The line abc is drawn through the point 7 and perpendicular to the direction of the wind pressure; hence with respect to the right support the arms of and resultant wind pressure are ac and be, and with respect to the left support, the arms of the resultant wind pressure are ac and ab. These different arms can be measured from a scale drawing of the truss or be computed as follows: The angle 17a equals the angle 417, and 417 was shown to be 30 degrees in Example 3, Page 26. Hence ab cos 30°, be = 28 cos 30°, ac = 42 cos 30°.
Since the algebraic sums of the moments of all the forces acting on the truss about the right and left supports equal zero, x 42 cos 30' = 10,800 X 28 cos 30°, and R, X 42 cos 30' = 10,800 X 14 cos 30°.
Adding the two reactions we find that their sum equals the load as it should.
Case (2) One end of the truss rests on rollers and the other is fixed to its support. The reaction at the roller end is always vertical, but the direction of the other is not known at the outset unless the loads are all vertical, in which case both reactions are vertical.
When the loads are not all vertical, the loads and the reactions constitute a non-concurrent nonparallel system and any one of the sets of conditions of equilibrium stated in Art. 35 may be used for determining the reactions. In general the fourth set is probably the simplest. In the first illustration we apply the four different sets for comparison.
Examples. 1. It is required to compute the reactions on the truss represented in Fig. 29 due to the wind pressures shown on the left side (3,100, 6,200 and 3,100 pounds), the truss resting on rollers at the right end and being fastened to its support at the left.
(a) Let and R2 denote the left and right reactions. The direction of R, (at the roller end) is vertical, but the direction of is unknown. Imagine resolved into and replaced by its horizontal and vertical components and call them and respectively (See Fig. 35.) The six forces, R2 and the
three wind pressures are in equilibrium, and we may apply any one of the sets of statements of equilibrium for this kind of a system (see Art. 35) to determine the reactions. If we choose to use the first set we find, resolving forces along a horizontal line, R,' + 3,100 cos 55° + 6,200 cos 55° + 3,100 cos 55° = 0; resolving forces along a vertical line, d-R," R, 3,100 cos 35° 6,200 cos 35° 3,100 cos 35° = 0 ; taking moments about the left end, + 6,200 X 12.2 + 3,100 X 24.4 R, x 40 = 0. From the first equation, R1= 3,100 cos 55° + 6,200 cos 55° + 3,100 cos 55°= 7,113 pounds, 6,200 x 12.2 ± 3,100 x 24.4 R2= 40 = 3,782 pounds.
Substituting this value of R2 in the second equation we find that R, " =3,100 cos 35° + 6,200 cos 35° + 3,100 cos 35° 3,782 = 10,156 3,782 = 6,374 pounds.
If desired, the reaction R, can now be found by compounding its two components and R1'.
(b) Using the second set of conditions of equilibrium stated in Art. 35 we obtain the following three "equilibrium equa tions" : As in (1), resolving forces along the horizontal gives - R,' + 3,100 cos 55° + 6,200 cos 55° + 3,100 cos 55° = 0, Taking moments about the right end gives R," X 40 - 3,100 X a6 - 6,200 X16 - 3,100 X c6 = 0 Just as in (a), we find from the first and second equations the values of R,' and To find R," we need values of the arms (A and c6. By measurement from a drawing we find that Substituting these values in the third equation and solving for R," we find that 3,100 x 32.7 + 6,200 x 20.5 +3,100 x 8.3 = 6,355 pounds. 1 40(c) Using the third set of conditions of equilibrium stated in Art. 35 we obtain the following three equilibrium equations : As in (b), taking moments about the right and left ends we get