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n: Question 1 Use Moment Distribution method to determine the reactions of the continuous beam shown in Figure 1. Modulus of

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given YEN 4K 2kn/m سے SYN am am foam um bo I hug 4 why fixed and moment MBC -2x42 12 - 2.667 khm mog २५५ 2.667 kom T2 Mcp coxDF • Jointe О.6 3у20 Ч listide се 0.4 ЧуэЕ) CD Joint О.Ч DC цу 20 (143) Е = 2.5E о. 6 чу за DE Joint E 04 ч 30 ED Re: Ии О. 66.6 0.4 MBA MBC mcol ०.५ moc 0.6 MOE 0.4, 0.6 MERMEE MEEMEGC mco 10.67-4 ५ --10.67 -4 -२.61 २.67 -17.33-8.665 4 -8 २६५ 8-8 4kim skilm B om LORG Ym ↑ 5.62 5.62 +२० R! Ro Rell Rg! RB TR = 8 Rp X4+567-20-4482 0 RBI = 7.595 KN Rel - +405kN { Rc46–24884Re- Ret Rell .405 + 7.42 Re = H4749) RC = 7,825149 RO= Rot Roll 8.587 1.667 RO TORUIKN RE RETRE 9339 + 935 RE 4.689 kH Refinally RB = 12.595kn Rc - 7.825 kn a RO RE 10.841 KN 4,689 KN 9.6sky RE > HC=0

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