Question 6 of 7: ASTM Grain Size Number and Hall–Petch Yield Strength
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — May 2018. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Notes on the paper state that
any five questions constitute a complete paper and only the first five questions appearing in the
answer book are marked, with all questions of equal value. All seven questions are solved below
for completeness.
Reference texts: Callister & Rethwisch, Materials Science and
Engineering: An Introduction, 9th ed. (crystal structure and packing, polymer molecular
weight, cold work and annealing, corrosion and diffusion, composites, ceramic glasses).
Question 6: ASTM Grain Size Number and Hall–Petch Yield Strength (20 marks)
Given. (a) Average grain diameter $d=30\,\mu$m. (b) Two grain
size–yield stress data pairs (table above); Hall–Petch relation (page 1)
$\sigma_y=\sigma_0+Kd^{-1/2}$.
Find. (a) ASTM grain size number $G$. (b) $\sigma_0$, $K$, and $\sigma_y$ at
$d=30\,\mu$m.
Approach
(a) By definition, $N=2^{G-1}$ grains occupy one square inch of a micrograph at $100\times$
magnification; converting the real area that one such square inch represents (at $100\times$, a
real length of $0.01''=254\,\mu$m per image inch) into an equivalent number of $d\times d$ grains
gives $N$, and hence $G$. (b) Two $(\,d^{-1/2},\sigma_y)$ points determine the line
$\sigma_y=\sigma_0+Kd^{-1/2}$ exactly (two unknowns, two equations); the fitted line is then
evaluated at $d=30\,\mu$m.
(a) Real area per image square inch. At $100\times$ magnification, $1''$ of
image corresponds to $1/100''=0.254$ mm of real specimen, so 1 square inch of image
represents a real area of
$$(0.254\ \text{mm})^2=0.064516\ \text{mm}^2$$
(a) Grains per image square inch. Treating grains as squares of side
$d=0.030$ mm:
$$N=\frac{0.064516}{(0.030)^2}\approx71.7\ \text{grains/in}^2\ \text{at}\ 100\times$$
(a) ASTM number. From $N=2^{G-1}$:
$$G=1+\log_2N=1+\frac{\ln71.7}{\ln2}$$
$$\boxed{G\approx7.2}$$
(A rounded ASTM No. 7 grain, whose nominal table diameter is $\approx32\,\mu$m, is consistent
with this $30\,\mu$m sample.)
(b) Solving the two-point Hall–Petch system. With
$x=d^{-1/2}$ ($d$ in $\mu$m): $x_1=(60.5)^{-1/2}=0.1286\,\mu\text{m}^{-1/2}$,
$x_2=(136)^{-1/2}=0.0858\,\mu\text{m}^{-1/2}$. From
$\sigma_0+Kx_1=160$ and $\sigma_0+Kx_2=130$:
$$K=\frac{160-130}{x_1-x_2}=\frac{30}{0.0429}$$
$$\boxed{K\approx700.7\ \text{MPa}\cdot\mu\text{m}^{1/2}}$$
$$\sigma_0=160-Kx_1\approx160-90.1$$
$$\boxed{\sigma_0\approx69.9\ \text{MPa}}$$
(b) Yield stress at $d=30\,\mu$m. $x_3=(30)^{-1/2}=0.1826\,\mu\text{m}^{-1/2}$:
$$\sigma_y=\sigma_0+Kx_3=69.9+(700.7)(0.1826)$$
$$\boxed{\sigma_y\approx197.8\ \text{MPa}}$$
This is higher than either tabulated point, as expected: $30\,\mu$m is the smallest grain size of
the three, and Hall–Petch strengthening increases as grain size decreases.
