Using the isothermal transformation diagram for a \(0.45 \mathrm{wt} \%\) C steel alloy (Figure 10.39), 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 \(845^{\circ} \mathrm{C}\left(1550^{\circ} \mathrm{F}\right)\) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Rapidly cool to \(250^{\circ} \mathrm{C}\left(480^{\circ} \mathrm{F}\right)\), hold for \(10^{3} \mathrm{~s}\), then quench to room temperature. (b) Rapidly cool to \(700^{\circ} \mathrm{C}\left(1290^{\circ} \mathrm{F}\right)\), hold for \(30 \mathrm{~s}\), then quench to room temperature. (c) Rapidly cool to \(400^{\circ} \mathrm{C}\left(750^{\circ} \mathrm{F}\right)\), hold for \(500 \mathrm{~s}\), then quench to room temperature. (d) Rapidly cool to \(700^{\circ} \mathrm{C}\left(1290^{\circ} \mathrm{F}\right)\), hold at this temperature for \(10^{5} \mathrm{~s}\), then quench to room temperature. (e) Rapidly cool to \(650^{\circ} \mathrm{C}\left(1200^{\circ} \mathrm{F}\right)\), hold at this temperature for 3 s, rapidly cool to \(400^{\circ} \mathrm{C}\left(750^{\circ} \mathrm{F}\right)\), hold for \(10 \mathrm{~s}\), then quench to room temperature. (f) Rapidly cool to \(450^{\circ} \mathrm{C}\left(840^{\circ} \mathrm{F}\right)\), hold for \(10 \mathrm{~s}\), then quench to room temperature. (g) Rapidly cool to \(625^{\circ} \mathrm{C}\left(1155^{\circ} \mathrm{F}\right)\), hold for \(1 \mathrm{~s}\), then quench to room temperature. (h) Rapidly cool to \(625^{\circ} \mathrm{C}\left(1155^{\circ} \mathrm{F}\right)\), hold at this temperature for \(10 \mathrm{~s}\), rapidly cool to \(400^{\circ} \mathrm{C}\left(750^{\circ} \mathrm{F}\right)\), hold at this temperature for \(5 \mathrm{~s}\), then quench to room temperature.

Step by step solution

01

Rapid cooling to 250°C

Before we start analyzing the transformation that occurs during this specific heat treatment, it is essential to make a rough estimate whether or not the cooling rate is high enough to cross the nose of the TTT diagram. If it does, we would end up with a martensitic structure that wouldn't have been involved in any isothermal transformation. In this case, however, given the description of the rapid cooling process, we can safely assume that we do not cross the nose and our analysis will be based on the IT diagram.

02

Holding at 250°C for 1000 seconds

The specimen is held at \(250^{\circ}\mathrm{C}\) for \(10^3\)s. On the IT diagram, locate the temperature and time curve intersection. This intersection reveals the transformation taking place, which should be the formation of pearlite (P) and proeutectoid ferrite (Fe).

03

Final microstructure

Due to the isothermal transformation at \(250^{\circ}\mathrm{C}\) for \(10^{3}s\), the final microstructure consists of pearlite (P) and proeutectoid ferrite (Fe). Repeat these steps for each of the time-temperature treatments (b) to (h) to find the final microstructure for each of them.

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