21_ch 13 Mechanical Design budynas_SM_ch13

21_ch 13 Mechanical Design budynas_SM_ch13 - . 5 k lbf Ans...

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Chapter 13 353 ± M D = R DC × F C + R DG × W + T = 0 (1) R DG × W =− 2404 i 1785 j + 2140 k R DC × F C =− 6 F z C j + 6 F y C k Substituting and solving Eq. (1) gives T = 2404 i lbf · in F z C =− 297 . 5 lbf F y C =− 356 . 7 lbf ± F = F D + F C + W = 0 Substituting and solving gives F x C =− 344 lbf F y D = 106 . 7 lbf F z D =− 297 . 5 lbf So F C =− 344 i 356 . 7 j 297 . 5 k lbf Ans . F D = 106 . 7 j 297
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Unformatted text preview: . 5 k lbf Ans . 13-39 P t = 8 cos 15 = 7 . 727 teeth/in y 2 z x a F a a 2 F t a 2 F r a 2 F a 32 F r 32 F t 32 G C D x z y W r W a W t 4.04" 3" 3" F y C F x C F z C F z T D F y 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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