Fixed End Moments:

There are no fixed end moments in any as the load in the members due to external loading of 30 kN is acting at one of the joints. However, there will be fixed end moments due to chord rotations of the members caused by sway. Figure 3.22(b) shows the deflected shape of the structure. Here, the joint B moves horizontally to B? and C moves to C?. Since B and C are connected by the rigid member BC the movement BB? = CC?= Δ (say). The vertical member AB then takes the new position AB? which is a clockwise rotation and will cause fixed end moment - 6EI Δ/6² = -EI Δ/6 at both ends

∴          M _FAB = M _FBA = -  EI Δ/6

Similarly, the vertical member DC rotates clockwise to DC′ and the fixed end moments will be

∴          M_FCD   = M_FDC = -- 6 EI Δ /3²= - 2 EI Δ/3 .

As there is no vertical movement of the joints B and C, there will not be any chord rotation and, therefore, no fixed end moments in the horizontal member BC.

Now, as we do not know the actual valued of E, I, or Δ so we cannot calculate the fixed end moments in absolute terms.

However, we can say that M_FAD/M_FCD = (- 6EI Δ/6)/(- 2 EI Δ/3) =1/4 this shows that the fixed end moment in member CD is four times that in member AB.

Here, we will assume some arbitrary values of fixed end moments in these members in this ratio and perform the moment distribution operation and later on we will calculate the actual values by considerations of horizontal equilibrium.

So we begin with fixed end moments of - 10 kNm in member AB and - 40 kNm in member CD. In member BC, the fixed end moments are 0 at each end.