Horizontal irregularity type 1a and type 1b: torsional and extreme torsional irregularities The following displacements are given for a two-story rigid diaphragm SSMF building shown in Figure 3.22. Horizontal irregularities—Type 1a and 1b only apply to buildings with rigid and semirigid diaphragms.

Question

Horizontal irregularity type 1a and type 1b: torsional and extreme torsional irregularities
The following displacements are given for a two-story rigid diaphragm SSMF building shown in Figure 3.22. Horizontal irregularities—Type 1a and 1b only apply to buildings with rigid and semirigid diaphragms. Determine if Type 1a or Type 1b torsional irregularity exists at the roof level.

Accepted Answer

The irregularity check is defined in terms of story drift.
[latex]\Delta_x=\left(\delta_x-\delta_{x-1}\right)[/latex] (3.30)
A torsional irregularity exists at level x when the maximum story drift is greater than 1.2 times the average story drift:
[latex]\Delta_{max}\gt 1.2\Delta_{Avg.}=\frac{1.2\left(\Delta_{L,x}\Delta_{R,x}\right)}{2}[/latex] (3.31)

Determining story drift at roof
[latex]∆_{L,roof} =\left(\delta_{L,roof}-\delta_{L,2}\right)=(1.3-1.1)=0.2[/latex] in.
[latex]∆_{R,roof} =\left(\delta_{R,roof}-\delta_{R,2}\right)=(1.98-1.2)=0.78[/latex] in.
[latex]∆_{avg.} =\frac{(0.2+0.78)}{2}=0.49[/latex] in.
[latex]∆_{max}[/latex] = 0.78 > 1.2 (0.49)
Thus a torsional irregularity exists—Type 1a
Check for extreme torsional irregularity—Type 1b
[latex]∆_{max}[/latex] = 0.78 > 1.4 (0.49)
Thus an extreme torsional irregularity exists—Type 1b
When a torsional irregularity exists at a level, the accidental torsional moment [latex]M_{ta}[/latex] must be increased by an amplification factor [latex]A_x[/latex]. A specific amplification factor is required at each level of the building.
The amplification factor is a function of displacements, not the story drifts.
[latex]A_x=\left[\frac{\delta_{max}}{1.2 \delta_{avg}}\right]^2[/latex] (3.32) (ASCE 7 Equation 12.8-14)
Then the amplification factor for the roof is calculated as shown here:
[latex]δ_{max}[/latex] = 1.98 in.
[latex]δ_{avg} = \frac{1.98+1.3}{2}=1.64[/latex]
Then,
[latex]A_x=\left[\frac{1.98}{1.2 (1.64)}\right]^2=1.012\gt 1.0[/latex]

Question 3.8: Horizontal irregularity type 1a and type 1b: torsional and e...

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