The weight of one girder = 61 700 - 2 = 30 850 lbs. " " 1 the floor = 5 X 6 000 = 30 000 lbs.
The dead-load shears are then computed by the methods given in Strength of Materials, and are given as follows, it being remembered that the concentrated load which comes at the end is one-half a panel load, or 3 000 pounds: where V, = the shear at the end; V, = the shear just to the left of point 1; V, = the shear just to the left of point 2; = the shear just to the right of point 2; and = the shear at the middle of the girder.
The determination of the wheel load position for maximum live-load shears is given in Table XVII. By comparing the Q and K, it will be seen that for the first panel Q = K, and q = k, as n = 1. The position of wheel loads for maximum moments at point 1 is the same as for maximum shear in the first panel. Accord ing to Table XII, wheels 3;4, and 5 at point 1 all give maximum shears in the first panel. In this case, as in previous ones, only the shear for the first position of the loading for any particular point will be determined, as the difference between this and the other eases is too small to warrant the additional labor necessary in com puting them. It is evidently unnecessary to go past panel 3, as only the maximum shears are required.
Wheel Positions, Shears in a Through Plate-Oirder For wheel 3 at point 1, the left reaction (see Table XIII) is 99.7. That portion of wheels 1 and 2 which is transferred to point 0 is 230 _ 15 = 15.33; and the shear in the first panel, therefore, is: V, = 99.70.— 15.33 = +84.37.
For wheel 2 at point 2, the left reaction is: R = (3 496 + 142 X 5) + 75 = 56.10; therefore, A computation with wheel 3 at point 2 will give a shear only 560 pounds greater, which difference would not influence the design to any appreciable extent.
For wheel 2 at point 3, the left reaction is: R = (2 155 + 116 X 1) _ 75 = 30.30; therefore, = 30.3 — 80 +24.97.
When the girder is a deck one, the computation of the dead shears is very much simplified, as all of the load is uniform.
Let it be required to determine the dead and live load shears at the tenth-points of the deck plate-girder of Article 57.
The total weight of one girder and track is 76 175 lbs.
= 76 175 2 = +38 088 pounds.
V, = 38 088 76 = +30 470 pounds.
The position of the wheel loads to produce the maximum shear cannot be determined by the same relation as that used in structures which have' a system of floor-beams and stringers, for here not a portion, but all of the load to the left of the section, must be sub tracted from the left reaction in order to give the shear.
The correct relation fbr the wheel load position will now be deduced.
Let Fig. 110 represent a beam of span 1 loaded with a series of wheel loads followed by a uniform load. Let P equal the weight of the first wheel, 1V equal the weight of all the loads, and g the distance from the center of gravity of all of the loads to the right abutment. The distance between the first and second wheel centers is a, and the first wheel is at the section b—b at a distance x, from the left support. Then, and = — (loads to left of section) _ Now, assume that the loads move forward the distance a. The wheel 2 will be at section b—b, and Fig. 111 will represent the position of the loads. Then, l and It is now evident that in order to get the greatest shear at section b—b, wheel 2 must be placed at the section whenever greater than _ b. Then, V"b—b > P >W9.
Now, canceling out the term appears on both sides of the equation, there results: Wa ] P.
For the engine under consideration, a = 8 feet, and P = 10 000 pounds, and the equation reduces to: 8 W > /, which is to say that when the load on the girder is greater than 14 times the span, then wheel 2 should be placed at the section in order the maximum shear.
For loading E 40, the following is true: For all sections up to and including the center of all spans, place wheel 2 W the section to give the maximum shear.
In Fig. 111 it is immaterial whether or not any additional loads come on the span at the right end when the loads move forward the distance a, as they would only tend to increase the left reaction and therefore the shear If the relation deduced is true for the case when no extra loads come on at the right end, it will be true when they do.