# Mechanical Metallurgy Questions and Answers – Stress Field & Energy of Dislocation

This set of Mechanical Metallurgy Multiple Choice Questions & Answers (MCQs) focuses on “Stress Field & Energy of Dislocation”.

1. In case of screw dislocation, the stresses around dislocation are ________
a) tensile in nature for right-hand screw dislocation
b) tensile in nature for left-hand screw dislocation
c) compressive in nature for right-hand screw dislocation
d) there are no tensile or compressive normal stresses

Explanation: In the case of screw dislocation there is no extra half plane associated, So the only shear component of stress exist.

2. One screw and one edge dislocation will have same strain energy if the burger vector is equal.
a) True
b) False

Explanation: Strain energy of edge dislocation will be higher compared to screw dislocation if both the dislocation has the same burgers vector.

3. The ratio of strain energy of edge dislocation to screw dislocation is equal to ___________
Where v is poison’s ratio, and b is the Burger’s vector.
a) 1/(1-v)
b) v
c) vb
d) b

Explanation: Strain energy of screw dislocation is equal to: $$\frac{(Gb^2)}{4π} ln⁡(\frac{r1}{ro})$$
The strain energy of edge dislocation is : $$\frac{(Gb^2)}{4π(1-v)} ln⁡(\frac{r1}{ro})$$
So the ratio will be 1/(1-v)

4. For a dislocation with burger’s vector equal to b, the strain energy per unit length is X. What will be the strain energy with burger’s vector is equal to 3b?
a) X
b) 3X
c) 9X
d) √3X

Explanation: Strain energy is propositional to Gb2.
So if the burger’s vector is increased by three times, the strain energy will be increased by nine times.
So total strain energy per unit length=9X

5. The strain energy of dislocation for each atom is in order of few __________
a) eV
b) mJ
c) J
d) V

Explanation: The strain energy of dislocation for each atom is around 8eV, while the core energy of dislocation is about 0.5eV (around 10-19J) per atom in the plane.
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6. Free energy of crystal __________ by introduction of dislocation in the structure.
a) increases
b) decreases
c) remain unchanged
d) sometimes increases and sometimes decreases

Explanation: Dislocation is associated with positive strain energy. When dislocation is generated in crystal, the positive strain energy adds to the total free energy. Hence, as a result, the free energy of the system increases.

7. The following diagram shows a positive edge dislocation. The AA’ shows the slip plane for the dislocation. Above the slip plane stress is _________ in nature, and below the slip plane stress is _______ in nature.

a) compressive, tensile
b) tensile, compressive
c) tensile, pure shear
d) compressive, pure shear

Explanation: Above the slip plane AA’ there is extra lattice plane introduced in the crystal, so it results in compressive stress in the lattice. In the plane below AA’ there is one lattice plane missing which result in tensile stress in the system.

8. Shear stress required to bend a dislocation of radius R and burger vector b is P. What will be shear stress needed to turn a dislocation of radius 2R and burger’s vector 2b to bend the dislocation?
a) P
b) 2P
c) 4P
d) 16P

Explanation: Shear stress required to turn a dislocation is:
τ=Gb/2R; where G is shear modulus, b is burger’s vector, R is the radius of dislocation.
P=Gb/2R
So if the radius is increased two times, and burger’s vector is also increased by two times, the net shear stress value will remain unchanged.
G(2b)/2(2R)=Gb/2R=P

9. Dislocation of same sign on the same plane will ___________ each other.
a) attract
b) repeal
c) not interact with
d) have stable equilibrium with

Explanation: The dislocation of the same sign will have compressive stress above the slip plane and tensile stress below the slip plane. So when two dislocations of the same sign on the same plane combines, they will create a region of higher compressive and tensile stress. This combination of two dislocations will result in an increase of free energy of the system. That is why they try to repel each other to minimize the energy of the system.

10. Dislocation of opposite sign on the same plane will ___________ each other.
a) attract
b) repeal
c) not interact with
d) have a stable equilibrium with

Explanation: The dislocation of the opposite sign will have compressive stress and tensile stress on the same side of the slip plane. So these two dislocations will try to attract each other to minimize the overall energy of the system. Also, they will annihilate each other.

11. Two edge dislocations of the same sign and same burger’s vector lie on above the same slip plane. If the energy of each dislocation is Q, what will be the difference in the total energy of two dislocations, when they are separated and when they are combined?

a) 0
b) Q
c) 2Q
d) 4Q

Explanation: The strain energy of dislocation is $$\frac{Gb^2}{4π(1-v)} ln⁡(\frac{r1}{ro})$$
$$\frac{Gb^2}{4π(1-v)} ln⁡(\frac{r1}{ro})$$=Q.
Total energy when dislocations are separated is 2Q.
Now when dislocation combines the burgers vector will become 2b.
So the total energy of combine of the two dislocations will be:
$$\frac{G(2b)^2}{4π(1-v)} ln⁡(\frac{r1}{ro})=\frac{4Gb^2}{4π(1-v)} ln⁡(\frac{r1}{ro})=4Q$$
So after combining total energy will be 4Q.
So the difference in the energy will be 4Q-2Q=2Q.

12. The force is _____________ between 2 parallel screw dislocation and ____________ for antiparallel screw dislocation.
a) attractive, repulsive
b) repulsive, attractive
c) attractive, attractive
d) repulsive, repulsive

Explanation: The force is attractive between 2 parallel screw dislocation and repulsive for antiparallel screw dislocation. Stress field in a screw dislocation is radially symmetrical.

13. Two opposite sign dislocations are separated by 100 nm in iron with grain size 4 mm. Determine the attractive force and total force on the two dislocations? Shear modulus of iron is 80 GPa, and the lattice parameter is 2.86 Ao
a) 312.59*10-3 N/m, 1.25*10-9 N
b) 112.59*10-3 N/m, 8.25*10-9 N
c) 312.59*10-9 N/m, 1.25*10-3 N
d) 312.59 N/m, 1.25 N

Explanation: The attractive force between the dislocation will be:
F=Gb2/2πr
80*109(b)2/2(3.14)(100*10-10)
The burger vector for iron=BCC=a0/2(111), So half the distance of ao direction.
Length of (111) = √3 a0 = √3*2.86*10-10 = 4.95*10-10
80*109(4.95*10-10)2/2(3.14)(100*10-10)
312.59*10-3 N/m
The total force on the dislocation will be:
312.59*10-3*4*10-3 = 1.25*10-9 N.

14. The number of dislocation of the same sign in the vertical plane is in stable equilibrium.
a) True
b) False

Explanation: A vertical array of edge dislocations of the same sign is always in stable equilibrium. This same arrangement of dislocations is followed in a low-angle grain boundary of the tilting variety.

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