25_ch 13 Mechanical Design budynas_SM_ch13

25_ch 13 Mechanical Design budynas_SM_ch13 - C k Putting it...

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Chapter 13 357 Or F B = [(613) 2 + (318 . 5) 2 ] 1 / 2 = 691 N radial ± F = F A + W + R B = 0 F A =− ( W + F B ) =− ( 637 i + 1333 j + 5119 k + 318 . 5 i + 613 j ) = 318 . 5 i 1946 j 5119 k Ans. Radial F r A = 318 . 5 i 1946 j N, F r A = [(318 . 5) 2 + ( 1946) 2 ] 1 / 2 = 1972 N Thrust F a A =− 5119 N 13-42 From Prob. 13-41 W G = 637 i 1333 j 5119 k N p t = p x So d G = N G p x π = 48(25) π = 382 mm Bearing D to take thrust load ± M D = R DG × W G + R DC × F C + T = 0 R DG =− 0 . 0725 i + 0 . 191 j R DC =− 0 . 1075 i The position vectors are in meters. R DG × W G =− 977 . 7 i 371 . 1 j 25 . 02 k R DC × F C = 0 . 1075 F z C j 0 . 1075 F y
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Unformatted text preview: C k Putting it together and solving Gives T = 977 . 7 N m Ans . F C = 233 j + 3450 k N, F C = 3460 N Ans . F = F C + W G + F D = F D = ( F C + W G ) = 637 i + 1566 j + 1669 k N Ans . G x y z F D F C W G D C 72.5 191 35 Not to scale...
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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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