Solutions Homework 8(Ch10) ME240 Spring 2020 PDF

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Chapter 10: Phase Transformations 1. What is more stable, the pearlite or spheroidite microstructure? Why? Spheroidite microstructures are more stable than pearlite ones. Since pearlite transforms to spheroidite, the latter is more stable. 2. Cite two major differences between martensitic and pearlitic transformations. Two major differences are: 1) atomic diffusion is necessary for the pearlitic transformation, whereas the martensitic transformation is diffusionless; and 2) relative to transformation rate, the martensitic transformation is virtually instantaneous, while the pearlitic transformation is time-dependent. 3. Rank the following iron-carbon alloys and associated microstructures from the highest to the lowest tensile strength: 0.25 wt% C with spheroidite; 0.25 wt% C with coarse pearlite; 0.60 wt% C with fine pearlite; 0.60 wt% C with coarse pearlite. Justify this ranking. The 0.25 wt% C, coarse pearlite is stronger than the 0.25 wt% C, spheroidite since coarse pearlite is stronger than spheroidite; the composition of the alloys is the same. The 0.60 wt% C, coarse pearlite is stronger than the 0.25 wt% C, coarse pearlite, since increasing the carbon content increases the strength (while maintained the same coarse pearlite microstructure). Finally, the 0.60 wt% C, fine pearlite is stronger than the 0.60 wt% C, coarse pearlite inasmuch as the strength of fine pearlite is greater than coarse pearlite because of the many more ferrite-cementite phase boundaries in fine pearlite. 4. For a eutectoid steel, describe an isothermal heat treatment that would be required to produce a specimen with 0.68 wt% of carbon to have a hardness of 93 HRB. You may want to consider the following two data as well as the Fe-C phase diagram (can be found in our textbook or lecture notes).

From animated data on the right side (see above), in order for a 0.68 wt% C alloy to have a Rockwell hardness of 93 HRB, the microstructure must be coarse pearlite. Thus, utilizing the isothermal transformation diagram (animated data on the left side, see above), this alloy, after 1

austenitizing at about 760°C, rapidly cool to a temperature at which coarse pearlite forms (i.e., to about 675°C), and allow the specimen to isothermally and completely transform to coarse pearlite. At this temperature, an isothermal heat treatment for at least 200 s is required. Then cool the specimen to room temperature (cooling rate is not important). 5. Name the two general stages that accompany a phase transformation. The first stage of a phase transformation is nucleation, which involves the formation of very small particles called nuclei. The second stage is growth, in which the nuclei increase in size. 6. Consider the following isothermal transformation diagram and the heat treatment designed by the line segments labeled a-b-c-d. From the list below, choose the microconstituents/phases that result from this heat treatment: proeutectoid cementite, tempered martensite, proeutectoid ferrite, spheroidite, martensite, bainite, pearlite.

Upon cooling to 725°C, the microstructure remains 100% austenite. While being held at this temperature for about 103 seconds, proeutectoid ferrite begins to form. During the rapid quench to room temperature, the remaining austenite begins to transform to martensite at about 300°C; all of this austenite will have transformed to martensite by the end of this quench. Furthermore, there are no changes with the proeutectoid ferrite that has already formed; thus the final composition will consist of proeutectoid ferrite and martensite. 7. Using the continuous-cooling transformation diagram for an iron-carbon alloy of eutectoid composition (see below), name the microstructural products of specimens having this eutectoid compositions that are first completely transformed to austenite, then cooled to room temperature at the following rates: 175 oC/s; 90 oC/s; 15 oC/s. For each cooling rate, choose the number corresponding to the following list of phases/microconstituents: 1. Martensite and pearlite. 2. Martensite. 3. Pearlite.

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High-resolution data can be found in Figure 10.27 in Page 383 in the textbook.

From the above diagram, the microstructural products that form for specimens of an ironcarbon alloy of eutectoid composition that are continuously cooled to room temperature at the various rates are as follows: (a) At a rate of 175 °C/s, only martensite forms since this rate is greater than the critical rate of 140°C. (b) At a rate of 90 °C/s, both martensite and pearlite form since this rate is less than the critical rate (140°C) yet greater the maximum rate for formation of a totally pearlitic structure (35°C). (c) At a rate of 15 °C/s, only pearlite forms since this rate is less than the maximum rate for the formation of a totally pearlitic structure (35°C). 8. Using the animated data below, the isothermal transformation diagram for a 0.45 wt% C steel alloy, specify the nature of the final microstructure (in terms of the microconstituents present) of a small specimen that has been subjected to the following temperature treatments: Rapidly cool to 700 oC, hold for 100,000 s, then quench to room temperature. It is assumed that the specimen begins at 845 °C and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure.

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First of all, we know from the above diagram that after cooling to and holding at 700°C for 30 s, a portion of the specimen has transformed to proeutectoid ferrite. While cooling to room temperature, the remainder of the specimen transforms to martensite. Hence, the final microstructure consists of proeutectoid ferrite and martensite. 9. Schematic room temperature microstructures for four iron-carbon alloys are shown below. Rank these alloy microstructures (by letter) from most ductile to the least ductile. You may want to refer to the Figure 10.30(b) in Page 386 in our textbook.

The following lists the microstructures for these four alloys: A = hypoeutectoid alloy (proeutectoid ferrite) and fine pearlite B = hypereutectoid alloy (proeutectoid cementite) and coarse pearlite C = spheroidite D = hypereutectoid alloy (proeutectoid cementite) and fine pearlite Alloy C will be the most ductile, because its microstructure is spheroidite, which more ductile than coarse and fine pearlites. The next most ductile alloy is A; its carbon content is lower than B and D (because it is a hypoeutectoid alloy)—as carbon content increases, ductility decreases. The carbon contents of B and D are approximately the same—the proeutectoid phase for both is cementite—however, the pearlite for B is coarse, while for D it is fine; coarse pearlite is more ductile than fine pearlite. Therefore, the ranking is C, A, B, D. 10. Match each transformation description below with name of a transformation of this type.

Diffusion-dependent with change(s) in phase composition(s)

Martensitic Transformation

Diffusionless

Eutectoid Reaction

Diffusion-dependent with NON change(s) in phase composition(s)

Recrystallization

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