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# Strength of Materials MCQs Strain Constants – 2

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

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Answer:d

Explanation: Poissons ratio = lateral strain / longitudinal strain

= δD/D x L/δL

= 0.0045/30 x 200/0.09

= 0.33.

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

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Answer: b

Explanation: As we know E = 3k(1- 2μ)

So E/K = 3(1-2×0.3) = 1.2.

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

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Answer: d

Explanation: 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.

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

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Answer: b

Explanation: The bulk modulus is given as K = E/3(1 – 2μ)

= 1.2 x 105/3(1 – 2/4)

= 0.8 x 105.

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

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Answer: b

Explanation: As we know, E = 2C(1 + μ)

μ= 0.25

K = E / 3(1 – 2μ)

= 8.

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

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Answer: a

Explanation: As we know E = 2G(1 + μ) so putting the values of E = 2.5G then we get μ= 0.25.

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)

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Answer: b

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

How many elastic constants does an isotropic, homogeneous and linearly elastic material have?

a) 1

b) 2

c) 3

d) 4

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Answer: d

Explanation: 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).

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

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Answer: a

Explanation: As E = 2G(1 + μ) putting E = 200 and G = 80 we get μ = 0.25.

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

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Answer: a

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

### Strength of Materials MCQs Elastic Constants Relationship – 1

How many elastic constants of a linear, elastic, isotropic material will be?

a) 2

b) 3

c) 1

d) 4

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Answer: aExplanation: 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.

How many elastic constants of a non homogeneous, non isotropic material will be?

a) 9

b) 15

c) 20

d) 21

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

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)

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Answer: c

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).

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

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Answer: c

Explanation: Modulus of resilience = f2/2E

= 200×2/2×2000

= 0.10.

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

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Answer: b

Explanation: As we know, μ= (3K – 2G) / (6K + 2G)

Given K = 2G

Then, μ = (6G – 2G) / (12G + 2G) = 4/14 = 2/7.

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

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Answer: c

Explanation: K = E / 3(1 – 2μ)

Since K = E

So (1-2μ) = 1/3

Therefore, μ = 1/3.

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)

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Answer: c

Explanation: The relation between the modulus of rigidity, modulus of elasticity and the Poissons ratio is given as

G = E / 2(1 + μ).

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)

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Answer: a

Explanation: The relationship between E, K, C is given by

E = 9KC / (3K + C).

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

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Answer: d

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

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)

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Answer: c

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).

### Strength of Materials MCQs Elastic Constants Relationship – 2

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)

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Answer: a

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

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

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Answer: a

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

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

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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.

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

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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.

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

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Answer: c

Explanation: As E = 2G(1 + μ)

1.25 x 102 = 2G(1 + 0.34)

G = 0.4664 x 102 MPa.

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

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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.

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

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Answer: b

Explanation: On using E = 2G(1 + μ) we can put the values of E and G to get the Poissons value.

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

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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.

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

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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.

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)

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Answer: a

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