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1 Problem Set 7 Materials101 1.) You wish to develop a gold alloy (mostly gold) that can be precipitation strengthened to provide high strength - high conductivity electrical leads to integrated circuits (pure gold as you may know is rather soft). The high conductivity requirement dictates that the Au matrix in which the precipitates are formed should be as pure Au as possible. The phase diagram of 4 different alloy systems, Au-Ag, Au-Ni, Au-Si and Au-Co, are given below. a) Which of these systems could you heat treat to produce precipitation from a high wt% Au solid solution? Give reasons why you rule out any of these systems. A: I would choose Au-Ni, or Au-Co where there can be significant amounts of the second component dissolved in a Au-rich solid solution but where that solubility decreases with decreasing T. Au-Si won t work because almost no Si can be dissolved in the Au solid solution. Au-Ag doesn t work because Au and Ag are completely soluble in the solid state just as they are in the liquid. There is thus no way to precipitate any particle from the solid solution. b) If there is more than one system in a) that can be heat treated to produce precipitates, which one would you pick to satisfy the high electrical conductivity criterion and why? A: I would Au-Co over Au-Ni, because from the phase diagram the solubility of Co in (Au) is less than the solubility of Ni in (Au) at any aging temperature. Thus after precipitation the (Au) matrix will be purer Au in the case of Au-Co. c) What would be the overall composition of the alloy you would try first? What would be the approximate composition of the precipitate you would expect to produce? A: Au-6 to7wt%co since a maximum of 8% can be dissolved at ºC, the precipitate would be ~100%Co (eco phase) at any temperature below 400 ºC. d) Describe the steps in the process you would use to produce precipitate particles. Give all pertinent information necessary for someone to duplicate your treatment and tell why each step is necessary. A: The first step is to solutionize by holding the Au-Co at about 990 ºC. The next step is to quench the alloy, that is, rapid cooling by putting it in water. Then the alloy is aged by raising the temperature so that the Co can diffuse. First forming a high number density of nuclei during the heating and then aging at a fixed temperature to allow the precipitates to grow and then coarsen until they are of optimum size and spacing (just large enough to resist cutting by dislocations but having the smallest spacing possible for that size. The exact time and temperature of the aging cannot be determined without a TTT curve for the specific composition.

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6 HW#7 Problem 2 a) An alloy knife blade of 14 at.% Be in Ni is held at 700 o C for several months. i) What is (are) the phase (s) present and its (their) composition? i) 14 at % Be in Ni = 86 at % 700 o C and 86 at% Ni in Ni!β and Ni phases are present Compositions: Drawing a tie line at 700 o C and 86 at% Ni between the β and Ni phases, we see that the β phase contains 51 % Ni and the Ni phase contains 95% Ni. ii) If there is more than one phase give the wt% of each phase present (at 700 o C) in the microstructure. ii) To calculate wt%, we can simply use the 2 nd phase diagram for Problem 2, which is as a function of wt% Ni. Drawing a tie line at 86 at % Ni (this is equivalent to 97 wt % Ni-see diagrams) and using the lever rule we obtain: wt % β = ( )/ ( ) = 13.0% wt % Ni = (97-87)/( ) = 87%

7 51 % 95 % 86 at% 97 wt b) Your boss suggests a precipitation hardening heat treatment to produce maximum hardness as follows. i) Heat rapidly from 700 o C to 1170 o C, hold for 1 hr. ii) Quench in ice water iii) Reheat to 1050 o C to age precipitation harden for 5 hours.

8 Do you agree? If not, write a diplomatic < 1 page memo discussing what is wrong with her suggestion and outlining what changes you would make to the heat treatment procedure and why. b) The precipitation hardening procedure that I would recommend would be to 1) heat rapidly from 700 o C to between o C. You want to heat above the solvus line at T~ 1080 o C but below the eutectic temperature of T=1150 o C. Above 1150 o C melting from the nonequilibrium microconstituent could occur. Therefore, we want to heat at a temperature between o C. At this point, the alloy would only contain Ni in its structure, which we call a solid solution. 2) I would then quench or cool rapidly which would prevent the atoms from diffusing to potential nucleation sites and therefore the β phase would not form. At this point the structure still only contains Ni, which is now a supersaturated solid solution containing excess Ni. This is not an equilibrium structure. 3) The last step would be to age or precipitate harden at an appropriate temperature and time to form finely dispersed β precipitates. The yield strength of the alloy is dependent on the aging temperature and time. From lecture, we saw that a copper alloy exhibited a maximum yield stress after aging for ~ 1-2 hours. For higher aging temperatures, shorter times are needed because diffusion is rapid and the precipitates form quicker. Thus, to age harden at 1050 o C for 5 hours as suggested in the problem is too excessive. I would recommend an aging time between 200 and 460 o C for ~1-2 hours.

9 Problem Set 7 3.) The compositions and isothermal transformation diagrams for two steels are given below. Steel A wt.% C, 0.29 wt.% Mn, balance Fe Steel B wt.% C, 0.56 wt.% Mn, 3.53 wt.% Ni, 0.74 wt.% Mo, balance Fe The isothermal transformation diagrams for the two steels are shown below. You may also want to consult the phase diagram for the Fe-C alloy system, which you can find in the lecture viewgraphs for lecture 15 or 16. (A word to the wise: I expect you to memorize the part of this phase diagram that is important for the heat treatment of steels and I will feel free to give you future problems on homeworks, quizzes and the final exam that require you to supply this information.)

10 a) How would you measure the hardenability of each steel? b) Which steel will have the highest hardenability? Why? c) Thin razor blades of each steel are heated to 950 C and then quenched in ice water. Describe the microstructure of each blade? Which blade will have the highest hardness? Why? d) How would you produce spheroidite (a morphology that consists of spheres of Fe 3 C in? - Fe) starting from the samples in c)? Which sample would have the higher percentage of cementite, if that phase forms?

11 A: a) Hardenability is the ability of an alloy to be hardened by the formation of martensite as a result of a given heat treatment. To test this, one should use the Jominy end quench test, which involves cooling a bar of steel from one end, and then measuring the hardness as a function of distance from the cooled end. b) Steel B should have the highest hardenability because the austenite to pearlite transformation is pushed back relative to Steel A, so that transformation can be more easily avoided. The result is that more martensite will be formed further from the quenched end in Steel B. c) Quenching in ice water produces very rapid quenching rates, fast enough ot avoid the austenite to F+C as it goes through this area of the TTT diagram. Thus the austenite of both steels will transform to martensite with a lenticular, plate-like morphology but the martensite of steel A will have a higher C content and will thus have a higher yield stress and hardness due to the strong interaction between the elastic dipoles formed by the C interstitial atoms in martensite and dislocations. Remember what makes martensite hard is carbon but whether martensite will form at all depends on the hardenability of the steel, which may be changed by changing the concentration of certain substitutional alloying elements. d) Spheroidite can be produced by heating to just below the eutectoid temperature (~ 600 C) and holding for a long time (~20 hrs). This allows the carbon to diffuse and precipitate from the martensite phase as the Fe 3 C or cementite phase. Once all the C is gone from the martensite, it is bcc Fe or ferrite. The Fe 3 C phase forms as sphere-like particles to reduce the high energy boundary area between phases. In this case, Steel A would have the higher percentage of cementite due to its higher carbon content.

12 Yes, martensite will form. The martensite will be found near the interface between the weld and the metal plate along the length of the weld where the metal plate has melted. The martensite that forms comes from the 0.8 wt% C metal plate of moderate harden-ability. The medal welding rod, of low hardenability, that is melted into the groove to form the weld will not form martensite The cooling rate at this interface should be greater than the 200 o C/s needed to miss the C- shape of the Isothermal Transformation Diagram to the left and reach the martensite formation temprature. Thus the melted portion of the eutectoid steel along the weld bead will be quenched into the martensite phase. The welding rod has a low carbon content (it has low hardenability). It is safe to assume that the cooling rate to room temperature will not be fast enough to quench the steel in the martensite phase. The cooling rate needed to quench pure steel into the martensite phase is about 10 3o C/s. The region further away from the weld is not heated to a temperature above the eutectoid temperature and does not form austenite. Because no austenite forms, martensite can not form here.

13 Question 4 Two batches of aluminum (99% pure) wire are produced from inch diameter cylindrical rod stock by drawing through a die at 25 o C. The rod stock has been annealed prior to drawing. Batch A was uniformly drawn to a diameter of inch whereas batch B was uniformly drawn to a diameter of inch. The yield stress σ y of batch A wire was 69 MPa, and σ y of batch B wire was 138 MPa, whereas σ y of the original rod stock was 55 MPa. The magnitude of the dislocation Burgers vector in Al is 0.29 nm and the shear modulus equals 28 GPa. a) (i) Estimate the dislocation density in the wire from each batch if the initial dislocation density of the rod stock was /m 3. Ans: ρ A = 7.4*10 13 m -3 and ρ B = 2.61*10 15 m -3 The initial dislocation density is ρ = /m 3. Now, we use the equation: σ o = σ y,i - α G b (ρ) 1/2 where we assume α to be 0.2. Using values provided, we calculate for σ o. σ o ~ 55 MPa Notice that the annealing has little effect on the yield strength of the material. We then solve for the new dislocation density for the two drawn samples: ρ A = [(σ t,a - σ o ) / (α G b)] 2 = [(69-55)/(0.2*(28E3)*(0.29E-9)] 2 = 7.4*10 13 m -3 5

14 ρ B = [(σ t,b - σ o ) / (α G b)] 2 = [(138-55)/(0.2*(28E3)*(0.29E-9)] 2 = 2.6*10 15 m -3 (ii) What would the dislocation arrangement in batch B wire look like under the electron microscope? (You may draw a picture). Comment on any significant features of this structure. Ans: see micrograph from lecture (slide 3, lecture 18) Because the dislocation density of sample B is extremely high, we expect the dislocations to arrange in a cell-like structure as shown in the micrograph above. b) Specimens from batches A and B are heated to 250 o C along with a piece of the original rod stock. The recrystallization temperature of Al is ~250 o C and the critical true strain is (i) Plot the changes you would observe in σ y (measure at 25 o C) as a function of time held at 250 o C for each of the specimens. Ans: True strain = ln (A o /A) = ln(d o 2 /d 2 ) = 2ln(d o /d) where A is the cross-sectional area of the rod and d is its diameter. The true strains of the samples are: ε original = 0 ε A = 2 ln(0.1/0.08) ~ 0.45 ε B = 2 ln(0.1/0.01) ~ 4.6 The original sample, having been annealed, is beneath the critical true strain. Therefore no recrystallization takes place. Samples A and B are strained above the critical value and we would expect recrystallization to take place. This means that in both cases, the dislocation density decreases as a function of time at 250 o C as new, dislocation-free grains nucleate and grow (The actual rate of recrystallization for the more highly strained sample B is faster than that of sample A so the yield stress of sample B should decrease before sample A the reverse of what is shown in the figure). When the microstructure of the two samples begin to look essentially like that of the original, we obtain our original yield stress (see plot).

15 yield stress, σ y original sample B sample A Annealing time

16 (ii) After holding at 250 C for 10 hours will there be any differences in the grain size of the two specimens of batch A and B? If you claim that differences will exist, explain why. Ans: Yes. Since regions of high dislocation density are starting points for the recrystallization process, we would expect the sample with higher ρ to initiate more new grains. Therefore sample B will have smaller grain size than sample A, which recrystallized with fewer nucleation centers. Fewer nucleation centers would result in larger grain size, as would be the case for sample A.

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