27_ch 20 Mechanical Design budynas_SM_ch20

27_ch 20 Mechanical Design budynas_SM_ch20 - 12/06/2006...

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Chapter 20 27 From Eqs. (20-18) and (20-19), µ y = ln[5 . 034(10 6 )] 0 . 528 2 / 2 = 15 . 292 ˆ σ y = p ln(1 + 0 . 528 2 ) = 0 . 496 From Eq. (20-17), defining g ( x ), g ( x ) = 1 x (0 . 496) 2 π exp ± 1 2 ² ln x 15 . 292 0 . 496 ³ 2 ´ x (Mrev) f / ( N w ) g ( x ) · (10 6 ) 0.5 0.00000 0.00011 0.5 0.04641 0.00011 1.5 0.04641 0.05204 1.5 0.09283 0.05204 2.5 0.09283 0.16992 2.5 0.16034 0.16992 3.5 0.16034 0.20754 3.5 0.24051 0.20754 4.5 0.24051 0.17848 4.5 0.13080 0.17848 5.5 0.13080 0.13158 5.5 0.08017 0.13158 6.5 0.08017 0.09011 6.5 0.06329 0.09011 7.5 0.06329 0.05953 7.5 0.05063 0.05953 8.5 0.05063 0.03869 8.5 0.04641 0.03869 9.5 0.04641 0.02501 9.5 0.03797 0.02501 10.5 0.03797 0.01618 10.5
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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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