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Phase diagrams and the iron-carbon diagram

Metallurgical and Materials Engineering (B.E./B.Tech) · Physical metallurgy · free study notes for exams, placements and interviews

In short

A phase diagram is a map of which phases are stable in an alloy at a given composition and temperature at equilibrium (usually 1 atm). The iron-iron carbide (Fe-Fe₃C) diagram is the single most important one in metallurgy because it explains why plain carbon steel exists in grades from mild steel to tool steel, and why cast irons behave differently. It is the backbone of heat-treatment practice at SAIL, Tata Steel, JSW, Jindal Stainless, foundries and forging shops, and is used daily by QA and process metallurgists. Phase diagrams carry heavy weight in GATE MT (lever rule, phase rule, invariant reactions, microstructure fractions) and in university Physical Metallurgy papers, and almost every campus technical round asks you to draw and label the Fe-Fe₃C diagram.

Explained

Basic definitions and the phase rule

Types of binary phase diagrams and invariant reactions

Reading a diagram: tie line and lever rule

Allotropy of iron and the Fe-Fe₃C diagram

Microstructures and how they develop

Limits of the diagram and Indian practice

Key points

Formulas and examples

Gibbs phase rule: P + F = C + 2 (pressure variable); P + F = C + 1 (constant pressure). P = number of phases, F = degrees of freedom, C = number of components.
Lever rule in a two-phase (α + β) field with overall composition C₀, phase compositions C_α and C_β (all wt%):
W_α = (C_β − C₀) / (C_β − C_α) ; W_β = (C₀ − C_α) / (C_β − C_α) ; W_α + W_β = 1 (mass fractions, dimensionless)
Volume fraction from mass fraction: V_α = (W_α/ρ_α) / (W_α/ρ_α + W_β/ρ_β), ρ in g/cm³ (ρ_ferrite ≈ 7.87, ρ_cementite ≈ 7.66)
Pearlite constituents at 727 °C: W_α = (6.67 − 0.76)/(6.67 − 0.022) = 0.889 ; W_Fe3C = (0.76 − 0.022)/(6.67 − 0.022) = 0.111
Hypoeutectoid steel just below 727 °C (overall C₀ < 0.76):
W_pearlite = (C₀ − 0.022)/(0.76 − 0.022) ; W_proeutectoid α = (0.76 − C₀)/(0.76 − 0.022)
Hypereutectoid steel just below 727 °C (0.76 < C₀ < 2.11):
W_pearlite = (6.67 − C₀)/(6.67 − 0.76) ; W_proeutectoid Fe3C = (C₀ − 0.76)/(6.67 − 0.76)

Example 1 — Pearlite constituents. Find the mass fractions of ferrite and cementite in fully pearlitic 0.76% C steel at 726 °C.

Example 2 — 0.4% C steel (a typical IS C40 / forging steel), just below 727 °C.

Example 3 — 1.2% C hypereutectoid steel at 726 °C.

Example 4 — Austenite at 900 °C in 0.4% C steel, and phase rule.

Example 5 — Pb-Sn solder, 40% Sn at 150 °C. Take C_α(Pb-rich) = 11% Sn and C_β(Sn-rich) = 98% Sn.

MicroconstituentPhases presentCarbon, wt%Typical hardnessFormed by
Ferrite (α)BCC Fe≤ 0.02280-100 HBSlow cooling, below A₃
Austenite (γ)FCC Fe≤ 2.11soft at temperatureAbove A₃/A_cm
CementiteFe₃C6.67~800 HVEutectoid / proeutectoid
Pearliteα + Fe₃C lamellar0.76200-350 HBEutectoid reaction at 727 °C
Ledeburiteγ (→ pearlite) + Fe₃C4.3very hardEutectic reaction at 1148 °C
ReactionTypeTemperatureEquation
PeritecticL + S → S1495 °CL(0.53) + δ(0.09) → γ(0.17)
EutecticL → S + S1148 °CL(4.3) → γ(2.11) + Fe₃C
EutectoidS → S + S727 °Cγ(0.76) → α(0.022) + Fe₃C

Where it is used

Interview questions

Common mistakes

Related topics

Heat treatment of steels · Crystal structure and defects · Mechanical behaviour: deformation and fracture · Mechanical testing of materials · Diffusion in solids · Solidification and casting of metals

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