Page 190 - Design of Reinforced Masonry Structures

4.54 CHAPTER FOUR

b
ε mu 0.80 fʹ m
a
2
a C = 0.80 fʹ m ab
c
h d a
NA d –
2
T = A s f s
A s
ε s
(f s < f y )
(a) (b) (c)
FIGURE 4.9 Stress and strain distribution diagram for an overeinforced masonry beam.

Substituting for e from Eq. (4.86) in Eq. (4.85), and noting that c = a/0.8 [Eq. (4.5)], we
s
obtain
−
08⎞
ρ
.
⎜
080 ′ f ab = bdE s ⎛ da /. ⎟ ε mu (4.87)
08 ⎠
m
⎝ a /.
Equation (4.87) can be simplified and expressed as
− )ε
ρ
080 ′ f ba 2 = bdE ( 08 . d a (4.88)
.
m s mu
2
Equation (4.88) can be expressed as a quadric in a (of the form Ax + Bx + C = 0):
⎛ ′ f ⎞
m
⎜ ρε ⎟ ⎠ a 2 + 125( . d a ) − d 2 = 0 (4.89)
⎝ E
smu
Equation (4.89) can be solved for a, the depth of compression block. Once the value of
a is known, the nominal strength of the beam can be determined by taking the moment of
the compression force resultant about the centroid of tension reinforcement (Fig. 4.3) as

given by Eq. (4.90):
⎛ a ⎞
M = C d −
⎝
n
⎠ 2 ⎛ a ⎞
′
= 080. fab d − ⎠ 2 (4.90)
⎝
m
Note that the value of M cannot be determined by taking the moment of T about C
n
because the value of f (and, hence, the value of T) is not known. Example 4.18 presents a
s
procedure for calculating the nominal strength of an overreinforced beam. Of course, such
a beam design is not permitted by the code.