Subject - Structural Analysis II

Title - Moment Distribution Method

Difficulty - High

1) Fixed end moments (FEM's) :

AB =>

$M_{ab} = \frac{-wl^2}{30} = \frac{-10*3^2}{30}$ = -3 KN.m

$M_{ba} = \frac{wl^2}{20} = \frac{10*3^2}{20}$ = 4.5 KN.m

BC =>

$M_{bc} = \frac{-M}{4} = \frac{5}{4}$ = -1.25 KN.m [Note : In couple either both (+ve) or (-ve) clockwise (+ve) anticlockwise (-ve) ]

$M_{bc} = \frac{-M}{4} = \frac{5}{4}$ = -1.25 KN.m

CD =>

$M_{cd} = \frac{-wl}{8} = \frac{-8*4}{8}$ = -4 KN.m

$M_{dc} = \frac{wl}{8} =\frac{8*4}{8}$ = 4 KN.m

2) Stiffness (K) :

$K_{BA} = \frac{HEI}{l} = \frac{4EI}{3}$ [$K_{B}$] [Joint B]

$K_{BC} = \frac{HEI}{l} = \frac{4EI}{6}$ [$K_{B}$] [Joint B]

$K_{CB} = \frac{HEI}{l} = \frac{4EI}{6}$ [$K_{C}$] [Joint C]

$K_{CD} = \frac{HEI}{l} = \frac{4EI}{4}$ = EI [$K_{C}$] [Joint C]

Total Stiffness :

$K_{B} = K_{BA} + K_{BC} = \frac{4EI}{3} + \frac{4EI}{6}$

$K_{B}$ = 2EI

$K_{C} = K_{CB} + K_{CD} = \frac{4EI}{6} + EI = \frac{5EI}{3}$

3) Distribution Factor :

$(D.F)_{BA} = \frac{K_{BA}}{K_{B}} = \frac{\frac{4EI}{3}}{2EI} = \frac{2}{3}$ [B]

$(D.F)_{BC} = \frac{K_{BC}}{K_{B}} = \frac{\frac{4EI}{6}}{2EI} = \frac{1}{3}$ [B]

$(D.F)_{CB} = \frac{K_{CB}}{K_{C}} = \frac{\frac{4EI}{6}}{\frac{5EI}{3}} = \frac{2}{5}$ [C]

$(D.F)_{CD} = \frac{K_{CD}}{K_{C}} = \frac{EI}{\frac{5EI}{3}} = \frac{3}{5}$ [C]

MDM Table :