Question 6 of 7: Diffusion and Grain Boundaries

Question VI: Diffusion 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 — Carburizing Time (Semi-Infinite Diffusion)

Given.

Find. The carburizing time $t$.

Approach. Apply the semi-infinite solid diffusion solution $\dfrac{C_s-C_x}{C_s-C_0}=\text{erf}\!\left(\dfrac{x}{2\sqrt{Dt}}\right)$ (appendix), read the inverse error function off the appendix's error-function table by interpolation, then solve for $t$.

1. Target error-function value. $$\frac{C_s-C_x}{C_s-C_0}=\frac{1.20-0.80}{1.20-0.25}=\frac{0.40}{0.95}=\boxed{0.4211}$$
2. Invert via the appendix erf table. The table gives $\text{erf}(0.35)=0.3794$ and $\text{erf}(0.40)=0.4284$; linearly interpolating for $\text{erf}(z)=0.4211$: $$z=0.35+0.05\left(\frac{0.4211-0.3794}{0.4284-0.3794}\right)=0.35+0.05(0.851)=\boxed{z=0.3925}$$
3. Solve for $t$. With $z=x/(2\sqrt{Dt})$: $$\sqrt{Dt}=\frac{x}{2z}=\frac{5\times10^{-4}}{2(0.3925)}=6.369\times10^{-4}\ \text{m}\ \Rightarrow\ Dt=4.055\times10^{-7}\ \text{m}^2$$ $$t=\frac{4.055\times10^{-7}}{1.6\times10^{-11}}=\boxed{2.536\times10^{4}\ \text{s} = 7.04\ \text{hours}}$$

VI.2 — FCC vs. HCP Ductility and Their Slip Systems

FCC is markedly more ductile than HCP under general loading. Ductility by slip requires a sufficient number of independent, easily-activated slip systems (slip plane × slip direction combinations) so that an arbitrary imposed strain can be accommodated by some combination of them (the von Mises criterion requires 5 independent systems for arbitrary polycrystal deformation without cracking). FCC slips on its close-packed $\{111\}$ planes along close-packed $\langle110\rangle$ directions, giving $4\times3=12$ slip systems, comfortably exceeding the von Mises minimum — FCC metals (Cu, Al, Ni, austenitic stainless steel) are reliably ductile at all temperatures. HCP slips most easily only on its single close-packed basal $(0001)$ plane along $\langle11\bar20\rangle$ directions, giving only $1\times3=3$ independent easy slip systems (well below 5); additional non-basal (prismatic, pyramidal) systems exist but require substantially higher stress to activate. HCP metals (Mg, Zn, Ti at room temperature) are consequently far more prone to brittle behaviour and limited formability than FCC metals of comparable bonding.

VI.3(a) — Percentage Decrease in Yield Strength on Annealing

Given. Hall–Petch relation $\sigma_y=\sigma_i+k_yd^{-1/2}$ with $\sigma_i=3$ MPa, $k_y=0.5\ \text{MPa}\cdot\text{m}^{1/2}$; grain size $d_1=4\ \mu\text{m}$ (as-received) → $d_2=104\ \mu\text{m}$ (annealed).

Find. The percentage decrease in yield strength caused by the grain growth.

1. Yield strength before annealing ($d_1=4\ \mu\text{m}=4\times10^{-6}$ m). $$\sigma_{y1}=3+0.5\left(4\times10^{-6}\right)^{-1/2}=3+0.5(500)=\boxed{253.0\ \text{MPa}}$$
2. Yield strength after annealing ($d_2=104\ \mu\text{m}=104\times10^{-6}$ m). $$\sigma_{y2}=3+0.5\left(104\times10^{-6}\right)^{-1/2}=3+0.5(98.06)=\boxed{52.03\ \text{MPa}}$$
3. Percentage decrease. $$\%\,\text{decrease}=\frac{\sigma_{y1}-\sigma_{y2}}{\sigma_{y1}}\times100\%=\frac{253.0-52.03}{253.0}\times100\%=\boxed{79.44\%}$$

VI.3(b) — Time for the Grain-Growth Anneal

Given. $d^n-d_0^n=Kt$ with $n=4$, $K=4.4\ \mu\text{m}^4\text{s}^{-1}$, $d_0=4\ \mu\text{m}$, $d=104\ \mu\text{m}$.

1. Solve for $t$. $$t=\frac{d^4-d_0^4}{K}=\frac{104^4-4^4}{4.4}=\frac{116{,}985{,}856-256}{4.4}=\boxed{2.659\times10^{7}\ \text{s}}$$ $$t=2.659\times10^{7}\ \text{s}\times\frac{1\ \text{hr}}{3600\ \text{s}}=\boxed{7{,}385\ \text{hours}\approx308\ \text{days}}$$