Structural Concrete - Hassoun

3.20 Additional Examples 141
Solution
1. The section behaves as a rectangular section with b = 14 in. and d = 21.5 in. Note that the width
b is that of the section on the compression side.
2. Check that ρ = A s /bd = 5/(14 × 21.5) = 0.01661, which is less than the maximum steel ratio of
0.018 for tension-controlled sections. Therefore, φ = 0.9. Also ρ>ρ min = 0.00333. Therefore,
ρ is within the limits of a tension-controlled section.
3. Calculate a ∶ a = A s f y ∕(0.85f c ′ b)=5 × 60∕(0.85 × 4 × 14) =6.3in.
(
φM n = φA s f y d − a )
(
= 0.9 × 5 × 60 21.5 − 6.3 )
= 4954.5K⋅ in = 412.9K⋅ ft
2
2
Example 3.16
A reinforced concrete beam was tested to failure and had a rectangular section, b = 14 in. and d = 18.5 in.
At failure moment, the strain in the tension steel was recorded and was equal to 0.004106. The strain in
the concrete at failure may be assumed to be 0.003. If f c ′ = 3ksi and f y = 60 ksi, it is required to:
1. Check if the tension steel has yielded.
2. Calculate the steel area provided in the section to develop the above strains. Then calculate the
applied moment.
3. Calculate the design moment strength based on the ACI Code provisions. (Refer to Fig. 3.40.)
Solution
1. Check the strain in the tension steel relative to the yield strain. The yield strain ε y = f y /E s =
60/29,000 = 0.00207. The measured strain in the tension steel is equal to 0.004106, which is
much greater than 0.00207, indicating that the steel bars have yielded and in the elastoplastic
range. The concrete strain was 0.003, indicating that the concrete has failed and started to crush.
Therefore, the tension steel has yielded.
2. Calculate the depth of the neutral axis c from the strain diagram (Fig. 3.40). From the triangles of
the strain diagram,
c
d = 0.003
0.003 + 0.004106
a = β 1 c = 0.85 × 7.81 = 6.64 in.
The compression force in the concrete, C c = 0.85,
( ) 3
and c = 18.5 = 7.81 in.
7.106
f c ′ ab = 0.85 × 3 × 6.64 × 14 = 237 K
The tension steel A s = C c /f y = 237/60 = 3.95 in. 2 (section has five no. 8 bars):
(
M n = A s f y d − a )
(
= 3.95 × 60 18.5 − 6.64 )
= 3597.6K⋅ in = 299.8K⋅ ft
2
2
3. Check ε t = 0.003(d t − c)/c.
c = 7.81 in. d t = h − 2.5in. = 22 − 2.5 = 19.5in.
Therefore, ε t = 0.003(19.5 −7.81)/7.81 = 0.0045, which is less than 0.005 for tension-controlled sections
but greater than 0.004. Section is in the transition region, and φ