Engineering Questions with Answers - Multiple Choice Questions

# MCQs on Simple Stress and Strain-2

1 - Question

MCQs on Strain Constants – 1
What will be the elastic modulus of a material if the Poisson’s ratio for that material is 0.5?
a) Equal to its shear modulus
b) Three times its shear modulus
c) Four times its shear modulus
d) Not determinable
Explanation: Explanation: Elastic modulus = E Shear modulus = G E = 2G ( 1 + μ ) Given, μ= 0.5, E = 2×1.5xG E = 3G.

2 - Question

A rigid beam ABCD is hinged at D and supported by two springs at A and B as shown in the given figure. The beam carries a vertical load P and C. the stiffness of spring at A is 2K and that of B is K.

What will be the ratio of forces of spring at A and that of spring at B?
a) 4
b) 3
c) 2
d) 1

Explanation: The rigid beam will rotate about point D, due to the load at C.

From similar triangle,
δa/2a = δb/3b
Force in spring A/Force in spring B = Pa/Pb
= 2k/k x 3/2 = 3.

3 - Question

A solid metal bat of uniform diameter D and length L is hung vertically from a ceiling. If the density of the material of the bar is 1 and the modulus of elasticity is E, then the total elongation of the bar due to its own weight will be ____________
a) L/2E
b) L2/2E
c) E/2L
d) E/2L2
Explanation: The elongation of bar due to its own weight is δ= WL/2AE Now W = ρAL There fore δ= L2 / 2E. advertisement

4 - Question

A bar of diameter 30mm is subjected to a tensile load such that the measured extension on a gauge length of 200mm is 0.09mm and the change in diameter is 0.0045mm. Calculate the Poissons ratio?
a) 1/3
b) 1/4
c) 1/5
d) 1/6
Explanation: Longitudinal strain = 0.09/200 Lateral strain = – 0.0045/30 Poissons ratio = – lateral strain/ longitudinal strain = 0.0045/30 x 200/0.09 = 1/3.

5 - Question

What will be the ratio of Youngs modulus to the modulus of rigidity of a material having Poissons ratio 0.25?
a) 3.75
b) 3.00
c) 1.5
d) 2.5
Explanation: Modulus of rigidity, G = E/2(1 + μ) Therefore, E/G = 2x(1+0.25) = 2.5.

6 - Question

An experiment was done and it was found that the bulk modulus of a material is equal to its shear modulus. Then what will be its Poissons ratio?
a) 0.125
b) 0.150
c) 0.200
d) 0.375
Explanation: We know that, μ = (3K – 2G) / (6K + 2G) Here K = G Therefore, μ = 3-2 / 6+2 = 0.125.

7 - Question

A bar of 40mm dia and 40cm length is subjected to an axial load of 100 kN. It elongates by 0.005mm. Calculate the Poissons ratio of the material of bar?
a) 0.25
b) 0.28
c) 0.30
d) 0.33
Explanation: Longitudinal strain = 0.150/400 = 0.000375 Lateral strain = – 0.005/40 = -0.000125 Poissons ratio = – lateral strain/longitudinal strain = 0.33.

8 - Question

What will be the approximate value of shear modulus of a material if the modulus of elasticity is 189.8 GN/m2 and its Poissons ratio is 0.30?
a) 73 GN/m2
b) 80 GN/m2
c) 93.3 GN/m2
d) 103.9 GN/m2
Explanation: The relationship between E, G, and μ is given by is given by E = 2G (1 + μ) G = 189.8 / 2(1 + 0.30) G = 73 GN/m2.

9 - Question

What will be the modulus of rigidity if the value of modulus of elasticity is 200 and Poissons ratio is 0.25?
a) 70
b) 80
c) 125
d) 250
Explanation: The relationship between E, G and μ is E = 2G (1 + μ) G = 200 / 2(1 + 0.25) G = 80.

10 - Question

MCQs on 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
Explanation: Poissons ratio = lateral strain / longitudinal strain = δD/D x L/δL = 0.0045/30 x 200/0.09 = 0.33.

11 - Question

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
Explanation: As we know E = 3k(1- 2μ) So E/K = 3(1-2×0.3) = 1.2.

12 - Question

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
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. advertisement

13 - Question

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
Explanation: The bulk modulus is given as K = E/3(1 – 2μ) = 1.2 x 105/3(1 – 2/4) = 0.8 x 105.

14 - Question

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
Explanation: As we know, E = 2C(1 + μ) μ= 0.25 K = E / 3(1 – 2μ) = 8.

15 - Question

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

16 - Question

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

17 - Question

How many elastic constants does an isotropic, homogeneous and linearly elastic material have?
a) 1
b) 2
c) 3
d) 4
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).

18 - Question

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

19 - Question

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

20 - Question

MCQs on Elastic Constants Relationship – 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.

21 - Question

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.

22 - Question

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). advertisement

23 - Question

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.

24 - Question

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.

25 - Question

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.

26 - Question

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).

27 - Question

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).

28 - Question

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.

29 - Question

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).

30 - Question

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

31 - Question

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

32 - Question

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
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. advertisement

33 - Question

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
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.

34 - Question

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

35 - Question

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
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.

36 - Question

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

37 - Question

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
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.

38 - Question

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
Explanation: We can use E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G) to get the relation between E, K and G.

39 - Question

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)