IMAGE EVALUATION TEST TARGET (MT-3) / o .w> Si' WP Wr M/ ^ 1.0 fM IIIIIM I.I 1.25 |r 140 12.0 1.8 \A. ill 1.6 (^ ""y.* /a >^M. e. '^W v: ^^^ / V'i^ .^ / V °m M fv 5V <^ ^ Cv V # e nationale du Canada Maps or plates too large to be entirely included in one exposure are filmed beginning in the upper Inft hand corner, left to right and top to bottom, as many frames as required. The following diagrams illustrate the method: Les cartes ou leb planches trop grandes pour dtr«i reproduites en un seul cliche sont film^es 6 partir de Tangle sup6rieure gauche, de gauche d droite et de haut en bas, en prenant le nombre d'images ndcessaire. Le diagramme suivant illustre la m^thode : 1 2 3 nt 1 2 3 4 5 6 r ■ITT I APPLIED MECHANICS. 'i BY HENRY T. BOVEY, M.A., ASS. Mem. Inst. C.E. ; Mem. Inst. Mech. E. ; Mem. Am. In.st. M.E. Professor of Civil En.ineerinr, and Applied Mechanics, MrGill University, Montreal. tMomoj Queens' Colkye, Camhridye, (Eng.). MONTREAL: PRLNTED BY JOHxN LOVELL & SON. 1882. 6^ /-?^ 207 a /3^ I :; ' , ^. ^A. in thp vear One Thousand Eight .• »trv Aot of Parliament of Canada, m tne year v»i.c Knterkd according to Act of rariia , . c„;„ j„ tho Office of the Minister of Hundn.d and Eighty-two. by John Lovkll & Son, m Agriculture. \ CONTENTS. r of CHAPTER I. FRAMES. PAfJE Definitions , Frame of two Members ^^ Frame of three or more Members 2 Method of Sections < ^'■^"^s !ll!V.''"l*l°!!'l ib. Bridge and Roof Trusses g Compound Roofs 25 Gothic Roof Truss Ig CHAPTER II. BOOFS. Types of Truss I g Principals, Purlins, etc 19 Roof Weights 21 Wind Pressure 22 Methods of Determining the Total Stresses in Roof Members 24 Tables 32 Examples ^ i CHAPTER III. BRIDGES. Classification 42 Depth of Truss ' j^^ Position of Platform i^ Number of Main Girders 44 Bridge Loads 4- Chord and Web Stresses 4,; Friction Rollers ^^ i IV CONTENTS. Trellis Girder Warren Girder Howe Tru.MS Pratt TruHH Fink TrusH Bollinan Truss • Quadrangular and Post Systems Bowstring Girder Bowstring Girder with Isosceles Bracing ^^ Lenticular Truss ' 70 PAOB 47 49 55 59 CI 62 63 ib. ber. =fi. Let iS',, St, be the stresses along OA, OB, respectively. ■'• P~ 00 ~ nina *" /' ~ 0'0~ siiui (3.) Franie of three or more inemhers. Let .'li^j^'lj ... be a polyf:;oiial franie jointed ut vl,, .ij, J3... Let ]\, 1\ 1\ ... be the resultant furce.s at the joints J,, A^, A3 ... respectively. Lit >S'„ jS'j, »S', be the forces alonS'i, and Os^ to S^. Similarly, F^, arts, and if each part is considered separately, ,.ne stresses in the hars (or memhers') intersected by the secant plane must balance the external forces upon the part in question. Hence, the algebraic sums of the horizontal components, of the vertical components, and of the moments of the forces with respect to any point, are severally zero, i.e., analytically, S(X)=o, I,(V)=o, and M=0 These equations are solvable if they do not involve more than three unknown quantities, i.e., if the secant plane does not cut more than three members, the corresponding stresses are determinate. (5) Jib-Crane. Fig. 7 is a skeleton diagram of an ordinary jib- crane. OA is the post fixed in the ground at 0. CRANES. OB is the jib, AB is the tie. The jib, tie aud gearing are sus- pended from the top of the post by a cross head, which admits of a free ro- tation round the axis of the post. Let the crane raise a weight W. Three forces in equilibrium meet at B, viz., W, the tension T in the tie, and the thrust C along the jib. They may be represented by the sides of the triangle AOB to which they are respectively parallel. .*. T = W. i^andC= W 12 AO AG The stress S in the chain tends to produce a compressive stress of equal magnitude in any member along which it passes. Thus, if it pass along AB the resultant stress in thj tie is T-S, while if it pass along BO, the resultant stress in the jib is C + S. W Also, neglecting friction, S= — , n being the number of falls of chain from B. Again, the post is a semi-girder fixed at one end and acted upon at the other by a force T. Let Mj. be the bending moment at any point E of the post distant x from A. M^ =T.CosBAD.x=W. AB ^^\.x^wM.x AG' AB AO The bending moment at is evidently W.AD, and is independent of the height of the pf^at. The vertical component of T, viz., T. — W. — , is transmitted AO' through the post. Hence, the total unit-stress in the layers of the transverse section at E, distant v from the axis, is — ^ ■^~ ± -7 -'"'' ' ^ 'A AO J A being the area of the section, and / its moment of inertia with respect to the axis. The total resultant pressure along the post at rp • T, An^n ' onn- w BD , .jr BO BD + AO = — T. sm BAD + C. sm BOF = — ^- -tt{-\- ^V. -_- — - — : AU AU MU w If the post revolves about its axis, the jib and gearing aie bolted to i, and the whole turns on a pivot at the toe G, In this case the frame, G CUANES. as a whole, is kept '"n equilibiiuin by the weifiht M\ the horizontal re- action //of the curb-plate at 0, and the reaction /£ at G. The first two forces meet in /', and therefore the reaction at G must also pass throuirh F. Hence, since OFG is a trianjjjle of forces, //= W. ^, and R = Tr. ^. OG GG (CO Derrick- Grave. Fij^. 8, shews a combination of a derrick and crane called a der- rick-crane. It is distiniiuished from the jib-crane, by having two bac/i- sf,(ii/K, which are u-iUiilly situated in planes at ritiht-aiigles to each other. One end of the jib is hinged at or near the foot of the post, and the other is held by a chain, which passes over pulleys to a winch on the post, so that the jib may be raised or lowered, as recjuired. The derrick-crane is gevierally of wood, is simple in construction, is easily erected, has a vertical as well as a lateral motion, and a range ccjual to a circle of from 10 to G(» ft. in radius. It is, therefore, very useful for temporary works, setting masonry, &c. The stresses in the jib and tie are obtained as in ^(5), and those in the back-stays and post may be calculated as follows : — Let Orr OK, OL. be the horizontal traces of the tie and back-stays. In //O produced, take ON^ to represent the horizontal component of the stress in the tie, i.^ ., T, sin f/,« being the inclin.ition of the tie to the vertical. Complete the parallelogram X Y. CRANES. very OX and 01' are the horizontal components of the stresses in the back- stays. Let the angle NOX = 6. .•. OX = ON. cos d = T. sin a. cos 6, and is a maximum when — 0" OY=OX. sin d= T. sin a. sin f), '• " '' t)=9{\o Hence, the stress in the back-stay is S'a and 7\ ni;iy be obtained in terms of »S'i and therefore of W; four forces >S'j„ f„ /S,, C,, act through (3), and the values of iS",, 6V being known, those of aS'^, T^, may be determined. Proceeding in this way, it is found that of the forces at each succeeding joint, only two are unknown, and the values of these are consequently determinate. il 8 TIMBER TRUSSES. The calculations may be checked by the method of moments and by the stress diaj^ram, (Fig, 11). e.(j., let W= 10 tons. Take moments about the point (7). .'. Tr. (y7)==10. (x?) or T, = 3. 25 1. 25' 10 = 26 tons. No other forces enter into the e(|uation o^ moments, as the portion of the crane above a plane intersecting 08 and passing through (7) is kept in ecjuilibrium by the weight of 10 tons and the stresses Tj, S^, Ce) the moments of S,-, and C^ abou [7) are evidently zero. .3. 9 Again, from the stress diagram, 7\ = QR-= ,-^- 10 = 26 to>ns. (8) Timhcr bridge ), respectively. P P are — . co< a, and _ 2 2 sin a If the girder is uniformly loaded, Pis one-half of tho whole load. {It). — In Fig. 18 a straining ciU, E F, is introduced, and the girder is supported at two points. Let P be the weight at each of the points E and F. The thrust in EC (or FD) = P- 1,^ and the horizontal thrust At in the straining piece = 7'. iiii- "* AC If a load is uniformly distributed over AB, it may be assumed that each strut carries one-half of the load upon AF (or BE), and that each abutment carries one-half of the load upon AE (or BF). i TIMBER TuUSSES. By means of straining cillsj the jjirdcrs may be supported at several points 1, 2, ... and the| weight concentrated at each may be assumed to be one-half of the I load between the two adjacent points of support. The calculations for the stresses in the struts, etc... are made precisely as above. If the struts are very long they are liable to bend, and countcrbraces AM, iiiV, are added to counteract this tendency. (ndicnlar to FC, the angle COB = /3 ; the angle A CF = W—ACO = 00"— 3 — "i Draw AM pcrpcndicnlar to the direction of /', and consider the rafter A(\ As before the thrust /j*, at .1. the resultant weight >r, at the middle point D of JC, and the thrust P at C, meet iu the point F. Take luouients about A, .*. P AM = ir,. AE But AM = AC. sinACM= AC. cnsT^a^ and AE= ^. cos. ax, W\ co. $ The vertical thrust upon the support A =-- W^ - P. cos 3 B =. W, + P. ens $ (d). — Kfng-pnsf truss. The simple triangular truss may be modified by introducing a king- post CO, (Fig. IG), which car ries a portion of the weight of the beam AB, and transfers it through the rafters so as to act upou the tic in the form of a tensile stress. d KING-POSTS. 11 *( r the It >r, point Let P bo the wvij^lit borne by tlu- kin<,'-po,4 ; represent it ))y (JO. Dr?,w OB parallel to BC and l)E parallel to AB. CF I' 1 DC= . '' ■ = TV — 1 is the thrust in CA due to /*, and is of course sm a I sin a, equal to DO, i. e., the thrust alonu; CB. P DE — CE.cot a = .-r . cot n, is the horizontal thrust on each rafter, and is also the tension in the tic due to /*. Let IF be the uniformly distributed load upon each rafter. The total horizontal thrust upon each rafter = ( ir f /*) r rot a IT The total vertical pressure upon each support = ^'+-:7 If the ape.x; (J is not vertically over the centre of the tie-beam, (Fi, together with z W' an \ipward pull of — at E, it which is the manner in which the beam AB neutralizes the efiFect of the weight W . The beam AB is therefore acted upon by two opposite forces, the one at F^ the other at 6?, ejich being equal Xa Z- m. magnitude. Let R be the reaction at A ; let AB ■=. I and FG — c. . p, W \l^c W' l-e ' W , „ W c • • R.l == — -o~ "T^" + — •' — 1-= — — • c, and R — — — . z. 2. I 2 2 2 2 1 The magnitude of the bending moment at F and G, at which pointe . . . 1 , . . W' c I r it 18 evidently a maximum, is . — ^ '212 The ordinates of the dotted lines, Fig. 24, are proportional to the bending moments due to the two equal and opposite forces — FF being the bending moment due to the upward pull at B, and GG' that due to the downward pull at E. Again, the rigidity of the truss may be increased by introducing diagonal braces BG and EF, as shewn by the dotted lines in Fig. 22. H" l-e : ■ COMPOUKD ROOFS, 15 El £ 2 'I points The shoaring force for points between /' and (i is ^ mAxinium whf n ^r> _ yy. _ II'. ^_^ the load is at /\)r G, itn vahie being ± — c -t- — = -+- y"" ~7~" ^ '^'^ force has to be transmitted to the horizontiil b^'ams throufjrh the uitidiuui of a diajj^onal, which is iJli when the load is at /' and KF when the load is at U. Thus, the nnijrnitude of the greati^'st thrust iu either of the diai^oiials is, —r ' — r-, ^ 2 I (I ' s bein^ the K'nirth of a diagonal, and d the depth of the truss. (10) (li'iicnil ruu'trks. — In the trasH«.» described in (), th^; vortical members are ties, i.e., are in tension, and the inclined members are struts, /.»■., are in compression, liy invertint^ the respective fiLrures another type of truss is obtained in which the verticals are struts while the inclined members are ties, lioth systems are widely ustd. and the mtiithod of calculatiuj,' the stresses is precisely the same in each. In desitniinir any particular member, allowance must be made for every kind of stress to which it nuiy be subjected. The collar-beam J)J'J in ri_u'. -1, for example, must be treated as a pillar subjected to a thrust in the direction of its length at er^ch end ; if it carry tv transverse load, its streiiirthas a beam, support<-Hi at the points DdwdJJ, must also be determined. Similarly, the rafters AC, BC, Fig. m, etc., must be designed to carry transverse loads and to act as pillars. But it must be remembered that struts and queen-posts provide additional points of support over which the rafters are continuous, and it is there- fore practically sufBcient to assume that the rai'ters are divided into a number of short lengths, each of which carries 07ie- /laZ/oi' the load betweeu the two adjacent supports. When a tie-beam is so long as to re- quire to be spliced, allowance must be made for the weakening effect of the splice. (11) Compound Roofs. — In mansard rov^fs and many other framings a number of rafters are made to abut against each other as in Fig 25. The relative position of the rafters is fixed by the condition that the horizontal pressure upon each must be the same, ■ ■ ,• /_;j;Vll. -^K, * ■■■"-' 'v> .. Vk ' ^A ^ /I/ ^-|f .^A/ ^^■^^■S^^E^H 1 ■ " •/ / / 'i , — In riozs 16 GOTHIC ROOF TRUSS. in order that the prcflsurc may be completely tranamittod from one rafter to the next. Let IK„ ir„ ir, ..., a„ a„ a,... be the wei}?ht3 and inclination.s to the horizontal of the rafters AJi, B,C, CD... respectively. Assume that the weights arc concentrr.ted at the points A, B, G in the proportions ^, -^-^-^\ ^^^^\ , respectively. The triangles BFA, BAG, CGII,... represent the forces at these points. Hence the horizontal pressure upon AB = -y- -^j, = -^-cotai = //» The resultant thrust along BC= — L_^ — ?. --- = — I- — . -^-- — and its horizontal component = ; . — ■. — — = Ha. 2 sin ia-i - «i) Similarly the horizontal component of the thrust along CD W'i + Wi ens a.2 cos a, ) 2 sill (fla -Ui) = //j, and so on. But III == Hi == //j . ... W Wi + T^ cos tti. cos a-i ''2. '"' "' - 2 • sin (a, - a,) TT,. or —7-" coi a\ = Wi -y W, 2 tan a-i — tan Uj W^ + W3 COS 052 COS flj 2 * COS (aa ~ 02) _ \V^ +_^,. 1_ 2 tan 03 — <«'i , C, E, and B, in the proportions -j-, -^, -^-j "2", d GOTHIC KOOF TKUSS. 17 w and -J-' rcspcctivoly. Hence the total weij^ht uj)on the pri- uiary truHs == -W. Trace the stress (liairraui ; the fine oflomh MN is evidently vertical ; take MN » -^ . W i draw OM, ON, parallel to DK, EL, respect- ively. The horizontal line Oil is the thrust aldiijj; L)E, and OJ/is the thrust aloni:; DK. Draw or parallel to AD; the sides of the trianule OTIl represent the stresses in the members AD, DF, and FA. Also, since DF is equal and parallel to AK, T is the middle point of Mil. The thrust along ^lA" =^ the thrust along DF + portion of weight concentrated at A •trr The secondary truss DCE carries a weight -^ at C W Take IIS ^= ^ and draw SO' parallel to CD ; SO' II is a triangle of forces for one-half of the secondary truss. W Also, OH - Oil . OH an IIS 4 2 : .J — = lu and the resultant thrust aloag DE is 1 OH. 1 CHAPTER II. Roofs. (1). — Ti/pcs of Truss. — A roof tru.«s may bo constructed of timber, of irctii, or of timber and iron combined. Timber is almo.'it invariably employed for small spans, but in tlie longer spans it has been larirely superseded by iron, in consequence of the couibined liuhtness, strength and durability of the latter. Attempts have been made to classify roofs according to the mode of construction, but the variety of form is so great as to render it imprac- ticable t — " and double ][ ) bars, bulb tee (T) and rolled (I) iron beams, are all excellent forms. Timber rafters are rectangular in section, and for the sake of economy and appearance, are often made to taper aniformly from heel to ptiak. Tlie heel is fitted intti a suitable cast-iron skew-back, or is fixed between wroughtriron angle brackets, (F'igs. 10, 11, 12), and rests either directly upon the wall or upon a wall plate. When the span exceeds GO-ft., allowance should be made for alterations i 20 PRIN'CIPALS, PURLINS, &C. of leniith due to chariijesof tempcraturo. This may be effoctctl by inter- posing'' a set of rollers between the skew-back and wall-plate at one heel, or by fixing one heel to the wall and allowinj;- the ojjposite skew-back to slide freely over a wall-jiiate. The junction at the peak is made by means of a casting or wrought iron plates, (Fig«. 13, 14, 15). Light iron and timber beams as well as angle-irons are employed as purlins. They are fixed to the top or sides of the rafters by brackets, or lie between them in cast-iron shoes, (Figs. 1(J to 21), and are usually held in place by rows of tie-rods, spaced at G or 8 ft. intervals between peak and heel, running the whole length of the roof. The sheathing boards and final metal or .^late covering is fa.stened upon the purlins. The nature of the covering regulates the spacing of the purlins, and the size of the purlins is governed by the distance between the main rafters, which may vary from 4-ft. t^) ui)wards of 25-ft. Hut when the interval between the rafters is so great as to cause an undue deflection of the purlins, the latter should be trus^ d. Each purlin may be trussed, or a light beam may be placed midway between the main rafters so as to form a supplementary rafter, and trussed as in Fig. 22. Struts are made of timber or iron. Timber struts are rectangular in section. Wrought iron struts may consist of L irons, T-bars, or light columns, while cast-iron may be emjtloyed for work of a more orncmental character. The strut-heads :;re attached to the rafters by means of cast caps, wrougbt-iron straps, brackets, &c., (Figs. 23 to 26), and the strut- I or ligbt ■ ROOF WEIGHTS. 21 iVet are easily desigiu'd both for pin and screw connections, (Fius. 27 to 30). Ties may be of flat or roniid bars attached either by eyes and pins or by !^es in tlie different mem- bers of a HMif truss two kinds of load have to be dealt with, the one jn'niiiniriif and the other „ „ normal wind pressure per sq. ft. of roof surface. d „ distance in feet from centre to centre of principals. /, and /jj be the lengths in ft. of two adjacent panels. The total ^>iT7««Ht'«Moad at the panel point on the windward side 9 'sr.W. d. 1' 24 EXAMPLES. The total accuhuttil load at the panel point on tlio windward side ^^ /'- „) maybe resolved into its vertieal (/>,) and horizontal (/'/,) eomponeiits ; j>,. may be eombined with the perman- ent load and />,, dealt with independently ; the respeetive efleets are to be added toiictiier. (I)). The )»ernianent and aceidental loads may be treated separately, and the respeetive effi'ets added toiiether. It' / is the Uiii:th (d'a ratter, and it' A" is tlie reaction at the windward sup|iort (hie to the horizontal jiressure of the wind, by takinu' moments about tile leeward sujiport, h .1 . /. i'ii>y II • y*„. .s/// (I. I. il . — =<>,orA = — /'„ — 7-' '. Hence. A'' is inifnttrr and acts downwards ; it must be le.ss than the reaction at the same su]>port due to the ju'ruuinent load and the vertieal wiiul pressure, or the root' will blow over. AV. 1. — Method ((/) aj»plietl to the roof tru.ss repre.sented in Fiir. 38- EXAMPLES. 25 (1 side itoly, Fi<^. 34 is the stress iliap:ram for the vertical load upon the roof. /> (j is the vertical reaction at H = -^- (>"«• '"«" + 2. cO- yz', />/•, /•«. yw, ifjjrcscnt tlu! stresses in the nienihers to \vhi<'li they are respectively parallel, viz., the rafter J /A and the ties HI), /)A. l>h\ Fi<;. 3,1 is the stress diai-rani for the load due to the liorizontal com- ponent oi'tlu^ wind jiressure when there are no rollers under the heel ol' either of the rafters. The horizonta! reaction at each heel may be assumed to be one-half of the total horizontal force of the wind, and e(jual tt) -■ "' ' .sin a. , , . , , , . ,, }},, I. (I sill'' a J) q IS the dowmimrd reaction at Ji= — . 4 ro.s ii The stres.se8 in the ineud)ers AB, BD, DA, I)E, arc represented by the lines q'r',j//, r's\ y/.s'. to which they are respectively parallel, and are evidently reversed in kind. Thus the total resultant compression in Ali = (jr - ij'r', the total resultant tension in HI) — ^n- p'r', in DA = rs - r's', in IJI'J = i>s-j,'s'. Fi^. 30 is the stress diagram for the load due to the horizontal com- ponent of the wind pressure, when rollers are placed at B. The total horizontal reaction is now at f, and equal to y)„. h <1. sin = — 4 ''".s' 1 1 t' y' is the Iidrizniital force of the wind at Ii. q' /•'. // /■', /•' s\ p s', are the stresses in AH, HP. DA, Dh\ respec- tively, and the n'sultant stres.seH in the dift'erent members are q r-q' / p i—j> /•', r s-r' s'. and /ts~p' s\ as ))efore. , Fig. 37 is tlie stress diagram for the load due to the horizontal com- ponent of the wind pressure, when rollers are placed at <\ The total horizontal reaction of the wiiul is now at B, and equal to ^>„. f. (h sin ii. The horizontal force of the wind at Bis . shni, so that the re. sultant horizontal force at B is ".,-" i^ln >' = t' q'. 2* t IS the duicnicnfd reaction at B - — r— . CUS u 26 EXAMPLES. The resultant stresses are obtained as above. The stress in I)E equilibrates the stresses in DA. BD, and also thoso in A\l, EC, so that the stresses in DA, DB arc respectively e(iual to those in EA, EC. The dotted lines in the diagrams represent the stresses in the mem- bers .1 C, E (\ E A, to which they are respeetivcly parallel. Ex. 2. — Method („. '•'>.s " + 2. ?'•)• qi\ OS, sr, pr. sf, pf. represent the stresses in the members to wliicli they are respectively parallel, viz.. BE. FA, DF, DB. DA, 1)E. Fig. 4(» is the stress diagram for the load due to the horizontal component of the wind pressure, when no rollers are placed either at B or C. p„. I. d .fin- a j>'/)i' is the downward reaction sit B = — ;j^ — cos^' pn. I. d m'(j' is the resultant horizontal force at B == . s*'«. a. The line througli •">' '< + '^. "")• g^'o' ia the weight at G =- -^. I. d. w. Ex. 3. — Metliod {(i) applied to the roof truss represented in Fig. 42. Fig. 43 is the stress diagram due to the vertical load upon the roof. pq is tlic vertical reaction at B r- P"-''"" " . (3. BIl + All), d + -.(4. BH + 2. AH ). d. 28 KXAMPLES. qm is the weight ^i F - IliiiSH^^JLl^ . BII --= the wi-i-lit at //-- mn Fiir. 41 i« the Htrcss cliaiinim duo to the hnrizoiital e<.»iui»onent of tlie wiiul pressure, when rollers are placi'd at li. />„• I- 'f •*•'"'' " //'i''. rcsjuctivdy. E.C. 4. — Mt'tliod i^ii) ap|(li('d to the roof truss represented in Fit:'. 45. Fig. 46 is the stress diagram due to the vertical load upon the roof. jiq is the vertical reaction at7i= — -. (10. p„. cos a + 14. n-). qm = mil = no, is the weight at 0, F, or Q, and is e(i[ual to — . [j),,. cost! + n-). Tf />„. rnn II =- y. tlie ;ioint o coincides with ^). To avoid ambiguity it is gem-rally assumed that the stresses in IIO, JIF, are ecjual to those in LQ, LF, respectively. KXAMI'LES. 29 Fi«r. 47 is the stress diaurain duo to tho horizontal couiponoiit of the wind pressure, wlieii rollers are placed at ('. , „ . , , , . „ />,, /. '/ «///* '/ J) I IS tlie downward reaction at xj = , — , . 4 ra.v II /' q' is the resultant horizontal force at B == -. j)„. I. „ cos a + 1(». ir), gm = 7)1 n = --— . (/>„. ros n + w), is the weight at F or //. o Fig. 50 is the stress diagram due to the horizont;il component of the wind pressure when rollers are placed at B. '£) I IS tlie downward reaction at li =- — - — V q' is the horizontal force of the wind at B == cos a p„. I. d . , sin (I. Ex. 7. — Method [h) applied to the roof truss represented in Fig. 51. Assuming the direction of the wind to be horizontal, th>' load at any joint of the bowstring roof in Fig. 51 due to wind may be taken to be 30 EXAMPLES. the pressure upon a surface o(jual in area to the portion of the roof supporttid by one panel or bay, and inclined at the same anf];le to the horizontal as the tanjirent to the curve of the roof at the joint in question. Let It be the resultant of the load /*, , 1\, I\, I\, at the joints B, d, c, /, and let it make an anj^'le tt with the horizontal. Fig. 5li is the stress diagram due to the normal wind pressure, when rollers are placed at C. The whole of the horizontal reaction (= K. cos f>) is at B, and is represented by mn. nt is the vertical reaction at B. ts, sr, rq, qp^ represent the loads />,, j9j, ^^3, p^^ respectively. «(?', rV §2', p3', are the stresses in Bd, de, c/,/A md',me',m/\ma', " '' 51,12,23,30 )-that the stress on BC Sit B is ir.-y and makes with the , (r).-that the stress on ^^ at B vertical an angle t).-the tension of the string AD is 2 ir, and determine the magnitudes and directions of the stresses at the joints, assuming that the strings are connected with pins distinct from the bars. (7). Shew that the stresses at C and F in the first case of Ex. 6 remain horizontal when the bars AF, FE, IU\ CI), are replaced by any others whieh are all ecjually inclined to the horizon. (8). An ordinary jib-crane (Kig. 7, Chap. I) is required to lift a weight of lO-tons at a horizontal distance of (i-ft. from the axis of the post. The post is a hollow cast-iron cylinder of 10-ins. external diar. ; find its thickne> principal members are two wrought-iron I rods ABC, supported at H upon two timber uprights BD, and connected at A and Cwithj two timber longitudinals ADE. The crane rests upon a truck, and the 36 EXAMPLES. centre of pressure is maiiitainecl at /> by iiieans of ;v movalile balanee- l)ox V)etweeii I) aial A'. 'I'lie crane is recjuired to lift a weight of 2-tons. deteruiine the stresses in the different members. (Data: — AK—M ft.. AI)^'.V,)-{\., D('='l-2-\\., BD^Mt., vertical distance between J and/;-7ift0 Point out a method by which the members may be efiectively braced together. (12). The inner flantre of a bent crane. (Fig. 10. chap. T), forms a (jnadrant of a circle of 2(>-ft. radius, and is divided into/oKr e(|ual bays. The iHifir flange forms the segment of a circle of 2.")-ft. radius. The two flanges are o-ft. apart at the foot, and are stru(,'k from centres in the same liorizuntal line. The bracing consists of a st'ries of isosceles triangles, of which the bases are the eipial l)ays of the inner flange. The crane is required to lift a weight of lO-tons, determine the stresses in all the members. (];}). A braced semi-arch is 10-ft. dei>p at the wall and ]>rojects 40-ft. The upper flange is horizontal, is divided into /'**//• eijual bays and carries a uniformly distributed load of 4n-tons. 'I'he lower flange form-! the segment of a circle of lU4-ft. radius. The bracing consists of a series of isosceles triangles of whica the bases are the e(|ual bays of the upper flang(>. |)etermine the stresses in all the meud).'rs. (14). Three bars, freely articulatt'd. form an e(piilati'ral trianglt! AJi(\ Tile system rests in a vertical plane upon suj»port< at /i ami (^ in the same horizontal line, and a weight IT is suspended from .1, determine the stress in Ii(\ neglecting the weight of the bars. (15). Three bars, freely articulated, iorm a triangle . I /^<^. and the system is kept in c((uilibrium by three forces acting on the joints, deter- mine the stress in each bar. What relation holds between the stresses when the lines of action of the forces meet. ('*).-in the centroid. (A).-in the orthoeentre of the triangle ? (16). A triangular truss of white pine consi.-^tsof two c(iual raftiTS ..J li, AC, and a tie-beam B(^ ; the span of the truss is l{l>-ft. and its rise is 7^-ft. ; the uniformly distribute I load upm each raft.M* is S4ers if ^ifl be subjected to a horizontal pres: ure of 156-lbs. per lineal ft. ? (23). Determine the dimensions of all the members of the truss in the preceding question, assuming the tie-beam to be also loaded with a weight of 12tM)-lbs. per lineal ft. (24). A roof truss consists of two equal rafters inclined at 60° to the vertical, of a horizontal tie-beam of length /, of a collar-beam of length n, and of a queen-post at each end of the collar-beam ; the truss is loaded with a weight of 2600-lbs. at tlie vertex, a weight of 4000-lbs. at one collar-beam joint, a weight of 1200-lb8. at the other, and a weight of 38 EXAMPLES. 13,500-lbs. at the foot of each quci'ii ; doteruiiiie tlic stress in each inoiuber. (25). Tho platform of a bridj^o for a cloar span of I 60-ft. is carried by two tiuibcr trusses 15-ft. deep, and of the form shewn in the Fij;. ; the load upon the I bridge is 5n-lbs. per sq. fi. of platform, which is 12- 1 ft. wide ; what should be the diuieiisions of tho ditiereut members, assumiiifz; the workiuij; stress to be 4('(l-lbs. per sq. in. ? If a single load of tiOdO-lbs. pass over the bridjje, and if its effect be equally divided between the trusses, find the greatest stress produced in the diagonals ^1.1, BB. (26). A frame is composed of a horizontal top beam 40-ft. long, two vertical struts 8-ft. long, and three tie-rods of which the middle one is horizontal and 15 ft. long; liiid the greatest stress produced in the several members when a single load of 12.(l0<)-lbs. passes ov.t th e truss. (27). The truss represented by the Fig. is conqioscd of two e(|ual ratters, A B.AC fu'.r equal tie-rods A I). BIJ, AE, C/'J, and the horizontal tie DE. If ir be the distributed load upon AB, IP that upon AC, s the span, $ the angle ABD, and/ the safe ineli-stre.-s in the 5 \V + W metal, shew that the total weight of the tie-rods is —. / s. cot $. GO" - 1 (28). If the rafters in the preceding Question are inclined to tho horizontal at an angle a, and if the tension on BD or EC is equal to that on DE, shew that /3 =- «, (t. e., D and E must fall in the hori- zontal line BC), or ^ ■ Petermine the stress in ^D corresponding to the latter value of/?. (29 1. The trusses for a roof of the same type as that in Question 27 are 12-ft. centre to centre, the span is 40 ft., the horizontal tie is 16-ft. long, the rafters are inclined at OlC'to the vertical, the deadweight of the roof including snow, is estimated at 10-lbs. per sq. ft. of roof surface, there are no rollers either at B or C ; determine the stress in each mem- ber wlien a wind blows horizontally o.i one side with a force of 40-lbs. per sq. ft. of vertical surface. (30). The accompanying truss is similar t(> that in question 27 with the addition of the struts l)F, EG. (a). — With the same data as in question 27 shew | f> a W+(W^ W). cos 'a that the weight of the metal in the tie rods is -• - • . , b y • s(« /3- c"s j3 and is a minimum when tan'' )3 == 2 + W EXAMPLES. 39 an d is a miuimuui when tan'^fi^ (h). If there arc no rollers either at B or at C, and if a wind blows horizontally upon .1 /i, tshew that the horizontal component of it^ pres- sure upon AB produces no stress in the rod Bl). (c). In a given roof, the rafters are of pitch-pine, the tie rods of wroughtr-irou, the sjiun is tiO-ft., the trusses are 12-i't. ecntre to centre, DF== 5-ft. == J'JG, tlie anj;le.4/i6' =- 30", the dead vvei-ht of the roof, including snow, is 9-lhs. per sq. ft. of roof surface, rollers are jilaced at C, a single weight of HllOO-lbs. is suspended from /', :ind the rool" is also designed to resist a normal wind pressure of U(i.4 lbs. per sq. it. of roof surface on one side; determine the dimensions of the several members. (31). The accompanying truss is similar to that in Question 27 with the addition of the/owr equal struts I)F. nil. AY/, EK. (((). — With th(; same data as in question 27 shew | that the weight of the metal in the tie-rods is 5_s_ 4 W+'d. (W+ W). cos'^ 18 /' cos fi. sin fi ' 7_ 3 W' 4 ^i"W' (6). — In a truss of this type adopted for the roof of a shed at Birk- enhead, the riifteri are of piich pine, are rectangular in section, have a constant breadtli of U|^-ins., and a depth which gradually diminishes from 13-ins. at the heel to It^-ins. at the peak ; the struts, of which 1>F and EG are vertical, are also of pitch pine, are rectangular in section, and have a scantlini: of Si-ins. by Gi-ins. ; the tit-rods are of wroui;ht-iron and AD == BJJ -= AE == CE -= 23-ft. ; BD and EC are round 'l}i-\n. bars, AD9.\\i\ AE are round H-in. bars, DE is a round 1^-in. bar; the angle ABC =- 30" ; the span == 7!)-ft. ; the trusses are 13 ft. centre to centre ; the heel B is free to slide on a smooth wall-plate ; the dead weight of the roof, including snow, is 8-lbs. per sq. ft. of roof surface ; determine the stress per unit of area to which each member is subjected when the wind blows horizontally witi a force of 40-lb8. per sq. ft. of verti'.'al surface, (1 — upon the side AB, (2) — upon the side^lf. (32). The Fig. represents the form of truss recent- ly ado})ted for the roof of a shed at Liverpool. The rafters are of pitch pine, are rectangular in section, have a depth of 14-ins., a breadth of 8-ins., are I inclined at GO" to the vertical, and are each supported at 5 equidistant points ; the six struts are also of pitch pine and reetau gular in section ; 1)F ViW A EC are each 10-ft. G-ins. long, and liave a scantling of 12-ins. by 8-ins. ; the remaining four struts have a .scant- lings of 8-ins. by 8-ins. ; the tie-rods are of wrough^iron ; BX and CO are round 2§-in. bars, ND and OE are round 2|-in. bars> Dl* and EQ are round Ig-in. bars, FA and QA are round 2;^ in. bars, FN, FF, GQ, and GO are round 1^-in. bars, DE is a round l.f in. bar ; the trusses are 20-11. centre to centre, and the weight of a bay of 40 EXAMPLES. the roof is 24,41 (i-lbf^ ; tho span is 90 ft.4-ins. ; the heel li is free to slide upon a sinonth wail-plat(^ ; dcteniiiiie the stress per unit of area in eaeli uunilur when the wind prodiiees a normal pressure of 2'>.4-lbs. pcrsq. ft. of roof surface, (l)-upon the side .1/^, (2)-upon the side AC. (.33). Make an alUsmative desiV'=- l(l()-ft., tiie an.ule A/U'--:W<\ a sinjrle weijiht of lOOO-lbs. is to be suspended from //. and one I of 2(HM(-lbs. from 0^ rollers are placed at (', the roof is| to be designed to resist a horizontal wind pressure on one side i>f40-lbs. per sq. ft. of vertical surface, the ties are to be of wrouudit-iron. the rafters and struts of timber or wrought-iron ; determine the dimensions of the different members. (35). The domiul roof of a ga^ holder ibr a clear span of HO ft. is I strengthened by secondary and primary trussing as in the Fig. The points /i\ and 6' are connected by the tie /^/V passing beneatli the central strut AP, which is 15 ft. long, and is also common to all the primary trusses; the rise of A above the horizontal is 5 ft. ; the secondarv truss ABEF, consists of the e(|ual bays All, HG, (Hi, the tven BE, Et\ FA, of which BE is horizontal and the struts d'E, FIf, which are each 2 ft. (5 in. long, and are parallel to the radius to the centre of Gil; the secondary truss ACLK'is similar to ABEF; when the holder is empty the weight support^'d by the truss is 3fi,fl00-lb>;., which may be asumed to be con- centrated at G, II, A, JA N, in the proportions.'HOOO, 4(100. 1000, 4000 and 8000-lbs., respectively; determine the stresses in the different mem- bers of the tru.ss. Should the braces HE and ML be introduced ? Why ? (3G). Dcisign the bowstring principals for the following roofs — (a). Span ■= 50-ft., rise == 10-ft., depth of truss at centre == 5-ft., distance between principals =r U-ft. (/>). Span=n 150-ft., rise =:, 30-ft., depth of truss at centre == 12-ft., distance between principals -= 20-ft. (c). Span == 200-ft., rise == 40-ft , depth of truss at centre == 20-ft., distance between principals == 20-ft. {]iort ; prove uenerally that tlie vertical ordinate inter- cepted between a point on the corresponding I'unicular polygon and a dos- ing lino wlio.se hori'/ontal projection is the span of the beam, represents on a certain determined scale the bending moment of a section at the ]ioint ; find its value by scale measurement f(«r a section at !>-it. from the left support, using the following .scales :-for hiKjths, ^-inch^ 1-foot ; \'oy forces, ^-inch == 1-ton ; the jxilar distance == 5 tons. Determine grapliieally by means of the same diagram the greatest bend- ing moment that can be ])roduced on the same section by the same series of loads travelling over the span at the stated distances apart. (89). The inclined bars of the trapezoidal trus.s repre- sented by the Fig. make angles of 45" with the horizontal ; a load of 10 tons is applied at the top joint of the left rafter in a direction of 45" with the vertical ; draw a frame diagram and determine graphically the two reactions, assuming' the one at the right to be vertical ; aKo, by means of a recipro- cal figure, find the stresses in all the pieces of the frame. Ex))lain in what manner the right hand reaction may be made approx- imately vertical. CHAPTER III. Bridges. (1). — CliisHijicatiun. — Bri(lij;i!a may be dividi'd intti tliroe general classes, viz. ; (I). — Bridge.s with horizontal girders, (2). — Suspension bridges, (3). — Arched bridges. The present chapter will treat of bridges of the 1st class oidy. In these the platforiu is supported by two or more main girders resting upon two or more supports, and exerting thereon pressures which are vertical or very nearly so. The neutral line (or surface) divides the girders into two parts, of which one is stretched and the other compressed. (2). — Comparative advuntages of curved and horizontal J(anges. — Jiittle. if any, advantage is gained by varying the depth of a girder, ana the practice is open to many grave objections. The curved or parabolic forui is not well suited to plate construction, and a diminution in depth lessens the resistance of the girder to distortion. Again, if the Vjottom flange is curved, the bracing lor the lower part of the girder is restricted within narrow limits, and the girder itself must be independent, so that in a bridge of several spans any advantage which might be derivable from continuity is necessarily lost. The depth is sometimes varied for the sake of appearance, but, gene- rally speaking, the best and most economical form of girder is that in which the depth is uniform througlunit, and in which the necessary thickness of flange at any point is obtained by increasing the number of plates. (3). — Depth of girder, or trnas. — The depth may vary from ^'^th to Tjth of the span, and even more, the tendency at present being to approx- imate to the higlier limit. If the depth fall below -j-'ath of the span, the deflection of the girder bec(unes a serious consideration. The depth should not be mon; than about Ih times the width of the bridge, and is therefore limited to ^-i-ft. for a single and to 40-ft, for a double track bridge. (4). — Position of platform. — The platform may be supported either at the top or bottom flanges, or in some intermediate position. It is POSITION OF PLATFORM. 43 cliiirncd in favour of the last tliat tho main ^jinlers may be braced to- gether below the iilatform ( Fii;. 1), while the upper portions serve a.s parapets or {guards, and also that tiie vibration comumuicated by a passing train is diminished. The position, hoW(!ver, is not conducive to rij^idity, and a large amount of metal is required to form the connections. The method of supporting the platform by the top flanges, (Fig. 2), renders the wh(>le depth of the girder available for braeing, and is best adapU'd to girders of shallow depth. Heavy cross girders njay be entirely dispensed with in the case of a single track bridge, and tlie load most effectively distributed, by laying the rails directly upon the flanges and vertically above the neutral line. Provision may be made for side spaces by employing sufliciently long cross-girders, or by means of sliort cantilevers fixed U) the flanges, the advantage of the former arrangement being that it increases the resistance to lateral flexure, and gives the platform more elasticity. Figs. 3, 4, 5, shew the cross girders attached to the bottom flanges, and the desirability of this mode of support increases with ths depth of the main girders, of which the centres of gravity should be as low as pos- sible. If the cross-girders are suspended by hangers or bolts below the flanges, (Fig. 5), the depth, and therefore the resistance to flexure, is increased. In order to stiffen the main girders, braces and verticals, consisting of angle or tee-iron, are introduced and connected with the cross-girders by gusset-pieces, etc. ; also, for the same purpose, the cros-t-girders may be 44 NUMBER OF MAIN GIRDEUR. prf)lon^'e(l on cacli side and the end jc»iiK'd to the top flanircs by suitable barH. When the depth of the main girders is more than about 5-ft., the top flanges should b • braced together. But the minimum clear head-way over the rails is 10-ft., so that some other method slumld be adopted for the support of the platform, when the depth of the main girders is more than 5-ft. and less tlian IG-ft. Assunu' that the depth of the platform below th(^ flanges is 2-ft., and that the dejith of the transverse bracing at the top is 1-ft. ; the total flmiti>it/ depths are 7-ft. andl9-ft., and if J is taken as a mean ratio of the depth to the span, the corresponding limiting spans are 5G-ft. and 152-ft. (5). — Comparative advantagps of two, three and four main girders. — A bridge is generally constructed with two main t^irders, but if it is crossed by a double track a third is occasionally added, and sometimes each track is carried by tiro independent girders. The tiro-girder system, however, is to be preferred, as the rails, by such an arrangement, may be continued over the bridge without devia- tion at the approaches, and a large amount of material is economized, even taking into consideration the increased weight of long cross-girders. The employment of four indepesident girders possesses the one great advantage of facilitating the maintenance of the bridge, as one-lialf may be closed for repairs withoi,t interrupting the trafiic. On the other liand, the rails at the approaclios umst deviate from the main lines in order to enter the bridge, the 'vidth of the bridge is much increased, and far more material is required in its construction. nillDGE LOADS. 45 Fow, if any, reasons can bo uiLri'tl in favour of tho introduction of ii third intonucdiato j^nrdcr, since it presents all the object ionahle features of the last system without any corresponding ncouunendatinn. (0). — Bridge hmds. — In order to deteruiino the stresses in the different members of a bridge truss, or main girder, it is necessary to ascertain the amount and character of the load to wliieh the bridjio may be subjected. The load is partly tlrml^ partly Hn\ and depends upon the type of truss, the span, the number of tracks, and a varii'ty of other conditions. The dead load increases with the s| an, and embraces the weight of the main girders (or trusses), cross-girders, platform, rails, ballast, and accumulations of snow. A table at the end of the chapter gives the dead load which may be adopted as a first approximation in calculating the strength and designing the parts of a bridge. (7). — Lirr load. — The live load due to a passing train is by no means unifnmdy distributed, but consists of a nund>er of diH'erent weights concentrated at definite points. This variable distribution is of little importsnce for spans of more than 100-ft., as the ratio of the dead to the live load then becomes very large, but it is nf vital moment when the span is short. The extreme weight of a locomotive upon its three drivers, with a wheel-base rarely exceeding 12-ft., variiis from 72,000 to ^^4,tl0t( lbs., and the maximum weight u]»on one pair of drivers may be taken at :-iO,()00-lbs. Thus, 30,000-lbs. may bo concentrated at the centre of spans of 12-ft. and under. This is etpiivalent to (50.(<00-lbs., uniformly distributed, /.*'., to 12,000-lbs. for a 5-11. span, and to 6000-lbs for a 10-ft. sj.an, per lineal ft. If the locomotive is placed centrally upon spans of 15, 20, 25, and 30-ft.. and if it is assumed that its weight ( == 84,000-lhs.) is equally distributed between the drivers, the bending moments at the centre are, 147,000, 252,000, 357,000, and 402,000 ft.-lbs., respectively. The g equivalent uniformly distributed loads per lineal ft. are, 147,000 X —^^ (==52205-lbs.), 252,000 x _(- 5040-lbs.), 357,000 x A.^(..4569a lbs), and 462,000 X -- ( == 4106§-lbs.) iifl 46 CHORD AND WEC STRESSES. Two locomotives at the head of a train will cover a span of GO-ft., and the load is practically uniform, or 2HO0-lbs. per lineal it. Two locomotives and tenders will cover a span of lOO-ft., and impose a practically uniform load of 2750 lbs per lineal ft. The couplinfjj of more than two locomotives is sometimes necessary upon snow-roads, but rarely occurs els«. where, and the case should not be included in a general rule for bridge design. It is a common practice, however, of the English Board of Trade, to prove a bridge by covering each track with as many locomotives and tenders as possible, and mea- suring the resulting deflection — an unreasonably severe proof for spans of more than lOii-ft. The weight of the heaviest train, exclusive of the engine, may be taken at 2250-lbs. per lineal ft. When a train crosses a bridge the conse(jueut strains are far more intense than if the train were at rest, and in France the government ri'gulations require a bridge to be proved under a moving as well as under a dead load. This additional intensity rapidly dimin- ishes as the span increases, and may be disregarded lor spans of more than 75-ft. For spans of less than 75-ft. the live load may be reduced to an equi- valent dead load by adding to tlie former certain arbitrary percentages, which may be taken at 10 for 75-i't., 20 for GO ft., 30 for 50-ft., 40 for 40ft., 50 for 30-ft., GO for 20-ft., and 70 for 10-ft. For spans of more than 75-ft. the total load -- ihad load + lice load. Provisit n may be made against accidents from high winds by the intro- duction of lateral bracing sufficiently strong to withstand a pressure of 40-lbs. per sq. ft. of truss and train. Tables at the end of the chapter give the uniformly distributed live loads upon railway and highw.iy bridges of diff"erent sjtans. (■g), — Chord and web stresses. — It may be assumed that the load upon a bridge is concentrated at the panel points of the main trusses. The stresses produced in the chords (flanges) by the passage of a train are to be calculated on the basis of a uniformly distributed load. In determining the strength of the web members allowance may be made for the irregularity in the dist'-ibution of the live load by sub- stituting for the nearest standard panel weight a weight equal in amount to that upon an independent sj)an of the same length as a panel. (9). — B'Mviiig, Friction (^also called expansion and hearirg) rollers, (.((•_ — The area of the bearing surface at the supports (abutments) must be sufficient to prevent any undue crushing. The length of the bearing for a cast iron girder is often made one-fifteenth of its central depth. BKAKING, FRICTION ROLLEUS. In order to provide for the lon(7v;('A»/'/(Moad concentrated at each apex in Fig. 6. Let be the inclination of the diagonals to the vertical. The reaction at .1 == 7^. v, and the shearing force at the section MX == 7J,. u- - 4. >r - U. ir. This shearing force must be transmitted through the diagonals. Hence, the stress in ah due to the permanent load -= U. w sec == -. u\ sec 0, Again, let w' be the flee load concentrated at an apex. The live load produces in ah the yrniffst stress of the same kind as that due to the permanent load, when it covers the longer segment up to and including b. WARUEN GIRDP:R. 49 55 Tlic sheariivj; force at J/.V ( == tlie n-actidu at .1) i,s thou '~.w', and 55 the correspond! II ii: stress in ah is ' '-.w'.sirO. ' ' ('.4 •'• the maxiniuiu stress in nh due to botli kinds of loads = /7 55 A The sheariiiiT t'oree ]irnduces its (jrmto^t eff, ct in rcrirsiiKj the strrss in (ih due to the piTuiaiii iii load, when the live IikmI covers the shurffr sejriueiit up to and iiieludlMi:' n. 5 The sheariiiij: force at AJX ( r- the rtaetioii at B) is then -. w', and 5 the corresponding stress in ah is —j-.ir'.sfc H; •*• the resultant stress in (i/> when the live load covers the sliorter sej:ment I .SVC IK = (^•'" - .To'"')-^' This stress may he. noirative. a»id must be provided ior by introducing a counter-brace or by proportioning the bar to bear (jitfli the greatest tensile and the greatest conij)ressive stress to which it may be subjected. The stress in any other bar may bi- obt.iinetl as above. The (7(o/-(/ stresses are greatest when the live load covers the whole of the girder, and may be obtaini'd by the method of moments, or in the manner described in the succeeding articles. (11^. — Witnrn Girder. — The Warren girder consists of two hori- zontal chords and a series of diagonal br 'es forming a single triangula- tiou, or zig-zag, Fig. 11. The principles which regulate the construction of trellis girders are equally applicable to tho.oc of the Warren type.. The cro.ss-girders ('loor-beams) are spaced so as to occur at the apex of each triangle. When the platform is supported at the top chords, the resistance of 50 WARREN GIRDER. the Btructure to lateral flexure may be increased by horizontal bracing between the cross-girders and by diagonal bracing between the main- girders. When the platform is supported at the bottom chords, additional cross- girders may be suspended from the apices in the upper chords, which also have the eflfect of adding to the rigidity of the main-girders. Let w be the dead load concentrated at an apex, or joint. w' " " live I *' " span of the girder. k " " depth " " s " ** length of each diagonal brace. " to the vertical ti " " inclination " N " " number of joints Two cases will be considered. Case I. — All the joints loaded. Chord Stresses. — These stresses are greatest when the live load covers the whole of the girder. Let *S^„ be the shearing force at a vertical section between the joints n and n + 1. Let H„ be the horizontal chord stress between the joints n-l and n + 1. The total load due to both dead and live loads = (w + w'). N- 1. w + w' The reaction at each abutment due to this total load = The shearing forces in the different bays are, 10 + w' 2 N-l. So = w + w' Si = — 5 . ^^3, S,= 2 w + w' . iV-1, between and 1, a 1 » 2, iV-5. S^ = w -f- w and S„ = . iV-7, w + w " 3 " 4. . (iV-2n-l) •T he corresponding diagonal stresses are, So-see 1^, Si.sec ^ , S^.$ec 9. WARREN GIRDER. 31 The last stresses multiplied by sin d give the increments of the chord stresses at each joint. Thus, I N.k H, = tension in 2 = ^o- <«« ^ = 9 — . iV-1, _ w + w' l_ N-\ 2 k' N ' II2 ~ compression in 13 = S^. tan d + Si. tan d _ to + w' I N- 1 w + w' l^ iV-3 _ tp-^w ' l_ 2.(iV-2) " 2 Ic N "*" 2 k' N ~ "Ik' N „ „ „ w^-xo' I 3.(.V-3) iZ, = tension in 2.^ = Hy\-Sy. tand + Sy tand=: ___._• w + w' I 4.(iV-4) B^ = comp'n in 36 = //, + S,. tan e + S3, tan e = ^ -r j^ and j^„ = horizontal stress in chord, between the joints n-1 and n + l - ^ —.— . "^ — 7= — , being a tension for a bay in the bottom chord, and a compression for a bay in the top chord. JVb^e. — The same results may be obtained by the method of momenta. e. g., find the chord stress between the joints n-1 and n + 1. Let a vertical plane divide the girder a little on the right of n. The portion of the girder on the left of the secant plane is kept in equilibrium by the reaction at the left abutment, the horizontal stresses in the chords, and the stress in the diagonal n, 7i + !• Take moments about the joint n, ■'• //„. k w + to' 2 * w + to' I n. I iV-l.n.^ -n-l.w-^w'.^^ .1. n. N-n. 2 N .'. //„ = &c. Diagonal Stresses due to dead load. Let d^ be the stress in the diagonal n, n + \ , due to the dead load. The shearing forces in the different bays due to the dead load are, -^.iV- 1, between and 1, ^.iV-3, between 1 and 2 Z-N-^ " 2 " 3,|.iNr-7, " 3 " 4 2 ^ and — .(iV-2 n-1), between n and n + l. 52 WARREN GIRDER. The corresponding diagonal stresses are, w "2 w 70 a compression ^.^- 1- «ec. = -^.iV- 1. — — (f, in 01, = (Z, in 12, c?j in 23, a tension --.iv- 3. «fc. 0= -c^Jy - o. , 1^ u li W W a compression — — .JV- 5. sec. =— .iV- 5.-.- = A; and the stress in the n-th diagonal between n and n + I is. ?tf /,.= -:5-. (.V-2»-l). /^ being a tension or compression according as the brace slopes down or up towards the centre. Diagonal stresses due to live load. The live load produces the greatest stress in any diagonal, («, n + 1), of the same kind as that due to the dead load, when it covers the longer of the segments into which the diagonal divides the girder; represent this maximum stress by Z)„. The live load produces the greatest stress in any diagonal {n, n + 1), of a kind opposite to that due to the dead load, when it covers the shorter of the segments into which the diagonal divides the girder; represent this maximum stress by D',,. The shearing force at any section due to the live load as it crosses the girder is the reaction at the end of the unloaded segment, and the cor- responding diagonal stress is the product of this shearing force by sec 6, or -. K The />» = values of the different diagonal stresses are, = compression in ^ when all the joints are loaded s w' N.N- 1 A = N ' tension in 1 2 when all the joints except one are loaded s w' N-l. N-2 ~k 2" N Dj = compression in 2 3 _ s ro' N-2.N-3 ^T2' N Di = tension in 3 4 _ s w^ N- 3.^-4 : "IT N " 1 and 2 <' 1,2, and 3 WARREN GIRDER. 63 D^ = stress in ?», n + 1 when all the joints except 1, 2, 3 . . . and n loaded _s w'. N-n. N- n-\ ~ /? 2 N ' //,= stress in 1 before the load comes upon the girder = D\ z= compression in 1 2 when the joint 1 is loaded =!_,—. 1 S Hi' i>'2 = tension "23 " joints 1 and 2 " =-I —.3 C 'If J D'a = compression " 34 " " 1, 2,&3 " = /•• "y'^ /)'„ = stress in 7i, n + 1, " l,2,...&?t" = -7 s w' H. n + I fc iV 2 The total maximum stress in the ?t-th diagonal of the sa7/if kind as that due to tlie dead load = '„. This resultant stress is of the same kind as that due to the dead load so long as Z>'„, and need not be considered since d„. it is necessary to provide for a stress in the given diago- nal of a kind opposite to that due to d„ + JJ„, and equal in amount to D'. - 'L- This is effected by counterbracing or by proportioning the bar to bear both the stresses r/, + /)„ and D'„ - d„. Case II. — Ordi/ joints denoted hy even numbers haded. Chord Stresses. — The stresses are greatest when the live load covers the whole of the girder. »'■ ! 64 WARREN GIRDER. iV-2 The total load due to both dead and live loads = (w> + «?'). — r — . w + w' -7— — The reaction at each abutment due to this total load= — -, .iV- 2. To find H,. take moments about 1, .'• H,.k = — j — -V- 2. ~ 2 Hk- '^•^"' iv^r9 o ^ I l_ N 4, - (to + w?'). ^. 4 iV -(«^ + ^«').2.^. " //« n, and^r*^ let n be even ; 5-.A; = w + ?o' 4 •^-2-]^ -(w +w'). j^ln-'I + 7t-4 + ... + 6 + 4 + 2| n 71. 71 - 2 1 4~ J iV- / 4 and, Zr„ == i« + 10' I 71. iV- n Next let 71 be odd. 4 • fc' ^ . . J?„.A; = — - — . N- 2. n. ^. 4 JV ~(w + w). -^.\n-2 ^ 71-4 + ... + 5 + 3 + 1| = (wj + w'VJl J N-2.n 71-1 ">4-{ and, i7„= -^— . ^ 4 4 W + 70' ? iV- 2.7i -71-1* i\r -AT, iVb^e. — If — be even, .'. /ijv^, the stress in the middle bay = — zrpi — T.iv r 16 ■ k HOWE TRCSS. 55 If ^ be odd, :.IIk, the Btress in the middle bay^ r » k 16 'k' N ' Diagonal stresses due to the dead load. The shearing forces in the difftrent bays due to the dead load arc, w 10 ■ic .N-2 between and 2,- . JV- 6 between 2 and 4, — . A"- 10 4 4 4 between 4 and 6, etc. The corresponding diagonal stresses are, rfo in 1= r- w . jV - 2 = , D', = D\ D, = D, D\ = //, A = A ^'. = ^'» If _ is odd, there is a single stress at the foot of each of these columns. The maximum resultant stress due to both dead and live loads is obtained as before. e. g., the maximum resultant stress in 3 4 when the longer segment is loaded = d^ + Dj = dj + Dj, and the maximum resultant stress in 3 4 when the shorter segment is loaded = d^ - D\ = d.. - D\. j^ote. — e is generally GO^, in which case s = ^~]^' (12). — Howe Truss. Fig. 14 is a skeleton diagram of a Howe truss. The truss may be of timber, of iron, or of timber and iron combined. The chords of a timber truss usually consist of three or more parallel 56 HOWE THLSS. nionibcrs, pliipcd a little tlistaiuH! apart so as to allow iron susponders with sercwt'd ends to pass bctwtvn tlirin (Kii;s. 15, 1<; ). Each meuibor is uiade up of a number of lengths scarfed or fished together (Figs. 17, 18). The main-braces, shewn by the full diagonal lines in Fig. 14 are composed of two or metre members. The counter-braees. which are introduced to withstand the v<^ffect of a live load, and are shewn by the dotted diagonal lines in Fig. 14, are either single or are composed of two or more members. They are set between the main braces, and are bolted to the latter at tlie points of intersection. The main braces and counters abut against solid hard wood or hollow cast-iron angle-blocks (Fig. 10). They are designed to with- stand compressive forces only, and arc kept in place by tightening up the nuts at the heads of the suspendi-rs. The angle-blocks extend over the whole width of the chords; if they are made of iron, they may be strengthened by ribs. If the bottom cliord is of iron it may be constructed on the same principles as those employed for other iron girders. It often consists of a number of links, set on edge, and connected by pins (Figs. 19, 20). HOWE TRUSS. 57 In Nucli a case tlie lower annK'-bloeks should have grooves to receive the burs. HO as to provoiit lateral flt-xure. It'tlio truss is iiiaile cntinly of iron, the top chord uiay be foruiod of K'li^ths of cast-iron provided with suital)le flanges by which they can be bolted together. Angle-blocks may also be cast in the same piece with the chord. To determine the stresses in the different members, the same data are assumed as i"or the Warren girder, except that N is now the number of panels. (.'Iiiird strrssi's. — These stresses are greatest when the live load covers tlie whole of the girder. Jict II„ be the chord stress in the 7j-th panel. The total load due to both dead and live loads == w + ?//. iV- 1. The reaction at each abutment due to this total load == — ^ — . A'-l. •■J Let a plane MM' divide the truss as in Fig. 14. The portion of the truss on the left of the secant plane is kept in e<|uilibrium by the load upon that portion, the reaction at the left abutment, the chord stresses in the ?i-th panels, and the tension in the Ji-th suspender. Jursf, let the load be on the top chord and take moments about the foot of the n-th suspender ; 7/1 4- 1/1 / I 71 . H - 1 Ar-"' + ''-N'—^- W + w' , 11. N'— 11 -. /, or //„ = jw + w' I n, N— n Next let the load be on the bottom chord and take moments about the head of the H-th suspender. .'. //„. k = IV + iv' , //. X- n I. , as before. 2 ■ X Thus, 7/„ is the same for corresponding panels, whether the load is on the top or bottom chord. DiagojKil stresses due to the dead load : — Let V'„ be the shearing force in the «-th panel, or the tension on the ?/-tli suspender due to the dead load. First, let the load be on the top chord, TV ^^' \ /^- 1 \ .-. y ,= -^.X- I - n. w = ^'''\— — »^ )• Next, let the load be on the bottom chord, ••• v'„= ^ . ix- i)-:n.w --^v. (^ -u). 58 HOWE TRUSS. The corresponding diagonal stresses are, Diagonal stresses due to the live load. Let F"„ be the shearing force in the «-th pan^l, or tension on the «-th suspender, when the live load covers the longer segment. First, let the load be on the top chord. The greatest stress in the H-th brace of the same kind as that produced by the dead load occurs when all the panful points on the right of MM' are loaded. .*. V"„, the shearing force on the left of MM' -■= the reaction at w' M— ?i H"' ■^~ n -1. — rj— , and the corresponding diagonal stress " k " k' 2 ^' N Hence, the resultant tension in the n-th suspender due to both dead and live loads = F„ = F„ + F', /iV-1 X ir ._ j\r-n iV T f and the resultant maximum compression on the n-th brace due to both dead and live loads _ * T7 * i /iV-l \ w' .V-H-l. N-n N }- d„+D„ The live load tends to produce the greatest stress in the 7i-th counter when it covers the shorter segment up to and including the nth panel point. Even then there will be no stress in the counter unless the effect of the live load exceeds that of the dead load in the (n + 1 )-th brace. The shearing force on the right of MM' = the reaction at N' I)'n the corresponding diagonal stress = — — ■ k 2 N and the resultant stress in the counter = !)'„ - (/„.^, s ( w' n.n + l /N-3 \ ) '=lc'l2-N--"'\-li V} Next let the load be on the bottom chord. ... F' =!!^7V-„^^±1 „ s w' ,_ N- 11 + 1 _ w' n.n 4- 1 \'~ir' PRATT, OR WHIPPLE TRUSS. Hence V„= ?',+ F'„ = t«. (^^ - n) + ~.N-n. and, < f /;„ - y| ^- (-^ - ") + y ^- "• —jr~ \ 59 N Also the fltresfl in the n-th counter is 71+1 JVbre=|:,and IF.= f. ' I, -i3 _ij- /; = ___. 5ec a, T^ T,= —■ sec 0, and T,=~-secy. If the truss represented by Fi^. 23 is inverted, another system is obtained m which the obliques are struts and the vertical ties. (l5).—J]oUman Truss. Fig. 24 represents a Bollman truss. The platform may be either at the top or bottom. The verticals are always struts and the obliques are always ties Each pair of tics, as AC, BC, is independent, and supports the load QUADRANGULAR AND POST SYSTEMS. 63 upon a vertical, {. e,, the weight of panel, or one-half of the weight be- tween the panel points A and E. Let T , Tj be the stresses in AC, BC, respectively. Let W\ be the weight upon the post DC Let a„ ttj, be the inclinations of AC, BC, respectively, to the vertical. sin ttj _j /T7 Ttr '^'^ ^1 f,= W. and 'R = W. sin (a, + ajj)' ' ' ' sin (a, + a.^) The stress in any other tie may be obtained in the same manner. The compression in the top chord is the sum of the horizontal compo- nents of the stresses in all the ties which meet at one end, (16). — Quadrangular and Post si/stems. — Fig. 25 represents the Quadrangular truss in which tiic bottom chord is supported at intermediate points half way between the verticals. Fig. 26 represents the Post system. In reality ;}- is the Pratt (Linville) truss shewn in Fig. 22, modified by inclining the compression members at an economical angle. The stresses in the different members are to be calculated in the manner described in the preceding articles. (17). — Bowstring girder, or truss. The bowstring girder in its simplest form is represented by Fig. 27, and is an excellent structure in point of strength and economy. The top chord is curved, and either springs from shoes (sockets) which are held together by a horizontal tie, or has its ends riveted to those of the tie. The strongest bow is one composed of iron or steel cylindrical tubes, but any suitable form may be adopted, and the inverted trough offers special facilities for the attachment of verticals and diagonals. 64 BOWSTRING GIRDEH, OR TRUSS. The tie is constructed on the same principles as those employed for other iron jrirders, but in its best form it consists of flat bars set on edge and connected with the shoes by gibs and cotters. The platform is suspended from the bow by means of vertical bars which arc usually of an I section, and are set with the greatest breadth transverse, so as to increase the resistance to lateral flexure. In large bridges the webs of verfieals and diagonals may be lattice-work. If the load upon the girder is uniformly distributi'd and stationary, verticals only are required for its suspension, and the neutral axis of the bow should be a parabola. An irregularly distributed load, such as tliat due to a passing train, tends to change the shape of the bow, and di- agonals are introduced to resist this tendency. A circular arc is often used instead of a parabola. To determine the stresses in the different members, assuming that the axis /lZ^(7of the top chord is a parabola : — Let w bo the dead load per lineal ft. " I " span of the girder. '' Je " greatest di'pth BD of the gird;T. Chord strcssfs. — These stresses are greatest when the live load covers the whole of the girder. The total load due to both dead and live loads = (ir + w'). I u- + ?r' The reaction at each abutment due to this total load -. I Let // be the horizontal thrust at the crown. Let T " " tension in the tie. Imagine the girder to be cut by a vertical plane a little on the right of Bl>. The portion ABD is kept in equilibrium by the reaction ut A the weight upon Al), and the forces // and 1\ BOWSTRING GIRDER, OR TRUSS. 65 Take moments about B and D ; 8 and • ' J =- — — . -j-==Jl Lot //' be the thrust alonj^ the cliord at any point /*. Let X be the liorizoiital distanee of 7' from li. The portion J'Ji is kept in i'(|uilibriuni ])y the thrust // at B, the thrust //' at /* and the weinlit {w + tr').x between /' and Ji ; ••• /r.siv' (•- //'•-'=. IP + (»• + »')■■../•-, i being tlie inelination of the tanjxent at P to the horizontal. Hence, the thrust at .1 = C'~r~-l) ( 777-7^ + 1 j Diiigrntnl strcssri^ diw to Urc fo'uf. Assume that the hiad is concentrated at the panel points, and let it move from .1 towards C l^m^m ri6.2L8 If the diau'onals sliipr as in the Fig. 28, thi'y are all ties, and the live load produces the greatest stress in any one ol'them. as QS, wlu'ii all the panel points from A up to and including (^ are loaded. Let ,r, y. be the horizontal and vertical co-ordinates, respectively, of any point on the parabola witii respect to B as origin. Tlu' e((uati(»n of the parabola is. // — - " ..r- (1) Li't tlu! tangent at the ap'X /* meet />>/) produced in L, and /-^'' jiro- duced in A'. Draw the horizontal line I'M. From the properties of the parabola, LM ^- 2. BM. Let I'M --= X, and BM ■■= >/. From the similar triangles L MP and LDl^, M r~ D E '"' ~ ~ xTt^E' 66 BOWSTRING GIRDER, OR TRUSS. 2.y 8.x . CE l-2x (i-2xy Sx 'QE^l + 2x (2) Draw EF perpendicular to QS produced, and imagine the girder to be cut by a vertical plane a little on the right oi' FQ. The portion of the girder between FQ and C is kept in equilibrium by the reaction li at C, the thrust in the bow at F, the tension in the tie at Q, and the stress in the diagonal QS. Denote the stress by Z>„. and let the panel OQ be the u-th. Let be the inclination of QS to the horizontal. Take moments about E, :. £>,, EF -- Jl CE orD„ = R0 cosec . Let N be the total number of panels, :. j^is & panel length, and w^ is a panel weight. (^) Also X = 71. ^r^ - ^^ , and . . , _ ^V 2' lr2x QE I -2.x CE N-n H R, the reaction at C when the // panel points preceding Tare loaded _?f^ , n.n T 1 Hence, equation (3) becomes. iV J^n = -^-f.n + 1. .coscc 4.k w Again, by equation (1), /.• - ST=~. DT'= 4./.-. ( ^Lll _ A' >ST ST Thus, finally, ...^/ iv^-„[^^+^-K^-'^-i)(«+oi^]* ^"~ 8 ^ -A^ This formula evidently applies to all the diagonals between B and C. (5) BOAVSTRIXG GIRDER WITH ISOSCELES BRACING. G7 (2) ' to um the ;^) 'i) r>) Similarly, it may bo easily shewn that the stress in any dinK -r -. «<« ('^ + ",)' ' sin (y, + W,) Hence, T, - >r. -;- Diagonal stresses due to the live lorn/. Let N be the number of half panels. 2.1 . 2.1 The lentrth of a panel == —rr-', the weight at a panel point = "•'. -~^- Let the load move from A towards C. All the braces indiiKil like 07' are ties, and all those inolinrd like (Jl' dw struts, girdir from A up to and including (). Tlie live load produces the greatest stress in OT when it covers the Denote this stress by />„ ; ()G is the /;-th half-panel. As before, I),. EF - A*. CE. ( 1 ) . rpv- 1 1 4/» ' ' ] • n '"'•"•*' 'i + 2 The load upon AO = n.w , — , and • • Ji = — -— . _^ The ratio of CE to iJF is denoted by the same expression as in the preceding article. .„_»_: 2 ,^ x-» [^ t5'^!i" -')"'^"iil' (2) ■■ " 8'//7i + l' iV ■ N-,>- 1 The live load produces the greatest stress iti OM when it covers the girder up to and including D. Denoti this stress by D'„ ; DG is now the 7i-i\i half panel. Let W be the reaction at (\ As before, D\r A". l]^p. cosecO, (3) e being the angle MOD. The weight upon AD= (h - l),!o'.— ,, and .'.li' = -^.l. -ttj— BO^VSTKING SUSPENSION lUUIMJE. 69 10 be It may be easily aliown, as in the precediiijr article, that, i oe' n+ 1 , and co8ccd = N. [/■+ !^fi(.v-„).<] • D' = / H- 1 4n.K'. (N-)i) (4) 8 k n iV ■ N-n Hence, when the load moves from A towards C, equation (2) gives the diagonal stress when n is even, and equation (4) gives the stress when u is odd. If the load moves from C towards A, the stresses are reversed in kind, so that the braces have to be designed to act both as struts and ties. Note. — By inverting Fig. 29, a bowstring girder is obtained with the horizontal chord in compression and the bow in tension, (19). — Jiou-striiig Suspension Bridge, {Lentie.ular Truss). — This bridge is a combination of the ordinary and inverted bowstrings. The most important example is that erected at Sultash, Cornwall, which has a clear span of 445— ft. The bow is a wrought-iron tube of an elliptical section, stiffened at intervals by diaphragms, and the tie is a pair of chains. A girder of this class may be made to resist the action of a passing load either by the stiffness of the bow, or by diagonal bracing. In Fig. aO, let BD^k. B'I) = k'. Let // be the horizontal thrust at B, and T the horizontal pull at B', when the live load covers the whole of the girder. //= w + w P 8 k + k' r^.= T. First, let k = k'. ponding stress in a bow-string girder of span I and depth k. •. 11= T— — :,— T) or one-half of the corres- lo k 70 CAMIiKU. Ono-hiilf of the total loud is .supportod by the bow and one-lialf is trausmittod through tlie verticals to the tie. .".The stress in each vortical = .^.0''' + "'"), A ^ ' w" being the portion of the dead weight per lineal ft. borne by the ver- ticals, and jY the number of panels. The diagonals are strained nnly under a passing load. Let /'/*' be a vertical through K, tlie point of intersection of any two diagonals in the same panel, and let the load move from A towards C. By drawing the tangent at i^and proceeding as in § (18), the expres- sion for the diagonal stress in i^S, becomes as before, ,. xv' . H.(//-l) /-2x ^^"^T-'— A^'TTii^-^"*""'- (1) Similarly the stress in the vertical QQ is „ I «•' , n.(n-\\l-2x \ = }(' — ^ — / — -^ '-• • \V 2 ^'^ l+'lx (^) '.under these loads,--- ■ — — II . o k ir= (3). Next, let A* and A*' be unequal. Let W be the weight of the bow, IF ' the weight of the tie. F' ""^ 7r " A;'" The verticals are not strained unless the platform is attached to them along the common chord ADC. In such a case, the weight of the plat- form is to be included in W '. The tangents at P and F* evidently meet AC produced in the same point 0', for EO' is independent of A or k'. Bence, the stresses in the verticals and diagonals due to the passing load may be obtained as before. (20). — Camber. — Owing to the play at the joints, a bridge truss, when first erected, will deflect to a much greater extent than is indi- cated by theory, and the material of the truss will receive a permanent set, which, however, will not prove detrimental to the stability of the structure, unless it is increased by subsequent loads. If the chords were made straight they would curve downwards, and, although it does not necessarily follow that the strength of the truss would be sensibly impaired, the appearance would not be pleasing. In practice it is usual to specify that the truss is to have such a camber, or upward convexity, that under ordinary loads the grade line will be true and straight. The camber may be given to the truss by lengthening the upper or CAMBER. 71 ( shortening the lower chord, and the difference o? length should beoqual- ly divided amongst all the panels. The lengths of the wt-b uifuiber.s in a cambered truss are not the same as if the chords were horizontal, and must be carefully calculatt-d, otherwise the several parts will not tit accurately together. Tojindnn opjiroximate value for the Camber, etc. Let d be the depth of the truss. Let .s-,. .v^, be the lengths of the upper and lower chords, respectively. J^<-'t/i,/i, be the unit stresses in " " « u Let rse dlnit'imlou. Sufe wcrklnB lo»il In Ibn. per iu\. In- Flat MidH. Hound i-udfi. Klftt unU. Uoiuicl (inila. 4,000 3,500 2.500 2,000 10 10 to 15 15 " 20 20 " 25 25 " 30 10,0000 9,000 8,000 7,500 6,800 7,000 6,500 6,000 5,500 5,o;-o 30 " 35 35 " 40 40 " 50 50 " 60 6,000 5,000 3,S(I0 3,000 Cast-iron compression members should not exceed 22-diam8. in length and should be subject to the same stresses as those prescribed for wrought-iron. TIMBER. Ratio of length to least transverse dimension. Safe load in lbs. per sq. in. OAK. PINE. 10 10 to 20 20 " 30 30 '' 40 1,000 800 600 400 9U0 700 500 300 RONDELET S RULE FOR OAK AND PINE. Batio of length to least transverse dimension. Extrt'uie load in lbs. per sq. in. borne by pillar without lateral tlexuri!. Safe Vforking load in lbs. per uq.in. with "as factor ot safety. 1 12 24 36 48 60 72 5,974 4,978 2,987 1,991 996 498 249 853 711 426 23i 112 71 35 In flexure, the greatest allowable stress per sq. in. may be taken at 1,200-lbs for oak, and 1,000-lbs. for pine. iV. J5. — The figures in the above tests and tables are by no means absolutely fixed, and are merely given to shew the standard practice of engineers. 74 TABLES. Table of loads for Highway Bridges. Sp.in in feet. City and Suburban Bridges liable to heavy t.rattic. Bridges in Manufac- turing Districts. Ballasted Koads. Bridges in Country Districts, rnballast- ed Roads. 100 and under. 100 ()200 .... 100 lbs. per sq. ft. HO " " 90 lbs. per sq. ft. 60 •' •' 70 lbs. per sq. ft. 60 " " 200 to 300 .... 70 " 50 '* " 50 " " 300 to 400 .... (JO " " .■)0 " " 45 " 400 and over... 50 " '' 50 " 45 " " i Table of loads for Railway Bridges. Dead load in lbs. per lineal ft. Span in ft. "!0 15 20 25 30 40 50 75 451) 500 550 625 625 65;> 700 750 800 Live load in lbs. Span in per lineal ft. ft. 12,000 100 5,500 125 5,000 150 4.750 175 4,500 200 4,000 250 3.500 3o0 3,250 350 3,000 400 Dead load in lbs per lineal ft. 900 1135 1225 1300 1500 2000 2400 3000 4000 Live load in lbs. per lineal ft. 2750 2600 2500 2500 2400 2400 2250 2250 2250 "try .Hast- .ft. a (I a u lbs, t. CHAPTER IV. Suspension Bridges. (1). CahJcs. — The modern suspension bridge consists of two or more cables, from whi^h the platform is suspended by iron or steel rods. The cables pass over lofty supports (piers), u' d are secured to anchorages upon which they exert a direct pull. Chain, or link cables are the most common in England and Europe, and consist of iron or steel links set on edge and pinned together. Formerly the links were made by welding the heads to a flat liar, hut they are now invariably rolled in one piece, and the proportional dimensions of the head, which in the old bridges are very imperfect, have been much im- proved. Hoop-iron cahh,. have been used in a few cases, but the practice is now abandoned, on account of the diifieulty attending the manufacture of endless hoop-iron. Win -rope cablis are the most common in America, and form the strongest ties in proportion to their weight. They consist of a number of parallel wire-ropes or strands, compactly bound together in a cylin- drical bundle by a wire wound round the outside. There are usually seven strands, one forming a core round which are placed the remain- ing six. It was found impossible to employ a serm-strnud cable in the construction of the East River Bridge, a.« tht; individual strands would have been far too bulky to manipulate. The same objection held against a thirteen-strand cable (13 is the next number giving an approximately cylindrical shape), and it was finally decided to make the cable with nineteen strands. Seven of these are pressed together so as to form a centre core around which are placed the remaining twelve, the whole being continuously wrapped with wire. In laying up a cable great care is required to distribute the tension uniformly amongst the wires. This may be effected either by giving each wire the same deflection or by using .s7ra/(j|/t^ wire, i.e., wire which when unrolled upon the floor from a coil remains straight, and shows no tendency to spring back. The distribution of stress is practically uniform in untwisted wire ropes. Such ropes are spun from the wires and strands without giving any twist to individual wires. 76 ANCHORAGE, ANCHORAGE CHAINS, SADDLES. The htchstay is the portion of the cable extending from an anchoraj^e to the nearest pier. The elevation of the cables should be sufTioient to allow for settling, wliich chiefly arises from the deflection due to the load and from changes of temperature. The cables may be prot/- (5). —Parameter, (f-c— Let /j„ /(,. be the elevations of A and B, re spcctively, above the horizontal line COl). Let OD rrz o„ 0C= oTjj, and let ((, + o.^ = u = CD. by equation (1), a\= ""./(„ and a;^= '^~ .h. w But a, -f- a.^ — a, The Parameter = (-t) «.V/'i (t.'^h. = -= , and (f.^ -- 2 //^ «r V / Again, fail 0: w .X, and sec 6 =v 1 + -^.x' CoroUuri/. — The length of any arc of the parabola measured from i8 approximately the same as that of the osculatory circle at measured from and having the same abscissa. Let R be the radius of this circle. R= Parameter x* 2 2.3/ "^^^= b-'(^)y= (^ + -7eO* = ^ + ^- approximately. -.x + X H./e*" 2 y^ --X + - . •— O X (13) (8\ — Weight of Cable. — The ultimate tenacity of iron wire is 90,000 lbs. per sq. in , while that of steel rises to 160,000-lbs, and even more. The strength and gauge of cable wire may be ensured by specifying that the wire is to have a certain ultimate teracity and elastic limit, and that a given number of lineal ft. of the wire is to wpigh o?ic pound. Each of the wires for the cables of the East River bridge was to have an ultimate tenacity of 3,400-lbs., an elastic limit of 1600-lbs., aad 14 lineal ft. of the wire were to weigh oiie pound. A very uniform wire, having a co- efficient of elasticity of 29,000,000-lbs., has been the result, and the process of straightening has raised the ultimate tenacity and elastic limit nearly 8 per cent. Let W be the weight of a length ^i (^ = 0D) of a cable of sufficient sectional area to bear safely the horizontal tension //. Let W2 be the weight of the length s, (=0-4) of the cable of a sec- tional area sufficient to bear safely the tension T^ at A. Let/ be the safe inch stress. Let p be the specific weight of the cable material. Wy= —.(ii-p, and W.i= — '—, — '-^i-P f f ^^\^ ir,.-^.sec< :0 = _±.( a, + -.-')( l+—-i + ...) or Tf,= >r,(l+|^i-), nearly, 6 (14) 82 DEFLECTION OF CABLE. A saving may be effected by proportioning any given section to the pull across that section. At any point («,//), the pull=//. sec0, and the corresponding sec- , Jf.secff ^, . , .^11 Il.sccd , , tional area= — ;; — . 1 he weight per unit oi lcngth= -, — .p, and the total weight of the length s, (^=0A) is, / W. / U.secB ds H.p/ ,., H.p I {. 2.y ^....y.^ Butx'-T^.y .■.F.=f!./-(U2.|.^....):..=f<(..i.'^),nea., 4 Ai Hence, W^=^^-(} + ^^ The weight of a cubic inch of steel averages .283-lb. " " wrought iron " .277-lb. The volume in inches of the cable of weight Tr, = 12.a,. (15; IV' .', ^— 77=. 283-lb. or .277-lb., according as the cable is made of 12. a,.:^ f steel or iron. Let the safe inch-stress of steel be taken at 33,960-lbs., of the best cable-iron at 14,958-lbs., and of the best chain-links at 9,972-lbs. •'• ^^=^' «'• ^^ ■'-^'^''3im= i^m ^'' ''''^ ''^^''' r,=JT,«,x.277x W,=ff.a,x.277x 12 14,958 "" 4,500 for iron cables. 12 H.a, , for link cables. (16) (17) (18) 9,972 3,000' Note. — About |^th may be added to the net weight of a chain cable for eyes and fastenings. (9). — Deflection of a cable due to an elementary change in its length. By the Corollary of § (7), the total length (S) of the cable AGB is, 2 hi 2 hi Now a, and a,, are constant; A, - /i, is also constant, and .'.dhi^dh.^. Hence, dS=t(^-^+^'A.dh, 6' \a. a J _^__. ■ k Cl'ltVK OF CABI.K. If /,,Z=/,,= /i. (i,~(i.,= 10 /. and (/aS'=^ o • ~- <^^* (V.)) rosp Draw tlu! ordinate /■'.V, and let the tangent at y-* uijct (J.X in JJ. As before, I*.VE is a triangh' of forces, and ^is the middle point of ON. •" - (20) the cquatiotj to a [larabola with its axis paralKl to OY, and its focus at a point aS' where 4. ^ "" , ■ CurnUir\j 1.— L't tlie axi'. meet the tangent at in 7", and h't its inclination to OX be ('. Let .1 be the vertex, and ON' a perpendicular to the axis. .•..SY>=,S'7"=*SM 4- AT'z=ii\ 4- AN' But, \.AS.A N'=ON"=N'T'\ tan^l =\. AS \ Vurl \ S ■ AS:=AN'. tat^i and SO- AS. (1 + C»t'i)= -,~Tv Hence, the parameter =4. AS=i. SO. .sin'l. (21) Corollari/2. — Let F be the obli({ueload upon tlie cable between &i F '• Q " " total thrust upon the plarform at E. " w " '* load per horizontal unit of length. '• q " " rate of increase of thrust along platform. '' t " " length of i'i;. w w '= - — -I and q=-w. cot i sin I ^ (22) 84 SUSPENSION BRIDGE LOADS. ' -» )/• ,r -' // H=",:L=2>r'.SO=2AS. 2.( w'.f.x sin^i =2. A S. -."-.- (23) r=iir^'= " •'- (24) .c // f=,r + ^ + x.t/.Cosi (25) CnroUiin/ 3.— Let s ha the lengtli of ()l\ and let f) bo the inclination .'.s=Ar—AO - ^•^^'l'l[tan (90-/?) ..wc (90-/0 t- Ai(7,.('"» 90^ + s^>c. 9(rr«) 2.W L -tnn (90-0.«''C (90-(")-/o(;, ( ^ni !KM + scr 0(Pi) | — , Acot iKcosec H — cot i.oscr I -^ dx]. , (20) It may b.' (>iisily shown as in tlio Cornllary of v^(7) that aj)proxiuiatoly, •> .y-. ^Z//- '• ^27) ,s=:.l' + y.<'".s' * H — .- 3 .'• //. ''ON i (11). — Siisjuuninii hriilijc lodds. — Tho heaviest distributtMl load to which a hiLihway bridire may be subject'd is that due to a dense crowd of people, and is fixed by modern French practice at S2-lbs. per sq. ft. Probably, however, it is unsafe to estimate tlie loud at less than iVoni 100 to 140-lbs. per s(|. ft., while allowance has also to be made for the concentration upon a single wheel of as nnich as HO.OOO-lbs., and per- haps more. A procession marching in step across a suspension bridge may strain it far more intensely than a dead load, and will wt up a synchronous vibration which may prove absolutely dangerous. The fucfor of sa/ctj/ for tlie dead load of a suspension bridge should not be less than 2^ or 3, and for the live load it is advisable to make it 6. With respect to this point it uuiy be remarked that the efficiency of a cable does not depjnd so niueh u[)oii its ultimate strength as upon its limit (>f elasticity, and .so long as the latter is not exceeded the cable remains uninjured. For example, the hnak'ng weujht of one of the 15-inch cables of the East River Bridge is estimated to be 12,000 tons, its limit of elasticity being 8,118-tons, so that with H only as a factor of safety, the stress would still fall below the ela.stic limit and MODIFICATIONS OK TIIK SIMPLE RUSPKNSION lUIIDOE. 8.') ,1 it us •:> liiive no iiijuriouH lifft-ct. Tlio riDitlinui/ applieation of such ii loail would doubtlt'SH iiltiiiiatcly K-ad to thi; dostria'tioii of the bridiio. The dip of tho cable of a suspoii.sioii hridLTC ttsiiully varies fi-oiu l-15th to 1-I2th of the span, and i.^ rarely as much as much as 1-lOth, except for small spans. (12). — Minlljifiifiotis of the simple sitxprusioii hr!ihji\ — The disadvantages connected with suspension bridt,'es are very <;reat. The po.sition of the platform is restrieted. massive anclioraijes and })iers are jrenerally re(|uired. and any change in the distribution of the load produces a sensible deformation in the struct\ire. Owinir to the want of rigidity, a considerable vertical and lutrizontal oscillatory motion may be caused, and many efforts have been made to modify the bridge in .such a manner as to neutralize the tendency to oscillation. (a). - The simplest, improvement >s that shewn in Fig. 13, where the point of the cable mo.st liable toi deformation is attach- ed to tlu' piers by short straight chains A B. (b). - A series of in- clined stays, or iron! ropes, radiating from the pier saddles, niay be made to support the platform at a number of equidistant points, (Fig. 14.) Such ropes were used in the Niagara Britlge, and still more recently in the East River Bridge. The lower ends of the ropes are generally made fast to the top or bottom chord of the briilge truss, so that the corresponding chord stress is increased and the neutral axis proportionati'ly displaced. To remedy this, it has been proposed to connect the rojH?s with a horizontal tie coincident in position with the neutral axis. Again, the cables of the Niagara and East Kiver bridges do not hang in vertical planes, but are inclined inwards, the distance between them being greatest at the j)iers and least at the centre ot the span. This draicing in adds greatly to the lateral stability, which may be still further increased by a series of horizontal ties. IMAGE EVALUATION TEST TARGET (MT-3) 10 ^ I.I 1.25 IIM 112.5 I: iiS 1.4 liiiU II 2.0 1.6 % ^3 ^" /J c^. (Tl ^'/ ^'^# :^ > V Qx w- \ 86 MODIFICATIONS OF THE SMPLE SUSPENSION BRIDGE. (c). -In F'v^. 15 two CJibles in the f^anio ver- tical [tlane are diagonal ly braced t =|'.^il.(^'^_,,.x'), (6) which is greatest when x'= „, i.e., at the centre of OB, its value then w Z — cc being — ^. — — , x^ Thus, the bending moment is an absolute maxi- mum when , Cl.x^ - x^") = 0, i.e., when x = ^.l, and its value is then dx X ; ^ ^ w 54" V The bending moment at any point between B and A distant x' from (?, =/?l.x + —^ .X " - — .(x - x)* =. -r-. .-.X - x.l - x' (7) d l + x which is greatest when — -, (x' -x. Z-x')=0, i.e., when .x'=— ^, or at dx: w x the centre of AB, its value then being q-.j-.(^ - x)*. Thus, the bend- ing moment is an absolute maximum when (x.l - x ) = 0, i.e., when dx X = _, and its value is then + — P. 3' 54 AUXILIARY OR STIFFENING TRUSS. 89 ! Hence, the maximum bending moments of the unloaded and loaded divisions of the truss are equal in viagnitude hut opposite in direction, and occur at the points of trisection ( D, C) of OA, when Ihe live load covers oners one-third (AC ) and tuw-thirds (AD) 0/ the bridge, respec- tively. Each chord must evidently be designed to resist both tension and compression, and in order to avoid unnecessary nicety of calcuhition, the section of the truss may be kept uniform throughout the middle half of its length. In order that the truss might act most efficiently, it should be made in two halves, (Fig. 20), con- nected at the centre by a bolt strong enough to re- sist the shear of u/.l The trusses can then deflect freely and are not sensibly strained by a rise or fall of the cable under a change of tem- perature. Case II. — A single concentrated load W at an 1/ point B 0/ the truss. W now takes the place of the live load of intensity w'. The remainder of the notation and the method of procedure being precisely the same as before, the corresponding equations are, Jii + Jii + {t-w).l- W={) (1') ^-^.P-W(l-x)=0 (2') RJ + P- W(l-x)=0 W t-^w= I -^=f(-4)=^' (3') (4') which shews that the reactions at and A are equal in magnitude but opposite in kind. They are greatest when x— and when x=l, i e., W when W is either at or at A, and the common value is then 2 The shearing force at any point between and B distant x' from 0, =R, + {t-w).x'=^.(x'-x+^-y (5') W which is a maximum when x' = x, and its value is then -^' 90 AUXILIARY OR STIFFENING TRUSS. w The web must theicftire be desijiiicd to bear a shear of throui^hout the whole leriL:th f)f the truss. Ajraiii. the bendinj; moment at any p(»int hifirccii (ind B distant x' from O, First le x< -• The bondinir moment is positive and is a maximum when x'=x, its value then being +-—- (?.x — x^j. iVcx/ let x>-^. The bending moment is then negative and is a maxi- ^ • , t- ,. . "' / /\' mum when x =x — -— , its value then being - • / x - -y ) . The bending moment at any point hetwecn B and A distance x' from = .S,.x' -. (< - ,.)/^ - ir.(x' - x)= -.(x' - ( ;^- - ^^ (7') which is a maximum when -, i x'- ^Y o- - -^ ) [ =0, /.f-., when 2 . W / l\^ — , and its value is then — 5— •( a; — :, ) . X =eX + Ahte. — The stiffening truss is most effective in its action, but adds considerably to the weight and cost of the whole structure. Provision has to be made both for the extra truss and for the extra material re- quired in the cable to carry this extra load. CHAPTER y. A R C II K D K I B S . (1). — Lincdr arrhcs. — When a cord han^rs from two points of sup- port, and is loaded with a iiunibcr of weights, it tends to assume the form of a /iinini/iir, or vquiUhriinn polyfron, and the horizontal pull at every point of the cord is the same, (^ (3). Chap. I). Let and .1 l)e the two points of support in the same horizontal plane. Let ./•. ij, be the co-ordin- ates of any point Q of the cord witli respect to 0. Let P be the resultant of the weights bt'tween and Q. Let X, bo the horizontal distance of /'from Q. Let Fand //, respectively, be the vertical aud horizontal components of the resultant force at 0. Take moments about Q. ILy-V.x + P.x, (1) Suppose the cord to be exactly invcrtal and to be stayed in such a manner that it retains its form unchanged. Also, let the weights be the same in magnitude and distribution. The stresses at different points of the cord are now reversed in /^lud but not in mapnitude, and equation (1) becomes, n.,/=V.x ~ P.x.^M (2) M being the bending moment at a distance x from \n sl horizontiil girder of the same span and similarly loaded. But JI is constant and .'. // oc M. 'Tence, the vertical ordinate^ at points of the inverted cord measured from the horizontal axis A are proportional to the bending moments for corresponding sectioits (fa girder of the same span and similarly loaded. 92 ARCHED RIBS. If the numbei" of the weights is increased indefinitely, and the distance between the points of application indefinit<.'ly diminished, the inverted cord becomes a curve, and forms a lincur arch, or cquilUrrated rib. An infinite number of linear arches, or curves of equilibrium, may be drawn through the points of support, as they are n.erely the bending moment curve plotted to different scales. These arches transmit a thrust only, and can never be constructed in practice, as the equilibrium is destroyed by the slightest variation in the distribution of the load from that for which the arch is designed, but the principles upon which they depend may be applied in the dt. termination of the lines <>/ resist- ance for arches and arched ribs. Corollitry 1. — If 7/ be unity, the ordinates (j/) are ecjuul to the bend- ing moments. Corolhry 2. — The linear arch for a load uniformly distributed along the horizontal is 2i, parabola, (§ (4), Chap. IV). (2). — Arched Ribs. — An arched rib is any truss in which both the chords are concave or convex in a vertical plane, and which gives rise to oblique reactions at the supports. Arches and suspension bridges are the two extremes in which the horizontal stresses developed by the load are met by horizontal forces applied at the points of support, instead of being made to neutralize each other within the structure itself, and are therefore the only systems in which the material should or might be subjected to the same kind of stress throughout. Wrought-iron and mild steel are very suitable for small arches in which parts of the rib are liable to tensile tresses. Cast-iron and cast- steel, having a greater compressive strength, might seem better adapted to large spans, as they admit of greater economy, but they are ill-suited to withstand shocks, and have been almost entirely superseded by the more elastic materials. The rib may consist of a number of tubes bolted together, as in the St. Louis steel bridge, but is usually of an /-section. The depth of the rib is not necessarily uniform from end to end, but may be advantageously increased in parts subjected to an excessive bending action. The space between the roadway and rib, i.e., the spandril, may be filled up with some kind of lattice work, which should be very carefully designed, as the artistic appearance of the arch is largely dependent thereon. ARCHED RIBS. 93 Generally, the ends of the rib are securely boltv>Ar(/j/« maximum stresses at the diiferent sections of an arched rib. The stresses cannot be found with mathematical accuracy, as they depend upon a variety of conditions which are altogether indeterminate. The total stress at any point is made up of a number of subsidiary stresses, of which the most important are : (1) a direci thrust, (2; a stress due to flexure, (3) a stress due to a change of temperature. Euoh of these may be iiivo^stigated separately. (4). — Bmclivg moment {M) at any jwint of an arched rth. Let A B c, the axis of an arched rib fiiiiged at the ends, coincide with a linear arch for a given load. The thrust at any cross-section of the rib is necessarily axial and uniformly distributed. For a diflferent load the linear arch assumes a new form, as a d c, and thf^ etrcBs is no longer axial. Let V and //, respectively, be the vertical and horizontal components of the thrust at A , Let P, the resultant of the load between A and any ordinate n E F, meet A c in G. ''•M at E,= V.af-P.qf-ILef But, = F. A p - P. a p - //. D p .'. i/= /f ( DF - EF) = //. D E (1), Hence, ihe bending moment at any point of the axis is proportional to the vertical distance of that point from the linear arch. The same is true if the ends of the rib are absolutely ^icec?. Let Ml be the moment due to the fixture at A. .". M= V. AP - P. GP - B. EP - Ml & = F. AP - P. GP - //. DP - Ml & M=H. DK, as befor3. (2). I' y I RIB WITH HINOED ENDS. 95 iV is zero at II, the point of intersection of the axis and linear arch, and is ncjrative, i.e., clianj^ca in direction, at points on the right of ll. Corollary. — Let T be the thrust along the axis at E; draw d'e a nor- mal at £. .-. r=//.s^cDED' = //.^ D E and •'*M= ZT. D e= T. d'e (5). Rih with hinged ends ; iuvnriability of span. Let ABC bo the axis of a rib supported at the ends on pins or on cylindrical bear, ings. The resultant thrusts at A and c must necessarily pass through the centres of rotation. The vertical components of the thrusts are e(jual to the cor- responding reactions at the ends of a girder of the same span and similarly loaded ; H remains to be found. Let ADC be a linear arch for any given distribution of 'he load, and let it intersect the axis of the rib at ll. The curvature of the mor* heavily loaded portion AEH will he flattened, while that of the remainder will be sharpened. The bending action at E tends to change the inclination of the rib at that point, and if the end a were free to move, the line E/ would take the position EL, al being of course very small ; draw the ver- tical LO. AO represents the horizontal displacement of the point E. Since LAE is approximately a right-angle, (1) ,AO = AL. AE M But the angle Ael = the change of curvature=-Tr-7 nearly, /being the moment of inertia of the section of the rib at E. .-. al_^,^ and .'. A0 = - AE, nearly, M „ DE.EF (2) But the length AC is assumed to be invariable, so that the total dis- placement between A and c is nil. 'Zr.DE.EP\ _ (■ E. I J 96 VALUE OF II. or, since 11 and E are constant, L\ ■;!! m 1 1!! (i^)=o C^) The actual linear arch may now be ascertained by drawinj; several linear arches and selecting the one which most nearly satisfies equation (3). Corollary 1. — Equation (li) may be written, /DP-EP\ /l»K. KK\ /EF'\ - ... l(^— p-j.EF = 0,ors(— ^-j-2(-y)=0 (4) Hence, the ordinates of the required arch may be directly calculated from the bending moments at the several sections of a girder of equal span and similarly loaded. Corollary 2. — If the section of the rib is uniform, / is constant, and equations (3) and (4) respectively become, 2. (de.ep) = and 2 (df.ep)-5; (ep*)— Corollary 3. — Let the axis of a rib of uniform section of span / and rise k, be a parabola, and lot a single weight be placed up- on the rib at a point of which the horizontal dis- tance from the centre of the span is x. The linear arch now consists of the two straight lines da, do, and it 32 k P is found, by applying the condition (4), that D P=— • - „'. ./ O 0.' ^TC.X Corollary 4. — If the axis of the rib is semi-circular, instead of a Ml parabola, it is found that dp is equal to . , i.e., is half the length of the rib. No linear arch, however, will even approximately coincide with the axis of a rib rising vertically at the springings, so that a semi-circular or semi-elliptical axis is not to be roconiniendcd. (6). — Value of II. — Let ADC be the actual linear arch of a rib of which the axis is ASC. Let DP, the maximum or- dinate of the arch,-=y Let x, be the horizontal distance from DP of P, the resultant of the load between A and dp. PROCESS OF DESIGNING A RIB. 97 Let AF = X Take uiomonts about D, v.x—p.x,=rr. V and V", th il npoiicntfi of tlic thrusts at A and c, are cfjual to tlio corresponding reactions at the ends of a fj;irdt'r of the eame span and simihirly h".adod, Tho resultant thrusts at A aud C are there- fore V V' + IP ^"'^ V Y'^ + II -, respectively. Tho resultant thrust at any other point of the rib is now easily com- puted, and is res,»lvahle into two coT»ij)onents, the one normal to tiie sec- tion of the rib at the u;iven point and the other in tho plane of the section. The latter i;ives rise to a shearing stress and the former (unless axial) to a uniformly varying stress, (7). — I'l'ofi'ss of ib'signing a rih. — For a fixed distribu- tion of the load the axis of the rib should be mado to coin" cide with a linear arch for that load. (generally, the rib lias to be designed to carry a live., in addition to a deatf load, so that provision must be made for the stresses which may arise from every possible distribution of the former. Let MN be a numbi',r of equi-distant crosfl-^ections of a given rib. Draw a linear arch for each of the following cases : — (l).-When the live load covers the whole of the bridge. (2),-When the live load covers the middle hal/ofiho bridge. (3). -When the live load covers one-half of the bridge from one end. (4). -When the live load covers three-fourths of the bridge from one end. Calculate the flange stresses in the diflferent sections from the thrust in the nearest of the four linear arches, and add or remove metal according as the stress is excessive or falls below the safe limit. If the form of the rib is much changed by this process, a second approximation must be made. (8). — ]{ib with ends ah- solutely fixed. — Let ABC be the axis of tlie rib. The fixture of the ends introduces two unknown moments, and since H is also unknown, three conditions must be satisfied before the strength of the rib can be determined. 98 RIB WITH ENDS ABSOLUTELY FIXED. :t The linear arch, as shewn by the dotted lines, may fall above or be- low the points A and C ; draw any ordinate df. Since the ends of the rib are absolutely fixed, the total change of in- clination of the rib between A and o must be zero. M But the change of inclination at any point Eis ~, approximately. ■■•'{in)='=/ (1)1 orchid n'h. — Li t the abut- ments be immovable. Let AHC be the axis of the rib in its normul ])nsition. Let ADC represent the posi- tion of the axis when the rib is loaded. Let BDK be the ordinat.^ at the centre of the span ; join ah, ad. ... I)F"-AD''- af'=.\ij-.( ) -af^ \(trc. \Ji/ But, "/•C.AB-'/rr.AD /being the intensity of stress due to the change in the length of the axis. .•.DF^=AB^ A-ZV- AF'=BF'''-An^ | 2. -^ -J lY ] .-. atjI I 2. j,-(C)' I =BF=' - i)f' = ( hf-dp) («f + df) — 2.nF.(Bj:))., approximately. ( -_, ) is also sufficiently le force at C may be resolved into a component T alonj» the neutral axis, and a component S in the plane; A A'. Tlif latter haa very little effect upon the curvature of the neutral axis, and may be dis- rejrarded as compared with M. Before deformation let the consecutive cross-sections lili' and AA' meet in R; R is the centre of curvature of the arc (.'(/" of the neutral axis. After deformation it may be assumed that the plane .LI' romain.s unchantrod. but that the jilane Bli' takes the position B"B"'. Let AA' and B"B"' meet in R' ; R is the centre of curvature of the arc GC after deformation. i I 102 ELEMENTARY DEFORMATION OF AN ARCIICD RIB. Let nJx'. be any layer at a distance z from C Let CC' = ^>i, CK=R, CR' r= li\ and let la be the seeiioual area of the layer tihc. T> • ., £ "c R' + z . iih R+z By siuular figures, = — „— , and - = — j^- Aii R Js R mi M .1 ,, f"' ., A.*.' /I 1 \ iho tensile Stress iiwf6r = 7:<.Aa.^-=:/i'.A^/.—- — ) (lb ah y R R/ = E.\<, -z. (^j^- - — ^ , very nearly. The moment of this stress with respect to C=E.l((.;.^.\ — j^ \ Hence, the moment of resistance at AA' .jA-..„..^(±-i)=i'.(;^- ;>jw, the intc'Tal extendin"- over the wliole of the section. •"=^-'(-/r- /t) (1) Airain, the effect of the force T is to lonjithen or shorten the element CC, so that the plane BB' will receive a motion of translation, but the position of R' is practically unaltered. Curulliirij 1. — ]jet A be the area of the section ^l.l'. T M.z The total unit stress in the; layer (thr=p=r -i____, (^2) the sign being 7>^^s■ or iiuiiks according as J/ acts towards or from the edge of the rib under consideration. From this expression may l)e deduced. (1). — the position of the point at which the intensity of the stress is a maxinium for any given distri- bution of the load, (-). — the distribution (if the load that makes the tensity an absolute maximum, (!>). — the value of the intensity. Corolla n/ 2. — Let ir be the total intensity of the vertic il load per horizontal unit of length. Let »", be the portion of w which produces only a direct compression. Let 7/ be the horizontal thrust of the arch. Let y^ be the total load between the crown and .LI' which jtroduces compression. Refer the rib to the horizontal OX and the vertical OPV as the axes of X and y, respectively. ) 1) i GENERAL EQUATIONS, Lot X, 1/, be the co-ordinates of C. :.P=Ij:^;hvLtdF=w,.dx d.c ' 103 also T=ill. d? ds dx (12). — Gcncnd equations Let X, y, be tlie co-ordiuates of the point C before deformation \ ^V/'- " " " " ^iftrr '' e be an-le between tr.M-ent at G and OX before deformation "'^'(='^-^^0 " <• " after '. ■ " ds be the length of tlie element CC before deformation. (3) (■i) ds' after <( Effect of flexure. ^L.^.^,,n^f=' ds W ds R ■ ■ £.1 R' R ~ ds' di- ''•AO:=:e-e'=^Ae„-r-^/t-dx, J L.J dx ds ve ry nearly. (5) ^(^o being the alteration of slope at F, which has yet to be determined. Again, cos e'=zeos (ft - so)=eos ft + Afi.sinO and siti 0'=zsi)iQ) - Aft) = sin ft - Aft.eos ft . dx' dx di/ di/' dij ds' ds 'ds ds' ds ■ AO. dx 17 (fi) Effect of T and of a change of t" in the temperature. — ds is compressed by T, and . • . its new length due to the compression In order to make alhiwunce for the change in temperature, this last expression must be multiplied by {l±a.t), a being the co-efficient of linear dilatation of the material of the rib. nt .■.ds' = ds.(l-^--^y(l±a.t). (7) Hence, by equations (G), dx'==(dx + A ft.di/). (j. - -Ly^i j.„.,) _ T — dx + A d.di/ - -p—-dx±^a.t.dx, approximately. 104 GENERAL EQUATIONS. and dy'={ihj - a fl.^/x). (l - -^^ (1 ± ,,.0 T = d\j- i^O.dx ~ p A 'dy ± a.t.dy, approximately dy T or, dx' - //x=d e,-d.Jx - -- .dx ± a.t.ilx (tx Jli.A and dy' -dy= -^ e.dx - -—'-■L^dr.-hn.t.'-^.dx E. A dx dx O O II and J, -y= - j ^".^x- J^_.^,,fa ±j ,,. (. -,f & (9) o Let / be the span of the arch. Let x„( = 0),y„, be tlie values oi'x.y, at 1\ Let A Ox, .''i,. .'/u be the values of Aft, x\ y\ respectively, at Q. C' M ds , ..A.. = Aft„-J^-^.--. be the dead loail per horizontal unit of length. Letwj' " live " " " Let the live load ccver a lenutli DIJ, - /'. /, of the span. Denote by (.1) formula) rclatinu' to the unloaded division OE, and by (7i) formuhv! relating to the loaded division DJJ Equations (14) and (1.5) respectively become, (A) s^-S,+ (^^-ir).x (26) (B) S=>%, -f (^--AZ^ - »•) . X - w'. ]x -(1 - r).l\ (27) (A) M^- M + .%,.x + (^dL _ ^o) -^ (28) (B) J/- ^f, + S..X + (^-^ - «) .^l^'ix-(i-o./r (29) PAUABOLIC RIB OF UNIFORM DEPTH AND STIFFNESS. Since the ends are fixed, _so„ = = Aff Hence, by wiuatiuiis (21) and (28), and by erjuations (21) and (29), When .r=A A^^rr^^, =. (». a,.,] therefore by tlie hist e(,uatinn, 107 (30) (31) o=^i/„ 4- >y Again, let :^fl^ , dx (32) (33) But A ff. = 0, and'i^^-= ^■^' By the conditions of the problem, x' - x and / - y are each zero at (J. Hence, equations (22) and (23) respectively becoue, O Substitute the value of A ^ in equation (32). and integrate between the limits; J -^(l.tj. which may be written, o=j/„ +§•.,+ (^/_ „.).f._ :::,.o,.+?./a_ -'.A'v, ^^T3i u 108 rAR\noiJc Rin of uniform dfpth and stiffness. Ecjuations (133), (37), ami (3S), arc the equations of condition. Stihtract (37) from (33), .Siil)tract (37) from (3S). uml iiiiilti]iK the result by 3, Sul)tr:ict (3!0 from (40). ■•^•==H-7^-'')-.io-'^'-'-U-i.r(r)-2\i.:A-^ 4 ' /r ^^^^ lleiico, 7/= (42) V 4 K.A-.^^ Elimiiiatin<,' .S'„ bctwct'ii (33) and (38), " \ p / 12 V 3 4-^ ^ ^ By equation (20), 3/ at Q is, .,,,,.,A,..v„./..C^-)f'p-'' W Eliniiiuvtinjz S^ between (33) and (44). ••• - ^^^' == - ^^^" + V-7- - "" ) 6 - "■ •'"• (2-3; ^^5) Substitute in this equation the value oi' M„ in (43), ' V r^ / 12 V2 3 4/ ^ ^ To find the greatest intensity 0/ stress, etc. — T II The intensity of the stres.s due to dire-^t compression — - = — The intejisity of the stress in the outside layers of the rib due to bending is the same as that in the outside layers of a horizontal beam of uniform section A^ acU'd upon by the same moments as act on the rib, for the deflections of the beam and rib are equal at every point, (Equa- tion 21). Also, since the rib is fixed at both ends, the bondinj:^ moment due to that portion of the load which produces flexure is a maximum at the loaded end, i.e., at Q. Hence, the maximum inten- sity of stress (p,) occurs at Q, and pi= — ± 3/,.-' (47) Zi being the distance of the layers from the neutral axis. rAUAnoLic Kin of unifohm depth and stiffness. 109 // and J/, aro both functions of ,-, and thcrofun. p, is an uhmlnte aximum when '^' = 0= A . '''^^- ^' 'Z''^' But, }.»y wjuation (42), and by equation (40), k.(\ l^^.\'^ \ " di: dr \i"-dr 1 , n-'./^llV. (,_,,. ■'^■/•.(i-f:*^.A \ A' ^ 4 ,l„/r7 (49) (50) (51) which may b(! written, The ro(jt.s of tliia equation are, 1 , ^^i /, •■— 1 and r=:-+-.2_ ^ vl|./.-' (52) (5:5) r= 1 makes Ul zero, so that the maximum value of p corresponds to one of thj remaining roots. llence, the maximum thnist= 1 ( II +^ .M^) ^p\ (54) the values of II and M^ being obtained by substituting iu Equations (42) and (4G), ^ i ^ (55) r = ~ I ^--^^ A ^ 1 -_^ A 2 1-/- and the maximum tension = —. ^ - H + ^^[ J^/^ =^y/' /'gg\ the values of // and M^ being obtained by substituting in Equations (42) and (4i;), no p\ii\noLic lun OF unifgum depth and stiffness. »• ==- 1 , 45 A 5,3 A (57) 1 + X- 2 vl,.2,.A' Efjuation (54) will give the jn-opi-r Hcctioruil area wlii'ii the depth ami i'lirui (if tlu- rib have been fixctl. If Eiinatioii (.')()) gives a negative result there is no tension at any point of tiK! rib. Curolliin/ 1. — The moment of inertia liiay bo expressed in the form, /, = g.2'f./l„ q being a eo-effic'eiit whieh depends upon the. figure of the section. .'. the vi'i.n'minn thrust = -•(II -f — ' ) (58) JiV q.z\/ and the corresponding value of /• is, 1 -1 45 3? 1 + _ .n. _' , ;} z, y (^ — O.J o ■< /. — -■-, suppose. (59) By (42;, 7/ = v'.l* ,15 .EI. t n '5 T 15 4 ,3 5\ 8 - H k V4 8 • 4 / k. b (GO) ,.2 9 By(4n),.,/,= -r.,.//,!^„„,,.(^l.-.,.,_;^) lb ^ A-'' 4 k Vb 4 'If, ,., //-^ 2 , r*\ = 1 +"-'^-(2-3'"4) •• '^ q.z, S.h \k 2 /tV"*" 4q.z^\ 2/^7 ^^(^,5 15 3,x /5 5 rS) ,^^ ^5..~,./>( /c W S-'^U- J Vb' 4' '^2Ji'^y..^, V-i 3 "^4/ 8.6 VA- 2 A-V 4 q.zyk h q.z, jb* 4 2) ( h j -^^;(2-3-''^r) But 1 - ;=• V=c and r=-- 2 a; c . T1 a. '^^> - "--^ /I 15 2, \ _ 1 a.^J?./, «,'.;» /r* ,.' r« x ^^^^ PAUABOLIC RIB OF UNIFORM l»i:i"TII AND STIKFXESS. HI Hence, ;/„ tho muxitnum thrust = ^ .V, 7 -f 15 r-k > I '> i.i 1 + 45 : rn -^\at.E.L ir',p Similarly./,",, the m;xhunn teminn = -^ ( - n,\- ^'^' ) 15 .:, 1 ^ ' 4 /i" 3 v^- /•=•+". (C3) t^>;W/..^ 2.-Lct tho depth of the rib be small compared with k. .-. ;■ is a small fraction, and the maximum value of ;>. both for tension and compression corresponds to r = |, approximately. Hence, equations (02) and (G3), respectively become P P A U U -^ 2 kO ^4 7^-+ aT25- ,T ! (64) ^..0..-, 3.-If 2.>5.. then . mustt'lv, '"^ ''^'^ ^^Lquations (42), (40), (C.2) and (G3), in thi^' case respectively be I'w + w' 15 a. t E.I, H^ 1 15 2, aA 8 '"-r-^VTrx +20-^.1 i'. = -r _ 1 15^, ^,( 8 (68) (69) ^ Lorof/,n-y 4.— A nearer approximation than is given by the preced- ing results may be obtained as follows :- « J^ « Pieced deLmatio '^^' ^ ^ ^^' ^^ ^^"^ co-ordinates of a point very near C before 112 PARABOLIC Rin OF UNIFORM STIFFNESS. Let x' + dx', y' + dy', be the co-ordinates of a point very near C after deforuiatlon. .-. ds"" = dx^ + dy-" and f/.s'^ = dx"' + dy'"" .-. ds'^ - di = dx!^ - dx? + dy" - dy" or, (ds' - ds). (ds' + ds) = (dx' - dx). {dx' + dx) + {dy' - dy)(dy' + dy) .• (ds - ds).ds:=(dx' - dx).dx + {dy' - dy)dy, approximately. ds , -, , , ^ dy .-. dx' - dx = ids' - ds). "- - (dy' - dy)/li dx 'ix da and dy' - dy - (ds' - ds)'!l /if - (rfx' - dx) . ^ dx dy Hence, by Equations (6) and (7), y%.. T /ds^^ dx dx'-dx=.M'^,dx - -£. Y f y,c7;c +«.<. C^y, E.A \dxj - \dxJ Jx T /ds\'dx J , /dsVdx .dx ^3«.^ . dy \dxJ dy and dy' - dy= - SO.dx - J. (% \ !^^,dx ±a.t.{'ll] ."^^.dx XL. A \dx/ df/ ■ < f^'^y :.x' -X— /Mr' '.dx- / __. _ . ((x-h 1 which may be written, Subtract (71) from (70), o)--- w'.l. ( V - - -r ~\ + 3 7/ ^' . ^ /5 V2 2"5;+^-^^x^ .-.0=/^ 7V Hence, Ii= PAio + ~^(5y-5.r* + 2.r')l (71) (72) (73) 8Jc.(l-,l^A.L) \ 8 A./c^J Eliminating *S'„ between (38') and (70), Also, by (27'), ^i = ^o+ {^—ji wj .1 - w'.rj= - P, suppose. (75) Eliminating ,S; between (70) and (75) ••■-^=^-('#'-)f»'K'-2-) W (70), (73). (74) and (76) give the values of //, ^„, ^„ and Agam, the maximum bending moment M' occurs at a point given by ^ = 0.n(29'). 8 (7V) M 114 PARABOLIC RIB OF UNIFORM STIFFNESS. Subtract (77) from (75), Hence, the distance from the loaded end of the point at which tlie bending moment is greatest is, F (78) l — x=- to + V) S.k:H_ Substitute this value of .r in (29'), and. for convenience, put 8./.-.// ii; + }C — P == 711. ^ ?/•' -- m , w' , , But by (70), 0=S„+-^,^—l-,yi^.l P' (79) Hence, M\ the maximum bendlug m<)nuiif. = - -~7 , S.I:.IJ\ As before the greatest stress (a fhriisf) = -j' ( H + ^j^M'j -7''i-(^^) and the value of r which niakrs //, an ohsnhitr niaxiniuui is given by ^=0. But by (79), M' ijivolvos /•'" in the numoraior, and r' in the dr denominator, so that 'p = r oonvenience, jiutting 1 4 - -j^ j^^ = n. By (73), -^=8:/;r7\""'-2.)' By (70), S.= ^-^{>r-,-^y~^-^\, (HI) (82) •8) PATIABOLIC RIB OF UNIFORM STIFFNESS. .,(74), -.=,^,{0-i;>'i^'-^'} B, (76), -S, = P=U(„:^'^)^^"' Ej (78), By (70) ( "-'N n-\ w' ,_8-{("'^ 2) — ^-| / w'\ n-\ w' M= 115 (83) (84) (85) (80) (15). — PuriihoUc Rib of uni/onu stiffness hinged at the crown and also at the ends.-h^ this case J/= at the crown, which introduces a fourth equation of condition. By(2n o=.„.^.(?^_.).(:.:^.(_i,,y which may be written, Eliminating S^ between (85) and (70), 8./.-.// .-. -J- - w =. w'.( - 2r' + 4r - 1) Hence, ^i=^^- \w- io'.(2r^ _ 4^ + 1 W By (87), By (76), By (74), S^=~{(3r'~4r+1) i.=s,=i^--',(,.-i)- A 6.,= 7C'.P " 24. E.I ■(l-4.r + 4.r'-r*) --(r-1) By (78) and (^QO),l-x=;^ _ tj.w (^r —- L ) By (79), ^f='!^Xr-ly _ _/ '~ 4 (88) (8;)) (90) (91) (92) (93) 116 MAXIMUM DEFLECTION OF AN ARCHED RIB. When r=\, 1 U'\P . ,„ ,r'./^ I (94) These results agree with those of (81) to (80), if?( = l. In gi'mnil, when » = 1, w + ^ (5.,-^ - b.r* + 1.r') = w - U-' (2./-^ - 4./- + 1), by (73) and (88) .•.2./-''- 5.r« + 9./-' + 8./- + 2 = 0= (2./- - !).(/• - 1 y.{r' - 2), 1 , - and the roots are, r=-, r=l, r- ± v2. Hf'iico, ?( = 1 only renders the expressions in (94) identical with the corresponding expressions of the preceding article when h=\ or 1. Again, the intensity of thrust is greatest at the outer flange of the loaded and the inner flange of the unloaded half (if the rib. and is =8-i;-i7,H-n-("'-^2J i 2,h/ 1 / V'\ . The intensity of tension is greatest at the inner flange of the loaded and the outer flange of the unloaded half »tf the rib, and is _ P (2, ■ O, The greatest total horizontal thrust occurs when r=l, and its value is -^.("■ + «0 (16). — Maximum (leJl"cti'on of an archrd rib The deflection must Dccessarily be a niaxiuiuui at a point given by A fl={). Solve for x and substitute in (9) to find the deflection y' ~ y ; the deflection is an abso hue maximum when — .(_y' - y)=i{). The resulting ecjuation involves r to a high power, and is too intricate to be of use. It has been found by trial, however, that in all ordinary cases the absolute maximum deflec- tion occurs at the middle of the rib, when the live load covers its whole length, I.e., when Jt^ = —y, and r=l. 45 /, =s Case I. — Rib o/ §(13). -For convenience, put 1 •♦"~T'~r7.2= /» w + w ' IbaAE.I^ 8jk r -¥« "1^ .-. By (42), U= (95) MAXIMUM DEFLECTION OF AN ARCHED lUB. II7 By (48) ana (4«),-,V„=^^.(„. ,. ,,ylzJ^l'^J,^ _ ,,^ (,,,, liy (3S; and (4:i), ,V„= - (1.''^" ^gj^ By m, (4:i,, (,,7,, .„= - ^!-^. (,,„.. _ ,,,,,;^, ,,.,,;j;') ^...^ IK'uoc, the niaxiimitii dofloctidn 384/;./,' s "^l^w'T'Z-"'^" ^"PP"'^^- (99) The central d.-flectmn ,/, of a uniform straight b.rizontal beam of same span, ofsamo section as the rib at the crown, and with its ends nxed, IS, I* w + w' (100) '^^"r84i^ Henco, neglecting the term involving the temperature, ^h=~^.L Case U.—jm, of^{U).~ By (78), //= P w + w' (101) (102) By (74) and (70), ., = ^-^,.. .o/^=^,(103) By(32),(102)and(103),..= ^.(;:.|!,i;) ^,,,^ Hence, the maximum deflection A^/■:/ V12 2 -3T/;-'^-^=3^47r/<'''-*-'^')-— ='^'. (1^5) If the ends of the beam in Case I are free, its central defection 384 EJ " 2) , . « , d n (106) Thus, the deflection of the arched rib in both cases is less than that of the beam. 118 HORIZONTAL DISTUIBUTING GIRDER. (17). — Archfd rlh of uniform slijf'm'ssfxid at the. ends and cnnnerted at the crown viith n horizonful distrlhufing girder (Sac ^ (2). — The load is transiiiitU'd U) the rib by vertical struts so that the vertical dis- plaeemeiits (if correspoiidiiijj; puiiits df the rib and ^irdiT are the same. The lutrizoiital thrust in tiie load^il is not lu'cossarily erjual to that in the uiiloadi'd division of the rib, but the excess of the tlirust in the loadt'd division will be borne by the distributinir trirdei, if the rib and frirder are connected in such a manner that the horizontal displacement of each at tlu' crown is the same. Till- forniulii3 of )^ (18) are applicable in the present case with tlie modification that /, is to include the uioment of tlie inertia of the girder. Tilt' maximum thrust and tension in the rib arc given by Equations ((11) and (05). Let ,: ' be the depth of the girder, A ' its sectional area, 11 M,.z' The ureatest thrust in the girder == (( tension .'l, + .l'"^ 2.A\/, (107) (lOfi) 7/ and J/, being given by Etjuations ((iO) and (t)7), respectively. The girder must luive its ends so supported as to be capable of trans- mitting a thrust. (18). — Semi-circulnr timber rlh. hinged at the ends. Let F C Q be the axis of such a rib. Consider a portion (X'" bounded by vertical planes at C and C. Let ;r, i/ be the co-ordinates of r' with resjM'ct to (\ Let /• be the radius of the semi- circle. Let w 1)0 the intensity of the load which is assumed to be uni- form/i/ distributed horizontalli/. The moment of resistance of the rib at 6" = the algebraic sum of the moments of the thrust at C, and of the weight upon CC, with respect to (7'. M=-U.ij + ir. x^ STRESSES IN SPANDRIL POSTS AND DIAGONALS. 119 v\ r >> At either end, x — i/-r. and Jf= 0. .", //= alsox* = ^. (2. >•--//) ?r J/ is a maximum when // = -^, its value then being — ^ In the horiz(tntal tan^-nt at (\ take CG = — ; Q G will be the direction of the resultant thrust at Q. (19). — Sfrcsst's in spiDufri/ j>(>stx itnd duujnnuh. Fip:. 1(5 ri'prosents an arch in which the spaiulril consists (»t' a series of vertical posts and diau'onal braces. Let the axis of the curved rib be a para- bola. The arch is tiuiii equilibrated under a unil'uruily distributed load, and the diagonals will be only called into play under a passing load. Letx, ^, be the co-ordinates of any point i^ of the parabola with re- ^f' 2 spect to the verte^ o. .'. y = -jy' -'^ Let the tangent at F meet C B in />, and the horizontal B E in G. Let BC=k' :. BL = BC-CL = BG-CN=k'-y. Let i.Vbe the total number of panels. Consider any diagonal E I) between the /<-th and (h 4- l)-th posts. Let u''be the greatest panel (ivc load. Tlie greatest compression in ED occurs when the passsng load is con- centrated at the first h-1 panel points. Imagine a vertical section a little on the left of EF. The portion of the frame on the right of this section is kept in wjuili- brium by the reaction R at I* and by the stresses in the three members met by the secant plane. Take moments about G. :. D.GE.cos e=R.AG, D being the stress in DE, and 6 the angle DEF. 120 METHOD OF DETERMININC RESULTANT THRUSTS. The greatest thru.st in EF = w' + " " tension " = D.coxd- Now,;;=";'.'^ , x+C B k' + y ,, „ k'.x - .T.y and • aE=GH^.c J''^;^^'!' ^ «,.d GA =U^L±Z^ •J,y 2 'l.y ,, T^ "'' « " - 1 /'/ + k'.x -xy Hence, Z>=—. r^ — .-^^ -.strd 2 N k ..I ^x.ij The stresses in the counter braces, (shewn by clotted lines in the Fig.), may be obtained in the same manner. w xus f - ?p, %o being the dead load upon EF, If the last expression is negative, EF is never in tension. (20) — CUrk MuxwvU s method 0/ determining the resultavf thrustn at the anjiporis of a framed arch. Lei .^ .s be the change in the length s of any meu-bcT of the frame under the action of a force P, and let u be tlie sectional area of the member. P ,•. ±-rr--*>' = •^•'*j the sign depending upon the character of the stress. E.a Assume that all the members, except the one under consideration, arc perfectly rigid, and let Sf be the alteration in the span / corresponding a/ to As. The ratio — is equal to a constant m, which depends only upon the geometrical form of the frame. .'.a/ =w. as = rh ni.P. ,\— E.a Again, F may be supposed to consist of two parts, viz., /, due to a horizontal force // between the springings, and /, due to a vertical force V applied at one springing, while the other is firmly secured to keep the frame from turning. By the principle of virtual velocities, f_A/ ^^^^ II AS A:. Similarly, ! is equal to some constant n, which depends only upon the form of the frame, P==f+A==m.JI+n.V .: Al=± (m\II +m.7i.V)-r^ METHOD OF DETERMINING RESULTANT THRUSTS. 121 Hence, the total clian^o in I for all the members is, I.sl=±^(in- /I. ■' ) 4- :l( m.ii.V. ,' ] If the abutments yield, let ^:s/=u.//, n being some co-offieient to be determined by experiment. If the abutments are iuunovable, 2..^/ is zero. (A) and 11= - (B) T^ is the same as the corresponding reaction at the end of a girder of the same span, and similarly loaded. The required thrust is the resultant of 77 and T', and the stress in each member may be computed grapliically or by the method of moments. In any particular case pro- ceed as follows : — (1 ).-Prepare tables of the values of /u and n for each menibcr. (2).-Assunie a cross-section for each member, based on a probable assumed value for the resultant of Fand //. (3) .-Prepare a table of the value of ;>r.^— for each member, and h.ii form the sum j; ( »r. 1,- ) V E.n I • (4). -Determine, separately, the horizontal thrust between the spring- ings due to the loads at the different joints. Thus, let c,, r.^ be the vertical reactions at the right and left supports due to any onv of these loads. Form the sum zim.n.V. — ^ using c, for all the members on the right of the load and r^ for all those on its left. The corresponding thrust may then be found by E'mperature, into a most irregular hole, but the exigencies of practical conditions often prevent its use, except for ordinary work, and its advan- tages cannot be obtained where they would be most appreciated as, e.g., in the riveting up oi' connexions. 130 STRENGTH OF PUNCHED OR DRILLED PLATER. (7.) — Dlinrnslnns of rivets — The diam(-ter {d) of a rivet in ordinary girder-work varies from ^-in, to 1-inch, and rarely exceeds l^in. The thickness (t) of a plate in ordinary girder-work should never be 5 less than i-in., and a thickness of fin., or even -r-;-in., is preferable. lb Let T be the total thickness through which a rivet passes. According to Fairbairn, when t < ^-in., d should be about 2.t " " •' " < > i-in. " " " \\.t ** '' Unwin, when t varies from ---in. to 1-in. and passes through two thicknesses of plate, d lies V . 3,5 ,7,8 between -t + 7-7 and 7:-t + ~. •i lb 8 8 T 5 " " " when the rivets join several plates, r bolts. The resistance must of course vary with the quality of the iron or steel, but the degree in which the tenacity of the plat«'s is affected by the indentation is most uncertain. The mean crushing intrnsiti/ may be defined to be the ratio of the total stress to the bearing area (the j)roduct of the diar. of the hole by tht- thicftness nf the plate), and has been fixed at 5 tons per stj. in. for icrought-iron, but on theoretical rather than practical grounds. If it is also assumed that the bearing and tearing strengths of a joint are the same, it is found that the crushing pressure rapidly increases with the ratio of the diar. of the rivet to the thickness of the plate. When the ratio exceeds 3, the crushing pressure in a double-cover joint is more than 40,000-lbs. per sq. in., i.e., it has reached its ultimate limit, and the joint will fail, although the tension in the plate is less than SGOO-lbs. per .sq. in. Hence, the diar. of the rivet should not be greater than three times the thickness of the plat«. In all probability the failure of the joint at such a low tensile stress is partly due to the deformation of the metal around the hole, which causes an unequal distribution of stress. Until further experiments give more definite information, the safe bearing intensity for wrought iron may be taken at 5-tons per sq. in. in ordinary work, and at 7^-tons per sq. in. in chain rivetted joints with well supportt>d rivets. The safe bearing intensity for steel may be taken at the same proportion of the ultimate stress as in the case of wrought- iron. Let S be the total stresses at a riveted joint. " /I'/s'./'a-/*! be the safe tensile, shearing, compressive, and bear- ing unit stresses, respectively. " t be the thickness of a plate, and w its width. " A' be the total number of rivets on one side of a joint. " n '' '' in one row. " p be the pitch of the rivets i.e., t!\e distance centre to centre. " (/ be the diar. of the rivets. " X be the distance between the centre line of the nearest row of rivets and the edge of the plate. 134 STRENGTH OF RIVETED JOINTS. It is impossible to apply the ordinary mathemafical rules to the determination of the ultimate . '' v-" v-f •-0 — — »=» — «=> — «::> — ^13 — iw)i ^_^ v.^ 1 FiG.23 in Fij;. 23. The advantiif^ea of the latter are that the plates may be cut in uniform lengths, and the flanges built up with a degree of accuracy which cannot be otherwise attained, while the short and awkward pieces accompanying l/roken joints are dispensed with. A good practical rule, and one saving much labour and expense, is to make the lengths of the plates, bars, etc., multiples of the pitch, and to design the covers, connectiona, etc., so as to interfere with the pitch as little as possible. The distance between two con.secutive joints of a group, (Fig. 22), is generally made equal to twice the pitch. (14). — Zig-zag and other riveted joints (Fig. 20), — A few experiments indicate that chain is a little stronger than zig- zng riveting. An excellent plan for lap and single cover joints is to arrange the rivets as shewn in Figs. 24 to 28. In reality, the strength COVEHS. 137 , 13 ind tch is '9- to th nf the plate at the joint is (>iily Wfukonod by nnr rivet Imlc. fur the pliii cannot tear at its weakest section, i.r., alonjr the central row of rivets {iiI).(j\~^'^.vi\/,, for the line 11 (H'-'J.l/)^7,= ^ .(m^-l).y'. '• " L'2 {w - 3d).tq, = .(/n» - :j ) /;. " •• 3.i 4 {w - 4.d).t.(j, = ^- .(m^ - t; )./;, •• 4-t ur - 1 Assume that /,=7„ .-. »•=(?«'+ 1).^/. Hence, by substituting this value of w in the first of the above I *• ^^ '^ /i • relations, ^ =j-.'--- since q,^,q^, are each less than /„ so that the as- sumption is justifiable. (1.5). — Covers. — In tension joints the strength of the covers must not be less than that of the plates to be united. Henco, a single cover should be at loa.st as thick as a single plate, and if there are two covers, each should be at least half as thick. When two covers are used in a tension pile it often happens that a joint occurs in the top or bottom plate, .so that the greater portion of the stress in that plate may have to be borne by the nearest cover. It is. tluTcfore, considered advi.^^able to make its thickness |-ths. that of the plate. The .lumber of the joints should be reduced to a minimum, as the introduction of covers adds a large percentage to the dead weight of the pile. Covers might be wholly dispen.sed with in per/ret y«w/* joints, and a great ecommiy of material effected, if the difficulty of forming such joints and the increased cost did not render them impracticable. Hence, it may be said that covers are required for all eomprrssion joints, and that they must be as strong as the plates, for, unless the plates butt closely, the whole of the thrust will be transmitted through the covers. In some 138 KIVET CONNEXION 15KTWKKN FLANOKS AND WEFl. of the best exiiniploH of bridyc construction, the tension and couipnssion joints are i(k'ntical. (IG). — liio(t ('i)nnexion hvtween Jl(ni(j< s and web. The web ia usually riveted to the angle-irons which form part of the fla" « .1/ be the bending niomont. and .S' the shearing force, at the jioint of >vnicli tlie abscissa is x, x being measured along the girder from any given origin. Let /( be the deptli of the girder, liCt/; be the safe shearing stress per unit of area. Two cases may be considered. Otisc I.-l'i'(te \Vt;h. — The shearing area of the rivets connecting the ■web with tlie angles must be e({ual to the theoretic horizontal section of the web, i.r., to .y, and the corresponding number of rivets per unit Jl-d[_ 11 nJ'/^ 1^ .'be the liorizontiil stress transmitted to the angles through the web, and tending to make them sUde over the flange face. ,'.F.h, the increment of the bending moment, = -^ = S,ot F=—. dx n s The requisite shearing area = -r-.- and number of rivets is therefore the same as before. Cuse II -Open M'eh. — One of the braces which meet at each apex is a strut, and the other a tie. The sectional area of the rivets con- necting each brace with the ann must be ecjual to the net section of that brace. Let /' be the horizontal component of the diagonal stresses. .•. F.h the increment of Uie bending moment, = aS' .". F =-j-, is the force which tends to make the angle slide over the flange face. Hence, the number of rivets between two consecutive apices. _ 14 1 S EXAMPLES. 139 iton the lint my ihe (.f lit he re le Examples (1)- — A trellis girder rests on two supports without exorting any lateral pressure upon them. A vertical section of the girder uieet^ two sets of H bars inclined in opposite directions, and it is assumed that in each set a mean stress maybe substituted for the different stresses in the bars. If />,/>', are the mean stresses and 7'C, the horizontal forces in the flanges at the section, shew that «.(/)- D') sin 0+ T- C must be zero at all points of the span, independently of any additional hypothesis, being the common inclination of the bars to the vertical. Note. — If T = C, D must necessarily be equal to D'. (2). — The tensile and compressive unit stresses in the chords of a bridge- truss of span I and uniform depth (/, are nowhere to exceed /', and/j, respectively ; shew that the greatest central deflection is approximately (/i+/2)- g ^, , • Hence, if the sp:in is eight times the depth, and if the grade line is to be truly horizontal when the bridge is loaded, shew that the length of the top chord should exceed that of the bottom chord by an amount etjual to the camber. (H) — A lattice girder 200-ft. long and 20-ft. deep, with two systems of right-angled triangles, carries a dead load of 800-lbs. per lineal ft. ; determine the greatest stresses in the diagonals of the fourth bay from one end, when a live load of 12()0-lbs. per lineal ft. passes over the girder. (4). — A lattice girder 80-ft. long and 8-ft. deep carries a uniformly distributed load of U-ijOOO-lbs. ; find the flange inch-stresses at the centre, the sectional area of the top flange being SG^sq. ins. gross, and of the bottom flange 45-sq. ins. net. What should be the camber of the girder, and what extra length should be given to the top flange, so that the bottom flange of the loaded girder may be truly horizontal ? (5). — A lattice girder 80-ft. long, and 10-ft. deep, with four systems of right-angled triangles carries a dead load of 1000 lbs. per lineal it. ; determine the greatest stresses in the diagonals met by a vertical plane ju the seventh bay from one end, when a live load of 2500-lbs. per lineal ft. passes over the girder. Design the flanges, which are to consist of plates riveted together. 140 EXAMPLIS. The lattice bars are riveted to angle irons ; find the number of ^ -in. o rivets required to connect the angle-irons with the flanges in the first bay, 10,000 lbs. per sq. in. being the safe shearing strength of the rivets. (6). — The bracing of a lattice girder consists of a single system of triangles in which one of the sides is a strut and the other a tie inclined to the horizontal at angles of a and ^, respectively ; in order to give the strut sufficient rigidity its section is made /;-times that indicated by theory, t e co-efficient /•; being > unity ; shew that the amount of material in the struts and ties is a minimum when - — = «. (6). — A bridge of n equal spans crosses a space of 7/-ft. ; w ^ and w, are respectively the live load and weight of the platform, permanent wry, etc., in tons per lineal ft.; shew that the weight ?" , of the main girders in tons per lineal ft. may be suppressed in the form L. Wj. {p. k 4- r) -f- w^.n n - L. (p-fc + q) k being the ratio of the span to the depth, and p, q, r, numerical co- efficients. Hence, determine the limiting span of a girder. If X is the cost of each pier, Y the cost per ton of the superstruc- ture, determine the value of ?i which will make the total coat per lineal ft., i.e.,- ■■{■w[.Y, a minimum, and prove that this is approxi- mately the case where the spans are so arranged that the cost of a span is equal to that of a pier. (8). — A warren girder with its bracing l formed of nine equilateral triangles, and with every joint loaded, is 90-ft. long; its dead weigl' 5001bs. per lineal ft.; prepare a table shewing the maximum stress in each bar and bay when a live- load of 1350-lbs.per lineal ft. crosses the girder, allowing for an engine- excess of 4,(l001bs. per panel, extending over two consecutive panels. The diagonals and verticals are riveted to angle-iront forming part of the flanges ; how many f in rivets are required for the connection of AB, AC, and AD, at A ? Also, how many are required between A and E to resist the tendency of the angle-irons to slip longitudinally ? EXAMPLES. 141 (9). — If a force of SOOOlbs. strike the bottom chord of the <:irdor in the prccedinfr quoi^tion at 20-ft. from one ' nd, and in a direction inclined at 30° to the horizontal, determine its effect upon the several members. (10.) — The platform of asinirle-track bridtre is supported upon the top chord> of two warren girders ; each girder i.^ 100-ft. loni.', and its bracing is formed of tin L'cjuilateral triangles; the dead weight of the bridge is 900-lbs. per lineal ft.; the greatest total stress in A A when a train crosses the bridge is 41,;-{94.8-lbs.; determine the weight of the live load per lineal ft. Prepare a table shewing the greatest stress in each bar and bay when a single load of 15,000-lbs. crosses the girder. (11). — The compression and tension bars in certain two bays of a lattice girder are required to bear stresses of 90-cwts. and 130-cwts respectively ; design the bars and specify the size and number of the rivets at the attachments, allowing 100-ewts. per sq. in. of net section in tension, 65-cwts. per sq. in. of gross section in compression, and a sliearing stress on the rivets of 100-cwts. per sq. in. (12).-The two trusses for a 16-ft. road- way are each 100-ft. in the clear, 17-ft. 3-in. deep, and of the type represented in the Fig.; under a live load of 1120-lbs. per lineal ft. the greatest total stress in Ali is 35,400-lbs., det<;rmine the permanent load. The diagonals and verticals are riveted to angle-irons forming part of the flanges ; how many |-in. rivets are required for the connection of AB and BC at B ? also, how many are required between B and D to resist the tendency of the angle-irons to slip longitudinally? (13). — Design a double flanged plate girder, 8-ft. deep and 80-ft. long, to carry per lineal ft. a dead load of 500-lbs. and a live load of 1600-lbs., the safe tensile and compressive iuch-stress being 10,000-lb8. and 8,000-lbs., respectively. Determine the number of rivets required per lineal unit of length at the end and centre of the girder to coanect the flange angle-irons with the web. (14). — Find the points in each chord of the girder in the preceding questioa at which the chord stress is equal to the vertical shearing stress. 142 EXAMPLES. (15). — A Howe truss, 80-ft. long and 8-ft. deep, has a single set of diagonals at 45"; find the chord and diagonal stresses in the ^rd hay : (a)-Whcn the truss carries a uniformly distributed load of 40-tons, (b)- Wheo the 40-tons are concentrated at the middle of the span. (^Neglect the weight of the truss). (16). — Design a 16-panelled Howe truss for a clear span of 200-ft., and of such a depth that '.,he chord stress in the 8th and 9th panels is one-half of the total distributed load (dead load per lineal ft.=:800-lbs. live load per lineal ft. = 1200-ibs.). (17). — What should be the depth of the truss in the preceding ques- tion 30 that the chord and diagonal stresses in the second panel may be ecjual ? (18). — A Howe truss with A^-panels and a span of ?ft. is designed for a dead load of ?{'-lbs. and a live load of i«'-lbs. at each of the panel points in the bottom chord ; assuming that the sectional nreas of the diflferent members are directly proportional to the stresses to which they are sub- jected, determine the depth of truss which will secure the greatest economy of material, (19). — The platform of a double-track railway bridge at Thamesville is supported from the bottom chords of two trusses bracid and counter-braced as in the Fig. The leng<'. of each truss is 184-ft. 2-in.. its depth is 34-ft., and it is designed to carry per lineal ft. a live load of 2250-lb3. and a dead load of 1 1 00-lbs. ; prepare a table shewino- the max, stress in each member. (20).— Design a cross-tie for the bridge in the presiding question, the live load for the floor-system being 800il-lbs. per lineal ft. (21).— A 16-panelled Muiphy-Whipple truss 200ft. long, and 18-ft. 9-ins. deep, is designed to carry per lineal ft. a dead load of 750-lbs. and a live load of 1200-lbs., the panel excess weight due to the engine being 4000-lbs., and that due to the tender 3000-lbs. ; prepare a table shewin*'- the max. stress in each member, and specify the bars which require to be counter-braced. (22). — Determine the greater weiglit that may be constructed at 50- ft. from one end of the truss iu the preceding question. (23). — Design a quadrangular truss (See ^^^^ Fig.) with square panels for a span of 91-ft., ^^^SlMMMMS the chords and posts being of timber and the ties of wrouKht-iron. EXAMI'LK-!. 143 (24). — Design a Fink truss for a 242-ft. span from the followini^ data : — dopth of truss = 30-ft. ; kntrth of panel = lo-ft H ins. ; dead load per lineal ft. of truss = 1282 lbs. live load per lineal ft. of truss = 1300-lbs. ; the en_i:;ine excess ■weij:^ht upon each of the end posts = 17U4-lbs. (25). — A^, A.,, and a,, a.^, are respectively the sectional areas and inclinations to the vertical of the two ties which meet at tlie foot of a post in a Bolhnan truss; if the sectional area of each tie is proportioned to the stress to which it is subjected, shew that >1,. cos'n.siti a = A .i.cos^aj.sin n^. (26). — Design an cight-panolled Bolhnan truss. 100-ft. long and 12iV- ft. deep, to carry a uniformly distributed load of 200-tons, together with a single load of 10-tons concentrated at 25-ft. from one end. (27). — In a Post truss for a 200-ft. span there are IH-panels, the posts incline towards the centre, and have a run of half a bay, the ties cross two panels, and are inclined at 45" to the vertical, the counter-ties cross one panel, and have the same inclination, the panel weight of engine = 17,600-lbs., of tender = 16,16(1 lbs., end of cars = 12,500-lbs.; the maxi- mum stress in the seventh tie from one end = 56,000-lbs.; determine the panel weight of truss, and specify the bars which require V, be counter-braced. (28). — A weight is placed at the centre of a truss of which the web consists of a single system of struts and ties; shew how to de.Vrmine the inclination of the web members to the vertical, so that the amount of material in them may be a minimum. (20). — Design a bowstring girder, (^^;)-with vertical posts and diagon- al bracing, (6)-with isosceles bracing, from the following data: — The span is 80-ft., the depth at the centre is 10 ft., the dead load per ft. run is ^-ton, the live load per ft. run is 1-ton. (80). — The bowstring girder repre- sented by the Fig. is designed to sup- port a load of 840-lbs. per ft. run ; its span is 100-ft., its central depth 14-ft., its depth at each end 5-ft., and its upper chord is a circular arc ; determine the stresses in all the members. (31). — The span of a suspension bridge is 200-ft.. the dip of the chains is 80-ft., and the weight of the roadway is 1-ton per ft. run ; find the tensions at the middle and ends of each chain. 144 EXAMI'LES. (Ii2). — Assuinin). — The dimensions and weight of the cables, (1) — if of uniform section throughout, (2) — if each section is proportioned to the pull across it. (c). — The alteration in the length of the cables and the corresponding depression of the platform at the centre of each span, due (1) — to a change of G0° F. from the mean temperature, (2) — to the proof load. (rf). — The crushing and bending efforts at the f«ot of a pier. (<'). — The bearing area at the top of the abutments, and the mass of masonry necessary to resist the tendency to overturning and to hori- zontal displacement. {Weight of masonry per cubic ft. — Vl'A lbs., Its safe comjjressive strength per sq.ft.~2ii^-lbs., its cuejficient uf friction = .76). EXAMPLES. 145 the as of lori- Its •tiun (37). — Solve the preceding question when the cahlcs for the several spans are independent, and have tlieir ends securely attached to the points of support. The piers are wrouirht-iron oscill'.lng columns, and in order to maintain the equilibrium of the structure under an unequally distributed loads the heads of the columns are connected with each other and with the abutments by iron wire stays ; determine the proper dimensions of the stays, assuming then to be approximately straiglit. (38). — The river span of a suspension bridge is 930-ft., and weigh.s 5976-tons, of which 1430-tons are borne by stays radiating from the summit of each pier, while the remaining weight is distributed between /our 15-in. steel wire cables producing in each at the piers a tension of 20G4-tons ; find the dip of the cables. The estimated maximum traffic upon the river span is 1311-tons, uniformly distributed ; determine the increased stress in the cables. To what extent might the traffic be safely increased, the limit of elasticity of a cable being 8116-tons, and its breaking stress 1^3,300- tons? (39).— A semi-circular rib, pivoted at the crown and springings, is loaded uniformly per horizontal unit of length ; determine the position and magnitude of the maximum bending moments, and shew that the horizontal thrust on the rib is one-fourth of the total load. (40). — Draw the equilibrium polygon for an arch of 100-ft. span and 20 ft. rise, when loaded with weights of 3, 2, 4, and 2 tons respectively, at the end of the 3rd, 6th, 8th, and 9th division from the left support, often e(|ual horizontal divisions. (Xeijlect the weight of the rib). If the rib consist of a web and of two flanges 2^ ft. from centre to centre, determine the maximum flange stress. (41), — A parabolic rib of span / and rise /.•, having its ends A and B fixed, supports a given load at a liorizontal distance x from the middle of the span ; the ecjuilibrium polygon consists of two straight lines CD CE, meeting the verticals through .1 and li in D an E respectively ; if ^•IZ>=//,, BE=ij.i, and if rj is the vertical distance of (J from AB shew that, 2 /+10..V, 2 /-10.;i- ^^ 6 , (42). — A parabolic double-flangod rib 2i-ft, deep, of 100-ft. span and 20-ft. rise, is fixed at both ends and loaded in the same maimer as the rib in Question 40 ; draw the equilibrium polygon, and det; — when hinged at the ends and centre, (c) — when fixed at ihe ends. (4(3). — Det^^-rmine the horizontal thru>t on a rib in tlie form of the scirment of a circle due to change of t" from the mean temperature. (47). — In CliMpfi-r X, determine the changes to be made in the analysis, (a) — of JJ i;}.14, 15, when the e(|uations deduced in Corollary 4, §(13), are employed instead of equations Id, 11 and 12. (fj). — Of ^ 13 and 14, when the abutments yield to the thrust ^o as to t'ohirge the sjian by an amount //.-//., // being a co-efficient to be determined by experiment. (r). — Of ^5 14 and 15, when the effect of a change of temperature is to be taken into account. (48). — Illustrate J^ 13, 14, 15, by the example of a rib in which h = ~=i\S.z, and q =■-;• h u (49). — A wrought-irou parabolic rib, of 90-ft. span and 10-ft. rise, is lunged at the two abutments ; it is of a double y-section, uniform throughout, and 24-ins. deep from centre to centre of the flanges; deter- mine the compression at the centre, and also the position and amount of the maximum bending moment, {<() — when a load of 48-tons is concen- trated at the centre, (/>). — when a load of 9G-tons is uniformly distributed per horizontal unit of length. IX'termine the deflection of the rib in each case. (50). — In the framed arch repre- sented by the Fig., the span is 120- ft.. the ri.ight of the bridge be equally distributed over the turntable. Shew how to find the stresses in the different members of the truss. (54).— The truss for a counterbalarced swing- bridge is of the I'orm and dimensions shewn by the Fig., the counterpoise consisting of concrete hung in a box from the points c and D ; the dead weight is 050-lbs. per lineal ft. of the bridge ; determine the stresses in the different mem- 148 EXAMI'LES, bern, ( load 280()-lbs.; tlio uhsymod wiM^dit of the pior is IHOO-llw. ])er ft. in heij:;ht; the cyliiidiT of tiw. locomotive is iniiis. in diani., tlm pri'ssurt; on the piston is l;{()-lbs. per sc^ in.: — (