Question 2 of 8

Question 2 (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. FCC lattice; close-packed slip plane (111); candidate slip directions are the twelve signed permutations of the <110> family.

Find. Which <110>-type directions $[uvw]$ actually lie in the (111) plane, i.e. form a valid FCC {111}<110> slip system.

Approach. A direction $[uvw]$ lies in a plane $(hkl)$ if and only if the zone law is satisfied: $hu+kv+lw=0$. Apply this with $(hkl)=(111)$ to every signed permutation of $\{1,1,0\}$.

1. Part (a) — apply the zone law. For (111), the condition is $u+v+w=0$. Testing the twelve signed <110> vectors: $[110]$ gives $1+1+0=2\neq0$ (out of plane); $[1\bar10]$ gives $1-1+0=0$ (in plane); and so on. Carrying this through the full family leaves exactly six vectors satisfying the condition, forming three independent lines (each line counted once, since $[uvw]$ and $[\bar u\bar v\bar w]$ describe the same crystallographic direction): $$\boxed{[\bar110],\quad[10\bar1],\quad[01\bar1]}$$
2. Part (a) — physical check. These three directions are mutually at $120^\circ$ in the (111) plane (each pair's dot product is $\tfrac12$ of the product of magnitudes, consistent with $\cos60^\circ$ between the positive vectors, i.e. $120^\circ$ between the slip-direction lines), matching the three edges of the close-packed triangular (111) net. This is the FCC {111}<110> slip system: 4 distinct {111} planes $\times$ 3 <110> directions each $=12$ independent slip systems, which is why FCC metals (Cu, Al, Ag, Ni, austenitic steel) are so ductile.
3. Part (b) — sketch. See the figure: the (111) plane forms an equilateral triangle of close-packed atoms, and the three valid <110> directions run along its three edges.