19_ch 05 Mechanical Design budynas_SM_ch05

19_ch 05 Mechanical Design budynas_SM_ch05 -...

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Chapter 5 133 Then σ A = 10 . 06 + 11 . 27 d 3 = 21 . 33 d 3 kpsi and σ B = 10 . 06 11 . 27 d 3 =− 1 . 21 d 3 kpsi For this state of stress, use the Brittle-Coulomb-Mohr theory for illustration. Here we use S ut (min) = 25 kpsi, S uc (min) = 97 kpsi, and Eq. (5-31 b ) to arrive at 21 . 33 25 d 3 1 . 21 97 d 3 = 1 2 . 8 Solving gives d = 1 . 34 in . So use d = 13 / 8 in Ans. Note that this has been solved as a statics problem. Fatigue will be considered in the next chapter. 5-24 As in Prob. 5-23, we will assume this to be statics problem. Since the proportions are un- changed, the bearing reactions will be the same as in Prob. 5-23. Thus xy plane: M B = 223(4) = 892 lbf · in xz plane: M B = 106(4) = 424 lbf · in So M max = [(892) 2 + (424) 2 ] 1 / 2 = 988 lbf · in σ x = 32 M B π d 3 = 32(988) π d 3 = 10 060 d 3 psi
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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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