# Strength of Materials Questions and Answers – Shear Force and Bending Moment diagram

This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Shear Force and Bending Moment diagram”.

1. What is the bending moment at end supports of a simply supported beam?
a) Maximum
b) Minimum
c) Zero
d) Uniform

Explanation: At the end supports, the moment (couple) developed is zero, because there is no distance to take the perpendicular acting load. As the distance is zero, the moment is obviously zero.

2. What is the maximum shear force, when a cantilever beam is loaded with udl throughout?
a) w×l
b) w
c) w/l
d) w+l

Explanation: In cantilever beams, the maximum shear force occurs at the fixed end. In the free end, there is zero shear force. As we need to convert the udl in to load, we multiply the length of the cantilever beam with udl acting upon. For maximum shear force to obtain we ought to multiply load and distance and it surely occurs at the fixed end (w×l).

3. Sagging, the bending moment occurs at the _____ of the beam.
a) At supports
b) Mid span
c) Point of contraflexure
d) Point of emergence

Explanation: The positive bending moment is considered when it causes convexity downward or concavity at top. This is sagging. In simply supported beams, it occurs at mid span because the bending moment at the supports obviously will be zero hence the positive bending moment occurs in the mid span.

4. What will be the variation in BMD for the diagram? [Assume l = 2m].

a) Rectangular
b) Trapezoidal
c) Triangular
d) Square

Explanation: At support B, the BM is zero. The beam undergoes maximum BM at fixed end.
By joining the base line, free end and maximum BM point. We obtain a right angled triangle.

5. What is the maximum bending moment for simply supported beam carrying a point load “W” kN at its centre?
a) W kNm
b) W/m kNm
c) W×l kNm
d) W×l/4 kNm

Explanation: We know that in simply supported beams the maximum BM occurs at the central span.
Moment at A = Moment at B = 0
Moment at C = W/2 × l/2 = Wl/ 4 kNm (Sagging).
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6. How do point loads and udl be represented in SFD?
a) Simple lines and curved lines
b) Curved lines and inclined lines
c) Simple lines and inclined lines
d) Cant represent any more

Explanation: According to BIS, the standard symbols used for sketching SFD are

7. ________ curve is formed due to bending of over hanging beams.
a) Elastic
b) Plastic
c) Flexural
d) Axial

Explanation: The line to which the longitudinal axis of a beam bends or deflects or deviates under given load is known as elastic curve on deflection curve. Elastic curve can also be known as elastic line or elastic axis.

8. The relation between slope and maximum bending moment is _________
a) Directly proportion
b) Inversely proportion
c) Relative proportion
d) Mutual incidence

Explanation: The relationship between slope and maximum bending moment is inversely proportional because, For example in simply supported beams slope is maximum at supports and zero at midspan of a symmetrically loaded beam where as bending moment is zero at supports and maximum at mid span. Hence we conclude that slope and maximum bending moment are inversely proportional to each other in a case of the simply supported beam.

9. What is the SF at support B?

a) 5 kN
b) 3 kN
c) 2 kN
d) 0 kN

Explanation: Total load = 2×2 = 4kN
Shear force at A = 4 kN ( same between A and C )
Shear force at C = 4 kN
Shear force at B = 0 kN
Maximum SF at A = 4 kN.

10. Where do the maximum BM occurs for the below diagram.

a) -54 kNm
b) -92 kNm
c) -105 kNm
d) – 65 kNm

Explanation: Moment at B = 0
Moment at C = – (10 × 3) × (3/2)
= – 45 kNm
Moment at A = – (10 × 3) × (1.5 + 2 )
Maximum BM at A = – 105 kNm
= 105 Nm (hogging).

Sanfoundry Global Education & Learning Series – Strength of Materials.

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