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-FeC diagram shows the phases of plain carbon steels and cast irons against carbon content (0 to 6.67 wt% C, where FeC 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 (): interstitial solid solution of C in BCC iron, max 0.022% C at 727 °C. Soft, ductile.
- Austenite (): C in FCC iron, max 2.14% C at 1147 °C. Non-magnetic, ductile, stable above 727 °C.
- -ferrite: C in BCC iron, up to 0.09% C at 1495 °C.
- Cementite (FeC): hard, brittle compound with 6.67% C.
- Liquid (L).
Invariant reactions
| Reaction | Temp. | C content | Equation |
|---|---|---|---|
| Peritectic | 1495 °C | 0.17% | |
| Eutectic | 1147 °C | 4.3% | (ledeburite) |
| Eutectoid | 727 °C | 0.76% | (pearlite) |
Critical lines: A (727 °C), A (GOS line, ferrite starts to form), A (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
| Constituent | Nature | Structure | Properties |
|---|---|---|---|
| Ferrite () | Interstitial solid solution of C in -Fe | BCC, max 0.022% C | Soft, ductile, magnetic below 768 °C, tensile strength about 280 MPa |
| Austenite () | Solid solution of C in -Fe | FCC, up to 2.14% C | Soft, ductile, tough, non-magnetic; stable above 727 °C (retained in austenitic stainless steel) |
| Cementite (FeC) | Intermetallic compound, 6.67% C | Orthorhombic | Very hard (about 800 HV), brittle |
| Pearlite | Eutectoid mixture of ferrite and cementite (0.76% C) | Alternate lamellae of ferrite and cementite | Hardness about 200 HB, tensile strength about 800 MPa; fine pearlite is harder than coarse |
| Martensite | Supersaturated solid solution of C in iron | Body-centred tetragonal (BCT), needle/plate shape | Extremely 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: . 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 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
- At 950 °C the structure is 100% austenite () of 0.4% C.
- On cooling to the A line (about 790 °C), pro-eutectoid (primary) ferrite nucleates at austenite grain boundaries. The remaining austenite is enriched in carbon.
- Between A and 727 °C more ferrite forms; austenite composition follows the GOS line toward 0.76% C.
- Just above 727 °C the austenite has 0.76% C. At 727 °C it transforms by the eutectoid reaction into pearlite.
- 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):
Pearlite (formed from the remaining austenite):
Total ferrite and cementite (phases, including those inside pearlite):
| Quantity | Fraction |
|---|---|
| Pro-eutectoid ferrite | 0.488 (48.8%) |
| Pearlite | 0.512 (51.2%) |
| Total ferrite | 0.943 |
| Total cementite | 0.057 |
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
- At high temperature the alloy is austenite of 1.2% C.
- On cooling to the A 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.
- The remaining austenite loses carbon and reaches 0.76% C at 727 °C.
- At 727 °C it transforms to pearlite.
- 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):
Pearlite:
(b) In 1 kg
Pro-eutectoid cementite = 0.0745 kg = 74.5 g; pearlite = 0.9255 kg = 925.5 g.
(c) Total phases
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
- Above the liquidus, the alloy is liquid.
- Between the liquidus and 1147 °C, primary austenite dendrites separate; the liquid becomes richer in carbon, reaching 4.3% C at 1147 °C.
- At 1147 °C the remaining liquid undergoes the eutectic reaction , giving ledeburite.
- Below 1147 °C, the austenite loses carbon as secondary cementite and at 727 °C reaches 0.76% C and transforms to pearlite.
- 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):
Liquid, which becomes ledeburite:
Phases just below 1147 °C
Total cementite (tie line 2.14 to 6.67):
| Stage | Result |
|---|---|
| Just above 1147 °C | 60.2% primary , 39.8% liquid |
| Just below 1147 °C | 81.0% 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
Volume change
The negative sign means a contraction of about 0.86% on heating through 912 °C (and an expansion on cooling).
Atomic radius
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 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 ; nm, 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.
| Process | Heating temperature | Cooling | Structure | Purpose |
|---|---|---|---|---|
| Full annealing | + 30-50 °C (hypoeutectoid); + 30-50 °C (hypereutectoid) | Very slow, in furnace | Coarse pearlite + ferrite | Soften, improve machinability, relieve stress |
| Normalising | (or ) + 30-50 °C | Air cooling | Fine pearlite + ferrite | Refine grain, uniform structure, moderate strength |
| Hardening | + 30-50 °C (hypoeutectoid); + 30-50 °C (hypereutectoid) | Rapid quench in water or oil | Martensite | High hardness and wear resistance |
| Tempering | Below : 150-650 °C | Any, after hardening | Tempered martensite | Reduce brittleness and quenching stress |
Notes
- Annealing gives the softest, most ductile state. Spheroidise annealing (just below 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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