 Fig. 54.

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) 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)/l. 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′, x 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 n/p. The reaction at B′ is {(p-n)x+n(x+p)}/pl. The shear at C′ is the reaction at B′ less the load at E′, that is, {p(x+n)-nl}/pl, which is the equation to the line DH (fig. 54). Clearly, the distribution of the load by the rail girder considerably alters the distribution of shear due to a load in the bay in which the section considered lies.

The total shear due to a series of loads P, P, ... at distances m, m, ... from the left abutment, y, y, ... being the ordinates of the influence curve under the loads, is S = Py+Py+.... 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 × 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 w × AGC and -w × CHB. But if the load is distributed to the bracing intersections by rail and cross girders, then the shear at C′ will be greatest when the load extends to N, and will have the values w × ADN and -w × NEB. 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 Fränkel, 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). Fig. 56.

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 l. Lay off kf, hg, horizontal and equal to l. Join f and g to h and k. Draw verticals at A, B, and join no. Obviously no is horizontal and equal to l. 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 of 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; C the cost of the main girders for one span, erected; n the number of spans; l 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 = al&SUP2;, where a is a constant. Hence the total cost of that part of the bridge which varies with the span adopted is -

C = (n-i)P+nal&SUP2;

= LP/l-P+Lal.

Differentiating and equating to zero, the cost is least when

 dC dl = -LP l&SUP2; +La = 0,

P = al&SUP2; = 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 of a 100-ft. span erected, and P the cost of one pier with its protection. Then the economic span is l = 100√P/√G.

30. Limiting Span. - If the weight of the main girders of a bridge, per ft. run in tons, is -

w = (w+w)lr/(K-lr)

according to a formula already given, then w becomes infinite if k-lr = 0, or if