Iron-carbon diagram, steels and cast irons

The Fe–Fe₃C diagram: phases, the three invariant reactions, critical lines, steels and cast irons, and lever-rule phase-fraction calculations.

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Why it matters

Crankshafts, gears, springs, brake drums and engine blocks are all iron–carbon alloys, and the Fe–Fe₃C diagram tells you what microstructure each one will have after slow cooling. It is the map you read before choosing a steel grade, specifying a heat treatment or deciding between grey and ductile iron. Phase-fraction calculations with the lever rule are among the most reliable scoring questions in this subject.

Key ideas

What the diagram is. The iron–iron carbide (Fe–Fe₃C) diagram plots temperature against carbon content (wt %) from pure iron to 6.67 % C (cementite, Fe₃C). It is a metastable diagram: cementite is not truly stable, but it forms in steels under normal cooling. The diagram is valid only for slow (near-equilibrium) cooling; fast cooling is described by TTT/CCT diagrams in the heat-treatment topic.

Phases.

  • Ferrite (α) – BCC, maximum 0.022 % C at 727 °C (about 0.008 % at room temperature); soft, ductile, magnetic below 768 °C.
  • Austenite (γ) – FCC, maximum 2.14 % C at 1147 °C; non-magnetic, the starting phase for all hardening treatments.
  • δ-ferrite – BCC, stable at very high temperature (above 1394 °C for pure iron).
  • Cementite (Fe₃C) – intermetallic compound with 6.67 % C; very hard and brittle. Pearlite and ledeburite are microconstituents (mixtures of phases), not phases.

Three invariant reactions.

  • Peritectic at 1493 °C (some books 1495 °C): δ (0.09 % C) + L (0.53 % C) → γ (0.17 % C).
  • Eutectic at 1147 °C, 4.3 % C: L → γ (2.14 % C) + Fe₃C. The eutectic mixture is ledeburite.
  • Eutectoid at 727 °C, 0.76 % C (some books 0.77 % or 0.8 %): γ → α (0.022 % C) + Fe₃C. The lamellar product is pearlite. Use the values given in the question; if none are given, use 0.022 / 0.76 / 2.14 / 4.3 / 6.67 % C.

Critical lines. A₁ is the eutectoid line (727 °C). A₃ is the γ/(α+γ) boundary for hypoeutectoid steels – it falls from 912 °C at 0 % C to 727 °C at 0.76 % C. A_cm is the γ/(γ+Fe₃C) boundary for hypereutectoid steels – it rises from 727 °C at 0.76 % C to 1147 °C at 2.14 % C. A₂ (768 °C) is the Curie temperature of ferrite.

Steels (≤ 2.14 % C).

  • Hypoeutectoid (< 0.76 % C): on slow cooling, proeutectoid ferrite forms between A₃ and A₁, the remaining austenite (now 0.76 % C) becomes pearlite. Room-temperature structure: ferrite + pearlite.
  • Eutectoid (0.76 % C): 100 % pearlite.
  • Hypereutectoid (0.76–2.14 % C): proeutectoid cementite forms along austenite grain boundaries between A_cm and A₁, then pearlite. Structure: pearlite + grain-boundary cementite network. As carbon rises, hardness and strength rise and ductility and weldability fall. Low-carbon steels (below about 0.25 % C) are used for body panels and are welded easily; medium-carbon (about 0.25–0.6 %) for shafts, axles and crankshafts; high-carbon (above about 0.6 %) for springs and cutting tools. Weldability is judged by a carbon equivalent – the IIW form is used most often.

Cast irons (2.14–6.67 % C, usually 2.5–4 %). They melt at lower temperature and cast well. What matters is the form of carbon:

  • White cast iron – carbon as cementite (fast cooling, low Si); hard, brittle, wear parts.
  • Grey cast iron – graphite flakes (Si promotes graphite); good damping, machinability and compressive strength, low tensile ductility. Engine blocks, brake drums, machine beds.
  • Malleable iron – white iron annealed so carbon forms compact temper-carbon rosettes; some ductility.
  • Ductile (spheroidal-graphite, SG) iron – Mg or Ce added to make graphite nodules; tensile behaviour approaches steel. Crankshafts, suspension parts.
  • Compacted-graphite iron – vermicular graphite; between grey and ductile iron (used in diesel blocks). For cast irons a different carbon equivalent, CE = C + (Si + P)/3, tells whether the iron is hypo- or hypereutectic (eutectic at CE ≈ 4.3).

Alloying elements. Ni and Mn stabilise austenite (lower A₁ and A₃); Cr, Mo, W, V, Si stabilise ferrite and many form carbides. Almost all alloying elements shift the eutectoid composition to lower carbon and raise hardenability.

Formulas

W_α = (C_β − C₀) / (C_β − C_α) W_β = (C₀ − C_α) / (C_β − C_α) (lever rule) W = mass fraction of each phase (dimensionless), C₀ = alloy carbon content (wt %), C_α and C_β = compositions at the ends of the tie line (wt %). Applies only inside a two-phase field at one temperature.

W_pearlite = (C₀ − 0.022) / (0.76 − 0.022) W_α,pro = (0.76 − C₀) / (0.76 − 0.022) (hypoeutectoid, just below 727 °C)

W_pearlite = (6.67 − C₀) / (6.67 − 0.76) W_Fe₃C,pro = (C₀ − 0.76) / (6.67 − 0.76) (hypereutectoid)

W_Fe₃C,total = (C₀ − 0.022) / (6.67 − 0.022) W_α,total = 1 − W_Fe₃C,total Total phase fractions, using the α + Fe₃C tie line below 727 °C.

CE (IIW) = C + Mn/6 + (Cr + Mo + V)/5 + (Ni + Cu)/15 All in wt %. Used for weldability of steels; higher CE means greater risk of hard, crack-prone heat-affected zones.

CE (cast iron) = C + (Si + P)/3 wt %; compare with 4.3 to classify a cast iron as hypo- or hypereutectic.

Worked examples

Example 1 (standard) – 0.45 % C steel, slowly cooled.

  1. Hypoeutectoid, so the microconstituents are proeutectoid ferrite + pearlite.
  2. W_pearlite = (C₀ − 0.022)/(0.76 − 0.022) = 0.428/0.738 = 0.580.
  3. W_α,pro = 1 − 0.580 = 0.420.
  4. Total phases: W_Fe₃C = (0.45 − 0.022)/(6.67 − 0.022) = 0.428/6.648 = 0.0644; W_α,total = 0.9356. Answer: 58.0 % pearlite and 42.0 % proeutectoid ferrite; as phases, 93.6 % ferrite and 6.4 % cementite.

Example 2 (GATE level) – 1.2 % C steel.

  1. Hypereutectoid: proeutectoid cementite + pearlite.
  2. W_pearlite = (6.67 − 1.2)/(6.67 − 0.76) = 5.47/5.91 = 0.9255.
  3. W_Fe₃C,pro = 1 − 0.9255 = 0.0745.
  4. Total cementite = (1.2 − 0.022)/6.648 = 0.1772. Answer: 92.6 % pearlite, 7.4 % proeutectoid cementite, 17.7 % total cementite. Check: the cementite inside pearlite is 0.9255 × (0.76 − 0.022)/6.648 = 0.1027, and 0.1027 + 0.0745 = 0.1772 ✓.

Example 3 – reverse problem. A slowly cooled plain-carbon steel shows 60 % pearlite and 40 % proeutectoid ferrite. Because the lever rule is linear, C₀ = 0.6 × 0.76 + 0.4 × 0.022 = 0.465 % C.

Common mistakes

  • Swapping the lever arms: the fraction of a phase is the length of the opposite arm divided by the whole tie line.
  • Calling pearlite or ledeburite a "phase". The phases below 727 °C are only ferrite and cementite.
  • Using the hypoeutectoid pearlite formula for a hypereutectoid steel (or vice versa).
  • Confusing eutectic (liquid → two solids, 1147 °C) with eutectoid (solid → two solids, 727 °C).
  • Assuming the equilibrium diagram predicts martensite or bainite – it cannot; those need cooling-rate diagrams.
  • Mixing the two different "carbon equivalent" formulas (weldability of steel vs eutectic position of cast iron).

For GATE ME

Expect identification questions on invariant reactions (temperature, composition, reactants and products), phases versus microconstituents, critical lines, and the structure of each cast iron. Numericals are almost always lever-rule fractions: proeutectoid phase, pearlite, total ferrite or total cementite for a given carbon content, or the reverse (carbon content from a measured fraction). Practise drawing the steel corner of the diagram from memory with all key numbers.

Quick check

  1. What reaction occurs at 727 °C and 0.76 % C?
  2. Which phase has the FCC structure, ferrite or austenite?
  3. What fraction of a slowly cooled 0.3 % C steel is pearlite?
  4. What makes the graphite spheroidal in ductile iron?
  5. Is pearlite a phase?

Answers: 1. Eutectoid, γ → α + Fe₃C (pearlite); 2. Austenite; 3. (0.3 − 0.022)/0.738 ≈ 37.7 %; 4. Small additions of Mg (or Ce); 5. No, it is a two-phase microconstituent of ferrite and cementite.

Interview questions

All Engineering Materials and Manufacturing Processes interview questions

Try answering each one aloud before you open it.

  1. 1.What is the iron-carbon phase diagram, and why is it important in metallurgy?Concept

    The iron-carbon phase diagram is a graphical representation of the phases present in iron-carbon alloys as a function of temperature and carbon content. It is important in metallurgy because it helps engineers understand the transformations that occur in steel and cast iron during heating and cooling, which in turn affects their mechanical properties.

  2. 2.Explain the difference between steel and cast iron based on the iron-carbon diagram.Concept

    Steels contain up to 2.14 % C, the maximum solubility of carbon in austenite, so they can be fully austenitised, hot worked and heat treated, and they solidify without the eutectic reaction. Cast irons contain more than 2.14 % C (usually 2.5–4 %), so the last liquid undergoes the eutectic reaction at 1147 °C; they melt at lower temperature and cast well but cannot be forged. In cast irons the excess carbon appears as cementite (white iron) or graphite (grey, malleable, ductile iron), and that carbon form governs their properties.

  3. 3.What is the eutectoid point in the iron-carbon diagram, and what transformation occurs at this point?Concept

    The eutectoid point in the iron-carbon diagram occurs at approximately 0.76% carbon and 727°C. At this point, austenite transforms into pearlite, a lamellar mixture of ferrite and cementite, during slow cooling. This transformation is crucial for the development of mechanical properties in steel.

  4. 4.Why is pearlite important in the context of steel microstructures?Application

    Pearlite is important because it provides a balance of strength and ductility in steel. It is a two-phase microstructure consisting of alternating layers of ferrite and cementite, which contributes to the toughness and wear resistance of steel.

  5. 5.What happens to the mechanical properties of steel if it is cooled rapidly from the austenite phase?Application

    If steel is cooled rapidly from the austenite phase, it can form martensite, a hard and brittle microstructure. This rapid cooling process is known as quenching. Martensite increases the hardness and strength of steel but reduces its ductility.

  6. 6.Why is cast iron often used in engine blocks and machinery bases?Application

    Cast iron is used in engine blocks and machinery bases because of its excellent castability, good wear resistance, and vibration damping properties. Its high carbon content makes it easy to cast into complex shapes, and its microstructure provides the necessary strength and durability for these applications.

  7. 7.What is the significance of the critical temperature lines (A1, A3, Acm) in the iron-carbon diagram?Concept

    The critical temperature lines in the iron-carbon diagram indicate the temperatures at which phase transformations occur. A1 is the eutectoid temperature, A3 is the temperature above which austenite is stable in hypoeutectoid steels, and Acm is the temperature above which austenite is stable in hypereutectoid steels. These lines are crucial for understanding heat treatment processes.

  8. 8.Calculate the amount of pearlite and proeutectoid ferrite in a 0.4% carbon steel at room temperature.Numerical

    Apply the lever rule just below 727 °C between ferrite (0.022 % C) and eutectoid austenite (0.76 % C). Pearlite fraction = (0.4 − 0.022)/(0.76 − 0.022) = 0.378/0.738 ≈ 0.51; proeutectoid ferrite = (0.76 − 0.4)/0.738 ≈ 0.49. So a slowly cooled 0.4 % C steel is about 51 % pearlite and 49 % proeutectoid ferrite.

  9. 9.What is the effect of adding alloying elements like chromium and nickel to steel?Application

    Adding alloying elements like chromium and nickel to steel can enhance its properties. Chromium increases hardness, corrosion resistance, and wear resistance, while nickel improves toughness and strength. These elements can also stabilize certain phases, affecting the steel's microstructure and performance.

  10. 10.Determine the carbon content of a steel that consists of 60% pearlite and 40% proeutectoid ferrite at room temperature.Numerical

    Because the lever rule is linear, the carbon content is the mass-weighted average of the two constituents' compositions: C = 0.60 × 0.76 + 0.40 × 0.022 = 0.456 + 0.0088 ≈ 0.465 % C. Check: (0.465 − 0.022)/(0.76 − 0.022) = 0.60 pearlite.

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