8_ch 07 Mechanical Design budynas_SM_ch07

# 8_ch 07 Mechanical Design budynas_SM_ch07 - 11.4 θ B = 1...

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Chapter 7 185 At gear mesh, B xy plane With I = I 1 in section OCA , y ± A =− 2153 . 1 / EI 1 Since y ± B / A is a cantilever, from Table A-9-1, with I = I 2 in section AB y ± B / A = Fx ( x 2 l ) 2 EI 2 = 46 . 6 2 EI 2 (2 . 75)[2 . 75 2(2 . 75)] =− 176 . 2 / EI 2 y ± B = y ± A + y ± B / A =− 2153 . 1 30(10 6 )( π/ 64)(1 . 25 4 ) 176 . 2 30(10 6 )( π/ 64)(0 . 875 4 ) =− 0 . 000 803 rad (magnitude greater than 0.0005 rad) xz plane z ± A =− 683 . 5 EI 1 , z ± B / A =− 128(2 . 75 2 ) 2 EI 2 =− 484 EI 2 z ± B =− 683 . 5 30(10 6 )( π/ 64)(1 . 25 4 ) 484 30(10 6 )( π/ 64)(0 . 875 4 ) =− 0 . 000 751 rad θ B = ± ( 0 . 000 803) 2 + (0 . 000 751) 2 = 0 . 001 10 rad Crowned teeth must be used. Finite element results: Error in simpliﬁed model θ O = 5 . 47(10 4 ) rad 3.0% θ A = 7 . 09(10 4 ) rad
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Unformatted text preview: 11.4% θ B = 1 . 10(10 − 3 ) rad 0.0% The simpliﬁed model yielded reasonable results. Strength S ut = 72 kpsi, S y = 39 . 5 kpsi At the shoulder at A , x = 10 . 75 in . From Prob. 7-4, M xy = − 209 . 3 lbf · in, M xz = − 293 . 0 lbf · in, T = 192 lbf · in M = ± ( − 209 . 3) 2 + ( − 293) 2 = 360 . 0 lbf · in S ± e = . 5(72) = 36 kpsi k a = 2 . 70(72) − . 265 = . 869 z A B x 128 lbf C O x y A B C O 46.6 lbf...
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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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