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Deflections of Trusses, Beams, and Frames: Work-Energy Methods 30 kN/m 50 KN + B T Hinge 4.5 m E=200 GPa I = 400(106)mm = 225

solve for horizontal deflection at point c using virtual work, please show work for reactions and all other steps

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Answer #1

Page. No-1 30 kN/m JI 50KN B Hinge 4.5m А Ax Dx 3m Ay -3m- IDY E = 200 G Pa = 200 x 109 Nim2 = 200x106 kN/m2 I - 400 x 106 mm

answered by: ANURANJAN SARSAM
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Answer #2

Page. No. 2 (Right side) -Dy x3 + Dx x 4.5+ 30x3; 3x = 0 Do = (Dy x3 – 30x3x 2 4,5 (127.5x3–30x3x3 = 55 4.5 Doc=55kN {My=0 TaPage No 3 k t C 1 KN B E 4.5m C A D dax Axt -3m- 3m- ay EMA=0 - dyx6 + 1x4.5=0 dy = 4.5 0 75 = |dy=( dy = 0.75 kn 6 ME=O (RigPage No.4 Bending Moment of each bart due to unit load al e in horizontal rightward directions Moc = -da X X = a - dac X k =Page No 5 6 4.5 Alte= 27.5 X 3 td EI El 0.75 [127.54.23 - 247.5x32 -152004 ]. [127.5 -2.25 16 - 247.5x - 1503 3 4,5 203 + 62.

answered by: ANURANJAN SARSAM
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