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32_ch 04 Mechanical Design budynas_SM_ch04

32_ch 04 Mechanical Design budynas_SM_ch04 -...

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Chapter 4 101 Since y = 0 at x = 6 in E I y | = 0 = − 250 3 (6 3 ) + F BE 6 (6 3) 3 + 6 C 1 + C 2 4 . 5 F BE + 6 C 1 + C 2 = 1 . 8(10 4 ) ( d ) y D = Fl AE DF = F DF (2 . 5) ( π/ 4)(5 / 16) 2 (30)(10 6 ) = 1 . 0865(10 6 ) F DF Substituting and evaluating at x = 9 in E I y D = 1 . 5(10 6 )[1 . 0865(10 6 ) F DF ] = − 250 3 (9 3 ) + F BE 6 (9 3) 3 + R C 6 (9 6) 3 + 9 C 1 + C 2 4 . 5 R C + 36 F BE 1 . 6297 F DF + 9 C 1 + C 2 = 6 . 075(10 4 ) ( e ) 1 1 1 0 0 3 6 0 0 0 0 1 . 3038 0 3 1 0 4 . 5 0 6 1 4 . 5 36 1 . 6297 9 1 R C F BE F DF C 1 C 2 = 500 4500 2250 1 . 8(10 4 ) 6 . 075(10 4 ) R C = − 590 . 4 lbf, F BE = 1045 . 2 lbf, F DF = − 45 . 2 lbf C 1 = 4136 . 4 lbf · in 2 , C 2 = − 11 522 lbf · in 3 σ BE = 1045 . 2 ( π/ 4)(5 / 16) 2 = 13 627 psi = 13 . 6 kpsi Ans. σ DF = − 45 . 2 ( π/ 4)(5 / 16) 2 = − 589 psi Ans. y A =
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