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Unformatted text preview: ME541: Fatigue of Materials Homework 6 Solutions 5.3) For a steel component that has a 1000 cycle fatigue strength of 90 ksi and an endurance limit (1,000,000 cycle) of 50 ksi. a) Using Basquin equation, N f at 60 ksi and 80 ksi are 117,341 and 3,992 respectively for b =  0.085 and C = 2.209 (N = 10C/b N 1/b ) b) Using PalmgrenMiner rule, ∑ = + = = 677 . 117341 50000 3992 1000 fj j N n D c) The curves are shifted by first subtracting n 1 from N f1 for S 1 , then calculating the new N f2 for S 2 from the constant slope, b, such that b = (logS 2  log S 1 )/ (log X  logN f1shift ) where X is N f2shift for S 2 after the shift. The final curve is then determined by subtracting n 2 from N f2shift for S 2 , then calculating the new N f1final for S 1 from the constant slope, b, such that b = (log S 2  log S 1 ) / (log N f2final  log X) where X is N f1final for S 1 after the second shift. Using this method, the order does not seem to be important. d) At 70 ksi, N f = 19172. Solving for n at 70 ksi after damage to this point of 0.677 gives n = 6200 cycles. 5.8) Use rainflow analysis to simplify the following loading history. Rain flow analysis from A to S=A The outside loop is the biggest (AF). Internal loops are BC, DE, GH, IJ, LM, QP, OR and possibly KN. The range and mean values for each loop are given Loop Peak Valley Range Mean AF 23 23 46 0 BC 15 10 5 12.5 DE 5 15 10 10 GH 10 20 10 15 IJ 10 5 5 7.5 LM 20 10 10 15 QP 2.5 7.5 5  5 OR 15 10 25 2.5 MN 20 15 35 2.5 2....
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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.
 Spring '10
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