Strength of Materials Questions and Answers – Elastic Constants Relationship – 2

This set of Strength of Materials Questions and Answers for Freshers focuses on “Elastic Constants Relationship – 2”.

1. What is the ratio of Youngs modulus E to shear modulus G in terms of Poissons ratio?
a) 2(1 + μ)
b) 2(1 – μ)
c) 1/2 (1 – μ)
d) 1/2 (1 + μ)
View Answer

Answer: a
Explanation: As we know G = E / 2(1 +μ) so this gives the ratio of E to G = 2(1 + μ).

2. The relationship between Youngs modulus E, bulk modulus K if the value of Poissons ratio is unity will be __________
a) E = -3K
b) K = -3E
c) E = 0
d) K = 0
View Answer

Answer: a
Explanation: As E = 2G(1 + μ) putting μ=1 we get E = -3K.

3. A rod of length L and diameter D is subjected to a tensile load P. which of the following is sufficient to calculate the resulting change in diameter?
a) Youngs modulus
b) Poissons ratio
c) Shear modulus
d) Both Youngs modulus and shear modulus
View Answer

Answer: a
Explanation: For longitudinal strain we need Youngs modulus and for calculating transverse strain we need Poisson’s ratio. We may calculate Poissons ratio from E = 2G(1 + μ) for that we need shear modulus.
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4. E, G, K and μ elastic modulus, shear modulus, bulk modulus and Poisson’s ratio respectively. To express the stress strain relations completely for this material, at least __________
a) E, G and μmust be known
b) E, K and μmust be known
c) Any two of the four must be known
d) All the four must be known
View Answer

Answer: c
Explanation: As E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G), if any two of these four are known, the other two can be calculated by the relations between them.

5. Youngs modulus of elasticity and Poissons ratio of a material are 1.25 x 102 MPa and 0.34 respectively. The modulus of rigidity of the material is __________
a) 0.9469 MPa
b) 0.8375 MPa
c) 0.4664 MPa
d) 0.4025 MPa
View Answer

Answer: c
Explanation: As E = 2G(1 + μ)
1.25 x 102 = 2G(1 + 0.34)
G = 0.4664 x 102 MPa.

6. If E,G and K have their usual meanings, for an elastic material, then which one of the following be possibly true?
a) G = 2K
b) G = K
c) K = E
d) G = E = K
View Answer

Answer: c
Explanation: As E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G)
The value of μ must be between 0 to 0.5, so as E never equal to G but if μ = 1/3, then E=K.

7. If a material had a modulus of elasticity of 2.1 kgf/cm2 and a modulus of rigidity of 0.8 kgf/cm2 then what will be the approximate value of the Poissons ratio?
a) 0.26
b) 0.31
c) 0.47
d) 0.43
View Answer

Answer: b
Explanation: On using E = 2G(1 + μ) we can put the values of E and G to get the Poissons value.
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8. Consider the following statements:
X. Two-dimensional stresses applied to a thin plater in its own plane represent the plane stress condition.
Y. Normal and shear stresses may occur simultaneously on a plane.
Z. Under plane stress condition, the strain in the direction perpendicular to the plane is zero.
Which of the above statements are correct?
a) 2 only
b) 1 and 2
c) 2 and 3
d) 1 and 3
View Answer

Answer: d
Explanation: Under plane stress condition, the strain in the direction perpendicular to the plane is not zero. It has been found experimentally that when a body is stressed within the elastic limit, the lateral strain bears a constant ratio to the linear strain.

9. What is the relationship between the linear elastic properties Youngs modulus, bulk modulus and rigidity modulus?
a) 1/E = 9/k + 3/G
b) 9/E = 3/K + 1/G
c) 3/E = 9/K + 1/G
d) 9/E = 1/K + 3/G
View Answer

Answer: d
Explanation: We can use E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G) to get the relation between E, K and G.
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10. Which of the relationship between E, G and K is true, where E, G and K have their usual meanings?
a) E = 9KC / (3K + C)
b) E = 9KC / (9K + C)
c) E = 3KC / (9K + C)
d) E = 3KC / (3K + C)
View Answer

Answer: a
Explanation: As we know E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G).

Sanfoundry Global Education & Learning Series – Strength of Materials.

To practice all areas of Strength of Materials for Freshers, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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