B-4 *Calculate Vertical Distribution of Forces.*

Fx = Cvx V

(EQ. 5.3.4-1 FEMA 302)

w xhk

x

C vx =

(EQ. 5.3.4-2 FEMA 302)

n

∑wh

k

ii

i=1

where; k = 1 in both directions for the building period is less than 0.5 seconds

The calculations are tabularized below*;

Story

wi

hi

wixhi

Cvx

CvxxVtrans

CvxxVlong

Level

=Fx,trans

=Fx,long

(kips)

(ft)

(ft-kips)

(kips)

(kips)

Roof

1283

33

42337

0.45

338

239

3rd

1573

22

34597

0.37

276

195

nd

2

1573

11

17299

0.18

138

98

SUM =

4428

94232

1.00

753

531

*Note: For metric equivalents; 1-ft = 0.30m, 1-kip = 4.448KN, 1-ft-kip = 1.36KN-m

Therefore;

Transverse direction;

Longitudinal direction;

B-5 *Perform Static Analysis.*

General;

Because the diaphragms are rigid, relative rigidities of the lateral load resisting elements must be

determined in order to establish the distribution of seismic loads. In the transverse direction, the shear

walls are analyzed based on their flexural and shear deformations of a cantilever wall using closed form

equations. In the longitudinal direction, moment frames are analyzed using a two-dimensional computer

analysis program (RISA-2D, version 4.0) to determine their rigidity. Increased flexibility due to cracking

for both the shear walls and the moment frames was accounted for by using cracked section properties in

accordance with Section 10.11.1 of ACI 318-95.

The following diagram shows the computer model geometry used to model the longitudinal moment

frames. Relative rigidities are determined for each floor level. For example, the stiffness at the roof is

determined by applying a 1000k load at the roof level (distributed uniformly along the length of the frame

at that level) with no other loads acting on the model. The deflection of the frame is then taken as the

average of all nodes at that level.

1-ft = 0.30m

1-kip = 4.448KN

H2-20

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