Question 7 of 8: Dislocations — Obstacle Bypass, Schmid Factor and Strain Rate

Question 7: Dislocations — Obstacle Bypass, Schmid Factor and Strain Rate (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.

7.1 — (a) Stress to bow a pinned dislocation past its obstacles

A dislocation segment pinned at two non-shearable obstacles a distance $L$ apart behaves, under an applied resolved shear stress $\tau$, like a flexible string under line tension. The line tension is $T\approx\tfrac12 Gb^2$, where $G$ is the shear modulus and $b$ the Burgers vector. As $\tau$ rises the segment bows into a circular arc of radius $R=T/(\tau b)$, curving further as $\tau$ increases because a smaller $R$ is needed to balance a larger driving force. The bow-out is stable while it can still find an equilibrium radius; it becomes unstable, and the dislocation sweeps past the obstacles (Orowan bypass), once the arc has bowed out to a semicircle pinned only at its two ends, at which point $R$ reaches its minimum possible value, $R_{\min}=L/2$.

Substituting the critical radius into the force balance $\tau b = T/R$ gives the stress at which bypass occurs:

1. Balance the line force against the applied shear force. $$\tau b = \frac{T}{R}$$
2. Insert the critical (semicircular) radius $R_{\min}=L/2$. $$\tau b = \frac{T}{L/2} = \frac{2T}{L}$$
3. Substitute $T=\tfrac12 Gb^2$ and solve for $\tau$. $$\tau = \frac{2(\tfrac12 Gb^2)}{bL} = \boxed{\dfrac{Gb}{L}}$$

This is the Orowan (bow-out/bypass) stress. It is the governing relation whenever a dislocation must get past obstacles it cannot shear or cut through — incoherent precipitates, dispersoid particles, or a forest of intersecting dislocations spaced $L$ apart — and it is the reason particle strengthening and forest hardening both scale as $1/L$: closer obstacle spacing means a tighter bow-out radius and a higher stress to break free.

7.2 — (b) Schmid factor of FCC, [100] loading

Given. An FCC single crystal loaded with its $[100]$ direction parallel to the tensile axis. FCC slip occurs on $\{111\}\langle110\rangle$ systems.

Find. The magnitude of the resolved-shear (Schmid) factor $m=\cos\phi\cos\lambda$, where $\phi$ is the angle between the loading axis and the slip-plane normal and $\lambda$ is the angle between the loading axis and the slip direction.

Approach. Take one representative slip plane, $(111)$, resolve the loading axis onto its normal and onto each of the three $\langle110\rangle$ directions that lie in that plane, then generalise by cubic symmetry.

1. Angle to the plane normal. For $\mathbf{n}_{\text{load}}=[100]$ and plane normal $\mathbf{n}=[111]$: $$\cos\phi=\frac{[100]\cdot[111]}{|[100]||[111]|}=\frac{1}{\sqrt3}$$
2. Angle to each in-plane $\langle110\rangle$ direction. The three directions satisfying $\mathbf{d}\cdot[111]=0$ are $[10\bar1]$, $[01\bar1]$ and $[\bar110]$. $$\cos\lambda_{[10\bar1]}=\frac{1}{\sqrt2},\qquad \cos\lambda_{[01\bar1]}=0,\qquad \cos\lambda_{[\bar110]}=-\frac{1}{\sqrt2}$$
3. Form $m=\cos\phi\cos\lambda$ for each and take the magnitude. $[10\bar1]$ and $[\bar110]$ both give $$m=\frac{1}{\sqrt3}\cdot\frac{1}{\sqrt2}=\boxed{\dfrac{1}{\sqrt6}\approx0.408}$$ while $[01\bar1]$ gives $m=0$ (it is perpendicular to the loading axis).

By the four-fold symmetry of $[100]$ about the cubic axes, every one of the four $\{111\}$ planes reproduces this same pattern — two directions at $m=0.408$ and one at $m=0$ — so eight slip systems share the maximum Schmid factor simultaneously. $[100]$ is therefore a high-symmetry, multiple-slip orientation: the eight equally favoured systems are activated together rather than a single one dominating, which is why single crystals pulled along $[100]$ work-harden faster than those pulled along a general orientation with a single m near the theoretical maximum of 0.5.

7.3 — (c) Shear strain rate from mobile dislocation density and velocity

Consider a crystal of unit volume containing a mobile dislocation density $\rho_m$ (total length of mobile dislocation line per unit volume). Let every mobile segment glide at the same average velocity $\bar v$ over a short time interval $dt$.

1. Area swept by one unit length of dislocation. A unit length of line moving a distance $\bar v\,dt$ sweeps an area $dA=\bar v\,dt$ on its slip plane.
2. Shear displacement produced. Sweeping an area $dA$ on the slip plane displaces the material above it relative to the material below by a Burgers vector $b$ per unit swept area, so the shear displacement contributed by that unit length is $b\,dA=b\bar v\,dt$.
3. Sum over all mobile line length per unit volume. With $\rho_m$ (length per volume) of mobile line moving, the total shear strain produced in $dt$ is $$d\gamma=\rho_m\,b\,\bar v\,dt$$ so $$\dot\gamma=\boxed{\rho_m\,b\,\bar v}$$

This is the Orowan equation. It is the kinematic bridge between microstructure (dislocation density, which strengthening mechanisms change) and macroscopic strain rate (an imposed test or process variable): for a fixed imposed $\dot\gamma$, any mechanism that raises $\rho_m$ (work hardening) must lower $\bar v$, and any mechanism that pins dislocations and lowers $\bar v$ (solute drag, obstacles) must raise the stress needed to keep $\dot\gamma$ constant — the microscopic root of strain-rate sensitivity.