Civil Engineering Reference
In-Depth Information
Solution:
4
8
1
2
dd
==−−−× =
18"1.5"
0.875
15.563"
t
4
8
1
2
4
8
d
=++×=
1.5"
2.25"
Assume
φ=
0.9
ε=
0.005
t
0.003
0.005
d
c
t
=
−=
11.67
c
dc
−
t
d
c
c
d
t
=
2.67
=
0.375 (see Figure 2.1)
t
ε
cu
=0.003
ε
s
c
=
0.375
d
t
=
0.375
×
15.563
=
5.84 in.
c
A
fb c
f
0.85
β
c
1
2.813 in.
2
=
=
s
y
d
t
fd
a
0.005
M
=φ −= =
1987 k-in.
165.58 k-ft
<
220 k-ft
ut
sy
2
Thus, enlarge section or use compression reinforcement.
Another solution approach is possible using Table 2.3 from Portland Cement
Association (2013).
R
nt
from Table 2.3 is 911 psi. This
R
nt
can be simply calculated
as follows:
M
bd
M
bd
165.58
×
××
12
nt
ut
R
=
=
φ
=
=
0.9115 ksi
=
911.5psi
nt
2
2
2
0.910
15.563
t
M
bd
M
bd
220
××
××
12
1000
n
t
u
R
=
=
φ
=
=
1211.1
>
911psi
n
2
2
0.910
15.563
2
t
TABLE 2.3
Design Parameters at Steel Strain of 0.005 for Tension-Controlled Sections
f
==
3, 000
f
==
4, 000
f
==
5, 000
f
==
6, 000
f
==
8, 000
f
==
10, 000
c
β
1
=
0.75
c
β
1
=
0.65
c
β
1
=
0.85
c
β
1
=
0.85
c
β
1
=
0.80
c
β
1
=
0.65
R
nt
683
911
1084
1233
1455
1819
615
820
975
1109
1310
1637
ϕ
R
nt
0.2709
0.2709
0.2550
0.2391
0.2072
0.2072
ω
t
ρ
t
Grade 40
0.02032
0.02709
0.03187
0.03586
0.04144
0.05180
Grade 60
0.01355
0.01806
0.02125
0.02391
0.02762
0.03453
Grade 75
0.01084
0.01445
0.01700
0.01912
0.02210
0.02762
Source:
Courtesy of Portland Cement Association (2013).
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