Calculating Torsional irregularity

[BAGW]

Hi,

I have couple of questions regarding torsional irregularity.

1) When one of the building deflects 10mm and other end deflects -2mm, how id average delta calculated. Is it done by making -2mm zero. So the deflection at the building ends become, 12mm and 0mm. Average delta = 6mm?

2) While calculating torsional irregularity, do you always use rigid diaphragm in analysis software, so that one end of the building deflects maximum and the other end minimum? If semi-rigid diaphragm is used, deflection at the mid span of the diaphragm may be more than the ends. See image attached.

Thanks

 

[Shu Jiang PEng SE]

To my understanding, the deflection should be like this. The deflection due to diaphragm rotation should be used to check the torsional irregularity.

 

[RobertHale]

Here are my thoughts on this:
(1) I think you need to preserve signs and not normalize to zero. Eg. δavg = 0.5*(10+-2) = 4 mm. Doing otherwise will impact the calculation of both the torsional irregularity check and the Amplification factor. In your example, you need to look at the ratio of δmax and δavg and compare to 1.2. If you use the first numbers (-2/10) then δmax/δavg = 2.5. If you normalize to 0 at one end then δmax/δavg = 2.0. This, in turn, leads to Ax = 4.34 (limited to 3.0) in the first case and Ax = 2.78 in the second case.
(2) I agree with Shu Jiang. You are trying to capture the rotational sensitivity of the structure. The calculation of δavg is only meant to give a notional base to measure the relative displacement of the ends so the in plane bending of the diaphragm itself is unimportant to the check.

Robert Hale, PE

 

[Shu Jiang PEng SE]

I agree with RoberHale except one: You may have a type error and I think the average deflection in the middle should be δavg = 0.5*(10-(-2)) = 6 mm.

 

[BAGW]

@RobertHale,

What happens if one end of the building is deflecting 10mm and the other end -10mm. Then the delta average is 0?

I think the delta average is how the building deflects at the center. So it has to be 10+abs(10)/2 = 10mm. Amplification factor will be (20/1.2*10)2 = 2.78.

@Shu Jiang,

With delta avg = 6mm, what will the accidental torsion? With what value of max displacement will this be compared?

Thanks

 

[wannabeSE]

BAGW,
In the case of +10" and -10", I would say δ[sub]max[/sub] / δ[sub]min[/sub] = ∞ and the torsional amplification factor would be 3 (maximum required amplification factor per ASCE 7-10 §12.8.4.3). However, this is based on interpreting the code literally without using engineering judgement.

I understand your reasoning because it is based on the actual rotation of the diaphragm. I just don't know if that is the intent. The code language hasn't changed since the 1997 UBC. (Previous UBC beginning in 88 are similar except use the maximum displacement and the average displacement are at extreme points rather than ends)

 

[RobertHale]

@Shu Jiang: δavg should be 4mm. The operation you propose is not an average; Link
@BAGW: I think the +/- 10mm situation is nonsensical. In that case you are suggesting the infinitely stiff in the translational direction yet torsionally flexible. It would be possible to produce this deflection if you applied a torsion to a symmetrical structure, but the nature of the problem is a lateral load applied at a point not coincident to the center of rigidity. There has to be some global translation. I cannot see a way to get the behavior you propose.

Robert Hale, PE

 

[wannabeSE]

RobertHale
I believe the +/-10mm is possible but not likely. There can be global translation at the center of mass while still having the extreme ends rotate so the displacement is +/-10 mm at the extreme ends

 

[Shu Jiang PEng SE]

wannabeSE. The possibility of deflection of +/-10mm is not possible unless the structure experiences a pure torsion. In most case, due to the eccentricity of weight center and stiffness center, the build ex experiences a horizontal force on weight center which is equivalent to a force and a moment on the stiffness center. Therefore, the maximum positive and negative deflection will not be the same.

 

[RobertHale]

@Shu Jiang. I have chewed over this problem for the past few days. wannabe is correct you need to have an offset center of rigidity with a center of mass offset even further from that. Even though the result of the calculation is undefined it indicates the behavior of an extreme torsional irregularity and you would need to use the Ax= 3.0 like wannabe said.
I have attached a diagram showing the configuration you would need to get the result. [URL unfurl="true"]https://res.cloudinary.com/engineering-com/image/upload/v1496160437/tips/ASCE_Torsional_Irregularity_bccsok.pdf[/url]

Robert Hale, PE

 

[Shu Jiang PEng SE]

I am not sure this picture will address your question