Page 517 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.79

Center of rigidity : Summing moments of rigidities of walls A and B (parallel to
the applied force) about Wall A:
⎛ R ⎞ ⎛ 200 ⎞
=
x = B () = ⎜ ⎟ () 244ft
96
96
R ⎜ ⎝ R + R ⎠ ⎟ ⎝ 600 + 200 ⎠
A B
Summing up moments of rigidities of walls C and D about Wall C:
⎛ R ⎞ ⎛ 300 ⎞
y = D (48 ) = ⎜ ⎟ (48 ) = 244ft
R ⎜ ⎝ R + R ⎠ ⎟ ⎝ 300 + 300 ⎠
C D
(or from symmetry, y = ½ (48) = 24 ft)
R
The center of mass is located at x = ½(96) = 48 ft from either wall.
M
Eccentricity e = x − x = 48 − 24 = 24 ft
M
R
Calculate the polar moment of inertia of shear walls, J, about the center of rigidity.
2
2
2
J = R (x ) + R (96 − x ) + R (y ) + R (48 − y ) 2
A
R
B
D
R
R
R
C
2
2
2
= 600 (24) + 200 (96 − 24) + 300 (24) + 300 (48 − 24) 2

= 1,728,000 (k/in.) ft 2
2. Direct shear in walls A and B: The rigidity of the building in the direction of the
force equals the sum of rigidities of walls parallel to the force. In this example, shear
walls A and B are parallel, and are oriented in the direction of applied lateral force.
Therefore, they would share the applied lateral force in proportion to their rigidities.
⎛ R ⎞ ⎛ 600 ⎞
=
V DA = ⎜ ⎝ R + A R ⎠ ⎟ V = ⎜ ⎝ 600 + 200 ⎟ 150 112..5 kips
⎠
,
B
A
⎛ R ⎞ ⎛ 200 ⎞
=
V DB = ⎜ ⎝ R + B R ⎠ ⎟ V = ⎜ ⎝ 600 + 200 ⎟ 150 37 .55kips
⎠
,
B
A
3. Plan irregularity considerations: The influence of torsional irregularity needs to
be determined as specified in Table 12.3-1 (ASCE 7-05). It requires determination
of story drifts in walls A and B. This evaluation must include accidental torsion
caused by an eccentricity of 5 percent of the building dimension perpendicular to
the applied force, 96 ft in this example (ASCE 7-05, Section 12.8.4.2).
acc
Additional eccentricity e = ± 0.05 (96) = ± 4.8 ft
Total eccentricity e total = 24 + 4.8 = 28.8 ft
Initial torsional shear: Based on V = 150 kips, and e = (24 + 4.8) ft
+ . )(24
0
R
A
acc
V ′ = Ve ( + e )( x )( R ) = 150 (24 4 8 )(600) = 36 kips
TA J 1 728 000
,
,
,
+
( + e
. )(72
B B
V ′ = Ve acc )( 96 − x R )(R ) = 150 (24 4 8 )(200 ) = 36 kips
, TB
J , 1 728 000
,
These shear forces are shown in Fig. E7.14B