21-Mat-A5 Phase Transformations and Thermal Treatment · December 2019
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, December 2019 — 10-Met-A5, Mechanical Behaviour and Fracture of Materials. Three hours, closed book, Casio/Sharp approved 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 sub-parts explicitly call for an essay-format answer, and the rubric rewards clarity and organisation, so those answers are written as structured prose rather than as note form.
Nothing on the paper is a phase-transformation or heat-treatment question in the TTT/CCT, hardenability or tempering sense; the syllabus actually examined is crystallography of slip and twinning, dislocation theory, creep, fatigue, toughness and fracture mechanics, and safe-life fatigue design.
Q4(a) states explicitly that the plate is semi-infinite and gives a service tensile stress (450 MPa), so it is solved using the free-surface geometry correction that detail calls for, alongside a cross-check against the simpler through-crack assumption.
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 |
|---|---|---|
| Cyclic stress amplitude | $\sigma_a$ | 450 MPa |
| Tensile mean stress | $\sigma_m$ | 200 MPa |
| Fatigue strength coefficient | $\sigma_f'$ | 1758 MPa |
| Fatigue strength exponent | $b$ | $-0.0977$ |
| Actual cycles to failure (8 years) | $N_{\text{actual}}$ | 228,000 |
| Design factor of safety (on life) | FOS | 2 |
Find. Whether the component was operated beyond its safe-life design retirement point, or whether it simply achieved an unusually long service life.
Approach. A metal's fully-reversed stress-life behaviour follows Basquin's equation $\sigma_a = \sigma_f'(2N_f)^b$. A non-zero mean stress is accounted for by Morrow's correction, which replaces $\sigma_f'$ with $(\sigma_f'-\sigma_m)$ — the correction of choice here because it needs only the fatigue strength coefficient the paper supplies, unlike a Goodman-type correction, which would need the alloy's ultimate tensile strength (not given). Solve Morrow's equation for the predicted life $N_f$ at the service stress state, then apply the aerospace “safe-life” design philosophy — the certified/retirement life is the predicted life divided by the design factor of safety — and compare that retirement point against the cycles actually accumulated before failure.
| Quantity | Symbol | Result |
|---|---|---|
| Predicted (mean) fatigue life at service stress | $N_f$ | 165,764 cycles |
| Certified safe-life retirement point (FOS = 2) | $N_f/\text{FOS}$ | 82,882 cycles |
| Actual cycles to failure | $N_{\text{actual}}$ | 228,000 cycles |
| Ratio, actual/safe-life limit | $N_{\text{actual}}/N_{\text{safe}}$ | 2.75$\times$ over limit |
| Governing conclusion | $\boxed{\text{Safe-life design limit was exceeded}}$ |
The paper supplies $\sigma_f'$ and $b$ but not the alloy's ultimate tensile strength, so a Goodman-type mean-stress correction (which needs $\sigma_{\text{UTS}}$) cannot be evaluated from the given data; Morrow's correction, which uses $\sigma_f'$ directly, is the one the supplied data set supports and is the standard textbook choice for exactly this situation.
Conclusion. The answer is (i): the component exceeded its safe-life design limit. The safe-life philosophy requires retirement at $N_f/\text{FOS}=82{,}882$ cycles specifically because scatter in fatigue testing means an individual part can fail anywhere in a statistical band around the mean predicted life $N_f=165{,}764$ cycles — the factor of safety exists to retire the part well before that band is ever reached. Instead the component remained in service to 228,000 cycles: not only past the 82,882-cycle certified limit, but even 38 per cent past the raw, unfactored mean-life prediction itself. Framed the other way, (ii) is not the correct characterisation — the outcome is not simply an “excessively long” lucky life, it is a direct consequence of operating the part well beyond the margin the safety factor was designed to provide.
Both features are records left on a fatigue fracture surface by the incremental advance of the crack front, but they record that advance at two entirely different scales and for two entirely different reasons.
(i) Origin. A striation is the fingerprint of a single load cycle: at the crack tip, each cycle blunts the tip on loading (by localised plastic flow) and re-sharpens it on unloading, and this plastic-blunting/re-sharpening sequence advances the crack front by one small increment and leaves one ripple on the fracture surface — the local striation spacing is, essentially, the local value of $da/dN$ from a Paris-law relation like the one in Question 3(b). A beach mark (also called an arrest mark or a concentric-ring mark), by contrast, is not tied to any single cycle at all: it marks a macroscopic change in growth conditions — a shutdown/restart of the machine or aircraft, a change in load spectrum or amplitude, a period of corrosive exposure during a dwell, or any other interruption — recorded as a visible ring showing where the crack front happened to be at that moment. A given fracture surface can show striations without ever showing beach marks (e.g. constant-amplitude laboratory testing) but essentially never the reverse, because beach marks are simply the visible trace of many thousands of striations accumulated between one operational interruption and the next.
(ii) Size. Striations are microscopic, with spacings set by the crack growth increment per cycle — typically sub-micron to a few microns, the same order of magnitude as the $da/dN$ values computed in Question 3(b) (there, growth rates of $2.9\times10^{-10}$ to $1.5\times10^{-8}$ m/cycle). Beach marks are macroscopic, typically millimetres to centimetres apart, because each one represents the cumulative crack advance over many thousands to millions of individual cycles between interruptions — several orders of magnitude larger than a single striation.
(iii) Detectability. Because striation spacing is comparable to the crack growth per cycle, striations are resolvable only under a scanning electron microscope at high magnification (typically $1{,}000\times$ or more); they are invisible to the naked eye and usually below the resolution of an optical microscope. Beach marks, being macroscopic, are visible to the naked eye or with a low-power hand lens, which is precisely why they are the first evidence a failure investigator examines on a recovered fracture surface — they map out the overall history and origin of the crack (where it started, how the front advanced, roughly how the loading varied) before any SEM work is undertaken to confirm the fine striation spacing and thereby cross-check the growth-rate law against the service load history.
