Iron Carbon Diagram Questions and Answers

Answer: A

The proeutectoid ferrite is about 44.9%.

Just above 723°C the steel is ferrite plus austenite of eutectoid composition, so apply the lever rule between 0.02 and 0.8 wt.% C:

\[ \begin{aligned} W_{\alpha} &= \frac{0.8 - 0.45}{0.8 - 0.02} \\ &= \frac{0.35}{0.78} = 0.449 \end{aligned} \]

All that austenite then becomes pearlite just below 723°C, while the ferrite already formed stays as proeutectoid ferrite. Taking the densities of the phases as nearly equal, as the question intends, the volume fraction equals the weight fraction, 44.9%. 55.1% is the pearlite fraction, the complement.

Answer: B

The pearlite is about 55.1%.

The pearlite forms from the austenite present just above the eutectoid temperature, so its fraction is the austenite fraction from the lever rule:

\[ \begin{aligned} W_{p} &= \frac{0.45 - 0.02}{0.8 - 0.02} \\ &= \frac{0.43}{0.78} = 0.551 \end{aligned} \]

With nearly equal densities, this is also the volume fraction, 55.1%. It is the complement of the 44.9% proeutectoid ferrite, which is the trap option. The lever rule here must use the eutectoid composition, 0.8 wt.% C, as its end point, not the 6.67 wt.% C of cementite, because pearlite is the product of austenite of eutectoid composition.

Answer: A

The fraction of pro-eutectoid ferrite is 0.387.

A 0.5% carbon steel is hypoeutectoid. On slow cooling, ferrite forms first; just above the eutectoid temperature the alloy is ferrite of 0.025% C and austenite of eutectoid composition, 0.8% C, which then turns to pearlite. The lever rule at that temperature gives

\[ \begin{aligned} f_{\alpha} &= \frac{0.8 - 0.5}{0.8 - 0.025} \\ &= \frac{0.3}{0.775} \approx 0.387 \end{aligned} \]

The rest, 0.613, becomes pearlite. 0.775 is the denominator alone, and 0.864 and 0.475 do not follow from the lever rule with these compositions.

Answer: C

Pearlite consists of ferrite and cementite.

Pearlite forms from austenite by the eutectoid reaction at about 727°C and 0.76-0.8% carbon in steel. The austenite transforms into alternating lamellae of ferrite (almost pure α-iron) and cementite (Fe₃C), which gives pearlite its mother-of-pearl appearance under the microscope.

Austenite is the high-temperature phase that pearlite forms from, so it is not present in pearlite. Martensite is a different product, formed by rapid quenching of austenite without diffusion.