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Chapter 10: Problem 27

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) \(10^{\circ} \mathrm{C} / \mathrm{s}\) (b) \(1^{\circ} \mathrm{C} / \mathrm{s}\) (c) \(0.1^{\circ} \mathrm{C} / \mathrm{s}\) (d) \(0.01^{\circ} \mathrm{C} / \mathrm{s}\)

Short Answer

Expert verified
Answer: The microstructural products are as follows: (a) Cooling rate: \(10^{\circ} \mathrm{C} / \mathrm{s}\) - Martensite (b) Cooling rate: \(1^{\circ} \mathrm{C} / \mathrm{s}\) - Bainite and Martensite (c) Cooling rate: \(0.1^{\circ} \mathrm{C} / \mathrm{s}\) - Fine-grained Pearlite, Bainite, and Martensite (d) Cooling rate: \(0.01^{\circ} \mathrm{C} / \mathrm{s}\) - Coarse-grained Pearlite, Fine-grained Pearlite, Bainite, and a small amount of Martensite.

Step by step solution

01

Transform to Austenite

To begin, we have to completely transform our 4340 alloy steel specimens to austenite. This can be done by heating the specimens to a temperature between 850 and 900\(^{\circ} \mathrm{C}\). This temperature range allows the specimens to transform into the desired austenite phase.
02

Apply the Time-Temperature Transformation diagram

The Time-Temperature Transformation (TTT) diagram is useful for understanding the effect of cooling rate on the microstructure of an alloy. We will use this diagram to determine the resulting microstructures for the given cooling rates.
03

(a) Cooling rate: \(10^{\circ} \mathrm{C} / \mathrm{s}\)

At this fast cooling rate, the microstructure transforms directly from austenite to martensite. This happens because there is insufficient time for the other intermediate transformations to occur. The final microstructural product is martensitic.
04

(b) Cooling rate: \(1^{\circ} \mathrm{C} / \mathrm{s}\)

At this cooling rate, there is enough time for an intermediate transformation: the formation of Bainite. Thus, the resulting microstructural product is a mixture of Bainite and Martensite. The amount of each phase depends on the specific cooling curve followed.
05

(c) Cooling rate: \(0.1^{\circ} \mathrm{C} / \mathrm{s}\)

By lowering the cooling rate further, we allow more time for transformations. The final microstructure will be a mixture of fine-grained Pearlite, Bainite, and Martensite. Again, the specific proportion of each phase in the microstructure depends on the cooling curve.
06

(d) Cooling rate: \(0.01^{\circ} \mathrm{C} / \mathrm{s}\)

At this slowest cooling rate, there is ample time for all the intermediate transformations to occur. The resulting microstructure will be a mixture of coarse-grained Pearlite, fine-grained Pearlite, Bainite, and a small amount of Martensite. In conclusion, the microstructural products of 4340 alloy steel specimens transformed to austenite and then cooled to room temperature at different rates are as follows: (a) Martensite (b) Bainite and Martensite (c) Fine-grained Pearlite, Bainite, and Martensite (d) Coarse-grained Pearlite, Fine-grained Pearlite, Bainite, and a small amount of Martensite.

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Most popular questions from this chapter

Cite two important differences between continuous cooling transformation diagrams for plain carbon and alloy steels.

On the basis of diffusion considerations, explain why fine pearlite forms for the moderate cooling of austenite through the eutectoid temperature, whereas coarse pearlite is the product for relatively slow cooling rates.

Briefly explain why the hardness of tempered martensite diminishes with tempering time (at constant temperature) and with increasing temperature (at constant tempering time).

The fraction recrystallized-time data for the recrystallization at \(600^{\circ} \mathrm{C}\) of a previously deformed steel are tabulated here. Assuming that the kinetics of this process obey the Avrami relationship, determine the fraction recrystallized after a total time of \(22.8 \mathrm{~min} .\) $$ \begin{array}{cc} \hline \text { Fraction Recrystallized } & \text { Time (min) } \\ \hline 0.20 & 13.1 \\ 0.70 & 29.1 \\ \hline \end{array} $$

Make a copy of the isothermal transformation diagram for a \(0.45 \mathrm{wt} \% \mathrm{C}\) iron-carbon alloy (Figure \(10.39\) ), and then sketch and label on this diagram the time-temperature paths to produce the following microstructures: (a) \(42 \%\) proeutectoid ferrite and \(58 \%\) coarse pearlite (b) \(50 \%\) fine pearlite and \(50 \%\) bainite (c) \(100 \%\) martensite (d) \(50 \%\) martensite and \(50 \%\) austenite
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