7_ch 07 Mechanical Design budynas_SM_ch07

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

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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 ± EIy ± =− 131 . 1 2 ( x 2 ) + 5 ² x 1 . 75 ³ 3 5 ² x 9 . 75 ³ 3 62 . 3 2 ² x 11 . 5 ³ 2 + C 1 (1) EIy =− 131 . 1 6 ( x 3 ) + 5 4 ² x 1 . 75 ³ 4 5 4 ² x 9 . 75 ³ 4 62 . 3 6 ² x 11 . 5 ³ 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: EIy ± = 1908 . 4 x = 11.5: EIy ± =− 2153 . 1 xz plane (treating z ↑+ ) EIz ± = 17 . 4 2 ( x 2 ) 2 ² x 1 . 75 ³ 3 + 2 ² x 9 . 75 ³ 3 + 206 . 6 2
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Unformatted text preview: x 11 . 5 2 + C 3 (2) E I z = 17 . 4 6 ( x 3 ) 1 2 x 1 . 75 4 + 1 2 x 9 . 75 4 + 206 . 6 6 x 11 . 5 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 = 1908 . 4 2 + 8 . 975 2 = 1908 . 4 lbf in 3 A : E I = ( 2153 . 1 ) 2 + ( 683 . 5 ) 2 = 2259 lbf in 3 (dictates size) = 2259 30(10 6 )( / 64)(1 . 25 4 ) = . 000 628 rad n = . 001 . 000 628 = 1 . 59 A z x y O C 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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