- Cantilever Method Theorem 20kn 20 N 300x300 400x400 300x300 300x300 3m 400x400 300x300 3m 40kn E X2 300x300 400x400 M 1 (29.15 KiB) Viewed 41 times
- Cantilever Method Theorem 20kn 20 N 300x300 400x400 300x300 300x300 3m 400x400 300x300 3m 40kn E X2 300x300 400x400 M 2 (24.59 KiB) Viewed 41 times
- Cantilever Method Theorem 20kn 20 N 300x300 400x400 300x300 300x300 3m 400x400 300x300 3m 40kn E X2 300x300 400x400 M 3 (20.55 KiB) Viewed 41 times
- Cantilever Method Theorem 20kn 20 N 300x300 400x400 300x300 300x300 3m 400x400 300x300 3m 40kn E X2 300x300 400x400 M 4 (22.73 KiB) Viewed 41 times
- Cantilever Method Theorem 20kn 20 N 300x300 400x400 300x300 300x300 3m 400x400 300x300 3m 40kn E X2 300x300 400x400 M 5 (19.45 KiB) Viewed 41 times
- Cantilever Method Theorem 20kn 20 N 300x300 400x400 300x300 300x300 3m 400x400 300x300 3m 40kn E X2 300x300 400x400 M 6 (29.03 KiB) Viewed 41 times
METHOD
CANTILEVER METHOD THEOREM 20kN 20 N 300x300 400x400 300x300 300x300 3m 400x400 300x300 - 3m 40KN E x2 300x300 400x400 m 300x300 3m X2 B c 5m 5m Step 1 For Second Floor Neutral Axis Af(x) = A (x0) + A2(x1) + A3(x2) 0.34() = (0.09)(0) + (0.16)(5m) + (0.09) (10m) x = 5.0m
CANTILEVER METHOD THEOREM 20KN 20kN NA H H Рpca Рен PF 40KN D E F = 5.0m 10m PAD Pas ТРce * = 5.0m 10m PpG PAD PEN PI PRE PcF
CANTILEVER METHOD THEOREM 20KN NA H Step 2 PDG PEH Determine the Relationship of the Column Forces Using Similar Triangles Ppg + Pei Pel 10m 5m Rewrite relationships in terms of Ppg Per = PDG * = 5.0m 10m Рос PEN PFI
CANTILEVER METHOD THEOREM 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 Per = PDG PAD PRE PCF X= 5.0m 10m PAD PRE PCE
CANTILEVER METHOD THEOREM 20KN G NA H Step 3 1.5m Determine the Forces at Columns EM = 0 (+) 20kN (1.5) - Ppc(5) - Pog(5) = 0 Рpc PER EN 1 Рос. PDG = 3.0KN * = 5.0m 10m Рpc PEN Ppt
CANTILEVER METHOD THEOREM 20KN PGH G VGH VpG Step 4 Calculating for Inflection Forces ΣF, = 0 ( +) VGH - Ppc = 0 VGH = 3kN Рps EM = 0 (+) 20kN (1.5m) - VGH (2.5m) - PG (1.5m) = 0 Pah = 15kN MGH 20kN PGH MDG VGN VOG ΣF, = 0(+) 20kN - Pou - Vpc = 0 VoG = 5kN Calculating for Moment at Joint from BEAM EMGH = 0 (+) EMGH = -VGH (2.5m) EMGH = -7.5kNm Calculating for Moment at Joint from COLUMN EMDG = 0 (+) EMDG = Vpc(1.5m) = 7.5kNm Рpa Complete Idealized Frame for JOINT G EMDG