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

1. A circular rod of dia 30 mm and length 200mm is extended to 0.09mm length and 0.0045 diameter through a tensile force. What will be its Poissons ratio ?

a) 0.30

b) 0.31

c) 0.32

d) 0.33

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explantion: 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 bulk modulus.

4. For a material, Youngs modulus is given as 1.2 x 10^{5} and Poissons ratio 1/4. Calculate the bulk modulus.

a) 0.7 x 10^{5}

b) 0.8 x 10^{5}

c) 1.2 x 10^{5}

d) 1.2 x 10^{5}

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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)

View Answer

Explanation: As E = 2G(1 + μ) = 3K(1 – 2μ).

8. How many elastic constants does an isotropic, homogeneous ans linearly elastic material have ?

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation: E, G, K and represents the elastic modulus, shear modulus, bulk modulus and poissons ratio respectively of a linearly elastic, isotropic and homogeneous material. To express the stress-strain relations completely for this material at lieat 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

View Answer

Explanation: 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 Poissons 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

View Answer

Explanation: If the Poissons ratio is zero then the material is rigid.

**Sanfoundry Global Education & Learning Series – Strength of Materials.**

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