Civil Engineering Reference
In-Depth Information
σ
−
σ
t
2
2
σ
+
σ
t
2
2
t
σ
1
=
+
+
τ
3
2
3
.
44
+
1
.
72
.
44
−
1
.
72
=
+
+
3
.
68
2
2
2
=
2
.
58
+
3
.
78
=
6
.
36 MPa
σ
2
=
2
.
58
−
3
.
78
=−
1
.
20 MPa
ε
d
=−
2. Select
0.0004
After several cycles of the trial-and-error process, assume
ε
r
=
0
.
0200
ζ
=
0
.
250
ε
d
ζε
o
=
0
.
800
<
1 ascending branch
800)
2
]
σ
d
=
0
.
250(
−
41
.
3)[2 (0
.
800)
−
(0
.
=−
9
.
92 psi
Solve the longitudinal steel strain
ε
:
Assume yielding of longitudinal steel,
f
=
413 MPa, and assume the strain of the longitu-
dinal steel to be
ε
=
0.00630
E
ps
(
ε
dec
+
ε
)
=
214 000(0
.
005
+
0
.
00630)
=
2420 MPa
E
ps
(
ε
dec
+
ε
)
2420
Equation 12
b
f
p
=
=
=
1726 MPa
1
4
1
4
1
4
1
4
2420
1860
E
ps
(
ε
dec
+
ε
)
f
pu
+
+
ε
r
−
ε
d
−
σ
d
0
.
0200
+
0
.
0004
10
−
3
/MPa
=
=
2
.
056
×
9
.
92
σ
−
ρ
f
−
ρ
p
f
p
ε
=
ε
r
+
ε
r
−
ε
d
−
σ
d
Equation
13
10
−
3
=
0
.
0200
+
2
.
056
×
[3
.
44
−
0
.
012(413)
−
0
.
003(1726)]
=
0
.
00623
≈
0
.
00630 OK
Solve the transverse steel strain
ε
t
:
Assume yielding of transverse steel,
f
t
=
413 MPa
σ
t
−
ρ
t
f
t
−
ρ
tp
f
tp
ε
t
=
ε
r
+
ε
r
−
ε
d
−
σ
d
Equation 14
10
−
3
=
0
.
0200
+
2
.
056
×
[1
.
72
−
0
.
012 (413)
−
0]
=
0
.
01334
>
0
.
00207 OK
