Question 7 of 8: ASTM Grain Size and the Effects of Cold Work and Low-Temperature Annealing

Question 7: ASTM Grain Size and the Effects of Cold Work and Low-Temperature Annealing (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.

Given. (a) 42 grains counted inside a 1.5 in. diameter circle on a micrograph at ×300 magnification. (b) Al alloy sheet, $\sigma_y$ rises 350→600 MPa after stretching (cold work). (c) The cold-worked sheet is heated to 155°C for 2 h, then cooled.

Find. (a) ASTM grain-size number and average grain diameter (mm). (b) Reason for the strength increase. (c) What happens on the 155°C/2 h anneal, and why.

Approach

For (a), convert the observed count and field size to grains per unit area at the ASTM reference magnification (100×) to get the ASTM number, then separately compute the true (unmagnified) grain density to get an actual average grain diameter. Parts (b) and (c) are qualitative applications of dislocation strengthening and recovery/recrystallization theory.

1. (a) Grain density at the working magnification. Field area $A_{300}=\tfrac{\pi}{4}(1.5)^2=1.767\ \text{in}^2$, so $$\left(\frac{\text{grains}}{\text{in}^2}\right)_{300\times}=\frac{42}{1.767}=23.77.$$
2. Convert to the ASTM 100× reference. Grain count per unit area scales with the square of magnification, so $$N_{100}=23.77\times\left(\frac{300}{100}\right)^2=\boxed{213.9\ \text{grains/in}^2}.$$ The ASTM grain-size number $n$ satisfies $N_{100}=2^{n-1}$: $$n=\log_2(213.9)+1=\boxed{8.7}\ \text{(ASTM grain size }\approx 8\text{–}9).$$
3. Actual (unmagnified) grain diameter. The true field diameter is $1.5/300=0.005$ in $=0.127$ mm, so the true field area is $A_{real}=\tfrac{\pi}{4}(0.127)^2=0.01267\ \text{mm}^2$, and the true grain density is $42/0.01267=3316\ \text{grains/mm}^2$. Treating each grain's footprint as roughly square/circular, $$d_{avg}=\sqrt{\frac{1}{3316}}=\boxed{0.0174\ \text{mm}}\ (17.4\ \mu\text{m}).$$

(b) Plastic stretching (cold work) drives a large increase in dislocation density; the resulting dense, tangled dislocation network impedes the motion of any individual dislocation (dislocations cutting through or piling up against one another), so a higher applied stress is required to continue plastic flow. This strain-hardening mechanism (Taylor relation, $\sigma\propto\sqrt{\rho_{dislocation}}$) raises the yield strength from 350 to 600 MPa, at the cost of remaining ductility.

(c) A 2-hour hold at 155°C is a low-temperature bake, well below the recrystallization temperature of a heavily cold-worked aluminum alloy (recrystallization typically needs a substantial fraction of the absolute melting point, and a longer time than 2 h at such a modest temperature). Only recovery occurs: point defects and some dislocations annihilate or rearrange into lower-energy configurations (polygonization/sub-grain formation), relieving internal (residual) stress and modestly softening the alloy and restoring some ductility — but the cold-worked, elongated grain structure is retained, since no new strain-free grains nucleate and no grain growth occurs at this temperature/time. The yield strength therefore drops somewhat from 600 MPa but does not return to the original 350 MPa.