Engineering Mechanics

(^270) „„„„„ A Textbook of Engineering Mechanics
Resolving the forces vertically and equating the same,
PAB sin 36.9° = 3
∴
33
5.0 kN (Compression)
AB sin 36.9 0.6
P ===
°
and now resolving the forces horizontally and equating the same,
PAD = 8 + PAB cos 36.9° = 8 + (5 × 0.8) = 12.0 kN (Tension)
Now consider the joint C. Let the directions of the forces PBC and PCD be assumed as shown
in Fig. 13.42 (b).
Resolving the forces vertically and equating the same,
PBC sin 36.9° = 9
99
15 kN (Compression)
sin 36.9 0.6
PBC===
°
and now resolving the forces horizontally and equating the same,
PCD = PBC cos 36.9° = 15 × 0.8 = 12.0 kN (Tension)
Now consider the joint D. A little consideration will show that the value of the force PBD will
be equal to the load 12 kN (Tension) as shown in Fig 13.42. (c). This will happen as the vertical
components of the forces PAD and PCD will be zero.
Now tabulate the results as given below :
S.No. Member Magnitude of force in kN Nature of force
1 AB 5.0 Compression
2 AD 12.0 Tension
3 BC 15.0 Compression
4 CD 12.0 Tension
5 BD 12.0 Tension
Example 13.12. 2 A truss of 8 metres span, is loaded as shown in Fig. 13.43.
Fig. 13.43.
Find the forces in the members CD, FD and FE of the truss.