+ U,L, 23.4 = 0 .. U,L, = + 23.4 To determine the minimum live-load stress in proceed as indicated on page 42. By loading points and to the left, the live load shear in the 4th panel will be 46.8, and in the 3d panel under this same loading it will be 46.8 + 62.4 = +15.6. The sign of the total shear in the two adjacent panels, and the members acting, are shown in Fig. 45. The stress in is then determined by using a circular section around and is simply the dead load at there being no live-load stress in the member when the bridge is loaded as has been done.
In finding the minimum live-load stress and also the minimum stress in the same method of procedure will be followed. Let L, and to the left be loaded. Then the shear in the 5th panel is 78.0, and under this same loading the shear in the 4th panel is 78.0 + 62.4 = 15.6. The sign of the total shear in each of the adjacent panels is given in Fig. 46. It should be remembered that a resultant shear with the same sign as the dead-load shear causes the main diagonal to act, while a resultant shear of opposite sign to that of the dead-load shear causes the counter to act. The members acting are shown, and a section 4 4 can be passed. The dead-load shear at this section is 105 3 X 20 4 X 10 = +5.0; and accordingly, Therefore, _ +5.0 = Dead-load stress in this case.
The live-load stress which acts at the same time is: the term 15.6 representing the live-load shear on the section 4 4. This is not the minimum stress, as will next be shown, but it illus trates the fact that the loading for minimum live-load shears does not always give the minimum live-load stress.
By loading the live-load shear in the second panel, and likewise all others from this to the right support, will be 7.8. The total shears, together with their sign, and also the members they cause to act, are given in Fig. 47. The minimum live-load stress in U,L, is found to be zero, and the dead-load stress is 10, as is derived by passing a circular section around U the equation being as follows: Live load at U, U,L, = 0 .'. U,L, 0 for live load.
Dead load at U,L, = 0 .'. UL = 10.0 for dead load.
A diagram of half the truss should now be made, and all dead and live load stresses placed upon it, and these should be combined so as to form the maximum and the minimum stresses. Such a dia gram, together with all stresses, is given in Fig. 48.
The stresses are written in the following order: Dead load, maximum live load, minimum live load, the maximum, and the minimum. In the chord and end-post stresses, there is no minimum live-load stress recorded, it being zero. Where pairs of stresses occur
simultaneously, a bent arrow connects them.
40. The Howe Truss. The physical make-up of the Howe truss differs from that of the Pratt in that the diagonals are made to stand com pression only, and the verticals can stand tension only. In the Pratt truss it was found that none of the inter mediate posts could be brought into tension by any loading. In the Howe truss it will be found that none of the verticals can be brought into compression.
Let it be required to determine the stresses in a Howe truss of the same span, height, and loading as the Pratt truss of Article 39. An outline diagram is given in Fig. 49.
The dead-load shears and the maximum and min imum live-load shears will be the same as for the Pratt truss, and they are:.
Inspection of these shows that counters are required in the 4th and 5th panels (see Article 37).
The dead-load lower chord stresses will be computed by the tangent method (see Article 31), the section being y y etc. The tangent of (A is 25 _ 30 = 0.8333. The stresses may be conven iently tabulated as follows: A simple method for the determination of the upper chord stresses, is to pass a section and to equate the sum of the horizontal forces. Pass section 1 1. The only horizontal forces are the stresses in and and as these are parallel, one must be equal and opposite to the other. In a like manner the stresses in the other sections of the top chord are found. The stresses are: U, U, = _ (+ 87.5) = 87.5 U, = L,L, = (+150.0) _ 150.0 U,U, = L,L, = = 187.5 A consideration of the Pratt truss shows that this method can be applied to it in determining the chord stresses.
As it is known that the diagonal web members are in compression under the dead load which produces a positive shear in the left half of the truss, it is evident that positive live-load shears will produce compressive stresses, and negative live-load shears tensile stresses, in the diagonals in the left half of the truss. Also, from Article 30, the stress in a diagonal is V sec (/). The stresses can now be written directly without the aid of the stress equation: Likewise the stresses in the verticals can be written directly, remem bering that here the secant is unity, and that the shear at the section cutting the member is to be used, not forgetting that 3 of the dead panel load is applied at the top panel points. The shears and stresses are: _ = +105.0 - 10 = +95.0 U,L, = +95.0 X10= +65.0 U,L,- +65.0