Strength of Materials Questions and Answers – Introduction to Shear Stress

This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Introduction to Shear Stress”.

1. At ________ the shearing stress in a beam are maximum.
a) Extreme fibres
b) Modulus of section
c) Neutral axis
d) Along the cross-sectional area
View Answer

Answer: c
Explanation: Shearing stress in a beam is maximum at the neutral axis. Shearing stress is defined as the resistance offered by the internal stress to the shear force.

2. Determine the shear stress at the level of neutral axis, if a beam has a triangle cross section having base “b” and altitude “h”. Let the shear force be subjected is F.
a) 3F/8bh
b) 4F/3bh
c) 8F/3bh
d) 3F/6bh
View Answer

Answer: c
Explanation: For a triangular section subjected to shear force the shear stress in neutral axis is
Shear stress at NA = 4/3 [Average shear stress].
= 4/3 [F/0.5 bh] = 8F / 3bh.

3. The maximum shear stress is ______ times the average shear stress [For rectangular beams].
a) 2.5
b) 3
c) 1.2
d) 1.5
View Answer

Answer: d
Explanation: The maximum shear stress occurs at neutral axis. Then y = 0.
Max shear stress = 3F/2bd = 3/2 [F/bd].
= 1.5 Average shear stress.
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4. Shear stress in a beam is zero at ______
a) Neutral axis
b) Extreme fibres
c) Cross section
d) Junctions
View Answer

Answer: b
Explanation: The resistance offered by the internal stress to shear is known as shearing stress. Shearing stress is zero at extreme fibres of the beam. The bending stresses are maximum at extreme fibres of the beam cross section.

5. Shear stress distribution over rectangular section will be _________
a) parabolic
b) elliptical
c) triangular
d) trapezoidal
View Answer

Answer: a
Explanation:
Maximum shear stress is 1.5 times that of average shear stress
Maximum shear stress is 1.5 times that of average shear stress.
The shear stress distribution is parabolic.

6. A round Steel rod of 100 mm diameter is bent into an arc of radius 100m. What is the maximum stress in the rod? Take E = 2×105 N/mm2.
a) 150 N/mm2
b) 200 N/mm2
c) 100 N/mm2
d) 300 N/mm2
View Answer

Answer: c
Explanation: D = 100m
y = 50 mm
R = 10 × 103mm
By equation of flexure; E/R = f/y
f=E/R ×y
= 2×105/ 100 × 103× 50
= 100 N/mm2.

7. For circular section, the maximum shear stress is equal to ____________ times of average shear stress.
a) 2/3
b) 3/2
c) 4/3
d) 3/4
View Answer

Answer: c
Explanation: Maximum shear stress occurs at neutral axis & y = 0.
Max. Shear stress = 4/3 [ F/A ].
F/A is average shear stress.
The maximum shear stress distribution is 33% more than average shear stress.
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8. A steel beam is 200 mm wide and 300 mm deep. The beam is simply supported and carries a concentrated load w. If the maximum stress are 2 N/mm2. What will be the corresponding load?
a) 50 kN
b) 80 kN
c) 40 kN
d) 85 kN
View Answer

Answer: b
Explanation: For a rectangular cross section
Max. Shear stress = 3/2 [F/A ] 2 = 3/2 [w/200 × 300].
w = 80 kN.

9. Maximum shear stress in thin cylindrical shell be ___________
a) pr/2t
b) pr/3t
c) pr/4t
d) pr/ 5t
View Answer

Answer: c
Explanation: Hoop stress P(h) = pr/t
Longitudinal stress P(l) = pr/2t
Thus, hoop stress is twice the longitudinal stress
Max. Shear stress = P(h) – P(l) / 2
= pr/4t.
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10. Circumferential stress is same as of _________
a) Hoop stress
b) Longitudinal stress
c) Transverse stress
d) Phreatic stress
View Answer

Answer: a
Explanation: In a thin cylindrical shell of internal radius r thickness t when subjected to internal fluid pressure P, the stress developed in the internal walls can be termed as circumferential stress or hoop stress.
P(h) = pr/t.

Sanfoundry Global Education & Learning Series – Strength of Materials.

To practice all areas of Strength of Materials, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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