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21-Mat-A4 Deformation Behaviour and Properties of Materials · May 2016

Question 5 of 8: Phase Diagram: Iron–Carbon (20 marks)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2016 — 10-Met-A4, Structure of Materials. Three hours, closed book, one approved calculator (Casio or Sharp). Eight questions of 20 marks each (Roman numerals I–VIII); the rubric asks for any five, with only the first five in the answer book marked. All eight are solved here, because this set is a study resource rather than an exam script. All necessary equations, constants and an error-function table are provided in the exam's own appendix (reproduced where used below).

Note on the exam title. The printed exam header reads 10-Met-A4, Structure of Materials. Only Question VIII (mechanical deformation) is genuinely deformation/mechanical-properties content; the paper as a whole is a broad introductory materials-science survey — bonding, crystal structure, crystallographic directions/planes, imperfections, phase diagrams, XRD and diffusion — and is answered as such below.

Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:


Question V — Phase Diagram: Iron–Carbon (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The printed Fe–Fe₃C diagram's own labelled values: pure-Fe melting 1538°C; peritectic 1493°C; $\delta\to\gamma$ transus (0%C) 1394°C; eutectic 1147°C at 2.14/4.30 wt%C; $\gamma\to\alpha$ transus (0%C) 912°C; eutectoid 727°C at 0.76/0.022 wt%C; cementite fixed at 6.70 wt%C.

[Figure not reproduced: Fe–Fe₃C metastable phase diagram redrawn from the paper's own printed labels (all values read directly from the printed diagram). See the official exam paper.]

V.1 — Invariant reactions

Approach. An invariant reaction in a binary system is a three-phase, zero-degree-of-freedom equilibrium (Gibbs phase rule: $F=C-P+1=2-3+1=0$ at fixed pressure), so it occurs at one fixed temperature and fixed compositions for all three phases. The Fe–Fe₃C system has three: a peritectic (low-C, high-T), a eutectic, and a eutectoid.

All three invariant reactions
ReactionTemperaturePhase reactionCompositions (wt%C)
Peritectic1493°C$\delta+L\rightleftharpoons\gamma$$\delta$(0.09) + L(0.53) → $\gamma$(0.17)
Eutectic1147°C$L\rightleftharpoons\gamma+\text{Fe}_3\text{C}$L(4.30) → $\gamma$(2.14) + Fe₃C(6.70)
Eutectoid727°C$\gamma\rightleftharpoons\alpha+\text{Fe}_3\text{C}$$\gamma$(0.76) → $\alpha$(0.022) + Fe₃C(6.70)
Check: the peritectic point (1493°C, $\delta$/L/$\gamma$ at 0.09/0.53/0.17 wt%C) sits in a very narrow, low-carbon corner of the diagram that is easy to compress to illegibility at small print scale; its temperature (1493°C) and the flanking transus (1394°C, 1538°C) ARE printed and legible on this paper's own diagram, so all three invariants are reported here even though the peritectic composition values themselves (0.09/0.53/0.17) are standard accepted Fe–Fe₃C data rather than individually re-printed at that scale.

V.2 — Maximum carbon solubility in α-ferrite and austenite

Approach. Both maxima occur where the relevant phase field is widest along the composition axis — for $\alpha$ this is exactly at the eutectoid temperature (727°C), and for $\gamma$ this is at the eutectic temperature (1147°C), both read directly off the diagram's own printed labels.

  1. α-ferrite. The $\alpha/(\alpha+\gamma)$ solvus reaches its rightmost extent exactly at the eutectoid point: $$\boxed{C_{\alpha,max}=0.022\ \text{wt\%C at }727^{\circ}\text{C}}$$
  2. γ-austenite. The $\gamma/(\gamma+L)$ and $\gamma/(\gamma+\text{Fe}_3\text{C})$ boundaries meet at the eutectic point, the widest extent of the $\gamma$ field: $$\boxed{C_{\gamma,max}=2.14\ \text{wt\%C at }1147^{\circ}\text{C}}$$

The huge difference (0.022 vs. 2.14 wt%C, a factor of ≈100) is the physical basis of steel heat treatment: austenite can dissolve nearly 100× more interstitial carbon than ferrite because the FCC $\gamma$ structure's octahedral interstices are larger than BCC $\alpha$'s — quenching austenite traps that carbon supersaturated in solution (forming martensite) because the low-temperature equilibrium phase (ferrite) cannot hold nearly as much.

V.3 — Lever rule: 99.65 wt%Fe–0.35 wt%C, just below 727°C

Given. $C_0=0.35$ wt%C; just below 727°C the two-phase field is $\alpha+\text{Fe}_3\text{C}$ with $C_\alpha=0.022$, $C_{Fe_3C}=6.70$ wt%C.

Find. $W_\alpha$, $W_{Fe_3C}$.

Approach. Apply the lever rule directly across the $\alpha+\text{Fe}_3\text{C}$ tie-line spanning the full 0.022–6.70 wt%C range.

  1. Lever-rule fractions. $$W_\alpha=\frac{C_{Fe_3C}-C_0}{C_{Fe_3C}-C_\alpha}=\frac{6.70-0.35}{6.70-0.022}=\frac{6.35}{6.678}=\boxed{0.951\ (95.1\%)}$$ $$W_{Fe_3C}=\frac{C_0-C_\alpha}{C_{Fe_3C}-C_\alpha}=\frac{0.35-0.022}{6.678}=\frac{0.328}{6.678}=\boxed{0.0491\ (4.91\%)}$$
  2. Check. $W_\alpha+W_{Fe_3C}=0.951+0.0491=1.000$. ✓

This is a hypoeutectoid steel far on the ferrite-rich side of the eutectoid composition (0.35 vs. 0.76 wt%C), so the result — overwhelmingly ferrite (95.1%) with only a small cementite fraction (4.9%) — is exactly the expected qualitative picture: most of this alloy's microstructure just below 727°C is proeutectoid ferrite plus a comparatively small volume of pearlite (whose own overall composition also averages to 0.35 wt%C once its internal $\alpha$/Fe₃C split is folded back in).

V.4 — Equilibrium microstructures of 1 wt%C steel

Approach. Locate $C_0=1$ wt%C (hypereutectoid, since 1 > 0.76 eutectoid) on the phase diagram at each of the three temperatures and read off which phase field it falls in.

single-phase γ (austenite) grains1000°C: 1 wt%C < 2.14 wt%C max solubilityγ grains + proeutectoid Fe₃C750°C: between Aₜm and 727°C, hypereutectoidpearlite (α+Fe₃C lamellae) + proeutectoid Fe₃C500°C: below 727°C eutectoid
Schematic equilibrium microstructures for 1 wt%C steel. 1000°C: single-phase austenite (1 < 2.14 wt%C max solubility, so fully dissolved, no proeutectoid phase). 750°C: between the A$_{cm}$ line and 727°C — hypereutectoid, so proeutectoid cementite forms as a network along the austenite grain boundaries while the grain interiors remain austenite. 500°C: below 727°C — the remaining austenite (at the eutectoid composition, 0.76 wt%C) has transformed to pearlite (alternating $\alpha$/Fe₃C lamellae), leaving a microstructure of proeutectoid cementite network + pearlite.
  1. (a) 1000°C. Since $C_0=1$ wt%C $< C_{\gamma,max}=2.14$ wt%C at this temperature (below the eutectic but the $\gamma$ field is still wide), the alloy is entirely single-phase $\boxed{\gamma\text{ (austenite)}}$ — no proeutectoid phase yet.
  2. (b) 750°C. This lies between the A$_{cm}$ (the $\gamma/(\gamma+\text{Fe}_3\text{C})$ boundary, which at 1 wt%C sits above 727°C) and the 727°C eutectoid line, so the alloy is hypereutectoid two-phase: $\boxed{\gamma+\text{proeutectoid Fe}_3\text{C (grain-boundary network)}}$.
  3. (c) 500°C. Below 727°C, all remaining austenite (which had been enriched back down to the eutectoid composition 0.76 wt%C by continued Fe₃C rejection between 750°C and 727°C) transforms via the eutectoid reaction to pearlite: $\boxed{\text{proeutectoid Fe}_3\text{C network}+\text{pearlite }(\alpha+\text{Fe}_3\text{C lamellae})}$.
Question V — final results
QuantityValue
Peritectic1493°C, $\delta$(0.09)+L(0.53)→$\gamma$(0.17)
Eutectic1147°C, L(4.30)→$\gamma$(2.14)+Fe₃C(6.70)
Eutectoid727°C, $\gamma$(0.76)→$\alpha$(0.022)+Fe₃C(6.70)
Max C in α-ferrite0.022 wt%C at 727°C
Max C in austenite2.14 wt%C at 1147°C
0.35 wt%C alloy, just below 727°C$W_\alpha$=95.1%, $W_{Fe_3C}$=4.91%
1 wt%C steel, 1000/750/500°C$\gamma$ / $\gamma$+Fe₃C net. / pearlite+Fe₃C net.