37. Counters. By reference to L, (Fig. 38), it is seen that when the live load is on the panel points L, and the total stress in the member is + 3.54 + ( 8.40) = 4.86, a compressive stress of 4 860 pounds; whereas, under dead load alone, the stress was + 3.54, a tensile stress of 3 540 pounds. If the member had been built of long, thin bars which could 'take only tension, and which con sequently would have doubled up under the resultant compression brought upon them by the combined action of the dead and minimum live load stresses, then this member could not be used in this case, but some other arrangement would be necessary in order to insure the stability of the truss.
In the Warren truss, no special ar rangement is necessary, as the web members are built so as to take either tension or compression; but with the Pratt and Ilowe trusses some special arrangement is necessary, as the diag onals are built to take one kind of stress only. The case of the Pratt will be considered first.
The Pratt truss has the diagonals made of long bars which take tension only, and the intermediate posts are constructed so as to be able to take compression only. It is not necessary to consider the intermediate posts, for the action of the web members is such that the resulting stresses are always compressive.
Let the 13-panel Pratt truss of Fig. 39 be considered. The panel length is 18 feet, the height 25 feet, the dead panel load 22 000 pounds, and the live panel load 58 500 pounds. The secant is + _ 25 = 1231. The dead-load shears and the maximum and minimum live-load shears are placed directly below their respective panels. Only those mem bers are shown full-lined in Fig. 39 which act under the dead load. Note that the dead-load shears in the center panel being zero, the dead-load stress in the diagonals in the center panel would be 0 X sec ’ = 0.
In the first four panels from either end, the live-load shear, which is of a different sign from that of the dead-load shear, is smaller than the dead-load shear, and therefore will not cause a reversal of stress in the member in that panel. For example, take L,; then, for dead-load stress, U,L, cos ’ + 66.0 = 0 ..U,L, = +66.0 X 1.231 = +81.20 For live-load stress, U,L, cos 27.0 = 0 .'.U,L, _ 27.0 X 1.231 = 33.25 The total stress = -F 81.20 33.25 = + 47.95, which is still The total stress, as before, is -1- 47.95, or a tension of 47 950 pounds.
An inspection of the center panel and the two panels on each side of it, shows that the live-load shear is of a different sign from the dead-load shear, and is also greater in value than the dead-load shear. If the members shown in Fig. 39 were the only ones in the panels, then the dead-load stresses would be: U,L, cos + 44.0 = 0 _ +54.20 U,L, cos 0 + 22.0 = 0 U,L, = +27.10 cos 22.0 = 0 L,U, _ cos 44.0 = 0 L,U, = +54.20 and the live-load stresses caused by the shear of opposite sign from that of the dead-load shear, are: U,L,, cos 45.0 = 0 = 55.40 U,L, cos 67.5 = 0 U,L, = 83.10 cos 4 + 67.5 = 0 = 83.10 +L,U, cos L,U,= 55.40 As no diagonal acts under dead load in the center panel, we may assume that L, acts under live load. The stresses which occur in this are: + cb + 94.5 = 0 U,L, = +116.30 U,L, cos 04.5 = 0 U,L, = The above shows that compressive stresses will occur in the diagonals which were built for tension only. These stresses are: = 55.40 = 1 200 pounds U,L, = +27.10 83.10 = 56 000 " L,U, = +27.10 83.10 = 56 000 " L,U, = 55.40 = 1 200 " U,L, _ - 0 116.30 = 116 300 " If some provision were not made for these stresses, they would cause the members to crumple up and the truss to fail. In order to allow for them, diagonals are placed in the panels, as shown by the dotted and dashed lines. These members will take up the above stress; and moreover, as they slope the opposite way from the main members, they will be in tension.
In order to prove this, assume to act when the live load is on points and L,. Now, will not be regarded, as its stress will be zero. Then the stresses will be: For dead load, and the total stress in will be 27.10 + 83.10 + 56.00.
In a similar manner, the stresses in the other members are: _ + 1.2; L,U, -I- 116.30; U,L, = + 56.00; and -1- 1.2. These diagonals are called counters or counter-bracing.
From a consideration of the foregoing, it is evident that: (a) If the live-load shear in any panel is of opposite sign and greater than the dead-load shear in the same panel, then a counter is required.
(b) The stress in a counter is equal to the algebraic sum of the dead-load shear and the live-load shear of opposite sign times the secant of the angle it makes with the vertical.