10_ch 04 Mechanical Design budynas_SM_ch04

# 10_ch 04 Mechanical Design budynas_SM_ch04 - x 2 + b 2 −...

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Chapter 4 79 z OB = Fbx 6 EIl ( x 2 + b 2 l 2 ), where b = 15", x = 12", l = 48", I = 0 . 2485 in 4 Then, z A = 502(15)(12)(144 + 225 2304) 6(30)(10 6 )(0 . 2485)(48) =− 0 . 081 44 in For z B use x = 33" z B = 502 ( 15 )( 33 )( 1089 + 225 2304 ) 6 ( 30 )( 10 6 )( 0 . 2485 )( 48 ) =− 0 . 1146 in Therefore, by superposition z A =+ 0 . 1182 0 . 0814 =+ 0 . 0368 in Ans. z B =+ 0 . 1103 0 . 1146 =− 0 . 0043 in Ans. 4-21 (a) Calculate torques and moment of inertia T = (400 50)(16 / 2) = 2800 lbf · in (8 T 2 T 2 )(10 / 2) = 2800 T 2 = 80 lbf, T 1 = 8(80) = 640 lbf I = π 64 (1 . 25 4 ) = 0 . 1198 in 4 Due to 720 lbf, ﬂip beam A-9-6 such that y AB b = 9 , x = 0 , l = 20 , F =− 720 lbf θ B = dy dx ± ± ± ± x = 0 =− Fb 6 EIl (3
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Unformatted text preview: x 2 + b 2 − l 2 ) = − − 720(9) 6(30)(10 6 )(0 . 1198)(20) (0 + 81 − 400) = − 4 . 793(10 − 3 ) rad y C = − 12 θ B = − . 057 52 in Due to 450 lbf, use beam A-9-10, y C = − Fa 2 3 E I ( l + a ) = − 450(144)(32) 3(30)(10 6 )(0 . 1198) = − . 1923 in 450 lbf 720 lbf 9" 11" 12" O y A B C R O R B A C B O R 1 R 2 12" 502 lbf 21" 15" z x...
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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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