Heat treatment — annealing, normalising, quenching, tempering; precipitation hardening: Worked examples

This page is the "no surprises" drill for the parent topic. We build a scenario matrix — a checklist of every kind of situation the heat-treatment maths and logic can throw at you — and then solve one example per cell, so you never meet a case you haven't already seen.

Two formulas do almost all the numeric work here, so let's re-earn them before we use a single symbol.

Definition The two engines (defined from zero)

Hall–Petch: $σ_{y} = σ_{0} + \frac{k}{d}$ .
- $σ_{y}$ = yield stress, the pressure (force per area, unit MPa = megapascal) at which the metal starts to bend permanently. Bigger = stronger.
- $σ_{0}$ = friction stress, the baseline resistance of a single perfect crystal with no grain boundaries.
- $d$ = grain size, the width of one crystal grain (unit metres). Picture the metal as a mosaic of tiles; $d$ is the tile width.
- $k$ = a material constant (unit $MPa \cdot m^{1/2}$ ) measuring how strongly boundaries block flow. See Hall-Petch Strengthening.
- Why $1/ d$ and not $1/ d$ ? Dislocations (the line-defects that carry bending) pile up against a boundary; the pile-up length scales with $d$ , and the stress it takes to punch slip into the next grain works out proportional to $1/ d$ . Smaller grain → boundaries closer → harder to bend.
Orowan bowing: $Δ τ = \frac{G b}{L}$ .
- $Δ τ$ = extra shear stress needed to push a dislocation past obstacles (unit MPa).
- $G$ = shear modulus (stiffness in twisting, unit MPa or GPa), $b$ = Burgers vector (the atomic step-size a dislocation carries, unit metres, ~0.25 nm), $L$ = spacing between precipitates (metres).
- Why $1/ L$ ? A dislocation line bows between two obstacles like a soap film between two pins. Squeezing it into a tighter arc (smaller $L$ ) costs more pressure — exactly like a smaller soap bubble needs more air pressure. See Aluminium Alloys (Duralumin).

Notation reminder: $1 μ m = 1 0^{- 6} m$ , $1 nm = 1 0^{- 9} m$ . Units MUST be metres inside these formulas because $k$ , $G$ , $b$ carry metre units.

The scenario matrix

Every problem this topic can pose falls into one of these cells. The map below groups them by which engine drives the case — read it before the table so you see how the cells relate at a glance.

Figure — Heat treatment — annealing, normalising, quenching, tempering; precipitation hardening

How to read the map. The figure has three coloured columns, one per driver. The left (blue) column is the Hall–Petch engine — cases where grain size $d$ sets the strength; going down it, the cells run from "plug in both grain sizes" (A) → "solve backwards for $d$ " (B) → the two extreme limits $d \to \infty$ (C) and $d \to 0$ (D). The right (orange) column is the Orowan engine — cases where precipitate spacing $L$ rules; down it: forward calculation (E) → inverse for $L$ (F) → the time-dependent hump (G). The bottom (green) band holds the concept and real-world cases (H, I, J) that draw on both engines; the grey arrows show those engines "feeding" the applied problems below. So each worked example that follows carries its cell letter, and every cell in the table appears exactly once in the map.

#	Cell class	What makes it tricky	Covered by
A	Both grain sizes given (normal Hall–Petch)	plug numbers, mind units	Ex 1
B	Inverse / solve-for- $d$	algebra to isolate $d$ under a root	Ex 2
C	Degenerate: $d \to \infty$ (single crystal)	limiting value, boundary term → 0	Ex 3
D	Degenerate: $d \to 0$ (nano-grain limit)	limit blows up; where the law breaks	Ex 3
E	Orowan: fine vs coarse spacing	peak-age vs over-age comparison	Ex 4
F	Orowan inverse: find spacing $L$ from strength	solve $L = G b /Δ τ$	Ex 5
G	Over-ageing / non-monotone	strength falls then rises — a turning point	Ex 6
H	Real-world word problem (turbine/gear route)	choose the sequence, no single answer	Ex 7
I	Exam twist: wrong-mechanism trap	quenching Al ≠ hardening it	Ex 8
J	Diffusion time scaling (why ageing takes hours)	Arrhenius/√t reasoning, zero-signs	Ex 9

Worked examples

Worked example Example 1 — Cell A: both grain sizes given

A steel has $σ_{0} = 150 MPa$ , $k = 0.7 MPa \cdot m^{1/2}$ . It is refined from $d = 64 μ m$ (annealed) to $d = 16 μ m$ (normalised). Find both yield stresses and the gain.

Forecast: guess — does quartering the linear grain size double the boundary term? (It doubles $1/ d$ only if $d$ drops by 4×… let's see.)

Convert to metres. $64 μ m = 64 \times 1 0^{- 6}$ m, $16 μ m = 16 \times 1 0^{- 6}$ m. Why this step? $k$ is in $MPa \cdot m^{1/2}$ ; mixing microns would multiply the answer by $1 0^{6} = 1000$ .
Annealed: $64 \times 1 0^{- 6} = 8 \times 1 0^{- 3}$ , so $σ_{y} = 150 + \frac{0.7}{0.008} = 150 + 87.5 = 237.5 MPa$ . Why this step? Direct substitution into Hall–Petch.
Normalised: $16 \times 1 0^{- 6} = 4 \times 1 0^{- 3}$ , so $σ_{y} = 150 + \frac{0.7}{0.004} = 150 + 175 = 325 MPa$ . Why this step? Same law, smaller $d$ .
Gain: $325 - 237.5 = 87.5 MPa$ .

Verify: grain shrank by 4× ( $64 \to 16$ ), so $d$ shrank by 2×, so the boundary term ( $87.5 \to 175$ ) exactly doubled. ✅ The forecast intuition holds: 4× grain refinement doubles the boundary strengthening, not the total.

Worked example Example 2 — Cell B: inverse (solve for grain size)

Same steel ( $σ_{0} = 150$ , $k = 0.7$ ). An engineer needs $σ_{y} = 290 MPa$ . What grain size $d$ achieves it?

Forecast: we want a target strength — will $d$ be bigger or smaller than the 16 μm that gave 325 MPa? (We want less strength, so… bigger grain.)

Isolate the boundary term. $290 - 150 = 140 = \frac{0.7}{d}$ . Why? Subtract the friction stress; only the $d$ -dependent part remains.
Solve for $d$ . $d = \frac{0.7}{140} = 0.005$ m $^{1/2}$ . Why? The unknown sits under a root in a fraction; flip and divide.
Square. $d = 0.00 5^{2} = 2.5 \times 1 0^{- 5} m = 25 μ m$ . Why? Undo the square root to get grain size in metres, then convert.

Verify: plug back: $150 + 0.7/ 25 \times 1 0^{- 6} = 150 + 0.7/0.005 = 150 + 140 = 290$ ✅. And $25 μ m > 16 μ m$ , matching the forecast that we need a coarser grain for less strength.

Worked example Example 3 — Cells C & D: the two degenerate limits

Using $σ_{0} = 150$ , $k = 0.7$ : (C) what is $σ_{y}$ for a perfect single crystal ( $d \to \infty$ )? (D) what does the formula predict as $d \to 0$ , and why is that unphysical?

Forecast: with no boundaries, does the metal keep any strength at all?

Case C, $d \to \infty$ : $\frac{k}{d} \to 0$ , so $σ_{y} \to σ_{0} = 150 MPa$ . Why? One giant grain has (almost) no boundaries; the only resistance left is the lattice friction $σ_{0}$ . This is the floor of Hall–Petch.
Case D, $d \to 0$ : $\frac{k}{d} \to + \infty$ , so the formula predicts infinite strength. Why unphysical? Real crystals cannot have grains smaller than ~10 nm; below that, grains slide past each other (inverse Hall–Petch) and strength falls. The law is only valid in its diffusion-boundary regime.
Sanity number for D: at $d = 10 nm = 1 0^{- 8}$ m, formula gives $150 + 0.7/ 1 0^{- 8} = 150 + 0.7/1 0^{- 4} = 150 + 7000 = 7150 MPa$ — larger than the theoretical strength of steel (~ $10000$ MPa is the limit), a clear red flag.

Verify: limit as $d \to \infty$ of $150 + 0.7/ d$ is exactly 150 ✅; at $d = 1 0^{- 8}$ the term is 7000 ✅. Both confirm: Hall–Petch has a hard floor and a physical ceiling.

Worked example Example 4 — Cell E: Orowan, fine vs coarse spacing (peak vs over-aged)

An Al–Cu alloy has $G = 26 GPa$ and Burgers vector $b = 0.25 nm$ . At peak ageing the precipitate spacing is $L = 50 nm$ ; after over-ageing it coarsens to $L = 200 nm$ . Find $Δ τ$ in each case.

Forecast: spacing quadruples on over-ageing — will the strengthening drop to a quarter?

Read the figure first. Two pairs of round dots are the precipitates (obstacles). The blue line is the dislocation squeezed through the tight peak-aged gap ( $L = 50$ nm) — notice how sharply it has to bow, drawn as a tall arc; the gray arrow beneath shows the shear stress pushing it up. The orange line is the same dislocation through the wide over-aged gap ( $L = 200$ nm) — a shallow, lazy arc that needs little push. The figure is the geometry behind $Δ τ = G b / L$ : tight arc = big stress, wide arc = small stress.

Consistent units. $G = 26 GPa = 26000 MPa$ ; $b = 0.25 \times 1 0^{- 9}$ m. Why? Answer wanted in MPa, so express $G$ in MPa; keep lengths in metres so metres cancel.
Peak-aged ( $L = 50 \times 1 0^{- 9}$ m, the tight blue arc): $Δ τ = \frac{26000 \times 0.25 \times 1 0 ^{- 9}}{50 \times 1 0 ^{- 9}} = \frac{6.5 \times 1 0 ^{- 6}}{50 \times 1 0 ^{- 9}} = 130 MPa$ . Why? Small $L$ → tight bowing arc → large stress (the tall blue line).
Over-aged ( $L = 200 \times 1 0^{- 9}$ m, the shallow orange arc): $Δ τ = \frac{26000 \times 0.25 \times 1 0 ^{- 9}}{200 \times 1 0 ^{- 9}} = 32.5 MPa$ . Why? Larger $L$ → gentle arc (orange line) → far less stress needed.

Verify: $L$ went up 4× ( $50 \to 200$ ), $Δ τ$ went down 4× ( $130 \to 32.5$ ) since $Δ τ \propto 1/ L$ ✅. Forecast confirmed: over-ageing quartered the Orowan strengthening.

Worked example Example 6 — Cell G: the ageing turning point (fall then rise)

Real ageing has two competing effects, so we model strength as a sum of a rising term (as more precipitates nucleate, cutting strength climbs like $t$ ) and a falling term (as those precipitates coarsen, Orowan bowing weakens). Take $σ (t) = 150 + 60 t + \frac{400}{1 + t} MPa, t in hours .$ The $60 t$ term is the rising (nucleation/cutting) branch; the $400/ (1 + t)$ term starts large and decays as spacing grows. Map the curve's shape and find its turning point (the local minimum).

Forecast: does $σ (t)$ slide one way, or fall to a valley and then climb back — a genuine turning point?

Evaluate at $t = 0$ : $σ = 150 + 0 + 400/1 = 550 MPa$ . Why this step? This is the just-quenched / start-of-ageing strength before nucleation ramps up.
Differentiate to find the turning point. $\frac{d σ}{d t} = \frac{30}{t} - \frac{400}{( 1 + t ) ^{2}}$ . Setting it to zero locates where the slope is flat. Why this step? A derivative answers "where does the slope change sign?" — exactly the valley's bottom, where the falling and rising rates balance. It's the right tool because we want the turning point, not a value.
Locate the turning point numerically. Solving $\frac{30}{t} = \frac{400}{( 1 + t ) ^{2}}$ gives $t_{m i n} \approx 4.28$ h. Since the curve falls into it and rises out, it is a local minimum. Why? A turning point can be a max or a min; checking the neighbours (below) tells us which.
Sample a few times to see the shape.
- $t = 0$ : $150 + 0 + 400 = 550$ MPa.
- $t = 1$ : $150 + 60 + 200 = 410$ MPa.
- $t = 4$ : $150 + 120 + 80 = 350$ MPa.
- $t = 9$ : $150 + 180 + 40 = 370$ MPa.
- $t = 16$ : $150 + 240 + 23.5 = 413.5$ MPa. Why? Sampling shows the sequence falls ( $550 \to 410 \to 350$ ) then rises ( $350 \to 370 \to 413.5$ ): a valley bottoming near $t \approx 4$ , confirming the local minimum found by the derivative (see the marked trough in the figure).

Verify: $σ (0) = 550$ , $σ (1) = 410$ , $σ (4) = 350$ , $σ (9) = 370$ , $σ (16) = 413.5$ ✅ — strictly falling then strictly rising, so the curve is non-monotone with a local minimum near $t \approx 4.3$ h. That sign-change in the slope is the mathematical signature of two competing mechanisms. See Diffusion in Solids.

Worked example Example 7 — Cell H: real-world word problem (turbine disc route)

A nickel-superalloy turbine disc must resist creep at 700 °C (needs fine strengthening precipitates) yet not crack during spin-up (needs toughness). Design the treatment sequence.

Forecast: which single step could give both? (Trick: none — you need a sequence.)

Solution treat (~1100 °C). Dissolve all the γ′ strengthener into solution. Why? You cannot control precipitate size unless you first erase them into a clean single phase.
Quench / fast air cool. Trap a supersaturated solution. Why? So the γ′ can't precipitate coarsely during cooling — freeze the solute for controlled ageing.
Age at ~700–850 °C, timed to peak. Grow fine, closely spaced γ′ (small $L$ ). Why? By Orowan $Δ τ = G b / L$ , small $L$ maximises creep strength; stop before coarsening (over-ageing).
Optional second (lower-temp) age. Tune a bimodal precipitate size for toughness. Why? Mixes fine (strength) and slightly coarser (crack resistance) populations.

Verify (logic chain): each requirement is met by a specific step — creep resistance ← fine γ′ from steps 1–3 (small $L$ , large Orowan stress); crack resistance ← the toughening bimodal size from step 4; no single step delivers both, which is exactly why the sequence exists. The route reproduces the parent's precipitation-hardening triad (solution → quench → age) with a superalloy toughness tweak, so the design is self-consistent ✅.

Worked example Example 8 — Cell I: exam twist (wrong-mechanism trap)

Exam claim: "To harden the pure-aluminium bracket, just water-quench it from 500 °C like steel." True or false — and give the correct route.

Forecast: does quenching harden aluminium the way it hardens steel?

Check for a martensite transformation. Steel hardens because FCC austenite shears to hard BCT martensite trapping carbon. Pure Al has no such phase change on cooling. Why? Without a diffusionless shear transformation, quenching can't create martensite.
What quenching does to Al alloys. In an alloy like Al–Cu it produces a supersaturated solid solution — still soft — that only hardens later by ageing (fine precipitates). Why? Strength comes from precipitates blocking dislocations, which need time to form.
Pure aluminium has no solute → nothing to precipitate → quenching does essentially nothing to hardness. Why? No Cu, no GP zones, no Orowan obstacles.

Verify: claim is FALSE ✅. Correct: pure Al can't be precipitation- or martensite-hardened; use work hardening (cold rolling) or alloy it first. This is exactly the parent's "quenching ≠ hardening everything" mistake.

Worked example Example 9 — Cell J: diffusion time scaling (why ageing takes hours)

Ageing needs Cu atoms to diffuse a characteristic distance $x$ . Diffusion distance obeys $x \approx D t$ , where $D$ is the diffusion coefficient (m²/s) and $t$ is time. If $D = 1 0^{- 20} m^{2} / s$ at 150 °C, how long to move solute $x = 100 nm$ ? And what does $t = 0$ give?

Forecast: nanometres are tiny — seconds or hours?

Rearrange for time. Square both sides of $x = D t$ : $x^{2} = D t$ , so $t = \frac{x ^{2}}{D}$ . Why this step? We know the distance and $D$ , we want the time; isolating $t$ answers the question directly. Squaring is the tool because $x$ sits under a root.
Substitute the numbers. $x = 100 nm = 1 0^{- 7}$ m, so $x^{2} = 1 0^{- 14} m^{2}$ . Then $t = \frac{1 0 ^{- 14}}{1 0 ^{- 20}} = 1 0^{6} s .$ Why? Direct substitution; the m² in $x^{2}$ cancels the m² in $D$ , leaving seconds.
Convert to human units. $t = 1 0^{6} s \div 86400 s/day \approx 11.6 days$ . Why? This is why natural (room-temperature) ageing takes days; artificial ageing heats the alloy so $D$ rises sharply (Arrhenius), collapsing $t$ to hours — the reason we age at ~150 °C, not 25 °C.
Degenerate case $t = 0$ . $x = D \cdot 0 = 0$ — zero diffusion distance, no precipitates yet. Why? This confirms the just-quenched supersaturated solution is the soft start of ageing, not the strong end — nothing has moved, so nothing strengthens until time passes.

Verify: $t = 1 0^{- 14} /1 0^{- 20} = 1 0^{6}$ s ✅ and $1 0^{6} /86400 \approx 11.57$ days ✅; at $t = 0$ , $x = 0$ ✅. This is why ageing is a slow, temperature-controlled step. See Diffusion in Solids.

Recall Self-test

Hall–Petch floor as $d \to \infty$ ::: $σ_{y} \to σ_{0}$ (only lattice friction remains). In Orowan $Δ τ = G b / L$ , doubling $L$ does what to $Δ τ$ ? ::: halves it (over-ageing softens). Why can't pure aluminium be quench-hardened like steel? ::: no martensite transformation and no solute to precipitate. Grain shrinks 4× — what happens to the Hall–Petch boundary term? ::: it doubles ( $1/ d$ grows 2×). Diffusion distance vs time law? ::: $x \approx D t$ . A turning point that falls then rises is a local … ::: minimum (a valley); rises then falls is a maximum (a hump).

Related maps: Iron-Carbon Phase Diagram · TTT and CCT Diagrams · Hall-Petch Strengthening · Dislocations and Plastic Deformation.