Wind Torsion Case - Flexible Diaphragm
Wind Torsion Case - Flexible Diaphragm
(OP)
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...
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...






RE: Wind Torsion Case - Flexible Diaphragm
Josh
RE: Wind Torsion Case - Flexible Diaphragm
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.
RE: Wind Torsion Case - Flexible Diaphragm
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.
RE: Wind Torsion Case - Flexible Diaphragm
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?
RE: Wind Torsion Case - Flexible Diaphragm
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...
RE: Wind Torsion Case - Flexible Diaphragm