Question

Which of the following is a possible slip plane for BCC iron?

Which of the following is a possible slip plane for BCC iron? O (110) O (211) O (100) • (111) Submit Request Answer

Answer 1

In materials science, a slip system describes the set of symmetrically identical slip planes and associated family of slip directions for which dislocation motion can easily occur and lead to plastic deformation.

Slip in face centered cubic (fcc) crystals occurs along the close packed plane. Specifically, the slip plane is of type {111}, and the direction is of type <110>. In the diagram on the right, the specific plane and direction are (111) and [110], respectively.

Given the permutations of the slip plane types and direction types, fcc crystals have 12 slip systems. In the fcc lattice, the norm of the Burgers vector, b, can be calculated using the following equation

${\displaystyle |b|={\frac {a}{2}}|\langle 110\rangle |={\frac {a{\sqrt {2}}}{2}}}$

Where a is the lattice constant of the unit cell.

Slip in body-centered cubic (bcc) crystals occurs along the plane of shortest Burgers vector as well; however, unlike fcc, there are no truly close-packed planes in the bcc crystal structure. Thus, a slip system in bcc requires heat to activate.

Some bcc materials (e.g. α-Fe) can contain up to 48 slip systems. There are six slip planes of type {110}, each with two <111> directions (12 systems). There are 24 {123} and 12 {112} planes each with one <111> direction (36 systems, for a total of 48). While the {123} and {112} planes are not exactly identical in activation energy to {110}, they are so close in energy that for all intents and purposes they can be treated as identical. In the diagram on the right the specific slip plane and direction are (110) and [111], respectively.

${\displaystyle |b|={\frac {a}{2}}|\langle 111\rangle |={\frac {{\sqrt {3}}a}{2}}}$

so the possible slip plane for BCC iron is (110)