The following points will suggest what position and what part of the load system are to be looked for to find the maximum moments.

1. For uniformly distributed load over the entire span, the maximum bending moment is in the middle of the span, and therefore at the center of gravity of the total load.

2. When the load is not uniformly distributed, the center of gravity, and with it the point of maximum moment, will approach the heaviest load.

3. For a load system of the second class, the absolute maximum of moments, however, does not occur at the middle point of the span, but under the load which is nearest to the center of gravity of the system, when they (that load and the center of gravity of the system) are equidistant from the middle point of the span.*

Now we proceed to an actual example:

As an example, let us take a single >track standard gauge deck girder of 50 ft. span, i. e., measured between the centers of bed plates.

The weight of the girder, according to the formula already given, with a=330 will then be equal to W=10(502+330) = 28,300 lbs. The weight of the floor system assumed to be 414 lbs. per lineal ft.=

414 Ã-- 50=20,700 Total dead load=49,000 lbs.

* For proof see DuBois' Elements of Graph. Statics, Chap, vii.

Since there are two girders, the load per lineal foot on one will be equal to 49,000/(2Ã--50) = 490 lbs.=w.

We will divide the entire span into say 10 equal parts to determine at these points of division the maximum stresses (Fig. 12).

Fig. 12.

Eq. (12) at once gives the moments at these points as follows.

These moments, when graphically represented, form the ordinates to a parabola with its vertex at the middle of the span (Fig. 13.)

The bending moment due to the moving load will be obtained graphically,* and the whole process is nothing more than what is to be seen in Figs. 1 and 2 of PI. I, and requires but little explanation.

In a horizontal line (Fig. 1, PL I,) lay down to any convenient scale; here taken at 20 feet to an inch (in practice a larger scale, say 5 feet to an inch, should be chosen), the system of load for length enough to cover the ranges of various positions of 50 feet girder , for which the bending moments are to be determined. As stresses need to be determined for one girder only, but half the weights are to be taken. Lay down in succession in a vertical line to a proportionate scale, here 30,000 lbs. to an inch (in practice take say 10,000 lbs. to an inch), the weights of the load system beginning at one end, as shown in Fig. 2. This is called the force diagram. In it choose any point P (called pale) at some convenient distance from the vertical line of weights, say 12/3 inches. This distance represents, according to the scale, the horizontal force of 12/3 Ã-- 30,000 = 50,000 lbs.

* Proofs of the method here used are found in any books on graphical Statics. See DuBois' Graph. Statics.

From the pole P draw lines PM, PN, PO, etc., to the end of each weight (Fig. 2). Then starting from any point l in Fig. 1, draw a line l M parallel to PM, and produce it until it cuts in point m, the perpendicular let fall from the load corresponding to the first weight in the force diagram. From this point m draw a line parallel to PN (of the force diagram), cutting the next perpendicular in n. In this way complete the polygon l m n o p . . . t u v w x, observing carefully that each side of the polygon always occupies the position corresponding to that of the line to which it is parallel in the force diagram. Thus the side n o running between perpendiculars, let fall from the loads 8,000 and 6,000, is parallel to the line PO (of force diagram) which is drawn to the point of junction of those weights.

The position of the locomotive, as shown in the figure, follows the rule already deduced.

The center of gravity, or more properly the resultant of the system of any number of loads, is at once found by prolonging the outermost sides of the polygon formed by so many loads as the number for which the resultant is required. Thus the resultant of the loads lying on the girder in the position I, viz.: 8,000, 6,000, 4 of 12,000 and 2 of 8,000, is found by producing the outermost sides m n and u v until they meet in Y, through which the resultant must pass.