Question 5 of 8: Dislocations, Slip and Grain Boundaries (20 marks)

Question V — Dislocations, Slip 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.

V.1 — Burgers vector; edge vs. screw dislocations

The Burgers vector $\mathbf{b}$ of a dislocation is defined by tracing a closed atom-to-atom circuit (a Burgers circuit) around the dislocation line in the real, defective crystal, using the same step sequence that would close perfectly in a perfect reference crystal; in the real crystal the circuit fails to close, and the vector needed to close it (from finish to start) is $\mathbf{b}$. It is a single, fixed vector for the whole length of a given dislocation line, however that line curves, and it quantifies the magnitude and direction of the lattice distortion the dislocation carries.

In an edge dislocation, an extra half-plane of atoms is inserted partway into the crystal; the Burgers vector is perpendicular to the dislocation line. In a screw dislocation, the crystal is instead sheared so that the atomic planes form a continuous helical ramp around the dislocation line, and the Burgers vector is parallel to the line. Most real dislocations are mixed, with edge and screw components that vary along a curved line, but the Burgers vector itself stays constant along the whole line in either case.

V.2 — Dislocations, slip and ductility

Plastic deformation in crystalline metals occurs almost entirely by slip: the coordinated glide of dislocations along close-packed planes in close-packed directions. A dislocation can move an entire plane of atoms past its neighbour by breaking and re-forming only the bonds immediately at its core, one row at a time, which requires vastly less stress than shearing every bond across the whole plane simultaneously (as a dislocation-free, "perfect" crystal would need to). Ductility — the capacity to sustain large plastic strain before fracture — is therefore fundamentally a consequence of dislocations existing and being mobile: a material with abundant slip systems and low lattice friction (e.g. FCC metals, with twelve {111}<110> systems) can accommodate large shape changes via slip and is ductile, while a material where dislocation motion is difficult (few slip systems, high Peierls stress, as in many ceramics) fractures with little plastic flow. Anything that impedes dislocation motion (solid-solution atoms, precipitates, grain boundaries, other dislocations) raises the strength but, by making slip harder, typically reduces ductility.

V.3 — FCC nickel single crystal loaded along [100]

Given. FCC nickel rod, radius $r=20\ \text{mm}$, loading axis $[100]$, axial load $F=50\ \text{kN}$.

Find. (a) the active slip system, (b) how many slip systems share the maximum Schmid factor for this orientation, (c) the Schmid factor, (d) the minimum resolved shear stress (= critical resolved shear stress condition) at this load.

1. (a) Slip system. FCC metals slip on the close-packed $\{111\}$ planes in the close-packed $\langle110\rangle$ directions — the $\{111\}\langle110\rangle$ system (twelve variants total: 4 planes × 3 directions each).
2. (b) Number of simultaneously-favoured systems. $[100]$ is a high-symmetry direction: it makes an identical angle with all four $\{111\}$ plane normals, and within each plane two of the three $\langle110\rangle$ slip directions are symmetric about $[100]$ and share the same (maximum) resolved shear stress. That gives $4\ \text{planes}\times2\ \text{directions} = \boxed{8\ \text{equally-favoured slip systems}}$, which is why single crystals loaded along $\langle100\rangle$ tend to show multiple, symmetric slip traces rather than single-system glide.
3. (c) Schmid factor. Take the representative system $(111)[10\bar1]$. The angle $\phi$ between the load axis $[100]$ and the slip-plane normal $[111]$, and the angle $\lambda$ between $[100]$ and the slip direction $[10\bar1]$: $$\cos\phi = \frac{[100]\cdot[111]}{|[100]||[111]|} = \frac{1}{\sqrt3},\qquad \cos\lambda = \frac{[100]\cdot[10\bar1]}{|[100]||[10\bar1]|} = \frac{1}{\sqrt2}$$ $$m = \cos\phi\cos\lambda = \frac{1}{\sqrt3}\cdot\frac{1}{\sqrt2} = \boxed{0.408}$$ (All 8 systems from part (b) share this same value by the symmetry noted above.)
4. (d) Minimum resolved shear stress. Normal (axial) stress from the applied load: $$\sigma = \frac{F}{A} = \frac{50000}{\pi(0.020)^2} = 39.79\ \text{MPa}$$ Schmid's law, $\tau_R = m\sigma$: $$\tau_R = (0.408)(39.79) = \boxed{16.2\ \text{MPa}}$$ This is the resolved shear stress delivered to the (equally-favoured) active slip systems by the 50 kN load; slip begins once $\tau_R$ reaches the crystal's critical resolved shear stress $\tau_{CRSS}$, so 16.2 MPa is the minimum shear stress this particular load state can drive onto the slip planes.