Question 1 of 7: Micrograph of a Hypoeutectoid Steel — Phase Identification, Carbon Estimate, and Weight Fraction of Austenite

Question I: Micrograph of a Hypoeutectoid Steel — Phase Identification, Carbon Estimate, and Weight Fraction of Austenite (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. An etched micrograph (5 μm scale bar) from a hypoeutectoid plain-carbon steel, with two arrows marking regions for phase identification. Fe–Fe₃C reference data: eutectoid composition $C_{eu}=0.77$ wt% C at 727 °C; ferrite solvus $C_\alpha=0.022$ wt% C; cementite $C_{\text{Fe}_3\text{C}}=6.70$ wt% C. For part (iv), a second, separate sample at $C_0=0.85$ wt% C is held a long time at a temperature just above the eutectoid temperature.

Find. The constituent at each arrow; an estimate of the micrograph sample's own carbon content; the heat-treatment route that produced this microstructure; and the equilibrium weight fraction of austenite in the second (0.85 wt% C) sample.

1.1 — (i) Phases at the two arrows

The steel is hypoeutectoid (below 0.77 wt% C), so on slow cooling from full austenite it must pass through a two-phase $\alpha+\gamma$ field before reaching the eutectoid, and the two constituents visible in the micrograph are exactly the two products of that route. The left arrow points into the fine, dark, mottled texture that occupies most of the field — at this magnification (5 μm scale bar) individual cementite lamellae are too closely spaced to resolve cleanly, so pearlite (alternating ferrite/cementite lamellae) reads as a uniformly dark, finely textured constituent rather than as distinct stripes. The right arrow points to the edge of a large, light, irregularly-shaped region with smooth internal contrast and no lamellar texture — this is proeutectoid ferrite, the soft, nearly carbon-free phase that forms first as the steel cools below $A_3$ and appears bright/featureless under this etch because, unlike pearlite, it contains no cementite to darken it. Proeutectoid ferrite of this kind nucleates preferentially at prior-austenite grain boundaries, which is consistent with the large constituent's blocky, boundary-following outline in the micrograph.

1.2 — (ii) Estimating the carbon content

Approach. For a hypoeutectoid steel slow-cooled to room temperature, the room-temperature proeutectoid-ferrite weight fraction is fixed by the lever rule on the $\alpha/\gamma$ tie line at 727 °C (just above the eutectoid reaction), because that is the last temperature at which the ferrite fraction is set before the remaining eutectoid austenite converts to pearlite; reading the ferrite AREA fraction off the micrograph and treating it as the ferrite WEIGHT fraction (reasonable since ferrite and pearlite have very similar density) inverts the lever rule to back out $C_0$.

1. Estimate the proeutectoid-ferrite area fraction from the micrograph. The light (ferrite) constituent visible in Fig. 1.1 — the large region at the right arrow plus the smaller light patches elsewhere in the field — covers roughly a quarter of the total field area, i.e. $W_\alpha\approx0.20$–$0.30$, central estimate $W_\alpha\approx0.25$.
2. Invert the $727$ °C lever rule for the $\alpha$ (ferrite, $C_\alpha=0.022$) / $\gamma$ (austenite, at the eutectoid composition $C_{eu}=0.77$) tie line: $$W_\alpha=\dfrac{C_{eu}-C_0}{C_{eu}-C_\alpha}\ \ \Rightarrow\ \ C_0=C_{eu}-W_\alpha\left(C_{eu}-C_\alpha\right)$$
3. Evaluate at the central estimate $W_\alpha=0.25$: $$C_0=0.77-0.25\times(0.77-0.022)=0.77-0.187=0.583$$ $$\boxed{C_0\approx0.6\ \text{wt\%\,C}}$$ — sweeping the visual read over its plausible range $W_\alpha=0.20$–$0.30$ moves the estimate only between $0.55$ and $0.62$ wt% C, comfortably below the eutectoid composition, consistent with the exam's own "hypo-eutectoid" statement.

1.3 — (iii) How the microstructure was produced

The ferrite–pearlite structure in Fig. 1.1 is the standard product of slow, near-equilibrium cooling of a hypoeutectoid steel from the austenite field — i.e. a full anneal or normalize, not a quench. The steel is first heated into the single-phase austenite field, above $A_3$, and held long enough to homogenize. On slow cooling (furnace cooling for a full anneal, or still-air cooling for a normalize) below $A_3$, the equilibrium phase diagram requires proeutectoid ferrite to begin nucleating and growing, preferentially at prior-austenite grain boundaries, because ferrite is the phase favoured on the low-carbon side of the $\alpha+\gamma$ two-phase field. As ferrite grows, it rejects carbon into the shrinking austenite, driving the remaining austenite's composition to increase along the $A_3$–$A_1$ boundary. This continues until the temperature reaches $727$ °C ($A_1$), by which point the remaining austenite has been enriched to exactly the eutectoid composition ($0.77$ wt% C); that remaining austenite then transforms by the eutectoid reaction $\gamma\rightarrow\alpha+\text{Fe}_3\text{C}$ into pearlite. The slow cooling rate leaves time for the pearlite lamellae to nucleate and grow at only modest undercooling below $A_1$, giving the finely but not exceptionally coarsely spaced lamellae read as a uniform dark texture at this magnification, alongside the blocky, grain-boundary-decorating proeutectoid ferrite formed in the earlier stage.

1.4 — (iv) Weight fraction of austenite at 0.85 wt% C, just above the eutectoid temperature

Approach. At $0.85$ wt% C the second sample is hypereutectoid, so a temperature just above $727$ °C places it in the $\gamma+\text{Fe}_3\text{C}$ two-phase field immediately above the eutectoid horizontal. Because the hold is "slightly higher" than $727$ °C, the austenite boundary ($A_{cm}$) has moved only infinitesimally away from the eutectoid point, so to an excellent approximation the austenite on the tie line still sits at the eutectoid composition, $C_\gamma\approx C_{eu}=0.77$ wt% C, while cementite remains at its fixed stoichiometric composition. Apply the lever rule on that tie line.

1. Identify the tie-line end compositions. $C_\gamma=0.77$ wt% C (austenite, eutectoid); $C_{\text{Fe}_3\text{C}}=6.70$ wt% C (cementite); alloy composition $C_0=0.85$ wt% C.
2. Apply the lever rule for the austenite fraction (the lever arm on the cementite side of $C_0$, divided by the full tie-line length): $$W_\gamma=\dfrac{C_{\text{Fe}_3\text{C}}-C_0}{C_{\text{Fe}_3\text{C}}-C_\gamma}=\dfrac{6.70-0.85}{6.70-0.77}=\dfrac{5.85}{5.93}$$
3. Evaluate: $$\boxed{W_\gamma\approx0.9865=98.65\%\ \text{by weight}}$$ with the balance, $W_{\text{Fe}_3\text{C}}=1-W_\gamma\approx0.0135=1.35\%$, the small amount of proeutectoid cementite that has already precipitated at the grain boundaries just above the eutectoid temperature.