Subject :- Structural Analysis II

Title :- Moment Distribution Method

Difficulty :- High

Free Body diagram :

1) F.E.M's [Fixed end Moment] :

AB =>

$M_{ab} = \frac{M}{4} = \frac{40}{4}$ = 10 KN.m

= $\frac{M}{4} - \frac{6EIδ}{l^2} = \frac{40}{4} - \frac{6*1785*0.012}{6^2}$

[ I=85$10^5$ $mm^4$ [ E=2.1$10^5$ mpa (N/$mm^4$)

I=85$10^5$$({10}^{-3})^4$ $m^4$ E=$\frac{2.1*10^5*{10}^{-3}}{{({10}^{-3})}^{2} m^2}$

I=85${10}^{-7} m^4$] E=2.1$10^8$ KN/$m^2$]

EI =85${10}^{-7} m^4$ * 2.1*$10^8$ KN/$m^2$ = 1785

= $\frac{40}{4} - \frac{6*1785*0.0012}{4}$ = 6.43 KN.m

$M_{ba} = \frac{M}{4} - \frac{6EIδ}{l^2}$ = 6.43 KN.m

BC =>

$M_{bc} = \frac{-wl^2}{30} + \frac{6EIδ}{l^2}$

= $\frac{-15*6^2}{30} + \frac{6*1785*0.012}{6^2}$ = -14.43 KN.m

$M_{cb} = \frac{+wl^2}{20} + \frac{6EIδ}{l^2}$

= $\frac{15*6^2}{20} + \frac{6*1785*0.012}{6^2}$ = 30.57 KN.m

CD =>

Note :- In overhanging beam we have to calculate direct final moment $M_{cd}$

$M_{cd}$ = 20*2 = 40 = -40 KN.m

2) Stiffness (K) :-

Consider joint B :

$K_{BA} = \frac{4EI}{l} = \frac{4EI}{6} [K_{B}$]

$K_{BC} = \frac{3EI}{l} = \frac{3EI}{6} [K_{B}$]

*Total stiffness :

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

=$\frac{4EI}{6} + \frac{3EI}{6} = \frac{7EI}{6}$

3) Distribution Factor :

$(D.F)_{BA} = \frac{K_{BA}}{K_{B}} = \frac{\frac{4EI}{6}}{\frac{7EI}{6}} = \frac{4}{7}$

$(D.F)_{BC} = \frac{K_{BC}}{K_{B}} = \frac{\frac{3EI}{6}}{\frac{7EI}{6}} = \frac{3}{7}$

4) Table :-