Question 5 of 7: ASTM Grain Size; Strain Hardening and Annealing of Aluminum

Question 5: ASTM Grain Size; Strain Hardening and Annealing of Aluminum (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 (a) — ASTM grain size and grain diameter

Given. $42$ grains counted inside a circle of diameter $1.5$ in on a micrograph taken at magnification $M=300\times$; $1$ in $=25.4$ mm.

Find. The ASTM grain-size number $n$ and the average grain diameter in mm.

Approach. Convert the count to grains per square inch at $100\times$ (the ASTM reference magnification), invert $N=2^{\,n-1}$ for $n$, then take the actual area per grain to get an average diameter.

1. Grains per in$^2$ on the print. Circle area $A=\tfrac{\pi}{4}(1.5)^2=1.767$ in$^2$, so $N_{300}=42/1.767=23.8$ grains/in$^2$ at $300\times$.
2. Rescale to 100×. Grains per print-area scale as $M^2$, so $$N_{100}=N_{300}\left(\frac{300}{100}\right)^2=23.8\times9=214\ \text{grains/in}^2.$$
3. ASTM number. From $N_{100}=2^{\,n-1}$, $$n=1+\log_2 N_{100}=1+\log_2 214=\boxed{8.7}.$$
4. Average grain diameter. The true field area is $A/M^2=1.767/300^2=1.963\times10^{-5}$ in$^2$, so the area per grain is $1.963\times10^{-5}/42=4.68\times10^{-7}$ in$^2$ and $$d=\sqrt{4.68\times10^{-7}}=6.84\times10^{-4}\ \text{in}=\boxed{0.0174\ \text{mm}}\ (\approx17.4\ \mu\text{m}).$$

Part (b) — why stretching strengthens the sheet

Stretching the sheet plastically is cold work (strain hardening). Plastic strain multiplies dislocations and drives their density up by orders of magnitude. The resulting dislocations tangle and pile up against one another and against grain boundaries, so each dislocation now has to push through the stress fields of many others before it can move. Because a higher applied stress is needed to keep dislocations gliding, the yield strength rises — here from 350 MPa to 600 MPa. The gain in strength comes at the cost of ductility, since the metal has used up much of its remaining capacity for plastic flow.

Part (c) — heating the stretched alloy to 155°C for 2 hours

155°C is a low-temperature bake, well below the recrystallization temperature of a cold-worked aluminum alloy. Pure metals recrystallize at roughly $0.4T_m$ (for aluminum, $T_m=933$ K, so $0.4T_m\approx373\ \text{K}=100^\circ$C, and Callister lists 80°C for 99.999% Al), but solute and second-phase particles raise that threshold sharply — up to about $0.7T_m\approx380^\circ$C for commercial alloys, which is why wrought aluminum alloys are annealed at 340–415°C. At $155^\circ\text{C}=428$ K the sheet is only at $0.46T_m$, and 2 hours is far too short at that temperature to nucleate new grains.

What occurs instead is recovery. Thermal energy lets point defects annihilate and lets dislocations climb, cross-slip and rearrange themselves into lower-energy arrays (polygonisation into sub-grain walls), so the residual (internal) stresses are relieved and electrical conductivity largely returns — but the dislocation density falls only modestly and the elongated, cold-worked grain structure is retained. The yield strength therefore drops a little from 600 MPa and a little ductility is regained; it does not return to the original 350 MPa, because that would require recrystallization. (Industrially this is exactly the stabilising bake applied to strain-hardened aluminum — the H3x tempers — whose purpose is to lock in the cold-worked strength, not remove it.)