Civil Engineering Reference
In-Depth Information
6.3.4.2
Frequency averaged radiation factor
It is not possible to give a simple formula to calculate the radiation factor for a plate
driven at a single frequency. The response will normally contain contributions from
several modes each with different amplitude, depending on the location of these modes
relative to the driving frequency and on the damping of the plate. The best one can do for
practical use is to find a frequency averaged radiation factor. One will then assume that
the excitation is relatively broadband as compared with the distance between the natural
frequencies, furthermore, that all modes inside the frequency band are equally excited.
Several expressions exist in the literature; see e.g. EN 12354-1, but it is not obvious that
one is better than another. We will show data given by Leppington et al. (1982), who
give the following expressions covering three frequency ranges
⎡
⎤
Uc
χ
+
12
χ
0
σ
=
⋅
ln
+
for
f
<
f
,
⎢
⎥
g
2
χ
−
1
2
2
χ
−
1
⎢
⎥
2
π
ff S
⋅
⋅
χ
−⎣
1
⎦
g
2
π
f
a
⎛
⎞
σ
=
⋅
a
0.5
−
0.15
for
f
≈
f
,
(6.48)
⎜
⎟
g
c
b
⎝
⎠
0
1
and
σ
=
for
f
>
f
.
g
f
f
g
1
−
The quantities
a
and
b
again give the dimensions of the rectangular plate, where it is
assumed that
a
<
b
. Further, the quantities
S
and
U
are the plate area and circumference,
respectively, i.e. S =
a
⋅
b
and
U =
2(
a
+
b
). The parameter χ is the square root of the ratio
f
c
/
f
. It should be noted that the critical frequency should be much higher than the first
eigenfrequency of the plate, due to the assumption that there should be resonant radiation
by an assembly of modes.
An example using these equations is shown in
Figure 6.15
. The radiation index is
calculated for plates of aluminium or steel where the length of one edge is 2 metres
whereas the other is varied between 0.5 and 2 metres.
Comparing the lowermost three curves in the diagram, representing plates having
identical thickness
h
, we observe that a long and narrow plate is a more efficient source
than a square one, of course provided the velocity is the same. Similarly, when
comparing the two uppermost curves we find that an increased thickness increases the
radiation. Increasing the thickness implies a reduced critical frequency, here by a factor
of two. This is the reason for the choice showing the radiation factor as a function of the
ratio of frequency to the critical frequency. Additionally, the curves are more general
than indicated by the examples. They may be applied to plates where the relationship
between circumference
U
, area
S
and thickness
h
, i.e. the quantity
U
⋅
h
/
S
is equal to the
ones given in the diagram. To calculate the critical frequency based on material
properties and thickness one should apply Equation
(6.41)
or
Figure 6.11
.
6.3.4.3
Radiation factor by acoustic excitation
The results given in the preceding section only apply to
resonant multimode
vibration of
a plate. It is presupposed that the plate is mechanically excited by a vibration source
having a given bandwidth, either directly excited or by vibrations transmitted from a
