Iron-carbon diagram, steels and cast irons

The Fe-Fe3C diagram's phases, invariant reactions and critical lines, slow-cooled microstructures of steels, classes of steel and cast iron, and lever-rule calculations of pearlite and cementite fractions.

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Why it matters

Steels and cast irons make up most of the mass of machines, from motor shafts and gears to machine-tool beds and engine blocks. The iron–carbon diagram tells you what microstructure a given carbon content produces on slow cooling, which in turn predicts strength, ductility, weldability and machinability. It is also the starting map for every heat treatment in the next topic.

Key ideas

The diagram. The engineering version is the metastable Fe–Fe₃C diagram, plotted from 0 to 6.67 wt% C (the carbon content of cementite, Fe₃C). It shows equilibrium phases for slow cooling only; fast cooling (quenching) produces non-equilibrium structures such as martensite that do not appear on it.

Phases.

  • Ferrite (α): BCC iron with very little dissolved carbon, maximum 0.022 % at 727 °C and about 0.008 % at room temperature. Soft, ductile, magnetic below 768 °C.
  • Austenite (γ): FCC iron, can dissolve up to 2.14 % C at 1147 °C because FCC octahedral holes are larger. Soft, ductile, non-magnetic; stable only above 727 °C in plain carbon steel.
  • δ-ferrite: BCC, stable at high temperature (above 1394 °C for pure iron), maximum 0.09 % C.
  • Cementite (Fe₃C): iron carbide with 6.67 % C. Very hard and brittle.
  • Liquid (L).

Three invariant reactions (fix the points as given here; some books quote 0.77 % and 2.11 % from slightly different data, and the method is unchanged):

  • Peritectic at 1495 °C, 0.17 % C: L + δ → γ.
  • 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 eutectoid product is pearlite, a lamellar mixture of ferrite and cementite. Pearlite is a micro-constituent, not a phase.

Critical lines. A₁ is the eutectoid line at 727 °C. A₃ is the boundary between γ and α + γ for hypoeutectoid steels; it falls from 912 °C at 0 % C to 727 °C at 0.76 % C. A_cm is the boundary between γ and γ + Fe₃C for hypereutectoid steels; it rises from 727 °C at 0.76 % C to 1147 °C at 2.14 % C. Heat-treatment temperatures are always quoted relative to these lines.

Slow-cooled steels at room temperature.

  • Hypoeutectoid (C < 0.76 %): proeutectoid ferrite forms between A₃ and A₁, then the remaining austenite (now 0.76 % C) becomes pearlite. Structure: ferrite + pearlite.
  • Eutectoid (0.76 %): 100 % pearlite.
  • Hypereutectoid (0.76–2.14 %): proeutectoid cementite forms along austenite grain boundaries below A_cm, then pearlite. Structure: pearlite + grain-boundary cementite network (brittle unless spheroidised).

As carbon rises from 0 to about 0.8 %, hardness and UTS rise and ductility, toughness and weldability fall. Common classes: low-carbon or mild steel (below 0.3 % C; sheet, structurals, weldable), medium-carbon (0.3–0.6 %; shafts, gears, axles, heat-treatable), high-carbon (0.6–1.4 %; springs, rails, cutting tools). Indian designations such as C45 or 45C8 indicate about 0.45 % C; take exact limits from IS 1570 or your data book. Alloying elements (Cr, Ni, Mo, Mn, V) raise hardenability, strength at temperature and corrosion resistance.

Cast irons (above 2.14 % C, usually 2.5–4 % C with 1–3 % Si). Silicon promotes graphite instead of cementite, and the form of carbon decides the properties:

  • Grey iron: graphite flakes. Excellent castability, damping, machinability and compressive strength, low tensile strength and nearly zero ductility (flake tips act as cracks). Machine-tool beds, engine blocks, brake drums.
  • White iron: carbon held as cementite (fast cooling, low Si). Very hard, wear resistant, brittle, nearly unmachinable. Grinding balls, and the starting material for malleable iron.
  • Malleable iron: white iron annealed for many hours so cementite decomposes into irregular graphite rosettes (temper carbon). Moderate ductility.
  • Ductile (spheroidal-graphite, SG or nodular) iron: Mg or Ce added to the melt makes graphite form as spheres directly on solidification. Highest strength and ductility of the cast irons. Crankshafts, gears, pipes.
  • Compacted-graphite iron: intermediate vermicular graphite.

Formulas

W_α = (C_Fe₃C − C₀) / (C_Fe₃C − C_α), W_Fe₃C = (C₀ − C_α) / (C_Fe₃C − C_α) (lever rule)

  • W: mass fraction of a phase (dimensionless); C₀: overall alloy carbon (wt%); C_α, C_Fe₃C: carbon in the two phases at the tie line ends (wt%). Applies only within a two-phase field at equilibrium.

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

W_pearlite = (6.67 − C₀) / (6.67 − 0.76) (hypereutectoid, just below 727 °C)

CE = C + (Si + P) / 3

  • CE: carbon equivalent of a cast iron (wt%); C, Si, P in wt%. CE ≈ 4.3 means eutectic behaviour; lower CE means hypoeutectic. For weldability of steels a different CE (IIW: C + Mn/6 + (Cr+Mo+V)/5 + (Ni+Cu)/15) is used.

Worked examples

Example 1 (standard): a 0.40 % C steel. Find the fractions of proeutectoid ferrite and pearlite just below 727 °C, and total ferrite and cementite at that temperature.

  1. Lever rule across α (0.022 %) and pearlite (0.76 %): W_pearlite = (C₀ − 0.022)/(0.76 − 0.022) = 0.378/0.738 = 0.512.
  2. W_proeutectoid ferrite = 1 − 0.512 = 0.488.
  3. Total phases, tie line α (0.022 %) to Fe₃C (6.67 %): W_Fe₃C = (0.40 − 0.022)/(6.67 − 0.022) = 0.378/6.648 = 0.057, so W_α total = 0.943.

So about half the structure is pearlite, but less than 6 % of the steel is cementite, which is why medium-carbon steels remain machinable.

Example 2 (GATE level): hypereutectoid steel. A 1.2 % C steel is slowly cooled. Find (a) the proeutectoid cementite fraction, (b) the pearlite fraction, (c) total cementite just below 727 °C.

  1. Proeutectoid Fe₃C forms between 0.76 % (pearlite) and 6.67 % (cementite): W = (1.2 − 0.76)/(6.67 − 0.76) = 0.44/5.91 = 0.0745.
  2. W_pearlite = (6.67 − 1.2)/5.91 = 5.47/5.91 = 0.926.
  3. Total Fe₃C with the α–Fe₃C tie line: (1.2 − 0.022)/(6.67 − 0.022) = 1.178/6.648 = 0.177.

Check: of the total 17.7 % cementite, 7.4 % is proeutectoid and 10.3 % lies inside pearlite (0.926 × 0.111 = 0.103, since pearlite itself is 11.1 % Fe₃C).

Example 3 (reverse problem). A slow-cooled plain carbon steel shows 60 % pearlite. Then C₀ = 0.022 + 0.60 × 0.738 = 0.46 % C (a C45-type steel).

Common mistakes

  • Calling pearlite a phase. It is a two-phase mixture of α and Fe₃C.
  • Confusing the eutectic (liquid → two solids, 1147 °C, 4.3 %) with the eutectoid (solid → two solids, 727 °C, 0.76 %).
  • Using the wrong tie line: proeutectoid fractions use pearlite (0.76 %) as one end; total phase fractions use α (0.022 %) and Fe₃C (6.67 %).
  • Putting martensite on the equilibrium diagram. It appears only on TTT/CCT diagrams.
  • Saying "cast iron is just brittle". Ductile and malleable irons have useful elongation; grey iron is brittle because of flake geometry, not carbon content alone.
  • Writing the steel limit as 2 % in a numerical; use the value your diagram gives (2.14 %).

For GATE ME

Expect conceptual questions on the three invariant reactions (temperature, composition, reaction type), which phases are FCC or BCC, and which cast iron has nodular, flake or rosette graphite. Numericals apply the lever rule: fraction of pearlite, proeutectoid ferrite or cementite for a given carbon content, or the reverse (carbon content from a pearlite fraction). Practise drawing the steel portion of the diagram from memory with all key coordinates.

Quick check

  1. What reaction occurs at 727 °C and 0.76 % C?
  2. What is the maximum solubility of carbon in austenite?
  3. Find the pearlite fraction in a 0.2 % C steel just below 727 °C.
  4. Which cast iron contains graphite spheroids formed directly on solidification?
  5. Why does grey cast iron have poor tensile strength but good damping?

Answers: 1. Eutectoid, γ → α + Fe₃C (pearlite). 2. 2.14 % at 1147 °C. 3. (0.2 − 0.022)/0.738 = 0.24. 4. Ductile (SG) iron. 5. Graphite flakes act as internal notches in tension but absorb vibration energy.

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 materials engineering?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 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 and suitability for different applications.

  2. 2.Explain the difference between steel and cast iron in terms of their carbon content and properties.Concept

    Steels contain less than about 2.1 % carbon (2.14 % on the usual Fe-Fe3C diagram) and can be fully austenitised, so they can be hot worked and heat treated; low and medium-carbon steels are ductile, tough and weldable. Cast irons contain more than 2.14 % carbon, typically 2.5 to 4 % with silicon, and solidify with a eutectic, which gives a low melting point and excellent castability. Most of their carbon is present as graphite or cementite, so they are strong in compression, damp vibration and machine well (grey iron) but have low ductility, except for ductile and malleable irons.

  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 consists of alternating layers of ferrite and cementite, which contribute to its mechanical properties. The presence of pearlite in steel can be controlled through heat treatment processes to achieve desired properties for specific applications.

  5. 5.What happens to the microstructure of steel when it is rapidly cooled from the austenite phase?Application

    When steel is rapidly cooled from the austenite phase, it undergoes a transformation to martensite, a hard and brittle microstructure. This process, known as quenching, increases the hardness and strength of the steel but reduces its ductility. Tempering is often performed afterward to improve toughness.

  6. 6.Why is grey cast iron preferred over steel for making engine blocks?Application

    An engine block is a complex hollow shape, and grey iron's low melting point and high fluidity let it be cast with thin walls and internal passages at low cost. Its graphite flakes give high damping capacity, good thermal conductivity, self-lubrication on cylinder bores and excellent machinability, and its compressive strength suits the loads involved. The flakes make it weak and brittle in tension, which is acceptable here because the block is stiffness- and compression-dominated; aluminium alloys are used where weight matters more.

  7. 7.Calculate the weight percent of carbon in a steel alloy that contains 0.5 kg of carbon and 99.5 kg of iron.Numerical

    Weight percent of carbon = (mass of carbon / total mass of alloy) × 100 = (0.5 kg / (0.5 kg + 99.5 kg)) × 100 = 0.5%

  8. 8.If a plain carbon steel is cooled slowly from 900°C to room temperature, what microstructures would you expect to find?Application

    It depends on carbon content relative to the eutectoid 0.76 % C. A hypoeutectoid steel gives proeutectoid ferrite plus pearlite, with more pearlite as carbon rises; a eutectoid steel gives 100 % pearlite; a hypereutectoid steel gives pearlite with a network of proeutectoid cementite on the former austenite grain boundaries. This assumes the steel was fully austenitic at 900 °C, which is true for carbon contents up to roughly 1 %; above that A_cm exceeds 900 °C and some cementite was already present.

  9. 9.What is the significance of the critical temperatures in the heat treatment of steel?Concept

    A1 (727 °C) is the eutectoid temperature, below which austenite cannot exist in plain carbon steel at equilibrium. A3 is the upper critical line for hypoeutectoid steels, above which the steel is fully austenitic, and Acm plays the same role for hypereutectoid steels. Annealing, normalising and hardening temperatures are specified relative to these lines, typically 30 to 50 °C above A3 for hypoeutectoid steels and above A1 for hypereutectoid steels, so the right amount of austenite forms before cooling.

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