Civil Engineering Reference
In-Depth Information
Figure 5.17
Example problem 5.3
5.4.4.2 Solution
1. Select
0.0002
After several cycles of the trial-and-error process we assume
ε
d
=−
ε
r
=
0
.
00385
0
.
9
0
.
9
Equation 8
ζ
=
√
1
ε
r
=
√
1
00385)
=
0
.
495
+
600
+
600(0
.
ε
d
ζε
o
=
−
.
0
0002
002)
=
.
<
0
202
1
ascending branch
0
.
495(
−
0
.
2
ε
d
ζε
o
2
ε
d
ζε
o
f
c
Equation
7
a
σ
d
=
ζ
−
202)
2
]
=
0
.
495(
−
41
.
3)[2(0
.
202)
−
(0
.
=−
7
.
43 MPa
