Past exam of the materials science course of the University of Cambridge / 2019 / Paper 1 / 5 / a / Solution

An FCC crystal slips on $\{111\}⟨110⟩$. For each slip system, Schmid's law gives $\tau =\sigma m$, with $m=|\hat{\mathbf{l}}\cdot \hat{\mathbf{n}}||\hat{\mathbf{l}}\cdot \hat{\mathbf{s}}|$. Enumerating the twelve systems for $\mathbf{l}=(5,1,4)$ gives the maximum atHere $\mathbf{n}=(1,-1,1)$ and $\mathbf{s}=(1,1,0)$ are perpendicular, while $\mathbf{l}\cdot \mathbf{n}=8$, $\mathbf{l}\cdot \mathbf{s}=6$ and $|\mathbf{l}{|}^2=42$. Hence $m=48/(42\sqrt{6})=0.46657$. Reversing both directed slip conventions describes the same physical system.