For a eutectoid steel describe isothermal heat treatments th

For a eutectoid steel, describe isothermal heat treatments that would be required to yield specimens having the following tensile strength-ductility (%RA) combinations: 1800 MPa and 30%RA 1700 MPa and 45%RA 1400 MPa and 50%RA

Solution

A: Thus, utilizing the isothermal transformation diagram for this alloy, we must rapidly cool to a temperature at which coarse pearlite forms (i.e., to about 675^oC), allow the specimen to isothermally and completely transform to coarse pearlite. At this temperature an isothermal heat treatment for at least 200 s is required.

B: For 1700Mpa and 45%RA

For case (1), after austenitizing, rapidly cool to about 580^oC , hold at this temperature for about 4 s (to obtain 75% fine pearlite), and then rapidly quench to room temperature.
For case (2), after austenitizing, rapidly cool to room temperature in order to achieve 100% martensite. Then temper this martensite for about 2000 s at 535⁰C .

C : According to this figure, in order to achieve a minimum yield strength of 1400 MPa a tempering temperature of less that about 410°C is required. On the other hand, tempering must be carried out at greater than about 360°C for a minimum ductility of 50%RA. Therefore, an Eutectiod steel possessing these characteristics is possible; tempering would be carried out at between 360°C and 410°C for 1 h.