21_ch 06 Mechanical Design budynas_SM_ch06

21_ch 06 Mechanical Design budynas_SM_ch06 -...

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Unformatted text preview: budynas_SM_ch06.qxd 11/29/2006 17:40 FIRST PAGES Page 167 167 Chapter 6 Eqs. (6-55) and (6-56): 2 σa = ( σa /0.85) 2 + 3τa 1/2 = ( 120.5/0.85) 2 + 3(134.5) 2 σm = ( −200.9/0.85) 2 + 3(224.1) 2 Goodman: ( σa ) e = 1/2 1/2 = 272.7 MPa = 454.5 MPa σa 272.7 = = 499.9 MPa 1 − σm / Sut 1 − 454.5/1000 Let f = 0.9 [0.9(1000)]2 a= = 2594 MPa 312.3 0.9(1000) 1 = −0.1532 b = − log 3 312.3 ( σa ) e N= a 6-29 1/b 499.9 = 2594 1/−0.1532 = 46 520 cycles Ans. S y = 490 MPa, Sut = 590 MPa, Se = 200 MPa σm = 420 + 140 420 − 140 = 280 MPa , σa = = 140 MPa 2 2 Goodman: ( σa ) e = σa 140 = = 266.5 MPa > Se ∴ finite life 1 − σm / Sut 1 − (280/590) [0.9(590)]2 = 1409.8 MPa 200 0.9(590) 1 = −0.141 355 b = − log 3 200 a= N= 266.5 1409.8 −1/0.143 55 = 131 200 cycles Nremaining = 131 200 − 50 000 = 81 200 cycles Second loading: 350 + ( −200) = 75 MPa 2 350 − ( −200) ( σa ) 2 = = 275 MPa 2 275 ( σa ) e2 = = 315.0 MPa 1 − (75/590) ( σm ) 2 = ...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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