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TIC

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Originally appearing in Volume V04, Page 557 of the 1911 Encyclopedia Britannica.
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TIC. T.J,C. TIC. T.1•C. 'T •1 TIC. 16-13 18-0 18 -0 12-17 I1-10 14- 10 14-18 Express Passenger Engine, G.N. Ry. 8=6- 5 1,f" 5=2#" 5 =2#12=63 ' 5=3" 5 =f0#~ _ _ _ . 8' 11over Buffers. _ } 12.13 14-08 14.14 12.45 10.11•.4.6 10.4.6 10•-•46 9.65 Goods Engine, L. & Y. Ry. _7 _ 56_ 6=9t; " ~I~ I 10=6"_ _ _ 7=_6 _ 7_=6" _ _ 14=5i ~_ _IO--O_'- 7 4# _ _ 65_6 oer _ Buffers - _ T- ' T T T T TJ 16.64- 12.25 16.0 17-18 27.5 27.5 Passenger Engine, Cal. Ry. In Austria the official regulations require that railway bridges shall be designed for at least the following live loads per foot run and per track: Span. Live Load in Tons. Metres. Ft. Per metre run. Per ft. run. t 3.3 20 6.1 2 6.6 15 4.6 5 16.4 0 3•t 20 65.6 5 I.5 30 98.4 4 1.2 It would be simpler and more convenient in designing short bridges if, instead of assuming an equivalent uniform rolling load, agreement could be come to as to a typical heavy locomotive which would produce stresses as great as any existing locomotive on each class of railway. Bridges would then be designed for these selected loads, and the process,would be safer in dealing with flooring girders and shearing forces than the assumption of a uniform load. Some American locomotives are very heavy. Thus a consolidation engine may weigh 126 tons with a length over buffers of 57 ft., corresponding to an average load of 2.55 tons per ft. run. Also long ore wagons are used which weigh loaded two tons per ft. run. J.A.L. Waddell (De Pontibus, New York, 1898) proposes to arrange railways in seven classes, according to the live loads which may be expected from the character of their traffic, and to construct bridges in accordance with this classification. For the lightest class, he takes a locomotive and tender of 93.5 tons, 52 ft. between buffers (average load 1.8 tons per ft. run), and for the heaviest a locomotive and tender weighing 144.5 tons, 52 ft. between buffers (average load 2.77 tons per ft. run). Wagons he assumes to weigh for the lightest class 1.3 tons per ft. run and for the heaviest 1.9 tons. He takes as the live load for a bridge two such engines, followed by a train of wagons covering the span. Waddell's tons are short tons of 2000 lb. ii. Impact.— If a vertical load is imposed suddenly, but without velocity, work is done during deflection, and the deformation and stress are, momentarily double those due to the same load at rest on the structure. No load of exactly this kind is ever applied to a bridge. But if a load is so applied that the deflection increases with speed, the stress is greater than that due to a very gradually applied load, and vibrations about a mean position are set up. The rails not being absolutely straight and smooth, centrifugal and lurching actions occur which alter the distribution of the loading. Again, rapidly changing forces, due to the moving parts of the engine which are unbalanced vertically, act on the bridge; and, lastly, inequalities of level at the rail ends give rise to shocks. For all these reasons the stresses due to the live load are greater than those due to the same load resting quietly on the bridge. This increment is larger on the flooring girders than on the main ones, and on short main girders than on long ones. The impact stresses depend so much on local conditions that it is difficult to fix what allowance should be made. E. H. stone '(Trans. Am. Soc. Of C. E. xli. p. 467) collated some measurements of deflection taken during official trials of Indian bridges, and found the increment of deflection due to impact to depend on the ratio of dead to live load. By plotting and averaging he obtained the following results: Excess of Deflection and straining Action of a moving Load over that due to a resting Load. Dead load in per cent 10 20 30 40 50 70 90 of total load. . . Live load in per cent 90 80 70 (0 50 30 10 of total load. . ` . 2.3 1.5 1.0 0.43 0.10 Ratio of . live to dead 9 4 load Excess of deflection and 23 13 8 5.5 4.0 t •6 0 3 stress due to moving load per cent .. These results are for the centre deflections of main girders, but Stolle infers `that the augmentation of stress for any member, due to causes included in impact allowance, will be the same percentage for the same ratios of live to dead load stresses. Valuable measurements of the deformations of girders and tension members due tamoving.trains have been made by S. W. Robinson (Trans. Am. Soc. C. E. xvi.) and by F. E. Turneaure (Trans. Am. Soc. C. E. xli.). The latter used a recording,deflectometer and two recording extensometers. The observations are difficult, and the inertia of the instrument is liable to cause error, but much care was taken. The most striking conclusions from the results are that the locomotive balance weights have a large effect in causing vibration, and next, that in certain cases the vibrations are cumulative, reaching a value greater than that due to any single impact action. Generally: (I) At speeds less than 25 M. an hour there is not much vibration. (2) The increase of deflection due to impact at 40 or 50 M. an hour is likely to reach40 to 50% for girder spans of less than 50 ft. (3) This pereeutage decreases rapidly for longer spans, becoming about 25 % for 75-ft. spans. (4) The increase per cent of boom stresses due to impact is about the same as that of deflection; that in web bracing bars is rather greater. (5) Speed of train produces no effect on the mean deflection, but only on the magnitude of the vibrations. A purely empirical allowance for impact stresses has been proposed, amounting to 20% of the live load stresses for floor stringers; 15% for floor cross girders; and for main girders, 1o% for 4o-ft. spans, and 5% for too-ft. spans. These percentages are added to the live load stresses. iii. Dead Load.—The dead load consists of the weight of niain girders, flooring and wind-bracing. It is generally reckoned to be uniformly distributed, but in large spans the distribution of weight in the main girders should be calculated and taken into account. The weight of the bridge flooring depends on the type adopted. Road bridges vary so much in the character of the flooring that no general rule can be given. In railway bridges the weight of sleepers, rails, &c., is o•2 to 0.25 tons per ft. run for each line of way, while the rail girders, cross girders, &c., weigh 0.15 to 0.2 tons. If a foot-way is added about 0.4 ton per ft. run may be allowed for this. The weight of main girders increases with the span, and there is for any type of bridge a limiting span beyond which the dead load stresses exceed the assigned limit of working stress. Let W1 be the total live load, Wf the total flooring load on a bridge of span 1, both being considered for the present purpose to be uniform per ft. run. Let k(Wi+Wf) be the weight of main girders designed to carry Wi-I-4Wf, but not their own weight in addition. Then Wo = (W1+Wf) (k+k2+ka ... ) will be the weight of main girders to carry WI+Wf and their own weight (Buck, Proc. Inst. C. E. lxvii. p. 331). Hence, Wo=(W1+Wf)kl(1-k) Since in designing a bridge W1+Wf is known, k(W1+Wf) can be found from a provisional design in which the weight W2 is neglected. The actual bridge must have the section of all members greater than those in the provisional design in the ratio kl (1-k). Waddell (De Pontibus) gives the following convenient empirical relations. Let wi, w2 be the weights of main girders per ft. run for a live load p per ft. run and spans 11, l2. Then w2/wl =1 [12/li + (12/11)2] Now let wi', w2' be the girder weights per ft. run for spans ii, l2, and live loads p' per ft. run. Then w2'/w2 = G (1 +4p'/p) w2'/1 = Is [12/11+ (12/li)2] (I +4p'/p) • A partially rational approximate formula for the weight of main girders is the following (Unwin, Wrought Iron Bridgss and Roofs, 1869, P. 40) :-- Let w=total live load per ft. run of girder; w2 the weight of platform per ft. run; w3 the weight of, main girders per ft. run, all in tons; 1=span in ft.; s= average stress in tons per sq. in. on gross section of metal; d =depth of girder at centre in ft.; r =ratio of span to depth of girder so that r=l/d. Then wa= (w,+w2)l2/(Cds—12) = (wi+w2)lr/(Cs—lr) , where C is a constant for any type of girder. It is not easy to fix the average stress s per sq. in. of gross section. Hence the formula is more useful in the form w = (wi+w2)l2/(Kd—12) = (w,+w2)lr/(K—lr) where k = (wi+w2+wa)lr/w3 is to be deduced from the data of some bridge previously designed with the same working stresses. From some known examples, C varies from 1500 to 1800 for iron braced parallel or bowstring girders, and from 1200 to 1500 for similar girders of steel. K=6000 to 7200 for iron and =7200 to 9000 for steel bridges. iv. Wind Pressure.—Much attention has been given to wind action since the disaster to the Tay bridge in 1879. As to the maximum wind pressure on small plates normal to the wind, there is not much doubt. Anemometer observations show that pressures of 3o lb per sq. ft. occur in storms annually in many localities, and that occasionally higher pressures are recorded in exposed positions. Thus at Bidstone, Liverpool, where the gauge has an exceptional exposure, a pressure of 8o lb per sq. ft. has been observed. In tornadoes, such as that at St Louis in 1896, it has been calculated, from the stability of structures overturned, that pressures of 45 to 90 lb per sq. ft. must have been reached. As to anemometer pressures, it should be observed that the recorded pressure is made up of a positive front and negative (vacuum) back pressure, but in structures the latter must be absent or only partially developed. Great difference of opinion exists as to whether on large surfaces the average pressure per sq. ft. is as great as on small,surfaces, such as anemometer plates. The experiments of Sir B. Baker at the Forth bridge showed that on a surface 30 ft. X 15 ft. the intensity of pressure was less than on a similarly exposed anemometer plate. In the case of bridges there is the further difficulty that some surfaces partially shield other surfaces; one girder, for instance, shields the girder behind it (see Brit. Assoc. Report, 1884). In 1881 a committee of the Board of Trade decided that the maximum wind pressure on a vertical surface in Great Britain should be assumed in designing structures to be 56 lb per sq. ft. For a plate girder bridge of less height than the train, the wind is to be taken to act on a surface equal to the projected area of one girder and the exposed part of a train covering the bridge. In the case of braced girder bridges, the wind pressure is taken as acting on a continuous surface extending from the rails to the top of the carriages, plus the vertical projected area of so much of one girder as is exposed above the train or below the rails. In addition, an allowance is made for pressure on the lee-ward gilder according to a scale. The committee recommended that a factor of safety of 4 should be taken for wind stresses. For safety against overturning they considered a factor of 2 sufficient. In the case of bridges not subject to Board of Trade inspection, the allowance for wind pressure varies in different cases. C. Shaler Smith allows 300 lb per ft. run for the pressure on the side of a train, and in addition 3o lb per sq. ft. on twice the vertical projected area of one girder, treating the pressure on the train as a travelling load. In the case of bridges of less than 5o ft. span he also provides strength to resist a pressure of 5o Ib per sq. ft. on twice the vertical projection of one truss, no train being supposed to be on the bridge. 19. Stresses Permitted.-For a long time engineers held the convenient opinion that, if the total dead and live load stress on any section of a structure (of iron) did not exceed 5 tons per sq. in., ample safety was secured. It is no longer possible to design by so simple a rule. In an interesting address to. the British Association in 1885, Sir B. Baker described the condition of opinion as to the safe limits of stress as chaotic: " The old foundations," he said, " are shaken, and engineers have not come to an agreement respecting the rebuilding of the structure. The variance in the strength of existing bridges is such as to be apparent to the educated eye without any calculation. In the present day engineers are in accord as to the principles of estimating the magnitude of the stresses on the members of a structure, but not so in proportioning the members to resist those stresses. The practical result is that a bridge which would be passed by the English Board of Trade would require to be strengthened 5% in some parts and 6o % in others, before it would be accepted by the German government, or by any of the leading rail-way companies in America." Sir B. Baker then described the results of experiments on repetition of stress, and added that " hundreds of existing bridges which carry twenty trains a day with perfect safety would break down quickly under twenty trains an hour. This fact was forced on my attention nearly twenty-five years ago by the fracture of a -number of girders of ordinary strength under a five-minutes' train service." It was pointed out as early as 1869 (Unwin, Wrought Iron Bridges and Roofs) that a rational method of fixing the working stress, so far as knowledge went at that time, would be to make it depend on the ratio of live to dead load, and in such a way that the factor of safety for the live load stresses was double that for the dead load stresses. Let A be the dead load and B the live load, producing stress in a bar; p=B/A the ratio of live to dead load; fi the safe working limit of stress for a bar subjected to a dead load only and f the safe working stress in any other case. Then fi(A+B)/(A+2B) =.fi(I+p)/(1+2p). The following table gives values of f so computed on the assumption that fi=7z tons per sq. in. for iron and 9 tons per sq. in. for steel. Working Stress for combined Dead and Live Load. Factor of Safety twice as great for Live Load as for Dead Load. Ratio 1 -{-p Values of f, tons per sq. in. P 1+2a Iron. Mild Steel. All dead load 0 1.00 7.5 9.0 •25 0.83 6.2 7.5 '33 0.78 5.8 7'0 .50 0'75 5.6 6.8 .66 011 5.3 5.9 Live load = Dead load 1 . o0 0.66 4'9 5.9 2•0o 0.60 4'5 5'4 4 00 0 56 4'2 50 All live load o•5o 3.7 4'5 Bridge sections designed by this rule differ little from those designed by formulae based directly on Wohler's experiments. This rule has been revived in America, and appears to be increasingly relied on in bridge-designing. (See Trans. Am. Soc. C.E. xli. p. 156.) The method of J. J. Weyrauch and W. Launhardt, based on an empirical expression for WOhler's law, has been much used in bridge designing (see Proc. Inst. C.E. lxiii. p. 275). Let t be the statical breaking strength of a bar, loaded once gradually up to fracture (t = breaking load divided by original area of section) ; u the breaking strength of a bar loaded and unloaded an indefinitely great number of times, the stress varying from u to o alternately (this is termed the primitive strength) ; and,. lastly, let s be the breaking strength of a bar subjected to an indefinitely great number of repetitions of stresses equal and opposite in sign (tension and thrust), se that the stress ranges alternately from s to -s. This is termed the vibration strength. Wohler's and Bauschinger's experiments give values of t, u, and s, for some materials. If a bar is subjected to alternations of stress having the range A=fads.-f,aie., then, by Wchler's law, the bar will ultimately break, if fma:. -FA, (I) where F is some unknown function. Launhardt found that, for stresses always of the same kind, F = (t-u)/(t-f,,,, .) approximately. agreed with experiment. For stresses of different kinds Weyrauch found F }_}•• (u-s)/)(2u-s-f,,,as.) to be similarly approximate. Now let f,aas-/fai,,.=0, where 4 is+or-according as the stresses are of the same or opposite signs. Putting the values of F in (1) and solving for f,,,,, ., we get for the breaking stress of a bar subjected to repetition of varying stress, fmax• = u(1 + (t -u)¢/u) [Stresses of same sign.] fads.=u(1+(u-s)(/u) [Stresses of opposite sign.] The working stress in any case is fa... divided by a factor of safety. Let that factor be 3. Then Wohler's results for iron and Bauschinger's for steel give the following equations for tension or thrust:- Iron, working stress, f =4.4 ('I +}0) Steel, ,, =5.87 (1+10). In these equations 4' is to have its + or - value according to the case considered. For shearing stresses the working stress may have o•8 of its value for tension. The following table gives values of the working stress calculated by these equations:- Working Stress for Tension or Thrust by Launhardt and Weyrauch Formula. . ¢ Working Stress f, 1 +-- tons per sq. in. 2 Iron. Steel. All dead load . . . . 1 •o 1 •g 6.6o 8.8o 0.75 1.375 6.05 8.07 0'50 1.25 5.50 7.34 All live load . . . . o•0 1.005 44'44o 6-60 5'87 -0.25 0.875 3.85 5.14 -0.50 0'75 3.3o 4.40 -0.75 0.625 2.75. 3.67 Equal stresses + and - . -1.00 0.500 2.2o 2.93 Practical experience taught engineers that though 5 tons per sq. in. for iron, or 62 tons per sq. in. for steel, was safe or more than safe for long bridges with large ratio of dead to live load, it was not safe for short ones in which the stresses are mainly due to live Load, the weight of the bridge being small. The experiments of A. Wohler, repeated by Johann Bauschinger, Sir B. Baker and others, show that the breaking stress of a bar is not a fixed quantity, but depends on the range of variation of stress to which it is subected, if that variation is repeated a very large number of times. Let K be the breaking strength of a bar per unit of section, when it is loaded once gradually to breaking. This may be termed the statical breaking strength. Let kma:. be the breaking strength of the same bar when subjected to stresses varying from k,, . to kmie. alternately and repeated an indefinitely great number of times; k,,,;a. is to be reckoned + if of the same kind as k,aas. and - if of the opposite kind (tension or thrust). The range of stress is there-fore k,,,as.-kale., if the stresses are both of the same kind, and kaas.+k„ie., if they are of opposite kinds. Let o = k,,,. .+=the range of stress, where 0 is always positive. Then Miller's results agree closely with the rule, k,, . = 1A+ d (K2-nzK), where n is a constant which varies from 1.3 to 2 in various qualities of iron and steel. For ductile iron or mild steel it may be taken as 1.5. For a statical load, range of stress nil, A =o, k,,,, . = K, the statical breaking stress. For a bar so placed that it is alternately loaded and the load removed, 0=/sass. and kma~. =o•6 K. For a bar subjected to alternate tension and compression of equal amount, 0=2 fees. and /sass. = 0.33 K. The safe working stress in these different cases is /sass. divided by the factor of safety. It is sometimes said that a bar is " fatigued " by repeated straining. The real nature of the action is not well understood, but the word fatigue may be used, if it is not considered to imply more than that the breaking stress under repetition of loading diminishes as the range of variation increases. To compare this with the previous table, (p =(A+B)/A=i+p. Except when the limiting stresses are of opposite sign, the two tables agree very well. In bridge work this occurs only in some of the bracing bars. It is a matter of discussion whether, if fatigue is allowed for by the Weyrauch method, an additional allowance should be made for impact. There was no impact in Wohler's experiments, and there-fore it would seem rational to add the impact allowance to that for fatigue; but in that case the bridge sections become larger than •:xperience shows to be necessary. Some engineers escape this difficulty by asserting that Wohler's results are not applicable to bridge work. They reject the allowance for fatigue (that is, the effect of repetition) and design bridge members for the total dead and live load, plus a large allowance for impact varied according to some purely empirical rule. (See Waddell, De Pontibus, p. 7.) Now in applying Wohler's law, fma~. for any bridge member is found for the maximum possible live load, a live load which though it may sometimes come on the bridge and must therefore be provided for, is not the usual live load to which the bridge is subjected. Hence the range of stress, f,,,az.—f,,,;,,., from which the working stress is deduced, is not the ordinary range of stress which is repeated a practically infinite number of times, but is a range of stress to which the bridge is subjected only at comparatively long intervals. Hence practically it appears probable that the allowance for fatigue made in either of the tables above is sufficient to cover the ordinary effects of impact also. English bridge-builders are somewhat hampered in adopting rational limits of working stress by the rules of the Board of Trade. ,Nor do they all accept the guidance of Wohler's law. The following are some examples of limits adopted. For the Dufferin bridge (steel) the working stress was taken at 6.5 tons per sq. in. in bottom booms and diagonals, 6•o tons in top booms, 5.o tons in verticals and long compression members. For the Stanley bridge at Brisbane the limits were 6.5 tons per sq. in. in compression boom, 7.o tons in tension boom, 5•o tons in vertical struts, 6.5 tons in diagonal ties, 8•o tons in wind bracing, and 6.5 tons in cross and rail girders. In the new Tay bridge the limit of stress is generally 5 tons per sq. in., but in members in which the stress changes sign 4 tons per sq. in. In the Forth bridge for members in which the stress varied from o to a maximum frequently, the limit was 5•o tons per sq. in., or if the stress varied rarely 5.6 tons per sq. in.; for members subjected to alternations of tension and thrust frequently 3.3 tons per sq. in. or 5 tons per sq. in. if the alternations were infrequent. The shearing area of rivets in tension members was made iZ times the useful section of plate in tension. For compression members the shearing area of rivets in butt-joints was made half the useful section of plate in compression. 20. Determination of Stresses in the Members of Bridges.—It is convenient to consider beam girder or truss bridges, and it is the stresses in the main girders which primarily require to be determined. A main girder consists of an upper and lower flange, boom or chord and a vertical web. The loading forces to be considered are vertical, the horizontal forces due to wind pressure are treated separately and provided for by a horizontal system of bracing. For practical purposes it is accurate enough to consider the booms or chords as carrying exclusively the horizontal tension and compression and the web as resisting the whole of the vertical and, in a plate web, the equal horizontal shearing forces. Let fig. 37 represent a beam with any system of loads Wi, W2, . . . W,.. R, t W,, Wz ° W, WM.. W, R= ~~ n C? x R2 = W1x1/l+W2x2/t-t- .. . That at the left abutment is R1= W1+W2+ ...-R2. Consider any section a b. The total shear at a b is S=R—E(W1+W2...) where the summation extends to all the loads to the left of the section. Let pi, p2 . . be the distances of the loads from a b, and p the distance of R1 from a b; then the bending moment at a b is M = Rlp—2(W1p1+W2p2 ...) where-the summation extends to all the loads to the left of a b. If the loads on the right of the section are considered the expressions are similar and give the same results. If Al Ac are the cross sections of the tension and compression flanges or chords, and h the distance between their mass centres, then on the assumption that they resist all the direct horizontal forces the total stress on each flange is H,=He=M/h and the intensity of stress of tension or compression is fa=M/A1h, f , = M/A,h. If A is the area of the plate web in a vertical section, the intensity of shearing stress is f.= S/A and the intensity on horizontal sections is the same. If the web is a braced web, then the vertical component of the stress in the web bars cut by the section must be equal to S. 21. Method of Sections. A. Ritter's Method.—In the case of braced structures the following method is convenient: When a section of a girder can be taken cutting only three bars, the stresses in the bars can be found by taking moments. In fig. 38 m n cuts three bars, and the forces in the three bars cut by the section are C, S and T. There are to the left of the section the external forces, R, W1, W2. / .\ / Y Let s be the perpendicular from 0, the join of C and T on the direction of S; t the perpendicular from A, the join of C and S on the direction of T; and c the perpendicular from B, the join of S and T on the direction of C. Taking moments about 0, Rx—Wl(x+a) —W2(x+2a) =Ss; taking moments about A, R3a—W12a—W2a = Tt; and taking moments about B, R2a—Wla = Cc. Or generally, if M1 M2 Ma are the moments of the external forces to the left of 0, A, and B respectively, and s, t and c the perpendiculars from 0, A and B on the directions of the forces cut by the section, then Ss=M1; Tt=M2 and Cc=Ma. Still more generally if H is the stress on any bar, h the perpendicular distance from the join of the other two bars cut by the section, and M is the moment of the forces on one side of that join, Hh=M. 22. Distribution of Bending Moment and Shearing Force.—Let a girder of span 1, fig. 39, supported at the ends, carry a fixed load. W at m from the right abutment. The reactions at the abutments are R1=Wm/l and R2=W(l—m)/l. The shears on vertical sections I•-m - . Shear. Ri. .T. Bending Moment. to the left and right of the load are RI and —R2, and the distribution of shearing force is given by two rectangles. Bending moment increases uniformly from either abutment to the load, at which the bending moment is M = Rem = Rt(l —m). The distribution of bending moment is given by the ordinates of a triangle. Next let the girder carry a uniform load w per ft. run (fig. 40). The total load n tRl ' / / / / % % % / % /O///% • I Shear . -Re N. Ri 1 Bending Moment. • IRz -r A = C 9 Old( c x. R2=w(c+x) Xc cx=w (c+x)2, 1•-•...1 Zc t --•----,I 4 4 i 1 which is also the shearing force at C 1 for that position of the load. As the load travels, the shear at the head of the train will be given by the ordinates of a parabola having its vertex at A, and a maximum F,,,as. = — Zwl at B. If the load of p. The web of a girder must resist the maximum shear, and, with a travelling load like a railway train, this is greater for partial than for complete loading. Generally a girder supports both a dead and a live load. The distribution of total shear, due to a dead load wi per ft. run and a travelling load wi per ft. run, is shown in fig. 42, arranged so that the dead load shear is added to the maximum travelling load shear of the same sign. 24. Counterbracing.—In the case of girders with braced webs, the tension bars of which are not adapted to resist a thrust, another circumstance due to the position of the live load must be considered. For a train advancing from the left, the travelling load shear in the left half of the span is of a different sign from that due to the dead load. Fig. 43 shows the maximum shear at vertical sections due to a dead and travelling load, the latter advancing (fig. 43, a) from the left and (fig. 43, b) from the right abutment. Comparing the figures it will be seen that over a distance x near the middle of the girder the shear changes sign, according as the load advances from the left or the right. The bracing bars, therefore, for this part of the girder must be adapted to resist either tension or thrust. Further, the range of stress to which they are subjected is the sum of the stresses due to the load advancing from the left or the right. 25. Greatest Shear when concentrated Loads travel over the Bridge.— To find the greatest shear with a set of concentrated loads at fixed distances, let the loads advance from the left abutment, and let C be the section at which the shear is required (fig. 44). The greatest shear at C may occur with Wi at C. If W1 passes beyond C, the shear at C will probably be greatest when W2 is at C. Let R be the resultant of the loads on the bridge when W; is at C. Then the reaction at B and shear at C is Rn/i. Next let the loads advance a distance a so that W2 comes to C. Then the shear at C is R(n+a)/l—W,, plus any reaction d at B, due to any additional load which has come on the girder during the movement. The shear will therefore be increased by bringing W2 to C, if Ra/l+d >W, and d is generally small and negligible. This result is modified if the action of the load near the section is distributed to the bracing intersections by rail and cross girders. In fig. 45 the action of W is distributed to A and B by the flooring. Then the loads at A and B are W(p—x)/p and Wx/p. Now let C (fig. 46) be the section at which the greatest shear is required, and let the loads advance from the left till Wi is at C. If R is the resultant of the loads then on the girder, the reaction at B and shear FIG. 45• at C is Rn/l. But the shear may be greater when W2 is at C. In that case the shear at C becomes R(n+a)/l+d—Wi, if a>p, and R(n+a)/l+d—W,a/p, if a WI in the first case, and if Rail> Wla/p in the second case. 26. Greatest Bending Moment due to travelling concentrated Loads.—For the greatest bending moment due to a travelling live load, let a load of w per ft. run advance from the left abutment (fig. 47), and let its centre be at x from the left abutment. The reaction at B is 2wx2/l and the bending moment at any section C, at m from the left abutment, is2wx2/(l—m)/l, which increases as x increases till the span is covered. Hence, for uniform travelling loads, the bending moments 4 >r i k 1 are greatest when the loading is complete. In that case the loads on either side of C are proportional to m and l—m. In the case of a series of travelling loads at fixed distances apart passing over the girder from the left, let W,, W2 (fig. 48), at distances x and x+a from the left abutment, be their resultants on either side of C. Then the reaction at B is Wix/l+W2(x+a)/l. The bending moment at C is M =W,x(l-m)/l+W2m {I — (x+a)/l f . If the loads are moved a iistance Ax to the right, the bending moment becomes M +AM = Wi (x+Ax) (1—m) /1+W2m { I — (x+Ax+a)/l{ Am=W1z x(l—m)/l-W20xm/l, and this is positive or Wi(l—m) >W2m, or if W,/m>W2/(l—m). But these are the average loads per ft. run to the left and right of C. Hence, if the average load to the left of a section is greater than that to the right, the bending moment at the section will be increased by moving the loads to the right, and vice versa. Hence the maximum bending moment at C for a series of travelling loads will occur when the average load is the same on either side of C. If one of the loads is at C, spread over a very small distance in the neighbourhood of C, then a very small displacement of the loads will permit the fulfilment of the condition. Hence the criterion for the position of the loads which makes the moment at C greatest is this: one load must be at C, and the other loads must be distributed, so that the average loads per ft. on either side of C (the load at C being neglected) are nearly equal. If the loads are very unequal in magnitude or distance this condition may be satisfied for more than one position of the loads, but it is not difficult to ascertain which position gives the maxi-mum moment. Generally one of the largest of the loads must be at C with as many others to right and left as is consistent with that condition. This criterion may be stated in another way. FIG. 48. The greatest bending moment will occur with one of the greatest loads at the section, and when this further condition is satisfied. Let fig. 49 represent a beam with the series of loads travelling from the right. Let a b be is wl; the reactions at abutments, R, = R2 = iwl. The distribution of shear on vertical sections is given by the ordinates of a sloping line. The greatest bending moment is at the centre and = Ms = ewl2. At any point x from the abutment, the bending moment is M = iwx(i—x), an equation to a parabola. 23. Shear due to Travelling Loads.—Let a uniform train weighing w per ft. run advance over a girder of span 2C, from the left abutment. When it covers the girder to a distance x from the centre (fig. 41) the total load is w(c+x) ; the reaction at B is /~////r//"'. R C Wi l Ws Ws B 1 - -•- m -- Im the bending moment increases, if D AXIOOCiliCK)OO b- ---• 1-2 e•--°•• 4+ s,. W, W2 A( ).0 CQi.QB ' i 1 yl..b--,i k m 1-m 4 1-a c the section considered, and let W. be the load at a b when the bending moment there is greatest, and W,, the last load to the right then on the bridge. Then the position of the loads must be that which satisfies the condition l greater than W1+W2+ ... WA-' x less than WI+W2+ ..• . l Wz Wi+W2+ Fig. 50 shows the curve of bending moment under one of a series of travelling loads at fixed distances. Let WI, W2, W3 traverse the girder from the left at fixed distances a, b. For the position shown the distribution of bending moment due to WI is given by ordinates of the triangle A'CB'; that due to W2 by ordin- ates of A'DB'; and that due to Ws by ordinates of A EB'. The a< x ` ~ total moment at -'-_... b L ---• WI, due to three loads, is the sum the intercepts which the triangle sides cut off from the vertical under WI. As the loads move over the girder, the points C, D, E describe the parabolas MI, M2, M2, the middle ordinates of which are ;WII, 4W21, and ;W3l. If these are first drawn it is easy, for any position of the loads, to draw the lines B'C, B'D, B'E, and to find the sum of the intercepts which is the total bending moment under a load. The lower portion of the figure is the curve of bending moments under the leading load. Till WI has advanced a distance a only one load is on the girder, and the curve A"F gives bending moments due to WI only; as WI advances to a distance a+b, two loads are on the girder, and the curve FG gives moments due to W1 and W2. GB" is the curve of moments for all three loads WI+W2+W2. Fig. 51 shows maximum bending moment curves for an extreme case of a short bridge with very unequal loads. The three lightly dotted parabolas are the curves of maximum moment for each of the loads taken separately. The three heavily dotted curves are curves of maximum moment under each of the loads, for the three loads passing over the bridge, at the given distances, from left to right. As might be expected, the moments are greatest in this case at the sections under the 15-ton load. The heavy continuous line gives the last-mentioned curve for the reverse direction of passage of the loads. With short bridges it is best to draw the curve of maximum bending moments for some assumed typical set of loads in the way Lust described, and to design the girder accordingly. For longer bridges the funicular polygon affords a method of determining maximum bending moments which is perhaps more convenient. But very great accuracy in drawing this curve is unnecessary, because the rolling stock of railways varies so much that the precise magnitude and distribution of the loads which will pass over a bridge cannot be known. All that can be done is to assume a set of loads likely to produce somewhat severer straining than any probable actual rolling loads. Now, except for very short bridges and very unequal loads, a parabola can be found which includes the curve of maximum moments, This parabola is the curve of maximum moments for a travelling load uniform per ft. run. Let we be the load per ft. run which would produce the maximum momentsrepresented by this parabola. Then we may be termed the uniform load per ft. equivalent to any assumed set of concentrated loads. Waddell has calculated tables of such equivalent uniform loads. But it is not difficult to find we, approximately enough for practical purposes, very simply. Experience shows that (a) a parabola having the same ordinate at the centre of the span, or (b) a parabola having 16 one the same ordinate at one-quarter span as the curve of maximum moments, agrees with it closely enough for practical designing. A criterion already given shows the position of any set of loads which will produce the greatest bending moment at the centre of the bridge, or at one-quarter span. Let Me and M. be those moments. At a section distant x from the centre of a girder of span 2c, the bending moment due to a uniform load we per ft run is M = awe(c-x) (c+x). Putting x =o, for the centre section Me = I --2'wee2; and putting x=ac, for section at quarter span Me=fwec2. From these equations a value of we can be obtained. Then the bridge is designed, so far as the direct stresses are concerned, for bending moments due to a uniform dead load and the uniform equivalent load we. 27. Influence Lines.—In dealing with the action of travelling loads much assistance maybe obtained by using a line termed an influence line. Such a line has for abscissa the distance of a load from one end of a girder, and for ordinate the bending moment or shear at any given section, or on any member, due to that load. Generally the influence line is drawn for unit load. In fig. 52 let A'B' be a girder supported at the ends and let it be required to investigate the bending moment at C' due to unit load in any position on the girder. When the load is at F', the reaction at B' is m/l and the moment at C' is m(l-x)/l, which will be reckoned positive, when it resists a tendency of the right-hand part of the girder to turn counter-clockwise. Projecting A'F'C'B' on to the horizontal AB, take Ff =m(l-x)ll, the moment at C of unit load at F. If this process is repeated for all positions of the load, we get the influence line AGB for the bending moment at C. The area AGB is termed the influence area. The greatest moment CG at C is x(1-x)/l. To use this line to investigate the maximum moment at C due to a series F. Mkc iC am of travelling loads at fixed distances, let PI, P2, P2, . . . be the loads which at the moment considered are at distances ml, m2, . from the left abutment. Set off • these distances along AB and let yI, y2, . . be the corresponding ordinates of the influence curve (y = Ff) on the verticals under the loads. Then the moment at C due to all the loads is M.=PIyr+Psy2+ .. . 0 W3 A B i I {h b •j• a• d ----- I • a,b • i X 1 ~ } t rfl 1 The position of the loads which gives the greatest moment at C may be settled by the criterion given above. For a uniform travel-ling load w per ft. of span, consider a small interval Fla=4m on which the load is wLm. The moment due to this, at C, is item (l-x) Am/l. But m(l-x)Om/l is the area of the strip Ffhk, that is yEm. Hence the moment of the load on Am at C is wyzm, and the moment of a uniform load over any portion of the girder is w X the area of the influence curve under that portiol, If the scales are so chosen that a inch represents I in. ton of moment, and b inch represents 1 ft. of span, and w is in tons per ft. run, then ab is the unit of area in measuring the influence curve. If the load is carried by a rail girder (stringer) with cross girders at the intersections of bracing and boom, its effect is distributed to the bracing intersections D'E' (fig. 53), and the part of the influence line for that bay (panel) is altered. With unit load in the position shown, the load at D' is (p-n)/p and that at E' is n/p. The moment of the load at C is m(l-x)/l-n(p-n)/p. This is the equation to the dotted line RS (fig. 52). If the unit load is at F', the reaction at B' and the shear at C' is m/l, positive if the shearing stress resists a tendency of the part of the girder on the right to move upwards; set up Ff =m/l (fig. 54) H on the vertical under the load- Repeating the process for other positions, we get the influence line AGHB, for the shear at C due to unit load anywhere on the girder. GC =x/l and CH =-(l-x)/i. The lines AG, HB are parallel. If the load is in the bay D'E' and is carried by a rail girder which distributes it to cross girders at D'E', the part of the influence line under this bay is altered. Let n (Fig. 55) be the distance of the load from D', xi the distance of D' from the left abutment, and p the length of a bay. The loads at D', E, due to unit weight on the rail girder are (p-n)/p and nip. The reaction at B' is I(p-n)x1+n(x1+P)}/pl. The shear at C is the reaction at B' less the load at E', that is, {p(xi+n)-ml} /pl, which is the equation to the line DH (fig. 54). Clearly, the distribution of the load by the rail girder considerably alters ID' the distribution of shear due to a load in the bay x, n. in which the section con-__ _•___ i sidered lies. The total l.-___ • _ ~_- 4_;i shear due to a series of loads PI, P2, . . . at dis- x -- t ,.i tances m1, m2, . . . from being the ordinates of the influence curve under the loads, is S = PIy1+P2y2+ Generally, the greatest shear S at C will occur when the longer of the segments into which C divides the girder is fully -loaded and the other is unloaded, the leading load being at C. If the loads are very unequal or unequally spaced, a trial or two will determine which position gives the greatest value of S. The greatest shear at C' of the opposite sign to that due to the loading of the longer segment occurs with the shorter segment loaded. For a uniformly distributed load w per ft. run the shear at C is w X the area of the influence curve under the segment covered by the load, attention being paid to the sign of the area of the curve. If the load rests directly on the main girder, the greatest + and - shears at C will be wXAGC and -wXCHB. But if the load is distributed to the bracing inter-sections by rail and cross girders, then the shear at C' will be greatest when the load extends to N, and will have the values wXADN and -wXNEB. An interesting paper by F. C. Lea, dealing with the determination of stress due to concentrated loads, by the method of influence lines will be found in Proc. Inst. C.E. clxi. p. 261. Influence lines were described by Frankel, Der Civilingenieur, 1876. See also Handbuch der Ingenieur-wissenschaften, vol. ii. ch. x. (1882), and Levy, La Statique graphique (1886). There is a useful paper by Prof. G. F. Swain (Trans. Am. Soc. C.E. xvii., 1887), and another by L. M. Hoskins (Proc. Am. Soc. C.E. xxv., 1899) 28. Eddy's Method.—Another method of investigating the maximum shear at a section due to any distribution of a travelling load has been given by Prof. H. T. Eddy (Trans. Am. Soc. C.E. xxii., 1890). Let hk (fig. 56) represent in magnitude and position a load W, at x from the left abutment, on a girder AB of span 1. Lay off kf, hg, horizontal and equal to 1. Join f and g to h and k. Draw verticals at A, B, and join no. Obviously no is horizontal and equal to 1. Also mn/mf=hk/kf or mn-W(l-x)/l, which is the reaction at A due to the load at C, and is the shear at any point cf AC. Similarly, po is the reaction at B and shear at any point of CB. The shaded rectangles represent the distribution of shear due to the load at C, while no may be termed the datum line of shear. Let the load move to D, so that its distance from the left abutment is x+a. Draw a vertical at D, intersecting fh, kg, in s and q. Then qr/ro=hk/hg or ro =W(l-x-a)/l, which is the reaction at A and shear at any point of AD, for the new position of the load. Similarly, rs=W(x+a)!l is the shear on DB. The distribution of shear is given by the partially shaded rectangles. For the application of this method to a series of loads Prof. Eddy's paper must be referred to. 29. Economic Span.—In the case of a bridge of many spans, there is a length of span which makes the cost of the bridge least. The cost of abutments and bridge flooring is practically independent of the length of span adopted. Let P be the cost of one pier; G the cost of the main girders for one span, erected; n the number of spans; 1 the length of one span, and L the length of the bridge between abutments. Then, n=L/l nearly. Cost of piers (n-1)P. Cost of main girders nG. The cost of a pier will not vary materially with the span adopted. It depends mainly on the character of the foundations and height at which the bridge is carried. The cost of the main girders for one span will vary nearly as the square' of the span for any given type of girder and intensity of live load. That is, G=ale, where a is a constant. Hence the total cost of that part of the bridge which varies with the span adopted is C = (n-t)P+,naP = LP/1- P +Lai. Differentiating and equating to zero, the cost is least when dC __ LP+La'=o, dl 12 P=a12=G; that is, when the cost of one pier is equal to the cost erected of the main girders of one span. Sir Guilford Molesworth puts this in a convenient but less exact form. Let G be the cost of superstructure W A; D !e k--.-x a---,! ...1 of a too-ft. span erected, and P the cost of one pier with its protection. Then the economic span is 1= tool P1- G. 30. Limiting Span.—If the weight of the main girders of a bridge, per ft. run in tons, is- wa=(w1+wz)lrl(K lr) according to a formula already given, then w8 becomes infinite if k-1r = o, or if l=K/r, ' x- ---=r 1 In * , ------- ----------- 11E' ----- ------- i I tI xl i - p~ li where 1 is the span in feet and r is the ratio of span to depth of girder at centre. Taking K for steel girders as 7200 to 9000, Limiting Span in Ft. r=12 l=600 to 750 =so =720 to 900 = 8 =900 to 1120 The practical limit of span would be less than this. Professor Claxton Fidler (Treatise on Bridge Construction, 1887) has made a very careful theoretical analysis of the weights of bridges of different types, and has obtained the following values for the limiting spans. For parallel girders when r = so, the limiting span is 1070 ft. For parabolic or bowstring girders, when r=8, the limiting span is 128o ft. For flexible suspension bridges with wrought iron link chains, and dip =i0th of the span, the limiting span is 2800 ft. For stiffened suspension bridges with wire cables, if the dip is 11,,th of the span the limiting span is 2700 to 3600 ft., and if the dip is 8th of the span, 3250 to 4250 ft., according to the factor of safety allowed. 31. Braced Girders.—A frame is a rigid structure composed of straight struts and ties. The struts and ties are called bracing bars. The frame as a whole may be subject to a bending moment, but each member is simply extended or compressed so that the total stress on a given member is the same at all its cross sections, while the intensity of stress is uniform for all the parts of any one cross section. This result must follow in any frame, the members of which are so connected that the joints offer little or no resistance to change in the relative angular position of the members. Thus if the members are pinned together, the joint consisting of a single circular pin, the centre of which lies in the axis of the piece, it is clear that the direction of the only stress which can be transmitted from pin to pin will coincide with this axis. The axis becomes, therefore, a line of resistance, and in reasoning of the stresses on frames we may treat the frame as consisting of simple straight lines from joint to joint. It is found in practice that the stresses on the several members do not differ sensibly whether these members are pinned together with a single pin or more rigidly jointed by several bolts or rivets. Frames are much used as girders, and they also give useful designs for suspension and arched bridges. A frame used to support a weight is often called a truss; the stresses on the various members of a truss can be computed for any given load with greater accuracy than the intensity of stress on the various parts of a continuous structure such as a tubular girder, or the rib of an arch. Many assumptions are made in treating of the flexure of a continuous structure which are not strictly true; no assumption is made in determining the stresses on a frame except that the joints are flexible, and that the frame shall be so stiff as not sensibly to alter in form under the load. Frames used as bridge trusses should never be designed so that the elongation or compression of one member can elongate or compress any other member. An example will serve to make the meaning of this limitation clearer. Let a frame consist of the five members AB, BD, DC, CA, CB (fig. 57), jointed at the A B points A, B, C and D, and all capable of resisting tension and compression. This frame will be rigid, i.e. it cannot be distorted without causing an alteration in the length of one or more of the members; but if from a change of temperature or any other cause b one or all of the members change their member, but will merely cause a change in the form of the frame. Such a frame as this cannot be self-strained. A workman, for instance, cannot produce a stress on one member by making some other member of a wrong length. Any error of this kind will merely affect the form of the frame; if, however, another member he introduced between A and I), then if BC be shortened Al) wit be strained so as to extend it, and the four other members will be compressed; if CB is lengthened AD will thereby be compressed, and the four other members extended; if the workman does not snake CB and AD of exactly the right length they and all the members will be permanently strained. These stresses will be unknown 1 the compression members being generally of timber and the tension quantities, which the designer cannot take into account, and such a combination should if possible be avoided. A frame of this second type is said to have one redundant member. 32. Types of Braced Girder Bridges.—Figs. 58, 59 and 6o show an independent girder, a cantilever, and a cantilever and suspended girder bridge. In a three-span bridge continuous girders are lighter than discontinuous ones by about 45% for the dead load and 15% for the live load, it no allowance is made for ambiguity due to uncertainty as to the level of the sup- ports. The cantilever and suspended girder types are as economical and free from uncertainty as to the stresses. In long-span bridges the cantilever system permits erection FIG. 58. by building out, which is economical and sometimes necessary. It is, however, unstable unless rigidly fixed at the piers. In the Forth bridge stability is obtained partly by the great excess of dead over live load, partly by the great width of the river piers. The majority of bridges not of great span have girders with parallel booms. This involves the fewest difficulties of workmanship and perhaps permits the closest approximation of actual to theoretical dimensions of the parts. In spans over zoo ft. it is economical to have one horizontal boom and one polygonal (approximately parabolic) boom. The hog-backed girder is a compromise between the two types, avoiding some difficulties of construction near the ends of the girder. Most braced girders may be considered as built up of two simple forms of truss, the king-post truss (fig. 61, a), or the queen-post truss (fig. 61, b). These may be used in either the upright or the inverted position. A multiple truss consists of a number of simple trusses, e.g. Bollman truss. Some timber bridges consist of queen-post trusses in the upright position, as shown diagrammatically in fig. 62, where the circles indicate points at which the flooring girders ''''\-41'r" /0 !x 11 V' transmit load to the main girders. Compound trusses consist of simple trusses used as primary, secondary and tertiary trusses, the secondary supported en the primary, and the tertiary on the secondary. Thus, the Fink truss consists of king-post trusses; the Pratt truss (fig. 63) and the Whipple truss (fig. 64) of queen-post trusses alternately upright and inverted. A combination bridge is built partly of timber, partly of steel, g. members of steel. On the Pacific coast, where excellent timber is obtainable and steel works are distant, combination bridges are still largely used (Ottewell, Trans. Am. Soc. C.E. xxvii. p. 467). The combination bridge at Roseburgh, Oregon, is a cantilever bridge, The shore arms are 147 ft. span, the river arms 105 ft, and the suspended girder 8o ft., the total distance between anchor piers being 584 ft. The floor beams, floor and railing are of timber. The compression members are of timber, except the struts and bottom chord panels next the river piers, which are of steel. The tension members are of iron and the pins of steel. The chord blocks and post shoes are of cast-iron. 33. Graphic Method of finding the Stresses in Braced Structures.—Fig. 65 shows a common form of bridge truss known as a Warren girder, with lines indicating external forces applied to the joints; 0 half the load carried between the two lower joints next the piers on either side is directly carried by the abutments. The sum of the two upward vertical reactions must clearly be equal to the sum of the loads. The lines in the diagram represent the directions of a series of forces which must all be in equilibrium; these lines may, for an object to be explained in the next paragraph, be conveniently named by the letters in the spaces which they separate instead of by the method usually employed in geometry. Thus we shall call the first inclined line on the left hand the line AG, the line representing the first force on the top left-hand joint AB, the first horizontal member at the top left hand the line BH, &c; similarly each point requires at least three letters to denote it; the top first left-hand joint may be called ABHG, being the point where these four spaces meet. In this method of lettering, every enclosed space must be designated by a letter; all external forces must be represented by lines outside the frame, and each space between any two forces must receive a distinctive letter; this method of lettering was first pro-posed by O. Henrici and R. H. Bow (Economics of Construction), and is convenient in applying the theory of reciprocal figures to the computation of stresses on frames. 34. Reciprocal Figures.—J. Clerk Maxwell gave (Phil. Mag. 1864) the following definition of reciprocal figures:—" Two plane figures are reciprocal when they consist of an equal number of lines so that corresponding lines in the two figures are parallel, and corresponding lines which converge to a point in one figure form a closed polygon in the other." Let a frame (without redundant members), and the external forces which keep it in equilibrium, be represented by a diagram constituting one of these two plane figures, then the lines in the other plane figure or the reciprocal will represent in direction and magnitude the forces between the joints of the frame, and, consequently, the stress on each member, as will now be explained. Reciprocal figures are easily drawn by following definite rules, and afford therefore a simple method of computing the stresses on members of a frame. The external forces on a frame or bridge in equilibrium under those forces may, by a well-known proposition in statics, be represented by a closed polygon, each side of which is parallel to one force, and represents the force in magnitude as well as in direction. The sides of the polygon may be arranged in any order, provided care is taken so to draw them that in passing round the polygon in one direction this direction may for each side correspond to the direction of the force which it represents. This polygon of forces may, by a slight extension of the above definition, be called the reciprocal figure of the external forces, if the sides are arranged in the same order as that of the joints on which they act, so that if the joints and forces be numbered 1, 2, 3, &c., passing round the outside of the frame in one direction, and returning at last to joint 1, then in the polygon the side representing the force 2 will be next the side representing the force 1, and will be followed by the side representing the force 3, and so forth. This polygon falls under the definition of a reciprocal figure. given by Clerk Maxwell, if we consider the frame as a point in equilibrium under the external forces. Fig. 66 shows a frame supported at the two end joints, and loaded at each top joint. The loads and the supporting forces are indicated by arrows. Fig. 67 a shows the reciprocal figure or polygon for the external forces on the assumption that the reactions are slightly inclined. The lines in fig. 67 a, lettered in the usual manner, correspond to the forces indicated by arrows in fig. 66, and lettered according to Bow's method. When all the forces are vertical, as will be the case in girders, the polygon of external forces will be reduced to two straight lines, fig. 67 b, superimposed and divided so that the length AX represents the load AX, the length AB the load AB, the length YX the reaction YX, and so forth. The line XZ consists of a series of lengths, as XA, AB . . . DZ, representing the loads taken in their order.' In subsequent diagrams the two reaction lines will, for the sake of clearness, be drawn as if slightly inclined to the vertical. If there are no redundant members in the frame there will be only two, members abutting at the point of support, for these two members will be sufficient to balance the reaction, whatever its direction may be; we can therefore draw two triangles, each having as one side the reaction YX, and having the two other sides parallel to these two members; each of these triangles will represent a polygon of forces in equilibrium at the point of support. Of. these two triangles, shown in fig. 67 c, select. that in which the letters X and Y are so placed that (naming the apex of the triangle E) the lines XE and YE are the lines parallel to the two members of the same name in the frame (fig. 66). Then the triangle YXE is the reciprocal a b c d FIG. 67. figure of the three lines YX, XE, EY in the frame, and represents the three forces in equilibrium at the point YXE of the frame. The direction of YX, being a thrust upwards, shows the direction in which we must go round the triangle YXE to find the direction of the two other forces; doing this we find that the force XE must act down towards the point YXE, and the force EY away from the same point. Putting arrows on the frame diagram to indicate the direction of the forces, we see that the member EY must pull and therefore act as a tie, and that the member XE must push and act as a strut. Passing to the point XEFA we find two known forces, the load XA acting downwards, and a push from the strut XE, which, being in compression, must push at both ends, as indicated by the arrow, fig. 66. The directions and magnitudes of these two forces are already drawn (fig. 67 a) in a fitting position to represent part of the polygon of forces at XEFA; beginning with the upward thrust EX, continuing down XA, and drawing AF parallel to AF in the frame we complete the polygon by drawing EF parallel to EF in the frame. The point F is determined by the intersection of the two lines, one beginning at A, and the other at E. We then have the polygon of forces EXAF, the reciprocal figure of the lines meeting at that point in the frame, and representing the forces at the point EXAF; the direction of the forces on EH and XA being known determines the direction of the forces due to the elastic reaction of the members AF and EF, showing AF to push as a strut, while EF is a tie. We have been guided in the selection of the particular quadrilateral adopted by the rule of arranging the order of the sides so that the same letters indicate corresponding sides in the diagram of the frame and its reciprocal. Continuing the construction of the diagram in the same way, we arrive at fig. 67 d as the complete reciprocal figure of the frame and forces upon it, and we see that each line in the reciprocal figure measures the stress on the corresponding member in the frame, and that the polygon of forces acting at any point, as IJKY, in the frame is represented by a polygon of the same name in the reciprocal figure. The direction of the force in each member is easily ascertained by proceeding in the manner above described. A single known force in a polygon determines the direction of all the others, as these must all correspond with arrows pointing the same way round the polygon. Let the arrows be placed on the frame round each joint, and so as to indicate the direction of each force on that joint; then when two arrows point to one another on the same piece, that piece is a tie; when they point from one another the piece is a strut. It is hardly necessary to say that the forces exerted by the two ends of any one member must be equal and opposite. This method is universally applicable where there are no redundant members. The reciprocal figure for any loaded frame is a complete formula for the stress on every member of a frame of that particular class with loads on given joints. Consider a Warren girder (fig. 68), loaded at the top and bottom joints. Fig. 69 b is the polygon of external forces, and 69 c is half the reciprocal figure. The complete reciprocal figure is shown in fig. 69 a. The method of sections already described is often more convenient than the method of reciprocal fgures, and the method of influence lines is also often the readiest way of dealing with braced girders. worm /i rs` r^iM1111a i ..~.mmul.7^amra11 r a Q U V w Al rsi . Ate_ X111 ®,~-®rSAiII VA II I I 1111^IFA r . r. a X 35. Chain Loaded uniformly along a Horizontal Line.—If the lengths of the links be assumed indefinitely short, the chain under given simple distributions of load will take the form of comparatively simple mathematical curves known as catenaries. The true catenary is that assumed by a chain of uniform weight per unit of length, but the form generally adopted for suspension bridges is that assumed by a chain under a weight uniformly distributed relatively to a horizontal line. This curve is a parabola. Remembering that in this case the centre bending moment Zwl will be equal to wL2/8, we see that the horizontal tension H at the vertex for a span L (the points of support being at equal heights) is given by the expression I. . .. H=wL2/8y or, calling x the distance from the vertex to the point of support, H =wx2/2 y. The value of H is equal to the maximum tension on the bottom flange, or compression on the top flange, of a girder of equal span, equally and similarly loaded, and having a depth equal to the dip of the suspension bridge. Consider any other point F of the curve, fig. 70, at a distance xfrom the vertex, the horizontal component of the resultant (tangent to the curve) will be unaltered; the vertical component V will be simply the sum of the loads between 0 and F, or wx. In the triangle FDC, let FD be tangent to the curve, FC vertical, and DChorizontal ; these three sides will necessarily be proportional respectively to the resultant tension along the chain at F, the vertical force V passing through the point D, and the horizontal tension at 0; hence H : V=DC : FC=wx2/2y : wx=x/2 : y, hence DC is the half of OC, proving the curve to be a parabola. The value of R, the tension at any point at a distance x from the vertex, is obtained from the equation R2 = H2+V2 = w2x4/4Y2 + w2x2, or, 2. . R=wx,l (I+xe/4y2). Let i be the angle between the tangent at any point having the co-ordinates x and y measured from the vertex, then 3 tan i=2y/x. Let the length.of half the parabolic chain be called s, then 4 s = x +2y2/3x. The following is the approximate expression for the relation between a change As in the length of the half chain and the corresponding change Ay in the dip:- s+As =x+(2/3x) ly2+2yLy+(Ay)2} =x+2y2/3xd-4yty/3x+20y2/3x, or, neglecting the last term, 5 As = 4yAy/3x, and 6 A y =3xAs/4y From these equations the deflection produced by any given stress on the chains or by a changt of temperature can be calculated. 36. Deflection of Girders.—Let fig. 71 represent a beam bent by external loads. Let the origin .0 betaken at the lowest point of the bent beam. Then the deviation y = DE of the neutral axis of the bent beam at any point D from the axis OX is given by the relation d2y MI dx2 ` EI' where M is the bending moment and I the amount of inertia of the beam at D, and E is the coefficient of elasticity. It is usually accurate enough in deflection calculations to take for I the moment of inertia at the centre of the beam and to consider it constant for the length of the beam. Then dy =El) I Mdx dxy=El f f Mdx2. The integration can be performed when M is expressed in terms of x. Thus for a beam supported at the ends and loaded with w per inch length M =w (a2-x2), where a is the half span. Then the deflection at the centre is the value of y for x=a, and is b= 5 wa4 24 The radius of curvature of the beam at D is given by the relation R=EI/M. 37. Graphic Method o finding Deflection.—Divide the span L into any convenient number n of equal parts of length 1, so that nl=L; compute the radii of curvature RI, R2, R2 for the several sections, Let measurements along the beam be represented according to any convenient scale, so that calling L, and li the lengths to be drawn on paper ,we have L = aLi ; now let ri, r2, r3 be a series of radii such that ri = RI/ab, r2 = Rz/ab, &c., where b is any convenient constant chosen of such magnitude as will allow arcs with the radii, ri, r„ &c., to be drawn with the means at the draughtsman's disposal. Draw a curve AiNti,WANAWANWo Y It 9 Q It x D X r U ----•1 L --a 1 , r as shown in fig. 72 with arcs of the length 11, l2i l3, &c., and with the radii rI, r2, &c. (note, for a length Ali at each end the radius will be infinite, and the curve must end with a straight line tangent to the last arc), then let v be the measured deflection of this curve from the straight line, and V the actual deflection of the bridge; we have V =av/b, approximately. This method distorts the curve, so that vertical ordinates of the curve are drawn to a scale b times greater than that of the horizontal ordinates. Thus if the horizontal scale be one-tenth of an inch to the foot, a= 120, and a beam loo ft. in length would be drawn equal to to in.; then if the true radius at the centre were Io,000 ft., this radius, if the curve were undistorted, 1, would be on paper r000 in., but making b =50 we can draw the curve with a radius of 20 in. The vertical distortion of the curve must not be so great that there is a very sensible difference between the length of the arc and its chord. This can be regulated by altering the value of b. In fig. 72 distortion is carried too far; this figure is merely used as an illustration. 38. Camber.—In order that a girder may become straight under its working load it should be constructed with a camber or upward convexity equal to the calculated deflection. Owing to the yielding of joints when a beam is first loaded a smaller modulus of elasticity should be taken than for a solid bar. For riveted girders E is about 17,500,000 lb per sq. in. for first loading. W. J. M. Rankine gives the approximate rule Working deflection = = l2/Io,000h, where l is the span and h the depth of the beam, the stresses being those usual in bridgework, due to the total dead and live load. (W. C. U.)
End of Article: TIC
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