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Chapter 6 · 10 hours

Iron – Iron Carbide diagram and Heat Treatment of Steels

Practice questions

Practice questions and answers

7 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 10 marks

Draw the iron-iron carbide equilibrium diagram, label the phases, temperatures and compositions, and explain the invariant reactions. State the applications and limitations of the diagram.

Answer

The Fe-Fe3_3C diagram shows the phases of plain carbon steels and cast irons against carbon content (0 to 6.67 wt% C, where Fe3_3C cementite forms) and temperature, under slow cooling.

 T(C)   (schematic)
 1538 +   Liquid (L)
 1495 +-- delta / peritectic (0.17%C)
 1147 +--- L + g ---- eutectic (4.3%C) -- L+Fe3C
      |  Austenite (g)   g+Fe3C   Ledeburite
  912 +-- a+g
  727 +--- eutectoid (0.76%C) ------------------
      |  Ferrite+pearlite | Pearlite+Fe3C | Cast
      +-----|-------|---------|-------|-----> %C
          0.022   0.76      2.14    4.3    6.67

Phases

  • Ferrite (α\alpha): interstitial solid solution of C in BCC iron, max 0.022% C at 727 °C. Soft, ductile.
  • Austenite (γ\gamma): C in FCC iron, max 2.14% C at 1147 °C. Non-magnetic, ductile, stable above 727 °C.
  • δ\delta-ferrite: C in BCC iron, up to 0.09% C at 1495 °C.
  • Cementite (Fe3_3C): hard, brittle compound with 6.67% C.
  • Liquid (L).

Invariant reactions

ReactionTemp.C contentEquation
Peritectic1495 °C0.17%L+δ⇌γL + \delta \rightleftharpoons \gamma
Eutectic1147 °C4.3%L⇌γ+Fe3CL \rightleftharpoons \gamma + \text{Fe}_3\text{C} (ledeburite)
Eutectoid727 °C0.76%γ⇌α+Fe3C\gamma \rightleftharpoons \alpha + \text{Fe}_3\text{C} (pearlite)

Critical lines: A1_1 (727 °C), A3_3 (GOS line, ferrite starts to form), Acm_{cm} (SE line, cementite starts to form from austenite).

Classification by carbon

  • Steel: up to 2.14% C (hypoeutectoid below 0.76, hypereutectoid 0.76-2.14).
  • Cast iron: 2.14-6.67% C (hypoeutectic below 4.3, hypereutectic above).

Applications

  • Selecting austenitising temperatures for annealing, normalising, hardening.
  • Predicting microstructure and phase fractions with the lever rule.
  • Understanding properties of steels and cast irons, and welding behaviour.

Limitations

  • It is an equilibrium diagram (very slow cooling); real heat treatment produces non-equilibrium phases such as martensite and bainite, which do not appear on it.
  • Alloying elements (Cr, Ni, Mn) shift transformation temperatures and the eutectoid composition; the diagram is valid only for plain carbon steel.
  • It gives no information about the rate of transformation (TTT/CCT diagrams are needed).
  • Cementite is actually metastable; graphite is the stable phase (Fe-C diagram).
  • Practice · 6 marks

Write short notes on ferrite, austenite, cementite, pearlite and martensite.

Answer

ConstituentNatureStructureProperties
Ferrite (α\alpha)Interstitial solid solution of C in α\alpha-FeBCC, max 0.022% CSoft, ductile, magnetic below 768 °C, tensile strength about 280 MPa
Austenite (γ\gamma)Solid solution of C in γ\gamma-FeFCC, up to 2.14% CSoft, ductile, tough, non-magnetic; stable above 727 °C (retained in austenitic stainless steel)
Cementite (Fe3_3C)Intermetallic compound, 6.67% COrthorhombicVery hard (about 800 HV), brittle
PearliteEutectoid mixture of ferrite and cementite (0.76% C)Alternate lamellae of ferrite and cementiteHardness about 200 HB, tensile strength about 800 MPa; fine pearlite is harder than coarse
MartensiteSupersaturated solid solution of C in ironBody-centred tetragonal (BCT), needle/plate shapeExtremely hard (up to 65 HRC) and brittle; formed by rapid quenching without diffusion

Notes

  • Pearlite forms when austenite of 0.76% C cools slowly through 727 °C: γ→α+Fe3C\gamma \rightarrow \alpha + \text{Fe}_3\text{C}. Lamellar spacing decreases with faster cooling.
  • Martensite forms by a diffusionless shear transformation when austenite is cooled faster than the critical cooling rate, below the MsM_s temperature. Carbon is trapped, distorting the lattice into BCT, which gives hardness. It is tempered to improve toughness.
  • Other constituent: bainite (fine ferrite + carbide formed between about 250 and 550 °C), ledeburite (eutectic of austenite and cementite in cast iron).
  • Practice · 6 marks

Describe the microstructural changes during slow cooling of a 0.4 wt% carbon steel from 950 °C to room temperature. Using the lever rule with ferrite 0.022 wt% C, eutectoid 0.76 wt% C and cementite 6.67 wt% C, calculate the fractions of pro-eutectoid ferrite and pearlite, and also the total ferrite and cementite just below 727 °C.

Answer

Cooling path

  1. At 950 °C the structure is 100% austenite (γ\gamma) of 0.4% C.
  2. On cooling to the A3_3 line (about 790 °C), pro-eutectoid (primary) ferrite nucleates at austenite grain boundaries. The remaining austenite is enriched in carbon.
  3. Between A3_3 and 727 °C more ferrite forms; austenite composition follows the GOS line toward 0.76% C.
  4. Just above 727 °C the austenite has 0.76% C. At 727 °C it transforms by the eutectoid reaction γ→α+Fe3C\gamma \rightarrow \alpha + \text{Fe}_3\text{C} into pearlite.
  5. Below 727 °C: pro-eutectoid ferrite + pearlite, little change on cooling to room temperature.
 950C: austenite     ->  790C: ferrite at grain boundaries
 +-----------+            +-----------+
 | gamma     |            |f gamma  f |
 +-----------+            +-----------+
 727C and below: ferrite (white) + pearlite (dark)

Phase fractions just below 727 °C

Pro-eutectoid ferrite (using tie line from 0.4 to 0.76 and 0.022):

Wα′=0.76−0.400.76−0.022=0.360.738=0.488W_{\alpha'} = \frac{0.76 - 0.40}{0.76 - 0.022} = \frac{0.36}{0.738} = 0.488

Pearlite (formed from the remaining austenite):

WP=0.40−0.0220.76−0.022=0.3780.738=0.512W_{P} = \frac{0.40 - 0.022}{0.76 - 0.022} = \frac{0.378}{0.738} = 0.512

Total ferrite and cementite (phases, including those inside pearlite):

Wα=6.67−0.406.67−0.022=6.276.648=0.943W_{\alpha} = \frac{6.67 - 0.40}{6.67 - 0.022} = \frac{6.27}{6.648} = 0.943 WFe3C=0.40−0.0226.67−0.022=0.3786.648=0.057W_{Fe_3C} = \frac{0.40 - 0.022}{6.67 - 0.022} = \frac{0.378}{6.648} = 0.057
QuantityFraction
Pro-eutectoid ferrite0.488 (48.8%)
Pearlite0.512 (51.2%)
Total ferrite0.943
Total cementite0.057

Answer: 48.8% pro-eutectoid ferrite and 51.2% pearlite; total phases 94.3% ferrite and 5.7% cementite.

  • Practice · 6 marks

A hypereutectoid steel contains 1.2 wt% carbon. Explain its microstructure development on slow cooling from the austenite region, and calculate (a) the mass fraction of pro-eutectoid cementite and pearlite just below the eutectoid temperature, (b) the amount of each in 1 kg of steel, and (c) the total mass of ferrite and cementite.

Answer

Microstructure development

  1. At high temperature the alloy is austenite of 1.2% C.
  2. On cooling to the Acm_{cm} line (about 900 °C for 1.2% C), austenite becomes saturated with carbon and pro-eutectoid cementite precipitates, mostly as a network along the austenite grain boundaries.
  3. The remaining austenite loses carbon and reaches 0.76% C at 727 °C.
  4. At 727 °C it transforms to pearlite.
  5. Room-temperature structure: cementite network + pearlite. The network makes the steel hard and brittle, which is why such steels are spheroidise-annealed.

Lever rule (just below 727 °C)

Pro-eutectoid cementite (tie line 0.76 to 6.67):

WFe3C′=1.20−0.766.67−0.76=0.445.91=0.0745W_{Fe_3C'} = \frac{1.20 - 0.76}{6.67 - 0.76} = \frac{0.44}{5.91} = 0.0745

Pearlite:

WP=6.67−1.206.67−0.76=5.475.91=0.9255W_{P} = \frac{6.67 - 1.20}{6.67 - 0.76} = \frac{5.47}{5.91} = 0.9255

(b) In 1 kg

Pro-eutectoid cementite = 0.0745 kg = 74.5 g; pearlite = 0.9255 kg = 925.5 g.

(c) Total phases

WFe3C=1.20−0.0226.67−0.022=1.1786.648=0.1772(177.2 g)W_{Fe_3C} = \frac{1.20 - 0.022}{6.67 - 0.022} = \frac{1.178}{6.648} = 0.1772 \quad (177.2\ \text{g}) Wα=1−0.1772=0.8228(822.8 g)W_{\alpha} = 1 - 0.1772 = 0.8228 \quad (822.8\ \text{g})

Answer: pro-eutectoid cementite 7.45%, pearlite 92.55%; per kg: 74.5 g cementite network + 925.5 g pearlite; total ferrite 822.8 g and total cementite 177.2 g.

  • Practice · 5 marks

Describe the solidification and cooling of a 3 wt% C hypoeutectic white cast iron from the liquid state. Calculate the fractions of primary austenite and eutectic (ledeburite) just above 1147 °C, and the total cementite and austenite just below 1147 °C. Take austenite 2.14% C, eutectic liquid 4.3% C, cementite 6.67% C.

Answer

Cooling path

  1. Above the liquidus, the alloy is liquid.
  2. Between the liquidus and 1147 °C, primary austenite dendrites separate; the liquid becomes richer in carbon, reaching 4.3% C at 1147 °C.
  3. At 1147 °C the remaining liquid undergoes the eutectic reaction L→γ+Fe3CL \rightarrow \gamma + \text{Fe}_3\text{C}, giving ledeburite.
  4. Below 1147 °C, the austenite loses carbon as secondary cementite and at 727 °C reaches 0.76% C and transforms to pearlite.
  5. At room temperature: pearlite (dendrites) + cementite + transformed ledeburite. The structure is hard and brittle white cast iron.

Fractions just above 1147 °C

Primary austenite (tie line 2.14 to 4.3):

Wγ′=4.3−3.04.3−2.14=1.32.16=0.602W_{\gamma'} = \frac{4.3 - 3.0}{4.3 - 2.14} = \frac{1.3}{2.16} = 0.602

Liquid, which becomes ledeburite:

WL=3.0−2.144.3−2.14=0.862.16=0.398W_{L} = \frac{3.0 - 2.14}{4.3 - 2.14} = \frac{0.86}{2.16} = 0.398

Phases just below 1147 °C

Total cementite (tie line 2.14 to 6.67):

WFe3C=3.0−2.146.67−2.14=0.864.53=0.190W_{Fe_3C} = \frac{3.0 - 2.14}{6.67 - 2.14} = \frac{0.86}{4.53} = 0.190 Wγ=1−0.190=0.810W_{\gamma} = 1 - 0.190 = 0.810
StageResult
Just above 1147 °C60.2% primary γ\gamma, 39.8% liquid
Just below 1147 °C81.0% austenite, 19.0% cementite

Answer: primary austenite 60.2% and ledeburite 39.8%; just below 1147 °C austenite 81.0% and cementite 19.0%.

  • Practice · 5 marks

Iron changes from BCC (alpha) to FCC (gamma) at 912 °C. The lattice constants at this temperature are 0.2903 nm for BCC and 0.3647 nm for FCC. Calculate the percentage volume change on transformation and the atomic radius in each structure. What is the practical effect of this change?

Answer

Method: compare volume per atom in the two structures.

BCC has 2 atoms per cell and FCC has 4 atoms per cell.

Volume per atom

VBCC=a32=(0.2903)32=0.024462=0.012232 nm3V_{BCC} = \frac{a^3}{2} = \frac{(0.2903)^3}{2} = \frac{0.02446}{2} = 0.012232\ \text{nm}^3 VFCC=a34=(0.3647)34=0.048514=0.012127 nm3V_{FCC} = \frac{a^3}{4} = \frac{(0.3647)^3}{4} = \frac{0.04851}{4} = 0.012127\ \text{nm}^3

Volume change

ΔV%=VFCC−VBCCVBCC×100=0.012127−0.0122320.012232×100=−0.86%\Delta V\% = \frac{V_{FCC} - V_{BCC}}{V_{BCC}}\times100 = \frac{0.012127 - 0.012232}{0.012232}\times 100 = -0.86\%

The negative sign means a contraction of about 0.86% on heating through 912 °C (and an expansion on cooling).

Atomic radius

RBCC=3 a4=1.732×0.29034=0.1257 nmR_{BCC} = \frac{\sqrt3\, a}{4} = \frac{1.732 \times 0.2903}{4} = 0.1257\ \text{nm} RFCC=2 a4=1.414×0.36474=0.1289 nmR_{FCC} = \frac{\sqrt2\, a}{4} = \frac{1.414 \times 0.3647}{4} = 0.1289\ \text{nm}

The FCC structure is denser (APF 0.74 against 0.68) but the atom appears larger because of coordination 12 against 8; the net effect is a small contraction.

Practical effect

When steel cools through the γ→α\gamma\rightarrow\alpha change, the volume change causes distortion and internal stresses, and in quenching can lead to cracking. Also, since FCC iron dissolves much more carbon (2.14% against 0.022%), steel is austenitised before hardening.

Answer: volume contraction of 0.86% on α→γ\alpha\rightarrow\gamma; RBCC=0.1257R_{BCC} = 0.1257 nm, RFCC=0.1289R_{FCC} = 0.1289 nm.

  • Practice · 8 marks

Explain the following heat-treatment processes of steel with the purpose and the resulting microstructure: (a) full annealing, (b) normalising, (c) hardening, (d) tempering. Give the heating temperatures with respect to the Fe-Fe3C diagram.

Answer

Heat treatment is controlled heating and cooling of a solid metal to obtain desired properties.

ProcessHeating temperatureCoolingStructurePurpose
Full annealingA3A_3 + 30-50 °C (hypoeutectoid); A1A_1 + 30-50 °C (hypereutectoid)Very slow, in furnaceCoarse pearlite + ferriteSoften, improve machinability, relieve stress
NormalisingA3A_3 (or AcmA_{cm}) + 30-50 °CAir coolingFine pearlite + ferriteRefine grain, uniform structure, moderate strength
HardeningA3A_3 + 30-50 °C (hypoeutectoid); A1A_1 + 30-50 °C (hypereutectoid)Rapid quench in water or oilMartensiteHigh hardness and wear resistance
TemperingBelow A1A_1: 150-650 °CAny, after hardeningTempered martensiteReduce brittleness and quenching stress

Notes

  • Annealing gives the softest, most ductile state. Spheroidise annealing (just below A1A_1 for a long time) turns lamellar cementite into spheres in high-carbon steel.
  • Normalised steel is stronger and harder than annealed, because air cooling is faster, giving finer pearlite.
  • Hardening needs the cooling rate to exceed the critical cooling rate so that austenite transforms to martensite without diffusion. Hardenability depends on carbon and alloy content. Retained austenite and cracking can occur in high-carbon steels.
  • Tempering allows carbon to precipitate as fine carbide. Low tempering (150-250 °C) keeps hardness for tools; medium (350-450 °C) gives spring steel; high (500-650 °C) gives tough, strong shafts and bolts (quench plus high temper is called hardening and tempering).
 T
 A3 ----- ---- hardening/annealing/normalising
 A1 ---------------------
         tempering (below A1)
 time -->

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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