Page 159 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.23

In Eq. (4.44), we substitute k for the bracketed quantity on the right-hand side:
n
0
f
k = ′ω(1 − .625 ω) (4.45)
m
n
Substitution of Eq. (4.45) in Eq. (4.44) yields
φM
2 n = φk n (4.46)
bd
A form of Eq. (4.46) that is useful for design can be obtained if we express the design
moment M in kip-ft units, and b and d in inches. With these substitutions, Eq. (4.46) can
u
be expressed as
φM n = bd 2 (4.47a)
,
φk 12 000
n
⎛ bd 2 ⎞
φM = φk n ⎜ ⎟ (4.47b)
n
⎝12 000, ⎠
where M = nominal strength, k-ft. For economical designs, φM = M , where M is
n
n
u
u
the design moment (or moment demand) calculated from the given loading conditions.
Therefore, for design, Eq. (4.47) can be expressed as
M bd 2
u = (4.48a)
φ k n 12 000
,
⎛ bd 2 ⎞
M = φ k n ⎜ ⎟ (4.48b)
u ⎠
⎝12 000,
Equations (4.45) and (4.48) are very useful for flexural design of masonry elements as
illustrated by Examples 4.8 and 4.9.
It is instructive to understand the significance of Eqs. (4.45) and (4.48) for design pur-
poses. For a given or assumed value of the reinforcement ratio w, the value of reinforcing
index, w, can be determined from Eq. (4.40) for the given values of ′ f and f . Knowing
m y
the value of w, the value of k can be determined from Eq. (4.45). Then, for a calculated
n
2
value of the design moment M , the value of the quantity bd can be determined from either
u
from Eq. (4.47) or (4.48). And finally, for a given or known value of the beam width b, the
required value of depth d is easily determined.
Note that k is a function of reinforcement ratio (r), masonry compressive strength ( ′ f ), and
n
m
the yield strength of reinforcing steel ( f ). Values of fk (f = 0.9) for f = 60 ksi and different
n
y
y
combinations of r and ′ f are listed in Table A.13. Interpolation may be used for intermediate

m
values of r to determine k without any appreciable error. See Examples 4.6 and 4.7.
n
Often in a design situation, the reinforcement ratio r is not known, in which case Eq. (4.45)
cannot be used. In such cases, one can resort to some reasonable approximation in analysis,
and use a trial-and-error procedure as discussed in Section 4.6.
Example 4.6 Solve Example 4.2 using Eq. (4.43).
Solution
2
A = 0.88 in. (two No. 6 bars), ′ f = 1.5 ksi, f = 60 ksi, b = 9.625 in. (10 in. nominal),
s
m
y
d = 34 in.