(12) 1. Just one True and False question and a couple of multiple choice calculations, circle one answer for each problem, no partial credit.

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1 (1) 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 [11]. Of the four slip systems listed below, choose the system along which you would expect slip should initiate. i) ( 111) [ 101] ii) 111 ( )! 110 " ( )! 011 # $ iii) 111 " ( ) 101 # $ iv) 111 [ ] 4 Schmid factors for i) and ii) are zero, Schmid factors for iii) and iv) are 18 1 and respectively. Since slip should initiate along the slip systems with the highest Schmid factor (largest resolved shear stress), iv) is the correct choice. (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 K 1c =76 MPa*m 0.5, σ 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 =.36 mm iii) a = 0.53 mm iv) a = 0.15 mm a c 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. K = σ πa, πa c = K 1c σ y, a c = 1 # K & 1c π % $ σ (, a c = 1 # 76 MPa m % y ' π $ 1310 MPa & ( ' =1.07 mm () c) Circle either TRUE or FALSE The charge imbalance created when Zr 4+ ions substitute TRUE FALSE for Ca + ions in CaO result in the formation of Ca + vacancies. Each Zr 4+ ion substituting for a Ca + ion in CaO results in two (+) charges too many for charge balance. To regain charge balance a Ca + vacancy must be formed.

2 (10). Recrystallization Before deformation, 3003 Aluminum (an aluminum alloy containing Al with small additions of manganese) has a yield strength of 40 MPa (σ 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 0% cold work resulting in a yield stress of 130 MPa. Sheet 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$ 130$ Sheet%% Sheet%1% The general shape of the curves is the same as found in Callister Figure 7.. 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 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: D 0 =6.x10-7 m /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. 3 " D = D o exp Q % $ ' D = m # RT & s exp * 80, 000 J mole, +, J mole K ( )( 103K) - /./ = m s (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). C x C o # x & =1 erf% ( C s C o $ Dt ' =1 erf " x % $ ' # Dt &! x $ = erf # & " Dt % if erf(z)=0.7166, z= by interpolation (closest round value is 0.75).! x $ # & = Dt = " Dt % x ( ) t = 1 ' x * ), D( ) ( ) +, 1 t = m s ' ) () m ( ) *, +, t = 4790 seconds or 490 seconds if you did not interpolate. Part c) is on the next page

4 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.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 5 (10) 4. Phases and Phase Diagrams I Please answer the following questions related to the Mg-Pb phase diagram below. C α " C 0 # C L # () a) At 300 C, the maximum solubility of lead (Pb) in magnesium (Mg) is approximately 17 wt. % lead. () 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 Phase(s) present in a 0 wt.% Pb- 80 wt.% Mg alloy Composition of each Phase Weight Fraction of each Phase 600 C α + L C α =1 wt.%pb C L = 35 wt.%pb W α = = 0.65 W L = = 0.35

6 (13) 5. Phases and Phase Diagrams II Refer to the phase diagrams on the next page to answer the questions below. (5) a) Below is part of a x-ray diffraction (XRD) pattern showing the positions of the (111) peaks for pure copper (Cu) and pure nickel (Ni). A mixture of 50 grams of Cu and 50 grams of Ni is heated to 1400 C and allowed to melt and mix completely. The mixture is subsequently cooled slowly to 1100 C. Directly on the XRD pattern sketch how you expect the pattern to appear when the alloy reaches 1100 C. If you expect no change in the pattern please state this. Cu(111) Ni(111) 6 I Θ 45 Due to their complete solid solubility the copper and the nickel will substitute for one another resulting in a single phase solid solution. This will result in a single x-ray diffraction peak representing a single lattice parameter. (8) b) Two alloys, one containing 10 wt.% Pb and 90 wt.% Mg (Alloy 1), and the other containing 70 wt.% Pb and 30 wt.% Mg (Alloy ) are cooled slowly from the liquid and held at 00 C. In the circles provided, make a qualitatively correct sketch of their expected microstructures, clearly labeling the phases present. Alloy 1 goes through the single phase α region before entering the two phase α and Mg Pb region. Therefore, there is no eutectic reaction (a horizontal line is not crossed) so there is no lamellar structure, but rather Mg Pb precipitates in a matrix of the α phase. Alloys crosses the horizontal line during cooling so one expects there to be a lamellar structure consisting of alternating layers of α and Mg Pb. Drawing a tie line just above the eutectic temperature suggests that most of the alloy is liquid with some primary Mg Pb (makes sense since we are near the eutectic composition). Upon cooling below the eutectic temperature the remaining liquid will undergo the eutectic reaction resulting in lamellae of α and Mg Pb.

7 7 Alloy%1% Alloy%%