21_ch 07 Mechanical Design budynas_SM_ch07

21_ch 07 Mechanical Design budynas_SM_ch07 - 1 Integrating...

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198 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 7-17 In Eq. (7-22) the term ± I / A appears. For a hollow unform diameter shaft, r I A = ² π ( d 4 o d 4 i ) / 64 π ( d 2 o d 2 i ) / 4 = ² 1 16 ( d 2 o + d 2 i )( d 2 o d 2 i ) d 2 o d 2 i = 1 4 ³ d 2 o + d 2 i This means that when a solid shaft is hollowed out, the critical speed increases beyond that of the solid shaft. By how much? 1 4 ³ d 2 o + d 2 i 1 4 ± d 2 o = ² 1 + ´ d i d o µ 2 The possible values of d i are 0 d i d o , so the range of critical speeds is ω s 1 + 0 to about ω s 1 + 1 or from ω s to 2 ω s . Ans. 7-18 All steps will be modeled using singularity functions with a spreadsheet. Programming both loads will enable the user to first set the left load to 1, the right load to 0 and calculate δ 11 and δ 21 . Then setting left load to 0 and the right to 1 to get δ 12 and δ 22 . The spreadsheet shown on the next page shows the δ 11 and δ 21 calculation. Table for M / I vs x is easy to make. The equation for M / I is: M / I = D 13 x + C 15 ± x 1 ² 0 + E 15 ± x 1 ² 1 + E 17 ± x 2 ² 1 + C 19 ± x 9 ² 0 + E 19 ± x 9 ² 1 + E 21 ± x 14 ² 1 + C 23 ± x 15 ² 0 + E 23 ± x 15
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Unformatted text preview: 1 Integrating twice gives the equation for Ey . Boundary conditions y = 0 at x = 0 and at x = 16 inches provide integration constants ( C 2 = 0) . Substitution back into the deection equation at x = 2, 14 inches provides the s. The results are: 11 2 . 917(10 7 ), 12 = 21 = 1 . 627(10 7 ), 22 = 2 . 231(10 7 ) . This can be veried by nite element analysis. y 1 = 20(2 . 917)(10 7 ) + 35(1 . 627)(10 7 ) = 1 . 153(10 5 ) y 2 = 20(1 . 627)(10 7 ) + 35(2 . 231)(10 7 ) = 1 . 106(10 5 ) y 2 1 = 1 . 329(10 10 ), y 2 2 = 1 . 224(10 10 ) w y = 6 . 177(10 4 ), w y 2 = 6 . 942(10 9 ) Neglecting the shaft, Eq. (7-23) gives 1 = 386 6 . 177(10 4 ) 6 . 942(10 9 ) = 5860 rad/s or 55 970 rev/min Ans ....
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