23_ch 05 Mechanical Design budynas_SM_ch05

23_ch 05 Mechanical Design budynas_SM_ch05 - − 25 333(...

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Chapter 5 137 The screw equation is F i = T 0 . 2 d (2) From Eqs. (1) and (2) p i = F w r i = T 0 . 2 d w r i dF x = fp i r i d θ F x = ± 2 π o fp i w r i d θ = fT w 0 . 2 d w r i r i ± 2 π o d θ = 2 π fT 0 . 2 d Ans. 5-32 (a) From Prob. 5-31, T = 0 . 2 F i d F i = T 0 . 2 d = 190 0 . 2(0 . 25) = 3800 lbf Ans. (b) From Prob. 5-31, F = w r i p i p i = F w r i = F i w r i = 3800 0 . 5(0 . 5) = 15 200 psi Ans . (c) σ t = r 2 i p i r 2 o r 2 i ² 1 + r 2 o r ³ r = r i = p i ( r 2 i + r 2 o ) r 2 o r 2 i = 15 200(0 . 5 2 + 1 2 ) 1 2 0 . 5 2 = 25 333 psi Ans . σ r =− p i =− 15 200 psi (d) τ max = σ 1 σ 3 2 = σ t σ r 2 = 25 333 ( 15 200) 2 = 20 267 psi Ans . σ ± = ( σ 2 A + σ 2 B σ A σ B ) 1 / 2 = [25 333 2 + ( 15 200) 2
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Unformatted text preview: − 25 333( − 15 200)] 1 / 2 = 35 466 psi Ans . (e) Maximum Shear hypothesis n = S sy τ max = . 5 S y τ max = . 5(63) 20 . 267 = 1 . 55 Ans. Distortion Energy theory n = S y σ ± = 63 35 466 = 1 . 78 Ans. dF x p i r i d ±...
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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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