21-Mat-A4 Deformation Behaviour and Properties of Materials · May 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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 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.]
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.
| Reaction | Temperature | Phase reaction | Compositions (wt%C) |
|---|---|---|---|
| Peritectic | 1493°C | $\delta+L\rightleftharpoons\gamma$ | $\delta$(0.09) + L(0.53) → $\gamma$(0.17) |
| Eutectic | 1147°C | $L\rightleftharpoons\gamma+\text{Fe}_3\text{C}$ | L(4.30) → $\gamma$(2.14) + Fe₃C(6.70) |
| Eutectoid | 727°C | $\gamma\rightleftharpoons\alpha+\text{Fe}_3\text{C}$ | $\gamma$(0.76) → $\alpha$(0.022) + Fe₃C(6.70) |
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.
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.
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.
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).
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.
| Quantity | Value |
|---|---|
| Peritectic | 1493°C, $\delta$(0.09)+L(0.53)→$\gamma$(0.17) |
| Eutectic | 1147°C, L(4.30)→$\gamma$(2.14)+Fe₃C(6.70) |
| Eutectoid | 727°C, $\gamma$(0.76)→$\alpha$(0.022)+Fe₃C(6.70) |
| Max C in α-ferrite | 0.022 wt%C at 727°C |
| Max C in austenite | 2.14 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. |