Iron-carbon equilibrium diagram

Why it matters

Steel and cast iron are the most used engineering metals, and almost everything about them — which can be hardened, how they are forged, why cast iron is brittle, what temperature to anneal at — is read from the iron–carbon diagram. Its key compositions and temperatures (0.022, 0.76, 2.14, 4.3, 6.7 % C; 727 and 1147 °C) must be known by heart.

Key ideas

Metastable Fe–Fe₃C diagram. The diagram used in practice runs from pure iron to cementite (Fe₃C, 6.67 ≈ 6.7 wt % C). It is strictly metastable: given very long times or silicon, cementite decomposes to iron and graphite (the basis of grey cast iron), but in steels cementite persists, so the Fe–Fe₃C diagram is treated as the equilibrium map.

Phases.

• α-ferrite: BCC interstitial solid solution of C in iron. Maximum solubility only 0.022 wt % C at 727 °C (about 0.008 % at room temperature). Soft, ductile, magnetic below 768 °C (Curie temperature, A2).
• γ-austenite: FCC solid solution; dissolves up to 2.14 wt % C at 1147 °C. Soft, ductile, non-magnetic; stable only above 727 °C in plain carbon steels. All hot working and hardening start from austenite.
• δ-ferrite: BCC, stable above 1394 °C, up to 0.09 wt % C.
• Cementite (Fe₃C): orthorhombic intermetallic, 6.7 % C; very hard and brittle.
• Liquid.

Invariant reactions.

• Peritectic at 1493 °C (about 0.17 % C): L + δ → γ. Matters mainly in welding and casting of low-carbon steel.
• Eutectic at 1147 °C, 4.3 % C: L → γ + Fe₃C. The eutectic mixture is called ledeburite.
• Eutectoid at 727 °C, 0.76 % C: γ → α + Fe₃C. The product is pearlite, alternate lamellae of ferrite and cementite (about 89 % ferrite, 11 % cementite by weight).

Critical lines (on slow heating and cooling).

• A1: the eutectoid horizontal at 727 °C.
• A3: the γ/(α + γ) boundary for hypoeutectoid steels, falling from 912 °C at 0 % C to 727 °C at 0.76 % C.
• Acm: the γ/(γ + Fe₃C) boundary for hypereutectoid steels, rising from 727 °C at 0.76 % C to 1147 °C at 2.14 % C. In practice heating lines (Ac) sit slightly above and cooling lines (Ar) slightly below these equilibrium values.

Classification by carbon content.

• Steels: up to 2.14 % C (commercially below about 1.5 %).
  • Hypoeutectoid (< 0.76 % C): on slow cooling, proeutectoid ferrite forms between A3 and A1 (along prior austenite grain boundaries); the remaining austenite, now 0.76 % C, becomes pearlite at 727 °C. Microstructure: ferrite + pearlite.
  • Eutectoid (0.76 % C): 100 % pearlite.
  • Hypereutectoid (0.76–2.14 % C): proeutectoid cementite forms between Acm and A1, often as a brittle grain-boundary network; the rest becomes pearlite.
• Cast irons: 2.14–6.7 % C (commercially 2.5–4 %). White cast iron follows the metastable diagram (ledeburite, very hard); grey, ductile and malleable irons contain graphite.

Microconstituent vs phase. Pearlite and ledeburite are microconstituents (mixtures with a recognisable structure), not phases. At room temperature every slowly cooled plain carbon steel contains only two phases, ferrite and cementite; the lever rule gives either the phase fractions (tie line 0.022 to 6.7 % C) or the microconstituent fractions just below 727 °C (tie line 0.022 to 0.76 % C for hypoeutectoid; 0.76 to 6.7 % C for hypereutectoid).

Property trends. In slowly cooled steels, more carbon means more pearlite (or cementite), raising hardness and UTS up to about 0.8 % C and lowering ductility and weldability. Above about 0.8 % C the brittle cementite network limits tensile strength even though hardness keeps rising.

Effect of alloying. Ni and Mn expand the austenite field (austenite stabilisers); Cr, Mo, Si, W shrink it (ferrite stabilisers). Most alloying elements shift the eutectoid to lower carbon contents, which is why the plain Fe–C numbers are approximate for alloy steels.

Formulas

Lever rule on the 727 °C line (just below A1), with Cα = 0.022, Ce = 0.76, Ccm = 6.7 wt % C and C0 = steel composition:

W_pearlite = (C0 − 0.022) / (0.76 − 0.022) and W_α,pro = (0.76 − C0) / (0.76 − 0.022) Hypoeutectoid steels (0.022 < C0 < 0.76).

W_pearlite = (6.7 − C0) / (6.7 − 0.76) and W_Fe₃C,pro = (C0 − 0.76) / (6.7 − 0.76) Hypereutectoid steels (0.76 < C0 < 2.14).

W_α,total = (6.7 − C0) / (6.7 − 0.022) and W_Fe₃C,total = (C0 − 0.022) / (6.7 − 0.022) Total phase fractions below 727 °C for any steel.

All W are weight fractions; compositions in wt % C. Some textbooks use 0.8 % C for the eutectoid and 0.025 or 0.02 % C for ferrite — answers then differ slightly, so follow the data given in the question.

Worked examples

Example 1 (standard): a 0.4 % C steel, slowly cooled.

1. Proeutectoid ferrite = (0.76 − 0.4)/(0.76 − 0.022) = 0.36/0.738 = 0.488 (48.8 %).
2. Pearlite = (0.4 − 0.022)/0.738 = 0.512 (51.2 %).
3. Total ferrite = (6.7 − 0.4)/(6.7 − 0.022) = 6.3/6.678 = 0.943; total cementite = 0.057.
4. Ferrite inside pearlite = 0.943 − 0.488 = 0.455, i.e. 0.455/0.512 = 89 % of the pearlite, as expected.

Example 2 (GATE level): a 1.2 % C steel, and reverse problem. (a) Just below 727 °C:

1. Proeutectoid cementite = (1.2 − 0.76)/(6.7 − 0.76) = 0.44/5.94 = 0.074 (7.4 %).
2. Pearlite = (6.7 − 1.2)/5.94 = 0.926 (92.6 %).
3. Total cementite = (1.2 − 0.022)/6.678 = 0.176. (b) A hypoeutectoid steel shows 60 % pearlite. Find its carbon content.
4. 0.60 = (C0 − 0.022)/0.738.
5. C0 = 0.022 + 0.60 × 0.738 = 0.465 % C (about a C45 steel).

Common mistakes

• Calling pearlite a phase; it is a two-phase microconstituent.
• Using the 0–6.7 tie line when the question asks for pearlite, or the 0.022–0.76 tie line when it asks for total ferrite.
• Mixing up A3 (hypoeutectoid) and Acm (hypereutectoid) lines.
• Saying a 0.8 % C steel is 100 % pearlite with the 0.76 % eutectoid; it contains about 0.7 % proeutectoid cementite.
• Thinking austenite exists at room temperature in plain carbon steels after slow cooling.
• Quoting the eutectic as L → α + Fe₃C; the eutectic product is austenite + cementite.

For GATE PI

Expect one-mark questions on the eutectoid, eutectic and peritectic compositions and temperatures, the crystal structures and carbon solubilities of the phases, and the names of microconstituents. Two-mark questions are lever-rule numericals — proeutectoid ferrite or cementite, pearlite fraction, total phase fractions — and reverse problems that find carbon content from a microstructure. Memorise the key numbers and always state which tie line you are using.

Quick check

1. What is the maximum solubility of carbon in ferrite and in austenite?
2. Write the eutectoid reaction with its temperature and composition.
3. What fraction of a 0.2 % C steel is pearlite after slow cooling?
4. What is ledeburite?
5. Name the line above which a hypereutectoid steel is fully austenitic.

Answers: 1. 0.022 % (727 °C) and 2.14 % (1147 °C); 2. γ (0.76 % C) → α + Fe₃C at 727 °C; 3. About 24 %; 4. The eutectic mixture of austenite and cementite formed at 1147 °C, 4.3 % C; 5. Acm.