Past exam of the materials science course of the University of Cambridge / 2019 / Paper 1 / 5

An FCC crystal slips on $\{111\}⟨110⟩$. For each slip system, Schmid's law gives $\tau =\sigma m$, with $m=|\hat{\mathbf{l}}\cdot \hat{\mathbf{n}}||\hat{\mathbf{l}}\cdot \hat{\mathbf{s}}|$. Enumerating the twelve systems for $\mathbf{l}=(5,1,4)$ gives the maximum atHere $\mathbf{n}=(1,-1,1)$ and $\mathbf{s}=(1,1,0)$ are perpendicular, while $\mathbf{l}\cdot \mathbf{n}=8$, $\mathbf{l}\cdot \mathbf{s}=6$ and $|\mathbf{l}{|}^2=42$. Hence $m=48/(42\sqrt{6})=0.46657$. Reversing both directed slip conventions describes the same physical system.

Applying Schmid's law at first yield,This infers the critical resolved shear stress for the initially active system, assuming the supplied tensile stress is the relevant uniaxial stress at yield and the twelve FCC systems have comparable initial slip resistance.

Initially, one slip system carries most of the deformation. Easy dislocation glide gives the nearly flat stage-I response, although lattice rotation and gradual dislocation interactions can already change the required tensile stress. In tension the loading direction expressed in the rotating crystal approaches the primary slip direction. Its path has the form $(5+t,1+t,4)$ until it reaches the symmetry boundary $y=z$, at $[844]=[211]$.

At that orientation, $(1\bar{1}1)[110]$ and $(11\bar{1})[101]$ have equal Schmid factors, $1/\sqrt{6}$. Conjugate slip therefore becomes competitive. Intersections between the two dislocation families generate obstacles, junctions and forest hardening, giving the sharp stage-II rise in flow stress. This is the ideal single-slip-to-conjugate-slip construction; real activation stresses can be shifted by latent hardening and imperfections.

The number and spacing of primary slip planes remain fixed, so the length projected on their normal is invariant: $l_0\cos {\varphi }_0=l_1\cos {\varphi }_1$. The initial and boundary loading directions are $[514]$ and $[211]$, respectively. With normal $[1\bar{1}1]$,ThereforeThis is engineering plastic strain in the ideal geometric model; its corresponding logarithmic strain is $\ln (4/\sqrt{7})$. Elastic extension and changes in slip resistance are neglected, as required by this construction.