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Materials Science · Chapter 9 of 10 · Advanced

Phase Transformations and Heat Treatment

The phase diagram says what should form at equilibrium. Cooling rate decides what actually forms. Control the rate, and the same steel can be soft and ductile or hard and strong.

Readiness check

This chapter adds time to the phase diagram and uses exponentials. Tick only what you can do closed-notes.

Read the iron-carbon eutectoid (0.76% C, 727 °C).
Recall ferrite, cementite, and pearlite.
Evaluate an exponential and a logarithm.
Read an Arrhenius (rate versus 1/T) relation.
Connect microstructure to hardness.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview the iron-carbon diagram in Chapter 8.

3 or more weak itemsRevisit diffusion in Chapter 4 first.

The core idea

Transformations take time, so cooling rate, not just temperature, decides the microstructure. Slow cooling gives soft pearlite; a fast quench traps hard martensite.

y = 1 − exp(−ktⁿ)rate = A exp(−Q/RT)austenite → pearlite, bainite, or martensite

Phase changes proceed by nucleation and growth, both needing time and atomic movement. The TTT diagram maps how long a transformation takes at each temperature, giving a C-shaped curve. Cool slowly through it and you get pearlite; cool fast enough to miss the nose and the austenite transforms without diffusion into martensite, hard but brittle, which tempering then toughens. Precipitation hardening uses the same kinetics to grow strengthening particles.

The skill works when: you treat microstructure as a function of cooling path, and match the heat treatment to the property wanted.

The skill breaks down when: the equilibrium diagram alone is used to predict a quenched structure, or kinetics (time) is ignored.

The concept. A TTT diagram. A slow cooling path crosses the C-curve high and forms pearlite; a fast quench passes to the left of the nose, misses it, and forms martensite below the M_s line.

The skills, taught in order

Heat treatment is kinetics put to work. Five skills run from transformation rate through the TTT diagram to martensite, tempering, and precipitation hardening.

9.1 Transformation kinetics

A new phase forms by nucleation then growth, giving an S-shaped fraction-versus-time curve described by the Avrami equation y = 1 − exp(−ktⁿ). The rate is fastest at intermediate temperatures: too hot and the driving force is small, too cold and diffusion is sluggish. This compromise is what creates the C-curve.

9.2 Isothermal transformation (TTT) diagrams

A TTT (time-temperature-transformation) diagram plots, for a fixed temperature, when transformation starts and finishes. Holding at high temperature gives coarse pearlite; lower gives fine pearlite then bainite. The nose marks the shortest time to begin transforming.

9.3 Continuous cooling and martensite

Real parts cool continuously, described by CCT diagrams. If cooling is fast enough to miss the nose, carbon has no time to diffuse and austenite shears into martensite, a hard, brittle, body-centred-tetragonal phase, below the martensite-start temperature M_s. Martensite formation is diffusionless and nearly instantaneous.

Product	Cooling	Hardness
Coarse pearlite	slow (furnace)	low
Fine pearlite	moderate (air)	medium
Bainite	faster, intermediate T	high
Martensite	rapid quench	very high, brittle

9.4 Tempering and heat-treatment cycles

As-quenched martensite is too brittle to use, so it is tempered (reheated below the eutectoid) to precipitate fine carbides, trading some hardness for much-needed toughness.

Treatment	Process	Result
Full anneal	slow furnace cool	soft, ductile (coarse pearlite)
Normalize	air cool	finer, stronger pearlite
Quench and temper	quench to martensite, reheat	strong and tough

9.5 Precipitation (age) hardening

For alloys like aluminium-copper, a three-step cycle (solution treat, quench to a supersaturated solid solution, then age) grows nanoscale precipitates that pin dislocations. Hardness rises to a peak then falls if overaged, so aging time and temperature must be tuned, faster at higher temperature but with a sharper peak.

Engineering connection: quench-and-temper makes tool and structural steels; precipitation hardening makes aerospace aluminium and nickel superalloys; both turn the same alloy into many materials.

Worked example 1: transformation kinetics

A transformation follows the Avrami equation with exponent n = 2.5 and reaches 50% completion in 100 s at a fixed temperature. Find the rate constant k, then the fraction transformed after 200 s.

Figure 1. The S-shaped Avrami curve. Once past 50%, transformation accelerates, so doubling the time from 100 to 200 s takes the fraction from one half to almost complete.

ProblemFind k and the fraction transformed at 200 s for the kinetics in Figure 1.
Given / findn = 2.5, y = 0.50 at t = 100 s. Find k and y(200 s).
AssumptionsIsothermal transformation following the Avrami equation.
ModelUse the 50% point to solve for k, then substitute t = 200 s.
Equationsy = 1 − exp(−ktⁿ) k = ln 2 / t₅₀ⁿ
SolveAt 50%, ktⁿ = ln 2, so k = 0.693/100^2.5 = 0.693/10⁵ = 6.93×10⁻⁶. At 200 s, ktⁿ = 6.93×10⁻⁶ × 200^2.5 = 6.93×10⁻⁶ × 5.66×10⁵ = 3.92, so y = 1 − e^−3.92 = 0.98 (98%).
CheckThe fraction stayed between 0 and 1 and rose with time, as it must. The steep middle of the S-curve explains why so much transformation happens in the second 100 s.
ConclusionThe Avrami equation quantifies how fast a structure forms at a given temperature. Repeating it at several temperatures traces out the TTT C-curve, the link to the cooling-path reasoning of this chapter.

Result. k = 6.93×10⁻⁶; 98% transformed after 200 s.

Worked example 2: aging time and temperature

An aluminium alloy reaches peak hardness after 10 hours of aging at 160 °C. The aging process has an activation energy of about 120 kJ/mol. Estimate the time to peak hardness if aged at 190 °C instead.

Figure 2. Precipitation hardening curves at two temperatures. Aging hotter reaches peak hardness much sooner, but the peak is lower and overaging sets in faster, so the window is narrower.

ProblemEstimate the time to peak hardness at 190 °C for the alloy in Figure 2.
Given / findt = 10 h at T₁ = 160 °C = 433 K, Q = 120 kJ/mol, T₂ = 190 °C = 463 K. Find t at 190 °C.
AssumptionsThe same precipitation reaction, Arrhenius kinetics, peak occurs at a fixed fraction transformed.
ModelThe rate follows an Arrhenius law; time to a fixed state is inversely proportional to rate.
Equationsrate = A exp(−Q/RT) t₂/t₁ = exp[(Q/R)(1/T₂ − 1/T₁)]
SolveThe rate ratio is exp[−(Q/R)(1/T₂ − 1/T₁)] = exp[−(120 000/8.314)(1/463 − 1/433)] = exp(2.16) = 8.7. So the time falls by that factor: t = 10/8.7 = 1.15 h.
CheckHigher temperature speeds the reaction, so the time must drop, which it does (from 10 h to about 1.2 h). A 30 °C rise gives almost a ninefold speed-up, the steep temperature sensitivity typical of diffusion-controlled aging.
ConclusionAging hotter is faster but less forgiving: the peak is lower and overaging arrives quickly, so the heat-treater trades speed against the width of the safe time window. This is the same kinetics as Example 1, now in temperature.

Result. Peak hardness in about 1.2 h at 190 °C, versus 10 h at 160 °C.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Equilibrium predicts quenched structure	Martensite missed from the phase diagram	"Was there time to diffuse?"	Fast cooling traps non-equilibrium phases; use TTT/CCT, not just the diagram.
Using martensite as-quenched	Brittle part cracks in service	"Has it been tempered?"	Temper martensite to restore toughness before use.
Hotter aging is always better	Overaged, soft precipitation alloy	"Did I pass the hardness peak?"	Aging has an optimum; beyond the peak, hardness falls.
Ignoring the nose	Slow quench expected to harden	"Does the cooling path miss the C-curve nose?"	Martensite needs a cooling rate fast enough to clear the nose.

Practice ladder

Level 1 · Direct skill

For the Worked Example 1 kinetics (k = 6.93×10⁻⁶, n = 2.5), find the fraction transformed at 150 s.

Show answer

ktⁿ = 6.93×10⁻⁶ × 150^2.5 = 6.93×10⁻⁶ × 2.76×10⁵ = 1.91, so y = 1 − e^−1.91 = 0.85 (85%). The curve is already well past its steep middle.

Level 2 · Mixed concept

Three identical eutectoid-steel parts are furnace-cooled, air-cooled, and water-quenched. Rank the resulting hardness and name the microstructure of each.

Show answer

Furnace cool gives coarse pearlite (softest), air cool gives fine pearlite (medium), water quench gives martensite (hardest, brittle). Faster cooling moves the structure down the TTT toward martensite.

Level 3 · Independent problem

A steel must be hard at the surface but tough overall. Outline a heat-treatment route and say what each step does.

Show answer

Austenitize, then quench to form martensite (hard but brittle), then temper to precipitate fine carbides and restore toughness. Optionally case-harden (carburize, Chapter 4) first so only the surface reaches high carbon and maximum quenched hardness.

Level 4 · Transfer to real engineering

Find a heat-treated product (a knife blade, an aluminium aircraft part, a gear). Identify the transformation or aging step and explain the structure-property outcome it targets.

What good work looks like

The treatment named (quench-and-temper or precipitation aging), the microstructure it produces, and the property goal (hardness, toughness, strength) tied to the cooling or aging path.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used kinetics (time), not just the equilibrium diagram, for this quenched part."

"Give me five cooling paths; I will name the microstructure each produces."

"Design the heat treatment." Reasoning from the TTT and the property goal is the skill.

"What is the aging time?" Applying the Arrhenius ratio yourself is the point.

Portfolio task

Design a heat-treatment cycle for one real part: state the cooling path or aging schedule, predict the microstructure with kinetics, and justify the resulting properties.

Must include: a TTT/CCT-based path or an aging schedule, a kinetics calculation, and a microstructure-property link.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the Avrami equation.

y = 1 − exp(−ktⁿ), the S-shaped fraction transformed versus time.

2. Why is the TTT curve C-shaped?

Transformation is slow when too hot (small driving force) and too cold (slow diffusion), fastest in between, the nose.

3. How does martensite form, and what is it like?

By a diffusionless shear on rapid quenching; it is very hard but brittle.

4. Why temper martensite?

To precipitate fine carbides that restore toughness, trading a little hardness.

5. What are the steps of precipitation hardening?

Solution treat, quench to a supersaturated solution, then age to grow strengthening precipitates.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the Avrami and aging calculations from a blank page.

+3 daysSketch a TTT and mark pearlite and martensite paths.

+7 daysConnect heat treatment to selection, Chapter 10.

+30 daysReview the structure-to-property chain across the whole course.

Textbook mapping

Item	Mapping
Primary source	Callister and Rethwisch, Materials Science and Engineering: An Introduction, Chapters 10 (Phase Transformations) and 11 (Processing of Metal Alloys)
Cross-reference	Askeland, Ch. 12 and 13 · Shackelford, Ch. 10
Core topics	9.1 Kinetics (Avrami) · 9.2 TTT diagrams · 9.3 Martensite · 9.4 Tempering · 9.5 Precipitation hardening
Engineering connection	Quench-and-temper steels and precipitation-hardened aerospace alloys.
Read next	Chapter 10: Material Classes and Selection.