22_ch 05 Mechanical Design budynas_SM_ch05

22_ch 05 Mechanical Design budynas_SM_ch05 - / 2 = [( 136 ....

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136 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design For the loading scheme depicted in Figure ( d ) M max = F 2 ± a + b 2 ² F 2 ± 1 2 ²± b 2 ² 2 = F 2 ± a 2 + b 4 ² This result is the same as that obtained for Figure ( c ). At point B , we also have a surface compression of σ y = F A = F bd 4 . 4(10 3 ) 18(12) =− 20 . 4MPa With σ x =− 136 . 2MPa . From a Mohrs circle diagram, τ max = 136 . 2 / 2 = 68 . 1MPa . n = 110 68 . 1 = 1 . 62 MPa Ans . 5-30 Based on Figure ( c ) and using Eq. (5-15) σ ± = ( σ 2 x ) 1 / 2 = (136 . 2 2 ) 1 / 2 = 136 . 2MPa n = S y σ ± = 220 136 . 2 = 1 . 62 Ans. Based on Figure ( d ) and using Eq. (5-15) and the solution of Prob. 5-29, σ ± = ( σ 2 x σ x σ y + σ 2 y ) 1
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Unformatted text preview: / 2 = [( 136 . 2) 2 ( 136 . 2)( 20 . 4) + ( 20 . 4) 2 ] 1 / 2 = 127 . 2 MPa n = S y = 220 127 . 2 = 1 . 73 Ans. 5-31 When the ring is set, the hoop tension in the ring is equal to the screw tension. t = r 2 i p i r 2 o r 2 i 1 + r 2 o r 2 We have the hoop tension at any radius. The differential hoop tension dF is dF = w t dr F = r o r i w t dr = w r 2 i p i r 2 o r 2 i r o r i 1 + r 2 o r 2 dr = w r i p i (1) dF r w...
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