Civil Engineering Reference
In-Depth Information
Design for doubly reinforced section.
MRbd
=
2
= ×× =
911
10
15.563
2
2, 206,505.5lb-in.
=
2206.5k-in.
=
183.88 k-ft
nt
nt
t
M
220
0.9
u
MMM
=−=
φ
−=−
M
183.88
=
60.56 k-ft
n
n
nt
nt
Strain in compression steel:
ε
−
0.003
5.84 2.25
5.84
−
60
29,000
s
=
ε =
×
0.003
=
0.001844
<
=
0.00207
s
cd
c
So compression steel does not yield
f E
MAfd d
=ε=
29,000
×
0.001844
=
53.48 ksi
s
s
s
(
)
=φ
−
u
s s
(
)
2
60.56
×=×
12
A
53.48
×
15.563
−
2.25
A
=
1.021 in.
s
s
f
f
53.48
60
s
y
or use Table 2.3 to determine
ρ
st
2
AAA
=+ =
2.813
+
1.021
×
=
3.723 in.
s
st
s
ρ
st
=
0.01806
AbdA
f
f
s
y
2
=ρ
+
=
3.721 in.
s
st
s
3.721
0.6
#7 bars (0.6 in.)
2
=
6.2 se seven #7 bars fortension
b
=×+ ×+× +×=
1.520.5270.875
6116.125"10"
>
Use two layers(ACI Section7.6)
b
=×+×+× +×=>
1.520.5240.875
3112
10 "
se threelayers(ACI Section7.6)
1.021
0.6
2
#7 bars (0.6 in.)
=
1.7
se two #7 bars forcompression (better fits the width)
It is evident that the bar arrangement in Figure 2.13 has caused a slightly higher bar area,
smaller effective steel depth (
d
), and larger compression steel depth (
d
′), all of which
may or may not furnish the moment capacity specified in Example 2.2. Accordingly,
the reader may check that by solving Problem 2.3 at the end of the chapter.
2#7
18"
1.0"
7#7
1.5"
10"
FIGURE 2.13
Cross-section design for Example 2.2.
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