+20 X U,U, + 40 X 15 833 20 X 6 333 = 0; whence = 25 333; +20 X U,U, + 60 X 15 833 40 X 6 333 20 X 6 333 = 0; whence = 28 500.
A diagram of half of the truss should now be made, and all the stresses placed upon it. The dimensions should also be put upon this diagram. The student should cultivate this habit, as it shows him at a glance the general relation of stresses and the general rules of their variations. Fig. 33 gives the half-truss, together with the stresses and dimensions. The stresses in the members of the right half of the truss are the same as those in the corresponding members of the left half.
From inspection of the above diagram, it is seen that the chord stresses increase from the end toward the center; that the web stresses decrease from the end toward the center; and that all members slant ing the same way as the end-post have stresses of that sign, while all that slant a different way have an opposite sign. These relations are true of all trusses with parallel chords and simple systems of webbings.
34. Position of Live Load for Maximum Positive and Negative Shears. The dead load, by reason of its nature, is an unchangeable load. The stresses due to it are the same at any and at all times.
With the live load, the case is different. The live load represents the movement of traffic upon the bridge. At certain times there may be none on the bridge, while at other times it may fill the bridge partially or entirely. In such cases the shears due to live load will vary.
Conventional Method. It has been found that the maximum positive shear at any section of a simple beam occurs when the beam is loaded from that section to the right support, and that the maximum negative shear occurs at the same section when this beam is loaded from the section to the left support. This can be proved as follows:
Let a beam be as in Fig. 34, and let a a be the section under consideration. The reaction R, is due to the load wy on the part y, and to the load wx on the part x. That is, Now the shear at the section a a is R, wx; or, From inspection of this last equation, it is seen that wx is the amount that is added to the reaction by loading the part x.
x Also, that 2 is less than unity, is evident. The amount in brackets in the last equation represents the effect of the loading of the segment x of the beam. As this is negative and will only reduce the positive valued term it is therefore proved that to get the largest positive shear the beam should be loaded from the section to the right support.
From further inspection of the equation, it will be seen that the term in brackets, which represents the effect of the load on the seg ment x on the shear, is always negative; and that the term represents the effect of the load on the segment y on the shear, is always positive. Hence, to get the largest negative shear at the section, the load should be on the segment x. That is, the loading should be from the section to the left support.
To get the maximum positive shear at a section or in a panel, load all panel points to the right of it.
To get the maximum negative shear at a section or in a panel, load all panel points to the left of it.
Example. Determine the maximum positive and the maximum negative shears in the panels of the 7-panel Pratt truss shown in Fig. 35, the live panel load being 40 000 pounds. (It will be noticed that the height of the truss is not required.) For maximum + V in 1st panel, load L L,, L,, L L, and L,.
" + " 2d " L L,, L,, L,, and L,.