Sound Insulation
The reduction of sound energy from
transmission of sound through materials
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As sound travels reaches a surface…there is
reflection, absorption and transmission
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Sound Insulation
The ability of a material to minimise the
transmission of acoustic energy through the
material.
Good sound insulation are obtained with
materials which are reasonable to heavy mass,
and are impervious.
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Sound Transmission
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An Analogy
If one wishes to contain water (sound)….
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An Analogy….
If one wishes to absorb water (sound)….
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Sound Insulation & Absorption
Noise treatment with sound insulation and sound absorption
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Sound Insulation
The difference in average sound pressure levels
between two rooms at any given frequency
Sound Insulation = L1 – L2 dB
where L1 = SPL in the sound source room
L2 = SPL in the adjacent receiving room
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Transmission Loss
A sound reduction index quantifying the sound insulation
performance of a material, building element or partition.
TL = L1 – L2 + 10 log10 (S/A2) dB
where TL = Transmission Loss
L1 = SPL in noise source room
L2 = SPL in receiving room
S = Area of partition
A2 = Absorption in receiving room.
Transmission loss performance varies with sound frequency.
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Factors Affecting Effectiveness of Sound
Insulation
Weight (mass)
Homogeneity and uniformity
Stiffness
Discontinuity or isolation
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Typical Sound Reduction Index
The performance is governed by mass and stiffness
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Transmission
Loss Data
Sheet
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Simplified Performance of Materials
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Calculations Involving Sound
Insulation
Sound Pressure Level in A Receiving Room
The resulting reverberant SPL in the adjacent room is given
by :
SPL 2 = SPL 1 – TL + 10 log Sp + 10 log (1/R)
where R = Room constant
Sp = Area of partition
TL = Transmission Loss
SPL1 = SPL in noise source room
SPL2 = SPL in receiving room
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Approximation
For a quick and simplified estimate, many practical
applications could be approximated as
SPL 2 = SPL 1 – TL
Example :
Plant room SPL = 100 dBA
TL of Brick wall = 45 dBA
Approx. SPL outside = 100 – 45 = 55 dBA
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Example
Foreman’s office to be built inside workshop.
Workshop SPL is 72 dBA. Office to be constructed with
metal of TL=35 dB.
Room dimensions as given. Use amean = 0.10.
Determine Room Constant and Room Correction
Rc = S amean /(1 - amean ) = 6.56
Rm correction = 10log (1/Rc) = -8 dB
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Example..…
Determine Partition Transmission Area
Sp = 10 m2
Noise radiation = 10log Sp = +10 dB
SPL in receiving Room
SPL 2 = SPL 1 – TL + 10 log Sp + 10 log (1/R)
SPL 2 = 72 – 35 + 10 + (-8) = 39 dBA.
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Composite Building Materials
In practice a partition may have a weaker
element (door, window) that would de-rate the
overall sound insulation.
In such case it would be necessary to obtain
the overall performance of the “composite
partition”.
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Nomogram
for Overall TL
Determination
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Example
A partition has a total area of 10 m2.
The partition is metal, with TL of 35 dB
Of the partition, the door is 2 m2 with a TL of 20 dB,
and glazing of 4 m2, and TL 27 dB.
Find the overall TL of the partition.
Material Area (m2) TL (dB)
Metal 4 35
Door 2 20
Glass 4 27
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Example…..
Ratio Metal : Glass 4:4 =1:1
Difference in TL 35 - 27 = 8 dB
From graph, loss of insulation = 6 dB
Combined Metal+Glass TL = 35 – 6 = 29 dB
Ratio Door : Metal+Glass 2 : 8 =1:4
Difference in TL 29 - 20 = 9 dB
From graph, loss of insulation = 4 dB
Combined Door+Metal+Glass = 29 – 4 = 25 dB
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Example….
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Example….
Consider previous example of the Foreman’s Office in the
factory……..
With metal partition only TL metal = 35 dB
With metal + door + glass TL overall = 25 dB
With metal only :
SPL inside office = 72 – 35 + 10 – 8 = 39 dBA
With metal plus door and glass :
SPL inside office = 72 – 5 + 10 – 8 = 49 dBA
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