Page 467 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.29

Considering the wall as fixed-ended (same as piers 2 and 3), the relative rigidity is

R = 12.053 + (12.053 − 11.602)(0.3)

= 11.92 > 7.424 (by interpolation from Table A.27)
Thus, the relative rigidity of the entire wall is greater than the sum of the relative
rigidities of piers 2 and 3 .

Example 7.6 Rigidity of a perforated wall with two openings.
Figure E7.6 shows an 8-in.-thick (nominal) reinforced concrete masonry wall.
2
Calculate its rigidity by the three methods discussed earlier. ′ f = 2000 lb/in. .
m
V
1

16'
3 4 6'
2

5 4'
8' 8' 6'–8' 8' 5'–4'
36'
FIGURE E7.6 Rigidity of a perforated shear wall.

Solution

2
′ f = 2000 lb/in E = 900 ′ f = 900(2.0)
m
m
m
2
= 1800 kips/in t = 7.625 in. (8-in. nominal)
Method A: The rigidity of the wall would be taken as the sum of the rigidities of
piers 2, 3, and 4 (assumed as fixed-ended).
Pier 2:
h = 10 = 125
.
d 8
⎛ ⎡
3
⎛
h ⎞
h ⎞
∆ = 1 = ⎜ ⎟ + ⎜ ⎟ ⎥ ⎤ ⎥
⎢
3
2 ⎝ d ⎠ ⎝ d ⎠
Et ⎣ ⎦
m
= 1 [( . 3 + 3 125)]
( .
125)
( 1800 7 625)
)( .
4 −
0
.
= 4 1553 1 ( 0 )in.
R = 1 = 1 = 2407 kips/in .
−
4
2 ∆ 4 1553 10 )
(
.
2