21-Mat-A4 Deformation Behaviour and Properties of Materials · May 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, May 2014 — 10-Met-A4, Structure of Materials. Three hours, closed book, one approved calculator (Casio or Sharp). Eight questions of 20 marks each; the rubric asks for any five, and only the first five in the answer book are marked. All eight are solved here, because this set is a study resource rather than an exam script. All necessary equations, constants and the error-function table are provided in the exam's own appendix and are used directly below.
Note on the exam title. The printed exam header reads 10-Met-A4, Structure of Materials. Only two of the eight questions (VI and VIII) are substantially deformation/mechanical-properties content; the paper as a whole is a broad introductory materials-science survey — bonding, crystallography, polymers, diffusion, XRD, phase diagrams, dislocations — and is answered as such below.
Check — figure-read values. Question VI.3's stress-strain curve and Question VII's Cu–Ag solvus/liquidus positions are read from the printed figures rather than given numerically. Graphically-read values carry a few percent uncertainty that closed-form calculations do not — this is flagged again at the point of use.
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. Cu–Ag eutectic diagram: $T_E=779^\circ\text{C}$, eutectic composition $C_E=71.9$ wt% Ag; $\alpha$-solvus limit at $T_E$, $C_{\alpha E}=8.0$ wt% Ag (point B); $\beta$-solvus limit at $T_E$, $C_{\beta E}=91.2$ wt% Ag (point G).
Approach. Read the $\alpha$-solvus (left boundary of the $\alpha$ field) and $\beta$-solvus (right boundary of the $\beta$ field) directly off the digitized diagram at the $600^\circ\text{C}$ isotherm.
Digitizing the printed curve at $600^\circ\text{C}$: the $\alpha$-solvus sits at about $2.7$ wt% Ag, and the $\beta$-solvus sits at about $97.0$ wt% Ag (i.e. $3.0$ wt% Cu dissolved in $\beta$).
(a) Maximum solubility of Cu in Ag ($\beta$ phase) at $600^\circ\text{C}$: $100-97.0 = \boxed{3.0\ \text{wt\% Cu}}$.
(b) Maximum solubility of Ag in Cu ($\alpha$ phase) at $600^\circ\text{C}$: $\boxed{2.7\ \text{wt\% Ag}}$.
Both solvus limits are well below their eutectic-temperature values (8.0 and 91.2 wt%), as expected — solid solubility falls as temperature drops below $T_E$.Check: read graphically from the digitized figure; the diagram gives no closed-form solvus equation away from the eutectic point, so this is inherently a "read the curve" answer.
Approach. At $800^\circ\text{C}$ (above $T_E=779^\circ\text{C}$), the two-phase field between the solidus ($\alpha$ boundary) and liquidus ($L$ boundary) is $\alpha+L$; read both boundary compositions at $800^\circ\text{C}$ and apply the lever rule about the alloy's own composition $C_0=20$ wt% Ag.
Digitizing the diagram at $800^\circ\text{C}$: solidus $C_\alpha \approx 8.0$ wt% Ag, liquidus $C_L \approx 70.6$ wt% Ag. Since $C_\alpha < C_0 < C_L$, the alloy is in the $\alpha+L$ field.
Approach. $775^\circ\text{C}$ is only $4^\circ\text{C}$ below $T_E$, so the $\alpha$ and $\beta$ solvus compositions have barely moved from their eutectic-temperature values B(8.0) and G(91.2); the alloy at 25% Ag lies in the $\alpha+\beta$ field, so apply the lever rule with those endpoints.
A eutectic reaction is an invariant (fixed-temperature, fixed-composition, three-phase) reaction in which, on cooling, a liquid transforms directly into two distinct solid phases simultaneously: $$L \rightleftharpoons \alpha + \beta$$ For the Cu–Ag system specifically: $$L\,(71.9\ \text{wt\%\ Ag}) \xrightarrow{\ 779^\circ\text{C}\ } \alpha\,(8.0\ \text{wt\%\ Ag}) + \beta\,(91.2\ \text{wt\%\ Ag})$$ On the phase diagram this is the point where the liquidus, both solidus branches, and the horizontal eutectic isotherm all meet (point E), and by the Gibbs phase rule three phases coexisting in a binary system fixes both temperature and all compositions — the reaction proceeds at constant temperature until all the liquid is consumed.
At the eutectic point, $\alpha$ (Ag-poor) and $\beta$ (Ag-rich) must both nucleate and grow simultaneously from a liquid whose composition (71.9% Ag) is intermediate between them. As each phase grows it rejects the solute it cannot accommodate (the growing $\alpha$ rejects Ag into the adjacent liquid; the growing $\beta$ rejects Cu), and the alternating-layer (lamellar) arrangement minimizes the diffusion distance each rejected species must travel sideways to feed the neighbouring lamella of the other phase — long-range diffusion through the liquid is far slower than the short lateral diffusion between adjacent lamellae. This solute redistribution by short-range lateral diffusion at the growth front is what selects a fine, alternating two-phase morphology rather than either phase growing as a single coarse blob.
| Quantity | Value |
|---|---|
| Cu solubility in Ag ($\beta$), 600°C | $3.0$ wt% |
| Ag solubility in Cu ($\alpha$), 600°C | $2.7$ wt% |
| 20%Ag/800°C phases | $\alpha+L$: $80.8\%\ \alpha$, $19.2\%\ L$ |
| 25%Ag/775°C phases | $\alpha+\beta$: $79.6\%\ \alpha$, $20.4\%\ \beta$ |
| Eutectic reaction | $L(71.9) \to \alpha(8.0)+\beta(91.2)$ at $779^\circ$C |
