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17-Phys-B7 Structure of Materials · December 2018

Question 6 of 7: Dislocations and Grain Boundaries

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 17-Phys-B7 Structure of Materials, National Examination December 2018 — a closed-book examination (Casio or Sharp approved calculators only; all necessary equations, constants and diagrams supplied in the paper's own appendix). Candidates attempt any five of the seven questions, each worth 20 marks; every question is nonetheless answered in full below so the paper remains a complete study resource.

Reference texts. W. D. Callister Jr. & D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed. (atomic bonding, crystal structure and packing, point defects, diffusion, dislocations and slip, mechanical properties, phase diagrams and the lever rule, precipitation hardening, X-ray diffraction).

Check — two points on the printed paper. (1) Question I.2 prints the ion as "Cl- (Z = 16)"; Z = 16 is sulfur, not chlorine (Cl is Z = 17) — a printed typo. The electronic structure below uses the correct Z = 17 for chlorine. (2) Question III.1(d) prints the hexagonal plane as $(2\bar{2}10)$, i.e. Miller–Bravais indices $h=2,\,k=-2,\,i=1$; the third index of a valid Miller–Bravais symbol is never independent — it is fixed by $i=-(h+k)$, here $i=-(2-2)=0$, not 1. Reducing the self-consistent index $(2\bar200)$ by its common factor of 2 gives $(1\bar100)$, a standard prism-type plane. The drawing below uses the symmetry-equivalent, non-degenerate face $(10\bar10)$ of the same $\{1\bar100\}$ family (chosen because it renders visibly in the cell projection used here) and states this substitution explicitly.

Question VI: Dislocations and Grain Boundaries (20 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.

Part 1(a) — screw vs. edge dislocations. An EDGE dislocation is geometrically an extra HALF-PLANE of atoms inserted into the lattice; its Burgers vector $\vec b$ is PERPENDICULAR to the dislocation line, and it moves (glides) in the direction of $\vec b$ under an applied shear stress. A SCREW dislocation instead corresponds to the lattice planes being sheared into a helical ("screw-thread") ramp around the dislocation line; its Burgers vector is PARALLEL to the dislocation line. Under TORSIONAL loading of an FCC solid, the applied stress state is pure shear acting parallel to the shaft (specimen) axis, which is exactly the geometry that drives dislocation lines running along that axis — SCREW dislocations are expected to dominate, because torsion twists one cross-section relative to the next about the shaft axis, the same helical-shear geometry a screw dislocation embodies, and because a screw dislocation's glide plane is not fixed (it can cross-slip onto any plane containing $\vec b$), which particularly favours screw character under a shear stress that rotates with position along the shaft.

Part 1(b) — Burgers vector. The Burgers vector $\vec b$ is the closure FAILURE of an atom-to-atom loop drawn around the dislocation line (a loop that would close perfectly in a perfect crystal): it quantifies both the magnitude and direction of the lattice distortion the dislocation carries, and it is conserved along the dislocation line (it cannot change partway along a single dislocation). Copper is FCC, so its perfect (full) dislocations have the shortest lattice translation vector, $\tfrac12\langle110\rangle$, i.e. $$\vec b = \frac{a}{2}[110]$$ (or any symmetry-equivalent $\tfrac a2\langle110\rangle$ direction) — this SAME Burgers vector applies to both the edge and the screw dislocation in Cu; what differs between the two is only the RELATIVE ORIENTATION of the dislocation LINE to $\vec b$ (perpendicular for edge, parallel for screw), not $\vec b$ itself.

Part 2(a) — deformation by twinning. Twinning is a deformation mechanism in which a region of the crystal shears homogeneously such that the atoms on one side of a mirror ("twin") plane are displaced into a mirror-image orientation of the atoms on the other side, by an amount PROPORTIONAL to each atom's distance from the twin plane (unlike slip, where atoms move whole numbers of lattice translations and the lattice orientation is unchanged across the slip plane). Twinning produces a distinct, differently-oriented region (the "twin") within the parent grain and is important at low temperature / high strain rate, or in metals (e.g. HCP metals, or BCC metals at low $T$) with too few easy slip systems to accommodate an arbitrary shape change by slip alone.

Part 2(b) — slip systems in FCC Cu. Slip occurs most easily on the close-packed PLANES, in the close-packed DIRECTIONS lying within them, since these maximize atomic packing (lowest energy barrier to sliding) and minimize $\vec b$ (Burgers vector length, $\propto$ dislocation energy). For FCC, the close-packed planes are the $\{111\}$ octahedral planes (4 non-parallel members of the family) and the close-packed directions are $\langle110\rangle$ (3 independent directions lying within each such plane), giving $4\times3=\boxed{12\ \{111\}\langle110\rangle\ \text{slip systems}}$ for pure FCC Cu.

FCC slip system {111}<110> [-110] [0-11] [10-1] (111)
One $(111)$ close-packed plane (shaded green) of the FCC cell, with its three close-packed $\langle110\rangle$ slip directions (blue arrows) — these ARE the three edges of the triangular $(111)$ plane traced through the cube corners. The same construction applies to the other three $\{111\}$-family planes, giving $4\times3=12$ slip systems in total.

Given. 70%Cu–30%Zn brass: grain size $d_1=100\ \mu\text{m}$, yield strength $\sigma_{y1}=64.5$ MPa; friction stress $\sigma_0=25$ MPa. New grain size $d_2=50$ nm.

Find. Yield strength $\sigma_{y2}$ at the refined grain size.

Approach. Use the Hall–Petch relation $\sigma_y=\sigma_0+kd^{-1/2}$: back out the material's Hall–Petch slope $k$ from the given $(d_1,\sigma_{y1})$ pair, then apply it at $d_2$.

  1. Hall–Petch slope $k$. $$k = (\sigma_{y1}-\sigma_0)\sqrt{d_1} = (64.5-25)\sqrt{100\times10^{-6}\ \text{m}} = (39.5)(0.01) = \boxed{0.395\ \text{MPa}\cdot\sqrt{\text{m}}}.$$
  2. Yield strength at $d_2=50$ nm. $$\sigma_{y2} = \sigma_0 + \frac{k}{\sqrt{d_2}} = 25 + \frac{0.395}{\sqrt{50\times10^{-9}}} = 25+1766.5 = \boxed{1791\ \text{MPa}}.$$
Check — this is a large extrapolation. The Hall–Petch relation is experimentally well established for grain sizes from roughly 1 µm up to a few hundred nanometres; below about 10–20 nm, real nanocrystalline metals typically show the "inverse Hall–Petch" softening effect (grain-boundary sliding takes over from dislocation pile-up), so the boxed $1791$ MPa is the value the FORMULA predicts, useful as an illustration of the strengthening TREND, but should not be read as a literal, achievable strength for a real 50 nm-grain brass.
Final results — Question VI
QuantityValue
Dominant dislocation type in torsion (FCC)Screw
Burgers vector, Cu (edge and screw)$\tfrac a2[110]$
Number of {111}<110> slip systems (FCC)12
Hall–Petch slope $k$0.395 MPa·m1/2
Yield strength at $d=50$ nm1791 MPa (extrapolated — see the check note)