Page 189 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.53

Substituting in Eq. (4.80), we obtain

2
A – 16.28A + 11.4 = 0
s
s
A = 0.73 in 2
s
Provide one No. 8 bar, A = 0.79 in 2
s
This is the same result that was obtained in Example 4.16. Other calculations should
2
follow as shown in Example 4.16, that is, check M with d = 20 in. and A = 0.79 in. ,
u
s
and that assumption e ≥ 1.5e is satisfied. These calculations would be exactly the same
y
s
as in Example 4.16. A check should also be made to ensure that r is not greater than
0.75r , as illustrated in Example 4.16.
b
4.9 OVERREINFORCED BEAMS
In some cases, a masonry beam may contain excessive tension reinforcement, which would
make it an overreinforced beam. The implication of a beam being overreinforced is that the
tension reinforcement would not yield, that is, the stress in the reinforcement would remain
in the elastic range. Under ultimate loads, such a beam may experience sudden (brittle)
failure initiated by crushing of the masonry in the compression zone of the beam. In such
cases, the analyses presented in preceding sections cannot be used because they are based
on the premise of tension reinforcement having yielded.
Overreinforced beams can be analyzed based on the same general equilibrium condition
(C = T) that was used for analyzing underreinforced beams except for the fact that the stress
in reinforcement, f , is not equal to the yield stress f (it is less than f ). Because the stress
s
y
y
in reinforcement is in the elastic range, it can be expressed as a function of elastic strain e s
and the modulus of elasticity of steel, E . The analysis follows.
s
Equilibrium of compressive and tensile force resultants in the beam gives
C = T (4.82)
where the value of C is given by Eq. (4.6), and T can be expressed as a product of the rein-
forcement area and stress in it. Thus,
′
C = 080 f ab (4.6 repeated)
.
m
T = A f (4.7 repeated)
s s
The stress in reinforcement can be expressed in terms of strain based on Hooke’s law:

f = e E (4.83)
s s
s
Substitution for f from Eq. (4.83) in Eq. (4.82) yields
s
T = A e E (4.84)
s s s
Substitution of Eqs. (4.6) and (4.84), and A = rbd [Eq. (4.38)] in Eq. (4.81) yields
s
.
080 ′ fab = Eρε bd (4.85)
m s s
From strain distribution diagram (Fig. 4.9), we obtain,
−
⎛ dc ⎞
ε = ε (4.86)
s mu ⎝ c ⎠