TABLE VI Determination of Position of Wheel Loads for Maximum Moments One should carefully note that in certain positions, as when wheels 11, 12, 13, and 14 are at some wheels are to the left of the left support; that is, they are not upon the bridge. In all such cases they are counted neither in the quantity L nor in W.
In the case of L, wheel 11, being the first large driver of the second engine, will give the greatest moment, as it is the first driver to come at the point when the maximum load of 304 000 pounds is on the truss. Fig. 93 shows the engine diagram on the truss in correct position to give the maximum moment at point 47. Pratt Truss under Engine Loads. In order to exemplify the use of the engine-load diagram, let it be required to determine the stresses in the Pratt truss of Article 45 due to E 40 loading, the height being 25 feet. The secant is (25t 20 )} =25 = 1.28.
The maximum positive shears in the various panels should first be computed. These, written in reverse or der, will be the maximum negative or minimum shears. Table V should now be re ferred to, and an outline dia gram drawn to the same scale as the engine used, on which to place the engine diagram in the correct position. The vari ous values can then be read off the diagram at the right hand end of the truss. It will be found convenient to lay off to scale the first ten feet of the lower chord of the truss from the right support, mak ing the divisions one foot apart. This will enable one to ascer tain the distance of the last wheel load from the right sup port, or the amount of uni form load upon the bridge, without scaling or further com putation. In case it is desired to have the wheel loads appear on the lower chord, as in Fig. 93, the outline of the truss should be on tracing cloth or transparent paper. This is not
to be advised, however, as er rors are likely to occur because of failure to distinguish clearly the various numerical values. It is far better to place the diagram as in Fig. 92, in which case the outline of both the truss and the dia gram can be drawn on good stiff paper.
For wheel 3 at point L, (see Articles 44 and 45), the left reaction is as follows, there tieing four feet of uniform load upon the truss: and the proportion of loads in the panel which is transferred to the point by the stringers is 230 _ 20 = 11.5. The shear is therefore V, _ +146.0 — 11.5 = +134.5. The computation for the shear when wheel 4 is at the point, will not be made; for, as has been noted before, the result will not be much different from the above.
For wheel 3 at wheel 16 comes over the right support. The left reaction is: R, = 12 041 = 120 =100.3; the proportional part of the loads which is transferred to L, is 11.5; and the shear is: ?, = (+100.3 - 11.5) _ +88.8.
For wheel 2 at L„ wheel 11 comes four feet from the right sup port. The left reaction is: R, = (5 848 + 172 X 4) - 120 = 54.5.
That part of wheel 1 which is transferred to is 80 _ 20 = 4.0, and the shear is therefore: ? +50.5.
For wheel 2 at L„ wheel 9 comes over the right support. The left reaction is: = 3 496 _ 120 = 29.1.
That part of wheel 1 which is transferred to L, is 4.0, and the shear is therefore: (+29.1 - 4.0) _ +25.1.
For wheel 2 at wheel 5 is five feet from the right support, and the left reaction is: If the dead panel load is 20 000 pounds, all the shears may now be written as follows: A comparison of the shears in the third and fourth panels shows that counters are required. The stress in these counters is: