30_ch 17 Mechanical Design budynas_SM_ch17

# 30_ch 17 Mechanical Design budynas_SM_ch17 - d 2 term. n f...

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Chapter 17 449 Design variables Choose 30-in D min . Table 17-27: w = 1 . 60 d 2 lbf/ft w l = 1 . 60 d 2 l = 1 . 60 d 2 (90) = 144 d 2 lbf, ea. Eq. (17-46): F t = ± W m + w l ²± 1 + a g ² = ± 5000 m + 144 d 2 ²± 1 + 4 32 . 2 ² = 5620 m + 162 d 2 lbf, each wire Eq. (17-47): F f = ( p / S u ) S u Dd 2 From Fig. 17-21 for 10 5 cycles, p / S u = 0 . 004;from p. 908, S u = 240 000 psi, based on metal area. F f = 0 . 004(240 000)(30 d ) 2 = 14 400 d lbf each wire Eq. (17-48) and Table 17-27: F b = E w d w A m D = 12(10 6 )(0 . 067 d )(0 . 4 d 2 ) 30 = 10 720 d 3 lbf, each wire Eq. (17-45): n f = F f F b F t = 14 400 d 10 720 d 3 (5620 / m ) + 162 d 2 We could use a computer program to build a table similar to that of Ex. 17-6. Alterna- tively, we could recognize that 162 d 2 is small compared to 5620 / m , and therefore elimi- nate the 162
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Unformatted text preview: d 2 term. n f ˙= 14 400 d − 10 720 d 3 5620 / m = m 5620 (14 400 d − 10 720 d 3 ) Maximize n f , ∂ n f ∂ d = = m 5620 [14 400 − 3(10 720) d 2 ] From which d * = ³ 14 400 32 160 = . 669 in Back-substituting n f = m 5620 [14 400(0 . 669) − 10 720(0 . 669 3 )] = 1 . 14 m Thus n f = 1 . 14, 2 . 28, 3 . 42, 4 . 56 for m = 1, 2, 3, 4 respectively. If we choose d = . 50 in, then m = 2. n f = 14 400(0 . 5) − 10 720(0 . 5 3 ) (5620 / 2) + 162(0 . 5) 2 = 2 . 06...
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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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