Structural Concrete - Hassoun

19.5 Design of Flexural Members 759
a = 0.377 β 1 d p (from Eq. 19.36). Substituting this value of a in Eq. 19.38,
(
M n = A ps f ps d p 1 − 0.32β )
1
1.7
for f ′ c = 5ksi,β 1 = 0.8. Then
=(ρ p bd p ) f ps d p (1 − 0.188β 1 )
= ω p f ′ c(1 − 0.188β 1 )bd 2 p
=(0.32β 1 − 0.06β 2 1 )f ′ cbd 2 p (19.42)
M n = 0.22 f ′ cbd 2 p = 1.09 bd 2 p
Similarly, for f ′ c = 4ksi,M n = 0.915 bd 2 p, and for f ′ c = 6ksi,M n = 1.238 bd 2 p.
19.5.3 Flanged Sections
For flanged sections (T- or I-sections), if the stress block depth a lies within the flange, it will be
treated as a rectangular section. If a lies within the web, then the web may be treated as a rectangular
section using the web width, b w , and the excess flange width (b − b w ) will be treated similarly to
that of reinforced concrete T-sections discussed in Chapters 3 and 4. The design moment strength
of a flanged section can be calculated as follows (see Fig. 19.7):
where
M n = M n1
(moment strength of web)+M n2
(moment strength of excess flange)
(
= A pw f ps d p − 1 )
2 a + A pf f ps
(d p − 1 )
2 h f
M u = φM n and a = A pw f ps
0.85 f ′ cb w
(19.43)
A pw
= A ps − A pf
A pf =[0.85 f c ′(b − b w )h f ]∕ f ps
h f = thickness of flange
Note that the total prestressed steel, A ps , is divided into two parts, A pw and A pf , developing the web
and flange moment capacity. For flanged sections, the reinforcement index, ω pw , must not exceed
0.32β 1 for tension-controlled sections, where
( )( )
( )
Apw fps
fps
ω pw =
b w d p f c
′ = prestressed web steel ratio ×
f c
′
19.5.4 Partial Prestressed Reinforcement
In some cases, nonprestressed reinforcing bars (A s ) are placed in the tension zone of a prestressed
concrete flexural member together with the prestressing steel (A ps ) to increase the moment strength
of the beam. In this case, the total steel (A ps and A s ) is considered in the moment analysis. For
rectangular sections containing prestressed and nonprestressed steel, the design moment strength,
φM n , may be computed as follows:
(
M n = A ps f ps d p − 1 ) (
2 a + A s f y d − 1 )
2 a (19.44)