6_ch 04 Mechanical Design budynas_SM_ch04

# 6_ch 04 Mechanical Design budynas_SM_ch04 - 2 400 3 1000 3...

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Chapter 4 75 4-13 R O = 7 10 (800) + 5 10 (600) = 860 lbf R C = 3 10 (800) + 5 10 (600) = 540 lbf M 1 = 860(3)(12) = 30 . 96(10 3 ) lbf · in M 2 = 30 . 96(10 3 ) + 60(2)(12) = 32 . 40(10 3 ) lbf · in σ max = M max Z 6 = 32 . 40 Z Z = 5 . 4in 3 y | x = 5ft = F 1 a [ l ( l / 2)] 6 EIl ±² l 2 ³ 2 + a 2 2 l l 2 ´ F 2 l 3 48 EI 1 16 = 800(36)(60) 6(30)(10 6 ) I (120) [60 2 + 36 2 120 2 ] 600(120 3 ) 48(30)(10 6 ) I I = 23 . 69 in 4 I / 2 = 11 . 84 in 4 Select two 6 in-8 . 2 lbf/ft channels; from Table A-7, I = 2(13 . 1) = 26 . 2in 4 , Z = 2(4 . 38) in 3 y max = 23 . 69 26 . 2 ² 1 16 ³ =− 0 . 0565 in σ max = 32 . 40 2(4 . 38) = 3 . 70 kpsi 4-14 I = π 64 (40 4 ) = 125 . 66(10 3 )mm 4 Superpose beams A-9-6 and A-9-7, y A = 1500(600)400 6(207)10 9 (125 . 66)10 3 (1000) (400 2 + 600 2 1000 2 )(10 3 ) 2 + 2000(400) 24(207)10 9 (125 . 66)10 3 [2(1000)400
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Unformatted text preview: 2 400 3 1000 3 ]10 3 y A = 2 . 061 mm Ans. y | x = 500 = 1500(400)500 24(207)10 9 (125 . 66)10 3 (1000) [500 2 + 400 2 2(1000)500](10 3 ) 2 5(2000)1000 4 384(207)10 9 (125 . 66)10 3 10 3 = 2 . 135 mm Ans. % difference = 2 . 135 2 . 061 2 . 061 (100) = 3 . 59% Ans. R C M 1 M 2 R O A O B C V (lbf) M (lbf in) 800 lbf 600 lbf 3 ft 860 60 O 540 2 ft 5 ft...
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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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