Civil Engineering Reference
In-Depth Information
slabs, characterized by their mass and bending stiffness and coupled together by an
elastic layer characterized by its stiffness only. The system is driven by a point force, a
falling hammer having a mass
m
h
. As the slabs are infinitely large there will only be a
bending wave near field propagating outwards from the driving point. This means that
there will be no reflections setting up a reverberant field, which implies that both slabs
are locally reacting.
Surprisingly, even if the slabs are characterized both by their mass and stiffness, the
latter property does not enter into the expression for the impact sound improvement Δ
L
n
.
This quantity is defined as the difference in the radiated sound power from the primary
construction applying the force directly on it and then to the floating slab. Cremer then
showed that we get
⎛⎞
f
j2
⋅
π
f m
h
Δ=⋅
L
40 lg
+⋅
20 lg 1
+
for
f
>
f
,
(8.40)
⎜
⎝⎠
n
0
f
Z
0
1
point impedance of the floating slab. For heavy floating slabs such as concrete we may
normally neglect the second term in the equation, thereby obtaining an improvement of
12 dB per octave. The assumption concerning local reaction for both slabs is however, as
pointed out above, not valid in practice for such floating floor constructions. One
therefore never experiences improvements as high as predicted by this equation. The
standard EN 12354-1 proposes a modified version of Equation
(8.40)
where the constant
40 is substituted by 30, thereby reducing the frequency dependence to 9 dB per octave.
a)
b)
F
F
Z
1
Z
d
Z
2
Figure 8.26
Model for calculating improvement in impact sound insulation.
It is also interesting to note that the Cremer equation is identical to the one arrived
at using a simple one-dimensional model, where each layer is characterized by its
mechanical impedance. Using the details found in
Figure 8.26
, we may calculate the
velocity amplitudes of the primary floor, first, when being driven directly by a force
F
and, second, when the same force is driving the top floor. Letting these velocities equal
u
2a
and
u
2b
, respectively, we get
F
FZ
⋅
d
u
=
and
u
=
.
(8.41)
2a
2b
(
)
Z
Z
⋅
Z
+
Z
d
Z
+
Z
2
1
2
1
2
The velocity ratio will be
u
Z
Z
Z
ZZZ
>> >>
2a
1
1
1
= +
1
+
⎯⎯⎯⎯⎯→
.
(8.42)
2
1
d
u
Z
Z
Z
2b
2
d
d
