54_ch 03 Mechanical Design budynas_SM_ch03

54_ch 03 Mechanical Design budynas_SM_ch03 -...

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Unformatted text preview: budynas_SM_ch03.qxd 11/28/2006 21:23 FIRST PAGES Page 67 67 Chapter 3 dA 2.6 = 1 ln − 2π 1.8 − r 1 1.82 − 0.42 = 0.672 723 4 A = 1(1.6) − π (0.42 ) = 1.097 345 2 in2 rn = 1.097 345 2 = 1.6312 in 0.672 723 4 e = 1.8 − rn = 0.1688 in ci = 1.6312 − 1 = 0.6312 in co = 2.6 − 1.6312 = 0.9688 in M = 3000(5.8) = 17 400 lbf · in σi = σo = 3-81 3 17.4(0.6312) + = 62.03 kpsi 1.0973 1.0973(0.1688)(1) 3 17.4(0.9688) − = −32.27 kpsi 1.0973 1.0973(0.1688)(2.6) From Eq. (3-68) a = K F 1/3 = F 1/3 3 2[(1 − ν 2 )/ E ] 8 2(1/d ) 1/3 Use ν = 0.292, F in newtons, E in N/mm2 and d in mm, then K= pmax = = 3 [(1 − 0.2922 ) /207 000] 8 1/25 1/3 3F 3F = 2π a 2 2π ( K F 1/3 ) 2 3 F 1/3 3 F 1/3 = 2π K 2 2π (0.0346) 2 = 399 F 1/3 MPa = |σmax | τmax = 0.3 pmax = 120 F 1/3 MPa 3-82 = 0.0346 Ans. Ans. From Prob. 3-81, 3 2[(1 − 0.2922 ) /207 000] K= 8 1/25 + 0 pmax = and so, 1/3 = 0.0436 3 F 1/3 3 F 1/3 = = 251 F 1/3 2π K 2 2π (0.0436) 2 σz = −251 F 1/3 MPa Ans. τmax = 0.3(251) F 1/3 = 75.3 F 1/3 MPa Ans . z = 0.48a = 0.48(0.0436)181/3 = 0.055 mm Ans . Ans. Ans. ...
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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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