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Metallurgical Engineering DISCUSSION
I am studying the iron-carbon phase diagram and get confused when applying the lever rule to a plain 0.40% C steel cooled slowly to room temperature. Some worked examples give about 49 percent ferrite and others about 94 percent ferrite for what looks like the same steel.
Which tie line am I supposed to use, which compositions go at its ends, and how do the results relate to what I would see under the microscope?
Both answers are right; they answer different questions. About 94 percent is the total amount of the ferrite phase, counting the ferrite inside the pearlite. About 49 percent is the proeutectoid ferrite, the separate ferrite grains that formed before the eutectoid reaction. The confusion comes from mixing up phases (ferrite, cementite) with microconstituents (proeutectoid ferrite, pearlite).
Under the microscope you see microconstituents: light ferrite grains and darker pearlite colonies in roughly equal areas. The 94 percent figure is not something you can see directly without resolving the lamellae.
For the microconstituents, draw the tie line just above the eutectoid temperature of 727 °C, across the ferrite plus austenite field. Its ends are ferrite at 0.022% C and austenite at 0.76% C. The rule is opposite arm over total length:
Proeutectoid ferrite = (0.76 - 0.40) / (0.76 - 0.022) = 0.36 / 0.738 = 0.49. Austenite = (0.40 - 0.022) / 0.738 = 0.51. On cooling through 727 °C all of that austenite becomes pearlite, so the steel is about 51 percent pearlite and 49 percent proeutectoid ferrite.
For the total phases, use the tie line just below 727 °C across the ferrite plus cementite field, with ends at 0.022% C (ferrite) and 6.70% C (cementite). Total ferrite = (6.70 - 0.40) / (6.70 - 0.022) = 6.30 / 6.678 = 0.943, and cementite = 0.057.
You can check the two results against each other. Pearlite itself is (6.70 - 0.76) / 6.678 = 0.889 ferrite. The ferrite inside the pearlite is then 0.51 × 0.889 = 0.455, and adding the 0.49 of proeutectoid ferrite gives 0.94, which agrees.
A shortcut worth remembering for hypoeutectoid steels: the pearlite fraction is roughly the carbon content divided by 0.76, since the ferrite solubility of 0.022% is nearly negligible. 0.40 / 0.76 = 0.53, close to the exact 0.51. It works in reverse too, so a slowly cooled plain carbon steel showing about 25 percent pearlite in a micrograph is near 0.2% C.
The reverse estimate only holds for plain carbon steels cooled slowly. Alloying elements move the eutectoid point to lower carbon, so an alloy steel shows more pearlite than the same carbon content would suggest.
Keep in mind that the diagram describes equilibrium, meaning very slow cooling. Air cooling (normalising) a 0.40% C steel gives less proeutectoid ferrite than the lever rule predicts and finer pearlite with a carbon content below 0.76%, because the transformation happens at a lower temperature than 727 °C. Faster cooling still produces bainite or martensite, which do not appear on the phase diagram at all.
For those cases you need the continuous cooling transformation (CCT) diagram of the specific steel. The lever rule remains the correct tool for equilibrium questions and for annealed structures.
