Page 493 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.55

Substituting the above values in Eq. (7.66) yields

=
in
∑ F x
=
F = ix w = ⎛ V ⎞ W = V (7.69)
=
px in px ⎝ W ⎠
∑ w i
=
ix
Thus, for a single-story building
V = F = C w px (7.70)
x
s
That is, the base shear, story shear, and the diaphragm force are all equal, with the limits of
0.2 S Iw ≤ F ≤ 0.4 S w
DS px x DS px
Example 7.11 presents calculations for this important concept. It was stated earlier that
F = C V (7.63 repeated)
x
vx
xx
and C = wh k (7.64 repeated)
vx Σ wh k
ii
For structures having periods of 0.5 s or less, k = 1 in Eq. (7.64). Therefore, for such one-
story buildings, Eq. (7.64) can be expressed as Eq. (7.69):
C = wh k = wh = 10.
11
xi
Σ wh wh
vx k
ii 11
Therefore, F = C V = V (7.71)
x
vx
Example 7.11 Determination of the diaphragm force for a single-story
building.
A single-story building having special reinforced masonry shear walls and a panel-
ized wood roof is shown in Fig. E7.11. The building is classified as Occupancy Category
1 and is assigned Seismic Design Category D. The site class for the building is D and
S and S for the site have been determined, respectively, as 55.2%g and 24.3%g. Dead
1
S
2
weight of the roof and the masonry walls may be assumed, respectively, as 16 lb/ft and
2
84 lb/ft . Calculate the diaphragm design force at the roof level.

Solution
Given information
S = 55.2%g = 0.552g
s
S = 24.3%g = 0.243g
1
I = 1.0
SDC = D
Site class = D
Bearing wall system with special reinforced masonry shear walls: R = 5 (Table 7.5)
r = 1.0 (assumed, explained later)
2
Diaphragm weight = 16 lb/ft
Wall weight = 84 lb/ft 2