Question

Find the forces in all the members of the truss shown in Fig.(a).

Answer 1

Due to symmetry, the reactions are equal.

RA = RE = 1/2 x Total load

= 1/2 (15 + 30 + 30 + 30 + 15) = 60 kN

Drop perpendicular CH on AF. 

In ∆ACH , ∠ACH = 45° 

∴ FC is inclined at 30° to vertical i.e., 60° to horizontal and CH = 5 m 

It is not possible to find a joint where there are only two unknowns. Hence, consider section (1)–(1). 

For left hand side part of the frame:

Substituting this value of F_FC in (1), we get

F_FC = 71.4042 × √2 - 60

= 40.98 kN (Tension)

Assumed directions of F_BC and F_FC are correct. 

Therefore, F_BC is in compression and F_FC is in tension. 

Now we can proceed with method of joints to find the forces in other members. Since it is a symmetric truss, analysis of half the truss is sufficient. Other values may be written down by making use of symmetry. 

Joint B: 

∑ forces normal to AC = 0, gives

F_BF – 30 cos 45° = 0 

F_BF = 21.21 kN (Comp.) 

∑ forces parallel to AC = 0, gives 

F_AB – F_BC – 30 sin 45° = 0 

F_AB = 71.40 + 21.21 

= 92.61 kN (Comp) 

Joint A: ∑V = 0, gives 

F_AF sin 30° – F_AB sin 45° – 15 + 60 = 0 

F_AF = 40.98 kN (Tension) 

The results are tabulated below:

+ means tension and – means compression