Page 487 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.49

(F = 1.35 found by interpolation for S = 0.4 and 0.5 from Table 7.3)
1
v
2
061 =
S = (. ) 041
.
D1
3
2. Calculate fundamental period T. Use approximate period, T , for T.
a
T = C (h ) x
n
t
a
C = 0.02 (all other structural systems) h = 60 ft x = 0.75
n
t
T = 0.02 (60) 0.75 = 0.43 s
a
3. Calculate the seismic response coefficient, C .
s
For a building frame system with special reinforced masonry shear walls, R = 5.5
(Table 7.5)
.
C = S DS = 060 = 011
.
⎛ R ⎞ ⎛ 55 . ⎞
s
⎝ I ⎠ ⎝ 10 . ⎠
C need not exceed
s
≤
C = S D1 = . 041 = . 017 f for TT
. ⎞
s,max ⎛ R ⎞ ⎛ 55 L
3
T (. )
04
. ⎠
⎝ I ⎠ ⎝ 10
T L, min = 4 s (ASCE 7-05 Fig. 22.15)
In this example T = 0.43 s < T L, min = 4 s. Therefore, C s, max = 0.17 [Eq. (7.52) does
not apply because T < T . C shall not be less than
L
s
C = 0.044S ≥ 0.01
DS
s
0.44S = 0.044(0.6) = 0.0264 > 0.01
DS
Therefore, C S, min = 0.0264
For the given structure, S = 0.45g < 0.60g; therefore Eq. (7.54) does not apply.
1
Therefore, C = 0.11 governs

s
V = C W = 0.11W
s
W = 400 + 450 + 500 + 600 + 650 = 2600 kips
V = 0.11W = 0.11(2600) = 286 kips
F = C V
x
vx
wh k
C = n xx
vx
∑ wh k
ii
i=1