Question

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

Answer 1

AB = 6 sin 60° = 3 m; 

BD = 6 cos 30° = 23.09 m 

In ∆ABC, AC = AB = 3 m 

∴ ∠ABC = ∠BCA = say θ 

∴ 2θ = 180 – 60 = 120 

or θ = 60°

Taking moment about A, we get 

R_D × 6 – 10 × 3 – 20 × 3 cos 60° = 0 

RD = 10 kN 

∴ R_A = 20 + 10 – 10 = 20 kN

[Note: H₄ = 0]

Joint A: 

Referring to Fig. (b) 

ΣV = 0, gives 

20 – F_AB sin 60° = 0 

∴ F_AB = 23.09 kN [Comp.] 

ΣH = 0, gives 

F_AC – F_AB cos 60° = 0 

∴ F_AC = 23.09 cos 60° = 11.55 kN [Tensile]

Joint D: 

ΣV = 0, gives 

F_DB sin 30° = 10 or F_DB = 20 kN [Comp.] 

ΣH = 0, gives 

– F_DC + F_DB cos 30° = 0

i.e., F_DC = 20 cos 30° = 17.32 kN [Tensile]

Joint C:

ΣV = 0, gives 

F_CB sin 60° = 10 

∴ F_CB = 11.55 kN 

check ΣH = 0, gives 

F_CB cos 60° = F_CD – F_CA = 17.32 – 11.55 

∴ F_CB = 11.55, Checked

These forces are marked in Fig. (e).