13_ch 20 Mechanical Design budynas_SM_ch20

# 13_ch 20 Mechanical Design budynas_SM_ch20 - = 30 . 3...

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Chapter 20 13 (b) F ( x 1 ) = F (0 . 748) = 0 F ( x 2 ) = (0 . 750 0 . 748)333 . 3 = 0 . 6667 If g ( x ) is truncated, PDF becomes g ( x ) = f ( x ) F ( x 2 ) F ( x 1 ) = 333 . 3 0 . 6667 0 = 500 in 1 µ x = a ± + b ± 2 = 0 . 748 + 0 . 750 2 = 0 . 749 in ˆ σ x = b ± a ± 2 3 = 0 . 750 0 . 748 2 3 = 0 . 000 577 in 20-18 From Table A-10, 8.1% corresponds to z 1 =− 1 . 4 and 5.5% corresponds to z 2 =+ 1 . 6. k 1 = µ + z 1 ˆ σ k 2 = µ + z 2 ˆ σ From which µ = z 2 k 1 z 1 k 2 z 2 z 1 = 1 . 6 ( 9 ) ( 1 . 4 ) 11 1 . 6 ( 1 . 4 ) = 9 . 933 ˆ σ = k 2 k 1 z 2 z 1 = 11 9 1 . 6 ( 1 . 4) = 0 . 6667 The original density function is f ( k ) = 1 0 . 6667 2 π exp ± 1 2 ² k 9 . 933 0 . 6667 ³ 2 ´ Ans. 20-19 From Prob. 20-1, µ = 122 . 9 kcycles and
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Unformatted text preview: = 30 . 3 kcycles. z 10 = x 10 = x 10 122 . 9 30 . 3 x 10 = 122 . 9 + 30 . 3 z 10 From Table A-10, for 10 percent failure, z 10 = 1 . 282 x 10 = 122 . 9 + 30 . 3 ( 1 . 282 ) = 84 . 1 kcycles Ans. 0.748 g ( x ) 500 x f ( x ) 333.3 0.749 0.750 0.751...
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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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