Page 194 - Design of Reinforced Masonry Structures

4.58 CHAPTER FOUR

7.63" (8" nominal) Example 4.19 A nominal 8- × 24-in. CMU beam
is reinforced with one No. 9 Grade 60 reinforcing bar
with d = 20 in. (Fig. E4.19). 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 1.0 k/ft over an
effective span of 12 ft. ′ f = 1500 psi. (a) Calculate the
m
d = 20"
24" shear resistance V provided by masonry, (b) what
m
would be the increase in the shear resistance provided
by the masonry if the specified strength of masonry
′ f is increased to 2000 psi from 1500 psi.
m
1#9
Solution
FIGURE E4.19 a. Since no axial force is present in the beam, the
shear strength provided by the masonry can be cal-
culated from Eq. (4.96):
b = 7.63 in. (8-in. nominal), d = 20 in., ′ f = 1500 psi
m
V = 2 25 A f ′ = 2 25 (7 63 20 1500) = 13 298 lbb ≈13 3 kips
.
.
.
.
)( )(
,
nm n m
b. When ′ f = 2000 psi, the value of V is
m nm
V = 2 25 A f ′ = 2 25 (7 63 20 2000) = 15 355 lbb ≈15 36 kips
.
.
.
)( )(
,
.
nm n m
The increase in shear strength of masonry is
∆V = (V ) – (V ) = 15.36 – 13.3 = 2.06 kips
m 2000 psi
m 1500 psi
−
5
% increase = 15 .36 13 .3 ≈15 .%
13 .3
Conclusion: Increasing the compressive strength of masonry ( ′ f m ) from 1500 psi
to 2000 psi, a one-third increase, results in only a 15.5 percent increase in the shear
strength of masonry.
4.10.2 Nominal Shear Strength of Reinforced Masonry
Shear reinforcement is not required in masonry beams when the reduced shear strength
provided by the masonry, fV , exceeds the factored shear V (i.e., when fV ≥ V ), where
u
nm
nm
u
f = strength reduction factor for shear (Table 4.1). See Example 4.20.
Example 4.20 A nominal 10- × 40-in. CMU beam is reinforced with two No. 6

Grade 60 reinforcing bars for tension. The centroid of reinforcement is located at 6 in.
from the bottom of the beam (Fig. E4.20). The beam carries a service load of 1600 lb/ft
over an effective span of 15 ft 8 in. in addition to its own weight. The grout unit weight
3
is 140 lb/ft . ′ f = 1500 psi. Check if shear reinforcement is required for this beam. A
m
∗
check for flexural calculations is not required for this problem .
Solution
Given: b = 9.63 in., d = 34 in., effective span = 15.67 ft, ′ f = 1500 psi, w = 2.932 k/ft
m u
(see Example 4.2 for details).
∗
( See Example 4.2 for flexural calculations for this problem.)