Engineering Mechanics

(^280) „„„„„ A Textbook of Engineering Mechanics
Pull in the chain
Let P = Pull in the chain.
Taking moments about the joint A and equating the same,
P × 0.9 = 2 cos 45° × 1.2 = 2 × 0.707 × 1.2 = 1.7
∴ 1.7 1.889 kN
0.9
P==^ Ans.
Force in each member
We know that horizontal reaction at A,
HA = 1.889 – (2 cos 45°) = 1.889 – (2 × 0.707) = 0.475 kN (→)
and vertical reaction at A,
VA = 2 sin 45° = 2 × 0.707 = 1.414 kN (↑)
Fig. 13.57.
First of all, consider the joint A. Let the directions of the forces PAB and PAD be assumed as
shown in Fig 13.57 (a). We have already found out that zthe horizontal and vertical reactions at A are
0.475 kN and 1.414 kN repectively as shown in the figure.
Resolving the forces vertically and equating the same,
PAB sin 30° = 1.414
1.414 1.414
2.828 kN (Compression)
AB sin 30 0.5
P ===
°
and now resolving the forces horizontally and equating the same,
PAD = PAB cos 30° – 0.475 = (2.828 × 0.866) – 0.475
= 1.974 kN (Tension)
Now consider the joint D. Let the directions of the forces PBD and PCD be assumed as shown in
Fig 13.57 (b). We have already found out that PAD = 1.974 kN (Tension) as shown in the figure.
Resolving the forces horizontally and equating the same,
PBD cos 60° = 1.974 – 2 cos 45° = 1.974 – (2 × 0.707) = 0.56 kN
∴
0.56 0.56
1.12 kN (Compression)
BD cos 60 0.5
P ===
°
and now resolving the forces vertically and equating the same,
PCD = PBD sin 60° + 2 sin 45°
= (1.12 × 0.866) + (2 × 0.707) = 2.384 kN (Tension)
Now consider the triangle BCD. From B, draw BE perpendicular to CD. Let the direction of
PBC be assumed as shown in Fig 13.57 (c).