An approximate method consists in applying the total of the static seismic force at each of the planar frames.
Since the forces are the same, the lateral floor rigidities at each level can be found dividing the (same) accumulated force (from top to level) by the displacement.
Alternatively, one can work with relative rigidities, taking the inverse of the lateral displacement, multiplied for some constant, to have more workable figures.
Knowing the rigidities, finding its centre of gravity, gives the centre of rigidity of the building at each floor level.
This procedure is useful when it is wanted to check the calculations, since the forces over each frame can be found, and even torsion can be included.
The center of rigidity of a given diaphragm is the point in the plane of the diaphragm that when a horizontal load is applied there, it will cause no rotation in the plane of the diaphragm, but only translation, not necessarily only in the direction of the applied force...
Assuming you have frames with stiffness Kxi in the x direction and frames with stiffness Kyi in the y direction.
The location of the center of rigidity (xr, yr) is given by:
The expressions presented by you to find the centre of rigidity are correct.
I think it is better to consider the frames as a whole.
When you apply a horizontal force at the centre of rigidity of a floor building, you get only horizontal displacement in the direction of the applied force; and, direct forces Hdi over each frame i parallel to such force:
Hdi = Sxn * Kxi/(sum Kxi), where
Sxn = sum of forces from top of building to top of columns of level n considered.
If you want, it is possible to consider torsion.
There are many books relating this topic. I just found the classical “Design of Multistory RC Buildings for Earthquake Motions” by Blume, Newmark and Corning. Also, give a look to the “Handbook of Concrete Engineering” by Fintel, pag. 411.
Don’t lose of sight that the main issue is to obtain the lateral rigidities and that this is supposed to be an expeditious method to answer the original question. If you have difficulties in the modelling, you can run the complete building and get the lateral rigidities of each element dividing the shear resisted by the differential displacement… but it is a lot of additional work.
