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
Froof = 338k
Therefore;
Transverse direction;
F3rd = 276k
F2nd = 138k
Froof = 239k
Longitudinal direction;
F3rd = 195k
F2nd = 98k
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
Previous Page
