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Chap 08 Solns-6E - Taking the natural logarithm of...

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CHAPTER 8 FAILURE PROBLEM SOLUTIONS 8.31 (a) The fatigue data for this alloy are plotted below. (b) The fatigue limit is the stress level at which the curve becomes horizontal, which is 290 MPa (42,200 psi). (c) From the plot, the fatigue lifetimes at a stress amplitude of 415 MPa (60,000 psi) is about 50,000 cycles (log N = 4.7). At 275 MPa (40,000 psi) the fatigue lifetime is essentially an infinite number of cycles since this stress amplitude is below the fatigue limit. (d) Also from the plot, the fatigue strengths at 2 x 10 4 cycles (log N = 4.30) and 6 x 10 5 cycles (log N = 5.78) are 440 MPa (64,000 psi) and 325 MPa (47,500 psi), respectively. 8.50 This problem gives Ý ε s values at two different temperatures and 140 MPa (20,000 psi), and the stress exponent n = 8.5, and asks that we determine the steady-state creep rate at a stress of 83
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Unformatted text preview: Taking the natural logarithm of Equation (8.20) yields 196 ln Ý ε s = ln K 2 + n ln σ -Q c RT With the given data there are two unknowns in this equation--namely K 2 and Q c . Using the data provided in the problem we can set up two independent equations as follows: ln 6.6 x 10-4 (h)-1 [ ] = ln K 2 + (8.5) ln(140 MPa)-Q c (8.31 J /mol - K)(1090 K) ln 8.8 x 10-2 (h)-1 [ ] = ln K 2 + (8.5) ln(140 MPa)-Q c (8.31 J /mol - K)(1200 K) Now, solving simultaneously for K 2 and Q c leads to K 2 = 57.5 (h)-1 and Q c = 483,500 J/mol. Thus, it is now possible to solve for Ý ε s at 83 MPa and 1300 K using Equation (8.20) as Ý ε s = K 2 σ n exp -Q c RT = 57.5 (h)-1 [ ] (83 MPa) 8.5 exp-483,500 J /mol (8.31 J/mol - K)(1300 K) 4.31 x 10-2 (h)-1 197...
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