Subject - Structural Analysis II

Title - Moment Distribution Method

Difficulty - High

1) Fixed end moments [FEM]:

AB :

$M_{ab} = \frac{-wl^2}{12} = \frac{-5*3^2}{12}$ = -3.75 KN.m

$M_{ba} = \frac{wl^2}{12} = \frac{5*3^2}{12}$ = +3.75 KN.m

BC :

$M_{bc} = \frac{-wl^2}{12} = \frac{-5*4^2}{12}$ = -6.67 KN.m

$M_{cb} = \frac{wl^2}{12} = \frac{5*4^2}{12}$ = 6.67 KN.m

2) Stiffness (K) :

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

$K_{BC} = \frac{3EI}{l} = \frac{3EI}{4}$ [$K_{B}$]

Total Stiffness :

$K_{B} = K_{BA} + K_{BC}$

= EI + $\frac{3 EI}{4}$

$K_{B} = \frac{7EI}{4}$

3) Distribution Factor :

$(D.F)_{BA} = \frac{K_{BA}}{K_{B}} = \frac{EI}{\frac{7EI}{4}} = \frac{4}{7}$ [B]

$(D.F)_{BC} = \frac{K_{BC}}{K_{B}} = \frac{\frac{3EI}{4}}{\frac{7EI}{4}} = \frac{3}{7}$ [B]

4) Distribution Table :