Engineering Mechanics

Chapter 13 : Analysis of Perfect Frames (Analytical Method) „„„„„ 279

50 5

()

30 3

B

WW

V ==↑

and

52

–()

A 33

WW

VW==↓

First of all, pass section (X-X) cutting the truss into two parts and consider the equilibrium of
the left part of the truss as shown in Fig. 13.55 (b). Now let the directions of the forces P 1 , P 2 , P 3 and
P 4 be assumed as shown in Fig 13.55 (b). First of all, let us consider the joint B. A little consideration
will show that the magnitude of the force P 4 will be equal and opposite to the reaction VB i.e., 5W/3
(Compression). This will happen as the vertical components of the horizontal members at B will be
zero.
Resolving the forces vertically and equating the same,

2

2

cos 45

3

W

P ×°=

or 2

21 2

3 cos 45 3 0.707

WW
P =× =
°× = 0.943 W (Compression)
Taking moments of the forces acting on the left part of the truss only about the joint E and
equating the same,

1

240

10 20

33

WW

P×= × =

∴ 1

40 1 4

(Compression)

3103

WW

P=×= Ans.

and now taking moments of the forces acting on the left part of the truss only about the joint F and
equating the same,

3

2

10 30 20

3

W

PW×= × =

∴ 3

20

2 (Tension)

10

W

PW== Ans.

Example 13.18. A pin-jointed frame shown in Fig 13.56 is hinged at A and loaded at D. A
horizontal chain is attached to C and pulled so that AD is horizontal.

Fig. 13.56.
Determine the pull in the chain and also the force in each member. Tabulate the results.
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 the force in each member.