20_ch 06 Mechanical Design budynas_SM_ch06

20_ch 06 Mechanical Design budynas_SM_ch06 - = 1080 N m T...

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166 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (b) If the shaft is not rotating, τ m = τ a = 0 . σ m = σ a =− 2009 P k b = 1 (axial) k c = 0 . 85 (Since there is no tension, k c = 1 might be more appropriate.) S e = 0 . 723(1)(0 . 85)(500) = 307 . 3MPa n f = 307 . 3(10 6 ) 2009 P P = 307 . 3(10 6 ) 3(2009) = 51 . 0(10 3 )N = 51 . 0kN Ans . Yield: n y = 800(10 6 ) 2(2009)(51 . 0)(10 3 ) = 3 . 90 Ans . 6-28 From Prob. 6-27, K f = 2 . 84, K fs = 1 . 76, S e = 312 . 3MPa σ max =− K f 4 P max π d 2 =− 2 . 84 ± (4)(80)(10 3 ) π (0 . 030) 2 ² =− 321 . 4MPa σ min = 20 80 ( 321 . 4) =− 80 . 4MPa T max = fP max ³ D + d 4 ´ = 0 . 3(80)(10 3 ) ³ 0 . 150 + 0 . 03 4 ´
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Unformatted text preview: = 1080 N m T min = 20 80 (1080) = 270 N m max = K f s 16 T max d 3 = 1 . 76 16(1080) (0 . 030) 3 (10 6 ) = 358 . 5 MPa min = 20 80 (358 . 5) = 89 . 6 MPa a = 321 . 4 80 . 4 2 = 120 . 5 MPa m = 321 . 4 80 . 4 2 = 200 . 9 MPa a = 358 . 5 89 . 6 2 = 134 . 5 MPa m = 358 . 5 + 89 . 6 2 = 224 . 1 MPa 307.3 m a 800 800...
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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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