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СЕ 3440 Homework #27 Due 11.22.2019 1A) Use lAB 890 in4 and lBc 1170 in4 A and C are fixed supports. Determine the internal m

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22ft 30k ABRE elt в с K2494 * loft D slope Deflection method o Fixed etd moments - MFAR= W2 = -2x92-96 kaft MFBAZ Co2 = 2432Mes = 60+ 3x8x0,656_ [204488= 39] &=s=0 MCB = 60+ 7x103 EOR 16 Joint- equilibrium equations 8– At joint B - Emozo MBA + MAC220 km 2 slope-Deflection methods Filed and merrents 8 km #snsm MFAB = MFBA=0 Maße = -0,2 = -20833=-iskrin es= constant. MFCBJoint- equilibsium equations- At joint B - Embro WBA + mec=0 O.SEIOB - 15+ 1.33ES OB+0167€ DO = 0 2113ESOB+0.67 €80c = 15 790 KK 80 sh V 3m 3 30 20 3) Ected eat morende s- MFAR = -22 = -2017= -54 tum team an HC MEBA= Code = 2xy = 8l kom Mifac -wo?Mga+ moc=0 Blot ou 44 EDOC -60+0.67 E208=0 loll ETOR = 21 EIOB = -18.92 Final momento From 90 MAB = -58.16 krim from eq ② MBA221444 30k A= 0.7in 0.40.08301074 LAR= 890 int = 890xC0.08934-0104444 kraft 1544 *c IBC = 1170 in = 1170x (0.08331-010544 OnMos = 60+ 2EX0105 5200+ 08- 3X0107 16 = 60 +6125x183 EOS – 8.2x103 MCB = 59199 +6.25*10% 60s -> Joint- equili blicam questionplss plss rate it

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