Mechanical Metallurgy Questions and Answers – Plastic Deformation of Single Crystal – Slip by Dislocation Movement

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This set of Mechanical Metallurgy Multiple Choice Questions & Answers (MCQs) focuses on “Plastic Deformation of Single Crystal – Slip by Dislocation Movement”.

1. The closed packed plane and close packed direction of hexagonal closed packed structure is _________ and _____________ respectively.
a) (0001) and <1120>
b) (0112) and <0001>
c) (1120) and <1120>
d) (0012) and <1212>
View Answer

Answer: a
Explanation: Generally, the slip plane is the plane of greatest atomic density and the slip direction is the closest-packed direction within the slip plane. In the case of HCP (0001) and <1120> is closed packed planes and closed packed directions.
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2. In the face-centered cubic structure, the (111) octahedral planes and the (110) directions are the close-packed systems. What is the total number of possible slip system in FCC?
a) 8
b) 12
c) 48
d) 16
View Answer

Answer: b
Explanation: There are eight (111) planes in the fcc unit cell. However, the planes at opposite faces of the octahedron are parallel to each other, so that there are only four sets of octahedral planes. Each (111) plane contains three (110) directions (the reverse directions being neglected). Therefore, the fcc lattice has 12 possible slip systems.

3. What is closed packed direction body-centered cubic system?
a) <111>
b) <110>
c) <100>
d) <122>
View Answer

Answer: a
Explanation: The <111> direction in the BCC unit cell is the most closed packed direction. Only along this direction, atoms will be touching each other edge to edge. The radius of atom in term of the unit cell is determined using this direction.

4. Which of the following is not a slip system in body centered cubic lattice?
a) {110} <111>
b) {112} <111>
c) {123} <111>
d) {112} <111>
View Answer

Answer: d
Explanation: The bcc structure is not a close-packed structure like the fcc or hcp structures. Accordingly, there is no one plane of predominant atomic density, as (111) in the fcc structure and (0001) in the hcp structure. Slip in bcc metals is found to occur on the {110}, {112}, and {123} planes, while the slip direction is always the [111) direction.

5. Which of the following is not a reason for the wavy appearance Slip lines in bcc metals?
a) The slip occurs on several planes, (110), (112), (123) but always in the close-packed (111) direction
b) Dislocations can readily move from one type of plane to another by cross sip
c) Higher shearing stresses are required to cross-slip
d) The ductile nature of BCC causes wavy nature
View Answer

Answer: d
Explanation: Slip lines in bcc metals have a wavy appearance. This is because slip occurs on several planes, (110), (112), (123) but always in the close-packed (111) direction, which is common to each of these planes. Dislocations can readily move from one type of plane to another by cross sip, giving rise to the irregular wavy slip bands. The planes are not so close-packed as in the fcc structure; higher shearing stresses are usually required to cause a slip. Generally, BCC material is brittle.
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6. The theoretical shear strength of perfect crystal is approximately equal to ___________
a) G/2π
b) G/4π
c) 2πG
d) G
View Answer

Answer: a
Explanation: Shear stress of perfect crystal is expressed as:
τ=τmsin(2πx/b)
Where τm is amplitude of sine wave and b is period
τ=Gγ=Gx/a
τ ≈ τm2πx/b
=> τm=Gb/2πa
=> τm=G/2π

7. Theoretical shear strength of metal range in _______
a) GPa
b) MPa
c) KPa
d) Pa
View Answer

Answer: a
Explanation: The shear modulus for metals is in the range of 20 to 150 GPa. Eq. τm=G/2π; predicts that the theoretical shear stress will be in the range (3 to 30 GPa), while actual values of the shear stress required to produce plastic deformation in single metal crystals are in the range 0.5 to 10 MPa. This large difference in values is due to defects and dislocation present in real crystals.

8. The force required to move a dislocation is known as ____________
a) peierls-nabarro force
b) hearing-nabarro force
c) drag force
d) frictional force
View Answer

Answer: a
Explanation: The dislocation width is important because it determines the force required to move a dislocation through the crystal lattice. This force is called the Peierls-Nabarro force. The Peierls stress is the shear stress required to move a dislocation through a crystal lattice in a particular direction. Hearing Nabarro is a phenomenon in the creep. Drag force is associated with fluid mechanics. The frictional force is the resistance offered by the material on movement in contact with the other surface.

9. The Peierls stress (stress required to move the dislocation) will be least in which type of crystal element?
a) Covalent compound
b) Metallic compound
c) Ionic compound
d) Intermetallic compound
View Answer

Answer: b
Explanation: Covalent solid have strong directional bonds, so the width of dislocation is low. Ionic crystal has large interatomic distance. In intermetallic compound, favorable slip system and directions are absent. So, in the metallic system, it is easiest to move a dislocation, that is why generally metal is ductile.
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10. What is the stress required to move a dislocation when the width of dislocation is equal to the Burger’s vector of the dislocation?
a) G
b) G/1027
c) G/400
d) G/2
View Answer

Answer: c
Explanation: Peierls stress is equal to:
\(\frac{2G}{1-v}e^{(-2\pi w/b)}\)
Ν=0.33, w=b
Substitute the values in the equation;
2G/0.66*(0.00187)
=> G/400.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn