6_ch 16 Mechanical Design budynas_SM_ch16

6_ch 16 Mechanical Design budynas_SM_ch16 -...

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Chapter 16 401 Eq. (16-2): M f = fp a br sin θ a ( rC aA ) = 0 . 25 p a (0 . 030)(0 . 150) sin 90° [0 . 15(1 . 1314) 0 . 25(0 . 48)] = 5 . 59(10 5 ) p a N · m Eq. (16-8): B = ± θ 2 1 4 sin 2 θ ² 98 . 13 π/ 180 rad 8 . 13 π/ 180 rad = 0 . 925 Eq. (16-3): M N = p a bra sin θ a B = p a (0 . 030)(0 . 150)(0 . 250) 1 (0 . 925) = 1 . 0406(10 3 ) p a N · m Using F = ( M N M f ) / c , we obtain 400 = 104 . 06 5 . 59 0 . 5(10 5 ) p a or p a = 203 kPa Ans. T = fp a br 2 C sin θ a = 0 . 25(203)(10 3 )(0 . 030)(0 . 150) 2 1 (1 . 1314) = 38 . 76 N · m Ans. 16-6 For + 3 ˆ σ f : f = ¯ f + 3 ˆ σ f = 0 . 25 + 3(0 . 025) = 0 . 325 M f = 5 . 59(10 5 ) p a ± 0 . 325 0 . 25 ² = 7 . 267(10 5 ) p a Eq. (16-4): 400 = 104 . 06 7 . 267 10 5 (0 . 500) p a p a = 207 kPa T = 38 . 75 ± 207 203 ²± 0 . 325 0 . 25 ² = 51 . 4N · m Ans. Similarly, for 3 ˆ σ f : f = ¯ f 3 ˆ σ f = 0 . 25 3(0 . 025) = 0 . 175 M f = 3 . 913(10 5 ) p a p a = 200 kPa T = 26
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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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