Question

Using Flexibility Method, please find:

a.Support reactions of the structure

b.Calculate and draw the internal force diagram of the structure

Note :

n = 2 , q = 3,84 , p1 = 38,4 , p2 = 3,84

P1 kΝ P1 kN qkN/m F A 3 E 2 E! A 2 EI 3nx 1 m nx 1 m nx 1 m 2 m

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Answer #1

own qkalm PEN BE I 6 (ЗЕТ am X X nxim k hxim 3nxIm n=2 9 = 3.84 P = 38.4 Flexibility PB=(3+2) -3 method (force method) - Dp=238.4 38.4 bole body diagram 361 3.84 kn/m mo 6m 9907.3 am RE A2 F G B (301. 2 153.6 4 76.8 B. 82714 A 99.84 1 38.4 v 38.4 476apply - + F SH B Am am c 01 moment egh due to unit load (m) + RB mide + Rclm, mzdy. in vertical downward dirh unit load at pos Mmdy (99,848 – 3,8422-837.1.2) (ex-C)dx + foto BEL Smmax 3893.76 0 El 10. ST 2 Mmady El S (99.548 – 3.89x2 «2_837.12) (7-103893.76+ 24 R$ + 48RC = 0 9223.35 + 114*667 Rc +48Rg - RB = -8.4038 RC= 76.918 38.4 38.4kn A 3.34 kn Im am D am 6m B am 4 f 7Sheee force desigram 38,9 KN 38.4 KN 14.5182 KN B 1118 -8.5218KN 38.518 KN shear force SA=14.5182 kN. SBeest -145187-3.84 X6shear force is at 3.78078 X= 14.5162 3.84 MA -17.5172 KOM 9.928knm -17.5172 + 14.5182X3.788 78 -- 3,84 X3 780782 2 •472 knm M

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