Dislocations and Grain Boundaries: 21-Mat-A4 Deformation Behaviour and Properties of Materials, December 2018, Q6

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.

VI.1(a) — Screw vs. Edge Dislocations; Torsional Deformation

An edge dislocation is the line defect bounding an extra half-plane of atoms inserted into the lattice; its Burgers vector is perpendicular to the dislocation line, and it moves by simple slip in the direction of $\mathbf{b}$ under a shear stress. A screw dislocation instead corresponds to a helical distortion of the lattice planes around the dislocation line (no extra half-plane); its Burgers vector is parallel to the dislocation line, and it can cross-slip onto any plane containing both the line and $\mathbf{b}$, unlike an edge dislocation which is confined to its own glide plane. Under torsional loading of an FCC solid, the imposed stress state is pure shear acting parallel to the specimen's long axis; because a screw dislocation's line and Burgers vector are both aligned along the direction of shear (and it is free to cross-slip to relieve local stress concentrations), screw dislocations are expected to dominate the resulting plastic deformation.

VI.1(b) — Burgers Vector; Screw and Edge in Cu

The Burgers vector $\mathbf{b}$ quantifies the magnitude and direction of lattice distortion associated with a dislocation: tracing a closed atom-to-atom loop (a Burgers circuit) around the dislocation line in the real, distorted crystal fails to close by exactly $\mathbf{b}$ when the identical circuit is drawn in a perfect reference lattice. In FCC copper, the perfect (full) dislocation glides on the close-packed $\{111\}$ plane along the close-packed $\langle110\rangle$ direction (Question VI.2(b) below), so for both the edge and the screw dislocation in Cu: $$\mathbf{b} = \frac{a}{2}[1\bar{1}0] \quad\text{(or any}\ \tfrac{a}{2}\langle110\rangle\ \text{family member),}\qquad |\mathbf{b}|=\frac{a\sqrt2}{2}$$ The two dislocation types share the identical Burgers vector magnitude and direction in FCC (both are $\tfrac{a}{2}\langle110\rangle$, the shortest lattice translation vector, which is what makes it energetically favoured); they differ only in how the dislocation line is oriented relative to that fixed vector — perpendicular for edge, parallel for screw.

VI.2(a) — Deformation by Twinning

Twinning is a plastic 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 positions that are the mirror image of the parent lattice on the other side — the sheared region ("twin") and the untwinned parent are related by a precise crystallographic reflection, rather than by the simple parallel-plane translation of slip. Twinning typically occurs at low temperature and/or high strain rate, and in materials (like HCP metals, or FCC metals with low stacking-fault energy) where the limited number of easy slip systems cannot otherwise accommodate the imposed strain.

VI.2(b) — Slip Systems in FCC Copper

Fig. VI.2(b) — the (111) close-packed plane (shaded) with its three <110> close-packed slip directions (red), the slip plane/direction pairing repeated on all 4 × {111} planes.

Slip in FCC metals occurs on the close-packed $\{111\}$ planes, along the close-packed $\langle110\rangle$ directions within those planes — the combination that minimizes the Burgers vector magnitude (Question VI.1(b)) and therefore the dislocation's strain energy ($\propto b^2$). There are 4 non-parallel $\{111\}$ planes in the FCC unit cell, and each contains 3 independent $\langle110\rangle$ directions, giving: $$4\ \{111\}\ \text{planes} \times 3\ \langle110\rangle\ \text{directions} = \boxed{12\ \text{slip systems}}$$ This large number of independent slip systems (satisfying the von Mises criterion of 5 independent systems for arbitrary polycrystal shape change) is exactly why FCC metals like Cu are characteristically ductile.

VI.2(c) — Hall–Petch Extrapolation to a Nanocrystalline Grain Size

Given. 70Cu-30Zn brass: $d_1=100\ \mu\text{m}$, $\sigma_{y1}=64.5$ MPa; $\sigma_0=25$ MPa (given); target $d_2=50$ nm.

Find. $\sigma_{y2}$ at $d_2=50$ nm.

Approach. Use the Hall–Petch relation $\sigma_y=\sigma_0+k\,d^{-1/2}$ (appendix) to back out $k$ from the first data point, then apply it at $d_2$.

Solve for the Hall–Petch constant $k$. With $d_1=100\ \mu\text{m}=1\times10^{-4}$ m, $d_1^{-1/2}=100\ \text{m}^{-1/2}$: $$k=\frac{\sigma_{y1}-\sigma_0}{d_1^{-1/2}}=\frac{64.5-25}{100}=0.395\ \text{MPa}\cdot\text{m}^{1/2}$$
Apply at $d_2=50$ nm. $d_2=5\times10^{-8}$ m, $d_2^{-1/2}=4472\ \text{m}^{-1/2}$: $$\sigma_{y2}=\sigma_0+k\,d_2^{-1/2}=25+(0.395)(4472)=\boxed{1791\ \text{MPa}}$$

Question VI.2(c) — summary
Quantity	Value
Hall–Petch constant $k$	0.395 MPa·m$^{1/2}$
Predicted $\sigma_y$ at $d=50$ nm	1791 MPa

Check

This is a straight extrapolation of the classic Hall–Petch law roughly 2000× below the grain size it was calibrated on (100 μm → 50 nm). Real nanocrystalline metals below $\sim$10–20 nm often show an inverse Hall–Petch effect (softening, not further strengthening) as grain-boundary sliding and rotation begin to compete with dislocation pile-up as the dominant deformation mechanism. The 1791 MPa figure above is the idealized textbook Hall–Petch prediction the formula gives when applied as instructed, not necessarily an experimentally achievable strength at that grain size — worth stating explicitly rather than presenting as unconditionally reliable.

Concept & Theory

Topic. Edge/screw dislocation geometry, Burgers vectors, FCC slip systems, twinning, and grain-size (Hall–Petch) strengthening.

Key relations. $\mathbf{b}=\tfrac{a}{2}\langle110\rangle$ (FCC); slip on $\{111\}\langle110\rangle$; $\sigma_y=\sigma_0+k\,d^{-1/2}$.

Why this works. Dislocations glide preferentially on close-packed planes/directions because that minimizes the Burgers vector and hence the elastic strain energy stored around the dislocation line; grain boundaries strengthen a polycrystal because they act as barriers to dislocation glide, forcing pile-ups whose stress concentration must reach a higher applied stress to trigger slip in the neighbouring grain as grain size shrinks.

Common pitfall. Applying Hall–Petch with $d$ left in $\mu$m or nm without converting to consistent SI units (m) — since $k$ carries units of MPa·m$^{1/2}$, mixing unit systems between the calibration point and the prediction point silently introduces large errors.

Source. Callister & Rethwisch, Materials Science and Engineering, Ch. 4, 7; Dieter, Mechanical Metallurgy, Ch. 4.