Page 196 - Design of Reinforced Masonry Structures

4.60 CHAPTER FOUR

′
to φ f ). This may very well be the case in a practical situation where a beam having short
m
spans may be adequate in flexure but not in shear. See Example 4.21.

7.63" (8" nominal) Example 4.21 A nominal 8- × 24-in. CMU beam
is reinforced with one No. 9 Grade 60 reinforcing bar
for tension with d = 20 in. (Fig. E4.21). The beam
is required to carry a service dead load of 1.0 k/ft
(including its self-weight) and a service live load of
1.75 k/ft over an effective span of 12 ft. ′ f = 1500 psi.
d = 20" m
24" Check if the beam section is adequate for shear. If
not, suggest alternatives for the beam to be adequate
in shear.
(Note: This example was presented earlier as
Example 4.13 for flexural calculations.)
1#9
FIGURE E4.21 Beam cross sec- Solution
tion for Example 4.21.
Given: b = 7.63 in., d = 20 in., effective span =
12.0 ft, ′ f = 1500 psi, w = 4.0 k/ft (see Example 4.13
u
m
for details).
Calculate the factored shear (or shear demand):
40 12)
V = wl = (.) ( = 24 0.kips
u
u
2 2
Calculate the shear strength of masonry (V ) from Eq. (4.96):
m
V = 2 25 A n f ′ = 2 25 7 63 20)( 1500) = 13 298 lbb = 13 3 kips
.
(
.
.
)(
.
,
nm
m
fV = (0.8)(13.3) = 10.64 kips < V = 24.0 kips
nm
u
Since fV < V , shear reinforcement is required for this beam. Check the maximum
nm
u
permissible value of V from Eq. (4.93).
n
63
V
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 = 24 kips
u
n max

Because fV (= 18.91 kips) < V (= 24 kips), the beam cross section is not acceptable.
u
n
Therefore, either (a) a larger beam size or (b) higher compressive strength of masonry
should be provided. Both options are explored as follows:
a. Larger beam size: The given beam is 24 in. deep (three 8-in. blocks). Try four 8-in.
blocks so that h = 32 in. and d = 32 − 4 = 28 in. Calculate the revised maximum
permissible value of V from Eq. (4.93).
n
() = 4 A f ′ = 4 ( . )(28 1500 = 33 ,097 lb = 33.1 kips
.
63
V
7
)
n max n m
(fV ) = (0.8)(33.1) = 26.48 kips > V = 24 kips OK
n max
u
Increase the beam cross section to nominal 8 × 32 in. to satisfy shear requirements.