43_ch 04 Mechanical Design budynas_SM_ch04

43_ch 04 Mechanical Design budynas_SM_ch04 - Then t ± = 1...

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112 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Now let x = y h ; then ˙ x y and ¨ x y . So the D.E. is ¨ x + ( k / m ) x = g with solution ω = ( k / m ) 1 / 2 and x = A cos ω t ± + B sin ω t ± + mg k At contact, t ± = 0, x = 0, and ˙ x = v 0 . Evaluating A and B then yields x =− mg k cos ω t ± + v 0 ω sin ω t ± + mg k or y =− W k cos ω t ± + v 0 ω sin ω t ± + W k + h and ˙ y = W ω k sin ω t ± + v 0 cos ω t ± To find y max set ˙ y = 0. Solving gives tan ω t ± =− v 0 k W ω or ( ω t ± )* = tan 1 ± v 0 k W ω ² The first value of ( ω t ± )* is a minimum and negative. So add π radians to it to find the maximum. Numerical example: h = 1in , W = 30 lbf, k = 100 lbf / in . Then ω = ( k / m ) 1 / 2 = [100(386) / 30] 1 / 2 = 35 . 87 rad/s W / k = 30 / 100 = 0 . 3 v 0 = (2 gh ) 1 / 2 = [2(386)(1)] 1 / 2 = 27 . 78 in/s Then y =− 0 . 3 cos 35 . 87 t ± + 27 . 78 35 . 87 sin 35 . 87 t ± + 0 . 3 + 1 For y max tan ω t ± =− v 0 k W ω =− 27 . 78(100) 30(35 . 87) =− 2 . 58 ( ω t ± )* =− 1 . 20 rad (minimum) ( ω t ± )* =− 1 . 20 + π = 1 .
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Unformatted text preview: Then t ± * = 1 . 940 / 35 . 87 = . 0541 s . This means that the spring bottoms out at t ± * seconds. Then ( ω t ± )* = 35 . 87(0 . 0541) = 1 . 94 rad So y max = − . 3 cos 1 . 94 + 27 . 78 35 . 87 sin 1 . 94 + . 3 + 1 = 2 . 130 in Ans . The maximum spring force is F max = k ( y max − h ) = 100(2 . 130 − 1) = 113 lbf Ans . The action is illustrated by the graph below. Applications: Impact, such as a dropped package or a pogo stick with a passive rider. The idea has also been used for a one-legged robotic walking machine....
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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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