Let X be the section at which the shear for various positions of g is to be found. When g is placed at any point between A and X the shear S at the latter point will be

but when the load is placed between B and X the shear becomes

Obviously these two values of the shear are equations of two parallel straight lines, that represented by equation (f) passing through A, and that represented by equation (h) passing through B, the constant vertical distance between them being g. Hence let BF be laid off negatively downward and AG positively upward, each being equal to g by any convenient scale. The ordinates drawn from the various positions 1, 2, 3 ... 6 of g on AB to AD and BC will be the shears at X produced by the load g at any point of the span, and determined by equations (f) and (h). The influence line, therefore, for the section X will be the broken line ADCB. When g is at X the sign of the shear changes, since the latter passes through a zero value.

If a train of weights W₁, W₂, W₃, etc., passes across the span, the total shear at X will be found by taking the sum of the vertical intercepts between AB and ADCB, drawn at the positions occupied by the various single weights of the train. If those single weights are expressed in terms of the unit load g, the shear S will have the value

y being the general value of the intercept between AB and the influence line. The latter shows that the greatest negative shear at X will exist when the greatest possible amount of loading is placed on AX only, while the greatest positive shear at the same section will exist when BX only is loaded. If BX is the smaller segment of span, the latter shear is called the “counter-shear,” and the former the “main shear.”

If the loads are applied at panel-points of the span only, the treatment is the same in general character as that employed for bending moments. In Fig. 25a let 4 and 5 be the panel-points between which the load g is found, and let the panel length be p. Also, let z′ be the distance of the weight g from panel-point 4. The reactions at A and 4 will then be

The shear at the section X for any position of the weight g will then be

As this is the equation of a straight line, with S and z or z′ for the coordinates, the influence line for the panel in which the section X is located will be the straight line represented by KL in Fig. 25a.

If z′ is placed equal to 0 and p successively, then will equation (k) become identical with equations (f) and (h) in succession. The shears at points 4 and 5 will therefore take the same values as if the loads were applied directly to the beam. For the reasons stated in connection with the consideration of bending moments, loads in other panels than that containing the section for which the influence line is drawn will have the same effect on that section as if they were applied directly to the beam or truss. Hence AKLB is the complete influence line for this case.

It is evident that there must be as many influence lines drawn as there are sections to be discussed. Also, if g is taken as some convenient unit, i.e., 1000 or 10,000 pounds, it is clear that the labors of computation will be much reduced.

102. Application of Influence-line Method to Trusses.—In considering both the bending moments and shears when the loads are applied at panel-points, it has been assumed, as would be the case in an ordinary beam, that the bending moments as well as the shears may vary in the panel; but this latter condition does not hold in a bridge-truss. Neither bending moment nor shear varies in any one panel. Yet the influence lines for moments and shears are to be drawn precisely as shown in Figs. 25 and 25a. The section X will always be found at a panel-point, and no intercept drawn within the limits of the panel adjacent to that section carrying the load g is to be used. This method will be illustrated by the aid of Fig. 25b.

The employment of influence lines may be illustrated by determining the moment and shear in a single section of the truss shown in Fig. 24, which is reproduced in Fig. 25c, when carrying the moving load exhibited in Fig. 25b, although its use may be much extended beyond this simple procedure.

The moving load shown in Fig. 25b is that of a railroad train consisting of a uniform train-load of 4000 pounds per linear foot drawn by two locomotives with the wheel concentrations shown; it is a train-load frequently used in the design of the heaviest class of railroad structures. If the criterion of equation (27) be applied to this moving load, passing along the truss shown in Fig. 25c, from left to right, it will be found that the greatest bending moment is produced at the section Q when the second driving-axle of the second locomotive is placed at the truss section in question, as shown in Fig. 25c.

The unit load to be used in connection with the influence lines will be taken at 10,000 pounds. Remembering that the panel lengths are each 30 feet, it will be seen that the panel-point Q is 150 feet from A. Hence the product gx will be 1,500,000 foot-pounds. Similarly the product g(l - x) will be 900,000 foot-pounds. Laying off the first of these quantities, as AD, at a scale of 1,000,000 foot-pounds per linear inch, and the second quantity, as BK, by the same scale, the influence line ACB can at once be completed. Vertical lines are next to be drawn through the positions of the various weights, including one through the centre of the uniform train-load 110 feet in length resting on the truss. The vertical line through the centre of the uniform train-load is shown at O. By carefully scaling the vertical intercepts between AB and ACB, and remembering that each of the loads on the truss must be divided by 10,000, the following tabulated statement will be obtained, the sum of the intercepts for each set of equal weights being added into one item, and all the items of intercepts being multiplied by 1,000,000:

The lever-arm of ef, i.e., the normal distance from Q to ef, is 39.7 feet. Hence the stress in ef is

All the chord stresses can obviously be found in the same manner.

In order to place the same moving load so as to produce the greatest shear at the same section Q, the criterion of equation (33) must be employed. The dimensions of the truss shown in connection with Fig. 29 give the following data to be used in that equation: i = 210 feet, m = 60 feet, and p = 30 feet.

Introducing these quantities into equation (33), and remembering that the train moves on to the bridge from A, it would be found that the second axle of the first locomotive must be placed at the section Q, as shown in Fig. 25d, which exhibits the lower-chord panel-points numbered from 1 to 7. The conventional unit load g will be taken in this case at 20,000 pounds. It is represented as AG and BF (Fig. 25d), laid off at a scale of 10,000 pounds per inch. K is immediately under panel-point 5 and L is immediately above panel-point 6, hence the broken line AKLB is the influence line desired. The vertical lines are then drawn from each train concentration in its proper position, all as shown, including the vertical line through the centre of the 54 feet of uniform train-load on the left. The summation of all the vertical intercepts between AB and the influence line AKL, having regard to the scale and to the ratio between the various loads and the unit load g, will give the following tabular statement:

These simple operations illustrate the main principles of the method of influence lines from which numerous and useful extensions may be made.