MASC 310

# MASC 310 - Midterm Exam II MASC 310 30 October 2008 Prof...

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Midterm Exam II MASC 310 30 October, 2008 Prof. Nutt CLOSED BOOK, CLOSED NOTES CALCULATORS ALLOWED High: 98% Low: 2% Average 70% MT 2 Distribution 0 2 4 6 8 10 12 14 16 18 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Score __________________________________ Name

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Equations 2 2 eff 1 erf 2 exp 8.31 J/mol 1 ln 1 xo so d o T i T dC JD dx CC D xx x Dt x Dt Q DD RT RK F A ab b a cos CRSS max 12 C cos cos %CW= 100% % RA 100 r o y y o y od o Ic of o t Et kd AA A K Y a A Impurity Diffusion Coefficients ( -Fe transforms to -Fe at 727°C) Material D o (m 2 /s) Q (kJ/mol) C in BCC Fe 2 x 10 -6 64 C in FCC Fe 2 x 10 -5 142
1. Diffusion . a. An FCC iron–carbon alloy initially containing 0.55 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1325 K (1052°C). Under these circumstances the carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere; that is, the carbon concentration at the surface position is maintained essentially at 0 wt% C. (This process of carbon depletion is termed decarburization.) At what position will the carbon concentration be 0.25 wt% after a 10-h treatment? The value of D at 1325 K is 4.3 × 10 –11 m 2 /s . b. A more common process is simple carburization of steel parts. Explain what effect this has on the part, and why it is performed. c. Explain how carburization might affect the fatigue resistance of a part. Solution Determine the position at which the carbon concentration is 0.25 wt% after a 10-h heat treatment at 1325 K when C 0 = 0.55 wt% C. From the solution to the diffusion eqn (Equation 6.5) C x C 0 C s C 0 = 0 . 25 0 . 55 0 0 . 55 = 0.5455 = 1 erf x 2 Dt Thus, erf x 2 Dt = 0.4545 Using data in the table and linear interpolation z erf ( z ) 0.40 0.4284 z 0.4545 0.45 0.4755

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z 0 . 40 0 . 45 0 . 40 = 0 . 4545 0 . 4284 0 . 4755 0 . 4284 And, z = 0.4277 Which means that x 2 Dt = 0.4277 And, finally x = 2(0.4277) Dt = (0.8554) ( 4 . 3 10 11 m 2 / s )( 3 . 6 10 4 s ) = 1.06 10 -3 m = 1.06 mm
2. Mechanical Properties. A cylindrical specimen of a brass alloy 10.0 mm in diameter and 120.0 mm long is pulled in tension with a force of 11,750 N; the force is subsequently released. (a) Compute the final length of the specimen at this time. The tensile stress– strain behavior for this alloy is shown below.

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MASC 310 - Midterm Exam II MASC 310 30 October 2008 Prof...

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