Civil Engineering Reference
In-Depth Information
and
2
f
ωπ
=
. As mentioned above, the calculations will inevitably demand iterations,
s
*
*
because
H
and
A
are functions of
and
. The iteration will take place on the
σ
σ
r
z z
rr
θθ
difference between
and
, which in general will be a small quantity.
ζ
ζ
i
ae
i
Example 6.4
Let us consider a simply supported horizontal beam type of bridge with span
L
=
L
=
500
m
exp
and set out to calculate the vortex shedding induced dynamic response at
x
=
L
2
which is
r
associated with the three mode shapes
⎡
⎤
0
0
⎡
⎤
⎡
⎤
⎢
⎥
⎡⎤
⎢
⎡⎤
⎢
⎡⎤
⎢
0
0
0
0
⎥
⎥
⎥
⎢⎥
⎛
⎞
x
⎢⎥
⎛
⎞
3
x
⎢⎥
⎢
π
π
⎢
⎥
⎢
⎥
sin
sin
0
0
⎥
φ
=
φ
=
φ
=
φ
=
and
φ
=
=
⎢⎥ ⎜ ⎟
⎢
⎢⎥ ⎜ ⎟
⎢
⎢⎥
⎢
1
z
2
z
3
1
⎥
2
⎥
⎝ ⎠
⎝ ⎠
⎥
⎢⎥
⎢
⎢⎥
⎢
⎢⎥
φ
x
L
⎥
⎥
⎛
⎞
π
0
0
⎢
⎥
⎣⎦
⎣
⎣⎦
⎣
⎣⎦
3
sin
0
0
⎦
⎦
⎜ ⎟
⎢
⎥
⎝ ⎠
⎣
⎦
with corresponding eigen-frequencies
0.8
,
1.6
and
2.5
rad/s. As can be seen,
φ
and
φ
contain only the displacement component in the across wind vertical direction while
φ
only
contains torsion. Let us adopt the following structural properties:
ρ
B
m
D
m
m
kg
m
m
θ
ζζ
%
ζ
ωω
=
rad
s
ωω
=
rad
s
ωω
=
rad
s
1
z
2
z
3
θ
1
3
2
kg
m
%
2
kgm
m
3
1.25
20
4
0.8
1.6
2.5
0.5
0.75
4
5
10
610
⋅
and the following vortex induced wind load properties:
St
ˆ
q
z
ˆ
q
b
b
θ
a
a
θ
K
K
σ
σ
λλ
=
z
az
a
θ
θ
θ
0
0
0.1
0.9
0.3
0.15
0.1
0.4
0.1
1.2
0.2
0.02
1
1
⎛
⎞
⎛
⎞
2
22
where:
ˆ
VB
⎠
and
ˆ
VB
⎠
σσ
=
ρ
σσ
=
ρ
⎜
⎟
⎜
⎟
q
q
q
q
z
z
2
θ
θ
2
⎝
⎝
Since
m
and
m
θ
are constant along the span, then the modally equivalent and evenly
distributed masses
mmm
==
and
3
mm
θ
=
.
1
2
z
