exam2fall2000 - ENSC 3313,Fall 2000 TEST 2 Name The width =...

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ENSC 3313, Fall 2000 TEST 2 Name The width = 1 ", the thickness = 0.25". Therefore the cross sectional area = 0.252 in. Hence, load x 4 = Stress. The gauge length = 4"; hence strain = elongation/4. a) E = 28 ooo/~ -11.2 x 106 psi , 4 b) YS -36.000 psi c) TS -56.000 psi d) % Elongation -(0.092/4)100, -~ 2. Please sketch a simple tetragonal crystal cell (c < a = b, a = [3 = y = 900) with atoms at each comer. Please determine a probable slip plane and a probable slip direction within the slip plane. The slip planes will have a high atomic density. By inspection, the plane with the greatest planar density is {010} . ..or { 100}. The slip directions will have the greatest linear density. Again, by inspection [001]. 3. A 3.25 w/o Si steel was cold worked 60%. The 50% recrystallization time was found to be 1.0 sec at 800°C and 2000 sec at 600°C. Please find the temperature corresponding to a 50% recrystallization time of 100 sec. Using the log form of the exponential equation, for illustration * In(tso)=B+% '1={ T~ T:O i rf 1 I= "l JOn IJ~ -+ c = 35,603 K J 1 -+ T = 669°C 73J 1 873 *~: Can write instead Rate = .!. = Ae-Q/RT t
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4. Hardness and Strength mandate <-22% CW ~ ~ ~ 'bi! c f! .. w 4) -= ..§ 5 Ductility mandates >-18% CW Choose ~ CW!
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This note was uploaded on 01/10/2012 for the course ENSC 3313 taught by Professor Staff during the Fall '11 term at Oklahoma State.

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exam2fall2000 - ENSC 3313,Fall 2000 TEST 2 Name The width =...

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