Engineering Mechanics

(^264) „„„„„ A Textbook of Engineering Mechanics
and now resolving the forces vertically and equating the same,
PBF = 1584 sin 18.4° = 1584 × 0.3156 = 500 N (Tension)
Now consider the joint B. Let the direction of PBG and PAB be assumed as shown in Fig 13.32.
Fig. 13.32.
From the geometry of the figure, we find that
1.5
tan 1.5
1
∠==GBF or ∠GBF = 56.3°
Resolving the forces horizontally at B and equating the same,
PAB cos 18.4º = PBG sin 56.3° + 3168 cos 18.4°
PAB × 0.9488 = PBG × 0.832 + 3168 × 0.9488
∴ 0.9488 PAB = 0.832 PBG + 3000 ....(ii)
Dividing the above equation by 3,
0.3156 PAB = 0.2773 PBG + 1000 ....(iii)
and now resolving the forces vertically at B and equating the same,
PAB sin 18.4° + PBG cos 56.3° = 1000 + 500 + 3168 sin 18.4°
= 1500 + (3168 × 0.3156)
PAB × 0.3156 + PBG × 0.5548 = 1500 + 1000
0.3156 PAB + 0.5548 PBG= 2500 ...(iv)
Substracting equation (iii) from equation (iv),
0.8321 PBG = 1500
or
1500
1801 N (Compression)
BG 0.8321
P ==
Substituting the value of PBG in equation (iii)
0.3156 PAB = (0.2773 × 1801) + 1000
0.3156 PAB = 500 + 1000 = 500
1500
4753 N (Tension)
AB 0.3156
P ==