Worked examples — Metals & alloys — Al alloys (2024, 7075), Ti alloys (Ti-6Al-4V), Ni superalloys (Inconel, Hastelloy), stainless steels

This page is the drill hall for the parent topic. The parent gave you the rules; here we run the rules across every kind of question the topic can throw at you — and we prove each number.

Before any formula appears, one promise: every symbol used below is defined the first time it shows up. If your eyes glaze at "$\sigma /\rho$", relax — we build it from "strength divided by heaviness."

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The scenario matrix

Think of this topic as a machine with a few "dials." Each cell below is one setting of the dials. Our worked examples must land on every cell.

#	Cell (case class)	What makes it distinct	Covered by
A	Ranking by specific strength	divide, compare ratios	Ex 1
B	Zero / degenerate input	pure metal, no strengthening ($d\to \infty$, $\sigma \to {\sigma }_0$)	Ex 2
C	Limiting value	grain size $\to 0$, precipitate spacing $\to 0$	Ex 3
D	Hall–Petch two-point solve	back out constants ${\sigma }_0,k$ from data	Ex 4
E	Orowan precipitate spacing	$\tau =Gb/L$, tiny units (nm)	Ex 5
F	Real-world word problem	choose a family from a spec	Ex 6
G	The "flip" case (temperature reverses the rule)	small grains help cold, hurt hot	Ex 7
H	Exam twist / trap	"stronger = better?" toughness trade-off	Ex 8
I	Unit-conversion landmine	MPa, g/cm³, SI mixups	Ex 9

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Ex 1 — Cell A: Rank three metals by specific strength

Forecast: which wins — the strongest (steel) or the lightest (Al)? Guess before reading.

1. Compute each ratio ${\sigma }_y/\rho$.
   • Why this step? A wing spar carries load while being flown, so we score strength per unit mass, not raw strength. $\frac{500}{2.8}=178.6,\frac{880}{4.43}=198.6,\frac{1000}{7.9}=126.6$
2. Order the numbers, largest first.
   • Why this step? Larger $\sigma /\rho$ = more strength you keep for every kilogram you're forced to fly. $\text{Ti }198.6>\text{Al }178.6>\text{Steel }126.6$

Verify: the steel is strongest in absolute MPa (1000 > 880 > 500) yet loses on the scoreboard — exactly the parent's claim that "titanium often beats steel even though steel is stronger." Units: $\frac{\text{MPa}}{{\text{g/cm}}^3}$ is consistent across all three, so the comparison is fair. See Precipitation Hardening for why 17-4PH reaches 1000 MPa at all.

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Ex 2 — Cell B: Degenerate input (pure metal, no obstacles)

Recall the parent's law:

Forecast: with the grain "infinitely big," should the added strength be large or nearly zero?

1. Take the limit $d\to \infty$.
   • Why this step? "No obstacles" literally means grain walls are so far apart a dislocation never meets one — Dislocations and Slip glide freely. $d^{-1/2}=\frac{1}{\sqrt{d}}\underset{d\to \infty }{\to }0$
2. Substitute.
   • Why this step? The boundary term vanishes, leaving only the base stress. ${\sigma }_y=20+k\cdot 0=20\text{MPa}$

Verify: a pure metal is weak — 20 MPa is tiny next to Ex 1's alloys (500 MPa). This is the degenerate floor of the formula: strip every strengthener and you fall back to ${\sigma }_0$. Sanity check on units: $k{d}^{-1/2}$ has ${\text{MPa⋅m}}^{1/2}\cdot {\text{m}}^{-1/2}=\text{MPa}$. ✓

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Ex 3 — Cell C: Limiting value (grains shrink toward zero)

Forecast: does strength climb without limit, or level off?

1. Convert each grain size to metres.
   • Why this step? $k$ is in ${\text{MPa⋅m}}^{1/2}$, so $d$ must be in metres or the units lie. $100\mu \text{m}=10^{-4}\text{m},10\mu \text{m}=10^{-5},1\mu \text{m}=10^{-6}$
2. Evaluate ${\sigma }_0+k{d}^{-1/2}$.
   • Why this step? Direct application; smaller $d$ → larger $d^{-1/2}$ → more strength. $d=10^{-4}:20+0.07\cdot 100=27\text{MPa}$ $d=10^{-5}:20+0.07\cdot 316.2=42.1\text{MPa}$ $d=10^{-6}:20+0.07\cdot 1000=90\text{MPa}$
3. Read the trend $d\to 0$.
   • Why this step? $d^{-1/2}\to \infty$, so the formula says ${\sigma }_y\to \infty$ — but physically grains can't shrink below a few atoms, so real strength saturates. The maths gives an idealized limit; the world caps it.

Verify: each 10× shrink in $d$ multiplies the boundary term by $\sqrt{10}\approx 3.16$ (100→316→1000). Strengths rise 27 → 42.1 → 90 MPa, monotonically — matching Hall–Petch Strengthening: "smaller grains → stronger."

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Ex 4 — Cell D: Solve for the constants from two data points

Forecast: Hall–Petch is a straight line in the variable $x=d^{-1/2}$. Two points → one line → unique ${\sigma }_0,k$.

1. Build $x=d^{-1/2}$ for each point (metres!).
   • Why this step? Plotting ${\sigma }_y$ against $d^{-1/2}$ turns the curve into a line $y={\sigma }_0+kx$. $x_1=(25\times 10^{-6}{)}^{-1/2}=200{\text{m}}^{-1/2},x_2=(4\times 10^{-6}{)}^{-1/2}=500$
2. Slope $k$ = rise/run.
   • Why this step? On a line, the slope is the constant $k$; it's the "strength gained per unit $d^{-1/2}$." $k=\frac{250-150}{500-200}=\frac{100}{300}=0.333{\text{MPa⋅m}}^{1/2}$
3. Intercept ${\sigma }_0$ = plug one point back.
   • Why this step? Knowing $k$, the base stress falls straight out of ${\sigma }_y={\sigma }_0+kx$. $150={\sigma }_0+0.333\cdot 200\Rightarrow {\sigma }_0=150-66.7=83.3\text{MPa}$

Verify: test the other point: $83.3+0.333\cdot 500=83.3+166.7=250$ MPa ✓. Both points sit on the line, so the fit is exact.

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Ex 5 — Cell E: Orowan spacing from precipitate strengthening

Recall:

Forecast: for a strong alloy do we expect $L$ in microns or nanometres?

1. Rearrange for $L$.
   • Why this step? We know the strengthening we want and solve for the microstructure that delivers it. $L=\frac{Gb}{{\tau }_{Orowan}}$
2. Match units — convert $b$ to metres, $G,\tau$ to Pa.
   • Why this step? Mixing nm with MPa gives nonsense; SI keeps it honest. $G=27\times 10^9\text{Pa},b=0.29\times 10^{-9}\text{m},\tau =100\times 10^6\text{Pa}$
3. Compute. $L=\frac{27\times 10^9\cdot 0.29\times 10^{-9}}{100\times 10^6}=\frac{7.83}{10^8}=7.83\times 10^{-8}\text{m}=78.3\text{nm}$

Verify: $L\approx 78$ nm — nanometre-scale, exactly the "fine second-phase particles" the parent describes for peak-aged tempers. If we halved $L$ to 39 nm, $\tau$ would double to 200 MPa (inverse law) ✓.

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Ex 6 — Cell F: Real-world word problem (choose the family)

Forecast: temperature is the loudest constraint — which family alone survives ~1000 °C?

1. List max service temperatures (parent table).
   • Why this step? Any material failing the temperature is disqualified before we even weigh it. $\text{Al }150^{\circ },\text{Ti }400^{\circ },\text{Stainless }600^{\circ },\text{Ni superalloy }1000^{\circ }+$
2. Eliminate by temperature.
   • Why this step? 950 °C exceeds Al, Ti, and stainless limits; only Ni superalloy remains.
3. Confirm the extra specs.
   • Why this step? A survivor must also match (b): creep resistance via γ′ precipitates and oxidation resistance via Cr — both are Inconel's specialty.

Answer: Nickel superalloy (Inconel-type). Weight being secondary is exactly why its high density ($\rho \approx 8.2$) is tolerable here.

Verify: cross-check the mnemonic ladder "All Tigers Snore Nightly" — the hottest rung is Ni. Only Ni clears 950 °C, so the choice is forced, not preferred. ✓

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Ex 7 — Cell G: The temperature "flip" (small grains help, then hurt)

Forecast: how can "add boundaries" and "remove all boundaries" both be right?

1. Cold regime: boundaries are obstacles.
   • Why this step? At low T, dislocation glide dominates; each boundary blocks slip → Hall–Petch ${\sigma }_y={\sigma }_0+k{d}^{-1/2}$ rewards small $d$. (Hall–Petch Strengthening.)
2. Hot regime: boundaries are highways for failure.
   • Why this step? Above ~0.7× melting T, atoms diffuse and grains slide past each other; boundaries crack via creep. More boundary length = more places to fail.
3. Take the limit that kills creep.
   • Why this step? Setting $d\to \infty$ (one crystal) removes every sliding surface → single-crystal blades.

Verify: both obey physics of their regime — the rule didn't break, the dominant deformation mode changed from slip (cold) to creep (hot). This is the parent's "strengthening rules flip with temperature." The Hall–Petch term $k{d}^{-1/2}$ simply becomes irrelevant when the failure mechanism is no longer dislocation slip.

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Ex 8 — Cell H: Exam trap ("stronger is always better")

Forecast: does the number 500 > 350 settle it?

1. Identify the governing failure mode.
   • Why this step? A lower fuselage skin is damage-tolerant / fatigue-driven, so the metric is fracture toughness, not peak yield.
2. Compare on the right metric.
   • Why this step? 7075 has higher yield but lower fracture toughness and worse stress-corrosion resistance; 2024 tolerates cracks longer before catastrophic failure.
3. Match load to family (parent's rule).
   • Why this step? Compression-loaded upper wing → 7075; fatigue-critical lower fuselage → 2024.

Answer: The reasoning is wrong. Correct pick is 2024-T3.

Verify: compute specific strengths — $500/2.8=178.6$ vs $350/2.8=125$. 7075 does win on $\sigma /\rho$, confirming the trap is seductive, yet the failure mode, not $\sigma /\rho$, decides here. Right tool, right load.

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Ex 9 — Cell I: Unit-conversion landmine

Forecast: same physics — why two different numbers?

1. Write out each with units.
   • Why this step? A ratio is meaningless without its units; the "0.199" and "198.6" differ only by the density's unit. $\frac{880\text{MPa}}{4430{\text{kg/m}}^3}=0.199\frac{\text{MPa}}{{\text{kg/m}}^3},\frac{880\text{MPa}}{4.43{\text{g/cm}}^3}=198.6\frac{\text{MPa}}{{\text{g/cm}}^3}$
2. Show they're the SAME quantity.
   • Why this step? $1{\text{g/cm}}^3=1000{\text{kg/m}}^3$, so dividing by a number 1000× larger gives a result 1000× smaller: $0.199\times 1000=199$.
3. Pick the engineering convention.
   • Why this step? Aerospace tables (parent table) quote $\sigma /\rho$ with $\rho$ in g/cm³, giving the familiar "~200."

Verify: $0.199\text{MPa}/({\text{kg/m}}^3)\times 1000=199\approx 198.6\text{MPa}/({\text{g/cm}}^3)$ ✓. Both classmates are right — they just reported different units. This matches the parent's "${\sigma }_y/\rho \approx 200$" for Ti-6Al-4V.

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Recall Quick self-test (reveal after guessing)

Which cell does "pure iron, no alloying, $d\to \infty$" belong to? ::: Cell B (degenerate input) — ${\sigma }_y\to {\sigma }_0$. Ti vs steel by specific strength — who wins and why? ::: Ti (~199 vs ~127); strength-per-mass favours the lighter metal even though steel is stronger absolutely. Why single-crystal blades, if boundaries make metals strong? ::: Cell G flip — at >0.7 T_melt boundaries slide/crack via creep, so removing them ($d\to \infty$) is stronger hot. Lower fuselage skin: 2024 or 7075, and why? ::: 2024 — fatigue/damage-tolerant, higher fracture toughness; yield strength is the wrong metric there.