Engineering Questions with Answers - Multiple Choice Questions
Strength of Materials MCQs Bending Equation
In simple bending, ______ is constant.
a) Shear force
d) Bending moment
Explanation: If a beam is undergone with simple bending, the beam deforms under the action of bending moment. If this bending moment is constant and does not affect by any shear force, then the beam is in state of simple bending.
If a beam is subjected to pure bending, then the deformation of the beam is_____
a) Arc of circle
Explanation: The beam being subjected to pure bending, there will be only bending moment and no shear force it results in the formation of an arc of circle with some radius known as radius of curvature.
When a beam is subjected to simple bending, ____________ is the same in both tension and compression for the material.
a) Modulus of rigidity
b) Modulus of elasticity
c) Poisson’s ratio
d) Modulus of section
Explanation: It is one of the most important assumptions made in the theory of simple bending that is the modulus of elasticity that is Young’s modulus [E] is same in both tension and compression for the material and the stress in a beam do not exceed the elastic limit.
E/R = M/I = f/y is a bending equation.
Explanation: The above-mentioned equation is absolutely correct.
E/R = M/I = f/y is a bending equation. It is also known as flexure equation (or) equation for theory of simple bending.
E stands for Young’s modulus or modulus of elasticity.
R stands for radius of curvature.
M stands for bending moment
I stand for moment of inertia
f stands for bending stress
y stands for neutral axis.
Maximum Shearing stress in a beam is at _____
a) Neutral axis
b) Extreme fibres
c) Mid span
d) Action of loading
Explanation: Shearing stress is defined as the resistance offered by the internal stress to the shear force. Shearing stress in a beam is maximum at a neutral axis.
At the neutral axis, bending stress is _____
Explanation: Neutral axis is defined as a line of intersection of neutral plane or neutral layer on a cross section at the neutral axis of that section. At the NA, bending stress or bending strain is zero. The first moment of area of a beam section about neutral axis is also zero. The layer of neutral axis neither contracts nor extends.
Curvature of the beam is __________ to bending moment.
b) Directly proportion
c) Inversely proportion
Explanation: From the flexural equation, we have 1/R is called as the “curvature of the beam”.
1 / R = M / EI
Hence the curvature of the beam is directly proportional to bending moment and inversely proportional to flexural rigidity (EI).
What are the units of flexural rigidity?
Explanation: The product of young’s modulus (E) of the material and moment of inertia (I) of the beam section about its neutral axis is called flexural rigidity.
Units for E are N/m2
Units for I are m4
Their product is Nm2.
What are the units for section modulus?
Explanation: The ratio of moment of inertia to the distance to the extreme fibre is called modulus of section or section modulus. It is generally denoted by the letter Z. Section modulus is expressed in m3
Z = I/y
= m4/ m
What are the units of axial stiffness?
c) N/ m
Explanation: Axial rigidity is a product of young’s modulus (E) and the cross-sectional area (A) of that section. Axial rigidity per unit length is known as axial stiffness the si units of axial stiffness are Newton per metre (N/m).
Calculate the modulus of section of rectangle beam of size 240 mm × 400 mm.
a) 5.4 × 106 mm3
b) 6.2 × 106 mm3
c) 5.5 × 106 mm3
d) 6.4 × 106 mm3
Explanation: b = 240 mm & d = 400 mm
Moment of inertia (I) = bd3/12; y = d/2
Section modulus (Z) = I/y = bd2/ 6
= 1/6 × 240 × 400 ×400
= 6.4 × 106 mm3.
What is the product of force and radius?
a) Twisting shear
b) Turning shear
c) Turning moment
d) Tilting moment
Explanation: Twisting moment will be equal to the product of the perpendicular force and existing radius. Denoted by letter T and SI units are Nm.
Determine section modulus for beam of 100mm diameter.<br/>
a) 785 × 103 mm3<br/>
b) 456 × 103 mm3<br/>
c) 87 × 103 mm3<br/>
d) 98 × 103 mm3
Explanation: d = 300mm
For circular sections; I = π / 64 × d4
Z = π/32 × d3 (d = 100 mm)
Z = 98.17 × 103mm3.
Strength of Materials MCQs Pure Bending Stress
In simply supported beams, the _____ stress distribution is not uniform.
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.
The maximum _________ stresses occur at top most fibre of a simply supported beam.
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.
The stress is directly proportional to _______
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).
At the extreme fibre, bending stress is______
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.
The curvature of a beam is equal to _____
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.
Skin stress is also called as ______
a) Shear stress
b) Bending stress
c) Lateral stress
d) Temperature 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”.
________ is the total Strain energy stored in a body.
a) modulus of resilience
b) impact energy
d) proof resilience
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.
In cantilever beams, there is _______ stress above 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.
The product of modulus of elasticity (E) and polar moment of inertia (J) is called torsional rigidity.
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.
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].<br/>
a) 2341.76 N/mm2<br/>
b) 1666.67 N/mm2<br/>
c) 5411.76 N/mm2<br/>
d) 4666.67 N/mm2
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.
The strength of beams depend merely on________
a) Modulus section
b) Moment of inertia
c) Flexural rigidity
d) Moment of resistance
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.
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
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.
Strength of Materials MCQs Section Modulus
What is the section modulus (Z) for a rectangular section?
Explanation: The modulus of section may be defined as the ratio of moment of inertia to the distance to the extreme fibre. It is denoted by Z.
Z= I/y ; For rectangular section, I = bd3/12 & y = d/2.
Find the modulus of section of square beam of size 300×300 mm.<br/>
a) 4.8 × 106 mm3<br/>
b) 4.5 × 106 mm3<br/>
c) 5.6 × 106 mm3<br/>
d) 4.2 × 106 mm3
Explanation: Here, a = side of square section = 300 mm.
I = a4/12. y= a/2.
Z = I/y = a3/6
= 4.5 × 106 mm3.
_________ of a beam is a measure of its resistance against deflection.
Explanation: A beam is said to be a strength when the maximum induced bending and shear stresses are within the safe permissible stresses stiffness of a beam is a measure of its resistance against deflection.
To what radius an Aluminium strip 300 mm wide and 40mm thick can be bent, if the maximum stress in a strip is not to exceed 40 N/mm2. Take young’s modulus for Aluminium is 7×105 N/mm2.
Explanation: Here, b = 300mm
d= 40mm. y= 20mm.
From the relation; E/R = f/y
=70×103 × 20 / 40
The bending stress in a beam is ______ to bending moment.
a) Less than
b) Directly proportionate
c) More than
Explanation: As we know, the bending stress is equal to bending moment per area. Hence, as the bending (flexure) moment increases/decreases the same is noticed in the bending stress too.
The Poisson’s ratio for concrete is __________
Explanation: The ratio of lateral strain to the corresponding longitudinal strain is called Poisson’s ratio. The value of poisons ratio for elastic materials usually lies between 0.25 and 0.33 and in no case exceeds 0.5. The Poisson’s ratio for concrete is 0.20.
The term “Tenacity” means __________
a) Working stress
b) Ultimate stress
c) Bulk modulus
d) Shear modulus
Explanation: The ultimate stress of a material is the greatest load required to fracture the material divided by the area of the original cross section in the point of fracture The ultimate stress is also known as tenacity.
A steel rod of 25 mm diameter and 600 mm long is subjected to an axial pull of 40000. The intensity of stress is?
a) 34.64 N/mm2
b) 46.22 N/mm2
c) 76.54 N/mm2
d) 81.49 N/mm2
Explanation: Cross sectional area of steel rod [Circular]be 490.87 mm2.
The intensity of stress = P/A = 40000/490.87
= 81.49 N/mm2.
The bending strain is zero at _______
a) Point of contraflexure
b) Neutral axis
d) Line of action of loading
Explanation: The neutral axis is a line of intersection of neutral plane or neutral layer on a cross section. The neutral axis of a beam passes through the centroid of the section. At the neutral axis bending stress and bending strain is zero.
Strength of the beam depends only on the cross section.
Explanation: The strength of two beams of the same material can be compared by the section modulus values. The strength of beam depends on the material, size and shape of cross section. The beam is stronger when section modulus is more, strength of the beam depends on Z.