Load distribution to skewed shear walls
Load distribution to skewed shear walls
(OP)
Normally the wind or seismic loads from flexible horizontal diaphragm structure are distributed to the shear walls based on the assumption that the roof or floor acts as simple beams spanning between the paralleled shearwalls in the direction of the loads. The distances between the paralleled shearwalls are the spans for the horizontal diaphragm. For some buildings, some shearwalls are either paralleled or perpendicular to each other, say at a skewed 45-degree with the others. The following are questions regarding to this type of building: (a) how should the skewed shear wall 'support' location for the horizontal diaphragm be defined, (b) how to break a long skewed shear wall into several supports for the diaphragm, and (c) how close in distance shall two shear walls be considered as one single shear line support.






RE: Load distribution to skewed shear walls
The force along the wall will, in actuality, vary from less to more depending on the width of the diaphram at its narrow and wider sides.
For a simplified analysis of the diaphram you could conservatively assume the widest width....or perhaps 67% of the width for design.
b)I'm not sure I'd break up the shearwall into several supports as it is truly acting as one unit.
c)Depends, I suppose, on the actual material you are dealing with. I'm not sure how to offer an opinion as it may depend on the actual layout/material/etc.
RE: Load distribution to skewed shear walls
RE: Load distribution to skewed shear walls
The loads distributed to the skewed and the right elevation walls will be different, since the spans between the right elevation wall and the first ‘support’ of the skewed shear wall in these two assumptions are different.
If no ‘break up’ shall be considered, for 56.56 feet long skewed shear wall the horizontal diaphragm will be considered as supported at 20 feet from the right elevation (the distributed loads to the right elevation wall will even be larger than if the skewed shear wall is just 28.28 feet).
RE: Load distribution to skewed shear walls
The diaprhragm is also acting as a single unit. Flexible, too. Thus, it will warp in plan as the lateral loads are applied to it. There is no pure method here to define the "span" of the trapezoidal diaphram. I would either conservatively design my diaphram based on the longest width, or some rationally derived reduced width...such as 2/3 of the full width. Treat the diaphram as a rectangle with the 2/3 span and design....then calculate the triangular (true) loading applied to the skewed shearwall - plf varying from less to more as the true diaphram span grows.
Hope this makes sense....and I hope I'm understanding your question and problem correctly.
RE: Load distribution to skewed shear walls
RE: Load distribution to skewed shear walls
Assuming:
1. The horizontal diaphragm deflects as a rigid body
2. The walls behave in a linear elastic fashion (doubtful) proportional to deflections in the plane of the wall.
3. Only the component of the developed racking force parallel to the racking load is effective.
Suppose the diaphragm deflects 10mm. The skewed wall will experience some of this as in plane deflection which will develop racking forces, and some as out of plane deflection which will cause the wall to tilt without developing racking force. The in plane deflection will be proportional to the cosine of the angle of the wall. Hence the racking force developed in the skewed wall will be less than if it was parallel to the load, and if linear elastic will be proportional to the cosine of the angle.
Of the racking force developed only the component parallel to the load is effective as resistance. The effective component will be proportional to the cosine of the angle of the wall.
Thus the angle of the wall must be used twice; once to find the deflection, and again to resolve the developed force into its components. Thus the effective bracing capacity of a skewed wall will be proportional to (cos Ø )^2, where Ø is the angle between the wall and the applied load. I.e. a wall at 45 degrees will provide 50% of the resistance of a similar wall parallel to the load.
I hope this explanation is clear and helps you.