The equation of vertical shearing force has already been given by eq. (3) viz.: x.

Sx =R-Ʃw o.

For a given position of load, and hence for a given value of R, therefore,

Sx will be maximum when:

For the dead load of girder we have Sx = R - w'x in which w' represents the load per unit of length. At the end of the girder where x=o:

Sx =R=490 lbs. X 50' x «=l 2,300 lbs.

At the middle of the girder where w' x =R the shearing force is zero. In Fig. 3, Plate I, A 6 represents the length of the girder, and C its middle point. If we now draw B D according to the scale of force, equal to 12,300 lbs., and connect CD, the amount of shearing action at every point between the end and the middle point of the girder will be equal to the ordinate from the line C B to this inclined line.

The maximum shearing action of the moving load can of course be determined by the use of equation (3) which is entirely general, but it requires too much work when the number of points for which the amount is required to be determined is large; and in such cases the graphical method is far preferable.

For the latter method, that part of Fig. 3 lying above the line A B is all that is necessary for our case.

If we look at equation (3) we see the shearing action of a moving load system is greatest at any section x when the system is in such a position that R shall be greatest possible and the least possible. Consequently for a single load it gives the maximum shear at any point when it lies over that point, and for a uniformly distributed load the shear is maximum at any point when the greater of the two parts into which the point divides the length of the girder is covered with the load.

To simplify the operation a little, we will take off the head load of the first locomotive and make the whole train advance from one end to the other. This gives evidently a somewhat greater amount of shear than when we take all the wheels into consideration, but the amount is very small, being at the middle of the girder, where the excess is greatest, equal to 6,000Ã--(25 - 7.5/50) =2,100 lbs.

In order now to determine the maximum shear at every point, lay off (Fig. 3) the locomotive weights along the line A B, with the first driver over the point A. In this position it gives the maximum shear at A. It is of course supposed that the first driver rests on the point at an infinitely small distance to the right of A. Beginning at the point B draw a vertical line and lay upward all the weights in succession, beginning at the first driver. Take the point A as the pole, and draw as before the radial lines from A, and complete the equilibrium curve A H. The process is exactly the same as in the case of Figs. 1 and 2. The line A E coincides with Ae, E F parallel to Af, F G to A g, and lastly O'H parallel to A O. The distance B H which, according to the scale, is equal to the force of 53,600 lbs., is the amount of shear at A, and to obtain the maximum shear for every other point between C and A we have but to measure in like manner the ordinate to the equilibrium polygon A E F G - - H from the line A B.* At the point B the shear is zero, as the locomotive is outside of the girder. For the total shear at any point we measure along the vertical line at that point between the equilibrium polygon and the line C D. Thus we obtain the maximum shears at the several sections into which we have previously divided the girder, the following amounts:

* For proof see DuBois' Graph. Statics, chap, vii.

According to what we have said in the foregoing pages, we have but to divide the shear eq. (9) at the end where it is maximum, by the allowable . stress per square inch for shear to obtain the necessary web section. Experiments have shown that the shearing strength of wrought iron across the fiber is equal to about 4 5 the tensile strength. The shearing strength along the fiber varies sometimes considerably, being, however, between 1/3 to 1 1/5 times that across the fiber, according to the kind of iron and the mode of treatment in the rolling mills. Consequently, where a piece of iron is subjected to shears in both directions, a somewhat smaller allowable stress should be given than merely across the fibers, say 2/3 the allowed tensile stress. The ratio of minimum to maximum shear at the end of the girder is about 1/5' and the same as in the case of flange stress; consequently we obtain at once the allowable shearing stress:

9,600 Ã--2/3,=6,400 lbs.

Adding 15% for impact to the end shear, we obtain 65,900Ã--1.15%= 75,800 lbs.

The section required is therefore equal to:

75,800/6,400 =11.9 square inches.

The depth of the web is 4 - 6". Supposing the number of rivet holes in a vertical row to be 14, and taking the thickness of the web at the least allowable dimension of ⅜", we have the net section:

(54" - 14") Ã--⅜=15 square inches,

which is about 3 square inches still on the side of safety.

Having thus found out that our web of ⅜ inch is strong enough so far as shearing stress is concerned, it now remains to be seen whether it can resist the stress as a column (see p. 24).

Taking the vertical distance between two rows of rivets at 4' - 2"=50" we have:

h/b=50"/⅜=133.

From the table on page 25 we see that the allowable stress per square inch is about 620 lbs., which is even less than the least shearing stress which we have at the middle of the web, and is equal to - 17,800/15=1,190 lbs. per square inch. Consequently the web must be stiffened throughout.

We have spaced stiffeners, as shown in Plate II. The spacing is, to a certain extent, governed by the positions of lateral and cross bracing connections; for at every point where we have cross-bracing it is convenient to have stiffeners to form connections and also to give stiffness at that section. Toward the end of the girder, stiffeners are placed much closer than toward the middle, for the evident reason of greatly increased shear, and hence of greater compression at the end.

Stiffeners must always be tightly fitted between the flange angles. In order to bring stiffeners in contact with web and vertical leg of angles, they are either bent, as shown in Fig. 14, or fillers maybe used, as in Fig. 15. The first method requires less material than the second, but requires more work in giving the angles required amount of bending, and will be economical only when there is a number of stiffeners to be bent in the same way, by using press and dies.

Fig. 14.

Fig.15.