Microstructures from the isothermal transformation diagram: 22-Mec-B8 Engineering Materials, December 2019, Q5

Question 5: Microstructures from the isothermal transformation diagram (20 marks)

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. A plain-carbon steel of eutectoid composition (0.76 wt% C), fully austenitised at 760 °C, and the isothermal transformation diagram for that composition reproduced on page 4. Values traced from the printed curves, all times measured from the instant the specimen reaches the holding temperature:

Values read from the page-4 isothermal transformation diagram
Quantity	Symbol	Value
Eutectoid temperature	$A_{1}$	727 °C
At 700 °C: start / 50% / finish	$t$	≈ 3 × 10² / 2.5 × 10³ / 1 × 10⁴ s
At 600 °C: start / 50% / finish	$t$	≈ 1.3 / 3.5 / 8.4 s
At 450 °C: start / 50% / finish	$t$	≈ 1.4 / 10 / 26 s
Martensite start	$M_{s}$	≈ 220 °C
Martensite 90%	$M_{90}$	≈ 145 °C

Find. The constituents present, and their approximate proportions, in the final room-temperature microstructure produced by each of the three treatments.

The three time–temperature paths superimposed on the page-4 diagram. Path (a) runs past the finish curve at 700 °C; path (c) stops between the start and finish curves at both holding temperatures, so it leaves untransformed austenite for the quench.

Approach. Read each holding temperature across to the transformation-start, 50% and finish curves, compare the holding time with those three times, and name the product from the temperature at which it forms — coarse pearlite just below the eutectoid, fine pearlite near the nose, bainite below the nose, martensite only on cooling through $M_{s}$. Any austenite still untransformed when the specimen is quenched becomes martensite; austenite that has already transformed cannot change into another product without being re-austenitised first.

Part (a) — establish where 10⁴ s at 700 °C lands. At 700 °C the specimen is only 27 °C below the eutectoid, so the driving force for nucleation is small and the reaction is slow: transformation does not begin until about 3 × 10² s and is not complete until about 10⁴ s. The prescribed hold of 10⁴ s therefore just reaches the finish curve, and the austenite transforms completely.
Part (a) — name the product and follow it to room temperature. Transformation so close to the eutectoid gives a large interlamellar spacing, because the carbon has time to diffuse a long way ahead of the growing colony. The product is therefore coarse pearlite — alternating lamellae of ferrite and cementite in the eutectoid proportions, roughly 88% ferrite to 12% cementite by mass. Cooling to room temperature afterwards changes nothing, because no austenite remains to transform. Final structure: 100% coarse pearlite, hardness of the order of 15 HRC.
Part (b) — recognise that 700 °C is below the eutectoid. The specimen entering this step is the fully pearlitic structure produced in part (a). Reheating it to 700 °C does not re-austenitise it, because 700 °C lies 27 °C below $A_{1}$; the isothermal diagram describes the decomposition of austenite and simply does not apply. What governs instead is the interfacial energy of the lamellar structure.
Part (b) — name the product. Twenty hours (7.2 × 10⁴ s) held just below the eutectoid is a classic spheroidising anneal. The thin cementite plates carry a very large ferrite–cementite interfacial area, and carbon diffuses down the resulting curvature gradient so that the plates pinch off and coalesce into roughly spherical particles. The final structure is spheroidite — coarse spheroidal Fe₃C particles dispersed in a continuous ferrite matrix. It is the softest and most ductile condition available to this steel, about 10 HRC and 68% RA, which is why the treatment is used before heavy machining or cold forming.
Part (c) — the hold at 600 °C. At 600 °C the reaction begins at about 1.3 s, is half complete at about 3.5 s and finishes at about 8.4 s. The prescribed 4 s therefore lies just past the 50% curve, so approximately$$f_{\text{pearlite}} \approx \boxed{50\,\%}$$of the austenite has transformed. Because 600 °C is much further below the eutectoid than in part (a), the lamellae are much closer together and the product is fine pearlite. The other 50% is still austenite when the specimen is moved.
Part (c) — the hold at 450 °C. The rapid cool to 450 °C is fast enough to miss the nose, so the remaining austenite arrives untransformed. At 450 °C, below the nose, the diffusion distance is short and the product is bainite — fine needles of ferrite with cementite precipitated within them rather than as continuous lamellae. Reading across, the reaction starts at about 1.4 s and is half complete at about 10 s, so the prescribed 10 s converts about half of the austenite still available:$$f_{\text{bainite}} \approx 0.50 \times 50\,\% = \boxed{25\,\%}\ \text{of the whole}$$
Part (c) — the quench and the final tally. Quenching from 450 °C takes the remaining austenite through $M_{s} \approx 220$ °C, and room temperature lies below $M_{90} \approx 145$ °C, so essentially all of it transforms by the diffusionless shear mechanism into martensite. Adding up,$$f_{\text{martensite}} = 100 - 50 - 25 = \boxed{25\,\%}$$giving a final structure of about 50% fine pearlite, 25% bainite and 25% martensite. Note that the fine pearlite and the bainite formed earlier are entirely unaffected by the quench — only austenite can transform.

Question 5 — final microstructures
Quantity	Symbol / basis	Value
(a) 700 °C, 10⁴ s, then cool	hold reaches the finish curve	100% coarse pearlite
(b) reheat to 700 °C, 20 h	below $A_{1}$: no austenite forms	spheroidite (spheroidal Fe₃C in ferrite)
(c) 600 °C for 4 s	just past the 50% curve	≈ 50% fine pearlite
(c) then 450 °C for 10 s	half of the remaining austenite	≈ 25% bainite
(c) then quenched	austenite left at $M_{s}$	≈ 25% martensite
(c) final structure	—	≈ 50% fine pearlite + 25% bainite + 25% martensite

Check: the 50% and 25% figures in part (c) are read from a printed diagram, and 4 s at 600 °C in fact sits slightly past the 50% curve, so a defensible answer is 50–60% fine pearlite with the balance split between bainite and martensite. What must be exact is the qualitative sequence and the reasoning, and one further point should be stated: strictly, the isothermal diagram applies to a specimen that is fully austenitic on arrival, so applying the 450 °C curve to austenite that is already 50% consumed is the standard textbook approximation rather than an exact treatment. A small amount of retained austenite also survives the quench in a eutectoid steel, typically a few per cent.

Concept & Theory

Topic. Isothermal (TTT) transformation of eutectoid steel: diffusional products above the nose, bainite below it, martensite only on cooling through $M_{s}$, and spheroidisation below the eutectoid.

Key relations. Eutectoid reaction $\gamma(0.76\ \text{wt}\%\ \mathrm{C}) \rightarrow \alpha(0.022) + \mathrm{Fe_{3}C}(6.70)$ at 727 °C; lever rule gives 88.7% ferrite and 11.3% cementite in pearlite; interlamellar spacing falls as the transformation temperature falls; martensite forms athermally between $M_{s}$ and $M_{f}$ with no time dependence.

Why this works. Two competing quantities set everything on a TTT diagram. Lowering the temperature raises the free-energy driving force, which speeds nucleation; it also slows carbon diffusion, which slows growth. Their product peaks at the nose near 540 °C, which is why the transformation curves are C-shaped and why the same steel takes hours to transform at 700 °C but seconds at 540 °C. The same competition sets the morphology: high in the diagram carbon can diffuse far, so the lamellae are widely spaced and the pearlite is coarse and soft; near the nose diffusion distances are short and the pearlite is fine and appreciably harder; below the nose diffusion is too slow to sustain cooperative lamellar growth at all and the reaction switches to the acicular bainitic morphology. Below $M_{s}$ diffusion stops mattering entirely, because martensite forms by a coordinated shear of the lattice; that is why the martensite lines are horizontal and why holding at a temperature between $M_{s}$ and $M_{f}$ does not increase the amount formed. Finally, the diagram describes the decomposition of austenite and nothing else, which is the whole content of part (b): a structure that is already pearlite has no austenite to decompose, so a sub-critical hold can only coarsen what is there.

Common pitfall. The most costly error is treating part (b) as another austenite transformation and reading the 700 °C curve again — the specimen never re-austenitises below 727 °C, and the answer is spheroidite, not more pearlite. The second is to apply the fractions in part (c) to the whole specimen at each step instead of to the austenite that actually remains, which double-counts. The third is to forget that products already formed are immune to the quench, and to write the answer as though the whole specimen became martensite.

Source. Callister & Rethwisch, 10th ed., §10.5 (isothermal transformation diagrams, Figs. 10.13–10.22), §10.6 and §10.7 (mechanical behaviour, spheroidite); Askeland & Wright, Ch. 12 and Ch. 13; Shackelford, Ch. 10.

Question 5 of 7: Microstructures from the isothermal transformation diagram

Question 5: Microstructures from the isothermal transformation diagram (20 marks)