Civil Engineering Reference
In-Depth Information
Figure 7.8
Box section for example problem 7.1
Assume
ε
r
=
0
.
00414 after several cycles of trial-and-error process:
0
.
9
0
.
9
Equation
[
14
]
ζ
=
√
1
ε
r
=
√
1
00414)
=
0
.
482
+
600
+
600(0
.
ε
ds
ζε
o
=
−
0
.
001
002)
=
1
.
037
>
1
descending branch
0
.
482(
−
0
.
2
482
2
ζ
0
.
=
=
0
.
1008
(2
−
ζ
)
2
(2
−
0
.
482)
2
1
1
1
2
2
ζ
1
3
ζε
o
ζ
)
2
ε
ds
1
3
ε
ds
Equation [13
b
]
k
1
=
−
−
+
−
)
2
ε
ds
ζε
o
ζε
o
(2
−
ζ
(2
−
ζ
1008]
1
037)
1
037
1
3
1
1
3
1
=
[1
−
0
.
−
+
0
.
1008(1
.
−
.
(1
.
037)
=
0
.
6102
−
0
.
0684
=
0
.
6786
k
1
, of course, can also be obtained from Table 7.1.
f
c
=
Equation [12]
σ
d
=
k
1
ζ
0
.
6786(0
.
482)(
−
41
.
3)
=−
13
.
51 MPa (
−
1963 psi)
.
Assume
t
d
=
96 mm after two or three trials
914(914
+
1220)
975 000 mm
2
A
c
=
=
2
2
914
2
p
c
=
(914
+
1220
+
+
152
2
=
3987 mm
