Page 202 - Design of Reinforced Masonry Structures

4.66 CHAPTER FOUR

Calculate the factored shear (or shear demand) V .
u
30 12)
V = wl = (. ) ( = 18.0 kips
u
u
2 2
Calculate the shear strength provided by the masonry (V ) from Eq. (4.96):
nm
V = 2 25 A f ′ = 2 25 7 63 20)( 1500) = 13 298 lbb = 13 3.kips
.
.
(
,
.
)(
nm n m
fV = (0.80)(13.3) = 10.64 kips < V = 24.0 kips
u
nm
Since fV < V , shear reinforcement is required for this beam. Check the maximum
u
m
permissible value of V from Eq. (4.93).
n
V
63
)
7
() max ≤ 4 A n f ′ = 4 ( . )(20 1500 = 23 ,641 lb = 23..64 kips
n
m
(fV ) = (0.8)(23.64) = 18.91 kips > V = 18 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 18 −10 64.
V = u nm = = 92.kips
φ 08
ns
.
Try 8 in. spacing for single vertical bars as shear reinforcement. Calculate A from
v
Eq. (4.103):
2 Vs () 29 2 8)
(.)(
A = ns = = 012 in. 2
.
v
)(
fd ( 60 20)
y
2
Provide No. 3 Grade 60 vertical bars as shear reinforcement, A = 0.11 in. OK
v
Check maximum permissible spacing of shear reinforcement:
s = d/2 = 20/2 = 10 in. > s = 8 in. (provided) OK
max
Check minimum reinforcement:
Minimum A per foot of beam = 0.0007bd = 0.0007(7.63)(20) = 0.11 in. 2
v
2
01
1
(. )( )
1
2
Actual A provided per foot length of beam = = . 017 in 2 > . 011 in. OK
v
8
Calculate distance from the center of the span in which shear reinforcement is not
< fV ). For a value of w = 3.0 k/ft and fV = 10.64 kips, this
required (i.e., where V u nm u nm
distance is calculated as x on either side of the midspan:
10
)( )
x = (.64 12 = 42 .56 in .
. 30
With cells in the CMU spaced at 8 in. on center, theoretically shear reinforcement is
not required for a distance 3 ft 6.56 in. on either side of the midspan. This may not result
in much of a saving, and some designers may choose to provide shear reinforcement in
every cell for ductility.
Provide No. 3 Grade 60 vertical bars @ 8 in. on center for shear reinforcement as
shown in Fig. E4.23B.