Problem Set 6 Due: Tuesday, March 6, 11:00 AM 1. TTT Diagrams Based on the transformation diagrams for eutectoid steel shown below, what microstructure would result from the following cooling histories? Assume the steel starts above the eutectoid temperature. Distiguish between coarse and fine pearlite when applicable. a) Rapidly cooled to 600 C, held for 7 seconds. Then rapidly cooled to 450 C and held for an additional 10 seconds. Quenched to room temperature. (2 pts) 80% Fine Pearlite, 10% Martensite, 10% Bainite. b) Rapidly cooled to 600 C, held for 1 minute, then rapidly cooled to 450 C and held for 10 seconds. Quenched to room temperature. (2 pts) 100% Fine Pearlite. Once this transformation has occurred, no further transformation will.. c) Rapidly cooled to 600 C, held for 7 seconds, then rapidly cooled to 170 C and held for 1 hour. Quenched to room temperature. (2 pts) 80% Fine Pearlite, 20% Martensite d) Rapidly cooled to 700 C, held for an hour and a half, then rapidly cooled to 570 C and held for 10 seconds. Quenched to room temperature. (2 pts) 50% Coarse Pearlite, 50% Fine Pearlite e) Cooled at a rate of 10 C/s. (2 pts) 100% Pearlite
f) Cooled at a rate 125 C/s. (2 pts) About 85% Martensite, 15% Pearlite. g) For part (f) draw what the microstructure might look like. (8 pts) 2. Lever Rule Please refer to the Cu-Ag phase diagram on the following page to answer the following questions. a) Label all phase fields present below 32 atomic percent Cu. (5 pts) b) Determine the phase composition at 500 o C in weight percent for an overall composition of C 0 =20at%Cu. (5 pts) The tie line is illustrated in the figure bellow. θ phase is 48%Al-52%Cu Al phase is 96%Al-4%Cu
Al+L θ+l θ+al A c) Determine the weight percent of each phase at 500 o C for an overall composition of C 0 =20at%Cu. (5 pts) Using weight percent Cu W θ =(C 0 -C Al )/(C θ -C Al )=(38%-4%)/(52%-4%)=71% W Al =100- W θ =29% d) Calculate the overall weight percent of Aluminum for C 0 =20 at%cu at 500 o C using parts b and c, and check to make sure this agrees with the phase diagram. (5 pts) Overall %Al = W θ (%Al in phase composition of θ)+ W Al (%Al in phase composition of Al) Overall %Al = (.71)(.48)+(.96)(.29)=.34+.28=.62 or 62% Al. The phase diagram shows 38% Cu. e) Identify as many invariant points as you can! (extra credit!)
3. Nucleation and Critical Radius The eutectoid temperature of iron/carbon alloy is 1000 K. This is the eutectoid between austenite, and pearlite (cementite and ferrite.) Assume that the latent heat of fusion and surface free energy for pearlite in the austenite phase are and, respectively. a) Compute the critical radius of nucleation for pearlite in austenite at the following temperatures of austenite: 950K, 850K, and 750K. State your answer in nanometers. (7 pts) b) Compute the free energy of activation for pearlite in the austenite phase at the same temperature undercoolings as above. (7 pts) c) From your answers in parts a and b, sketch (qualitatively) schematics of the pearlite nucleation process (one at each temperature stated above). Your drawing does not need to be up to scale as long as it captures the average number of stable pearlite nuclei and their average critical radius at all three temperatures relative to one another. Recall that the number of stable nuclei is exponentially proportional to the term (equation 10.8). Since this is the beginning of the nucleation process, the pearlite regions may be represented as spheres instead of the equilibrium lamella structure. Be sure to label the different phases in your drawings. (6 pts)
4. Phase Diagram Fun Please refer to the X-Y phase diagram below to answer the following questions. a) Label all regions on the phase diagram. Notice the two line compounds already labeled. Distinguish the single-phase regions with α, β, γ and L and the two-phase regions between them as appropriate (7 pts) b) Use the Gibb s phase rule to solve for the degree(s) of freedom in the two phase, single line compound phase and single phase regions. (4 pts) Gibb s Phase Rule F = C - P + N P is # of phases, F is degrees of freedom, C is # of components (which can be elements, or stable compounds), N is the number of non-compositional variables. Since this phase diagram is at a fixed pressure, the only non-compositional variable is temperature. Single phase field (eg. α) C=2 (Ag + Mg = 2 components), P=1, N=1 F = 2-1+1 = 2 Two phase field (eg. α + β) C=2 (Ag + Mg = 2 components), P=2, N=1 F = 2-2+1 = 1 Single phase line compound (eg. Ag 3 Mg) C=1, P=1, N=1 F=1-1+1=1
c) Label all invariant points on the phase diagram (eg. eutectics, peritectics, eutectiods, peritectiods, and congruent transformations). (7 pts) Eutectic- start with liquid, end with 2 solid phases. (eg. L α + β ) Eutectoid- start with 1 solid phase, end with 2 solid phases (eg. α β + Ag 3 Mg ) Congruent transformation- start 1 liquid or solid phase, end 1 solid phase without a change in composition. (eg. L β) d) Find the number of degrees of freedom for an eutectic point. (2 pts) Eutectic point C=2, P=3, N=1 F = 2-3 + 1 = 0 5. Making a Medieval Sword You want to reproduce an authentic Medieval European sword. In order for the sword to maintain a sharp edge the steel used must have sufficient hardness. However, increasing hardness also will increase brittleness. A brittle sword will fracture under the heavy stresses of battle. a) You decide to use AISI 1060 steel. What is the composition of this steel? (4 pts) The first two digitis (10) indicate that this is a plain carbon steel, so it contains predominantly iron with carbon, a little manganese and only residual concentrations of other elements. The carbon concentration is given by the second two digits (60) as one hundred times the weight percent of carbon, so this steel contains 0.60 wt% C. b) How would the mechanical properties of the material be different if you used 1040 steel and 1080 steel as compared to the 1060 steel? (2 pts) The tensile strength, yield strength and hardness increase with carbon concentration while the ductility decreases. 1040 steel would be of lower strength and hardness and of higher ductility compared to the 1060. The 1080 steel would have increased strength and hardness while being more brittle (lower ductility). c) After forging the blade with a hammer and anvil into the desired shape, you perform a normalizing process. Briefly describe the grain structure of the sword before and after the process of normalizing. (4 pts) After the steel is shaped via cold working, it contains large irregularly shaped grains which vary significantly in size. The normalizing process decreases the grain size while making them more regular in shape with a smaller distribution of grain sizes. d) In order to add finishing details to the sword you would like to make it softer and more ductile. This allows for easier grinding of the blade edge as well as any decoration you may wish to engrave in the steel. What heat treatment process will result in a coarse pearlite grain structure which is soft and ductile? How is the cooling of this process different than that of the normalizing process? (4 pts)
The process described is a Full Annealing of the steel. The cooling step of a full anneal is to turn off the furnace and let the furnace and steel cool together slowly. The cooling step of the normalizing process is performed faster by removing the steel from the furnace and letting it cool in air. e) Finally, you have your sword shaped and decorated as desired. However, it is now too soft to hold a sharp edge. The steel must be tempered to the desired hardness. This process converts some of the microstructure to what microconstituent? (2 pts) The microconstituent formed when temper hardening the steel is martensite. The temper process makes the steel more martensitic. f) When performing the cooling step of the tempering process, sword makers have used material such as clay to cover portions of the blade to slow the cooling rate of the covered steel. Why might this technique be beneficial for making a sword which holds a sharp edge while remaining flexible enough to withstand the forces subjected to it in battle? (4 pts) When tempering the steel, a faster cooling rate will have the result of making the steel more martensitic which means that it will be harder and more brittle. By cooling selective areas more slowly such as the middle of the blade running the length of the sword, those areas will be more ductile and less brittle. This will result with the sword having harder edges and more durable core. The blade will be more flexible while maintaining hard sharp edges. This happens somewhat naturally without using a material to alter the cooling rate due to the cooling rate profile throughout the volume of the steel (i.e. the middle of the steel will cool more slowly than the outer areas).