Subject :- Structural Analysis II

Title :- Moment Distribution Method

Difficulty :- High

1) Fixed end moments :-

AB =>

$M_{ab} = \frac{-wab^2}{l^2} = \frac{-80*3*2^2}{5^2}$ = -38.4 KN.m

$M_{ba} = \frac{wa^2b}{l^2} = \frac{80*3^2*2}{5^2}$ = 57.6 KN.m

BC =>

$M_{bc} = \frac{-wab^2}{l^2} = \frac{-60*1*2^2}{3^2}$ = -26.67 KN.m

$M_{bc} = \frac{wa^2b}{l^2} = \frac{60*1^2*2}{3^2}$ = 13.33 KN.m

BD =>

$M_{bd}$ = 0 [Because there is no load in member BD]

$M_{db}$ = 0 [Because there is no load in member BD]

2) Stiffness (K) :-

Consider Joint B

$K_{BA} = \frac{4EI}{l} = \frac{4*2EI}{5} = \frac{8EI}{5}$ [B]

$K_{BD} = \frac{4EI}{l} = \frac{4*2EI}{4}$ = 2EI [B]

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

Total Stiffness :-

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

=$\frac{8EI}{5} + 2EI + \frac{4EI}{3}$

$K_{B}$ = 4.933EI

3) Distribution factor (DF) :-

$(DF)_{BA} = \frac{K_{BA}}{K_{B}} = \frac{\frac{8EI}{5}}{4.933EI} = 0.32$

$(DF)_{BD} = \frac{K_{BD}}{K_{B}} = \frac{2EI}{4.933EI} = 0.41$

$(DF)_{BC} = \frac{K_{BC}}{K_{B}} = \frac{\frac{4EI}{3}}{4.933EI} = 0.27$

4) Distribution Table :-

5) B.M.D :-