ANALYSIS OF HIGH FREQUENCY BEHAVIOR OF PLATE AND …DAVRANIŞLARININ İSTATİSTİKSEL ENERJİ...

ANALYSIS OF HIGH FREQUENCY BEHAVIOR OF PLATE AND BEAM STRUCTURES BY STATISTICAL ENERGY ANALYSIS METHOD

A THESIS SUBMITTED TO THE GRADUATE SCHOOL OF NATURAL AND APPLIED SCIENCES

OF THE MIDDLE EAST TECHNICAL UNIVERSITY

BY

CANAN YILMAZEL

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR

THE DEGREE OF DOCTOR OF PHILOSOPY IN

MECHANICAL ENGINEERING

JUNE 2004

Approval of the Graduate School of Natural and Applied Sciences

_____________________

Prof. Dr. Canan ÖZGEN

Director

I certify that this thesis satisfies all the requirements as a thesis for the degree of

Doctor of Philosophy.

_____________________

Prof. Dr. Kemal İDER

Head of the Department

This is to certify that we have read this thesis and that in our opinion it is fully

adequate in scope and quality, as a thesis for the degree of Doctor of Philosophy.

____________________

Prof. Dr.Y. Samim ÜNLÜSOY

Supervisor

Examining Committee Members:

Prof. Dr. Bülent E. PLATİN (METU, ME) ___________________

Prof. Dr.Y. Samim ÜNLÜSOY (METU, ME) ___________________

Prof. Dr.T. Mehmet ÇALIŞKAN (METU, ME) ___________________

Prof. Dr. Yavuz YAMAN (METU, AEE) ___________________

Assoc.Prof. Dr. Müfit GÜLGEÇ (Gazi Üniv.) ___________________

iii

PLAGIARISM

I hereby declare that all information in this document has been obtained and presented in accordance with academic rules and ethical conduct. I also declare that, as required by these rules and conduct, I have fully cited and referenced all material and results that are not original to this work. Name, Last name : Canan, YILMAZEL

Signature :

iv

ABSTRACT

ANALYSIS OF HIGH FREQUENCY BEHAVIOR OF PLATE AND BEAM

STRUCTURES BY STATISTICAL ENERGY ANALYSIS METHOD

YILMAZEL, Canan

Ph.D., Department of Mechanical Engineering

Supervisor: Prof. Dr.Y. Samim ÜNLÜSOY

June 2004, 179 pages

Statistical Energy Analysis (SEA) is one of the methods in literature to

estimate high frequency vibrations. The inputs required for the SEA power balance

equations are damping and coupling loss factors, input powers to the subsystems. In

this study, the coupling loss factors are derived for two and three plates joined with a

stiffener system. Simple formulas given in the literature for coupling loss factors of

basic junctions are not used and the factors are calculated from the expressions

derived in this study. The stiffener is modelled as line mass, Euler beam, and open

section channel having double and triple coupling. Plate is modelled as Kirchoff

plate. In the classical SEA approach the joint beam is modelled as another

subsystem. In this study, the beam is not a separate subsystem but is used as the

characteristics of the joint and to calculate the coupling loss factor between coupled

plates. Sensitivity of coupling loss factors to system parameters is studied for

different beam approaches.

v

The derived coupling loss factors and input powers are used to calculate the

subsystem energies by SEA. The last plate is joined to the first one to simulate the

fuselage structure. A plate representing floor structure and acoustic volume are also

added. The different modelling types are assessed by applying pressure wave

excitation. It is shown that deriving the parameters as given in this study increases

the efficiency of the SEA method.

Keywords: Statistical Energy Analysis, SEA, Coupling Loss Factor, High Frequency

Vibrations, Wave Propagation, Stiffened Plate, Fuselage

vi

ÖZ

PLAKA VE KİRİŞ YAPILARININ YÜKSEK FREKANS

DAVRANIŞLARININ İSTATİSTİKSEL ENERJİ ANALİZİ METODU

İLE ANALİZİ

YILMAZEL, Canan

Doktora, Makina Mühendisliği Bölümü

Tez Yöneticisi: Prof. Dr.Y. Samim ÜNLÜSOY

Haziran 2004, 179 sayfa

İstatistiksel Enerji Analizi (İEA) yüksek frekanslarda titreşim davranışlarını

tahminde kullanılan metotlardan biridir. İEA güç denge denklemlerine girdi olarak

sönümleme ve etkileşim kayıp faktörleri gerekmektedir. Bu çalışmada bir kiriş ile

birleştirilmiş iki ve üç plaka için etkileşim kayıp faktörleri çıkarılmıştır. Yayınlarda

bu faktörler için verilen basit denklemler yerine, bu çalışmada elde edilen ifadeler

kullanılmıştır. Ara bağlantı elemanı doğrusal kütle, Euler kirişi, ikili ve üçlü

etkileşimli açık kesit kiriş olarak modellenmiştir. Plaka Kirchoff plaka olarak

modellenmiştir. Genel İEA yaklaşımı ara kirişi ayrı bir altsistem olarak

modellemektir. Bu çalışmada kiriş ayrı bir altsistem değil, bağlantının karakteristiği

olarak alınmış ve etkileşim kayıp faktörlerini hesaplamak için kullanılmıştır.

Etkileşim kayıp faktörlerinin sistem parametrelerine duyarlılığı tüm modelleme

tipleri için yapılmıştır.

vii

Sayısal olarak hesaplanan güç girdisi ve etkileşim kayıp faktörleri, altsistem

enerjilerini İEA ile hesaplamada kullanılmıştır. Son plakayı ilkine bağlayıp kapalı bir

sistem oluşturularak uçak gövde yapısı simüle edilmiştir. Ayrıca akustik hacim

eklenmiş ve taban yapısı da bir plaka ile modellenmiştir. Değişik modelleme

tiplerinin sonuçlara etkisi gösterilmiştir. Parametreleri bu çalışmada verildiği şekilde

almanın, İEA metodunun verimliliğini arttırdığı gösterilmiştir.

Anahtar Sözcükler: İstatistiksel Enerji Analizi, Etkileşim Kayıp Faktörü, Yüksek

Frekans Titreşimi, Dalga Yayılımı, Takviye Edilmiş Plaka, Uçak Gövdesi

viii

to me

and the last 8 years

ix

ACKNOWLEDGEMENTS

I express sincere appreciation to Prof. Dr. Y. Samim Ünlüsoy for his

guidance and insight throughout the research. Special thanks go to the other Thesis

Progress Committee members, Prof. Dr. E. Bülent Platin and Prof. Dr. Yavuz

Yaman, for their valuable suggestions and comments.

I offer sincere thanks to my colleagues for their endless encouragement and

support. Thanks go to Dr. Akif Erşahin, Dr. Fatih Cıbır, Emel Aslan and İlke Dikici

for their helps in writing the thesis.

Thanks to my family to understand my madness and to leave me by myself

during the last painful months.

x

TABLE OF CONTENTS

PLAGIARISM ................................................................................................................................. III

ABSTRACT ..................................................................................................................................... IV

ÖZ ..................................................................................................................................................... VI

ACKNOWLEDGEMENTS ............................................................................................................ IX

TABLE OF CONTENTS ..................................................................................................................X

LIST OF TABLES........................................................................................................................ XIII

LIST OF FIGURES...................................................................................................................... XIV

LIST OF SYMBOLS AND ABBREVIATIONS ........................................................................ XIX

CHAPTER

1 INTRODUCTION.................................................................................................................... 1

1.1 LITERATURE SURVEY........................................................................................................ 3 1.2 OBJECTIVE AND SCOPE OF THE PRESENT STUDY .............................................................. 7

2 BASICS OF SEA.................................................................................................................... 10

2.1 CLASSICAL STATISTICAL ENERGY ANALYSIS ................................................................. 11 2.2 SUBSYSTEMS................................................................................................................... 13 2.3 MODAL DENSITIES .......................................................................................................... 14 2.4 INTERNAL LOSS FACTORS ............................................................................................... 15 2.5 COUPLING LOSS FACTORS............................................................................................... 16 2.6 DERIVATION OF COUPLING LOSS FACTORS..................................................................... 18

2.6.1 Wave Approach...........................................................................................................19

xi

2.6.2 Modal Approach .........................................................................................................23

3 DERIVATION OF SEA PARAMETERS............................................................................ 28

3.1 FORMULATIONS OF COUPLING PARAMETERS τ AND η12.................................................. 28 3.2 ASSUMPTIONS OF LINE JOINT SYSTEM ............................................................................ 30 3.3 DERIVATIONS FOR TWO PLATES - LINE MASS JOINT ...................................................... 32

3.3.1 Oblique Incidence.......................................................................................................33 3.3.2 Normal Incidence........................................................................................................39

3.4 DERIVATIONS FOR TWO PLATES - BEAM JOINT............................................................... 41 3.4.1 Motion Equations of Beam Types ...............................................................................42 3.4.2 Oblique Propagation ..................................................................................................46 3.4.3 Normal Incidence........................................................................................................55

3.5 POWER INPUT.................................................................................................................. 58 3.6 DERIVATIONS FOR THREE PLATES - BEAM JOINT............................................................ 59

3.6.1 Motion Equations of Beam Types ...............................................................................60 3.6.2 Oblique Propagation ..................................................................................................63 3.6.3 Normal Incidence........................................................................................................74

4 COUPLING LOSS FACTOR SENSITIVITY..................................................................... 79

4.1 COUPLING LOSS FACTOR FOR TWO PLATES COUPLING................................................... 79 4.2 SENSITIVITY TO SYSTEM PARAMETERS........................................................................... 89

4.2.1 Sensitivity to Density.................................................................................................100 4.2.2 Sensitivity to Bending Stiffness .................................................................................103 4.2.3 Sensitivity to Lateral Bending Stiffness.....................................................................105 4.2.4 Sensitivity to Torsional Stiffness ...............................................................................107 4.2.5 Sensitivity to Vertical Shear Centre Offset from Plate Surface ................................109 4.2.6 Sensitivity to Horizontal Shear Centre Offset from Plate Surface ............................111 4.2.7 Sensitivity to Warping Coefficient ............................................................................112

4.3 COUPLING LOSS FACTORS FOR THREE PLATES COUPLING ............................................ 115

5 SEA APPLICATIONS......................................................................................................... 123

5.1 TWO PLATES ................................................................................................................. 123 5.2 SIX PLATES ................................................................................................................... 125 5.3 CLOSED STRUCTURE ..................................................................................................... 131 5.4 FUSELAGE STRUCTURE ................................................................................................. 136 5.5 CLOSED STRUCTURE WITH FLOOR PANEL..................................................................... 138 5.6 FUSELAGE STRUCTURE WITH FLOOR PANEL ................................................................. 141

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5.7 CLOSED STRUCTURE WITH ACOUSTIC CAVITY.............................................................. 145 5.8 FUSELAGE STRUCTURE WITH ACOUSTIC CAVITY.......................................................... 147 5.9 CLOSED STRUCTURE WITH FLOOR PANEL AND TWO ACOUSTIC CAVITIES.................... 149 5.10 FUSELAGE STRUCTURE WITH FLOOR AND TWO ACOUSTIC CAVITIES ........................... 151

6 CONCLUSIONS .................................................................................................................. 154

REFERENCES .............................................................................................................................. 158

APPENDIX

A. DYNAMIC STIFFNESS METHOD................................................................................... 165

A.1 DYNAMIC STIFFNESS SOLUTION.................................................................................... 166 A.2 POWER INPUT................................................................................................................ 171 A.3 EXACT CALCULATION OF AVERAGE ENERGY ............................................................... 172

B. SEA PARAMETERS OF PLATE – ACOUSTIC CAVITY............................................. 173

C. BEAM AND PLATE PROPERTIES ................................................................................. 175

CURRICULUM VITAE ............................................................................................................... 179

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LIST OF TABLES

2-1 Number of modes, modal densities. .............................................................................................15

2-2 Coupling loss factors ....................................................................................................................18

C-1 Example line mass and Euler Beam coupling system properties...............................................175

C-2 Example double and triple coupling beam properties................................................................176

C-3 Example triple coupling system properties................................................................................177

C-4 Example frame properties..........................................................................................................178

xiv

LIST OF FIGURES

2-1 SEA Power flow model. ...............................................................................................................12

2-2 Two rods coupled by a linear spring. ...........................................................................................19

2-3 Model of coupled simple oscillators.............................................................................................23

3-1 SEA power flow model. ...............................................................................................................28

3-2 Two plates connected via a line junction. ....................................................................................31

3-3 Line Mass Joint ............................................................................................................................32

3-4 Oblique Incidence ........................................................................................................................33

3-5 Open Section Channel ..................................................................................................................42

3-6 Three plates connected via a line junction....................................................................................59

4-1 Transmission loss of normal incidence. .......................................................................................79

4-2 Transmission loss of normal incidence for mass (green) and open section beam (blue)..............80

4-3 Transmission loss for 30° incidence angle ...................................................................................81



4-6 Transmission loss versus frequency and oblique angle for line mass modelling. ........................84

4-7 Transmission loss versus frequency and oblique angle for Euler beam modelling. .....................85

4-8 Transmission loss vs. frequency and oblique angle for triple coupling modelling.......................85

4-9 CLF vs. frequency and oblique angle for line mass modelling ....................................................86

4-10 CLF vs. frequency and oblique angle for Euler beam modelling...............................................87

4-11 CLF vs. frequency and oblique angle for triple coupling modelling..........................................87

4-12 Coupling loss factor for line mass (green), beam (red) and open section channel (blue)

modelling.................................................................................................................................88

4-13 Coupling loss factor for line mass (green), beam (red) and classical SEA (blue). .....................89

4-14 Transmission loss vs. frequency and oblique angle for Euler beam modelling..........................90

4-15 Transmission loss vs. frequency and oblique angle for double coupling modelling ..................90

4-16 Transmission loss vs. frequency and oblique angle for triple coupling modelling.....................91

xv

4-17 CLF for line mass (green), beam (red), double coupling (black) and triple coupling (blue)

modelling.................................................................................................................................91

4-18 Transmission loss at 10000 Hz vs. angle for line mass (green), beam (red), double coupling

(black) and triple coupling (blue) modelling...........................................................................92

4-19 Transmission loss at 30° vs. frequency for line mass (green), beam (red), double coupling




4-21 Transmission loss vs. frequency and oblique angle for line mass modelling .............................95

4-22 Transmission loss vs. frequency and oblique angle for Euler beam modelling..........................95

4-23 Transmission loss vs. frequency and oblique angle for double coupling modelling ..................96

4-24 Transmission loss vs. frequency and oblique angle for triple coupling modelling.....................96

4-25 Transmission loss at 10000 Hz vs. angle for line mass (green), beam (red), double coupling






4-28 CLF for line mass (green), beam (red), double coupling (black) and triple coupling (blue)

modelling.................................................................................................................................99

4-29 Transmission loss of normal incidence for line mass modelling, change in density. ...............100

4-30 Transmission loss of 30° incidence for Euler beam modelling, change in density ..................101

4-31 Transmission loss of 30° incidence for double coupling modelling, change in density ...........102

4-32 Transmission loss of normal incidence for line mass modelling vs. density. ...........................102

4-33 Transmission loss of 30° oblique angle for Euler Beam, change in Iζ. ....................................103

4-34 Transmission loss of 30° oblique angle for triple coupling, change in Iζ. ................................104

4-35 Transmission loss of 30° oblique angle for Euler Beam vs. Iζ vs. frequency...........................104

4-36 Transmission loss of 30° oblique angle for double coupling vs. Iη vs. frequency....................105

4-37 Transmission loss of 30° oblique angle for double coupling, change in Iη. .............................106

4-38 Transmission loss of 30° oblique angle for triple coupling vs. Iη vs. frequency ......................106

4-39 Transmission loss of 30° oblique angle for triple coupling, change in Iη.................................107

4-40 Transmission loss of 30° oblique angle for Euler beam, change in J. ......................................108

4-41 Transmission loss of 30° oblique angle triple coupling, change in J........................................108

4-42 Transmission loss of 30° oblique angle for Euler Beam vs. J vs. frequency............................109

4-43 Transmission loss of 30° oblique angle triple coupling, change in sz.......................................110

xvi

4-44 Transmission loss of 30° oblique angle for triple coupling vs. sz vs. frequency ......................110

4-45 Transmission loss of 30° incidence for triple coupling modelling, change in sx ......................111

4-46 Transmission loss of 30° oblique angle for triple coupling vs. sx vs. frequency ......................112

4-47 Transmission loss of 30° incidence for double coupling modelling, change in Γo...................113

4-48 Transmission loss of 30° incidence for triple coupling modelling, change in Γo .....................113

4-49 Transmission loss of 30° oblique angle for double coupling vs. Γo vs. frequency...................114

4-50 Transmission loss of 30° oblique angle for triple coupling vs. Γo vs. frequency .....................114

4-51 Transmission loss between plate 1 and 2 vs. frequency and oblique angle for line mass

modelling...............................................................................................................................115

4-52 Transmission loss between plate 1 and 2 vs. frequency and oblique angle for Euler beam

modelling...............................................................................................................................116

4-53 Transmission loss between plate 1 and 2 vs. frequency and oblique angle for double

coupling.................................................................................................................................116


modelling...............................................................................................................................117


modelling...............................................................................................................................117


coupling.................................................................................................................................118


modelling...............................................................................................................................119


modelling...............................................................................................................................119


coupling.................................................................................................................................120

4-60 CLF between first and second plates for mass (green), beam joint (blue) and open section

beam (red) .............................................................................................................................120

4-61 CLF between first and third plates for mass (green), beam joint (blue) and open section

beam (red) .............................................................................................................................121

4-62 CLF between second and third plates for mass (green), beam joint (blue) and open section

beam (red) .............................................................................................................................122

5-1 SEA solution of two plates system for line mass and beam joints. ............................................123

5-2 Comparison with mean energy for diffuse wave field CLF, η12(ω). ..........................................124

5-3 Comparison with mean energy for normal incidence CLF, η12(0,ω). ........................................124

5-4 SEA results for 6 plates array, line mass joint............................................................................126

xvii

5-5 SEA results for 6 plates array, Euler beam joint. .......................................................................126

5-6 SEA results for 6 plates array, double coupling. ........................................................................127

5-7 Comparison with dynamic stiffness results for 6 plates array, line mass joint, diffuse

wave field CLF - η12(ω). .......................................................................................................128

5-8 Comparison with dynamic stiffness results for 6 plates array, Euler beam, diffuse

wave field CLF - η12(ω). .......................................................................................................129

5-9 Comparison with dynamic stiffness results for 6 plates array, Euler beam, normal

incidence CLF - η12(ω,0).......................................................................................................129

5-10 Comparison with dynamic stiffness results for 6 plates array, double coupling, diffuse

wave field CLF - η12(ω). .......................................................................................................130

5-11 Comparison with dynamic stiffness results for 6 plates array, double coupling, normal

incidence CLF - η12(ω,0).......................................................................................................130

5-12 Closed structure........................................................................................................................131

5-13 SEA results for 6 plates array, line mass joint..........................................................................132

5-14 SEA results for 6 plates array, Euler beam joint. .....................................................................132

5-15 SEA results for 6 plates array, double coupling. ......................................................................133

5-16 Comparison with dynamic stiffness results for 1st and 6th plates, line mass...........................133

5-17 Comparison with dynamic stiffness results for 1st and 6th plates, Euler beam .......................134

5-18 Comparison with dynamic stiffness results for 1st and 6th plates, open section channel ........134

5-19 Comparison with AutoSEA and dynamic stiffness results for 6th plate ..................................135

5-20 Fuselage structure.....................................................................................................................136

omparison with AutoSEA results for 6th plate ..................................................................................137

5-22 Comparison of energies of 1st, 2nd and 13th plate......................................................................138

5-23 Closed section with floor panel ................................................................................................138

5-24 Comparison of energies of 6th plate..........................................................................................139

5-25 Comparison of energies of floor panel (13th plate)...................................................................140

5-26 Comparison of energies of 6th plate for closed section with and without floor panel...............140

5-27 Fuselage structure with floor panel ..........................................................................................141

5-28 Comparison of energies of 6th plate..........................................................................................142

5-29 Comparison of energies of floor panel of first section (37th plate)...........................................142

5-30 Comparison of energies of 30th plate........................................................................................143

5-31 Comparison of energies of floor panel of third section (39th plate)..........................................143

5-32 Comparison of energies of 6th plate for closed section with and without floor panel...............144

5-33 Closed structure with acoustic cavity .......................................................................................145

5-34 Comparison of energies of 6th plate for closed section with acoustic cavity ............................146

5-35 Comparison of energies of 6th plate for closed section with and without acoustic cavity .......146

xviii

5-36 Fuselage structure with acoustic cavity ....................................................................................147

5-37 Comparison of energies of 6th plate for fuselage section with acoustic cavity .........................148

5-38 Comparison of energies of 6th plate for fuselage section with and without acoustic cavity ....148

5-39 Closed structure with floor panel and two acoustic cavities.....................................................149

5-40 Comparison of energies of 6th plate for closed section with floor panel and two acoustic

cavities...................................................................................................................................150


cavities...................................................................................................................................150

5-42 Comparison of energies of floor panel (13th plate) for closed section with floor panel and

two acoustic cavities..............................................................................................................151

5-43 Fuselage structure with two acoustic cavities...........................................................................151


cavities...................................................................................................................................152


cavities...................................................................................................................................153

5-46 Comparison of energies of floor panel (37th plate) for closed section with floor panel and

two acoustic cavities..............................................................................................................153

xix

LIST OF SYMBOLS AND ABBREVIATIONS

A Area

Aw Area of coupling aperture

Abi Cross-sectional area of beam i

cg Group velocity

ci Longitudinal wave speed in beam i

cx Horizontal distance between beam centroid and shear centre

cz Vertical distance between beam centroid and shear centre

C Centroid

e Exponential function symbol

ei Modal energy

f Frequency, Hz

ft Total force on joint beam in z direction

ftx Total force on joint beam in x direction

F Force

h Thickness of plate

i = 1−

Ip Polar second moment of area

Iη Second moment of area about η-axis

Iζ Second moment of area about ζ-axis

Iηζ Product moment of area about η and ζ-axes

J Torsion constant

kL, cL Wave number and wave speed of longitudinal waves

xx

ko, co Wave number and wave speed of sound in air

kB, cB Wave number and wave speed of bending waves

Ki Radius of gyration of beam i

l Spacing of points at junction

L Length

Lj Length of junction

m Mass

mp Mass per unit area

mt Total y moment on joint beam

M Moment

Mi Mass of beam i

n Modal density

N Number of modes with resonance frequencies below ω

O Shear centre

P Beam and plate connection point

r Viscous damping coefficient

R Radius

sx Horizontal distance between beam shear centre and point P

sz Vertical distance between beam shear centre and point P

u Displacement in x direction

v Displacement in y direction

V Volume

w Displacement in z direction

z Moment impedance

Z Point impedance

α Radius

λb Flexural wavelength

σ Radiation efficiency

ηij Coupling loss factor

ηi Internal (damping) loss factor

xxi

τij Diffuse field wave transmission coefficient

Γ o Warping constant about shear centre

ρ Density

ρo Density of air

ω Circular frequency, rad/s

∆N Number of modes in frequency band ∆ω

SEA Statistical Energy Analysis

CLF Coupling Loss Factor

Equation Section (Next)

1

CHAPTER I

1. INTRODUCTION The lowest few vibration modes are generally the ones which are associated

with the greatest deflections, highest stresses, and gross structural failures. This is

why vibration engineers have focused their attention on low frequency oscillations.

Although the classical methods are valid in principle at all frequencies, their use is

very often impractical for high frequencies, particularly for randomly excited

complex structures.

The statistical energy analysis approach to structural vibration problems was

developed in response to the need for a simple means for understanding and

estimating significant properties of multimodal vibrations of complex systems. SEA

began to develop in 1959, when Lyon and Smith independently made their first

calculations concerning power and response of linearly coupled resonators, [1].

There are two basic ideas underlying all SEA work, [2]. The first concerns the

primary variables used to describe problems; these are long term averages of (kinetic)

energy flows (or levels). Since these energy flows are time invariant and also because

all dissipative flows are detailed explicitly, energy balance equations may be set up

and solved to find the long term average energy levels in all parts of the system under

investigation. These equations encompass details of the known forcing functions and

system parameters, and allow a designer to see how vibrational energy is distributed

around a system. This information can then be related to the various average motions,

stress levels, etc.

2

Secondly, SEA is based on the concept of structures with random parameters

in which responses are calculated as averages across ensembles of similar systems.

This leads to the idea of random natural frequencies, i.e., the analysis requires that

the number of natural frequencies occurring in a given frequency band is known, but

not their exact positions. As a combination of these two ideas SEA is thus concerned

with the calculation of average response spectra, e.g., the spectra are those that would

be found if very many similar, but not identical systems were examined and the

individual spectra averaged, frequency by frequency, across all the results.

In the SEA method, systems are considered to be divided into subsystems,

which are linearly coupled together, and which exchange energy via resonant

vibration modes. Subsystems are structural or acoustical entities that have modes

which are similar in nature. In many applications geometrical structural elements are

subsystems, while sometimes different wave types, e.g. bending or longitudinal

waves, form subsystems. The primary variable of interest in SEA is energy. For

steady state conditions, a power balance is derived in each analysis band in which the

input power to the subsystem(s) is either dissipated within the subsystem(s) or

coupled to other subsystems where it is dissipated or radiated to the acoustic farfield.

For a complex system the theoretical justification of the SEA equations is

normally based on either a diffuse wave field approach or a modal approach to the

system dynamics. A number of studies have compared the SEA approach with the

exact analytical results which can be obtained for relatively simple systems, including

two coupled rods and two coupled beams. The method has also been compared with

experimental results for a wide range of systems including assemblies of plates,

shells, beams and acoustical spaces. In general it has been found that the SEA

approach is most applicable to structures composed of parts containing a significant

number of resonant modes and has a high degree of modal overlap, although it is

difficult to produce definitive guidelines.

3

1.1 Literature Survey

There are other methods in the literature developed for studying high

frequency vibrations. A potential alternative to SEA that is gaining increasing interest

in recent years is the power flow approach, [3-14]. Its advantage is represented by the

possibility of modelling the spatial distribution of energy density at high frequencies,

thus yielding a more effective estimate of the system behaviour than the average

constant value given by SEA. The method is also known as the “thermal analogy”

because of the similarity of the differential energy equation with that for heat

conduction in thermal problems. By considering the physics of power transmission in

three-dimensional structures, it was shown [13] that an exact time-averaged energy

density equation determined for the power balance can be obtained only for particular

structures such as beams and plates. Yet, even in these simple cases, the power flow

does not have a thermal-like behaviour and the equations depend non-linearly on the

energy density. Therefore, it is not appropriate to use the thermal analogy to describe

the time-averaged energy density, especially for complex structures.

A wave intensity technique for the analysis of high frequency vibrations is

presented by Langley [15] in which the vibration of each component of the system is

represented in terms of a homogeneous random wave field. The directional

dependency of the wave intensity is represented by a Fourier series. The use of a

single component in the intensity series is equivalent to the assumption that the wave

field is diffuse, in which case the method reduces to conventional SEA. The method

is applied to a number of panel arrays for which exact results may be obtained by

using the dynamic stiffness method, and it is found that a significant improvement

over conventional SEA may be achieved for relatively little additional effort.

Dynamic Stiffness Technique is one of the methods used to study the

vibrations of complex structures, [16-18]. Langley [16] has used this technique for

the analysis of stiffened shell structures. The method is based on a singly curved

orthogonally stiffened shell element which has a constant radius of curvature and

4

which is simply supported along the curved edges. The stiffeners are taken to be

smeared over the surface of the element, and the appropriate modifications to the

shell differential equations and boundary conditions are performed. The resulting

differential equations are solved exactly to yield the dynamic stiffness matrix and the

loading vector for the element. Any number of elements may be assembled to model

the cross section of a built up structure such as an aircraft fuselage.

Langley and Bremner have presented a hybrid method for the vibration

analysis of complex structural-acoustic systems, [19]. The method is based on

partitioning the system degrees of freedom into a global set and a local set. The

solution method consists of a deterministic model of the global response and a

statistical energy analysis model of the local response with due allowance for the

coupling which exists between two types of response.

Fahy and Yuan [20] has shown that the time-averaged power flow between

two oscillators coupled by spring and damping elements, subjected to white noise

force sources, are not simply proportional to the time averaged oscillator energy

difference, but also to the time averaged energy of the individual oscillators. The

proportionality constants are functions only of the oscillator and coupling parameters.

Wave approach in SEA is used by several authors, [21-28]. Wester and Mace

[21] was studied on two edge-coupled, simply supported, rectangular plates exposed

to rain-on-the-roof excitation of one of the plates. Two parameters quantifying the

strength of coupling between plates are found, and four distinct regimes of wave

component energy flow and storage are observed, involving weak and strong

coupling. The traditional SEA hypothesis of proportionality between the coupling

power and the difference in subsystem mean modal energies is found to hold for the

ensemble average response, for all coupling strengths, but not generally for the

responses of individual ensemble member systems. The traditional estimate of

coupling loss factor, found by a wave approach in which semi-infinite subsystems

and diffuse fields are assumed, is seen generally to be an over-estimate of the true

value for the present system, except when the coupling is weak. It is also found that

modal overlap, which has been proposed as an indicator of coupling strength and of

5

the accuracy of the traditional coupling loss factor estimate, is inappropriate in this

role for the rectangular plate systems considered.

Sound transmission in buildings is an important area of SEA application, [29-

33]. Wöhle et al present a method to calculate coupling loss factor at the rectangular

slab joints for incident bending, longitudinal and transverse waves, [32, 33]. Springs

existing at the coupling point as well as potential losses are taken into consideration.

They studied both the excitation by free bending waves and forced bending waves.

For structural slab systems, they found that the structure-borne sound transmission

that is caused by forced bending waves is almost negligible compared with the

structure-borne sound transmission due to free bending waves.

L junction of plates is studied by several authors, [34-37]. McCollum and

Cuschieri, [22] includes the shear and rotary inertia effects and the in-plane waves

effects. First, they performed the analysis for the transmission of waves through the

junction between the plates pinned to remove the in-plane waves transmission but

including the influence of transverse shear deformation and rotary inertia. Second,

the constraints of a pinned junction are removed and the influence of in-plane waves

through an unsupported junction is considered. The in-plane shear and longitudinal

waves are described using the generalized plane stress theory. From the results

presented, it is concluded that in terms of the vibrational power flow through the

junction of an L-shaped plate the effects of shear and rotary inertia and that of in-

plane wave generation are important and, the contribution to the power transmission

of the coupling between in-plane vibration and the thickness modes of the structure

can be included to the analysis in an enhanced SEA model.

Not much work has been done in the use of SEA on complex, heavy machine

structures. The reason for this is the difficulty in estimating or measuring the modal

densities and the internal dissipation and coupling loss factors for the “heavy”

structure. Cuschieri and Sun [38-40] have a study on experimental determination of

these three parameters of a rotating machine structure. The technique they have used

is suited to both conservatively and non-conservatively coupled systems and effect of

indirect coupling. Lim and Singh have worked on gear noise, [41]. They modelled

the gearbox using ideal simply supported geometries and calculated the coupling loss

6

factors accordingly. Experimental results are in reasonable agreement with the SEA

results.

Cuschieri and Sun [38-40] presented a method for determining the dissipation

and coupling loss factors of a fully assembled machinery structure. The presented

experimental approach is suitable for both conservative and non-conservative

coupling. Coupling loss factors for directly coupled subsystems and indirectly

coupled structures are determined. Indirect coupling exists when the subsystems of

the structure are small compared to the wavelength of vibration and all coupling

interfaces are in the near field of each other. The method is very time consuming and

the subsystems capable of energy transmission, storage and dissipation can not be

considered.

The equations of total loss factors of complex structures are very difficult or

sometimes can not be obtained by theoretical calculations. In recent years, SEA

offers a very powerful means of estimating the energy throughout the structure and

total loss factors. Sun and Richards, [42], has a study on total loss factor of a welded

steel box with and without additive damping treatment. A formula for estimating

total loss factors of a structure has been derived from the linear steady state energy

balance equations. The formula was simplified by assuming that the structures,

except for the measured substructure, are coupled weakly. The results are compared

with the experimental measurements. The agreement is quite good. However, there is

a need for further investigation into the loss factors of substructures, such as plate-

like, beam-like, box, shell and so on, of which a complex structure is composed.

There are different models of stiffened panels studied by several researchers,

[43-49]. Langley and Smith proposed that the panel can be modelled as a damped

coupling element between two adjoining structural components, and the transmission

and absorption coefficients calculated on the basis of periodic structure theory, [49].

The method is applied to the forced response of two panels which are coupled by a

stiffened panel. Although the study shows that this approach is a feasible technique

for the type of structure used, further work is needed to consider the panels in more

complex structures.

7

Petyt et al [50] has worked on a rectangular singly curved, finite strip shell

element and a compatible thin walled, open section beam element are derived. They

use a finite strip method to find natural frequencies and mode shapes.

Mace [51, 52] has studied the vibration of and sound radiation from an

infinite fluid-loaded plate stiffened periodically by line supports. The response to a

convicted harmonic pressure has been found by using Fourier transforms. Then, the

response of fluid-loaded periodically stiffened plates has been found for line and

point force excitations. It was seen that at low frequencies, where the separation of

the stiffeners was less than one third of the wavelength in the plate, the behaviour of

the stiffened plate can be approximated by that of an equivalent orthographic plate.

1.2 Objective and Scope of The Present Study

The problem of calculating the response of a complex engineering structure to

dynamic loading may generally be approached using the finite element method. If,

however the concern is with very high frequency excitation which produces short

wavelength deformations the finite element method may require an excessive number

of elements to model the system. The problem of predicting the response of a

complex system to high frequency loading may be approached by using SEA. The

system is modelled as a collection of subsystems, the mean energies and external

power inputs of which are related via a set of linear equations, the coefficients of

which are expressed in terms of quantities known as the damping loss factors and the

coupling loss factors. Several studies exist in the literature to overcome the

limitations of the method. There are works on strong coupling, non-conservative

coupling, prediction of transient vibration envelopes, etc. Another important subject

is the determination of SEA parameters like coupling loss factors, damping loss

factors and modal densities. There are experimental and theoretical methods

presented to obtain these parameters.

8

Several organisations have prepared SEA computer codes. However, the

subsystem library is limited to basic shapes and there is a need to add the structural

elements and junction types that are widely used.

Stiffened panels are used widely in aerospace and marine vehicles, and it is

often necessary to predict high frequency noise and vibration levels in structures of

this type. This thesis study is on high frequency vibratory characteristics of stiffened

panel structures.

At low frequencies, when the wavelength in the plate is much greater than the

stiffener separation, the stiffened structure can be approximated by an equivalent

orthographic plate. At higher frequencies, when the plate wavelength is comparable

with the stiffener separation, other methods of analysis, in which use is made of the

spatial periodicity of the structure, have been applied to determine the forced

response, free wave propagation and acoustic radiation, [52].

There are a number of studies in the literature on low frequency vibration

analysis of the stiffened panel structures. In the high frequency analysis of stiffened

structures, there are two basic approaches. The first one is “smearing” the stiffener

properties to produce an equivalent orthotropic plate. In SEA terms, the other

approach is to model each stiffener element as a subsystem. The associated coupling

loss factors can be found by considering wave transmission across stiffener. The

inputs required for the power balance equations are damping and coupling loss

factors and input powers. In this thesis, the equations for coupling loss factors are

derived for a stiffener and an infinite panel first. Then stiffener is modelled as a line

mass, beam and open section channel beam having double and triple coupling. Three

plates with a line junction are also studied. Sensitivity of transmission loss to the

modelling technique and the system parameters is examined. The analysis is

continued by placing more stiffeners on the panel which are equally spaced.

The last panel is attached to the first one and a closed structure is obtained.

Next step is to compose a structure by combining several of this closed structure via

beams and add a floor structure. This is to simulate a classical fuselage structure

consists of skin, stringer and frames. At the conclusion of the study, the high

9

frequency vibration analysis by SEA of this structure is performed and compared

with the results calculated with Dynamic Stiffness Method and AutoSEA.

10

CHAPTER II

2 BASICS OF SEA

SEA is used to predict the response of a dynamic system to external power

input. A system will refer to the entire complement of coupled structures and

acoustic spaces under consideration. The system is then divided into subsystems

which consist of a collection of similar resonant modes within a structure or acoustic

space. For example, a bounded acoustic space is usually treated as a subsystem

containing acoustic modes. Bending waves in a plate and longitudinal waves in the

same plate may each be treated as separate sets of modes and therefore as separate

subsystems. Subsystems are then coupled via coupling loss factors in the

development of power balance equations.

The basic concepts of SEA are [53]:

1- Power flow between subsystems is proportional to the differences in the modal

energies of the coupled subsystems. This is supported by mathematical modelling

that begins with the coupling of two simple oscillators. The coupling of two

oscillators becomes the building block for coupling of multimodal resonant

subsystems where each oscillator serves as a modal model.

2- Power input or transmitted to a subsystem is either dissipated in the subsystem or

transmitted to adjacent subsystems via junctions of structures or interfaces

between structures and acoustic spaces. Thus a complete account of all of the

power is taken.

3- Energy resides only in resonance modes, so that the more modes a subsystem

have in the analysis band, the greater the capacity of the subsystem to accept and

store energy. Within each analysis band, the energy in a subsystem is uniformly

11

divided among the modes. The net power coupled between subsystems is

proportional to the difference in their modal energies and only passes from the

subsystem with higher modal energy to those of lower modal energy. When

modes have equal energy, it is referred to as an equipartion of energy.

There are several assumptions that are generally made in the development of

SEA models.

a- The input power spectrum is broadband, i.e. there are no strong pure tones in the

input power spectra.

b- Energy is not created in the couplings between subsystems. Energy may be

dissipated in junctions between subsystems, such as in isolation mounts, but

generally this effect is added to subsystem damping loss factors.

c- The damping loss factor is equal for all modes within a subsystem and analysis

band.

d- Modes within a subsystem do not interact except to share an equipartition of

energy. The coherent effects between modes are ignored so that power sums

apply.

2.1 Classical Statistical Energy Analysis

SEA is an analytic method for calculating power transmission between

connected substructures. It is based on a calculation by Lyon of the power flow

between two weekly coupled linear oscillators excited by independent white noise

sources, [2]. Lyon found the power flow between the oscillators to be related to the

uncoupled energies of the oscillators; the power flow always went from the oscillator

of higher energy to that of lower energy, and the power flow was proportional to the

difference in the uncoupled energies.

Pik = β( Wi - Wk ) (1)

This basic idea can be extended to a complex structure by dividing it into

distinct substructures, leading to the SEA formulation. It is then assumed that, in a

12

frequency band of interest, the power transmitted between any two substructures is

proportional to the difference of the average dynamical energies of their (uncoupled)

resonant modes. Note that it is exact only for a two-oscillator system. Thus the power

flow, Pik, from an arbitrary substructure i to another arbitrary substructure k may be

written as,

Pik = ω( ηik Ei - ηki Ek ) (2)

where ω is a representative frequency, ηik and ηki are the “coupling loss factors”,

and Ei and Ek are the blocked (uncoupled) total energies. SEA makes the

fundamental assumption that, within narrow frequency bands, the energies in all

independent modes equalise at steady state. The total energy in each element, Ei, in a

frequency band centred at ω is derived from the product of the element’s modal

energy ei , and the element’s modal density, ni , at that frequency.

Ei = ni ei (3)

In addition, the power dissipated by a substructure is proportional to ω times

the substructure’s “damping loss factor” or “internal loss factor”, ηi.

Pidiss = ω Ei ηi (4)

Then the power balance equation for a subsystem i is the following.

Piin = Pidiss + Pik (5)

Piin = ω Ei ηi + ω( ηik Ei - ηki Ek ) (6)

Figure 2-1 SEA Power flow model

P12

P21

Pin1

Pdiss1

Subsystem 1

Subsystem 2

Pdiss2

Pin2

13

Thus the power balance equations may be expressed in the following matrix

form.

P = ω [C] E (7)

P is a vector of input powers from external sources, [C] is the matrix of

coupling loss factors and damping loss factors, and E is the vector of blocked

energies.

An important relationship in SEA is the reciprocity relationship;

ηik ni = ηki nk (8)

where ηik is the coupling loss factor from element i to element k, and ni is the modal

density of element i, respectively for ηki and nk .

The inputs required for the power balance equations are damping and

coupling loss factors, subsystem masses, input powers, and modal densities. With

these inputs, the power balance equations become algebraic equations with known

coefficients which can be solved for the unknown energies in each of the subsystems.

The average vibration and sound pressures are then computed from predictions of the

subsystem energies. It is assumed that the input powers are known or can be

computed based on force or displacement inputs and subsystem admittances.

2.2 Subsystems

A subsystem is a part of the system which stores and/or dissipates energy.

Physically, a subsystem may be 0, 1, 2 or 3 dimensional; for example a point mass, a

beam, a plate or an acoustic cavity. A continuous subsystem is one with distributed

mass and elasticity, while a discrete one has lumped mass and stiffness; for example,

a set of oscillators. In a practical case a discrete subsystem will be an idealisation of a

continuous one.

14

Energy propagates through a continuous subsystem as waves, although the

subsystem may be analysed in terms of modal or wave-based model. The complexity

of the subsystem depends on how many distinct and independent energy-propagating

wave modes exist, and this number can be identified with the dynamic

dimensionality of the subsystem. Thus a subsystem is defined to be one dimensional

dynamically if energy is propagating by just one wave mode. Examples of such

systems are a rod in torsion or a beam in bending for which near field effects can be

neglected. If there is motion in more than one direction, then the energy associated

with each direction can be computed separately. If the beam having motion in each

of two directions is to be modelled as a single subsystem then the two energies can

be summed. Subsystems with a higher dynamic dimensionality are, for example, a

beam undergoing both in-plane bending and axial vibration (two dimensional) while

a general structural member is dynamically four-dimensional (four wave modes,

namely torsion, axial tension and in or out of plane bending). In modal terms each

dimension can be loosely identified with a group of modes of a similar kind.

2.3 Modal Densities

The mode count N, which is the fundamental quantity, is, for a subsystem, the

number of modes that subsystem resonates in the frequency band ∆f considered. It

may sometimes be estimated as a product of modal density, n(f), and the frequency

bandwidth ∆f. As a derived quantity, the modal density is defined as the number of

modes per unit frequency. Sometimes it is given per unit angular frequency as n(ω),

these being related by the following equation.

n(f) = 2 π n (ω) (9)

Theoretically derived modal densities are available in the literature for

idealised structural elements [54].

15

Table 2-1 Number of modes, modal densities.

System Number of Modes Modal Densities

Beam, longitudinal N=kLL/π=ωL/cLπ ∆N/∆ω=L/cLπ

Beam, bending N=kBL/π = ω L/1.7 c hL

∆N/∆ω=kBL/2πω =L/3.4 c hL ω

Plate, bending N=kB2A/4π

=ωA/3.6cLh ∆N/∆ω=kB

2A/4πω =A/3.6cLh

Volume, airborne sound N=ko3V/6π2

=ω3V/6π2co3

∆N/∆ω=k02V/2π2c0

3

=ω2V/2π2c0

Ring, radially excited N=2kB R =3.7 LR / c hω

∆N/∆ω=kB R/ω

=1.9R / c hL ω Thin walled tube, for v<1 N≅ 3 ( )3/ 2

L3 l R/cω /2πh ∆N/∆ω=2 ω R3/2 =L/1.6hcL

3/2 Thin walled tube, for v>1 N≅ 3 LRω/cLh ∆N/∆ω≅ 3 LR/cLh

2.4 Internal Loss Factors

The internal loss factor represents the amount of damping present. For a

simple oscillator, ηi , is defined as below.

ηπi = ⋅1

2Energy dissipated per cycle of oscillationMaximum energy stored during the cycle

(10)

The frequency band averaged value of ηi, for a particular subsystem, is

defined as,

ηωi

dissPE

= (11)

where

Pdiss = Total time averaged power dissipated in the frequency band

E = Total time averaged energy in the frequency band

16

ω = Band centre frequency (rad/s)

The loss factor is proportional to the ratio of energy dissipated per cycle to

the energy stored. Sometimes it is defined as the phase angle of a complex Young’s

modulus of elasticity. The internal loss factor of a structural element includes several

different damping or energy-loss mechanisms. Commonly accepted forms of linear

damping are structural (hysteric or viscoelastic) damping and acoustic radiation

damping. In practice, other non-linear damping mechanisms are also present at the

structural joints. These include gas pumping, squeeze-film damping and frictional

forces. The internal loss factor of a structural element forming part of a built-up

structure is given by,

η = ηs + ηrad + ηj (12)

where ηs is the structural loss factor, ηrad is the radiation loss factor and ηj is the loss

factor associated with energy dissipation at the boundaries of the structural element.

Typically, engineering structures are lightly damped and 2.5x10-4 < η < 5.0x10-2 .

For most structures η tends to decrease with frequency. Theoretical estimates of loss

factors are not generally available for structural elements. In practice, measured

values are used. It is very important to know the measurement conditions. This

means that one must know what components are included in the above equation. For

many engineering structures ηj is zero and ηrad may or may not be included. When

only ηs is required, it is measured in a vacuum. This point is not always clearly

stated, although for very thin plates the radiation loss factor is of the same order as

the internal loss factor.

2.5 Coupling Loss Factors

The coupling loss factor ηij is related to energy flow from subsystem i to

subsystem j. This indicates the efficiency of vibrational power transmission from one

17

subsystem to another. Thus the coupling loss factor from subsystem i to subsystem j

is defined as below.

ηωij

ij

i Ej

PE

==0

(13)

There are two approaches for deriving expressions for coupling loss factors

for structures; the modal and wave approaches. In the modal approach, the coupling

between individual modes is computed and an average taken over modes in each

frequency band. In the wave approach, the coupling loss factor can be related to the

power transmissibility for semi-infinite structures, which is often easier to estimate

than the average of the couplings between modes of finite structures. The power

transmitted from the first to the second structure through the junction is then energy

lost by the first structure via the coupling. Since the coupling loss factor has been

defined as the energy lost per radian of motion relative to the total energy in the

structure, and the source of energy loss is transmission through the junction at the

boundary of the first structure, it is possible to relate the coupling loss factor to the

power transmission coefficient τ12 as follows,

τ12 = PP

EE c l

lc

trans

inc

tot

tot g f

f

g= =

ωη ωη12 12

( / ) (14)

where Ptrans is the power transmitted through the junction, Pinc is the power incident

on the junction, lf is the mean-free path length between incidences on the junction,

and cg is the group velocity, i.e. the velocity of energy propagation. For beams, cg = 2

cb where cb is the wave phase speed and lf = L, the length of the source beam.

ηω

τ12 12

2=

cLB (15)

For plates, lf = π A/ L12 , where L12 is the length of the junction, so that;

ηπ

τ1212

12

2=

Lk Ap

(16)

where kp is the wave number for freely propagating bending waves in the source

plate.

18

For beams or plates, expressions for the power transmissibility can be

obtained by matching the transverse and angular velocities, and the shear forces and

bending moments at the junction, or relating the transmissibility to the blocked forces

generated by the incident wave, the impedances relating the blocked forces to the

velocities at the junction, and the transmitted waves generated by the velocities at the

junction.

Because of the complexities involved in deriving meaningful expressions for

coupling loss factors for structures, coupling loss factors have been the subject of a

number of studies and incorporated into SEA computer codes. Table 2.2 on the

following page gives the coupling loss factors for the basic subsystem and junction

types, [55].

Table 2-2 Coupling loss factors

Subsystem i

Subsystem j

Type of Junction

Formulae for ηηηηij

Plate Plate Line cgiLjτij/πωAi

Cantilever Beam

Plate Point (2ρiciKiAbi)2(ωMi)-1Re(zj-1)|zj/(zj+zi)|-2

Plate Plate Stiff Bridges

2ZiZj/(πωni)/(Zi+Zj)2

Plate or cylindrical shell

Acoustic cavity

Acoustic ρocoσ/(ωmpi)

Acoustic cavity

Acoustic cavity

Aperture or common partition

coAwτij/(8πfVi)

Plate Plate N points 2 2i ji i

2 2 2i i j

h hh c4NA (h h )3 ω +

for λb<l

3/ 2 3/ 21/ 4 1/ 2j i ji i

3/ 2 3/ 2 2i i j

L h h2 h c3 A (h h )

ω +

for

λb>l

19

2.6 Derivation of Coupling Loss Factors

There are two approaches to theoretical derivation of coupling loss factors

which are wave approach and modal approach, [2, 20, 56-62]. In the following

paragraphs, these approaches are explained.

2.6.1 Wave Approach

Wave approach is used to derive the coupling loss factor of coupled

multiresonant subsystems, [61]. The mechanical system considered is two long thin

rods vibrating longitudinally and coupled together with a linear massless spring. The

ends of rods are connected to linear dashpots to simulate the edge damping often

encountered in real structures. Only rod 1 is forced with a point force as shown in the

figure.

F

r0

r0

r0

r0

Rod 1

Rod 2

Kc

Figure 2-2 Two rods coupled by a linear spring

For the case of many oscillators coupled together as in the case of two

coupled multimodal systems, one generally assumes that all modes in a frequency

band in a given multimodal system have the same total energy. This result in the

20

following expression for the time averaged power in a given frequency band flowing

between the two rods of the coupled multimodal system.

P nEn

En12 12 1

1

1

2

2= −

ωη (17)

A technique that has come to be called the “wave transmission approach” or

“impedance approach” is used to calculate an average η12 for the system. The wave

approach is based on the observation that the impedance of a finite dynamic system

becomes the impedance of the infinite systems when that impedance is measured in a

sufficiently broad frequency band. For the system here, this simply means that the

impedances of the rods are taken to be impedances of rods infinitely long.

The time average power flow from rod 1 to rod 2 may be written as,

{ }P Fc12 2

12

= Re ! *ξ (18)

where Fc is the force amplitude in the coupling spring and ! *ξ 2 is the complex

conjugate of the amplitude of the velocity in rod 2 at the point of attachment of the

coupling spring. Using the point impedance of rod 2, Z2 , P12 is given below.

PF

Zc

12

2

221

=

Re * (19)

It is desirable to express Fc in terms of a quantity easily relatable to the

energy in rod 1. For this reason, Fc is expressed as the sum of the force FBL that

would be required to hold the point of attachment of the coupling spring to rod 1

rigidly and the force Fm due to the motion of that point.

Fc = FBL + Fm (20)

The force Fm may be written as,

F Zm = − 1 1!ξ (21)

where Z1 is the point impedance of rod 1 and ! *ξ 1 is the velocity in rod 1 at the point

of attachment of the coupling spring. The total coupling force Fc applied to rod 1 is

applied equally and oppositely to the coupling spring such that,

21

F Zc = 1 1' !ξ (22)

where Z’2 is the impedance of the coupling spring attached to rod 2 but separated

from rod 1.

Z Zj K

Z j Kc

c2 2

2

' ( / )( / )

=+ +

ωω

(23)

Combining the above equations leads to an expression for the power flow in

terms of FBL:

PF Z

Z Z ZBL

12

22

1 2

2

221

=+

'

' *Re (24)

Now model the vibration in rod 1 by assuming that a right running travelling

wave of amplitude and an uncorrelated left running wave of the same amplitude are

incident on the coupling point in rod 1. By assuming that, the point is held rigidly,

one can easily show that,

F ZBL2

12

022= !ξ (25)

The energy contained in rod 1 due to the action of these two uncorrelated

travelling waves may be written as,

E A LZ Z

Z Z1 1 1 1

02

2

2

1

2

1 2

221= +

+

+

ρ

ξ! '

' (26)

where ρ1 is the density of rod 1, A1 its area, and L1 its length. In this equation, the

effect of the coupling spring on waves transmitted and reflected from the coupling

point has been taken into account. In the case of light coupling, the equation becomes

the following.

E A L1 1 1 1 0

2= ρ ξ! (27)

Since the correction amounts to (at most) 2 dB, above equation is used for

simplicity.

22

Ordinarily, at this point one assumes that the energy in the receiving system

(rod 2) is negligible compared to that in the deriving system and writes the following.

P E12 12 1= ωη (28)

Substitution of energy equations into above equation yields the below

equation, for the coupling loss factor.

ωηρ12

1 1 1

1 2

1 2

2

2

1 1=

+

A L

Z ZZ Z Z

'

*Re (29)

For weak coupling P12 equation is a good approximation, but for strong

coupling, where the modal energy in rod 1 and rod 2 may be comparable, the validity

of that equation becomes doubtful. Then calculate the coupling loss factor using P12

equation based on an estimate of the energy in rod 2.

For the velocity at the coupling point in rod 2 one may write,

!ξ 22

=FZ

c (30)

or, in terms of the blocked force.

!'

'ξ 22

1 2 2

1=

+

F

ZZ Z ZBL (31)

Because of the symmetry of the forcing of rod 2 by the coupling spring there

are two waves of amplitude !ξ 2 /2 in that rod travelling away from the coupling point.

Hence, the total energy in rod 2 may be written as,

EA L

22 2 2 2

2

2 4=

ρ ξ! (32)

or, in terms of the blocked force.

EA L F Z

Z Z ZBL

22 2 2

2

2

1 2

2

2

22 41

=+

ρ '

' (33)

23

Substituting these equations into power balance equation, one obtains below

equationfor the coupling loss factor.

ωηρ ρ

ρ

121 1 1

1 2

1 2

2

2 1 1

2 2

2

1 2

2

1 1 1

114

=+

−+

A LZ Z

Z Z Z AA

ZZ Z

'

' * '

'

Re (34)

2.6.2 Modal Approach

Two simple oscillators are shown in the figure below coupled via a spring

and a gyroscope, [20]. The coupling forces transmitted through the spring are

proportional to the differences in the displacements, ξ1 and ξ2, of the two masses in

the oscillators. Coupling forces transmitted through the gyroscope are proportional to

the velocities, !ξ 1 and !ξ 2 , of the two oscillator masses. Arms connecting the

oscillator masses to the top of the gyroscope slide in a fixed trough and are assumed

to be inflexible along their axes. The bottom of the gyroscope is hinged in a ball

socket. The displacement of the mass away from the base of the oscillator is taken as

positive in each oscillator. A positive velocity for oscillator 1 results in a positive

force on oscillator 2, and a positive velocity for oscillator 2 results in a negative force

on oscillator 1.

r1 r2

k1 k2

m1 m2

ξ1 ξ2

Figure 2-3 Model of coupled simple oscillators

B

24

Adding the coupling forces by the spring and gyroscope between the

oscillators to the equations of motion for the two oscillators yields;

m r k B s F1 1 1 1 1 1 2 12 2 1 1!! ! ! ( )ξ ξ ξ ξ ξ ξ+ + + + − = (35a)

m r k B s F2 2 2 2 2 2 1 12 1 2 2!! ! ! ( )ξ ξ ξ ξ ξ ξ+ + − + − = (35b)

where m1 and m2 are masses of the oscillators, r1 and r2 are the viscous damping

coefficients, k1 and k2 are the oscillator spring constant, and F1 and F2 are the

magnitudes of the harmonic forces applied to the oscillator masses. Group the terms

and define below equations.

s1 = k1 - s12 s2 = k2 - s12 gives

m r s B s F1 1 1 1 1 1 2 12 2 1!! ! !ξ ξ ξ ξ ξ+ + + + = (36a)

m r s B s F2 2 2 2 2 2 1 12 1 2!! ! !ξ ξ ξ ξ ξ+ + − + = (36b)

With harmonic time dependence, eiωt, of the applied forces, the equations

above reduce to algebraic equations which can be solved for the velocities of the

oscillator masses:

{ }ν ωξω

ωω ωδ ω ω1 1

1 22

22 2

21 12 22= =

−− − + +i

im m D

m i F i B s F( )

( ) ( ) (37)

{ }ν ωξω

ωω ω ωδ ω2 2

1 212 1 1

21 1

222= =

−− + + − −i

im m D

i B s F m i F( )

( ) ( ) (38)

where

24 3 2 2 2

1 2 1 2 1 21 2

2 2 2 2121 2 2 1 1 2

1 2

BD( ) 2i ( ) ( 4 )m m

s 2i ( )m m

ω = ω − ω δ + δ − ω ω + ω + δ δ +

+ ω δ ω + δ ω − + ω ω (39)

25

In the equation, ωi = (si / mi )1/2 are the natural frequencies of the uncoupled

oscillators, and δi = ri / 2mi are related to the half power bandwidths of the oscillators

about their natural frequencies. Of interest is the power flow between the oscillators,

which can be described as,

P12 = ½ Re{ F12 ν*12 } (40)

where F12 is the force applied by oscillator 1 to oscillator 2, and * denotes the

complex conjugate. Combining with the velocity equation,

Ps B

m m Dm F m F12

2122 4 2

12

22 2 2 2 1

2

1 1 2

2=

+−

ω ω

ωδ δ

( )( ) (41)

This is the power flow at the single frequency ω. For broadband excitation,

the power flow given by above equation must be integrated over the frequency

bandwidth ∆ω, of interest. If it is assumed that the natural frequencies of the

oscillators are far from both limits of the frequency band, then the integration can be

extended to ±∞ without significant loss in the accuracy.

2 2 2 2 2 21 2 1 2 2 1 1 2 12

12

1 1 2 2 1 2

F F ( )B ( )sP2 m m m m Q

π δ ω + δ ω + δ + δ= − ∆ω δ δ

(42)

where

2 2 2 2 21 2 1 2 1 2 2 1

2 2 2 2 21 2 1 2 12

1 22 1 1 2 1 2 1 2

Q ( ) 4( )( )

B ( ) s ( )m m m m

= ω − ω + δ + δ δ ω + δ ω

ω ω δ + δ+ δ + δ + + δ δ δ δ

(43)

In the derivation of power equation above, it was assumed that the input

forces are independent of frequency, i.e. the excitations are broadband with flat

frequency spectra. Next, relate the power flow between the oscillators to the

differences in the energies in two oscillators. The energy, Wi , in the ith oscillator in

∆ω bandwidth is given below.

Wm

dii

i i=−∞

+∞

∫2∆ων ω ν ω ω( ) ( )* (44)

26

Find the energies and take the difference, divide the found equation to power

equation (1) where,

βδ ω δ ω δ δ

ω ω δ δ δ ω δ ω=

+ + +− + + +

221 2

1 22

2 12 2

1 2 122

12

22 2

1 2 1 22

2 12m m

B s( ) ( )( ) ( )( ) (45)

This is a function of the parameters of the oscillators (δ1, δ2, ω1 and ω2) and

coupling factors (B and s12) and not the energies in the oscillators.

When two coupled multiresonant subsystems are considered, subsystem 1 has

N1 modes and subsystem 2 has N2 modes in a given analysis band. The power flow

from the m1th mode in subsystem 1 to that for two coupled oscillators, since each of

the two modes act like a simple oscillator with inertial (mass), elastic (spring), and

dissipative (dashpot) elements. Thus,

P W Wm m m m m m1 2 1 2 1 2= −β ( ) (46)

where Pm m1 2 is the power flow from the m1th mode in subsystem 1 to the m2

th mode

in subsystem 2, βm1m2 is the coupling parameter analogous to β equation, and

W Wm m1 2( ) is the modal energy in the m1th (m2

th) mode in subsystem 1 (subsystem

2). The bars over the variables denote an average over a frequency band. If we

assume that all of the modes in subsystem 1 are strongly coupled, then the energy in

subsystem 1 is equally divided among all of the modes within the analysis band.

Strongly coupled modes can be thought of as modes that are all directly responsive to

the same broadband excitation, such as the acoustic modes in a room responding to a

broadband acoustic source in the room. The energy is equally divided among the

modes when the damping loss factor is the same for all modes.

The coupling of N1 modes in subsystem 1 to the m2th mode in subsystem 2

can be expressed as,

P N W Wm m m N m m1 12 1 2 1 1 2=< > −β ( ) (47)

where <>N1 indicates an average over the N1 modes in subsystem 1. Extending this to

include the N2 modes in subsystem 2,

27

P N N W Wm m N N12 1 2 1 21 2 1 2

=< > −β ( ) (48)

where P12 is the net power flow from subsystem 1 to subsystem 2 in a frequency

band. Defining,

ηβ

ω1221 2 1 2=

< >m m N N N

(49)

and

ηη

211 12

2=

NN (50)

where ω is the centre frequency of the analysis band, then the equation can be written

as,

[ ]P W W12 12 1 21 2= −ω η η (51)

where W N Wm1 1 1= and W N Wm2 2 2= are total energies in subsystems 1 and 2,

respectively, and η12 and η21 are the coupling loss factors. With the modal densities

in subsystems 1 and 2 given by n1 = N1 / ∆ω and n2= N2 / ∆ω, one can write the

equation as below.

η η211 12

2

1

212= =

nn

nn

∆ωη∆ω

(52)

Thus the coupling loss factor for subsystem 2 to subsystem 1 is greater than

for subsystem 1 to 2 when the modal density in subsystem 1 is greater than the modal

density in subsystem 2. The higher the modal density, the more modes there are to

store energy within a fixed frequency band. Since modal densities are often easier to

estimate than coupling loss factors, above equation can be used to reduce the effort

required in obtaining estimates for the coupling loss factors, since only one of the

two coupling loss factors for two coupled subsystems need to be calculated.

28

CHAPTER III

3. DERIVATION OF SEA PARAMETERS

3.1 Formulations of Coupling Parameters ττττ And ηηηη12

Figure 3-1 SEA power flow model

The basic equation of the SEA theory is the power transmission equation

which is the equality of the input power to the addition of the transmitted power and

dissipated power, [2, 63].

Pin1 = P12 + Pdiss1 (53)

Transmitted power is the following.

12 12 1 21 2P E E= ωη − ωη (54)

Dissipated power is the following equation.

Pdiss = ωη1E1 (55)

P12

P21

Pin1

Pdiss1

Subsystem 1 Subsystem 2

Pdiss2

Pin2

29

Transmitted and reflected power terms are given below.

Ptra = P12 = τ . Pinc (56)

Pref = r 2 Pinc = (1 – τ ).Pinc (57)

τ is the transmission and r 2 is the reflection coefficient. Since no energy is

dissipated in the joint, sum of these coefficients is equal to unity.

r 2 + τ = 1 (58)

The power flow per unit width, in the direction of wave propagation, is the

structural intensity, I, is equal to the multiplication of energy density (average kinetic

energy per unit area) and the group velocity by definition.

I = ε . cg (59)

where ε = ρ [kg/m2]. v2

If a wave is incident on a boundary at an angle θ then the power incident on a

1m length of boundary will be reduced by cosθ so that the power incident is Icosθ.

Total power of subsystem 1 is the following.

Pinc + Pref = [(E1 / A1) . cg ]. Lj (60)

Then the power flow equation is written and the transmission coefficient is

calculated with the below equation.

tra12

inc

PPower transmitted across the joint to system 2Power incident on the joint from system 1 P

τ = = (61)

Transmission efficiency calculated with above equations depends on angle θ.

2Pinc( ) tPtra

τ θ = = (62)

30

Transmission Loss R, in dB, is defined as follows.

R = -10.log(τ) (63)

Ptra = P1→2 = ω.η12.E1 = ω.η12.( Pinc + Pref). A1 / (cg .Lj) (64)

τ . Pinc = ω.η12.( Pinc + r 2 Pinc). A1 /(cg .Lj) (65)

Equation for coupling loss coefficient, η12 , can be derived by using the

relation between r and τ.

( ) 112

1

cg Lj,A 2

⋅ τη θ ω = ⋅ω − τ

(66)

Since τ depends on angle θ, η12 is also angle dependent. The following

integral is used to find an average value.

( ) ( )/ 2

12 120

2 , dπ

η ω = ⋅ η θ ω ⋅ θπ ∫ (67)

3.2 Assumptions of Line Joint System

The formulas derived up to this point will be used in the following sections to

calculate the coupling parameters of two plates connected via a line junction. The

junction is modelled by a line mass and a beam.

Line mass is assumed having no bending and torsional stiffness. The beam on

the junction is studied by an Euler beam assumption and a general open section

channel. Double coupling and triple coupling are studied for open section channel,

[63-67]. Plates are modelled as Kirchoff plate, [68, 69]. It is assumed that the

junction line has no lateral motion.

31

Figure 3-2 Two plates connected via a line junction

Net force and moment on the beam resulting from both plates are ft and mt

defined with the following equations.

ft + F– – F+ = 0 (68a)

mt + M+ – M– = 0 (68b)

An input wave is applied to the first plate and the response of the second plate

is found. Since the transmission ratio is the ratio of velocities, the system response is

solved by assuming wave velocities proportional to the input wave velocity, v1+.

Velocities of plates are given below.

ikx ikx kx i t11 1

w v v (e r e rj e ) et

− ω+

∂ = = + ⋅ + ⋅ ⋅∂

(69)

ikx kx i t22 1

w v v (t e tj e ) et

− − ω+

∂ = = ⋅ + ⋅ ⋅∂

(70)

The ratio of reflected wave velocity to the input wave velocity is r and rj is

the same for reflected near field wave velocity. The ratio of transmitted wave

velocity on plate 2 to the input wave velocity on plate 1 is t and tj is the same for

transmitted near field wave velocity.

F−

F+

ft

mt

M− M+

Plate 1 Plate 2

32

Plates are assumed semi-infinite so that no wave reflection occurs from the

other end of both plates.

The angle of input wave is considered by calculating the coupling parameters

for both normal and oblique incidence cases.

3.3 Derivations For Two Plates - Line Mass Joint

Figure 3-3 Line Mass Joint

Two plates are connected via a line mass having no bending or torsional

stiffness. Equation of plate is,

4 4 4 2

p4 2 2 4 2

w w w wD 2 m 0x x y y t

∂ ∂ ∂ ∂+ + + = ∂ ∂ ∂ ∂ ∂ (71)

mp is mass per unit area.

mb, Ip

Plate 2 Plate 1

33

3.3.1 Oblique Incidence

Figure 3-4 Oblique Incidence

A bending wave is assumed incident on the joint from plate 1 at an angle θ

with an amplitude w1+, [63]. The wavenumbers are,

kx = k cosθ (72a)

ky = k sinθ (72b)

The displacement of this incident wave is,

ikcos x iksin y i t1w(x, y, t) w e e e− θ − θ ω+= ⋅ ⋅ ⋅ (73)

The wavelength in y direction must be the same for all waves on all plates.

1 1 2 2k sin k sinθ = θ

12 1

2

ksin sin

kθ = θ

(74)

θ2, the angle at which the wave leaves the joint is calculated from this

equation. For some cases sinθ2 may be found greater than 1. This means that there

are no real angles at which the waves leave the joint.

θ

34

Homogeneous solution of plate bending:

xk x ik sin y i tw(x, y, t) W e e e− θ ω= ⋅ ⋅ ⋅ (75)

2

p4B

mk

Dω

= (76)

1,2

3,4

4 2 2 2 4 4 4x x B

2 2 2x B

2 2 2x B

k 2k k sin k sin k

k k k sin

k k k sin

− θ + θ =

= ± + θ

= ± − + θ

(77)

Velocities of plates for sinθ2<1:

( ) ( ) ( )1 1 1 1 1 1n1ik cos x ik cos x ik sin yk x i t11 1

w v v (e r e rj e ) e et

− θ θ − θ ω+

∂ = = + ⋅ + ⋅ ⋅ ⋅∂ (78)

( ) ( )2 2 1 1n2ik cos x ik sin yk x i t22 1

w v v (t e tj e ) e et

− θ − θ− ω+

∂ = = ⋅ + ⋅ ⋅ ⋅∂ (79)

where

2 2 2n1 B 1 1k k k sin= + + θ (80a)

2 2 2n2 B 1 1k k k sin= − + θ (80b)

for sinθ2>1, there will be no travelling wave and the velocity of plate 2 is:

( )2 2 21 11 1 2 n2 ik sin yk sin k x k x i t2

2 1w v v (t e tj e ) e et

− θ− θ − − ω+

∂ = = ⋅ + ⋅ ⋅ ⋅∂ (81)

35

Under this condition no power is radiated from the joint. Instead there is a

second nearfield wave resulting in local deformation at the joint.

Boundary conditions:

• Translational velocities at x = 0 are equal: v1 = v2 at x = 0

1 + r + rj = t + tj

r + rj – t – tj = -1 (83)

• Angular velocities at x = 0 are equal: 1 2v v

x x∂ ∂=∂ ∂

1 1 1 1 n1 2 2 n2

1 1 n1 2 2 n2 1 1

( ik cos ) r (ik cos ) rj (k ) t ( ik cos ) tj ( k )

r (ik cos ) rj (k ) t (ik cos ) tj (k ) ik cos

− θ + ⋅ θ + ⋅ = ⋅ − θ + ⋅ −

⋅ θ + ⋅ + ⋅ θ + ⋅ = θ (84)

• Total force at x = 0 is equal to the inertial force of the mass:

2b

3 3 3 32 2 1 1

3 2 3 2

vm ft Seff Seff

tv v v vD D (2 ) (2 )

i ix x y x x y

+ −∂= = −

∂ ∂ ∂ ∂ ∂

= − + − ν + + − ν ω ω∂ ∂ ∂ ∂ ∂ ∂

(85)

mb is mass per unit length of junction line mass.

36

( )

( ) [ ]

( )

3 3b 2 2 n2

21 1 2 2 n2

33 31 1 1 1 n1

1

Dm i t tj t ( ik cos ) tj ( k )i

(2 ) ik sin t ( ik cos ) tj ( k )

D ( ik cos ) r ik cos rj (k )i

(2 ) ik sin

⋅ ω⋅ + = − ⋅ ⋅ − θ + ⋅ −ω

+ − ν − θ ⋅ − θ + ⋅ −

+ ⋅ − θ + θ + ⋅ω

+ − ν − θ( ) ( ) ( )21 1 1 1 1 n1ik cos r ik cos rj k − θ + θ + ⋅

(86)

( ) ( ) ( )( )

( )( )

( )( ) ( )

( )( ) ( )

( ) ( ) ( )

2 21 1 1 1 1 1

22n1 n1 1 1

22 b2

2 2 2 2 1 1

22 b2

n2 n2 1 1

21 1 1 1

r ik cos ik cos 2 ik sin

rj k k 2 ik sin

m it (ik cos ) ( ik cos ) 2 ik sin

D

m itj k ( k ) 2 ik sin

D

ik cos ik cos 2

⋅ θ ⋅ θ + − ν − θ

+ ⋅ ⋅ + − ν − θ

⋅ ω + ⋅ θ ⋅ − θ + − ν − θ −

⋅ ω + ⋅ ⋅ − + − ν − θ −

= θ − θ + − ν ( )21 1ik sin − θ (87)

• Total moment at x = 0 is equal to the inertial moment of the mass:

22

b

2 2 2 22 2 1 1

2 2 2 2

vIp mt M Mt x

v v v vD D i ix y x y

+ −∂ρ = = − +∂ ∂

∂ ∂ ∂ ∂= + ν − + ν ω ω∂ ∂ ∂ ∂

(88)

37

[ ]

( ) ( )

( ) ( ) [ ]

2 2 n2

22 22 2 n2 1 1

2 22 21 1 1 1 n1 1 1

Ip i t( ik cos ) tj( k )

D t( ik cos ) tj ( k ) ik sin t tji

D ( ik cos ) r ik cos rj (k ) ik sin 1 r rji

ρ ⋅ ω⋅ − θ + − =

⋅ − θ + ⋅ − + ν − θ + ω

− ⋅ − θ + θ + ⋅ + ν − θ + + ω

(89)

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

2 2 221 1 1 1 n1 1 1

222

2 2 1 1 2 2

222

n2 1 1 n2

2 21 1 1 1

r ik cos ik sin rj (k ) ik sin

Ip it ( ik cos ) ik sin ik cos

D

Ip itj ( k ) ik sin k

D

ik cos ik sin

⋅ − θ − ν − θ + ⋅ − − ν − θ

ρ ⋅ ω + ⋅ − θ + ν − θ − ⋅ − θ

ρ ⋅ ω + ⋅ − + ν − θ − ⋅ −

= − θ + ν − θ

(90)

Matrix relation for line mass joint to solve oblique incidence wave

transmission is the following.

( ) ( ) ( )( )

( ) ( )

1 12 2

1 1 1 1 1 1

2 21 1 1 1

1rik cosrj

R ik cos ik cos 2 ik sinttj ik cos ik sin

− θ ⋅ = θ − θ + − ν − θ − θ + ν − θ

(91)

R11 = 1 (92)

R12 = 1 (93)

R13 = -1 (94)

38

R14 = -1 (95)

R21 = i k1 cosθ1 (96)

R22 = kn1 (97)

R23 = i k2 cosθ2 (98)

R24 = kn2 (99)

R31 = ( ) ( ) ( )( )2 21 1 1 1 1 1ik cos ik cos 2 ik sin θ ⋅ θ + − ν − θ (100)

R32 = ( )( )22n1 n1 1 1k k 2 ik sin ⋅ + − ν − θ (101)

R33 = ( )( ) ( )22 b2

2 2 2 2 1 1m i

(ik cos ) ( ik cos ) 2 ik sinD

ω θ ⋅ − θ + − ν − θ − (102)

R34 = ( )( ) ( )22 b2

n2 n2 1 1m i

k ( k ) 2 ik sinD

ω ⋅ − + − ν − θ − (103)

( ) ( )2 241 1 1 1 1R ik cos ik sin= − θ − ν − θ (104)

( )2242 n1 1 1R (k ) ik sin= − − ν − θ (105)

( ) ( ) ( )2

2243 2 2 1 1 2 2

Ip iR ( ik cos ) ik sin ik cos

Dρ ⋅ ω

= − θ + ν − θ − ⋅ − θ (106)

( ) ( ) ( )2

2244 n2 1 1 n2

Ip iR ( k ) ik sin k

D

ρ ⋅ ω = − + ν − θ − ⋅ −

(107)

Transmission efficiency, τ, depends on angle θ as given in equation (62) and

transmission loss R in dB was given as equation (63).

39

3.3.2 Normal Incidence

For normal incidence substitute sinθ = 0 & cosθ = 1.


kx i tw(x, y, t) W e e ω= ⋅ ⋅ (108)

From plate equation,

k4 = mp . ω2 / D (109)

2 2p p4 4

m mk i ,

D Dω ω

= ± ⋅ ± (110)

Velocities of plates:

ikx ikx kx i t11 1

w v v (e r e rj e ) et

− ω+

∂ = = + ⋅ + ⋅ ⋅∂

(111)

ikx kx i t22 1

w v v (t e tj e ) et

− − ω+

∂ = = ⋅ + ⋅ ⋅∂

(112)



1 + r + rj = t + tj

r + rj – t – tj = -1 (113)

• Angular velocities at x = 0 are equal: 1 2v vx x

∂ ∂=∂ ∂

( ik) r (ik) rj (k) t ( ik) tj ( k)− + ⋅ + ⋅ = ⋅ − + ⋅ −

40

r i rj t i tj i⋅ + + ⋅ + = (114)

• Total force at x = 0 is equal to the inertial force of the mass:

2b

3 32 1

3 3

vm ft F F

tv vD D

i ix x

+ −∂= = −

∂ ∂ ∂

= − + ω ω∂ ∂

(115)

mb is mass per unit length.

( ) 3 3 3b

3 3

Dm i t tj ( ik) r (ik) rj ki

D t ( ik) tj ( k)i

⋅ ω⋅ + = ⋅ − + ⋅ + ⋅ ω

− ⋅ ⋅ − + ⋅ − ω

2 2

b b3 3

m mi r rj t i tj 1 i

k D k D

ω ω− ⋅ + + ⋅ − + + ⋅ + = −

(116)

• Total moment at x = 0 is equal to the inertial moment of the mass:

22

b

2 22 1

2 2

vIp mt M M

t xv vD D

i ix x

+ −∂ρ = = − +∂ ∂

∂ ∂= − ω ω∂ ∂

2 2Ip Ipr rj t 1 i tj 1 1

k D k D ρ ω ρ ω− + ⋅ − − + ⋅ − = − ⋅ ⋅ (117)

41

Matrix relation: 1

r 1 r 1rj i rj i

R Rt i t itj 1 tj 1

−

− − ⋅ = ⇒ = ⋅ − − − −

(118)

2 2b b3 3

2 2

1 1 1 1i 1 i 1

m mi 1 i 1R

k D k D

i Ip Ip1 1 1 1k D k D

− − ⋅ ω ⋅ ω − − − + =

⋅ ⋅

⋅ρ ⋅ ω ρ ⋅ ω − − − − ⋅ ⋅

(119)

The above equations are solved for r, rj, t and tj. From these, transmission

coefficient is found as below.

' 21 B1 1' 22 B2 2

2m c VPincPtra 2m c V

τ = =!! (120)

2tτ = (121)

Transmission loss R in dB was given as equation (63).

3.4 Derivations for Two Plates - Beam Joint

Equation of plate:

4 4 4 2

p4 2 2 4 2

w w w wD 2 m 0x x y y t

∂ ∂ ∂ ∂+ + + = ∂ ∂ ∂ ∂ ∂ (122)

mp is mass per unit area.

42

3.4.1 Motion Equations of Beam Types

3.4.1.1 Open Section Channel

Figure 3-5 Open Section Channel

Junction beam is assumed an open section channel with non-coincident shear

centre and centroid. Equations of beam with triple coupling as defined in Figure 3-5

are given below. The horizontal distance between shear centre and beam-plate

connection point is assumed zero in literature. In order to see the effect of this

assumption in the sensitivity analysis, the distance is assumed nonzero as sx.

43

4 4 2 2

b b x4 4 2 2w` u` wÈI EI m m c ft

y y t tζ ηζ∂ ∂ ∂ ∂ φ+ + + =∂ ∂ ∂ ∂

(123)

4 4 2 2

b b z x4 4 2 2u` w` uÈI EI m m c ft

y y t tη ηζ∂ ∂ ∂ ∂ φ+ + + =∂ ∂ ∂ ∂

(124)

4 2 2 2 2

o b o b x b z4 2 2 2 2

x x z

w` uÈ GJ Ip m c m cy y t t t

mt ft s ft s

∂ φ ∂ φ ∂ φ ∂ ∂Γ − + ρ + +∂ ∂ ∂ ∂ ∂

= + ⋅ − ⋅ (125)

The term mb is mass per unit length of line junction beam. Since the motions

of the plates are defined in a coordinate system at the level of plates, the beam

equations are transferred to the same coordinate system with the following relations.

u = up = u` – sz φ (126)

w = wp = w` + sx φ (127)

Then the equations of beam motion are the followings.

( )

( )

4 4 4 2

x z b4 4 4 2

2

b x x 2

w u wEI EI E s I s I my y y t

m s c ftt

ζ ηζ ζ ηζ∂ ∂ ∂ φ ∂+ + ⋅ − + +∂ ∂ ∂ ∂

∂ φ+ − + =∂

(128a)

( ) ( )4 4 4 2

z x b z z x4 4 4 2u wEI E I s s I EI m s c ft

y y y tη η ηζ ηζ∂ ∂ φ ∂ ∂ φ+ ⋅ − + + + =∂ ∂ ∂ ∂

(128b)

( )

4 2 2 2 2

o b o b x b z4 2 2 2 2

2

b z z x x x x z2

w uE GJ Ip m c m cy y t t t

m c s c s mt ft s ft st

∂ φ ∂ φ ∂ φ ∂ ∂Γ − + ρ + +∂ ∂ ∂ ∂ ∂

∂ φ+ − = + ⋅ − ⋅∂

(128c)

44

It is assumed that the point P has no x motion. Therefore the u term above is

zero. Since u is already specified, the differential motion equations can be reduced to

two.

( ) ( )4 4 2 2

x z b b x x4 4 2 2w wEI E s I s I m m s c ft

y y t tζ ζ ηζ∂ ∂ φ ∂ ∂ φ+ ⋅ − + + + − + =∂ ∂ ∂ ∂

(129a)

( ) ( )4 4 2

z x b z z x4 4 2wE I s s I EI m s c ft

y y tη ηζ ηζ∂ φ ∂ ∂ φ− + + + =∂ ∂ ∂

(129b)

( )

4 2 2 2

o b o b x4 2 2 2

2

b z z x x x x z2

wE GJ Ip m cy y t t

m c s c s mt ft s ft st

∂ φ ∂ φ ∂ φ ∂Γ − + ρ +∂ ∂ ∂ ∂

∂ φ+ − = + ⋅ − ⋅∂

(129c)

After some manipulations an equation for mt is found as below.

( ) ( )

( ) ( ) ( )

4 4 22 2

z x o x z x z4 4 2

2 22 2

b x x b c b x x z z2 2

2 2 22 2

2 2

wE s I s I E s I s I 2s s I GJy y y

w m c s Ip A c s c s mtt t

M M

w w w D Dx y

ηζ ζ ζ η ηζ

+ −

∂ ∂ φ ∂ φ− + ⋅ Γ + + − −∂ ∂ ∂

∂ ∂ φ + − + ρ + − + + = ∂ ∂

= − +

∂ ∂ ∂= + ν − ∂ ∂

21 1

2 2w

x y

∂+ ν ∂ ∂

(130)

45

( ) ( )4 4 2 2

x z b b x x4 4 2 2

3 3 3 32 2 1 1

3 2 3 2

w wEI E s I s I m m c s fty y t t

Seff Seff

w w w w D (2 ) D (2 )

x x y x x y

ζ ζ ηζ

+ −

∂ ∂ φ ∂ ∂ φ+ ⋅ − + + + − =∂ ∂ ∂ ∂

= −

∂ ∂ ∂ ∂= − + − ν + + − ν

∂ ∂ ∂ ∂ ∂ ∂

(131)

3.4.1.1.1 Double Coupling Case

The equations of motion for beam with double coupling case are,

4 2

b4 2

3 3 3 32 2 1 1

3 2 3 2

w wEI m fty t

Seff Seff

w w w w D (2 ) D (2 )

x x y x x y

ζ

+ −

∂ ∂+ =∂ ∂

= −

∂ ∂ ∂ ∂= − + − ν + + − ν

∂ ∂ ∂ ∂ ∂ ∂

(132a)

( ) ( )4 2 2

2 2o z b o b z z z4 2 2

2 2 2 22 2 1 1

2 2 2 2

E s I GJ Ip m s 2c s mty y t

M M

w w w w D D

x y x y

η

+ −

∂ φ ∂ φ ∂ φ ⋅ Γ + − + ρ + + = ∂ ∂ ∂

= − +

∂ ∂ ∂ ∂= + ν − + ν

∂ ∂ ∂ ∂

(132b)

46

3.4.1.2 Euler Beam

Equation of Euler beam in bending is,

4 2

z b4 2

w wEI ft my t

∂ ∂− + =∂ ∂

(133)

mb is mass per unit length of line junction beam.

4 2

z b4 2

3 3 3 32 2 1 1

3 2 3 2

w wEI m ft Seff Seffy t

w w w w D (2 v) D (2 v)

x x y x x y

+ −∂ ∂+ = = −∂ ∂

∂ ∂ ∂ ∂= − + − + + − ∂ ∂ ∂ ∂ ∂ ∂

(134)

Equation of Euler beam in torsion is,

2 2

b2 2GJ mt Ipy t

∂ φ ∂ φ+ = ρ∂ ∂

(135)

2 2

b2 2

2 2 2 22 2 1 1

2 2 2 2

GJ Ip mt M My t

w w w w D v D vx y x y

+ −∂ φ ∂ φ− + ρ = = − +∂ ∂

∂ ∂ ∂ ∂= + − + ∂ ∂ ∂ ∂

(136)

3.4.2 Oblique Propagation

3.4.2.1 Open Section Beam


with an amplitude w1+. The wavenumbers are kx and ky.

kx = k cosθ (137a)

47

ky = k sinθ (137b) The displacement of this incident wave is given as equation (73). The wavelength in y direction must be the same for all waves on all plates.

1 1 2 2k sin k sinθ = θ

12 1

2

ksin sink

θ = θ (138)

θ2, the angle at which the wave leaves the joint is calculated from this





2p4

B

mk

Dω

= (140)

1,2

3,4

4 2 2 2 4 4 4x x B

2 2 2x B

2 2 2x B

k 2k k sin k sin k

k k k sin

k k k sin

− θ + θ =

= ± + θ

= ± − + θ

(141)

Velocities of plates for sinθ2<1:

( ) ( ) ( )1 1 1 1 1 1n1ik cos x ik cos x ik sin yk x i t11 1


− θ θ − θ ω+

∂ = = + ⋅ + ⋅ ⋅ ⋅∂ (142)

48

( ) ( )2 2 1 1n2ik cos x ik sin yk x i t22 1


− θ − θ− ω+

∂ = = ⋅ + ⋅ ⋅ ⋅∂ (143)

where,

2 2 2n1 B 1 1k k k sin= + θ (144a)

2 2 2n2 B2 1 1k k k sin= + θ (144b)

For sinθ2>1, there will be no travelling wave and the velocity of plate 2 is:

( )2 2 21 11 1 2 n2 ik sin yk sin k x k x i t2


− θ− θ − − ω+

∂ = = ⋅ + ⋅ ⋅ ⋅∂ (145)



Beam solution:

1 1i( t k sin y)b b 1 2w W e w (0, y, t) w (0, y, t)ω − θ= ⋅ = = (146)

1 1i( t k sin y) 1 2b

w (0, y, t) w (0, y, t)ex x

ω − θ ∂ ∂φ = φ ⋅ = =∂ ∂

(147)



1 + r + rj = t + tj

r + rj – t – tj = -1 (148)

• Angular velocities at x = 0 are equal: 1 2v v

x x∂ ∂=∂ ∂

49

1 1 1 1 n1 2 2 n2

1 1 n1 2 2 n2 1 1



− θ + ⋅ θ + ⋅ = ⋅ − θ + ⋅ −

⋅ θ + ⋅ + ⋅ θ + ⋅ = θ (149)

• Total z force at x = 0 is equal to the vertical force on the beam:

( ) ( )4 4 2 2

x z b b x x4 4 2 2

3 3 3 32 2 1 1

3 2 3 2


Seff Seff

w w w w D (2 ) D (2 )

x x y x x y

ζ ζ ηζ

+ −

∂ ∂ φ ∂ ∂ φ+ ⋅ − + + + − =∂ ∂ ∂ ∂

= −

∂ ∂ ∂ ∂= − + − ν + + − ν

∂ ∂ ∂ ∂ ∂ ∂

(150)

( )

( ) ( )

( ) ( )

( ) [ ]

24 b

1 1

2z x b x x4

1 1

2 2 n2

3 32 2 n2

21 1 2 2 n2

31 1 1

E I m( ik sin ) t tj

D D

E s I s I m c s( ik sin )

D D

t ik cos tj k

t ( ik cos ) tj ( k )


( ik cos ) r ik c

ζ

ηζ ζ

⋅ ω− θ − ⋅ + +

⋅ − − ω + − θ −

⋅ ⋅ − θ + ⋅ − =

− ⋅ − θ + ⋅ −

+ − ν − θ ⋅ − θ + ⋅ −

+ − θ + ( )

( ) ( ) ( )

3 31 n1

21 1 1 1 1 1 n1

os rj (k )

(2 ) ik sin ik cos r ik cos rj k

θ + ⋅

+ − ν − θ − θ + θ + ⋅ (151)

50

Define Λw and Λφ as

24 b

w 1 1E I m

( ik sin )D D

ζ ⋅ ωΛ = − θ −

(152)

( ) ( ) 2z x b x x4

1 1E s I s I m c s

( ik sin )D D

ηζ ζφ

⋅ − − ω Λ = − θ −

(153)

( ) ( ) ( )( )

( )( )

( )( )

( )( )

( ) ( )

2 21 1 1 1 1 1

22n1 n1 1 1

222 2 2 2 1 1 w 2 2

22n2 n2 1 1 w n2

21 1 1 1


rj k k 2 ik sin

t (ik cos ) ( ik cos ) 2 ik sin ik cos

tj k ( k ) 2 ik sin k

ik cos ik cos

φ

φ

⋅ θ ⋅ θ + − ν − θ

+ ⋅ ⋅ + − ν − θ

+ ⋅ θ ⋅ − θ + − ν − θ − Λ + Λ θ

+ ⋅ ⋅ − + − ν − θ − Λ + Λ

= θ − θ +( )( )21 12 ik sin − ν − θ

(154)

• Total moment at x = 0 is equal to the moment on the beam:

( ) ( )

( ) ( ) ( )

4 4 22 2

z x o x z x z4 4 2

2 22 2


2 2 22 2

2 2



M M

w w w D Dx y

ηζ ζ ζ η ηζ

+ −

∂ ∂ φ ∂ φ− + ⋅ Γ + + − −∂ ∂ ∂

∂ ∂ φ + − + ρ + − + + = ∂ ∂

= − +

∂ ∂ ∂= + ν − ∂ ∂

21 1

2 2w

x y

∂+ ν ∂ ∂

(155)

51

Define Ipp as the following

( ) ( )2 2p c b x x z zIp Ip A c s c s = + − + +

(156)

Define Γpp as the following

( )2 2p o x z x zp s I s I 2s s Iζ η ηζΓ = Γ + + − (157)

( ) ( ) ( )

[ ]

( ) ( ) ( ) ( )

( ) ( )

24 2p b p

1 1 1 1

2 2 n2

2z x 4 b x x

1 1

22 22 2 n2 1 1

E p Ip iGJik sin ik sinD D D


E s I s I m c s i + ik sin (t tj)

D D

t( ik cos ) tj ( k ) ik sin t tj

( ik

ηζ ζ

Γ ρ ⋅ ω − θ − − θ +

⋅ ⋅ − θ + ⋅ −

− − ⋅ ω ⋅ − θ + ⋅ + =

− θ + ⋅ − + ν − θ +

− − ( ) ( ) [ ]2 22 21 1 1 1 n1 1 1cos ) r ik cos rj (k ) ik sin 1 r rj θ + θ + ⋅ + ν − θ + +

(158)

Define Υw and Υφ as

( ) ( ) ( ) ( )2z x 4 b x x

w 1 1E s I s I m c s i

ik sinD D

ηζ ζ − − ⋅ ω ϒ = ⋅ − θ +

(159)

( ) ( ) ( )24 2p b p

1 1 1 1E p Ip iGJik sin ik sin

D D Dφ

Γ ρ ⋅ ω ϒ = − θ − − θ +

(160)

52

( ) ( )

( ) ( )

( ) ( ) ( )

( )

( ) ( )

2 21 1 1 1

2 2n1 1 1

2 22 2 1 1 w 2 2

22n2 1 1 w n2

2 21 1 1 1

r ik cos ik sin

rj k ik sin

t ik cos ik sin ik cos

tj ( k ) ik sin k

ik cos ik sin

φ

φ

⋅ − θ − ν − θ

+ ⋅ − − ν − θ

+ ⋅ − θ + ν − θ − ϒ + ϒ ⋅ θ

+ ⋅ − + ν − θ − ϒ + ϒ ⋅

= − θ + ν − θ

(161)

Matrix relation for open section beam to solve oblique incidence wave


( )1 1

2 21 1 1 1 1 1

2 21 1 1 1

1 rik cos rj

Rik cos ( ik cos ) (2 )( ik sin ) ttj ( ik cos ) ( ik sin )

− θ = ⋅ θ − θ + − ν − θ − θ + ν ⋅ − θ

(162)

The first two rows of matrix R are the same with line mass and given as

equations from (92) to (99). The expressions of third and fourth rows are given

below.


R32 = ( )( )22n1 n1 1 1k k 2 ik sin ⋅ + − ν − θ (164)

R33 = ( )( )222 2 2 2 1 1 w 2 2(ik cos ) ( ik cos ) 2 ik sin ik cosφ

θ − θ + − ν − θ − Λ + Λ θ (165)

53

R34 = ( )( )22n2 n2 1 1 w n2k ( k ) 2 ik sin kφ

⋅ − + − ν − θ − Λ + Λ (166)

( ) ( )2 241 1 1 1 1R ik cos ik sin= − θ − ν − θ (167)

( )2242 n1 1 1R (k ) ik sin= − − ν − θ (168)

( ) ( )2243 2 2 1 1 w 2 2R ( ik cos ) ik sin ik cosφ= − θ + ν − θ − ϒ + ϒ ⋅ θ (169)

( )2244 n2 1 1 w n2R ( k ) ik sin kφ

= − + ν − θ − ϒ + ϒ ⋅ (170)


First and second rows of R matrix for double coupling case are identical to R

matrix of open section beam. The term sx will be zero for the third boundary

condition and this will change the following terms.

24 b

w 1 1E I m

( ik sin )D D

ζ ⋅ ωΛ = − θ −

(171)

2z 4 b x

1 1Es I m c

( ik sin )D D

ηζφ

ωΛ = − θ −

(172)

The fourth row of R matrix will be used with the new definitions of Υw and

Υφ as below.

( ) ( )24z b x

w 1 1Es I m c i

ik sinD D

ηζ ⋅ ω ϒ = ⋅ − θ +

(173)

( )( ) ( ) ( )2 2

o z 4 2 b p21 1 1 1

E s I Ip iGJik sin ik sinD D D

ηφ

Γ + ρ ⋅ ω ϒ = − θ − − θ +

(174)

54

where

( )2p2 o b z z zIp Ip A s 2c s= + + (175)

3.4.2.2 Euler Beam

Matrix relation for Euler beam to solve oblique incidence wave transmission,

( )1 1

2 21 1 1 1 1 1

2 21 1 1 1

1 rik cos rj

Rik cos ( ik cos ) (2 )( ik sin ) ttj ( ik cos ) ( ik sin )

− θ = ⋅ θ − θ + − ν − θ − θ + ν ⋅ − θ

(176)

First and second rows of R matrix for Euler beam are identical to R matrix of

open section beam.

Third row of R matrix for Euler beam to solve oblique incidence wave

transmission is given below.


R32 = ( )( )22n1 n1 1 1k k 2 ik sin ⋅ + − ν − θ (178)

( )( )

( )

2233 2 2 2 2 1 1

24 bz

1 1

R (ik cos ) ( ik cos ) 2 ik sin

mEI ik sin

D D

= θ ⋅ − θ + − ν − θ

ω− − θ +

(179)

55

R34 = ( )( ) ( )2

2 42 bzn2 n2 1 1 1 1

mEIk ( k ) 2 ik sin ik sinD D

ω ⋅ − + − ν − θ − − θ + (180)

Fourth row of R matrix for Euler beam to solve oblique incidence wave


( ) ( )2 241 1 1 1 1R ik cos ik sin= − θ − ν − θ (181)

( )2242 n1 1 1R (k ) ik sin= − − ν − θ (182)

( )

( ) ( ) ( )

2243 2 2 1 1

22

1 1 2 2

R ( ik cos ) ik sin

Ip iGJ ik sin ik cosD D

= − θ + ν − θ

ρ ⋅ ω + − θ − ⋅ − θ

(183)

( ) ( ) ( ) ( )2

2 2244 n2 1 1 1 1 n2

Ip iGJR ( k ) ik sin ik sin kD D

ρ ⋅ ω = − + ν − θ + − θ − ⋅ −

(184)


For normal incidence case substitute sinθ = 0 & cosθ = 1


Matrix relation for open section beam to solve normally incident wave


56

( )1

31

21

1 rik rj

Rtiktj ( ik )

− = ⋅ −

(185)

The first row of matrix R is the same with oblique transmission and given as

equations from (92) to (95). The expressions of the remaining rows are given below.

R21 = i k1 (186)

R22 = kn1 (187)

R23 = i k2 (188)

R24 = kn2 (189)

R31 = ( )31ik (190)

R32 = 3n1k (191)

R33 = 32 w 2(ik ) ikφ− Λ + Λ (192)

R34 = 3n2 w n2k kφ− Λ + Λ (193)

( )241 1R ik= − (194)

242 n1R (k )= − (195)

( )243 2 w 2R ( ik ) ikφ= − − ϒ + ϒ ⋅ (196)

244 n2 w n2R ( k ) kφ= − − ϒ + ϒ ⋅ (197)

57


First and second rows of R matrix for double coupling case are identical to R

matrix of open section beam. The third and fourth rows will be used with the

definitions of Λw, Λφ, Υw and Υφ as given in section 3.4.2.1.1.

3.4.3.2 Euler Beam

Matrix relation for Euler beam to solve normally incident wave transmission,

( )1

31

21

1 rik rj

Rtiktj ( ik )

− = ⋅ −

(198)

First and second rows of R matrix for Euler beam are identical to R matrix of

open section beam.

Third row of R matrix for Euler beam to solve normally incident wave

transmission,

R31 = ( )31ik (199)

R32 = 3n1k (200)

23 b

33 2m

R ( ik )Dω

= − − + (201)

R34 = 2

3 bn2

m( k )

Dω

− − + (202)

58

Fourth row of R matrix for Euler beam to solve normally incident wave

transmission,

( )241 1R ik= − (203)

242 n1R (k )= − (204)

( ) ( )2

243 2 2

Ip iR ( ik ) ik

Dρ ⋅ ω

= − − ⋅ − (205)

( ) ( )2

244 n2 n2

Ip iR ( k ) k

Dρ ⋅ ω

= − − ⋅ − (206)

Transmission loss and coupling loss factors will be calculated from the

previously derived formulas.

3.5 Power Input

Power input for pressure wave excitation is calculated with formulation given

in Appendix A.

59

3.6 Derivations for Three Plates - Beam Joint

The coupling parameters of three plates connected via a line junction will be

derived in this section. The assumptions are the same as two plates system.

Figure 3-6 Three plates connected via a line junction

Net force and moment on the beam resulting from all plates are ft and mt

defined with the following equations.

ft + F1– – F2

+ – F3+cosβ = 0 (207a)

mt – M1– + M2

+ + M3+ = 0 (207b)

An input wave is applied to the first plate and the response of the second plate

is found. Since the transmission ratio is the ratio of velocities, the system response is

solved by assuming wave velocities proportional to the input wave velocity, v1+.

Equation of plate is the same with equation (71).

Plate 1 F1

- M1

-

Plate 2 F2

+ M2

+

Plate 3 F3

+ M3

+ β

Beam

60

3.6.1 Motion Equations of Beam Types

3.6.1.1 Open Section Channel

Junction beam is assumed an open section channel with non-coincident shear

centre and centroid. Equations of beam for triple coupling are given in the previous

section. For three plates case, forces on the beam come from three plates.

Then the equations of beam motion are,

( ) ( )4 4 2 2

x z b b x x4 4 2 2

1 2 3

3 3 3 31 1 2 2

3 2 3 2


Seff Seff Seff cos

w w w w D (2 ) D (2 )

x x y x x y

ζ ζ ηζ

+− +

∂ ∂ φ ∂ ∂ φ+ ⋅ − + + + − =∂ ∂ ∂ ∂

= − + + β

∂ ∂ ∂ ∂= + − ν − + − ν

∂ ∂ ∂ ∂ ∂ ∂

3 33 3

3 2w w

D cos (2 )x x y

∂ ∂− β + − ν

∂ ∂ ∂

(208)

61

( ) ( )

( ) ( ) ( )

4 4 22 2

z x o x z x z4 4 2

2 22 2


1 2 3

2 21 1

2 2



M M M

w w Dx y

ηζ ζ ζ η ηζ

− − −

∂ ∂ φ ∂ φ− + ⋅ Γ + + − −∂ ∂ ∂

∂ ∂ φ + − + ρ + − + + = ∂ ∂

= − −

∂ ∂= − + ν ∂ ∂

2 22 2

2 2

2 23 3

2 2

w wDx y

w w D

x y

∂ ∂+ + ν ∂ ∂

∂ ∂+ + ν

∂ ∂

(209)


The equations of beam motion for double coupling case are,

4 2

b4 2

1 2 3

3 3 3 31 1 2 2

3 2 3 2

3 33 3

3

w wEI m fty t

Seff Seff Seff cos

w w w w D (2 ) D (2 )x x y x x y

w w D cos (2 )x x y

ζ

+− +

∂ ∂+ =∂ ∂

= − + + β

∂ ∂ ∂ ∂= + − ν − + − ν ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂− β + − ν∂ ∂ ∂ 2

(210)

62

( ) ( )4 2 2

2 2o z b o b z z z4 2 2

1 2 3

2 2 2 21 1 2 2

2 2 2 2

2 23

2

E s I GJ Ip m s 2c s mty y t

M M M

w w w w D Dx y x y

w w Dx

η

− − −

∂ φ ∂ φ ∂ φ ⋅ Γ + − + ρ + + = ∂ ∂ ∂

= − −

∂ ∂ ∂ ∂= − + ν + + ν ∂ ∂ ∂ ∂

∂ ∂+ + ν∂

32y

∂

(211)

3.6.1.2 Euler Beam

Equation of Euler beam in bending is, 4 2

z b4 2

w wEI ft my t

∂ ∂− + =∂ ∂

(212)

mb is mass per unit length of line junction beam.

4 2

z b 1 2 34 2

3 3 3 31 1 2 2

3 2 3 2

3 33 3

3 2

w wEI m ft Seff Seff Seff cosy t

w w w w D (2 ) D (2 )x x y x x y

w w D cos (2 )

x x y

+− +∂ ∂+ = = − + + β∂ ∂

∂ ∂ ∂ ∂= + − ν − + − ν ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂− β + − ν

∂ ∂ ∂

(213)

63

Equation of Euler beam in torsion is,

2 2

b2 2GJ mt Ipx t

∂ θ ∂ θ+ = ρ∂ ∂

(214)

2 2

b 1 2 32 2

2 2 2 2 2 23 3 2 2 1 1

2 2 2 2 2 2

GJ Ip mt M M Mx t

w w w w w w D D D

x y x y x y

− − −∂ θ ∂ θ− + ρ = = − −∂ ∂

∂ ∂ ∂ ∂ ∂ ∂= + ν + + ν − + ν ∂ ∂ ∂ ∂ ∂ ∂

(215)

3.6.2 Oblique Propagation



with an amplitude w1+. The wavenumbers are

kx = k cosθ (216a)

ky = k sinθ (216b)

The displacement of this incident wave is,

ik cos x ik sin y i t1w(x, y, t) w e e e− θ − θ ω+= ⋅ ⋅ ⋅ (217)

The wavelength in y direction must be the same for all waves on all plates.

1 1 2 2 3 3k sin k sin k sinθ = θ = θ

64

12 1

2

ksin sin

kθ = θ

(218a)

13 1

3

ksin sin

kθ = θ (218b)

θm, the angle at which the wave leaves the joint is calculated from this





2pi4Bi

i

mk

Dω

= (220)

1,2

3,4

4 2 2 2 4 4 4x x Bi

2 2 2x Bi

2 2 2x Bi

k 2k k sin k sin k

k k k sin

k k k sin

− θ + θ =

= ± + θ

= ± − + θ

(221)

Velocities of plates for sinθ2 & sinθ3<1:

1 1 1 1 n1 1 1ik cos x ik cos x k x ik sin y i t11 1


− θ θ − θ ω+

∂ = = + ⋅ + ⋅ ⋅ ⋅∂

(222)

2 2 n 2 1 1ik cos x k x ik sin y i t22 1


− θ − − θ ω+

∂ = = ⋅ + ⋅ ⋅ ⋅∂

(223)

3 3 n3 1 1ik cos x k x ik sin y i t33 1

wv v (t e tj e ) e e

t− θ − − θ ω

+∂

= = ⋅ + ⋅ ⋅ ⋅∂

(224)

65

where,

2 2 2n1 B1 1 1k k k sin= + + θ (225a)

2 2 2n2 B2 1 1k k k sin= − + θ (225b)

2 2 2n3 B3 1 1k k k sin= − + θ (225c)

for sinθ2 & sinθ3 >1, there will be no travelling wave and the velocity of the

plate 2 and 3 are:

2 2 21 1 2 n2 1 1k sin k x k x ik sin y i t2


− θ − − − θ ω+

∂ = = ⋅ + ⋅ ⋅ ⋅∂

(226)

2 2 21 1 3 n3 1 1k sin k x k x ik sin y i t3

3 1w

v v (t e tj e ) e et

− θ − − − θ ω+

∂= = ⋅ + ⋅ ⋅ ⋅

∂ (227)



Beam solution:

1 1i( t k sin y)b b

1 2 3

w W e

w (0, y, t) w (0, y, t) w (0, y, t) cos

ω − θ= ⋅

= = = β (228)

1 1i( t k sin y) 31 2b

w (0, y, t)w (0, y, t) w (0, y, t)ex x x

ω − θ ∂∂ ∂φ = φ ⋅ = = =∂ ∂ ∂

(229)


• Translational velocities at x = 0 are equal: v1 = v2 = v3 .cosβ at x = 0

66

1 + r + rj = t2 + tj2 = (t3 + tj3).cosβ then,

r + rj – t2 – tj2 = -1 (230)

r + rj – (t3 + tj3).cosβ= -1 (231)

• Angular velocities at x = 0 are equal: 31 2 vv vx x x

∂∂ ∂= =∂ ∂ ∂

1 1 1 1 n1 2 2 2 2 n2

1 1 n1 2 2 2 2 n2 1 1



− θ + ⋅ θ + ⋅ = ⋅ − θ + ⋅ −

⋅ θ + ⋅ + ⋅ θ + ⋅ = θ (232)

1 1 1 1 n1 3 3 3 3 n3

1 1 n1 3 3 3 3 n3 1 1



− θ + ⋅ θ + ⋅ = ⋅ − θ + ⋅ −

⋅ θ + ⋅ + ⋅ θ + ⋅ = θ (233)

• Total z force at x = 0 is equal to the vertical force on the beam:

( ) ( )4 4 2 2

x z b b x x4 4 2 2

1 2 3

3 3 3 31 1 2 2

3 2 3 2


Seff Seff Seff cos

w w w w D (2 ) D (2 )

x x y x x y

ζ ζ ηζ

+− +

∂ ∂ φ ∂ ∂ φ+ ⋅ − + + + − =∂ ∂ ∂ ∂

= − + + β

∂ ∂ ∂ ∂= + − ν − + − ν

∂ ∂ ∂ ∂ ∂ ∂

3 33 3

3 2w w

D cos (2 )x x y

∂ ∂− β + − ν

∂ ∂ ∂

(234)

67

( )

( ) ( )

( ) ( )

( ) [ ]

24 b

1 1 2 2

2z x b x x4

1 1

2 2 2 2 n2

3 32 2 2 2 n2

21 1 2 2 2 2 n2

3

E I m( ik sin ) t tj

D D

E s I s I m c s( ik sin )

D D

t ik cos tj k



cos t ( i

ζ

ηζ ζ

⋅ ω− θ − ⋅ + +

⋅ − − ω + − θ −

⋅ ⋅ − θ + ⋅ − =

− ⋅ − θ + ⋅ −

+ − ν − θ ⋅ − θ + ⋅ −

− β ⋅ −

( ) [ ]

( )

( ) ( ) ( )

3 33 3 3 n3

21 1 3 3 3 3 n3

33 31 1 1 1 n1

21 1 1 1 1 1 n1

k cos ) tj ( k )


( ik cos ) r ik cos rj (k )

(2 ) ik sin ik cos r ik cos rj k

θ + ⋅ −

+ − ν − θ ⋅ − θ + ⋅ −

+ − θ + θ + ⋅

+ − ν − θ − θ + θ + ⋅

(235)

Define Λw and Λφ as

24 b

w 1 1E I m

( ik sin )D D

ζ ⋅ ωΛ = − θ −

(236)

( ) ( ) 2z x b x x4

1 1E s I s I m c s

( ik sin )D D

ηζ ζφ

⋅ − − ω Λ = − θ −

(237)

68

( ) ( ) ( )( )

( )( )

( )( )

( )( )

2 21 1 1 1 1 1

22n1 n1 1 1

222 2 2 2 2 1 1 w 2 2

222 n2 n2 1 1 w n2

3 3 3


rj k k 2 ik sin

t (ik cos ) ( ik cos ) 2 ik sin ik cos

tj k ( k ) 2 ik sin k

t cos (ik cos )

φ

φ

⋅ θ ⋅ θ + − ν − θ

+ ⋅ ⋅ + − ν − θ

+ ⋅ θ − θ + − ν − θ − Λ + Λ θ

+ ⋅ ⋅ − + − ν − θ − Λ + Λ

+ ⋅ β ⋅ θ ( )( )

( )( )

( ) ( ) ( )( )

223 3 1 1

223 n3 n3 1 1

2 21 1 1 1 1 1

( ik cos ) 2 ik sin

tj cos k ( k ) 2 ik sin

ik cos ik cos 2 ik sin

⋅ − θ + − ν − θ

+ ⋅ β ⋅ ⋅ − + − ν − θ

= θ − θ + − ν − θ

(238)

• Total moment at x = 0 is equal to the moment on the beam:

( ) ( )

( ) ( ) ( )

4 4 22 2

z x o x z x z4 4 2

2 22 2


1 2 3

2 21 1

2 2



M M M

w w Dx y

ηζ ζ ζ η ηζ

− − −

∂ ∂ φ ∂ φ− + ⋅ Γ + + − −∂ ∂ ∂

∂ ∂ φ + − + ρ + − + + = ∂ ∂

= − −

∂ ∂= − + ν ∂ ∂

2 22 2

2 2

2 23 3

2 2

w wDx y

w w D

x y

∂ ∂+ + ν ∂ ∂

∂ ∂+ + ν

∂ ∂

(239)

69

( ) ( ) ( )

[ ]

( ) ( ) ( ) ( )

( ) ( )

24 2p b p

1 1 1 1

2 2 2 2 n2

2z x 4 b x x

1 1 2 2

22 22 2 2 2 n2 1 1 2 2

E p Ip iGJik sin ik sinD D D


E s I s I m c s i + ik sin (t tj )

D D

t ( ik cos ) tj ( k ) ik sin t tj

ηζ ζ

Γ ρ ⋅ ω − θ − − θ +

⋅ ⋅ − θ + ⋅ −

− − ⋅ ω ⋅ − θ + ⋅ + =

⋅ − θ + ⋅ − + ν − θ +

( ) ( )

( ) ( ) [ ]

22 23 3 3 3 n3 1 1 3 3

2 22 21 1 1 1 n1 1 1

+ t ( ik cos ) tj ( k ) ik sin t tj

( ik cos ) r ik cos rj (k ) ik sin 1 r rj

⋅ − θ + ⋅ − + ν − θ +

− − θ + θ + ⋅ + ν − θ + +

(240)

Define Υw and Υφ as the following by using the same Ipp and Γpp given in

paragraph 3.4.2.1.

( ) ( ) ( ) ( )2z x 4 b x x

w 1 1E s I s I m c s i

ik sinD D

ηζ ζ − − ⋅ ω ϒ = ⋅ − θ +

(241)

( ) ( ) ( )24 2p b p

1 1 1 1E p Ip iGJik sin ik sin

D D Dφ

Γ ρ ⋅ ω ϒ = − θ − − θ +

(242)

70

( ) ( )

( )

( ) ( )

( )

( )

( )

2 21 1 1 1

22n1 1 1

222 2 2 1 1 w 2 2

222 n2 1 1 w n2

223 3 3 1 1

223 n3 1 1

r ik cos ik sin

rj (k ) ik sin

t ( ik cos ) ik sin ik cos

tj ( k ) ik sin k

t ( ik cos ) ik sin

tj ( k ) ik sin

φ

φ

⋅ − θ − ν − θ

+ ⋅ − − ν − θ

+ ⋅ − θ + ν − θ − ϒ + ϒ ⋅ θ

+ ⋅ − + ν − θ − ϒ + ϒ ⋅

+ ⋅ − θ + ν − θ

+ ⋅ − + ν − θ

( ) ( )2 21 1 1 1 ik cos ik sin= − θ + ν − θ

(243)

Matrix relation for open section beam to solve oblique incidence wave


( )

1 1 21 1 2

2 21 1 1 1 1 1 3

2 2 31 1 1 1

1 r 1 rjik cos t

Rik cos tjik cos ( ik cos ) (2 )( ik sin ) t

tj ( ik cos ) ( ik sin )

− − θ = ⋅ θ θ − θ + − ν − θ − θ + ν ⋅ − θ

(244)

Matrix R for open section beam,

R11 = 1 (245)

R12 = 1 (246)

R13 = -1 (247)

71

R14 = -1 (248)

R15 = 0 (249)

R16 = 0 (250)

R21 = 1 (251)

R22 = 1 (252)

R23 = 0 (253)

R24 = 0 (254)

R25 = -cosβ (255)

R26 = - cosβ (256)

R31 = i k1 cosθ1 (257)

R32 = kn1 (258)

R33 = i k2 cosθ2 (259)

R34 = kn2 (260)

R35 = 0 (261)

R36 = 0 (262)

R41 = i k1 cosθ1 (263)

R42 = kn1 (264)

R43 = 0 (265)

R44 = 0 (266)

R45 = i k3 cosθ3 (267)

R46 = kn3 (268)


72

R52 = ( )( )22n1 n1 1 1k k 2 ik sin ⋅ + − ν − θ (270)

R53 = ( )( )222 2 2 2 1 1 w 2 2(ik cos ) ( ik cos ) 2 ik sin ik cosφ

θ − θ + − ν − θ − Λ + Λ θ (271)

R54 = ( )( )22n2 n2 1 1 w n2k ( k ) 2 ik sin kφ

⋅ − + − ν − θ − Λ + Λ (272)

R55 = ( )( )223 3 3 3 1 1cos (ik cos ) ( ik cos ) 2 ik sin β ⋅ θ − θ + − ν − θ (273)

R56 = ( )( )22n3 n3 1 1cos k ( k ) 2 ik sin β ⋅ ⋅ − + − ν − θ (274)

( ) ( )2 261 1 1 1 1R ik cos ik sin= − θ − ν − θ (275)

( )2262 n1 1 1R (k ) ik sin= − − ν − θ (276)

( ) ( )2263 2 2 1 1 w 2 2R ( ik cos ) ik sin ik cosφ= − θ + ν − θ − ϒ + ϒ ⋅ θ (277)

( )2264 n2 1 1 w n2R ( k ) ik sin kφ

= − + ν − θ − ϒ + ϒ ⋅ (278)

( )2265 3 3 1 1R ( ik cos ) ik sin= − θ + ν − θ (279)

( )2266 n3 1 1R ( k ) ik sin= − + ν − θ (280)


First four rows of R matrix for double coupling case are identical to R matrix

of open section beam triple coupling formulation above. For the fifth and sixth rows,

the elements are the same but the definitions of Λw, Λφ, Υw and Υφ are different. The

equations given in paragraph 3.4.2.1.1 are used for these parameters.

73

3.6.2.2 Euler Beam

Matrix relation for Euler beam to solve oblique incidence wave transmission

is the following.

( )

1 1 21 1 2

2 21 1 1 1 1 1 3

2 2 31 1 1 1

1 r 1 rjik cos t

Rik cos tjik cos ( ik cos ) (2 )( ik sin ) t

tj ( ik cos ) ( ik sin )

− − θ = ⋅ θ θ − θ + − ν − θ − θ + ν ⋅ − θ

(281)

First four rows of R matrix for Euler beam are identical to R matrix of open

section beam.

Fifth row of R matrix for Euler beam to solve oblique incidence wave

transmission,


R52 = ( )( )22n1 n1 1 1k k 2 ik sin ⋅ + − ν − θ (283)

( )( )

( )

2253 2 2 2 2 1 1

24 bz

1 1

R (ik cos ) ( ik cos ) 2 ik sin

mEI ik sin

D D

= θ ⋅ − θ + − ν − θ

ω− − θ +

(284)

R54 = ( )( ) ( )2

2 42 bzn2 n2 1 1 1 1

mEIk ( k ) 2 ik sin ik sinD D

ω ⋅ − + − ν − θ − − θ + (285)

74

R55 = ( )( )223 3 3 3 1 1cos (ik cos ) ( ik cos ) 2 ik sin β ⋅ θ − θ + − ν − θ (286)

R56 = ( )( )22n3 n3 1 1cos k ( k ) 2 ik sin β ⋅ ⋅ − + − ν − θ (287)

Sixth row of R matrix for Euler beam to solve oblique incidence wave

transmission,

( ) ( )2 261 1 1 1 1R ik cos ik sin= − θ − ν − θ (288)

( )2262 n1 1 1R (k ) ik sin= − − ν − θ (289)

( )

( ) ( ) ( )

2263 2 2 1 1

22 P

1 1 2 2

R ( ik cos ) ik sin

Ip iGJ ik sin ik cosD D

= − θ + ν − θ

ρ ⋅ ω + − θ − ⋅ − θ

(290)

( ) ( ) ( ) ( )2

2 2 P264 n2 1 1 1 1 n2

Ip iGJR ( k ) ik sin ik sin kD D

ρ ⋅ ω = − + ν − θ + − θ − ⋅ −

(291)

( )2265 3 3 1 1R ( ik cos ) ik sin= − θ + ν − θ (292)

( )2266 n3 1 1R ( k ) ik sin= − + ν − θ (293)


For normal incidence case substitute sinθ = 0 & cosθ = 1.

75


Matrix relation for open section beam to solve normally incident wave


( )

1 2

1 23

312 31

1 r1 rj

ik tRik tj

tiktj ( ik )

− − = ⋅ −

(294)

First and second rows are the same with oblique incidence.

R31 = i k1 (295)

R32 = kn1 (296)

R33 = i k2 (297)

R34 = kn2 (298)

R35 = 0 (299)

R36 = 0 (300)

R41 = i k1 (301)

R42 = kn1 (302)

R43 = 0 (303)

R44 = 0 (304)

R45 = i k3 (305)

R46 = kn3 (306)

76

R51 = ( )31ik (307)

R52 = 3n1k (308)

R53 = 32 w 2(ik ) ikφ− Λ + Λ (309)

R54 = 3n2 w n2k kφ− Λ + Λ (310)

R55 = 33cos (ik )β ⋅ (311)

R56 = 3n3cos kβ ⋅ (312)

( )261 1R ik= − (313)

262 n1R (k )= − (314)

( )263 2 w 2R ( ik ) ikφ= − − ϒ + ϒ ⋅ (315)

264 n2 w n2R ( k ) kφ= − − ϒ + ϒ ⋅ (316)

265 3R ( ik )= − (317)

266 n3R ( k )= − (318)

3.6.3.2 Euler Beam

Matrix relation for Euler beam to solve normally incident wave transmission

is the following.

77

( )

1 2

1 23

312 31

1 r1 rj

ik tRik tj

tiktj ( ik )

− − = ⋅ −

(319)

First four rows of R matrix for Euler beam are identical to R matrix of open

section beam. Fifth row of R matrix for Euler beam to solve normally incident wave


R51 = ( )31ik (320)

R52 = 3n1k (321)

23 b

53 2m

R (ik )Dω

= + (322)

R54 = 2

3 bn2

m( k )

Dω

− − + (323)

R55 = 33 3cos (ik cos )β⋅ θ (324)

R56 = 3n3cos (k )β ⋅ (325)

Sixth row of R matrix for Euler beam to solve normally incident wave


( )261 1R ik= − (326)

262 n1R (k )= − (327)

( ) ( )2

P263 2 2

Ip iR ( ik ) ik

Dρ ⋅ ω

= − − ⋅ − (328)

78

( ) ( )2

P264 n2 n2

Ip iR ( k ) k

Dρ ⋅ ω

= − − ⋅ − (329)

265 3R ( ik )= − (330)

266 n3R ( k )= − (331)

Transmission loss and coupling loss factors will be calculated from the

previously derived formulas.

79

CHAPTER IV

4. COUPLING LOSS FACTOR SENSITIVITY

4.1 Coupling Loss Factor for Two Plates Coupling

101 102 103 1040

10

20

30

40

50

60

R [d

B]

Frequency [Hz]

Figure 4-1 Transmission loss of normal incidence

Transmission loss of two plates system is calculated by using the equations

derived in the previous chapter. Transmission loss of normal incidence is the same

for line mass and beam modelling techniques since bending and torsional stiffness of

80

the beam does not have any effect on the transmission. The beam and plate properties

given in Appendix C Table C-1 are the same as Ref [54] to check the calculations;

the resulting Figure 4-1 is also the same as Ref [54].

The graph shows that the joint acts like a low pass filter. There is a total

transmission point at around 300 Hz, where the transmission loss is zero. At a

frequency near 2000 Hz, R takes its maximum value. A local minima occurs around

4500 Hz and after this point transmission loss is continuously increasing.

The third joint modelling technique is the junction via an open section thin

walled beam with double and triple coupling. Since the example beam used here has

a perpendicular cross section, there is no difference between double and triple

coupling results. The following figure shows the transmission loss of normal

incidence calculated for line mass (or Euler beam) and open section beam.

101 102 103 1040

10

20

30

40

50

60

Frequency [Hz]

R [d

B]

Figure 4-2 Transmission loss of normal incidence for mass (green) and open section beam (blue)

81

Transmission loss calculated with open section beam modelling starts to

increase at a lower frequency than line mass modelling. Total transmission point at

around 300 Hz is not a definite point for open section beam results like line mass

modelling. R reaches its maximum value for open section beam modelling at a lower

frequency that means line mass modelling allows transmitting a wider range of low

frequency waves. Transmission loss of open section beam is lower than line mass

joint at higher frequencies. This means if an open section beam is modelled as a line

mass, the calculated transmitted energy will be lower than open section case.

Following figures give the transmission loss calculated for oblique wave

propagation. As the incidence angle increases to 30°, R for line mass does not change

much; however beam joints shows different behaviour than normal incidence. The

most interesting section is that the maximum loss point of line mass joint

corresponds to the total transmission point of Euler beam modelling. Minimum

transmission loss point of open section beam modelling slides to a lower frequency

than Euler beam at this incidence angle.

101 102 103 1040

10

20

30

40

50

60

Frequency [Hz]

R [d

B]

Figure 4-3 Transmission loss for 30° incidence angle

(line mass -green, Euler beam -red, triple coupling -blue)

82

101 102 103 1040

10

20

30

40

50

60

Frequency [Hz]

R [d

B]

Figure 4-4 Transmission loss for 45° incidence angle (line mass -green, Euler beam -red, triple coupling -blue)

As the angle increases, the peak at low frequency region at low frequencies

for Euler beam and open section beam modelling types disappears. The total

transmission points for both slides to higher frequencies. At 60° incidence angle,

there is no peak left in open section beam results and the line mass transmission loss

at low frequencies increases. Beam joints are affected more from the angle changes

than line mass joint.

83

101 102 103 1040

10

20

30

40

50

60

R [d

B]

Frequency [Hz]

Figure 4-5 Transmission loss for 60° incidence angle (line mass -green, Euler beam -red, triple coupling -blue)

It is shown that the transmission loss is changing with the oblique angle.

Surface plots Figure 4-6 and Figure 4-7 with oblique angle on one axis would be

more helpful to show this change. The plots do not cover 90°, because the loss goes

to very high values and it becomes difficult to interpret the changes at lower angles.

84

Figure 4-6 Transmission loss versus frequency and oblique angle for line mass modelling

Oblique angle does not result in much change in transmission loss for line

mass modelling. However, transmission loss for beam joint differs much with angle.

After 20°, the behaviour changes completely. For the angles less than 20°, maximum

transmission loss is at about 1000 Hz. However, the maximum transmission loss

point becomes the minimum transmission loss at 20° and the minimum point moves

from 1000 Hz to higher frequencies with the increasing angle.

85

Figure 4-7 Transmission loss versus frequency and oblique angle for Euler beam modelling

Figure 4-8 Transmission loss vs. frequency and oblique angle for triple coupling modelling

86

Euler beam calculations do not include the vertical offset of beam centroid

from plate surface and flexibility of beam in x direction. Triple and double coupling

formulations include these items and the results given in Figure 4-8. It is shown that

the transmission behaviour changes after 30° incidence angle. The dip about 1000 Hz

for Euler Beam changes the path after this angle and ends at 50°. Transmission loss

is much higher for triple coupling.

The coupling loss factor (CLF) is calculated with the following formula and

the results are given with surface plots Figure 4-9, Figure 4-10 and Figure 4-11.

( ) 112

1

cg Lj,A 2

⋅ τη θ ω = ⋅ω − τ

Figure 4-9 CLF vs. frequency and oblique angle for line mass modelling

87

Figure 4-10 CLF vs. frequency and oblique angle for Euler beam modelling

Figure 4-11 CLF vs. frequency and oblique angle for triple coupling modelling

88

Coupling loss factor of line mass joint does not change much with angle.

Therefore averaging the CLF over angle may be an acceptable assumption. However,

CLF of beam joint differs much with angle and the result of averaging does not

represent the actual joint characteristic well.

Coupling loss factors of studied joint models are calculated and averaged

over propagation angle, and the resulting curves are given in Figure 4-12. It is seen

that for frequencies lower than 200 Hz, triple coupling modelling does not produce

different results then Euler beam. However, the behaviour differs after this frequency

and there is 700 Hz shift between beam and triple coupling modelling local minima.

Besides line mass and beam joints results, the result of classical SEA for line

mass joint is also plotted in the Figure 4-13. It is seen that classical SEA takes

approximately the average of line mass and beam assumptions for low frequencies

and assumes that this is also valid for high frequencies.

101 102 103 10410-5

10-4

10-3

10-2

10-1

100

Frequency [Hz]

Cou

plin

g Lo

ss F

acto

r

Figure 4-12 Coupling loss factor for line mass (green), beam (red) and open section channel (blue) modelling

89

1.E-06

1.E-05

1.E-04

1.E-03

1.E-02

1.E-01

1.E+00

10 100 1000 10000Frequency [Hz]

Cou

plin

g Lo

ss F

acto

r

CLF (SEA)

CLF(Beam)

CLF(Line Mass)

Figure 4-13 Coupling loss factor for line mass (green), beam (red) and classical SEA (blue)

4.2 Sensitivity to System Parameters

The beam properties used above belongs to a rectangular cross section beam.

The following figures from Figure 4-14 to Figure 4-20 are obtained by using a C

cross section with double coupling and the properties are given in Appendix C Table

C-2.

The horizontal offset between shear centre, centroid and the point P, which

has no x movement, is not included in double coupling formulation. Since the C

beam has zero product moment of area, this horizontal offset is the only difference

between triple and double coupling. The vertical offset of beam centroid from plate

surface, warping and flexibility of beam in x direction are the differences between

Euler beam and double coupling modelling.

Since the mass modelling parameters are chosen similar to the previous

rectangular beam example, the line mass modelling results are approximately the

same as Figure 4-6, therefore its figure does not given here.

90

Figure 4-14 Transmission loss vs. frequency and oblique angle for Euler beam modelling

Figure 4-15 Transmission loss vs. frequency and oblique angle for double coupling modelling

91


101 102 103 10410-5

10-4

10-3

10-2

10-1

100

Frequency [Hz]

Cou

plin

g Lo

ss F

acto

r

Figure 4-17 CLF for line mass (green), beam (red), double coupling (black) and triple coupling (blue) modelling

92

Euler beam results are similar to rectangular beam results but higher as seen

from Figure 4-14. In addition, total transmission point slides to lower frequencies

and maximum transmission points are reached at a lower angle of incidence.

The difference between double coupling of rectangular beam and C beam is

the warping coefficient and larger second moment of area about z axis for C beam. It

is found that the reason of the peak at 30° is the new combination of second moment

of area values and this situation will be shown in the related paragraph of sensitivity

study for the second moment of area about z axis. The addition of sx to the

calculations with triple coupling, this peak moves to a lower angle of 20°. The

positions of peaks are well represented with the following figure showing the

transmission loss values at 10000 Hz for all joint modelling types.

0 10 20 30 40 50 60 70 80 900

10

20

30

40

50

60

70

80

90

100

Angle [Deg]

R [d

B]

Figure 4-18 Transmission loss at 10000 Hz vs. angle for line mass (green), beam (red), double coupling (black) and triple coupling (blue) modelling

93

Coupling loss factors of studied joint models for C beam are calculated and

averaged over propagation angle, and the resulting curves are given in Figure 4-17. It

is seen that for frequencies lower than 100 Hz, triple and double coupling modelling

do not produce different results then Euler beam. However, the behaviour differs

after this frequency and double coupling results are in between Euler beam and triple

coupling modelling results.

The following figures gives transmission loss values for 30° and 45° angles.

Differences between modelling technigues are well shown at 30°. Triple coupling

has two peaks at 400 Hz and 1000 Hz while double coupling has one peak at 5000

Hz. As incidence angle increases, triple coupling reduces to one peak and the peak of

double coupling slides to lower frequencies.

Figure 4-19 Transmission loss at 30° vs. frequency for line mass (green), beam (red), double coupling (black) and triple coupling (blue) modelling

94

101 102 103 1040

10

20

30

40

50

60

70

80

90

100

Frequency [Hz]

R [d

B]


The beam properties used for the above figures belong to a C beam which has

zero product moment of area and zero vertical distance between shear centre and

centroid. In order to show the transmission loss values calculated with the addition of

these parameters and hence having triply coupled cross section, an L beam with

triple coupling is used and the properties are given in Appendix C Table C-2. The

results for L beam are given in figures from Figure 4-21 to Figure 4-24.

95

Figure 4-21 Transmission loss vs. frequency and oblique angle for line mass modelling

Figure 4-22 Transmission loss vs. frequency and oblique angle for Euler beam modelling

96

Figure 4-23 Transmission loss vs. frequency and oblique angle for double coupling modelling


97

The L beam Euler and line mass modelling transmission loss values are lower

than C beam, but the characteristics of both surfaces are the same. However, double

coupling results are different. The peak observed at 30° incidence angle in C beam

results slides to higher angles for L beam. The differences between C and L beam in

terms of double coupling formulation are the vertical distance between shear centre

and centroid for L beam, smaller area moment of inertia and warping values. It will

be shown in the related paragraph that as area moment of inertia decreases the peaks

slide to higher frequencies and as the vertical distance increases the dips slides to

lower frequencies. The differences between double and triple coupling transmission

losses of C and L beam are the results of these parameters.

0 10 20 30 40 50 60 70 80 900

10

20

30

40

50

60

70

80

90

100

Angle [Deg]

R [d

B]

Figure 4-25 Transmission loss at 10000 Hz vs. angle for line mass (green), beam (red), double coupling (black) and triple coupling (blue) modelling

98

101 102 103 1040

10

20

30

40

50

60

70

80

90

100

Frequency [Hz]

R [d

B]


101 102 103 1040

10

20

30

40

50

60

70

80

90

100

Frequency [Hz]

R [d

B]


99

101 102 103 10410-5

10-4

10-3

10-2

10-1

100

Frequency [Hz]

Cou

plin

g Lo

ss F

acto

r

Figure 4-28 CLF for line mass (green), beam (red), double coupling (black) and triple coupling (blue) modelling

The coupling loss factors of L beam for examined modelling techniques are

not as different as C beam. Euler and open section beam CLF are the same up to

1000Hz and getting closer again at 10000Hz.

In order to see the effect of the joint beam properties on the transmission

characteristics, the parameters of beam has changed and the results are discussed by

considering the modelling technique in the following paragraphs.

100

4.2.1 Sensitivity to Density

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-29 Transmission loss of normal incidence for line mass modelling, change in density

(blue-ρ = 2000 kg/m3, green-ρ = 6000 kg/m3, magenta-ρ = 10000 kg/m3) Properties of Appendix C Table C-1 are used. This means the joint beam is a

rectangular one and therefore double coupling and triple coupling calculations give

the same results.

As the density of the joint line mass is increasing, the transmission between

plates is squeezed to lower frequencies. The maximum reflection frequency is

decreasing with increasing mass. Only low frequencies can be transmitted with high

joint mass. At high frequencies the transmission loss is increasing with increasing

mass. The Figure 4-32 surface plot shows the behaviour much better.

101

Similar conclusions can be done for Euler beam and double (or triple)

coupling results of frequencies larger than 1000 Hz. The rate of maximum reflection

frequency decrease with increasing mass is higher in double coupling than Euler

beam. The maximum transmission point around 200 Hz slides to higher frequencies

slowly with the same rate for both beam models.

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-30 Transmission loss of 30° incidence for Euler beam modelling, change in density

(blue-ρ = 2000 kg/m3, green-ρ = 6000 kg/m3, magenta-ρ = 10000 kg/m3)

102

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-31 Transmission loss of 30° incidence for double coupling modelling, change in density

(blue-ρ = 2000 kg/m3, green-ρ = 6000 kg/m3, magenta-ρ = 10000 kg/m3)

Figure 4-32 Transmission loss of normal incidence for line mass modelling vs. density

103

4.2.2 Sensitivity to Bending Stiffness

The effect of joint beam bending stiffness is studied below. The beam joint

transmission loss is plotted for 30° incidence angle on the following figures by

changing Iζ. It is shown that transmission loss changes significantly with inertia.

Total transmission frequency does not change with inertia except a transmission

region. The transmission loss at very low inertia is equivalent to line mass modelling.

As the inertia increases maximum transmission loss frequency gets lower which

means if the joint beam has higher inertia, much lower frequency waves can be

transmitted. At high frequencies, transmission loss increases as inertia increases.

However, at an inertia value, zero transmission loss is obtained at frequencies above

1000 Hz.

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-33 Transmission loss of 30° oblique angle for Euler Beam, change in Iζ

(blue - Iζ = 2.10-6 m4, green - Iζ = 4.10-5 m4, magenta - Iζ = 1.10-4 m4)

104

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-34 Transmission loss of 30° oblique angle for triple coupling, change in Iζ

(blue - Iζ = 2.10-6 m4, green - Iζ = 4.10-5 m4, magenta - Iζ = 1.10-4 m4)

Figure 4-35 Transmission loss of 30° oblique angle for Euler Beam vs. Iζ vs. frequency

105

4.2.3 Sensitivity to Lateral Bending Stiffness

The effect of joint beam lateral bending stiffness is studied below. The beam

joint transmission loss is plotted for 30° incidence angle on the following figures by

changing Iη. It is shown that transmission loss changes significantly with inertia.

Total transmission frequency changes rapidly with inertia. The transmission loss at

very low inertia is equivalent to line mass modelling. As the inertia increases

maximum transmission loss frequency gets lower which means if the joint beam has

higher lateral bending stiffness, much lower frequency waves can be transmitted. At

high frequencies, transmission loss increases as inertia increases.

Figure 4-36 Transmission loss of 30° oblique angle for double coupling vs. Iη vs. frequency

106

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-37 Transmission loss of 30° oblique angle for double coupling, change in Iη

(blue - Iη = 2.10-6 m4, green - Iη = 4.10-5 m4, magenta - Iη = 1.10-4 m4)

Figure 4-38 Transmission loss of 30° oblique angle for triple coupling vs. Iη vs. frequency

107

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-39 Transmission loss of 30° oblique angle for triple coupling, change in Iη

(blue - Iη = 2.10-6 m4, green - Iη = 4.10-5 m4, magenta - Iη = 1.10-4 m4)

4.2.4 Sensitivity to Torsional Stiffness

The effect of joint beam torsional stiffness is studied below. Increase in

torsional stiffness increases the total transmission frequency for all frequencies. Both

the total and maximum transmission points are moving to higher frequencies as

torsional stiffness increases. At 10000Hz, the change in torsional constant does not

change the transmission loss much. Transmission loss is more sensitive to changes in

torsional stiffness than changes in bending stiffness.

108

101 102 103 1040

10

20

30

40

50

60

70

80

R [d

B]

Frequency [Hz] Figure 4-40 Transmission loss of 30° oblique angle for Euler beam, change in J

(blue - J = 2.10-7 m4, green - J = 4.10-6 m4, magenta - J = 1.10-5 m4)

101 102 103 1040

10

20

30

40

50

60

70

80

R [d

B]

Frequency [Hz] Figure 4-41 Transmission loss of 30° oblique angle triple coupling, change in J

(blue - J = 2.10-7 m4, green - J = 4.10-6 m4, magenta - J = 1.10-5 m4)

109

Figure 4-42 Transmission loss of 30° oblique angle for Euler Beam vs. J vs. frequency

4.2.5 Sensitivity to Vertical Shear Centre Offset from Plate Surface

The effect of joint beam shear centre vertical offset from plate surface (sz), is

studied below. Line mass and Euler beam modelling do not include this parameter.

Rectangular beam properties used up to here reflect this effect in the same way for

triple and double coupling modelling.

110

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-43 Transmission loss of 30° oblique angle triple coupling, change in sz

(blue - sz = 0 m, green - sz = 0.08 m, magenta - sz = 0.2 m)

Figure 4-44 Transmission loss of 30° oblique angle for triple coupling vs. sz vs. frequency

111

The total transmission point is moving to lower frequencies as torsional

stiffness increases and the maximum transmission point moves also slightly to lower

frequencies. It is seen from the above figures that if this parameter is taken as zero,

the calculated transmission loss is smaller than the real value. This means that the

transmitted energy will be calculated higher than the exact one.

4.2.6 Sensitivity to Horizontal Shear Centre Offset from Plate Surface

The effect of joint beam shear centre horizontal offset from the point with no

x movement on the plate surface is studied in this paragraph. Line mass, Euler beam

and double coupling modelling do not include this parameter. C beam properties

given in Appendix C Table C-2 are used here to reflect the effect for triple coupling

modelling.

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-45 Transmission loss of 30° incidence for triple coupling modelling, change in sx

(blue - sx = 0 m, green - sx = 0.025 m, magenta - sx = 0.05 m)

112

Figure 4-46 Transmission loss of 30° oblique angle for triple coupling vs. sx vs. frequency

The effect of horizontal offset is not seen at low frequencies. It is seen from

the above figures that if this parameter is taken as zero, the calculated transmission

loss is lower. This means that the transmitted energy will be higher with the zero

horizontal distance assumption.

4.2.7 Sensitivity to Warping Coefficient

The sensitivity to warping coefficient is studied below in order to show the

effect on transmission loss if warping is neglected. Line mass and Euler beam do not

include this parameter. C beam properties are used here to reflect this effect for

double and triple coupling modelling.

113

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-47 Transmission loss of 30° incidence for double coupling modelling, change in Γo

(blue - Γo = 1.10-10 m4, green - Γo = 5.10-9 m4, magenta - Γo = 1.10-8 m4)

101 102 103 1040

10

20

30

40

50

60

70

80

Frequency [Hz]

R [d

B]

Figure 4-48 Transmission loss of 30° incidence for triple coupling modelling, change in Γo

(blue - Γo = 1.10-10 m4, green - Γo = 5.10-9 m4, magenta - Γo = 1.10-8 m4)

114

Figure 4-49 Transmission loss of 30° oblique angle for double coupling vs. Γo vs. frequency

Figure 4-50 Transmission loss of 30° oblique angle for triple coupling vs. Γo vs. frequency

115

The results shows that warping does not have a significant effect for

frequencies less than 100 Hz. Transmission loss is decreasing after this frequency as

warping coefficient increases. The total transmission point seen in double coupling

modelling results is moving to higher frequencies with the increase in warping

coefficient. However, the maximum transmission point seen in triple coupling

modelling results does not change much.

4.3 Coupling Loss Factors for Three Plates Coupling

Transmission loss factors calculated for the three plate coupling joint are

given below with surface plots. Properties are the ones in Appendix C Table C-1.The

general behaviour of the transmission loss between facing plates is similar with the

two plates coupling as shown with the following figures.

Figure 4-51 Transmission loss between plate 1 and 2 vs. frequency and oblique angle for line mass modelling

116

Figure 4-52 Transmission loss between plate 1 and 2 vs. frequency and oblique angle for Euler beam modelling

Figure 4-53 Transmission loss between plate 1 and 2 vs. frequency and oblique angle for double coupling

117



118


It is seen that the transmission loss between plates 1-2 and 1-3 are showing

the same characteristics, but the peak values of transmission loss between plates 1-2

are higher.

The following figures show the transmission loss between second plate and

the tilted plate for three joint modelling techniques. However, consideration of

flexibility in x direction and joint beam shear centre vertical offset from plate

surface, sz, does not result in the same deviation in transmission loss as two plates

coupling. The surface plots shows that the transmission characteristics are different

than the transmission between plates 1-2 or 1-3.

119



120


Figure 4-60 CLF between first and second plates for mass (green), beam joint (blue) and open section beam (red)

121

Coupling loss factors calculated for the properties given in Appendix C Table

C-1 for three plates coupling case is given in this section. Figure 4-60 shows that

CLF between first and second plates for beam and triple coupling modelling are

approximately the same at low frequencies. However, as the frequency gets higher,

beam modelling CLF results are closer to line mass modelling. It is seen that line

mass modelling results in higher CLF values for frequencies lower than 1000 Hz but

after this frequency triple coupling modelling CLF are higher.

The general behaviour explained above is the same for CLF between first and

third plates also. The triple coupling modelling has some peaks which are not

encountered in the first and second plate CLF.

Figure 4-61 CLF between first and third plates for mass (green), beam joint (blue) and open section beam (red)

122

Figure 4-62 CLF between second and third plates for mass (green), beam joint (blue) and open section beam (red)

The CLF between second and third plates shows different behaviour than the

previous plots for coupling between plates 1-2 and 1-3. The line mass modelling has

a peak at 1000 Hz while on the other plots there is a decrease at this point. Triple

coupling and Euler beam modelling gives higher CLF values up to 600 Hz and after

1500 Hz. The line mass modelling CLF is higher in between these frequencies.

Triple coupling and Euler beam CLF are approximately the same up to 100 Hz and

getting closer to each other again at 6000 Hz.

123

CHAPTER V

5. SEA APPLICATIONS

5.1 Two Plates

The coupling loss factors calculated as given in previous chapters are used to

solve a two plate with line junction system with SEA. Properties are given in

Appendix B Table B-1. Excitation on the first plate is plane wave acoustic excitation

ini (wt-k x)yPoe sink y ⋅ with wavenumbers ky = mπ/Lj kin=ω/Uo where m=1, Po=1 Pa

and Uo=190 m/s. The following figures are showing the results.

1.E-22

1.E-211.E-20

1.E-191.E-18

1.E-171.E-16

1.E-15

1.E-141.E-13

1.E-121.E-11

1.E-101.E-09

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy SEA PL1 (L.Mass)

SEA PL2 (L.Mass)SEA PL1 (E.Beam)SEA PL2 (E.Beam)

Figure 5-1 SEA solution of two plates system for line mass and beam joints

124

1.E-221.E-21

1.E-201.E-191.E-18

1.E-171.E-161.E-151.E-14

1.E-131.E-121.E-11

1.E-101.E-091.E-08

1.E-071.E-06

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

SEA PL1 (L.Mass)

SEA PL2 (L.Mass)

E1 (L.Mass)

E2 (L.Mass)

SEA PL1 (E.Beam)

SEA PL2 (E.Beam)

E1 (E.Beam)

E2 (E.Beam)

Figure 5-2 Comparison with mean energy for diffuse wave field CLF, η12(ω)

1.E-221.E-211.E-201.E-191.E-181.E-171.E-161.E-151.E-141.E-131.E-121.E-111.E-101.E-091.E-081.E-071.E-06

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

SEA PL1 (L.Mass)

SEA PL2 (L.Mass)

E1 (L.Mass)

E2 (L.Mass)

SEA PL1 (E.Beam)

SEA PL2 (E.Beam)

E1 (E.Beam)

E2 (E.Beam)

Figure 5-3 Comparison with mean energy for normal incidence CLF, η12(0,ω)

125

The mean energy of plate 1 is approximately the same for both modelling.

However, the response calculated for plate 2 differs around 1000 Hz. The energy

calculated for line mass joint case is lower than the one for Euler beam modelling.

This means with beam assumption more energy is transformed to the second plate.

The SEA results for three joint beam modelling case are compared with the

mean energy calculated with the solution method used for transmission loss factor

derivation in Figure 5-2. The plates are taken very big to simulate semi-infinite

plates. It is seen that SEA with beam joint does not catch the dip around 1000 Hz but

line mass modelling does. This is because of input power characteristics. The mean

energy is calculated for normally incident input wave, however SEA assumes

random input in all incidence angles. As it was shown in the previous chapter, beam

modelling is sensitive to the angle. In order to see the situation clearly, the diffuse

transmission loss is not taken for the SEA calculation and the subsystem energies are

calculated by using transmission loss for normal incidence. As seen in Figure 5-3,

beam modelling results in this case catch the dip like line mass modelling. Therefore,

if proper transmission loss is used, SEA may give good results also for excitations

other than random.

5.2 Six Plates

The calculated coupling loss factors are used to solve six plates connected

end to end by line joints. The open ends of the plates are free. The SEA results are

compared with Dynamic Stiffness method results for 6 plate case. The input wave is

on the first plate and normal incidence is assumed in dynamic stiffness calculations.

Properties of Appendix B Table B-1 are used for the line mass and beam joint

models.

126

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-4 SEA results for 6 plates array, line mass joint. (blue - plate 1, green - plate 2, cyan - plate 3, red - plate 4, yellow - plate 5, black - plate 6)

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-5 SEA results for 6 plates array, Euler beam joint. (blue - plate 1, green - plate 2, cyan - plate 3, red - plate 4, yellow - plate 5, black - plate 6)

127

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-6 SEA results for 6 plates array, double coupling. (blue - plate 1, green - plate 2, cyan - plate 3, red - plate 4, yellow - plate 5, black - plate 6)

The above figures Figure 5-4 to Figure 5-6 show the mean energy values for

six plates system. It is seen that less energy is transmitted to the next plate by line

mass modelling. Euler beam and double (or triple) coupling results are the same up

to 200 Hz and they are reaching to same values again at 10000Hz. In between, two

modelling techniques produce different mean energy values.

SEA results are compared with dynamic stiffness method results on the below

figures from Figure 5-7 to Figure 5-11. The same joint beam modelling technique is

used in dynamic stiffness calculations for comparison. The first and the last panel

results are used in the figures.

Energy of plate 1 is not much different in all methods. However, the

difference between methods can be seen in the energy of plate 6. SEA can follow the

general trend of dynamic stiffness method. It is seen that double coupling modelling

results are closer to the dynamic stiffness for diffuse wave field (average) coupling

loss factor η12(ω).

128

The idea of performing SEA calculations with normal incidence coupling loss

factor is tried for six plate case also. It is seen in Figure 5-8 that the results of Euler

beam joint do not follow the dip around 1000 Hz. Figure 5-9 gives the results of

same SEA calculation with transmission loss for normal incidence and it is shown

that beam joint gives better results with this replacement. Comparison of dynamic

stiffness results and triple coupling modelling results shows a similar situation as

seen in Figure 5-10 and Figure 5-11. supports this conclusion as seen in Figure 5-8

and Figure 5-9. SEA calculation with triple coupling transmission loss for normal

incidence results are closer to dynamic stiffness method results for frequencies less

than 2000 Hz. For higher frequencies than 2000 Hz, SEA results calculated with

diffuse wave field coupling loss factor are better.

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-7 Comparison with dynamic stiffness results for 6 plates array, line mass joint, diffuse wave field CLF - η12(ω).

(blue - plate 1 SEA, green - plate 6 SEA, cyan - plate 1 Dyn.Stf, red - plate 6 Dyn.Stf)

129

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-8 Comparison with dynamic stiffness results for 6 plates array, Euler beam, diffuse wave field CLF - η12(ω). (blue - plate 1 SEA, green - plate 6 SEA, cyan - plate 1 Dyn.Stf, red - plate 6 Dyn.Stf)

102 103 10410-35

10-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-9 Comparison with dynamic stiffness results for 6 plates array, Euler beam, normal incidence CLF - η12(ω,0). (blue - plate 1 SEA, green - plate 6 SEA, cyan - plate 1 Dyn.Stf, red - plate 6 Dyn.Stf)

130

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-10 Comparison with dynamic stiffness results for 6 plates array, double coupling, diffuse wave field CLF - η12(ω). (blue - plate 1 SEA, green - plate 6 SEA, cyan - plate 1 Dyn.Stf, red - plate 6 Dyn.Stf)

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-11 Comparison with dynamic stiffness results for 6 plates array, double coupling, normal incidence CLF - η12(ω,0). (blue - plate 1 SEA, green - plate 6 SEA, cyan - plate 1 Dyn.Stf, red - plate 6 Dyn.Stf)

131

5.3 Closed Structure

Figure 5-12 Closed structure

A closed structure is obtained by joining the last plate to the first one. This is

to simulate the classical skin-stringer fuselage section between two frames. The

results are given for 12 plate section.

The power input is given to Panel 1 and it is identical to the previous

paragraphs. The mean energy values for all plates are calculated by SEA and

dynamic stiffness method.

1st Plate

6th Plate

132

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-13 SEA results for 6 plates array, line mass joint. (blue - plate 1, green - plate 2, cyan - plate 3, red - plate 4, yellow - plate 5, black - plate 6)

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-14 SEA results for 6 plates array, Euler beam joint. (blue - plate 1, green - plate 2, cyan - plate 3, red - plate 4, yellow - plate 5, black - plate 6)

133

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-15 SEA results for 6 plates array, double coupling. (blue - plate 1, green - plate 2, cyan - plate 3, red - plate 4, yellow - plate 5, black - plate 6)

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-16 Comparison with dynamic stiffness results for 1st and 6th plates, line mass (blue - plate 1 SEA, green - plate 6 SEA, cyan - plate 1 Dyn.Stf, red - plate 6 Dyn.Stf)

134

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-17 Comparison with dynamic stiffness results for 1st and 6th plates, Euler beam (blue - plate 1 SEA, green - plate 6 SEA, cyan - plate 1 Dyn.Stf, red - plate 6 Dyn.Stf)

102 103 10410-30

10-25

10-20

10-15

10-10

10-5

Frequency [Hz]

Mea

n E

nerg

y

Figure 5-18 Comparison with dynamic stiffness results for 1st and 6th plates, open section channel (blue - plate 1 SEA, green - plate 6 SEA, cyan - plate 1 Dyn.Stf, red - plate 6 Dyn.Stf)

135

1.E-30

1.E-28

1.E-26

1.E-24

1.E-22

1.E-20

1.E-18

1.E-16

1.E-14

1.E-12

1.E-10

1.E-08

1.E-06

100 1000 10000

Frequency [Hz]

Mea

n E

nerg

y

PL6 AutoSEASEA PL6 (L.Mass)SEA PL6 (E.Beam)SEA PL6 (3c)Dyn Stiff. PL6

Figure 5-19 Comparison with AutoSEA and dynamic stiffness results for 6th plate

It is shown in the figures given in this section; triple coupling gives better

results, which means closer to dynamic stiffness results, for closed section structure.

The Figure 5-19 above combines the results for 6th plate, where the power input is on

the 1st plate, for all joint modelling types, dynamic stiffness and the commercial

program AutoSEA. It is seen that AutoSEA results are higher than this studies

results. The modelling technique that AutoSEA use is not well known. The only hint

in hand is that it does not ask for the coupling beam product of inertia, offset of beam

shear centre from plate surface and warping constant which means that it does not

include warping, coupling between rotational and translational motions. AutoSEA is

used to validate the results calculated in this study for fuselage structure.

136

5.4 Fuselage Structure

Figure 5-20 Fuselage structure

A fuselage structure is obtained by joining the closed skin-stringer fuselage

sections via beam joints. These beams are representing the frames of the fuselage.

The power input is given to Panel 1 and it is identical to the previous

paragraphs. The mean energy values for all plates are calculated by SEA and

compared with AutoSEA results.

1st Plate

6th Plate

13th Plate

137

1.E-27

1.E-25

1.E-23

1.E-21

1.E-19

1.E-17

1.E-15

1.E-13

1.E-11

1.E-09

100 1000 10000

Frequency [Hz]

Mea

n E

nerg

yPL6 AutoSEA

SEA PL6 (L.Mass)

SEA PL6 (E.Beam)

SEA PL6 (3c)

Figure 5-21 Comparison with AutoSEA results for 6th plate

The results are similar in characteristics with the closed section case. This is

because of the properties of the frame structure. It is stiffer than the beam used

between panels of closed section, which can be called as stringers. Transmission via

frame is lower than stringer. Therefore the more energy flows in the section between

frames.

The next figure shows the mean energy values calculated with SEA for the

panel 1 which has the input power, the panel 2 connected to it within the section, and

panel 13, which is the neighbouring panel after frame. It is seen that energy of Plate

13 is lower than Plate 2.

138

1.E-191.E-181.E-171.E-161.E-151.E-141.E-131.E-121.E-111.E-101.E-091.E-08

100 1000 10000

Frequency [Hz]

Mea

n E

nerg

ySEA PL1SEA PL2SEA PL13

Figure 5-22 Comparison of energies of 1st, 2nd and 13th plate

5.5 Closed Structure with Floor Panel

Figure 5-23 Closed section with floor panel

1st Plate

6th Plate

Floor Panel (13th Plate)

139

A floor panel is introduced in the closed section and the previously derived 3

panel joint transmission loss equations are used in SEA calculations. The results are

given for 6th panel and the floor panel.

It is seen that as the structure becomes complex the results of different

modelling techniques are getting closer. The results of 6th panel for closed section

with and without floor panel is compared in Figure 5-26. Addition of floor panel

increases the stiffness of the structure and this modification increases the mean

energy of the panel.

1.E-211.E-201.E-191.E-181.E-171.E-161.E-151.E-141.E-131.E-121.E-111.E-101.E-091.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

PL6 AutoSEASEA PL6 (L.Mass)SEA PL6 (E.Beam)SEA PL6 (3c)

Figure 5-24 Comparison of energies of 6th plate

140

1.E-211.E-201.E-191.E-181.E-171.E-161.E-151.E-141.E-131.E-121.E-111.E-101.E-091.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

PL13 AutoSEA

SEA PL13 (L.Mass)

SEA PL13 (E.Beam)SEA PL13 (3c)

Figure 5-25 Comparison of energies of floor panel (13th plate)

1.E-27

1.E-25

1.E-23

1.E-21

1.E-19

1.E-17

1.E-15

1.E-13

1.E-11

1.E-09

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

PL6 AutoSEA w /o Floor

PL6 AutoSEA w ith Floor

SEA PL6 w /o Floor (L.Mass)

SEA PL6 w ith Floor (L.Mass)

SEA PL6 w /o Floor (E.Beam)

SEA PL6 w ith Floor (E.Beam)

SEA PL6 w /o Floor (3c)

SEA PL6 w ith Floor (3c)

Figure 5-26 Comparison of energies of 6th plate for closed section with and without floor panel

141

5.6 Fuselage Structure with Floor Panel

Figure 5-27 Fuselage structure with floor panel

Floor panel is introduced to the three section fuselage structure and the

previously derived 3 panel joint transmission loss equations are used in SEA

calculations. The results are given for 6th panel and the floor panel, 37th panel, for the

first section and the corresponding panels, 30 and 39, in the third section.

Three section fuselage panels mean energy values are less than the one

section case at low frequencies. The difference is decreasing as the frequency

increases. The general response characteristics of corresponding panels at the first

and last section are the same. However, the mean energy values of last section are

considerable small from the first section results.

1st Plate

6th Plate

37th Plate

30th Plate

39th Plate

142

1.E-21

1.E-201.E-19

1.E-18

1.E-171.E-16

1.E-15

1.E-14

1.E-131.E-12

1.E-11

1.E-101.E-09

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy PL6 AutoSEA

SEA PL6 (L.Mass)SEA PL6 (E.Beam)

SEA PL6 (3c)


1.E-211.E-201.E-191.E-181.E-171.E-161.E-151.E-141.E-131.E-121.E-111.E-101.E-091.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

PL37 AutoSEA

SEA PL37 (L.Mass)

SEA PL37 (E.Beam)

SEA PL37 (3c)

Figure 5-29 Comparison of energies of floor panel of first section (37th plate)

143

1.E-26

1.E-24

1.E-22

1.E-20

1.E-18

1.E-16

1.E-14

1.E-12

1.E-10

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n E

nerg

y

PL30 AutoSEA

SEA PL30 (L.Mass)

SEA PL30 (E.Beam)

SEA PL30 (3c)


1.E-26

1.E-24

1.E-22

1.E-20

1.E-18

1.E-16

1.E-14

1.E-12

1.E-10

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

PL39 AutoSEA

SEA PL39 (L.Mass)

SEA PL39 (E.Beam)

SEA PL39 (3c)

Figure 5-31 Comparison of energies of floor panel of third section (39th plate)

144

1.E-27

1.E-25

1.E-23

1.E-21

1.E-19

1.E-17

1.E-15

1.E-13

1.E-11

1.E-09

100 1000 10000

Frequency [Hz]

Mea

n E

nerg

yPL6 AutoSEA w /o Floor

PL6 AutoSEA w ith Floor

SEA PL6 w /o Floor (L.Mass)

SEA PL6 w ith Floor (L.Mass)

SEA PL6 w /o Floor (E.Beam)

SEA PL6 w ith Floor (E.Beam)

SEA PL6 w /o Floor (3c)

SEA PL6 w ith Floor (3c)

Figure 5-32 Comparison of energies of 6th plate for closed section with and without floor panel

The results calculated for the fuselage structure with and without floor are

compared in the figure above. Mean energy of the 6th panel increases with the

addition of floor panel. The differences between results are larger at high frequency

region.

145

5.7 Closed Structure with Acoustic Cavity

Figure 5-33 Closed structure with acoustic cavity

An acoustic cavity is defined in the closed section representing the inner

volume of air. The differences in the results with this model change are studied with

the below figures.

Addition of acoustic volume closes the difference between modelling

techniques results after 2500 Hz. Since the modal density is very low at lower

frequencies, the effect of acoustic cavity is not sensed by SEA. At 6000 Hz, all

curves have a peak and this frequency corresponds to the critical frequency of plates.

Results of closed section with and without acoustic cavity are compared in

Figure 5-35. Acoustic cavity does not change the low frequency results. After

1000Hz, the results become different and the mean energy of the structure with

acoustic volume gets higher than the structure without acoustic volume.

146

1.E-22

1.E-21

1.E-20

1.E-19

1.E-18

1.E-17

1.E-16

1.E-15

1.E-14

1.E-13

1.E-12

1.E-11

1.E-10

1.E-09

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

PL6 AutoSEA + V

SEA PL6+V (L.Mass)SEA PL6+V (E.Beam)

SEA PL6+V (3c)

Figure 5-34 Comparison of energies of 6th plate for closed section with acoustic cavity

1.E-28

1.E-26

1.E-24

1.E-22

1.E-20

1.E-18

1.E-16

1.E-14

1.E-12

1.E-10

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n E

nerg

y

PL6 AutoSEA

PL6 AutoSEA + V

SEA PL6 (L.Mass)

SEA PL6+V (L.Mass)

SEA PL6 (E.Beam)

SEA PL6+V (E.Beam)

SEA PL6 (3c)

SEA PL6+V (3c)

Figure 5-35 Comparison of energies of 6th plate for closed section with and without acoustic cavity

147

5.8 Fuselage Structure with Acoustic Cavity

Figure 5-36 Fuselage structure with acoustic cavity

An acoustic cavity is defined in the fuselage section representing the inner

volume of air. The differences in the results with this model change are studied with

the below figures.

Addition of acoustic volume closes the difference between modelling

techniques results after 3000 Hz. At 6000 Hz, all curves have a peak and this

frequency corresponds to the critical frequency of plates.

Results of fuselage section with and without acoustic cavity are compared in

Figure 5-38. Acoustic cavity does not change the low frequency results. After

1000Hz, the results become different and the mean energy of the structure with

acoustic volume gets higher than the structure without acoustic volume.

148

1.E-23

1.E-21

1.E-19

1.E-17

1.E-15

1.E-13

1.E-11

1.E-09

100 1000 10000

Frequency [Hz]

Mea

n En

ergy PL6 AutoSEA + V

SEA PL6+V (L.Mass)

SEA PL6+V (E.Beam)

SEA PL6+V (3c)

Figure 5-37 Comparison of energies of 6th plate for fuselage section with acoustic cavity

1.E-28

1.E-26

1.E-24

1.E-22

1.E-20

1.E-18

1.E-16

1.E-14

1.E-12

1.E-10

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

PL6 AutoSEA

PL6 AutoSEA + V

SEA PL6 (L.Mass)

SEA PL6+V (L.Mass)

SEA PL6 (E.Beam)

SEA PL6+V (E.Beam)

SEA PL6 (3c)

SEA PL6+V (3c)

Figure 5-38 Comparison of energies of 6th plate for fuselage section with and without acoustic cavity

149

5.9 Closed Structure with Floor Panel and Two Acoustic Cavities

Figure 5-39 Closed structure with floor panel and two acoustic cavities

Two acoustic cavities are defined in the closed section representing the inner

volume of air below and under the floor. The differences in the results with this

model change are studied with the below figures.

Addition of acoustic volumes does not change the results much but increases

slightly. Results of closed section for 6th plate with and without acoustic cavities are

compared in Figure 5-41.

The results for floor panel given in Figure 5-42 shows that addition of

acoustic volumes decrease the mean energy. Since there are acoustic volumes on

both sides of the floor panel, the energy loss from the panel increases and the mean

energy of the panel decreases.

150

1.E-211.E-20

1.E-191.E-18

1.E-171.E-16

1.E-15

1.E-14

1.E-131.E-12

1.E-111.E-10

1.E-09

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n E

nerg

yPL6 AutoSEA + V

SEA PL6+V (L.Mass)

SEA PL6+V (E.Beam)

SEA PL6+V (3c)

Figure 5-40 Comparison of energies of 6th plate for closed section with floor panel and two acoustic cavities

1.E-211.E-20

1.E-191.E-18

1.E-171.E-161.E-15

1.E-141.E-131.E-12

1.E-111.E-10

1.E-091.E-08

100 1000 10000

Frequency [Hz]

Mea

n E

nerg

y

PL6 AutoSEA

PL6 AutoSEA + V

SEA PL6 (L.Mass)

SEA PL6+V (L.Mass)

SEA PL6 (E.Beam)

SEA PL6+V (E.Beam)

SEA PL6 (3c)

SEA PL6+V (3c)


151

1.E-211.E-201.E-191.E-181.E-171.E-161.E-15

1.E-141.E-131.E-121.E-111.E-10

1.E-091.E-08

100 1000 10000

Frequency [Hz]

Mea

n E

nerg

yPL13 AutoSEAPL13 AutoSEA + VSEA PL13 (L.Mass)SEA PL13+V (L.Mass)SEA PL13 (E.Beam)SEA PL13+V (E.Beam)SEA PL13 (3c)SEA PL13+V (3c)

Figure 5-42 Comparison of energies of floor panel (13th plate) for closed section with floor panel and two acoustic cavities

5.10 Fuselage Structure with Floor and Two Acoustic Cavities

Figure 5-43 Fuselage structure with two acoustic cavities

152

Two acoustic cavities are defined in the closed section representing the inner

volume of air below and under the floor. The differences in the results with this

model change are studied with the below figures.

Addition of acoustic volumes does not change the results much but increases

slightly at low frequencies and decreases at high frequencies. Results of closed

section for 6th plate with and without acoustic cavities are compared in Figure 5-45.

The results for floor panel given in Figure 5-46 shows that addition of

acoustic volumes decrease the mean energy at high frequencies. Since there are

acoustic volumes on both sides of the floor panel, the energy loss from the panel

increases and the mean energy of the panel decreases.

1.E-22

1.E-20

1.E-18

1.E-16

1.E-14

1.E-12

1.E-10

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

PL6 AutoSEA + V

SEA PL6+V (L.Mass)

SEA PL6+V (E.Beam)

SEA PL6+V (3c)


153

1.E-22

1.E-20

1.E-18

1.E-16

1.E-14

1.E-12

1.E-10

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

PL6 AutoSEA

PL6 AutoSEA + V

SEA PL6 (L.Mass)

SEA PL6+V (L.Mass)

SEA PL6 (E.Beam)

SEA PL6+V (E.Beam)

SEA PL6 (3c)

SEA PL6+V (3c)


1.E-22

1.E-20

1.E-18

1.E-16

1.E-14

1.E-12

1.E-10

1.E-08

100 1000 10000

Frequency [Hz]

Mea

n En

ergy

PL37 AutoSEA

PL37 AutoSEA + V

SEA PL37 (L.Mass)

SEA PL37+V (L.Mass)

SEA PL37 (E.Beam)

SEA PL37+V (E.Beam)

SEA PL37 (3c)

SEA PL37+V (3c)

Figure 5-46 Comparison of energies of floor panel (37th plate) for closed section with floor panel and two acoustic cavities

154

CHAPTER VI

CONCLUSIONS The analysis of the dynamic behaviour of complex structures by the

Statistical Energy Analysis Method is the subject of this thesis. The most important

steps in SEA Analysis are the determination of the transmission loss and the coupling

loss factors.

The coupling loss factors required for the power balance equations of SEA

are found by the wave transmission technique. The equations for coupling loss

factors are derived for two semi-infinite plates connected first via a line joint. The

joint is modelled by a line mass, an Euler beam, and an open section channel beam

having double and triple coupling. The transmission parameters were calculated for

all these joint types and used for SEA calculations. The differences between coupling

parameters calculated for all joint types are studied. The sensitivity of the coupling

parameters to system characteristics is also examined. The results indicate that open

section beam joint is not as efficient low pass filter as a line mass joint. Transmission

loss of the open section beam is lower than the line mass joint at higher frequencies.

Therefore, it is concluded that if an open section beam is modelled as a line mass, the

calculated transmitted energy will be lower than that for the actual open section case;

a conservative estimate will be obtained.

The transmission loss for oblique propagation is also studied. The change in

transmission loss with oblique incidence angle for beam and line mass joints is given

in separate surface plots. The angle dependency of transmission loss is presented in

literature by 2-D curves for some selected frequencies. Surface plots given in this

study presented, more clearly, the dependency on incidence angle and frequency with

155

one graph. It is shown that Euler and open section beam joints are affected more by

the incidence angle changes unlike the line mass joint.

Coupling loss factors are calculated by using the transmission loss for each

incidence angle. In SEA, coupling loss factors averaged over incidence angle are

used because of the random behaviour. Coupling loss of line mass joint does not

change significantly with incidence angle. Therefore, averaging the coupling loss

factor over incidence angle may be an acceptable assumption. However, CLF of

beam joint differs appreciably with incidence angle and the result of averaging does

not represent the actual joint characteristic well.

The calculated coupling loss factors averaged over all incidence angles are

compared with the classical SEA approach. The low frequency behaviour is

assumed, as the dynamic behaviour does not change in the high frequency range in

classical SEA. It is illustrated; however, that in reality the low frequency behaviour

of coupling losses is not the same as that in the high frequency region. Therefore, it

is proposed that more realistic behaviour can be obtained, if loss factors are

calculated for each frequency with the expressions derived in this study.

Sensitivity of transmission loss to density, bending stiffness, torsional

stiffness, and the beam shear centre offset from plate surface is studied. The distance

between shear centre and beam to plate connection point in x-direction, which is

assumed to be zero in literature, is included in the analysis. It is shown that if this

parameter is not taken as zero, low frequency behaviour will not effect besides the

high frequency transmission loss turns out to be larger. This means that the

transmitted energy will be overestimated with zero horizontal offset assumption. It is

also shown that transmission loss does not change with the direction of horizontal

offset. The reason is that transmission loss is a ratio of energy and the direction of

rotation, which is determined by the sign of sx, does not affect the amount of energy.

A new joint type is introduced as a three-plate-joint with a line junction and it

is also studied for three stiffener modelling types. The general behaviour of the

transmission losses between the first – second and the first - third plates do not show

significant differences when compared with just two plates coupled together.

However, transmission loss values for the second and third plates show peaks and

dips at different incidence angles and frequencies.

156

The coupling loss factor results are used to solve a two plate with line

junction system with SEA. The SEA results for two plates joint are compared with

the mean energy calculated with the solution method used for transmission loss

factor derivation. The mean energy is calculated for normally incident input wave;

however, SEA assumes random inputs as explained before. It is shown that if the

subsystem energies are calculated by using transmission loss for normal incidence

instead of diffuse wave field transmission loss, beam modelling gives better results.

It is concluded that if proper transmission loss is used, SEA may give good results

also for excitations other than random.

The analysis is continued by placing more stiffeners on the panel, which are

equally spaced. The calculated coupling loss factors are used to solve six plates

connected end to end by line joints. The SEA results are compared with Dynamic

Stiffness method results for this case. The results support the above conclusion for

six plate system also.

Further, a closed structure is obtained by joining the last plate to the first one.

This is to simulate the classical skin-stringer fuselage section between two frames.

The results are given for a 12-plate section. It is shown that triple coupling gives

better results for closed section structure at high frequencies. The results from all

joint modelling types, the dynamic stiffness method, and the commercial program

AutoSEA are compared.

The substructures are assembled to obtain a fuselage structure by joining the

closed skin-stringer fuselage sections via beam joints. These beams are assumed to

model the frames. Since dynamic stiffness formulation cannot be applied for this

assembly, it can not be used here for validation. Therefore, the calculated results for

fuselage structure are compared with AutoSEA results.

Generally the fuselage structure is studied with SEA for noise transmission

applications by using very coarse models. The application of SEA to the fuselage

structure, modelling the plate and stringers individually is presented for the first time

in literature. It is shown that the method can be used for structural vibration

transmission and this means SEA can be used for estimating the vibrational response

of fuselage panels.

157

The next step of assembly is adding a floor panel to the fuselage structure. It

is seen from the results that adding a floor panel changes the transmission

characteristics. Addition of floor panel increases the stiffness of the structure and this

modification increases the mean energy of the fuselage panels.

The closed structures studied are modified by adding an acoustic volume

inside representing the volume of air. Acoustic cavity does not change the low

frequency results of fuselage structure without floor panel. At high frequency region,

differences in the results become apparent and the mean energy of the structure with

acoustic volume gets higher than the structure without the acoustic volume. The

fuselage structures with floor panel require two acoustic volumes above and under

the floor. Addition of acoustic volumes does not appreciably change the results in the

case of this somewhat stiffer assembly.

SEA is developed for calculating mean energy values and energy flow in

complex structures. It is not an exact method. However, in this thesis it is shown that

calculating the parameters as given in this study increases the efficiency of the

method. Unlike the main assumption of the classical SEA, one does not have to limit

the analysis to random excitation. It is shown that if the coupling losses are

calculated for normal incidence only, SEA can give acceptable results for excitations

other than random.

For complex structures like fuselage structure, SEA is a very easy and

efficient method for the calculation of mean energies and for the observation of the

energy flow in the structure. It is shown that closer to exact results can be obtained

by using more advanced modelling techniques such as double and triple coupling

open section formulations for coupling between subsystems.

By following this idea, the study can be extended by investigating the effect

of in-plane waves, and shear deformation in plates on the transmission

characteristics. Timoshenko beam theory can be used for beam modelling by adding

shear deformation.

The further step of the study may be the investigation of the effects of

fuselage radius on the transmission characteristics by performing the formulation in

which the plate is replaced by a curved shell. The effects of the modelling technique

and the subsystem parameters can be studied.

158

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[55] N. Lalor, “An Introduction to Automotive NVH Seminar Notes”, İstanbul,

November 1998.

[56] R.S. Langley, Journal of Sound and Vibration, “A Derivation of the Coupling

Loss Factors Used in Statistical Energy Analysis”, 1990, 141 (2), 207-219.

[57] Y.K. Tso, C.H. Hansen, Journal of Sound and Vibration, “An Investigation of

the Coupling Loss Factor for A Cylinder/Plate Structure”, 1999, 41 (7), 831-843.

[58] D.A. Bies, S. Hamid, Journal of Sound and Vibration, “In Situ Determination

of Loss and Coupling Loss Factors by the Power Injection Method”, 1980, 70

(2), 187-204.

[59] S.H. Crandall, R. Lotz, Journal of Acoustical Society of America, “On the

Coupling Loss Factor in Statistical Energy Analysis”, 1971, 49 (1), 352-356.

[60] G.R. Khabbaz, Journal of Acoustical Society of America, “Comparison of

Mechanical Coupling Factor By Two Methods”, 1970, 47 (1), 392-393.

[61] A.J. Keane, Journal of Sound and Vibration, “A Note on Modal Summations

and Averaging Methods as Applied to Statistical Energy Analysis (SEA)”, 1993,

164 (1), 143-156.

[62] M.L. Lai, T.T. Soong, Journal of Engineering Mechanics, “Statistical Energy

Analysis of Primary-Secondary Structural Systems”, 1990, 116(11), 2400-2413.

164

[63] R. J. Craik, “Sound Transmission Through Buildings Using Statistical Energy

Analysis”, Gower Publishing, 1996.

[64] Y. Yaman, Journal of Sound and Vibration, “Vibrations of Open-Section

Channels: A Coupled Flexural and Torsional Wave Analysis”, 1997, 204 (1),

131-158.

[65] Y. Yaman, 6. Ulusal Makina Teorisi Sempozyumu, Eylül, Trabzon, “Kirişlerde

Birlikte Var Olan Eğilme-Burulma Titreşimlerinin Analizi”, 1993, 423-430.

[66] Y. Yaman, 8. Ulusal Makina Teorisi Sempozyumu, Gaziantep Üniversitesi,

Gaziantep, “Elastik Sınır Koşullarına Sahip Bir Açık Kesitli Kirişte Zorlanmış,

Bağlaşık Titreşimlerin Analitik Modellenmesi", 1997, 133-142.

[67] Y. Yaman, Makina Tasarım ve İmalat Dergisi, “Bir Arada Oluşan, Birbirine

Dik İki Eksendeki Eğilme Titreşimleri Ve Burulma Titreşimlerinin Analitik

Modellemesi", 1997, 3 (3), 111-120.

[68] T. Çalışkan, Y. Yaman, I. Uluslararası Havacılık ve İleri Teknolojiler

Sempozyumu, Maslak, İstanbul, “Kombine Zorlama Altındaki Sonlu Plakların

Analitik Modellenmesi”, 1995, 492-500.

[69] S. P. Timoshenko, S. W. Krieger, “Theory of Plates and Shells”, Mcgraw-Hill

Inc., Second Edition, 1959.

[70] F. Fahy, “Sound and Structural Vibration, Radiation, Transmission and

Response”, Academic Press, 1985.

165

APPENDIX A

A. DYNAMIC STIFFNESS METHOD

Dynamic Stiffness Technique is one of the methods used to study the

vibrations of complex structures. Langley [17] has used this technique for the

analysis of stiffened shell structures. The resulting differential equations are solved

exactly to yield the dynamic stiffness matrix and the loading vector for the element.

Any number of elements may be assembled to model the cross section of a built up

structure such as an aircraft fuselage.

Figure A- 1 Panel array

166

The panels are simply supported along longitudinal edges. The stiffeners are

open section channel beams with triple coupling.

A.1. Dynamic Stiffness Solution

4 4 4 2

p4 2 2 4 2

w w w wD 2 m p(x, y, t)x x y y t

∂ ∂ ∂ ∂+ + + = ∂ ∂ ∂ ∂ ∂ (A.1)

where

yx ikin yikin x i top(x, y, t) P e e e−− ω= ⋅ ⋅ ⋅ (A.2)

Then the following form is assumed for the panel deflection in z direction.

m m mm m

w(x, y, t) X (x, t)sin(m y / Lj) X (x, t)sin(k y)= π =∑ ∑ (A.3)

Substitute this formula to the differential equation of panel vibrations.

Multiplying both sides by sin(my/Lj) and integrating over y results in the following.

Lj'''' 2 '' 4m m m m m m m0

2DX 2k DX Dk X X p(x, y, t) sin(k y) dyLj

− + +ρ = ⋅ ⋅∫!! (A.4)

Name the integral as Hm.

yLj ikin ym y 0

2H (kin ) e sin(m y / Lj) dyLj

−= ⋅ π ⋅∫ (A.5)

Solution to this equation may be written as a sum of a complementary

function and a particular integral.

nm4

k xm nm mp m y

n 1X (x) A e X (x) H (kin )

== + ⋅∑ (A.6)

167

where Anm are the integration constants and knm terms are the four roots of the

following equation in k.

4 2 2 4 2m mDk 2k Dk Dk 0− + −ρω = (A.7)

2p4

Bm

k Dω

= (A.8)

4 2 2 4 4m m B

2 21 B m

2 22 B m

k 2k k k k

k k k

k k k

− + =

= ± +

= ± − +

(A.9)

Define Ain as the following.

22 2 2in x yA D kin kin = ⋅ + −ρω (A.10)

Particular solution is the below equation.

xx

ikin xikin xo o

mp 22 2 2 inx y

P e PX (x) eAD kin kin

−−⋅= = ⋅

⋅ + −ρω

(A.11)

To satisfy the boundary conditions in x direction, the following vector is

defined.

T ' 'm m m m mu X (0) X (0) X (L) X (L) = (A.12)

The solution equation of Xm(x) can be written as below.

m m m np yu P A u H(kin )= + ⋅ (A.13)

where

168

[ ]Tm 1m 2m 3m 4mA A A A A= (A.14)

3m1m 2m 4m

3m1m 2m 4m

1m 2m 3m 4mT1m k Lk L k L k L

k Lk L k L k L1m 2m 3m 4m

1 1 1 1k k k k

Pe e e e

k e k e k e k e

=

(A.15)

x xikin L ikin LT omp x x

in

Pu 1 i kin e i kin e

A− − = ⋅ − ⋅ − ⋅ ⋅ (A.16)

The shear force and the bending moment for the deflection Xm(x) are given

below.

''' 2 'm m y mV (x) D X (2 ) kin X = − − − ν ⋅ ⋅ (A.17)

'' 2m m y mM (x) D X kin X = − − ν ⋅ ⋅ (A.18)

nm

nm

4k x 2 2

m nm nm nm yn 1

k x 2 2om x x y

in

V (x) D A (k ) e (k ) (2 )kin

P H ( ikin ) e ( ikin ) (2 )(kin )

A

=

= − ⋅ ⋅ − − ν + ⋅ ⋅ − ⋅ − − − ν

∑ (A.19)

nm

nm

4k x 2 2

m nm nm yn 1

k x 2 2om x y

in

M (x) D A e (k ) kin

P H e ( ikin ) (kin )

A

=

= − − ν ⋅ + ⋅ ⋅ − − ν ⋅

∑ (A.20)

The force vector at both ends of plate is defined as Fm.

[ ]Tm m m m mF V (0) M (0) V (L) M (L)= − − (A.21)

169

m 2m m mp yF P A F H(kin )= + ⋅ (A.22)

2 22m 1n nm nm y(P ) D k (k ) (2 )kin = ⋅ ⋅ − − ν (A.23)

2 22m 2n nm y(P ) D (k ) kin = − ⋅ − ν (A.24)

nmk L2m 3n 2m 1n(P ) (P ) e= − ⋅ (A.25)

nmk L2m 4n 2m 2n(P ) (P ) e= − ⋅ (A.26)

2 2omp 1 m x x y

in

P(F ) D H ( ikin ) ( ikin ) (2 )(kin )

A = ⋅ ⋅ ⋅ − ⋅ − − − ν (A.27)

2 2omp 2 m x y

in

P(F ) D H ( ikin ) (kin )

A = − ⋅ ⋅ ⋅ − − ν ⋅ (A.28)

nmk L 2 2omp 3 m x x y

in

P(F ) D H ( ikin ) e ( ikin ) (2 )(kin )

A = − ⋅ ⋅ ⋅ − ⋅ − − − ν (A.29)

nmk L 2 2omp 4 m x y

in

P(F ) D H e ( ikin ) (kin )

A = ⋅ ⋅ ⋅ − − ν ⋅ (A.30)

Combining equations um and Fm,

1 1m 2m 1m m mp 2m 1m mp y

m m m y

F P P u F P P u H(kin )

Q u q H(kin )

− − = + − ⋅

= − (A.31)

where the panel dynamic stiffness matrix Qm and the panel force vector qn is defined.

This equation gives the dynamic properties of one single panel element. The response

of the complete row of elements is analysed by using the finite element method.

170

As derived before the equations of open section channel beam motion are

given below.

( ) ( )4 4 2 2

x z b b x x4 4 2 2

3 3 3 32 2 1 1

3 2 3 2


Seff Seff

w w w w D (2 ) D (2 )

x x y x x y

ζ ζ ηζ

+ −

∂ ∂ φ ∂ ∂ φ+ ⋅ − + + + − =∂ ∂ ∂ ∂

= −

∂ ∂ ∂ ∂= − + − ν + + − ν

∂ ∂ ∂ ∂ ∂ ∂

(A.32)

( ) ( )

( ) ( ) ( )

4 4 22 2

z x o x z x z4 4 2

2 22 2


2 2 22 2

2 2



M M

w w w D Dx y

ηζ ζ ζ η ηζ

+ −

∂ ∂ φ ∂ φ− + ⋅ Γ + + − −∂ ∂ ∂

∂ ∂ φ + − + ρ + − + + = ∂ ∂

= − +

∂ ∂ ∂= + ν − ∂ ∂

21 1

2 2w

x y

∂+ ν ∂ ∂

(A.33)

Arrange the equation as matrix multiplication.

D.ub = Fb (A.34)

where

[ ]Tbu w= φ (A.35)

[ ]bF V M= (A.36)

Perform the assembly of panel and stiffener equations as finite element

method and obtain one matrix equation.

171

Gm.Ψm = gm.Hm(kiny) (A.37)

where Ψm is a vector containing the displacement and longitudinal slope of each

panel/stiffener attachment point, and Gm and gm is the assembled dynamic stiffness

matrix and force vector. This equation is solved by standard techniques and Ψm is

calculated. The response at any point within a particular panel can be calculated with

the below equation.

T 1m m 1m m mp m y mp m yX (x) h P u u H (kin ) X (x)H (kin )− = ⋅ ⋅ − + (A.38)

The vector h is defined below.

mik xmih e= (A.39)

A.2. Power Input

The total power input by pressure wave excitation is calculated with an

integral over panel area, [17].

2 /

inp0 A

P ( ) p(x, y, t) v(x, y, t)dAdtπ ω

ω = ⋅∫ ∫ (A.40)

The excitation is taken in the following form.

ini (wt-k x)o yp(x,y,t)=P e sink y ⋅ (A.41)

with wavenumber

ky = mπ/Lj (A.42)

Assume that the damping is low. Then the equation of power input is the

following.

172

2 /2 2

inp o x y0 A

P ( ) P V cos ( t kin x) sin (k y)dAdtπ ω

ω = ⋅ ⋅ ω − ⋅∫ ∫ (A.43)

A.3. Exact Calculation of Average Energy

SEA results are compared with the results obtained by solving the systems

numerically with dynamic stiffness method and calculate the mean subsystem

energies by taking average over the plate area, angle and time.

2 / / 2 20

0 Area

2E( ) h v (x, y, t, , )d dxdydt2

π ω πωω = ρ⋅ ⋅ ⋅ ω θ θπ π ∫ ∫ ∫ (A.44)

173

APPENDIX B

B. SEA PARAMETERS OF PLATE – ACOUSTIC CAVITY

Coupling loss factor from a wall, 1, to a room, 2, is computed with the

following equation, [54, 63, 70].

0 012

p

cm

ρ ση =

ω (B.45)

The radiation efficiency, σ, is defined as the power radiated by a structure

compared to that of a piston with the same area and with the same velocity. The

following expressions are used to calculate the radiation efficiency for three cases

where the frequency of interest is less than, equal to, or greater than, the critical

frequency.

0c2 1/ 2 1/ 2 2 1/ 2 2

c

Uc 1 2ln , f < f14 f f A( 1) 1

µ + µσ = ⋅ + µ −π µ − µ − (B.46)

1/ 2x

x c0 y

L2 f L 0.5 0.15 , f = fc L

πσ = − (B.47)

1/ 2c

cf

1 , f > ff

− σ = −

(B.48)

The plate dimensions are Lx and Ly where Ly ≥Lx. U is the perimeter of the

radiating area, usually 2(Lx + Ly), and µ is defined as (fc / f)1/2. Critical frequency for

a homogeneous plate is calculated with the equation below.

174

1/ 222p0

c 3

3m (1 )cf

Eh

− µ= π

(B.49)

The following equation is used to calculate the coupling loss factor from a

room, 1, to a wall, 2.

20 0 c

12 3p

c Af8 Vm fρ σ

η =π

(B.50)

175

APPENDIX C

C. BEAM AND PLATE PROPERTIES

The following beam and plate properties are used to calculate transmission

ratio and coupling loss factor.

Properties given in Table C-1 are the same with Ref [54].

Table C- 1 Example line mass and Euler Beam coupling system properties.

Beam Properties Plate Properties

ρb = 7810 kg/m3

Ab = 8.51x10-3 m2

mb = ρb .Ab = 66.38 kg/m

Ip = Ix + Iz = 3.85x10-5 m4

Ix = 3.75x10-5 m4

J = 4.5x10-8 m4

G = 7.93x1010 MPa

E = 20.7x1010 (1+i.0.01) MPa

ν = 0.29

Lz = 0.23 m

Lx = 0.037 m

ρp = 7810 kg/m3

h = 0.02 m

mp = ρp .h = 4.32 kg/m2

G = 7.93.1010 MPa

E = 20.7.1010 (1+i.0.01) MPa

ν = 0.29

D = E.h3 / [12.(1 – v2)]

Lx

Lz x ξ

z η

176

Table C- 2 Example double and triple coupling beam properties.

Double Coupling Beam Properties Triple Coupling Beam Properties

ρb = 7810 kg/m3

Ab = 8.56x10-3 m2

mb = ρb .Ab = 66.85x10-3 kg/m

Iξ = 5.80x10-5 m4

Iη = 6.63x10-6 m4

Iηξ = 0

J = 1.44x10-6 m3

IpC = 6.46x10-5 m4

Γo = 5.19x10-8 m6

G = 7.93x1010 MPa

E = 20.7x1010 (1+i.0.01) MPa

ν = 0.29

cx = 0.046 m

cz = 0

sx = 0.066 m

sz = 0.115 m

ρb = 7810 kg/m3

Ab = 8.65x10-3 m2

mb = ρb .Ab = 67.56x10-3 kg/m

Iξ = 4.52x10-5 m4

Iη = 4.72x10-6 m4

Iηξ = 7.15x10-6 m4

J = 2.40x10-6 m3

IpC = 4.99x10-5 m4

Γo = 7.98x10-9 m6

G = 7.93x1010 MPa

E = 20.7x1010 (1+i.0.01) MPa

ν = 0.29

cx = 0.010 m

cz = 0.082 m

sx = 0.035 m

sz = 0.115 m

Plate properties are the same with Table C-1.

x ξ

z η

177

Table C- 3 Example triple coupling system properties.


ρb = 2700 kg/m3

Ab = 9.68x10-5 m

hb = 38.1x10-3 m

mb = ρb .Ab = 261.36x10-3 kg/m

Iξ = 2.24x10-8 m4

Iη = 5.08x10-9 m4

Iηξ = 4.25x10-9 m4

J = 5.20x10-11 m3

Ipo = 4.60x10-8 m4

Γo = 7.11x10-12 m6

G = 2.6x1010 Pa

E = 7x1010 (1+i.0.01) Pa

ν = 0.29

cx = 10.43x10-3 m

cz = 9.09x10-3 m

ρp = 2700 kg/m3

h = 0.002 m

mp = ρp .h = 5.40 kg/m2

G = 2.6x1010 Pa

E = 7x1010 (1+i.0.01) Pa

ν = 0.29

D = E.h3 / [12.(1 – v2)]

x ξ

z η

178

Table C- 4 Example frame properties.


ρb = 2700 kg/m3

Ab = 3.3x10-4 m

hb = 0.10 m

mb = ρb .Ab = 891x10-3 kg/m

Iξ = 4.81x10-7 m4

Iη = 3.91x10-8 m4

Iηξ = 2.64x10-8 m4

J = 4.46x10-10 m3

Γo = 5.66x10-11 m6

G = 2.6x1010 Pa

E = 7x1010 (1+i.0.01) Pa

ν = 0.29

cx = 0.013 m

cz = 0.018 m

ρp = 2700 kg/m3

h = 0.002 m

mp = ρp .h = 5.40 kg/m2

G = 2.6x1010 Pa

E = 7x1010 (1+i.0.01) Pa

ν = 0.29

D = E.h3 / [12.(1 – v2)]

x ξ

z η

179

CURRICULUM VITAE PERSONAL INFORMATION Surname, Name: Yılmazel, Canan Nationality: Turkish (TC) Date and Place of Birth: 15 October 1970 , Gölcük/Kocaeli Marital Status: Single Phone: +90 312 811 18 00 Fax: +90 312 811 14 25 email: [email protected] EDUCATION

Degree Institution Year of Graduation MS METU Mechanical Engineering 1996 BS METU Mechanical Engineering 1992 High School Barbaros Hayrettin High School,

Gölcük 1987

WORK EXPERIENCE

Year Place Enrollment 1998- Present TUSAŞ Turkish Aeronautical Industries Design Engineer 1995-1999 MEKSA AutoCAD and AMD Instructor 1994 August Chamber of Civil Engineers AutoCAD & Microsoft Office

Instructor 1993-1998 METU Department of Mechanical

Engineering Research Assistant

FOREIGN LANGUAGES Advanced English PUBLICATIONS 1. Yılmazel, C., Tümer, S.T., and Platin, B.E., “An Investigation On The Wheelchair Propulsion During Take-off”, The Third Biennial World Conference On Integrated Design & Process Technology, July 6-9, 1998, Berlin, Germany 2. Yılmazel, C., Tümer, S.T., and Platin, B.E., “Tekerlekli Sandalye ile Harekete Kalkış Üzerine Bir İnceleme”, The Eighth National Machine Theory Symposium, September 17-19, 1997, Elazığ, Turkey HOBBIES

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