Rigid Diaphramn Lateral Analysis

[mjohan]

When performing a rigid diaphramn lateral analysis utilizing only 3 exterior walls, what is the eccentricity when considering lateral restraint in the direction of the one wall? Is it the width of the building?

How can this eccentricity and now a calculated torsional force be applied into the perpendicular walls? I have only seen examples where the torsional force calculated was applied to shear walls acting in the same direction. Can this torsional force be restrained in any direction? Are their additional considerations when applying the torsional force to the perpendicular walls?

Also, any literature would be of great help.

Thanks
Mark

 

[whyun]

In a simplified case where you have a square shaped floor and walls on north, south and west ends, and lateral load is towards the north, the west wall takes the entire shear (direct shear).

Torsional moment is lateral force times the half the east-west dimension of the building plus 5% times the east-west dimension for accidental torsion. This counterclockwise moment is resisted by torsional shear in all three walls.

Detailed examples with procedure and equations can be found in many text and exam review manuals. I am not sure if this was what you were looking for...

 

[mjohan]

Thanks whun,

Can you also consider this type of rotational analysis with a flexible, or semi flexible diaghramn?

Regards,
Mark

 

[whyun]

For a pure flexible diaphragm (although there is no such thing), you may use a simple supported beam method or alternately (slightly more clumsy way, but...) a tributary area method. Using the former method, in the process, you obtain diaphragm shear values as well. In a purely flexible diaphragm, the rotational analysis is not used.

In a semi-flexible diaphragm (UBC isn't too specific about the design method), an envelope solution from flexible and rigid diaphragm design may be used.

This method was introduced when it was determined that horizontal diaphragms in a small wood framed houses with relatively small diaphragm with many exterior/interior shear walls (by code definition) will calc out to be rigid. Traditionally, it had been considered as a pure flexible diaphragm. Despite this finding, since small wood houses performed relatively well through many earthquakes. Thus, SEAOC recommend using an envelope solution.

To answer your question:
1) Flexible Diaphragm - no rotation
2) Semi Flexible - Perform Ridig and Flexible diaphragm analysis and use the maximum envelope.

That's what I would do... Regards,

 

[rlflower]

Mark:

Before you get carried away with this analysis, step back and think about it for a minute. There is a difference between field experiance and theory in this case:

The most common case where we have three perimeter walls is the typical stand-alone 2-car garage. Yes, in theory, you can analyze the stucture as a three-walled system, depending upon the roof as a rigid or semi-rigid diaphragm; and subsequently, we can banter about until our ears turn blue on whether or not a conventional roof framing system would provide a rigid or semi-rigid diaphragm.

But practical experience tells us that such analysis is not effective - it does not produce satisfactory results. Consider what happens to the non-structural side of the garage - i.e., the side with the garage door - during a seismic event. This non-structural side will undergo pernament distortions with unsightly cracking in the stucco wall finish. I have seen plenty examples of this very thing in sunny Southern California; many cases which would be considered un-salvagable.

In this case, I say enough with theory.

Let's place some steel pipe or tube columns on either side of the gagage door, embedded into a continuous concrete grade beam below the concrete slab. These columns will cantilever the tributary siesmic load. Design these columns for bending stress, but also for drift limitation. Limit your drift to eliminate the cracking in the stucco.

 

[whyun]

Very good suggestion rlflower. If it is a garage chances are, it will consist of a light wood-framed roof. If we introduce a lateral resisting element such as tube columns embeded in wood stud wall or pre-fab short shear wall system, it will carry half of the total seismic from the wood roof. Now, no need for rigid diaph analysis.

If this is not a garage and indeed has a concrete roof or any other rigid diaphragm, the rigidity of the element on the open oprtion will be small so it will behave as a three sided structure for the most part.

mjohan, what is the horizontal diaphragm construction? Also, what is the seismic system? Is it a concrete wall or plywood sheathed wall?

 

[mjohan]

The building I am working on is four stories, with a pretensioned slab construction. The client's question to me was how much cmu will be required in the front of the building? The front of the building is very open, about 90% windows and doors. My first approach is to analyize the lateral restraint of the building using only three walls. All the exterior walls are cmu, I have the ability to use any size block. The rear wall is 70' long, no openings. The left side is 26' long with no openings and the right side is 26' with openings.

The following is a description of how I am analyzing these three walls. Please feel free to comment on anything I have assumed. I used a perforated shear wall analysis on the right side wall. My first assumption was to analyize the cmu on a floor to floor basis. I came up with an effective rigidity per floor, and found a center of rigidity on each level. (The seismec governed this building) The placement of openings in this wall is as such: the top of the wall has no openings and each level down has more openings, the first floor has many openings. What I am not sure is if I should consider the force into each wall as laterally restrained into the wall, and retains the force down the entire height of the building. Each level of the building will have a different center of rigidity. If the force from the 4th floor is 40 kips, 3rd is 30, 2nd 25, 1st is 20, how should the force be restrained on the first floor? Should I assume the cummulative force is restrained by three walls according to their rigidity, and then back into the diaphramn to analyze again? If I follow that analysis, I can sum up the forces on each level, 115 kips, and use the center of rigidity on the first floor, designing the walls and footings accordingly. If I consider the force to stay in the wall, not get transfered back into the diaghram, the force on the first floor would be much higher on the wall with many openings. I believe their is enough cmu in this building to use a shear wall analysis.

Another question I have is how to apply the torsional moment when considering the seismic event in the direction of the one wall. The center of rigidity is at the wall so when I calc the torsional force, I get values for the perpendicular walls only. Is that correct? Also, does this torsional force applied to the perpendicular walls get added to the force obtained from the values obtained when considering the seismic in the direction of the two walls? It seems like it shouldn't, since seismic, like wind, I assume to only happen in one direction. It seems as if assuming the building has only three sides is better, per analysis, then assuming a small amount of restraint in the front of the building.

I know I have a lot of questions, any comments would be greatly appreciated.

If you know of any similar examples utilizing three walls with openings that would be of great help.

Thanks,
Mark

 

[whyun]

First of your question, I would be inclined to say the total shear generated down to a particular level shall be re-distributed through the diaphragm based on the center of rigidity of the seismic system below the diaphragm and a cumulative center of mass of all floors above the diaphragm. Usually, for regular structures, this has almost no impact (i.e. cumulative center of mass is very close to the center of mass of a particular level in question), but in irregular structures, "stacking" method produced wrong results!!!

As for torsional shear, I agree with you that since the center of rigidity is at the 70 ft wall, it shouldn't have any torsional shear. Thus, design the 70 ft wall to carry the entire base shear and design the orthogonal walls for the greater of the direct shear due to load parallel to the 26 ft wall or torsional shear due to load perpendicular to the 26 ft wall.

 

[mjohan]

Thanks for your help.

One interesting result was the torsional force was not even close to the force calculated for that direction. That is why it seemed better to me to assume no restraint in the front of the building, because the torsional force will not govern the orthogonal walls.

Rlflowers point regarding tube steel columns at garages is a great suggestion. Sometimes I will design a cmu pilaster to provide stability at the garage front. Expecially on split level stacking townhouses with cmu foundations.

 

[whyun]

Torsional shear was not controlling due to the length of the building. The torsional moment is approximately base shear times half the width(~25 feet) and coupling arm is ~75 feet.

Stiffening up the open side by using a thicker CMU is a good idea.