Engineering Mechanics

Chapter 13 : Analysis of Perfect Frames (Analytical Method) „„„„„ 255

Now consider the equilibrium of the right part of the truss. Let the directions of P 1 , P 2 and P 3
be assumed as shown in Fig. 13.14.

Fig. 13.14.

Taking moments about joint M and equating the same,

P 1 × 2a sin θ = 375 × 2a

∴ 1

375 375

625 N

sin 0.6

P===

θ

(Compression)

Similarly, taking moments about joint A and equating the same,

P 2 × 2a = 375 × 4a = 1500a

∴ 2

1500

750 N

2

a

P

a

== (Tension)

and now taking moments about the joint L, and equating the same,

3

3

375 2 750

2

a

Paa×= ×=

∴ 3 750 500 N

1.5

P == (Tension)

Example 13.4. An inclined truss shown in Fig 13.15 is loaded as shown.

Fig. 13.15.
Determine the nature and magnitude of the forces in the members BC, GC and GF of the truss.
Solution. From the geometry of the figure, we find that the load 8 kN at B is acting at a
distance of 1.5 m from the joint A. Taking moments about A and equating the same,

RE × 6 = (8 × 1.5) + (6 × 2) + (12 × 4) = 72

∴

72

12 kN

E 6

R ==

RA = (8 + 6 + 12) – 12 = 14 kN