Question 5 of 8: Dislocations, Slip and Grain Boundaries

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.

Part 1. The Burgers vector $\mathbf{b}$ quantifies the magnitude and direction of lattice distortion caused by a dislocation: trace a closed rectangular loop of equal atom-to-atom steps around the dislocation line in the real (defective) crystal, then trace the same step sequence in a perfect reference lattice — the loop fails to close, and the vector needed to close it is $\mathbf{b}$. For an FCC metal, $\mathbf{b}=\tfrac{a}{2}\langle 110\rangle$, the shortest full lattice translation. An edge dislocation is formed by inserting an extra half-plane of atoms into the lattice; its Burgers vector is perpendicular to the dislocation line, and the line marks where the extra half-plane terminates (left figure below). A screw dislocation instead forms when the lattice is sheared parallel to the dislocation line, turning successive atomic planes into a continuous helical ramp around the line; its Burgers vector is parallel to the dislocation line (right figure below). Both are line defects carrying the same Burgers vector magnitude for a given slip system, but they glide and interact with stress fields differently because $\mathbf{b}$ sits at a different angle to the line in each case.

Part 2. Dislocations are the line defects whose motion (glide) through the crystal constitutes slip — the shearing of one atomic plane over another along a close-packed plane and direction. Because moving a dislocation requires breaking and re-forming only the bonds immediately along its core (rather than shearing an entire plane of atoms simultaneously), the shear stress needed to move a dislocation is orders of magnitude below the theoretical shear strength of a perfect lattice; this is precisely why real crystals deform plastically at practical stresses at all. Ductility is the macroscopic consequence: a metal with many available slip systems and mobile dislocations can accommodate large plastic strain by sequential slip before fracturing, whereas a material where dislocation motion is difficult (few slip systems, strong lattice friction, as in most ceramics) is brittle. Anything that impedes dislocation motion — grain boundaries, solute atoms, precipitates, other dislocations — raises the yield strength but, past a point, reduces the ductility left available before fracture, the classic strength–ductility trade-off.

Part 3 — Given. FCC nickel rod, radius $r=20$ mm, loading axis $[100]$, axial load $F=50$ kN.

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

Approach. FCC metals slip on the close-packed $\{111\}\langle110\rangle$ system; evaluate the Schmid factor $m=\cos\phi\cos\lambda$ for the $[100]$ loading axis by vector dot products, then scale the applied axial stress by $m$.

1. (a) Slip system. FCC nickel slips on the close-packed octahedral planes and close-packed directions, $\boxed{\{111\}\langle110\rangle}$ — 4 distinct $\{111\}$ plane orientations, each with 3 $\langle110\rangle$ directions lying in it, giving 12 slip systems in total.
2. (b) Multiple slip for [100] loading. The $[100]$ axis is a 4-fold symmetry axis of the cubic lattice, and by that symmetry the resolved shear stress works out identical on 8 of the 12 $\{111\}\langle110\rangle$ systems (the 4 remaining systems have zero resolved shear, since their slip direction is perpendicular to $[100]$) — this is the standard "$\langle100\rangle$-corner" multiple-slip orientation of the FCC stereographic triangle, so $\boxed{8}$ slip systems are simultaneously most highly stressed.
3. (c) Schmid factor. Take plane normal $\mathbf n=[111]/\sqrt3$ and an in-plane $\langle110\rangle$ direction $\mathbf d=[10\bar1]/\sqrt2$ (valid since $\mathbf n\cdot\mathbf d=0$), against loading axis $[100]$: $$\cos\phi=\mathbf n\cdot[100]=\frac{1}{\sqrt3}=0.5774,\qquad \cos\lambda=\mathbf d\cdot[100]=\frac{1}{\sqrt2}=0.7071,$$ $$m=\cos\phi\cos\lambda=\frac{1}{\sqrt3}\times\frac{1}{\sqrt2}=\frac{1}{\sqrt6} =\boxed{0.408}.$$
4. (d) Minimum shear stress. Axial stress from the applied load, $$\sigma=\frac{F}{A}=\frac{50\times10^3}{\pi(0.020)^2}=\boxed{39.79\ \text{MPa}}.$$ Resolving onto the most-favoured slip system via Schmid's law, $$\tau=\sigma\,m=39.79\times0.408=\boxed{16.24\ \text{MPa}}.$$ Because 8 systems share this same maximum Schmid factor for $[100]$ loading, slip on all 8 initiates essentially simultaneously once $\tau$ reaches the critical resolved shear stress of the material.