Consider 3 kg. of austenite containing 0.50 wt. % C, cooled to below 727 degree

Question

Consider 3 kg. of austenite containing 0.50 wt. % C, cooled to below 727 degree C What is the proeutectoid phase? How many kilograms each of total ferrite and cementite form? How many kilograms each of pearlite and the proeutectoid phase form? Schematically sketch and label the resulting microstructure. Proeutevtiod Phase alpha + Fec MFc C = 30 9 kg m alpha = 691 kg alpha 0.022 gamma = 0.74 The critical resolved shear stress for copper is 0.41 MPa (70 psi). Determine the maximum possible yield strength for a single crystal of Cu pulled in tension. Table 7.1 Slip Systems for Face-Centered Cubic Body-Centered cubic, and mexagonal Close-Packed Metals TR = 0.48 M pal (70 psi) Plane {1,1,3} Direction lamda [u, v, w] = (110) [u2, v2 q2] = [111] lamda = cos -1[(1)(1) + (1)(1) + (0)(1)]/ lamda = cos -1 2/ lamda = 24.6 phi = [u, v, w]=[1 1 0][u2 v2 w2] = [111] phi = cos-1[(1)(1) + (1)(1) + (0)(1)/ ] phi = cos -12/ 24.4 degree A 6.4 mm (.25 in) diameter cylindrical rod fabricated form a 2014 -T6 aluminum alloy is subjected to reserved tension compression load cycling along its axis. If the maximum tensile and compressive loads are +5340 N (+1200 lbs) and -5340 N (-1200 lbs) respectively, determine its fatigue life. Assume that the stress plotted in figure is stress amplitude. A 6.4 mm (.25 in) diameter cylindrical rod fabricated form a 2014 -T6 aluminum alloy is subjected to reserved tension compression load cycling along its axis. If the maximum tensile and compressive loads are +5340 N (+1200 lbs) and -5340 N (-1200 lbs) respectively, determine its fatigue life. Assume that the stress plotted in figure is stress amplitude. sigma a = sigma r/2 = sigma max-sigma min/2 = 5340(1200 lbs)-(5340)(-1200)/2 sigma 1 = 10680 (2400 lbs)/2 [sigma a = 5340(1200 lbs)].

Explanation / Answer

1) Ferrite is the proeutectoid phase since 0.5 wt% C is less than 0.76 wt% C 2)Using lever rule we get 0.92*1.5kg=1.39 kg of ferrite will form 0.08*1.5kg=0.12kg of cementite will form 3) Now consider the amounts of pearlite and proeutectoid ferrite. The fraction of pearlite is 0.64 This corresponds to (0.64)(1.5 kg) = 0.97 kg of pearlite. Thus, amount of proeutectoid ferrite is given as: Wa' = 1 – WP = 0.36 or, (0.36)(1.5 kg) = 0.54 kg of proeutectoid ferrite

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