# Structural Analysis Questions and Answers – Analysis of Externally Redundant Trusses

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This set of Structural Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Analysis of Externally Redundant Trusses”.

1. Externally redundant truss can be solved by three equilibrium equations.
a) True
b) False

Explanation: Redundant truss is solved by using equilibrium equations along with the displacement equilibrium.

2. Calculate the P value of the member AC for the given external redundant truss, considering the horizontal reaction of support B as a redundant force.

a) 1kN
b) 3kN
c) 5.77kN
d) 2.89kN

Explanation:

Va=Vb (Due to symmetry)
Va=Vb=5kN
At joint A,

∑V=Fac sin⁡ θ=5kN
Fac=$$\frac{5}{sin60}$$=5.77kN.

3. Calculate the P value of the member AC for the given external redundant truss, considering the horizontal reaction of support B as a redundant force.

a) 1kN
b) 3kN
c) 5.77kN
d) 2.89kN

Explanation:

Va=Vb (Due to symmetry)
Va=Vb=5kN
At joint A,

∑V=Fac sin⁡θ=5kN
Fac=$$\frac{5}{sin60}$$=5.77kN
Fac cos⁡θ=Fab=5.77*cos⁡60=2.89kN.

4. Calculate the P value of the member BC for the given external redundant truss, considering the horizontal reaction of support B as a redundant force.

a) 1kN
b) 3kN
c) 5.77kN
d) 2.89kN

Explanation:

Va=Vb (Due to symmetry)
Va=Vb=5kN
At joint B,

∑V=Fbc sin⁡θ=5kN
Fbc=$$\frac{5}{sin60}$$=5.77kN.

5. Calculate the k value of the member BC for the given external redundant truss, considering the horizontal reaction of support B as a redundant force.

a) 0kN
b) 1kN
c) 2.89kN
d) 3kN

Explanation: If the support reaction at support is considered 1kN rightward, the horizontal force at the joint will be balanced by horizontal force in member AB. Thus, member BC becomes a zero-force member.

6. Calculate the k value of the member AC for the given external redundant truss, considering the horizontal reaction of support B as a redundant force.

a) 0kN
b) 1kN
c) 2.89kN
d) 3kN

Explanation: If the support reaction at support A is considered 1kN leftward, the horizontal force at the joint will be balanced by horizontal force in member AB. Thus, member AC becomes a zero-force member.

7. Calculate the k value of the member AB for the given external redundant truss, considering the horizontal reaction of support B as a redundant force.

a) 0kN
b) 1kN
c) 2.89kN
d) 3kN

Explanation: If the support reaction at support is considered 1kN leftward, the horizontal force at the joint will be balanced by horizontal force in member AB. Thus, member AC becomes a zero-force member.

8. Calculate the redundant horizontal forces at support B, for the given truss considering the horizontal reaction of support B as a redundant force.

a) 1kN
b) 5kN
c) 3kN
d) 2.89kN

Explanation:

Member P values k values Length of Member $$\frac{P k L}{AE}$$ $$\frac{∑k^2 L}{AE}$$
AB 2.89 1 3 8.67 3
BC -5.77 0 3 0 0
AC -5.77 0 3 0 0

Redundant horizontal force at support B=$$\frac{- ∑\frac{P k L}{AE}}{\frac{∑k^2 L}{AE}}$$ = -2.89kN

9. Calculate the force in the member AB for the given truss.

a) 0kN
b) 2.89kN
c) 5.78kN
d) 10kN

Explanation:

Member P values k values Length of Member $$\frac{P k L}{AE}$$ $$\frac{∑k^2 L}{AE}$$
AB 2.89 1 3 8.67 3
BC -5.77 0 3 0 0
AC -5.77 0 3 0 0

Redundant horizontal force at support B=$$\frac{- ∑\frac{P k L}{AE}}{\frac{∑k^2 L}{AE}}$$ = 2.89kN(leftward)
Considering, joint B

∑H=0
2.89-2.89+Fab=0
Fab=0 kN.

10. Externally redundant trusses are always internally determinate.
a) True
b) False