40_ch 03 Mechanical Design budynas_SM_ch03

40_ch 03 Mechanical Design budynas_SM_ch03 -...

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Chapter 3 53 3-48 F t = 1000 2 . 5 = 400 lbf F n = 400 tan 20 = 145.6 lbf Torque at C T C = 400(5) = 2000 lbf · in P = 2000 3 = 666 . 7 lbf ± ( M A ) z = 0 18 R Dy 145 . 6(13) 666 . 7(3) = 0 R Dy = 216 . 3 lbf ± ( M A ) y = 0 ⇒− 18 R Dz + 400(13) = 0 R Dz = 288 . 9 lbf ± F y = 0 R Ay + 216 . 3 666 . 7 145 . 6 = 0 R Ay = 596 . 0 lbf ± F z = 0 R Az + 288 . 9 400 = 0 R Az = 111 . 1 lbf M B = 3 ² 596 2 + 111 . 1 2 = 1819 lbf · in M C = 5 ² 216 . 3 2 + 288 . 9 2 = 1805 lbf · in Maximum stresses occur at B . Ans. σ B = 32 M B π d 3 = 32(1819) π (1 . 25 3 ) = 9486 psi τ B = 16 T B π d 3 = 16(2000) π (1 . 25 3 ) = 5215 psi σ max = σ B 2 + ³ ´ σ B 2 µ 2 + τ 2 B = 9486 2 + · 9486 2 ¸ 2 + 5215 2 = 11 792 psi Ans. τ max = ³ ´ σ B
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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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