12 1 Just one True and False question and a couple of multiple choice calculations circle one answer for each problem no partial credit The next page is left blank for your use but no partial will be given for anything written there 5 a A single crystal of a FCC metal is loaded in tension along the 112 Of the four slip systems listed below choose the system along which you would expect slip should initiate i 11 1 101 ii 111 1 10 iii 111 0 11 iv 111 101 4 6 and 18 12 18 12 respectively Since slip should initiate along the slip systems with the highest Schmid factor largest resolved shear stress iv is the correct choice Schmid factors for i and ii are zero Schmid factors for iii and iv are 5 b Aircraft components are often made from alloy 17 7 PH 70 wt Fe 17 wt Cr 7 wt Ni along with smaller amounts of C Al and Mn Alloy 17 7 PH has a fracture toughness and yield stress of K1c 76 MPa m0 5 y 1310 MPa respectively When a component of this material is subject to tensile loading what is the flaw size a above which the component is expected fail by fracture and below which the component is expected to exhibit plastic deformation For your calculation assume Y 1 and that any flaws are oriented perpendicular to the tensile loading direction i a 1 07 mm ii a 2 36 mm iii a 0 53 mm iv a 0 15 mm ac refers to the critical crack size above which the component is expected fail by fracture and below which the component is expected to exhibit plastic deformation 2 K a 2 K1c 1 K1c 1 76 MPa m ac a c a c 1 07 mm y y 1310 MPa 2 c The charge imbalance created when Zr4 ions substitute for Ca2 ions in CaO result in the formation of Ca2 vacancies Circle either TRUE or FALSE TRUE FALSE Each Zr4 ion substituting for a Ca2 ion in CaO results in two charges too many for charge balance To regain charge balance a Ca2 vacancy must be formed 2 10 2 Recrystallization Before deformation 3003 Aluminum an aluminum alloy containing Al with small additions of manganese has a yield strength of 40 MPa y 40 MPa You begin with two identical thick sheets of 3003 aluminum and each sheet is reduced in thickness using a rolling mill operating at room temperature Sheet 1 is reduced in thickness such that it experiences 20 cold work resulting in a yield stress of 130 MPa Sheet 2 is reduced in thickness such that it experiences 50 cold work resulting in a yield stress of 175 MPa After deformation the two sheets are heated from room temperature to above their recrystallization temperature For each sheet make a plot in the space below showing how you expect the yield stress of the material to change during heating from room temperature to above the temperature at which recrystallization is complete The beginning and ending values for yield stress should be quantitatively correct The shape of the graph in between the beginning and ending values should be qualitatively correct Clearly label which plot belongs to which sheet 175 Sheet 2 130 Sheet 1 The general shape of the curves is the same as found in Callister Figure 7 22 Both materials start at the yield stress associated with their amount of cold work and then drop down to their annealed yield stress of 40 MPa as they are heated past their respective recrystallization temperatures At lower temperatures the yield stress remains at the deformed value because at room temperature and lower temperatures diffusion is very slow and the atoms do not rearrange despite the driving force provided by the cold work Sheet 2 begins to recrystallize at a lower temperature because it has a greater amount of cold work and therefore has a greater driving force to recrystallize compared to Sheet 1 3 15 3 Diffusion A gear made from 1040 steel iron with 0 40 wt carbon is caburized to harden the gear surface Given that carburization is performed at 750 C please answer the following questions For carbon diffusing through BCC iron D0 6 2x10 7 m2 s Q 80 kJ mole R 8 314 J mole K 3 a Calculate the diffusivity of carbon through iron at 750 C Make sure your answer has appropriate units and please show your work Q D Do exp RT D 6 2 10 7 m 2 2 80 000 J mole exp 5 1 10 11 m s s 8 314 J mole K 1023K 8 b Given a surface concentration of 1 0 wt C calculate the time in seconds required for carburization to result in a carbon concentration of 0 57 wt C at a depth of 0 75 mm below the gear surface Please show your work use the top of the next page if you need more room x Cx Co 1 erf Cs Co 2 Dt x 0 57 0 4 1 erf 2 Dt 1 0 0 4 x 0 7166 erf 2 Dt if erf z 0 7166 z 0 7587 by interpolation closest round value is 0 75 x 0 7587 2 Dt x Dt 2 0 7587 2 1 x t D 2 0 7587 2 7 5 10 4 m t 2 5 1 10 11 m 2 0 7587 s 1 t 4790 seconds or 4902 seconds if you did not interpolate Part c is on the next page 4 4 c The plot below represents the approximate carbon concentration profile in the gear after the completion of carburization If the carburizing gas was turned off such that no new carbon entered the gear but the elevated temperature remained such that carbon diffusion within the gear continued sketch a qualitatively correct concentration profile after heating for approximately one hour after the gas was turned off Assume that none of the carbon leaves the sample and that after approximately one hour the carbon concentration equals 0 40 wt at a depth of 2 5 mm Since the gas is turned off and no new carbon enters the gear the total amount of carbon in the gear will not change with increasing time However since the temperature is still elevated the carbon will continue to diffuse through the gear with the concentration gradient decreasing as Fick s first law says it should The area under the curve representing the amount of carbon in the gear should stay roughly the same over time 5 10 4 Phases and Phase Diagrams I Please answer the following questions related to the Mg Pb phase diagram below C C0 CL 2 a At 300 C the maximum solubility of lead Pb in magnesium Mg is approximately 17 wt lead 2 b The liquidus temperature of an alloy containing 50 wt Pb and 50 wt Mg is approximately 565 C 6 c Please complete the table below Temperature 600 C Phase s present in a 20 wt Pb 80 wt …
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