20KN 300x300 400x400 = 300x300 - 3m 40KN 40KN E E 300x300 400x400 300x300 300x300 400x400 67 3m 300x300 3m А В с X1 5m 5

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20KN 300x300 400x400 = 300x300 - 3m 40KN 40KN E E 300x300 400x400 300x300 300x300 400x400 67 3m 300x300 3m А В с X1 5m 5m X2 Step 1 For Ground Floor Neutral Axis A7() = A1(x0) + A2(x1) + A3(x2) 0.34(x) = (0.09)(0) + (0.16) (5m) + (0.09) (10m) x = 5.0m
20kN G NA H I Step 2 Determine the Relationship of the Column Forces Using Similar Triangles PDG PEH ТРЕ PEL PDG + PEL 10m 5m X = 5.0m Rewrite relationships in terms of Ppg 10m PeL = PDG PDG РЕН PFI
20kN H 1 Step 2 Determine the Relationship of the Column Forces Using Similar Triangles PAD + PCF PCF 10m 5m 40kN D E F Rewrite relationships in terms of Ppg Pce = PDG PAD PBE ↑ Per ž= 5.0m 10m PAD PBE PCF
20kN G NA H 1 Step 3 1.5m Determine the Forces at Columns PpG ТРЕн ↑ Poo EM = 0 (0 +) 20kN (1.5) - Ppg(5) - PpG(5) = 0 Ppg = 3.0kN ž= 5.0m 10m Рpc РЕН PFI
20kN PGH G Step 4 VGH 《一 VPG Calculating for Inflection Forces EF, = 0 (1 +) VGH - PDG = 0 VGH = 3kN PDG EM = 0 (0 +) 20kN (1.5m) - VGH (2.5m) - PG (1.5m) = 0 PGH = 15kN MGH EFx = 0 (+) 20kN – PGH – Vpg = 0 VDG = 5kN 20kN PGH MDG VGH VPG Calculating for Moment at Joint from BEAM EMGH = 0 (0 +) EMGH = -VGH (2.5m) EMGH = -7.5kNm PDG Complete Idealized Frame for JOINT G Calculating for Moment at Joint from COLUMN EMDG = 0 (0 +) EMDG = V0G(1.5m) EMDG = 7.5kNm