Subject :- Structural Analysis II

Title :- Moment Distribution Method

Difficulty :- High

We know that point B is settles down by 0.01 mm

1) Fixed end moments :-

AB =>

$M_{ab} = \frac{-wl^2}{30} - \frac{6EIδ}{l^2} = \frac{-18*6^2}{30} - \frac{6*2EI*0.01}{6^2}$

= -21.6 - 66.67 = -88.27 KN.m

$M_{ba} = \frac{wl^2}{20} - \frac{6EIδ}{l^2} = \frac{18*6^2}{20} - \frac{6*2EI*0.01}{6^2}$

= 32.4 - 66.67 = -34.267 KN.m

BC =>

$M_{bc} = \frac{-wab^2}{l^2} + \frac{6EIδ}{l^2} = \frac{-40*4*2^2}{6^2} + \frac{6*20000*0.01}{6^2}$ = 15.55 KN.m

$M_{bc} = \frac{wab^2}{l^2} + \frac{6EIδ}{l^2} = \frac{40*2*4^2}{6^2} + \frac{6*20000*0.01}{6^2}$ = 68.89 KN.m

Note :- In overhanging part CD direct come the final moment $M_{cd}$ as shown :

$M_{cd}$ = 30*3 = 90 KN.m = -90 KN.m

Final moment :

2) Stiffness (K) :

$K_{BA} = \frac{4EI}{l} = \frac{4*2*EI}{6} = \frac{4EI}{3}$ [$K_{B}$]

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

Total Stiffness :

$K_{B} = K_{BA} + K_{BC} = \frac{4EI}{3} + \frac{3EI}{6} = \frac{11EI}{6}$

Distribution Factor :

$(D.F)_{BA} = \frac{K_{BA}}{K_{B}} = \frac{\frac{4EI}{3}}{\frac{11EI}{6}} = \frac{8}{11}$ [B]

$(D.F)_{BC} = \frac{K_{BC}}{K_{B}} = \frac{\frac{3EI}{6}}{\frac{11EI}{6}} = \frac{3}{11}$ [B]

4) Table :