Engineering Mechanics

(^260) „„„„„ A Textbook of Engineering Mechanics
Resolving the forces vertically and equating the same,
PAB sin 60° = 10
∴
10 10
11.5 kN (Tension)
AB sin 60 0.866
P ===
°
and now resolving the forces horizontally and equating the same,
PAD = PAB cos 60° = 11.5 × 0.5 = 5.75 kN (Compression)
Fig. 13.25.
Now consider the joint B. Let the directions of PBD and PBC be assumed as shown in Fig
13.25 (b). We have already found out that the force in member AB is 11.5 kN (Tension) as shown
in the figure 13.25 (b). Resolving the forces vertically and equating the same,
PBD sin 60° = PAB sin 60° = 11.5 sin 60°
∴ PBD = PAB = 11.5 kN (Compression)
and now resolving the forces horizontally and equating the same,
PBC = PAB cos 60° + PBD cos 60°
= (11.5 × 0.5) + (11.5 × 0.5) = 11.5 kN (Tension)
Method of sections
First of all, pass section (1-1) cutting the truss through the members AB and AD. Now consider
the equilibrium of the right part of the truss. Let the directions of the forces PAB and PAD be assumed
as shown in Fig 13.26 (a).
Fig. 13.26.