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ENGR302Wk5-2 - F= ° σA = ° σπ d 2 2 = 40 x 10 6 N/m 2...

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California National University ENGR 302 Kathryn Toth Week #5 Assignment #2 3/14/2011 Chapter 8 Questions:
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8.9) Ti-6Al-4V loaded to 1/2 σy = ½ (910MPa) K Ic = 55 MPa m = ( ) ac 1π KIcσY 2 = ( ( . )) 1π 55MPam455Mpa 1 50 2 = (0.08 m ) 2 = 0.002m 8.23) Cite 5 factors that could lead to scatter in fatigue life data. Some factors are: specimen fabrication, surface preparation, metallurgical variables, specimen alignment in the apparatus, mean stress, and test frequency. Basically, there are bound to be some variables that are uncontrollable and produce unpredictable results. 8.29) It is necessary to find the steady state creep rate to determine tensile stress. ∆ls = 6.44mm – 1.8mm = 4.64mm ϵs = ∆ϵ∆t = ∆lsl∘ t = ( 4.64 mm/635mm)/5,000 h = 1.46 x 10 -6 h -1 Assume: a steady state creep rate of 1.46 x 10 -6 h -1 corresponds to a stress σ of about 40 MPa
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Unformatted text preview: F= ° σA = ( ° ) σπ d 2 2 = ( 40 x 10 6 N/m 2 ) π ( .-19 0x10 3m ) 2 2 = 11,300 N 8.34) ϵs = K 2 σn e ^ (-QcRT ) 2.5x10-3 h-1 = K 2 (55MPa) n e^ (-, / . ( ) 140 000J mol8 31 473K ) => . 2 5x10- - . ×-/ 3 h 1 3 4 10 16Jmol K = K 2 (55MPa) n = . × / 7 35 1012K mol J h K 2 = . × / ( ) 7 35 1012K mol J h 55MPa ^ n And: 2.4x10-2 h-1 = K 2 (69MPa) n e^ (-, / . ( ) 140 000J mol8 31 473K ) K 2 = . × / ( ) 7 05 1013K mol J h 69MPa ^ n . × / ( ) 7 35 1012K mol J h 55MPa ^ = . × / ( ) n 7 05 1013K mol J h 69MPa ^n . × = . × (( ) ) 55MPan 7 05 1013KmolJh 7 35 1012KmolJh 69MPa ^n Take logs of both sides and get n= 9.974. K 2 =3.22 ×-10 5 h-1 MPa-n ϵs = K 2 σn e ^ (-QcRT ) = ϵs (3.22 ×-10 5 h-1 MPa-n ) ( ) . 48Mpa 9 974 e^ (-, / . ( ) 140 000J mol8 31 523K ) = ϵs 0.019h-1...
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