21-Mat-A4 Deformation Behaviour and Properties of Materials · December 2013
Question 7 of 8: Question VII — Mechanical Behaviour (20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2013 — 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. All necessary equations and constants are provided in the exam's own appendix (reproduced where used below).
The printed exam header reads Met-A4, Structure of Materials. Only two of the eight questions (VI and VII) are genuinely diffusion/mechanical-properties content; the paper as a whole is a broad introductory materials-science survey — bonding, crystal defects, crystallography, XRD and phase diagrams — and is answered as such below.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
W. D. Callister and D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed. — atomic bonding, crystal structure, imperfections, diffusion, mechanical properties, phase diagrams.
G. E. Dieter, Mechanical Metallurgy, 3rd ed. — stress–strain analysis, resilience and toughness.
Ductility is a measure of plastic deformability — how much a material can permanently stretch or reduce in cross-section before fracturing, reported as %EL or %RA from a simple uniaxial tensile test. Fracture toughness is a distinct material property (units $\text{MPa}\sqrt{\text{m}}$, symbol $K_{Ic}$) that measures resistance to crack propagation in the presence of a pre-existing flaw, combining the applied stress and flaw size via $K_{Ic}=Y\sigma\sqrt{\pi a}$. A material can be ductile yet still have a moderate fracture toughness, or vice versa: ductility describes bulk deformation capacity of a flaw-free specimen, while fracture toughness describes crack-tip behaviour in a flawed one, and the two are measured by entirely different tests (uniaxial tension vs. compact-tension/notched-specimen fracture tests).
VII.2 — Four representative stress-strain curve archetypes
(a) A typical metallic alloy (e.g. steel) shows a linear elastic region to a distinct yield point, strain hardening to an ultimate tensile strength (UTS) peak, then necking and softening to fracture — exactly the shape analysed quantitatively in VII.3 below. (b) A brittle material (e.g. a ceramic or cast iron in tension) deforms elastically along a single straight line with no yielding or plastic region at all, fracturing abruptly at a relatively low strain. (c) A linearly elastic–perfectly plastic material is elastic up to a sharp yield stress $\sigma_y$, then deforms at constant stress (a flat plateau) under continued straining, with no strain hardening — an idealisation used in limit-analysis and simple plasticity models. (d) A nonlinearly elastic material (e.g. natural rubber) follows a smooth, concave-upward curve with no distinct yield point at all; unlike the other three, deformation here is fully reversible — unloading retraces (approximately) the same path back to the origin rather than leaving permanent set.
Four representative stress–strain archetypes: strain-hardening ductile alloy, brittle fracture with no yield, ideal elastic–perfectly plastic, and reversible nonlinear elastic (e.g. rubber).
VII.3 — Properties from the printed stress–strain curve
Given. The printed $\sigma$–$\varepsilon$ curve for the specimen: origin at $(0,0)$, rising steeply and continuously to a peak of ≈510 MPa near $\varepsilon\approx0.03$, then descending smoothly through the marked fracture point at exactly $(\varepsilon,\sigma)=(0.10,\,300\ \text{MPa})$ (the printed curve's own endpoint marker).
Find. Yield strength, UTS, $E$, %EL, modulus of resilience, and toughness.
Check: the printed curve has no gridded data table, only axis tick marks (0.02 strain / 100 MPa spacing) and the labelled origin/peak/fracture region — all six values below. The proportional limit reads ≈400 MPa directly off the curve, consistent with the source figure's own annotation.
Approach. Identify the linear (elastic) region for $E$, apply the 0.2%-offset construction for a defensible yield point on this smoothly-curving alloy, read UTS at the peak, take %EL directly from the fracture strain (appendix $\%EL=100\varepsilon_f$), then compute resilience from $E$ and yield, and toughness by numerically integrating the full digitised curve.
Digitised source curve with the elastic (0.2%-offset) yield, UTS peak and fracture point marked; the dashed line is the offset construction used for VII.3(a).
(c) Young's modulus — from the linear region. The digitised curve is linear from the origin to the proportional limit at $(\varepsilon,\sigma)\approx(0.012,\,400\ \text{MPa})$:
$$E \approx \frac{\Delta\sigma}{\Delta\varepsilon} = \frac{400\ \text{MPa}}{0.012} = \boxed{33.3\ \text{GPa}}$$
(a) Yield strength — 0.2% offset. Constructing a line of slope $E$ through $\varepsilon=0.002$ and intersecting it with the digitised curve (dashed line in the figure) gives the offset yield point at $\varepsilon\approx0.0147$:
$$\sigma_y = E(\varepsilon-0.002) = 33{,}300(0.0147-0.002) = \boxed{420\ \text{MPa}}$$
(This sits just above the ≈400 MPa proportional limit read directly off the curve, as expected — the 0.2% offset always reports slightly above the true proportional limit for a smoothly-curving alloy with no sharp yield point.)
(d) % elongation to failure. Reading the fracture strain directly off the horizontal axis, $\varepsilon_f=0.10$, and applying the appendix relation $\%EL=100\varepsilon_f$:
$$\%EL = 100(0.10) = \boxed{10\%}$$
(e) Modulus of resilience. Assuming linear-elastic behaviour up to yield, the elastic strain energy per unit volume is the area of the elastic triangle, $U_r=\sigma_y^2/(2E)$:
$$U_r = \frac{(420)^2}{2(33{,}300)} = \boxed{2.65\ \text{MJ/m}^3}$$
(f) Toughness (total energy absorbed to fracture). Numerically integrating the full digitised curve from $\varepsilon=0$ to $\varepsilon_f=0.10$:
$$U_T = \int_0^{0.10}\sigma\,d\varepsilon \approx \boxed{40\ \text{MJ/m}^3}$$
This is the area under the entire curve (elastic + plastic, including the post-UTS softening branch) — it is what Question VII.1 above distinguishes from $K_{Ic}$: this is the tensile-test "toughness" (total absorbed energy per unit volume), not the crack-tip $K_{Ic}$ fracture toughness, which this smooth curve alone cannot supply (no notch/crack geometry is given).
