Page 514 - Design of Reinforced Masonry Structures

7.76 CHAPTER SEVEN

V T 1 D
D
D
e = 6'
B T = V(e + e acc ) B
A CR CM A CM
V T 1 A
CR
V D 1 A V T 1 B
24' V 36' V D 1 B
C C
V T 1 C
FIGURE E7.13D Direct and torsional shear in all shear walls.

torsional shears in walls A and B only, which are parallel to the applied shear
force V.
4. Plan irregularity considerations: The influence of torsional irregularity needs to
be determined as specified in Table 12.3-1 (ASCE 7-05). It requires determina-
tion 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 per-
pendicular to the applied force (96 ft in this example) each way (ASCE 7-05,
Section 12.8.4.2).

Additional eccentricity e = ± 0.05 (60) = ± 3 ft
acc
Total eccentricity e total = (6 ± 3) ft

Initial torsional shear: Calculate from Eq. (7.75).
For the determination of torsional irregularity, the initial most severe torsional
shears, V′ and corresponding story drifts (so as to produce the lowest value of
average story drift) would result from the largest eccentricity, e + e . Thus,
acc
based on V = 100 kips, and e = (6 + 3) ft, M = Ve , the torsional shear are
ta
acc
+ )(24 6
′ =
R
acc
A
.
V TA Ve ( + e )( x )( R ) = 100 (6 3 )( ) = 10 95 kips
,
J 11 ,8840
−
( +
(
Ve + e )( 60 − x )(R ) 1000 6 3 60 24 4)
)(
)(
.
V ′ = acc R B = = 10 95 kips
, TB
,
J 11 840

These shear forces are shown in Fig. E7.13D.
Initial total shears in walls A and B are:
−
V
V ′ = V − ′ = 60 10 .95 = 49 .05 kips
,
,
A D A T A
=
+
V V
V B ′ = B D B + ′ = 40 10 .95 50 .95 kips
,
,
TB
V
(Check: total shear, ′ + ′ = 49 05 50 95 100. + . = kips)
V
B
A
Note: V = 49.05 kips in Wall A, which is smaller than the direct shear value of
60 kips, is not the design shear for this wall as the accidental eccentricity, and has
been used here to reduce the force.