22-Mec-B8 Engineering Materials · May 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, May 2017 — 16-Mec-B8 Engineering Materials. Three hours, open book; any non-communicating calculator permitted. Eight problems, all of equal value; any FIVE constitute a complete paper, so each problem is worth 20 marks. Candidates are urged to submit a clear statement of any assumptions made. All eight problems are solved below, because the set as a whole is the study resource.
Reference texts (22-Mec-B8 Engineering Materials).
Note on this sitting. Seven of the eight problems restate archetypes seen in 2013–2016 with fresh data, and every calculation below has been re-worked from this paper's numbers. Question 8 is the genuinely new one: the familiar aluminium–lithium substitution has been expanded into a three-alloy selection problem that also asks for the beam volume and for a strength-to-density ranking, and the two halves of it do not pick the same alloy.
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. AISI 1080 is a plain-carbon steel of eutectoid composition (0.80 wt % C), so its isothermal time–temperature–transformation (TTT) diagram is the classic single-C-curve form: eutectoid temperature A1 = 727 ℃, nose at roughly 540 ℃ and about 1 s, martensite start Ms ≈ 220 ℃ and martensite finish Mf ≈ −50 ℃.
Find. For each of the five required changes, the austenitising step, the quench path, any isothermal hold, and the final cooling — each traced on the TTT diagram. Part (e) needs the hold carried a little beyond the 50 % curve the diagram already draws, which is what fixes its hold time.
One principle governs every part of this question and is worth stating before any individual answer: the TTT diagram describes the decomposition of austenite, and nothing else. Pearlite, bainite and martensite are all products of austenite. None of them transforms directly into another at ordinary temperatures. Consequently every part of this question except the tempering leg of (c) begins with the same step — re-austenitise — and a candidate who tries to convert pearlite straight into martensite by quenching from room temperature has missed the point of the diagram entirely. A second consequence matters for 1080 in particular: because Mf lies well below room temperature, any treatment demanding 100 % martensite requires a sub-zero (cryogenic) finish, otherwise 5–10 % retained austenite is left behind.
(a) Pearlite → martensite. Austenitise at about 760 ℃ (some 30 ℃ above A1) and hold to dissolve the pearlite completely into homogeneous austenite, typically 30 min for a section of modest thickness. Then quench continuously and severely so that the cooling curve misses the nose — for eutectoid plain-carbon steel that means clearing about 540 ℃ in well under a second, which in practice demands agitated brine and a thin section, since 1080 has poor hardenability and a thick bar will form pearlite at its centre no matter how the surface is cooled. Continue below Ms = 220 ℃; the transformation is athermal, so the martensite fraction depends only on how far below Ms the steel has been taken, not on how long it is held. To reach a fully martensitic structure the part must be carried below Mf ≈ −50 ℃, so a sub-zero step is required. As-quenched martensite is normally tempered immediately, if only to relieve the quench stresses that otherwise crack the part.
(b) Pearlite → bainite. Austenitise at about 760 ℃ and hold long enough to dissolve all the pearlite into homogeneous austenite. Then austemper: quench rapidly into a salt or lead bath held between the nose and Ms — about 350 ℃ is a good choice for lower bainite — fast enough that the cooling curve passes to the left of the nose and no pearlite forms on the way down. Hold isothermally at that temperature until the transformation-finish curve is crossed, which the diagram puts at roughly 103 s at 350 ℃. Then cool to room temperature in air; the rate no longer matters because there is no austenite left to transform. Austempering has a practical bonus: because the part is held at a uniform temperature while it transforms, distortion and quench cracking are far lower than in a quench-and-temper route to comparable hardness.
(c) 65 % pearlite + 35 % martensite → 100 % tempered martensite. The starting mixture is irrelevant once it is re-austenitised, which is exactly what must be done: heat to about 760 ℃ and hold until both the pearlite and the existing martensite have dissolved into uniform austenite. Note that the martensite reverts to austenite on this heating leg like any other structure, so no separate step is needed for it. Now quench severely — brine or agitated water for a plain-carbon steel of this hardenability — so that the cooling curve passes to the left of the nose without touching it, and continue below Ms and then below Mf with a sub-zero treatment to eliminate retained austenite. That gives 100 % (untempered) martensite, which is hard and brittle and in a state of high residual stress. Finally temper: reheat to somewhere between 250 and 650 ℃ depending on the strength–toughness balance wanted, hold about one hour, and air cool. Tempering precipitates the supersaturated carbon as fine cementite in a ferrite matrix. Note that the tempering leg cannot be drawn on the TTT diagram: it starts from martensite, not austenite.
(d) Martensite → fine pearlite. Again the route is through austenite. Re-austenitise at 760 ℃, holding long enough for full homogenisation — this erases the martensite completely, which is the only way to reach a diffusional product from it. Then quench to just above the nose — about 580 ℃ — and hold isothermally past the transformation-finish curve, of the order of 30 s at that temperature, before air cooling. The choice of hold temperature is what selects the pearlite spacing: transformation just below A1 (650–700 ℃) gives coarse pearlite because carbon diffuses far before the lamellae impinge, whereas transformation nearer the nose (540–600 ℃) gives fine pearlite, since the larger undercooling raises the nucleation rate and shortens the diffusion distance. Fine pearlite is both harder and stronger than coarse pearlite of the same composition, which is why the question specifies it.
(e) 100 % pearlite → 55 % pearlite + 45 % martensite. Austenitise at about 760 ℃ and hold until all the pearlite has dissolved into homogeneous austenite. Quench rapidly to about 620 ℃, which is above the nose, and hold isothermally until 55 % of the austenite has transformed — that is, marginally past the 50 % curve drawn on the diagram, of the order of 25 s at that temperature. At that instant 55 parts in a hundred of the austenite have become pearlite. Immediately quench again, fast enough to miss the nose, past Ms and down below Mf: the remaining 45 % of the austenite transforms athermally to martensite. The two-stage quench is the whole trick — the hold sets the pearlite fraction, the second quench freezes whatever austenite is left. Because the target sits just past the printed 50 % curve rather than on it, the hold time has to be interpolated between the 50 % and finish curves, and a slightly conservative hold is the safer error: interrupting the hold late gives too much pearlite and cannot be undone except by starting again.
| Part | Austenitise | Quench to / hold | Finish | Product |
|---|---|---|---|---|
| (a) | 760 ℃ | Severe quench, miss the nose | Below Mf (sub-zero) | 100 % martensite |
| (b) | 760 ℃ | 350 ℃, hold past the finish curve (≈103 s) | Air cool | 100 % bainite (austempered) |
| (c) | 760 ℃ | Severe quench, miss the nose | Below Mf, then temper 250–650 ℃ for 1 h | 100 % tempered martensite |
| (d) | 760 ℃ | 580 ℃, hold past the finish curve (≈30 s) | Air cool | 100 % fine pearlite |
| (e) | 760 ℃ | 620 ℃, hold just past the 50 % curve (≈25 s) | Quench below Mf | 55 % pearlite + 45 % martensite |
Check: the hold times quoted are read from a standard published TTT diagram for eutectoid steel and will shift with austenitising temperature, prior austenite grain size and residual alloy content; the 25 s for part (e) in particular is an interpolation between the 50 % and finish curves rather than a value the diagram prints. The exam expects the correct path and the correct region of the diagram; the times are indicative. Note also that a TTT diagram is strictly valid only for isothermal transformation — for the continuous-cooling legs of (a), (c) and (e) the CCT diagram, whose curves are shifted down and to the right, is formally the correct tool.