18_ch 04 Mechanical Design budynas_SM_ch04

18_ch 04 Mechanical Design budynas_SM_ch04 -...

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Chapter 4 87 4-37 I = π 64 (40 4 ) = 125 . 66(10 3 )mm 4 R O = 2(500) + 600 1000 1500 = 1900 N M = 1900 x 2000 2 x 2 1500 ± x 0 . 4 ² 1 where x is in meters EI dy dx = 950 x 2 1000 3 x 3 750 ± x 0 . 4 ² 2 + C 1 EIy = 900 3 x 3 250 3 x 4 250 ± x 0 . 4 ² 3 + C 1 x + C 2 y = 0at x = 0 C 2 = 0 y = 0at x = 1m C 1 =− 179 . 33 N · m 2 Substituting C 1 and C 2 and evaluating y at x = 0 . 4m, y A = 1 207(10 9 )125 . 66(10 9 ) ± 950 3 (0 . 4 3 ) 250 3 (0 . 4 4 ) 179 . 33(0 . 4) ² 10 3 =− 2 . 061 mm Ans. y | x = 500 = 1 207(10 9 )125 . 66(10 9 ) ± 950 3 (0 . 5 3 ) 250 3 (0 . 5 4 ) 250(0 . 5 0 . 4) 3 179 . 33(0 . 5) ² 10 3 =− 2 . 135 mm Ans . % difference = 2 . 135 2 . 061 2 . 061 (100) = 3 . 59% Ans. 4-38 R 1 = w ( l + a )[( l a ) / 2)] l = w 2 l ( l 2 a 2 ) R 2 = w ( l + a ) w 2 l ( l 2 a 2 ) = w 2 l ( l + a ) 2 M = w 2 l ( l 2 a 2 ) x w
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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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