18_ch 13 Mechanical Design budynas_SM_ch13

18_ch 13 Mechanical Design budynas_SM_ch13 - F C =(11 4 2...

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350 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design So F B = [( 678 . 8) 2 + ( 178 . 4) 2 ] 1 / 2 = 702 lbf Ans . F A =− ( F B + W ) =− ( 178 . 4 i 678 . 8 k + 106 . 9 i 53 . 4 j + 328 . 3 k ) = 71 . 5 i + 53 . 4 j + 350 . 5 k F A (radial) = (71 . 5 2 + 350 . 5 2 ) 1 / 2 = 358 lbf Ans . F A (thrust) = 53 . 4 lbf Ans . 13-34 d 2 = 15 / 10 = 1 . 5in , W t = 30 lbf, d 3 = 25 10 = 2 . 5in γ = tan 1 0 . 75 1 . 25 = 30 . 96°, 0 = 59 . 04° DE = 9 16 + 0 . 5 cos 59 . 04° = 0 . 8197 in W r = 30 tan 20° cos 59 . 04° = 5 . 617 lbf W a = 30 tan 20° sin 59 . 04° = 9 . 363 lbf W =− 5 . 617 i 9 . 363 j + 30 k R DG = 0 . 8197 j + 1 . 25 i R DC =− 0 . 625 j ± M D = R DG × W + R DC × F C + T = 0 R DG × W = 24 . 591 i 37 . 5 j 7 . 099 k R DC × F C =− 0 . 625 F z C i + 0 . 625 F x C k T = 37 . 5 lbf · in Ans . F C = 11 . 4 i + 39 . 3 k lbf Ans
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Unformatted text preview: F C = (11 . 4 2 + 39 . 3 2 ) 1 / 2 = 40 . 9 lbf Ans . ± F = F D = − 5 . 78 i + 9 . 363 j − 69 . 3 k lbf F D (radial) = [( − 5 . 78) 2 + ( − 69 . 3) 2 ] 1 / 2 = 69 . 5 lbf Ans . F D (thrust) = W a = 9 . 363 lbf Ans . W r W a W t z C D E G x y 5" 8 0.8197" 1.25" Not to scale F x D F z D F x C F z C F y D 1.25 0.75 g...
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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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