22_ch 20 Mechanical Design budynas_SM_ch20

22_ch 20 Mechanical Design budynas_SM_ch20 -...

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22 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design y = S y y 0 = 64 . 1, θ = 81 . 0, b = 3 . 77 ¯ y = 64 . 1 + (81 . 0 64 . 1) ± (1 + 1 / 3 . 77) = 64 . 1 + 16 . 9 ± (1 . 27) = 64 . 1 + 16 . 9(0 . 902 50) = 79 . 35 kpsi σ y = (81 64 . 1)[ ± (1 + 2 / 3 . 77) ± (1 + 1 / 3 . 77)] 1 / 2 σ y = 16 . 9[(0 . 887 57) 0 . 902 50 2 ] 1 / 2 = 4 . 57 kpsi p = 1 exp ± ² y y 0 θ y 0 ³ 3 . 77 ´ p = 1 exp ± ² 70 64 . 1 81 64 . 1 ³ 3 . 77 ´ = 0 . 019 Ans. 20-31 x = S ut = W [122 . 3, 134 . 6, 3 . 64] kpsi, p ( x > 120) = 1 = 100% since x 0 > 120 kpsi p ( x > 133) = exp ± ² 133 122 . 3 134 . 6 122 . 3 ³ 3 . 64 ´ = 0 . 548 = 54 . 8% Ans. 20-32 Using Eqs. (20-28) and (20-29) and Table A-34, µ n = n 0 + ( θ n 0 ) ± (1 + 1 / b ) = 36 . 9 + (133 . 6 36 . 9) ± (1 + 1 / 2 . 66) = 122 . 85 kcycles ˆ σ n = ( θ n 0 )[ ± (1 + 2 / b ) ± 2 (1 + 1 / b )] = 34 . 79 kcycles For the Weibull density function, Eq. (2-27), f W ( n ) = 2 . 66 133 . 6 36 . 9 ² n 36 . 9 133 . 6 36 . 9 ³ 2 . 66
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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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