27_ch 04 Mechanical Design budynas_SM_ch04

27_ch 04 Mechanical Design budynas_SM_ch04 - s ² T s T s =...

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96 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Since δ c = δ b , P c L ( π/ 4)(4 . 5 2 4 2 ) E = P b L 6( π/ 4)(0 . 375 2 ) E P c = 5 . 037 P b (2) Substituting into Eq. (1) 6 . 037 P b = 7 . 54 P b = 1 . 249 kip ; and from Eq. (2), P c = 6 . 291 kip Using the results of (a) above, the total bolt and cylinder stresses are σ b = 76 . 5 + 1 . 249 6( π/ 4)(0 . 375 2 ) = 78 . 4 kpsi Ans. σ c =− 15 . 19 + 6 . 291 ( π/ 4)(4 . 5 2 4 2 ) =− 13 . 3 kpsi Ans. 4-53 T = T c + T s and θ c = θ s Also, T c L ( GJ ) c = T s L ( GJ ) s T c = ( GJ ) c ( GJ ) s T s Substituting into equation for T , T = ± 1 + ( GJ ) c ( GJ
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Unformatted text preview: ) s ² T s % T s = T s T = ( G J ) s ( G J ) s + ( G J ) c Ans. 4-54 R O + R B = W (1) δ O A = δ AB (2) 500 R O AE = 750 R B AE , R O = 3 2 R B 3 2 R B + R B = 3 . 5 R B = 7 5 = 1 . 4 kN Ans. R O = 3 . 5 − 1 . 4 = 2 . 1 kN Ans. σ O = − 2100 12(50) = − 3 . 50 MPa Ans. σ B = 1400 12(50) = 2 . 33 MPa Ans. R B W ± 3.5 750 mm 500 mm R O A B O...
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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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