31_ch 08 Mechanical Design budynas_SM_ch08

# 31_ch 08 Mechanical Design budynas_SM_ch08 - T / d = 716....

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234 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 8-46 2 P (200) = 12(50) P = 12(50) 2(200) = 1 . 5 kN per bolt F s = 6 kN/bolt S p = 380 MPa A t = 245 mm 2 , A d = π 4 (20 2 ) = 314 . 2mm 2 F i = 0 . 75(245)(380)(10 3 ) = 69 . 83 kN σ i = 69 . 83(10 3 ) 245 = 285 MPa σ b = CP + F i A t = ± 0 . 30(1 . 5) + 69 . 83 245 ² (10 3 ) = 287 MPa τ = F s A d = 6(10 3 ) 314 . 2 = 19 . 1MPa σ ± = [287 2 + 3(19 . 1 2 )] 1 / 2 = 289 MPa m = S p σ ± = 380 289 = 91 MPa Thus the bolt will not exceed the proof stress. Ans. 8-47 Using the result of Prob. 5-31 for lubricated assembly F x = 2 π fT 0 . 18 d With a design factor of n d gives T = 0 . 18 n d F x d 2 π f = 0 . 18(3)(1000) d 2 π (0 . 12) = 716 d or
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Unformatted text preview: T / d = 716. Also T d = K (0 . 75 S p A t ) = . 18(0 . 75)(85 000) A t = 11 475 A t Form a table Size A t T / d = 11 475 A t n 1 4 28 0.0364 417.7 1.75 5 16 24 0.058 665.55 2.8 3 8 24 0.0878 1007.5 4.23 The factor of safety in the last column of the table comes from n = 2 f ( T / d ) . 18 F x = 2 (0 . 12)( T / d ) . 18(1000) = . 0042( T / d ) 12 kN 2 F s 2 P O 200 50...
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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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