Question 6 of 7: Dislocations and Grain Boundaries

Question 6: 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.

6.1(a) — Dislocations, Slip, Yield Strength and Ductility

A dislocation is a line defect whose motion through the crystal lattice, one atomic bond-shuffle at a time along a close-packed plane and direction, constitutes slip — the mechanism of plastic deformation in crystalline metals. Slip is dramatically easier than breaking every bond across a plane simultaneously (the theoretical shear strength), which is why real metals yield at stresses orders of magnitude below the theoretical value. Yield strength is therefore, physically, the applied stress at which dislocations begin to move (and multiply) macroscopically on their easiest available slip systems — anything that impedes dislocation motion (solute atoms, grain boundaries, precipitates, other dislocations) raises the yield strength. Ductility is the capacity for slip to continue accumulating (dislocations gliding, multiplying, interacting) before fracture; a material with many available slip systems (e.g. FCC's 12) and mobile dislocations is ductile, while one that cannot easily generate/move dislocations (e.g. covalent ceramics, or BCC/HCP at low temperature) fractures with little slip and is brittle.

6.1(b) — Cross Slip

Cross slip is the process by which a screw-dislocation segment, which (unlike an edge dislocation) is not confined to a single slip plane because its line and Burgers vector are parallel, switches from gliding on one crystallographic slip plane to a different plane that shares the same slip direction (same Burgers vector) — typically to bypass an obstacle (precipitate, forest dislocation, pile-up) blocking its original glide plane. Cross slip is confirmed experimentally by transmission electron microscopy (TEM) of thin foils, which reveals characteristic wavy or zig-zag slip traces (versus the perfectly straight, planar traces of single-plane glide), and by etch-pit or slip-line analysis on a polished, deformed surface showing slip bands that change plane partway along their length. Cross slip plays a central role in plasticity: it allows dislocations to circumvent obstacles that would otherwise pile up and stop deformation, it is the mechanism behind stage III (dynamic-recovery) work softening in the stress–strain curve as annihilation of opposite-sign screw dislocations becomes possible, and materials with low stacking-fault energy (where partial dislocations are widely separated and must first constrict before cross-slipping) show suppressed cross slip and correspondingly more planar, less recoverable slip.

6.2 — Resolved Shear Stress in BCC Iron

Given. Tensile axis $[010]$; slip plane $(110)$; slip direction $[\bar111]$; applied stress $\sigma=52$ MPa (part a); critical resolved shear stress $\tau_{\text{CRSS}}=30$ MPa (part b).

Find. (a) $\tau_R$ at $\sigma=52$ MPa; (b) $\sigma$ at yielding.

Approach. Schmid's law, $\tau_R=\sigma\cos\phi\cos\lambda$, with $\phi$ the angle between the stress axis and the slip-plane normal and $\lambda$ the angle between the stress axis and the slip direction, both from dot products of the direction/normal index vectors.

1. Direction cosines. Plane normal $\mathbf{n}=[110]$, slip direction $\mathbf{d}=[\bar111]$, stress axis $\mathbf{t}=[010]$: $$\cos\phi=\frac{\mathbf{t}\cdot\mathbf{n}}{|\mathbf{t}||\mathbf{n}|}=\frac{1}{\sqrt2}=0.7071\qquad \cos\lambda=\frac{\mathbf{t}\cdot\mathbf{d}}{|\mathbf{t}||\mathbf{d}|}=\frac{1}{\sqrt3}=0.5774$$
2. (a) Resolved shear stress at $\sigma=52$ MPa. $$\tau_R=\sigma\cos\phi\cos\lambda=52(0.7071)(0.5774)=\boxed{21.2\ \text{MPa}}$$
3. (b) Applied stress at yielding, $\tau_R=\tau_{\text{CRSS}}$. $$\sigma_y=\frac{\tau_{\text{CRSS}}}{\cos\phi\cos\lambda}=\frac{30}{0.4082}=\boxed{73.5\ \text{MPa}}$$

6.2(c) — Slip Systems: BCC Iron vs. FCC Copper

FCC copper slips on its close-packed $\{111\}$ octahedral planes in the close-packed $\langle110\rangle$ directions: the $\{111\}\langle110\rangle$ family gives 4 distinct $\{111\}$ planes × 3 $\langle110\rangle$ directions each = 12 slip systems, e.g. $(111)[\bar101]$, $(111)[0\bar11]$, $(\bar111)[101]$, etc. BCC has no truly close-packed plane, so slip is observed on several plane families that all share the close-packed $\langle111\rangle$ direction: primarily $\{110\}\langle111\rangle$ (48/2 = the most common, 6 planes × 2 directions each = 12 systems), with $\{112\}\langle111\rangle$ (12 systems) and $\{123\}\langle111\rangle$ (24 systems) also active at higher temperature/stress — e.g. $(110)[\bar111]$, $(112)[\bar111]$, $(123)[\bar111]$ all share the $[\bar111]$ direction used in part 6.2(a)/(b) above.