20_ch 20 Mechanical Design budynas_SM_ch20

# 20_ch 20 Mechanical Design budynas_SM_ch20 - + 18 . 5 (1 ....

This preview shows page 1. Sign up to view the full content.

20 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Weibull parameters see Shigley, J. E., and C. R. Mischke, Mechanical Engineering Design, McGraw-Hill, 5th ed., 1989, Sec. 4-12. 20-27 Let x = ( S ± fe ) 10 4 x 0 = 79 kpsi, θ = 86 . 2 kpsi, b = 2 . 6 Eq. (20-28) ¯ x = x 0 + ( θ x 0 ) ± (1 + 1 / b ) ¯ x = 79 + (86 . 2 79) ± (1 + 1 / 2 . 6) = 79 + 7 . 2 ± (1 . 38) From Table A-34, ±( 1 . 38 ) = 0 . 88854 ¯ x = 79 + 7 . 2 ( 0 . 888 54 ) = 85 . 4 kpsi Ans. Eq. (20-29) ˆ σ x = ( θ x 0 )[ ± (1 + 2 / b ) ± 2 (1 + 1 / b )] 1 / 2 = (86 . 2 79)[ ± (1 + 2 / 2 . 6) ± 2 (1 + 1 / 2 . 6)] 1 / 2 = 7 . 2[0 . 923 76 0 . 888 54 2 ] 1 / 2 = 2 . 64 kpsi Ans. C x = ˆ σ x ¯ x = 2 . 64 85 . 4 = 0 . 031 Ans. 20-28 x = S ut x 0 = 27 . 7, θ = 46 . 2, b = 4 . 38 µ x = 27 . 7 + (46 . 2 27 . 7) ± (1 + 1 / 4 . 38) = 27 . 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: + 18 . 5 (1 . 23) = 27 . 7 + 18 . 5(0 . 910 75) = 44 . 55 kpsi Ans. x = (46 . 2 27 . 7)[ (1 + 2 / 4 . 38) 2 (1 + 1 / 4 . 38)] 1 / 2 = 18 . 5[ (1 . 46) 2 (1 . 23)] 1 / 2 = 18 . 5[0 . 8856 . 910 75 2 ] 1 / 2 = 4 . 38 kpsi Ans. C x = 4 . 38 44 . 55 = . 098 Ans. From the Weibull survival equation R = exp x x x b = 1 p...
View Full Document

## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online