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HW10+solution (2)

# HW10+solution (2) - MechEng 382 Winter 2010 Homework#10...

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Unformatted text preview: MechEng 382 Winter 2010 Homework #10 Solutions 1. Dowling 9.4 σ a Mpa N f Cycles log(σa) 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 , σ a ) 1 = (257, 524 MPa) (N f , σ 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 log-log plot log(Nf) MechEng 382 Winter 2010 Homework#10 Solutions 2 Using Basquin’s 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) Re-arranging 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 ª = 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 Basquin’s 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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HW10+solution (2) - MechEng 382 Winter 2010 Homework#10...

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