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'««wi»*5#iesiias*'* ■ ails du tdifier une nage rrata :o pelure, id D 32X The copy fiimed hern has been reproduced thanks to the generosity of: Library of Congress Photoduplicaiion Service The images appearing here are the best quality possible considering the condition and legibility of the original copy and in keeping with the filming contract specifications. Original copies in printed paper covers are filmed beginning with the front cover and ending on the last page with a printed or illustrated impres- sion, or the back cover when appropriate. All other original espies are filmed beginning on the first page with a printed or illustrated impres- sion, and ending on the last page with a printed or illustrated impression. The last recorded frame on each microfiche shall contain the symbol -^ (meaning "CON- TINUED"), or the symbol V Imeanmg "END"), whichever applies. f^aps. plates, charts, etc.. may be filmed at different reduction ratios. 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Un des symboles suivants apparaitra sur la dernidre image de cheque microfiche, selon le cas: le symbole — »>signifie "A SUIVRE", le symbole V signifie "FiN ". Les cartes, planches, tableaux, etc., peuvent dtre fiim6s A des taux de reduction diffdrents. Lor»«que le document est trop grand pour dtre reproduit en un seul clich6, il est film6 d partir de Tangle supiriieur gauche, de gauche d droite, e^ de haut en bas, en prenant le nombre d'images ndcessaire. Les diagrammes suivants illustrent la m6thode. 5 6 ^^0^s^^m^^~ ^^^.^"^i^.M''^'^''^'' 2nd COPY, 1898. HEMATICAL WORKS. ESSOR EDWARD A. BOWSER. ACADEMIC ALGEBRA. With Numerous Examples. COLLEGE ALGEBRA. With Numerous Examples. PLANE AND SOLID GEOMETRY. With Numerous Exercises. ELEMENTS OF PLANE AND SPHERICAL TRIG- ONOMETRY. With Numerous Examples. A TREATISE ON PL/INE AND SPHERICAL TRIG- ONOMETR^ , AND ITS Applications to Astronomy AND Geodbsv. With Numerous Examples. AN ELEMENTARY TREATISE ON ANALYTIC GEOMETRY, Embracing Plans Geometry and AN Introduction to Geometry of Three Dimen- sions. With Numerous Examples. AN ELEMENTARY TREATISE ON THE DIFFER- ENTIAL AND INTEGRAL CALCULUS. With Numerous Examples. AN ELEMENTARY TREATISE ON ANALYTIC MECHANICS. With Numerous Examples. AN ELEMENTARY TREATISE ON HYDROME- CHmNICS. With Numerous Examples. LOGARITHMIC AND TRIGONOMETRIC TABLES. A TREATISE ON ROOFS AND BRIDGES. With Numerous Exercises. J; ''' II iiri(ii r»iiiiii«ia»»!P''l'K"mtuij n il II I .nii i ^ if i.i i i 1, 1 H I I , ui nil u% i f \i i ri0iiif)im^iiiriifgimrmiifmi'>^ 20013 OoPTBisnT, 1898, , Bt e, a. bowser. OF DEC7-1CG8 J * I Xartnaati )^\m 1. 8. Ctuhinx Ii Co llprwick .1 Smith Nurwoud Mua8. II.8.A f PREFACE. \^ r I Thk present treatise on Roofs and Bridges is designed as a textrbook for the use of schools. The object of tliis work is ^o develop the principles and explain the methods em- ployed in finding the forces in Roofs and Bridges, and to train the student to compute the stresses, due to the dead, live, snow, and wind loads, in the different members of any of the simple roof and bridge trusses that are in common use. The aim has b iVBi Iv PREFACE. Chapter III is devoted to IJridge Trusses with Unequal Distribution of the Loads. This is divided into three parts, as follows : (1) The use of a unifomihj distributed excess load covering one or mcjre panels, followed by a uniform train load cover- ing the whole span. (2) The use of one or two concentrated excess loads, with a uniform train load covering the span. (3) The use of the actual specijied locomotive ivheel loads, followed by a uniform train load. Chapter IV treats of Miscellaneous Trusses, including the Crescent Roof Truss, the Pegram and Parabolic Bowstring Uridge Trusses, and Skew Bridges. The stresses in this work are nearly all given in tons, the word " ton " meaning a ton of 2000 pounds. Any ocher unit of load and of stress might be used as well. My best thanks are due to my friend and former pupil, Mr. George H. Blakeley, C.E., of the class of '84, now * Chief Engineer of the Passaic Rolling Mill Company, for reading the manuscript and for valuable suggestions. E. A. B. RUTOBRS COLLBOE, Nbw Brunswick, N.J., October, 1898. k *•»«> ii.M*«>»iSSa»^ - ».i3£4&S&^;ri^^ ABT. 1. 2. 3. 4. 5. (J. 7. 8. 9. 10. 11. 12. TABLE OF CONTENTS. CHAPTER I. Roof Tkuhses. Definitions .... 'I'lie Dead Load Tlio Live I-oad .... Tlie Apex F.oud.'i and Reactions Relations between External Forces Aletliods of Calculation . Lever Arms — Indeterminate Cases Snow Load Stresses . . . Wind Loads .... Wind A}iex Loads and Reactions \Vind Stresses .... Complete Calculation cf a Roof Truss and Internal Stresses 1 3 4 4 11 14 19 20 22 25 29 aW CHAPTER n. Bridge Trusses with Uniform Loads. 13. Definitions . . . . 14. Different Forms of Trusses 15. The Dead Load 10. The Live Load .... 17. Shear — Shearing Stress . 18. Web Stresses due to Dead Loads 19. vbord Stresses due to Dead Loads 38 89 42 44 40 47 50 A"iWl".' Vi AST. 'JO. 'Jl. 22. 23. 24. 2."). 20. 27. 28. •jn. m. .12. :t:t. :tl. :<(!. ;(7. 38. CONTENTS. Position of IJiiifonii Live Loail causiiif; (.'liord Sti'0.ssc'N .... Mftxiinum Stresses in tiie Clionls . I'ositiuii of Uniform Io W'.ieu the Uniform Train Load is preceded by One or More Heavy Excess Panel Loads .... 120 When One Concentrated Excess Load accompanies a Uniform Train I^cad 127 When Two Ecjual Concentrated Excess Loads accompany a Uniform Train Load l!J3 The Baltimore Truss 142 True Maximum Shears for Uniform Live Load . . 154 Locotnotive Wlieel Loads 157 ~4l'M-f, '■■^■K~ M)>fHWMHH>>~.~,. 51 5fi R7 60 70 74 81 84 87 00 »1 08 100 101 lo:) 108 101) no 118 i..r ^■f 40. PoHition of Wlusel Loads for Mivxi'iium Slicnr 47. I'.)sitioii of Wheel I.oitdH fur Maxiiiiuiii Moini'iit at •loiiit ill Loudc'l Chord 48. Posilioii of Wheel I.oadH for Miixiiiiiiin Moineiit al Joint ill I'liloailrd Chord 40. Tabulation of Mouiuuts of Wlieel Loads. CIIAPTEll IV. MiSCKLLAJJEOt'S TllUSSKH. Ron/ Truttts. 50. The King and Queen Truss — The Fink Truss •A. The Crescent Truss 1(10 108 Bridge Trusses. 52. 'he Pegram Tniss — The Parabolic Bowstrinjr Truss 53. tjKew Bridges 177 182 187 102 i;>o 120 127 'I? 1!5:j 142 154 157 1\ .^ -^iS^Wi'^- ROOFS AND BRIDGES. o»{o CHAFTER I. ROOF TRUSSES, Art. 1. Definitions. — A Tramed Scructurs is a collection of pieces, either of wood or ii-etal, or both combined, so joined together a.s to causn the structure to act as one rigid body. The points at which the pieces are joined together are usually c&V ""d joints. A Truss 1? a structure des .^ned to transfer loads on it to the supports at each end, while each member of the truss is subject only to longitudinal stress, either tension or com- pression. The simplest of all trusses is a triangle; and all tmsses, however complicated, containing no superfluous members, must be composed of an assemblage of triangles, since a triangle is the only polygon whose form cannot be changed without changing the lengths of its sides. A Strut is a member which takes compression. Struts are sometimes called j9os Art. 2. The Dead Load. — The dead or iixcd load sup- ported by a roof truss consists of the weight of the truss itself and the weight of the roof. The weight of the truss can .only be approximated. The following formula* gives approximately correct results for short spans : Let I = span in feet; b = distance between trusses in feet; and W= ap- proximate weight of one truss in pounds. Then The roof includes the roof-coverinf, the sheeting, the rafters, and the purlins. Its total weight will vary from 6 to 30 lbs. per square foot of roof surface. The purlins are • Modern Framed Structures, by Johnson, Bryan, and Turneaure. -^v« lbs. to 12 lbs. per cubic foot, although snow which is saturated with water weighs nnich more. Tlie snow load need not be considered when the inclination of the roof to the horizontal is 60° or more, as the snow would slide off. The luind load is variable in direction and intensity, and often injurious in its effects ; especially is this the case with large trusses pkced at considerable intervals apart. The maximum wincl pressure on a surface normal to its direction is variously estimated at from 30 lbs. to HG lbs. per square foot. The calculation, therefore, of the stresses cansed by the wind forces is often of considerable importance, and should never be left out of account in designing iron roofs of large span. Art. 4. The Apex Loads and Reactions. — Both the dead and live loads are transferred, by means of the purlins, to the joints or apexes of the upper chords of the truss ; and these loails at the apexes are called the apex loads. They are also called the panel loads. The truss, roof, and snow loaiu, being vertical, and uni- formly distributed over each panel, the apex loads are each equal to one half the sum of tl.e adjacent panel loads. Thus, the load at h, Fig. 1, is eoaal to one half the panel load on Ac plus one half the panel load on ah. The wind • . -',' ROOF TliUSSES. '•-.•'■> load at h is also 0(iiial to one lialf the wind load on Ac plus one half the wind load on ah, the load on eaoh being normal to the surface. If the truss bo synnuetri- cal, and the panels be of equal length, the apex loads are d(;termined by dividing the total load by the number of panels in the upper chords. Thus in Fig. 1, the apex loads ^''*" ^ at h and c are each one fourth of the total load. At the suii- ports, a and l, the loads are only one half those at h and c. Reactions. — For dead and snow loads both reactions of the supports of the truss are vertical, and each is one half of the total load, if the truss is symmetrical. For wind load the reactions depend upon the manner of supporting the truss. ■_ : ■ ,; Pi'ob. 1. A roof truss, like Fig. 1, has its span 80 feet and its rise 40 feet; the distance between trusses is 12 feet, center to center. Find (1) the weight of the truss, (2) the weight of the roof, (3) the snow load, (4) the apex loatls, and (5) the reactions. For these problems, take 20 lbs. per square foot of roof sin-face for weiglit of roof, and 20 lbs. per square foot of horizontal projection for snow load. Here ab = 80 feet, dc = 40 feet, and ac = 5G.5C feet. Weight of truss, \V, = -,^br- = 3200 lbs. Weight of roof, TF, r^ 56.5G x 20 x 12 x 2 = 27148.8 lbs. Weight of snow, W, = 80 x 12 x 20 = 19200 lbs. Each apex load = J x 49.548.8 = 12.SS7.2 lbs. Each reaction, ii = J x 49548.8 = 24774.4 lbs. i ■'^' ^ .. «-- >^ *-.- i , * ,. •*•." ■ M S & ' i-:%*M Mti •aiwi' '^^m 6 ROOFS AND li JUDGES. Proh. 2. A roof truss, like Fig. 2, has its si)au 00 feet, its rise ;«) feet, and distance between trusses la feet: find the Via. a '■■ , " ^' :' total apex loads and the reactions for the weights of the truss, roof, and snow. Ans. Apex loads = G989 and 13977 lbs.; reaction = 27954 lbs. Pfob. 3. In the roof truss of Fig. 3, the span is 100 feet, the rise is 25 feet, and the distance between trusses 12.5 feet: find the total apex loads and the reactions for the truss, roof, and snow loads. k d FiK. 3 Ans. Loads = 7270 and 14540 lbs. ; reaction = 29079 lbs. Art. 5. Relations between External Forces and -^intemal Stresses. — The external forces acting upon a tvui^f-^j-fl in equilibrium with the internal stresses in the members'iS r^^e truss. Let mn be a'stvi/^'n passed through the truss, Fig. 4, cut- #^ nnOF TRUSSEfi. feet, its find the ts of the eaction = 1 100 feet, isses 12.5 >s for the 29079 lbs. ces and ig upon a ies in the i'ig. 4, cut- ting tlie members whcse stresses are desired, and let tliese stresses be replaced by equal external forces. Then it is Fig. 4 clear that the equilibrium is undisturbed. Therefore we have the principle: The internal stresses in any section hold in equilibrium the external forces acting upon either side of that section. If we remove either portion of the truss, as the one on the right of tlie sectiovi, then the external forces on the remaining part of the truss, together with the internal stresses, form a sy.stem of forces in static equilibrium. And from Anal. Mechanics, Art. Gl, the conditions of equilibrium for a system of forces acting in any direction in one plane on a rigid body are: 2 horizontal components = (1) S vertical components =0 (2) S moments =0 (3) These three equations of condition state the relations between the internal stresses in any section, and the ex- ternal forces on either side of that section. If only three of these internal stresf^es are unknown, they can therefore be determined. For example, in Fig. 4, let R, P^, Pi be the reaction and apex loads, found as in Art. 4; let .Sj, .1.2, .% be the stresses in the members he, kc, and kd, that are (!ut by the section mn, and let fl, and 63 be the angles made by .s, and s^ with Jit' ■' <"- 8 fiooF.9 AND ntiinnEsi. the vertical. Applying our throe equations of equilibrium, wo have : for horizontal components, », sin $i + a, sin Oj + Sg — 0, for vertical components, R - 1\— 1\ -\- s, cos fl, + Sj cos 0^=0. In applying our third equation of condition, the ct-nter of moments may l)e chosen at any point, Anal. Mechanics, Art. 40. If we take it at c, the moments of s, and Sj, are zero, and the equation is — (R - I\)\ ab-PiX \ab - Sg x cd = 0. These three equations enable us to find the unknown stresses. ■'"■- " . ' ;[.,^;'- ■■■'^>;. ' NoTK 1. —In all cases, in this work, a tensile stress is denoted by the positive sign, and a compressive stress, by the negative sign.* In stating the equations, it will bo convenient to represent the unknown stresses as tensile, pulling away from the section, as in Fig. 4. Then, if the numerical values of these stresse* are found to be positive, it will show that tliey were iissuuied in the right direction, and are ten- sile ; but if they are negative, they were assumed in tlie wrong direc- tion, and are compressive. Prob. 4. In the truss of Fig. 5 the span is 90 feet, the rise is 35 feet, hk is perpendicular to the rafter at its mid- point, and the loads are as shown : let it be required to find the stresses in all the members of the truss. Representing the stresses by Si, .% s^, S4, Sj, and a^, we proceed to apply our three equations of equilibrium. To find the stress s„ pass a section cutting s, and Sg, sepa- rating the portion to the left, take moments about h, and regard .5 lbs. To liiiil .Vfl, pass tln! section cuttinj,' n.j, .%, and s^, and take moments about a. Tlion, 10000 x22.5-8«x 35=0 (since the lever arm of S8=c(«=35), .-. .s„ = (; tL'.S lbs. Since the truss is symmetrical a.id syuimetrieally loaded, it is evident that the stresses in all the pieces of the right half are cMpial to those just found in the left. NoTK 2. — In tlio ivbovo Hulution we Imvo called forces acting up- wanls iin, and tluwe in the oppn.sitc! direction, negative. Tho oiiposito coiivtuition would do a« well ; we have only to introduce opposite forces and opposite nionients with unlike HJgns. Prob. 5. A roof truss like Fif,'. 1 has its span 40 feet, its rise 20 feet, and the apex loads 2000 lbs. : find the stresses in ad, ah, he, find hd. Ans. (id = + 1.5 tons, ah = — 2.12 tcnis, he = — 1.41 tons, hd = - 0.71 tons. Pri>b. 6. A truss like Fig. 2 has its span (50 feet, its rise 20 feet, and tlm panel loads 4000 lbs. : find the stresses in ad, ah, he, and hd. Aiis. ad = 4- 4.5 tons, ah = — 5.40 tons, /jo = — 3.G4 tons, hd = - 1.82 tons. Prnh. 7. A truss like Fig. 3 has its span 80 feet, its rise 20 feet, and the panel load? 10,000 lbs. : find the stresses in all the memligrs. Ana. ak = + 15 tons, kd = + 10 tons, ah — ~ 10.75 tons, /«*= — 1 1 5 tons, /(A;=— 4.5 tons, fcc= +5 trms, rf?=00 tons. -t.. t,-iRre55wj^.:^'5«?tTS:'*«iE«^-ai g,J ! !' ! 'yw-.t!l^ ' - ' ' ■ »T?ff^"' TnracBn^sacv.^ -11.-- ItoOF TltUSSEfl. 11 Alt. 6. Methods of Calculaf on. — Our tliipo oq\ia- tioiis III' cipiililirimn, Art. r», I'liniisli iis witli two nit'lliiMis uf ciilciil.'iliou : //((' iiK'tliod hi/ irniihtliiiii nf fiti'ci'n, iiiul tlif iiii'thdil of mniiu'iil.i. 'VUv priiiciiik) of llin lirst im'tliod is L'lubriiciul in tho first two iMnuitioiiM of ('(luilil)iiuin ; tlio Iiriiicii)U' of tlio second iiKitliod is enilnaced in tlie third tM|niition of ('({iiililiriiini. Kitlun- of tiit'sc! nu'tJKjd.s may be used in tlie solution of ftny KiviMi case; but in general there will be one, the em- jiloynn'Mt of whi(!h in an}' sjiccial ease; will bo found easier and simpler than the other. Sometimes a eumbiimtion of both methods furnishes a readier solution. Kkmaiik. — A section may bo imsseil tlir(m;,'li iv tni.ss in any dirnc- ti(in, sciianilinK it into any two p'lvtions. 'Ilnis. in Fi^. Ti, a Hi'ctinn may In- i)a.sHial arouuil /(, cntlinj; he. hk\ and hn. 'I'licii llie inttTiial Ntii'HHt'H »i, Hi, «8, and the apex h)ad of 10,(KX) lbs. at h foi'in ii HyHtcui (if fiircus in t'(|ullibriuiii, to wliicii our ctiuntiouH are applicabli'. A judiciiiiis .selieliiin (if dircctidns for tlie ri'solutioii of tbc forrcs often Kimplities Uic dcti'rniinatiou of tlie Hi rcasci. Tluis, to iind .15 in Fi^. 5, if wo rcHolve tlie forces into a direction periiendicular to tli(! rafter nr. we sliall oblain an ei|nation free from the forces S:i and S4 ; wliereius if the directioiiH are talteii at randoni, all of tiie forces will enter the O(ination. This principle is a very useful one. Thus, we have at onee in Fifj;. /», ealling the angle between the load and the rafter, 10000 sin e + .Si = 0. .-. s, = - 10000 cos cwl = - 7.S95 lbs. as before. Similarly, if a section be passed around d in Fig. T), cut- ting (Ik, ik, and /= (only supports part of the tie rod), cflf = + 0.75, ah = + 1.5, ck=+ 4.5, bg = - 1.68, e/t = - 2.15, dk=- 2.7. Prob. 14. In Fig. 8 the span 's 60 feet, the rise of truss is 15 feet, the rise of tie rod at point ^ is 2 feet, the panel Viit. loads are 1.6 tons, the struts 3-4, 5-6, 7-8 are vertical, dividing the rafter 2-7 into three equal parts: find the stresses in all the members. Ans. Stress in 2-3 = - 9.75, in 3-6 = ~ 7.8, in 5-7 = -5.85, in 2-4=4-8.73, in 4-6 = + 8.73, in 6-8 = + 7, in 3-4 = (only supports part of tie rod), in 5-6 = -|- 0.75, in 7-8 = -H 3.7, in 3-6 = - 1.85, in 5-« = - 2.24. Proh. 16. In Fig. 9 the span is 90 feet, the rise of truss is 22.6 fe«t, the rise of tie rod is 3 feet, the struts, 3-4, 5-6, I'Ms. O 7-8, and 9-10 are vertical, dividing the rafter into four ecpial parts; the apex loads are 2 tons: find the stresses in all the members. 18 HOOFS A Nit llliinaEN. Aiix. Sticbs in 2-3 = - 18, in .'{-5 = - 15.4, in 5-7 --= - lli.S, in 7-1) = - 10.2, in 2-1 = + IG.l, in 4-(} = + U).l, in (•,-8 = + 13.8, in 8-10-r 4- ll.u, in 3-4^=0 (this is not necessary to the staliility of tlie truss), in 5-() = + 1.0, in 7-8 =+2.0, in 9-10 =+7.2, in 3-6 =-2.46, in 5-8 = - 2.96, in 7-10 = •- 3.68. Pmb. 16. In Fig. 10 the span is 120 fc't, the rise is 30 feet, the struts 3-4, 6-6, 7-8 are drawn normal to the ! rafter, dividing it into four equal parts, the apex loads are 2.5 tons : find the stresses in all the members. Atis. Stress in 2-3 = - 19.5, 3-5 = - 18.38, 5-7 = - 17.25, 7-9 = - 16.13, 2-4 = + 17.5, 4-6 = + 15.0, 6-10 = + 10.0, 3-A = - 2.23, 5-6= -4.45, 7-8= -2.23; 4-5= +2.5, 5-8 = + 2.5, 6-8 = + 5.0, 8-9 = + 7.5, 9-10 = (only sup- ports part of the tie rod, and not necessary to the stability of the truss). It will be observed fhat the members 3-4 and 7-8 are symmetrical with respect to their loads, and therefore their stresses arc equal, and also that 5-4 and 6-8 are symmetrical, and hence their stresses are equal. Prob. 17. In Fig. 11 the span is 90 feet, the rise of truss is 22.5 feet, the rise of tie rod is 3 feet, the struts 3—4, 5-6, 7-8 are drawn normal to the rafter, dividing it into four equal parts ; the apex loads are 2 tons ; lind the stresses in all the members. • 4k ■?f- in r»-7 .-= - = 4- Ki.l, in (this is not I = + 1.0, in 6, in 5-8 = he rise is 30 riual to the pex loads are 8.38, 5-7 = 15.0, G-10 = ; 4-5= +2.5, ; (only sup- the stability re symmetrical . arc eciual, and eir sti-esses are I rise of truss •II ts 3-4, 5-6, it into four lie stresses in ROOF TIIUSSES. Ans. Stress in 2-3= -20.34, 3-n= - 19.44, 5-7= -18.54, 7-9 = --17.(J4, 2-4 = + 18.3, 4-0 = + lo.GH, 0-10 = + 9.3, 3^-4 = - 1.78, 5-0 = -3.5(;, 7-8 = -1.78, 4-5 = + 2.64, 5-8 = + 2.64, 6-8 = + ().84, 8-9 = + 9.48, 9-10 = (not necessary to stability of truss). Art 8. Snow Load Stresses. — The snow load is estimated per stpiare foot of horizontal projection (Art. 3). If the main rafters of the truss are straight — as in all of our previous publems — the snow load is uniformly dis- tributed over the whole roof, and the apex paow loads are all equal (see Art. 4); therefore tlie dead load and snow load stresses are proportional to the corresponding apex loads. Thus, in Prob. 13, the dead panel load was 3083 lbs. and the snow panel load = 10 x 20 x 12 = 2400 lbs. Therefore \2 we multiply each dead load stress by ^^§^3 (=.778), we shall have the corresponding snow load stress. IJut if the main rafters are not straight, the snow load is not uniformly distributed over tlie whole roof, and the apex snow loads are not all equal; in such case, the snow load stresses have to be determined independently. Thus, with the dimensions given in Fig. 12, for 9 feet between trusses, we have the apex snow load at b and at 6' each equal to one half the ]»anel load on ub plus one half the panel load on 6c ' = 6 X 20 X 9 = 1080 lbs. = 0.54 ton, '-■ « " ' g . ji3;r " m 20 nOOFH AND nUIDGES. The apex snow load at c equals one half the panel load on cb plus one half the panel loiui on cb' = 8 X 20 X 9 = 1440 lbs. = 0.72 ton. The stresses duo to these apex snow loads may now be found in the same way as the deail load stresses were found in the preceding problems. Via. I'-a It is possible for one side only of a roof to be loaded with snow. This possibility is recognized in desi" ing roofs of very large span, such as the roof of the Jersey City train shed of the Pennsylvania R.R. In special forms of trusses such a distribution of snow load may produce a maximum stress in some members. Problem. With the dimensions and apex snow loads above given, find the snow load stresses in all the members of Fig. 12. Am. ui'* = -1.44, be = -1.3, oti = + 0.86, W = + 0.68, dh = + i:.i, feet; let tlie liistiuu'e between trusses be lli •eet, and suppose the wind to be on the left side. Then the /'tclination of the rafter ae to the horizon = tair' J=L'«;".'M'. Heuw, ?>i>iu our table, the normal wind pressure per s(piaru Vie. 13 foot = .30.2 lbs. The total normal wind pressure on the si-'s of the roof ae is therefore = 30.2 X 12 X Vfi(F+W= 30.2 X 12 X 56.9 = 20258 lbs., one fourth of which, or 5004.5 lbs., is the pressure on cacli panel. Jlenee the normal wind load at each apex b, e, d, is 5004.5 lbs., or say, in round numbers, 5000 lbs., or 2.5 tons, and at each apex a, and <>, it is 2500 lbs., or 1.25 tons (Art. 4). The Reactions caused by the wind pressure are inclined; the horizontal component of the wind tends to slide the entire truss ot? its supports. The weight of the truss and the roof are usually sufficient to cause friction enough to hold it in place. But if it is necessary, the truss should be fastened at its ends to the wall. Roof trusses of short span, and especially wooden trusses, have generally both ends fixed to the supporting walls. But la'-ge iron trusses have only one end fixed, while the other end is free, and resting upon friction rollers, so that it may move horizontally, 7i-^m-'v!^! mmmmk tions. r.ft 110 I'.'ft. ami ^ be lli •Vt't, Tlien the ■' J =!-'(;" .'M'. 5 per t)(|tiiiiu lure on the J02o8 lbs., lire on eacli 3x b, e, (I, is or 2.0 tons, V 1.25 tons re inclined; slide the i truss and 1 enough to s should be short span, both ends I'lisses have md resting urizontally, nOOF TRUSSES. under changes of teniporaturo. Wo have then two cases: till! first, when both ends are fixed; the .scm oiid, when one end only is fixed, and the other end is free to move upon rollers. Case I. When both ends are /.mi. — Let Fig. 13 represent a roof truss with both ends fixed, its span being 100 feet, its rise 25 feet, and the wind ajicx loads 1.2r>, 2.5, 2.5, 2.5, and 1.25 tons, as found above. In this case, the two reac- tions Ri and ii, are jiarallel to the normal wind loads, and may easily be found, as follows : Let 6 be the angle between the rafter and the lower ohoid ak, and take moments about the left end a. We have then < i?j X ah - (? 5 X ab + 2.5 x ac + 2.5 x ail + 1.25 x ae) = 0, or, we may take the ic.'jidtant of all the loads, or 10 tons, acting r.t c, .-. R, X 100 cos ^ - 10 X 27.05 = 0, .-. Rt = 3.13 tons (since cos = ^S). The reaction Ri may be found by subtracting T?, from the total wind load, giving us /i, = 6.87 tons. Or, we may find Ri by taking moments about the right end k. Thus, RiXah — 10 (ah — ac) = 0, from which we get Ri — 6.87 tons, as before. Case II. When one end is fixed and the other is free. — (1) Suppose the right end of the truss to be free, and the wind blowing on the fixed side, as in Fig. 14. The right end of the truss is supposed to rest upon rollers, the support at o taking all the horizontal thrust due to the wind. This being the case, the reaction iij at the free end will be verti- cal, and the reaction at the fixed end will be inclined. #» 24 HOOFS AND II It I DOBS. 'A y T(i find R.2 take nionunts about n ; let the (liinciisioiiR nii.i wind loads ho the siiiiic as in Ki;;. 1.'5. Thus, Jij X 100 - 10 X 27.05 = 0. .-. Ji^ = 2.8 toua. ♦ 11/ II n J 71 /■ 4 •B. . Ib. Klg. 14 Resolve the left reaction into its horizontal and vertical components, // and /?„ thus, 2 hor. comp. = gives 10 8infl-//=0. .-. H=iA6. S ver. comp. = gives -B, -I- 2.8 -10 cos ^ = 0. .-. /?, = 6.13. Or, we might find 72, by taking moments about the right support; thus, /?, X 100 - 10 (100 cos e - 28) = 0. .: Ri = 6.13, as before. (2) Suppose the wind to blow on the free side ek of the truss, Fig. 14. The reaction iJ, is vertical, as before, and the reaction at the fixed end a may be resolved into its horizontal and vertical components, in the same way as above. Thus, we find, //=4.'G, i?, = 2.8, i2, = 6.13; that is, when the wind changes from one side of a roof to the other, the horizontal component // has the same value as before, but acts in the opposite direction, and the re- actions Ri and li^ interchange their values. rm tuut). nd vertical 3. ■ the right , as before. ek of the )efore, and )d into its le way as a roof to ame value id the re- mbtmumm ROOF TRVSSKS. 86 rri>h. 18. A trims like Fi^,'. 14 lias its span 40 feet, rise Kl IVcl. mill t\u' lotul noi'iiial wind load on tin* Hxed side li.'J tons: liiid tlio wind load reactions. Ann. /;, = 1.90, Itt = 0.90, //= 1.43 tons. Proh. 10. A truss like Fi>j. 3 has its span fiO feet, its rise 12.5 feet, and tlu' distance between the trusses 8 feet: find the reactions when both e- :L are fixed. Ana. /f, = 2.32, /i, = l.Ofi tons. Prob. 20. A truss like Fif?. 4, with one end free, has its t.Mi 80 feet, it,s rise 20 feet, and the total normal wind load on the fixed sic'ic 5 tons : find the wind load reactions. Au8. R^ = 3.07, /;, = 1.4, //= 2.23 tons. Proh. 21. A truss like Fig. 4, with one end free, has its span (50 feet, rise of truss 12 feet, rise of tie rod 2 feet, and the total normal wind load on the fixed side 4.5 tons : find the wind load reactions. Ana. Hi = 2.97, li, = 1.21, H= 1.67 tons. Prob. 22. A truss like Fig. 6, with one end free, has its span 90 feet, its rise 18 feet, and the total normal wind load on the fixed side 6 tons : find the wind load reactions. Ana. Ill = 3.94, R^ = 1.62, //= 2.25 tons. Art. 11. Wind Stresses. — (1) If the truss have both ends fixed, we must consider the wind blowing normally to the principal rafters on one side only (either side indiffer- ently) ; and as the wind load is unsymmetrical to the roof, the wind stresses in the members oi. one side of the truss are different >om those in the corresponding members on the other side, and hence they must be computed for every member in the truss. (2) In trusses with one end fixed and the other free, we I J^ r » '^ I 26 IKxn-S AM) It II 11)11 KS. uwut coiisii! T Mio wind lilo'viiii,' lirHt on onn siiln of tho ti'iiHS, and lilt < on tin- (itlicr; and llii' Ht.ri'.s.scs inodiu'cd in tlio twr) casi's will liiivn to ho, (jomimti'd. Prof). 23. A truss liko Fig. 15, with both ends fixed, has its span HO feet, its riso 12,5 feet, and tho wind loads and reactions as sliown: lind all the wind stresses. Wo find «c = 27.95 feet, or, in round numbers, 28 feet : ak = 15.()S feet = kc ; hk = l feet. Ilopresentiiif^ the stresses by «„ Sj, Sj, s^, Sj, and s^, and applying the printuples of Arts. 5 and G, we have for the left half of the truss : 2.(52 X 14 - .s, X fi.25 = 0, 2.G2 X 28 - 3 X 14 - h., x 12.5 = 0, 2.62 X 14 + s,, X 7 =0, 3 + », =0, - 3 X 14 + «8 X 12.5 = 0, ,s, = 4- 5.87 tons. Sj, = + 2.51 tons. Sg = — 5.24 tons = s^. sj = - 3.00 tons. Sfl = + 3.36 tons. For the right half it is better to resolve the right hand rcaetion 1.88 into its horizontal and vertical components, thus, //= 1.88 sin e = 0.84, and V = 1.88 cos d = 1 .68 tons, and to state the e(piation of each piece including the ex- ternal forces on the right of the section rather than on the left. Thus, ' ■•'i;iJii51!i>i*Sniym»JJ-'!MW«'at*ll!alte- ■ tmmmtm ItnoF TitirssEs, 27 I 4 i.OSx t2.n-.HI x(5.1»r. -Vxr..2n=:(), .-. h,'^ fli.r.2 tons. i.(Wx2r.-.si X \'j.r>^.s,'x\'j.r>^(), .-. «;=4--w mns. l.«W X 1 <-),(;« f.v,,' X 7^0, .-. *,'= -.'{ "(> tons -:=«;. I'nih. 24. A truss likn Fi(,'. lo, with oiio end frco, liiis its span 40 leot, its riso 10 feet, ami tlio total normal wind load on till! lixcd siilo .'{.U tons; tind all tlio wind stresses. Wo must lirst find tho reactions and horizontal eonipo- nent, as in Case II. Thus, /?, = 1.90, Ji, = 0.9, // = 1.43. Ans. A<, = + ;{.«, »,= + 1.8, /»a=:-2.8, «4=-2.8, s,= -1.6, a„= + i.8, .V=+1.8, s;=-2, ,V=-2, ^=0, «„'=0. 8i o. — It will often bo best to state tlib equation, uhIiik the ex- ternal forces on the riylU of the section. Thus, to C'.nl tlii' stress In «2- It' wo use tho forces on the right of the section, the eciuation is .Ox20-»aX 10 = 0. .-. »2 = 1.H. Hut If wo use tho forces on tho left of the section, iho eijuallon la 1.00 X 20 + 1.4.} X 10-32 x 11.18 -»j x 10 = 0. .-. «, = 1.8. Of course, to find tho 8tres.ses in members near the loft end, not so much is gained by using tho forces on tho right of the section. Prob. 26. A truss like l*'ig. 4, with one end free, has its span (50 feet, its rise lit feet, rise of tie rod .3 feet, and the total normal wind loud on the fixed side .'i ton : tind the wind stresses in all the members. Ana. Stress in ak = + S.7, kd = + 3.5, «/t = -7.78, he = - 7.78, hk = - 2."), kc = + r>Ar>, bk' = + 4..3(i, bh' = - 4.8, h'c = - 4.8, h'k' = 0, ck' = + 1.0, cd = (n(»t necessary to stability of structure). Prob. 26. A truss like Fig. 14, with one end free, has its span 80 feet, its rise 20 feet, and the total normal wind load >f ^^1 . w i K i mn i ■^Tvr-r •^•SC~-f**"**- ># ill 28 ROOFS AND nUWGES. on the fixed side 8 tons : find the wind stresses in all the members Ans. Stress in «/= + 11.16 =yj/, gh = + S.92, /(j = + G.()8, bg = - 2.5, c/t = - 3.10, dj = -4.02, eg = +1.12, dh = + 2.24, e; = + 3.35, ab = -9, be = -7.5, cd = -(j.O, tie = -4.5, kf =:f'g' = g'h' = h'J = + 4.47, Vg' = e'h' = d'j = 0, b'f = e'g' = d'h' = 0, kb' = b'c' = e'C = d'e = - 5.0 tons. Prob. 27. A truss like Fig. 16, with one end free, has its span 60 feet, its rise 12 feet, the struts hk and h'k' normal to the rafters at their middle points, and the total normal wind load on the fixed sida ac 4 tons: find the wind stresses in all the members. Ans. Stress in ak = + 5.4, kd = + 2.7, ah = — 4.6 = he, hk = -2, cfc = + 2.7, bk' = + 2.7 = dk', bh' = -2.91 = ch', h'k' = = ek'. Prob. 28. In the same truss Fig. 16, with the same dimen- sions as given in Prob. 27, let the same normal wind load of 4 tons blow on the free side be: find the wind stresses in all the members. In this problem the values of the reactions are the same as those in Prob. 27, but interchanged, and th ' horizontal component has the same value, but aiitf. in the opposite direction. See Case II. of Art. 10 ; therefore, here Hi = 1.08. Ri - 2.63, //= 1.48, ^Jl^JiSS^-lggf I ROOF riiUSSES. 29 II all the = + G.()8, = + 2.24, = -4.5, //' = cy ), has its ;' normal 1 normal he wind 4.6 = he, .91 = eft', le dircen- d load of resses in 18 those in [t has the of Art. 10 ; Ans. Stress in bk' = + 3.88, dk'= + 1.10, 6ft' = -4.58 = eft', h'k' = + 2, ck' = + 2.7, uk = -^ 1.22, dk = + 1.22, aft = - 2.91 = eft, hk = = ck. Art. 12. Complete Calculation of a Roof Truss.— By comparison of the values of tlie stresses in Prob. 27 uith those in Prob. 28, we see that the stresses are quite different, and generally greater when the wind blows on the fixed side of the roof than when it blows on the free side. When the wind blows on the fixed side it tends to " flatten " the truss ; and when it blows on the free side it tends to " shut up " the truss, or " double it up." In the complete calculation of a roof truss, we must find the stresses due to the greatest dead load, and combine them with the greatest stresses due to the live load, so as to get the greatest possible tension and compression in each member. If the dead load and live load stresses in any piece are of the same character, both compressive or both tensile, we must add them to obtain the greatest stress in the piece. But if these stresses are of opposite characters, one compression and the other tension, their difference will be the resulting stress die to the combination of live and dead loads, and if the live load stress is less in amount than the dead load stress, it will only tend, when the wind blows, to relieve the stress due to the dead load by that amount, and the dead load stress is the maximum stress \r> the mem- ber. But if the live load stress is greater than the dead load stress and of an opposite character, it will cause a reversal of stress and this piece will then need to be counterbraced. It is the customary American practice to determine the greatest stresses in each member of the fixed side of the roof truss which could be caused by the wind force acting t ■ t ' it •2H noOFS AND nnWGES. on the fixed side 8 tons : find the wind stresses in all the members Ans. Stress in «/= + 1 1 .16 = ft/, gli = + 8.92, lij = + C.()8, bg = - 2.5, ch = - 3.10, dj = -4.02, eg = +1.12, dh = + 2.24, ej = + 3.35, ab = -9, be = -7.5, ed = -'6.0, de = -i.5, kf =:f'g' = g'h' = h'J = + 4.47, b'g' = e'h' = d'j = 0, b'f = c'g' = d'h' = 0, kb' = b'e' = c'C = d'e = - 5.0 tons. Prob. 27. A truss like Fig. 16, with one end free, has its span GO feet, its rise 12 feet, the struts hk and h'k' normal to the rafters at their middle points, and the total normal wind load on the fixed sida ac 4 tons: find the wind stresses in all the members. Ans. Stress in ak = + 5.4, kd = + 2.7, ah = — 4.6 = he, hk = -2, c& = + 2.7, bk' = + 2.7 = dk', bh' = - 2.91 = ch' , h'k' = = cA:'. Prob. 28. Tn the same truss Fig. 16, with the same dimen- sions as given in Prob. 27, let the same normal wind load of 4 tons blow on the free side be : find the wind stresses in all the members. In this problem the values of the reactions are the same as those in Prob. 27, but interchanged, and th ' horizontal component has the same value, but aiitf. in the opposite direction. See Case II. of Art. 10 ; therefore, here Hi = 1.08. i?2 - 2.fl.S, //= 1.48, a,BS!i»3«'.v«#^i&v«?SE*?a»*as^-;sair€ ■ ■ ^msmmm^m^^^r Irv9 ROOF rUUSSES. 29 Ans. Stress in bk'=+ 3.88, dk'= + 1.10, 6/t' = - 4.58 =r-. ch', h'k' = + 2, ck' = + 2.7, ctA,' = -^ 1.22, dk = + 1.22, ah = - 2.91 = ch, hk = = cfc. Art. 12. Complete Calculation of a Roof Truss.— By comparison of the values of the .stresses in Prob. 27 'vith those in Prob. 28, we see that the stresses are quite different, and generally greater when the wind blows on the fixed side of the roof than when it blows on the free side. When the wind blows on the fixed side it tends to " flatten " the truss ; and when it blows on the free side it tends to " shut up " the truss, or " double it up." In the complete calculation of a roof truss, we must find the stresses due to the greatest dead load, and combine them with the greatest stresses due to the live load, so as to get the greatest possible tension and compression in each member. If the dead load and live load stresses in any piece are of the same character, both compressive or both tensile, we must add them to obtain the greatest stress in the piece. But if these stresses are of opposite characters, one compression and the other tension, their difference will be the resulting stress due to the combination of live and dead loads, and if the live load stress is less in amount than the dead lead stress, it will only tend, when the wind blows, to relieve the stress due to the dead load by that amount, and the dead load stress is the maximum stress in the mem- ber. But if the live load stress is greater than the dead load stress and of an opposite character, it will cause a reversal of stress and this piece will then need to be counterbraced. It is the customary American practice to determine the greatest stresses in each member of the fixed side of the roof truss which could \y) caused by the wind force acting I mmi f' n- i i i 80 ROOFS AND niHbUES. normally to the truss on the fixed side only, and then to build the members of the two sides of the truss of the saiuf size. Since the stresses caused by the wind blowing on the fixed side of the roof are at least as great as those caused by the wind blowing on the free side, this arrangement gives the maximum stresses, and is on the safe side; and for reasons of economical manufacture both sides of the truss are constructed alike. Prob. 29. A trust, iikc : - 14, with one end free, has its span 80 feet, its ris' i<; ,<'jt, distance apart of trusses 1(3 feet, rafter divided into four equal parts, struts vertical, dead load of roof 20 lbs. per sciuare foot of roof surface, snow load 20 lbs. per square foot of horizontal projection, normal wind load on fixed side by table of Art. '.; : find the dead load, snow load, wind load, and maximum stresses in all the members. Fnmi ^he given rise and span we have the length of one half of rooi i»« = VIO" + 16* = 8V21) = 43.08 feet. Weight of truss = j'^ bP (Art. 2) = ^^' ^ ^^^^)' = 42GG lbs. 24 Dead panel load 420fi , 4.S.()8 X 1(5 X 20 3977 lbs. 8 4 = 1,9885 tons, ' . ij, 2 tons. Snow load per panel = 10 x 16 ; ." - 200 lbs. =1.6 tons. Inclin. of roof = tan-' ■ 4 = 2lM^* 11(1 then to if tlic .saiiif 'ing on the ose caused 'i'angenieu„ side ; and les of the ree, has its trusses 10 is vertical, [■face, snow on, normal I tlie dead 3 in all the j;th of one t. 42GG lbs. [)77 lbs. =1.6 tons. 1=25.2 lbs. !, 2.2 tons. wind load computed, m \ ROOF TliUSSKS. 31 and the maximum stresses found, for all the members of the truss, and tabulated, as follows : StRK88K(* in tub L0W£lt ClIORI). Mkmiikkh. "/ fa (/A M Dead lo.ad stresses . . Snow load stiTSiiis . . Wind 1( ad stresses . . + 17.50 + 14.00 + 14.74 + 17.50 + 14.00 + 14.74 + 14.96 + 11.07 + 11.77 + 12.44 + 0.06 + 8.80 Ma:;i.uuu) stresses . . +46.24 + 46.24 + 38.70 +31.10 Stkesses in the Upper CnoHu. Mkmueiui. ah be C'(/ de Tlead load stresses . . Snow load stresses . . Wind load stresses . . -18.80 -15.04 -12.86 -16.10 - 12.88 -10,52 - 13.40 -10.?;; - 8.21 -10.70 - 8.56 - 6.34 Maximum stresses . . -46.69 -39.60 -32.33 -25.60 Stressks in the Web Membeiis. ME.MIIEK8. «C dh ej hg ch dj Dead load stre.sses Snow load stresses Wind load stresses + 1.00 +0.80 + 1.10 +2.00 ] + 6.00 + 1.60 1 + 4.80 +3.96 1 + 3.56 -2.68 -2.14 -3.19 -3.20 -2.56 -3.78 - 3.00 - 3.12 - 5.04 Maximum stresses +2.99 + 7.56 1+14.36 -8.01 -9.64 -12.00 (6/ is not necessary to the stability of the structure.) Here the maximum stress of each kind for each member ill the windward side of the truss is found by adding the J~lg^ 82 HOOFS AM) lilt IDG EN. (leiid load, snow load, and wind load stresses giving the Rrcatost total tension and the greatest total compression. Of course, the wind stresses in the nienibors of the other half of the truss will be less than those above given for the members of the half on the windward side, and therefore, the maximum stresses in these members will be less than those in the above table, since the dead and snow load stresses in the corresponding members of the fixed and free sides of the truss are the same. If there is no wind blowing, the maximum stresses are found by adding together the dead load and snow load stresses. If there is neither wind nor snow the dead load stresses are also the maximum stresses. If the wind blows on the free side of the truss, the maximum stresses cannot exceed those found above. Proh. 30. A truss like Fig. 9, with one end free, has its span 90 feet, rise of truss 18 feet, rise of tie rod 3 feet, rafter divided into four equal parts, struts vertical, dead load per panel 2 tons, snow load [)er panel 1.5 tons, normal wind load per panel, wind on fixed side, 2.25 tons: find the dead load, snow load, wind load, and maximum stresses in all the members of the half of the truss on the windward side.. Here the effective reaction for dead load = 2 X 3.5 = 7 tons. To find the dead load stress in 2-4 or 4-6, take moments around the point 3. 7 X 11.25 dead load stress in 2^ : 3.75 : 21.00 tons. Similarly the dead load, snow load, and wind load stresses may be computed for all the members of the truss. v&v.-^-^f ,\¥e.-.. :?;«■ cjiviiig the nipression. the other /en for the therefore, I less than snow load (1 and free tresses are snow load dead load kind blows ses cannot •ee, has its od 3 feet, tical, dead ns, normal 1 : find the stresses in windward ? moments ons. id stresses Ans. ROOF TRUS,SES. Stuesses in the Lower Ciionn. Stresses in the Uiter Chord. 88 Memiiirs. 2-» + (! (.-•< S-IO Dead load 8t.re.s.se8 . . Snow loiul sti-L'sses . . Willi' load stresses . . + 21.00 + 15.75 + 18.l:j + 21.00 + 16.75 + U,.i3 + 18.00 + 13.50 + 14.40 + 15.00 + 11.25 + 10.84 Maxiniuin stresses . . + .-)4.8a + r,4.S8 + 45.00 + 37.00 Memiikrk. 2-3 3-1 5-7 7-9 Dead load stresses . . Snow load stresses . . Wind load stresses . . -22.00 -16.05 -16.20 -10.30 -14.52 - 13.28 -10.12 -12.00 - 10.20 -12.88 - 0.06 - 7.74 Mivximu 1 stress, s . . -55.84 -47.16 -38.47 -30.28 Stresses in THE W 'ed Memdehs. Memberr. 6-fl 7-3 9-10 8-6 6-8 7-1 n Deail load stre.s.ses Snow load stresses Wind load stresses + 1.00 +0.76 + 1.21 + 2.00 + 1.50 + 2.43 + 7.60 + 5.70 + 4.54 -3.10 -2.32 -3.7e - 3.50 - 2.62 - 4.23 - 4.10 - 3.08 - 4.'M Maximum stresses +2.06 + 5.03 + 17.84 -9.20 -10.35 -12.17 (h/is not necessaiy to the stability of the truss.) Prob. 31. A truss like Fig. 10, with one end free, has its span 100 feet, its rise 20 feet, the rafter divided into four equal parts by struts drawn normal to it, dead load per panel 2.5 tons, snow load per panel 1.5 tons, normal wind load per panel, wiad on fixed side, 2.3 tons : find all the ii II f 84 HOOFS AND nilllHUCS. stn>ssn.s in all tl'.e members of the liiilf of the truss on the wimlwuril side. Ana. Sthkssus in the Lowe It Chord. Mkmiikkh. a- 4 4-6 fi-IO Dciiil load stve8.sea . . Snow load Htressus . . Wind load stres-ses . . + 21.83 + 13.10 + 15.64 + 18.70 + ll.i;2 + 12.64 + 12.50 + 7.50 + 0.21 Maximum stresses . . + 50.67 + 42.40 + 20.21 Sthksses in the UrrEit Choisii. Mkmiikkh. 2-8 3-5 n-T 7-9 Dcvd load stre.s.se.s . . Show load streasea . . Wind load stresses . . -23..")0 -14.10 -13.57 -22.57 — 13.55 -13.67 -21.66 -12.00 -13.67 -20.73 -12.44 -13.57 Maxiuiuin stresses . . -61.17 -40.09 -48.21 -40.74 Stuesses in the Wep MeML'"U8. Mrmiiebs. 3-4 ttiid l-S ,')-•(■) 4-fi niKl 0-8 8-9 Dead load stresses Snow load stresses Wind load stresses -2.33 -1.40 -2.30 - 4.65 - 2.79 - 4.G0 + 3.10 + 1.80 +3.11 + 6.20 + 3.72 + 0.21 + 9.30 + 6.68 + 0.i>l Maximum stresses -0.03 -12.04 + 8.07 + 10.13 +24.19 (0-10 is not necessary to the stability of the truss.) Prob. 32. A truss like Fig. 11, with one end fi-ee, has its spau 100 feet, the rise of truss 20 feet, the rise cf tie rod 2.5 feet, the rafter divided into four equal parts ly struts drawn normal to it, the dead, snow, and wind loads 2.5 'SWRM tmmm BwiMimnnswigJWW''"'"'*'*-' russ on the ft-I(» + 12.50 + 7.50 + 0.21 + 20.21 8-9 20 + 9.30 72 + 5.58 il + O.iil 13 +24.19 S8.) "ree, has its cf tie rod ts ly struts L louds 2.5 BOOF TRUSSES. 35 tons, 1.r» toi»s, and 2.3 tons respectively, or the same as in I'rol). 31 : tind all the stres.'es in all the members of the half of the truss on the windward side. Ann. Stkbahks in the Lowkk CiIOKI). .Mkmhrrh. L'-l •IB fi-in Dead load stre-sscs . . Snow load Htrt's.si's . . Wind load Htresses . . + 28.55 + 17.1;{ + 19.92 + 21.60 + 11.70 + 17.30 + 14.35 + 8.01 + 7.00 Maxiiuum stresses . . + 05.00 + 50.60 + 29.96 Stuessks in the Urrmt Chohd. ME.UIIKR8. 2-8 8-5 6-T T-9 Dead load stresses . . Snow load stresses . . Wind load stresses . . -30.00 -18.12 -19.08 -17.81 -18.12 -28.75 -17.25 -18.12 -27.83 -10.70 -18.12 Maximum stresses . . -67.08 -66.61 -64.12 -62.65 Stresses in the Web Members. Membebb, 8-4 and 7-8 o-« 4-5 and 5-8 6-8 8-9 Dead load stre.sses Snow load stresses Wind load stresses -2.33 -1.40 -2.30 - 4.66 - 2.79 - 4.00 + 4.00 + 2.40 + 4.02 + 10.05 + 6.39 + 9.15 + 14.65 + 8.79 + 1.".10 Maxii.mm stresses -0.03 -12.04 + 10.42 + 20.19 +30.00 (9-10 is not neceasary to the stability of the truss.) Prob. 33. A truss like Fig. 11, with one end free, has its span 120 feet, rise of truss 20 feet, rise of tie rod 3 feet, rafter divided into four equal parts by struts drawn uornuil IIP" 86 HOOFS AND nUIDGES. to it, distance lietwcon trusses 20 feet, dead load of roof 15 Ib.s. per square foot of roof surface, siow load 20 lbs. per sijuare foot of horizontal projection, normal wind load on fixed side by table of Art. 9 : find all the stresses in all the members in the half of the truss on the fixed side. Ans. Dead load per panel = ,'5.1 tons; snow load per panel = 3 tons ; wind load per panel = 3.3 tons. Stiiessks in the Lower Ciioud. .Mkuiicbd. 2-4 4-0 8-10 Dead load Htresscs . . Snow load stresses . . Wind load stresses . . + 45.26 + 43.80 + 37.35 + 38.01 + 37.65 + 20.02 + 22.17 +21.46 + 12.56 Maximum stresses . . + 126.41 + 106.48 + 66.18 Stkerses in the Uri'ER Chord. MiMIIERS. 2-:i S-.'i 5-7 T-9 Dead load stresses . . Snow load stresses . . Wind load stresses . . - 47.52 - 45.00 - 35.31 - 46.56 - 46.06 - 35.31 - 45.60 - 44.13 ■- 35.31 - 44.64 - 43.20 - S\3l Maximum stresses . . -128.82 -126.03 -125.04 j -123.16 Stresses in the Web Memheks. Members. 3-4 and 7-S 5-6 4-5 and 6-S 6-8 8-9 Dead load stresses Snow load stresses Wind load stresses -2.05 -2.86 -3.35 - 5.80 - 5.70 - 6.70 + 6.38 + 6.15 + 7.44 + 17.36 + 16.80 + 17.76 +23.72 + 22.05 +25.10 Maximum stresses -0.15 -18.20 + 10.05 j +51.02 +71.86 •^'i^&^sSl^l^^^fi,^ '^V^-V^ "" MMMMI load of roof load 20 11)8. lal wind load itresses in all ced side. ow lo• an' rarricd up tho tics to r and d, and tlu>n by tliti two inclined struts m and til, di.wu to tht! abntnit'uta, a and h. The Howe Truss, Fi^. 11), has its voi-tic-al niombors in tension and the intdincil ones in coniprcssion ; the dia),'onal counter struts aro in I ken lines. The cliouls and diagonal /. v/ Kic. lU web members are of wood, and the vertical tics of iron. This truss was patented in the United States in 1840 by William Howe. It has proved the most useful style of bridge truss ever devised for use in a new and tind)ered country. It is still very largely used where timber is eheap,.for both highway and railway bridges. The Pratt Truss, Fig. 20, }ms its vertical members in compression and the inclined ones in tension. All the members of this truss are of iron or steel ; though it was formerly built all of wood, except the diagonal ties, which .mVi'.mkdmmiSmm-i'iifi'M -'■ 1 I 1 1 UUIUnK TIlUSSRS. 41 with 1(111!,' Tim |iaii('l 1, iiul then abntinuata, lonibors in (' (liii>,'()iial III diagonal B8 of iron, n 1840 by il stylo of 1 timbered timber is lendiers in , All the ngli it was ;ies, which worn of wrought iron. The I'ratt truss is a favorito tyjie, and is used inme tlian any otiicr kin.l. I'"i^'iire '_'(> hIihwh a f/ccA'-bridgc. The deck or ilirnugli I'ratt truss is tlie stand- ard form of triisH for botli higiiway and railway bridges of nio(b'rate sjians, though it is not generally nse(i for railway bridges in whieh the span is much less than lOO feet. The Warren Truss, or Warren Oirder, iuis all its web mend)ers inelino;! at eqnal angles, some of them being in tension and some in eompression. This truss is an example of the pine triangular type. Its web nuMubers consist always of ('(jiiilntenil Iriuiiylcn. When the triangles are not ecjuilateral, the truss is simply a ^' tridiifjiilnr truss." AAAAAAAA/ ^ KiB. ai Fignre 21 shows a Warren truss as a deck truss. The War- ren truss is generally built all of iron or steel, and is used for comparatively short spans; it is of more frequent oceurrenee in England than in this country. The Double Triangular, or Double Warren Truss (or Oirder), Fig. 2'J, has each panel braced with two diagonals intersecting each other, and forming a single lattice. wmymm m Ktg. 39 Other forms of trusses will be explained as we proceed. All these forms of trusses, and bridge trusses generally, may Im; arranged so as to be used either for deck-bridges or through bridges. I ^^S^Jk . w *H 42 noOFS AND It JUDGES. Art. 15. The Dead Load, or permanent load, consists (if tlio entire weij^ht oi' llie hrnlge. It ineliitlos the weight of the trusses, the lateral bracing, and the floor system. In highivai/ biidgen, the floor system consists of the floor beams, which are supported by the chords at the panel points, the stringers, which are supported by the floor beams, and the planks, which are supported by the stringers. In railway briclyes the stringers support the cross ties, rails, guard- rails, spikes, etc. The dead load depends upon the length of the bridge, its width, its style, and upon the live loads it is intended to carry. For highway bridges with plank floors the total dead load jier foot may be found approxi- mately by the following formula: 10 = 150 -f cbl + 4 bt, where iv = weight in pounds per linear foot of bridge. I = length of bridge in feet, b = width of roadway in feet, t = thickness of planking in inches, c = ^ for heavy city bridges, = ^ for ordinary city or suburban bridges, = I for light country bridges. For single track railroad bridges the dead load per linear foot is given very closely by the following formula (taken from "Modern Framed Structures," by Johnson, Bryan, and Turneaure, p. 44). For deck-plate girders, to = 9^4- 520 (1) For lattice girders, w = 7l + mO (2) 5SSS^*S^S1^'^^3|^ ilKlE TRUatiES. 43 d, consists the weight 'stem. Ill oor beams, points, the s, and the [n railway ils, guard- tlie knigth ve loads it ith plank i approxi- dge. ;e8, per linear ilae (taken )n, Bryan, ■ • (1) . . (2) For through pin-connected bridges, v,^lH-\- 750 ...*.... (3) For Howe trusses, iv = G.5 1 -\- G75 (4) where I is the span in feet, and lo the dvjad load in pounds per linear foot. These four formulte give dead loads of iron i-ailway bridges, including an allowance of 400 lbs. per foot for the weight of track material, for bridges designed to (iarry lOO-ton locomotives. For lighter locomotives the dead load would be less, and for heavier locomotives it would be more. For double track bridges add 90 per cent to the above values;* for the load on each truss take one half the above values. Prob. 34. What is the weight of a Vv'arreu truss bridge 105 feet long ? Here we may find the dead load from formula (2), as follows: Dead load per linear foot = ly = 7 ^ + COO = 1336 lbs. .-. weight of bridge = 1335 x 105 = 140,175 lbs. and weight of bridge to be carried by one truss = 70087.5 lbs. Proh. 35. What is the weight of a through pin-connected bridge of 100 feet span ? Ans. 125,000 lbs. Proh. 36. What is the weight of a Howe truss bridge cf 144 feet span ? Ans. 231,984 lbs. ; or 115,992 lbs. per truss. ♦For double track brulges the weight of the me/a/ toorA; is alKiutfK)% greater than for a single track; Imt tlie weight of tlie track materiat is just double that for a single track bridge. i! ill i • -"-""rKBaSe*. 44 HOOFS AND nillDGES. Art. 16. The Live Load, or moving load, is that which moves over thft bridjjo, ami consists of \vaiited live loads per linear foot of track, i)roduce approximately the same stresses as the above concentrated loads. Span, 10, 20, 30, 40, 50, 100, 200, 300 feet. Load, 10,000, 6600, 5500, 4900, 4600, 4000, 3700, 3500 lbs. This loading is very nearly that used by the rcnnsyl- vania Railroad. The Erie Railroad nses a live load one fifth greater; and the Lehigh Valley Railroad a live load one third greater. The live load is taken greater for short spans than for long ones, because if the span is short one t)r two locomo- tives may cover the whole bridge, while if the span is long, the whole bridge would not often be loaded with more than a train drawn by one or two locomotives. 'J'he calculations of the stresses are made in precisely the same way for railway as for highway bridges, so long as the live load is uniform. Ill the use of equivalent uniform loads, the following precautions are to be observed : In obtaining the stresses in the chords hrA the main web members, the equivalent load corresponding to the length of the truss is to be used. But in such members as receive a maxi- mum stress from a single panel load, another equivalent load must be used. Thus, in the "hip verticals" of a Pratt truss, Fig. 30, or in the "vertical suspenders" of a Warren truss. Fig. 35, the maximum stress is obtained by using the equivalent load corresponding to a span of two panel lengths. Prob. 37. A bridge for a city has its roadway 20 feet wide in the clear, and also two sidewalks, each 6 feet wide in the clear. The span is 200 feet and there are 10 panels. Find the live panel load per truss. Ans. 12.8 tons. HI i H I i I V'i mm 46 ROOFfI AND niilDGKS. Fig. S3 O, Pinb. 38. A country l):i(lj,'C, .50 feet long, lias its roadway 1(5 feet wide in the dear, and a sidewalk 8 feet wide in tlie clear. There are 5 panels. Find the live panel load per truss. Ans. 6.4 tons. Art. 17. Shear — Shearing Stress. — Let Fig. 23 repre- sent a beam fixed horizontally at one end and sustaining a load P at the other end. Imagine the beam divided into vertical slices or transverse sections of small thick- ness. The weight /* tends to sepa- rate or shear the section or slice on which it immediately rests from the adjoining one. The lateral con- nection of the sections prevents this separation, and the second section or Siice is drawn by a vertical force ecpial to the weight P which tends to slide or shear it from the third section, and so on. Thus, a vertical force equal to the weight P is transmitted from section to section throughout the length of the beam to the point of support. This vertical force is called the " shearing force," or ^' shear" ; and the equal and opposite internal force or stress in the section that balances it is called the " shearing stress." The shear then at any section is that force ivhich tends to make that section slide upon the one imtnediatel;/ following. The vertical shear at any section is the total vertical force at that section, and it is equal to the sum of the vertical com- ponents of all tl external forces acting upon the beam on either side of the section. Thus, let Fig. 24 represent a beam I feet long, resting horizontallj' on supports at its extremities; and let the 'CS:r7SrvS(Sr^T^-fi %i BIIIDGE TRUSSEi.. 47 beam have a uiiifonn l(ni(l of to lbs. i)or foot. Consider a section ab at any distance a; from the left end. Then, the reaction at each siii)i)ort is J wl; the slicar in the section ab is the weiglit that is between that section and tiic center Hiw. yilvi. Fig. S4 of the beam, or (| I — x)w, since this is the total vertical force at ab, wliich is the shear bi' definition; i)ut tliis is the same as the algebraic sum of the vertical components on the left side of ab. For the algebraic sum of the vertical components on the left of the section ab = ^ Iw — xw, or (^ I — x)w. Art. 18. Web Stresses due to Dead Loads — Hori- zontal Chords. — Let ABCD be a truss with horizontal c cT /» fc h' e' D -4 « b y, « d c ' jt>' a 7-^B P. Fig. SB chords* AB, CZ), and let a section mn be drawn in any panel be cutting the diagonal bh and the two chords, and let 6 be the angle which the diagonal nuvkes with the ver- tical. Then (Art. 5) the stresses in these three members mist be in equilibrium with the external forces on the left of the section; and therefore the algebraic sum of the . * Called also Ihui^bs. K T? 48 nooFs AND liiiinaKs. w m vertical components of theso forces (Mjiials zero. Hence, calling S the stress in the diagonal bh, wt; have R-Pi-I\ + Scos6 = (1) But It— Pi — Pj, is the vertical shear for the given socticjn Miw (Art. 17). Hence (1) becomes Shear + 5008(9=0; .". S = — shciiT sec (2) Therefore, for horizontal chords and* vertical loads, the stress in any web member in etjmd to the vertical shear multi- plied by the secant of the angle which the member makes tvith the veHical. Since the shear in the section cutting he is the same as that in the section mn, no load being at h, therefore the stress in Ch is equal to the shear in ch, that is, equal to the shear in the panel be. Since there are no loads between the joints h and c, the shear is constant lluoughout the panel be; and we nsually speak of it as the shear in the panel be. It will be observed from (2) that, for the diagonal bh, the stress is negative, or compressive, provided that the shear is positive; but for the dead load the shear is always positive in sections left of che middle of the truss. Hence the stress in b' is compressive, and so for all the other diagonals in th . left half of the truss, since they are all inclined in the same direction, that is, downwards toward the left. Conversely, for members inclining downwards toward , he right, in the left half of the truss, the stresses axe 2>ositive or tensile. Prob. 39. A throngh Howe truss, like Fig. 2-'i, has 8 panels, each 15 feet long, aud 20 feet deep: find all the i i BltlDGE TliUSSES. 49 web stresses due to a dead load of 450 lbs. per linear foot per truss. Panel load = 1^0^15 = 27 ^ 3 gg ^^^^ 2000 8 27 7 Reaction =-- y.^ = ^ = 11.8 tons, 16 which is also the shear in panel Aa. The shear in each of the other panels is found by subtracting from the reaction the loads on the left of the panel. sec lilillXiES. /'loh. 41. A throujrh Warroii truss, Fig. liCi, lias 10 piincls, (uu'li 10 feet long, its web uiuinbeis all loruiiiig I'tiuilatoral t s r, 7 [t ;i' 7' r' s' f /\/.\/.\/\/\/\AA/\A 10 It 10' Ki«, SO 4' t' triangles (Art. 14): find the stresses in all the web mcnihers dno to a cleatl load of 400 lbs. per linear foot per truss. Ans. Stresses in 1-2, ,">-4, 5-0, 7-8, 9-10 are - 10..'{«, - 8.08, ~r>.7(i, - 3.4(5, - 1.14 tons. Stresses in 1-4, 3-0, 5-8, 7-10, 9-12, are + 10.38, + 8.08, + .5.70, + 3.40, + 1.14 tons ; that is, the signs alternate in the web members. Art. 19. Chord Strasses due to D«>.ad Loads — Horizontal Flanges. (1) To find the chord stresses by the method of moments. Pass a section cutting the chord member whose stress is required, a web member, and the other chord, and take the center of moments at the intersection of the web member and the other chord. Then, supposing the right part of the truss removed, state the equation of moments between the unknown stress and the exterior forces on the left of the section. For stresses in the npper chord members the centers of moments are at the lower panel points; and for stresses in the lower chord members the centers are at the upper chord points. Thus, in Vrob. 41, each panel load is 2 tons, and each reaction is therefore 9 tons; the depth of the truss is 10 sin 00° = 8.66 feet. Hence for the lower chord stresses, we have : -«v IS 10 piint'l.s, i i'(liiilatoral s' 1' 4' f kX'b inciiihers jr truss. we - 1().,'}«, ).38, + 8.08, alternate in 1 Loads — moments. 086 stress is and take tlie veb member t part of tlie between the tlie left of nembers the nts; and for I's are at the IS, and each the truss is ord stresses, BRIDGE riiUSSES. 61 9 X 5 - stress in '2-4 x 8.(i(; — 0; .-. Stress in 2-4 = + r».li() tons. 9x15-2x5- stress in 4-0 x H.m = 0; .-. Stress in 4-G = + 14.4.'! tons. 9 X 25 - 2(15 + 5) - stress in O-S x H.GG = 0; .-. Stress in 0-8 = + 21. .'50 tons. 9 X 35 - 2 X 45 - stress i!i 8-10 X 8.00 = 0; .-. Stress in 8-10 .= + 25.98 tons. 9 X 45 - 2 X SO - stress in 10-12 x 8.00 = 0; .■• Stress in 10-12 = + 28.28 tons ; and for the upper chord stresses, we have : 9 X 10 - stress in 1-3 x 8.00 = ; .-. Stress in 1-3 = - 10.38 tons. 9 X 20 - 2 X 10 - stress in 3-5 x 8.06 = ; .-. Stress in 3-5 = - 18.48 tons. 9 X 30 - 2(20 + 10) - stress in 5-7 x 8.06 = 0; . .-. Stress in 5-7 = — 24.24 tons. 9 X 40 - 2 X 00 - stress in 7-9 x 8.06 = 0; .-. Stress in 7-9 = — 27.70 tons. 9 X 60 - 2 X 100 - stress in 9-9 X 8.60 = 0; .-. Stress in 9-9 = - 28.86 tons. (2) By the method of chord increments. Let it be required to find the stress in the chord member 0-8, ¥\^. 20. Denc^te the vertical shears in the web mem- bers 2-1, 1-4, 4-^}, etc., by ?'i, r^, v^, etc., and the angles these members make with the vertical by fl„ 0^, 6^, etc. Now pass a curved section cutting the chord member 6-8 and all the web members on the left. Then from the first i ar" -^raai»^»r/8»«teTf?^V'«E!s?jg5p«fvs:aE^^ 62 ROOFS AND HlUbGEs. coiulition of «Mi\iilibri»iiii the sum of the hoiizontal rom- poncnts is zero. Hut the horizontal eoinpoiiont of tlio 8tresH in any wob nicnilu'r is 0(iual to tlio vertical shear in that nioniher iniiltii)lied by the tangent of its angle with the vertical. Hence we have: Stress in G-S = , , tan 0, + ih tan 61 4- 1\ tan 6^ + ('« tan ^4 + I'j tan d.,. 'riicirjhn; to Jind the stn-itu in uny chord member, imiss a riirri'il ncHiou ckUiiiij the member ami all the web membeh on the left, mnllliihl the uertinU shear in vin'h ireh member by the taiiijeid of its amjle with the vertical, and take the sum of the jtroihtcts. From an inspection of Fig. 26 it is seen that the stress in any chord member, as 4-6 for example, is greater than the stress in the immediately preceding chord member, 2-4, by the sum of the horizontal components of the stresses in the web members, 1-4 and 4-3, intersecting at the panel point between these two members ; That is, the increment of chord stress at any panel point is equal to the sum of the hqrizontal com2>onents of the web stresses intersecting at that jmint ; or, expial to the sum of the jn-oducts of the vertical shears of these web members by the tangents of their respective arujles ivith the vertical. It is well to test the stress found by this method by the method of n.oments, as the results obtained by the two methods should agree ; and this affords a check on the work. The method by chord incremants applies only to hori- zontal chords, while the method by moments applies to trusses of any form, that is, when the chords are not hori- zontal as well as when they are. Prob. 42. Let it be required to find the chord stresses in Prob. 41 by the method of chord increments. •izontal <'<>iu- lUPMt. of tlio iciil Hhciir in 8 angle with + 1-4 tan 6y ember, }Htns a h members on :„ember b;/ the. he sum of the t the stress in [vter than the niber, 2-1, by itresses in the e panel point ime)U nf chord the horizontai hat 2)oint ; or, shears of these angles ivith the nethod by the tl by the two k on the work. , only to hori- its applies to ! are not hori- 3rd stresses in i\ t lUilhOK TUUSSEfi. 58 Hero each panel load is 2 tons, and all the web members are inclined at an aiiK'le of ',\0°; .-. tan Ox = tan 0^ = etc. = tan IW = .5773. Hence for the lower chord stresses we have: Stress in 2^ = 9 x .r.773 = + r>.2(). Stress in 4-<5 =(9 + 1) + 7) x ..".773 = + 14.43. Stress in G-8 =(9+9 + 7+7 + 5) x.fi773= +21.36. Stress in 8-10 =(32+5+5+3) X. 5773= +25.98. Stre,ss in 10-12 =(42+3 +3 + l)x. 5773= +28.29; and for the upper chord stresses we have: Stress in 1-3 = -(9 + 9) X .5773 = - 10.39. Stress in li-o = -(18+2 x 7) X .5773= -18.47. Stress in 5-7 = -(32 + 2 x 5) x .5773= -24.25. Stress in 7-9 = -(42 + 2 x3) x .5773 = -27.71. Stress in 9-9 = - (48+2 x 1 ) x .5773 = - 28.80. I^ob. 43. Find all the chord stresses in the deck Tratt truss of I'rob. 40. Ana. Stresses in lower chords = + 8.25, +15.0, +20.25, +24.0, +20.25 tons. Stresses in upper chords = -8.25, -15, -20.75, -24, -20.25, -27.0 tons. Prob. 44. A through Howe truss, like Fig. 25, has 12 panels, each 10 feet long, and 10 feet deep : find the stresses in r.ll the members due to a dead loa^l of 400 lbs. per linear foot per truss. Ans. Lower chords=11.0, 20.0, 27.0, 32.0, 35.0, 30.0 tons. Verticals =11.0, 9.0, 7.0, 5.0, 3.0, 2.0 tons. Diagonals =15.5, 12.08, 9.80, 7.04, 4.22, 1.4 tons. Of course the upper chord stresses can be writtpu directly from those of the lower chord by (1) of Art. 5. Wi»»si?!pct.;Kj»7aR(? HUWaES. Art. 20. Position of Uniform Live Load Causing Maximum Chord Stresses.— Let Kig. U7 bo a truss sup- I .1 s 1\ 1 n tt N FiB. 07 ii Ik III. ported at the oiuls. Tl.en t(. fiiitl the stress in any chord ineuiber, we i>iiss a section cutting that member, a web member, and the other chord, and take the center of moments at (he iniersectiim of the web mcmbvr and the other chord; and since tiie hn-er arm for the chord is con- stant, tlie stress in any chord mend)er will be greatest when the live load is so arranged as to give the greatest bending moment. Now siippose we have a uniformly distributed moving load coming on the truss from the right, till it produces the left abutment reaction Ru then any increase in the load on the right of the section .V will artVct the forces on the left only by increasing the reaction U„ and consequently the bending moment. Hence, as R, increases with every load added to the right of .V, the bending moment increases, aiul therefore the chord stress also increases. Also, suppose we have a uniformly distributed moving load covering the truss on the left of the section producing the right abutment reaction R.t\ then any increase in the load on the left of the section N will affect the forces on the right only by increas- ing the reaction It,, and consequently the bending moment. Hence every load, whether on the right or left of the sec- tion, increases ths bending moment, and therefore the chord stress. t-!^^- id Causing ! ;i truss Slip- in any chord uber, 11 wcl) 10 centt'i' of i\\hv aail the flioid is con- jreatt'st wlicii itest, bendiii;,' lutnd moving prodiicos tho 11 the h)ad on 9 on tho left jeqnently the th every h)ad increases, and 0, suppose we ring the truss ;ht abutment :he left of the ly by increas- ing moment, ft of the sec- Fore the chord I hlillXlK TliVSHFU. 66 Thnrfnre, for a uniform had, thv maximum hpmli»o ,„nmnd at 'au'i paiut, a>ul cum'r'uHn thv maximum rlnml Hlrr^x in «»// ' mrmhvr, » irx whru Ihv live Imd corers the whole lenyth of the truss. To determine tho chord stresses then due to a uniform live load, we have only to sui.pose th.* live load to cover tlu* whole truss, just as the .lead h.a.l does, au.l compute tho cliord stresses in exactly the sauie way as wo compute tho dead load stresses. Prob. 46. A through Warren truss, like Fig. 2C>, has 8 panels, each 8 feet long: find the stresses in all tho chord mend)ers duo to a live load .)f 1000 lbs. per linear foot per truss. Ans Upper chord stresses=ir,.lG, 27.72, 34.04, 30.0(> tons. Lower chord strcsses= 8.08, 21.92, 31.20, 35.80 tons. Prob 46. A deck Tratt truss has 11 panels, each 11 feet long, atul 11 feet deep: find all tho chord stresses due to a live load of 800 lbs. per linear foot per truss. Ans. Upper chord stresses = 22.0, 39.6, 52.8, 61.6, 66.0, 66.0 tons. Art 21. Maximum SttesseB in the Chords.— Accord- in" to "the principles of tho preceding Article the stresses in the chords will be greatest when both dead and live loads cover tho whole truss. We have then only to determine the stress in each chord merabDr due to the dead and live loads, as in Arts. 19 and 20, and take their sum; or, we may add together at first the dead and live panel loads, and determine tho maximum stress in each chord member directly. This method is the simplest and shortest; but it is not the one which in practice is generally employed. •'^mmssssmET --..-.■vj^'mm*. fi.-5SGS1^P" 66 nOOFS AND BRIDGES. .■.-. J, = 2J sq. in. All modern specifications require a separation of dead and live load stresses. One type of specilication pernjits twice the allowable stress per s(puivo incli of niital for dead load that it permits for live load; tluis 1(),0()0 lbs. per S(iuare inch lor live load stresses and 20,000 lbs. per square inch foi (lea J f AAAAA/\/\/\/\/\ t h 6 s 10 a 10' s If ' Kig. so all forming equilateral triangles (Art. 14) ; the dead load is given by formula (2), Art. 15, the live load is 1500 lbs. per foot per truss : find the maximum and minimum stresses in all the members. The dead panel load per truss from formula (2) _ ^7^0 20 + 600)12 ^ 8g4o j^s. = 4.32 tons <> = say, 4 tons, for convenience in computation. Live panel load per truss = -^^^jjjjjj — = 9 to^^s- •A. % 1 5 counter- lel 10-12, itresses in I of chord 53 by the id stresses the raaxi- and mini- l directly, live loads the deck are found ad bridge, > members ead load is DO lbs. per stresses in ) ans ition. BlilDUE TltUSfiEH. 61 Since the loads are distributed uniformly along the joints of the upper chord, .3 and 3' will receive three fourths of a panel load each. The dead and live load stresses will be fou:id separately, tan 6 = .577 ; sec d = 1.1547. Dead Load Stresses. Left reaction = 4.75 x 4 = 19 tons. This is also the shear in panel 1-3. The dead load shear in each of the other panels is found by subtracting the loads on the left of the panel from the abutment reaction. We have then the fol- lowing chord stresses by chord increments : Upper Chord Stresses. Stress in 3-5 = (19 + 16) X .577 = - 20.20 tons. Stress in 5-7 = (19 + 2 x 16 -1- 12) x .677 = - 36.36 tons. Stress in 7-9 = (19 + 2 x 16 4- 2 X 12 + 8) x .577 = - 47.90 tons. Stress in 9-11 = (75 + 2 X 8 + 4) X .577 =-54.82 tons. Stress in 11-11'= (91 + 2 x 4) X .577 = - 57.12 tons. Stress in 1-3 = 00.00 tons. Lower Chord Stresses. Stress in 2-4 = 19 x .577 = 10.96 tons. Stress in 4-6 = (19 + 2 x 16) x .577 = 29.42 tons. Stress in 6-8 = (51 + 2 X 12) x .577 = 43.28 tons. Stress in 8-10 = (75 + 2 x 8) x .577 = 52.51 tons. Stress in 10-12 = (91 + 2 x 4) x .577 = 57.12 tons. -»-«wBnw!wggS';' ■ ipi 6*! ROOFH AM) nuiDaiis. Weh Stuhsseh (iiY Akt. 18). Stress in 2-3 = 1 1) x 1 . 1 '>i = - 21 .85 tons. Stress in 3-4=10 x 1.154:^^ +18.40 tons= -stress in 4-5. Stress in 5-6=12x1.154= +13.84 tons= -stress in L ". Stress in 7-8= 8x1.154=+ 9.23 tons =- stress iu 8-9. Stress iu 9-10= 4x1.154=+ 4.G1 tons = —stress in 10-11. Stress in 11-12=0.00 = -stress in 12-12'. Stress in 1-2=^ of a panel load=l ton. Live Load Stresses. The maximum live load chord stresses occur when the live load covers the whole length of the truss (Art. 20); hence they are found in precisely the same way as the dead load chord stresses above; or we may find them by multi- plying the above dead load chord stresses by the ratio of live to dead panel load, which = 2^ here, giving us the following chord stresses : Uppke Chord Stresses. 1-3 = 00.00 tons. 3-5 = — 45.45 tons. 5-7 = - 81.81 tons. 7-9 = - 107.77 tons. 9-11 = - 123.34 tons. U-ll'=- 128.52 tons. Lower Chord Stresses. 2-4 = 24.66 tons. 4-6= 66.19 tons. 6-8= 97.38 tons. 8-10 = 118.15 tons. 10-12 = 128.52 tons. BRIDGE TKUSSES. m ss in 4-5. S3 in t r. 88 iu 8-9. when the (Art. 20); s the (lead by multi- lie ratio of ng us the • tons. > tons. i tons. ) tons. ] tons. Wkh Stkk.sses. The stress in any web member is equal to the shear in the section whicli cuts that member and two liorizontal chord members, multiplied by the secant of tlie angle which the web member makes with the vertical (Art. 18) ; the stress is therefore a maximum when the shear is a maximum. The maximum positive live load shear in any panel occurs when all joints on the right are loaded and the joints on the left are unloaded (Art. 22). The maximum jjositive live load shear for 2-3 will occur when the truss is fully loaded. We have then Stress in 2-3 = 9 X 4.76 x 1.154 = - 49.30 tons. The maximum positive shear for 3^ and 4-5 will occur when all joints except 3 are loaded (Art. 22). Taking moments about the right end, and multiplying by sec 6, we have Stress in 3-4 = A(f + 3-f- = Ax^|^x Stress in 5-6 = /Tra + 3 + = 33.13 tons : Stress in 7-8 5 + 7 4- 9 + 11 + 13 + 15 -I- 17) X 1.1547 1.1547 = 41.96 tons — — stress in 4-5. 5 + 7 + 9 + 11 + 13 -!- 15) X 1.1547 = — stress in 6-7. = Ad + 3 + 5+7+9+11 + 13) X 1.1547 = 25.33 tons = — stress in 8-9. Stress in 9-10 = ,_,\ (| + .3 + 5 + 7 + 9 + 1 1) x 1.1547 = 18.58 tons = - stress in 10-11. Stress in 11-12 = /^(J + 3 + 5 + 7 + 9) x 1.1547 = 12.87 tons = - stress iu 12-11'. :'>'«ite»itti^y«>iii«aj@^tBa!8W»MHeffii«K«r^ ^ 04 HOOFS AND IIIIIDGKS. Stress in ll'-lO' = /o (1 + -^ + •'> + ") x l.ir)47 = 8.11) tons = - stress in lO'-U'. Stress in 9'-8' = .'V (3 + .*i + 5) x 1.1517 = 4.55 tons = — stress in 8'-7'. Stress in 7'-G' = ^'V (f + •^) ^ l-^''*-*^ = l.DC tons = — stress in (5-6'. Stress in 5'-4' = ^\ (|) x 1.1547 = 0.40 tons = — stress in 4'-3'. Stress in 3'-2' = 0.00 tons. Collecting the above results we may enter them in as follows : TABLE OF STRESSES IN ONE TKUSS. Upper Chord Stuessbs. a table Memhihs. 1-3 8-5 5-7 7-» 1) 11 11-11' Dead load Live load 00.00 00.00 -20.20 -46.45 - 30.30 - 81.81 - 47.00 -107.77 - 54.82 -123.34 - 57.12 -128.52 -185.04 - 67.12 Max. stress Mill, stress 00.00 00.00 -05.05 -20.20 -118.17 - 30.30 -165.07 - 47.00 -178.10 - 54.82 Lower Chord Stresses. Mkmiieks. i-'-i 4-6 C-S 8-10 10-12 Dead load . . Live load . . . + 10.0(J + 24.C,0 +29.42 + 00.19 + 43.28 + 97.:5H + 52.51 + 118,15 + 57.12 + 128.62 Max. stress . . Min. stress . . + 35.02 + 10.00 +95.01 + 29.42 + 140.00 + 43.28 + 170.00 + 62.51 + 186.64 + 67.12 -y. V. 5'. 3'. )m in a table 3S. I a 11 11-11' 4.82 3.34 - 57.12 -128.52 8.10 )4.82 -185.04 - 67.12 ) 10-12 .51 15 + 57.12 + 128.62 .0(i .51 + 186.64 + 57.12 BlilDQE TliU88E8. 66 Weu Stresses. Mkmhkkn. 2-3 it 4 4-5 frfl 8-7 ^ Dead load .... J j From rif?lit . . . i ( From left . . . . -21.8.1 -40,;to 00.00 + 18.40 + 41.110 - 0.40 -18.40 -41.00 + 0.40 + 13.84 + 33.13 - 1.90 -13.84 -33.13 + 1.90 Max. 8tri!S8 Mill, stresa -71.15 -21.86 + 60.42 + 18.00 -00.42 - 18.00 +40.07 + 11.88 -46.07 -11.88 M KM HERS. T-9 8-9 9-10 10-11 11-12 ^ Dead load .... § j From right . . . 1 1 From left .... + 9.23 +25.33 - 4.56 - 9.23 -26.33 + 4.55 + 4.01 + 18.58 - 8.10 - 4.01 -18.68 + 8.10 + 0.00 + 12.87 -12.87 Max. stress Min. stress + 34.66 + 4.08 -34..'i0 - 4.08 +23.19 - 3.58 -23.19 + 3.58 + 12.87 -12.87 We see from this table of tueb stresses that, (1) when the live load comes on from the right, all the web members in the left half of the truss are subjected to but one kind of stress, that is, the members 2-3, 4-5, 6-7, 8-9, 10-11 are subjected to compressive stresses, and the members 3-4, 5-6, 7-8, 9-10, 11-12 are subjected to tensile stresses ; (2) when the live load comes on from the left, all the web members to the left of 9-10 are subjected to but one kind of stress, but the members 9-10, 10-11, 11-12 have their stresses changed, that is, the stresses in 9-10 and 11-12 are changed from tensile to compressive, while that in 10-11 is changed from compressive to tensile. Hence, the members 2-3, 4-5, 6-7, 8-9, should be struts to carry compression only, and the members 3-4, 6-6, 7-8, 'iiaxiiBjaiWMia I? i liOUFS AM) ltl!lJ)(li:S. ■■■, !i It !i should be tit^s to carry tension only, while 9-10, 10-11, 11-lL' shouUl lie nionibcrH I'apabli' of iv,sistin(< both coni- l)rcs,sion and tcn.siini, and should thnieforo be amiiter hritcea (Art. 1): and the same is true for the ri.Ljht half of the truss. In this solution for web stresses the live load was brought on from the n'(jlit, and the niaxiuuna /losiUce shear was found in eaeh pantd of the rifiht half of the truss, and then the resulting,' stress ; if preferred, the live load may \w brought on from the lif}, and the maximum wyative shear bo found in each iianel of the Jeft half of the truss (Art. 22), and then the resulting stress. Thus, the maximum jjositive shear in any paii'l in the right half of the truss as (J'-8' has the same numerical value as the maximum negative shear in the corresponding lueiubor C-8 in the left half; and the resulting stress in any member 7-8' lias the same value as that in the corresponding nuMuber 7-8. The maximum and miniminn stresses may be determined directly from a single equation by placing the dead and live loads in proper position. Thus, for the nuiximum stress in 8-9, wo pass a section cutting it, place the live load on the right, and have Max. stress in 8-9==-[2x4-f /^(|+3+5 + 7+9+ll+13)] X 1.1547= —34.57 tons, as before. For the niinimuni stress tlie live load is reversed and covers the truss on the left of the section. Thus Min. stress in 8-9=-[2x4-^\(f+3+5)]xl.l547 = — 4.()9 tons, as before. Prob. 53. A deck Warren truss, like Fig. 29, has 10 panels, each 10 feet long, its web mendiers all forming equi- lateral triangles; the dead load is 800 lbs. per foot per truss, and the live load is 1000 lbs. per foot per truss ; the f'^f' ..i Bnil'HE TRUSSEfi. fi7 9-10, 10-11, \\; both coiu- •oitnter hraccti I of the truss. I was hiDUglit [•e slioiar was US8, and then load may he wyutivc shear •uss (Art. 22), iiuini i)()sitive s as (5-8' has egative shear lalf ; and the same vahie as le determined he dead and ass a section have -9+11 + 13)] 1 before, reversed and lUS < 1.1547 ;. 20, has 10 foriiiin^ equi- per foot i)Pr er truss: the ■^.'i. J5^ joints 3 and 3' receive three fourths of a i)anol h>ad each: iind the maximum and minimum strosHi".-) in M the members. At)s, Ui-i'Kit CiioKi> Sthksse Mkmhkrh. 1 a 8-5 -60.60 -20.20 r>7 TO 9-11 11-11' Max, Ht n'H.st'8 iMin.Mtri'KNr.s 00 00 00.00 - 100.08 ;iO.:j« - 14:(.70 - 47.02 - ltt4..V.' - 64.84 -17I.4M - .17.1(1 LOWKK !^iioiti) .Strksses. MKMIIKR!), 9-» 4-a 9 f-in KI-12 Max. stresses . . . Mill, HtrcsHPS . , . +32.88 1 10.06 + 88.20 + 29.40 + 120.8.4 + 4;J.28 + 1.')7.50 + 62.52 + 171.48 + 57. 16 Web Strbsses. Mkuiikkh. li-3 8-4 4-8 6-0 6-1 Max. 8trt's.se8 . . . Mill. stroascH . . . -65..52 -21.H4 + 65.80 + 18.12 -65.80 -18.12 + r.i.2!i + i2.o;i -4.'5.20 -12.00 MKMIIIiRK. 7-8 8-9 9-10 i«-ii u-ia Max. strcsaes . . . Mill, stresses . . . +;{1.78 + 5.18 -31.78 - 6.18 +21.ia - 2.70 -21.1.3 + 2.70 + 11.46 -11.46 Prob. 54. A deck Warren truss, like Fiw^,^«(^£gusaiSi^i^^^^sgi@l$a6^^»^aiaaiista9£^&tie^-^^'?MttiMfi»t@if»^ > ]i(JOFS AM) nil I DUES. 1 ■ Aug. Max. stresses in upimr dionl = 0.00, -()7.()8, -117.97, -UH.L'o, -158.17 tons. Mill. HtiTH.se.s in uppt'i' chonl = 0.00, -L'l.OO, -.'50.01, -Ki.OI, -lO.O'J tons. Max. sti'CMsi's in lower choid = + n7.r>s, +07.00, +i;w.i5, -\ ir,HA~ touH. Min. HtrcHscs in lower cliord = 4-1 100, +'M.W, +'lli.87, +40.00 tona. Max. web Htresses = _ 74.00, +00.08, -00.08, +48.-1.3, - 4.'U.'J, + 27.57, - '27.07, + 1;{.08 tons. Min. web Htrcssea = -U3.L'7, +18.20, -18.20, +0.23, -9.23, -1.35, + 1..35, - 13.08 tons. Pivb. 65. A throiiKli Warren truss, Fij,'. 2(5, has 10 panels, each 12 feet loiif,', its web nienilier.s all forniinK eciuilateral triangles; the dead load is 500 lbs. per loot per truss, and the live load is 8.'i4 lbs. per foot per truss : find the maxi- nulla and niiniiuum stresses in all the members. Alts. Upper ehord max. = -41.52, -73.92, -90.90, -110.80, -115.44 tons. Ui»per ehord min. = _15.57, -27.72, -3G.30, -41.55, -43.29 tons. Lower chord max. = +20.80, +57.70, +85.44, +103.92, +113.12 tons. Lower cliord min. = +7.80, + 21.00, + 32.04, + 38.97, + 42.42 tons. Web stresses max. = -41.55, +41.55, -32.91, +32.01, -24.81, +24.81, - 17.31, + 17.31, - 10.37, + 10.37 tons. %t^f m^^'- ITtona. 9 tuns. >ltH. .'J.I.'J, + 27.57, 9.23, - 1.35, has 10 i)aiipls, 1^,' cciuilatcral per trus.s, and intl the maxi- s. .16.44 tons. 1.29 tons. 113.12 tons. B.42 tons. !4.81, +24.81, s. 'W^^^- UUWUE Tit U SUES. ti'J Web sticsses niin. =:_ ir,.57, + 15.57, -11.55, +11.55, -(5.91, +0.91, - 1.73, + 1.73, + 4.()('., - 4.0(5 tons. Prob. 66. A through Wanen tni,ss, like Kig. 2fi, lias K panels, each 15 feet long, its web nuMiibt'is all turniing equi- lateral triangles; the dead load is SOO lbs. per foot per truss, aiul the live load 1(500 lbs. per fool per truss; find the uiaxi- muni and luininuim stresses in all the nieud)er8. Ans. Upper ehord max. = -48.48, - 83.10, - 103.92, - 110.88 tons. Upper chord min. = - 10.10, - 27.72, - 34.04, - 30.90 tons. Lower ehord max. = + 24.24, + 05.70, + 93,00, + 107.40 tons. Lower chord min. = + «.()S, + 21.92, + 31.20, + 35.80 tw.is. Web stresses max. =.-48.48, +48.48, -35.08, +35.08, -24.20, +24.20, - 13.80, + 13.80 tons. Web stresses min. = - 10.10, + 48.48, - 10.40, + 10.40, - 3.48, + 3.48, + 4.08, - 4.08 tons. Prob. 57. A deck Warren truss, like Fig. 29, has 7 ])anels, eacdi 15 feet long and 15 feet deep, its web nuMubers all forming isosceles triangles; the dead load is 4 tons, and the live load is 9 tons, per panel per truss, the first and last joints 3 and 3' receiving three cpiavters of a panel load each, as in Vrobs. 51, 52, and 53: find the maximum and mini- mum stres.ses in all the members. Ans. Upper ehord max. = - 0.00, - 37.38, - 03.38, - 70.38 tons. *.^i'!ft5&^3,!K^i3fclt3iil0S^5'Sft**?'''****^*^ ■*j*w»/st-'«*»*''waR*sss«wts*w.»«*efi»we«si9B^^ ROOFti AND liRIDGES. Upper clionl iiiin. = _ ().(K), _ 11.50, - 19.60, - 23.50 tons. Lower chord max. = 4- -1.1.'5, + 5y,0i~, ,- TlJ.ia, + 79.6.'} tons. Lower cliur'l niiii. = + (i.-'ilS + 10.50, + 22.50, + 24.50 tons. Web stresses max. = -•47.2."., +;}(j.87, -30.87, +24.49, -24.49, +13.50, — 13.5(5 tons. Web stresses niin. = - 14.63, + 10.04, - 10.64, + 4.01, - 4.01, - 4.05, + 4.05 tons. Art. 24. Mains and Counters. — In the Pratt truss, Fig. 30, all the verticals, except 1-4 and l'-4', are to take compression only, and all the inclined members, except 1-2, 1 .1 5 5' 5' 1' 1-2' are to take tension only. The two inclined members, 1-2 and l'-2', are usually called the " inclined end posts." The verticals t-4 and l'-4' do not form any part of the truss proper, since they serve only to carry the loads at 4 and 4' np to the hip joints 1 and 1'; they are called hip verticals. The members 1-0, 3-8, 3-8', and I'-O' are mains, or main ties. They are the onlj^ inclined ties that are pnt under stress by the action of the dead load, or by a uniform dead and live load extending over the whole truss. The mem- bers 5-6, 5-8', 5'-8, and 5'-6', in dotted lines, are counters, "o^^^HE^^ BRIDGE TliVSSES. 71 .49, + 13.50, l.Ol, -4.or>, Pratt truss, , are to take , except 1-2, ed members, I end posts." part of the e loads at 4 e called hip tins, or main e put under uiform dead The mem- are counters, or counter ties. In the I'ratt truss there are no l-ounter braces (Art. 1). These four counters are called into j)lay ', it is carried by the truss to the abutments at 2 and 2'. The part of this load whi(;h goes to 2 may be conceived as being carried up to o', down to 8', up to 5, down to 8, up to 3, down to o, up to 1, down to the alnitment at 2. The other part of this load, which goes to the right, passes up to 1', then down to 2'. For this loading, the counters 5 -G', 5-8', and the mains 3-8, l-(5, and I'-O' are put under stress, while the counters 5'-8, r>-{), and the main tie 3-8' are idle, and might be removed without endangering the truss. If the two middle joints, 8 and 8', are loaded equally, the part of the load at 8' going to 2' is just balanced by the part of the load at 8 going to 2, and hence there is no stress in the intermediate web members 5-8', 5'-8, 5-8, and 5'-8', nor in the counters 5-6 and 5'-6'. If the live load going from 6' to the left abut- ment is greater than th'> lieaci load going from 8' to the right abutment, the countc; 5-6' must be inserted, if this counter 5'-C' were not inserted, the panel G'-8' would be distorted. The load at G' would bring the opposite corners 3' and 8' nearer together, and the opposite corners 5' and G' farther apart, because the main tie 3'-8' cannot take com- pression. Whenever the live load would tend to cause com- pression in any main tie, a counter must be inserted uniting the other corners of the panel. Both diagimals in any panel cannot have compression or tension at the same time. If the action of the dead and live loads tend to sul)ject a member to stresses of op])osite kiiuls, the resultant stress in the member will be equal to the numerical diffe-enec of the ojvposite stresses, and will be of the same kind as the greater. Tims, if the dead load, g(nng from 8 to the left •■<»«-=«»f«*o»,;jaiSiSg,c^j;,l55|r,-.>nj,,rj.ij,.i r,v^siiift««-^/-*c>«.v~v.-»4is«(.i»v-'. ■ tiS»i3«*^*«nHiayj;- 72 nOOFS AND nUlDGES. m t I- I 1 abutment, shoukl subject the main tie 3-8 to a tension of 8 tons, and if the live load, going from to the right abut- ment, should tend to subject the same tie to a compression of ;') tons, there would bo a resulting tension of 3 tons in the member ,'J-8 ; and the counter 5-0 would not be needed for this loading. Ihit, if the dead load should subject the tie 3-8 to a tension of only 2 tons, while the the live load should tend to sidiject the same piece to a compression of 5 tons, there would be a resulting force of 3 tons, tending to bring the joints 3 aiul 8 nearer together, or to separate the joints o and (j from each other. In order t.) provide for this compression or tension, either a brace would be needed alongside the main tie 3-8, to take the compression of 3 tons, or the counter tie fl-G would be needed to take the tension of 3 tons. In the Howe truss. Fig. 31, the verticals are to take ten- sion only, and the inclined members are to take compression Kit.'. 31 only. The mains, or main braces, are represented by the full lines ; the dotted lines de lote counters, or counter braces. Under the action of the dead load, or of a uniform dead and live load covering the whole truss, the main braces are the only diagonals that are put under stress. The counter braces are called into play only when the live load covers a part of the truss, as iu the case of the Pr itt truss. Thus, if a load be placed at the joint G, the part of it which goes to the right abutment L" may be conceived as |isl'#,^*^6Sfesv tension of 8 i right abiit- compression of 3 tons iu [)t be needed I subject the bhe live load inpression of tons, tending f to separate ) provide for Id be needed ion of 3 tons, J the tension to take ten- I compression 3nted by the I, or counter of a uniform 3S, the main under stress, fhen the live of the I'r.tt he part of it conceived as niilDGE riiUSSES. 78 being carried up to 3, down to 8, up to 5, down to 8', up to 5', down to G', up to 3', i;owu to 4', up to 1', down to 2'. If the live load going from 6 to the right abutment is greater than the dead load going from 8 to the left abut- ment, the counter brace 3-8 must be inserted. If this counter 3-8 were not inserted, the panel 6-8 would be dis- torted. The load at G would bring the op])osite corners 3 and 8 nearer together, and the opposite corners 6 and 6 farther apart, because the main brace 5-0 cannot take tension. But if the live load going from G to the right abutment is less than the dead load going from 8 to the left abutment, the counter brace 3-8 will not be needed for this loading. The main and counter 'n any panel cannot boih take stress at the same time by any system of loading. We may determine where the counters are to begin, as follows: Consider the left half of the truss, and let the live load come on from the left ; then the dead and live load shears, or stresses, are of opposite kinds. It results from this at once that counters — counter ties in the Pratt, and counter braces in the Howe — must begin in that panel in which the stress or shear caused by the live load is greater than that of the opposite kind caused by the dead load. Thus, in Fig. 30, let the dead panel load be 3 tons and the live panel load 14 tons. Then we have the following maximum negative shears : Shear in 2-4 = 4-9- -0 = + 9. Shear in 4-G = 4-6- -1 xl4 = 4-4. Shear in 6-8 = 4-3- -14 (1 + f) = -3. Hence the counters must begin in the third panel, and therefore the third, fourth, and fifth panels must have counter ties for this loading. ■' -■•««'**«w;«»i«»SK*«AV»*;#««R'iT^<**t'*^Jft"MaimE?t.MBa«!^^ 74 UOOFS AND niilDGES. Prol). 58. Tn the Howe truss of Fig. 28, with 12 panels, the (h'iul [)iUiol h);i(l liciiij,' 4 tons ami tlic live 1) tons, find the unnil)ei' of panels to he conuterbracetl. Alls. The 5th, Gth, 7th, and 8th panels must have counter braces. Art. 25. The Howe Truss. — The niaxinnini and mini- mum stresses, both in the chords and the web menibers of the Howe truss, may now be computed by the principles of Arts. 18, 19, 22, and 24. The dead and live load stresses may be found separately, and their sum taken for the maxi- mum and minimum stresses; or, the maximum and mini- mum stresses of any inember may be determined directly, from a single equation, by placing the dead and live loads : ptoper position. In the following solution of the through ilowe t"uss, tliis method is the one employed : J^rub. 59. A through Howe truss, as a through railroad bridge, Fig. 32, has 10 panels, each 12 feet long and 12 feet 3 5 7 .1 ;; fi' ?'____ 5' s' i'l;;. 33 deei); the dead load is given by formula (4), Art. 15, the live load is loO(» lbs. per foot per truss: liud tlie maxiintua and mininuini stresses in all t)ie members. From tormula (4), Art. 15, we find the dead panel load per truss = (0.5 X J20 ±i!mi2 ^ 3^30 1^^ ^ 4 3^5 ^^^^ 2 = say, 4 tons, for convenience of computation. T • 111 4. l''>00 X 12 Q . Live panel load ]»er truss = — ^,7-;: — = ^ tons. tan0= 1, sec ^ = 1.41. lUv Bam itli 12 panels, e 1) tons, find ; have counter until and niini- ib ineiiibers of iJiiiiciples of load stresses for the iiiaxi- nni and niini- liiied directly, ind live loads of the through ongh railroad iig and 12 feet S' :), Art. 15, the the maximtuii id panel load 505 tons itation. ous. [.% L nUIDGE TRUSSES. 76 Tlic niaxinunn and iniiiiinnni chord ntresnes may he written mit at once (Arts. I'.) and 21). They are as follows: Max. stresses in lower chord = 58.5, 104.0, l;j().5, 150.0, 102.5 tons. Min. .stresses in lower chords 18.0,32.0,42.0,48.0,50.0 tons. The upper cliord stresses may be written from the lower at once. Thus, stress in 3-5 = — stress iu 2-4 ; and so on. Weu Stresses. The maximum stress in any web member is equal to the maximum shear in the section which cuts that member and two horizontal (diord members, multiplied by the secant of the angle which the Aveb member makes with tlie vertical (Art. 18). Tlie maximum positive live load shear in any panel in the Icfi. half of the truss occurs when all joints on the right an; loaded and the joints on thc^ left are unloaded (Art. 22). For the maximum stress in 2-3, we pass a sec- tion cutting it, i»lace the live load on the right, in this case covering the whole truss, and have Max. stress in 2-3 = - [4 X 4.6 + ^!V(1 + 2 + 3 + 4 +5 + 6 + 7+ 8 + 9)] 1.41 = - 82.48 tons. For the max. stress in 4-6 all the joints on the right of 4 are loaded. Hence, Max. stress in 4-5 = -[4x3.5+V'(i(l + 2+3+44- Max. stress iu 0-7 = -[4x2.5 + J-o (1 + 2 + 3 4-. Max. stress in 8-9 = -[4xl.5 + -f-a-(l+2 + 3+. Max. stress in 10-11 = - [4 X 0.5 + T-9^ (1 + 2 + ... + 5)] 1.41 .+8)] 1.41=- -05.42 tons. + 7)] 1.41=- 49.03 tora. + 0)] 1.41= -35.10 tons. 21.85 tons. I |*¥«Mi?i^;SBfo''»iAtf**'4«m»«-- tons to ivered all the )vered all the J be subjocted aum St 1 ess in lO-1 1 becomes m 9-12. The coiiipressiou. Hence, theoretically, there is only one counter brace needed in the h'ft half of this truss; but, practically, a counter brace would be put in the fourth panel also, so that the tniss would have counter braces in its four middle panels. The maximuui stress in any vertical, except the middle one, is e(pial to the nuiximum positive shear in the section cutting it and two chord members, or it is equal tt> the maximum positive shear in the panel toward the abutment from the vertical. Thus, Mi X. stress in 3-4 = 13 X 4.5 = 58.5 tons. Max. stress in 5-() = [4 X 3.5 + .9(1 + 2 + ••• + 8)j = 4G.4 tons. Max. stress in 7-8 = [4 X 2.5 + .9 (1 + 2 + - + 7)] = ao.a tnna. Max. stress in 9-10 = [4 X 1.5 + .9(1 + 2 + ••• + 6)] = 24.9 tons. The maximum stress in the middle vertical 11-12 is equal, either to the maximum positive shear in panel 10-12, or to a full panel dead and live load, whichever is the greater. For, if this maximum positive shear is greater than a panel load, then the shear in the next right hand panel under the same loading is also positive, and the counter Imice in that panel will he in action, thus making the shear in panel 10-12 the same as that in the vertical 11-12 ; but. when th« coun- ters are not in action, the stress in 11-12 is always equal to the load at 12. Max. positive shear in 10-12 = 4 X 0.5 + ^(1 + 2 + ••• + 5) = 15.5 tons. Full panel load = 4 -f- 9 = 13 tons. .-, Max. stress in 11-12 = 15.5 tons. W«WPtfi'Z"-^*'V>^- '-■■—- .'->-3l*e?'59^fiaWWWSK».^^^;/;!W^-t.' ( ! 78 HOOFS A\n miiixsKs. Tlic miiiiiiuiin strras in any vertical is e(|iial to tlic maxi- mum iK'gativf shear in the stiction cutting it and two chord luonihers, or it is equal to the maximum negative shear in the panel toward the cahutment from the vertical, or to a dead jHinel load, whichever is the greater. Thus, Min. stress in 3-4 = 4 x 4,r» - = 18.0 tons. Mill, stress in 5-G = 4 x 3.5 — ^^ x 1 = 13.1 tons. Min. stress in 7-8 = 4 x 2.5 - ^s (1 + 2) = 7.3 tons. Min. stress in U-10 = 4 x 1.5 - /^(l + 2 + 3) = 0.6 tons. IVit the stress in a vertical of this truss cannot be less than the weight of a dead panel load; therefore Min. stress in 0-10 — 4 tons. Min. stress in 11-12 = 4 tons. We may now collect these web stresses in a table as follows ; Wkh Stresses. Mkmiikrh. 2-3 Max, Htrc8,ses Min, 8truiM(!H -82.48 -25.38 4-5 -05.42 -18.47 fi-7 -49.03 -10.29 > 9 -35.10 - 0.85 10-11 -21.85 0.00 9-li -U.87 0.00 VERTiCALS. Mkmiikm. 8-1 ;,-'•, T '* 9 111 11 [i Max. stresses . . , , Min. stresses .... 68.5 18.0 40.4 I'M 36.2 7..T 24.0 !.0 15.6 4.0 -''Si-.'.^'t -^J^^T^ ■ i^*- "' I to tlic maxi- 11(1 two chord tive shear in 'tical, or to a us, 5. 1 tons. 7.',i tons. J) = 0.0 tons. iniiot b(! less re n a tal>le as 10-11 9-1/ -^ 11.87 0.00 -21.85 0.00 1 1(1 II I'i u.n 10 16.5 4.0 nitlDOE TliUStWS. i!» Til building Howe trus.ses of tiiubor, it is usually the practice to juit in all the dotted diagonals to keep the tru.ss rigid. This is not done in metallic trusses. I'rob. 60. A through Howe truss has 11 panels, each 10 feet long and 10 feet deep; the dead load is SOO Ihs., and the live load is 1(500 1!,.h. per foot per truss: tind the iiiaxi- iiiuni and minimuin stresses in all the members. Ans. Max. stresses in lower chord = GO, 108, 144, 1G8, ISO, and 180 tons. Mill, stresses in lower chord -= 20, 3(;, 48, 5(5, GO, and 00 tons. Max. stresses in main braces = _84.G0, -08.71, -53.83, -40.0, -27.18, — 15.37 tons. Mill, .stresses in main braces = -28.28, -21.5.",, -13.85, -5.13, -0.00, -0.00 tons. Max. stresses in counters = — 4.31, — 15.38 tons. Max. stresses in verticals = + 00.0, +48.73, +38.18, +28.30, +19.28 tons. Mill, stresses iu vcvtioala = -I- yO.O, 1 15 '.'7, + 9.82, + 4.00, + 4.00 tons. I'rnh. 61. A deck Howe trimN liiis 12 [lanels, each 10 feet long and 10 feet deep; the dead load is 000 lbs., and the live load is 1200 llis. per foot per truss: find the maxi- mum and minimum stresses in all the members. Ans. Wax. stresses in lower chord = 49.5, 90.0, 121.5, 144.0, 157.5, 102 tons. Max. stresses in main braces = -09.79, -57.81, -40.53, -36.95, -20.08, -lO.Otons. 9't iron or steel, the verticals bciiiL; ennijircHsion members, except the hip verticals, and the diiij^onals ten- sion nicnibers. The niaxinmni and niininuiin stresses of an}' member jniiy be (k'tcrnuned directly from a single Cipui- tion, as in the case of the Howe trnss (Art. 2o). I'i-dIi. 64. A tliroii^di Pratt trnss, as a through railroad bridge, Fig. 33, has 10 panels, each 14 toet long and 14 feet 10 It 10' 8' Vlit. 33 deep; the dead load from formula (1), Art. 15, the live load is 1800 lbs. per foot per truss : find the maximum and minimum Htressos :i all the members. From formula (I;, Art. 15, the dead panel load per truss = (0x140 + 520)14 ^ 12,460 lbs. -= G.23 tons = say, 6 tons. Live panel load per truss = 16.2 tons = say, 16 tons. tan = 1, sec 6* = 1.414. The max. ar.d min. chord stresses may be written out at once (Arts. 10 and 'il). They are the following: Max. stresses in lower chord = !)<>.0, 90.0, 176.0, 231.0, 264.0 tons. Min. stresses in lower chord = 27.0, 27.0, 48.0, 63.0, 72.0 tons. •£fin»iM»?l3H£M6Ses«3;;#.i': i <,■.-■>.- rf,\p:..,. I ■^ il»l)l-T«W-— »S»f^"»P. I P- '/ IMAGE EVALUATION TEST TARGET (MT-3) 1.0 'r V» I.I |5A S3 1^ 2.5 2.2 110 1.8 Photographic Sciences Corporation '<' ^f /. fA 1.25 L4 1.6 ^ 6" ^ -^•\ #% gb iV :\ \ ^\^ 23 VVEST MAIN STREET WEbSTER.N.Y. 1 4.^90 (716) 872-4503 | | lilli . . , i y i|W p ili j| i| ^^l i ;L i l i j; i |i< li y;jpyjyi i [u^ ^ ^^^^^ 4E?- CIHM/ICMH Microfiche Series. CIHM/ICMH Collection de microfiches. Canadian Institute for Historical Microreproductions / Institut Canadian de microreproductions historiques -•..--..-...*», ..w^*»6«-.»afc.j.,,.^.^jj,^S^,,,^,^.^,,^,,„,.„^^,,,.,;i^ •^y -"""-'" 1 M II I l; 82 HOOFS A.\h lililbGES. Wi;ii Stukssks. Tlic wfl) stro.sscs may he fuiiiul exactly as in Prob. 60. Th(! I'olluwiiig arc (lie o X () -f- l.O^l + 2 + ■.. +. S)]1.41 = 110.8 to'is. Max. stress in 0-8 = [2.5 X 6 + l.G(l +2 + ... +- 7)11.41 =~. 84.3 tons Max. stress in 7-8 -: -[i.5 X + 1.0(1 +. 2 + ... +- ())] = -42.0 tons. Max. stress in 11-12 - - [- 0.5 X + 1.0(1 +. 2 + ... + 4)]= - 13.0 tons. Min. stress in 5-8 = [2^ X - 1.6(1 + 2)]1.41 = 14.4 tons. Max. stress in 10-11 = [- i X + 1.0(1 + 2 + ... 4- 4)]1.41 = 18.5 tons, Thus we find the stresses in the following table : Weii Stuesses. \!'.mp;krm. •:-'! S-fi 1 5-S 1-10 9-12 S 9 10-11 Max. .strtsses Min. stresses -l:!9.'. - .-Js.l + 110.8 + 27.4 +81.3 1 +(10.0 + 14.4 0.0 -f .S8.0 0.0 + 0.8 00 + 18.6 0.0 Max. stresses in Ihe vertieals =+-22.0, -50.8, -42.0, - 27.0, - iii.b tons. HRWUE TRUSSES. m 1 ill I'loi). (;o. Ik! iiu'iiihers: Proh. 65. A tliio.ij,']. I'liitt truss lias H i.anels, ea'-I. 10 fffi Inn- and 10 feet (l.-cp; tilt; dead load is IL'OO Iks., and tlh' iiv.. lead is L'OOO lbs. per fout per truus : find the stresses in all the 'neiuhers. Ans. Max. sticsses in lower eliord = S0, SO, Ml, liH,', L'L'-l,lM(M,oi.s. Max. in main diagonals = -ll;U, -i-OI..'-,, +7l.r,, +50,8, 4..'W.4, +19.2 tons. Min. in main diaj^onals = - IL'.;}, -I- 1^2.{^, + 2\j>, + 9.2, 0.0, 0.0 tons. Max. in eounters ^ -f- 4.4, + J9.L'l tons. Max. in verticals = + IG.O, - r>0.7, - 37.4, - LT).], - 13.6 tons. /Vo/>. 66. A deck Pratt icuss, Fij,'. 34. has 10 panels, each lli feet long and 12 feet deei>; the dead load is 833 18.5 tons, ible: S 9 lii-tl +0.8 0.0 + 18.5 0.0 59.8, -42.0, ll>s., and the live load is 1667 lbs. per foot per truss: find the stresses in all the members. Ans. Afax. stresses in lower chords = 0.0, 67.5, 120.0, 157.5, 180.0 tons. Max. in main ties = 95.1, 75.4, 57.1, 40.2, 24.6 tons. Min. in main ties - 31.7, 23.2, 13.4, 2.1, 0.0 tons. Max. in eonnters = 10.6 tons. Max. in verticals = - 75.0, - 67.5, - 53.5, - 40.5, - 28.5, - 20.0 tons. ■ k I f ( 84 HOOFS A XI) lilllDGKs. I'fth. 67. A ik'ck I'latt truss liiis i» panels, caoh IS feet long mid L'l Icct dccj); tlio dead load is .'.(MMlis., and flu- live load is !()(»» His. per foot per truss: Hud the stresses in all the nienibcrs. Ans. Max. stresses iu lower chords = 0.0, JO.-), 70!).<)1, 1)1.1, lOl.;} tons. Ma.\. stresses iii main lies = (•.:.:., .-.1.0, -.Mli, L'4. 1, 12..-. tons. Mill, stresses in main ties ^ L'L'..-., l.-..r., 7..-., 0.(», 0.0 tons. Max. stresses in counters — 1.0, lL'..j tons. Mi'.x. stresses in verticals = _ no.s, - 54.0, - 41..-., - .30.0, - 10.5 tons. The chord stresses are the same in the Howe trusses, and also in the Tratt trusses, whether the load be jilaced on the top or the bottom chord. Art. 27. The Warren Truss with Vertical Sus- penders. — The left half of this truss is represented in Fig. 35, in which the web bracing consists of equilateral or isosceles triangles and vertical ties or suspenders. In a through truss, each of the verticals .3, 5, 7, 9. simply carries a paiud load from the lower chord to the upjier, while the verticals I, (>, 8, 10, are only to siipjiort and stiffen the upper chord. If the truss is a deck, the last named ver- ticals become posts. ■ ''slSSl*^'" BlilDGE TliUSSEfi. 85 caoh is feet 11)8., anil tlir- the stresses - lO./i tons. ? trusses, ami [)lacetl on the irtical Sus- 'l)resent('(l in 3quilateral or nders. In a imply carries er, while tlie I stiffen the t named ver- Prnh. 68. A through Warren tnbTi with vertical ties, one half of which is shown in Fig. .};■>, has 16 panels, each S feet long, its braces all forming equilatenil triangles; the dead load is given by formula (2) of Art. 15, the live load is loCM) lbs. per foot per truss: find the maximum and minimum stresses in all the members. The dead panel load per truss _/7x 128 + 600\y ^oeiiu •> * =■' — - 18=5984 lbs. = 3 tons. Live panel load per truss = — - ^-? = G tons. tan e = .577 ; sec d = 1.1547. The chord stresses may be written out at once by chord incremc rts. For the verticals the maximum r.tress is evidently the dead and live panel load, or 9 tons; the minimum stress is the dead panel load, or :i tons. The following are the equations for iletermining the stresses in a few of the members: Max. stress in 2-t = 7.50 x 9 x .677 = 38.9 tons. Max. stres3 in 4-6 = [7.50 + G.50 + 5.50]9 x .577 = 101..3 tons. Max. stress in 3-5 -= -[7.50 + (5.50] X 9 X .577= - 72.7 tons. Max. stress in 2-3 = -[7.50 x 9 x 1.154] = - 77.9 ton.s. Max. stress in ,5-() = [4.5 X 3 + /, (1 + 2 4- 3 + ... 4- 12)] 1.154 = 42.75 X 1.154 = + 49.3 tons. Min. stress in 7-8 = [2.5 X 3 - T-\ (J + 2 + 3 + - + 5)] 1.154 = + 2.2 tons. lu this way all the stresses may be found. m ft I — ^ 86 HOOFS A\h llllIDaEs. riiiiiiii SriiKKSKs. Memukim. «-6 .V7 I 9 9-9 2-4 4-8 6-8 S-H) Max. strpssi's Mln. Htrir.HM'.H -72.7 -24.2 i24.r. - 41.;. ~ IW.*. - 51.9 - lfifi.2 - .V..4 + 3^.9 + l.'-.O f 101.3 + 8S.S + 142. S + 47.6 + 111;.'- Wkii Stbksses. Mrmiiikk. Max. rtr«»e!i . Mln. ttrttaM<8 , 2 8 -77.9 -2110 3-4 + 117.9 ■^ 22. 1 -^>*.4 -17.7 + 49.3 + 13.0 C-7 -40.6 - 7.S .S-9 -(-•32.6 -24.7 + 2.2 I -I- 3.9 -»■ 17.3 10.1 We see fn)iii tlii.s tiiblo of wel) stre.sses tliat tlie niciiibors 8-9 and 9-10 should lie cai)al)lo of resisting both tension and eoiiipression, and hence siiould be eouiiter braces (A, . 1^; and, of course, tlie same is true for the right half of the truss. J'rob. 69. A through Warren truss with vertical ties, like Fig. ;jr>, has 20 panels, each 10 fe.'t long, its braces all forming equilateral triangles; the dead load is given by formula (li) of Art. lo, the live load is lOCK) lbs. per foot l)er truss : find the niaximum and minimum stresses in all the membeis. Aus. Max. stresses in upper chord = -i;{r>.o, -2-10.0, -3ir,.o, -sco.o, -375.0 tons. Max. stresses in lower chord = + 71.;i, +191.3, -f- 281.3, +341.3, +371.3 tons. Max. web stresses = -142.;-., +128.0, -113.9, +100.2, -87.1, +74.4, - (52.2, + 50.4, ~ 39.1, + 28.3 tons. Min. web stresses = -54.8, +48.G, -41.9, +34.7, -27.1, +19.0, -10.5, + 1.5, + 8.0, - 17.9 tons. m 6-8 8-11) + 1«.S + 47.6 + Kill.'- + M.- S S(-9 ' + -- .6 2 -24.7 + 8.8 the nicnibors li tension and ices (A, .1^; f of the truss. tieal ties, like ts braces all is given by 11)8. per foot tresses in all 375.0 tons. ?71.3 tons. ■87.1, -1-74.4, -19.0, -10.5, HHlltaK TRUSSES. 87 Prnh. 70. A tliroii;j;li Warren triiss with vertical ties, lik<' V'y^. ."r>. has 10 panels, each 12 feet long and VJ feet tlcep, its braces all forn.ing is().s<'eles triangles; th(^ dead and live loads are 100(( lbs., and I'OOO lbs. per loot per trn.ss: tind the niaxiniuni and niinininni .stresses in all the meiubers. Art. 28. The Double Wairen Truss, or Double Tri- aitfjiilar Truss, or Ginlcr. - This truss, shown in Fig. ofi, has two systems of triangular bracing, the one system represented in full lines and tlie other in dotted Hues, the J s r, 7 ." ti ?' 7' s' .1' r chords being common to both systems. The roadway may be carried either npon tho upper or the lower chord. In all the trusses hitlierto considered, a vertical section taken at any point of the truss will cut only three members ; but in the double Warren truss this is not the case, since a vertical section will in general cut fimr mend)ers, causing the problem at first to seem to be indeterminate (Art. 7). This is obviated, however, by the fourth condition that: the truss is ecpiivaleut to the two trus.ses, shown in full and in dotted lines, welded into one. Thus, the loads at fi, 10, 10', G' are carried to the abutments by the diagonals drawn in full, while the loads at 4, 8, 12, 8', 4' are carried by the dotted diagonals. The chord and web stresses are retulily found as in a simple triangular truss, assuming each system as independent. This truss is used both as a riveted, and as a pin-eonuccted bridge. .-^"uriAiMr^^ii .jMMH -^TJTflT •f I } i aaMH 88 nnoFs AND unnxms. Pmh. 71. A (louhlo Wan-on trus.s, Fig. m, as a dock rail- road hiid^'o, li.xs Id paiiols, oach 14 foot lonj,' and 14 foot dooii; tlic d,.;i(l load is Lfivcii l.y foniiida (L') of Art. ir», the livo load is L'lMIO ilw. per loot per truss: tind the stresses in all the members. The dead panel load per truss V 2 ) "^ l^OGO lbs. = 5.5 tons. Live panel load i)er truss = 14 tons. tan 5 = 1, sec 5 .= 1.414. The followiiif,' aro the etpiations for determining the stros.sos in a few of the niend)ers: Max. stress in l-H = - 2 x 1!).;") x 1 = - 39 tons. Max. stress in .3-5 = - [2.0 + 2.5 4- 1.5] 19.5 = - 117 tons. Max. stress in (Mi = [2.5 + 4.0 -|- 3.0] 19.5 = + 185.2 tons. Max. stress in .'M! ^ [1.5 X 5.5 +- 1.4 (1 + 8 +5 +- 7)] 1.414= + 43.3 tons. Min. stress in 5-.S ^ [5.5 - 1.4 x 2] 1.414 = + 3.8 tons. Thus the following stresses are determined : Upper Chord Stresses. Mkmheiw. 13 3-5 .^•7 7-9 9-1! Max. stroRses . . . Min. stresses . . . -3i».0 -117.0 -11.0 - .3.S.0 -175.5 - 49.5 -214.5 - 00.5 -234.0 - 00.0 Lower niioRi) St KESSRS. .Mkuiiers. 2-1 4-r. fi-S s-in 1(1-12 Max. 8lre.s.ses . . . Mill, stri'sscs . . . +48.8 + 1.3.8 + 120.8 + 35.8 + 18.5.2 + 52.3 + 224.3 1- 03.3 +24.3.8 + 08. S a a (lock rail- and 14 foot :' Alt. IT), the lio stresses in 5 tons. nnining the ma. = - 117 tons. -f- 185.2 tons. - 43.3 tons. - 3.8 tons. 9 9-11 4.5 0.5 -234.0 - 00.0 n 1(1-12 1.3 5.3 +243.8 + 08.8 BlUDdK riiUSSES. 89 Wkh Strksshs (Dottki) "ivSTEM). MKMItKltf*. i-i 1 .•. .'i H s-K !t-|-.' .Max. stresses .Mill. .Htreases + 55.2 + 15.0 -55.2 - 15.0 + 31 6 + 3.8 -31.5 - ;i.8 + n.» -ll.O Wkii Strksskh [FiM, System). .Mkmiikk". ■i-:\ :!-■! <;-T 7-10 10-11 Max. stresses -00.0 -10.4 + 43.3 + 0.7 -43.3 - 0.7 + 21.7 - 4.1 -21.7 + 4.1 Mill, stresses Prob. 72. A double deck Warron truss, like Fig. 3f), has 12 panels, each 10 feet lon<,' and 10 foet deep; the dead load is SOO lbs., and the live load is 1(500 lbs, {ler foot per truss : tiiid the stresses in all the nieinbers. Aus. Max. stresses in upper chord = - 30, - 90, - 138, - 174, - 198, - 210 tons. Max, stresses in lower chord = 30, 9(), 144, 180, 204, 21G tons. Max. stresses in dotted diagonals = + 42.4, -42.4, -j-27.3, -27.3, -f 14.1, -14.1 tons. Max. stresses in full diagonals = -60.9, +34.9, -'M.9, +20.7, -20.7, +8.5 tons. Min. stresses in dotted diagonals = + 14.1, - 14.1, + G.G, - 6.6, - 2.9, + 2.9 tons. Min. stresses in full diagonals = -17.0, +10.4. -10.4, +1.7, -1.7, -8.5 tons. I 90 HOOFS AMt ItllllX/KS. Pnth. 73. A (lowltlc tliruii^'li Wjiin-ii Iniss, like Fij». .'{f., IliVS 1(1 |i,llirls, iMcIl 10 IVct, li.lij,' iiikI 10 r,(.|, .ItTl-; tlic .Icml and liv." loiids arc SOO ll.s., and L'OCM* \h». per foot per Uiisa: find tho stresses in all tlie inendx-rs. Ana. Max. stresses in npiier chord = - ;jr>. - i>i, - i;;.!, - u\\, - ito tons. Max. stresses in lower chord = + L'S, + St, f IL'C, + loj, -»- ICS tons. Max. stresses in dia.i,'onal ties = + -l!)..|, +■■!'.).(•., +.'{1.1, 4-lil.'.r,. +ir>.r., +H4ton8. Min. stresses in dia<,'onal ties = + 14.1, +11.;!, +7.1, +L'.8, -IM), -8.4 tons. J'rnh. 74. A donlile thr.)ni,'h Warren trn.ss, like Fi;L'. .'!(), has IL' panels, each 1". feet Ion;,' and 1". feet deep; tlie dciul load is given hy fornmla (.'i) of Art. 15, the livi^ load is L'OOO lbs. per foot per truss : find the stresses in all the mend)er8. Art. 29. Tho Whipple Truss. -This truss, shown in Fig. ;{7, consists of two simple Tratt trusses combined. ;5* 10' KiB. 37 The ties in the web system extend over two panels, and it is therefore often called a " double intersection Pratt truss."* The advantage over the I'ratt for long spans is that it has short panels, wliile keeping the inclination of the diagonals* at about 45°. • Called also the Linville Truss. mam till' ilciid [M'l truss: uuiixu: TurssKs. 91 Tliis tnisH WHS at one tinu- iiKtre coiiiiiioii lliaii any otiicr III Aiiicricaii liriih^c |ii;iili( c. I), is slill iiscil Ili(iii;,'li vi-ry ran-ly tor lii^,'liwii_v hiidgrs; lor railway luiilKfs it. is as a rule ivvoidnl by tin- licst prai'tico. It is assiiiiu'il that tliis truss is oiiuivalciit to tli«> two trusses shown in full and in dotted linos, and that each system aets indei)endeiitly and carries only its own loads to the abiitinents; but in the case of the web nienibers, this is not stn'rth/ true. If, however, the truss is built witli an even miiiiber of panels, the error in the web stresses ( btained on the assumption of independent systems is prolialily very small. With the chord stresses there is no anibij,'Mity ; the idiord stresses are a maximum for a full load, and may bo found as usual by chord increments, or by monu'iits. The vertical shear will be tlie same whether the load is applied at the top or at the bottom chord. Pn>h. 75. A througli Whipple truss, Fig. J^", has 12 j)anel8, each 10 feet long and I'O feet deep; the dead load is ToO lbs., and th(> live load is 12(M> lbs. per foot per truss: find the stresses in all the meml)er.s. The dead panel load per truss = S.T.T tons. The live panel load per truss = CO tons. tane = l: seed = 1.414: tan 0' = 0.5: secfl' = 1.118. The following are the cipiations for finding tlie stresses in a few of tlie members : Max. stress in 4-G = 3 x 9.75 x 0.5 = 14.(» tons. Max. stress in 0-8 = 14.G + 2}^x 9.75 x 1 = 39.0 tons = — stress in 1-3. Max. stress in 9-11 = - [39.0 + (2 + 1^1 + i)9-75]= - 87.8 tons = stress in 11-13, •-'i r i 1 92 ittutFs .i.v/> itni'xiKs. sinro for a iinifdnn Icnul the (li.i^fonals niocMiii},' lln' ii|i|«M' olionl Im'I wtfll ',) iillil '.>' ;ili' lint ill iirtidil. Mux. Htrt'HS ill l-l — .'{ X 'Xlii X l.llK j^i'W." tons. Mftx. Htri'sfl in MJ = [l'.J X ;J.7r. + ,«^ (L' -f. 4 + + 8 -j 10)] X 1.414 = lil.;t7r. X l.JM-^.'M.ntons. Mux. stress in 1 -L' = - •.i'J.'J't - LM.STfi = - 53.(5 tons. Thus tho following stresses are (letorniiiied . {'iii>ni> Stiikshes. Mkmiikkk. Max. . .Mill. . •:-i 4 a ll.ll r..ti 0* »-IO ii;-i!j 7:i.l 2H.I IMJ ll-l!l H7.8 :).'{.8 ll.tM) II.IMI :M).o 15.0 58.5 22.6 «2.0 ;ii.8 Stukhhi:* i; 1 niA ■-1(1 Ii)-l-i 111-14 M-ll> IB-IS 1» IT Max. . Mill. . . (HI. 102. i:J2. 44. IM. r.2. 174. OH. IHtl. 02. 1',I2. «14. HrnKssKS iv Fri.i, DiAfioxAi.s. Mknhkbh. Ml .'.-in n-u l:t-ls n-14' Max. . . . Mill. . . . r.0.4 10.8 4;i.8 12.7 20.7 4.2 17.0 0.0 f).7 0.0 STiu:asK8 IN DoTTKit Diauonals. MRMHKB!). i-« :t ■< 7 12 11-16 15-1«' 15-12 Max. . . Mill. . . r.:t.7 22.0 51.0 10..-! 30.8 8.4 2.1.3 0.0 11.3 0.0 0.7 0.0 Stukssks in I'osTH. Mkmkkkh. 1-2 !14 .'. « 7-» 20.0 0.0 U-10 11-12 lit 14 l.Vl« n-i» Max. . . Mill. . . 1K).0 ;io.o 30. f) 11.5 ;ii.o 0.0 21.0 3.0 10.5 0.0 12.0 ' 0.0 8.0 0.0 4.0 0.0 r I < ri \ 94 HOOFS Ash nniunKs. Proh. 11. A (leek Wliipplc truss, like Fi^. 38, has Ifi panels, each 1L' feet long ami L'l feet »l<'cji; tiic dead load is THO ihs. por foot per truss, and tlif liv ■ load is -(HH) lbs. per foot por truss: Hiid the stresses in all the member^'. Art. 30. The Lattice Truss, or Quadruple Warren TrusiS.* — This truss, Fig. 31), contains four web sysloinb 15 g 7 !) fl 19 /5 n 10 «/ l» Ci 8 iu U Ik W 18 to in i»' 18' 10' 11,' i,' lu' 8' 0' U' ^ j' welded into one. It is built only as a short-span riveted structure. In determining the .stresses in the different nit-iubers of this tiass, an ambiguity arises from two eanses : (1) the web members being riveted together at each intersection, the different systems cannot act independently ; and (2) two of the systeinr, are not symmetrically placed in reference to the center of the truss. Thus, if etpial loads 1m, placed at each joint, the abutment at 2 will carry more than half of the load on the system 4; 12, 20, 16', 8', and less than Ijalf of the load on tli" system 8, 13, 20', 12', 4'. I?nt it simplifies the solution toassunu' that each system acts independently, and is symmetrical in reference to the center of the truss. The chord and web stresses are then readily found, as in a simple triangular truss. * A'l rtimble sy teiiis, such .is this aii-10 be- longs is ayniinetncal with reference t the center of the truss ; but the shear In 7-12 Ir not the weight of the two panel loads at the iolnts 12 and 20, because this system Is not synniietrlcal in reference to the center. If the joint 20 were at the center 22, the shear for 7-12 woidd be 1 J panel loads ; but If It were at the joint is, llu' shear would be 2 p.int'l loads. Being midway between the joint.s 18 and 22, the shear for 7-12 is the mean of l.J and 2 panel loads, or 1.75 panel loads ; and so for the shear in any other web member. By chord increments, we have Max. stress in 2-4 = 2 x 15 x 1 = 30 tons. Max. stress in 4-0 = (2 + 1.75 + 2.75) X 15 = 97.5 tons. Max. stress in 0-8 = 97.5 + (1.5 + 2.5) X 15 = 157.5 tons. Max. stress in l-*5 = 2.5 x 15 = 37.5 tons. ■ Max. stress in 3-5 — (2.5 + 2 x 2.25) x 15 = 105.0 tons. ■■ i i , » np ^ fi g»HWWi— »» I i 96 ROOFS AND nuiDGKS. Maximum Web Stues.sk.s. Tho maximum stress in ;\-S will bo when the dead and live loads cover the whole sy.st,eni to which 3-8 belongs. Likewise for the menilmr o-lO. The maximum live load stress in 7-lL' will be when the live pane! loads are at the joints IL', 'JO, 1(5', 8', since these are the only loads which a.;t in the .system to which 7-12 belongs, on the right of 7-12. For the dead load stress, see Note. Max. stress in 3-8 = 2.25 X 15 X 1.414 = 47.7 tons = - stress in .1-3. Max. stiass in 5-10 = 2 X 15 X 1.414 = 42.4 tons = - stress in 2-5. Max. stress in 7-12 = [1.75x5+ .JO (3+7^11^.iri)-]l 414^3-. g^^^g = — stress in 4-7. Max. stress in 9-14 = [1.5 x 5 + ^ (2 + + 10 + 14)] 1.414 = 33.2 tons = — stress in (J-l). Thus the following stresses are determined: Ui'PKit Ciioni) Stresses. Memhkrs. 1-3 S-A .V7 T-9 9-11 Max. . . . Min. . . . 37.5 12.5 106.0 35.0 106.0 55.0 217.5 72.6 202.6 G7.5 MiMIIKUS. Max. . . . Min. . . . 11-13 l.S-l,'> 15 IT 17 111 19-21 375.0 125.0 300.0 100.0 .330.0 110.0 352.5 117.5 .'!07.5 122.5 m^ ! dead and S l»elong,s. 1 live load are at the latls which le right of a ^-3. tons tons 9-11 202.5 19-21 375.0 125.0 U JUDGE TRUSSES. Lower Ciiorp Stresses. 97 .Memiip.iis. 2-t ■)-■'. 0-1 ^-m lU-13 Max. . . . Mill. . . . 30.0 fl.O 07.5 ;{2.5 157.5 52.5 210.0 70.0 255.0 85.0 Mkmiikus. IJ-U 11 111 IC-li H-2() 300.0 120.0 •M 22 Max. . . . Mia. . . . 202.5 07.5 322.5 107.5 345.0 115.0 307.5 122.5 Web Strkh,se.s. Max. stre.ss in A-4 = 58.4, in l-C = M.O, in 3-8 = 47.7, in 5-10=42.5, in 7~1L'=.'{7.8, in \)-U=3:i.2, in 11-1G=l'8.7, in K5-18 = 24.0, in 15-20 = 20.2, in 17-22 = 1G.3, in 19-20' = 12.3, in 21-18' = 8.5, in 19'-1G' = 5.4 tons. Min. stress in A-4 = 10.5, in 1-6 = 17.7, in 3-8 = 15.9, in 5-10 = 14.2, in 7-12 = 11.8, in 9-14 = 9.2, in 11-16 = 6.8, ill i;}-18 = 4.3, in lJ-20 = 1.3, in 1.7-22 = - 2.1, in 19-20' = - 5.3, in 21-18' = - 8.5, in 19'-16' = - 8.9 tons. Max. stress in 1-.>:1 = 37.5, in ^1-2 = 112.5 tons. Min. stress in 1-^1 -■ 12.5, in ^1-2 = 37.5 tons. We see from these results that the 10 diagonals 14-17 and 17-22, 16-19 and 19-20', 18-21 and 21-18', 20-19' and 19'-16', 22-17' and 17'-14' require to be counterbraced. Prob. 79. A deck lattice truss, containing four web sys- tems, like Fig. 39, has 20 panels, each 12 feet long, the depth of t'-uss is 24 feet; the dead a.ul live loads are 500 lbs., and 1000 lbs. per foot per truss : find the stresses in all the members. Ill 98 HOOFS AM) ItniDGKS. V Alls. Max. stresses in upper clionl =z 18.0, 58.5, 94.5, 12G.0, iny.O, ITo.u, VJS.n, L'Or.O, lilO.O, L'l'O.n tons. Ma.\. stresses in lower cliord = lil'."), (53.0, UD.O, i'M5 157.5, 180.0, 198.0, liU.S, 220.5, 225.0 tons. ' Max. stresses in struts A-li, 2-5, ete. = 35.0, 31.8, 28.G, 25.5, 22.7, 19.9, 17.2, 14.4, 12.1, 9.8, 7.4, 5.1, 3.2 tons. Mill. stre.sses in struts .1-3, 2-5, etc. =-11.7, -10.(5, -9.5, -S.5, -7.0, -5.5, -4.1, -2.0, -0.8, -f 1.2, +3.2^ 4- 5.1, 4- 7.4 tons. Max. in .1-1 = 22.5; niin. in .1-1 = 7.5. Max. in ^1-2 = <>7.5 ; mill, in .1-2 = 22.5 tons. I'rob. 80. xV through lattice truss, containing four web systems, like Fig. 39, has 1»} i)aiiels, each 10 feet long, the (l.'].th of truss is 20 feet; the dead and live loads are 1000 lbs., and 2000 lbs. per foot per truss: liud the stresses in all the members. Art. 31. The Post Truss.* — This truss, shown in iMg. 40, is a special form of the double triangular truss (Art. 28). The members 3-4, 5-0, etc., are struts, and all the other diagonals are ties; the counters are shown in dotted lines. The distinctive feature of this truss is that the web struts, or posts, instead of standing vertically, are • Tlxis truss has becoiuj ubsulote. BRIDGE TliUSSES. 99 )4.6, 12G.0, !).0, 1^0.5, 31.8, •M.a, )ns. .7, - 10.(5, 1.2, + li.2, X. in ^1-2 four web : long, tlie loads are le stresses shown in liar truss s, and all shown in ss is that cally, are inclined by one half of a panel length, and the ties by one and one hulf panel lengths. All the panels of Ixilh chords aif of ctpud length, except the two end panels oi" the lower chord, which are each one half a panel in length. There arc and)iguities in the character of the stresses in iliis truss, as there are in all double systems of bracing. It is impossible to separate the two systems as they are (connected at the center. But if we assume that, under a full load, the members 7-10' and 7-10 are not acting, if it is a deck truss, or that the members 7-10', 7'-10, 9-10, and 9-10' are not acting, if it is a tlirough truss, tlien the chord stresses are readily found. Also, if we assunu' that the systems 2, 1, C, 5, 10, 7', 8', 3', 4', 1', 2', and 2, 1, 4, 3, 8, 7, 10', ')', 0', r, 2' are independent, the web stresses may be readily found. Prob. 81. A deck Post truss, Fig. 40, has 8 panels in the ujiper chord, each 10 feet long, and 20 feet deep; the dead and live loads are 400 lbs., and IGOO lbs. per foot per truss: lind the stresses in all the members. Dead panel load per truss = 2 tons. Live panel load per truss = 8 tons. tan e = 0.25, and sec d = 1.0308 for the posts. tan fl = 0.75, and sec 6 = 1.25 for the ties. Max. stress in 1-3= [2 x ^ + 1^ X J] 10 = 16.3 tons. Max. stress in 4-6= [2 x J + 2 x i^] 10 = 10.0 tons. Max. stress in 5-6 = [Hx2+(2+44-6)]x 1.03=16.5 tons. Max. stress in 3-8 = [2 + (1 4- 3 + 5)] X 1.25 = 13.8 tons. Min. stress in 3-8=[2 - 8 x i] x 1.25 = 1.3 tons. iilf ■ iii f im m i m if AT KK KT % Ml "^'' nOOF& AND niilDGES. Thus the following stresses are determined: Chord StHK!*8KS. Meuiikhs. i-a ,S .1 28.7 &-I ;}rt.2 7.2 T-l) ;i8.7 7.7 4-1 10.0 2.0 c-t S 10 lU 10' Max. . . Mill. . . 25.0 6.0 a.-i.o 7.11 40.0 8.0 Sthkh.sivS IX Ties. Mkmiikks. 14 Ifi 1 8-8 &-10 7-10' 7.6 0.0 S-'J I'-T 4-:> Max. . . Mill. . . 20.0 4.2 1S.7 ;!.8 13.8 8.7 0.0 5.0 0.0 2.-) 0.0 1..1 0,0 Stuk.sses IV Posts. Memiikrn. \-'i u-J .Ml 7-8 li-lo 2-4 Max. . , Mill. . . 40.0 8.0 22.0 4.2 15.4 .•i.l 11.4 2.1 7.2 1.0 0.0 0.0 Prob. 82. A through Post truss, like Fig. 40, lias 12 panels in the lower cliord, each 10 feet long and 20 feet deep ; the dead and live loads are 1000 ll,.s., and 2000 ll)s. per foot per truss : find the stresses in all the members. Art. 32. The sollman Truss. — This truss, shown in I'Mg. 41, was the earliest type of iron truss built in the United States. It eonsists of a series of inverted king post trusses with une«pially inclined ties, which carry the load at each joint directly to the ends of the upper chord. The vertical pieces are struts; the upper chord is in compres- sion ; and there is uo stress in the lower chord. The short ■"■ *nJ>V.W**■"•^i^9«ll*fe^ii■.' fc, 'Jt^<^:^ij:^u- l, and was regarded as an improvement on the Bollman truss. It consists of a com- A J f ; n K D' ( ;' B' A' ^ 5^ >. t \ ^ ^ :::^m a h e (1 eb' a' Fig. 43 bination of inverted king post trusses with equally inclined ties, as the primary .system AdA', the secondary systems AhK and A'b'E, and the tertiary systems AuC, CcE, etc. The vertical pieces are struts, the upper chord is in com- pression, and there is no stress in the lower chord ; all the diagonals are ties. Tiie load may be carried either upon the upper or the lower chord, though it has been used more often for deck V)2 HOOFS AM) nnrnGEs. th.in for tluougli l>ii(l(,'('3. Tliesc inisscs were built from ISJI down to about l.S7<5; but they arc nut now generally regarded with favor by bridge builders. I'riifi. 84. A deek Kink truss, Fig. 41', has 8 panels, earh Ili feet long, ami Ki feet deep; the d.-atl and live loads are ")(»() lbs., and 2(>0() lbs. per foot per truss: find the stre.sses in all the mend)ers. The dead panel load per truss = .'? tojis. The live panel load per truss = 12 tons. Wo see at onee, from Fig. 42, that the maximum stresses in all the members oeeur when there is a full panel load at every apex. Hence the maximum stresses will be when 1/5 tons arc jdaeed at every apex, and the minimum stresses will be one fifth of the maximum. The maxinnim stress in ea<'h of the posts lia, Dc is — 15 tons. Tin; maximum stress in the post (Jb is the panel load of lo tons at C, plus one half the panel load of 15 tons at />, plus one half the panel load of 15 tons at B, = — 30 tons. Similarly the stress in Ed = — 60 tons. Also, Stress in An =-- ] x 15 sec; = {).4 tons = On = Cc = Ec. Stress in Ah = | x ."iO see ", = 27.0 tons = Eh. Stress in Ad = ^ X 00 see 0^ = 5)5.0 tons. Stress in AA' = - [,J- x 15 x 5 + .J X ."0 X 2 + J X CO x J^] = -118.1 tons. Prob. 85. A deck Fink tntss has 8 panels, eaeh 10 feet long and 15 feet deep; the deid and live loads are 400 lbs. and 2000 lbs. per foot per truss: tiad the stresses in all the members. ""^k*::ii,i^^A!%HMi&-J*'^i«i.^i i-i^»ir.^w.^\s*fc»#-,;j«rt^l!*«sa**t5i«tMjrt«ik.jM«Ui5afla'^ built from ff goneriilly )aneli), each c loads are tho striisst's nil strosses lu'l load at l>o when 1.5 mi stresses nil stress in maxiiiiuiii I at C, plus le half the luilarly th« = Ec. X 60 X J^] aeh 10 feet lie 400 lbs. s ill all the HUWQK TIlirssES. Rrid*!!' Tkhshks with TNci.t\Ki» OiinKns. 108 Art. 34. The Parabolic Bowstring Truss. -lu this truss, Fig. l.'{, tiie lower ulioiil is horizontal, ami the uiijier ^-'rr^ A EI) C yilu Via. *3 D chord joints lie on the arc of a parabola ; the verticals may be in compression and the diagonals in tension, or tlie ver- ticals may be in tension and the diagonals in compression. Let I = the length of the span vl/i, d = the height of the are (>C at the center of the sjian, and w = the uiiiforiii load in lbs. ]ier foot of truss; let (.r, y) be any joint /'of the ujiper chord referred t'- the axes ATi and AY, and S the stress in the lower chord panel opposite P. Then we have (1) The equation of the parabola referred to its vertex O as origin is {\l-xf^2p{a-v) (2) 72 and for the point A this becomes — = 2pd\ Sub.stituting this in (2) and solving for y, we get y=^^Hx-aP) . . . . 2/) = All which in (1) gives 5 = (3) (4) IIpncp,f(>r a uniform loud on ii pnrdhoUc truss the stress in the lower chord is the same in every panel. ■ ' ' u:iavtm^f^i. S"'^* "' 104 IKXIFS A.\h nitlDCKs. Sim-e llu- liorizoiital compoiiPiit of flip stress in any iliaj,'<)ii:il, iis /'/> for cxainiilr, is f(|iial to the (lillfiviiti- ot tlio diord iitri'sscH in the adjiiceiit iiaiiols, /<;/> and IX'. Thvi-cf()ri>, fitf (I Kiiijhnn loiul there ix uo .slretrn in an if dkiyuiKil. It fuUowH at once tliat, /or ti iDiifhrm load tho hitrizautnl comjinni'Ht of the .s/ms.s i„ the n/iiter churd in the name in erery imnel, and is expri'sstul by ('(iiiation (4j. Therefore, for a intifnnn Indd the stress in nvij inrlined jntnel of the np/ier chord is eiimd to the stress in a fninel of the lower chord mnltijtiieil lij/ the secant of the inclination. I'roli. 86. A paralKjlic! liowstrini,', Kij,'. 11, as a tlir(ni},'li biidgt;, has 8 panels, oaeli 1(» feet loiiy, and lU feet eenler Kin. 44. depth, the verticals t;>ke either tensioji or eompression, the diai,'onals are ties; the dead and live loads are 4(10 lbs., and S(M> lbs. per foot per truss: hnd the stre.s.ses in all the niendters. The max. .stre.ss in any panel of the lower chord, by (4), 1200 X .so-' ,„ ^ .-. Min. stress =1 x 48 = IG tons. Dead panel load = li tons. Live ])anel load = 4 tons. ^^.^^.^^<^Mw^A.ca»^saw^^,^i^;y^:.^^a=lrit^ ■Jk T II It I in; K TltUHS.ZH. lor) % (.">) we iiiivc lfiit,'tli.s of ;{-J, ;"■)(;, iind 7 S = L.'JTr., 7.r>, ami l)..'{7r» ft't't. Max. stress in m_;, = 48_x y (1 ()) ^ f (1;{7;^^ II) = fi2.4 tons. Sinco the (load load pnidiicis no .stroHscs in tlic diaKonals, the max. stress in any diagonal is found liy |iiittinH only tin- live load on tho hrid^,'o in the position to },'ive the max. shear in that memlier (Art. '2'J,). Thus, for the maxinuim stress in H-S the live load is l»liu;ed on the right, the center of moments is on tho lower chord at L'(» feet to the left of the i)oint 2. The lever arm of r. 8 is ;!(► feet. The reaction for this loading = 1(1 + 2 + 3 + 4 + 5) = 7.fi tons. .-. Stress in 5-8 = ^^^ = 5 tons. till For live load on the left, the stress in 5-8 = 0, and the connter 0-7 comes into action. The lever arm of (5-7 is 27.4 feet ; tlie reaction for this loading = (i.a tons. There- fore the stress -S' for G-7 is found from the eciuatiou G.5 X 20 - 4 (30 4- 40)+ .S' x 27.4 = 0. .-. S = stress in 0-7 = 5 /» tons. If we cut a vertical as 5-(i by a section, place the live load on tho right, and take the center of moments at tho intersection of .'}-,'> and 4-0, which is 4 feet to the left of the point 2, we shall obtain tlie maximvim eompresslon iu it caused liy the live load. Thus, The reaction for this loading = 71) tons. ■ •. Max. stress in r)-(> due to live load 7.5 X 4 24 ■ = — 1 .25 tons. Min. stress in 5-0 = 2.0 - 1.25 = 0.75 tons. I! 11 i 100 /•OOFS Axif iiiiihaKS. Th« tiiii.xiiimm HtrcsH in cm-h vorticul is omo of tciisidii, aiitl will (MTiir wlifii \\w livo Iduil fovors tlio hritlgu, ami is u (loiul ami livo piiiicl l((iul, nr () tons. It llicrc bt! IK) live Idiul dm tlit' tiiiss, fho strtss in any vcrtiful is u dead imni'l load, or a tt'nHhm of '* toii.s. Now let till' Iivt( load conn, on from tli« rij,'lit, a iiart of it which KOCH to the left support will cuii.sc ruiiiprr.s.sion in each vertical, or will diminish the tension in that vertical duts to the .load load. Thus, when the live load reaches the joint 8', it will cause in the vertical U-IO a compression of 'J.'J tons, which will neutralize the dead loatl tension of 2 tons, and leave as the result a compression of 0.1' tons. Also, when it reaches the ape.K S, it will cause in the vortical fi-(; a compression of l.L'o tons, as we .saw above, which will neutralize that much of the dead loail tension of li tons, .-.nd leave as the result a tetision of 0.7ri tons. 'i'h(i following stresses are found in a manner similar to the above ; Ciioni) Stiikhsks. Mkmiikkh. Max. . . . -62.4 - 17.5 .'t ft f-7 •-9 -48.1 - irt.o •i-i, 4-1!, otc. + 48,0 + 10.0 -60.:. -16.8 -48.8 - WM Strk«se8 in Diaoonals, MKMIl'Jh.S. 3-0 .vs 7-10 ■1-,^ 8 7 S-9 Max. . . . Min. . . . +4.;} 0.0 + 5.0 0.0 + 5.6 0.0 +6.0 0.0 + 5.5 0.0 -^5.7 0.0 ■■>»-a«.j»,s**.5ii»9*Mi-te'»»M,.t>.wSjSi«i;jj(isj(Miwiiiwaiai^ >f t, +14.0, +-14.5 tons. Mill, stresses in verticals^ +2.0, +2.0, +-2.0, +2.0 tons. Mill, stres.ses in diagonals = 0. J'liil). 88. A thidiigli paraboli(! bowstring truss, Fig. 44, has 8 panels, eaidi 12 b-et long and IG feet center depth ; the verticals arctics and the diagonals are braces, the dead and live loads are 500 lbs. and 2000 lbs. per foot per truss: liiid the stresses in all the menibers. Alls. Max. stress in eacdi panel of lower chord = + JM) tons. Min. stress in each pamd of lower chord = +- 18 tons. Max. stres.ses in upper chord =5 - 104.1, - D7.5, - 1)2.7, - 90.3 tons. wfffWBlK^T'y**^ mmm iJk 108 HOOFS AND liltlDGES. r.Tiii. stresses in upper cliord = - L'()..S, - V.).r>, - IS.n, - 18.1 tons. Max. stresses in verticals = + ir).0, +18.7, f-21.0, +21.7 tons. Min. stresses i" verticals = +3.0, +3.0, +3.0, +3.0 tons. Max. stress in 3-(> = - 10.4, m 5-^ = - 12.7, in 7-10 = - M.4, in 4-5 = - 12.7, in G-7 = - 14.4, in 8-9 = - 15.0 tons. Min. stresses in diagonals = 0. Pmb. 89. A tliroii.L,']] iiarabolic bowstring truss has 12 panels, eac'i S Jeet long, ami 12 feet center depth; Ihe verticals are ties and the diagonals are braces; the dead and live loads aie oOO lbs. and 2000 lbs. per foot per truss: tind the stresses in all the members. Art. 35 The Circular Bowstring Truss. — This form of tiuss, Fig. 45, is often used for highway bridges. 'l"he joints of the upper chord lie upon the arc of a circle, each joint being directly over the middle of the panel below. The diagonals are built to take either tension or com- pression. Proh. 90. A through circular bowstring truss, Fig. 45, has G panels, each 12 feet long, and 11.7 feet center depth ; the joints of the upper chord lie on the arc of a I'ircle of 00 feet radius; the dead and live ionils mv 150 li)s. and I.'tOO lbs. per foot per truss: find the stresses in all '1 e uiend)ers. •*i««aiW««wiw*«i7W«a»«t!i&-i«e^^ 1 HI! ma!': thusnes. 10!) A)tH. Max. stress in '_'-.'{=.- 40.4, in 3-5 = - 54.3, in 5-7 = - 51.1, in 7-7' = - 50.1 tons. Ma.\. stress in 2-4 = + 41.1, in 4-6 = + 45.8, in fi-8 = + 47.3 tons. Max. stress in 3-4 = + 9.0, in 4-5 = + 9.0, in 5-(i = + 10.4, in G-7 = + 10.0, in 7-S .■= + 10.8 tons. Min. stress in 4-5 =— 2.2, in 5-6=— 1.3, in 0-7 = -3.0, in 7-8 = -3.1 tons. This pnibi'Mii is lakcn, with slight changes, from Burr's Stresses in nnilj;e and IJoof Triussos. Proh. 91. A throu,t,'h circular Oow.striiif; truss has 8 panels, eaoh 15 feet long, the center depth is 10.74 feet; the joints of the upper chord lie on the are of a circle of 100 feet radius, each joint being directly over the center of the panel below; the dead and live loads are KKKt lbs., and 2000 lbs. per foot per truss: find the stresses in all the members. Art. 36. Snow Load Stresses. — For railway bridges the snow load is not taUen into account, since the floor is opeUi so that but little is retained. For highway bridges the snow load is taken from to 15 lbs. per square foot of floor surface, depending upon the climate where the bridge is situated.* The snow load is tcaken lower than for roofs, since the full live load is not likely to come upon the bridge v.'hile it is heavily loaded with snow. Since the snow load is uniform, the stresses due to it are computed in the same way as the dead load stresses ; or the snow load and dead load stresses are proportional to the corresponding apex loads (Art. 8). * In bnilding highway bridgen iu England and F rauce the snow load is not generally considered. ii !; i ; !!) if ,./■ no HOOFS A\J} lUUDdES. i Pruh. 92. A thioufijli llowo truss, Fig. 32, him 10 icuiels, each 12 feet long ami 12 tVct deoj); tlie width of roadway is 20 feet, and the width . eiudi sidewallc i.s ;"» feet; tlic snow h>ad is 10 lbs. per stjiiare foot of floor : find the snow load stresses in all the niendiers. Ann. Stresses in lower chord = 4.1, 9.5, lO.S, 1 l.;j tons. Stresses in the verticals = 4.1, 3.2, 2.3, 1.4, 0.9 tons. Stresses in the diagonals = -S.7, -4.4, -3.2, -.9, -lO.Gtons. P)-(>b. 93. A (leek Pratt truss. Fig. 31, has 10 jtanels, eaeh 10 feet long and 1(5 feet deep; the width of roadway, in- cluding sidewalks, is 3o feet; the snow load is 15 lbs. per square foot of floor: find the snow load stresses in all tlie members. Art. 37. Stresses due to Wind Pressure. — In bridges of large sjjan, it is ofleu found that the stresses produced in some of the members by a gale of wind are almost as great as those caused by both the dead and live loads. In the principal members of the Forth bridge, Scotland, Sir 15en- jamin Maker estimated that tlu? maximum stresses due to these t'iiree separate forces were as follows : Stress due to dead load = 2282 tons. Stress due to live load = 1022 tons. Stress due to wind Joad = 2920 tons. In estimating the wiiul pressure, ami the resulting stresses in the mend)ers of a bridge, the practice of engineers varies greatly. Different railroads have their own speciiications for wind i)rcssurc. The standard wind ])ressure i)er scpiarc foot ranges in this country from 30 to 50 lbs. It is assumed by many engineers that there will be no train on the bridge ^?*3»l^i,^»^,•«»4iWMM«Ba»**K■<*. - lilllDGE TltUSSkS. Ill whiiii the wind hli.ws with greater pressure than .30 lbs. per S(|nare foot. A wind pressure of 30 lbs. per square foot will overturn an empty freigiit car. In sucli a gale it would be hardly possible for a train to reach the bridge. It might, however, be eaught there by a sudden s(piall ; and this is just what ajipi-ars to hav(! happeiu'd in the case of the Tay bri. A maximum pressure of oO lbs. per square foot is taken as applying to the bridge alone. The surface exposed to the wind action is commonly taken as ilouble the side elevation of one truss, for the vca.son that the windward truss cannot afford much siielter to the leeward truss, whatever may be the direction of the wind. In a heavy gale of wind there is not much shelter to be fouiul ujider the lee of a lamjvpost at a distance of lit) feet from it, even if the post be directly to windward. Experiments show that the wind pressure against the two trusses of a bridge is more than 1.8 times that on the exjwsed surface of one truss. For Highway Bridges the wind i)ressure is fretpiently taken at about 30 lbs. per square foot of exposed surface of both trusses. It is assumed that there will be no live load upon the bridge when the wind is blowing at this maximum pressure of 30 lbs. i)er squar" foot. For Railroad Bridges tlie wind pressure is taken at about 30 lbs. per scpiare foot of exposed su face of bv)th trusses, and about 3(K> lbs. per linear foot due to the train surface. The train surface is about 10 square feet for each linear foot of the bridge. The 30 lbs. per square foot of the trusF^s is treated usually as a dead load, and the 300 lbs. per linear foot flue to the train surface as a live load. This regards ejich truss as fully exposed even when the train is on, though it partially shelters one t; uss. 112 UOOFS Ay I) niilDGES. Ti) estimate the .'50 lbs. pressure \iev s L'OOO This load acts as a dead load at each apex of the wind- ward and leeward chords in the direction of the arrows, shown in Fij^. H'r, hence there is no stress in the dotted diai^onals. tantf=l; sec 5 = 1.414. Stress in hh' = — 0.6 tons. Stress in //'' =: - O.G x 2 = - 1.2 tons. Stres.s in ce' = — 0.(> x 4 = — 2.4 tons. Stress in he — — ().<» x 1) x 1 = — 0.4 ton Stress in al = - [o.4 -f- 4.2] x 1 = - 9.0 tons. Stress in fh' = -f- 0.0 x 1.414 = 4- O.S tons. Stress in c/' = + 3 x O.G x 1.414 = -f 2.5 tons. (2) Lower Lateral System (12 Panels). Panel load for both chords ino X 10 , 300 X 10 ■ + = 1.2 tons + 2.4 tons. 2000 2000 The former we treat as a dead load, the latter as a live load, both acting at each ajicx of the windward chord in the direction of the arrows, shown in Fig. 48. (See Remark.) Stress in AB = - 6.6 x 3.0 = - 19.8 tons. Stress in C't" :-r4.6xl.2+|i(l-f2+3-h - +10)1= -16.4 tons. Stress in DE' = [2.ox 1.2+0.2(1+2+3+ -. +8)] 1.414= 14.4 tons. L8). : ().(■» tons. f the wind- ;lio arrows, the dotted HIillK.K TIlUiiSKS. TliiLs the I'ullowiiij; strcssos are (M)iiiimted: ITl'I'KIt liATKKAl- SVSTKM. Stresses in chords = - r>A, - <».(;, - V2.(\, - It. 4, - m.o tons. Stresses in struts = - r>.7, - l.H, - :U\, - 'lA, - l.L', - 0.0 tons. Stresses in dia^jonals = -f 7.(i, +").•.», + 1.1,', -+!-'.'», +0.8 tons. 11 ■> )ns. ons. IS). ns. I" as a live liord in the Remark.) 16.4 tons. L4.4 tons. LoWEK LATKIlAr, Sv.STEM. Stres.ses in chords -^ 10.8, 30.0, 48.G, .57.6, G.3.0, G4.8 tons. Stresses in struts = -'2\.(\, -1J).8, -10.4, -i;i.2, -10.2, -7.4, -ratons. Stresses in diagonals = 27.9, 23.1, IS.O, 11.4, 10.4, 0.8 tons. Tiie cliord bciUi is in eompression under the action of the dead and live loads; this conii»ression is increased by that due to the wind pressure, as just found. The chord A'WC'Il' is in tension under the action of the dead and live loads ; this tension is increased by that due to the wind pressure, as just found. When the wind Idows on the opposite side of the bridge the diagonal and chord stresses are to be interchanged, while the strut tresses remain the same. This there will be the same stresses in b'c\ c'd', etc., as in Ix', cd, etc., and the same in A'li', Ti'C, etc., as in Ali, li<\ etc., and the dotted system of braces will act in eiuOi lateral. Tlie wind huvds on the left hidf of the upper lateral are transicrred to the abutment at A by means of the portal hrac,':'tj AA'b'b in the transver.se plane of Ab. no /?OOF.S AST) niilDQES. Proh. 96, A tliruugli IIowo truss niilwuy briiltjo -<> f't'ct wide has I'J paiu'ls, cacli '-'() U'ci loiif,'; llic upper anil lover lateral systems aro Howe trusses: iind the stresses in Ix.th lateral systems tlue to the same wind pressure as in tiie previous problem. Hy taking the dotted diagonals, Fij,'s. 47, M!, and 48 will represcMit tlie left half of the Howe truss and the uppe- and lower lateral systems. Then by riil<' : i'aiiel load for one chord of upper lateral = 0.75 tons, to he taken as a dead load. I'anel load for both ehords of lower lateral = 1.5 tons + .'{.0 tons ; tho former to be taken as a dead load, the latter as a live loa?^; all acting in tho direction of the arro vs. Stress in b'c = - 9 X .75 X 1.414 = - 9.5 tons. Stress in CD = - [3.5 X 1.5 4- -rSj(l + 2 + ■■• + 9)] 1.414 = - 23.3 tons. Stress in B'C = 10 x 4.5 x 1 = 45 tons. Upper Lateral System. Stresses in chor.ls = G.S, 12.0, 15.8, 18.0, 18.8 tons. Stresses in struts = 0.4, G.O, 4.5, 3.0, 1.5, 0.8 tons. Stresses in diagonals = _ 9.5, _ 7.4, - 5.3, - 3.2, - 1.1 tons. LowKK Lateral System. Stresses in chords = 24.8, 45.0, 60.8, 72.0. 78.8, 81.0 tons. Stresses in struts = 2..3, 20.5, 1G.5, 12.8, 9.2, 6.0, 3.8 tons. Stresses in diagonals = _ 34,9, _ 28.9, - 23.3, - 18.0, - 13.0, - 8,5 tons. ii iiIk'o L'O fVct r anil lover ;ses ill l)(>t)i as ill tlio and 48 will e vippe- ami ).75 tons, to cr as a live - 23.3 tons. tons. ;on8. I, 81.0 tons. 5.0, 3.8 tons. ■ 8.5 tons. hltlDGE riiVssKs. \V I'rttli. 96. A ilock Pratt truss railroad briilj,'*- -<> fi't't wido has 10 panels, each i'O feet long ; the upixT and lower lateral systems are Pratt trusses: 'ind the stresses in both lateral systems dno to a wind pressnre of 40 Ihs. |.er stpiare foot of expo.sed surface of lioth trusses and 3(M) Ihs. i)er linear foot of train surface. Hy taking only the fidl diagonals, Fig. 4('» will represent the left half of the deck Pratt truss and also the upper and lower lateral systems. Since this is a deck bridge, tlie upper Iriterals are on the loaded, and the lower laterals are on the unloaded chord. Hence by the rule: I'anel load for one eliord on lower lateral system 20()0 ' 1 ton. to be taken as a dead load at each apex of the windward and leeward chords (Rem.). Panel load for both chords on upper lateral system 200 X 20 , 300 X 20 „ , , ., , = "2000- + ~-2mr = ^ ^''"^ + '' ^""« ' the former to be taken as a dead load, the latter as a live load, both acting at each apex of the >vitidward chord (Item.), and all acting in the direction of the airows in Fig. 40. Uri'KU Lateral System. Stresses in chords = - 22.5, - 40.0, - 52.5, - GO.O, - 62.5 tons. Stresses in struts = - 25.0, - 22.6, - 17.8, - 13.4, - 9.3, - 5.0 tons. Stresses in diagonals = + 31.7, +26.1, +18.9, + 13.1, +7.7 tons. cr 118 HOOFS AM) niilDOES. LoWKH liATUKAL SvHTKM. Stresses in chonls = -9.0, -10.0, -21.0, -24.0, -25.0 tons. Stri's.st's in struts = - 9.5, - HA), - CO, - 4.0, - 2.0, - 1.0 tons. Stresses in (Ha^jonals = + 12.7, + 1»-1». + 7.1, + 4.2, 4- 1.4 tons. Prof). 97. A throiigli I'ratt tni.ss railroad bridge IfiJ fett wido lias paticl.s, cacli 17 f»'ot long; the upper and lower lateral systems are Tratt trusses: find the stresses in both lateral systems due to the same wind pressures as in l»rob. 94. Art. 38. The Factor of Safety for a body is tlie ratio of its breaking to its wDikiiij; stress; or it is tlu' ratio of the load which will just erusli the body to the assumed load, or load whieh it is intended to earry. Tiuis, if the tearing unit-stress of an iron jilate be 20 tons and the work- ing luiit-stress be 4 tons, the faetor of safety will be 5. The value of the factor of safety is generally assumed by the engineer; different engineers assume different factors of safety, depending somewhat upon the manner of applying the loads, the character of the structure, and the nature and quality of tlie material. Thus, for steady loads and slowly varying stresses, tin factor of safety may be low; but when the load is applied with shocks and sudden stresses, tlie faetor ought to be large. In a building the stresses on the walls are steady, and hence the factor of safety may be low. In a l)ridge the stresses on the different members are more or less varying, and hence the factor of safety must be higher. ma. tons. Ikc ir.j f.'ft r and litwrr jses in both iures us in is ilic ratio the ratio of ho asHiinu'd rinis, if th(! il the work- 1 be .'). assumed by it factors of of apply in;^ nature and and slowly ; but when tresses, the ?sses on the nay be low. rs are more ty must be nniDOE TRUSSES. ll'J Also, it has been seen (Art. 1(5) that the live load for a short span is mueh more tiian that for a loii^' span. For thi.s r(>ason the variation of stiess in pa.ssin;.; froni a loaded to an unloaded state is iiiuth j,'reater in the mend)ers of a short span than in those of a long one. Conseciuently, the material in a short span .vill suffer what is termed fathjue more than in a hmg oiu'. And although the fatigue of metals is a subject not yet well i;nderstans. In American practieo the values of the factor of safety, for steiuly and for varying stresses, are: from 3 to 7 f<»r wrought iroti, from rt to lo for cast iron, from 4 to S for steel, from S to 12 for tind)er, and from lli to iiO for stono or brick. At times it may be necessary to enploy a much larger factor of safety than either of these, owing to local circum- stances. If the risk attending failure (such as loss of life or property) is small, the factor of safety may be sm-iU. But if the risk is large, the factor of safety must be large in proportion. With a bridge in perfect conditiim, a very small factor would be sufficient. Hut no bridge is in per- fect condition; all rough places, such as rail joints, more or less open, produce shocks which cause sinlden stresses. These stresses cannot be measured. A very large allowance has to be made for these uncertainties and for the imper- fect state of nr knowledge ; and therefore there must be a large factor of safety to cover all uncertainties. if CHAPTKK HI. imiDcjK TitUKSKH wrrii unkqual distuiuution OF THE LOADS. Art. 39. Preliminary Statemonfc — Tlio preceding <'ha|)tor hiis troatod (jf HtroHHos iirodiiccd by dead loads and uniformly distrihntt'd live loads. While this is the general method of treatment for hiijhwaij bridges, and in English practice for railway bridges, it has become the general jinictico "f American engineers to ealcnlate these stresses lor railway bridges by one of the following three methods: (1) The vi.se of a uuiformJii (listi'ihiUed excasH had covering one or more panels ft)llowed by a uniform train load cover- ing the whole span. (2) The use of one or two commtnUed excess loads with a uniform train load covering the span. (3) The use of the actutd specijled locomotive tcheel loads followed by a uniform train load. It is proposed in this chapter to .show how to find the maximum stress in each member of a truss by each of these three methods. Art. 40. Method of Calculating Stresses when the Uniform Train Load is preceded by one or more Heavy Excess Panel Loads. — This method of linding maximum stresses is sometimes used to avoid the laborious practice of finding the stresses due to the actual locomotive wheel loads to be described later. '»^r^^mx^^^mmmu^m!ii^^£. niilDCE TItUSSKS. \\>\ JUTION preceding loads and :.hi' general in Knglish lio general 80 strcHyes niethoils : id covering oad cover- loada with vheel luadu find the !h of these 68 when r one or method of avoid the the actual rrt)h. 98. A through double Warren truss, •''ig. 41), lias Id |Kinels, eaeh VI feet long iiud \2 fe<*t deep; the deml load is KMM) lbs. per foot per truss, and the train load is 1' »'/ 1 .» r r U u y • ' ,;' .r x> \/ /\ \/ )] 1.414 = 67.7 tons. Min. stress in 7-10 = [l x()-30x I?,] 1.414 = 0.0. That is, the dead load passing through 7-10 to the left abutui.ut is neutralized l)y the part of the locomotive load passing through 7-10 to the right abutment. The following .stresses are found in a manner similar to the above: UiTER Chord Stresses. Mkmbkkk. 1-8 8-5 M 7-9 a-ii Max. ... -61.2 -16.0 -133.2 - 39.0 -183.0 - 67.0 -210.0 - 00.0 -2.34.0 - 76.0 Min Lower ( SiioRD Stresses Meubirs, 2-4 ■i-i) fi-8 s-io 10-12 Max .Mill + 60.4 + 12.0 + 122.4 ■; 30.0 + 176.4 + 64.0 + 208.8 + 60.0 + 226.8 + 72.0 Diagonal Stresses. Membibs. i-i «-6 .vs 7-10 9-12 11-10' Max Min + 86.5 + 21.2 + 71.3 + 17.0 + 57.7 + 8.5 +44.1 + 0.0 + 32.2 -10.3 + 20.3 -20.3 Miuc. stress iu 1-2 = - 61.2 tons. •■s»FBai^«i»E,:' loiitl alone. 10 live loa lbs. per foot per truss, and the train load is 1000 lbs. per foot per truss preceded by two locomotive i)anel loads i.f ;!(> tons each per truss: find the stresses in all the members. Consider three fifths of the dead load as ai>plied at the upper chord, and two fifths at the lower chord.* Dead panel load ])er truss = 5 tons. Train panel load per truss = 10 tons. Excess panel load per truss = 20 tons. For maximum chord stresses put locomotive panel loads at joints 3 and 5, and the train panel loads at all the other joints. Thus, Max. stress in 1-3 = 4^ X 15 + 20 (-j\ -f- ^%) = 101.5 tons .-= stress in 4-6. Max. stress in 3-5 Max. stress in 5-7 = 10.V X 15 + f^ (17 -I- 7 - 3) = 199.5 tons. For ma.ximum stress in any diagonal the live load is placed on the right, by the usual method. * In railroad bridges it is customary to take two tliird.s of the dead load as applied at the loaded ihord ; that is, the chord which carries the live load, aud cne third at the uuloadeJ chord. 11' 124 PGOFS A XI) nniDGES. CflonD Strkssbs, .MllMIIKKH. 1 t) ;■-*) .'.-7 ■ -» 9-11 Max Mill -101.5 - 22.6 - MJ8.0 - 40.0 - iim.5 - 52.5 -210.0 - 00.0 -217.6 - 02.5 StRK'.SES 1" THE DiAGON.VI.S. I! ^ li 'i I' i' 11 Mrviiikhh. 1-4 » l> .) !* ■ 10 i»-12 8-9 10-1 1 Max.. . Miu. . . + 14:!.5 f ;!i.7 + 118.0 + 20.4 + 04.0 + 4.0 + 71..'J 0.0 + 60.2 0.0 + 12.1 0.0 + 30.4 0.0 Max. compression in the i)().st.)= 102.0, 00.6, 81.5, 04.5, 48.6, iJfl.O tons. Proh. 100. A through Howe truss, Fig. 32, has 10 panels, oach 12 feet lonj; and 12 feet deep; the dead lojid is 1000 Ihs. per foot per truss, and the train load is 16(i7 lbs. per foot per truss, preceded by two locomotive panel loads of 30 tons each per truss : lind the stresses in all the members. Dead panel load = 6 tons. Train panel load — I'Vtons. Excess panel load - :'V \>os. Consider one third of the dead ii d ^ applied at the upper chord. CiioRi) Stresses. Mkmhkrh. 2-4 4-8 fi-S 8-10 10-12 Max Min H 100.0 + 27.0 + 170.0 + 48.0 + 210.0 + (!:{.0 + 228.0 + 72.0 + 2.'10.0 + 75.0 9-11 1.0 ).0 -217.6 - 02.6 8-9 IC-1 1 12.1 0.0 + .'50.4 0.0 48.5, 30.0 tons. IS 10 panels, lojod is 1000 I6(i7 lbs. per nel loads of lie members. plied at the 10-12 8.0 2.0 + 230.0 + 76.0 ItltllKiE TRUSSES. I2r, Strkh»kh IN TIIK DlAOONALS. Mkmiikiu. .Max. . . Mill. . . 2-3 4-5 «-7 &-9 1(1-11 7-10 « 12 -140.0 - 38.0 -123.0 - 25.6 -97.6 - 8.5 -73.0 0.0 -50.9 0.0 -9.0 0.0 -29.7 0.0 Stkkssks in the Verticai.b. Mkmiikbk. Mill. Mill. !?-« + 104.0 + 25.0 6-8 + 85.0 + i0.0 +07.0 + 4.0 !)-10 + 60.0 + 4.0 II +37.0 + 4.0 I*)-oh. 101. A through Pratt truss, Fig. 33, has 10 panels, ea'-h 15 feet long and In feet deep ; the dead load is 1200 lbs. per foot per truss, and the train load is 2000 lbs. per foot per truss, preceded by two locomotive panel loads of 30 tons each per truss : find the stresses in all the members. (Consider one third of the dead load as applied at the upper chord. Chord Stresses. Mkmiikks. 2-4 4-6 c-s 8-10 10-12 g-n Max. . . Mill. . . + 133.5 + 40.6 + 133.5 + 40.6 +228.0 + 72.0 +28.3.6 + 94.5 + 316.0 + 108.0 -322.6 -112.6 Stresses in THE Dl AOONALS. Mrmiirks. 2-'l !i-tt 5-S 7-10 »-!2 8-9 10-11 +29,7 0.0 Max Min -188.8 - 57.3 + 162.7 + 40.3 + 118.7 + 19.1 +87.0 0.0 + 67.3 0.0 + 4.2 0.0 : ; \ r " -? • Ii ii.f »■ ii 126 RooF.s AND niiiuaKs. Stkkssks in tiik Vkiiticals. Mkuiikks. 3-1 .vo 7-S 9-11) 11-12 -28.5 - 3.0 Max Mill + 30.0 + 0.0 -87.0 -10.5 -«4.6 - 3.0 -43.5 - 3.0 Proh. 102. A through Whijiple truss, Fig. 37, has VJ panels, each 12 feet long and 24 feet deep; the dead load per foot ])er truss is 10(M) lbs., and tht; train load is2()()() lbs. per foot per truss, preeeded by one locomotive panel load of 30 tons per truss : find the stresses in all the members. For max. chord stresses put locomotive panel load at joint 4 and the train panel loads at uU the other joints. * Dead panel load = 6 tons. Train panel load = 12 tons. Excess panel load = 18 tons. Max. stress in 4-6 = (3 x 18 -f- 1 J x 18 ) x .5 = 35.3 tons. Max. stress in G-8 = 35.3 + 2| x 18 = 80.3 tons. Max. stress in 8-10 = 80.3 + 2 x 18 - xV X 18 = 114.8 tons. Chord Stresses. MUMBIIBfl. 2-4 4-0 6-S 8-10 10-1 2 12-14 9-11 Max. . . . Min. . . . 0.0 0.0 + 35.3 + 9.0 + 80.3 + 24.C + 111.8 + 30.0 + 141.8 + 45.0 + 158.3 + 51.0 -107.3 - 54.0 * Put all tlie dt-ad loftd un the loaded (^liord, unless otberwiso stated. 11-12 -28.5 - 3.0 37, has IL' 10 (load load 1 i.s 2000 Iks. lanel load of embers, load at joint ts.* ? tons. :18 -14 9-11 .58.:! 51.0 -167..S - 54.0 vf\nv stated. n RIDGE TinrssKs. STRESSBS in TIIK l)|Aflfl.V.\LS. 127 Mkmhkks. i-i i-fi 3-S .'-10 T-l'.' !l-ll 11-12' 13-10 ll-s Max. . . Mill. . . + 78.8 + 20.1 + 84.8 + 21.2 + 71.^ + l.V I +.-)0.0 l:+ 5.7 1 +40.0 o.r +3;!.o 0.0 + 2.3..3 0.0 + 12.7 0.0 +.3.5 0.(1 Sthkssk.s in TIIK Vehticai.s. MKMHF.ItH. 1-2 l-J .Vfi "-S '•-'" 11-12 i;i-u Max. . . Min. . . -1!.^6 - .S.'J.O -50.5 - 0.5 -41.0 - 4.0 -.32.5 0.0 -24.0 0.0 -10.6 0.0 -9.0 0.0 Pmb. 103. A through Whipple truss, Fig. 38, has IC. panels, each 10 feet long and L'O feet deep; the dead and train loads are 1000 lbs. and 3000 lbs. per foot jier truss, the train being preceded by one locomotive i»auci . -id of 30 tons per truss : find *;lie stresses in all the members. ArL 41. Method of Calculating Stresses when one Concentrated Excess Load accompanies a Uniform Train Load. — Tliis method is sonu'timt'S iisi'd, like tlie one ill Art. 40, to avoid the jiractice of finding the stresses due to the locomotive wlieel loads. IWit we h.ave seen tliat the method in Aru. 40 does not give strictly the maximum stresses in all the chord members. Let Fig. 51 be a truss sujiporting a nniform train load covering the span, and a concentrated excess load P. We snppose the excess load /' to be the ditference between tlie locomotive ])anel load and the uniform train ]tanel load as in Art. 40. Then (lie stress in any chord member, us lid, caused by the concentiated loud /*, is eipial to the tending moment at c, the renter of moments for hil, divided by the lever arm for the chord ; and hence the chord stress will be H £;^ 128 nnoFs AMt itiaixiEs. a iiiaxiinuiii wlu-ii tlic roiicviitratfil loiul /' is ho phwed as to uviiko tlin ImmhUh.u; immitMit a iiiaxiiimm. >Jiiw, lor ii single concentrated load, the niaxiiiiinii liendiiig inoniciit at any point occurs when the load is at that poii.L, for, if tlm load be niovtMl to either side of the point, tlie reaction of the opposite abutment will be diminished, and hence the mo- ment will be diminished. Therefore, for a concentrated excess load and a uniform train load, the maxiniuni hendinij moment at amj point, and consp(juentl;i the maximum chord stress in anij member, occurs when the concuitrated loud is at the center of moments for that member, or at the vertical section thromjh the center of moments, and the uniform train lo id covnj the tchole span (Art. 22). Thus, for the maximum stres > in bd, the concentrated load P is at c, the center of mom ants for bd; and so for any other panel in the lower chord, or unloaded chord j,'t'nerally. For any panel in the upper, or loaded chord, as he, of a like the Warren, the concentrated load /* acts at o, oi at that end of tiie panel which is to the right of the ver- tical section through 6, the center of moments for he, while the uniform train load covers the whole span (Art. 22). For any panel in the loaded chord of a truss like the I'rait or //owe, where the apexes of the upper chord ■"■" vertically above thos(! of the lower chord, the concent n. ted load /' is at the apex directly over or under the cente- of mo- ments for that member. ) pliwi'd iia to ', for ii siiit,'le inu'tit ill any ^ )!*, if l\w load [letioii of the eiice the luo- nd a nuifunn Mij point, (ind uember, (tcenra mo me II tH for \ the. venter of he tvhvle sjiun concentrated uid so for any ord j^cncrally. d, as //(■, of a /* acts at 0, lit of the ver- s for he, while Art. L'li). For 'i the Pralt or "•■'1 vertically onlr:.tcd load cento.- of uio- BllIDGE TRUSSES. 129 For a single concentrated load, the niaxinuiin positive shear at any section will occur when the load is just to the right of the section ; for tin- left reaction is then a niaxiniuiii. Thi'ri'fon', for a cniwcntrnlcil ovcs.s Imul and a vniform train loud, tlw maxininni ttlirss in iini/ lirair. occnrn irheii the cnnvcntriited hail in at thi' jhiih'I point iinniediatelif on the riijht of the sertion, anil thi- uniform train load covers the xjian from the riijht ahxtment to this samr jiani'l point. Thus, the greatest positive shear, and therefore the niaxinnun stress, in be or in hh, occurs when the con- centrated loiul is at the apex c and the uniform train load extends from this apex to the right abutment. Tlie greatest negative shear for he or for hh would occur when the concentrated load is at tlie apex h and the uni- form train load reaches from this apex to the left aljutment. For maximum chord stresses, cars both precede and follow the locomotive. This does not often happen. For maxi- mum stresses in the braces the locomotive precedes the cars. It follows therefore, that maximum chord stresses are of less frequent occurrence than maximum web stresses, which occur for every passage of the train. Proh. 104. A through Howe truss, Fig. 32, has 10 panels, each 12 feet long and 12 feet deep; the dead ami train loads are 1000 lbs. and 2000 lbs. per foot per truss, and the loconu)tive panel load is 30 tons per truss: find the stresses in al' the members. Consider one third of the dead load as applied at the upper chord. L*'ead panel load = G tons. Train panel load =12 tons. Excess panel load = 18 tons. ii I I i;jo HOOFS .\M> niniKiKs. i I For the niivxinmin strfss in any cluml ni.'nibor, fuj 0-8, tin- fxcfss load is iiliK't'd at the joint H and the tniiu load covi'is the wliole span. Then, (U) of Alt. 19, Max. stress in O-H = lO.r, X 18 + T^j X 18 X 3 = 220.8 tons. Otherwise by moments, (1) of Art. 11), thns: Left leaetion = 4.J x 18 + ^'^ x 18 = '.Ki.O tons. The equation of moments about the point 7 is ".KIO x ;<0 - 18(12 + ^4) - stress in 0-8 x 12 = 0. .-. max. stress in (]-H = f),'}.0 X ;{ - 18 X 3 = 220.8 t feet dci'i); the dead and train loads art' IL'OO ilis. uiid L'lMK) His. per foot per truss, and the excess loud is 1*0 tons per tri;«s : tuid tiie stresses in ail tiie inenihers. Consider one third of tlie dead load as applied at the upper chord. Dead jiane) load = 9 tons. Train jianel load = 15 ions. Excess panel load = 20 to:is. Max. stress in 8-10 = lOJ x 24 -f- f^ X -0 X 3 = 294 tons. Max. stress in 3-G = m X 9 + U X 36 + f J X 8] 1.414 = 143.0 tons. Max. stress in 7-8 = - [1| X 9 + 3 -f 1.5 X 21 -I- 2 X 6] = - 60 tons. Chord Stresses. Mbhbkiu. 8-» *s 6-S 8-10 10-12 9-11 Max. . . Min. . , + 120.0 + 40.6 + 120.0 + 40.6 + 224.0 + 72.0 + 294.0 + 94.5 + 330.0 + 108.0 -350.0 -112,5 Stressks IN THE Diagonals. Mkmhkhh, Max. . Mill. . 2-) c-c 5-8 7-10 9-12 t-9 10-11 -178.2 - 40.5 + 14.3.5 + 39.0 + 111.0 + 19.8 + 80.0 0.0 +62,3 0.0 + 2.1 0.0 +20.2 0.0 *4 t ■ym i^ ■««p • i! Vi2 HOOFS AS I) nil I DUES. t SrilBSHKH IN TIIK VERTirALi. Mkmukim. S-t 6-6 7-S W-IO ll-ii Max. .Mill. 41.0 + -81.6 -fllt.O - 40,0 - 3.0 -20.0 -3.0 -17.0 - 3.0 I'nih. 106. A double Warren truss, Fig. 49, used as a dock bridge, lias 10 pauels, each 14 feet loug and 11 feet deep; the dead and train loads are 1000 lbs. and I'OOO lbs. per foot per truss, and tiie excess load is 20 tons per truss: liud the stresses in all the members. L'pPKii C'litniit Strk»se.«. .Mkuhkkk. 13 it-.'i -154.0 - 42.0 .V * M» '.MI Max -00.0 -14.0 -213.0 - O.i.O -2.') 1.0 - 77.0 -208.0 - 84.0 Mill Lower Chord St KES.<ns. nitihcK ritrssEH. 13:{ rri>h. 107. A ilf.k Trutt truss. Fig. W), has 10 paiu'ls, caili l.-» tVct loiiK and IT* feet di-fp; tlie d.'ad and train loads lire SOO ll»s. and L'OOO llis. jicr foot jit-r truss, ami the oxccss load is L'(» tons per truss: tind the stresses in all the inend)ers. Art. 42. Method of Calculating Stresses when two equal Concentrated Excess Loads, placed about 50 feet apart, accompany a Uniform Train Load. — This method is often used, like the two in .\rts. In ami 41. to avoid the praetiet! of iinding the stresses due to the hx'O- motive wheel loads, and is a nearer approximation to the aetual loads than either of the i)ther two. i^et Fig. *■»- lie a truss siipporting a uniform tr.ain load eovering the s])an, and two equ.al eoneeiitrated exeess loails, (^„A— ^fi Vitf. Oid /', and Pj. As in Arts. 40 and 41, we suppose ea^h excess load to he the ditference hetween the loeomotive jianel load and the iniiforni train panel lo.ad. Then the stress in any ehord nieud)er, as bd, eaused by the concentrateil loads J', and I\ is eciual to the bending moment at <; the center of moments for M, divided by the lever arm of bd ; and hence the chord stress will be a maxinnim when the two con- centrated lo.ads are so jHaeed as to make the bending moment a maximum. Now, for two ecpial loiids, a tixed distance apart, the maximum moment at any point occurs when one load is at i, » r : [' H ?': ; 184 HOOFS ASH nun (iki.fi. tlit> |i(iiiit and tli(> (>tlit Iuiik'i'I' M>g;nont uf tlui tniHS. 'I'liiH may 1m' shown iw follows: Ltit /= iiMi^'lli t)l' Hpan in Fi),'. ol', (pial concentrriteil excess loiuh, and a nni/orm tntiii loin\ wlinr the upper apexes are vertiialh above file ecirrespondin); lower ones, the concenlrated loail /', is at the apex directly over or under the eenter of uiouieuts for that MUMuher. l<\)r two e section and /', is as near to it as possible; for the left reaction is then a nuuimum. Therefore, far two eqiinl concent rated exeen» hmla am\ a uniform train load, the maximum atreas in any Imire on-nrx tehen both loads are on the riijht of the sertion anil one of them is at the panel point immediidelij on the riijht of the sertion, and the uniform train load covers the sjnin from this point lu the riijht abutment. The second excess load P^ sliould be placed at a panel pt.int at an interval of about at) ft'ct to the right of /'„ to simplify the computation. Thus, should the panel length be IL', ir», or 18 feet, the distance between the loads would be 48, 45, or 54 feet. Prob. 108. A through Warren truss, '^ig. 53, has 8 panels, each 10 feet long and 10 feet deep, its web members all 7' jt' 1' H 10 * 4' '«' forming isosceles triangles; the dead ami train loads are 1(MM( lbs. aiul 2000 lbs. uer foot i)er truss, and there are two m 136 uo()F< Axn liUihCKs. pxcoss loads 50 fopt apart, each of 3.'] tons : liud tlie stresses in all the inenihers. Dead panel load = T) tons. Train panel load = 10 tuns. Each excess load — 'M tons. tanfl=: \] seed? = 1.117. For the maxinuini stress in any chord menil er, as 4-(), one excess load is placed at (> and the other one at 4', or .10 feet to llie ri'^dit of llie first, and tiie dead and train loads cover the whole span. Then I'J) of Art. li), Max. stress in 4-G ^ [0.\ x 1") + Y 0- + 0)3] x .5 =:114.() tons. Otlierwise by nionuMits, (1) of Art. 10, as follows: Left reaction = 3,^ x 1") + 33 (| + S) =-■ 81.375 tons. The equation of moments abont the point 3 is 81.375 X 15 - 15 x 5 — stress in 4-0 y 10 = 0. .-. max. stress in 4-G = 114.0 tons, as before. For the maximum stress in any dia;,'onal, as 3-0, one excess load is placed at <> and the otlier 50 feet to the right of it at 4', and the train loads cover the span fronx G to the right abutment. Tiien, Max. stress in 3-G = [2. J x 5 + V (1 4- 2 + - + G) 4- y (G 4- 1) ] 1-117 = 75.5 tons = - stress in 3-4. Min. stress in 5-6 = - [1| X 5 - f X 10 - | x 33] 1.117. = 5.0 tons. I'i'PEU CiioiiD Stkesses. hT MrMIIEKS. 1-$ i]~'i 5-7 7-'t Max. -80.« -17.5 -147.8 - :}o.O -174,4 - :J7.5 -180.0 - 40.0 Min . il tlie stresses 111 or, us 4-0, 10 at 4', or HO 1(1 train loads ] X.5 lows : '5 tons. is L0 = 0. ofore. , as w-C), one t to the riglit from to the ...+C) in 3-4. X 33] 1.117. 5-7 7-1 14.4 37.5 -180.0 - 40.0 mU^ E TRUSSES. LowKH CiioHD Stresses. Diagonal Stkesses. 13? Memhkkx. '2-1 4-6 C-4 f-U) Max +44.8 + 8.8 + 114.0 + 23.8 + 152.8 + 33.8 + 174.0 + 38.8 Mill .M KM UKHfl. 1-2 1-1 8-0 v-s T-io Max. Mill. -100.1 - 19.5 + 100.1 + 10.5 + 75.5 + 8.0 +52.4 - 5.0 + 35.2 -19.4 Prob. 109. A deck Pratt truss, Fig. 50, has 10 panels, each 12 feet long and 12 feet deep; the dea«l and train loads are 1000 ibs. and 2000 lbs. per foot per truss, and then^ are two excess loads 48 feet apart, each of 30 tons : find the stresses in all the members. Cuonn Stresses. y K.MHKKS. l-i .?-") f.-7 7-9 9-U Max. . . . .Mill -123.0 - 27.0 -210.0 - 48.0 -270.0 - 03.0 -312.0 - 72.0 -315.0 - 75.0 RriiESSEs IN Diagonals. Mkmiik)- Max. Mill. 1-1 + 17.^0 + .38.0 + 141.7 + 2.'!. 8 f-S + 111.1 + 7.0 7-10 + 82.3 0.0 »-li s-9 + .16.1 0.0 + 7.2 0.0 lo-ll + 20.7 0.0 fT 188 ItOOFS AND lililDGES. Sthessea in Vekticals. McMltEIM. l-i n-i 6-15 7-8 »-l« U-12 Max Mill -138.0 - 30.0 -123.0 - 27.0 -100.2 - 16.8 -78.0 - 0.0 -58.2 - 0.0 -48.0 - 0.0 Prob. 110. A deck Whipple truss, Y\<^. 37, has 12 panels, each 12 foi't long and 24 feet (i('('i>; the (load and train loads are 10(M> lbs. and 3000 lbs. per foot ]»er truss, and there are two excess loiuls 48 feet apart, each of 30 tons : find the stresses in all the members. Dead panel load = 6 tons. Train panel load = 18 tons. Each excess load = 30 tons. tan d = 0.5; tane' = l; sec w = 1.117; sec 6' = 1.414. Max. stress in 4-6 = [3 x 24 + f§ (11 + 7) J x. 5 =58.6 tons. Max. stress in 6-8=4x24 + 2.5(10+0)1^ = 150.0 tons. Max. stress in 3-8 r=[2x6 + 1.5x 25+2.5x14] 1.414=119.5 tons. Min. stress in 5-10=[9-1.5x2-2.5x2] 1.414=1.4 tons. Chord Stresses. ft, i Mp.MBKIIli. 'i-\ 4-0 c-> t-lO ll!-l'i I'J-14 »-il 1 1 n Max. . Mil). . ^no.o 0.0 + .'•.8.6 + 0.0 + 166.0 + 24.0 +211.6 + 30.0 +286.0 + 46.0 + 316.6 + 61.0 -326.0 - 54.0 SA^ rnKm. 9-10 11-12 -58.2 -48.0 - 0.0 - 0.0 as 12 panels, id train loads uid there are )ns: t ind the MM W^j 6' = 1.414. •=58.6 tons. 15G.0 tons. tons. H4 = i.4 tons. I '.'-14 K-ll -310.6 - 61.0 -320.0 - 54.0 niiWGK TRUSSES. Stiiksses in Diaoonals. 139 .Mkmuekh. I-l 1-8 i!-l .vio 7-1-2 9-14 11-1-2' 13-10 ll--< Max MIn + 130.7 + -20. 1 + 141.4 + •l\.i + 119..'! + 11. !i +97.a + 1.4 +77.S 0.0 +.\s.n 0.(1 +40.8 o.u +W.6 0.0 + l(l.ti 0.0 Stresses i.n Verticals. Mkmiikks. 1-2 3-1 .V3 7-S 9-10 11-1-2 lH-14 Max. . . Min. . . -170.0 - 30.0 -117.0 - 18.0 -100.0 - 15.0 -84.6 - 8.0 -00.0 - 0.0 -56.0 ~- 0.0 -64.0 - 0.0 Pmh. 111. A through parabolic bowstring truss, Fig. 44, has 8 jxuicl.s, eacii 10 feet long, and 10 feet center depth; the verticals are ties, and the diagonals are struts; the dead and train loadj are 1000 lbs. and 3000 lbs. per foot per truss, and there are two excess loads .50 feet apart, each of 30 tons : tiud the ! tresses in all the ineinbers. Dead panel load = 5 tons. Train panel load = 15 tons. Each excess load = 30 tons. Length of 3-4 =: 4.375 feet ; length of 5-0 = 7.5 feet. Length of 7-8 = 9.375 feet; length of 0-10 = 10.0 feet. Then, for the inaxiinuin chord stresses, a full dead and train panel load must be at each lower apex. For any ineinber of the lower chord, as L*-4, one excess load of 30 tons is placed at 4 and the other at (V, or 50 feet to the right of the fir.it, and the dead and train loads cover the whole span, as just stated. Here we cannot use the "method of chord iiici-eiuents," since the chords are not j)arallel, but must use the "method of moments," (1) of .Vrt. I'J. if ; m 140 ROOFS AND nillDGES. m Reju'tion at the left end = 3^ x 20 + 30 ( J + g) - 103.75 tons. Mux. stie.ss in 2-4 = ^^^'^•"'"' ^ '*' = 237.1 tons. 4.375 This tensile stress in 2-4 is eijual to the horizontal com- ponent of the eoiapressive stress in 2-t3, by (1) of Art. 5. Therefore the stress in 2-3 is ecjual to the stress in 2-4 nuiltii)lie(l by the .secant of the angle between 2-^5 and 2-4. Thns, Max. stress in 2-3 = - 237.1 x 1.092 ^ - 25S.9 tons. For the maximuni stress in 3-5, one excess load is pnt at and the other at 4', while the deail and train loads (iover the whole span ; then take center of moments at intersection of 4-5 and 4-() or at 4. And so on for the stresses in 5-7 and 7-',). The diagonal .stresses are fonnd by jmtting only the train and excess loads on the truss in the proper position for each mend)er, since the dead load jiriMluces no stress in the diaj;?- oiials (Art. 34). Thius, to find the nmximiim stress in 4-5, one excess load is placed at and the other 50 feet to the right of it at 4', and the train loads cover the span from to the right abutment. Now cut 3-5, 4-5, and 4-6, and take center of moments at the intersection of 3-5 and 4-0, which is 4 feet to the left of 2; the diagonal 3-0 is not in aetion for this loading. Lever arm of 4-5 = 14 sin = 14 x -^p = 8.4 feet. Left reaction for this loading = ^•'* (1 + 2 + •■•+(;) -f ■ V (G +- 1) = 05.025 tons. s '. .Mkmiikkh. 4-.') C-T s-i» 3-0 '.-t 7-111 Max Mill -31.3 0.0 -34.2 0.0 -38.2 0.0 -40.0 0.0 -43.75 0.0 -41.1 0.0 Max. stresses in the verticals = -|- 50.0 tons. Min. stresses in the verticals = -|- 5 tons. Prah. 112. \ throut,di circular bowstring truss, Fig. 45, has panels, each 15 feet long, its web members all form- ing isosceles triangles, with the upper apexes on the cir- cumference of a circle whose radius is 75 feet, the center depth of the truss being 14| feet; the dead and train loads arc 720 lbs. and 217«) lbs. per foot per triiss, and there are two excess loads 45 feet apart, each of 24 tons: fiud the stresses in all the members. Dead panel load = 5.4 tons. Train pa«el load = 1G.32 tons. Kach excess load — 24 tons. if ; ■MM BMM 142 HOOFS AM) li lit DUES. Ciioiti) Stresses. . Mkmbeks. 2-8 8-6 5-7 7-T 2 4 4 6 6-8 Max. . . . Mill. . . . -140.0 - 24.7 - 103.0 - 27.2 - 140.2 - 25.0 -137.1 - 25.0 + 125.0 + 20.0 + 130.8 + 22.0 + 125.4 + 23.7 Wur. STIIE8.SES. Mkmiieks. ii-4 & « "-■> 7'-«' 5' 4' Max Mir + 20.0 + 4.9 + 33.4 -11.0 + 34.8 -15.1 + 37.2 -13.0 +43.2 - 7.1 J'rob. 113. A tlirout^h Vratt truss, ¥\g. 33, has 10 panels, eatih 12 feet lon^ and 12 feet deep; tlie dead and train loads ar« 1000 lbs. and 3000 lbs. j>er foot per truss, and there are two excess loads 4S feet apart, each of 30 tons: find the stresses in all the uiendjers. Art. 13. The Baltimore Truss. — This truss. Fig. 54, or some modification of it, is now used very generally for long si)ans when it is desired to avoid lont; panel lengths. Each large pane' is divided into two smaller ones by insert- Vig. 04 ing half ties and subvertical struts, or half stmts and sub- vertical ties, according as it is a deck or a tiirough truss. In the deck form, Fig. 54, the s»il)verticals ai, hj, ck, etc., are strained only by the panel loads which they directly 4 6 6-8 + 130.8 + 22.0 + 125.4 + 23.7 -«' 5' 4' V.2 13.0 +43.2 - 7.1 hivs 10 pan el .s, ,iul train loads and there are :ous: find the triLss, Fig. 54, generally for [)anel leni,'tli,s. jnes hv insert- * t' h l' / \ / M 1 V" trnts and sub- Ihrnngli tr'.iKs. ai, hj, ck, etc., they directly noOF TllUSSES. 113 siipjinrt; the dotted diagonals 4k, CI, G'm, 4'n are counter.s, not licing in acition foj; full load. Proh. 114. .\ deck Haltiniore truss, Fig. /i4, ha.s 10 panels, each 10 feet long and 20 feet deep; all the verticals are po.sts, an ve 'iVS tons at or of uioiiit'iits IS. X *-'0 = 0. ch subvertical, id, 4 + 8 + 30 \, bj, (k, etc., is h' tie J ».'3, j5, '. the half ties 4 =29.7 tons.' any main tie, (• and at 7', r»0 are at all the k for this loud- 1)= 78.875. 1.5 tons. 3al, as 5-4, is greatest shear Hter!>. Thus, take ill loads at all the the sireate.st slioar . As /-8 is not in t uf the stres!> iu ituns. in 5-1 occurs wlien the excess loads act at 5 and at c, fiO fict l<> tlie ri^,'ht of 5, and the train loads are at all the apexes from the ri^'ht abutment to 5 inclusive. Then the greatest shear in 5-4 is that which is due to this loading, together with the four dead loads at 5, c, 7, and (/, half of tlu! dead load at '.), and half of the dead load at b, which is transmitted to 5 through the half tie j5. Hence, Max. stress in 5-4 = -[5 X 4 + fo(l- + ")+ A(12 + - + 1)] = - 94.0 tons. The maximnm stress in the lower part of any main tie, as ./I, is equal to the maximum stress in 4-5 intdtiplied by the secant of the angle which the tie makes with the vertical. Hence, Max. stress in > 4 = 94.6 X 1.414 = 133.8 tons. Tlie maximum stress in 6-Z occurs when the excess loads are at a and 7, and the train loads are at all the joints from the left abutment to 7 inclusive. Then the shear in the section cutting rf-9, /-9, and Z-8 is — 21.G25 tons, which is the vertical comp(ment of the stress in M), since l-H is not in action. The load at d produces in /-9 a negative shear of 2 tons, so that the difference, or 19.625, must come from the member 6-^ .-. max. stress in C>-1 = + 19.625 x 1.414 =-. + 27.7 tons. Otherwise as follows: The live loads going to the right abutment through the panel 6-8 = 23.63 tons. The dead loads crossing this at the section cutting f/-9, /-9, and l-H are I of load at 9 + .[ of load at d - 4 tons. The difference is 19.()3 tons, for which the panel 6-8 must be counterbraced. Thus are found the following maximum stresses : 1M5 HOOFS AMt iiiniiCKS. Maximi^ Clloltli SrllKHWKH. 1 il ii-'i 6-7 7 9 ii-i 4-9 + 201.1 -130.0 -228.8 -281.0 -205.0 + 127.1 f2l.-..;5 M.v.xiMiM SniKHHKs IN riM'KU I'akt OF Main Tik)* ani> IIai.1- Tiks. 1 t + 11(3.5 H b-k ;-; •A-i 6-J 7* 4-40.8 + 151.1 + 111.6 + 74.H + 20.7 + 20.7 +20.7 .Mavimim Stiikhsks in LowKii I'.Mti <>i .Main Tiks. i i ./-» k i\ ;-H (>^i + 0.7 + 174.8 + 133.8 + 05.0 + 00.3 +27.7 Maximum Sthe'^sks in tiik VKnTicAi.s. ft'^^i i-A 4-5 B 7 S-9 «/, h}, ck, (I!. -123.0 -94.0 -07.0 -59.0 -42.0 Note. — In the above determination of the niaxiinum Ktresses in ttie long verticals and in the lower ends ', f the long diauon.-ils, a.s, for exiiniiilc, in 5-4 and j-4, it was assumed thiit the larjrest p(>s.sibk' shear in 5-4 or in J-4 occurs when the excess loads act at 5 and at f, and the train loads are at all the apexes from the rij;ht ahutinent to the joint 5 inclusive. It is evident, however, from ;i little inspection of KIg. 54, that, if a train pane! load be placed at llie joint /(, one half of this train load at b will be transndlted to the joint 5 through the half tie j5, just as one half of the dead load at 6 is tri-nsuiitled too through the half tie j5, and that this half train load transmitted to 5, minus 'S^, *««:.,,. •1-0 <^s 21. -..;'. + 2(11.1 AND II.M.I- TiKH. T-* ii-i 4-40.8 +20.7 N 'I'lKS. 4-t .7 +n.7 (II, fij, ck; ,1!. -42.0 imuiu Htresses in ilingoiials, as, for est possible sliear u ami at f, and , abutment to tlio ;tle inspection of inl h, om; half of tliroiifrli tlif liiilf tti'd toi") tliroiifili litted to 5, ininiis nitllXlK THUSSES. 11" tlip part that is tran.iinittcd to tlic ri«lit abninuni, wliicli is ,'^ of tin- train load at h, will form part of tlit> shear in 5-4 or in j-\ ; thai is, the shear will l)u iiicrciuti'il hy J x 8 — ,9g x H = ,"4 x 8 = 2 5. Ik'nec, Max. strfsaln &-» = - [5 x4 + JX(12 + 7) + .'4(l2 + ••• + lj+2.5] = -07.1 tons. Max. 8tn>88 in j-\ - 1)7.1 x 1.11 » = l.'H.:) tons. Similarly, Max. Blri'HH in 2-^ = - [7 x 12 f };i( U ) !•) ] - 127. 1 tons. Max. Htrt'HS in i-2 - 127.1 x I. II I -- 1711.7 tons, and HO on for U-7 and k-i\, 8-1), and /-8. Thus, stn.ss in «-7 - - (i8.(5, in A-fl - 07.0, in 8-!» = - 4.1.1, in /-8 - (11 tons. It is cvidunt that by putting; n train panel load on the apex just to the left of the section considered, the shear in that section is inere-i-sed the most when near the left end of the triisH ; and that it liecreases on approaching thu ndddic of the tru88 without beuondng zero. l',"'i. 115. A deck liultiinort; truss, Fi^'. .W, has 14 panels, each 10 feet loiij.,' aiul L'(» feet (leep, the verticals being posts and the diagonals ties ; the dead and train loads are 1(M)0 lbs. 7 a 3 h ,'! e 7 a 7' e J' f fa V w \/ \/ \l/ H^ \/ \/ \/m \ \ /\ A A\ / / a «' 4' and 2000 lbs. ])er foot per trnss, and there are two excess loads 50 feet apart, each of 30 tons: find the inaxiniiiin stresses in all the members. (See note to last problem.) Maximum C-iord Stuf.sses. - l-s ;!-5 r» -7 7 7' •i-i 4-fi + 214.:i t;-i'.' + 260.7 -142.6 -2.30.4 -271.1 -204.7 + 130.7 •«.'?•' I. ' iii' MH hoot's A.M) iiiiiiiaHs. Maximum Sthknhki* in Upi-ku Part <»k Main 'I'ikh ani> IIai.k Tik.*. + 201.6 »-J 6-* 7-V +ni.H +:ji.8 7-* + 31.8 + ltlO,0 + 102.6 + 00.1 Maximi M Stuksskn in Lowi.h I'aht or Main Tn.n. l-i J-i i~n W 4-* + 184.8 + v.n.n + 85.8 + 42.4 + 8.1 Maximi'M SmKOfiKH IN riiK Vkuticals. ^» 4-.'i ft-T ,ii, hj. ,1, ,11. -i:jo.7 -MM -00.7 -4:).o /'/•(»/>. 116. A through Biiltimore truss, Fig. WJ, has Ifi jianels, eacli Hi feet h)ng and .'51-' feet (Um'Ji ; the dead load is given liy formula (.'{), Art. ir», tin; train load is luUO lbs. .t fi 7' X- S' JUT foot per trus'-., and there are two excess loads 4S feet apart, each of .'><• tuns- find the inaxiinuni .stresses in all the nienibcrs. Head panel load - (^Ar^:^~!^^^ = « tons. 1,' X L'(HM) Train iiancl load — 12 tons. Each excess U)ad -- oO tons. HS^, ■teS, ANi> IIai.i-- 'I'ik*. b-J + 31.8 h:n.8 UN TlK*. 4-* + 8.1 L». ((, /,J. iX, '//. -4.').0 'ijj. Tifi, hiis 1(5 the (load load •ad is luUO Ib-s. s' u k' h ^ t' s loads 4S feet .stresses in all ^ 8 tons. nit 1 1)111-: iiirssics. ll'.» Tlit^ niaxinuini cliord stiPs.srs iK-ciir when tlio dead and train loads att at every lower apex (Art. '_'(>), and the excess loads act at the proper apexes (,\rt. I-). Tims for /><>, \vc have :W tons at I iind at c, •« feet to the ri^'lit of (>. The center of nionients is at .'$, the iiiterseetion of li~."» and .'{-0. Hen(!0 the left reactiuM for dead and live loads is, i? = 7i X liO + ,'!! (I t -f 11) = 190.875 ton.s. Calling s the max. stress in iO, we have, I0().87r. X .'!•-' - 20 X IT) 4- 20 x U\ - s x 32 = 0. .•. s = max. stress in hO = llMi.l) tons = max. stress in 46. For any upper chord meinhcr, as 5-7, we have ,".0 tons at 8 and at c, IH feel to the right of 8. The center of moments is at 8. Thus, ii = 7,\ X 20 + ^H (10 4 7) = 181.87.-. tons. .-. 181.875 xOx 16-20(1(5 + 32 + 48 + (U-f-80) + .sx32=0. .•. 8 = max. stress in 5-7 = - 3;)5.(» tons. Similarly, the other chord stresses are found. It is evident that the m,'<.xinnim stress in each snhvertical, ai, bk, d, dm, etc., is etpial to a full panel load 8 -|- 12 + .'>0 = 50 tons tension. It is also evi x 1.414 = .35.4 tons, o The maximum stress in the lower part of any main tie, as /8, occurs wlien the excess loads act at 8 and at f, 48 feet to the riglit of 8, and the train loads are at all the apexes "? ■ i i "' i I f k %{ 150 HOOFS AND liRlDGES. from the right abutment to 8 inclusive. The shear in /8 lor this loading is, 2i X 8 + -Jg (10 + 7) + H (10 + ... 4- 1) = 03.125. .-. max. stress in 18 = 93.125 x 1.414 = 131.7 tons. The maximum stress in the upper part of any main tie, as 3 A;, occurs when tlio excess loads act at G and at (/, 48 feet to the riglit of G, am tliu train loads are at all the apexes from the right abutment to b inclusive : for half of the train load at b is transmitted to the joint G through the half strut kG, and f^^r of the train load at b is transmitted ft) the right abutment; therefore the difference of these, or ^\ of the train load at b, goes to the left abutment and forms part of the shear in 3/i'; that is, the train panel load at b will increase the sliear in '6k by yV X 12 = 3| tons, it is evident also that half of the dead load at b, which is trans- mitted to G through the half stmt kC>, goes to the left abut- ment and forms part of the shear in 3k. Similarly for 5/ and 7 m. Hence, Max. stress in 3 A; = ["> X 8 4- fga- + 9) + H (12 + - + 1) + 3.75] 1.414 = 200.3 tons. The maximum stress in 3-4 is ecjual to the full panel load at 4 plus half the sum of the panel loads at a and b = ,30 -j- 12 -I- 8 -f I (24 + IG) = 70 tons. The maxiuium stress in either of the other long verticals, as 7-8, is equal to the shear in 7 m plus whatever load is applied to the upper apex 7. Hence, Max. stress 7-8 = -[8 + M(8 + «''>)+1|(8 + - + 1)+tVx12] ."=-60.1 tons. ns», iliear in /S lor = 93.125. .31.7 tons. any main tie, ! and at (/, 48 ire at all the ; : for half of G throut,'h the transmitted to )f these, or j^^r !nt and forms mel load at /* I tons. It is ,liich is tran.s- the left abut- niilarly for 5/ + 3.75] 1.414 full panel load and b )ns. long verticals, atever load is rV X 12] JililhGK TRUSSES. mi The maximum stress in 3 (" occurs when the excess loads act at 4 and at c, and the train loads cover the whole tr i 's ; it is easily se«;n that lialf of the dead train loads .( (■, which are transmitted to 4, form part of the shear r 3;'. Hence, Max. stress in 3 1 = - [7 x 20 + f f (14 + 11)] 1.414 = -- 2W.2 tons. The following stresses are found in a manner similar to the above : Maximum Ciiokd Stuesses. S 6 r,-7 T-!) 2-» 4-6 o-s 8-10 -318.8 -395.6 -417.6 + 100.9 -106.9 + 328.8 +405.0 Maximum Stresses in Lower Paiit of Main and Half I/iaconals. i-i A--8 /-3 HI-IO (-4 i-4 1-6 m S -283.7 + 189.3 + 131.7 + 78.3 -35.4 -3>.4 -35.4 -35.4 Maxi.mcm Stresses in Upper Part of Main Diagonals. 8-i 3-* 6-1 i-m 9-m t-l -204.2 +200.3 + 140.5 + 85.0 + 53.2 +6.2 Maximum Strf-sses tn the Verticals. S-4 5-« T-S 8-10 ai, bk, ol, dm. + 70.0 -90.4 -00.1 -23.9 + 50.0 IS i I f. il ■ w 152 HOOFS AND nillDOES. The stress in 0-)ii •-= 9-u occurs whoa the excess loads act at »' 1111(1 iV, and the train loads act at all the joints from the riyht ahulnicnt to c. 'rhen the shear in the sec- tion cutting y-7', '■)-((, and !(»-/( ^^ lOJ.OL'o-S x .S= -{-o'-iVJo tons; and as lO-ii cannot take compression, this is the vertical component of stress in 9-)i. .-. max. stress in 9-u = -f 37.025 x 1.414 = + 53.2 tons. If 10-u were constructed to take compression, this would ho greatly modified. In this case the excess loads would act at 8' and ij, and the train loails ut all the joints from the right abutment to 8'. Then the left reaction = 92.G25 ton.s and the shear in the section cutting l)-7', 9-)i, and 10-jt = 9'_'.(525 - CA = -+- 28.025 tons. The member 10->i takes a positive shear of 4 tons, so that the vertical component of stress in 9-?i = + 28.(>25 - 4 = + 24.025 tons. .-. max. stress in \)-n = 4- 24.025 x 1.414 = + 34.8 tons. The stress in 7'-0 occurs when the excess loads act at f and 4', and the train loads at all the joints from the right abutment to/. Then the left reaction = -f "^4.375 tons and the shear in the section cutting 7-5', 7-0, and 8-0 = 81.375-8 x 10= -f 4.375 tons. As 8'-0 does not act in compression, this shear must be resisted by ^'-O alone. .-. max. stress in 7'-0 = + 4.375 x 1.414 = + 0.2 tons* Prob. 117. A deck Baltimore truss. Fig. 54, has 10 panels, each 10 feet long and 20 feet tleep, the verticals being post^ and the diagonals ties; the dead and train loads are 1000 lbs. and 2000 lbs. per foot per truss, and there are two excess loads 50 feet apart, each of 33 tons : find the stresses in all the members. •It should be noted that the members m-8 and ;i-8' are also tension members for couiiipr stresses. The teusiuu stress iu each of these members = 20.025 x 1.414 := 29.2 tons. mi Jumt excess loads ill the joiiiLs ir ill tlic scc- ;S= +;>7.G2u , this is the + 53.2 tons. »n, this would i loads would lints from the = 92.G25 tons , and 10-jt = 10-?i takes a component of + 34.8 tons, loads act at nts from the + "^A.Tib tons 0, and 8'-0 = 33 not act in -0 alone. + 6.2 tons* lias IG panels, Is being post^ )ads are 1000 here are two d the stresses are also tension 1.414 = 29.2 tons. nUIDGE TRUSSES. Maximum CnoKD Stbesses. 15:} l-l S-,'> . '' ' T it 2-1 4 6 fi ^ -104.1 -274.1 -337.7 -354.7 + 152.4 +258.1 + 317.8 Maximum Stresses in Uppeii Pakt of Main and Half Ties. l-i »-J r, i- :-/ 37 +34.0 .'■'-; --<: +47.4 + 232.0 + 180.0 + 132.7 + 88.3 + 34.0 + 34.0 Maximum Stuesses in Lower Paht of Main Ties. 1-2 J~i 1-6 /-S 6-/ + 215.6 + 164.1 + 116.2 + 09.1' + 31.8 The counter 4-k is found not to be required; but in actual practice a counter would bo used to provide for loads differ- ing from the ones above assumed. Maximum Stresses in the Verticals. 2-3 4-5 6-7 S-9 fii, hi, ei; ,11. -152.4 -110.1 -82.2 -09.2 + 48.0 This problem is taken from "Strains in Framed Strnctnres" by A. J. Du Bois. It will Ik- seen that the answers above given for the maximum stresses in the lower part of the nuiin ties and in the long verticals differ soinrwhat from those given by Prof. Du Rois. "n f ' h n 154 ItOOFS AND BllIDGES. Art. 44. True Maximum Shears for Uniform Live Load.— Tliiis iar, it h;is been assmiu'd Uiat the live panel loads were all coiici'iitiated at the panel points, and in order to find the niaxiniiun stress in any diagonal dne to a nniforni live load per linear foot, the niaxinuini live load shear in that diagcnial has been found by putting a live panel load on every joint on the right of the seetion cutting the diagonal (Art, 22). For exaniide, it has been assumed that the first apex on the right of the section in Fig. 57 has a full live ])anel load aiul lilt' first one on the left has no live load. Now the joint n on the right of the section receives the half panel load A7\ KA7gg\ , | t„ ;■'". — >M J Kip. 07 between the joints n and n - 1, but only a small part of the load on the left of n ; the other part of the load on the left of n is received by the joint n + 1. When the uniform load extends from the right abutment to the joint n + 1, the joint }) receives a full panel load ; but then the joint n-\-\ receives a half panel load. Let us therefore ascertain at what dis- tance X to the left of the joint ?i the load must extend to produce the greatest shear in the n + 1th panel. I A>t N= the number of panels in the truss, p = the panel length, w = the uniform live load i)er linear foot, n = the number cf whole panels covered by the load, li = the reac- tion at the left or unloaded end. Then, : the part of the load ivx which is carried by the joint n'+l; and wx — ^ = the part 2p carried by the joint n. of the load wx which is rnif orm Live the live panel ;s, ami in order le to a uniform load shear in ive panel load ig the diagonal 3 first apex on live i)anel load Now the joint lalf panel load lall part of the oad on the left le uniform load n + 1, the joint it n + l receives in at what dis- nust extend to lel. i, p = the panel ir foot, n = the i, It = the reac- ied by the joint id wx which is ItlllDflK TliUssES. We have then for the reaction li ^'-je:^" + ') 'jp wxn '"IT + IV):' wn^p '2pN- • '2N Hence the shear in the n + 1th panel is ~~N'^2pX '■"" 2p 2N 155 \ 2p)N 2N (1) (2) Equating the derivative of Fto zero, we have, N^pX~J~' X = n N-1 P, which is the distance to the left of tlie joint n that the live load must extend in order to produce the niaximuni siiear in the n + 1th panel from the right end. Thus, let the truss have 8 panels; then for the true maximum shear in the sixth panel from the right end X = ^p, for the seventh panel x = S-p, and for tlie eighth or last panel x=lp = p, or tlie whole truss is covered. The total live load on the truss = mcp+J^^ = ^!lIEP N-1 N-1 = ^ times the load on the n + 1th panel. Thm'forp, the lirp had on the n + 1th pvicl As ^ th of the total lire load on the triins. '^ Substituting the above value of x in (2) and reducing, we Maximum shear = — ^-^^ 2 (N-1) ft i i I t ir,»! HOOFS AM) lUllhClEfi. These valiips of a* and the inaxinmin sliear are iii(U'j>eii(lent of the form of the tniss^ wlifthrr llie weliliiiij,' be iiicliiietl, as in llie Warren, or inelined and vtMliciil. as in tlie Tratt and Howe. Pi-oh. 118. A trns.s, Kit,'. r>7, has 7 jiaiiels, eaeh 10 feet lonK', tlie live h)ad is 2 tons per foot: iind (1) the true niaxinnim live load shears in every i)anel and (2) the niaxinmm live load shears by the usual method (Art. 22). Here N=l,p = 10 feet, and to = 2 tons. For the maximum shear in the 1st panel from the left n — (5. ... .T = ;. = IC. feet, and shear = ^ili^l*'=:9(J tons, ' 2 X (> which is half of the effective load, or the effective reaction. For max. shear in the 4lh ])anel from left, n — 'A. .•. x= I X !(■)=: S feet, or the load reaches to the middle of the 4th panel. 8hear = g. xj<> ^ '^' = 24.0 tons. 2 X (> By the usual way of taking a full panel load as con- centrated at the third apex from the right n - 1, and none at the 4th apex n, we have Shear = -^i (1 + 2 + 3) = 27.4 tons, which is greater than that obtained by the strictly correct method. Thus the following live load shears are computed (1) by the Strictly Correct Method and (2) hy the Method in Common Practice : True Method. 90.0, 00.7, 42.7, 24.0, 10.7, 2.7, 0.0 tons. Common Method. 90.0, 08.0, 4r>.7. 27.4. 1.3.7, 4.0, 0.0 tons. -i)Oii(lent \\i lie iiicliiu'ti, s in tlic Vvixil i, oiuili K) I'eet I (1) the tine 1 and (2) the od (Art. 22). tons. from the left H2 - =:9() tons, ective reaction. 'ft, n r= 3. i to the middle il load as con- i — 1, and none ns, strictly correct !omputed (1) by hod in Common 7, 2.7, 0.0 tons. 7, 4.C, 0.0 tons. •PWI iinihuK ritussKs. 157 U is seen that Hit' .sliciirs dlilaiiit'il liy tlie ?(.s»(((/ nn'tlioil, that is, liy siipitosinji; the [lani'l loads to Ixs coneentrati'il at the panel |)oints, are larger than tlio.se obtained hy the true mclliiiil. For liie, reason that the shi'ais obtained by the \isnal method iihvays err on the safe side, and that the error is always small, it is the common practice to snppose all the load from middle to middle of panel to be concentrated at the panel point, as in Art. 4. Art. 45. Locomotive Wheel Loads. — In compntin^ the stresses in railway bridges in America, the live load which is now generally taken, consists of two of the heaviest engines in nse on the line, at the head of the heaviest known train load. The weights of the engines and tenders are assumed to 1)0 concentrated at the wheel-bearings, giving definite loads at these {)oints, while the train load is taken as a uniform load of about 3000 pounds per linear foot of single track. (a.) ^ nonho noP.o ^ hnnnn o o o o{^^>!l (b.) (ItO/eel = il inch tcole) Fig. 58. The first diagram of Fig. r)8 represents two 88-ton pas- senger locomotives, as specified by the Pennsylvania Rail- road. The second diagram shows two 112-ton decapod engines, used on the Atlantic Coast Line. \ \ ( t I i m 158 HOOFS AM) nilUHiKS. The tiiinibers aliovc tli«' wImmOs sliow their weights in tons for hotli rails of a .siii.L;lc triu-k. Tlie numbers between the wlieels show their distaiiees apart in feet. Art. 46. Position of Wheel Loads for Maximum Shear. — Let Fi};. iV.) represent a truss with a (leea|io(i engine and train upon it. Let N= the number of panels in the truss, ]i = tlie ])anel length, 7'= /', 4 /'a +•••+/'«= the sum of the wheel loads, ?« = the uniforju train load per linear foot, U'— the total live load on the truss, d = the distanee of the center of gravity of /* from the front of the train, x = the distanee of the front of the train from the right abutment, a — the dis- tanee of /', from the front of the train, y = its distance at the left of the nth panel point, and li = the reaction at the left end. Then J\ •' = the part of the load ]\ that is carried P by the n -f 1th panel jjoint. We have then for the reaction iV> '>Np Hence the shear in the u 4 lili panel is r — -r-z 1- — I I ■ Up 2 Nu p (1) (^) n litithGE TIlUSSES. 15!) figlits in toiiH i betwei;a the ' Maximum Lli a (li'ca|io(l p z= tlip ])aiu'i t' the wIh'I'I fdot, W= the ' the center of k 16 distance of b, u = tlie (lis- ts distance at Raction at the that is carried (1) (-',) But from the fifjure, x + « = tip ■\- y ; which in (-') gives y ^ (I -f X - np, Np • 2Np /" + ^-"^')- Equating the first derivative of Fto zero, we have, P + 1VX X -y'. = 0; or, since P + ivx= W, we have, Ilencr, the shear in an>i panel ix a maximum when the load on the imnel is —th of the entire live load on the truss. In practice it is convenient to put one of the U)ads at tlie nth panel pijint, so that the above condition cannot, in gen- eral, be exactly satisfied. We must have in general, P, = orAiWA0MMMKmv>rMMi« iT H\ 1«',0 II OOFS ANFt UlUltnKfi fict it !>(< n'i|ini'i'i| In timl tin- iiiaxiiiuiin slicar in the 2d piuifl tn.iii tlic Icll. llciK « == S; S 1<». Hy the aliovi' rule for tlio iniixiimnn sliciir in this iniiicl, the first driver must W put ;it llif joint h, since H < '1',^-, and In tliis position of the loads, tlie left reaction is /;= A. X 1(»4 + -f^ C.»(;.(» f 91 .75 + 87.r. -f m.'l't +■ 70.0) 4 JiL(71.5 + GG.83 + C1.23 + 5G.5f))=74.0 tons. /. max. .shear = 74.0 -y- = G9.C tons. Let it 1)0 rcpiired to find the niaxinuini shear in the 8th panel from tlie left. J lore 7* = 1.'. To iind the jHisition of the wheels for maximum shear in this jianel, try iho first driver at the joint u, as lieforo. For this position the total live load on tlie truss = 8 + 5 X 12.8 = 72 tons, since the tender wheels are oft' the truss. T\v the rule, therefore, this is not the correctt position, since 8 > j'g x 72. If we put the pilot wheel at 71, the load on the truss = 8 + 4 X 12.8 = 50.2 tons, since the last driver is off the truss. Hence for maximum shear in this panel this is the correct luadini?. With the live load in this position, the left reaction is iJ = A X 24 + ^|:y IG + 11.75 + 7.5(» + 3.25)= 5.7 tons, which is also the maximum shear, since in this case there is 110 load to he suhtraeted. Thus the fullowin}; shears are determined: Max. shears = 80.8, GO.G, 58.4, 47.2, 3G.0, 24.8, 13.0, 5.7. T^'''^«r**:s ^^x,"*? liciir ill the '2d • ill tlii.s luiufl, le 8 < ',7, aud ion is t.25 + 70.0) 50) =74.'.) tons. liear in the 8th (iiiiiini shear in as lii'l'ore. truss Tiy tlie rule, the truss 5 for maximum r left reaction is 25)= 5.7 tons, lis case there is {4.8, 13.9, 5.7. nillDGE llii'.'SES. tni Pfoh. 120, A deck I'ratt truss, Fij,'. 50, has 10 jiani'ls, each 20 feet long and 24 feet deep: find the inaxiiuuiu web str(^sst's in e;u^h panel caused by a sinnU; pas.senger locomo- tive and tendt-r, as shown in {(() of Fij,'. 58. \\i) tiist lind tiie niaximnin shear in the left ])anel. I>y the i-\\\{' the second pilot wheel must be put at the joint 4, since H< jp„ and 8 -f 8> flJ. With the live load in this position, the left reaction is Ji = , ^ (185.5 4- I'^O + IG.'J.r* + 148.5 + 14.S -f- 138) + 11^-^(171 + 163)= 71.34 tons; aud the max. shear = 71.34 — 8 x 5J) 20 : 09.14. .-.stress in l-l = 09.14 sec 6 = 09.14 x 1.302 = 90.0 tons. For 3-t we have the same loading as for 1-4; and the maximum compression in 3-4 is equal to the shear just found for 1-4. .-. stress in 3-4 = — 09.1 tons. The maximum compression for the end vertical 1-2 is found by placing the wheels so as to bring the greatest load to the joint 1, and is found to be 71.9 tons. Thus we find the following maximum stresses: Diagonals = -I- 90.0, +78.0, +07.1, +55.0, +44.2, +.32.7, + 21.2, + 9.8 tons. Posts = -71.9, -09.1, -00.3, -51.5, -42.7, -33.9 tons. If the second driver be placed at the middle panel point II, the Ktress in 11-12 will be found to be - 39.0 tona. Prob. 121. A deck I'ratt truss, Fig. .50. has 10 panehs, each 12 feet long and 12 feet deep ; the dead load is 1000 ;->«tftfc*» 'ia < T - j i r ... 1 162 llOoFf^ AM' ItUlhUKS. llis. \M'v liiu'iir fiiDt iif track, tlic livr load is a i>a,ss(MiK'(>r lociniKitivo ami tt'inlcv, as hIiowii in (n) of Kig. M: timl the iiiaxiiiiuin web stresses in oach paiu'l. Tho (Icatl and live load stresses may lie comimted sepa- rately. HtRKHHK* in DudliNM.S. Mkmiikkh. 1-4 H-ll &-■« 7-l() + 12.7 + 61.(1 +03.7 ii-li 11-10' !''->' Dead strpftHcs. + 3H.0 + HH.3 + 21t.fl + 76.0 + 106.6 + 21.1 +03.4 + H4.6 + 4.2 + 3H.6 + 42.7 - 4.2 +-20.1 -12.7 +-13.7 Max. BtresaoH . + 120.;! +-21.9 +- 1.0 Kthksskm in Tiir. I'owts. MfMltKliS. 1-.' .-. .',-« T-^ !l-ll> U-l'J Dead htrcswcs . Live KtrosHPS . - 30,0 - 70.0 -27.0 -02.6 -21.0 -63.7 -16.0 -44.0 - 9.0 -30.1 - fl.O -27.:i Max. Htroiwes . -100.0 -80,6 -74.7 -69.9 -46.1 -33.3 If tlie first (liiviT !»• jiUvcd at llii' iiiiililli' imihI point 11, tlio stress in 11-12 will lie fimiul |.) be - 2H,7 tens, I; will In- noticiMl tliat the tension of 4.2 tons in tin- iliasonal 0'-12 is iMiiiivaicnt to a c'oinpre.ssioii of 4.2 tons in the diagonal 11-10' ; and siinilavly for the next diagonal 0'-8'. Proh. 122. A through Pratt truss, Fig. GO, has 8 panels, each 18 feet long and 24 feet deep : find the niaxinuun web ; 3' * Ji^ , 0,; ,..* '-■-§1- imiiKiE mvstiKs. 108 ,s a i>a.sH(Mijjpr [. r*« : timl i\w oinpiued sepa- 11-10 !''->' 2 7 - 4.2 +20.1 +21.9 -12.7 + i;i.7 + 1.0 (1-1(1 U-l'i - 9.0 -30.1 - 0.0 -27.:) -45.1 — M>..» )iiit 11, tlio stress noticud tliiit the to a c'oinim'ssioii the next diagonal ), has 8 panels, maximum web X.. Hlifsscs ill each iiaiicl due to a |titss('ii^,'t'r locoimitivc and It'iidiT, as .shown in (/() of Kit;, aS. followed by a uniform train load of .'UXK) lbs. per linear foot. (1) To find the niaximuni stress in 2S. Try the first driver at the panel })oiiit4. In this position the total live loa X l.r. = 22.3 tons. Sinee S+.S<23\ and S + 8 + 20>JJ^ tliis is the eorrect position for the maximum shear in the panel 2-4. The reaction is /f = -^^ aaS + 140.6 -f- 108.5 + 10.'i.r> -l- 0» + 93) 144 + ,tt(^-''' + 11«)+ ?. X -T^,~ = 113.77 tons. 144 Z 144 .-. max. shear = 113.77 - ~- (0 -f- 14.5)= 103.33. 18 .-. stress in 2-3 = - 103.33 x 1.25 = - 129.2 tons. (2) To find the maximum stress in 9-8'. Try the first driver at the joint 8'. In this position the total live load on the truss ■ =88 + 18 X 1.5 = 115 tons. But 8-f 8> Lp; hence this is not the correct position for the maximum shear in the panel 10-8'. We therefore try the second pilot wheel at 8'. We then have, for the total live load on the truss, Tr= 88 + 9 X 1.5 = 101.5 tons. Since 8 < — — -> this is the correct position for the maxi- o mum shear in the panel 10-8'. f.-i?t«i««*vi*M«»«/F*!iT.»»«T^Mm:-' T3ii^---it i^vnxikfi'i Im Ml n m lot HOOFS AND DllIDGES. The reaction is 1540 4- IfilO 4- 13.r> X 4.5 11 = = 22.51 tons. 144 .-. max. shear = 22.51 - ?-^-i'-''^ = 20.1 tons. 18 = — stress ill 9-10. .-. stress in 9-8' ::= 20.1 x 1.25 = 25.1 tons. The max. tension in the hip vertical 3—1 is fouinl by puttiiii,' tlie wlioels so as to bring the greatest load at 4. Tims iho following uiaximum stresses are found: Diagonals = -129.2, +9G.4, +07.9, +43.6, +25.1, +11.3 tons. Verticals = + 36.9, - 54.8, ~ 34.9, - 20.1 tons. Art. 47. Position of Wheel Loads for Maximum Moment at Joint in Loaded Chord.— In addition to lh(! notation of Art. 4(!, let P' — the load on the left of the panel point n, Fig. 59, x' = the distance of its center of gravity from the point n, and n' = the number of panels between the left abntuient and the point n. Then for the moment at the panel point 7i, Fig. 59, wc have ■P(a; + (r) K'j;- Xp li~Xj>_ n'p - Fx', [by (1) of Art. 46]. Equating the first derivative of M to zero, we have, since N ' or since P+ ivx— TF(Art. 40), we have N "" ■ I niilDGE TliUSSES. 165 1 tons, tons. ;ons. -4 is found l)y itpst load at l. found : 5.1, +11.3 tons, tons. or Maximum - In addition to 1 tlio left of the if its center of mhcr of panel.s Then for the ve (1) of Art. 4G]. , \vc have, since Jlence the moment nt (tni/ paiui pniitt in the hailed chord is v' a maximxm when the UkuI on the lejl of the point is -ths of the entire lice lo(((l on the triiH.^. In practice it is convenient to put one of the loads at the Hti. panel point, so that the above condition can very seldom l)e exactly satisfied. We must have in general P' or < - W. N Hence, in general, the moment at any panel point of the loaded chord in a maximuni ifhen the load on the left, of the point has to the entire load on the trusn a ratio which is equal to or J II xt less than the ratio irhich the nnmher of panels on the left of the point bears to the entire number of panels in the truss. By tliis rule the maximum chord stress in any member of the upli)adcd clii rd may be determined, since we have only to divide the maximum moment at the panel point opposite the chord mendier by the depth of truss. Prob. 123. A deck Pratt truss, like Fig. 50, lias 8 panels, each 18 feet long and 24 feet deep: find the maximum chord stresses due to a single passenger locomotive and tc'uder. Let it be required to find the maximum stress in the second panel 3-5. Here n' = 2, X=S, ll'=S,S; therefore, by the rule, J" must be equal to or less than 5 x 88, or 22 tons. Hence, the lir,st driver must be put at 5 since 8 + 8 < 22, and 8 + 8 + 20 > 22. With the live load in this position, the left reaction is H ^ ^- (122.5 + 117 4- 90.5 + 85.5 + 80 + 76) 144 + -'i (108 -I- 100) = G0.58 tons. 144 ^ '-^, -.tim/itMttt/'M i----.^. 1G(; ROOFS AND li HI DOES. .-.stress in 3-:> = ~l^^^^^^^^^^^^^±^^ = - 83 ton. = — stress in C-8. Thus aro foinul the following stresses: Max. stress in l-.'?:^ —47.7, in 3-5= -83, in 5-7= —103.8, in 7-".) = - 110.5 tons. I'i'cli. 124. A (li'ck I'ratt truss, Fig. 50, has 10 iiancls, each 20 feei hnvj; and L'4 feet deep: find llic nia.\iniiini chord stiesses in each panel due to a passenger loeouiotive and tender. ..Ills. Max. stresses in upper chord = — 57.(5, — 103.0, - 13G.4, - 155.2, - 161.9 tons. Proh. 125. A through Warren truss has 10 panels, each 12 feet long and 12 feet deep: find the niaxiiiinni stresses in tlie ui)])er chord due to a decapod engine and tender. Ann. Max. stresses = - 80.8, -145.0, - I'JO.U, -217.3, - 225.2 tons. Art. 48. Position of Wlieel Loads for Maximum Moment at Joiat in Unloaded Chord. — I'-y the nUe deduced in Art. 47, the niaxiiuum chord stresses may be determined in the nnlnaded chord of any siuijjle truss, and also in the loaded chord of s\ich trusses as the Pratt and Howe, where the web members are vrtiad and inclined so that the panel points of the upper chord are directly over those of the lower chord. For trusses like the Warren and lattice, Avhere all the web members are inclined, it applies only to the unloaded chord. For the haded choid of suci trusses a modification of the formula or rule is necessary, which may be deduced as follows : Let it be required to fiiul the maximum moment at the panel point c of the unloaded chord, Fig. 59. ii ttlllDGE TItUSSES. 167 i = -83 tons ir)-7=-10;{.8, lias 10 jiaiit'ls, 1li(> niiixiiiniiii iger locomotive 57.G, -103.0. 10 i);uiels, each xiiinnii stresses uiil tender. I'JO.y, -217.3, or Maximum — I'.y the vn\v tre.sses may be in])k'. truss, and i tlie I'ratt and (iiid inclined so re directly over the Warren and ined, it applies [ choid of suci le is necessary, moment at the Lot W^ the total live loa«l on the truss, Q - the load (.11 the (/(. - l)th panel, and /• = the Inul .m Mie left of the (n - l)ti> piiiiel. Let I' = the distance of c from the left support, 7 ^ the distance be, p = the panel length, and I = the length of the span. Let a; = the distance from the center of gravity of W to the right support, .f,= the distance from the center of gravity .jf F to the panel point n, and x\ = the distance from Q to the panel j/oint n — 1. The part of Q ihat is carried by the nth panel point 18 -£-^. Hence the moment at c is I V Equating the first derivative of M to zero, we have, since dx = dxx = dx.^, J 1> V ^ (1) If the braces are cipially inclined, that is, if the center (,f moments c is directly over or under the center of the opposite member, as is usually the case, we have | = ^, and (1; becomes P'+iQ = pF. (2) For the Pratt and Howe trusses 7 = 0, and (1) becomes which is the same as the formula that was found in Art. 47. BMMll 1G8 nooFs AM) imiDaFS. Ft is convenient in practice to put one of tlie loads at the n — 1th panel point, so that, In (jviwral, wo imist have /^-f-i-Q = or <^jW. (3) Proh. 126. A thron<,'h Warron truss has 10 panels, each 12 fort luui-j and 11' feet deep: tind the niaxiinuni stresses in tl'.c lower chord due tu a decajiod eu^'ine and tender. To tind the niaxitniini stress in the first panel 2-4. Here V:=y,l^\i), and )r=112. Therefore by fonnula (8), F + I Q must 1x1 erpial to or less than o',, X 1 12 or 5.0 tons. Hence the Ist driver must be put at 4, since I X 8 < 5.G, and ^ + \x 12.8 > 5.6. The left reaction then := 8G.14 tons. The moment at the point 1 is i»/= 8G.14 X 6 - 8 X ^»^ X G = 484.84. 484.S4 stress in 2—1 ■ 12 : 40.4 tons. Similarly the maximum stresses in the other lower chord .members are found. Ma.K. stresses = 484.8, 111.9, 1G7.1, 203.4, 220.7 tons. iS'h<7. To lind stress in 4-(i, put i!d driver at (! ; in 0-8, put T.d driver at 8; in 8-U), put 4lh driver at 10; in 10-12, put 0th driver at 12. Art. 49. Tabulation of Moments of Wheel LoadB. — A diafj;niin such as is shown in l''i|,' '51, diminishes con- siderably the work of coinp\iting stresses due to actual wheel loads. The Gist diagram is for an 88-ton passenger loco- motive and its tender: the se(^ond is fur a 112-ton decajtod engine and its tender. Aay locomotive can l/e o i o 1 o to o 1 1 o 1 tt> o 1^ iiionperlinlt. a&oi 8'0 ■ -^ III ^'>: I »'.§ ^!^ . . its §1^ i< 5b7oni,?2:s >! 7: ==4: U 64Tons;3P:o >! 1 k 72Ton5i 37.'0 >! 1 k eOTon5,4?:5 >, , , K 88Ton5,47.'S '■ >i j j< - eSTonSiSa'S "•>! Si^ ^ SI I I (a) I I »i iS ^ IS ^ «» Co cvj tvi SI; I I IS t9 o OQQOQ 00 6 6[ES I.-.- K— ! 1 U ^/?»!« '♦.ie ogie i i ! ! ! 1 «33.67&«v?*'^* ^4: S^'^g # 5l| --is •!§ I'l K 59 ?7o/75, ?0'75 ~A I --^ iSlf Si!i H ISTonS; ?5:0 A i ! ! i< SZTons, 32'.5 J I ! gBTons,37:i7 ; >i ! l02Ton5:42J7 ■■ •••• >« , . /IZTonsj 47.4} >l J //em5,-4St'69 - ••'•■•>) (b.) Fig.ei. (20 feet = 1 inch scale. ) Each diagram shows the weights and distances aj)art of the wheels. Kelow each Avheel, on the horiznntnl line, is sliown the weight of that wheel togetlier wijii that <*f all the preceding ones, and its distance from the front wheel. ., Ma 170 nnoFH Ay I) II HI DOES. Below ouch wheel on the vertu al line is given the moment of all the preeoding wheels, witli reference to tluit wheel. Thus, for the third driver of the decapod engine, we have 4().4 tons for the weight of it and the three preceding wheels, 10.5 feet for its distance from the front wheel, and liy5.2 foot-tons for the moment of the preceding wheels with reference to the third driver. At the beginning of the imiforni load we hav(> 112 tons for the weight of all the preceding loads, 49.08 feet for the distance of tliis point from the front wheel, and 2910 foot-tons for the moment of all the [)receding loads with reference to this point. Each diagram shows also that, the itwment at any ivlieel is equal to the moment at the next preceding wheel on the left, plus the mm of all the preceding ivheel loads multiplied by the dixtance from the next left preceding irheel to the wheel in question. Thus, in (a) the moment of the first three wheels about the third is 188, and about the fourth it is 476. But 470 is equal to 188 plus 30 multiplied by 8; and similarly for the moment with reference to any other point. (See Du Bois's Strains in Framed Structures.) This diagram may be used for finding reactions, shears, and monuntx. Thus Let I = the length of the truss, and the other notation as in Art. 4G. Then, from (1) of Art. 4G, the left reaction is R = hPd + I'x + ^tva?), which for the passenger locomotive is /f=.:^('2388-|-88«-h and for the decapod engine is 1 i^"^) (1) ij- I '^no + U2x + hc3i^l («) '**•* -m-is - '^A^ -■ ^h/c <#^iSi, ; "•-■^"'"TfS,/*! n the moment to that wheel, giue, we have ree preceding mt wheel, and ig wheels with inning of the j;ht of all the of this point ;he moment of point. Each heel is equal to e left, plus the by the dititunce el in (juestiun. leels about the ut 470 is equal ilarly for the (See Du Bois's actions, shears, ler notation as ft reaction is (1) (2) niilDGK TliUSSKd. 171 Suppose, for example, that the third tender wheel of the decapod engine is 2 feet to the left of the right ubutiiiciit. We take out the numbers 2181.8 and 102 from the diagram, and the reaction is 7i = ~ (2181.8 + 102 + 2). In practice it is convenient to draw a skeleton outline of the truss to the same scale as tiie diagram, to be placed directly above it in the proper position for the maximnm stress in each member. Prob. 127. If the span \h 120 feet, find the reactions for the passenger Uxvinotive, (1) when liie last tender wlieel l« ."i feet to the left of the right abtttment, (2) when the UlHt (fewer is 50 feet to the right ..f the left abulniolil, and (.^) when the st>eond pilot wheel is 40 feet to the lidt of the right abutiiwMit. Ahs. (1)31.4 tons; (2) 52.1 tons; (3) 1G.4 tons. I\ob. 19&. If the span be 100 feet, find the moments for the decapod engine, (1) at an apex .'SO feet to the left of the right abutment when the second driver is at that apex, and (2) at the (tenter of the span when the first driver is there. Ans. (1) 1.341.4 fooUons; (2) 1883 foot-tons. Prob. 129. A through Pratt truss, Fig. 00, has 8 jxiMels, each 18 feet long and 24 f et deep: find the niaxinium chord stresses due to a passenger hxomotive ami tender followed l)y a uniform train load of .30(M) lbs. per linear foot. To find the position for the maximum moment in 2 4 caused by the live load, we try the first driver at the [lanel point 4. Then for tbe uniform train kjad we have .c = 7 X 18 - 36 = DO feet. In this i)Osition, the total live load on the truss = 88 + 1.5 X 90 = 223 tons. / 172 HOOFS AND UniUGKS. , ( Since 8 + 8 < J x 22.S, aii.l S + 8 + 20 > J x 223, this is the ciincct position for niaxintiun nionicnt in 2-4. For tl'is loud tlie reaction, hy formula (1), is n=^\i (2388 + 88 X 'JO + .75 X Wi') = 1 1 3.8 tons ; and the moment at 3 is .V= 113.8 X 18- 188 = 1800.4. 1800.4 •. max. stress in 2-4 24 = 111) tons = stress in 4-6. Similarly, the maximum stresses in the other three chord nieml)ers arc found to be the following : Max. stress in 3-5 = - 128.3, in 5-7 = - 157.0, in 7-9 = - 165.8 tons. Note. — To determine the maxiiiiurn strisis in 7-0 put 10 ffPt of {lie iriiiii to the lift of the point 10 ; tliiit i.«, put tlic third tender wheel at H. Then ir= 88 + 1.5x82 -211 tons. Rut 88 + 10 X 1.5 < ^ X 211 ; therefore this is about the correct loading. !(=■• s\i Ci'MH f 88 X 82 + .75 x 82') = 101 .7 tons. M = 101.7 x 7y (2388 f 88 x 10 + .75 x lo') = 3U70.4 foot-toiis. max. stress In 7-9 = — 3979.4 = - lfi5,8 tons. Proli. 130. A tlirough Tratt truss, Fig. (">(», has 8 panels, each 20 feet long and 24 feet deep; find (1) the maximum chord stresses, and (2) the maximum wob stresses, in pach jKiMcl, diw to a decapod eni^ine and tender followed bj u uniform train load of 3000 lbs. per linear foot. MM lUUhr.K TliUStiES. na 23, tliis is tons ; 88 in 4-G. luee chord 7.0, 1) fppt of tlie der wheel at about the u»8. ns. 5 8 panels, nuixituuiil B8, ill ench twed \\y a i (1) To find the position for niaxinunn stress in 2-4 wo try tho IM driver at 4 ; then x -^ W2X> feet; and W= 11-* + l.r) X 102.0 = '2m tons. Since i X L'Cr, = 3.';', > 8 + IL'.S, and < 8 + 12.8 + 12.8, this is the eorrect position for inaxinuun .stress ni 2-1, althonj^'h if the third driver should be put at 4 we woidd Ket alH)nt the same stress, as 8 + 12.8 -h 12.8 would »v just less than j^ of the load on the truss.* The resR-tion is U =. rJ,5 (20O'.».'.) -I- 112 X 102.G + .75 X i02.(>') = i:W.;5 tons. r.i9.3 x20-ir.2.4 •. max. stress in 2-4 = 24 = stress in 4-G. l-^ = 10').7 tons Similarly, the maxitnuin stresses in the other three chord members are found to be - 181.5, -218.9, and - 22(;. I tons. (•J) To find the position for maxinmm shear in 2-i} put the 3d driver at 4; then * = 100.8 feet, and W= 112 4 1.5 X 100.8 = 272.2 tons. Since H f 12.8 -t- 12.1 < It X 272.2, this is the correct position for maximum shear in 2-^, 'Iho reaction is R =- J- (2909.9 -f 112 X 106.8 + .75 x 100.8') - 140.4 tons ; 100^ and shear in 2-4 = 140.4 - ^2f (4.25 + 8.5) - 1 x 10.5 = 131.04 tons. .-. max. stress in 2-3 - - 131.64 x 1.302 = - 171.4 tons. • Tho conditions of Arts. 4f and 48 may sometimes be satisfied by diHuruiil iH'sllions of the load. 171 noOFS AM) IlI'lOaES. II I lip I'd driver lie put at 1 we sliall obtain about tlio .same result. Tlio following stresses are found in u manner similar to the above : Muxliiiiiiii Strcsupn In the f^iitgonalg. 2-3 =- 171.4 tons. 3-0 = + 130.7 tons. 6-8 =+ !»4.7 Inns. 7-10 = + 0:!.(1 tens. 0-8' = + 3H.3 tons. TAV-+ 18.0 tons. MaxiiiMiin Sin<«ii<>it In the Vrrliiuilt. 3-4 =+51.1 tons. 6-0 =- 72.7 tons. 7-8 = - 48.9 tons. 0-10 = - 29.4 tons. Prob. 131. A through ''ratt truss has 7 panels, each L'O feet long and 20 feet deep; the dead load is Jr>(»(> lbs. jier linear foot, the live load is a i)as.seng('r locomotive and ti-mh'r followed by a uniform train load of .'iOOO lbs. jier Jinear foot : find the maximum stresses in all the mendiers. CiioKo Stressks. MKMlltRS. •.!-l, -l-fi 8-5 5-T T-7' Dead load stnsses .... Live loatl stri'sscs .... + 48.0 + 08.3 - 80.0 -167.3 - 96.0 -186.4 - 96.0 -185.4 Ma.xiiiuiin stresses .... + 140.3 -237.3 -i;hi.4 -281.4 Stresses in the Diaoonals. MEUIIF.U-ki '2-3 8-6 +22.6 +64.6 "->' T-6' Dead Itjad stresses . Live load stresses . - 67.9 -138.9 + 45.2 + 08.7 0.0 + 30.5 0.0 + 16.7 Maxiiniun stresses . -206.8 + U.I. 9 + 87.2 +.%.. - 5.9 Ate nitlDGE riWSSES. ITf) 1 abuut tho r Hiiniliir to the Vtrtie.iiU. tons. tolLS. Ions. tona. »ls, eacli -'() [JOO Ills. i>or molivn ami 00 U.S. per e lucnibers. 7-7' 4 - 96.0 -185.4 4 -281.4 7-6' ).0 5.5 0.0 + 10.7 i.w 1 - 5.0 The livr load stn>ss in T-^V is + 10.7, wliilo tho (load load stress in a s' or 'y-it> i. -}• !,"_'.»>. Siiic»( tlic nifiubor 7'-(>' cannot takv (■ouij.iv.ssion, this losult shows that tho ("onnter 7'-(5' is not uoeded f<t IN Till'. Vl.llTICAJ -. MKMtirHK. 8-4 ,'i-C, - 0.0 -25.8 Deatl load stresflcs Live load strt'Sfit's + 10.0 +;io.o + 55.0 -Ifl.O -45.7 -01.7 Miiviinuin streast'S -25.8 Pinh. 132. A through Pratt truss, V\^. .'!.'>, has lO panels, each L'O feet loiij,' and 20 feet deep; the dead load is L'OOO Ills, per foot, the livt' load is a decapod engine and tender followed l)y a luiiforin Irain load of .'iOOO lbs. per loot: tind the maxinuim stress* s iu all the members. Chord Strewi '^, Mkmhf-rs. 2-4, l-Ci 0-8 ^-10 10-12 u-11 -250.0 -|0(i.O Dead load stresses . Live lord stresses . Maximum stresses . + 00.0 + i«;!.o + 100.0 + 280,*! + 210,0 + ;557.7 + 240.0 4 :508.0 '-253.0 + 440.0 + 507.7 + 0;!8.0 -650.0 Stressks N THE DlAOON.Vl-S. Mf.hiifbs. 2-3 8-C 5-9 7-10 U-12 n-10' «■->' Dcnd load etrestos .... Live loiMl stresses .... -127.3 -2.30.6 + 98.9 + 18fl..'i + 70.7 +146.9 + 42.4 +111.4 + 14.1 +S0.2 0.0 +54.1 (1.(1 +33.1 Mn.xlinmn stresops ... -357.S 42W.4 +217.6 +i.-ins +<>' 3 +40.0 -9.3 \ mvi* IMAGE EVALUATION TEST TARGET (MT-3) k{0 // ^^, Ik? MJi .<' M? <[<■ -k' %"*" i<'^ Ui f/. 1.0 I.I 1.25 PIIIIIM 12.5 Z 1112 IM m 1.4 20 i.8 1.6 i Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, NY. 14580 (/16) 872-4503 '' ^'^ '" ' -^i .: ' ' -'1^'i^^n;>:>; : --mJiJPP|^4^p^lil|if V^ ' ^ 4j ,% W^ /#- CIHM/ICMH Microfiche Series. CIHM/ICMH Collection de microfiches. Canadian Institute for Historical Microreproductions / Institut Canadian de microreproductions historiques O^ -^,^..-. ...-.• 1 -,^.- ..i;,: 17m»iiiwiiiiwffw»imiiii 1^«- Tile, on 3" fireproof blocks 30 to 36 lbs. The weight of the steel roof construction must be added to the above. For ordinary light roofs without ceilings, the weight of the steel construction may be taken at 5-lbs. per square foot for spans up to 60 feet, and 1 lb. additional for each 10 feet increase of span. (Manual of useful infor- mation and tables appertaining to the use of Structural Steel, as manufactured by the Tassaic Rolling Mill Co., Paterson, N.J. By George H. Blakeley, C.E.) Prob. 135. A truss. Fig. 02, with one end free, has its span 120 feet, its rise 30 feet, the ties, 3-4, 5-6, 7-8, 9-10, M ISC EL I A N i:0 US TR USSE8. 179 Corrugated 1 to 3; for ro liable to e not tight. lto2. weight per onstructiou. CovuniNos. 1 to 3 lbs. 2 lbs. 8 to 10 lbs. 7 to 9 lbs. 1 tolill^s- 1 to 11 lb.s. 2^ lbs. 4 to 10 lbs. 3 lbs. 4 lbs. 2 1b,s. 8 to 10 lbs. 16 to 20 lbs. 8 to 10 lbs. 30 to 35 lbs. ist be added out ceilings, len at 5- lbs. b. additional useful infor- f Structural ng Mill Co., free, has its .6, 7-8, 9-10, 11-12, vprtical, dividing the rafter into (5 ecjual parts; the dead load i)or panel is 2.5 tons, snow load per panel l.fi tons, normal wind load per i)anel, wind on fixed side, 2 tons : find all the stresses in all the members of the half of the truss on the windward side. ^ Stresses is the Loweu Chord. Mf.mukrh. 2 6 c-s 8- in 10-12 12-14 Dead load stresses . . . Snow load strcssi's . . . Wind load stresses . . . + 27.5 -16.5 + 17.8 + 25.0 + 15.0 + 15.6 + 22.5 + 1.3.5 + 13.4 +20.0 + 12.0 + 11.1 + 17.5 + 10.5 + 8.9 Maximum stresses . . . +61.8 +56.6 + 49.4 +43.1 +36.9 Stresses in the Upp ;r Chord. Mrmiiera. 2-3 3-6 6-7 7-0 8-11 11-18 Dead load stresses Snow load stresses Wind load stresses -.30.75 -18.45 -14.62 -27.05 -16.77 -13.00 -21.15 -16.09 -11.48 -22.35 -l.'Ul - 9.96 -10.55 -11.73 - 8.44 -16.75 -10.05 - 7.44 Maximum stresses -63.72 -5772 -61.72 -45.72 -.30.72 -.34.24 ii'^' »*ffir JM I tilfc gya^g'rT ■ I 180 ROOFS AND n HI DOES. fTBKHSES IN TIIK VeHTICALS. Memhkks, Dead load Hlresflo.s . . Snow load slressfS . . Wind load stresses . Maximum streases f>-« + 1.25 +0.76 + 1.12 + 3.12 7-8 + 2.50 + 1.60 + 2.24 +0.24 9-10 + 3.76 +2.26 + 3.30 + 9.36 11-12 + 5.00 + 3.00 + 4.46 + 12.46 13-14 + 12.60 + 7.60 + 6.56 +25.56 (3-4 is not necessary to tlie stability of the truss.) Stresses in the Diagonals. MRMItRKK. 8-6 6-8 7-10 0-12 11-14 Dead load stre-sses . Snow load stresses . Wind load stresses . -2.80 -1.68 -2.48 -3.5r -2.11 -3.16 - 4.50 - 2.70 - 4.04 - 5.66 - 3.33 - 5.00 - 6.70 - 4.02 - 0.00 Maximum stresses . -0.96 -8.79 -11.24 -13.88 -16.72 Prob. 136. A truss of the type oi fig. 62, with one end free, has its span 150 feet, its rise 25 feet, the ties, 3-4, 5-6, etc., vertical, dividing the rafter into six equal parts; the dead load per panel is 2.8 tons, snow load per panal 1.5 tons, normal wind load per panel, wind on fixed side, 1.8 tons : find all the stresses in all the members of the half of the truss on the windward side. Stresses in the Lower Chord. Mkmhkks. 2-n fl-8 8-10 1(V12 12 14 Dead load stresses . . Snow load slres.ses . . Wind load stresses . . +40.20 + 24.76 + 22.79 +42.00 +22.50 + 19.93 +37.80 + 20.26 + 17.00 + 3.S.00 + 18.00 + 14.20 +29.40 + 16.76 + 11.34 Maximum 8t.re^^ses . . j +93.74 + 84.43 + 75.11 +06.80 + 66.49 mmm i 13-U 00 00 46 + 12.60 + 7.50 + 6.56 46 +25.56 •i 11-14 66 33 .00 - 6.70 - 4.02 - 6.00 .88 -16.72 ith one end lie ties, 3-4, Bqual parts; d per par.al 1 fixed side, s of the half V12 12 14 53.60 8.00 4.20 +29.40 + 16.76 + 11.34 16.80 + 56.49 WittM MISCKLLA NKOUS Tit USSES. Strbssbs in the UrPEU Choki). 181 Mkmiikuh. Duiwl load slresseH . Snow load stresses . Wind load st'esses. 2-i !Wi B-T T-it 0-11 11-18 -48.97 -26.24 -20.74 -44.62 -23.86 -18.32 -80.69 -40.07 -21.47 -15.91 -86.62 -19.08 - 13.50 -31.16 -16.70 -11.09 -26.71 -14.31 - 8.96 Maximum stresses . -95.96 -77.46 -68.20 58.96 -49.98 Stresses in the Verticals. Mkuiikks. B-fi t-'» r-io 11-12 18-14 Dead load stresses . . . Snow load stresses . . Wind load stresses . . . + 1.40 + 0.76 -J-0.04 +2.80 + 1.50 + 1.89 + 4.20 +2.25 +2.84 + 6.60 + 3.00 + 3.78 + 17.02 + 9.12 + 6.76 Maximum stresses . . . +3.09 +6.19 +9.29 + 12..38 +31.90 Stresses ^s the Diaoonai-s. Memherh. 8-0 .'.-S 7-10 9-12 11-14 Dead load stresses . . . Snow load stresses. . . Wind load stresses. . . - 5.16 - 2.76 - 3.38 - 5.60 - 3.00 - 3.82 - 0.27 - 3.30 - 4.27 - 7.28 - 3.90 - 4.97 - 8. .34 - 4.47 - 6.65 Maximum stresses . . . -11.29 -12.42 -13.90 -16.15 -18.40 Prob. 137. A Fink truss, Fig. 10, with one end free, has its span 150 feet, its rise 25 feet, the rafter divided into four equal parts by struts drawn novii'.al to it, the dead load per panel 3 tons, the snow load per panel 1.5 tons, the normal wind load per panel, wind on fixed side, 2 tons: find all the stresses in all the members of the half of the truss on the windward side. ■; ^^sm MaSitWMili^ J 182 R00F8 AMJ BHWaES. Sthkshks in tiik Lower Chord. Mehbkiui. '.'-I 4-6 ft-m Dead load streHses .... Snow load »trc88e» .... Wind load stressos .... + 31.50 + 15.76 + 16.84 + 27.00 + 13.60 + 12.68 + 18.00 + 9.(H) + o.:J4 Maximum stressuH .... +03.09 + 53.18 +;}3.34 Stiikssks in the Ui'Pkr Ciiouu. Mkuderh. 2-3 «-a 6-T 7-9 Dead load atresseH .... Snow load stresses .... Wind load stresses .... -33.15 -10.58 -14.30 -32.22 -10.11 -14.30 -31.29 -15.06 -14.30 -30.30 -16.18 -14.30 Maximum stresses .... -04.09 -62.09 -01.30 -69.90 Stressks in tiik VVkii Mkmhers. Mkmiikks. 8-t, 7-3 £•-!) 4-5, 6-S c-s e-9 Dead load stresses . . . Snow load stresses . . . Wind load stresses . . . -2.85 -1.43 -2.00 - 5.70 - 2.86 - 4.00 +4.60 + 2.25 + 3.12 +- 9.00 + 4.50 +- 6.32 + 13.50 + 0.75 + 48 Maximum stresses . -0.28 -12.66 + 9.87 + 19.82 + 2ft 73 Art. 51. The Crescent Truss, Figs. 63 and M, is a good form for comparatively large spans. Riveted iron- work is used throughout. Prob. 138. A circular Crescent truss, Fig. 0.3, with one end free, lias its span 160 feet, the rise of the upper chord 32 feet, the rise of the lower chord 20 feet, the number of 6-10 + 18.00 + 0.(H) + tt.U +;w.34 7-9 9 -3(>.3(» 6 -10.18 -14.30 -59.90 -s fi-9 9.00 4.50 8.32 + 13.50 + 0.75 + 48 9.82 + 29.73 md M, is a iveted iron- 53, with one upper cliortl ! numher of I MI8CKLLANK0U8 TRUSSES. 1S3 l)aiiels 8, the apexes of both the ui)per and lowpr chords lying on arcs of circles, dividing thcin into 8 eqna,\ parts ; the dead load is 3 tons per panel, the wind load is 2 tons iTiu.oa per panel, the wind being supposed to act vertically on the fixed side only : ' find all the stresses in all the members of the half of the truss on the windward side. The computation of the stresses is effected by the appli- cation of the principles of Chapter I. The radius of the upper chord = IIG feet. The radius of the lower chord = 170 feet. The lengths of 2-3, 3-5, etc. = 22.04 feet. The lengths of 2-4, 4-6, etc. = 20.81 feet. The horizontal distance of each apex from the left abut- ment 2, and its vertical distance above the line 2-2', are as follows : (S) (5) (7) (9) 17.33, 36.94, 68.06, 80 feet. 13.01, 23.70, 29.91, 32 feet. (4) (6) (8) (10) 18.93, 38.76, 69.22, 80 feet. , 8.66, 14.92, 18.72, 20 feet. Hor. distance from 2, Height above line 2-2', Hor. distance from 2, Height above line 2-2', 1 n the normal wind pressure were taken, it would have a different value for each panel, owing to the curved surface of the rafters, which wonid increase the difficulty of the computation. 1H4 ItooFS AM) nniDGES. Tlie dead load reiu-tion = 10.fi tons. Tho stresses in tlie elunds are best found by inonientH; for the other nienihers the method of resolution of forces may be used. Thus, to compute the dead load stress in 2-il, we find the lever arm of 2-3 to be 4.89 feet. .-. dead load stress in 2-^1 = - lM>^«i?§ = _ 40.7 tons. 4.oy Similarly, the stress in 2-4 = IM >ill:33 ^ g^ ^ ^^^^ To find the stresses in 3-4 and 3-5 pass a secition cutting 2-4, ;i-4, and 3-5. Then, denoting the stresses in 3-4 and .'5-5 by s, and .% we have, for horizontal and vertical com- ponents respectively, 35.1 X .9095 + A-^ s, + .889 », = 0, yjlS' + 4.96* and 35.1 x .4150 - ^ ». + .4579 s, + 7.5 = 0. .•. s, = + 5.0 tons = stress in 3-4 ; and Sj = — 37.7 tons = stress in 3-6. The wind load reaction = 5,1 tons. 6.1 X 18.93 wind load stress in 2-3 = - 4.89 = - 19.7 tons. In this way all the stresses may be found. Stresses in the Top Chord. MiMHKRH. 2-8 8-5 5-7 T-9 Dead load stres-ses . . . Wind load stresses . . . -40.7 -19.7 -37.7 -17.0 -37.8 -16.6 —37.8 -14.8 Maximum stres-ses . . . -60.4 -54.7 -54.4 -52.6 ■,..... ^ etc., is o(iual lo one-eighth of the lower cliord, while the two end puiu'lH, L'-l luul 2-4', arc eacii e(|iial to IJ of the other panels, tliat 's, v(\\\n\ to 1 \ ei{,'lith,s of the Iowit chord ; liie (lead load is .'t tons per panel inid the wind load is 12 tons per panel, the wind hei'ig snpposed to act vertically on the fixed side: lind all the .stresses ii. all the menihers of the lialt of the trnss on the windward side. Here the radii of the npjter and lower chords are 110 and 170 feet, respectively, as in i'rob. 138, and the eoiVrdinates of the apexes of the upper chord are also the same as in I'roW. l;i8. The length of 2-1 is found to be 31.21 feet, and the cottr- di nates of 4 to be 28.77 feet and 12.0'.) feet. The reactions are the same as in I'roVj. 138. Di'ad load stress in 2-3 = — Dead load stress in 2-4 = Wind load stress in 2-3 = — Wind load stress in 2-4 = lO.fi X 28.77 8.26 ' " lO.f) X 17.33 6.83 6.1 X 28.77 8.26 6.1 X 17.33 5.83 - 3().G tons. 31.2 tons. - 17.7 tons. 16.1 tons. Strebseh in tub Tor Chord. Memiiiim. ^-A 8-8 M T-» Dead load stresses Wind load stresses -30.0 -17.7 -40.5 -18.4 -30.3 -10.8 —38.7 -14,4 Maxinuun stresses -54.3 -58.9 -56.1 -53.1 - •^• m aff rn m 7-» 3 8 -38.7 -14,4 1 -53.1 MlSVFf. I. .1 S Kors Til I ' ^ SA'.S. 187 Sl'IH-'.^HKH IN TIIK liorTOM ('lll)ltll MlMHRIW. !i-l 4-(i ft-s H-h' + ::rt.o + !•-'. 4 + 10.0 Dnul lliilll Htl'I'MMCH Wind Iciail utri'HMiH + 31.2 + 16.1 + 35.1 + 10.2 + :wi.o + 14.7 .Miixiiiiuin KtrcHHts + 46.3 + 51.3 + 51.3 Strksuks in TIIK Wkii Mkmiikiih. Mkhhkkh. 11-1 4-.'. K-ll <:-7 "-i s-!> Doail load stresses . . Wind loHil streBst'H ■ . + 7 2 + 3.0 +3.2 + 1.4 +4.1 + 2.2 + 2.7 + 2.0 + 3.1 + 2.0 +0.0 H3.0 +2.7 + 5.7 Maxinniiii stmnes . . . + 10.2 +4.0 +0.3 + 6. ft BiiiDOK Tbussks. Art. 52. The Fegram Truss - The Parabolic Bow- string Truss. — Tli« I'ognim truss, Fig. (i."i, (ionsists of U»e same nuiuber of panels in each cliord. Ml the panels of oiich chord are of equal Ipngth, the upi)or chord panels being shorter than the lower. The apexe.s of the upper chord lie on an arc of a cinrle, the chord of which, .V.'V, is about one and one half panel lengths shorbu' than the span. The rise of the <.ord members of this paiiel are horizontal, the stress in 9-8' is found by Art. 18. .-. max. stress in 9-8' = ^ x ^^- = 18.5 tons. 7 oo.oo 190 ROOFS AND li HI DUES. Sthkssks IV THE Upper Ciiohd. Mkmbkkh. !>.') 5-7 7-9 9-U' Dead lofwl stresseH Live load stresses -S3.6 -00.4 - 42.4 - 70.2 - 45.1 - 81.1 - 44.3 - 70.7 Maximum stresses -94.0 -118.6 - 120.2 - 124.0 STnESSBS IN TUB LoWEK ClIOIlD. Memiiekh. 2-4 4-0 C-i 8-8' Dead load stressep Live, loiul stresses +25.0 + 45.0 + 30.1 + 65.0 + 41.6 -t 74.0 + 44.0 + 80.3 Maxi.iuun stresses + 70.0 + 101.1 + 116.6 + 124.0 II Stresses in the Web Ties. Members. 8-4 e-« 7-3 9-8' »'-0' Deiid load stresses . . Live load stresses . . +20.6 +37.2 + 12.2 + 27.4 + 6.2 + 22.7 + 0.0 + 18.6 + 0.0 + 14.1 Muximum stresses . . +67.8 +30.0 +28.0 + 18.5 + 14.1 Stresses in the Web Struts. Mkmbebs. 2-8 4-6 6-7 8-9 Dead load stresses Live ioad stresses . . . - 38.8 - 09.8 -10.0 -27.8 - 1.3 -17.3 - 0.0 - 12.0 Maximum stresses . . -108.6 -38.4 -18.0 -12.0 m^Km + 0.0 + 14.1 + 14.1 8-9 -12.0 MISCELLANEOUS THUSfitS. 191 See Framed Structures, by Johnson, Bryan, iiinl Turneaure j also Engineering News, Dec. 10 and 17, 1887, and Feb. 11, 1801, where the answers differ very slightly from the above on account of the center member 9-9' being t^ less. Prob. 141. A through parabolic bowstring triisfj, Fig. (Jfi, has 8 pauels, each 24 feet long, and 32 feet center depth, the verticals are ties and the diagonals are struts ; the dead and live loads are 10 tons and 15 tons per panel per truss : find the stresses in all the members. SxRr^SSES IN THK TOP ClfORD. Memiiers. 2-3 - 65.0 - 97.6 ft T 7 9 Dead load stresses .... Live load stresses - 69.4 -104.1 - 61.8 - 92.7 - 60.2 Maximum stresses .... -173.6 -102.5 -154.6 -160.5 Dead load stress in each panel of lower chord = +■ 60 tons. Live load stress in each panel of lower chord = -|- 90 tons. Maximum stress in each panel of lower chord = -1- 150 tons. Stresses IX THE Diagonals. Mkmiikrs. r-a 5-S 7-10 4-8 6-7 0.0 -18.1 Dead load stresses . Live load stressi-s . 0.0 -13.0 0.0 -16.9 0.0 -18.0 0.0 -1.5.9 0.0 -18.0 Maximum stresses . -13.0 -15.9 -18.0 -16.0 -18.0 -18.1 uf fSS^- I, i 192 liOOFS AND UHIDGES. 8tke88k» in tub Vkuticals. ! Mkmiikkh. 3-1 5-8 T-9 9-10 Dead load strcssi'S .... Live load stresses .... + 10.0 + 15.0 + 10.0 + 19.7 + 10.0 +22.6 + 10.0 +23.4 Maximum stresses .... +26.0 + i!0.7 +32.6 +33.4 Art. 53. Skew Bridges ave those in which the end supports of one truss are not directly opposite to those of the other. Tho trusses of a skew bridge are usually placed so that the intermediate panel points are directly opposite in the two trusses, and the floor beams arc at right angles to the trusses. When the skew is the same at each end the trusses are symmefriml ; otherwise they are inis>immetriccd. In the analysis of unsymmetrical trusses, each truss must be treated separately ; and the stresses are to be computed for all the members of the truss. Frob. 142. Fig. C8 is a plan and Fig. 67 is tlie elevation of one of the two trusses of an unsymmetrical through ^ m it satwjiaiBBBwanBnw MISCELLANEOUS TRUSSES. 193 9-10 + 10.0 +23.4 +33.4 I the end ,o tliose of illy placed y opposite gilt angles eh end the immetriccU. truss must I computed e elevation il through 16 Pratt bridge; the span 2-18 is 120 feet^ *^«/«Pf ^^^ feet the panels 3-5 and 2-4 are each 18 feet, the panels ltl5 and 16-18 are each 12 feet, the other panels are ea.h 15 feet, the inclination of the - i. nmm il f ( 194 HOOFS AND ItKIDGES. Also the following greatest live load stresses: Stress in 3-6 =-■ ^i, [13.6 x 12 +- 15 x 285] 1.345 = + 49.7 tons. Stress in 5-G = -j^ll 02 + 15 x 198] = - 26.1 tons. Stress in 12-13 = ^^^ [297 + 15 x 222] 1.166 = + 35.2 tons. Chord Strebseh • MsMnEiw. 2-4 4-6 +20.8 +40.0 6-S 8-10 10-12 12-14 U-16 10-18 7-11 Demi load atreasos . . . Ltvo load Btressea . . . +l».l +:JSi.2 + 85. H + 71.fi + 43.1 + 88.2 + 41.0 + 82.0 +31.5 +88.0 +19.S +38.0 +20.8 +40.0 - 44.8 - 89.fi Maxliiiuin strcsans . . . +57.8 +00.9 + 107.4 + 129.3 +12.S.II +94..^ +57.9 +00.9 -184.4 Stresses in tub Diagonals. MKMBBR8. 8-1 8-0 5-8 7-10 9-12 11-14 Dead load stresses . Live load st,resses . + 8.3 + 16.6 +23.2 +49.7 + 12.2 +35.1 + 2.8 +23.0 - 6.6 + 13.5 -15.9 + 6.3 Maximum stresses +24.0 + 72.9 +47.3 +25.8 + 6.9 - 9.6 Mkmukbs. 15-10 14-15 12-18 lO-ll 8-9 0-7 " Dead load stresses . Live load stresseo . + 6.8 + 13.6 + 23.6 +48.8 + 15.9 +35.2 + rt,6 +23,9 - 2.8 + 14.7 -12.2 + 7.7 Maximum stresses . +20.4 + 72.4 + 51.1 +30.6 + 11.9 - 4.r. We see from the above that, theoretically, the diagonals 6-7 and 11-14 are not needed. I MKBter )45 tons. ;5.2 tons. lft-18 7-11 +20.8 +40.6 - 44.8 - 89.6 +60.9 -184.4 11-14 6 6 -16.9 + 6.3 .9 - 9.6 6-7 ■ .8 .7 -12.2 + 7.7 .9 - 4.f> . i | i ^^ j|| i ij^^ ii j^ I . w ^ i jujj < J »|Fi ' ^ i ' I- ■ ■ .14 i ^ ^ I ! ". . -V" ■ ' . ■■ ' . ■ ^ '^ * ^. 1 ' W! " [ ■' iipwi L i t y i. ii . i' i y'^ppj MISCELLANEOUS TliUSSES. Stresses in the Posts. 196 MIMBKK8. 2-3 6-6 7-S 9-10 11-12 11-14 16-18 Dead load stresses . Live load stresses . -31.0 -63.8 - 0.8 -26.1 - 2.3 -17.1 - 0.0 -12.6 - 5.M -20.5 -12.8 -30.2 - 33.8 - 67.6 Maximum stresses . -05.7 -35.0 -10.4 -12.6 -25.8 -43.0 -101.4 Prob. 143. Let the dimensions of Fig. G7 be as follows: span = 144 feet, depth = 24 feet, the panels S-5 and 2-4 = 21 feet, the panels 13-15 and lG-18 = 15 feet, all the other panels = 18 feet ; let the dead and live loads be 800 lbs. and 2000 lbs. per foot per truss: find all the stresses in all the members. diagonals i