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7_ch 07 Mechanical Design budynas_SM_ch07

# 7_ch 07 Mechanical Design budynas_SM_ch07 - ² x − 11 5...

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184 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design 7-6 If students have access to finite element or beam analysis software, have them model the shaft to check deflections. If not, solve a simpler version of shaft. The 1" diameter sections will not affect the results much, so model the 1 " diameter as 1 . 25 " . Also, ignore the step in AB . From Prob. 18-10, integrate M xy and M xz xy plane , with dy / dx = y E I y = − 131 . 1 2 ( x 2 ) + 5 x 1 . 75 3 5 x 9 . 75 3 62 . 3 2 x 11 . 5 2 + C 1 (1) E I y = − 131 . 1 6 ( x 3 ) + 5 4 x 1 . 75 4 5 4 x 9 . 75 4 62 . 3 6 x 11 . 5 3 + C 1 x + C 2 y = 0 at x = 0 C 2 = 0 y = 0 at x = 11 . 5 C 1 = 1908 . 4 lbf · in 3 From (1) x = 0: E I y = 1908 . 4 x = 11.5: E I y = − 2153 . 1 xz plane (treating z ↑+ ) E Iz = 17 . 4 2 ( x 2 ) 2 x 1 . 75 3 + 2 x 9 . 75 3
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Unformatted text preview: ² x − 11 . 5 ³ 2 + C 3 (2) E I z = 17 . 4 6 ( x 3 ) − 1 2 ² x − 1 . 75 ³ 4 + 1 2 ² x − 9 . 75 ³ 4 + 206 . 6 6 ² x − 11 . 5 ³ 3 + C 3 x + C 4 z = 0 at x = ⇒ C 4 = z = 0 at x = 11 . 5 ⇒ C 3 = 8 . 975 lbf · in 3 From (2) x = 0: E I z ± = 8 . 975 x = 11 . 5: E I z ± = − 683 . 5 At O : E I θ = ± 1908 . 4 2 + 8 . 975 2 = 1908 . 4 lbf · in 3 A : E I θ = ± ( − 2153 . 1 ) 2 + ( − 683 . 5 ) 2 = 2259 lbf · in 3 (dictates size) θ = 2259 30(10 6 )( π/ 64)(1 . 25 4 ) = . 000 628 rad n = . 001 . 000 628 = 1 . 59 A z x y O C B...
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