, 37, Aug be part of a curved flange. Let the value of 
 
 /Moment of Rupture at Bx. i*j * t , .u- 
 
 ( DepthofGl^er.tB )^ represented to scale by the 
 horiioiUal line DE, measaring it firom some point 
 (B!) on Oie tatigent (f £) to ABC at B. If D7 
 be drawn vertical, >. e , perpendicnlar to AE, then 
 
 F B wiU = "°'°""°'^'"^'""*° X seoant DEF, 
 r a wiu— uepth of Girder at B * ««»"' ui^n. 
 
 (DEF being the angle of Inclination of the flange 
 atB.) 
 
 78. The strains in the flanges are either direct compression (168), or 
 direct tension (167). 
 
 Wei. 
 
 7A. In girders with obli^ne or <!urved flanges, the strips in the web are 
 affected by the incUnations of the flanges. 
 
 • As in "box" or "tubular" girders. 
 
 t This is but a close approximation to the truth. Bee also fttot-hote (t), p, 61, 
 
 t Which will generally be at the centre of whole girders. 
 
FULL-WEBBKD OURVJCD AND OBLUJUE GIRUERS, 29 
 
 SO.* For ihv Stmiu o» tli<* M'rb nt nnr VrDivsil Srrtioii, — FiiiJ 
 the shearing force (I) developed at that section (;t4), and modify it as 
 directed below. 
 
 If at the vertical section the flange in compression (US, JS) be inclined 
 
 down j , ' > , or the flange in tension be inclined down j '°™ ! , the 
 
 nearest point of support, j '^" ??° f the vertical component of the strain 
 
 in that flange j ' j the shearing force. The sign of the result need not 
 here be regarded. 
 
 If S = strain in flange, 
 
 6 = angle which the flange or a tangent to it makes with a 
 horizontal line, as D E P (fig. 37) ; then, 
 S X Sin 6 = the " vertical component referred to above." 
 
 81. By «ion»ti'iicti«u.— Drawavertical A B = sheaT- '^^ ^^' 
 
 ing force. If the flange in tension («.'», IS) be inclined 
 down from, or the flange in compress-ion be inclined 
 down to the nearest point of support (we will suppose 
 the latter case), — Draw CD representing both in direc- 
 tion and amount the strain in that flange, so that a 
 horizontal drawn through D shall cut A. Then CA 
 will be the vertical component of the strain in D. 
 
 Again, if the flange in compression be inclined dovn 
 from, or the flange in tension be inclined down to the 
 nearest point of support (we will suppose the latter case),— -Let it be re- 
 presented byEF, Obtain its vertical component EB, which should be 
 added to A B, in the same way that A C was subtracted from it. Then 
 C E will be the total resulting strain on the web. Should the amount to 
 be subtracted exceed the sum of the original and that added to it, the 
 difference must still be taken. 
 
 SI A. The ."train taken must be that nbtainiug when the shearing force being 
 considered is developed. For instance, with a load gradually advancfnt: across 
 a beam supported at both ends, the maximnm shearing? foroe will be developed at 
 the centre, when the load covers only half the span (45), at wliich time the moment 
 of rupture, from which the 3afif?e strain is fpund, vriii be that given by (20), and 
 not by (19), Which latter would be used when isaking the calculation for the 
 flanges. 
 
 88. If the girder have more webs than one, the strain as found by (80 or 
 81) must be divided by the number of webs for the strain in each, 
 
 83. NoTK. It will obviously follow, from the abov^ rules, that in 
 girders with curved or oblique flanges, the maximum strain in the web does 
 not necessarily occur when the maximum shearing strain is produced. 
 
 * The rules, &c., given here are not advanced as mathematically accurate, as 
 there is much connected with the effect of curvature in the flanges on the strains 
 in the web that hits not been satisfactorily determined. 
 
so CALCOLATIOK OF STKAIHS. 
 
 GiRDBKS WITH Webs or Opek EiiAoiNa. 
 Fig. 3Sa. 
 
 OEBERAIi KULES FOK OFEN-ITEBBIID GIRDEBS. 
 
 JBoomg. 
 
 84. At an; vertical section of a girder the strains in the two booms are 
 opposite in kind. (See 65, which also applies here.) 
 
 85. When the girder is loaded at the joints (87), the strain in an; hay 
 of either boom is constant throughout its length, and acts only in the 
 direction of its length. 
 
 86. A bay cannot be in direct compression and tendon simnltaneously 
 (SI). 
 
 87. Wheneaer the load or part of the load upon a girder is situated 
 betweenihe two extremities of any bay, that bay must be considered as a 
 loaded beam, and the strains in it calculated and provided for accord- 
 ingly. 
 
 Web. 
 
 88.. The strain on any brace is constant throughout its length, and acts 
 only in the direction of its length. 
 
 89. A brace cannot suffer compression and tension simnltaneously. 
 
 90. If two or more strains, not all of the same kind, be acting upon a 
 brace at the same time, the total actual strain in the brace will be equal 
 to the algebraical sum of those strains.* 
 
 91. The Nature of the Strains in the diagonal braces of girders 
 symmetrically loaded will be — 
 
 (a) in all bars inclined down to the nearest support — compression (168). 
 (6) in all bars inclined down from the nearest support— («Ksio» (IC7). 
 
 98. If the girder be not symmetrically loaded, some of the bars will be 
 Eu\)ject to strains of a nature-contrary to that stated in (91). 
 
 9.t. Counter stralnH.-^With a, moving load, some of the bars will be 
 subject to strains both of compression and tension, depending on the 
 position of the load and the proportion it bears to the weight of the girder. 
 The strain not according with (91) is known as the "counter strain." 
 
 Note. (OS) and (93) do not apply to semi-beams or cantilevers. 
 
 • Equal to tbe difference between the sum of the tensile and the sum of the 
 
 cutnpressive strains. 
 
SIBAISHI WAKBEN SIKDERS. 31 
 
 94. If the load be brought only upon one boom, * an; two bars forming 
 an apex on the unloaded boom may be termed a "pair ;"f if these make 
 equal angles with the boom they are equally strained. 
 
 95. If both booms be loaded, then — 
 
 (a) in a warren (zig-zag or single triangle) girder there will be no 
 " pairs." + 
 
 (i) in a straight latticet girder, if the load be equally divided between 
 the two booms, bars iiitersec]bing at the mid-depth of the girder, and 
 making equal angles with the booms, may be termed a '-'pair," and are 
 equally strained. 
 
 CALCpLATIO,^ BY MEADS OF THS MOMENTS OP BUPTURB (pp. 2 to 16) 
 AND TUB SHEARISG FORCES (pp. 16 to 23). 
 
 9C. lonilii regularly concentrated at the apices of the diagonall, 
 may be considered as itniformiy distributed loads (w, in % 
 and 34) when the moments of rapture (1) and sbciiring Fig^^S^ 
 forces are being determined, so long as the concentrated 
 load at any apex is equal tp half the euiq of f,h^ siipposed 
 uniformly distributed load on the two adjoining bays. ' Thus 
 the weif^ht at A (Fig. 39J must be =^ ^ that at C, before the load ban 
 tie considered as upifqrmly distributed. If the uniformly rlistri- 
 bnt'eii fooj} "were on tl)e "bottom Ijdom in the flg., the weiRhts concentrated at 
 jD and I^jyoiild be equal. 
 
 OIRSEKS ^IIH FASALLEL SliiAIGHT BOOMS. 
 
 97. Note. As the depth of the girder (a constant divisor in calculating 
 the strains from moments) is uniiorm, the diagrams (4, 5) may be con- 
 sidered to give strains instead of moments, if the vertical scale be multiplied 
 by the nun^ber of unit^ of length in the depth of the girder. 
 
 Thus, suppose the diagram bad been drawn to a vertical scale of six 
 tons to the inch, and that the depth of the girder was two feet (the |^pt 
 being the lineal unit used in the case), then the ordinates in the diagram 
 may be considered to give actual strains instead of moments if they be rea4 
 off on a scale of three tons to the inch ; three tons to the inch being a sc^le 
 twice as large as six tons to the inch. 
 
 98. ICoiE. The Uepth of the Girder is Ike disla^e between the 
 centres of yravity of the cross sections of the booms, and must always be 
 expressed in the same" units of measurement as the length of spap (S, 34). 
 
 99. Warren Clrder (single triangle), isosceles Bracine, Loaded oh 
 Que Boom Only — for any case given under Moments of Rupture, 
 pp. 2 to 15. 
 
 BooMS.g— For the strain in any bay (84 — 87) of the unloaded booni,;^ 
 Divide the moment of rupture (^x in the formula!, or the ordinate in the 
 
 ^ Verticals from loaded bays to opposite apices may be considered to distri- 
 bute the load between both the booms. ' ' "" 
 
 t When two bars are said to form a pair, it is meant that the Mame amount o/ 
 tJte vertical pressure of the load is transmitted through them both. 
 
 X Any girder' whose web consists of more than one system of trian^ulation is 
 considered a " latBce." § See also 91. 
 
32 OALOuiiATioir or straisb. 
 
 disgram, for the case T to 33) at the opposite apex, by the depth (98) of 
 the girder. For any bay (84—87) of the loaded boom, take the arithmetical 
 mean (half the sum) of the strains in the two opposite bays of the unloaded 
 boom. 
 
 Web. — For the strain on any pair (94) of diagonals forming an apex on 
 the unloaded boom, — Multiply the shearing force (S Hx in the formula, or 
 the ordinate in the diagram, for the case 34 to Sa), developed at the apex, by 
 the secant of the angle which the brace makes with a vertical ; or increase 
 the shearing force^ (as above) in the proportion of the inclined length of 
 the br%ce to its vntical depth. For the counter strains (93) from jioving 
 lo^ds, sf e (IflO). 
 
 100. The Connter 8t»lng (93) in girders not continnons will be given 
 by the smaller value of the two shearing forces given in 42 or 4S, 
 or by the ordinates (35) to the lines (AK^'K B. §^. 80,' or A E, £ B, 
 fiig. 82) in the diagrams which accompany those ^verid cases ; ,tho values 
 thns obtained being, of course, multiplied by the 8eet.'(n, ISft) of the 
 angle between the brace and a vertical, and subject ta indeed all tiui other 
 stipnlations made (under " Web ") for the strains normal. 
 
 101. In continuous girders with moving loads the connter strains are 
 indefinite, but may be supposed to act equal in intensity to the values 
 allowed for the regular strains, and for some distance on either side of the 
 point of minimum shearing force * (51, SS). 
 
 lOi. yranf^ Girder, Isoseeles or scalene BmcIiiK, wltb the I«ad 
 bTonRbt danally upon Both pooms't*— for any case given under 
 moments of rui^UT% pp. 2 to 1 6. 
 
 Bpojw. J— Foir the strain in any bay (84 — 87) of either boom, — Divide 
 the moment of ruptur^ ^Miin the fonr ite, or the ordinate in the diagram, for 
 the ca8$ 3 tp 33) at the opposite apex, by the depth (98) of the girder. 
 
 Wbb. — For the strain in any brace, multiply the shearing force (8 Hi 
 in the formijls, or the ordinate in the diagram, for the case 34 to S9), developed 
 at t^e mid'iength of the brace, by the secant (126) of the angle which the 
 brace inaikes with a vertical ; or increase the shearing force (as above) in 
 the proportion of the inclined length of the brace to its vertical depth. 
 
 For the counter strains (93) from moving loads, see (100^, 101). 
 
 103, Gltrder with One Syst^jn of Vertical ^tinti gf^i Inclined 
 
 Tlje», laajJ.ed either on the top or 
 
 gig. 40. bottom, oi both — for any case given 
 
 under moments of mptnre, pp. 2 to 
 15. 
 
 BoouB.:^ — For the strain in any 
 bay (C D for instance) of either boom 
 (84—87), — Divide the moment of 
 rupture (^x in the formnlee, or the ordinate in the diagram, for the cose 7 to 
 
 * Actually there will be counter stains for a greater length of the several spans 
 than there will be necessity practically to provide for them in, as their amount 
 will always be exceedingly small near the piers compared with the other strains. 
 
 f This is accomplished when there are verticals at the unloaded apices, 
 
 X See also 97. 
 
SISAIGHT WARBES SIRSEHS. 33 
 
 33) at the opposite apex B, by the depth (98) of the girder. The strain in any 
 bay of the upper boom in a whole girder (A B, fig. 40^ for instance,) will be equal 
 to the strain m the bay (C D) pf the lower boom on the isiiiltspw ei^^ of t^e oppo- 
 site apex ; for the moment of rapture will be taken at the seme p6m1.' (horizon^ 
 tally considered) for both the bays. The bays G E and F G are unnecessary for 
 an mdependent whole girder. 
 
 Wkb. — Stntts. If the load he only on the upper boom, the compression 
 on apy strut will equal the shearing force (S Hi iif the formute, or the 
 ordinate in the diagram, for the case 34 to 62), developed at the middle 
 of the adjoining bay on the far abutment side ; * if only on the lower, 
 then that at the middle of the adjoining bay on the near dhvimeni side ; 
 and if equally on the upper a/nd lower booms, then that dev0oped at the 
 strut itself. 
 
 Ties. For the tension on any inclined tie, — Multiply the shearing force 
 (34) developed at its mid-length by the secant of the angle it wakes with 
 a vertical, or increase that shearing force in the proportion of the length of 
 the tie to the length of a strut. 
 
 Counter Strains (93 — 101). In a girder vith the bracing disposed 
 similar to that in the figure, a ni!iTin|,;leadj,- or a,.,3jp»J^^isering less 
 than either half of the girder would "jfoifuop ten§op' p some of the 
 verticals and compressiqri in some of the diagonals (to eatjmate which see 
 100). This can be obviSfed by introducing other ties (as H C ajid H F) wliioh will 
 Buffer the tension otherwise bronght on the verticals. These ties would, however, 
 be useless with a st^tipnary symmetrical load. 
 
 104. Warren Clrder witli scalene Bracing, Loaded on one Boom 
 
 — for any case given under moment of rupture, pp. 2 to 15. 
 
 Booms, t — iPor any'tftj {84 — 8 J) in the unloaded boom (CD,, fig. 41, for 
 instance), — Divide the moiiient qf i;uptnre (Mz in the formulis, or the ordi- 
 nate in the diagram, for the case T to 3S) at the"oppoaite apex (G), by the depth 
 (98) of the girder. 
 
 The strain on any hay of the loaded boom (as E G, fig. 41) will bear the 
 same relation to the strains in the opposite bays (H C ^d C D) as its apex 
 does to their apices horizontally considered (that is, as the point F does to 
 the ppillts E and Ci). Therefore — 
 
 F F 
 Strain in E (J = Str. in H C -I- --- (str. in C D — str. in H C). (See 105). 
 Jii ur 
 
 Web. — For the strain in a pair (94) of liraees forming an apex on the 
 unloaded boom, — Multiply the shearing force (SHa-inthe forraulaj, or the 
 ordinate in the diagram, for the case 34 to S!!), developed at the middle of the 
 loaded bay included^ between the bars, by the secants of the respective 
 angles wfiirch they foi'm with, a vei-tical ; or increase the shearing force (as 
 above) in the proportion of the inclined length of the respective braces to 
 their vertical depth. 
 
 For the counter strains (93) &om moving loads, see 100. 
 
 105. method of a|;plyine tbe Dlasramg for tbe Sloments of 
 Rupture to the foregoing cases.t 
 
 * Or ilie " no abutment" side in a semi-beam, 
 t See also 97. 
 
S4 
 
 CALCVLATIpII Qf SIRAIS3, 
 
 As before stated (96), leads w}iieh are in reality concentrated at the apices may 
 be considered as unitonnly distributed if they be such as a uniformly distributed 
 load would bring on the apices in question. 
 
 Fig.«. 
 
 Q 
 
 Place the diagmiu of the moments of ruptoire (1) for the case of loadinpt, 
 support, 4o, (T to 33) immediately above (or under), and dvawn to the same hori- 
 zontal scale as the skeleton elevation (88, 50) of the girder. (See flg. 41, in which 
 the load is supposed to be uniformly diaiributed (00, 19) nlong the upper boom.) 
 Draw verticals right IhroORh the diagram at each of the Uiackd apices, whether 
 on one or both booms{as IS Jl' and & G fig. 41.) If the lines to or from wluch the 
 .<pr4!natei in tbP diagram are supposed to be taken be curved, draw chords to 
 ^tihai onrva af the intersections of the verticiils just drawn with that curve (bs in 
 fig. 4t.) A polygoiKJvjll be thus formed, and the strain on any bay of either a loaded 
 or unloaded boon) will be given (97) by the ordinate to that polygon taken at the 
 opposite apex. (See also 100.) 
 
 For the diagonals the shearing forces may be taken direptly from the diagrain 
 (ST th§ case (.t4 to 89) dr^wn to a large scale (SKi), sulnecl to ^e modifications 
 given under '' Web ■' io the cases above. 
 
 106. Nof^. If all tbe line; in the diagram for the moments of rupture, 
 to or from whish the ordinates are directed to b« j;aken, be etraight, thw 
 the strain in any bay of eitlier boom, whether loaded or imloaded, ni»y be 
 found frpBi the ordinate tftken »t the opposite ^pejc, By dividing the vftlu^ 
 of that prdinate by4b9 detitl) of the girder (unless 97 be complied with, 
 when the latter operation musl^ of course, be omitted). 
 
 1Q7. InMlce (iiirdec ^dthgul a central lay, and if of more than two 
 systems p/ triangles then with a complete 
 Sj/atem of triungulation in the half 
 spari (o> A B C! D K F Q), bMt an;/ num- 
 ber of intermediate systems losulcd 
 miirorinl}- on on« bonni for an vtiiinl 
 flistniiee on eitlivr side of flic mill 
 span (see no); may be applied also to 
 SM^s of continuous girders (30— 3.?). 
 
 BsuWS.-'^Flsce the digram (5, 6, 97) for 
 the monieptg of luptpH (1) for the case 
 Qboye (or below) the outline sketch of the 
 girder (59) »s in fig. it. 
 
 Draw ebords mi t^ng^nts (as shown 
 in fig. ) at points (0' E' Q') eowesponding 
 with the diTinoQi made by the "primary 
 triiiDgles." Tyro polygons will thqs be formed, one by (or including*) the 
 
 Wbei)(?l) is applied the load being oentrally situated,. tjie chordg and tan. 
 ifpnte will of coarse extend only along tbe parabola C B D (8g. 15). 
 
 Fig. 43, 
 
STRAIQHT LATTICE SEMI-GIRUER. 
 
 35. 
 
 ebards, another by (or including*) the tangents. For the strain in any- 
 bay of the unloaded boom take the ordinate to the nrcumscribed polygon,* 
 at the centre of the bay ; and for any bay (as E H) of the loaded boom take 
 the ordinate to the in»criied polygon* at the centre of the bay. If the ordi- 
 nates represent the moments of rapture (4, 5, 97), divide this value by the 
 depth (98) of the girder. 
 
 Web. — Divide the load per unit of length (34) by the number of sys- 
 tems of triangles (^ the number of bays in Che base of one of the primary tri- 
 angles) for a new unit of weight, w, ; thus- = Wi or - = »",. Then for 
 
 the strain on any pair (95) of lattice bars, — Multiply the shearing force 
 (S H^ in the formulse, or the ordinate in the diagram— .S4, 35,— calculated or 
 constructed with the new unit u, , or tc'i ,) developed at their apex by the 
 secant of the angle they make with a vertical, or increase that shearing 
 force in the proportion of the inclined length of the bar to its vertical 
 depth. 
 
 108. Any lattice Hemi-Kirder loaded on One 
 
 only. 
 
 If distributed, place the 
 Pig. 43. 
 
 Bogus. — if the loud be concentrated, see IIO. 
 diagram for the moments of rupture (4, 5) 
 immediately above (or below) the outline 
 sketch of the girder (.19). 
 
 Trace out the system of triangulation 
 which terminates fairly, at the extremity 
 of the girder (see the thick lines in the 
 fig.) To the curve of moments (4, 5) 
 draw chords and tangents at points 
 ( E', D') corresponding to the divisions 
 made '.l)y the triangles just traced out 
 (D, E, fig. 43). If necessary for the con- 
 struction, the diagram may be continued 
 into the abutment. Two polygons will 
 thus be formed, when, whatever be the 
 number of systems of triangulation, — 
 For the strain in apj bay (S^-r»j;i of 
 the unloaded boom take tlie ordinate, 
 at the centre of that bay, to the circumscribed polygon ; and for apy bay 
 of the loaded boom, to the inscribed polygon. If the ordinates represent 
 the moments of rupture (*, 5, 97), divide their value by the depth of the 
 girder (98). 
 
 Web. — If the load be concentrated, see (110). If distributed, divide 
 the load per unit of length (w) by the numSer of systems of triangles 
 (= the number of bays in the base of a primary triangle (4 in fig. 43) fbranew unit 
 
 of load ( - = Wi ). Then for the strain on any pair (95) of braces forming 
 
 an apex on the unloaded boom, — Multiply the shearing force (s Hx in the 
 formulae, or the ordinate in the diagram for the case, calculated or constructed 
 with the nf u' imit i;,— 34, 35), developed at their apex, by the secant of the 
 angle which they (eaeh) make with a vertical, or increase the shearing 
 force (as above) in the proportion of the inclined length of the brace to its 
 vertical depth. 
 
 ' See note, p. 34. 
 
 D 2 
 
36 
 
 CALOOLATIOS OF STRAINS. 
 
 109. Kattlce'Glrder, if whole, and of more than tvm aysiema of tri- 
 angles, then not withoiil a etmiral apex on one or other of the booms (see 
 4g. 44), loailed nnlformly and equally on both the Boomn for an 
 eaual Distance on either side of the Slid-span, may be applied to con- 
 tinnoua girders (30 — 33); 
 
 Boous. —Place the diagram (4, 4) of the moments of rupture (1) immeaiately 
 
 above (or below) the outline sketch (6:t 39) 
 
 Vig- 44. of the girder (as in the fig., where tlie load 
 
 is 'supposed to extend over the whole 
 
 length of the girder). 
 
 Where the line to or from which the or- 
 dinates are taken is curved, draw chords 
 to it at points (E' P'G'C) correspond- 
 ing to the several apices (on both booms) 
 of tb% system of triangulation which has 
 an apex at the mid span (E F G in fig.) 
 For the strain in any bay (84—81) of 
 either boom, take the ordinate to the 
 polygon,*. thus formed, at the centre of 
 that bay. if the ordinates represent the 
 moments of rupture (4, 5, 9T), divide 
 their value by the depth of the girder, 
 
 WsB, — Divide the load per unit of length (w) by the number of systems 
 of triangulation (= the number of bays in the base of one of the primary 
 
 ('w \ 
 
 - = till ). 
 
 strain in any lattice bar, or pair t of bars. Multiply the shearing force 
 (S Hi in the formulse, or the ordinate in the diagram for the case — 34, .15), 
 developed at its or their mid-length, by the secant of the angle it or they 
 (each) make with a vertical, or increase the shearing force (as above) in 
 t]^ proportion of the inclined length of brace to the vertical depth. 
 
 I09A. If there be only Two Systems of Triangvlation, Then for the 
 strain in any bay (84 — 87) of either boom. Take the arithmetical mean 
 between the moments of rupture at the two ends of the bay (p and ;), and 
 divide it by the depth of the girder. 
 
 „^ . . ■ , Mp + M, ' 
 
 Stram in any bay = . ■ ■. 
 
 2u 
 The ohords in the diagram irill, however, give the same result. 
 
 110. Concentrated loads on Ijtttice Girders. If a lattice girder bo 
 subjected to theTaction of a concentrated load at any apex, as at D (fig. 42), 
 the system of triangles uptn one apex of which the load is situated 
 (ABODEFG, im.) should be oonsidered as constituting the sole web of 
 the girder — ^that is, as far as the concentrated load is concerned, for there 
 is also the distributed weight of the girder (61). The strains from the con- 
 centrated load might be calculated as if the girder were a warren, and as if the 
 intermediate braomg did not exist. Indeed, to introduce other systems of trian- 
 gles for a stationaiy concentrated load would be an error in design, unless tihe 
 boom upon which the load is placed be made sufficiently rigid to distribute part 
 of the pressure upon adjacent intermediate apices, in wmch case the strains 
 would be very indefinite. 
 
 Then for the 
 
 * If it be an application of case (SI) the lines A F and O B (fig. 16) must be con- 
 sidered to form .part of the polygon, 
 t There will be no pairs (95), cmless the number of systems of triangles be even ■ 
 
ODRVED OR OBHQUE OPEN-WBBBED GIBDEKS. 87 
 
 OIRSEBS WITH CDRVEO OB OBLIQUE BOOMS. 
 
 111. Any Carved of Abllaue wltole or Seml-elrder witli a Slncle- 
 trlauKUlar IVeb, loaded on One Boom.* (This will include many roof 
 principalB, bow-strinsf girders, bent cranes, die.) 
 
 BoouB.— If alternate braces be Tertical, then for- the strain in an; 
 bay of either boom; if not, then for the strain in any bay of the im- 
 loaded ]Mom only, — Divide the moment of rupture (m< in the formule, or 
 the ordinate in the diagram (4,5), for (be case 7 to 33) at the opposite apex by 
 the length of the perpendicular let fall from that apex on to the bay. For 
 the strains in the loaded bays, when all the braces are inclined, the best 
 way is by the diagram as follows. Place the diagram of the moments of 
 mpture, immediately above (or below) the outline sketch (53, 59) of the girde^^ 
 If the lines in the diagram to or from which the ordinatea are directed to' 
 be taken be carved, draw chords at points corresponding to the posi- 
 tion of the apices in the loaded boom. Then for the strain in any bay of 
 the loaded boom, divide the value of the ordinate to the just-formed 
 polygon, taken at the opposite apex, by the length of a perpendicular let 
 &U from that apex on to the bay (121). (See 154.) 
 
 Web. — For any brace, first determine the shearing force (34), acting at 
 the middle of the bay on the loaded boom, which forms part of the same 
 triangle with the brace in qaestiou. Next, find the vertical components (80, 
 81) of the strains in those bays which are opposite sides of a quadrilateral 
 figure, whose diagonal is the brace in questisnr. ' If either of these bays be 
 part of a boom in tension, and sloped as a strat (91), or part of a boom in 
 compreosion and sloped as a tie (91), add the vertical component' of the 
 strain in it (alr^y found) to the shearing force first obtained. Again, if 
 either of the two same bays be part of a boom in compression and sloped as 
 a strut, or part of a boom in tension and sloped as a tie, tuibtraet its 
 vertical component (already found) from the'i'esult of the last operation 
 (the addition, if any). The total resulting quantity must then be multi- 
 plied by the secant of the angle the brace makes with a vertical, or 
 increased in the proportion of the inclined length of the brace to the ver- 
 tical distance between its ends. If the sign of the reiuU be negaiwie (— ), 
 it shows that the nature of the strain on the brace is apposite to that 
 which its position would have indicated according to the general rules (91). 
 If the brace be horizowliU, the shearing force must be disregarded, and the 
 horizontal components of the bays substituted for their vertical components 
 in the process detailed abotlt. The strain in a horizontal bay can have 
 no vertical component. 
 
 112. Any Curved or Oblinne whole or Seml-Elrder with a slncle> 
 triangmlar Web. loaded equally on Both Booms. 
 
 Booms. — For the strain in any bay of either boom, — Divide the moment 
 of rupture (Mz in the formulee, or the ordinate in the diagram (4, 5) for the oa£e 
 7 to 33), taken at the opposite apex, by the length of a perpendicular let 
 fall from that apex on to the bay. (See 154.) 
 
 Web. — As in the last case, excepting that the shearing force must be 
 taken at the mid-length of a line joining the centres of the bays, which are 
 two opposite sides of a quadrilateral whose diagonal is the brace. 
 
 * SeefootnoteC"), p. 31. 
 
09 CALOULATIOH OF STBIIHS. 
 
 llSa For Continuons Girdera, and drdem flxed at the Ends, tlie 
 foregoing methods may be used in connection with cases 24 to 33 and 4$ tii 
 53, provided that the conditions stated in the latter and belov (113A) are 
 complied with. 
 
 II3A. Wherever there is a negative moment of rupture at (or for 
 safety in practice near) the abutments, the girder should be anchored down 
 at its extremities. (See foot notes, pp. 13 and IS.) 
 
 r 
 114. Flxlns the Points of Inflexion of Continuous Girders. — The 
 
 points of inflexion (23) may be practically fixed at any part of a continuous 
 girder subject to a movinjg load, by severing either boom at that part ; 
 if the upper boom, the parts thus severed must be prevented ft'om coming in 
 contact. The structure is thui resolved into a_feriea of independent girders, 
 the strains upon which can then bi most T^ecSlMy calcvlated. 
 
 Fig. 46. 
 
 ExAHFliE. — In the accompanying fig.f4S), by aBvering(or really removmg) the 
 bays of the tipper boom opposite and D, AC and D B become virtually senn- 
 giiders, each having to sustain in addition to the distributed load upon its length, 
 half the totta load on G D, suspended at the extremity (11, 39). C U is smiply an 
 independent girder supported at both ends. 
 
 115. Thepointa of nation may ie conndered at fixed in those con- 
 
 Fig. 46. 
 
 on them may be obtained trom 155 or 118 
 
 tinuons girders, and girders fixed (or 
 tied back) at the ends, whose depths 
 vary as (or nearly as) the moments of 
 mptnre. The strains in these also 
 may be calculated as if the several 
 divisions were independent girders. 
 
 If theiteiision members running down 
 from thS'fefrera (as in flg. 46) be made to 
 act simplyw suspension chaiiiB, the strains 
 
 CAlCUlATIOir BT THE COMPOSITION AND RESOLUTION OF FORCES. 
 (applicable to all oases of OPEN-WBBBED INDBPBHDENT GIRDEB8.) 
 
 116. The following principles should be applied to the calculation of the 
 strains on the various members of an open girder (discontinuous) by (1st) 
 finding the reaction of the supports from any weight in the girder, and 
 (2n<J) tradiig this reactionary pressure throughout the various parti. 
 
COMPOSIIIOH AND KXSOLnilON OF FORCES. 
 
 IIT. Reaction of 8apport«. Let a and b equal the distances from the 
 sapports A and B, respectively, of the 
 centre of gravity of a load (W) on a Fig. 47. 
 
 beam. Then, 
 
 Pressure on A = Seaction of A = W ~ 
 
 Pressure on B = Beaction of B = W -. 
 
 t 
 In girders, ptoperly so called, the sup- 
 ports are supposed to be capable of resisting. vertical pressure only. Their 
 reaction can then only be vertical, and this must be borne in mind.* 
 
 If two forces {ab and 
 Fig. 48. 
 
 118. ComposlUan and Resolntton of Forces. 
 
 a e) acting at a given point (a), be represented both 
 in direction amd amount by the two adjacent sides 
 (a b, a c) of a parallelograin, an equivalent force 
 will be represented in iath direction and amount 
 by the diagonal (a d) of that parallelogram. 
 The converse of this is also true. 
 
 119. If three forces, acting at a given point, be in 
 equilibrinm (balance), their direction and amount 
 will correspond to the three sides of a triangle, if any two of which be 
 given, the third may be found. This trianele is nothine more tban abdar 
 aed,ag.4B. 
 
 130. ir there be more than one concentrated load on a clrder, two 
 
 courses may be followed : 
 
 The reactions and the strains produced by each weight separately may 
 be found, and the algebraical sum (see foot note, p. 30) taken ; or. 
 
 The reaction of the piers from the whole load may be worked towards 
 the mid-span, the downward, pressures of the several weigl^ts (when they 
 are met with), being compounded with the other pressures during the 
 progress. See Example (1«8). 
 
 The first should be adopted when the load is variable — for then the 
 maximum resulting strains of either kind on any member may be found 
 (136) ; the second when the load is stationary. 
 
 SXAMPIE— 
 Sraina in a Beat Girder {Roof PrincipaJ). — Plate I. 
 
 I%1. CoJcuJation ty the Mimmtt of Rupturt (1 (o33) and Sluaring: Fona (M 
 to 58). 
 
 AFKishalf of a bent girder whose clear span is 80 feet. The lower boom 
 consists of chords to a curve of 98 feet radius ; and the upper, chords to a curve of 
 £7 feet radius. The load is considered as uniformly distributed (regarded 
 horizontally) on the upper boom, and:= 8 tons = *1 ton (w) per foot of span. The. 
 braces are alternately vertical and inclined. (See 111.) 
 
 * Where the structure almtt againtt the supports, and those supports are 
 supposed to be capable of resisting the lateral pressure, the structure is virtually 
 an arch, and the reaction of abutments must not be considered solely as ver- 
 tioal, (See 146.) 
 
10 0ALCin,A.TIOS OF SI8AIK8. 
 
 Strains in the Boomt. — The strain in the upper boom is compreuire, 
 and in the lower tensile (66). Aa the braces are lUtematielj vertical and inclined 
 (111)» the strains in any bay of either boom will equal the moment of rupture (1) 
 for the case (19) taken ai the opposite apex, and divided by the depth of the girder 
 taken as a perpendicular from the apex on to the bay. (See Plate I.) 
 
 The moments of rapture (M) may be obcained either iVom the formula 
 
 ^M=^i?(2 — x) (8)(19) V or, itom a properly constructed diagram (4, S). 
 
 The depths (as above) must be scaled from the outline sketch (S3— 59). 
 
 The diagram for the moments as given in 19 apply properly to the unloaded 
 boom only ; while foir the loaded boom, an inscribed polygon is required (111), as 
 shown in Plate I. ' But wherever the apices on one boom are vertically under (as 
 in the present instance) or over tltose on the other boom, it will be evident that 
 tSie only ordinates required from the. polygon will be those taken to its angles, 
 that is, to the very paints where it coincides with the parabola: so that the 
 polygon is unnecessary. 
 
 ' It wUl be seen that by drawing the lines of construction for the parabola, as 
 dotted in on the Diagram, there is really no necessity to draw the actual curve. 
 (See also note to 238, ni.) 
 
 _»_. . . M !»«((— «) .Diagram Ordinate 
 
 Stram m any bay = -;- = — ;r^ — ' = -*?• ; • 
 
 a Z a a 
 
 JBy Formvlae, Sy Dktgrcan. 
 »^- , . „ •! X 6 X 76 1876 
 
 Stram in AB « -j-^^-JT^" =''ve' = ""^ '°»»- 
 
 • 1 X 13-7S X ee-M 4863 
 B0= 2x3-9 ~ 8-9 ="'8tOM. 
 
 •1 X 22-26 X 67-6 64-68 
 0D = j^^-jty =-5:^=ll-3ton». 
 
 •1 X 31'2S X 48-76 76-17 
 
 DEo- 
 
 2x6-9 
 
 - 6-9 
 
 = 11-039 tons. 
 
 EF=- 
 
 1x40x40 
 2 X 7-25 
 
 80 
 -7-26 
 
 = 11-084 tons. 
 
 AQ = 
 
 18-76* 
 1-8 1 
 
 
 = 9-8 tons. 
 
 GH = 
 
 18-76 • 
 2-0 
 
 
 = 9-3 tone. 
 
 HI = 
 
 46-63* 
 4-6 
 
 
 = 10-1 tons. 
 
 IJ = 
 
 64-68* 
 6-1 
 
 
 = 10-6 tons. 
 
 JK = 
 
 78-17 • 
 7-08 
 
 
 = 10-7 tons. 
 
 Strains in the Web. — The values of the Shearing (brces (1, 34) may be 
 obtained either from the formula for the case (44) or from a diagram (35), the 
 
 • The moments of rupture are the same as for the upper boom, as the apiees 
 ore at the same horizontal distance from the supports. 
 
 t This "depth" is a'peipendionlar.let fkU from B on to AG produced. (See 
 Plate.) 
 
XXiMPLKS. 41 
 
 triangle K"BC (Plate I.) oQivtoponding to A C E Ifig. 31). The vertical compo- 
 nents of theetxaihs iif thtibpome mnsi be obtained (1 1 1) either by (80) moltiplymg 
 such strains by the sines of the angles of inclination of the several bays (as the 
 angle i j D for the bay D E, Plate I.), or by a geoinetrical construction, as shown 
 for the baysDE and! Jj fbr (he former, D^ is the strain in the bay, and D i the 
 vertical component. 
 
 Strain in any brace = | S H + (vert, oomp.) + (vert, comp.) | see. t. 
 
 Strain in PK = -IS?— -85 =— -414 tension.* 
 
 KB = (-437 + -48 — -86) X 17 = "lU tension.* 
 „ B J = 1'3 + -48 — 2-6 = — '82 tension. 
 
 „ J D = (1'3 + 1-4 — 2-6) X 2 = -218 tension. 
 „ D I = 2-2 + 1-4 — 4-3 = — -7 tension. 
 
 i, 10 = (2*2 + 2-22 — 4-3) X 3-6 = '42 tension. 
 „ C H = 307 + 2-22 — 6'2 = — "93 tension. 
 
 „ H B = (3-07 + 3-0— 6-2) X 9-6 = — 1-21 tension. 
 „ B6 s: Diff. bet, vert. comp. of AGandGH. 
 = S'O — 37 = — •? tension. 
 
 1S!9. CSiIcuIation ly tht Compoikwm and Xaolvtim of Flma (116— iSO). 
 For the date see 181. the weights (125) on the several apices oi the tipper 
 boom moist be considered as half the load on the adjacent bays, the load being 
 
 ■6 ton being snppbrted directly by the two piers (°26 ton on each, being half th^ 
 load on the outer bay). The reaction (111) of eiUier pier from the total load oa 
 
 is 
 
 the girder ifOl be -j^ e ^-75, and this mnst be worked up to the centre by the 
 
 application of (118 or 119). , 
 
 At pier A (Plate I.) draw a perpendicnlar (A b) s the reaction of A as above. 
 FroduceAB toa, andtbtonghftdrawtapaiallel to AG. A6a is the triangle 
 of forces at A. Then (119) 6 a is the tension in A O ; and A a the compres- 
 sion in A B. Through a draw a c parallel to G H ; then a c 6 is the triangle of 
 forces at G; c a ia the tension in GH, ande6 that in G B. AtB there are acting 
 the compression in A B, the tension in G B, and the vertical pressure of the 
 weight (°e87 tons) at B, idl of which are held in equilibriohi by B G and B H, Let 
 A d = me snin of tlie tension (e b) in GB and the weight at Si Joiil d to o ; and 
 dais the resultant of the three forces just nainedi To avoid coiitlision, repeat d a 
 at ad'. Through d* draw d' < parallel toCB; and through a, a c parallel to B H. 
 Then «d' is the compression mBO; and a e the tension in B H. Transfer a«to 
 ef, andjoin/toa; tnen/a is the resultant of the strains in B Hand GH; which 
 resultant may be resolved in the directions of H C and HI; and so on to juie 
 centre. 
 
 The mexperieneed praetiiioner wUl find U more convmvmi am safe to 
 worh by the paraUdogramt (118) imtead of the tritmglei offorcei, as the 
 natur£ of the resolting strains is more oUarly shown by the former, and 
 there is also less liability to oonfasion. 
 
 KXIHODS or CALODLAHON FOttlrDZD on THU FASALLIIiOaitUI OF 
 
 FoiiaiiEl; 
 
 123. General Eaw of the Strains In the Booms of Horizontal 
 BtralBht etrdexs. The increment of strain developed at any apex in the 
 
 » See 111, "Web." 
 
42 CALCOLATIOir OF STBAIITS. 
 
 boom, is eqnal to the resultant of the horizontal components of the strains 
 in the two diagonal bars forming the apex. 
 
 Let ab andac represent in direction and amount the stefdns in tivo bars 
 forming an apex, one bar .beins in compreBsion, the 
 Sig. 40. other in tension. Let the verticals b d and e be drawn, 
 
 and through a, the horizontal de. Then bd and ce will 
 correspond to the load on the diagonals 6 a and ae respec- 
 tively; da and ae will be their horizontal componenta ; and 
 de will be the increment of strain developed at a. 
 
 When, however, tta strains in the bar; are both tension 
 or both compression, their horizontal components will be 
 antagonistic, and the increment will then be the difTetenoe 
 between them instead of their sum (as d e), 
 
 A vertical braee hat no horizontal component. 
 
 124. Tha general RtUes for booms a/nd bracing (84 to 95) liold good 
 here alio. 
 
 12s. If the load be not concentrated at the apices, each apex must 
 he considered to sustain part of the load on the two adjacent bays. If the 
 load be uniformly distributed, each apex will sustain half the load on the 
 two adjacent bays. For other distributions of load, the pressure or eqnivslent 
 weight on each apex must be determined firom IIT. The total loads thus 
 allotted to the several apices will be called the " weights." 
 
 US. In this division of the work — 
 Let = Angle made by any brace with a vertical line. Then — . 
 
 o^. a Inclined length of brace ^ ^o^ 
 
 oee. u— YjrtlcmliUitlmce between lt»en(U ' 2 %» 
 
 m A — J^<"^*oP**^ dfatance between iU endt S o^ 
 
 lam u— Yertlcal dirtance between Iti enda * 5 * 
 
 
 Jin. 9 = 
 
 Horlnmtal distance between lie endi 
 
 Inclined lengtlt. > 
 
 ^ ; ...Horizontal length"*' 
 
 18 T. Straight 8emi-Blrder loaded In any manner. • 
 
 Wbb. — Etery weight (125) on the girder it trantinittfd through ill own 
 tyitem of trianglee to the abutment. The had on any brace is equal to 
 tiie sum of all the weights upon its system, between it and the nnsupported 
 end of the girder. For the tlrain on any brace multiply the load on it (aa 
 above) by sec. 9. 
 
 Books. — ^By (183), — ^The horizontal component of the strain in any 
 brace = (strain in brace) x sin. 0. (1S«.) 
 
 128, — EzAXPLS. —Let fig. SO represent a parallel straight girder of otherwise 
 irregular construction loaded at three points (a/c). The weight of the load at a 
 isoonductedalong the braces (tied e,thatat(! along ode. If I ii! represent Uie 
 
STBAISHI WABREN SIBOERS. 
 
 43 
 
 weight Wi) ti a mil be the tension (01) on a b, and I a = increment (123) 
 
 at a. If m n also equal Wi.ni will eqaal the 
 
 compreasion in c &, and m b the increment at 6. Fif;. 60. 
 
 The load on i: d is equal to Wi andWs together. 
 
 IS op — this load, a e will equal the tension on 
 
 d c, and p c the increment at c, and so on ; »t 
 
 being equal to Wi, «/will equal the stntiu in 
 
 fOt and £/the increment at/. 
 
 Then (123) the tension (66) in bg = mb + la; 
 that in <(0 = ii(r + (/+ strain in 6 ;; that in 
 ii2=^g<i + pc + that in d^. So also the com- 
 pression(6S)ina/=a Z; tfaatin/c = t/+ that 
 ina/; thatinch = cp4'mft+thatinc/+that 
 
 in fa; and so on. If the triangles for the strains be drawn to a large scale, the 
 strains may be obtained quickly and accurate. 
 
 12JK StrolBht Warren Seml-Kirder, Isosceles Braeinc. loaded at 
 tbe Extremity. 
 
 Web. — Strain (Z) in any brace = W sec. 0. . . , (126.) 
 Booiis. — Strain (S) in any bay of .either boom, 
 
 8 = »Wtan. «, , . . (126,) 
 
 where W=the weight, and m= number of diagonala between the centre of 
 the bay and W. 
 
 130. gtralclit urarren Beml-sirder, Isosceles Bracins, loaded uni- 
 formly on One Boom. The weight on the end apex = half that on the 
 others (06). 
 
 Web. — Strain (2)>in any pair of diagonals forming an apex on the un- 
 loaded boom. 
 
 2 = w « sec. 0, . . . (126,) 
 
 where w = load per unit of length, and as = number of units between 
 the apex and the unsupported end of the girder. 
 
 BoouB. — ^The strain (S) in any bay of the loaded boom, 
 
 S = I m(in— l) + i |wy tan. 9 . . . (126.) 
 
 In sny bay of the unloaded boom, 
 
 S = tn' w y tan. d, . . . (I26,) 
 
 where m =■ number of tbe bay counted from the outer end of the girder, 
 and y = length of bay. 
 
 The fallowing cases (viz., 131 to 138 inclusive) do not apply to con- 
 tinuous girders or whole girders fixed at the ends. 
 
 131. Stralsbt Warren Girder, Isosceles Braelns, loaded at Any 
 Point. 
 
 Wbb. — Strain (2) in any diagonal be- 
 ween A and W, ^K- "• 
 
 I 
 2 = W 5" sect. ». 
 
 In any diagonal between B and W, 
 
 2 = Wjsec. «... (ia«.) 
 
 BooxB. — strain in either end bay of tbe longer boom = (reaction of 
 
^^ 0AL01TLAII01I or SIKAIIIB. 
 
 Bnpport) X ten. (117, 186) ; and if this ralne be considered a unit, tlien 
 the strains in the' bays of the longer boom will be proportional to the series 
 !> 3> p, 7, &c. ; and in the shorter boom to the series 2, 4, 6, 8, &c., 
 counting from tke supports. Thns the Btrain in tbe second bay of the longer 
 boom, from pier B will equal (3 W -^ tan. < V 
 
 Strain in the bay opposite the loaded apex= . . ; — d being the depth 
 (98) of the gilder. ' " 
 
 138, If W Ibe In the centre of the clnler, the sttwn (2) on any 
 diagonal, 
 
 2 = ^ see. 9 . . . (186.) 
 And the strain (S) on the centre bay, 
 
 133. Btralcht Warren Girder, iMMweles Braelnc, with h cob- 
 centmted HoTlnc load. 
 
 WsB. — Every diagonal except the two end ones will be subject to counter 
 strains (93). The maximum strain normal on any pair of braces forming 
 an apex on the unloaded boom, will occur when IJie load is at the inner 
 (mid- span) end of the inclosed bay. The maximum cqunter strain on any 
 pair will obtain when the weight is at the outer (near abutment) end of 
 the enclosed bay. The values can be obtained irom (131). 
 
 Boons. — Tbe maximum strain on any bay (84-T) of the unloaded boom 
 is when the weight (W) is at the opposite apex ; and on the loaded boom 
 when the weight is at the next inner apex. The values can be obtained 
 from (131). 
 
 134. Any Straight Warren or lattice Cirder (isosceles of scalene 
 ^bracing), with any load gynunetrically disposed aboat the Centre, 
 
 either on one or both booms. 
 Suppose, the load collected at'the apices, according to (125). 
 Web. — The load on any bar mil he equal to the nun of <al tht viaghte 
 (125) on it! tyitem of trianglei, ietiiieea 
 ^e-S2. j-( g^ fj^ eenlre of the girder. The itrain 
 
 '===^=> ^jj ^jjg ^^ ^jj]^ jqmi thg j,^ on j^ multi- 
 plied by the secant of the angle it makes 
 with a vertical (126). 
 
 Let the aocompanyine fig. represent half a 
 girder, with loads on both booms symmetiioally 
 disposed atmut the centre <t. Then the load on 
 b a will be half the weight (125) at a, that on 6e half the weight at a, and tbe 
 whole weight at h ; that on c d, the weights at 1; and & and half the weight at a, and 
 so on ; fg will take the whole weight at /; ij the whole w^bt at f 3 «/ Kit&'i t 
 will notabe necessary. (See foot note, p. 46.) 
 
 Boohs. — The strains on the several boys vnU be best obtained ^m 
 the web, by the application uf (123), by sumsning the seyenl i^rements 
 (commencing at the outer ends of the ^rder) as follows, 
 
 The strain in on is the increment at «.• But ttie str^ in e » perrades 1^ 
 whole length of tbe bottom boom, and therefore must be added to all .tb? other 
 increments inuardt (towards tbe centre of the girder). The saine may be said of 
 the strains in the other baysj so that, strain in » m — (iec. at 0) ■(■ (ipg, at n)i 
 
STRAIGHT aiBDERS WITH MOVING LOADS. 
 
 45 
 
 Strain in m ! = (inc. St o) + (inc. at n) + (inc. »tm) = (strain in m«) + (inc. at m). 
 ^h^ff for«, the strain in any bay is equal to the strain in the next outer 
 Jbay { + )jiluf the increment at the intermediate apex. Note. This applies 
 afso ip ssmi-ipr^ers. 
 
 135. Any 8f|^tlplit IVnrren nr Kntticc CSinler (isosceles or scalene 
 bracing), wll.li |» iiniforiiily illMtrlbiitcil TMovlnff Kotul.* 
 
 Wkb. — The simplest method of obtaining the strains and connter strains 
 ()^!$) is b; tabulating the strains produced by each weight separately, 
 using (.l.'H, "Web"), which will hold good for other than isosceles bracing, 
 provided the .sec. 8 (I'iii} is corrected for the varying angles. The co- 
 efficient, sec. d, shoultl'not, however, be employed until after the summary is 
 made. See example (I :«;). 
 
 Booms. — The strains in the boo.m^ are greatest when the girder is fully 
 loaded, and may be most easily obtained from the web by (1%3 and 134), 
 
 136. jR]r^jr/-£s.— AB.fii;. 53,isagirder60ft.Ion|;,10ft. deep. The lattice bars 
 are inclined at angle of 45° with a vertical. 
 
 The moving loadis equal to 1 ton per lin»al Fiff. .'53. . , 
 
 toot, = 10 tons at each apex (on the lower 
 ' boom) (ItSS). In tabulating the sU^ns, 
 only those bars which incline in ono direction 
 need be considered, for the remainder are 
 stFgined similarly and equally. (Thus a, 
 Sg^ 6S, corresponds exactly with the bar 
 intersecting e, and so on). In the table 
 below, + indicates compression, and — ten- 
 sion. The nunnbers are the loadu on the bars. For Mic ftmiiix, the loads must 
 be multiplied by sect. 45° = 1 "4. The columns (maximum + ) and (mfudmnm — ) 
 are obtamed by adding the + values together and the — values tofrother.respcc- 
 tively in each -horizontal row. The columu (uniform load) is the algebraical sum 
 of the values in the horizontal rows. 
 Tracing the action i^W^, it is seen (from i^i) that in a,, fig. 53, there is from it 
 
 aload = W- = 10 x * =8 tons; evidently Iproducing tension ( — ); and in c 
 
 I , SO 
 and e there is a load = W— = 10 x — = 2 tons, as evidently producing compres- 
 jjion (+). And so on for th« ptber wpi^hfs. • 
 
 »"»• 
 
 w. 
 
 w» 
 
 TVj 
 
 w, 
 
 Maxi- 
 mum 
 
 + 
 
 M«x|. 
 mum 
 
 Uiuform 
 Load. 
 
 a 
 
 — 8 
 
 -4 
 
 
 
 -^12 
 
 — 12 
 
 b 
 
 
 — 6 
 
 
 ^i 
 
 
 -r S 
 
 — 8 
 
 e ■ 
 
 + 2 
 
 
 — 4 
 
 
 + 2 
 
 ~. 4 
 
 — 2 
 
 • d 
 
 . 
 
 + 4 
 
 
 — 2 
 
 + 4 
 
 — 2 
 
 + 2 
 
 e 
 
 + 3 
 
 
 + 8 
 
 ■ 
 
 
 + 8 
 
 
 + 8 
 
 It will be seen from the above Table that bars e and d (and of coarse the bars 
 inclined the other way which ooirespond to them) will suSisr strains of both 
 tension and coqtpression. 
 
 * C^se of a railway train passiJtg over. 
 
46 
 
 CALCULATION OF STRAINS. 
 
 137. Note. — The strains in the web resalting from the -weighty of the 
 girder itself, must not be calculated simnltaneously and in combination 
 with those resulting from the moving load. They shonld be calculated as 
 for a stationary load (either by 134, or by tabulating as above, to form the ' 
 column "uniform load ").* The values thus obtained for the several bars 
 shonld then be algebraically added to each of the columns ("maximum +", 
 "maximum—", and "uniform load"); and it will be found that the 
 modification in the first two will be more advantageous as the weight of the 
 girder becomes large compared with that of the moving load ; for the 
 counter strains (93) will be less, and the amount of counter bracing 
 necessary will be diminished also. 
 
 138. Anr Stralsrlit Opcn-bntccil nirder, witli a load nnRymme- 
 trlrally lUspoMCd alioiit (be <:cn(ro. 
 
 Web. — a. By tabulating the strains produced by the several weights, and 
 taking their algebraical sum ;. 
 
 (A, i. By (1st) abstracting the unsymmetrical parts of the load, and 
 proceeding with the r^ainder by (134), (2nd), calcu- 
 lating for the unsymmetrical part by (135) ; and (3rd), then 
 taking their algebraical sum of (1) and (2). 
 
 Boous. — From the web by (1S3). 
 
 130. lAttiee Ulnlen havlnc their Diagonal Bitm llxcd tosrrtlicr 
 at tliclr Intenieetiun may be calculated as if the bars were not so fixed ; 
 for this mode of constructing a girder " can never, add to the longitudinal 
 strain upon it ; but by calling into play the resistance'^of the bars tC cur- 
 vature adds to the stiffness of the bridge, "f 
 
 140. Simple Traas.- 
 
 Fig-S*. 
 
 -Let A B = 2, W C = d, and 9 the an|;le between the 
 inclined tie and tke horizontal. Then, — 
 
 Compression in AB = -j-y 
 
 Tension in A C and C B : 
 
 W? 
 
 If WC represent 
 
 W 
 
 Compresmon on C W = W. 
 then AW or BW will equal strain along AB ; and 
 
 AG or CB that in the ties AC, CB. 
 
 * The two methoflH will not f^ve equal results at and about the very point (the 
 centre of the girder) where the values arc of the e;*cnteBt conEC(iuence. Thiti is 
 quite UTiavoidable ; and the Kiifcst course (which need be pursued only in works 
 of importance) is lo employ both methods, and to provide against tiie greater 
 strains. 
 
 t " Wrought Iron Bridges," by J. H. Latham, M.A. 
 
SIMPLK AHD COHFODND TBCSSE3. 
 
 47 
 
 141. When W is not in the Centre, 
 
 Let A W = a, 
 •BW = i, 
 
 (p= angle CAW, 
 ^i' = angle CBW. 
 The other notations as before. 
 
 Compression along AB = W 
 
 Fig. 5B. 
 
 ab 
 Id' 
 ab 
 
 Tension in AC = W ^ sect, (f, 
 Id 
 
 Tension in CB = W — sect. ?' 
 Id 
 
 (Vie.) 
 
 Compression on W C = W. 
 For the pressures on the supports, see (ill). 
 
 14a. If the truss in either Jig. 64 or Jig. 55 were inverted, the strains ou 
 the Tarious parts would change in kind but not ia amount. 
 
 143. In both the above cases, the tension on the ties and the direct 
 (ST) compression on A B would remain the same as at present, if instead of 
 being loaded with the concentrated weight W, a. load equal to (2 W) were 
 %read uniformly along AB, 
 
 144. Coinponnd Truss. — A structure of this kind may be regarded 
 as compounded of a number -of simple trusses (140). The simple truss 
 ACBD has to sustain at its centre (C) half the entire weight of the structure 
 and its' load. The truss AECF has to sustain at £ half the total load - 
 
 Fig. S6. 
 
 between A and C. AGEH must likewise be considered as supportin); at 
 G half the total load between A and B, and so on with the others. With 
 a uniformly distributed load on A B, A H B and the three other similar 
 trusses would each take > ; AF C and its fellow, each \ ; and A D B, ' of 
 the total load. The strains on the Tarious parts can then be obtained from 
 (140). The compression along AB is uniform throughout 'the whole 
 length. 
 
48 OALOUIiATtON OF SIRAIKS. 
 
 Aboh BnicaES. 
 
 ABOHEB WITH SFAHDKIL BRAOIHO. 
 
 145. HVlth a Vnlform Horizontal toad. 
 
 Let I = length of span, 
 V = rise, or versine, 
 
 y — horizontal distance of any point 
 from the crown, 
 
 to = load per unit of lengtli. 
 
 Abched Bib. — Compression at the crown (C), 
 
 C = ■ 
 
 8v 
 
 Compression at any other point (C) 
 
 "-VW^ 
 
 The exprea^n for C is stricUy accurate only when the aroh* is a parabola,— 
 the curve of equilibrium for the load is. question. It may, however, be safely 
 used iu most cases of practice for arcs of dnsles. 
 
 SpandbH. — If the arch* he parahoUc, the only strain on the spandril 
 will be the vertical pressure of the load. (See foot note t, pi 49.) . 
 The spandril may then condst of a number of pillars or struts (as in 
 
 fig. 57), each sustaining at vertical pressure = nearly j N being the 
 
 N — 1 
 number of spaces into which the pillars divide the span. 
 
 If it consist of a continuofl^ web, the compression of it per unit of length 
 will be equal to w. 
 
 If the areh* be not a parabola, the strains in the spandril bracing 
 may be obtained from 80, 81, (for continuous or "plate bracing"), or 111 
 (for diagonal open work), the lib being considered as the " compression 
 boom." 
 
 tot HoBlzoHTAL Mbhbbb. — WUh d parabolic (trek this member 
 (B E fig. 67) simply acts as an immediate support for the load (87). With 
 an ardi, not a parabola, there will be a strain on it acting in the direction 
 of its length, the nature and amount of which may be determined from 
 the Increments at the apices of the diagonals in the spandril, by the appli- 
 cation of the law in 1S3 as illustrated in 134. 
 
 146. With a MovIdk load. 
 
 For obbtlninf; the strains oil the various parts of a braced arch. Subject to a 
 distributed moving load, the following method may be employed.t 
 
 • More properly the neutral axis, or the line traced through the centres of 
 gravity of the cross sections of the nb. 
 t Given by B. StOney, B.A. (in " Theory of Strains ") j and others. 
 
BRACED AND IINEAB AKCHES. 49 
 
 * Conceive tbe load to be collected into weights at the several apices on the 
 horizontal member, each apex sustain- 
 ing half the load on the two adjacent ' ■ ^'K- *^' 
 
 bays. The strains produced by each 
 weight separately must then be found 
 and tabulated, as in 135. Suppose 
 the strains from weight W (fig. 58) are 
 to be considered. At W draw the 
 vertical J7 D. From the abutment 
 A draw A C D through the crown of 
 
 the arch (C) till it. intersects the perpendicular at D. Join D to 
 B. If the weight W be resolved (118) in the directions A D and B S, the 
 amounts and directions of the reactions of the two abutments will be found. 
 This may be readily done by producing A D to E, and erecting a perpen- 
 dicular at B. Then, if E B equal the weight W, D E will be the reaction 
 of A, and D B the reaction of B. If the fbi'mer be traced up &om A 
 towards the weight, and the latter from B, by the resolution of forces 
 (118, 119) the strains on the various parts may be thus found ; and the 
 same being done for the other weights in succession, the maximum strains 
 produced by any position of the load may be derived from a table similar to 
 that on page 4S. 
 
 If the weight of the structure be small compared with that of the rolling 
 load, it will be found that some of the end bays of the horizontal member, 
 and of the middle bays in the ai-ch, will occasionally suffer tension. See 
 also 137. 
 
 UBBRAOED AKCHKS, 
 
 Or arches whose stability depends upon the stiffness of the rib itself. 
 
 147. Tlie Neutral Snrfaee, or neutral curve of an arched rib, is a line 
 traced through the centres of gravity (SSO) of the cross sections of the rib. 
 
 148. The Une of Pressures is a Une the ordinatet to which vary at 
 iAe moments of rupture (1) for tlix load* 
 
 The line of pressures is given at once in those diagrams (pp. 4 to 6), in 
 which the ordinates for the moments of rupture are directed to be taken 
 from and on the same side of one horizontal line. Where this is not the 
 case (as in 20 or 21, for instance), the ordinates must be transferred to 
 some new horizontal datum. 
 
 The ordliiates may be taken to afiy scale for ready comparison with the 
 neutral curve of the rib. 
 
 149. Wnenever in any arch the line of pressures eoineidei tott/t the 
 neutral surface, the arch is in equilibrium, and the strain upon it ii 
 everywhere compressive.f 
 
 • For BIiuanr]ranclinanyottaerAreheBcasel4wUlbefoundexceedingIyoon- 
 Tcnient. Tbe structure and its load should be conaideced as made up of several 
 small portions, each collected at its centre of gravity'. The line of pressures can 
 then be obtained by summing the ordinates for uiew outline, as there directed. 
 
 t Strictly speaking, the pressure on an arch ofother than a thoroughly incom- 
 pressible matsrial alters the form of the arch; and this alteration of form, or 
 tmiUng action, induces strains similar to those found in beams (slee 1 and 93). 
 The tension on equilibriatedribs (ISO, ISl) is in practice so small (even wl^ere it it 
 developed,— for this does not occur till the tension induced by the bmdiag exceeds 
 the direct compression on the rib) that it need not be regarded. 
 
 •K 
 
50 OALODLATIOS OP STKAISS, 
 
 150. The Stability of an Arch incawablc of Rcaistins Tension i|^ 
 
 secure so long as the line of pressures does not at any point deviate from 
 the neutral curre by an amount, the proportion of which to the depth of 
 the rib is given by the values of q (153) for various forms of cross section, 
 
 151. Wherever the line of pressures deviates beyond this limit, there is 
 a temion on the nb on the other, side of the neutral surface, incrSising with 
 the deviation. 
 
 153. Pressure •along one edge of an arched rib invai'iablyaproduces 
 tension on the other edge, whatever be the form of cross section. 
 
 153. Limits of the deviation of the line of pressures from the neutral 
 surface consistent with there being no tension on ihe rib ^— = 
 
 Form of Cross Sectiim, Value of q. * 
 
 Rectangle . . ' -J- 
 
 Ellipse and Circle ..'.... \ 
 Hollow Rectangle (area = hh — b' h') \ 
 
 also I formed section, i' being the ( , / b' h'^\ _ / b' h' \ 
 
 sum of the breadths of the. lateral i ^ Kj; ~ b h? / "^ \ bh )' 
 hollows. ) 
 
 (V \ 
 ^ + la /■ 
 
 Ho'low Ellipse j(l.*_^)^(l_^'). 
 
 Hollow Circle . 
 
 i(>*F> 
 
 b and V — external and internal breadths ; and 
 
 h and A' — the external and internal heights or depths. 
 
 I, section alike above and below . J- ( 1 + -77 — r~r )• 
 
 A = area of each table or flange ;"' 
 
 A' = area of the connecting web. 
 
 1 54. The above values of j should be applied as' a test for tension in 
 cases of braced arches, and also where the spau'dril 
 Fig, fig. consists of columns ; for the line of pressures* is 
 
 th«n in reality !l polygon, with the angles at 
 the apices on the rib. -(See fig. 59.) 
 
 154 A. When an Arched BrldBc consists of 
 several unequal Spans, the neutral surfaces of 
 aU*the ribs should be parts of the same figure-^ 
 (generally arcs of the same circfe, or the sam« parabola). 
 
 * These values or$ are from Professor Bankine'g " Civil Engineerinf;.' 
 
8DSPBNSI0K BRIDSES. 51 
 
 SnsPEHSiON Bridois. 
 
 ISS. ORDIHART SUSPENBION BSIDSB OP ONE BPAK, WITH A DNIPOBM 
 HORIZONTAL LOAD. 
 
 Maih CHilss.— The curve .which the main chains will as3«m« will be 
 very nisarly a parabola, 
 liet IS = load on each chain per nnit of length ; 
 
 I = length of span ; 
 
 V = Tersine, or degfession of the chain ; 
 
 f = angle which a tangent to the cliain at any point makes with a 
 horizontal ; 
 
 y — horizontal distance of em; point from the mid-span. 
 
 Tension at Centre, T = |^ ; ^.^ g„ 
 
 At any ether point, 
 
 «"=V(sy^""(^~^>=' 
 
 
 SusPKUMNO Rods. — Let N =■ the 
 number of spaces into which they divide the span, then, 
 
 Tension an each^K Jf 
 
 Towers and Cocnter-ohains. — The tension on the counter-chains, 
 and the pressures on the towers may be readily found as follows : — Pro- 
 duce the tangent to the main-chain at the tower (C) till its length (CD) 
 on a scale of parts, equals the tension at that point (found from 
 (F aboye). Through D draw D E parallel to the direction of the 
 counter-chain (C F). Produce the centre line of the tower tillit inter- 
 sects DE. Then (IIS) CE will give the pressure on the tower, and D E 
 the tension on the counter-chain. 
 
 156. OBSINART SUSPENSION BBIDQE OF UORE THAN ONE SPAN, WITli 
 A DNirOBM HORIZONTAL LOAD. 
 
 When a suspenaioa bridge consists of several spans, the chains of all of 
 them must form portions of one and the same parabola. 
 
 The strains on the whole spans — as B C — will be the same as in (ISS). 
 
 E 2 
 
52 OALCULATION OV STRAINS. 
 
 The Btrains on the outer spans (A B) are identical with those in B D, 
 Tig. 81. 
 
 as the parts A B ^d B D are equal. But where the abutment spans are 
 not equal to half the inner ones, the first conditions made in this paragraph 
 
 must be observed, and then T = ( — — j sec. (155), I bmg still the 
 
 inner ipan. 
 
 The compression on the towers will equal the disgonnl .of a parallelogram, 
 whose sides oorrespond in direction and length to the tensions ip the obams on. 
 either side (as at iT is &g. 61). 
 
 15T. SnSPENSIOH BKIDOB WITH SLOFIira BODS, WITH A VVIFOBK BOBI- 
 ZOHTAL LOAD. 
 
 MAm-0EAiR8.-^The curve which each half of the chains will assume 
 will be a parabola with its axis parallel 
 "^•*^' to the direction of the sloping rods. 
 
 Tension at mid-span, T = -g^ } wnere 
 
 (ml') is the total vertical load on the 
 rods ; the other notations as in 155. 
 Tension at any other pointy 
 T sjs wy cosec. <p; 
 
 y being the distance from the mid-span to the bottom of the sloping rod, 
 at the top of which T' is required ; and 4> the angle which a tangent to the 
 chain makes with the hccrizontaL^ak 
 
 SLOPina KoDS.— Tension on a^rod, 
 
 « = Wsec. /8; 
 where W is the vertical load on the rod (half the load on the two adjacent 
 bays) ; and J3 the an^ which the rods make with a vertical. 
 
 HoBizoKTAii Meubeb. — ^The eompressioii (e) on the horizontal platform 
 at any point distant (]f) from the mid-span, is 
 e = wy tan. fi. 
 For the towen, see 155 or 156. *' 
 
 158. SusrinaiOH BBinaEs with Moviko Loads are subject to mneh 
 disfigurement, to prevent or modify which several means have been devised. 
 
 159. (1.) An auxiliary girder from pier to pier, anchored down to the 
 
 abutments. If this girder be continuous for each span, its booms (fig. 38 a) 
 
 for abont the middle half of its length must be able to resist a strain 
 
 «'{' 1^2 
 
 = r-T-j, and the web a shearing force of about -r-; v/ bang the intensity 
 
 of the moving load per unit of length ; I, the length of span, and d the 
 depth (V8) of the girder. * 
 
ABUTMENTS AND PIERS. a3 
 
 IGO. (2.) When the brUge consists of several spans, — Fixing the chaim 
 to the top of the, toweii, and considering the latter as semi-girder!:. 
 
 each one loaded at the extremity with - — , notation as before (IS9). Wbll^ 
 
 8 V 
 the tower is suffering the strains consequent on the application o'' 
 .this force to its extremity, there is also a direct compression on it 
 
 ~(w+ 7- )'• This latter, it must be remembered, modifies the ten - 
 
 sional (4) strains produced by the former. 
 
 161. (3.) Inaerting diagonal bracing between the roadway and th^^ 
 chains. The strains on the various parts may in this case be obtained aij 
 iu 146 ; they will be altered in Icind only, not in intensity. 
 
 161 A. (4.) A pair of chains of identical curvature, placed one above 
 the other, and having diagonal bracing between, the greatest shearing forcS 
 
 on which would be — r- • 
 
 161 K. (5.) Counter chaim attached to the main chains at about | spa'i 
 from the abutments or piers, and running down to the latter. They 
 
 should be made to resist a strain of ( -^ x sec. 9); 9 being the angli? 
 
 between the counter chains and a vertical. 
 
 16IC. (6.) Inclined straight chains, for oanying the platform ari-i 
 the moving load. They extend from the towers, and meet or intersect 
 each other. They are sustained in the required straight lines by rodS; 
 which are connected to onrved chains, the latter having to carry th.-- 
 weight of the straight chains only.* The tension on the latter may bs 
 found from 155. The tension on the straight chains may be most readilv 
 found by a parallelogram of forces (IIS). 
 
 162. Abntments and Picn. Girders, properly so called, viz., thosV. 
 structures which simply rest upon the supports, bring upon those supporlt 
 a vertical pressure equal to the shearing force developed there. (See last 
 paragraph in 46. ) 
 
 AliiUments of Arches. The thrust at the abutments of anarch is exactly 
 equal to the compression in the arch rib at the springing, the valne of 
 which may be determined from 145, 146. . ' 
 
 For the towers and piers of suspension bridges, see 1S5, 156. 
 
 Whenever the piers of a bridge consist of columns, their strength as 
 such — their liability to flexure, &c. — must not be overlooked. (See 168.) 
 
 SECTION m. 
 
 DlSTRIBUIIOH Or'MATEBIAL 10 BliSIST THE CALCULATED SiRAIHS. 
 
 (Embodying Frocesseses VI, VII, p. 24.) 
 
 .. 163. Tlie Strenictli of a Stmctnre, or of any part of it, is its ability 
 to resist the external forces tending to cause its rupture. 
 
 * Mi. B. M, Ordisli's system. 
 
54 DISTRIBUTION 07 TEE UATERUL. 
 
 164, Axiom. No whole is stronger than its wedkesl part. 
 
 105. Uniform Streiieth. AStructure is said to be nf nniform strength 
 ■when no one part would yield bwi*"e another, supposing the structure to be 
 subjected to the load, or a multiple of that load for which it was designed. 
 
 In structures not of .uniform Btt^gth, all the material in excess of that neces- 
 saryfor uniformity of strenghis redundant.* ' 
 
 To secure wniformity of strength, a constant co efScient of safety ixii)- 
 must be used for the same material strained in the same way. For beams 
 of uniform strength, see SIS — 918. 
 
 16G. Vntts. It is necessary to adopt : 
 
 I. Unit of Strain or Stress ; generally 1 lb. avoirdnpois.' 
 9. Unit of Sectional Area ; generally 1 superficial inch. 
 3. Compouvded Unit of Strain and Area ; 1 lb. per sq. inch. 
 Let A = area of a section in units (sq. inches). 
 
 S = calculated strain in units (pounds avoirdupois). 
 n.,= ultimate strength, or breaking weight of the material — 
 in -lbs. per sq. inch of section, (fm: numerical values, 
 see 9Zl.) 
 Co = coefSoient of safety (178). 
 W S = working strain (119). 
 
 FniNciPAL Strains to be met with in Bbidses, Girders, &c. 
 
 IG7. Tension, causes or tends to cause the fracture of the material 
 iipou which it acts by tearing asunder its particles. 
 
 The resistance to Tension is directly as the area of the cross section of 
 the material, takeik perpendicular to the direction of the strain (164). 
 
 S X Co S 
 Area necessary to safely resist a strain, A = — jt — = tF5' 
 
 168. Compression, causes or tends to cause the failure of the mate- 
 rial, by crashing, buckling, or both combintfd. 
 
 Crushvag. Materials in compression ( " struts ") can be considered liable 
 to crushing alone, only when their least ^ameter (taken perpendicular to 
 the direction of , the strain) is not more than about \ of their unsupported 
 length,-)' The resistance is then directly as the sectional area. X 
 
 Area necessary to safely resist a strain, A = = 
 
 ■ f-r u ws 
 
 Backling. When struts have an unsupported length equal to about 25 § 
 times their least diameter, they may be considered to suffer almost en- 
 
 * It cannot be said that the excess is entirely useless ; but more on this point 
 would be out of place here. 
 
 t Hodgkinson. 
 
 t The resistance to crushing of a body whose diameter normal to the pres- 
 sure far exceeds^ its dimensions* in a line with the pressure, is very great, but 
 , equally indefinite. 
 
 9 For wi'ought-iron struts with riveted joints, from 40 to 50 times. 
 
BREAKING WEIOHT OP COHIMKS. 55 
 
 tirely ty being buckled, ». e., crumpled up.* Struts shorter than this fail, 
 partly by crushing, and partly by .buckling. 
 
 169. bueakingI WEiaHT OF ooniHNS. — Prof. HodgkimotCs fotirmlce 
 for cast-iron columns. When of more than from 25 to 80 diameters in length, 
 
 Break, wt. = -^ for solid pillars ; and 
 
 Break, wt. = 41-3 (i^ '» - d'^ ■ ) 
 
 for hollow cylindrical pillars, flat, or firmly fixed at both ends, d being 
 the external diameter, and d' the i/iternal diameter in inches, and I the 
 length in feet. Columns rounded or moveable at both ends have but J the 
 strength of those flat or fixed ; and the strength of those with one end fixed 
 or flat, and the other rounded or moveable, is about an arithmetical mean 
 between these two cases. 
 
 When of less than from 25 to 30 diameters in length, let h be the value 
 obtained by the above formulce, and c the crushing load of a short block 
 (231) of the same sectional area as the column|iyuu the corrected breaking 
 
 '*^^' " l + ic 
 
 ■ OeneroU formula for the ireaMng meight of Cast onij Wrought Iron ■ 
 Colamni. 
 Let C= compressive resistance of a short block of the same sectional area. 
 
 »• = T-= length of column divided by the greatest diameter. 
 
 n 
 For cast-iron, B. W. 
 
 For wrought iron, B. W. = ^. 
 
 68 + -1 r 
 C 
 
 •85 + -Oir 
 
 Breaking Weight in tons per sgwire inch of section. 
 
 I ,. 13,600 
 
 I cast iron, „ ' — ;• 
 
 T> J ) 330 + r' 
 
 Bound 
 
 ] •, ^ . 34,000 
 
 (wrought iron, ^^jjj,^ 
 
 ciuare timber. 
 
 2000 + r" 
 850 
 
 350 + »■» 
 
 Breaking Weight of timber pillars, taking the ttrength of a cube as 
 unity. 
 
 Value of r . 
 Breaking weight . 
 
 * Professor Hadgkinson. 
 
 1 
 
 1 
 
 12 
 
 i. 
 
 24 
 
 i 
 
 88 
 
 48 
 * 
 
 60 
 
58 
 
 DISTBIBDIION OF HAIERIAL. 
 
 110. LOKO {IIBUI8 should b« made with a cross section, which will 
 
 Figr. 63. 
 
 insure a certain amount of rigidity or stiffnesSi 4Bd 
 thus resist the tendency to bucl(ie : or they should 
 be braced (either externally or internally), and 
 thus divided into a number of shorter lengths, 
 each of which (and, therefore, the braced strut as 
 a whole) may be considered as suffering crushing alone. 
 
 ITI. Staearlnc causes, or tends to cause, contiguous sections of the 
 material to slide over each pig. ei. other (as at A B, sup? 
 
 posing that D S is part of a 
 and B C part of the load), 
 is directly as the area that 
 
 Area necessary ts safely 
 
 A = 
 
 loaded beam, A one support, 
 The resistance to shearing 
 wotdd be sheared. 
 
 resist a shearing strain. 
 
 S X Co 
 U 
 
 WS 
 
 Shearing strains will b^aund to act on the vertical web * of continncjis 
 webbed girders, and in{||p|ra generally, which see. 
 
 For Bending, bending intenaitiet, and the resistance of materialt to 
 
 them, see I, and 191 et eeq. 
 
 » 
 
 ITa.CoemclentB of Safety are numbers representing the proportions of 
 the ultimate strength pf materials to the strains that can safely be brought 
 upon them. Coefficients of safety may be variously estimated. The fol- 
 lowing May,, however, be taken as a fair average of the factors at present in 
 use where the materials employed and the workmanship are ordinarily 
 good, 
 
 Metals. Timber. Masonry. 
 
 For a dead load . . 3 4 to 5 4 
 
 For a live load . . 5 8 to 10 8 
 
 Under dead load may be included all permanent or stationary loads, 
 and loads very gradually applied. 
 
 Under Hue load, all rapidly moving, and suddenly applied loads. 
 
 ' Ultimate reaistance 
 
 - Working strain. 
 
 Coefficient of ufety 
 
 1T3. The Sfodnlns of Blasttelty (E) (in pounds per square inch as 
 given in SSI) is> the weight (in pounds) required to elongate or shorten, by 
 an amount equal to its original length, a bar of material (of one square 
 inch of cross-section), and is on the supposition that the elasticity of the 
 material would remain constant throughout the operation. 
 
 strain (In Vn.) per gq. inch on a bar 
 Mod.of elfti. (in lbs,} per iq. .Inch 
 
 Increage or diminution of lenyth 
 Original iengtli of tUe bar ' 
 
 Joiiris. 
 IT4. Jciintt thould always ie as strong as the porta they serve to join. 
 
 • In the paragraphs 09 and 74. the flanges have not been supposed to ta,ke 
 part in resistinfr the shearing farces (so called from the shearing strains above), 
 any more than the web in taking its share of the horizont^ strains. (66.) 
 
IKON JOINTS AMD FASTEKINOS. 
 
 67 
 
 ITS. The variom partt of a joint, and the leveral pmrti in a fattening, 
 thould be of uniform ttrengtk. 
 
 IRON JOINTS AND VASIENINaS. 
 
 176. Rivets may fail in leveral different vays, depending on their 
 office in a joint. The head may be shorn off (as at G D, j,. ^ 
 
 EF) ; or the rivet nay be ruptured at any section (as 
 A B) when the rivet is in tension. 
 
 Let S = the tensional strain ; then d (diam. of tivet) 
 
 should not be less than 
 
 V 8 X 
 Ux • 
 
 ^0 
 
 7864 
 
 XT being the 
 
 ultimate resistance of the material to tension (SSI) ; and C D or B F should 
 
 not be less than f= — --.,„, (118), U being here the nltimate resistance 
 
 U X 8'141o a ' 
 to shearing (SSI). 
 When the rivet has to resist a shearing strain (S) at AB, d must not be 
 
 less than A/fr- — .-„-., U being the ultimate shearing resistance. 
 
 Oenerdl Sulc—The height of the head (A) should tiever 6(! fc»» (htm 
 half the diameter of the rivet. 
 
 ITT. BoKs. — The diameter of a bolt liable to shearing at the fpiiidDe 
 (as at A B, fig. 65) must be determi^^ from (ITS). 
 
 If the Wt be in tension it will fail either^! st, by shearing 9ff the 
 thr^ ; 2nd, or by shearing off the heftd i 9rd, qr ij tegsiffij^ Tt>p(an of 
 the spindle. For a perfect thr«^d the height cf the qut and of tft« head 
 should b? equal : but to alloir ^r inaccuracies of workinanship, the h^ht 
 of the nut should b? about twice that of the head. The height sf &e 
 put should not be less than the diameter of the spindle ; in praatiiN it 
 is frequently made mnoh more than this, 
 
 178. The diameter of a nut or head ofaAoU, prqfihe head of a rtve^ 
 ihffald he not maeh lem than tviee thSitifil^ ^indle. 
 
 ITO. Plm Joints In Tension Bars. — (Such as in some suspension 
 chuns, triangular girders, and trusses). 
 
 Pin. — Let n = the least number of sections at 
 which the pin must be divided before the joint 
 can fail (4 in. fig. 66), <; = sectifmal area of 
 pin, and S the tension on ike joint, then 
 
 «=TmSSf^- TlietoefBcient(lT8)Bhoul!jbe 
 larf^e, as any inaccuraoies in the workmanship will 
 tend to concentrate the strain in certain parts. 
 
 IMc. — The section of any link-head taken 
 through the centre of the pin-hole (A B) should 
 
 equal about half as muoh again as that b^ken through the body of the liuls 
 (as ]>), in consequence of the inequality in the distribution of the strain. 
 
 Fig, 86. 
 
 JH 
 
58 
 
 DISTBIBUTIOH OF UATEBIAL. 
 
 Let I = the length of overlap, E F or G H (fig. 66) ; t= thiokneas of 
 all the overlaps, in one series, taken together (2 or 3, fig. 66), then 
 
 I = 
 
 S X Co 
 
 where S is the strain oi the joint, and U the ultimate resistance to 
 shearing. 
 General Sule. — Diameter of pin may be f of the width of the links. 
 
 Riveted JoiiUa in Tension. 
 
 180. The Effective or Available Section of a plate with rivet hA<B 
 in it depends upon the disposition of the 
 I'ig. 67. rivets. Thus, in fig. 67, the least section 
 
 of plate that could he taken is that at 
 A A, But before the joint can fail here 
 by the rupture of (say) the upper plate 
 at A A, the three rivets marked a a a 
 must be shorn. And it will be found that 
 in a joint arranged in this way, the efiiective 
 section will be equal to that talcen through 
 the fint rivet, or lint of rivets, as at B B. 
 
 ISl. The sectional area of aU the rivets in a joint taken together should 
 he equal to the effeCtiiie section of the plate. 
 
 The distance between the centres of rivets which stand in a liAe (perpen- 
 dicular to the tension on the joint) should be made = d + - ('7854<i'n), 
 
 in which n is the number of lines of rivets as above (5 in fig. 67) ; t, the 
 thickness of the plate ; and d diameter of rivets. 
 
 Fig. 88. 
 
 182. lap Joints may fail, — Ist, from the tenmonal rupture of the 
 effective section (180) of the plate ; 2nd, by the 
 shearing of the rivets ; 3rd, by the shearing out 
 of the overlaps (A A A A, iig. 68). The jtram 
 
 en each rivet = u„^°°t'^ ' »nd they have each to 
 
 be sham at one section only (170). The distance 
 between the lines of rivets (181), (of which there 
 are 3 in fig. 68) must not be less than the over- 
 lap required for the rivets in the first row (as at 
 A). The latter may be determined from the 
 following equation. 
 
 1 = 
 
 S X Co 
 2ntxf 
 
 in which S = tension on the whole joint ; n, total number of rivets ; t, 
 thickness of plate ; U, ultimate resistance of the plate to shearing ; Co, a 
 suitable coefBcient of safety (172). 
 
PLATE II. 
 
JOIBia IN IIMBEB STRUCTURES. 
 
 59 
 
 183. Fish Joints. — ^Where only one cover or fish-plate is used (as in 
 fig. 69), the case is virtually identical with two 
 
 suosessive lap-joints, and can be calculated as 'Pig- ^• 
 
 such (I8S). Where two cover-plates are em- 
 ployed, it is to he borne in mind, that before the 
 joint can f4il, each rivet must be shorn at two 
 sections, so tiiat the section of each need be but 
 half that necessary with a single cover-plate (164). The thickness of each 
 cover-plate must never be less thiui half that of the main plates. 
 
 Riveted JoiiUi i/n Compretsion, 
 
 184. In lap-Joints the effective sectional area of the plate is eqnal 
 to the loss of section by rivet-holes, not counting 
 
 those which arc behind any others in the Fig. 70. 
 
 direction of the pressure. Thus, in fig. 70, the 
 effective bearing section of the plate is that from 
 AtoB. 
 
 For the shearing strain (171) on each rivet 
 (lT6), divide the strain on the joint by the total 
 number of rivets. A rather Urge coefficient of 
 safety (lis) should be used, as inaccuracies of 
 worlimanship will materially concentrate the 
 strain at certain points. 
 
 185. Bntt Joints (same as fish-joints in tension) may be considered as 
 having an effective section equal to the total section of the plate ; for if the 
 rivets fill the holes as they should, there is hardly any loss of strength 
 from them. 
 
 The cover-plates are required simply to keep the main plates in their 
 proper positions. 
 
 186. Joints formed by kITm and cotters may be calculated from ITO. 
 It is advisable that the obliquity of the surfaces to the direction of the 
 strain should not exceed 4° or 5°. 
 
 A few cast and wrought iron joints are given in Plate II. 
 
 JOIRIS IH TIUBES 8TKU0TUKXS. 
 
 18T. A form of joint having been determined, the parts at which it is 
 liable to fail must be traced out, and sufBcient strength be given to them, 
 using (165 — ITI), and parjiicularly observing (164). 
 
 188. Joints l|i Tension. 
 
 pesent forms offish-joints. 
 In fig. 71 the tension 
 may cause rupture at A B, 
 or C B ; or at E F and 
 OH. (JBotA of the latter 
 must &il before the joint 
 gives way by the rnpture 
 of the fish-pieces.) And 
 further, the joint may &il 
 by the shearing off of Q F 
 andBQ; orof FSandQT; orofEQandLM; orof SNandOF. The 
 
 -Pithed cmd scwrfed joints. Figs. 71, 72 re- 
 Kg.n. 
 
60 
 
 BEAM8 OF TARIorS SEOTIOHS. 
 
 above is duregarding the shearing resistances of the fonr holts sbOvn in fig. 
 
 71, which must he taken into account 
 Fi^. 72. in a somewbat similar way. 
 
 Fig. 72 represents a joiqt fished with 
 iron plates, and also " scarfed." Fig. 
 73 shows another form of scarfing. 
 
 189. Joints In Compression. — 
 
 Surfaces abutting against each other 
 should be as nearly as possible per- 
 pendicular to the direction of the 
 thrust. Where convenient, fish pieces 
 may be used to retain the main- 
 pieces in their proper pdsition; or 
 
 the latter may abut into cast-iron sockets suitably designed. 
 
 On Plate II. are given several forms of joints for timber structures, Which 
 
 can profess only to be suggestive. 
 
 190. Shouldered Tenon tor attacbiiiir Cross to Main Beamk..— 
 
 Fie 74 '''''^ weight on the end of the cross beam AB is borne 
 
 ' — '■ by the shoiildet C, which is let into the main beam for & 
 
 diMsnce equal to about one-sixth of the depth of the 
 former. The length of the tenon, J), is about twice its 
 depth. 
 
 BEAMS OF VARIOUS SECTIONS. 
 
 191. For those beams, Orders, and other similar structures, in which 
 certain parts are 8ut>posed to resist certain definite strains (68), and other 
 parts, otber sirains (for instance, flanged girders '«^ith thin, contibnous webs, 
 all open-webbed girden, trusses, ftCi) see 63 to Xi9i A inodd of. pro- 
 cedure is there adopted which w6uld not be thoroughly appUcaUe to 
 those beains in which evel-y fibr^ oi' ;^articl6 is cohlsidered io take part in 
 r^isting the bending actiod of the moment of rupture (1), and where the 
 whole secijon is liable to the action of the shearing force (1). 
 
 7» Designing a Beam — 
 
 198. Determine the nature of its cross section. 
 
 If the exact proportions of the section are to he adhered to, 
 and the area alone required, — express all the dimensions of 
 the section in terms of one of them, that there may he but 
 one unknown quantity. 
 
 Thus— suppose a beam to support a given load is to be rectan- 
 gular, with the deptti twice tue breadth .then let t = breadth, 
 and 2 5 = depth. 
 
 If all the dimetssions of the section, except one, he given, that 
 
 Otte will of course be the unknown quantity. Then, — 
 
 193. Suhstitnte for M, in the equations given hereafter, its value as 
 
 found from the span, manner of loading and supporting, £c.j pp. 2 to 15. 
 
 The dimension, oi dimeiisions, reqtiired may then he obtained. Lastly, — 
 
8TABIUIT or LOADBO BIA.U. 61 
 
 194. If at any Terdcal geetion, tbere be not sufficient material to resist 
 the shearing force (1..34, et leq. and ITI), the necessary addition must be 
 made to the section. 
 
 This will seldom be required at other places tban near the snpporle in dts- 
 continuoua beams, and near the points of contraiy flexure (193) m continuoas 
 beams. 
 
 106. The vielght of the beam ittelf must alirays* be added to the 
 estraneouB load, upon it ; and may be approximately estimated 1? a pro- 
 cess similar to that in 60. *' 
 
 196. The Siabiuit of a Loaded Beau depends on the equation 
 
 M = R. (1.) 
 lOT. Abbreriations — 
 
 Let M = moment of rupture (1, a), the values of which may be de- 
 termined from the several oases 7 to 33. 
 
 C I 
 S == — = moment of resistance (1) of the section. 
 
 I = moment 'of inertia of the section. 
 
 t — distance of the neutral axis (198) from the farthest edge of 
 
 the section. 
 a — total area of the section. 
 C = modulus of rupture (803). 
 
 198. TUe Keutral Axis (N — A in the sections, figs. 82—84) is a 
 section of the neutral surface, — a layer in the beam (and the only one) 
 which is neither extended nor shortened by the action of the load (4). 
 
 199. Provided that the limits of elasticity of the material of the beam 
 be not exceeded, the neviral axa vnll ptut through the eetUre of gravity of 
 
 the lectim (n^o). 
 
 goo. There most be but one lineal unit need in obtaining the values of 
 M, B, I, < ; and the superficial unit used for a must correspond to that 
 lineal unit. 
 
 201. The section at which R and I (107) are taken may be made 
 parallel to the reaction of the supports of the beam, i 
 
 Fig. 76. 
 
 S02. Wherever either the upper or lower surface 
 of the beam is not perpendicular to the action 
 of the load, then C must he modified to (C cos.' B), 
 6 being the inclination of the most inclined surface 
 to that perpendicular. 
 
 * Except in small girders, or beams of minor importance. 
 
 t The section at which the moments of resistance should really be taken is a 
 curved surface cutting the upper and lower edges of the beam, and the neutral 
 surface all at rif^ht angles, whatever be the form of the beam ; and the moment of 
 rupture to be equated with it should be taken at the intersection of this curved 
 surface with the neutral surface of the beam. 
 
62 
 
 BEAU8 or TARIODS SECTtOSS. 
 
 803, Hodnlm of RnptitFe.— The ilumttical-n,lne of C is thereeiatance of the 
 material to direct compreBsion or teneion. But it is found from ekperimentB on 
 cross breaking that this value is not sufficiently high. Amongst the reasons that 
 have been assigned for this, are— 1st, that in addition to the resistances of the 
 particles of the beam to a direct Strain, there is another resistance arising from 
 the lateral adhesion of the fibres to each other, termed the " Resistance of 
 Flexure." (See Barlow on the "Strength of Materials," eth edition.*) And 
 2nd, that in most metallic beams (especially when cast) the outer skin, which is 
 strained more than any other part of the section, is very much stronger (from 
 many well-known causes) than the nvernge section ; whereas if the direct tensile 
 or compressive resistance of the same beam, in, the direction of ite length, were 
 being experimentally ascertained, it would be i&t average section at least, and 
 perhaps me centre (weaker) portion especially, from which the strength would be 
 determined. However, there is evidently a necessity to employ a higher value 
 than that for the direct resistance ; and Professor Hankine has adopted a modulus 
 of rupture which is 18 times the load required to break a bar of 1 sq. inch section, 
 supported on two pdnts one foot apart, and loaded in the middle between the 
 -supports (831). 
 
 MoMiKTS OP Inertia and Rksistanok of Beams of Vabiocs Seomohs. 
 804. Beam of a solid rectangnlar section. 
 
 Fig. 76. 
 
 1 = 
 
 Id' 
 
 12 ■ 
 
 _ Cirfs Cad „ 
 
 20S. Beam of a hollow rectangnUir section. 
 
 Fig. 77. 
 
 I = 
 
 bd^—b'd'" 
 12 
 
 - c(&d»-yd'3) „ 
 
 K = — = M. 
 
 806. Beam of a solid circnlar section.. 
 Fig. 78. 
 
 I = -7854 r* = — '. 
 4 
 
 R = C -7854 r» = ^H = ji. 
 
 i 
 
 * Bdited by W.Hnmber. London : Lockwood & Co. 
 
K0XSHI8. or niBBIIA AHS BESISTAHCE. 
 
 XOT. Beam of a hollow ciieular section. 
 *I = -7854 (r* - j'*). 
 
 _ •7854Cfr«- r'«) „ 
 
 es 
 
 Fig. 79. 
 
 20S. Beam of a solid elliptical section. 
 
 I = -7854 b d\ 
 
 K = -7854 C 6 <P = M. 
 
 rig. 80. 
 
 209. Beam of a hollov elliptical section. 
 
 I = -7854 (6 cP - V d-").' 
 „ •7854 0(6«P-yd'>) 
 
 Fig. 81. 
 
 %I0. Beam irith one flange. 
 
 \ 
 
 I = i { 5cP + i'd"- (y-6)d'" } 
 
 K=2i=M. 
 
 Fig. 82. 
 
61 BEAKS or TABIOVS SEOTIOHS. 
 
 811, Beam witb tiro equal flanges. 
 Fig. 83. 
 
 I = 
 
 12 
 
 
 SIS. Beam vith two nnequid flanges. 
 Fir. 84. 
 
 I = J { 5 tP - (6 - *) (rf - c)» + 5' d" 
 (l/-]c) id' - (/)' }• 
 
 E = — = M. 
 
 813, To And tUe Homentg of Inertia and Rcslntsnce of anr 
 CroHs Seetion made up of a number of simple figures. 
 
 Find the moment of inertia of each of the simple figures about an axis 
 traversing its centre of gravity parallel to the neutral axis of the complex 
 figure. 
 
 Multiply the area of each of the simple figures by the square of the 
 distance between its centre of gravity and the nentnil axis of the whole 
 figure. 
 
 Add all the results together for the moment of inertia of the whole 
 figure. 
 
 Let I, = moment of inertia df one of the simple figures about its own 
 neutral axis ; A its area ; v the distance &om its centre of gravity to 
 that of the whole section (880) ; and I, moment of inertia of the whole 
 section ; then, 
 
 I = (Ii + v^ A) + &C. 
 
 C I 
 Moment of resistance, B = — 
 
 (w.) 
 
 814. IWoiaeitts of Inertia dnd Resistance, Ultimate Strenattli, 
 and Deflections of Similar Beams. 
 
 Moments of inertia \ . .•-.•i»t ( *''* **'' power J 
 
 Moments of resistance f ggij.l,- ) the 3rd power f of their linear 
 Strengths I ) the 2nd power l dimensions. 
 
 Deflections ) ^'J ** ( the 1st power ) 
 
nnsuAVtO BMAIts or vnivorh stkehqth. 
 
 65 
 
 BlAKB OF RKCTAttdtrtlA SKOTIOK AND OF UnIFOKH StBEKOTH (16R). 
 SUvatiotu of Bmrta ^ GiHildent irioMh, and Plant of Smmt of Conttant l>epth. 
 
 Fig. 85.— Elevation. 
 
 915. A tin!- 
 beam (T) losbclgd 
 Iritii a caiMKnir&tedl 
 veigiit at iU iit' 
 
 . tireMity. 
 
 21T. Beam sup- 
 ported at both 
 ends, loaded at any 
 'point (13) with 
 a concentrated 
 weight. 
 
 SIB. SemUbeam 
 (A) loaded uui' 
 ibrmly over its en- 
 tire length. 
 
 ' A ^, aparabola, with 
 its vertex at B. 
 
 ^ A triangle. 
 
 AB, a straight lUe. 
 
 A'BandCB, a peir 
 ofj^arabolaB, with 
 their apices at B> 
 
 AOandBC, pair of 
 pELrabolag) with' . 
 their vertices at 
 A and B respec- 
 tively.* 
 
 A C D, B t), a 
 pair of triangles, 
 having a coimmon 
 base, C D.» 
 
 318. Beam sup' 
 ported at both ends, 
 loaded uniformly . 
 (ID) over its en 
 tlfe length. 
 
 'A semi-ellipse, AB 
 major ajuis.^ 
 
 AOB. ADB, pairof 
 parabolas, having 
 their Vertices at 
 mid'^pan.* 
 
 * ¥he additional material dotted iA at the supports is necessary to resist the 
 fehearmg force (104>. 
 
66 
 
 SBTUEOTIOH. 
 
 919. How to cut the. (1.) Stroncu* and> (^<) BtilTut Reetansnlnr 
 Beam firoin a Cylindrical loc 
 
 Let the accompanying figs. 93, 94 represent sections of the log. - 
 
 Pis. 93. I^fT. «*. 
 
 strongest. 
 
 Stlffest. 
 
 Draw a diameter. For the beam whose ultimate strength will be the 
 greatest^ trisect the diameter. For the beam which will deflect the least, 
 divide the diameter into fonr equal parts. 
 
 Draw perpendiculars to the diameter as shown, and their intersection 
 with the circumference will determine the inscribed rectangle^ which is 
 the section of the required beam. 
 
 MO. To find the Centre of Gravity (199) of any Section. 
 
 Let a, a', a", &o., represent the sectional area of the several elemen- 
 tary parts into which the section may be decomposed ; g, g', gf', &c., the 
 known distances of their respective centres of gravity from any fixed axis — 
 say the lower edge of the beam — and S, the distance from the latter to 
 the centre of gravity of the total section ; then, 
 
 a _ "g + "'g' + <»"!/' + ^»- 
 a + a! + a" + &0> 
 
 DEFLECTION. 
 
 9X1. Deflection is the "displacement of any point of a loaded beam 
 ftrom its position when the beam is unloaded." 
 
 MS. Camber is an upward curvature, similar and equal to the maxi- 
 mum calculated deflectiMi, given to a beam or girder or some line in it, in 
 order to ensnre its hoiizontality when fiilly loaded. 
 
 aiRDXM'¥aosB oaoss asoiibira are imiroiiM am> equal 
 
 THBOUaHODT THKIK IiXireiB8. 
 
 3X3. The maximum deflections for several cases are given with the 
 moments of rupture (7 to XS), fay the values for Def. 
 
 ILAHOED OIBDEBS OP UHIFOBM SlBXMeTH. 
 
 SS4. Girder sapported at both ends. 
 
 Let D = central deflection. 
 d s central depth. 
 I = length of span. 
 
 K = sum of the extension of one flange or boom, and the short- 
 ening of the other by the strains upon them. 
 
(I05II1IV0V8 eiBDKBS. (S7 
 
 K may be found aa fultowa : Ii«t S = itrain in Iba. per sq. in. on either 
 boom when the load imdniang the deflection is on the beam ; B = nlodulns 
 
 of eUaUeity (»IS)j I = length of boom j then i-^I, + ^) = length 
 
 of b|pm after the atrain ia on ; and if li coireapond aimilarly for the other 
 boom, then k + Ji =li.. If the booms be of equal length and section, 
 thenK = (2i). 
 
 Let d = depth at support ; the other notations as before. 
 The deflection at the unsupported extremity, — 
 
 2d 
 
 tSt. ContiBnom Glrden, and einlers flxed »t One or Both Knds 
 
 («4— 88). 
 
 When any whole span ia aiulysed it will be seen (M to SO) that it ia equivalent 
 to a whole giider supported at ue ends, and one or ma aemi-girdera, as the case 
 may he. To UieBe the above fbrmnlee 
 (MS— S) mn be applied, and the nuud- Fig. K. 
 
 mum deflection obtained. Thus— 
 
 Forasrtrder (aa AB);!^ at lalh 
 ends {iA—M), the deflection of the 
 semi-beama A C and D B at and D ( II 
 or H5), added to the. deflection of C D 
 below its ends, aa olibuned from (IS, 
 19, or »S4), will be the total maximum deflection of E, belnw A. B. 
 
 Again, for eoMinuoiw girdtn with moving JoadttSX, SS), the maximum 
 dtfitctian at the middle of a span (B C, flg. 26) vill aeeur <imti2(aneoiM{y 
 mtk the maximum potttivtaumeM of rupture ()[» page 14), at which time 
 the points of oontrary flexure will be at M, M^ (fig. 25), wboae positions 
 may be determined either from the diagnm, or formulie. In the outer 
 ipant of continuoue girdere, and in girdere Jiieed at one end and tupported 
 only at the otiier, the deflection at the midcUe of the part oorreaponding to 
 a whole girder aimply anpported (aee »8, SO), may be found by adding the 
 cential dofleetion of the latter aa anch (19, 9M) to half the deflection of 
 the remaining (semi-beam) portion (11 or S9S). 
 
 BBBAEINQ AND SAFE LOADS FOB BRIDGES, QIBDEBS, ETC. 
 
 m. In the whole of the foregoing pages it is supposed that the span, 
 load, and other datit as &r as necessary ai'e given, in order to find the 
 resulting strains, and the quantity of material to resist them. 
 
 X88. To And the iMad (breaking or aafe) when the Qnantltj of 
 Haterinl, 8|mn, Ac., are given, is simply an inversion of the former 
 calculationa. 
 
 For uetional area of material, eubttitute an equivalent itrain, breaking 
 or safe acconliog as breaking or safe load is required. 
 
68 
 
 BRSAKIBO AHD SAFX U>AD8. 
 
 , Then cuca-tain the load that would product that strain, and that will 
 M the load requii«d. 
 
 Remembering that if the itructwe he not of uniform itrengtA, the 
 strength of the veakest part determines the strength of the whole (184). 
 
 HM. BXAUTLS I.— What load, at th« centre of a wronght-iron, single-webbed 
 or plate girder of uniform strength (165), and of the following dimensions, wonld 
 cause the rupture of the girder,— Length of span, 20 ft. ; central depth (OTi IS), 
 1-6 it. ; effeouve section (180) of the lower flange at the centre, 4 sq. in. ; t|jp>n of 
 average quality (SSI). ^ 
 
 As the girder is of unifbrm strength, it matters not what part ia considered, for 
 the same result would be obtained. 
 Lower flange will be in tension (6S— 1). 
 
 intimate strength of average wrought iron plate (>31), 66,000 lbs. per sq. in. 
 iq.in. lbs, lbs. 
 4 X 66,000 = 220,000 = brealdng strength of lower flange. 
 
 W i W X 20 
 
 Then from (19 and •!) strain on lower flange = — = 220,000 lb. = ^., ' 
 
 Therftfore W = ^^^ ^ iXJ^ _ ^j^ y^ ^ ^.^ ^^^ 
 
 230. Bi-^MTLx n.— Required the greatest safe (172) load nniformly distributed 
 on a rectangular beam of British oak projecting from a wall. 
 
 Length of beams 6 ft. = 72 in. . . . (200.) 
 Breadth „ =6 in. 
 Depth „ s=9in. 
 
 Taking coef, (1T2) as 6 ; O (203) as 10,000 (231.) 
 
 M 8 B. . . . (1, 196.) 
 
 "'• (a) E = ,£±L . . . (,01., 
 
 V. 
 
 2 
 r(w(}38. 
 
 36 (w I) = 162,000. 
 
 w I = 4,600 lbs. a 3 torn, 
 
 6 X Co 
 
 10,000 X 64X9 
 
 6x6 
 162,000 
 
 231. TaBL> of the SikKNOTH, &0., of MlTERIAtS IN FOUNDS AvOIBBVIOU 
 
 fKB Square Incb of Seoiior, 
 
 Jbtn-foii. 
 
 
 Ifoihiluio/ 
 JEliutieity. 
 
 Aot 
 •n lU.' 
 
 '-"• 
 
 '*x"- 
 
 OuaHng. 
 
 Cnu 
 Bnating. 
 
 MlTALS: 
 Brass, Cast 
 
 18,000 
 
 49,000 
 60.000) 
 36,000 f 
 
 40,000 
 16,000 
 
 10,300 
 
 uOiOoo 
 
 110,000 
 
 27,700 
 
 40.000 
 
 9,000,000 
 
 14,230,000 
 17,000,000 
 
 18,000,000 
 
 1 487 
 
 ' to 
 
 624-4 
 
 633 
 
 1649 
 
 1666 
 
 444 
 
 „ Wire 
 
 Copper, Rolled 
 
 „ _ Bolts 
 
 Iron, Cast 
 „ American best 
 „ „ average 
 
 * The specific gravities may be readily found when it is known that a cubic foot 
 of water weighs 62*6 lbs. 
 
BtnSKaiH, Ice, OF MATERIALS. 
 
 69 
 
 Tabli or THE Stmhgih, bto,, or MatsriaiiS — eontimied. 
 
 
 mtlmateX'tManctto 
 
 
 
 MaUriali. 
 
 
 Xoduliu at 
 OiuHeUy. 
 
 Foot 
 miti. 
 
 T«uim. <*X"' 
 
 Bhearim. 
 
 Crou 
 BnakinQ, 
 XoMmof 
 
 
 
 
 Suptw*. 
 
 
 
 MOTAM:— «««. 
 
 
 
 
 
 
 Iron, Wroaght .. 
 
 ( 73,000 
 
 
 
 
 
 
 „ „ „ iverage 
 
 66,000 
 f 41,000 
 
 100 000> 
 36,000 
 
 60,000 
 
 43V60O) 
 
 29,000,000 
 
 480 
 
 „ .. Plates .... 
 
 I 71,000 
 
 to 
 40,000 
 
 
 ' 
 
 16,000,000 
 
 
 » M •■ •▼wage 
 
 cs.ooo 
 
 
 
 
 
 
 „ ,, Jcunta— 
 ■iDRle riveted . 
 doubleriretecH- 
 
 ■ 
 
 
 
 
 
 
 40,000 
 
 
 
 
 
 
 62,000 
 
 
 
 
 
 
 •■ It Wir© ■«•(■• 
 
 84,0001 
 
 88.000 r 
 
 • •■ 
 
 • •* 
 
 *•• 
 
 16,000,000 
 
 
 .. „ ,.CBbI«.. 
 
 
 
 
 
 
 
 1 134,000 i 
 
 
 
 
 
 ( 487 
 
 SteelBan 
 
 160,000 
 
 
 
 
 to 
 
 
 
 120,000 
 
 >■• 
 
 80,000,000 
 
 K «8 
 
 ;; hii^f:,.:. 
 
 9S«» 
 
 ao,«o 
 
 
 
 
 
 
 TiMBu : 
 
 
 
 
 f 12,000) 
 
 - 
 
 
 Asb 
 
 16,H)0 
 
 9,000 
 
 1,400 
 
 \ to 
 
 1,600,000 
 
 47 
 
 
 
 
 
 ( 14.000) 
 
 
 
 
 
 
 
 \ w,oooi 
 
 
 
 Beech 
 
 11.600 
 
 9,000 
 
 
 i,3{o,oao 
 
 
 
 
 Birch 
 
 15,000 
 
 6,600 
 
 >.* 
 
 11,700 
 
 1,600,000 
 
 44-4 
 
 Box 
 
 20,000 
 
 10,000 
 
 .., 
 
 
 1,000000 
 
 60 
 
 Cedm 
 
 ll.MO 
 12,000 
 
 6,800 
 10,000 
 
 1,400 
 
 7^400 
 
 486.000 
 1,000,000 
 
 30-4 
 84 
 
 aam 
 
 ^ ■" ••«•••....,•» 
 
 
 ( 600 
 
 6,000 
 
 700,000 
 
 80 
 
 nr- 
 
 12,000 
 
 6,600 
 
 . to 
 I 1,600 
 
 to 
 
 1,000,000 
 
 to 
 4« 
 
 
 Uihogany 
 
 f 8,000) 
 
 to \ 
 
 20,000 ) 
 
 8,0fl0 
 
 
 f 7.«<iO) 
 ■< to 
 
 1,600,000 
 
 f 36 
 to 
 68 
 
 
 
 
 (11,600 
 
 
 
 10,000 
 
 
 
 (10,000 
 
 900,000 
 
 43 
 
 Oalr,BBgUah .... 
 
 to 
 
 WOO 
 
 2,300 
 
 ■i to 
 
 to 
 
 to 
 
 
 I 18,000 
 
 
 
 (18,600 
 
 1,700,000 
 
 62 
 
 „ American ., 
 
 12,000 
 ( 10,000) 
 
 6,000 
 
 ... 
 
 10,600 
 ( 12,000) 
 
 1,000,000 
 
 M 
 \ 4 
 
 Teak 
 
 to 
 
 12,000 
 
 
 to 
 
 1,300^000 
 
 ■< to 
 
 
 I 16,000 
 
 
 l,i».ooo) 
 
 1 61 
 
 Bmiomh CABtU 
 
 6,000 
 
 
 
 
 
 
 • It 18 difficult to estimate the cnnipreMi ve reeietance of ahoct bloOke o( wronght 
 iron, SB the material bulges very much under pressure, 
 
 f One and two rows of nveta (181). Joints of ec|nal lacUon to the plates titkeit 
 through the line of rivets, — From numerous experiments by W. Fairoaim. 
 
70- OOBSIRnOXIOH 0» JARABOLAS. 
 
 Tabu of ihb Stmsoth, rao., o» a^rssuis— continued. 
 
 XateriaU. 
 
 
 
 Foot 
 inltt. 
 
 Xtntim. 
 
 Oomprea' 
 Mien. 
 
 Shmim- 
 
 Bnatlin. 
 
 SH»m,Cn>iii, 
 
 sec.: 
 Bride Fire 
 
 „ Btnmg Ked 
 „ Weak Bad 
 
 Cbalk 
 
 Gnalta 
 
 f 276 
 i to • 
 ( 300 ) 
 
 CO 
 ••• 
 
 1,700 
 
 i.ipo 
 
 6S0) 
 to > 
 
 I sooj 
 
 400 
 11,000 
 
 4,600 
 
 ».000 
 ■ to 
 (6,600 
 
 ••* 
 
 1.100 ) 
 
 to \ 
 
 2,360 1 
 
 2,600,000 
 
 115 
 ' to. 
 I 136 
 
 rm 
 
 '■•■to : 
 I m 
 
 PS 
 
 I m 
 no 
 
 t 170 
 to 
 ISO 
 
 ISO 
 
 to 
 I 167 
 
 Mortar, ord. «■.. 
 LlmeBtono *•■».. 
 
 Saadatoqe 
 
 S3!i. YaSIOUB VBtHODS OF DBAWIlta FAKABOLAa, SBB BaSIC AHD HuSBI 
 BEINa aiVXN.* 
 
 I. (Fhte ni., fig. 1.) Bt obdihatxs on oitseis niox a iakocdi (E 0) 
 
 ID IBS PASABOLA AI ITS TIKIXX (S). 
 
 Tbioagh h diav I> E parallel and equal to A C. The onlinatei or 
 ofiets £rom any points in D E to the parabola will be proportional to the 
 sqnarei of the distances of those points from S, Thus, if the ordinate at a 
 be 1, then the ordinate at 6, tviee the distance of a bom. D.dnnst be 4 ; 
 thSit at Ct three times the distance, most be 9 ; and so on. To proceed 
 pra«tic^7 : DiTide E D into a number of equal parts (») as at a, i, ev he., 
 fig, 1 ; then if B A be divided into (n') parts, each of these parts will' be' 
 the required nnit, 1 of which is the o^t at a, 4 at &, 9 at e,. and so on. 
 TiiTough the points «.' V. il, &o., thus detenuined, the required enrre can 
 be' drawn. 
 
 II. (Plate III.J fig. 3.) Bt OBDnrAncs fBOH ibb basi. 
 
 Divide the base (half of which is repreeented by A 0) into an even 
 number of equal parts ;> then if the haght or ordinate at centre D corre- 
 qwnd to the square of. half the number-of those parts (8x8 = 64), the 
 
 * The terms height (or ordinate at oentre) and beee have been need 
 instead of absCiBBa for the former, and donble-ordinate for the latter, that tlie 
 
 garabola might appear in a moie almple light than perbape it otherwiee would 
 ave done. Any height con be adopted itar the parabola; oonvenience for 
 scaling off the momenta (S), Ac., being alone studied. 
 
CONSTRUCTION OF PARABOLAS. 
 
 Fiy 1. 
 
 Fiy ;.'. 
 
 / 
 
 / ^ '-^ ~^' 
 
 /' 
 
 
 / ::* 
 
 / ■ -, 
 
 A 
 
 C 
 
 Fiy.3. 
 
 ";- ': ' ^'' : ' 
 
 - / ■ 
 1, / ■ ' 
 
 c' ' / 
 
 : / ■ ■ 
 ,/■ / . 
 ' / , 
 
 Fit;- 4 
 E 
 
 /■ \ 
 
 /■■' 
 
 a'" C 6 
 
 F 1 q ; 
 
 C 
 
 "H 
 D 
 
 K 
 C 
 
 /;, ,/,„„ „ r.,„.,,r,r 
 
 ,..aP..,-,.M„:.r , 
 
 „„y ynf„pr.,nt | 
 
 />'"" 1 ^^" 
 
 p/I Jc 
 
 / 1 
 
 //'■' ! 
 
 
 \ 
 
 / \ 
 
 P 
 
 A M 
 
OONSTKCOIIOir OF PAKABOIiAS. 71 
 
 ordinate xt any other point (d for instance) will be represented by the 
 product of the numbers of parts in the two segments into which d divides 
 the base (4x12 = 48), The paraboia may then be drawn through the 
 extremities of the ordinates. 
 
 III. (Plate III., iig, 3.) Br the constki^toh of a duqbah. 
 Draw D E parallel and equal to AG; divide D E and E A similarly ; 
 
 the end B of E A corresponding to the end D of E D. Through a, b, &c., 
 in £ D, draw a a, bb, &e„ parallel to D C Join D to the several points 
 a' b', &c , in E A. The parabola will pass through the intersection of a (r 
 with Va'ibb with D V, &c, ■ 
 
 Note. If this mode of construction be adopted when the ordinates are 
 I i at certain paints only (and this will generally be the case 4n 
 
 piaciice), the actual curve need not be drawn, after the pointi it patati 
 through have been determined. 
 
 IV. (Plate III., fig. 4,) B7 the constritctiok of a siaorah. 
 
 On the base AB describe an isosceles tiiangle, whose height C E is 
 double that of the required parabola. • Divide the two sides A E, E B of 
 the triangle into an even number of equal parts, and draw lines as in the 
 figure. These lines will be tangents to the parabola, which may therefore 
 be readily drawn. 
 
 v. (Plate III., fig. S.) Bt means of a striso. 
 
 Draw E D equal and parallel to AC. iToin C to F at the bisection of 
 E D. Make G F 6 a right angle. Let F Q intersect the production of 
 C D. MakeDK*= DG, parallel to F D ; H H is a "straightedge," 
 against which slides the " set-square " S. A piece of thread or fine string 
 equal in length to the distance A E is fixed, one end at E and the other 
 at the point M, in the set square which will traverse the base A C as 
 the set square slides along. A pencil, P, by which the string is kept 
 tight, and close to the edge of the set square, will describe a true 
 parabola. 
 
 To DRAW A tAKOtST TO A FARABOIiA AT AKT POIKT P, (Plate III., fig. 6.) 
 
 Draw P C perpendicular to the axis S G. Make D E = D C, Join 
 E to P, and P B will be the required tangent. 
 
 * K is the "focus "of the parabola. GG' the directrix. 
 
SCIEN-flFIC BOOKS 
 
 , #t;te,isHXl} St 
 
 D. VAN NOSTRAND, 
 
 Publisher and Ittiporter. 
 
 -o ■■ « ■■ » 
 
 LOWELL HTDSAITLIC EZPEEIMEKTS— being a 
 
 Selection from Experiments on Hydraulic Motors, 
 Oh the Flow of Water over Weirs, and in Open 
 Canals of Uniform Rectanguki" Section, made ii 
 LoWisllj Mass. By J. B. Francis^ Civil Engififeeif. 
 Secofid editiorij rfevised and . enlarged, including 
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