Wind Torsion Case - Flexible Diaphragm

[jgVA]

ASCE-7 (05) figure 6-9 gives two wind load cases with torsion (Case 2 and Case 4). What is the appropriate way to apply these cases to a flexible diaphragm? Adjust the load intensity to one side to achieve the required eccentricity, or just calculate torsional shears as if a rigid diaphragm?

I used section 6.5.12.2.1 Rigid Buildings of All Heights (Figure 6-6) to determine my wind pressures. The building is a four-story wood structure but does not qualify for the exception in 6.5.12.3...

 

[JoshPlumSE]

Take a look at the 1998 version of ASCE-7. That version of the code shows a slightly different load pattern. For rigid diaphragms, it is equivalent to the 2005 version. But, for flexible diaphragms, I think the 1998 version would be easier to apply.

Josh

 

[WillisV]

JoshPlum - I disagree with your comment regarding equivalency with ASCE 7-05. The required torsional eccentricities were increased significantly between ASCE 7-98 and ASCE 7-05.

For example ASCE 7-98 Figure 6-9 has an equivalent eccentricity of e = 0.0625B (when normalized against 100%P) which is 1.8 times less than the equivalent eccentricity of Case 2 in ASCE 7-05 for the same load (100%P). See also ASCE 7-05 Commentary C6.5.12.3.

I do agree with your proposed method; however, as the cases in ASCE 7-05 can be broken down into equivalent off-balanced loading as suggested by the O.P. for flexible diaphragms.

 

[JoshPlumSE]

Guilty as charged! I did not actually look at the magnitudes of the torsion. I was going purely off of memory of something I had looked at 5 years ago or so.

Clearly my memory isn't what it used to be!

But, at least the old diagram provides a codified method of handling the torque through un-equal diaphragm distribued loads rather than applying a point torque.

Another method (perhaps a simpler one) would be to first apply your distributed loads to your walls and frames as if there were NO torque. Then apply a set of point loads (or distributed loads) to each frame that creates the torque. The sum of these forces would have to be zero. That's actually probably a lot quicker for most structures.

 

[Stillerz]

I have a question here too!!!

In ASCE 7-05 section 6.5.12.3 they give a long equation for "e" for flexible buildings but in figure 6-9 they simply show e=+/- 0.15B.

Is figure 6-9 for rigid structures only?

 

[jgVA]

Stillerz - I think you are correct. It appears that the intent is to use 15%B for rigid structures and equation 6-21 for flexible structures. Note the definition of eQ under equation 6-21 where it specifically states that the e per Figure 6-9 is for rigid structures...

Thank you all for your responses. It makes sense to me to apply a uniform 0.75P over the entire width and then calculate additional shears directly to the shear walls per the torsional moment. I am going to assume that the wall shears due to torsion will be proportional to the distance from the geometric center...

 

[Stillerz]

There are ways to model this if you are using a modeling software.