19_ch 20 Mechanical Design budynas_SM_ch20

19_ch 20 Mechanical Design budynas_SM_ch20 - 0.00286 100...

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Chapter 20 19 From Table A-10, for 1% failure, z 0 . 01 =− 2 . 326 . Thus, ln x 0 . 01 = 4 . 587 + 0 . 043 74( 2 . 326) = 4 . 485 x 0 . 01 = 88 . 7 kpsi Ans. The normal PDF is given by Eq. (20-14) as f ( x ) = 1 4 . 30 2 π exp ± 1 2 ² x 98 . 26 4 . 30 ³ 2 ´ For the lognormal distribution, from Eq. (20-17), defining g ( x ), g ( x ) = 1 x (0 . 043 74) 2 π exp ± 1 2 ² ln x 4 . 587 0 . 043 74 ³ 2 ´ x (kpsi) f / ( N w ) f ( x ) g ( x ) x (kpsi) f / ( N w ) f ( x ) g ( x ) 92 0.00000 0.03215 0.03263 102 0.03676 0.06356 0.06134 92 0.06985 0.03215 0.03263 104 0.03676 0.03806 0.03708 94 0.06985 0.05680 0.05890 104 0.01838 0.03806 0.03708 94 0.09191 0.05680 0.05890 106 0.01838 0.01836 0.01869 96 0.09191 0.08081 0.08308 106 0.01471 0.01836 0.01869 96 0.13971 0.08081 0.08308 108 0.01471 0.00713 0.00793 98 0.13971 0.09261 0.09297 108 0.01471 0.00713 0.00793 98 0.06250 0.09261 0.09297 110 0.01471 0.00223
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Unformatted text preview: 0.00286 100 0.06250 0.08548 0.08367 110 0.00735 0.00223 0.00286 100 0.04412 0.08548 0.08367 112 0.00735 0.00056 0.00089 102 0.04412 0.06356 0.06134 112 0.00000 0.00056 0.00089 Note: rows are repeated to draw histogram The normal and lognormal are almost the same. However the data is quite skewed and perhaps a Weibull distribution should be explored. For a method of establishing the f ( x ) g ( x ) Histogram 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 90 92 94 96 98 100 102 104 106 108 x (kpsi) Probability density 110 112...
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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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