Illustration .2 Hypoeutetoid steel that was cooled slowly from g-state to room temperature was determined to contain 10% eutectoid ferrite. Suppose no change in structure arises on cooling from just underneath the eutectoid temperature to room temperature. Estimate the carbon content of steel.

Solution

Submit as to phase diagram of Figure of Iron-Carbon Phase Diagram and assume the vertical line xx cross the isotherm at 5 such as 5 is at a distance x¢ from temperatures axis. After that by lever rule:

% total ferrite = (6.67 - x')/(6.67 - 0.025)

 = (6.67 - x')/6.645

 % pro-eutectoid ferrite =

= (0.80 - x')/(0.80 - 0.025)

= (0.80 - x')/0.775

% eutectoid ferrite = % total ferrite - % proeutectoid ferrite

 or        10/100 = (6.67- x')/ 6.645

=(0.80 - x')/ 0.775

= 0.51, 498

=5.169

= 0.775 x' - 5.316 + 6.645 x'

= 0.51, 645

= 5.87 x'

X'= 0.51645 / 3.87

= 0.088%

The steel has 0.088% C.