19_ch 07 Mechanical Design budynas_SM_ch07

# 19_ch 07 Mechanical Design budynas_SM_ch07 - (0 . 7% low)...

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196 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design One element : Eq. (7-24) δ 11 = 12(12)(24 2 12 2 12 2 ) 6(30)(10 6 )(0 . 049 09)(24) = 1 . 956(10 4 ) in/lbf y 1 = w 1 δ 11 = 5 . 316(1 . 956)(10 4 ) = 1 . 0398(10 3 )in y 2 1 = 1 . 0812(10 6 ) ± w y = 5 . 316(1 . 0398)(10 3 ) = 5 . 528(10 3 ) ± w y 2 = 5 . 316(1 . 0812)(10 6 ) = 5 . 748(10 6 ) ω 1 = ² g w y w y 2 = ² 386 ³ 5 . 528(10 3 ) 5 . 748(10 6 ) ´ = 609 rad/s (30% low) Two elements : δ 11 = δ 22 = 18(6)(24 2 18 2 6 2 ) 6(30)(10 6 )(0 . 049 09)(24) = 1 . 100(10 4 ) in/lbf δ 12 = δ 21 = 6(6)(24 2 6 2 6 2 ) 6(30)(10 6 )(0 . 049 09)(24) = 8 . 556(10 5 ) in/lbf y 1 = w 1 δ 11 + w 2 δ 12 = 2 . 658(1 . 100)(10 4 ) + 2 . 658(8 . 556)(10 5 ) = 5 . 198(10 4 )in = y 2 , y 2 1 = y 2 2 = 2 . 702(10 7 )in 2 ± w y = 2(2 . 658)(5 . 198)(10 4 ) = 2 . 763(10 3 ) ± w y 2 = 2(2 . 658)(2 . 702)(10 7 ) = 1 . 436(10 6 ) ω 1 = ² 386 ³ 2 . 763(10 3 ) 1 . 436(10 6 ) ´ = 862 rad/s
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Unformatted text preview: (0 . 7% low) Three elements : δ 11 = δ 33 = 20(4)(24 2 − 20 2 − 4 2 ) 6(30)(10 6 )(0 . 049 09)(24) = 6 . 036(10 − 5 ) in/lbf δ 22 = 12(12)(24 2 − 12 2 − 12 2 ) 6(30)(10 6 )(0 . 049 09)(24) = 1 . 956(10 − 4 ) in/lbf δ 12 = δ 32 = 12(4)(24 2 − 12 2 − 4 2 ) 6(30)(10 6 )(0 . 049 09)(24) = 9 . 416(10 − 5 ) in/lbf δ 13 = 4(4)(24 2 − 4 2 − 4 2 ) 6(30)(10 6 )(0 . 049 09)(24) = 4 . 104(10 − 5 ) in/lbf 1.772 lbf 1.772 lbf 1.772 lbf 4" 4" 8" 8" 2.658 lbf 2.658 lbf 6" 6" 6" 6"...
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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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