Engineering Mechanics

Chapter 13 : Analysis of Perfect Frames (Analytical Method) „„„„„ 253

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

PBE × 3 sin 60° = (PBC × 3 sin 60°) – (2 × 1.5) = (1.732 × 3 × 0.866) – 3.0 = 1.5

1.5 1.5

0.577 kN (Tension)

BE 3 sin 60 3 0.866

P == =

°×

Now pass section (3-3) cutting the truss through the members BC, CE and ED. Now consider
the equilibrium of the right part of the truss. Let the directions of the forces PCE and PDE be assumed
as shown in Fig. 13.12 (a) Taking moments of the forces in the right part of the truss only, about the
joint D and equating the same,

PCE × 3 sin 60° = (4 × 1.5) – (PBC × 3 sin 60°)

= 6.0 – (1.732 × 3 × 0.866) = 1.5

∴

1.5 1.5

0.577 kN

CE 3 sin 60 3 0.866

P == =

°×

(Compression)

and now taking moments of the forces in the right part of the truss only about the joint C and equating
the same,

PDE × 3 sin 60° = 3.5 × 1.5 = 5.25

∴

5.25 5.25

2.021 kN

DE 3 sin 60 3 0.866

P == =

°×

(Tension)

(a) Section (3–3) (b) Section (4–4)
Fig. 13.12.
Now pass section (4-4) cutting the truss through the members CD and DE. Let the directions of
the forces PCD be assumed as shown in Fig 13.12 (b). Taking moments of the forces acting in the right
part of the truss only about the joint E and equating the same,

PCD × 3 sin 60° = 3.5 × 3

3.5 3.5

4.042 kN

sin 60 0.866

PCD===

°^

(Compression)