Page 476 - Design of Reinforced Masonry Structures

7.38 CHAPTER SEVEN

For the entire wall,
∆ net, wall = ∆ gross − ∆ solid strip + ∆ piers = 0.08125 − 0.0377 + 0.08628 = 0.12983

R = 1 = 1 = 770
.
∆ 0 12983
wall
.
net
The relative rigidity of the perforated shear wall is 7.70.
Method C: Piers 3, 4, and 5 are in parallel. Therefore, their combined relative
rigidity equals the sum of their relative rigidities:
R 3+4+5 = R + R + R
3
4
5
= 3(4.96) = 14.88
Piers 3-4-5 and 6 are in series. Therefore, the reciprocal of the relative rigidity
of Pier 3-4-5-6 equals the sum of the reciprocals of the relative rigidities of
piers 3-4-5 and 6:
1 1 1
= +
R 34 5 6 R 34 5 R 6
+
++
++
= 1 + 1
.
14 88 33 223
.
0 0973
= .

.
R R = 1 = 10 28
++
+
34 5 6
.
0 0973
Now, piers 2, 3-4-5-6, and 7 are in parallel. Therefore, their combined relative
rigidity is given by the sum of their relative rigidities.
R 2+3+4+5+6+7 = 0.714 + 10.28 + 0.714 = 11.71
Pier 1 is in series with all other piers combined. Therefore, the reciprocal of the
relative rigidity of the entire wall equals the sum of the reciprocals of the relative
rigidity of Pier 1 and the pier group 2-3-4-5-6-7:
1 1 1 1 1
= + = + = .
0 1237
R R R 26 122 11 71
.
.
−
wall 1 2 7
R = = 1 = 808.
wall
0 1237
.

The relative rigidity of the perforated shear wall is 8.08.
b. Distribution of shear to various piers:
Method 3:
V = 100 kips
1
0 714 ⎞
⎛ .
V = ⎜ ⎟ ( 100 = .
6 1 kips
)
2 ⎝ 11 71 ⎠
.
⎛ 10 28 ⎞
.
V 3 34 5 = ⎜ ⎝ 11 71 ⎠ ⎟ ( 100 = 87 8 . kips
)
++
.
⎛ 07 . 114 ⎞
V 7 = ⎜ ⎝ 11 71 ⎠ ( ⎟ 100) = 6 1 kips
.
.