Given. A hypoeutectoid Fe–C steel, slowly cooled from the austenite region, shows a room-temperature microstructure of 40 wt% pearlite and 60 wt% proeutectoid ferrite. On the standard Fe–Fe3C diagram, the eutectoid composition is $C_{eut}=0.77\,\text{wt\%C}$ and the $\alpha$-ferrite solubility limit just above the eutectoid temperature (727°C) is $C_\alpha=0.022\,\text{wt\%C}$.
Given data
Quantity
Value
Weight fraction pearlite, $W_p$
0.40
Weight fraction proeutectoid ferrite, $W_\alpha$
0.60
Eutectoid composition, $C_{eut}$
0.77 wt%C
$\alpha$-ferrite solvus at 727°C, $C_\alpha$
0.022 wt%C
Find. (i) the alloy's overall carbon concentration $C_0$; (ii) the equilibrium phase constitution at 730°C after a long hold; (iii) the equilibrium phase constitution at 900°C.
Approach. Apply the lever rule on the tie line just above the eutectoid temperature (where the untransformed austenite fraction equals the pearlite fraction that will form on further cooling) to back out $C_0$, then locate 730°C and 900°C relative to the steel's own $A_1$ and $A_3$ boundaries.
Set up the lever rule at the eutectoid tie line. Just above 727°C the steel is two-phase ($\alpha$ + austenite); the austenite fraction present there is exactly the fraction that will become pearlite on cooling through 727°C, and the proeutectoid $\alpha$ fraction is fixed below that temperature. So:
$$W_p=\frac{C_0-C_\alpha}{C_{eut}-C_\alpha}$$
Solve for $C_0$. Substituting $W_p=0.40$:
$$C_0=C_\alpha+W_p(C_{eut}-C_\alpha)=0.022+0.40(0.77-0.022)=\boxed{0.32\ \text{wt\%C}}$$
Cross-check with the ferrite fraction. $W_\alpha=\dfrac{C_{eut}-C_0}{C_{eut}-C_\alpha}=\dfrac{0.77-0.32}{0.748}=0.60$ — matches the given 60% exactly, confirming $C_0\approx0.32\,\text{wt\%C}$.
Locate 730°C relative to this steel's own critical temperatures. The eutectoid ($A_1$) temperature is 727°C regardless of composition; the upper critical temperature $A_3$ for a hypoeutectoid steel (Andrews' empirical relation) is
$$A_3=912-203\sqrt{\text{wt\%C}}=912-203\sqrt{0.32}\approx797^\circ\text{C}$$
Since $727^\circ\text{C} < 730^\circ\text{C} < 797^\circ\text{C}$, holding at 730°C places the steel just inside the two-phase $\alpha+\gamma$ (ferrite + austenite) field — NOT in the pearlite-stable region, because pearlite is a transformation product that can only exist below $A_1$.
Describe the 730°C equilibrium structure. A long hold lets the steel reach its lever-rule equilibrium at 730°C: proeutectoid $\alpha$ grains coexisting with austenite grains, in proportions set by the (very short) 727–730°C tie line — essentially the same $\approx$60/40 $\alpha$/$\gamma$ split as at 727°C, since 730°C is only 3°C above it. Any pearlite present before reheating is destroyed: on crossing back above $A_1$ the eutectoid ferrite and cementite of the pearlite immediately revert to $\gamma$, and only the proeutectoid $\alpha$ grains survive as a separate phase. A long hold coarsens the two-phase grain structure but does not change this phase constitution while the steel remains above $A_1$.
Locate and describe the 900°C structure. $900^\circ\text{C} > A_3\approx797^\circ\text{C}$, so the entire steel lies above its own upper critical temperature: the equilibrium structure is a single phase, 100% austenite ($\gamma$), with no ferrite present at all.
Fig. 1.1 — equilibrium structure at 730°C (two-phase $\alpha+\gamma$, no pearlite present at temperature) versus 900°C (single-phase $\gamma$, above $A_3$).
Final results
Quantity
Result
Overall carbon concentration, $C_0$
0.32 wt%C
Equilibrium structure at 730°C
Two-phase $\alpha$ (proeutectoid ferrite) + $\gamma$ (austenite), no pearlite
