29_ch 04 Mechanical Design budynas_SM_ch04

29_ch 04 Mechanical Design budynas_SM_ch04 -...

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98 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (2) ± F x = 0, 10(10 3 ) 5(10 3 ) R O R C = 0 R O + R C = 5(10 3 ) lbf (3) δ C = [10(10 3 ) 5(10 3 ) R C ]20 AE [5(10 3 ) + R C ] (10) R C (15) = 0 45 R C + 5 ( 10 4 ) = 0 R C = 1111 lbf Ans. R O = 5000 1111 = 3889 lbf Ans. 4-59 (1) Choose R B as redundant reaction (2) R B + R C = w l ( a ) R B ( l a ) w l 2 2 + M C = 0( b ) (3) y B = R B ( l a ) 3 3 EI + w ( l a ) 2 24 [4 l ( l a ) ( l a ) 2 6 l 2 ] = 0 R B = w 8 ( l a ) [6 l 2 4 l ( l a ) + ( l a ) 2 ] = w 8 ( l a ) ( 3 l 2 + 2 al + a 2 ) Ans. Substituting, Eq. ( a ) R C = w l R B = w 8 ( l a ) ( 5 l 2 10 a 2 ) Ans. Eq. ( b ) M C = w l 2 2 R B ( l a ) = w 8 ( l 2 2 a 2 ) Ans. 4-60 M =− w x 2 2 + R B ± x a ² 1 , M R B x a ² 1 U
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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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