Page 209 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.73

Equation (4.119) can be simplified and written as Eq. (4.120):

.
064 ′ fbc 2 + ′ A ε E c − ′ A ε E d ′ − A f c = 0 (4.120)
m s m s s m s s y
Equation (4.120) can be expressed as a quadric in c:
) − ′ A ε
(.064 ′ fb )c 2 + ( ′ A ε E − A f c E d ′ = 0 (4.121)
m s m s s y s m s
Equation (4.121) can be solved for c, and the magnitude of M can be determined by sum-
n
ming up moments due to C and C about T as given by Eq. (4.117).
s
m
Example 4.25 Nominal strength of a doubly reinforced beam.
A nominal 8 × 24 in. simply supported concrete masonry beam is reinforced with
one No. 8 Grade 60 bar for tension reinforcement and one No. 8 Grade 60 bar for com-
pression reinforcement as shown in Fig. E4.25. Assuming ′ f = 2000 psi, calculate the
m
nominal strength of the beam as (a) singly reinforced beam, (b) a doubly reinforced
beam. What is the percentage increase in the nominal strength of the beam as a doubly
reinforced beam?
7.63'' (8'' nominal)
d′ = 2'' Solution
a. Nominal strength as a singly reinforced beam:
The nominal strength of this beam was determined
#8 earlier in Example 4.15 (calculations not repeated
h = 24'' d = 20'' here). fM = 64.2 k-ft

#8 n
b. Doubly reinforced beam
2
Given: b = 7.625 in., d = 20 in., d′ = 2 in., A = 0.79 in.
s
2
FIGURE E4.25 Cross section (one No. 8 bar), ′ = 079. in. (one No. 8 bar), ′ f =
A
m
s
of beam in Example 4.25. 2000 psi, f = 60 ksi.
y
The location of neutral axis for a doubly reinforced
beam is determined from Eq. (4.115):
)c
(.064 ′ fb )c 2 + ( ′ A ε E − A f − . 080 ′′ − ′ A εε Ed′ = 0 (4.115 repeated)
f A
m s m s s y m s s m s
2
Equation (4.115) is a quadratic of the form: Ax + Bx + C = 0, which can be solved
for x [= c in Eq. (4.115)]:
where
A = 064. f b ′ = 064 20 7625. ( . )( . ) = 976. k/in.
m
B = A′ε E − A f − 080. f A ′
ε
s m s s y m s y
−
−
20
(.
9
(.
)
)
(
9
= 0 79 0 0025 29 000)) (. )(60 ) (.080 )(.)(. )
07
07
,
= . 861 kips
0
,
C = ′′ AE ε d ′ = (. )(0 79 29 000 )(.0025 )(.) = 114 .555 k-in.
0
2
s m
s
Substituting in Eq. (4.115), we obtain
2
9.76c + 8.61c −114.55 = 0