Page 495 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.57

<
C = S D1 for TT (= 4 s minimum)
s max L
(
TR I)/
n
T = C (h ) = 0.02(20) 0.75 = 0.19 s < T = 4 s (minimum)
t
L
a
x
>
C = S D1 = . 031 = . 0 326 0 .10
S,max
TR I) 019 . )
.(5 10/
(/
C s, min = 0.01
Therefore, C = 0.1 (governs)
s
V = C W = 0.1W
s
3. Calculate the tributary weight W for a 1-ft wide strip of wall and the roof
diaphragm.
Tributary wall height = 20 = 10 ft
2
Tributary length of roof diaphragm = 80 ft
W = weight of tributary walls + weight of diaphragm
p1
= (2 walls)(10)(84) + 80(16) = 2960 lb
For a one-story building,

F = V = 0.1W = 0.1(2960) = 296 lb/ft
p1
Check limits,
0.2 S Iw = 0.2(0.5)(2960) = 296 lb/ft
DS
px
0.4S Iw = 0.4(0.5)(2960) = 592 lb/ft
px
DS
Therefore, F = 296 lb/ft
px
Commentary: The redundancy factor r = 1.0 can be used in this example as this
building (assigned to Seismic Design Category D) satisfies requirements of ASCE
7-05 Section 12.3.4 2a (story resists 100 percent of base shear, more than 35 percent
required).

7.8 INFLUENCE OF BUILDING CONFIGURATION
ON LATERAL FORCE DISTRIBUTION
IN SHEAR WALLS

7.8.1 General Considerations
Building configuration can be described as regular or irregular. Regular buildings are those
that are symmetrical or almost symmetrical about both axes in plan, and having load-resisting