Civil Engineering Reference
In-Depth Information
a
a
a θ
⎡⎤ ⎡ ⎤
⎢⎥ ⎢ ⎥
=
1
0.5
0.1
y
Let us consider the case that
L
==
L
500
m
,
4 Vms
=
and
and adopt the
⎢⎥ ⎢ ⎥
⎢⎥ ⎣ ⎦
exp
z
⎣⎦
numerical values given in the following tables:
B
m
D
m
m
Kg/m
ω
Rad/s
EI
2
Nm
EI
2
Nm
GI
2
Nm
EI
ζ
i
y
z
t
w
%
4
Nm
20
4
0.8
0.5
5
10
11
10
12
210
10
510
12
210
ρ
x f
x f
L
m
L
C
C
C
C
C
I
I
kg
uy f
wy f
D
M
u
w
m
3
m
1.25
0.7
5
1.5
1.5
1.0
0.2
0.1
162
13.5
Thus,
(
)
L
ˆ
ω
ˆ
1exp
+
ω
ω
i
exp
()
u
2
u
ˆ
C
16
4
2
0.24
ω
=
=
ψω
=⋅
+
π
=
u
uy f
ui
V
2
2
(
)
2
ˆ
ωπ
+
2
2
ˆ
ωπ
+
u
u
(
)
L
1exp
ˆ
ω
ˆ
+
ω
ω
i
exp
()
w
2
w
ˆ
C
10
4
2
0.37
ω
=
=
ψω
=⋅
+
π
=
w
y f
wi
2
2
2
V
(
)
ˆ
ωπ
+
2
2
ˆ
ωπ
+
w
w
Let us adopt the typical Kaimal type of turbulence spectra
()
x f
()
x f
S
S
ω
1.08
LV
/
ω
1.5
LV
/
u
w
u
w
=
and
=
(
)
(
)
2
5 3
2
5 3
σ
σ
x f
x f
u
11.62
+
ω
LV
/
w
12.25
+
ω
LV
/
u
w
()
2
()
2
S
ω
S
ω
ui
u
wi
w
=
0.206
and
=
0.229
σ
σ
The joint acceptance is then
2
()
()
S
S
D CaI
ω
ω
2
ui
() (
)
wi
()
2
Ca
BC a I
ψω
+
+
ψ ω
Dyu
u
i
Lz
M
w
w
i
θ
B
2
2
σ
σ
ˆ
2
u
w
J
0.127
=
=
ii
2
(
)
2
2
2
aaa
++
y
z
θ
Let us for simplicity also adopt quasi-steady values to the aerodynamic derivatives:
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