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54_ch 03 Mechanical Design budynas_SM_ch03

# 54_ch 03 Mechanical Design budynas_SM_ch03 -...

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Chapter 3 67 d A r = 1 ln 2 . 6 1 2 π 1 . 8 1 . 8 2 0 . 4 2 = 0 . 672 723 4 A = 1(1 . 6) π (0 . 4 2 ) = 1 . 097 345 2 in 2 r n = 1 . 097 345 2 0 . 672 723 4 = 1 . 6312 in e = 1 . 8 r n = 0 . 1688 in c i = 1 . 6312 1 = 0 . 6312 in c o = 2 . 6 1 . 6312 = 0 . 9688 in M = 3000 ( 5 . 8 ) = 17 400 lbf · in σ i = 3 1 . 0973 + 17 . 4 ( 0 . 6312 ) 1 . 0973 ( 0 . 1688 )( 1 ) = 62 . 03 kpsi Ans. σ o = 3 1 . 0973 17 . 4 ( 0 . 9688 ) 1 . 0973 ( 0 . 1688 )( 2 . 6 ) = − 32 . 27 kpsi Ans. 3-81 From Eq. (3-68) a = K F 1 / 3 = F 1 / 3 3 8 2[ ( 1 ν 2 )/ E ] 2 ( 1 / d ) 1 / 3 Use ν = 0 . 292, F in newtons, E in N/mm 2 and d in mm, then K = 3 8 [(1 0 . 292 2 ) / 207 000] 1 / 25 1 / 3 = 0 . 0346 p max = 3 F 2 π a 2 = 3 F 2 π ( K F 1 / 3 ) 2 = 3 F 1 / 3 2 π K 2 = 3 F 1 / 3 2 π (0 . 0346) 2 = 399 F 1 / 3 MPa = | σ max | Ans. τ max = 0 . 3 p max = 120 F 1 / 3 MPa Ans. 3-82 From Prob. 3-81, K = 3 8 2[(1
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