Mechanical Metallurgy Questions and Answers – Elastic Stress and Strain Relation

This set of Mechanical Metallurgy Multiple Choice Questions & Answers (MCQs) focuses on “Elastic Stress and Strain Relation”.

1. The tension force along x-direction produces extension along the x-axis; it also produces contraction in transverse y and z-directions.
a) True
b) False
View Answer

Answer: a
Explanation: Volume of the material always remains constant. If one of the directions gets elongated, then to keep the volume constant the other two direction shrinks. Strain along the x-direction is called longitudinal strain and along y and z-direction is called transverse strain.

2. The transverse strain is a constant fraction of the strain in the longitudinal direction. This ratio is known as __________
a) poisson’s ratio
b) pilling-Bedworth ratio
c) concentration ratio
d) reduction ratio
View Answer

Answer: a
Explanation: Poisson’s ratio defines the relation between longitudinal strain and transverse strain.
Mathematically, it is expressed as εy = εz = -νεx = -ν σx/E; here ν is the Poisson’s ratio.
Pilling-Bedworth ratio is associated with a thickness of corrosion layer. The concentration ratio is used in extractive metallurgy. The reduction ratio is a change in the diameter of material during the manufacturing process.

3. For most isotropic metal the value of Poisson’s ratio is equal to ___________
a) 0.5
b) 1
c) 0.33
d) 0
View Answer

Answer: c
Explanation: Metals are crystalline and therefore, atomic arrangement is well defined. If this ordered structure of the atom is pulled in one direction, the atomic shift causes the contraction in the other two directions in well-defined pattern and contraction in a transverse direction. Generally, the calculated result comes around 1/3=0.33.

4. Consider a unit cube subjected to normal stress σx, σy, σz. Assume normal stress does not produce any shear stress on planes x, y, z. What will be the net strain along x direction?
a) 1/E [σx-ν(σy+σz)]
b) 1/E [σy-ν(σx+σz)]
c) 1/E [σz-ν(σy+σx)]
d) 1/E [σx]
View Answer

Answer: a
Explanation: The stress along y and z-directions will produce an equivalent strain of -νσy and -νσz in the x-direction. So, these strain values will get subtracted from the original strain produced by σx and net strain in the x-direction will be 1/E [σx- ν(σy+σz)].

5. Modulus of elasticity defines the relationship between shear stress and shear strain.
a) True
b) False
View Answer

Answer: a
Explanation: Shearing stress acting along a cube edge produces a shearing strain. τxy=G*γxy
The torsion test determines value of G.
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6. Bulk modulus or volumetric modulus of elasticity(K) is equal to ____________
a) τxy / σx
b) γxy / σx
c) σm / Δ
d) Δ / σm
View Answer

Answer: c
Explanation: Bulk modulus is defined as the ratio of hydrostatic pressure to dilation.
Mathematically σm/Δ = -p/Δ = 1/β
Where σm is hydrostatic stress, Δ volumetric strain, p hydrostatic pressure, β is defined as compressibility.

7. Which of the following relation is not correct?
a) Δ={(1-2ν)/E}3σm
b) K = E/3(1-2ν)
c) G = E/2(1+ν)
d) ν = {1+(2G/3K)} / {1- (2G/3K)}
View Answer

Answer: d
Explanation: The expression ν = \(\frac{1+(2G/3K)}{1- (2G/3K)}\) is not correct. The correct relationship will be ν = \(\frac{1-(2G/3K)}{1+ (2G/3K)}\).
The expression Δ={(1-2ν)/E}3σm, K = E/3(1-2ν), G = E/2(1+ν) are correct and well-defined relation in mechanical metallurgy.
Easier way to remember these formulae is:
=> E=2G (1+ν)
=> E=3K (1-2ν).

8. In case of plane strain, the strain(ε3=0) in one of the principal directions is zero, but the stress is non-zero (σ3 is nonzero) in this direction. σ3= _____________
a) ν (σ1+ σ2)
b) E (1- ν2) (σ1+ σ2)
c) σ1+σ2
d) {E/(1- ν2)} (σ1+ σ2)
View Answer

Answer: a
Explanation: Strain in the direction 3:
=> ε3 = 1/E[σ3 – ν(σ1+ σ2)] = 0
=> σ3 = ν(σ1+ σ2)
The stress in the direction 1 will be equal to:
=> σ1 = {E/(1- ν2)} (ε1 + ν ε2).

9. A strain gauge measurement made on the free aluminum surface indicates the principle strains 0.003 and 0.002. What are the principal stresses? (Given that poison’s ratio for aluminum=0.33, young modulus = 70 GPa)
a) σ1=965 MPa, σ2=235 MPa
b) σ1=857.8 MPa, σ2=125.86 MPa
c) σ1=305 MPa, σ2=458 MPa
d) σ1=287.5 MPa, σ2=233.96 MPa
View Answer

Answer: d
Explanation: The value of principle stresses in case plane strain condition is given as:
\(\sigma 1=\frac{E}{1-V^2}(\epsilon 1+v \epsilon 2)\)
\(\sigma 1=\frac{70}{1-0.33^2}(0.003+0.33*0.002)\)
σ1 = 287.5 MPa
Similarly, for σ2:
\(\sigma 2=\frac{70}{1-0.33^2}(0.002+0.33*0.003)\)
σ2 = 233.96 MPa.

10. Calculate the shear modulus for steel in GPa, if given that young modulus is 215 GPa and possion’s ratio is 0.33?
a) 40 GPa
b) 50 GPa
c) 80 GPa
d) 100 GPa
View Answer

Answer: c
Explanation: The relation between young modulus and poison’s ratio is:
=> E=2G(1+ν)
E=215 GPa, ν=0.33
=> 215=2*G(1+0.33)
=> G=80.82 GPa.

Sanfoundry Global Education & Learning Series – Mechanical Metallurgy.

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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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