Question 6 of 7: Dislocation Theory and Grain Boundaries

Question VI: Dislocation Theory 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 — Given. Slip plane oriented with its normal parallel to the tensile axis (i.e. the slip plane itself is normal/perpendicular to the applied load).

Find. The Schmid law statement, and whether slip occurs in this orientation.

Schmid's law: the resolved shear stress on a slip system is $$\tau_R=\sigma\cos\phi\cos\lambda,$$ where $\sigma$ is the applied uniaxial tensile stress, $\phi$ is the angle between the tensile axis and the slip plane normal, and $\lambda$ is the angle between the tensile axis and the slip direction (which lies IN the slip plane). Slip begins once $\tau_R$ reaches the critical resolved shear stress, $\tau_{crss}$.

1. Geometry. If the slip plane is normal to the tensile axis, the plane's normal is parallel to the axis, so $\phi=0^\circ$. Any slip direction lies within the slip plane, hence perpendicular to the plane's normal and to the tensile axis: $\lambda=90^\circ$.
2. Resolved shear stress. $$\tau_R=\sigma\cos(0^\circ)\cos(90^\circ) =\sigma\times1\times0=\boxed{0}.$$
3. Conclusion. Because $\tau_R=0$ for every possible slip direction in this orientation regardless of how large $\sigma$ is, $$\boxed{\text{no slip occurs}}$$—this is the classic zero-Schmid-factor orientation (pure tension/compression normal to the slip plane resolves no shear onto it at all).

Part 2 — Find. The primary and secondary slip systems of the HCP lattice.

HCP metals slip most easily on the close-packed basal plane, $\{0001\}\langle11\bar20\rangle$ (3 slip systems, along the 3 close-packed $\langle11\bar20\rangle$ directions) — this is the primary system for ideal/high $c/a$ HCP metals (e.g. Zn, Cd, Mg). Because basal slip alone supplies too few independent systems for general (von Mises) ductility, secondary systems activate at higher stress or temperature: prismatic slip $\{10\bar10\}\langle11\bar20\rangle$ and pyramidal slip $\{10\bar11\}\langle11\bar20\rangle$ (and, in some metals, the $\langle11\bar23\rangle$ pyramidal system, which adds a $c$-axis component). Which secondary system activates, and how readily, depends strongly on the metal's $c/a$ ratio.

Part 3(a) — Given. Copper (FCC), lattice constant $a=3.615\ \text{\AA}=3.615\times10^{-10}\ \text{m}$.

Find. The Burgers vector of an edge dislocation in the (close-packed) slip plane, and its magnitude.

1. Burgers vector. In FCC metals, the perfect (full) slip dislocation's Burgers vector connects nearest neighbours along a close-packed $\langle110\rangle$ direction: $$\vec b=\boxed{\tfrac{a}{2}\langle110\rangle}.$$
2. Magnitude. $$|\vec b|=\frac{a}{2}\sqrt{1^2+1^2+0^2}=\frac{a}{\sqrt2} =\frac{3.615\times10^{-10}}{\sqrt2}=\boxed{2.556\times10^{-10}\ \text{m}=2.556\ \text{\AA}}.$$

Part 3(b) — Given. $\tau_{crss,1}=2.10$ MPa at $\rho_1=10^5/\text{mm}^2$; $G=48$ GPa, $\alpha=0.2$, $b=2.556\times10^{-10}$ m (Part 3a). Find $\tau_{crss,2}$ at $\rho_2=10^7/\text{mm}^2$.

Find. $\tau_0$, then $\tau_{crss}$ at the higher dislocation density.

1. Convert densities to /$\text{m}^{2}$. $\rho_1=10^5/\text{mm}^2=10^{11}/\text{m}^2$, $\rho_2=10^7/\text{mm}^2=10^{13}/\text{m}^2$.
2. Solve for the intrinsic strength $\tau_0$. $\alpha Gb\sqrt{\rho_1}=0.2\times48\times10^9\times2.556\times10^{-10}\times\sqrt{10^{11}} =0.776\ \text{MPa}$, so $$\tau_0=\tau_{crss,1}-\alpha Gb\sqrt{\rho_1}=2.10-0.776=\boxed{1.324\ \text{MPa}}.$$
3. Hardening term at $\rho_2$. Since $\sqrt{\rho_2/\rho_1}=\sqrt{100}=10$, the hardening term scales by exactly $10\times$: $\alpha Gb\sqrt{\rho_2}=0.776\times10 =7.76\ \text{MPa}$.
4. CRSS at the higher density. $$\tau_{crss,2}=\tau_0+\alpha Gb\sqrt{\rho_2} =1.324+7.76=\boxed{9.08\ \text{MPa}}.$$ Strain hardening (rising dislocation density) roughly quadruples–to–order-of-magnitude increases the flow stress here, consistent with the $\sqrt\rho$ (Taylor) hardening law.