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.
Full (perfect) dislocation. Its Burgers vector connects one lattice point to another equivalent lattice point, so slip leaves the crystal structure unchanged after the dislocation passes — e.g. in FCC metals (Cu, Al, Ni) the perfect Burgers vector is $\mathbf{b}=\tfrac{a}{2}\langle110\rangle$, the shortest full lattice translation. Partial dislocation. Its Burgers vector does not connect equivalent lattice points, so its passage leaves a stacking fault (a local region of incorrect stacking sequence) in its wake; a perfect dislocation can dissociate into two Shockley partials bounding a stacking-fault ribbon, e.g. in FCC metals $\tfrac{a}{2}[\bar{1}01] \to \tfrac{a}{6}[\bar{2}11]+\tfrac{a}{6}[\bar{1}\bar{1}2]$ — low-stacking-fault-energy metals such as austenitic stainless steel and brass widely dissociate this way, which widens the stacking-fault ribbon and suppresses cross-slip.
Tilt boundary. A low-angle grain boundary produced when two adjacent grains are misoriented by a small rotation about an axis lying within the boundary plane; it can be modelled as a simple vertical array of parallel edge dislocations spaced $D=b/\theta$ apart ($\theta$ = tilt angle, $b$ = Burgers vector magnitude). Twist boundary. Produced by a small rotation about an axis perpendicular to the boundary plane; it is modelled as a cross-grid (mesh) of screw dislocations rather than a single array, since the misorientation has no in-plane rotation axis to align a single dislocation family with.
Given. $\sigma_y = \sigma_i + k_yd^{-1/2}$, $\sigma_i=3$ MPa, $k_y=0.5\ \text{MPa}\cdot\text{m}^{1/2}$; $d_0=4\ \mu\text{m} \to d=103.82\ \mu\text{m}$ after annealing. Grain growth: $d^n-d_0^n=Kt$, $n=4$, $K=4.4\ \mu\text{m}^4/\text{s}$.
Find. (a) % decrease in yield strength on annealing; (b) time $t$ for the anneal.
Approach. (a) Evaluate Hall-Petch before and after (converting $d$ to metres). (b) Solve the grain-growth law directly for $t$ (keeping $d,d_0$ in $\mu$m to match $K$'s units).
| Quantity | Value |
|---|---|
| $\sigma_y$, as-received ($d=4\ \mu$m) | $253$ MPa |
| $\sigma_y$, annealed ($d=103.82\ \mu$m) | $52.1$ MPa |
| % decrease in yield strength | $79.4\%$ |
| Annealing time, $t$ | $2.64\times10^7$ s $\approx 305.6$ days |
