# Strength of Materials Questions and Answers – Elastic Constants Relationship – 1

This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Elastic Constants Relationship – 1”.

1. How many elastic constants of a linear, elastic, isotropic material will be?
a) 2
b) 3
c) 1
d) 4

Explanation: Isotropic materials have the same properties in all directions. The number of independent elastic constants for such materials is 2. out of E, G, K, and μ, if any two constants are known for any linear elastic and isotropic material than rest two can be derived. Examples are steel, aluminium, copper, gold.
Orthotropic materials refer to layered structure such as wood or plywood. The number of independent elastic constants for such materials is 9.
Non isotropic or anisotropic materials have different properties in different directions. They show non- homogeneous behaviour. The number of elastic constants is 21.

2. How many elastic constants of a non homogeneous, non isotropic material will be?
a) 9
b) 15
c) 20
d) 21

Explanation: Non isotropic or anisotropic materials have different properties in different directions. They show non- homogeneous behaviour. The number of elastic constants is 21.

3. How can be the Poissons ratio be expressed in terms of bulk modulus(K) and modulus of rigidity(G)?
a) (3K – 4G) / (6K + 4G)
b) (3K + 4G) /( 6K – 4G)
c) (3K – 2G) / (6K + 2G)
d) (3K + 2G) / (6K – 2G)

Explanation: There are four elastic modulus relationships. the relation between Poissons ration, bulk modulus and modulus of rigidity is given as
μ = (3K – 2G) / (6K + 2G).

4. Calculate the modulus of resilience for a 2m long bar which extends 2mm under limiting axial stress of 200 N/mm2?
a) 0.01
b) 0.20
c) 0.10
d) 0.02

Explanation: Modulus of resilience = f2/2E
= 200×2/2×2000
= 0.10.

5. In an experiment, the bulk modulus of elasticity of a material is twice its modulus of rigidity. The Poissons ratio of the material is ___________
a) 1/7
b) 2/7
c) 3/7
d) 4/7

Explanation: As we know, μ= (3K – 2G) / (6K + 2G)
Given K = 2G
Then, μ = (6G – 2G) / (12G + 2G) = 4/14 = 2/7.
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6. What will be the value of the Poisson’s ratio if the Youngs modulus E is equal to the bulk modulus K?
a) 1/2
b) 1/4
c) 1/3
d) 3/4

Explanation: K = E / 3(1 – 2μ)
Since K = E
So (1-2μ) = 1/3
Therefore, μ = 1/3.

7. What is the expression for modulus of rigidity in terms of modulus of elasticity and the Poissons ratio?
a) G = 3E / 2(1 + μ)
b) G = 5E / (1 + μ)
c) G = E / 2(1 + μ)
d) G = E/ (1 + 2μ)

Explanation: The relation between the modulus of rigidity, modulus of elasticity and the Poissons ratio is given as
G = E / 2(1 + μ).

8. What is the relationship between Youngs modulus E, modulus of rigidity C, and bulk modulus K?
a) E = 9KC / (3K + C)
b) E = 9KC / (9K + C)
c) E = 3KC / (3K + C)
d) E = 3KC / (9K + C)

Explanation: The relationship between E, K, C is given by
E = 9KC / (3K + C).

9. What is the limiting values of Poisson’s ratio?
a) -1 and 0.5
b) -1 and -0.5
c) -1 and -0.5
d) 0 and 0.5

Explanation: The value of Poissons ratio varies from 0 to 0.5. For rubber, its value ranges from.45 to 0.50.

10. What is the relationship between modulus of elasticity and modulus of rigidity?
a) C = E / 2(1 + μ)
b) C = E / (1 + μ)
c) C = 2E / (1 + μ)
d) C = 2E / 2(1 + μ)

Explanation: The relation is given by calculating the tensile strain of square block is given by taking tensile strain in a diagonal. On equating that stains we get the relation,
C = E / 2(1 + μ).

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