Mechanical Metallurgy Questions and Answers – Plastic Deformation of Single Crystal – Deformation by Twinning


This set of Mechanical Metallurgy online quiz focuses on “Plastic Deformation of Single Crystal – Deformation by Twinning”.

1. Among the given metal, which has the highest stacking fault energy?
a) Aluminum
b) Copper
c) Brass
d) Stainless steel
View Answer

Answer: a
Explanation: The material with high stacking fault energy is not strain-hardenable material. So among all the given metals, aluminum is the not metal which is not strained hardened material.

Metal Stacking fault energy (mJ m-2)
Stainless steel 20
Copper 80
brass 10
Aluminum 200

2. Inhomogeneous deformation of crystal results in regions of different orientation called deformation bands. Deformation bands have been observed in both fcc and bcc metals, but not in hcp metals.
a) True
b) False
View Answer

Answer: a
Explanation: Consideration of the equation for critical resolved shear stress shows that it will be difficult to deform a hexagonal crystal when the basal plane is nearly Parallel to the crystal axis. Orowan found that if a cadmium crystal of this orientation were loaded in compression, it would deform by a localized region of the crystal suddenly snapping into a tilted position with a sudden shortening of the crystal.

3. Which of the following is not a source of dislocation multiplication?
a) Condensation of the vacancy
b) Frank read mechanism
c) Emission from high angle grain boundary
d) Annealing
View Answer

Answer: d
Explanation: Dislocation multiplication can arise from condensation of vacancies, by regeneration under applied stress from existing dislocations by either the Frank-Read mechanism or multiple cross-slip mechanisms or by the emission of dislocations from a high-angle grain boundary. But the annealing generally reduces the dislocation density.

4. A zinc crystal is strained to point A, unloaded, and then reloaded in the direction opposite to the original slip direction:
What is the reason for the drop in shear stress when loaded in the opposite direction?
a) Back stress
b) Strain hardening
c) Increase in the number of slip system
d) Slip plane reorientation
View Answer

Answer: a
Explanation: The back stress developed as a result of dislocations piling up at barriers during the first loading cycle is aiding dislocation movement when the direction of slip is reversed. Furthermore, when the slip direction is reversed, dislocations of opposite sign could be created at the same sources that produced the dislocations responsible for strain in the first slip direction. Since dislocations of opposite sign attract and annihilate each other, the net effect would be a further softening of the lattice.

5. The dislocation of low mobility that is produced by a dislocation reaction is called a ______ dislocation.
a) sessile
b) glissile
c) screw
d) edge
View Answer

Answer: a
Explanation: The reaction of 2 moving dislocations can result in new dislocation, which does not lie in low shear stress. Hence they are not mobile and called sessile dislocation.

6. Which of the following phenomena is responsible for the formation of sessile dislocation?
a) Lomer-Cottrell barriers
b) Frank read source
c) Age hardening
d) Dislocation jog and kink
View Answer

Answer: a
Explanation: The sessile dislocations do not lie on the slip plane of low shear stress; they act as a barrier to dislocation motion until the stress is increased to a high enough level to break down the barrier. The most critical dislocation reaction, which leads to the formation of sessile dislocations, is the formation of Lomer-Cottrell barriers in FCC metals by slip on intersecting {111} planes.

7. The interaction of dislocation does not result in ______________
a) jog
b) kink
c) dislocation threading
d) slip
View Answer

Answer: d
Explanation: Jog and kink are formed when dislocation on two different planes cut each other. Dislocation threading is the result of the interaction of many dislocations at an obstacle. Slip result from shear, not from dislocation interaction.

8. Cross slip can occur in both the edge dislocation as well as in screw dislocation.
a) True
b) False
View Answer

Answer: b
Explanation: The phenomenon of cross slip is restricted to screw dislocations. Since the line of screw dislocation and its Burgers vector are parallel, this does not define a specific plane as with an edge dislocation (where b is perpendicular to the dislocation line). So, in edge dislocation, cross slip is not possible.

9. The formation of jogs in screw dislocations _________ its motion. Jogs in edge dislocations _________ its motion.
a) restricts, do not restricts
b) do not restricts, restricts
c) restricts, restricts
d) do not restrict, do not restrict
View Answer

Answer: a
Explanation: The formation of jogs in screw dislocations impedes their motion and may even lead to the formation of vacancies and interstitials if the jogs are forced to move non-conservatively. Jogs in edge dislocations do not impede their motion.

10. Strain hardening due to ___________ process arises from short range force occurring over a distance of 5 to 10 atomic distance. This hardening can be overcome at finite temperatures with the help of thermal fluctuations, and therefore, it is temperature and strain-rate-dependent.
a) dislocation cutting
b) dislocation pile-up
c) second phase particle
d) alloying
View Answer

Answer: a
Explanation: Strain hardening due to dislocation cutting process arises from short range force occurring over a distance of 5 to 10 atomic distance. On the other hand, strain hardening arising from dislocation pile-up at barriers occurs over longer distances, and therefore, it is relatively independent of temperature and strain rate.

11. What is the resolved shear stress, when dislocation density in metal is equal to 1012m-2? Given that shear stress needed in the absence of dislocation is 500MPa. Numerical constant is 0.3 and shear modulus is 80GPa, burger vector is 1 nm.
a) 500MPa
b) 550MPa
c) 600 MPa
d) 524 MPa
View Answer

Answer: d
Explanation: Change in resolved shear stress value with dislocation density is given as:
=> τ = τo+αGbρ1/2
=> τ = 500*106+(0.3)(80*109)(106)(10-9)
=> τ = 500+24=524 MPa.

Sanfoundry Global Education & Learning Series – Mechanical Metallurgy.

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