Subject: Structure Analysis 2

Topic: Moment Distribution Method

Difficulty: Hard

1) Fixed End Moments (FEM's)

$M_{ab} = \frac{-wab^2}{l^2} = \frac{-100*2*3^2}{5^2} = -72 KN.m$

$M_{ba} = \frac{-wa^2b}{l^2} = \frac{-100*2^2*3}{5^2} = 48 KN.m$

$M_{bc} = \frac{-wl^2}{12} = \frac{-20*3^2}{12} = -15 KN.m$

$M_{cb} = \frac{wl^2}{12} = \frac{20*3^2}{12} = 15 KN.m$

$M_{bd} =0$ (because of no load)

$M_{db} =0$ (because of no load)

2) Stiffness:

Considering Joint B:

$K_{BA} = \frac{4EI}{l} = \frac{4E2I}{5} = \frac{8}{5}EI$ [B]

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

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

Total Stiffness:

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

$=\frac{8}{5}EI + \frac{4}{3}EI + EI = 3.933EI$

3) Distribution Factor:

$(D.F)_{BA} = \frac{K_{BA}}{K_B} = \frac{\frac{8EI}{5}}{3.93EI} = 0.40$

$(D.F)_{BD} = \frac{K_{BD}}{K_B} = \frac{\frac{4EI}{3}}{3.93EI} = 0.33$

$(D.F)_{BC} = \frac{K_{BC}}{K_B} = \frac{EI}{3.93EI} = 0.27$

4) Distribution Factor: