F. K. Kong MA, MSc, PhD, CEng, FICE, FIStructE, R. H. Evans CBE, DSc, D ès Sc, DTech, PhD, CEng, FICE, FIMechE, FIStructE (auth.)-Reinforced and Prestressed Concrete-Springer US (1987)

106 Reinforced concrete beams-the ultimate limit state
0.45d). Example 4.6-2 shows that the compression reinforcement will
reach the design strength of0.87[y provided d' lx does not exceed 0.43; that
is (for xld = 0.5), provided d' /d does not exceed 0.21. In other words, the
force in the compression reinforcement can normally be taken as 0.87[yA~,
and this has a lever arm of (d - d') about the tension reinforcement.
Equating this additional resistance moment to the excess moment,
0.81[yA~(d - d') = M - Mu (4.6-6)
where Mu = K'fcubtf- (eqn 4.6-5). Equation (4.6-6) gives the required
area A~ of the compression reinforcement. An area As of tension
reinforcement must then be provided to balance the total compressive
force in the concrete and the compression reinforcement. Referring to Fig.
4.6-2(b),
(4.6-7)
Noting from eqn (4.6-5) that 0.2fcubd = Mulz, we can write eqn (4.6-7)
as
where
A Mu A'
s = 0.87[yz + s
z = 0.775d (from Fig. 4.6-l(c)); and
Mu = K'fcubd 2 (from eqn 4.6-5).
(4.6-8)
The term 'balanced section' has been defined in Section 4.3, and again in
a slightly different way in Section 4.5. In connection with the use of the
simplified stress block here, a balanced section is defined as a singly
reinforced section having such an area As of tension reinforcement that the
x/d ratio is equal to 0.5. From eqn (4.6-2)
0.5 = 2.15/.[y :d
cu
or
e(balanced) = 0.233Jyu
(4.6-9)
where e = As! bd.
Example 4.6-l
Determine the limiting value of the neutral axis depth x for which tension
reinforcement, of [y = 460 N/mm 2 , will reach the design strength of 0.87[y
at the ultimate limit state.
SOLUTION
The strain distribution at the ultimate limit state is shown in Fig. 4.6-3.
For [y = 460 N/mm 2 , the design strength 0.87[y = 400 N/mm 2 , and ~s =
400/Es = 0.002. Therefore, from the geometry of Fig. 4.6-3,
0.0035 0.002
-x- = d-x
X= 0.64d