250+ TOP MCQs on Strain Constants – 2 and Answers

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This set of Strength of Materials Interview Questions and Answers on “Strain Constants – 2”.

1. A circular rod of dia 30 mm and length 200mm is extended to 0.09mm length and 0.0045 diameters through a tensile force. What will be its Poissons ratio?
a) 0.30
b) 0.31
c) 0.32
d) 0.33
Answer:d
Clarification: Poissons ratio = lateral strain / longitudinal strain
= δD/D x L/δL
= 0.0045/30 x 200/0.09
= 0.33.

2. The Poissons ratio of a material is 0.3. what will be the ratio of Youngs modulus to bulk modulus?
a) 1.4
b) 1.2
c) 0.8
d) 0.6
Answer: b
Clarification: As we know E = 3k(1- 2μ)
So E/K = 3(1-2×0.3) = 1.2.

3. What is the bulk modulus of elasticity?
a) The ratio of shear stress to shear strain
b) The ratio of direct stress to direct strain
c) The ratio of volumetric stress to volumetric strain
d) The ratio of direct stress to volumetric strain
Answer: d
Clarification: When a body is subjected to the mutually perpendicular like and equal direct stresses, the ratio of direct stress to the corresponding volumetric strain strain is found to be constant for a given material when the deformation is within a certain limit. This ratio is known as the bulk modulus.

4. For a material, Youngs modulus is given as 1.2 x 105 and Poissons ratio 1/4. Calculate the bulk modulus.
a) 0.7 x 105
b) 0.8 x 105
c) 1.2 x 105
d) 1.2 x 105
Answer: b
Clarification: The bulk modulus is given as K = E/3(1 – 2μ)
= 1.2 x 105/3(1 – 2/4)
= 0.8 x 105.

5. Determine the Poissons ratio and bulk modulus of a material, for which Youngs modulus is 1.2 and modulus of rigidity is 4.8.
a) 7
b) 8
c) 9
d) 10
Answer: b
Clarification: As we know, E = 2C(1 + μ)
μ= 0.25
K = E / 3(1 – 2μ)
= 8.

6. The Youngs modulus of elasticity of a material is 2.5 times its modulus of rigidity. Then what will be its Poissons ratio?
a) 0.25
b) 0.33
c) 0.50
d) 0.60
Answer: a
Clarification: As we know E = 2G(1 + μ) so putting the values of E = 2.5G then we get μ= 0.25.

7. How the elastic constants E and K are related?
a) E = 2K(1 – 2μ)
b) E = 3K(1 – 2μ)
c) E = 2K(1 – μ)
d) E = K(1 – 2μ)
Answer: b
Clarification: As E = 2G(1 + μ) = 3K(1 – 2μ).

8. How many elastic constants does an isotropic, homogeneous and linearly elastic material have?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: E, G, K represents the elastic modulus, shear modulus, bulk modulus and poisson’s ratio respectively of a linearly elastic, isotropic and homogeneous material. To express the stress-strain relations completely for this material at least any two of the four must be known, E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G).

9. The modulus of rigidity and the modulus of elasticity of a material are 80 GPa and 200 GPa. What will be the Poissons ratio of the material?
a) 0.25
b) 0.30
c) 0.40
d) 0.50
Answer: a
Clarification: As E = 2G(1 + μ) putting E = 200 and G = 80 we get μ = 0.25.

10. Which of the following is true if the value of Poisson’s ratio is zero?
a) The material is rigid
b) The material is perfectly plastic
c) The longitudinal strain in the material is infinite
d) There is no longitudinal strain in the material
Answer: a
Clarification: If the Poissons ratio is zero then the material is rigid.

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