(^286) A Textbook of Engineering Mechanics
and ∠FDB = ∠ACG = 180° – (46.1° + 60°) = 73.9°
Now at D, resolving the forces along BD and equating the same,
PBD = PED cos 83.4° + PFD cos 73.9°
....(The component of force PAD about BD is zero)
= (1.68 P × 0.1146) + (2.78 P × 0.2773)
= 0.963 P (Compression) Ans.
and at C resolving the forces along AC and equating the same,
PAC = PCH cos 83.4° + PCG cos 73.9°
....(The component of force PBC about AC is zero)
= (3.36 P × 0.1146) + (1.39 P × 0.2773)
= 0.772 P (Compression) Ans.
Taking moments about B and equating the same,
RA × 6 = (P × 5) + (2 P × 4) + (P × 2) + (2 P × 1) = 17 P
17
2.83
6
A
P
R ==P
Fig. 13.65.
Now pass section (1-1) cutting the truss into two parts as shown in Fig 13.65. Let us extend the
line AC and through D draw DL perpendicular to AC. From the geometry of the figure, we find that
DL=AD sin 30° = 5.196 × 0.5 = 2.598 m
Taking moments of the forces in the left part of the truss about D and equating the same,
2.83 P × 4.5 = (0.772 P × 2.598) + (P × 3.5)
- (2 P × 2.5) + (PFG × 2.598)
12.74 P= 10.5 P + (PFG × 2.598)
∴ 2.598 PIG = 12.74 P – 10.5 P = 2.24 P
or PFG=
2.24
2.598
P
=0.862 P (Tension) Ans.