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Unformatted text preview: MechEng 382 Winter 2010 Homework #10 Solutions 1. Dowling 9.4 a Mpa N f Cycles log(a) log(N f ) 524 257 2.72 2.41 459 1494 2.66 3.17 410 6749 2.61 3.83 352 19090 2.55 4.28 315 36930 2.50 4.57 270 321500 2.43 5.51 241 2451000 2.38 6.39 (a) Taking two points from the graph, (N f , a ) 1 = (257, 524 MPa) (N f , a ) 2 = (2451000, 241 MPa) 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 2.72 2.66 2.61 2.55 2.50 2.43 2.38 loglog plot log(Nf) MechEng 382 Winter 2010 Homework#10 Solutions 2 Using Basquins law, [Page20 Data Handbook] ? 1 ? 2 = ? ? 1 ? ? 2 = ? ? 1 ? ? 2 ?? ? 1 ?? ? 2 = ? 524 ? 241 ? 257 ? 2451000 = 0.0848 = ? 1 ? ? 1 = 524 257 0.0848 = 839 ? (b) Rearranging the equation in to the linear form y = mx + c ? = ? ? ?? ? = 1 ? ? ? Fitting the data from Table P9.4 to this model, we get m = 10.916 = 1 ? = 1 10.916 = 0.0916 c = 74.02 = exp = exp 0.0916 74.02 = 881 ? b = B =0.0916 ? = 2 = 881 2 0.0916 = 939 ? MechEng 382 Winter 2010 Homework#10 Solutions 3 2. Dowling 9.10 (a) Safety factor in life = 3 Safety factor in stress can be found using Basquins law again: ? 1 ? 2 = ? ? 1 ? ? 2 Or ? 1 500 = ( 1 3 ) 0.0977 ? 1 = 556.7 ? Safety factor = 556.7/500 = 1.11 (b) The suggestion is not a good one because although the safety factor in life seems high, the corresponding safety factor in stress is low. Fatigue lives are quite sensitive to stress, so relatively large safety factors in life are needed to achieve reasonable safety factors in stress. In this case, for a safety factor of 1.5 in stress, a safety factor of 63 on life would be required. MechEng 382 Winter 2010 Homework#10 Solutions 4 3. Dowling 9.13 Type of load Source Static Bicycle weight Working Rider weight, pedaling forces, centripetal forces while turning Vibratory Uneven road surface Accidental Crash MechEng 382...
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This note was uploaded on 05/12/2010 for the course MECHENG 382 taught by Professor Thouless during the Winter '08 term at University of Michigan.
 Winter '08
 Thouless

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