Page 515 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.77

Calculate displacements of shear walls A and B due to shear forces calculated above.
The resulting displacements are
V ′ 49 05
.
δ = A = = 818
.
A
R A 6
V ′ 50 95
.
δ = B = = 12 74.
B
R B 4
+
δ avvg = 05.(δ A +δ B ) = 05 818 1274) = 1046
.
.(.
.
δ max = 12..74
δ 12 .74
>
max = − . 122 12 but < . 14
.
δ 10 .46
avg
Note: Units of shear wall displacement, d, are not given here because relative
rigidity rather than actual rigidity of the wall has been used to for its calculation.
Because we use ratio of wall displacements in this example, actual units of d are
not required here.
The above calculation shows that torsional irregularity Type 1a (ASCE 7-05
Table 12.3-1, Item 1a) exists. Because the seismic design category (SDC) for
this building is D, structural modeling would require a three-dimensional model
(ASCE 7-05 Section 12.7.3). In addition, the diaphragm shear forces to collectors
must be increased 25 percent (ASCE 7-05 Section 12.3.3.4).
ASCE 7-05 Section 12.8.4.3 requires the evaluation and application of torsional
amplification factor, A .
x
⎛ δ ⎞ 2 ⎛ 12 74 ⎞ 2
.
=
A = ⎜ max ⎟ = ⎜ ⎟ = 103 30. < .
X
⎝ 12. δ avg ⎠ ⎝ 1 2. ×10 46. ⎠
Therefore, use A = 1.03
x
5. Most severe torsional shears: To account for the effects of torsional irregularity,
ASCE 7-05 Section 12.8.4.3 requires that accidental torsional moment, Ve , be
acc
multiplied by the torsional amplification factor, A .
x
The most severe total torsional shears result from the use of torsional moment
equal to V(e – A e ) for wall A (minus sign is used for A e as this force acts
x
x
acc
acc
opposite to the direct shear V D, A ), and V(e + A e ) (plus sign is used for A e
x
x
acc
acc
as this force acts with the direct shear V D, B ) for wall B. Thus,
Ve ( − A e )( x )( R ) 100 6 .03 3 )(24 )(()6
×
( −1
V TA = X acc R A = 11 ,840 = . 11 kips
,
J
×
( + A e
)( )( )
B B
V = Ve X acc )(60 − x R )(R ) = 150 ( +6 1 .03 3 36 4 = 11 .1 kips
, TB
J 11 ,840
6. Total shear in walls: Total shear in walls A and B equals the sum of direct and
torsional shears.
=
.9
V A ′ = V D A − V T A = 60 − .1 1 58 kips
,
,
V ′ = V + V = 440 5 11 1 51 1. + . = . kips
B D B T B
,
,