3. Using the isothermal transformation diagram for a 1.13 wt%csteel alloy (Figur

Question

3. Using the isothermal transformation diagram for a 1.13 wt%csteel alloy (Figure 11.49 in your book), determine the final microstructure (in terms of just the microconstituents present) of a small specimen that has been subjected to the following time-temperature treatments. In each case assume that the specimen begins at 920 C 1690 F) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Rapidly cool to 775°C (1430 F), hold for 500 s, then quench to room temperature. temperature. (c) Rapidly cool to 350°C (660°F), hold for 300 s, then quench to room temperature. (d) Rapidly cool to 600 C (1110°F), hold at this temperature for 7s, rapidly cool to 450°C (840 F), hold at this temperature for 4 s, then quench to room temperature.

Explanation / Answer

(All temp. in deg celsius)

A. On cooling to 775 and standing for 500sec, approx 50%A + 50% C is formed. On quenching to RT, the A converts to M(martensite). Thus, the finalcomposition is 50% M & 50% C ( ProeutectoidCemetite)

B. On standing for a very long time, all the Austentite has converted into Pearlite. Thus, even after quenching to RT, we get 100% Pearlite.

C. In this process, after standing for 300s, we get approx. 40% bainite formation. Thus on quenching, we get 60% martensite and 40% bainite.

D. After waiting for7s, all A converts to P. This P cannot be transformed to B or M. Thus, we get 100 % pearlite.

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