20_ch 12 Mechanical Design budynas_SM_ch12

# 20_ch 12 Mechanical Design budynas_SM_ch12 -...

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Chapter 12 323 Solving Eq. (12-32) for t t = π DL w 4 f 1 f 2 KVF = π (1)(1)(0 . 005) 4(1 . 8)(1)(0 . 6)(10 10 )(52 . 4)(500) = 1388 h = 83 270 min Cycles = Nt = 200(83 270) = 16 . 7rev Ans. 12-23 Estimate bushing length with f 1 = f 2 = 1, and K = 0 . 6(10 10 )in 3 · min/(lbf · ft · h) Eq. (12-32): L = 1(1)(0 . 6)(10 10 )(2)(100)(400)(1000) 3(0 . 002) = 0 . 80 in From Eq. (12-38), with f s = 0 . 03 from Table 12-9 applying n d = 2 to F and ¯ h CR = 2 . 7 Btu/(h · ft 2 · °F) L . = 720(0 . 03)(2)(100)(400) 778(2 . 7)(300 70) = 3 . 58 in 0 . 80 L 3 . 58 in Trial 1 : Let L = 1in , D = 1in P max = 4(2)(100) π (1)(1) = 255 psi < 3560 psi O.K. P = 2(100) 1(1) = 200 psi V = π (1)(400) 12 = 104 .
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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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