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Chapter 10: Problem 31

Cite two reasons why martensite is so hard and brittle.

Short Answer

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Answer: The two main reasons for martensite's hardness and brittleness are its body-centered tetragonal (BCT) crystal structure and rapid transformation rate from austenite. The BCT structure introduces a high level of stress within the crystal lattice, and the rapid transformation rate leaves a high number of dislocations locked in the martensite.

Step by step solution

01

1. Crystal Structure of Martensite

One of the reasons why martensite is so hard and brittle is due to its crystal structure. Martensite has a body-centered tetragonal (BCT) crystal structure. The BCT structure has elongated unit cells along one crystallographic direction, which makes it less symmetric and introduces a high level of stress within the crystal lattice. This increased stress and distorted lattice with strong atomic bonds result in increased hardness. The high stress within the structure also makes it difficult for dislocation movement, preventing it from undergoing plastic deformation and resulting in brittleness.
02

2. Rapid Transformation Rate

The second reason for martensite's hardness and brittleness is the rapid transformation rate of martensite from the parent phase, austenite. The transformation process, known as martensitic transformation, involves a sudden shear motion of the crystal lattice that happens at very fast cooling rates. This rapid movement doesn't allow enough time for dislocations to rearrange and for vacancies to be eliminated. Consequently, a high number of dislocations remain locked in the martensite, leading to the material being hard and brittle. In summary, the hardness and brittleness of martensite are primarily caused by its body-centered tetragonal crystal structure and rapid transformation rate from austenite.

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Most popular questions from this chapter

Figure \(10.40\) shows the continuous cooling transformation diagram for a \(1.13 \mathrm{wt} \%\) C iron-carbon alloy. Make a copy of this figure and then sketch and label continuous cooling curves to yield the following microstructures: (a) Fine pearlite and proeutectoid cementite (b) Martensite (c) Martensite and proeutectoid cementite (d) Coarse pearlite and proeutectoid cementite (e) Martensite, fine pearlite, and proeutectoid cementite

It is known that the kinetics of recrystallization for some alloy obey the Avrami equation and that the value of \(n\) in the exponential is \(2.5\). If, at some temperature, the fraction recrystallized is \(0.40\) after \(200 \mathrm{~min}\), determine the rate of recrystallization at this temperature.

(a) Rewrite the expression for the total free energy change for nucleation (Equation 10.1) for the case of a cubic nucleus of edge length \(a\) (instead of a sphere of radius \(r\) ). Now differentiate this expression with respect to \(a\) (per Equation 10.2) and solve for both the critical cube edge length, \(a^{*}\), and also \(\Delta G^{*}\). (b) Is \(\Delta G^{*}\) greater for a cube or a sphere? Why?

In terms of heat treatment and the development of microstructure, what are two major limitations of the iron-iron carbide phase diagram?

What is the driving force for the formation of spheroidite?
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