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44.1 Theory of Sound Insulation

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44.1.1 Expressions of Sound Insulation [1]

44.1.1.1 Transmission Coefficient

Let us denote by Ii the acoustic energy incident on a wall per unit area and unit time. Some energy is

dissipated in the wall, and, apart from the energy that is reflected by the wall, the rest is transmitted

through the wall. Using It to denote the transmitted acoustic energy, the transmission coefficient of the

wall is defined as

t ¼

Ii ð44:1Þ

44.1.1.2 Transmission Loss

As an expression for sound insulation performance, we may use transmission loss (TL), which is defined

as (also see Chapter 42 and Chapter 43)

TL ¼ 10 log

􀀏 􀀐

¼ 10 log

􀀏 􀀐

ð44:2Þ

44.1.2 Transmission Loss of a Single Wall

Consider a plane sound wave incident on a impermeable infinite plate at angle u, which is placed in a

uniform air space as shown in Figure 44.1. The sound pressure of the incident, reflected, and transmitted

44-1

waves, denoted by pi; pr; and pt; respectively, are

given by

pi ¼ Pie jvt2jkðx cos u þ y sin uÞ

pr ¼ Pre jvt2jkð2x cos u þ y sin uÞ ð44:3Þ

pt ¼ Pte jvt2jkðx cos u þ y sin uÞ

where Pi; Pr; and Pt are the sound pressure

amplitudes of incident, reflected, and transmitted

waves, respectively; v is angular frequency; k is the

wave number of the sound wave; c is the speed

of sound, respectively in the air. Assuming

that the plate is sufficiently thin compared

with the wavelength of the incident sound wave,

the vibration velocities on the incident and

transmitted surfaces of the plate are equal. Then vibration velocity, u, of the plate in the x direction is

equal to the particle velocity of the incident and transmitted sound waves, and we obtain relations

u ¼ 2

jvr

›ðpi þ prÞ

›x ¼ 2

jvr

›pt

›x ð44:4Þ

pi þ pr 2 pt

u ¼ Zm ð44:5Þ

where r is the air density and Zm is the mechanical impedance of the plate per unit area. From these

equations, the transmission coefficient, tu ; and then the transmission loss, TLu ; at the incident angle, u; are

obtained according to

TLu ¼ 10 log

tu ¼ 10 log

􀀈 􀀈 􀀈 􀀈 􀀈

¼ 10 log 1 þ

Zm cos u

2rc

􀀈 􀀈 􀀈 􀀈

ð44:6Þ

44.1.2.1 Coincidence Effect

Consider the vibration of the plate in the x – y plane shown in Figure 44.1. Denoting by m the surface

density, and by B the bending stiffness per unit length of the plate, the equation of motion of the plate is

given by

›2j

›t2 þ Bð1 þ jhÞ

›4j

›y4 ¼ pi þ pr 2 pt; B ¼

Eh3

12ð1 2 n 2Þ

ø Eh3

12 ð44:7Þ

where

j ¼ displacement in the x direction

E ¼ Young’s modulus of the plate

h ¼ thickness of the plate

h ¼ loss factor of the plate

n ¼ Poisson’s ratio of the plate

The plane sound wave of angular frequency, v; and of incidence angle, u; causes a bending wave in the

plate where displacement is assumed to be j ¼ j0e jðvt2k1 yÞ; as a solution of Equation 44.7. Hence, the

mechanical impedance per unit area is obtained:

Zm ¼

pi þ pr 2 pt

›j=›t ¼ h

Bk41

v þ j vm 2

Bk41

􀁻 !

ð44:8Þ

where k1 ¼ k sin u ðk ¼ v=cÞ is the wave number of the bending wave in y direction caused by the

incident sound wave. Propagation speed of the forced bending wave, c1; and a free bending wave of

0 x

pr pt

FIGURE 44.1 Plane sound wave incidence on an

infinite plate.

44-2 Vibration and Shock Handbook

the plate, cB; are given by

c1 ¼ v =k1; cB ¼

v2B

􀁻 !1=4

ð44:9Þ

Equation 44.9 reduces Equation 44.8 to

Zm ¼ hv m

􀀏 􀀐4

þ jvm 1 2

􀀏 􀀐4 􀀒 􀀓

ð44:10Þ

When the speed of forced bending wave, c1; and the speed of free bending wave, cB; are equal in Equation

44.10, the imaginary part of Zm becomes 0, and a form of “resonance” occurs. Then the transmission loss

decreases rapidly. This phenomenon is called the coincidence effect, and the resonant frequency dependent

on the incident angle is given by

f ¼

2p sin2u

ffiffiffiffi

ð44:11Þ

The minimum of the resonant frequency is called coincidence critical frequency, or critical frequency for

short, and it reduces to

fc ¼

ffiffiffiffi

ø c2

1:8cLh ð44:12Þ

where cL ¼

ffiffiffiffiffiffi

E=rP p is the speed of longitudinal wave in the plate, and rP denotes the density of the plate.

Let us show the relations of the critical frequency and the plate thickness of typical material

of sound insulation shown in Figure 44.2. Using the relation cB=c1 ¼

ffiffiffiffiffi

f =fc

sin u; Equation 44.10

104

103

102

0.1 2 3 4 5 6 7 81.0 2 3 4 5 6 7 8 10

Thickness h (cm)

Critical frequency fc (Hz)

2 3 4 5 6 7 8100

104

103

102

2 (CL = 4000m/s)

5 (CL = 1000m/s)

3 (CL = 3000m/s)

4 (CL = 2000m/s)

1 (CL = 5000m/s)

FIGURE 44.2 Critical frequency vs. plate thickness of typical sound insulation materials: (1) aluminum, steel, or

glass; (2) hardboard or copper; (3) dense concrete, plywood, or brick; (4) gypsum board; (5) lead or light weight

concrete. (Source: Beranek, L.L. 1988. Noise and Vibration Control, INCE/USA. With permission.)

Design of Sound Insulation 44-3

becomes

Zm ¼ hvm

􀀏 􀀐2

sin4u þ jvm 1 2

􀀏 􀀐2

sin4u

" #

ð44:13Þ

44.1.2.2 Mass Law of Transmission Loss

When f ,, fc; Equation 44.13 becomes Zm ø jvm: Then, the transmission loss depends on the

incident angle, the frequency and the surface density of the plate. This is called the mass law of

transmission loss.

Mass law of normal incidence represents the transmission loss at the incident angle u ¼ 0; as given by

TL0 ¼ 10 log 1 þ

2rc

􀀏 􀀐2 􀀒 􀀓

ð44:14Þ

For vm .. 2rc; it becomes

TL0 ø 10 log

2rc

􀀏 􀀐2

¼20 log mf 2 42:5; for air ð44:15Þ

Mass law of random incidence represents the transmission loss at the angle averaged over a range of u

from 0 to 908, which is realized for perfectly diffused sound field. We have

TLr ¼ 10 logð1=trÞ ø TL0 2 10 logð0:23TL0Þ ð44:16Þ

where the random incident transmission coefficient, tr; is defined as

tr ¼

ðp=2

tu cos u sin u du

􀀝 ðp=2

cos u sin u du ð44:17Þ

An approximation for Equation 44.16, as given below, is generally used for a practical use and this is often

useful.

TLr ¼ 18 log mf 2 44 ð44:18Þ

Mass law of field incidence represents the transmission loss at the angle averaged over a range of u from

0 to about 788, which is said to agree with actual sound field. We have

TLf ¼ TL0 2 5 ð44:19Þ

The three types of transmission loss presented above are compared in Figure 44.3.

44.1.2.3 Stiffness Law of Transmission Loss [2]

The plate described above is assumed to be infinite. However, an actual plate is always supported by some

structures at its boundaries and the plate size is finite. Transmission loss of a finite plate is considered to

be related to the nature of excitation of vibration in the plate, for example, sound wave incidence, modes

of vibration and characteristics of sound radiation. Therefore, the governing relationships become very

complex. However, in the following frequency range, it is known that the transmission loss conforms to

the mass law

, f ,, fc ð44:20Þ

where a is length of shorter edge for rectangular plate.

When f , c=2a; the whole plate is excited in phase, and stiffness effects from the supports of its

edges will appear. If we denote the equivalent stiffness of the plate as K and assume a loss factor of 0,

44-4 Vibration and Shock Handbook

the mechanical impedance of the plate is obtained by using Equation 44.8; thus

Zm ¼ j vm 2

􀀏 􀀐

ð44:21Þ

The frequency at Zm ¼ 0 corresponds to the first mode natural frequency, f11; of the plate (see Chapter 4),

and consequently, the equivalent stiffness of rectangular plate with simple edge-support is given by

f11 ¼

ffiffiffiffi

; p

ffiffiffiffi

􀀏 ��2

􀀏 􀀐2 􀀒 􀀓

ð44:22Þ

Then,

K ¼ Bp4 1

a2 þ

􀀏 􀀐2

where a and b are the length of the short and long edges for the rectangular plate, respectively. When

f ,, f11 is assumed in Equation 44.21, the mass term can be neglected, and from Equation 44.6 the

normal incidence transmission loss, TLS0, is given by

TLS0 ¼ 10 log 1 2 j

2vrc

􀀈 􀀈 􀀈 􀀈

¼ 10 log 1 þ

2vrc

􀀏 􀀐2 􀀒 􀀓

ø 20 logðK=f Þ 2 74:5 ð44:23Þ

This is called the stiffness law of the transmission

loss, and it shows a 6 dB decay per octave.

The characteristics mentioned above for single

wall transmission loss are shown in Figure 44.4

and summarized below.

FIGURE 44.3 Theoretical transmission loss based on mass law.

Region I

Region II

Region III Region IV

Frequency (Hz)

f11 fc

− 6dB/octave

5.4 dB/octave

Transmission loss TL (dB)

FIGURE 44.4 Transmission loss characteristics of a

single wall.

Design of Sound Insulation 44-5

1. Region I ð f ,, f11Þ: Transmission loss is controlled by the stiffness of the panel:

TL ¼ TL0 2 40 log

f11

􀀏 􀀐

ð44:24Þ

2. Region II ð f < f11Þ: Transmission loss is controlled by the lower-mode natural frequencies of the

panel, and the estimation becomes very complex.

3. Region III ð f11 ,, f # fc=2Þ: Transmission loss is controlled by the mass (surface density) of the

panel:

TL ¼ 18 log mf 2 44 ð44:25Þ

4. Region IV ðf . fc=2Þ: Transmission loss is controlled by the mass and the damping of the panel,

and it is reduced by coincidence effects.

For fc=2 , f # fc:

TL is represented by a straight line connecting the value at f ¼ fc=2 of Equation 44.25 and the value at

f ¼ fc of Equation 44.26.

For f . fc:

TL ¼ TL0 þ 10 log

􀀏 􀀐

ð44:26Þ

44.1.3 Transmission Loss of Multiple Panels

To realize sound insulation of high performance, we often use a double wall or a multiple panel

composed of insulation materials like steel plates and absorbing materials like fiber-glass. In this

subsection, transmission loss of a multiple panel is described [3].

44.1.3.1 Calculation Method

Consider a multiple panel of infinite lateral extent

as shown in Figure 44.5, which is composed of n

acoustic elements, each element consisting of three

basic materials, an impermeable plate, air space,

and an absorption layer. Furthermore, consider a

plane wave incident on the left-hand side surface

of the nth element at angle u: Let the sound

pressure of the incident wave be pi; and of the

reflected wave be pr; and the wave be propagating

through the structure, and then radiating from

the right-hand side of the first element as a plane

wave of pressure pt into a free field at transmission

angle u:

In the analysis, we append the subscript

kð¼ 1; 2; …; nÞ to the physical parameters of the

kth element, and “2” and “1” to the left- and righthand

side values of these parameters, respectively, as shown in Figure 44.5. The ratio of the sound

pressure at the incident surface, pn2; to the incident wave, pi; is given by

pn2

pi ¼

pi þ pr

pi ¼

2Zn2

Zn2 þ rc=cos u ð44:27Þ

where Zn2 is the acoustic impedance of the left-hand side normal to the surface of the nth element and

rc=cos u is the acoustic impedance normal to the surface, which is equal to the radiation impedance of the

q q

n k 1

pr pt

Zn2 Zk2 Zk1 Z11

pn2 pk2 pk1 p11

FIGURE 44.5 Calculation model of n-element multiple

panel.

44-6 Vibration and Shock Handbook

first element, Z11; shown in Figure 44.5. Using the usual condition of pressure matching at each interface,

we can write the expression for the oblique incidence transmission coefficient as

t ðuÞ ¼

􀀈 􀀈 􀀈 􀀈 􀀈

; p11

􀀈 􀀈 􀀈 􀀈 􀀈

pn2

􀀈 􀀈 􀀈 􀀈 􀀈

pn1

pn2

· · ·

pk1

pk2

· · ·

p11

p12

􀀈 􀀈 􀀈 􀀈 􀀈

ð44:28Þ

Hence, we obtain the following expression for the random incidence transmission loss:

TL ¼ 10 log

ðul

cos u sin u du

ðul

t ðuÞ cos u sin u du

BBB@

CCCA

ð44:29Þ

where u l is the limiting angle above which no sound is assumed to be received, and it varies between

788 and 858.

If we know Zn2 in Equation 44.27 and the pressure ratio across each of the single elements in

Equation 44.28, we can calculate the TL using Equation 44.29. We can obtain Zn2 by using the conditions

of impedance matching at each interface from the rightmost to the leftmost element in order, if we know

the impedance relations across each of the single elements.

Now, we present the pressure ratios and the acoustic impedance relations across three basic elements.

44.1.3.2 Impermeable Plate

Consider the vibration of an infinite impermeable

plate of thickness, h, induced by the sound

pressure difference on each side of the plate, as

illustrated in Figure 44.6. In this case, the particle

velocity on both sides of the plate must be the same

as the plate vibration velocity. Then, from

Equation 44.8, the following expressions are

obtained:

Z2 ¼ Z1 þ Zm ð44:30Þ

p1 ¼

Z1 ð44:31Þ

where p2; p1 are the sound pressure at the incident

surface x ¼ 0 and at the transmitted surface x ¼ h;

respectively, Zm is the mechanical impedance of the plate, and Z2; Z1 are the acoustic impedance normal

to the incident surface at x ¼ 0 and the transmitted one at x ¼ h; respectively.

44.1.3.3 Sound Absorbing Material

For a sound absorbing material layer of thickness d and infinite lateral extent, consider a plane wave

incident at an angle u to the normal, as shown in Figure 44.7. Deriving the wave equation in the sound

absorbing material and applying the continuity conditions of the sound pressure across the surface at

x ¼ 0 and x ¼ d; with some mathematical manipulation we get the following results:

Z2 ¼

gZ0

coth ðqd þ wÞ ð44:32Þ

p1 ¼

cosh ðqd þ wÞ

cosh w ð44:33Þ

q ¼ g

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 þ

􀀏 􀀐2

sin2u

; w ¼ coth21 qZ1

gZ0

􀀏 􀀐

ð44:34Þ

q q

q h

Pr Pt

Pi P2

Z2 Z1

FIGURE 44.6 Excitation of infinite plate by a plane

sound wave.

Design of Sound Insulation 44-7

where g is a propagation constant and Z0 is a

characteristic impedance of a homogeneous, isotropic

absorbing material.

If porous material is used as the absorbing

material, the following relations are applicable for

g and Z0 [4]:

Z0 ¼ R þ jX

R =rc ¼ 1 þ 0:0571ðr f =RlÞ20:754

X=rc ¼ 20:0870ðr f =RlÞ20:732

ð44:35Þ

g ¼ a þ jb

a =k ¼ 0:189ðr f =RlÞ20:595

b=k ¼ 1 þ 0:0978ðr f =RlÞ20:700

ð44:36Þ

ð0:01 # r f =Rl # 1Þ

where r is the air density, f is the frequency, and Rl

is the flow resistivity, respectively. Specifically, note

that Rl is defined as the flow resistance of the

porous absorbing material per unit thickness.

With data measured with a measuring tube of

flow resistance, we can write

Rl ¼

l·u ð44:37Þ

where Dp is pressure difference between the inlet

and the outlet of the absorbing material in the tube,

u is the mean flow velocity in the tube, and l is the

thickness of the absorbing material. It is known

that the flow resistivity of porous absorbing

material such as fiber-glass or rock wool is related

to the bulk density, as shown in Figure 44.8.

44.1.3.4 Air Space

For an air space, Z0 ¼ rc and g ¼ jk: Hence, Equation 44.32 to Equation 44.34 reduce to

Z2 ¼

cos u

coth ð jkd cos u þ dÞ ð44:38Þ

p1 ¼

cosh ð jkd cos u þ dÞ

cosh d ð44:39Þ

d ¼ coth21 Z1 cos u

􀀏 􀀐

ð44:40Þ

44.1.3.5 Double Wall [2]

Applying the theory formulated above, we can easily obtain the transmission loss of a double

wall composed of the three elements: impermeable plate, air space, and impermeable plate, as shown

in Figure 44.9. Assume that the two impermeable plates have the same surface density, m, and

the mechanical impedance of the plates is jvm: Then, we can obtain following equations for element

105

104

10 5 102 5

Fiber glass

Rock wool

Flow resistivity R1 (MKS rayls/m)

Bulk density rm (Kg/m2)

FIGURE 44.8 Flow resistivity vs. bulk density for

porous, sound absorbing materials.

0 d x

Z2 Z1

Pi Pi

FIGURE 44.7 Schematic relation of sound wave

directions.

44-8 Vibration and Shock Handbook

one and element three:

Z12 ¼ Z11 þ Zm ¼

cos u þ jvm;

p12

p11 ¼

Z12

Z11 ¼ 1 þ j

vm cos u

ð44:41Þ

Z32 ¼ Z31 þ Zm ¼ Z22 þ jvm;

p32

p31 ¼

Z32

Z22 ¼ 1 þ j

Z22

ð44:42Þ

For element two:

Z22 ¼

cos u

coth ð jkd cos u þ d 0Þ;

p22

p21 ¼

cosh ð jkd cos u þ d 0Þ

cosh d 0

ð44:43Þ

where, by applying impedance matching

conditions at the interface of element one and element two, the following definition is introduced:

d 0 ¼ coth21 Z21 cos u

􀀏 􀀐

¼ coth21 Z12 cos u

􀀏 􀀐

¼ coth21 1 þ j

vm cos u

􀀏 􀀐

ð44:44Þ

In this case, Equation 44.27 reduces to

p32 ¼

Z32 þ rc=cos u

2Z32 ¼

Z22 þ jvm þ rc=cos u

2ðZ22 þ jvmÞ ð44:45Þ

Substituting Equation 44.41 to Equation 44.45 into Equation 44.28, we obtain the transmission loss of

the double wall:

TLu ¼ 10 log½1=t ðuÞ􀀉 ¼ 10 log½1 þ 4a2 cos2u ðcos b 2 a cos u sin bÞ2􀀉 ð44:46Þ

a ¼ vm=2rc; b ¼ kd cos u

In Equation 44.46, the transmission loss is zero, and full passage (i.e., “all-pass” in the filter

terminology) of sound occurs when the following equation holds:

cos b 2 a cos u sin b ¼ 0 ð44:47Þ

When b ,, 1ðkd ,, 1Þ; the frequency of full passage for normal incidence is given by

fr ¼

ffiffiffiffiffiffiffi

2rc2

ð44:48Þ

This is the natural frequency of a vibrating system consisting of two masses, m, connected by a spring of

spring constant, rc2=d:

When b .. 1ðkd .. 1Þ; the solution of Equation 44.47 for b is b ø np; and the frequency of all

passage for normal incidence is given by

fn ¼

2d ðn ¼ 1; 2; 3; …Þ ð44:49Þ

These are the acoustic resonant frequencies of the air space d.

Characteristics of the transmission loss given by Equation 44.46, in case of normal incidence ðu ¼ 0Þ;

are as follows:

1. f , fr

􀀍

b ,

ffiffiffiffiffiffiffiffi

2rd=m

p 􀀎

TL ø 10 logð4a2Þ ¼ TL0 þ 6 ð44:50Þ

This is equal to the transmission loss of a single wall of surface density 2m.

pr pt

m m

3 2 1

q q

FIGURE 44.9 Calculation model of a double wall with

air space.

Design of Sound Insulation 44-9

2. fr # f , f1=p

􀀍 ffiffiffiffiffiffiffiffi

2rd=m

# b , 1

􀀎

TL ø 10 logð4a4b2Þ ¼ 2TL0 þ 20 logð2kdÞ ð44:51Þ

This transmission loss indicates an 18 dB increase per octave.

3. f ¼ ð2n 2 1Þc =4dðb ¼ np 2 p=2Þ

TL ø 10 logð4a4Þ ¼ 2TL0 þ 6 ð44:52Þ

A straight line connecting the transmission losses at these frequencies in Figure 44.10 indicates a 12 dB

increase per octave. When the two impermeable plates have different surface densities, m1 and m2;

Equation 44.41 to Equation 44.52 reduce to

1. f , fr

􀀍

b ,

ffiffiffiffiffiffiffiffi

2rd=m

p 􀀎

TL ¼ 20 log½vðm1 þ m2Þ=2rc􀀉 ð44:53Þ

2. fr # f , f1=p

􀀍 ffiffiffiffiffiffiffiffi

2rd=m

# b , 1

􀀎

TL ¼ TL1 þ TL2 þ 20 logð2kdÞ ð44:54Þ

3. f ¼ ð2n 2 1Þc=4dðb ¼ np 2 p=2Þ

TL ¼ TL1 þ TL2 þ 6 ð44:55Þ

In these equations, TL1 and TL2 are the transmission losses of each plate, which are given by

Equation 44.15.

The transmission loss of a double wall, as mentioned above, is shown schematically in Figure 44.10.

An actual double wall, however, is finite in size and the air space forms a closed acoustic field, which

f1/π f1

Frequency (Hz)

TL = 20 log[(m1 + m2) f ] − 42.5

TL = TL1 + TL2 + 20 log(2kd)

TL = TL1 + TL2 + 6

Transmission loss TL (dB)

FIGURE 44.10 Transmission loss of a double wall with air space.

44-10 Vibration and Shock Handbook

makes the transmission loss deviate from the theoretical value. Figure 44.11 gives a design chart of an

actual double wall, which is based on theory and experiments.

44.1.4 Transmission Loss of Double Wall with Sound Bridge [5]

In the previously presented theory, each plate of the multiple panel is considered to be structurally

independent. In actual multiple panels, such as partitions of a building or sound insulation laggings of a

duct, however, each plate is connected with steel sections, stud bolts, and the like, which are called sound

bridges. This is illustrated in Figure 44.12.

The sound pressure of the transmitted wave through a double wall with sound bridges is given by the

summation of radiated sound pressure from the vibration of the transmitted side plate excited by the

sound in the air space and that mechanically excited by the sound bridges.

Double wall with air space

√2 fr 2 fr

3 fc

Frequency (Hz)

Mass LawTL

18 dB/octave

12 dB/octave

Double wall with sandwiched

porous material

m1 m1

Transmission loss TL (dB)

FIGURE 44.11 Design chart for estimating the transmission loss of a double wall with sandwiched porous material

or air space.

FIGURE 44.12 Examples of actual double wall with sound bridges.

Design of Sound Insulation 44-11

The acoustic power radiated from the area, S, of an infinite plate excited by sound pressure is given by

WP ¼ rcSv2

2 ð44:56Þ

where v2

2 is the space averaged mean square vibration velocity over the plate. The acoustic power radiated

from the plate mechanically excited by a point force or a line force is

WB ¼ rcxv2ð f ,, fcÞ ð44:57Þ

x ¼

p3 l2c

ðpoint force excitationÞ

llc ðline force excitationÞ

ð44:58Þ

where v2 is the mean square vibration velocity of the plate at the excitation point, lc ¼ c=fc is the

wavelength of the bending wave at the critical frequency, and l is the length of the line force. By

comparing Equation 44.56 and Equation 44.57, it is noted that x is the effective area of the acoustic power

radiated from the infinite plate excited by the point or line force. Acoustic power, WB; is the power

radiated from a small area near the excitation point, because a free bending wave propagating in an

infinite plate can radiate little sound when f , fc:

From the equations given above, the total acoustic power radiated from the transmitted side plate is

obtained as

WT ¼ WP þ WB ¼ rcSv2

2 1 þ

�� 􀀐2 􀀒 􀀓

ð44:59Þ

where n is the number of excitation forces applied to the area, S. Then, transmission loss TL T of the

double wall with sound bridges is given by

TL T ¼ 10 log

􀀏 􀀐

¼ 10 log

􀀏 􀀐

¼ TL 2 TLB ð44:60Þ

where WI is the acoustic power incident on the double wall, TL is the transmission loss of the double wall

without a sound bridge, and TLB denotes the transmission loss reduction by the sound bridges, and is

given by

TLB ¼ 10 log

􀀏 􀀐

¼ 10 log 1 þ

􀀏 􀀐2 􀀒 􀀓

ð44:61Þ

We assume the following:

1. The vibration velocity of the incident side plate is not affected by the sound bridges.

2. The vibration velocity of the transmitted side plate at the excitation points (connecting points) is

equal to the velocity v1 of the incident side plate, and consequently, the next equation holds

ø v1

With these assumptions, we apply the method presented in section 44.1.3, to determine v=v2 as

ø v1

v2 ¼

v2m2d

rc2 ð fr , f , f1=pÞ ¼

vm2

rc ð f . f1=pÞ ð44:62Þ

Using Equation 44.60 and Equation 44.53, we can obtain the increase in transmission loss, DTL;

from the transmission loss of mass law based on the total mass of the double wall, as presented

below.

44-12 Vibration and Shock Handbook

1. Point connection

DTL ¼ TL 2 TLB 2 20 log

vðm1 þ m2Þ

2rc

􀀒 􀀓

¼ 20 logðefcÞ þ 20 log

m1 þ m2

􀀏 􀀐

þ 10 log

8c2

􀁻 !

ð44:63Þ

2. Line connection

DTL ¼ TL 2 TLB 2 20 log

vðm1 þ m2Þ

2rc

􀀒 􀀓

¼ 10 logðbfcÞ þ 20 log

m1 þ m2

􀀏 􀀐

þ 10 log

􀀏 􀀐

ð44:64Þ

where e ¼

ffiffiffiffi

S=n p is the distance between point forces and b ¼ S=nl is the distance between line

forces.

Figure 44.13 presents a practical and useful design chart of the transmission loss for a double wall with

sound bridges, which is based on Figure 44.11 and Equation 44.63 and 44.64.

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44.1 Theory of Sound Insulation

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