PS: I'm searchig for general expressions for rigidity center. Think that I found them:
[tab][tab][tab]Xr=[Σ]i(Kyi*Xi)/[Σ]iKyi
[tab][tab][tab]Yr=[Σ]i(Kxi*Yi)/[Σ]iKxi
I just have to add inclination koeficient?
Rigidities are vectorial quantities (thinking in finding a rigidity center). Hence you should be able to treat them as vectors, the components can be treated separately at their own eccentricity, i.e., substituting the resultant inclined rigidity by its x and y components at corresponding eccentricities at the center of the wall element.
I agree with ishvaaag that shearwall rigidities are vector quantities. Each of them can be represented in vector notation by a.i + b.j where i and j are unit vectors in the x and y direction. But, assuming the rigidity is the same in the opposite direction, which is usually true, then each of them can also be represented as -a.i -b.j.
Is there necessarily a single center of rigidity for an arbitrary group of walls? I would guess that the answer is no because some can act in the opposite direction. It depends upon the forces and the location of forces acting on them.
However, I have not thought it through, so do not want to commit to a particular answer at this moment. It is nearly midnight and I am going to bed. Nighty night.
