# Structural Analysis Questions and Answers – Analysis of Beams-5

This set of Structural Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Analysis of Beams-5”.

A and B are fixed supports. B and C are roller supports. Roller support at point C is pushed downward by 0.1 m. AB = 24m, BC = 20m and CD = 15m.
All moment options are in KNM.
Take EI as constant.

1. How many separate parts will be required for this question?
a) 0
b) 1
c) 2
d) 3

Explanation: Since, 4 supports are there we will divide it into 2 separate parts to solve.

2. What is FEM of point B in beam BC?
a) 7.2
b) -7.2
c) 10.8
d) 0

Explanation: No force is acting on beam BC.

3. What is FEM of point C in beam BC?
a) 7.2
b) 0
c) 10.8
d) -10.8

Explanation: No force is acting on beam BC.

4. What will be the FEM at point A in beam AB?
a) 453.1
b) -453.1
c) 72
d) -72

Explanation: Formula for moment for unit load applied at mid point is WL2.12 and direction will be anti clockwise.
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5. What will be the FEM at point B in beam AB?
a) 453.1
b) -453.1
c) 72
d) -72

Explanation: Formula for moment for unit load applied at mid point is WL^2.12 and direction will be clockwise.

6. What will be the deflection of beam AB?
a) 0
b) 1/EI
c) 2/EI
d) can’t say

Explanation: As, point A is a fixed support, it won’t allow any deflection to take place.

7. What will be the deflection of beam BC?
a) -.1
b) .1
c) 0.005
d) -0.005

Explanation: Deflection will be 0.1 and it will be +ve as it will be rotating clockwise.

8. What will be the rotation of beam AB at point A?
a) 0.2
b) -0.2
c) 0
d) can’t say

Explanation: There won’t be any rotation at point A as it is a fixed support.

9. What will be the rotation of beam CD at point D?
a) 0.2
b) -0.2
c) 0
d) can’t say

Explanation: There won’t be any rotation at point C as it is a fixed support.

10. After using all the joint conditions, how many unknowns are still left?
a) 0
b) 1
c) 2
d) 3

Explanation: Rotation at point B and at point C in either of beam is not known (they are both equal).

11. What will be one of the extra condition, which we will get if we conserve moment near joint B?
a) mBA + mCA = 0
b) mBA + mCB = 0
c) mBA + mBC = 0
d) mAB + mBC = 0

Explanation: Just cut a small part near joint B and conserve moment around that joint.

12. Total how many equations will be generated?
a) 3
b) 4
c) 5
d) 6

Explanation: Three separate fixed beams are considered. So, there will be a total of 6 equations.

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