Question

Determine the forces in all the members of the truss shown in Fig. (a) and indicate the magnitude and nature of forces on the diagram of the truss. All inclined members are at 60° to horizontal and length of each member is 2 m.

Answer 1

Now, we cannot find a joint with only two unknown forces without finding reactions. Consider the equilibrium of the entire frame. 

∑M_A = 0, gives

R_D × 4 – 40 × 1 – 60 × 2 – 50 × 3 = 0 

∴ R_D = 77.5 kN 

∑_H = 0, gives 

∴ H_A = 0

∴ Reaction at A is vertical only 

∑V = 0, gives 

R_A + 77.5 = 40 + 60 + 50 

∴ R_A = 72.5 kN Joint A: 

∑V = 0, gives 

F_AB sin 60° = R_A = 72.5

F_AB = 83.7158 kN (Comp.) 

∑ H = 0, gives 

F_AE – 83.7158 cos 60° = 0 

F_AE = 41.8579 kN (Tension)

∑V = 0, gives 

F_DC sin 60° = R_D = 77.5 

∴ F_DC = 89.4893 kN (Comp.) 

∑H = 0, gives 

F_DE – 87.4893 cos 60° = 0 

∴ F_DE = 44.7446 kN (Tension)

Joint B: ∑V = 0, gives 

F_BE sin 60° – F_AB sin 60° + 40 = 0

F_BE = \(\frac{72.5-40}{sin\,60^\circ}\) = 37 5278(Tension)

∑ H = 0, gives 

F_BC – F_AB cos 60° – F_BE cos 60° = 0 

F_BC = (83.7158 + 37.5274) × 0.5 

F_BC = 60.6218 kN (Comp.)

Joint C: ∑V = 0, gives 

F_CE sin 60° + 50 – F_DC sin 60° = 0

F_CE = \(\frac{72.5-50}{sin\,60^\circ}\) = 31.7543 kN (Tension)

Now the forces in all the members are known. If joint E is analysed it will give the check for the analysis. The results are shown on the diagram of the truss in Fig. (f).