Torsional Analysis for Flexible Diaphragms

Case 2 Wind Torsions

ASCE 7 Commentary C 27.4.6 says that the torsional load cases are due to non-uniform wind loading, which is applicable to flexible diaphragm buildings as well as rigid diaphragms. Therefore ASCE 7 Directional Case 2 loading is applied to flexible diaphragm design. This case involves 75% of the wind load in one direction plus an accidental eccentricity of 15% of the building width. AS Commentary

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_i²

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.

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 for this simple case, adding 15% eccentricity increases total force by 15%, consider a 40 ft wide building with 100 lb/ft force on the diaphragm.

F = 4000 lb; F_d1 = F_d2 = 2000 lb; K₁ = K₂ = 2000 lb; e_a = 6 ft; di = 20 ft; T = 24,000 lb-ft; J = 1,600,000 lb-ft2

F_t1 = F_t2 = 24,000 lb-ft x 2000 lb x 20 ft / 2000 kN-m2 = 600 lb

= 15% F