MSE 230 : Solutions for Assignment 10 Fall 2014

*Problems with an * by them will not be graded, but solutions will be provided.

Problem 1* First we cool the alloys rapidly to 0°C such that no diffusion can occur. At this point, the microstructure just consists of a polycrystalline single phase α microstructure. The α phase is supersaturated with copper. The copper would like to diffuse to form the θ phase, but the temperature is too low to accomplish this. Heating to 50°C and holding for 1 hour won't change things because there hasn't been enough of a temperature increase to get diffusion going. The microstructure would look the same as when the material was quenched. Heating to 200°C for one hour will result in a fine dispersion of θ precipitates in an α matrix, while heating to 350°C will result in a smaller number of larger θ precipitates. The ordering in increasing strength will be: 50°C < 350°C < 200°C The alloy treated at 50°C is the lowest because the θ particles did not form (although you should expect some solid solution strengthening). The strength of the alloy at 200°C will be the largest because it contains the finest θ particles and therefore the largest number of θ particles per unit volume of alloy. Many small particles are more effective at inhibiting dislocation motion. Problem 2 Two samples of an alloy of Al containing 4 wt. % Cu are heat treated at 540°C until all of the copper is in solid solution (see phase diagram below). One sample of the alloy is quenched rapidly to room temperature and then heated back up to 300°C (in the α+θ two phase region) until the θ particles have just completed precipitating. A second sample is cooled very slowly from 540°C to room temperature. (5) a) Assuming that 300°C is below the nose of the TTT diagram, modify the microstructures below to show the microstructures you would expect to observe for the different heat treatments.

Slow cooling leads to large precipitates because the material is in the two phase region at higher

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temperatures where the driving force for nucleation (undercooling) is small and the diffusivity is large (allowing rapid mass transport that favors precipitate growth). Also, given the rapid mass transport one expects some of the particles to grow on grain boundaries to minimize their interfacial energy. Quenching and reheating to below the noise creates situation where the driving force for nucleation is large and the diffusivity is relatively small. This favors the growth of many more fine precipitates compared to slow cooling. b) Which microstructure do you expect to have the higher yield stress? Briefly explain your answer. Quenched and heated to 300°C should have the higher yield stress for two main reasons:

a) For a given volume of second phase a smaller particle size leads to a greater number of particles per unit volume.

b) Smaller particles maintain a coherent interface and the coherency strain associated with the interface is very effective at inhibiting dislocation motion. At some point particle growth results in an incoherent interface that is much less effective at inhibiting dislocation motion.

Problem 3* a) After holding for 100 seconds at 600°C all of the austenite has transformed to

pearlite. Quenching to room temperature does not change the microstructure. b) After holding for 100 seconds at 700°C none of the austenite has transformed to

pearlite. So we still have 100% austenite. Upon quenching to room temperature virtually all of the austenite transforms to martensite via a displacive or athermal transformation. Usually not all of the austenite is able to transform (Fig. 11.23) so we are left with 90+% martensite and <10% austenite.

c) After holding for 100 seconds at 350°C approximately half of the austenite has

transformed to bainite. Upon quenching to room temperature most of the remaining austenite will be transformed to martensite. The bainite formed at 350°C will not be affected by the quench. So we are left with ≈50% bainite, ≈45% martensite, and ≈5% retained austenite.

Problem 4 The key here is to recognize that if you heat your iron-carbon alloy of eutectoid composition above the eutectoid temperature (727°C) the alloy will consist of single phase austenite. Then you can make a variety of microstructures following the same approach as problem 1. a) Making martensite from coarse pearlite: heat the coarse pearlite above 727°C (say to

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800°C) and hold it there for sufficient time to transform the pearlite to austenite. Then quench the austenite to room temperature. This will result in a microstructure that is ≈90+% martensite with the remainder retained austenite.

b) Coarse pearlite from fine pearlite: again heat to ≈800°C until the alloy is all

austenite. Quickly cool to ≈650°C and hold it there until the austenite transform completely to coarse pearlite (approximately 2 minutes). Holding above 650°C (but below the eutectoid temperature!) will produce pearlite that is more coarse, but will also take longer.

c) Same thing as part b. Problem 5* We are called upon to name the microstructural products that form for specimens of an iron-carbon alloy

of eutectoid composition that are continuously cooled to room temperature at a variety of rates.

Figure 10.27 is used in these determinations.

(a) At a rate of 200°C/s, only martensite forms.

(b) At a rate of 100°C/s, both martensite and pearlite form.

(c) At a rate of 20°C/s, only fine pearlite forms. Problem 6* This problem asks us to consider the tempering of a water-quenched 1080 steel to achieve a

hardness of 50 HRC. It is necessary to use Figure 10.35. (a) The time necessary at 425°C is about 650 s. (b) At 315°C, the time required (by extrapolation) is approximately 4 x 106 s (about 50

days). Problem 7* a) Extended heating of pearlite near the eutectoid temperature will cause the plates of

ferrite and cementite to break up into spheres of cementite in a ferrite matrix (see Fig. 10.19). The formation of spheroidite from pearlite is driven by the reduction in interfacial energy between the ferrite and cementite phases. This process results in relatively coarse (≈1-3 µm diameter) cementite particles.

Tempered martensite is formed when martensite is heated to a temperature

sufficiently large to allow carbon to diffuse. If we stay in the α+Fe3C two phase region this will result in the formation of cementite particles in a ferrite matrix. In contrast to spheroidite, the cementite particles in tempered martensite will be much smaller (see Fig. 10.33), in this case 10 times smaller. This is because they arise from the nucleation and growth of new cementite particles from carbon that was in solid solution rather than the breakup of existing cementite plates.

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b) Tempered martensite will have a greater strength because of the finer dispersion of

cementite particles. Referring to Table B.4 on page 891 you can see that the strength of "oil-quenched and tempered (@315°C)" steel is significantly larger than the strength of annealed steel for a number of different steel alloys.

Problem 8 In this problem we will make use of Figures 10.32 and 10.35. Forming a fine pearlite

structure in 1080 steel (0.8 wt.%C) will result in a Brinell hardness of no greater than 300 which is too low (unfortunately, there is no data for bainite here). The data for martensite tempered at 371°C has a value of approximately 475, also too low. The data for martensite indicates a Brinell hardness of over 600, too large.

However, we can make martensite and then temper the martensite to reduce the hardness.

Figure 10.35 indicates that tempering the martensite at 315°C for approximately one hour will result in a steel with a Brinell hardness of approximately 550. To make the martensite in the first place one would likely need to heat above the eutectoid temperature to allow the original microstructure to convert to all austenite before quenching to form martensite.

Problem 9* We are asked to specify a practical heat treatment for a 2014 aluminum alloy that will produce a minimum

tensile strength of 380 MPa, and a minimum ductility of 15%EL. From Figure 11.28(a), the following

heat treating temperatures and time ranges are possible to the give the required tensile strength.

Temperature (°C) Time Range (h)

204 0-12 h

149 0-1000 h

121 0-?

With regard to temperatures and times to give the desired ductility:

Temperature (°C) Time Range (h)

204 <0.2 h

149 <10 h

121 <500 h

From these tabulations, the following may be concluded:

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At 204°C, the time would be roughly 10 minutes which is a bit impractical when considering

heating and cooling times.

At 149°C, the time would be 10 h

Finally, at 121°C, the time range is about 500 h. Heat treatments for such long time periods

are not practical.

Problem 10 a) In the manner of Example Problem 11.1, the equivalent distances and hardnesses tabulated below were determined from Figures 11.15 and 11.18b. Radial Equivalent HRC Position Distance, mm Hardness Surface 7 54 3/4 R 11 50 Midradius 14 45 Center 16 44 The resulting hardness profile is plotted below.

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(b) A 70-mm (2

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34-in.) diameter cylindrical specimen of a 1040 steel alloy that has been

quenched in moderately agitated water.

Solution

In the manner of Example Problem 11.1, the equivalent distances and hardnesses tabulated below were determined from Figures 11.15 and 11.18a. Radial Equivalent HRC Position Distance, mm Hardness Surface 3 48 3/4 R 8 30 Midradius 13 23 Center 15 22 The resulting hardness profile is plotted below.

c) The alloying elements of Ni, Cr and Mo in 8640 steel suppress the diffusive transformation of austenite to pearlite enabling the formation of martensite (and perhaps bainite). This leads to greater hardness throughout the sample although the middle is somewhat softer due to the slower cooling rate. 1040 steel, being a plain carbon steel and lacking the additives to suppress diffusive transformations, forms martensite only close to the surface where the cooling rates are the fastest, and pearlite once one gets below the surface.

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Problem 11* A ninety-millimeter (three and one-half inch) diameter cylindrical steel specimen is to be quenched in moderately agitated water. We are to decide which of eight different steels will have surface and center hardnesses of at least 55 and 40 HRC, respectively. In moderately agitated water, the equivalent distances from the quenched end for a 90 mm diameter bar for surface and center positions are 3 mm (1/8 in.) and 22 mm (7/8 in.), respectively [Figure 11.18a]. The hardnesses at these two positions for the alloys cited are given below. The hardnesses at these two positions for the alloys cited are given below (as determined from Figures 11.15 and 11.16). Surface Center Alloy Hardness (HRC) Hardness (HRC) 1040 50 < 20 5140 56 34 4340 57 53 4140 57 45 8620 42 < 20 8630 51 28 8640 56 38 8660 64 55 Thus, alloys 4340, 4140, and 8660 will satisfy the criteria for both surface hardness (minimum 55 HRC) and center hardness (minimum 40 HRC).