Engineering Mechanics

(^252) „„„„„ A Textbook of Engineering Mechanics
Resolve the forces vertically and equating the same,
PBE sin 60° = PAB sin 60° – 2.0 = 2.887 × 0.866 – 2.0 = 0.5 kN
∴
0.5 0.5
0.577 kN
sin 60 0.866
PBE===
°
(Tension)
and now resolving the forces horizontally and equating the same,
PBC = 2.887 cos 60° + 0.577 cos 60° kN
= (2.887 × 0.5) + (0.577 × 0.5) kN = 1.732 kN (Compression)
Now consider joint C. We have already found out that the forces in the members BC and CD
(i.e. PBC and PCD) are 1.732 kN (Compression) and 4.042 kN (Compression) respectively. Let the
directions of PCE be assumed as shown in Fig. 13.10 (b). Resolving the forces vertically and equating
the same,
PCE sin 60° = 4 – PCD sin 60° = 4 – (4.042 × 0.866) = 0.5
∴
0.5 0.5
0.577 kN (Compression)
CE sin 60 0.866
P ===
°
Method of sections
First of all, pass section (1-1) cutting the truss through the members AB and AE. Now consider
equilibrium of the left part of the truss. Let the directions of the forces PAB and PAE be assumed as
shown in Fig. 13.11 (a).
(a) Section (1-1) (b) Section (2-2)
Fig. 13.11.
Taking moments of the forces acting in the left part of the truss only, about the joint E and
equating the same,
PAB × 3 sin 60° = 2.5 × 3
2.5 2.5
2.887 kN (Compression)
sin 60 0.866
PAB===
°
Now pass section (2-2) cutting the truss through the members BC, BE and AE. Now consider
equilibrium of the left of the truss. Let the directions of the forces PBC and PBE be assumed as shown
in Fig. 13.11 (b). Taking moments of the forces acting in left part of the truss only, about the joint E
and equating the same,
PBC × 3 sin 60° = (2.5 × 3) – (2 × 1.5) = 4.5
∴
4.5 4.5
1.732 kN (Compression)
BC 3 sin 60 3 0.866
P == =
°×