Load distribution to skewed shear walls 4

[LKT]

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.

 

[JAE]

a) with a skewed shearwall, the lateral force coming into the wall at an angle can be resolved into two orthoganal reactions along the shearwall. The reaction parallel to the shearwall, will be that used to design the diaphragm. The perpendicular reaction (perp. to the shearwall) would have to be resisted by some external element bracing the skewed wall against falling over.

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.

 

[Kramer]

As far as question #3 goes, there is a wide range of opinions on how far shear walls can be offset and still be assumed to act in unison. Some engineers have felt that if you are assuming flexible diaphragms, the shear walls can not be offset at all. To my way of thinking however, if you have a pair of shear walls two feet apart, and they are ten feet tall, it's pretty obvious they are going to share the load. You have to look at the stiffness of the shear wall elements compared to the stiffness of the diaphragm. I'm not saying that you should try to put numbers to this, use your engineering judgement. By the way, if you are interested, there is a pretty good discussion paper on rigid versus flexible diaphragm assumptions at

http://www.seaint.org/seaocconvention/convention1999/Proceedings/panelreport.htm

 

[LKT]

The main point of my first question is where should the ‘support’ location be. To describe the question more clearly, I assign some numeric values for discussion. Say the skewed 45-degree shear wall is 28.28 feet long and one end of this skewed wall meets the right elevation wall of the building. The skewed wall has an effective shear wall length of 20 feet (component parallel to the right elevation, as described by JAE) and its rigidity center is 10 feet from the right elevation shear wall. Should the horizontal diaphragm be considered as supported at 10 feet from the right elevation, or ‘break up’ into 2 supports (just for discussion as if the skewed wall is separated in two shear walls), each has an effective shear wall length of 10 feet and the ‘supports’ located at 5 and 15 feet from the right elevation shear wall, respectively.

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).

 

[JAE]

I would not "break up" the skewed shearwall. It is acting, in reality, as a single unit....simply receiving a varying amount of lateral force because the diaprhagm connecting it is varying in width. Thus, you would have a triangular distributed load along the skewed wall. And: that lateral force is coming into the wall at a 45 degree angle...so you have a parallel and perpendicular component applied to the wall....the parallel works into the wall as a shearwall. The perpendicular component must be taken out by some other means.

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.

 

[LKT]

I agree there shall not be ‘break up’ is relative short skewed shear wall. But for a long one or if there are actually several separated (by windows and doors etc.) short shear walls in the skewed wall line, then the use of several ‘supports’ seems more realistic. Selection of the ‘support locations’ is more of an engineering judgment than anything else.

 

[sdz]

My analysis of the load carried by a skewed shear wall is as follows.
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.