Page 453 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.15

∆ = 1 [( . 3 + 3 0 4545)]
( .
4 0 4545)
c
( 1800 7 625)
)( .
4 −
= 12 . 6671 10( ) in.
R = 1 = 1 = 7892 k/in.
4 −
c ∆ 1 2671 10 )
(
.
c
The rigidity of the wall is 7892 kips/in., that is, a lateral force of V = 7892 kips would
be required to deflect the wall at the top by 1 in. in the plane of the wall.
Example 7.2 Relative rigidity of a cantilevered shear wall.
Calculate the relative rigidity of the cantilevered shear wall described in Example 7.1.

Solution
The relative rigidity of a cantilevered shear wall can be determined from Eq. (7.9):

R = 1
r
⎡ ⎛ h ⎞ 3 ⎛ h ⎞ ⎤
⎢ 4 ⎝ d ⎠ + 3 ⎝ d ⎠ ⎥
⎣ ⎦
h 20
= = 0 4545
.
d 44
Substituting for h/d in the above equation, we obtain
1 1
R = = = . 0 575
⎡ ⎛ h ⎞ 3 ⎛ h ⎞ ⎤ [( . 3 + 3 0 ( ..4545 )]
r
4 0 4545)
⎢ 4 ⎝ + 3 ⎝ ⎥
⎣ d ⎠ d ⎠ ⎦
Alternatively, we can determine the relative rigidity of a cantilevered shear wall from
Eq. (7.10) if its rigidity is known a priori:

R = R r = 7892 = 0 575.
r
)
(
.
Et ( 1800 7 625)
m

For a shear wall with an h/d ratio of 0.4545, the relative rigidity is 0.575.
Commentary: The relative rigidity of a cantilevered shear wall can also be deter-
mined from Table A.26. For an h/d ratio of 0.4545, the value obtained from Table A.26
is 5.75 (by interpolation). As explained earlier, this value represents 10 times the actual
value of relative rigidity, so that the relative rigidity is 0.575.
Example 7.3 Rigidity and relative rigidity of a fixed-ended masonry shear
wall.
Calculate the rigidity and relative rigidity of the shear wall described in Example 7.1
assuming it to be fixed at both top and bottom. (Fig. E7.3.)