Question 1
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: Option 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.
Question 2
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: Option 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.
Question 3
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: Option 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.
Question 4
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: Option 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.
Question 5
The theoretical shear strength of perfect crystal is approximately equal to ___________
A. G/2π
B. G/4π
C. 2πG
D. G
View Answer
Answer: Option A
Explanation:
Shear stress of perfect crystal is expressed as:τ=τmsin(2πx/b)Where τm is amplitude of sine wave and b is periodτ=Gγ=Gx/aτ ≈ τm2πx/b=> τm=Gb/2πa=> τm=G/2π
Question 6
Theoretical shear strength of metal range in _______
A. GPa
B. MPa
C. KPa
D. Pa
View Answer
Answer: Option 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.
Question 7
What is closed packed direction body-centered cubic system?
A. <111>
B. <110>
C. <100>
D. <122>
View Answer
Answer: Option 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.
Question 8
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
Question 9
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: Option 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.
Question 10
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: Option 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.