Engineering Mechanics

(^262) „„„„„ A Textbook of Engineering Mechanics
Solution. The example may be solved either by the method of joints or method of sections.
But we shall solve it by the method of section only as we have to find out the force in member AB
only.
First of all, let us find out the force in the member AB
of the truss in terms of W. Now pass section (1-1) cutting the
truss through the members AB, BE and ED as shown in Fig.
13.28.
Now consider the equilibrium of the right part of the
truss. Let the direction PAB be assumed as shown in Fig 13.28.
Taking moments of the forces in the right part of the truss
only, about the joint E and equating the same,
PAB × 2 = (W × 1.5) + (W × 4.5) = 6 W
6
3
AB 2
W
PW==
Thus the value of W, which would produce the force of 15 kN in the member AB
15 5 kN
3
W
W
=×= Ans.
Example 13.8. Figure 13.29 shows a cantilever truss having a span of 4.5 meters. It is
hinged at two joints to a wall and is loaded as shown.
Fig. 13.29.
Find the forces in all the member of the truss.
Solution. 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 forces in all members of the truss.
Force in all the members of the truss
Fig. 13.30.
First of all, consider the joint D. Let the directions of PCD and PDE be assumed as shown in Fig.
13.30 (a).
Fig. 13.28.