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Equivalent Static And Accidental Torsion Load

Many seismic codes have procedures for calculation of equivalent static or accidental torsion loads.

GSA provides a method of calculating these loads.

The first stage in the calculation is to establish the nodal masses. This includes the mass of elements plus the additional mass derived from any loads. If a modal analysis has previously been carried out this information is picked up from that analysis task. For each storey we can calculate the storey mass MsM_{s} by summing the mass of all nodes in that storey.

If a response spectrum case has been selected the base shear, VV, is extracted from that calculation, otherwise the base shear is calculated using the code equations (e.g. UBC, IBC or FEMA).

At this point different codes have different requirements. In UBC 1997, depending on the period, an additional force, FtF_{t}, is added to the top storey.

Ft=0T<0.7sFt=0.25×V0.07>T>0.25sFt=0.07×TV\begin{aligned}F_{t} =& 0 & T < 0.7s\\ F_{t} =& 0.25 \times V & 0.07 > T > 0.25s\\ F_{t} =& 0.07 \times TV\end{aligned}

For IBC and FEMA an exponent KK on the distribution function is required

K=1T<0.5sK=2T>2.5sK=1+(T−0.5)2\begin{aligned}K &= 1&T < 0.5s \\ K &= 2&T > 2.5s \\ K &= 1 + \frac{(T - 0.5)}{2}\end{aligned}

For other codes KK is set to 1.

Equivalent Static​

We then calculate the force to be applied to each storey, ss, at height, hh

Fs=MsHsK∑MsHsK(V−Ft)F_{s} = \frac{M_{s}{H_{s}}^{K}}{\sum_{}^{}{M_{s}{H_{s}}^{K}}}\left( V - F_{t} \right)

And for the top storey

Fs=MsHsK∑MsHsK(V−Ft)+FtF_{s} = \frac{M_{s}{H_{s}}^{K}}{\sum_{}^{}{M_{s}{H_{s}}^{K}}}\left( V - F_{t} \right) + F_{t}

This storey force is then distributed to the nodes, nn, in proportion to their mass

Fn=MnMsFsF_{n} = \frac{M_{n}}{M_{s}}F_{s}

Accidental Torsion​

For the accidental torsion we calculate the storey masses as for the equivalent static and we calculate the centre of mass of each storey. Storey calculations are relative to the centre of mass. We also need the width of the storey which is calculated by the difference in the extreme coordinates in the direction of interest.

We have an offset, oo, which is based on the width of the storey. The accidental torsion moment for the storey Mzz,sM_{zz,s} is then

Mzz,s=Fs⋅oM_{zz,s} = F_{s} \cdot o

This is then applied as forces to the nodes in the storeys

F^n=MnMsFs⋅o{\widehat{F}}_{n} = \frac{M_{n}}{M_{s}}F_{s} \cdot o

As well as a resulting torsion on the storey this may lead to a force

F^=∑nodesF^n\widehat{F} = \sum_{nodes}^{}{\widehat{F}}_{n}

So we correct for this by adjusting these forces by

F~n=F^n−MnMsF^s{\widetilde{F}}_{n} = {\widehat{F}}_{n} - \frac{M_{n}}{M_{s}}{\widehat{F}}_{s}

And sum the moment on the storey is

M~zz,s=∑nodesFn⋅on{\widetilde{M}}_{zz,s} = \sum_{nodes}^{}{F_{n} \cdot o_{n}}

Finally we adjust these force values so that we have the correct moment on the storey

Fn=F~nMzz,sM~zz,sF_{n} = {\widetilde{F}}_{n}\frac{M_{zz,s}}{{\widetilde{M}}_{zz,s}}