Engineering Mechanics

(^256) „„„„„ A Textbook of Engineering Mechanics
The example may be solved by any method. But we shall solve it by the method of sections, as
one section line can cut the members BC, GC, and GF in which the forces are required to be found
out. Now let us pass section (1-1) cutting the truss into two parts as shown in Fig. 13.16
Fig. 13.16.
Now consider equilibrium of the left part of the truss. Let the directions of the force PBC, PGC and
PGF be assumed as shown in Fig 13.16. Taking moments about the joint G and equating the same,
PBC × 2 sin 30° = (14 × 2) – (8 × 0.5) = 24
∴
24 24
24 kN
BC 2sin30 2 0.5
P ===
°×
(Compression)
Similarly, taking moments about the joint B and equating the same,
PGC × 1 cos 30° = (14 × 1.5) + (6 × 0.5) = 24 kN
24 24
27.7 kN
GC cos 30 0.866
P ===
°
(Compression)
and now taking moments about the joint C and equating the same,
PGF × 3 tan 30° = (14 × 3) – (6 × 1) = 36
∴
36 12
20.8 kN
GF 3 tan 30 0.5774
P ===
°
(Tension)
Example 13.5. A framed of 6 m span is carrying a central load of 10 kN as shown in
Fig. 13.17.
Fig. 13.17.
Find by any method, the magnitude and nature of forces in all members of the structure and
tabulate the results.