1. SOUND INSULATION

2. Transmission Loss, R
When sound energy is transmitted through a panel (wall or floor) one way of expressing the proportion of energy transmitted is by giving the reduction in energy flow in decibels.
This is the Transmission Loss, R
Sometimes also given as the “Sound Reduction Index”, SRI

3. Diffuse Field Sound Transmission Loss, R
“Coincidence
Controlled “ - flexural wave resonance
LOW FREQUENCIES
“Stiffness
Controlled”
HIGHER FREQUENCIES
“Mass Controlled”
TRANSMISSION LOSS. R (dB)
FREQUENCY (Hz)
Slope 6 dB per
doubling of frequency
First panel resonance
Coincidence
Critical frequency

4. Equation of Motion for a Panel
Force on wall = Incident sound pressure
(pressure = force / area)
Force = P0 sin(2ft)
Force required to vibrate the wall
F = Mass x Acceleration
F = Damping x Velocity
F = Stiffness x Displacement
Force = mx + dx +kx
m is mass, d is damping factor, k is stiffness
x is displacement, x is velocity, x is acceleration +  .. . . ..

5. Simple theoretical model - mass region only .. .
Equating we get,
Po sin(2ft) = mx + dx + kx
This 2nd order differential equation can be solved to find the vibration of the wall and hence sound transmitted from the opposite side into an adjoining space.
Difficulties
Based on single angle of incidence
Stiffness and damping are difficult to predict
Simple theoretical model - mass region only
Luckily this is usually the most important .. .

6. Transmission Loss at Angle θ
Solving the equation of motion for a single angle and converting this into a decibel reduction we get:-  P A R T I O N

7. Theoretical Mass Law Rearranging etc. we get, R0 = 20lg(mf) – 42
For normal incidence, θ = 0 so Cosθ = 1
we can replace:- ω by 2πf,
ρc by 135 rayls, and the
superficial density ρs by m (mass/m2)
Rearranging etc. we get, R0 = 20lg(mf) – 42
Since RField = R0 – 5dB the THEORETICAL MASS LAW is
RField = 20lg(mf) – 47 dB

8. Example
A thin panel has a surface mass of 20 kg.m-2 find the transmission loss, a frequency of 500Hz
R = 20 lg(mf) – 47 dB
substitute in for m and f we get,
R = 20 lg(20x500) – 47 dB
R = 33 dB
Repeating this for other frequencies gives,
27 250Hz
39 1kHz
Note that this is a 6 dB improvement for each doubling of frequency
and also a 6 dB improvement for each doubling of mass
R (dB)
Slope 6 dB/ doubling
of frequency
Frequency (Hz)

9. Mass .v. Layers From the previous example Double the mass
surface mass 20 kg.m-2
R = 500Hz
Double the mass
surface mass 40 kg.m-2
R = 500Hz
Create two separate isolated layers
surface mass 20 kg.m-2 per layer
potential R = = 500Hz
Two thin layers are better than one thick layer
R=33dB
R=39dB
R=66dB

10. PRACTICAL SOUND TRANSMISSION
A wall (or floor) is a barrier to airborne sound energy
Sound Transmission Coefficient,  (Tau)
 = transmitted energy
incident energy
For instance, if 1% of the energy is transmitted then  = 0.01
5% transmitted  = 0.05
100% transmitted  = 1.0
air
Incident
Sound energy
wall
Source room
Receiving room
transmitted

11. Transmission Loss, R
In practical terms this is the inverse of the transmission coefficient and expressed in dB
For instance
if 1% of the energy is transmitted then R = 20 dB
if 5% of the energy is transmitted then R = 13 dB
if 10% of the energy is transmitted then R = 10 dB
if 100% of the energy is transmitted then R = 0 dB

12. SOUND TRANSMISSION EQUATION
Mic L1 L2
The level, L2 in the receiving room depends on:-
Transmission Loss, (R) of the wall being tested
The area, (Se) of the separating wall
The amount of absorption, (A) in the receiving room

13. Measurement of R in a Laboratory
Spring isolation to prevent flanking sound transmission L1 L2
Sending Room
Receiving Room, volume V
Loudspeaker
Mic
Wall construction being tested with area, Se
Isolation pads
Sound Transmission Suite
Measure L1 & L2, and the reverberation time,T in room 2
Find the room absorption from Sabine formula
Note: No flanking sound

14. MEASUREMENT OF ROOM ABSORPTION
The reverberation time can be measured if the room exists
Absorption area found by transforming the Sabine formula
What if the building hasn’t been built – the design stage!
PREDICTION OF ROOM ABSORPTION
Room absorption can be calculated by adding all the room surface areas x absorption coefficients.
A = s
A = s11 + s22 + 

15. Example Calculation
It is proposed to convert an old warehouse building into sheltered accommodation for the elderly.
One of the flats will be next to the boiler room and you have been asked to determine whether the proposed partition wall between the boiler room and the flat will provide sufficient sound reduction.
Flat
Boiler
room

16. The target level in the receiving room is NR35.
Flat
Boiler
room
Data
The noise level in the boiler room is 87 dB in the dominant 500 Hz frequency band. The separating wall is to be of plastered timber stud construction 3m high x 5m long.
For a wall 50mm x 100mm studs, 9mm plasterboard, 12mm plaster the R values are (Woods Practical Guide to Noise Control):-
The absorption in the receiving 500 Hz has been calculated by adding together the all the surface areas x absorption coefficients giving a total absorption of 11 m2 Sabine units.
The target level in the receiving room is NR35. 63
125
250
500 1k 2k 4k 8k Hz 20 25 28 34 47 39 50 56 R
39 dB

17. Solution The level in the boiler room L1 = 87 dB @ 500 Hz
Transmission Loss, R at 500 Hz is 34 dB (table)
The areas of the wall, Se is 3x5 = 15 m2
Total absorption, A is 11m2 sabins
Substituting these values into the transmission equation we get,
L2 = L1 - R + 10lg(Se/A)
L2 = lg(15/11)
L2 = 54 dB
NR35 requires a level of 39 Hz, therefore the target has not been achieved.
Action -- Reduce noise level in boiler room by 15 dB
Choose a better wall (double masonry wall)

18. Composite Panels (walls)
Let’s consider the sound transmission through the wall
between the performance space and the control room

19. Triple leaf concrete block wall
Partition Details
Triple leaf concrete block wall
Double glazed unit
Wall 3m x 3m has R values of:-
Glazing 1.5m x 1m has R values of:-
What is the composite R say, 500 Hz?
Triple leaf cavity wall, 112/215/112mm with 50mm cavities between leaves
Freq 63
125
250
500 1k 2k 4k 8k R 53 56 62 69 78 84
6/200/10 double glazing
Freq 63
125
250
500 1k 2k 4k 8k R 35 46 56 65

20. Composite R for panels with different values of R in different areas
This is based on calculating the composite sound transmission coefficient, composite
composite = 1x Area1 + 2 x Area2 + +n x Arean
Total area of the panel
Individual ,s are found from R = 10lg(1/ )
 = 1/10(R/10)
Rcomposite = 10lg(1/ composite)

21. similarly for the wall (R=69dB)
Solution (for 500 Hz)
 = 1/10(R/10)
glazed panel (R=46dB)
 Glazing = 1/10(46/10) =
similarly for the wall (R=69dB)
Wall = 1/10(69/10) =
composite = 1x Area1 + 2 x Area2 + +n x Arean
Total area of the panel
Total area = 9m2 , Glazed area = 1.5m2 , Wall area = 7.5m2
composite = x x = 9
R composite = 10lg(1/ composite ) = 53.7 dB (54 dB)
Note comparison with --- wall = 69 dB, Glaze = 46 dB

22. NEXT WEEK
The Building Regulations & sound insulation tests for walls and floors
Typical constructions for residential and commercial buildings & remedial treatments
Design Process