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 EIy | = 0 =− 250 3 (6 3 ) + F BE 6 3) 3 + 6 C 1 + C 2 4 . 5 F + 6 C 1 + C 2 = 1 . 8(10 4 ) ( d ) y D = ± Fl AE ² DF = F (2 . 5) ( π/ 4)(5 / 16) 2 (30)(10 6 ) = 1 . 0865(10 6 ) F Substituting and evaluating at x = 9in D = 1 . 5(10 6 )[1 . 0865(10 6 ) F ] 250 3 (9 3 ) + F 6 3) 3 + R C 6 6) 3 + 9 C 1 + C 2 4 . 5 R C + 36 F 1 . 6297 F + 9 C 1 + C 2 = 6 . 075(10 4 ) ( e ) 11 10 0 36 0 0 0 01 . 3038 0 3 1 04 . 50 6 1 4 . 53 6 1 . 6297 9 1 R C F F C 1 C 2 = 500 4500 2250 1 . 8(10 4 ) 6 . 075(10 4 ) R C 590 . 4 lbf, F = 1045 . 2 lbf, F 45 . 2 lbf C 1 = 4136 . 4 lbf · in 2 , C 2 11 522 lbf · in 3 σ = 1045 . 2 ( 4)(5 / 16) 2 = 13 627 psi = 13 . 6 kpsi Ans. σ 45 . 2 ( 4)(5 / 16) 2 589 psi Ans. y A = 1 1 . 5(10 6 ) ( 11 522) 0 . 007 68 in Ans. y B = 1 1 . 5(10 6 ) ³ 250 3 (3 3 ) + 4136 . 4(3)
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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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