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23_ch 10 Mechanical Design budynas_SM_ch10

# 23_ch 10 Mechanical Design budynas_SM_ch10 -...

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Unformatted text preview: budynas_SM_ch10.qxd 12/01/2006 17:23 FIRST PAGES Page 283 283 Chapter 10 The useable root for C is 2 2 2S 2S πd a πd a π d Sa − + + 2 C = 0.5 144 144 36 π (0.081) 2 (42.2)(103 ) π (0.081) 2 (42.2)(103 ) = 0.5 + 144 144 = 4.91 2 − π (0.081) 2 (42.2)(103 ) 36 D = Cd = 0.398 in Fi = 33 500 π d 3 τi π d3 C −3 = ± 1000 4 − 8D 8 D exp(0.105C ) 6.5 Use the lowest Fi in the preferred range. Fi = π (0.081) 3 8(0.398) 33 500 4.91 − 3 − 1000 4 − exp[0.105(4.91)] 6.5 = 8.55 lbf For simplicity we will round up to next 1/4 integer. Fi = 8.75 lbf k= 18 − 9 = 36 lbf/in 0.25 d4G (0.081) 4 (10)(106 ) = = 23.7 turns 8k D 3 8(36)(0.398) 3 G 10 Nb = Na − = 23.7 − = 23.3 turns E 28 L 0 = (2C − 1 + Nb ) d = [2(4.91) − 1 + 23.3](0.081) = 2.602 in L max = L 0 + ( Fmax − Fi ) / k = 2.602 + (18 − 8.75) /36 = 2.859 in Na = ( σa ) A = = (n f ) A = Body: 4.5(4) π d2 4C 2 − C − 1 +1 C −1 18(10−3 ) 4(4.912 ) − 4.91 − 1 + 1 = 21.1 kpsi π (0.0812 ) 4.91 − 1 Sa 42.2 = = 2 checks ( σa ) A 21.1 KB = 4C + 2 4(4.91) + 2 = = 1.300 4C − 3 4(4.91) − 3 τa = 8(1.300)(4.5)(0.398) (10−3 ) = 11.16 kpsi π (0.081) 3 τm = Fm 13.5 τa = (11.16) = 33.47 kpsi Fa 4.5 +2 ...
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