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5_ch 20 Mechanical Design budynas_SM_ch20

# 5_ch 20 Mechanical Design budynas_SM_ch20 - 008 929 From...

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Chapter 20 5 20-7 F ( x ) = 0 . 555 x 33 mm (a) Since F ( x ) is linear, the distribution is uniform at x = a F ( a ) = 0 = 0 . 555( a ) 33 a = 59.46 mm. Therefore, at x = b F ( b ) = 1 = 0 . 555 b 33 b = 61.26 mm. Therefore, F ( x ) = 0 x < 59 . 46 mm 0 . 555 x 33 59 . 46 x 61 . 26 mm 1 x > 61 . 26 mm The PDF is dF / dx , thus the range numbers are: f ( x ) = ± 0 . 555 59 . 46 x 61 . 26 mm 0 otherwise Ans. From the range numbers, µ x = 59 . 46 + 61 . 26 2 = 60 . 36 mm Ans. ˆ σ x = 61 . 26 59 . 46 2 3 = 0 . 520 mm Ans. (b) σ is an uncorrelated quotient ¯ F = 3600 lbf, ¯ A = 0 . 112 in 2 C F = 300 / 3600 = 0 . 083 33, C A = 0 . 001 / 0 . 112 = 0 .
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Unformatted text preview: 008 929 From Table 20-6, for σ ¯ σ = µ F µ A = 3600 . 112 = 32 143 psi Ans. ˆ σ σ = 32 143 ² (0 . 08333 2 + . 008929 2 ) (1 + . 008929 2 ) ³ 1 / 2 = 2694 psi Ans. C σ = 2694 / 32 143 = . 0838 Ans. Since F and A are lognormal, division is closed and σ is lognormal too. σ = LN (32 143, 2694) psi Ans....
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