27. Influence Lines.—In dealing with the action of travelling loads much assistance may be 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)/l, 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(l-x)/l. To use this line to investigate the maximum moment at C due to a series of travelling loads at fixed distances, let P1, P2, P3, ... be the loads which at the moment considered are at distances m1, m2, ... from the left abutment. Set off these distances along AB and let y1, 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 = P1y1+P2y2+...
The position of the loads which gives the greatest moment at C may be settled by the criterion given above. For a uniform travelling load w per ft. of span, consider a small interval Fk = ∆m on which the load is w∆m. The moment due to this, at C, is wm(l-x)∆m/l. But m(l-x)∆m/l is the area of the strip Ffhk, that is y∆m. Hence the moment of the load on ∆m at C is wy∆m, and the moment of a uniform load over any portion of the girder is w × the area of the influence curve under that portion. If the scales are so chosen that a inch represents 1 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) 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′, x1 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)x1+n(x1+p)}/pl. The shear at C′ is the reaction at B′ less the load at E′, that is, {p(x1+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 P1, P2, ... at distances m1, m2, ... from the left abutment, y1, y2, ... being the ordinates of the influence curve under the loads, is S = P1y1+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 × 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.