Sound attenuation in building elements like massive concrete walls or floors
The sound transmission through a massive wall or floor depends primarily on the mass of the construction.
Mean attenuation through a massive construction is indicated in the diagram below:
Note! - more mass - more attenuation!
The attenuation for a specific frequency can be calculated by adding the value in the table below to the mean value indicated in the chart above.
|Attenuation Correction |
Note! - higher frequency - more attenuation!
Example - A Concrete Floor and Sound Attenuation
The mass m of a concrete floor with density 2300 kg/m3 and thickness 0.2 m can be calculated as
m = (2300 kg/m3) (0.2 m)
= 460 kg/m2
From the chart above the mean sound attenuation for the floor can be estimated to
The attenuation dL at 250 Hz can be calculated as
dL = (52 db) + (- 5 dB)
= 47 dB
The attenuation at 2000 Hz can be calculated as
dL = (52 db) + (7 dB)
= 59 dB
Sound Transmission Loss - or Attenuation - for some typical Building Elements
|Building Element||Sound Transmission Loss|
|230 mm brickwork, plastered both sides||55|
|230 mm brickwork, plastered one side||48|
|115 mm brickwork, plastered both sides||47|
|100 mm timber studs, plasterboard both sides, quilt in cavity||46|
|6 mm double glazing, 100 mm air gap||44|
|75 mm clinker concrete block, plastered both sides||44|
|115 mm brickwork, plastered one side||43|
|75 mm timber suds, plasterboard both sides||36|
|6 mm single glazing||29|
|one layer plasterboard||25|