Engineering Questions with Answers - Multiple Choice Questions
MCQs on Pure Bending Stress
1 - Question
In simply supported beams, the _____ stress distribution is not uniform.
a) Bending
b) Shearing
c) Tensile
d) Compressive
View Answer
Answer: a
Explanation: In a simply supported beam, there is compressive stress above the neutral axis and tensile stress below it. It bends with concavity upwards. Hence the bending stress distribution is not uniform over the section.
2 - Question
The maximum _________ stresses occur at top most fibre of a simply supported beam.
a) Tensile
b) Compressive
c) Shear
d) Bending
View Answer
Answer: b
Explanation: As bending stress distribution is not uniform over the section in simply supported beams, the maximum compressive stress lies above the neutral axis. Obviously, top most fibre of beam. The maximum tensile stress occurs at bottom most fibre.
Explanation: As bending stress distribution is not uniform over the section in simply supported beams, the maximum compressive stress lies above the neutral axis. Obviously, top most fibre of beam. The maximum tensile stress occurs at bottom most fibre.
3 - Question
The stress is directly proportional to _______
a) E
b) u
c) y
d) R
View Answer
Answer: c
Explanation: By two equations; we have e = y/R & e = f/E
Equating both equations; we get e = f/E = y/R
Hence stress (f) is directly proportional to the distance from neutral axis(y).
4 - Question
At the extreme fibre, bending stress is______
a) Minimum
b) Zero
c) Constant
d) Maximum
View Answer
Answer: d
Explanation: Bending stress is defined as the resistance offered by internal stress to bending. In beams, stresses occurs above or below the neutral axis i.e at the extreme fibres. Hence bending stress is maximum at the extreme fibres.
Explanation: Bending stress is defined as the resistance offered by internal stress to bending. In beams, stresses occurs above or below the neutral axis i.e at the extreme fibres. Hence bending stress is maximum at the extreme fibres.
5 - Question
The curvature of a beam is equal to _____
a) EI/M
b) M/E
c) M/EI
d) E/MI
View Answer
Answer: c
Explanation: From the bending equation, E/R = M/I = f/y.
Where R is called “radius of curvature “
1/R is called “curvature of the beam “.
So, 1/R = M/EI.
So curvature of the beam is directly proportional to bending moment.
6 - Question
Skin stress is also called as ______
a) Shear stress
b) Bending stress
c) Lateral stress
d) Temperature stress
View Answer
Answer: b
Explanation: The bending moment leads to deform or deflect the beam and internal stress resists bending. The resistance offered by internal stress to bending is called bending stress or “fibre stress” or “skin stress” or “longitudinal stress”.
Explanation: The bending moment leads to deform or deflect the beam and internal stress resists bending. The resistance offered by internal stress to bending is called bending stress or “fibre stress” or “skin stress” or “longitudinal stress”.
7 - Question
_________ is the total Strain energy stored in a body.
a) modulus of resilience
b) impact energy
c) resilience
d) proof resilience
View Answer
Answer: c
Explanation: When a load acts on a body, there is deformation of the body which causes movement of the applied load. Thus work is done is stored in the body as energy and the load is removed this stored energy which is by virtue of strain is called resilience.
8 - Question
In cantilever beams, there is _______ stress above neutral axis.
a) Compressive
b) Tensile
c) Temperature
d) Shear
View Answer
Answer: b
Explanation: In a cantilever beam maximum compressive stress occurs at bottom most fibre and maximum tensile stress occurs at the top most fibre and zero at neutral axis hence the tensile stresses lies above the neutral axis.
Explanation: In a cantilever beam maximum compressive stress occurs at bottom most fibre and maximum tensile stress occurs at the top most fibre and zero at neutral axis hence the tensile stresses lies above the neutral axis.
9 - Question
The product of modulus of elasticity (E) and polar moment of inertia (J) is called torsional rigidity.
a) True
b) False
View Answer
Answer: b
Explanation: The product of the modulus of rigidity (C) and polar moment of inertia (J) is called torsional rigidity and it produces a twist of one radian in a shaft of unit length.
10 - Question
Calculate the maximum stress due to Bending in a steel strip of 30 mm thick and 60 mm wide is bent around a circular drum of 3.6 m diameter [Take Young’s modulus = 200kN/m2].
a) 2341.76 N/mm2
b) 1666.67 N/mm2
c) 5411.76 N/mm2
d) 4666.67 N/mm2
View Answer
Answer: b
Explanation: Thickness of steel strip = 30 mm; b = 60 mm; d = 3.6m
R = 3.6/2 = 1.8 m
E = 200 kN/m2
y = 30/2 = 15 mm
E/R = f/y ; f = 200000×15/1800
= 1666.67 N/mm2.
11 - Question
The strength of beams depend merely on________
a) Modulus section
b) Moment of inertia
c) Flexural rigidity
d) Moment of resistance
View Answer
Answer: a
Explanation: The ratio of moment of inertia to the distance to the extreme fibre is called modulus of section. The Beam is stronger when section modulus is more. The strength of beam depends on section modulus. The beams of same strength mean section modulus is same for the beams.
Explanation: The ratio of moment of inertia to the distance to the extreme fibre is called modulus of section. The Beam is stronger when section modulus is more. The strength of beam depends on section modulus. The beams of same strength mean section modulus is same for the beams.
12 - Question
The steel plate is bent into a circular path of radius 10 metres. If the plate section be 120 mm wide and 20 mm thick, then calculate the maximum bending stress. [Consider Young’s modulus = 200000 N/mm2].
a) 350 N/mm2
b) 400 N/mm2
c) 200 N/mm2
d) 500 N/mm2
View Answer
Answer: c
Explanation: R = 10000 mm; y = 20/2 = 10 mm; E = 200000 N/mm2
By bending equation we have E/R = f/y
f = 200000×10 / 10000
= 200 N/mm2.