B-8 *Determine need for redundancy factor, ρ.*

The building has a seismic design category of D, and therefore, per Paragraph 4-4, the redundancy factor is

calculated as follows;

20

(EQ. 4-1)

rmax A x

Evaluation of rmax;

For the longitudinal moment frames, rmax is taken as the maximum of the sum of the shears in any two

adjacent columns in the plane of the frame divided by the story shear. The portal method is used as an

approximation for the distribution of shear in the columns.

Therefore, since each story has 7 columns per frame, 2 frames per story, and the frames are each equally

loaded;

2

rmax =

= 0.17 at each floor level

12

For the transverse shear walls, rmax is taken as the shear in the most heavily loaded wall divided by 10/lw.

Where lw is the wall length in feet divided by the story shear. At every story level, the most heavily loaded

wall occurs at either grid lines 1 or 9.

Therefore, at each story level;

10

10

=

= 0.339

l w 29.5'

At the third story level;

34.1k

rmax =

(0.339) = 0.034

338k

At the second story level;

27.6k

rmax =

(0.339) = 0.034

k

276

At the first story level;

13.7 k

rmax =

(0.339) = 0.034

138k

Calculations to determine the redundancy factor are tabularized below;

Story

Earthquake

rmax

Ax

rx

2

(ft )

Level

Direction

Third

Transverse

0.034

8,181

-4.5 < 1

Longitudinal

0.170

8,181

0.70 < 1.0

Second

Transverse

0.034

8,181

-4.50 < 1.0

Longitudinal

0.170

8,181

0.70 < 1.0

First

Transverse

0.034

8,181

-4.50 < 1.0

Longitudinal

0.170

8,181

0.70 < 1.0

B-9 *Determine need for overstrength factor *Ω o .

Per Paragraph 5.2.6.3.2 of FEMA 302, collector elements, splices, and their connections to resisting

elements shall be designed for the load combinations of Section 5.2.7.1 of FEMA 302. Therefore, for these

force controlled elements the following seismic load will be used;

For both the transverse or the longitudinal direction;

H2-26

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