This set of Advanced Mechanical Behaviour Questions and Answers focuses on “Dislocation Multiplication”.
1. The stress required to activate the longest Frank-Read source would be _____ and _____ on the crystal structure.
a) large, depends
b) large, doesn’t depend
c) small, depends
d) small, doesn’t depend
View Answer
Explanation: σc ≈ 2T/lb, here l is free length, T is tension and b is Burgers vector. The stress required to activate the largest Frank-Read sources would be small. It varies from crystal to crystal.
2. The density of dislocations increases by an order of ____ in magnitude during deformation.
a) 2-4
b) 6-7
c) 8-9
d) 12-13
View Answer
Explanation: Density of dislocations increases during plastic deformation. It is due to the Frank-Read source. It increases by an order of 2 to 4 in magnitude depending on the degree of deformation.
3. What is density of dislocations in annealed crystal?
a) 108-1010 m-2
b) 1010-1012 m-2
c) 1012-1014 m-2
d) 1014-1016 m-2
View Answer
Explanation: Annealing heat treatment reduces dislocation density in crystal. It is due to annihilation of dislocations on high temperature. Dislocation density ranges from 1010-1012 m-2.
4. What is the density of dislocations in heavily deformed crystal?
a) 108-1010 m-2
b) 1010-1012 m-2
c) 1012-1014 m-2
d) 1014-1016 m-2
View Answer
Explanation: Heavy plastic deformation causes rapid increase in dislocation density in crystal. It is in range 1014-1016 m-2 . It causes strain hardening effect in crystal.
5. Which metals can be cold worked to the largest extent?
a) FCC
b) BCC
c) HCP
d) SC
View Answer
Explanation: BCC metals show some covalent characteristic due to ‘d’-orbital bonding. Therefore, these metals show less ductility. FCC metals show high ductility and wide dislocations so these can be cold worked to the largest extent.
6. Higher dislocation density ____ yield strength of the metal.
a) increases
b) decreases
c) doesn’t affect
d) can increase or decrease
View Answer
Explanation: Cold working of materials causes an increase in dislocation density. It is due to the formation of Frank-Read source. Higher dislocation density enhances yield strength due to work hardening.
7. Dislocations generate on _____ under applied stress.
a) every plane
b) slip planes
c) planes with Frank-Read source
d) {1 1 1} planes
View Answer
Explanation: Dislocation generation takes place on planes with Frank-Read source. These sources produce dislocation loops on application of stress. It can be understood by Frank-Read source mechanism.
8. The stress required for the generation of dislocation through Frank-Read source is inversely proportional to _____
a) shear modulus
b) burgers vector
c) distance between the pinning sites
d) p-n stress
View Answer
Explanation: The stress required is given by τ = 2Gb/x. Here G is shear modulus, b is Burgers vector and x is a distance between pinning sites. Hence it is inversely proportional to x.
9. Which factor is not a reason of Frank-Read source immobile pinning sites?
a) Other solute atoms
b) Second phase particles
c) Solvent atoms
d) Intersection with other dislocations
View Answer
Explanation: Frank read source forms by immobile pinning sites. These sites may form by other solute atoms, second phase particles or intersection with other dislocations. These generate dislocation loops on applying stress.
10. Which of the following term is used to refer to the immobile pinning sites in the Frank-Read source?
a) Nodal points
b) Pin points
c) Frank points
d) Shear points
View Answer
Explanation: The points say X and Y, remain immobile in applying stress. These are generally pinned by second phase particles. Hence, they are known as nodal points.
11. Frank-Read source is a mechanism of dislocation multiplication.
a) True
b) False
View Answer
Explanation: Frank-Read source is a mechanism which explains dislocation multiplication. It explains dislocation multiplication in well spaced slip planes on deformation.
12. The dislocation bowing decreases with increasing applied stress.
a) True
b) False
View Answer
Explanation: Bowing (R) is given by R ∝ Gb/τ. Here G is shear modulus, b is Burgers vector and τ is applied stress. Hence bowing increases with increase in applied stress.
Sanfoundry Global Education & Learning Series – Mechanical Behaviour & Testing of Materials.
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