21-Mat-A5 Phase Transformations and Thermal Treatment · May 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, May 2017 — 10-Met-A5, Mechanical Behaviour and Fracture of Materials. Three hours, closed book, any non-communicating calculator permitted. Eight questions of 20 marks each; the rubric states that five questions constitute a complete paper and that only the first five appearing in the answer book are marked. All eight are answered here, because this set is a study resource rather than an exam script. Several questions ask explicitly for essay-format answers, and the marking scheme rewards clarity and organisation, so the discursive answers below are written as structured prose rather than as note form.
The printed exam header reads 10-Met-A5, Mechanical Behaviour and Fracture of Materials. The paper has no phase-transformation or heat-treatment question in the classical (TTT/CCT diagram, hardenability, tempering-curve) sense; the syllabus actually examined is fracture mechanics and fatigue-crack-growth life, strengthening and toughening of engineering materials, creep and fatigue testing, deformation processing, and elastic–plastic forming behaviour.
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.
| Quantity | Symbol | Value |
|---|---|---|
| Fracture toughness | $K_{Ic}$ | 25 MPa$\sqrt{\text{m}}$ |
| Cyclic (range) tensile stress | $\Delta\sigma$ | 100 MPa |
| Initial surface crack length | $a_0$ | 2.0 mm |
| Paris-law coefficient | $A$ | $1\times10^{-12}\ \text{MPa}^{-3}\text{m}^{-1/2}$ |
| Paris-law exponent | $n$ | 3 |
Find. The number of load cycles $N_f$ to grow the crack from $a_0$ to the critical size at which fast fracture takes over.
Approach. First find the critical crack length $a_c$ at which the stress-intensity range reaches $K_{Ic}$ (a "relatively large sheet" is modelled as a through-thickness crack in an infinite plate, $Y=1$, per the given data); then integrate the Paris law between $a_0$ and $a_c$.
| Quantity | Result |
|---|---|
| Critical crack length, $a_c$ ($Y=1$) | 19.9 mm |
| Initial $\Delta K$ at $a_0$ | 7.93 MPa$\sqrt{\text{m}}$ |
| Fatigue life, $N_f$ ($Y=1$) | $\boxed{5.49\times10^{6}\ \text{cycles}}$ |
No compliance function $Y(a/W)$ is given, so the through-crack, large-plate value $Y=1$ is used above, matching "a relatively large sheet." Because the flaw is explicitly described as a surface crack, the more conservative free-surface correction $Y\approx1.12$ reduces $a_c$ to about 15.9 mm and the estimated life to about $3.69\times10^{6}$ cycles — a 33% reduction — and should be treated as the design-appropriate figure, with the $Y=1$ result above as an upper-bound estimate.
A nominal tensile test measures the strength of defect-free (or nearly defect-free) material under a single, slowly applied load. Two entirely different loading histories let a material fail well below the $\sigma_y$ that same test reports, because in each case the local driving force for fracture is amplified far above the applied nominal stress.
Condition 1 — fast fracture at a pre-existing crack (linear elastic fracture mechanics). If the component contains a crack or crack-like flaw of length $a$, the crack tip sees a stress intensity $K=Y\sigma\sqrt{\pi a}$ that can reach the material's fracture toughness $K_{Ic}$ at a nominal stress $\sigma$ far below $\sigma_y$, provided the flaw is larger than the transition flaw size $a_t=(1/\pi)(K_{Ic}/\sigma_y)^2$ (worked explicitly in Question 5(a) of this paper). Once $K=K_{Ic}$, the crack propagates unstably with essentially no macroscopic warning, even though a small crack-tip plastic zone exists. This is exactly the mechanism quantified in part (a) above: the sheet fails catastrophically once the fatigue crack reaches $a_c$, at a nominal stress of only 100 MPa — far below any reasonable yield strength for structural steel.
Condition 2 — fatigue (sub-critical cyclic crack growth). As part (a) demonstrates quantitatively, a stress that never once exceeds $\sigma_y$ can still drive a crack from a small, sub-critical size to the critical size $a_c$ through millions of load cycles, via $da/dN=A(\Delta K)^n$. The microscopic mechanism is repeated, localised plastic slip right at the crack tip on each loading cycle — irreversible glide that opens the crack tip a small increment (a striation) each cycle even though the bulk component is elastic and the nominal stress is comfortably below yield. Fatigue is therefore also a sub-yield, low-apparent-ductility failure when viewed only through the nominal stress–strain curve, because the accumulated crack extension is entirely invisible to a single-cycle tensile test.
A third, related family worth noting (developed further in Question 8 of this paper) is environmentally assisted cracking — stress-corrosion cracking and hydrogen embrittlement — where a sustained sub-yield stress combined with a specific chemical environment again produces slow, brittle-appearing crack growth to failure with none of the ductility a tensile test alone would predict.
