HW18 - __ PROBLEM 6.43 )m 1—° ” 9 A Warren bridge...

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Unformatted text preview: __ PROBLEM 6.43 )m 1—° ” 9 A Warren bridge truss is loaded as shown. Determine the force in :7 1'1": 3:” 'rm 7 m h m members EG, PG, and F11. L1; 1 I i ‘ [‘7 7°} IU 'I‘lv "Em 5m ")m j SOLUTION ‘ FBD Truss: —- ZR =O:K\ =0 / K 24-144 2 0: (25 m]K_‘ 410 m)(3 kN) — (5 m)(3 kN) : 0 1921,8ka {ZMC : o; {10111)[LS kN) + (6 m)fi,H = 0 Section FED: F”, = 73 kN F“, : 3.00 kN C 4 Liza/1,, = 0:(12.5 m}(1.8kN) —(6m)(Fm) : 0 En H J’ - FM; 2 3.75 kN Fm : 3.75 kN T 4 _ 12 ‘ TEE, = 0: Bay +1.8kN : 0 Fm = —1.95kN Fm =1,95(}kN C4 PROBLEM 6.51 SOLUTION FBI) Truss: FED Secfion: Notes: y, = so yu‘)’/:1m PROBLEM 6.55 *: I 1 ‘ A rooftruss is loaded as shown. Determine the force in members FH, ‘ 4-1 1“ f7 ._ 6.], and 1 H 1 l hir‘ \ H a , _ l :n . at I‘ n 9 41h . .t \v 'r . ‘ k "LL A, E .. PMHL M --~Hl~l~Wv—7m -- «nu SOLUTION By symmetry: A1. 2 M). = 2.45 kipsT 72550: AY=0 = (£le 7 5 3(03 kip) — 12 fi(2.35 kips) + 7 fiLfiFE-H] = 0 5' F0, : 3.887 kips Fm = 3.89 kips T 4 I :1 2M“ 7 (6 my,” 7 (4 fi)(1.0 kips} 7 (10 fi){0.3 kips) 7 (16 fi)(0.l kips) + (16 fi)(2.45 kips) = 0 Fm : 75.10 = 5.10kips C By inspection: Fm: Fm = 5.10 kipsC 4 I 2!; = 0: fiasm kips) + 245 kips 7 1.4 kips 7 fig, 2 0 Fa : 2.40 kips T 4 ...
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This note was uploaded on 05/18/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

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HW18 - __ PROBLEM 6.43 )m 1—° ” 9 A Warren bridge...

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