50_ch 03 Mechanical Design budynas_SM_ch03

# 50_ch 03 Mechanical Design budynas_SM_ch03 -...

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Chapter 3 63 Using Eq. (3-65) σ i = F A + Mc i Aer i = 1000 0 . 7854 + 2000(0 . 4682) 0 . 7854(0 . 031 754)(1 . 5) = 26 300 psi Ans . σ o = F A Mc o Aer o = 1000 0 . 7854 2000(0 . 5318) 0 . 7854(0 . 031 754)(2 . 5) =− 15 800 psi Ans . 3-74 Section AA: D = 0 . 75 in, r i = 0 . 75 / 2 = 0 . 375 in, r o = 0 . 75 / 2 + 0 . 25 = 0.625 in From Table 3-4, for R = 0 . 125 in, r c = ( 0 . 75 + 0 . 25 )/ 2 = 0 . 500 in r n = 0 . 125 2 2 ( 0 . 5 0 . 5 2 0 . 125 2 ) = 0 . 492 061 5 in e = 0 . 5 r n = 0 . 007 939 in c o = r o r n = 0 . 625 0 . 492 06 = 0 . 132 94 in c i = r n r i = 0 . 492 06 0 . 375 = 0 . 117 06 in A = π (0 . 25) 2 / 4 = 0 . 049 087 M = Fr c = 100(0 . 5) = 50 lbf · in σ i = 100 0 . 049 09 + 50(0 . 117 06) 0 . 049 09(0 . 007 939)(0 . 375) = 42 100 psi Ans . σ o = 100 0 . 049 09 50(0 . 132 94) 0 . 049 09(0 . 007 939)(0 . 625) =− 25 250 psi Ans . Section BB: Abscissa angle θ of line of radius centers is θ = cos 1 ± r 2 + d / 2 r 2 + d + D / 2 ² = cos 1 ± 0 . 375 + 0 . 25 / 2 0 . 375 + 0 . 25 + 0 . 75 / 2 ² = 60 M = F D + d 2 cos θ = 100(0 . 5) cos 60 = 25 lbf · in
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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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