Subject - Structural Analysis II

Title - Moment Distribution Method

Difficulty - High

1) Fixed end moments (FEM) :

AB =>
$M_{ab}$ =$\frac {-wl^2}{12}$ = $\frac{-10*3^2}{12}$ = -7.5 KN.m

$M_{ba}$ = $\frac{wl^2}{12}$ = $\frac{10*3^2}{12}$ = 7.5 KN.m

BC =>

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

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

2) Stiffness (K) :

$K_{ba}$ = $\frac{HEI}{l}$ = $\frac{4EI}{3}$

$K_{bc}$ = $\frac{HEI}{l}$ = $\frac{4EI}{4}$ = EI

Total Stiffness :

K = $K_{ba}$ + $K_{bc}$

= $\frac{4EI}{3}$ + EI =$\frac{7EI}{3}$

3) Distribution Factor :

$(DF)_{ba}$ = $\frac{K_{ba}}{K}$ = $\frac{\frac{4EI}{3}}{\frac{7EI}{3}}$ =$\frac{4}{7}$ [B]

$(DF)_{bc}$ = $\frac{K_{bc}}{K}$ = $\frac{EI}{\frac{7EI}{3}} =\frac{3}{7}$ [B]

4) Table :

B.M.D