written 7.8 years ago by | modified 2.8 years ago by |
Mumbai University > Civil Engineering > Sem 5 > Structural Analysis – II
Marks: 10 Marks
Year: May 2015
written 7.8 years ago by | modified 2.8 years ago by |
Mumbai University > Civil Engineering > Sem 5 > Structural Analysis – II
Marks: 10 Marks
Year: May 2015
written 7.8 years ago by |
It is a Non – Sway Frame.
Fixed End Moments :
FE MAB = Wab² / L² = 20 * 2 * 3² / 5² = 14.4 KN-m
FE MBA = - Wa²b / L² = 20 * 2² * 3 / 5² = -9.6 KN-m
FE MBC = FE MCB = WL² / 12 = 10 * 4² / 12 = 13.33 KN-m
FE MBD = FE MDB = WL / 8 = 30 * 3 / 8 = 11.25 KN-m
Distribution Factor & K :
Joint A ,
AB = 0 = K = DF
Joint B ,
BA = 1 / L = 1 / 5 = 0.2 I = K , DF = 0.28
BC = 3 I / 4 L = 3 I / 4 * 4 = 0.1875 I = K , DF = 0.26
BD = 1 / L = 1 / 3 = 0.33 I = K , DF = 0.46
ΣK = 0.71715 I
Joint C ,
1 = K , DF = 1
Joint D ,
DB =0 = K = DF
Moment Distribution Table :