Structures · MCQ practice Bending Moment and Shear Force MCQs – Part 3
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Part 3 of 5 Question 31 Easy
Pick the correct statement from the following:
A For a uniformly distributed load, the shear force varies linearly B For a uniformly distributed load, the bending moment curve is a parabola C For a load varying linearly, the shear force curve is a parabola D All options are correct
Answer: D. All options are correct
Each integration raises the degree by one: UDL (constant) → linear SF → parabolic BM; linear load → parabolic SF → cubic BM.
Question 32 Easy
Impact load results from which type of effect of the loads applied?
A Static B Dynamic C Static and dynamic D Neither static nor dynamic
Answer: B. Dynamic
Impact is the extra effect of a load applied suddenly or with velocity — a dynamic effect.
Question 33 Hard
A pile of length L with weight W per metre is lifted by two suspension points placed so that the bending moment at the centre equals that at the points of suspension. The maximum bending moment is:
A WL/8 B WL²/24 C WL²/47 D WL²/26
Answer: C. WL²/47
Suspension points at 0.207L from each end. Hogging there = W(0.207L)²/2 ≈ WL²/46.6 ≈ WL²/47, equal to the sagging at the centre.
Question 34 Easy
The general expression for the bending moment of a beam of length l is M = (wl/2)x − wx²/2. The beam carries:
A a uniformly distributed load w per unit length B a load varying linearly from zero at one end to w at the other C an isolated load at mid-span D None of these
Answer: A. a uniformly distributed load w per unit length
(wl/2)x is the moment of the reaction wl/2; wx²/2 is the moment of a UDL of length x. So it is a simply supported beam with a full UDL.
Question 35 Easy
A beam curved in plan is designed for:
A bending moment and shear B bending moment and torsion C shear and torsion D bending moment, shear and torsion
Answer: D. bending moment, shear and torsion
The load on a curved beam does not lie in the line joining its supports, so besides bending and shear it also twists the beam: design for all three.
Question 36 Medium
A simply supported beam of length a carries a load varying from zero at one end to w per unit length at the other. The maximum bending moment is:
A wa/27 B wa²/27 C w²a/√27 D wa²/(9√3)
Answer: D. wa²/(9√3)
R_A = wa/6. Shear is zero at x = a/√3; M_max = (wa/6)(a/√3) − (w/6a)(a/√3)³ = wa²/(9√3).
Question 37 Easy
A simply supported beam of span 3 m carrying a UDL of 10 kN/m has a shear force at mid-span of:
A 15 kN B 30 kN C 7.5 kN D Zero
Answer: D. Zero
R_A = 15 kN. V(1.5) = 15 − 10 × 1.5 = 0.
Question 38 Medium
The bending moment diagram of a beam ABCDE is zero over AB and DE, rises linearly from zero at B to a peak at C, and falls linearly back to zero at D (a symmetric triangle). The corresponding shear force diagram is:
A zero over AB and DE; a constant positive value over BC and an equal constant negative value over CD B a straight line falling from a positive value at B to a negative value at D, crossing zero at C C a constant positive value over the whole length AE D zero over BC and CD, with non-zero values only over AB and DE
Answer: A. zero over AB and DE; a constant positive value over BC and an equal constant negative value over CD
V = dM/dx. Flat BMD (AB, DE) ⇒ V = 0; constant rising slope (BC) ⇒ constant positive V; constant falling slope (CD) ⇒ constant negative V.
Question 39 Medium
A beam CAB carries a 20 kN downward load at its free end C. It is pinned at A, 100 mm from C, and rests on a roller at B, a further 100 mm from A. The reaction at support A is:
A 40 kN downward B 40 kN upward C 20 kN upward D 20 kN downward
Answer: B. 40 kN upward
ΣM about B: R_A × 100 = 20 × 200 ⇒ R_A = 40 kN upward. Then R_B = 20 − 40 = −20 kN, i.e. 20 kN downward. The book key prints (a).
Question 40 Easy
The point of contraflexure is the point where:
A bending moment changes sign B bending moment is maximum C bending moment is minimum D shear force is zero
Answer: A. bending moment changes sign
Contraflexure: the BM passes through zero and changes sign (sagging ↔ hogging); the curvature of the beam reverses there.
Question 41 Easy
Which of the following beams is likely to have a point of contraflexure?
A Cantilever beam B Simply supported beam C Beam with overhangs D None of these
Answer: C. Beam with overhangs
Overhangs cause hogging over the supports while the span sags, so the BM changes sign between them.
Question 42 Easy
In the design of a simply supported beam under a uniformly distributed load:
A check the section at the centre for both bending moment and shear B check the section at the centre for bending moment and at the support for shear C check at the centre for shear and at the support for bending moment D None of these
Answer: B. check the section at the centre for bending moment and at the support for shear
BM is maximum (wL²/8) at mid-span; shear is maximum (wL/2) at the supports.
Question 43 Easy
The slope of the shear force diagram at any section equals:
A the slope of the loading at the section B the ordinate of the loading diagram at the section C the area of the loading diagram from the end to that section D None of these
Answer: B. the ordinate of the loading diagram at the section
dV/dx = w: the slope of the SFD at a section equals the load intensity (ordinate of the load diagram) there.
Question 44 Medium
A simply supported beam of length a carries a load varying from zero at one end to w at the other. The shear force is zero at a distance x from the least loaded end, where x is:
A a/2 B a/3 C a/√3 D a√3/2
Answer: C. a/√3
R_A = wa/6. V = wa/6 − wx²/(2a) = 0 ⇒ x = a/√3 ≈ 0.577a.
Question 45 Easy
The maximum bending moment for a simply supported beam with a uniformly distributed load w per unit length is:
A wL/2 B wL²/4 C wL²/8 D wL²/12
Answer: C. wL²/8
M_max = (wL/2)(L/2) − (wL/2)(L/4) = wL²/8, at mid-span.
Part 3 of 5