Engineering Mechanics

Chapter 13 : Analysis of Perfect Frames (Analytical Method) „„„„„ 261

Taking moments of the forces acting on right part of the truss only, about the joint D and
equating the same,
PAB × 3 sin 60° = 10 × 3

∴

10 10

11.5 kN (Tension)

sin 60 0.866

PAB===
°
and now taking moments of the forces in the right part of the truss only about the joint B and equating
the same,
PAD × 3 sin 60° = 10 × 1.5 = 15

∴

15 15
5.75 kN (Compression)
AD 3sin 60 3 0.866
P == =
°×
Now pass section (2-2) cutting the truss through the members BC, BD and AD. Now consider
the equilibrium of the right part of the truss. Let the directions of the forces PBC and PBD be assumed
as shown in Fig. 13.26 (b)

Taking moments of the forces acting on the right part of the truss only, about the joint D and
equating the same,

PBC × 3 sin 60° = 10 × 3

∴

10 10
11.5 kN (Tension)
BC sin 60 0.866
P ===
°
and now taking moments of the forces in the right part of the truss only, about the joint C and equating
the same,
PBD × 1.5 sin 60° = (10 × 3) – PAD × 3 sin 60° = 30 – (5.75 × 3 × 0.866) = 15
15 15
11.5 kN (Compression)
BD 1.5 sin 60 1.5 0.866
P == =
°×
Now tabulate the results as given below :

S.No. Members Magnitude of force in kN Nature of force

1 AB 11.5 Tension

2 AD 5.75 Compression

3 BD 11.5 Compression

4 BC 11.5 Tension

Example 13.7. A cantilever truss is loaded as shown in Fig 13.27.

Fig. 13.27.

Find the value W, which would produce the force of magnitude 15 kN in the member AB.