Structural Concrete - Hassoun

116 Chapter 3 Flexural Analysis of Reinforced Concrete Beams
Solution
1. For tension-controlled condition, ε t = 0.005, ρ max = 0.01806 (Table 3.2), and ρ b = 0.0285. Check
ρ = A s
bd = 6
12 × 23.5 = 0.02128 >ρ max
Section is in the transition region.
2. Calculate a, c,andε t :
3. Calculate:
a =
A s f y
0.85f c ′ b = 6 × 60
= 8.824 in.
0.85 × 4 × 12
c =
a
0.85 = 10.38 in. d t = h − 2.5 = 27 − 2.5 = 24.5
c
= 0.424 > 0.375
d t
( )
dt − c ( )
24.5 − 10.38
ε t = 0.003 =
0.003 = 0.00408 > 0.004
c
10.38
( ) 250
φ = 0.65 +(ε t − 0.002) = 0.823
3
φM n = φA s f y
(
d − 1 2 a )
23.5 − 8.824∕2
= 0.823(6)(60) = 471 K ⋅ ft
12
Discussion
For a tension-controlled section limitation, ρ max = 0.01806 and R u = 820 psi,
φM n = R u bd 2 = 0.82(12) (23.5)2 = 452.8K⋅ ft
12
This is a conservative value: It is advisable to choose adequate reinforcement to produce
tension-controlled condition with φ = 0.9.
3.14 RECTANGULAR SECTIONS WITH COMPRESSION REINFORCEMENT
In concrete sections proportioned to resist the bending moments resulting from external loading
on a structural member, the internal moment is equal to or greater than the external moment, but a
concrete section of a given width and effective depth has a minimum capacity when ρ max is used.
If the external factored moment is greater than the design moment strength, more compressive and
tensile reinforcement must be added.
Compression reinforcement is used when a section is limited to specific dimensions due
to architectural reasons, such as a need for limited headroom in multistory buildings. Another
advantage of compression reinforcement is that long-time deflection is reduced, as is explained
in Chapter 6. A third use of bars in the compression zone is to hold stirrups, which are used to
resist shear forces.
Two cases of doubly reinforced concrete sections will be considered, depending on whether
compression steel yields or does not yield.