0.20(76.6 - 0.259P) - 15.32 - 0.052P. Substituting for Rx from step 4 yields 64.3 -
0.966P - 15.32 - 0.052P; so P = 53.6 Ib (238.4 N).
- Draw a second free-body diagram
In Fig. 2c, draw a free-body diagram of the bar, with Rx being directed downward.
- Solve as In steps 1 through 5
As before, Ry = 76.6 - 0.259P. Also the absolute value of Rx = 0.966P - 64.3. But Rx =
0.20^, = 15.32 x 0.052P. Then 0.966P - 64.3 = 15.32 - 0.052P; so P = 78 2 Ib (347 6N).
ANALYSIS OFA STRUCTURAL FRAME
The frame in Fig. 30 consists of two inclined members and a tie rod. What is the tension
in the rod when a load of 1000 Ib (4448.0 N) is applied at the hinged apex? Neglect the
weight of the frame and consider the supports to be smooth.
Calculation Procedure:
- Draw a free-body diagram of
the frame
Since friction is absent in this frame, the
reactions at the supports are vertical.
Draw a free-body diagram as in Fig. 3b.
With the free-body diagram shown,
compute the distances X 1 and Jc 2. Since
the frame forms a 3-4-5 right triangle, X 1
= 16(4/5) = 12.8 ft (3.9 m) and X 2 =
12(3/5) -7.2 ft (2.2m).
- Determine the reactions on
the frame
Take moments with respect to A and B to
obtain the reactions:
^MB = 20/^-1000(7.2) = O
SM 4 = 1000(12.8) - 20RR = O
R 1 = 360 Ib (1601.2 N)
RR = 640 Ib (2646.7 N)
- Determine the distance y in
Fig. 3c
Draw a free-body diagram of member A C
in Fig. 3c. Compute y = 13(3/5) = 7.8 ft
(2.4 m).
4. Compute the tension in the
tie rod
Take moments with respect to C to find
the tension Tin the tie rod:
^Mc= 360(12.8) -7.8f= O
T= 591 Ib (2628.8 N) FIGURE 3
Tie rod
