Test: Trusses -1 - Civil Engineering (CE) MCQ

At joint D, there is no external force so force in FD and BD is zero.
Taking moment about A, the reaction at B,

Considering joint B, force in member BF,

force in member BE,
F_BE=1t (tensile)

Considering joint equilibrium at D. The force equilibrium equation in vertical direction gives FD = 50 kN tensile

Only a horizontal reaction can exist at A.

Taking moments about hinge.
H_A x 5 = 10 x 15 
H_A = 30t

By considering joint equilibrium at B,

Force in EF is zero if joint equilibrium at E is considered force in SE will be tensile, and its value will be 10 kN by considering joint equilibrium at B.

The given truss is symmetrical. So force in member BO and EO will be same in magnitude and nature both. Joint equilibrium at O.

Cutting a section through 1-2; 2-5; and 5-6. Taking moment of left part about 5,

Cut a section through 1-2, 1-5 and 4-5 and balance vertical force for left part only,

Cutting a section through AC, CD and DF and taking moment of upper portion about C.The forces in the members AC and CD meet at C so they do not produce any moment. The force in member DF,