Question

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

Answer 1

Σ Forces in horizontal direction = 0, gives,

Joint A: ΣV = 0 → F_AB = 60 kN (Comp.) 

ΣH = 0 → F_AF = 20 kN (Tensile)

Joint E: ΣV = 0, gives, F_ED = 90 kN (Comp.) 

ΣH = 0, gives, F_EF = 0

Joint B: Noting that inclined member is at 45°

ΣV = 0; gives, 

– F_BF sin 45° – 30 + 60 = 0

Joint C: ΣV = 0, gives F_CF = 50 kN [Comp.] 

ΣH = 0, gives 30 – F_CD = 0 

or F_CD = 30 kN [Comp.]

Joint D: Noting that DF is at 45° as shown in Fig. (f) 

ΣV = 0 → 

– F_DF cos 45° + 90 – 40 = 0

F_DF = \(\frac{50}{cos\,45^\circ}\) = 70.71 kN [Tensile]

Check ΣH = 0, gives 

– F_DF cos 45° + 30 + 20 = 0 or F_DF = 70.71 kN checked. 

Final result is shown in Fig.(g)