tensile. Equate the sum of the forces along each axis to zero. For instance, if the truss
members are in tension, the x components of these forces are directed to the left, and ^Fx
= 4000 -ABx -ACx -ADx = O.
Express each component in terms of the total force to obtain ^Fx = 4000 - 0.26&45 -
0.249^C - 0.707'AD = O; 2Fy = -0.894,45 - 0.83IAC - 0.707'AD = O; 2FZ = 0.358,45
-0.498,4C = O.- Solve the simultaneous equations in step 4 to evaluate the
forces in the truss members
A positive result in the solution signifies tension; a negative result, compression. Thus,
AB = 3830-lb (17,036-N) compression; AC = 2750-lb (12,232-N) compression; and AD =
8080-lb (35,940-N) tension. To verify these results, it is necessary to select moment axes
yielding equations independent of those previously developed. - Resolve the reactions into their components
In Fig. 1 Ib, show the reactions at the supports B, C, and D, each reaction being numeri-
cally equal to and collinear with the force in the member at that support. Resolve these re-
actions into their components. - Take moments about a selected axis
Take moments with respect to the axis through C parallel to the x axis. (Since the x com-
ponents of the forces are parallel to this axis, their moments are zero.) Then 2MCjc =
IQABy - 6ADy = 10(0.894)(3830) - 6(0.707)(8080) = O. - Take moments about another axis
Take moments with respect to the axis through D parallel to the x axis. So ^M0x = 4ABy —
6ACy = 4(0.894)(3830) - 6(0.831)(2750) = O.
The computed results are thus substantiated.
ANALYSIS OFA COMPOUND SPACE TRUSS
The compound space truss in Fig. Via has the dimensions shown in the orthographic pro-
jections, Fig. I2b and c. A load of 5000 Ib (22,240 N), which lies in the xy plane and
makes an angle of 30° with the vertical, is applied at A. Determine the force induced in
each member, and verify the results.
Calculation Procedure:- Compute the true length of each truss member
Since the truss and load system are symmetric with respect to the xy plane, the internal
forces are also symmetric. As one component of an internal force becomes known, it will
be convenient to calculate the other components at once, as well as the total force.
Record in Table 4 the length of each member as projected on the coordinate axes. Cal-
culate the true length of each member, using geometric relations. - Resolve the applied load into Its x and y components
Use only the absolute values of the forces. Thus Px = 5000 sin 30° = 2500 Ib (11,120 N);
Py = 5000 cos 30° = 4330 Ib (19,260 N). - Compute the horizontal reactions
Compute the horizontal reactions at D and at line CC (Fig. 126). Thus 2MCC = 4330(12)
- 2500(7) - 10//! = O; H 1 = 3446 Ib (15,328 N); H 2 = 3446 - 2500 = 946 Ib (4208 N).