Civil Engineering Reference
In-Depth Information
Figure 5.25
Normalized Mohr circle for proportional loading in example problem 5.4
.
.
0
9
0
9
ζ
=
√
1
ε
r
=
√
1
00345)
=
.
Equation 8
0
514
+
600
+
600(0
.
ε
d
ζε
o
=
−
0
.
000275
002)
=
0
.
268
<
1 ascending branch
0
.
514(
−
0
.
2
ε
d
ζε
o
2
ε
d
ζε
o
f
c
Equation 7
a
σ
d
=
ζ
−
268)
2
]
=
0
.
514(
−
27
.
6) [2(0
.
268)
−
(0
.
=−
6
.
57 MPa
Solve the longitudinal steel strain
ε
:
ε
r
−
ε
d
−
σ
d
0
.
00345
+
0
.
000275
10
−
3
=
=
0
.
567
×
/
MPa
6
.
57
m
σ
1
−
ρ
f
−
ρ
p
f
p
ε
=
ε
r
+
ε
r
−
ε
d
−
σ
d
Equation
13
P
10
−
3
=
0
.
00345
+
0
.
567
×
(0
.
5(3
.
65)
−
0
.
0103
f
−
0)
10
3
Assume
f
=
200
×
ε
before yielding
Then
ε
+
1
.
168
ε
=
0
.
00345
+
0
.
001035
0
.
004485
2
ε
=
=
0
.
00207
=
0
.
00207 (
ε
y
)OK
.
168
