For values of moments at points which fall in between wheels, or at positions in the uniform load where the value of the moment is not given, a very important principle of applied mechanics is used. It is: At' 2.
in which, M. — Moment at section desired; M' = Value of moment at preceding vertical line; W — Sum total of all loads to the left of and at the point where M' is taken; x = Distance from section under consideration to vertical line to which M' is referred; to = Uniform load on the distance x.
Let it be desired, for example, to determine the moment at a point c, 3 feet to the right of wheel 13. The position of the loads is given in Fig. 86. The moment is: = 7 668 + 212 X 3 = 7 668 + 636 = 8 304 = 8 304 000 pound feet, there being no uni form load.
To illustrate the method when applied to points in the uniform load, assume the point to be 7 feet to the right of line 2. The po sition is illustrated in Fig. 87. The moment is: The use of the moment diagram is now apparent. Reactions due to any position of the engines may be determined by dividing the span into the value obtained for the moment at the right end of the span. Likewise, if the moment of the reaction about any panel point is determined and from it the moment of the wheel loads about that same panel point are subtracted, then the result, divided by the height of the truss, will give the chord stress. For example, if the right end of an 8-panel 196-foot span truss, height 25 feet, came 7 feet to the right of the vertical line 2, then the moment at this point (see Fig. 87) would be 21 481 000, and the reaction would be 21 481 000 _ 196 = 109 600. This position of the loads would cause the panel point L, to come 3 feet to the right of wheel 13. The moment of the reaction about L. is 109 600 X 6 X 24.5 = 16 111 200; In using the engine to determine the shear in any particular panel, it must be remembered that the shear is not the left reaction less all the loads to the left of the panel point on the right of the section, as the loads in the panel under consideration are carried on stringers, and these stringers transfer a portion of the loads to the panel point on the left of the panel, and a portion to the panel point on the right of the panel. Only that portion of the loads in the panel
which is transferred to the left panel point should be subtracted from the reaction, as should all of the loads to the left of the panel under consideration. If, in a 6-panel 120-foot span Pratt truss, the wheel 6 comes at the left reaction will be: R1 120 (16 364 + 3 X 284 + 3° = 143.6; and the loads in the first two panels will be in position as indicated by Fig. 88, the wheel 3 being 1 foot to the right of point Let it be required to determine the shear in the panel when the loads are in this position. It will be the reaction 143.6 minus loads 1 and 2 and also that portion of the loads 3, 4, and 5 which will be trans ferred by the stringers to- the point L. As the stringers are simple beams, the amount transferred to L, will be the reaction of the stringer Re ferring to Fig. 89, the reaction is : 20 = 42.0 • The shear in the second panel is now found to be: = 143.6 -(10 + 20 + 42.0) _ +71.6 In the majority of cases where it is necessary to determine the shear in a panel, none of the loads will be in the panel to the left of the one under consid eration. In this case the operation is somewhat simplified, as the engine dia gram can be used directly. If the engine be placed so that the third wheel is at wheel 16 will be just over the right support, and the left reaction will be: R,=12 041 = 120 = 100.3.
As there are no wheel loads in the first panel, the amount to be sub tracted from the reaction will be that proportion of the loads l and 2 which is transferred to L,; and this (see Fig. 90) is 230 - 20 = 11.5. The shear in the second panel is then 100.3 — 11.5 = +88.8.
From inspection of the resulting shear in the second panel when wheel 6 is at and when wheel 3 is at it is seen that different wheels at will give different shears in the panel to the left. Evi dently there is some wheel which will give the greatest shear possible. The same is true of the relation between wheels and moments The next two articles are devoted to subject-matter which will enable one to tell which of several wheels is the correct wheel at the point, without the necessity of solving for the shear each time every wheel is at the point.