[

]

(Eq D-14)

= 4(20.5' )2(30)*k*siθ

The shear panel furthest from the center of rigidity provides the greatest torsional resistance.

However, the end panels in the short direction against the exterior walls will not be loaded as heavily

as the panels one bay in from the end because the end panels have only half the tributary area as the

panel one bay in. Therefore, the panels one bay in from the end will be the most critically loaded

because of lateral loads in the short axis and the full width of that bay, and because of its large

contribution to torsional resistance. The torsional resistance of the two shear panels that make up the

critical short-direction frame, Mtrc, may be expressed as follows:

[

]

n

Mtrc = ∑ ρi2k siθ = 2 (3x20.5)2 k siθ = 2(20.5' )2 (9)k siθ

(Eq D-15)

i=1

Equation D-15 shows that the critical short-direction frame provides 3/20 of the total building torsional

resistance (Equation D-15 divided by Equation D-14). This ratio can be used to calculate the applied

torsion in the critical frame by equating the total accidental torsion, Mta, and torsional resistance from

all shear panels, Mtr, as follows:

Mtrc

3

Mtrc =

Mta =

Mta

(Eq D-16)

Mtr

20

The additional shear force due to accidental torsion for the critical frame is now calculated by solving

Equation 3-3 for Qsic,as follows:

Mtrc

Qsic =

(Eq D-17)

ρc

Values of this additional shear force are given in Table D-3 for each floor level.

Values of seismic story shear in the short direction, VxS, are calculated by modifying Equation C-27 to

include the effects of accidental torsion as follows:

(Eq D-18)

D8. LONG-DIRECTION EARTHQUAKE FORCE DEFINITION. The same process is repeated for

the definition of earthquake forces in the long direction of the building. These results are summarized

in Table D-4. The effects from accidental torsion are not added to the frames in the long direction of

the building.

D-6

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