# Mechanical Metallurgy Questions and Answers – Dislocation Climb & Intersection of Dislocation

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This set of Mechanical Metallurgy Multiple Choice Questions & Answers (MCQs) focuses on “Dislocation Climb & Intersection of Dislocation”.

1. The stacking fault energy of aluminium is 200 mJ m-2. Find the equilibrium spacing between partial dislocation of aluminium. If the shear modulus of the material is 27 GPa and lattice parameter is 3.57 A°.
a) 20 A°
b) 1000 A°
c) 1 A°
d) 20 micron

Explanation: Equilibrium spacing between partial dislocation is given by:
d=Gb2/2π(1-v)ϒ
First, find out the burger’s vector for the aluminum
b=ao/2(110); the slip direction in FCC material is 110
(3.57*√2)/2= 2.53 A°
Substitute the values in the equation:
27*109(2.53*10-10)2/[(2*3.14)(1-0.33)*200*10-3)
0.208*10-8
20.8 A°.

2. A dislocation always feels the attractive force from the surface of the material.
a) True
b) False

Explanation: In general, dislocation is always attracted to the surface of the material, but if the surface is corroded, then the thin oxide layer tends to repeals the dislocation from the surface.

3. Which of the following is a non-conservative type of movement of dislocation?
a) The glide of positive edge dislocation on the slip plane
b) The glide of positive edge dislocation on the slip plane
c) The glide of positive screw dislocation on the slip plane
d) The climb of an edge dislocation

Explanation: The conservative type of movement is the one which occurs for the same slip plane. But in case of climb of edge dislocation, it leaves the slip plane to move on parallel slip plane, hence termed as a non-conservative movement.

4. The dislocation climb is least affected by ___________
a) movement of the atom to dislocation
b) movement of vacancy to the dislocation
c) shear stress
d) high Temperature

Explanation: The climb of dislocation is generally associated with the movement of the atom to dislocation end for the negative climb and removal atom from slip plane for a positive climb. This movement of the atom will be easier at higher temperatures. But shear stress plays a critical role in the glide of dislocation but not in the climb of dislocation.

5. It is commonly observed that climb generate jog in dislocation line because ________
a) dislocation climb creates stress in dislocation line to bend it
b) the extra plane of atoms is highly unlikely to remove from the entire width of dislocation
c) a jog is a form of dislocation climb
d) statement is incorrect

Explanation: It is highly unlikely that complete rows of atoms are removed or added at the extra half plane in the climbing process. Individual vacancies diffuse to the dislocation and climb occurs over a short segment of the dislocation line. This incomplete removal or addition of atoms results in the formation of steps or jogs along the dislocation line.

6. The activation for the climb is equal to _____
where;
Uj = Activation for nucleation of jog
Uv = Activation for formation of the vacancy
Um = Activation for formation of the self diffusion
Ui = Activation for formation of the interstitial diffusion
a) Uj+Uv+Um
b) Uj
c) Uj+Uv
d) Uj+Uv+Um+Ui

Explanation: In the crystal, there is thermodynamically equilibrium number of jog present. The number of jogs present is a crystal is governed by the rate of formation of vacancy and diffusion of an atom to dislocation site.

7. Dislocation climb is temperature dependent phenomena and present in both edge and screw dislocation.
a) True
b) False

Explanation: Climb is only possible in edge dislocation, not in screw dislocation because there is no extra half plane present in screw dislocation.

8. No Diffusion of atom is needed to allow ____________ to move from one plane to another plane.
a) edge dislocation
b) screw dislocation
c) jog
d) kink

Explanation: The Burgers vector of a screw dislocation is parallel to the dislocation, so it is free to slip from one plane to another plane which contains the dislocation line and the Burgers vector. No diffusion of atoms is needed to allow the screw dislocation to move on to another slip plane.

9. Two edge dislocation of same sign lies in parallel slip angle with an angle of 30 degrees with each other. What will the nature of force between these two dislocations?

a) Attractive
b) Repulsive
c) Neither attractive nor repulsive
d) Cannot predict with given information

Explanation: The force between interacting dislocation can be given as:
$$\frac{Gb^{2x}(x^2-y^2)}{2π(1-v) (x^2-y^2)^2}$$

The plot of this curve will look like:

The solid line represents the two edge dislocation of the same sign, and the dashed line represents the two dislocations of opposite sign. So if (x>y) means the angle is less than 45 degrees, the nature of force is repulsive.

10. Two edge dislocation of opposite sign lies in parallel slip angle with an angle of 55 degrees with each other. What will the nature of force between these two dislocations?

a) Attractive
b) Repulsive
c) Neither attractive nor repulsive
d) Cannot predict with given information

Explanation: The force between interacting dislocation can be given as:
$$\frac{Gb^{2x}(x^2-y^2)}{2π(1-v) (x^2-y^2)^2}$$

The plot of this curve will look like:

The dashed line represents the two edge dislocation of opposite sign, and the dashed line represents the two dislocations of opposite sign. So if (x<y) means the angle is more than 45 degrees the nature of force is repulsive.

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