MASC 310

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

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 2
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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 4
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.
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 13

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online