27_ch 10 Mechanical Design budynas_SM_ch10

27_ch 10 Mechanical Design budynas_SM_ch10 - 458 = . 078...

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Chapter 10 287 This means (2 . 5 2 . 417)(360 ) or 29 . 9 from closed. Treating the hand force as in the middle of the grip r = 1 + 3 . 5 2 = 2 . 75 in F = M y r = 57 . 2 2 . 75 = 20 . 8 lbf Ans. 10-33 The spring material and condition are unknown. Given d = 0 . 081 in and OD = 0 . 500, (a) D = 0 . 500 0 . 081 = 0 . 419 in Using E = 28 . 6 Mpsi for an estimate k ± = d 4 E 10 . 8 DN = (0 . 081) 4 (28 . 6)(10 6 ) 10 . 8(0 . 419)(11) = 24 . 7 lbf · in/turn for each spring. The moment corresponding to a force of 8 lbf Fr = (8 / 2)(3 . 3125) = 13 . 25 lbf · in/spring The fraction windup turn is n = Fr k ± = 13 . 25 24 . 7 = 0 . 536 turns The arm swings through an arc of slightly less than 180 , say 165 . This uses up 165 / 360 or 0.458 turns. So n = 0 . 536 0 .
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Unformatted text preview: 458 = . 078 turns are left (or . 078(360 ◦ ) = 28 . 1 ◦ ). The original configuration of the spring was Ans. (b) C = . 419 . 081 = 5 . 17 K i = 4(5 . 17) 2 − 5 . 17 − 1 4(5 . 17)(5 . 17 − 1) = 1 . 168 σ = K i 32 M π d 3 = 1 . 168 ± 32(13 . 25) π (0 . 081) 3 ² = 296 623 psi Ans. To achieve this stress level, the spring had to have set removed. 10-34 Consider half and double results Straight section: M = 3 F R , ∂ M ∂ P = 3 R F 3 FR L ± 2 28.1 ±...
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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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