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Chapter 10: Problem 24

Name the microstructural products of eutectoid iron-carbon alloy \((0.76\) wt \(\%\) C) specimens that are first completely transformed to austenite, then cooled to room temperature at the following rates: (a) \(1^{\circ} \mathrm{C} / \mathrm{s}\) (c) \(50^{\circ} \mathrm{C} / \mathrm{s}\) (b) \(20^{\circ} \mathrm{C} / \mathrm{s}\) (d) \(175^{\circ} \mathrm{C} / \mathrm{s}\)

Short Answer

Expert verified
Answer: The final microstructure of the eutectoid alloy will depend on the cooling rate as follows: - At 1°C/s cooling rate, the microstructure will be a pearlite mixture. - At 50°C/s cooling rate, the microstructure will be a mixture of pearlite and bainite. - At 20°C/s cooling rate, the microstructure will be mainly bainitic with some residual pearlite. - At 175°C/s cooling rate, the microstructure will be martensite.

Step by step solution

01

Understand the cooling process

We must first understand that when an iron-carbon alloy is completely transformed into austenite and then cooled, various microstructures will form depending on the cooling rate. The iron-carbon phase diagram helps us identify the transformations that occur during the cooling process. The eutectoid reaction, which involves austenite transforming into ferrite and cementite, happens at 727°C for a 0.76 wt% C alloy.
02

Analyze microstructure for the cooling rate of 1°C/s

For a cooling rate of 1°C/s, the alloy will have enough time for the eutectoid reaction to happen entirely during the cooling process. The final microstructure will be a pearlite mixture.
03

Analyze microstructure for the cooling rate of 50°C/s

At a cooling rate of 50°C/s, the alloy will cool rapidly, but not fast enough to allow a complete martensitic transformation. The final microstructure will be a mixture of pearlite and bainite.
04

Analyze microstructure for the cooling rate of 20°C/s

For a cooling rate of 20°C/s, the rate at which the alloy cools is faster than in the previous case. The cooling rate is enough to result in a predominant bainitic transformation. The final microstructure will be mainly bainitic with some residual pearlite.
05

Analyze microstructure for the cooling rate of 175°C/s

The cooling rate of 175°C/s is extremely fast, causing the austenite to transform into martensite without enough time for diffusion-controlled transformations like the formation of pearlite or bainite. The final microstructure will be martensite.

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Most popular questions from this chapter

Make a copy of the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition (Figure \(10.22\) ) and then sketch and label time- temperature paths on this diagram to produce the following microstructures: (a) \(100 \%\) coarse pearlite (b) \(50 \%\) martensite and \(50 \%\) austenite (c) \(50 \%\) coarse pearlite, \(25 \%\) bainite, and \(25 \%\) martensite

Briefly explain why there is no bainite transfor- mation region on the continuous-cooling trans- formation diagram for an iron–carbon alloy of eutectoid composition.

Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at the following rates: (a) \(0.005^{\circ} \mathrm{C} / \mathrm{s}\) (b) \(0.05^{\circ} \mathrm{C} / \mathrm{s}\) (c) \(0.5^{\circ} \mathrm{C} / \mathrm{s}\) (d) \(5^{\circ} \mathrm{C} / \mathrm{s}\)

Compute the rate of some reaction that obeys Avrami kinetics, assuming that the constants \(n\) and \(k\) have values of \(2.0\) and \(5 \times 10^{-4}\), respectively, for time expressed in seconds.

Briefly describe the simplest continuouscooling heat treatment procedure that would be used in converting a 4340 steel from one microstructure to another. (a) (Martensite \(+\) ferrite \(+\) bainite) to \((\) martensite \(+\) ferrite \(+\) pearlite \(+\) bainite \()\) (b) (Martensite + ferrite + bainite) to spheroidite (c) (Martensite \(+\) bainite \(+\) ferrite) to tempered martensite
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