Engineering Mechanics

(^274) „„„„„ A Textbook of Engineering Mechanics
The example may be solved either by the method of joints or method of sections. But we shall
solve it by the method of joints, as we have to find out the forces in all the members of the truss.
First of all, consider the joint P. Let the directions of the forces PPR and PPT be assumed as
shown in Fig 13.48(a). We know that a horizontal force of 20 kN is acting at Q as shown in
Fig. 13.48 (b).
Fig. 13.48.
Resolving the forces vertically and equating the same,
PPR sin 60° = 17.3
∴
17.3 17.3
20 kN (Compression)
PR sin 60 0.866
P ===
°
and now resolving the forces horizontally and equating the same,
PPT = PPR cos 60° = 20 × 0.5 = 10 kN (Tension)
Now consider the joint Q. Let the directions of the forces PSQ and PQT be assumed as shown in
Fig. 13.48 (b). We know that a horizontal force of 20 kN is acting at Q as shown in Fig 13.48 (b).
Resolving the forces vertically and equating the same,
PSQ sin 30° = 17.3 – 10 cos 30° = 17.3 – (10 × 0.866) = 8.64
∴
8.64 8.64
17.3 kN (Compression)
SQ sin 30 0.5
P ===
°
and now resolving the forces horizontally and equating the same,
PQT = PSQ cos 30° + 20 – 10 sin 30°
= (17.3 × 0.866) + 20 – (10 × 0.5) = 30 kN (Tension)
Fig. 13.49.
Now consider the joint S. We have already found out that PSQ = 17.3 kN (Compression). A
little consideration will show that the value of the force PTS will be equal to the force 20 kN
(Compression). Similarly, the value of the force PRS will be equal to PSQ i.e., 17.3 kN (Compression)
as shown in Fig. 13.49 (a).