Heat Treatment of Steels and Aluminium Alloys: TTT and CCT, Annealing to Tempering, Precipitation Hardening

This chapter answers the first two paragraphs of Section XE2.5 (Processing of Materials) of XE2 Materials Science, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam. The syllabus text reads: “Heat Treatment of steels: TTT and CCT diagram: coarse and fine pearlite, martensite and bainite; Annealing, normalizing, quenching, tempering. Heat treatment of aluminium alloys: Precipitation hardening: Solutionising, quenching and ageing. Hardness vs. aging time and its dependence on aging temperature.” It teaches the Fe–Fe₃C diagram as the map of steel heat treatment, the TTT and CCT diagrams, the products pearlite, bainite and martensite with the Ms and Mf temperatures, the four classical treatments and hardenability in outline, and then the solutionise–quench–age route of aluminium alloys with its precipitation sequence and hardness–time curves. The numericals are lever-rule fractions of ferrite, cementite and pearlite, carbon content from phase fractions, a critical cooling rate read from a simplified TTT diagram, and the effect of ageing temperature on the time to peak hardness.

1. The Fe–Fe₃C diagram as the map of steel heat treatment

Heat treatment changes microstructure, and so properties, by controlled heating and cooling without changing composition. For steel the map is the Fe–Fe₃C diagram. Its landmarks, in wt% C: austenite (γ, FCC) dissolves up to 2.14 wt% C at 1147 °C; ferrite (α, BCC) dissolves at most 0.022 wt% C (at 727 °C); cementite (Fe₃C) contains 6.67 wt% C. At the eutectoid point, 727 °C and 0.76 wt% C, austenite transforms to a lamellar mixture of ferrite and cementite called pearlite: γ (0.76) → α (0.022) + Fe₃C (6.67). A steel with less than 0.76 wt% C is hypoeutectoid and cools to proeutectoid ferrite plus pearlite; one with more is hypereutectoid and cools to proeutectoid cementite (on the austenite grain boundaries) plus pearlite.

Three critical lines carry symbols. A₁ is the eutectoid line (727 °C). A₃ is the upper boundary of the ferrite + austenite field, falling from 912 °C for pure iron to 727 °C at 0.76 wt% C. Acm is the boundary of the austenite + cementite field. Measured on heating they are written Ac₁, Ac₃, Accm and on cooling Ar₁, Ar₃, Arcm; the real transformation lags the equilibrium line (Ac is above it, Ar below). Austenitising for hardening means heating into the austenite field: above A₃ for a hypoeutectoid steel, but only just above A₁ for a hypereutectoid steel, which keeps some fine undissolved cementite for wear resistance and avoids coarse austenite grains.

Lever rule just below 727 °C (take 0.022, 0.76 and 6.67 wt% C). Hypoeutectoid steel of carbon C₀: proeutectoid ferrite = (0.76 − C₀)/(0.76 − 0.022) and pearlite = (C₀ − 0.022)/(0.76 − 0.022). Hypereutectoid: proeutectoid cementite = (C₀ − 0.76)/(6.67 − 0.76) and pearlite = (6.67 − C₀)/(6.67 − 0.76). Worked: for 0.40 wt% C, proeutectoid ferrite = 0.36/0.738 = 0.488 and pearlite = 0.512. Total ferrite (free plus in pearlite) uses the end of the tie line at 6.67: (6.67 − 0.40)/(6.67 − 0.022) = 0.943, so total cementite is 0.057. Inside pearlite the mass fractions are 0.889 ferrite and 0.111 cementite, so the ferrite lamellae are about eight times thicker than the cementite ones.

⚠️ Pearlite fraction versus total ferrite
“Fraction of pearlite” uses the eutectoid composition 0.76 as one end of the tie line, because the austenite that remains at 727 °C has exactly that composition. “Fraction of ferrite” (all of it) uses 6.67 as the other end. Mixing the two tie lines is the commonest slip: a 0.40 wt% C steel is 51% pearlite but 94% ferrite.

2. TTT diagram: pearlite, bainite and martensite

The TTT (time–temperature–transformation) diagram is drawn for one composition. Austenite is cooled rapidly to a chosen temperature, held there, and the times at which transformation starts and finishes are plotted against temperature on a log-time axis, giving C-shaped curves. The shape has a reason. Just below A₁ the driving force is small, so nucleation is slow; far below, diffusion is slow. The rate is highest at the nose (near 550 °C and about 1 s in a plain eutectoid carbon steel). The product depends on where the specimen is held, not on how it got there.

Products of austenite in a plain-carbon eutectoid steel (approximate ranges; textbooks differ slightly)
ProductFormedMechanism and structureRelative hardness
Coarse pearliteJust below A₁, high temperatureDiffusional; widely spaced ferrite and Fe₃C lamellaeLowest
Fine pearliteLower, towards the noseDiffusional; lamellar spacing ∝ 1/(A₁ − T), so it falls as undercooling risesHigher than coarse
Upper bainiteAbout 550 to 350 °CFerrite laths with carbides between the laths; featheryHigher than pearlite
Lower bainiteAbout 350 °C down to MsFerrite plates with very fine carbides inside; needle-likeHigher, and tougher than upper bainite
MartensiteBelow Ms on rapid coolingDiffusionless shear; carbon trapped in a body-centred tetragonal (BCT) latticeHighest; brittle

Martensite forms by a diffusionless shear of FCC austenite into a supersaturated BCT solid solution with the same composition as the parent. Because no diffusion is needed it is athermal: the amount depends only on how far below Ms (start) the steel is cooled, reaching zero further growth if held, and finishing at Mf. Ms and Mf fall as carbon and most alloying elements rise. In high-carbon steels (roughly above 0.6 wt% C), Mf lies below room temperature, so a room-temperature quench leaves retained austenite, which sub-zero treatment can reduce. Martensite is hard because of trapped carbon, a very high dislocation density and fine laths or plates; it is brittle and carries large residual stress.

🎯 Why the C-curve has a nose
Transformation rate is a product of two opposing factors. Nucleation rate rises with undercooling (larger driving force), while diffusion, which feeds growth, falls as temperature drops. The product peaks in between, so time-to-start is shortest at the nose. Alloying elements such as Mo, Cr, Mn and Ni slow nucleation and diffusion and push the whole curve to the right.

3. CCT diagram and cooling-rate reasoning

Real heat treatments cool continuously, not isothermally. The CCT (continuous-cooling-transformation) diagram shows where cooling curves T(t) meet the start and finish of transformation. Compared with the TTT diagram of the same steel its curves lie at longer times and lower temperatures, because the austenite spends part of the cooling time at high temperatures where nucleation is slow. Slow furnace cooling gives coarse pearlite, still-air cooling fine pearlite, oil quenching a mixture that may include bainite and martensite, and water quenching martensite. For a plain eutectoid carbon steel the CCT diagram has no bainite region: a cooling curve either transforms to pearlite or reaches Ms and forms martensite. Bainite appears in the CCT diagram of low-alloy steels.

The critical cooling rate is the slowest rate that just misses the nose and so gives a fully martensitic structure. From a TTT diagram a quick estimate treats the cooling curve as a straight line from A₁ to the nose. Worked example (simplified data): a eutectoid steel at 727 °C must reach the nose at 547 °C within 0.90 s. The rate is (727 − 547)/0.90 = 200 °C/s. Because the true CCT nose lies at longer time and lower temperature, this TTT-based estimate over-estimates the required rate slightly. Cooling slower than critical gives pearlite or bainite plus martensite; hardness then falls with depth below the surface, because the core of a thick section cools slowly.

  • Austempering: quench to a temperature just above Ms (in the bainite range), hold until the austenite has become bainite, then cool in air. Gives bainite with good toughness and little distortion or cracking.
  • Martempering: quench to just above Ms, hold briefly so that surface and core reach the same temperature, then cool slowly through the Ms–Mf range so that martensite forms nearly together everywhere. Residual stress and cracking are lower; tempering follows.

4. Annealing, normalising, quenching, tempering and hardenability

The four classical treatments of steel
TreatmentHeatingCoolingResult
Full annealingAustenitise: above A₃ (hypoeutectoid) or between A₁ and Acm (hypereutectoid)Slow, in the furnaceCoarse pearlite (plus proeutectoid phase); softest, most ductile, stress-free
NormalisingAbove A₃ (hypoeutectoid) or above Acm (hypereutectoid)Still airFiner pearlite and finer grains; harder and stronger than annealed
Quenching (hardening)Austenitise as aboveFaster than the critical rate: brine, water, oil, in falling order of severityMartensite; hard, brittle, high residual stress
TemperingReheat the quenched steel below A₁ (about 150 to 650 °C)Any, usually airTempered martensite: fine Fe₃C in ferrite; hardness down, toughness up

Spheroidising (prolonged holding just below A₁) turns lamellar cementite into spheres in a ferrite matrix, the softest condition and the best for machining high-carbon steel. Process annealing below A₁ restores ductility to cold-worked low-carbon steel by recrystallisation without any phase change. Normalising and full annealing both give ferrite and pearlite; they differ only in cooling rate, and the faster air cool of normalising gives finer pearlite and therefore higher hardness.

Tempering is diffusional. As temperature or time rises, carbon leaves the BCT lattice, which relaxes towards BCC, fine carbides precipitate and later coarsen, and retained austenite decomposes. Hardness and strength fall smoothly, while toughness and ductility rise; temperature matters more than time. Hardenability is the ability of a steel to form martensite to depth on quenching; it is not the same as hardness. It is measured by the Jominy end-quench test: a standard bar is austenitised and water-sprayed on one end only, and hardness is plotted against distance from the quenched end. A steel of high hardenability keeps its hardness far from the end. Mn, Cr, Mo and Ni raise hardenability by moving the nose to the right and lowering the critical rate; a coarser austenite grain size does the same. The maximum hardness at the quenched end is fixed by carbon content, not by alloying.

⚠️ Hardenability is not hardness
Adding a hardenability element such as Mo to a 0.4 wt% C steel lets thicker sections go fully martensitic but does not make the martensite itself harder. A 1 % C plain steel quenched at its surface can be harder than a 0.2 % C alloy steel, yet the alloy steel hardens deeper.

5. Precipitation hardening of aluminium alloys

Precipitation hardening needs an alloy system whose solid solubility falls with temperature (a solvus), so that a single-phase solid solution at high temperature becomes supersaturated on cooling. Al–Cu is the model: the Al-rich α phase dissolves up to about 5.65 wt% Cu at the eutectic temperature 548 °C and very little at room temperature, and the equilibrium precipitate is θ = CuAl₂ (about 54 wt% Cu). The treatment has three steps. Solutionising: heat to a temperature above the solvus but below the eutectic (to avoid melting at grain boundaries) and hold until all Cu is dissolved. Quenching: cool rapidly, usually in water, so that the Cu has no time to precipitate; the result is a soft, unstable supersaturated solid solution with many retained vacancies. Ageing: hold at room temperature (natural ageing) or at about 100 to 200 °C (artificial ageing) so that fine precipitates form.

The precipitation sequence is: supersaturated α → GP zones → θ″ → θ′ → θ (CuAl₂). Guinier–Preston (GP) zones are Cu-rich discs one to two atomic layers thick, fully coherent with the matrix; θ″ is a coherent, larger, ordered phase; θ′ is semi-coherent; θ is the incoherent equilibrium phase. Each step is closer to equilibrium but is reached through a smaller nucleation barrier than going straight to θ, because coherent interfaces have low interfacial energy.

🎯 Why the metastable phases come first
The nucleation barrier scales as ΔG* ∝ σ³/ΔGv². The stable θ phase has a high-energy incoherent interface (large σ), so its barrier is large. GP zones have an almost zero-energy coherent interface, so they nucleate homogeneously and in huge numbers even at room temperature, with the help of quenched-in vacancies that speed Cu diffusion. The system takes the path of fastest kinetics, not the path of lowest final energy.

The strength comes from the strain fields around coherent particles, which obstruct dislocations. Small particles are sheared by dislocations, so strength rises with particle size and volume fraction. Beyond a critical size dislocations bypass the particles by bowing between them (Orowan looping) at a stress of about Gb/L, with G the shear modulus, b the Burgers vector and L the interparticle spacing, so strength falls as L grows. Hardness is maximal at the changeover.

6. Hardness versus ageing time and ageing temperature

At any ageing temperature, hardness rises, reaches a peak and then falls. The rise reflects GP zones and θ″ forming and growing; the peak is the optimum of size and spacing (the peak-aged condition, temper T6 for artificially aged alloys); the fall, over-ageing, is due to θ′ and θ forming as larger, more widely spaced and less coherent particles, together with coarsening (Ostwald ripening, in which large particles grow at the expense of small ones). The same applies to other heat-treatable alloy families such as the 6xxx (Al–Mg–Si) and 7xxx (Al–Zn–Mg) series; the 1xxx, 3xxx and 5xxx series are not hardened this way. Temper T4 means solution treated and naturally aged.

Effect of ageing temperature on the hardness–time curve (artificial ageing range)
Higher ageing temperatureLower ageing temperature
Peak is reached sooner (faster diffusion)Peak takes much longer
Peak hardness is lower (fewer, coarser precipitates; GP zones unstable)Peak hardness is higher (finer, denser precipitates)
Over-ageing starts sooner and the fall is quickerWider time window near the peak; slow, gentle fall

The time to reach peak hardness follows an Arrhenius law, t_peak ∝ exp(Q/RT) with T in kelvin, because precipitation is limited by solute diffusion. Worked example (R = 8.314 J mol⁻¹ K⁻¹, Q = 100 kJ/mol assumed): an alloy peaks after 48 h at 150 °C. At 180 °C, t₂/t₁ = exp[(Q/R)(1/T₂ − 1/T₁)] with T₁ = 423.15 K and T₂ = 453.15 K: Q/R = 12 028 K and 1/T₂ − 1/T₁ = −1.5646 × 10⁻⁴ K⁻¹, so t₂/t₁ = exp(−1.882) = 0.1523 and t₂ = 7.3 h. A 30 °C rise cuts the ageing time by a factor of about 6.6, at the cost of lower peak hardness.

ℹ️ Why over-ageing is sometimes chosen
Temper T7 deliberately over-ages high-strength 7xxx alloys beyond the peak. Strength falls a little, but resistance to stress-corrosion cracking and dimensional stability improve. Peak ageing is therefore not always the engineering optimum.

Key takeaways

  • Heat treatment is read off the Fe–Fe₃C diagram: eutectoid 0.76 wt% C at 727 °C, ferrite up to 0.022 wt%, cementite 6.67 wt%. Use the lever rule with the right tie line: pearlite fraction (C₀ − 0.022)/0.738 for hypoeutectoid steel, but total ferrite uses the 6.67 end.
  • On a TTT diagram the nose separates slow diffusional products (coarse then fine pearlite) from bainite and, below Ms, martensite. Pearlite spacing falls with undercooling; martensite is diffusionless, athermal, BCT, set by temperature reached rather than time, with retained austenite if Mf is below room temperature.
  • CCT curves lie to longer times and lower temperatures than TTT. The critical cooling rate just misses the nose; a TTT-based straight-line estimate slightly over-states it. Plain eutectoid steel shows no bainite on continuous cooling.
  • Annealing (furnace cool) gives coarse pearlite and the softest ductile steel; normalising (air cool) gives finer pearlite, harder and stronger; quenching gives hard brittle martensite; tempering below A₁ trades hardness for toughness. Alloying raises hardenability (nose moves right) without raising the maximum hardness, which is set by carbon.
  • Precipitation hardening is solutionise, quench, age. In Al–Cu the sequence is supersaturated α → GP zones → θ″ → θ′ → θ; coherent metastable phases form first because their interfacial energy and nucleation barrier are low. Particles are sheared when small and bypassed when large, so strength peaks and then falls.
  • Raising the ageing temperature brings the hardness peak sooner (t_peak ∝ exp(Q/RT)) but lowers the peak value and speeds over-ageing; lowering it does the reverse. Over-ageing (T7) is sometimes chosen for stress-corrosion resistance.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Austenite of eutectoid composition (0.76 wt% C) is cooled very slowly through 727 °C, close to equilibrium. The product formed is:

    1. Bainite, with ferrite laths and carbide particles lying between the laths
    2. Martensite, a body-centred tetragonal supersaturated solid solution of carbon in iron
    3. Pearlite, a lamellar mixture of ferrite and cementite formed by diffusion of carbon
    4. Proeutectoid cementite along the austenite grain boundaries, with no ferrite present
    Show answer

    Answer: C — Pearlite, a lamellar mixture of ferrite and cementite formed by diffusion of carbon

    Slow cooling lets carbon diffuse and the eutectoid reaction γ → α + Fe₃C run, giving pearlite. Bainite needs a much larger undercooling (isothermal holding below the nose), and martensite needs a quench fast enough to avoid diffusion altogether and to reach Ms. Proeutectoid cementite appears only in hypereutectoid steel; at exactly 0.76 wt% C there is no proeutectoid phase.

  2. In a plain-carbon steel, which statement about the austenite-to-martensite transformation is correct?

    1. It is diffusionless, so the amount formed depends on the temperature reached below Ms and not on the holding time
    2. It needs carbon diffusion, so holding the steel just below Ms for longer transforms steadily more austenite
    3. It produces a product whose carbon content differs from that of the parent austenite, as in the pearlite reaction
    4. It appears on the TTT diagram as a C-shaped curve with a nose at about 550 °C
    Show answer

    Answer: A — It is diffusionless, so the amount formed depends on the temperature reached below Ms and not on the holding time

    Martensite forms by a cooperative shear with no atomic diffusion, so it has the composition of the parent austenite and is athermal: the fraction grows only as the temperature falls from Ms towards Mf, and stops growing if the temperature is held. The tempting distractor, that a longer hold transforms more, describes diffusional products such as pearlite and bainite. On a TTT diagram Ms and Mf are horizontal lines, not a C-curve.

  3. In the aged Al–Cu alloy, the precipitation sequence from the supersaturated solid solution α, in order of appearance, is:

    1. α → θ → θ′ → θ″ → GP zones
    2. α → θ′ → GP zones → θ″ → θ
    3. α → θ″ → GP zones → θ′ → θ
    4. α → GP zones → θ″ → θ′ → θ
    Show answer

    Answer: D — α → GP zones → θ″ → θ′ → θ

    The sequence runs from fully coherent Cu-rich zones (GP zones), through the coherent θ″ and semi-coherent θ′, to the incoherent equilibrium phase θ = CuAl₂. The other orders put the equilibrium phase or a larger, less coherent phase before the finer zones, which reverses the progression of size and coherency loss.

  4. Two pieces of the same solution-treated and quenched Al–Cu alloy are aged artificially, one at 150 °C and the other at 190 °C, and hardness is measured against ageing time for each. Which description of the two hardness–time curves is correct?

    1. The 190 °C curve peaks sooner and the peak is lower, because the precipitates are fewer and coarser
    2. The 190 °C curve peaks sooner and the peak is higher, because faster diffusion produces more precipitates
    3. The 190 °C curve peaks later and lower, because diffusion of copper is slower at the higher temperature
    4. Both curves peak at the same time, because the order of precipitation is fixed by the alloy
    Show answer

    Answer: A — The 190 °C curve peaks sooner and the peak is lower, because the precipitates are fewer and coarser

    Diffusion is faster at 190 °C (Arrhenius), so the peak comes sooner. But the supersaturation, and hence the driving force and the number density of nuclei, is smaller at the higher temperature, and the GP zones, which give the finest dispersion, are not stable there. The precipitates are fewer and coarser, so the peak hardness is lower. Faster diffusion alone would suggest a higher peak, which is the tempting but wrong reading, and a fixed sequence does not fix the rate.

  5. A 0.45 wt% C steel is austenitised above A₃. One piece is then cooled slowly in the furnace and another in still air. Compared with the furnace-cooled piece, the air-cooled piece has:

    1. coarser pearlite and lower hardness, because transformation occurs closer to A₁
    2. finer pearlite and higher hardness, because transformation occurs at a larger undercooling
    3. a fully martensitic structure, because air cooling exceeds the critical cooling rate of this steel
    4. the same microstructure and hardness, because both start from the same austenite
    Show answer

    Answer: B — finer pearlite and higher hardness, because transformation occurs at a larger undercooling

    Faster air cooling shifts the pearlite reaction to a lower temperature, a larger undercooling, where the lamellar spacing is smaller (spacing ∝ 1/(A₁ − T)) and the hardness higher; this is why normalised steel is stronger than annealed steel. Air cooling of a plain medium-carbon steel is far slower than the critical rate for martensite, so the claim of a fully martensitic structure is wrong.

  6. A simplified TTT diagram of a plain eutectoid steel is described as follows. Pearlite starts at the nose (550 °C) after 1 s. At 350 °C bainite starts after 20 s and finishes after 500 s. Ms is 230 °C. A specimen is austenitised and cooled in 0.2 s to a salt bath at 350 °C, held there for 1 hour, and then quenched in water to room temperature. Its microstructure is:

    1. fully bainite, because the austenite finished transforming at 350 °C before the final quench
    2. bainite plus martensite, because the final water quench transforms the remaining austenite
    3. fine pearlite, because the specimen was in the pearlite temperature range for part of the cooling
    4. fully martensite, because water quenching takes any steel below Ms
    Show answer

    Answer: A — fully bainite, because the austenite finished transforming at 350 °C before the final quench

    The specimen reaches 350 °C in 0.2 s, well before the 1 s pearlite start at the nose, so no pearlite forms. Holding for 3600 s exceeds the 500 s needed to finish bainite, so no austenite remains. The later water quench then has nothing to transform, because martensite forms only from austenite; the quench to Ms cannot touch bainite. Hence the structure is fully bainitic.

  7. Adding about 0.2 wt% Mo to a medium-carbon steel increases its hardenability. The metallurgical reason is that Mo:

    1. raises the carbon content of the martensite, so that the martensite formed near the surface is harder, in the surface layers of the quenched part
    2. raises Ms, so that martensite begins to form at a higher temperature in the slower-cooled regions of the core, during the quench of a thick section
    3. raises the thermal conductivity of the steel, so that the core cools as fast as the surface during the quench
    4. delays the pearlite and bainite reactions, moving the nose to longer times and lowering the critical cooling rate
    Show answer

    Answer: D — delays the pearlite and bainite reactions, moving the nose to longer times and lowering the critical cooling rate

    Hardenability is governed by how slowly steel can be cooled and still miss the nose. Mo is a strong retarder of the diffusional reactions, so the C-curves move to the right and the critical rate falls, letting thick sections reach martensite. The hardness of martensite itself is fixed by carbon, not Mo. Mo slightly lowers Ms rather than raising it, and alloying lowers the thermal conductivity of steel, so the conductivity and Ms explanations are wrong.

  8. A quenched, fully martensitic steel is reheated to 500 °C, below A₁, and held. Which change in properties results, and why?

    1. Hardness rises, because carbon diffuses together to form more martensite
    2. Hardness and toughness both rise, because the body-centred tetragonal lattice becomes face-centred cubic
    3. Hardness falls and toughness rises, because carbon leaves the tetragonal lattice as fine carbide particles
    4. Hardness is unchanged and only residual stress rises, because the transformation is diffusionless
    Show answer

    Answer: C — Hardness falls and toughness rises, because carbon leaves the tetragonal lattice as fine carbide particles

    Tempering is thermally activated and diffusional: carbon leaves the supersaturated BCT lattice, the lattice relaxes towards BCC ferrite and fine cementite precipitates. Hardness and strength fall while toughness and ductility rise, and residual stress is relieved rather than increased. Reheating below A₁ cannot form FCC austenite or new martensite; martensite only forms on cooling from austenite.

  9. An Al–4 wt% Cu alloy is solution treated, but cooled very slowly in the furnace instead of being quenched. On later ageing it stays soft. The best explanation is that slow cooling:

    1. dissolves the CuAl₂ phase even more completely, so that no copper remains available to precipitate
    2. lets coarse incoherent θ precipitate, largely at grain boundaries, during cooling, leaving little supersaturated Cu to form fine coherent zones later
    3. makes the alloy amorphous, so that no precipitate can nucleate during the later ageing treatment
    4. raises the number of retained vacancies so much that the GP zones redissolve at room temperature
    Show answer

    Answer: B — lets coarse incoherent θ precipitate, largely at grain boundaries, during cooling, leaving little supersaturated Cu to form fine coherent zones later

    Hardening needs the Cu to stay in a supersaturated solution until ageing. Slow cooling passes through the (α + θ) field slowly, so equilibrium θ nucleates heterogeneously at grain boundaries and grows coarse, consuming solute. Little is left for fine coherent precipitates, and the coarse particles harden poorly. Slow cooling does not make the alloy amorphous, and it reduces rather than raises the retained vacancy concentration. Dissolution is completed at the solution-treatment temperature, not during cooling.

  10. When a peak-aged Al–Cu alloy is held much longer at the ageing temperature (below the solvus), its hardness falls. The main cause is:

    1. complete redissolution of all precipitates into the aluminium matrix, as happens when an alloy is heated above its solvus temperature
    2. an increase in coherency strain as θ becomes fully coherent with the matrix, which would make the particles harder for dislocations to cut
    3. recrystallisation of the aluminium matrix, which removes the dislocations, which would soften the metal at the ageing temperature
    4. coarsening of θ′ and θ, which widens the particle spacing so that dislocations bypass the particles by Orowan looping
    Show answer

    Answer: D — coarsening of θ′ and θ, which widens the particle spacing so that dislocations bypass the particles by Orowan looping

    On prolonged ageing the precipitates coarsen (Ostwald ripening) and lose coherency, so the interparticle spacing L grows and the Orowan bypass stress, about Gb/L, falls. Below the solvus the precipitates do not redissolve, which would need a higher temperature. θ becomes less, not more, coherent, and the matrix does not recrystallise at ageing temperatures without prior cold work.

  11. Which of the following statements about the TTT and CCT diagrams of steel are correct?

    1. A TTT diagram is obtained by holding austenitised specimens at constant temperatures, whereas a CCT diagram follows continuous cooling curves
    2. For the same steel, the transformation curves of the CCT diagram lie at longer times and lower temperatures than those of the TTT diagram
    3. For a plain eutectoid carbon steel, a bainite region appears in the CCT diagram between the pearlite and martensite regions
    4. Raising the carbon content of a steel raises its Ms temperature
    Show answer

    Answer: A — A TTT diagram is obtained by holding austenitised specimens at constant temperatures, whereas a CCT diagram follows continuous cooling curves; B — For the same steel, the transformation curves of the CCT diagram lie at longer times and lower temperatures than those of the TTT diagram

    The statements on how the two diagrams are obtained and on the displacement of CCT curves relative to TTT curves are standard. In a plain eutectoid carbon steel continuous cooling gives only pearlite or martensite; bainite appears on continuous cooling in alloy steels, so the claim of a bainite region is wrong. Ms falls, not rises, as carbon increases, so the claim about raising Ms is wrong.

  12. Which of the following statements about the heat treatment of a 0.45 wt% C plain-carbon steel are correct?

    1. Tempering is carried out just above A₁ so that the martensite changes back to austenite and then to pearlite
    2. Normalising heats the steel above A₃ and cools it in air, giving finer pearlite than full annealing
    3. Full annealing ends with a water quench, so that the residual stress locked into the martensite strengthens the steel
    4. Spheroidising holds the steel for a long time just below A₁, turning lamellar cementite into spheres and softening the steel
    Show answer

    Answer: B — Normalising heats the steel above A₃ and cools it in air, giving finer pearlite than full annealing; D — Spheroidising holds the steel for a long time just below A₁, turning lamellar cementite into spheres and softening the steel

    Normalising (air cool from above A₃) gives finer pearlite than furnace-cooled full annealing, and spheroidising (long hold just below A₁) converts lamellar cementite to spheres and softens the steel. Tempering is done below A₁, never above it, so no austenite forms. Full annealing ends with slow furnace cooling to a soft, stress-free structure; a quench would produce martensite, not annealed steel.

  13. Which of the following statements about precipitation hardening of an Al–Cu alloy are correct?

    1. Quenched-in vacancies slow the diffusion of copper and so delay ageing at room temperature
    2. The solution-treatment temperature must lie above the solvus and below the eutectic temperature
    3. An over-aged alloy is harder than a peak-aged alloy, because the equilibrium phase θ is the most effective obstacle to dislocations
    4. Coherent GP zones have a lower interfacial energy than incoherent θ, so they nucleate more easily and appear first
    Show answer

    Answer: B — The solution-treatment temperature must lie above the solvus and below the eutectic temperature; D — Coherent GP zones have a lower interfacial energy than incoherent θ, so they nucleate more easily and appear first

    Solutionising must dissolve all Cu (above the solvus) without incipient melting at the 548 °C eutectic. Coherent zones have low interfacial energy, hence a low nucleation barrier, so they form first. Quenched-in vacancies speed Cu diffusion, which is why ageing occurs at room temperature at all, so slowing is wrong. Over-aged alloys are softer than peak-aged ones, because the coarse incoherent particles are bypassed rather than cut.

  14. Take the eutectoid composition as 0.76 wt% C and the maximum solubility of carbon in ferrite as 0.022 wt% C. A slowly cooled plain-carbon steel with 0.40 wt% C is examined just below the eutectoid temperature. The mass percentage of proeutectoid ferrite is ____ (answer to one decimal place).

    Numerical answer — type the value.

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    Answer: 48.8

    Lever rule with the tie line from 0.022 to 0.76: W(proeutectoid α) = (0.76 − 0.40)/(0.76 − 0.022) = 0.36/0.738 = 0.4878, which is 48.8 %. The remaining 51.2 % is pearlite. Using the 6.67 end of the tie line would instead give total ferrite, 94.3 %, which is a different quantity.

  15. A slowly cooled plain-carbon steel is found to contain 70 % pearlite and 30 % proeutectoid ferrite by mass. Take the eutectoid composition as 0.76 wt% C and the maximum solubility of carbon in ferrite as 0.022 wt% C. The carbon content of the steel is ____ wt% (answer to two decimal places).

    Numerical answer — type the value.

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    Answer: 0.54

    Pearlite fraction W_p = (C₀ − 0.022)/(0.76 − 0.022) = 0.70, so C₀ = 0.022 + 0.70 × 0.738 = 0.022 + 0.5166 = 0.5386, which is 0.54 wt%. Check: the proeutectoid ferrite fraction (0.76 − 0.5386)/0.738 = 0.30, as given. A steel with 70 % pearlite is therefore a medium-carbon steel of about 0.5 wt% C.

  16. A simplified TTT diagram of a eutectoid steel has the pearlite-start nose at 547 °C and 0.90 s, with time measured from the start of cooling at 727 °C. Treat the cooling curve as a straight line on a linear temperature–time plot, and use the TTT diagram directly without any CCT correction. The minimum constant cooling rate that just avoids the formation of pearlite is ____ °C/s (answer as a whole number).

    Numerical answer — type the value.

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    Answer: 200

    The cooling line from 727 °C must pass through the nose point (547 °C, 0.90 s). The temperature drop is 727 − 547 = 180 °C in 0.90 s, so the rate is 180/0.90 = 200 °C/s. Any slower rate crosses the pearlite-start curve to the right of the nose and forms pearlite. A real CCT diagram, with its curves moved to longer times, would permit a somewhat slower rate.

  17. An Al–Cu alloy reaches peak hardness after 48 h of ageing at 150 °C. Assume the time to peak hardness varies as t ∝ exp(Q/RT), with Q = 100 kJ/mol, R = 8.314 J mol⁻¹ K⁻¹ and T in kelvin (0 °C = 273.15 K). The time to peak hardness at 180 °C is ____ h (answer to one decimal place).

    Numerical answer — type the value.

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    Answer: 7.3

    t₂/t₁ = exp[(Q/R)(1/T₂ − 1/T₁)], with T₁ = 423.15 K and T₂ = 453.15 K. Q/R = 100000/8.314 = 12028 K. 1/T₂ − 1/T₁ = 0.0022067 − 0.0023632 = −1.5646 × 10⁻⁴ K⁻¹. The exponent is −1.882, so t₂/t₁ = 0.1523 and t₂ = 48 × 0.1523 = 7.3 h. Using Celsius temperatures directly in the exponent would give a wrong answer.

  18. Assume that the interlamellar spacing of pearlite is inversely proportional to the undercooling below the eutectoid temperature A₁ = 727 °C. Pearlite formed isothermally at 690 °C has spacing S₁ and pearlite formed isothermally at 650 °C has spacing S₂. The ratio S₁/S₂ is ____ (answer to one decimal place).

    Numerical answer — type the value.

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    Answer: 2.1

    Undercooling at 690 °C is 727 − 690 = 37 °C, and at 650 °C it is 727 − 650 = 77 °C. With S ∝ 1/ΔT, S₁/S₂ = ΔT₂/ΔT₁ = 77/37 = 2.08, which is 2.1. The pearlite formed at the lower temperature is about twice as fine, which is why it is harder.