11_ch 20 Mechanical Design budynas_SM_ch20

11_ch 20 Mechanical Design budynas_SM_ch20 -...

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Chapter 20 11 From Table 20-6, ¯ δ = ¯ F ¯ l (1 / ¯ A )(1 / ¯ E ) ¯ δ = 14 700 ( 1 . 5 ) 1 0 . 226 1 29 . 5 ( 10 6 ) = 0 . 003 31 in Ans. For the standard deviation, using the first-order terms in Table 20-6, ˆ σ δ . = ¯ F ¯ l ¯ A ¯ E ( C 2 F + C 2 l + C 2 A + C 2 E ) 1 / 2 = ¯ δ ( C 2 F + C 2 l + C 2 A + C 2 E ) 1 / 2 ˆ σ δ = 0 . 003 31(0 . 0884 2 + 0 . 00267 2 + 0 . 0133 2 + 0 . 03 2 ) 1 / 2 = 0 . 000 313 in Ans. COV C δ = 0 . 000 313 / 0 . 003 31 = 0 . 0945 Ans. Force COV dominates. There is no distributional information on δ . 20-15 M = (15000, 1350) lbf · in, distribution unspecified; d = (2.00, 0.005) in distribution unspecified. σ = 32 M π d 3 , C M = 1350 15 000 = 0 . 09, C d = 0 . 005 2 . 00 = 0 . 0025 σ is of the form x / y , Table 20-6. Mean: ¯ σ = 32 ¯ M π ± d 3 . = 32 ¯ M π ¯ d 3 = 32(15 000) π (2 3 ) = 19 099 psi
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