BRIDGES
375
tribution of shearing force is given by two rectangles (Fig. 1). Bending moment increases uniformly from either abutment to the load, at which the bending moment is M = R2?/i=R! (£ - to). The distribution of bending moment is given by the ordinates of a triangle. Next let the girder carry a Straining uniform load w per foot run (Fig. 2). The total load action. is wl; the reactions at abutments, R1 = R2—The distribution of shear on vertical sections is given by the ordinates of a sloping line. The greatest bending moment is at the centre Steel, = 5-87 i+ ( Hand = JVL = At any point In these equations <p is to have its + or - value according to the x from the abutment, the bendcase considered. For shearing stresses the working stress may ing moment is — have 0'8 of its value for tension. The following table gives an equation to a parabola. Let a uniform train weighing values of the working stress calculated by these equations :— w per foot run advance over a Working Stress for Tension or Thrust by Launhardt and girder of span 2c, from the left Weyrauch Formula. abutment. When it covers the girder to a distance x from the Working Stress/, centre (Fig. 3) the total load is tons per sq. in. l+ w[c+x); the reaction at B is Steel. Iron. R, = w(c+x)xcJ^=™(C + xf, 4c 4c 1-0 1-5 6-60 8-80 All dead load. 0-75 1-375 6-05 8-07 which is also the shearing force jqg, 2. 0-50 1-25 7-34 5-50 at C for that position of the 0'25 1T25 6-60 4-95 load. As the load travels, the shear at the head of the train will 0-00 1-00 All live load . 4-40 5-87 be given by the ordinates of a parabola having its vertex at A, •0-25 0-875 5-14 3-85 and a maximum Fmoa:. = — ^wl ■0-50 0-75 3-30 4-40 at B. If the load travels the ■0-75 0-625 2-75 3-67 reverse way, the shearing force •1-00 0-500 Equal stresses + and 2-20 2-93 at the head of the train is given by the ordinates of the dotted To compare this with the previous table, 0 = (A + B)/A = 1 +p. parabola. The greatest shear Except when the limiting stresses are of opposite sign, the two at C for any position of the tables agree very well. In bridge work this occurs only in some of load occurs when the head of the bracing bars. the train is at C. For any It is a matter of discussion whether, if fatigue is allowed for by load/) between C and B will the Weyrauch method, an additional allowance should be made for increase the reaction at B and impact. There was no impact in Wohler’s experiments, and theretherefore the shear at G by fore it would seem rational to add the impact allowance to that part of p, but at the same for fatigue; but in that case the bridge sections become larger than time will diminish the shear experience shows' to be necessary. Some engineers escape this at C by the whole of p. The difficulty by asserting that Wohler’s results are not applicable to web of a girder must resist the bridge work. They reject the allowance for fatigue (that is, the maximum shear, and, with a travelling load like a railway train, effect of repetition) and design bridge members for the total dead this is greater for partial than for complete loading. Generally a and live load, plus a large allowance for impact varied according to girder supports both a dead and a live load. The distribution some purely empirical rule. (See Waddell, Zte p. 7.) Now of total shear, due to a dead load v)d per foot run and a travelling in applying Wohler’s law, fmax- for any bridge member is found for load Wi per foot run, is shown in Fig. 4, arranged so that the the maximum possible live load, a live load which though it may dead load shear is added to the maximum travelling load shear of sometimes come on the bridge and must therefore be provided for, the same sign. is not the usual live load to which the bridge is subjected. Hence In the case of girders with braced webs, the tension bars of the range of stress, fmax. - fmin; from which the working stress is which are not adapted to resist a deduced, is not the ordinary range of stress which is repeated a thrust, another circumstance due practically infinite number of times, but is a range of stress to which to the position of the live load the bridge is subjected only at comparatively long intervals. must be considered. For a train Hence practically it appears pro- advancing from the left, the travelbable that the allowance for ling load shear in the left half fatigue made in either of the of the span is of a different sign tables above is sufficient to cover from that due to the dead load. the ordinary effects of impact Fig. 5 shows the maximum shear Fig. 4. also. at vertical sections due to a dead English bridge - builders are and travelling load, the latter adsomewhat hampered in adopting vancing (Fig. 5, a) from the left and (Fig. 5, b) from the right rational limits of working stress abutment. Comparing the figures it will be seen that over a dis-R2 by the rules of the Board of tance x near the middle of the girder the shear changes sign, Trade. Nor do they all accept according as the load advances the guidance of Wohler’s law. from the left or the right. The The following are some examples bracing bars therefore for this of limits adopted. For the Dufpart of the girder must be ferin Bridge (steel), the workadapted to resist either tension ing stress was taken at 6 "5 tons or thrust. Further, the range per sq. in. in bottom booms and of stress to which they are subdiagonals, 6 "0 tons in top booms, jected is the sum of the stresses 5"0 tons in verticals and long due to the load advancing from compression members. For the the left or the right. To find Stanley Bridge at Brisbane, the Fig. 1. the greatest shear with a set of limits were 6-5 tons per sq. in. concentrated loads at fixed dism compression boom, 7 '0 tons in tension boom, 5 "0 tons in vertical tances, let the loads advance struts, 6 -5 tons in diagonal ties, 8 "0 tons in wind bracing, and 6 "5 from the left abutment, and let tons in cross and rail girders. In the new Tay Bridge the limit of C be the section at which the stress is generally 5 tons per sq. in., but in members in which the shear is required (Fig. 6). The stress changes sign 4 tons per sq. in. greatest shear at C may occur Let a girder, supported at the ends, carry a fixed load W at to with Wj at C. If Wx passes from the right abutment. The reactions at the abutments are beyond C, the shear at C will Fig. 5. and R2 —W(Z - to)/?. The shears on vertical sections be greatest when W2 is to the left and right of the load are Rj and - R2, and the dis- at O. Let R be the resultant ofprobably the loads on the bridge when
fmax- = M ^ 1 + ^ u S Q ^ [Stresses of opposite sign. ] The working stress in any case is fmax. 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 (l'"l*)
