22_ch 05 Mechanical Design budynas_SM_ch05

# 22_ch 05 Mechanical Design budynas_SM_ch05 - 2 = − 136 2...

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: / 2 = [( − 136 . 2) 2 − ( − 136 . 2)( − 20 . 4) + ( − 20 . 4) 2 ] 1 / 2 = 127 . 2 MPa n = S y σ ± = 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. σ t = r 2 i p i r 2 o − r 2 i ± 1 + r 2 o r 2 ² We have the hoop tension at any radius. The differential hoop tension dF is dF = wσ t dr F = ³ r o r i wσ t dr = w r 2 i p i r 2 o − r 2 i ³ r o r i ± 1 + r 2 o r 2 ² dr = w r i p i (1) dF r w...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online