Question

[7Pts] B2. (a) Aluminum is a FCC metal. Identify families slip planes and directions in Al (write their Miller indices). On what basis did you select the planes and direction? Explain. (b) Consider the figure given below. Draw the Burger's circuit to identify and mark the Burgers vector b. What is the angle between the Burgers vector band the dislocation line? Based on b, classify this dislocation as edge or screw type. Dislocation Motion IT

Answer 1

1. The Slip Planes in Al (FCC) are planes with miller indices {111} and the Slip directions in this are <110>.

The Slip planes are the planes with highest planar density, while the slip directions are the directions with highest linear density. In other words, the energy of a dislocation according to Frank's rule will be maximum along these planes and directions. It can also be said that along these directions and planes, the work requirement for movement of dislocation will be minimum. The Planar and Linear densities can be calculated by:

Planar density = (No. of atoms in the plane)/(area of the plane)

Linear density = (No. of atoms in the plane)/(length of the line or direction)

2. The Burger's circuit has been drawn in the figure below, with the assumption that the circuit is Right handed with thumb pointing in direction of the dislocation motion (unit vector t). Since there is a defect in the crystal, the citcuit is not closed. We started the circuit at point s and finished it at point f. Hence our Burger's vector (b) is a vector whose start point is f and end point is s. it has been shown in the figure. It is clear that the Burger's vector is Paralell to the dislocation line, and hence the dislocation is of Screw type.

Dislocation Motion NA HA