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

Intuition How to use this page

Work each problem before opening its solution. The problems climb a ladder: L1 Recognition (know the words) → L2 Application (plug the formula) → L3 Analysis (compare, reason) → L4 Synthesis (design a route) → L5 Mastery (defend a subtle claim). Every solution lives inside a collapsible [!recall]- block so the page tests you, not just tells you. The two master formulas you will reuse: $σ_{y} = σ_{0} + \frac{k}{d} Δ τ = \frac{G b}{L}$ where $σ_{y}$ is yield strength in MPa (the stress at which the metal starts to permanently bend), $σ_{0}$ is the lattice friction in MPa (resistance of a perfect crystal), $k$ is the Hall–Petch coefficient with units $MPa\cdotm^{1/2}$ (it must carry $m^{1/2}$ so that dividing by $d$ in metres gives MPa), $d$ is the grain size in metres, $G$ is the shear modulus in Pa, $b$ is the Burgers vector in metres (the size of one atomic slip step), and $L$ is the precipitate spacing in metres. See Hall-Petch Strengthening and Dislocations and Plastic Deformation if any of these words feel new.

Intuition WHERE these two formulas come from (the mechanism, not just the algebra)

Plastic bending happens when line-defects called dislocations glide through the crystal. Anything that blocks their glide raises strength. The two formulas are two different blockers:

Hall–Petch ( $k / d$ ): a grain boundary stops a dislocation dead. Behind it, more dislocations arrive and pile up like cars at a wall — see the pile-up figure below. The pile-up concentrates stress on the boundary; a longer grain fits a longer pile-up (bigger $d$ ), which reaches the wall more easily, so small grains resist more. The $1/ d$ falls out of how pile-up length scales with grain size.
Orowan ( $G b / L$ ): a dislocation cannot cut hard precipitates, so it must bow between them like a soap film pushed through a gap — see the bowing figure. A narrower gap $L$ forces a tighter arc, which costs more stress ( $\propto 1/ L$ ), leaving a residual loop around each particle.

The parent note this drills is Heat Treatment.

Level 1 — Recognition

Recall Solution

Recall the master idea: cooling rate decides which phase you trap.

(a) Furnace cool (slowest) → coarse pearlite (soft, ductile). This is annealing.
(b) Still-air cool (medium) → fine pearlite + small grains (tougher). This is normalising.
(c) Intermediate rate → bainite — feathery ferrite + fine carbides that forms between the pearlite and martensite ranges. Upper bainite (higher hold ~400–550 °C) is coarser; lower bainite (~250–400 °C) is finer, harder and tougher. Bainite is the phase students forget — it fills the whole intermediate window on the TTT and CCT Diagrams.
(d) Water quench (fastest) → martensite (BCT, hard, brittle). This is quenching. Slow = atoms diffuse to equilibrium; fast = no time to diffuse, lattice shears instead. The intermediate window (bainite) is partial diffusion. See TTT and CCT Diagrams for the curves.

Recall Solution

Pearlite — diffusion-controlled. Carbon must migrate to build alternating ferrite/cementite layers → needs time + temperature.
Bainite — mostly diffusion-controlled (carbon still diffuses to form carbides) but at low enough temperature that the ferrite forms by a shear-like step — an intermediate between pearlite and martensite.
Martensite — diffusionless (martensitic shear). The FCC lattice snaps to BCT with no atom hopping.
GP zones — diffusion-controlled. Copper atoms must slowly cluster, which is why ageing takes hours, not milliseconds. See Diffusion in Solids.

Level 2 — Application

Recall Solution

Convert to metres: $d = 36 \times 1 0^{- 6} m$ , so $d = 36 \times 1 0^{- 6} = 6 \times 1 0^{- 3}$ . $σ_{y} = 120 + \frac{0.6}{6 \times 1 0 ^{- 3}} = 120 + 100 = 220 MPa .$ The unit trick: because $k$ carries $m^{1/2}$ , $d$ must be in metres or the answer is nonsense.

Recall Solution

Keep units consistent (SI, everything in metres and pascals). $G = 26 \times 1 0^{9} Pa$ , $b = 0.286 \times 1 0^{- 9} m$ , $L = 50 \times 1 0^{- 9} m$ . $Δ τ = \frac{G b}{L} = \frac{( 26 \times 1 0 ^{9} ) ( 0.286 \times 1 0 ^{- 9} )}{50 \times 1 0 ^{- 9}} = 1.487 \times 1 0^{8} Pa \approx 149 MPa .$ Look at the bowing figure (s01): tighter spacing $L$ forces a smaller bowing radius on the yellow dislocation, so more shear stress (blue arrow) is needed to push it through the gap.

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

Level 3 — Analysis

Recall Solution

$d_{1} = 100 \times 1 0^{- 6} = 1 0^{- 2}$ , $d_{2} = 25 \times 1 0^{- 6} = 5 \times 1 0^{- 3}$ .

(a) $σ_{y 1} = 120 + \frac{0.6}{1 0 ^{- 2}} = 120 + 60 = 180$ MPa. $σ_{y 2} = 120 + \frac{0.6}{5 \times 1 0 ^{- 3}} = 120 + 120 = 240$ MPa.
(b) Rise $= 240 - 180 = 60$ MPa.
(c) Boundary term at $d_{2}$ is $120$ MPa out of $240$ → $120/240 = 0.5 = 50%$ . Quartering $d$ halved $d$ , so the boundary term doubled (60→120). This is why normalising strengthens: it shrinks $d$ .

Recall Solution

Peak: $Δ τ = \frac{( 26 \times 1 0 ^{9} ) ( 0.286 \times 1 0 ^{- 9} )}{40 \times 1 0 ^{- 9}} = 1.859 \times 1 0^{8} Pa \approx 186 MPa .$
Over-aged: $Δ τ = \frac{( 26 \times 1 0 ^{9} ) ( 0.286 \times 1 0 ^{- 9} )}{160 \times 1 0 ^{- 9}} = 4.648 \times 1 0^{7} Pa \approx 46 MPa .$ Coarsening ( $L$ : 40→160 nm) dropped the Orowan stress by a factor of 4. Read it off the hump figure (s02): early on, ageing creates precipitates so the yellow curve rises; past the pink peak, precipitates coarsen and spacing $L$ grows, so $G b / L$ falls and the curve drops. Rising then falling = a hump, not a ramp.

Level 4 — Synthesis

Recall Solution

Austenitise — heat above the steel's $A_{c 3}$ temperature to form FCC austenite (γ). Note: $A_{c 3}$ is not a fixed 723 °C — that number is only the eutectoid line $A_{1}$ . $A_{c 3}$ falls as carbon rises, so a medium-carbon steel (~0.4 %C) austenitises fully only around 800–850 °C. Why: the quench can only trap carbon that is dissolved. See Iron-Carbon Phase Diagram.
Case-harden the surface — the standard way to get a hard skin on a tough core is carburising: hold the part in a carbon-rich atmosphere so carbon diffuses into the surface layer only, raising its carbon content. Why: only the high-carbon skin will quench to hard martensite; the low-carbon core stays tougher. (See Diffusion in Solids — case depth grows as $time$ .)
Quench (oil, not water, to limit cracking) — the carbon-rich case transforms to hard martensite for wear resistance while the core stays softer/tougher.
Temper at ~180–300 °C — raw martensite is glass-brittle and the quench leaves large residual tensile stresses and distortion. Tempering precipitates fine carbides, relaxes those stresses, and restores toughness with little hardness loss. ✅ Result: a carburised, quench-and-temper gear — hard wear surface, tough shatter-resistant core, and (bonus) the surface ends in useful compressive residual stress that resists fatigue cracking. Skipping the temper would leave a distorted, crack-prone part.

Recall Solution

Solution treat ~500 °C — dissolve all Cu into single-phase α. Why: need a single solid solution to trap.
Quench (water) — freeze a supersaturated solid solution (SSSS); Cu has no time to leave. Why fast: to prevent premature coarse precipitation.
Age ~150 °C for hours — Cu forms fine coherent GP zones → θ″. Why moderate: enough atom mobility to nucleate many, fine precipitates (small $L$ , large $G b / L$ ), not so hot they coarsen. Skip the quench? Slow cooling lets Cu precipitate coarsely at grain boundaries → large $L$ , tiny $G b / L$ → the fitting stays soft. The quench is what makes fine ageing possible. See Nickel Superalloys for the same logic at high temperature.

Level 5 — Mastery

Recall Solution

Grain-boundary term: $\frac{k}{d} = \frac{0.07}{50 \times 1 0 ^{- 6}} = \frac{0.07}{7.071 \times 1 0 ^{- 3}} = 9.9$ MPa.
Precipitate term: $Δ σ = 3 \times 149 = 447$ MPa.
Lattice friction: $40$ MPa. Ranking: precipitates ( $\approx 447$ ) ≫ friction ( $40$ ) > grain boundary ( $\approx 10$ ). Total $\approx 40 + 10 + 447 = 497$ MPa. Lesson: in age-hardenable aluminium, precipitation is the workhorse — this is exactly why Duralumin exists and why "just refine the grains" is not enough for aircraft-grade strength.

Recall Solution

For peak-aged: peak ageing gives smallest $L$ , largest $G b / L$ , so maximum room-temperature strength — good for a static airframe fitting. Against (why over-ageing can be chosen deliberately):

Thermal stability: a peak-aged part run hot (engine bay, Nickel Superalloys regime) will coarsen in service and drift off-peak unpredictably. A slightly over-aged (stabilised) temper coarsens more slowly, keeping properties steady.
Toughness / stress-corrosion: peak-aged Al can be more prone to stress-corrosion cracking; a controlled over-age (e.g. T73 temper) trades a little strength for much better crack resistance. Verdict: the claim is too strong. "Best" depends on service temperature and the strength-vs-durability trade, not on peak hardness alone.

Recall Self-check summary

Each line below is Question ::: Answer — cover the right of the ::: and try to recall it. Cooling rate sets the phase ::: slow → equilibrium/soft, fast → trapped/hard Phase at intermediate cooling rate ::: bainite (upper coarser, lower finer/tougher) Hall–Petch says smaller grains do what ::: raise yield strength via $k / d$ Where Hall–Petch reverses ::: below ~10–20 nm grains (inverse Hall–Petch, boundary sliding) Orowan says wider precipitate spacing does what ::: lowers $G b / L$ → over-ageing softens Steel reheat after quench is called ::: tempering (softens, toughens, relieves residual stress) Aluminium reheat after quench is called ::: ageing (hardens)