Page 174 - Design of Reinforced Masonry Structures

4.38 CHAPTER FOUR

Combining Eqs. (4.62) and (4.71), and substituting d = d (where d = the distance from
t
t
the extreme compression fibers to the tensile reinforcing bar closest to the tension face of
the beam), we obtain
ρ fd ⎛ ε ⎞
y t = ⎜ mu ⎟ d (4.72)
.
080 080 f ) ′ m ⎝ ε + 15 ε ⎠ t
.
.
(
mu
y
Substitution of r = r max in Eq. (4.72) gives a general expression for the reinforcement ratio
corresponding to 1.5 times the yield strain e in steel:
y
⎛ ε ⎞ ⎛ ′ ⎞
ρ = ⎜ mu ⎟ ⎜ . 064 f m ⎟ (4.73)
max
⎝ ε + 15 ε ⎠ ⎝ f y ⎠
.
mu
y
Values of r max depend on the values of strain in masonry, e (= 0.0025 for concrete
mu
masonry and 0.0035 for clay masonry). For concrete masonry, substitution of e = 0.0025
mu
6
and e = f /E , and E = 29 × 10 psi in Eq. (4.73) yields
s
y
y
s
⎛ ⎞
⎜ . 0 0025 ⎟ ⎛ . 064 ′′ ⎞
f
ρ = ⎜ ⎟ ⎜ m ⎟ (4.74)
max . 15 f ⎝ f ⎠
⎜ . 0 0025 + y ⎟ y
⎝ 29 ×10 6 ⎠
Equation (4.74) can be simplified and expressed as Eq. (4.75):
⎛ 72 ,500 ⎞ ⎛ . 064 f ′ ⎞
ρ max = ⎜ ⎟ ⎜ m ⎟ ⎟ (4.75)
+
⎝ 72 ,500 1 f y ⎠ ⎝ f y ⎠
.5
An expression similar to Eq. (4.75) can be obtained for clay masonry by substituting e =
mu
0.0035 in Eq. (4.73):
⎛ ⎞ ⎛ . 064 f ⎞ ⎞
′
ρ = ⎜ 101 ,500 ⎟ ⎜ m ⎟ (4.76)
+
max
.5
⎝ 101 ,500 1 f y ⎠ ⎝ f y ⎠
Tables 4.5 and 4.6 give the values of r determined form Eqs. (4.75) and (4.76) for con-
max
crete and clay masonry, respectively, for various combinations of ′ f and f .
y
m
It is important to understand the significance of Eqs. (4.75) and (4.76). They give the
maximum reinforcement ratios for concrete and clay masonry, respectively. These limits
represent threshold values that may not be exceeded if a beam is to remain ductile according
the code. In cases where a beam is reinforced excessively (i.e., r > r max ), the design would

not qualify as acceptable according the MSJC Code even though its nominal strength may
be considerable. The reason: such a beam does not posses ductility; the masonry would
fail by premature crushing without sufficient yielding of reinforcement. The value of the
strength reduction factor, f = 0.9, associated with flexure, is intended to be used for ductile
members, that is, only when r < r max , which ensures that e ≥ 1.5e . Example 4.13 illustrates
y
s
the intent of this code requirement.
Example 4.13 A nominal 8 × 24 in. concrete masonry beam is reinforced with one
No. 9 Grade 60 reinforcing bar placed at d = 20 in. for tension (Fig. E4.13). The beam is
required to carry a uniform 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.
m