Torsional Analysis for Flexible Diaphragms

Seismic Torsions

The program implements NBC Structural Commentary J 174, which says.

Structures with flexible diaphragms are designed so that their loads, including the effects of accidental torsion, are distributed to the vertical elements using the tributary area concept.  Accidental torsion should be taken into account by moving the center of mass by +/- 0.05D_nx and using the largest seismic loads for the design of each vertical element.

Note that the Wood Design Manual Example 1, Seismic Design Considerations includes this force by adding 5% to the total shear line force.

Calculation Procedure

Noting that to direct (non-torsional) component of the shear line force should be that determined by tributary area distribution, this can be achieved by setting the rigidities K to the forces determined using the flexible procedure, seeing that

F_di = F * K_i / Σ K_i ,

using the notation in the Rigid Diaphragm topic, F being the total force. In that case, the center if mass CM =  Centre of rigidity CR, and we are including just the accidental eccentricity ae_a and not the eccentricity of the structure or loads. We also do not consider the torsional moment J in the other direction, as none of the loads are in the other direction.   The torsional component on each line is then

F_ti = T * K_i * d_i / (J_x)

where d_i is the distance of the shear line from the center of load, and

J_x = Σ K_i * d_i ²   

T = F * ae_a

Shear lines already heavily loaded get higher contributions of accidental torsion, rather than those that are stiffer as in the case of rigid analysis.

The distribution of load is proportional to the distribution of the mass of building material, which itself is proportional to the “area” that the NBC commentary is referring to. So with this method, the torsional component is distributed using the tributary area concept as the NBC mandates.

Verification of Calculation Procedure

For a simple case of a uniform load on a rectangular building, adding a certain percentage of torsional eccentricity will add that percentage of total force on the structure. For more complicated situations, the amount of force will vary due to the effect of the moment arms of the shear line locations.

To show that this simple case is consistent with the seismic example from the Wood Design Manual, consider a 20 m wide building with 1 kN/m force the diaphragm.

F = 20 kN; F_d1 = F_d2 = 10 kN; K₁ = K₂  = 10 kN; e_a = 1 m; di = 10 m; T  = 20 kN-m; J = 2000 kN-m2

F_t1 =  F_t2 = 20 kN-m x 10 kN x 10 m / 2000 kN-m2  = 1kN

= 5% F