Engineering Questions with Answers - Multiple Choice Questions
Strength of Materials MCQs Properties of Strain
The property of a material by which it can be drawn into thin wires is?
Explanation: The ductile material can be drawn into wires because it can resist large deformation. Malleability is the property by which it can be made into thin sheets.
If the material has identical elastic properties in all directions, it is called ____________
Explanation: An homogeneous material is that with uniform composition. An elastic and plastic are different on the criteria.
Why is the strain the fundamental property but not the stress?
a) Because it is dimensionless
b) Because it is a ratio
c) Because it’s value is calculated in the laboratory
d) No stress is the fundamental property
Explanation: The stress is the fundamental property because it is calculated in the laboratory. It is a non dependable value.
The material in which large deformation is possible before absolute failure by rupture is called ____________
Explanation: The ductile material can be drawn into wires because it can resist large deformation before it fails.
What is a creep?
a) Gradual increase of plastic strain with time at constant load
b) Gradual increase of elastic strain with time at constant load
c) Gradual increase of plastic strain with time at varying load
d) Gradual increase of elastic strain with time at varying load
Explanation: Creep is the property by virtue of which a metal specimen undergoes additional deformation with the passage of time under sustained loading within elastic limit. It is permanent in nature and cannot be recovered after removal of load, hence is plastic in nature.
If the material has different elastic properties in perpendicular directions, it is called ____________
Explanation: Isotropic material has the same elastic properties but ortho tropic material has the same.
Which one of the following pairs is NOT correctly matched?
a) Visco-elastic – small plastic zone
b) Orthotropic material – different properties in three perpendicular directions
c) Strain hardening material – stiffening effect felt at some stage
d) Isotropic material – same physical property in all directions at a point
Explanation: Visco-elastic material exhibit a mixture of creep and elastic after effects at room temperature. Thus their behaviour is time dependent. Materials with different properties in different directions are called anisotropic. Orthotropic material is a special case of an anisotropic material in three mutually perpendicular directions. However, these are symmetric about any axis.
The phenomenon of slow extension of materials having a constant load, I.e. increasing with the time is called
d) None of the mentioned
Explanation: The creeping is the phenomenon of deformation in materials which have been under load for several time. When the load is put on the material, initially it deforms but when the load is not removed, it causes a small amount of deformation which increases with time.
Strength of Materials MCQs Strain Constants – 1
What will be the elastic modulus of a material if the Poisson’s ratio for that material is 0.5?
a) Equal to its shear modulus
b) Three times its shear modulus
c) Four times its shear modulus
d) Not determinable
Explanation: Elastic modulus = E
Shear modulus = G
E = 2G ( 1 + μ )
Given, μ= 0.5, E = 2×1.5xG
E = 3G.
A rigid beam ABCD is hinged at D and supported by two springs at A and B as shown in the given figure. The beam carries a vertical load P and C. the stiffness of spring at A is 2K and that of B is K.<br/>
What will be the ratio of forces of spring at A and that of spring at B?
Explanation: The rigid beam will rotate about point D, due to the load at C.
From similar triangle,
δa/2a = δb/3b
Force in spring A/Force in spring B = Pa/Pb
= 2k/k x 3/2 = 3.
A solid metal bat of uniform diameter D and length L is hung vertically from a ceiling. If the density of the material of the bar is 1 and the modulus of elasticity is E, then the total elongation of the bar due to its own weight will be ____________
Explanation: The elongation of bar due to its own weight is δ= WL/2AE
Now W = ρAL
There fore δ= L2 / 2E.
A bar of diameter 30mm is subjected to a tensile load such that the measured extension on a gauge length of 200mm is 0.09mm and the change in diameter is 0.0045mm. Calculate the Poissons ratio?
Explanation: Longitudinal strain = 0.09/200
Lateral strain = – 0.0045/30
Poissons ratio = – lateral strain/ longitudinal strain
= 0.0045/30 x 200/0.09
What will be the ratio of Youngs modulus to the modulus of rigidity of a material having Poissons ratio 0.25?
Explanation: Modulus of rigidity, G = E/2(1 + μ)
Therefore, E/G = 2x(1+0.25) = 2.5.
An experiment was done and it was found that the bulk modulus of a material is equal to its shear modulus. Then what will be its Poissons ratio?
Explanation: We know that, μ = (3K – 2G) / (6K + 2G)
Here K = G
Therefore, μ = 3-2 / 6+2 = 0.125.
A bar of 40mm dia and 40cm length is subjected to an axial load of 100 kN. It elongates by 0.005mm. Calculate the Poissons ratio of the material of bar?
Explanation: Longitudinal strain = 0.150/400 = 0.000375
Lateral strain = – 0.005/40 = -0.000125
Poissons ratio = – lateral strain/longitudinal strain
What will be the approximate value of shear modulus of a material if the modulus of elasticity is 189.8 GN/m2 and its Poissons ratio is 0.30?
a) 73 GN/m2
b) 80 GN/m2
c) 93.3 GN/m2
d) 103.9 GN/m2
Explanation: The relationship between E, G, and μ is given by
is given by
E = 2G (1 + μ)
G = 189.8 / 2(1 + 0.30)
G = 73 GN/m2.
What will be the modulus of rigidity if the value of modulus of elasticity is 200 and Poissons ratio is 0.25?
Explanation: The relationship between E, G and μ is E = 2G (1 + μ)
G = 200 / 2(1 + 0.25)
G = 80.