6_ch 06 Mechanical Design budynas_SM_ch06

6_ch 06 Mechanical Design budynas_SM_ch06 -...

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152 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Table A-20: S ut = 440 MPa, S y = 370 MPa S ± e = 0 . 5(440) = 220 MPa Table 6-2: k a = 4 . 51 ( 440 ) 0 . 265 = 0 . 899 k b = 1( axial loading) Eq. (6-26): k c = 0 . 85 S e = 0 . 899(1)(0 . 85)(220) = 168 . 1MPa Table A-15-1: d /w = 12 / 60 = 0 . 2, K t = 2 . 5 From Fig. 6-20, q ˙= 0 . 82 Eq. (6-32): K f = 1 + 0 . 82(2 . 5 1) = 2 . 23 σ a = K f F a A S e n f = 2 . 23 F a 10(60 12) = 168 . 1 1 . 8 F a = 20 100 N = 20 . 1kN Ans . F a A = S y n y F a 10(60 12) = 370 1 . 8 F a = 98 700 N = 98 . 7kN Ans . Largest force amplitude is 20.1 kN. Ans. 6-11 A priori design decisions: The design decision will be: d Material and condition: 1095 HR and from Table A-20 S = 120, S y = 66 kpsi. Design factor: n f = 1 . 6 per problem statement. Life: (1150)(3)
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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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