Page 204 - Design of Reinforced Masonry Structures

4.68 CHAPTER FOUR

Calculate the shear strength provided by the masonry, V , from Eq. (4.96):
nm

V = 2 25 A f ′ = 2 25 9 20)( 2500) = 20 250 lb = 20..25 kips
(
,
(
)
.
.
m n m
fV = (0.80)(20.25) = 16.2 kips < V = 17.09 kips
nm
u
Since fV < V , shear reinforcement is required for this beam. Check the maximum
nm
u
permissible value of nominal shear strength, V , from Eq. (4.93).
n
V
() ≤ 4 A f ′ = 4 ( )(20 ) 2500 = 36 ,000 lb = 36 .0 kkips
9
n max n m
(fV ) = (0.8)(36.0) = 28.8 kips > V = 17.09 kips OK
u
n max
Beam size is adequate. Calculate shear from Eq. (4.100) that must be carried by
shear reinforcement:
V − φ V 17 09 −16 2.
.
V = u nm = = 111. kips
φ 08
ns
.
Try No. 3 Grade 60 vertical bars as shear reinforcement. Calculate the required spac-
ing of bars from Eq. (4.102):
Af d ( 011 60 20)
)(
)(
.
s = vy = = 59 46 in .
.
2 V 21 11)
(.
ns
Check maximum permissible spacing of shear reinforcement.
s max = d/2 = 20/2 = 10 in. < 48 in.
Therefore, use a spacing of 10 in. Check minimum reinforcement:
Minimum A per foot of beam = 0.0007bd = 0.0007(9)(20) = 0.13 in. 2
v
01
(. )( )
2
1
1
Actual A provided per foot length of beam = = 0.13 in. 2 OK
v
10
Calculate distance from the center of the span in which shear reinforcement is not
required (i.e., where V < fV . For a value of w = 2.136 k/ft and fV = 16.2 kips, this
u
nm
u
nm
distance is calculated as x on either side of the midspan:
16 2 12
x = (. )( ) = 91 in .
. 2 136
Therefore, shear reinforcement is required for only a distance of 8(12) – 91 = 5 in.
from each support. This would need only one stirrup. However, provide No. 3 stirrups
@ 10 in. on center for construction purposes.
Provide No. 3 Grade 60 one-legged stirrups @ 10 in. on center for shear reinforce-
ment for the entire beam as shown in Fig. E4.24b. The first stirrup is to be placed at
d/4 = 20/4 = 5 in. from the support.