# SEAOC Structural/Seismic Design Manual 2009 IBC Vol 3: Example 1A Question



## Formula1251 (Mar 21, 2013)

Hopefully this is a quick one. (I feel I need to mention I'm a bridge guy, because I feel the answer should be relatively straightforward for someone more familiar with the material.)

On p. 13, Table 1A-3, in the _SEAOC Structural/Seismic Design Manual 2009 IBC Vol 3: Building Design Examples for Steel and Concrete_, Example 1A, they are calculating the accidental torsion for each of the braced frames as a percentage of the total shear load. Due to symmetry (rectangular building), there is no inherent torsion load, only accidental. For design of the braced frame in question (BF-2 oriented in the building-transverse direction, for seismic loads in the same building-transverse direction), the additional shear due to accidental torsion, calculated as %-total shear, is calculated using the equation: |e_acc * R_i * d_i / sum(R_i * d_i^2)|. I can follow the calculation for the transverse frames, where:


e_acc_trans=10.6' (n-s)

d=105'

R=1.25

sum(R_i * d_i^2) = J = 88,980

and get 1.6%.

However when calculating for the percentage to the longitudinal braced frames, I cannot arrive at the number shown in the table (0.7%). I am using the same formula to calculate, and replacing only the longitudinal frame stiffness, R=1.00, and the offset, d=75' (I am keeping the same eccentricity, since it is asking for the accidental torsion due to the same moment), and I get 0.9%. I'm not sure if I'm looking at this incorrectly or not.

If I try also replacing with the accidental eccentricity for forces in the longitudinal direction (e_acc_long=7.6'), I get 0.64% -- and I do not believe this to be correct, because it is due to a different loading.

Can anyone please shed any light on this and confirm if I'm on the right track leading to 0.9%?

Thank you!


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## ajk244 (Mar 21, 2013)

I don't have it in front of me to look, but I wasn't getting my answers to match on a couple of the SEAOC rigid diaphragm distributions. I believe on one I figured out they were multiplying by the eccentricity twice or something and on another they were adding the rotational shear to the direct shear when they should have been subtracting. On one, they even had the answer correct in a previous version, but changed it for some reason.

Anyway, I'll take a look at this one today and see if I'm having getting the same answers you are.


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## ajk244 (Mar 23, 2013)

I agree with your approach and numbers. The table explicitly says "Vy" and "Ty" and the footnotes are even clearer about it, so the north-south eccentricity should be used (10.6'), leading to 0.9%. Error-filled references have been the most frustrating part of studying for this.


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## Formula1251 (Mar 24, 2013)

Thanks for your responses. I completely agree about the sense of frustration in regards to the errors in the study material. I try to reassure myself that if I can identify that it is, or is likely an error, then I tend to understand the concepts. In that regard it seems to help with the ability to weed out the misleading multiple choice answers during the test!


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