# hmwk2b - ME541 Fatigue of Materials Homework 2 Solutions...

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ME541 Fatigue of Materials Homework 2 Solutions 1.1) This is an application of the modified Basquin Equation shown in Eq. 1.3 such that S = 10 c N b between N = 10 3 and N = 10 6 cycles. Assuming S 1000 at N = 1,000 cycles is 0.9 S u and S e at N=1,000,000 is 0.5 S u (for S u < 200 ksi) or 100 ksi (for S u >200 ksi) and solving for C and b leads to Eq 1.7 such that S N = 1.62S u (N) _0085 for steels with S u < 200 ksi and S N =10 c N b for steels with S u >200 ksi. For S u = 100 ksi N S N (ksi) 10 3 90 10 4 74 10 5 61 10 6 50 For S u = 220 ksi N S N (ksi) 10 3 198 10 4 158 10 5 126 10 6 100 1.4) Eq 1.7 is S N = 1.62S u (N) _0.085 for steels assuming S 1000 at N=1,000 cycles is 0.9 S u and S e at N=1,000,000 is 0.5 S u . Material A, S u = 42 ksi Exp N Exp S (ksi) Curve Fit A (ksi) Curve Fit b Basquin A = 1.62 S u (ksi) Basquin b 4.5 x 10 4 32.3 61.27 -0.058 68.04 -0.085 2.4 x 10 5 30.3 8.0 x 10 5 27.9 1.5 x 10 6 25.9 2.7 x 10 6 25.4 7.8 x 10 6 24.4 There is general agreement between the curve fit and the Basquin Equation with the assumptions for steel. Note however that the slope, n is less for the curve fit as is the intercept, A.

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Material B, S u =102.6 ksi Exp N Exp S (ksi) Curve Fit A (ksi) Curve Fit b Basquin A = 1.62 S u (ksi) Basquin b 4.4 x 10 4 81.4 175.26 -0.075 166.21 -0.085 8.5 x 10 4 74.7 1.4 x 10 5 71.6 6.3 x 10
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## This note was uploaded on 11/28/2010 for the course ENES enes100 taught by Professor Staff during the Spring '10 term at Maryland.

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hmwk2b - ME541 Fatigue of Materials Homework 2 Solutions...

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