FUNDAMENTAL STUDIES IN THE LAMB-WAVE...

FUNDAMENTAL STUDIES IN THE LAMB-WAVE INTERACTION

BETWEEN PIEZOELECTRIC WAFER ACTIVE SENSOR AND HOST

STRUCTURE DURING STRUCTURAL HEALTH MONITORING

by

Giola Santoni

Bachelor of Science

University of Pisa, Pisa, Italy, 1999

Submitted in Partial Fulfillment of the Requirements

For the Degree of Doctor of Philosophy

Mechanical Engineering

College of Engineering and Information Technology

University of South Carolina

2010

Accepted by:

Victor Giurgiutiu, Major Professor

Sarah Baxter, Committee Member

Yuh Chao, Committee Member

Sarah Gassman, Committee Member

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ACKNOWLEDGMENTS

The author is grateful to many to accomplish this goal. The author would like to

expresses her sincere gratitude for her advisor, Prof. Victor Giurgiutiu, for his continuous

support, encouragement, motivation and guidance throughout all phases of her Ph.D.

study. The author would like to thank her defense committee members, Prof. Sarah

Baxter, Prof. Yuh Chao, Prof. Sarah Gassman, for their comments, suggestions and time

for reviewing this work. The author is thankful to former graduate director Prof. Xiaomin

Deng and current graduate director Prof. Tony Reynolds for their countless helps during

the years.

The author would like to thank LAMSS research group members: Lucy Yu, Bin Lin,

Buli Xu, Tom Behling, Patrick Pollock, Weiping Liu, Adrian Cuc, James Kendall, Greg

Crachiolo, for their invaluable suggestions, comments, and friendship.

The author wishes to dedicate this dissertation to her husband for his continuous

support, patience and understanding and to her children for being so nice and for thinking

that her work consists in making toys.

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ABSTRACT

The scope of my research was to develop a better understanding of the engineering

variables that influence the interaction of PWAS with structure during activation of the

transducer. This is a key feature needed to develop more power/energy efficient structural

health monitoring (SHM) systems. SHM is the field of engineering that determines the

health of a structure while it is in service. Active SHM can be performed through

piezoelectric transducers such as piezoelectric wafer active sensors (PWAS) that can be

permanently attached to the structure through a bonding layer. PWAS transducers can

actively interrogate the structure by exciting and receiving Lamb waves propagating in

the structure or by passively listen to changes in the structure. PWAS-structure

interaction modeling is fundamental in order to achieve single mode excitation, i.e.,

tuning, a requirement for most of the SHM algorithms (time reversal, phase-array, and

imaging).

To achieve our research goal, we had to go beyond the current state of the art in

modeling and understanding the load transfer from PWAS to the structure. The existing

modeling methods rely on the low frequency assumption of axial/flexural waves only.

This assumption is not true in the high frequency range of ultrasonic SHM applications.

We derived, through the normal mode expansion methods (NME), the interfacial shear

stress, hence, the load transfer from the PWAS to the structure through the bond layer,

without limitations on the frequency and the number of modes. This allowed us to derive

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more accurate predictions of the tuning between PWAS and Lamb waves which

compared very well with experimental measurements.

This dissertation is constructed in three major parts.

In Part I, we developed a generic formulation for ultrasonic guided waves in thin wall

structures. The formulation is generic because, unlike many authors, in many parts of our

derivation (power flow, reciprocity theorem, orthogonality, etc.) we stayed away from

specifying the actual mathematical expressions of the guided wave modes and maintained

a generic formulation throughout.

In Part II, we addressed some unresolved issues of the PWAS SHM predictive

modeling. We extended the NME theory to the case of PWAS bonded to or embedded in

the structure. We developed the shear layer coupling between PWAS and structure using

N generic guided wave modes and solving the resulting integro-differential equation for

shear lag transfer. We applied these results to predicting the tuning between guided

waves and PWAS and obtained excellent agreement with experimental results.

Another novel aspect covered in this dissertation is that of guided waves in composite

materials. The increasing use of composites in aeronautical and space applications makes

it important to extend SHM theory to such materials. For this reason, the NME theory is

extended to the case of composites. We developed a generic formulation for the tuning

curves that was not directly dependent on the composite layup and can be easily extended

to various composite formulations. We conducted carefully-planned experiments on

composites with different orientations. The comparison between our predictions and

experiments was quite good.

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In Part III, SHM applications and related issues are addressed. We discussed the

reliability of SHM systems and the lack of specifications for quality SHM inspections

with particular focus on the case of composites SHM. We determined experimentally the

ability of PWAS to detect damage in various composite specimens. We tested the

performance of the PWAS for damage detection on composite plates, on unidirectional

composite strips, on quasi-isotropic plates, on lap-joints junctions, and composite tank

sections. We also tested the ability of PWAS transducers to operate under extreme

environments and high stress conditions, i.e. the survivability of PWAS-based SHM. We

proved the durability of the entire PWAS-based SHM system under various different load

conditions. We also tested the influence of bond degradation on PWAS electrical

capacitance as installed on the structure, which gives a measure of the quality of the

PWAS installation, a key feature in PWAS-based SHM. We developed theoretical

models for shear horizontal waves scattering from a crack and Lamb waves scattering

from change in material properties. We studied the acoustic emission (AE) in infinite

plate and we used NME to model AE phenomena.

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TABLE OF CONTENTS

ACKNOWLEDGMENTS............................................................................................................................II

ABSTRACT ................................................................................................................................................ III

LIST OF TABLES ..................................................................................................................................... XI

LIST OF FIGURES ................................................................................................................................ XIII

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

1.1 MOTIVATION.................................................................................................................................1

1.2 RESEARCH GOAL, SCOPE, AND OBJECTIVES ...................................................................................3

1.3 DISSERTATION LAYOUT ................................................................................................................5

PART I ELASTIC WAVES FOR STRUCTURAL HEALTH MONITORING .................................9

2 ACOUSTIC FIELD EQUATIONS.....................................................................................................12

2.1 EQUATION OF MOTION ................................................................................................................12

2.2 STRAIN-DISPLACEMENT EQUATION .............................................................................................15

2.3 HOOKE’S LAW.............................................................................................................................16

2.4 ACOUSTIC FIELD EQUATIONS SUMMARY .....................................................................................17

3 GUIDED WAVES IN PLATES ..........................................................................................................18

3.1 STRAIGHT-CRESTED GUIDED WAVES IN RECTANGULAR COORDINATES .......................................18

3.2 CIRCULAR-CRESTED GUIDED WAVES IN CYLINDRICAL COORDINATES .........................................37

4 POWER FLOW AND ENERGY CONSERVATION – THE ACOUSTIC POYNTING

THEOREM..................................................................................................................................................61

4.1 POWER FLOW IN RECTANGULAR COORDINATES ..........................................................................67

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4.2 POWER FLOW IN CYLINDRICAL COORDINATES.............................................................................90

5 RECIPROCITY RELATION ...........................................................................................................113

5.1 REAL RECIPROCITY RELATION...................................................................................................115

5.2 COMPLEX RECIPROCITY RELATION............................................................................................117

5.3 REAL RECIPROCITY RELATION IN RECTANGULAR COORDINATES...............................................122

5.4 COMPLEX RECIPROCITY RELATION IN RECTANGULAR COORDINATES........................................126

5.5 REAL RECIPROCITY RELATION IN CYLINDRICAL COORDINATES.................................................129

5.6 COMPLEX RECIPROCITY RELATION IN CYLINDRICAL COORDINATES ..........................................133

6 ORTHOGONALITY RELATION...................................................................................................137

6.1 ORTHOGONALITY RELATION WITHOUT ASSUMPTIONS ON THE SOLUTION..................................137

6.2 ORTHOGONALITY RELATION IN RECTANGULAR COORDINATES .................................................157

6.3 ORTHOGONALITY RELATION IN CYLINDRICAL COORDINATES ...................................................167

PART II PWAS-BASED STRUCTURAL HEALTH MONITORING .............................................174

7 PWAS EXCITATION OF GUIDED WAVES.................................................................................177

7.1 PIEZOELECTRIC WAFER ACTIVE SENSORS CHARACTERISTICS ....................................................177

7.2 PWAS EXCITATION OF STRAIGHT-CRESTED GUIDED WAVES.....................................................182

7.3 PWAS EXCITATION OF CIRCULAR-CRESTED GUIDED WAVES ....................................................191

7.4 NORMAL MODE EXPANSION MODEL WITH SURFACES FORCES....................................................197

7.5 NORMAL MODE EXPANSION MODEL WITH VOLUME FORCES .....................................................210

8 SHEAR LAYER COUPLING BETWEEN PWAS AND STRUCTURE......................................213

8.1 PROBLEM DEFINITION ...............................................................................................................215

8.2 SHEAR-LAG SOLUTION FOR AXIAL AND FLEXURAL MODES........................................................217

8.3 SHEAR-LAG SOLUTION FOR TWO MODES, ONE SYMMETRIC AND THE OTHER ANTISYMMETRIC ..232

8.4 SHEAR-LAG SOLUTION FOR N GENERIC MODES ......................................................................238

8.5 STRESS DISTRIBUTION IN THE BONDING LAYER .........................................................................260

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9 TUNED GUIDED WAVES EXCITED BY PWAS .........................................................................264

9.1 SHEAR TRANSFER THROUGH BOND LAYER ................................................................................265

9.2 PWAS – LAMB WAVES TUNING ................................................................................................271

10 TUNED GUIDED WAVES IN COMPOSITE PLATES ................................................................288

10.1 DISPERSION CURVES IN COMPOSITE PLATES ..............................................................................289

10.2 PWAS – GUIDED WAVES TUNING..............................................................................................313

PART III STRUCTURAL HEALTH MONITORING ISSUES AND APPLICATIONS .................326

11 RELIABILITY OF STRUCTURAL HEALTH MONITORING..................................................329

11.1 SPECIFICATIONS FOR QUALITY STRUCTURAL HEALTH MONITORING INSPECTION ......................329

11.2 PROBABILITY OF DETECTION CURVES .......................................................................................333

12 SPACE QUALIFIED NON-DESTRUCTIVE EVALUATION & STRUCTURAL HEALTH

MONITORING .........................................................................................................................................340

12.1 INTRODUCTION .........................................................................................................................340

12.2 SUBSYSTEM/COMPONENT SPECIFICATION .................................................................................341

12.3 DAMAGE DETECTION EXPERIMENTS ON TEST SPECIMENS..........................................................347

13 SURVIVABILITY OF SHM SYSTEMS .........................................................................................362

13.1 TEST SPECIFICATIONS................................................................................................................362

13.2 TEST PROCEDURE......................................................................................................................363

13.3 RESULTS ...................................................................................................................................365

13.4 CONCLUSION.............................................................................................................................373

14 DURABILITY OF SHM SYSTEMS ................................................................................................374

14.1 REQUIREMENTS.........................................................................................................................374

14.2 MISSION PROFILE ......................................................................................................................375

14.3 SHOCK TEST..............................................................................................................................377

14.4 RANDOM VIBRATION TEST .......................................................................................................379

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14.5 THERMAL TEST .........................................................................................................................379

14.6 ACOUSTIC ENVIRONMENT TEST.................................................................................................380

15 EFFECT OF PARTIAL BONDING BETWEEN TRANSDUCER AND STRUCTURE ON

CAPACITANCE .......................................................................................................................................381

15.1 POWER ANALYSIS AND SAMPLE SIZE .........................................................................................383

15.2 POPULATION VARIANCE ............................................................................................................384

15.3 EXPERIMENTAL SETUP ..............................................................................................................385

15.4 EXPERIMENT .............................................................................................................................386

15.5 ANALYSIS .................................................................................................................................389

16 GUIDED WAVES SCATTERING FROM DAMAGE...................................................................393

16.1 MODE DECOMPOSITION OF INCIDENT, REFLECTED, AND TRANSMITTED WAVES: SH WAVES

SCATTERING FROM A CRACK ........................................................................................................................396

16.2 MODE DECOMPOSITION OF INCIDENT, REFLECTED, AND TRANSMITTED WAVES: LAMB WAVES

SCATTERING FROM CHANGE IN MATERIAL PROPERTIES ................................................................................402

17 ACOUSTIC EMISSION IN INFINITE PLATE .............................................................................407

17.1 ACOUSTIC EMISSION THROUGH INTEGRAL DISPLACEMENT .......................................................411

17.2 ACOUSTIC EMISSION THROUGH NORMAL MODE EXPANSION......................................................416

18 CONCLUSIONS AND FUTURE WORK........................................................................................421

18.1 RESEARCH CONCLUSIONS .........................................................................................................423

18.2 MAJOR CONTRIBUTIONS ...........................................................................................................429

18.3 RECOMMENDATION FOR FUTURE WORK ....................................................................................430

19 REFERENCES...................................................................................................................................432

APPENDIX ................................................................................................................................................441

A EQUATION OF MOTION IN CYLINDRICAL COORDINATES ..............................................442

B MATHEMATIC CONCEPTS ..........................................................................................................451

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C POWER AND ENERGY...................................................................................................................464

D ORTHOGONALITY FOR VIBRATION AND WAVE PROBLEMS..........................................475

D.1 ORTHOGONALITY PROOF FOR SOME VIBRATION PROBLEMS ......................................................475

D.2 STURM-LIOUVILLE PROBLEM ...................................................................................................489

D.3 ORTHOGONALITY RELATION FROM THE REAL RECIPROCITY RELATION .....................................490

E NORMALIZATION FACTOR ........................................................................................................498

E.1 SHEAR HORIZONTAL WAVES .....................................................................................................498

E.2 LAMB WAVES............................................................................................................................502

F STRAIN DERIVATION THROUGH NME AND FOURIER TRANSFORMATION ...............510

G STRUCTURE EXCITED BY TWO PWAS ....................................................................................514

G.1 SYMMETRIC MODE ....................................................................................................................514

G.2 ANTISYMMETRIC MODE ............................................................................................................517

H STATISTICAL DATA ANALYSIS..................................................................................................519

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LIST OF TABLES

Table 7.1 Material properties .................................................................................. 179

Table 8.1 Shear stress parameters ........................................................................... 224

Table 9.1 Actual and effective PWAS length ......................................................... 284

Table 10.1 Ply material properties (Herakovich 1998)............................................. 311

Table 11.1 Health monitoring reliability needs. ....................................................... 332

Table 11.2 Summary of the specimen configurations (Note: A, B, C, D, E: seed

location) .................................................................................................. 336

Table 11.3 95% CI amplitude for different sample sizes (n) and probabilities ........ 337

Table 12.1 Summary of PWAS damage detection methods. (Cuc et al., 2005) ....... 341

Table 12.2 Summary of experiments discussed in this paper. .................................. 347

Table 12.3 Hole sizes for corresponding readings in the unidirectional composite

strip experiments. .................................................................................... 348

Table 12.4 Hole diameters corresponding to the quasi-isotropic plate damage

detection experiment............................................................................... 350

Table 12.5 Summary of impact test parameters on quasi-isotropic plate specimen. 354

Table 13.1 Full-scan, 12 min (for 1000 sample at 200 Hz) (T=transmitter,

R=receivers)............................................................................................ 364

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Table 13.2 Test sequence for impedance. ................................................................. 364

Table 14.1 Notional test plan for space certification of NDE system....................... 375

Table 15.1 Capacitance variance. ............................................................................ 385

Table 15.2 Power as a function of n for 3r = , 0.05α = ............................................ 385

Table 15.3 Experiment setup. 1 – 3: specimen identification number; a – c: type

of bonding ............................................................................................... 385

Table 15.4 Capacitance (nF) of the PWAS............................................................... 388

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LIST OF FIGURES

Figure 3.1 Plane wave notations................................................................................. 18

Figure 3.2 Dispersion curves of SH waves propagating in an aluminum plate.

Dash lines: antisymmetric modes; solid lines: symmetric modes. ........... 23

Figure 3.3 Dispersion curves of Lamb waves propagating in an aluminum plate.

Dash lines: antisymmetric modes; solid lines: symmetric modes. a)

Dispersion curves for the frequency range 0-4000 kHz-mm b)

Dispersion curves below the first cut-off frequency (fd<780 kHz-

mm). .......................................................................................................... 27

Figure 3.4 S0 and A0 wave propagation at the low frequencies. a) Wave

propagation of non-dispersive S0 mode. b) Wave propagation of

dispersive A0 mode................................................................................... 27

Figure 3.5 An element of plate subjected to forces (after Giurgiutiu, 2008) ............. 30

Figure 3.6 An element of plate subjected to forces and moments (after Graff,

1991) ......................................................................................................... 34

Figure 3.7 Cylindrical wave notations ....................................................................... 38

Figure 4.1 Coordinate notation. Power flow from 1 to 2 is ˆ ˆdS dS− ⋅ ⋅ = ⋅v T n P n

(after Auld 1990)....................................................................................... 64

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Figure 4.2 Rectangular section dxdydz of a plate of thickness 2d. a) Section

notations; b) Power flow through surface with normal nx. ....................... 67

Figure 4.3 Average power flow apparent variation with x. The abscissa is equal

to cos(2 )XX xξ′ = . ....................................................................................... 76

Figure 4.4 Circular section rdzdθ of a plate of thickness 2d. a) Section notations;

b) Power flow through surface with normal nr. ........................................ 90

Figure 4.5 Power flow in the r direction as a function of the radius (Symmetric

SH0 mode for an Aluminum with wave propagating at 100 kHz). .......... 94

Figure 4.6 Variation of RR′ (solid red line) and XX ′ (dashed blue line) with

respect to frequency-radius product.......................................................... 95

Figure 4.7 Bessel function ( )1J rξ approximated with the sum of two sine

functions (forward and backward propagating waves)........................... 102

Figure 4.8 Average power flow as a function of rξ . ............................................... 105

Figure 5.1 Reciprocity relation................................................................................. 113

Figure 7.1 Lead Zirconite titanate PWAS. a) atomic structure of PZT for

temperature below Curie temperature. (www.piezo-kinetics.com). b)

PWAS transducer notations .................................................................... 178

Figure 7.2 GaPO4 PWAS. a) GaPO4 crystal structure. (www.roditi.com), b)

Transducer deformation for an electric field in direction 3. ................... 181

Figure 7.3 Lamb waves wave front and external load Trz applied on the surface

of the structure ........................................................................................ 196

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Figure 7.4 Surface forces due to a PWAS bonded on the top surface of the

structure................................................................................................... 198

Figure 7.5 Surface forces due to a PWAS bonded on the top surface and a second

on the bottom surface of the structure..................................................... 203

Figure 7.6 Volume forces due to a PWAS embedded in the structure..................... 210

Figure 8.1 Interaction between the PWAS and the structure through the bonding

layer......................................................................................................... 215

Figure 8.2 Forces and moments acting in the plate. a) Stress distribution of the

axial and flexural modes. a) Equilibrium of an infinitesimal element.... 218

Figure 8.3 Normalized shear strain as a function of the normalized PWAS

position and bond layer thickness (All other parameters are defined in

Table 8.1). ............................................................................................... 224

Figure 8.4 Effect of the different parameters on the shear stress transmission.

The abscissa is the normalized position of the PWAS length (in the

graph is shown only the portion close to the actuator tip:0.8 to 1.)........ 226

Figure 8.5 Stress distribution of the first symmetric and antisymmetric modes at

frequency-thickness product of 780 kHz-mm......................................... 231

Figure 8.6 Stress distribution of the first three Lamb wave modes (A0, S0, and

A1) at frequency-thickness product of 1600 kHzmm............................. 236

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Figure 8.7 Repartition mode number as a function of frequency. (I): classic

solution 4α = ; (II): 2 modes solution Equation (8.73) for S0 and A0;

(III): N generic modes solution Equation (8.98) for S0, A0, and A1;

(IV): N generic modes solution Equation (8.98) for S0, A0, and A1

and contribution form shear stress equal to zero in the power flow. a)

Total repartition number. b) Repartition number divided between α

for the antisymmetric modes and α for the symmetric modes. ............. 244

Figure 8.8 Dispersion curves for an aluminum plate. a) Frequency-phase velocity

plane; b) Wavenumber-radial frequency plane. Left plane imaginary

wavenumbers, right plane real wavenumbers. ........................................ 245

Figure 8.9 Repartition mode number as a function of frequency with imaginary

A1. (I): classic solution 4α = ; (II): two modes solution, Equation

(8.73), for S0 and A0; (III): N generic modes solution Equation (8.98)

for S0, A0, and A1. a) Total repartition number. b) Repartition

number divided between α for the antisymmetric modes and α for

the symmetric modes. (IV): contribution from A1. ................................ 247

Figure 8.10 Differential element of length dx. a) normal stress due to bending; b)

Shear force due to the presence of stress. ............................................... 248

Figure 8.11 Forces and moments on a plate. a) Shear stress sign convention. b)

Moment balance...................................................................................... 248

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Figure 8.12 Normal stress distribution and shear stress distribution across the plate

thickness for the first antisymmetric mode for three different

frequencies. Solid black line: normal stress ( , )x x yσ as in Equation

(8.121); Dot line: shear stress ( , )xy x yτ as in Equation (8.122); Dash-

dot line: normal stress distribution of A0 as from Equation (3.31);

Dash line: shear stress distribution of A0 as from Equation (3.31). ....... 250

Figure 8.13 Repartition mode number due to the A0 mode as a function of

frequency. (a): classic solution with Bernoulli-Euler assumption,

3α = ; (b): 2 modes solution Equation (8.73) for A0; (c): N generic

mode solution Equation (8.98) for A0; (d): 2 modes solution Equation

(8.142) for A0 (normal + shear stress).................................................... 254

Figure 8.14 Shear stress variation with frequency. a) Shear stress transmitted by

the PWAS to the structure through a bond layer. (I): shear stress

derived for low frequency approximation ( 4α = ), Equation (8.44); All

other curves: shear stress derived in the generic N mode formulation

( fα = (frequency,no. of modes)), Equation (8.162). (II) fd=1 kHz-

mm; (III) fd=783 kHz-mm before the cut-off frequency; (IV) fd=784

kHz-mm after the cut-off frequency). b) Percentage difference in load

transmitted at different frequency-thickness products............................ 262

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Figure 8.15 Interfacial shear stress distribution and percentage of change a) Shear

stress transmitted by the PWAS to the structure through a bond layer

with imaginary A1. (I): shear stress derived for low frequency

approximation ( 4α = ); (II): shear stress derived in the generic N

mode formulation ( fα = (frequency, no. of modes)) for different

frequencies (fd=1; 200; 781; 850 (solid line); 1000 kHz-mm (dash-dot

] line)). b) Percentage difference in load transmitted at different

frequency-thickness products.................................................................. 263

Figure 9.1 S0 and A0 particle displacement and interaction of PWAS with Lamb

waves. (Bao 2003) .................................................................................. 272

Figure 9.2 Comparison of tuning curves for the strain excited by a PWAS

derived through the Fourier transformation model and the normal

mode expansion method. ........................................................................ 273

Figure 9.3 Aluminum plate 2024-T3 1-mm thick with square, rectangular and

round PWAS ........................................................................................... 275

Figure 9.4 Tuning for aluminum 2024-T3, 1-mm thickness, 7-mm square

PWAS; experimental A0 (cross) and S0 (circle) data; theoretical

values (solid lines) for 6.4 mm PWAS. .................................................. 276

Figure 9.5 Group velocity: Aluminum 2024-T3, 3-mm thick, 7-mm square

PWAS ..................................................................................................... 277

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Figure 9.6 Aluminum 2024-T3, 3-mm thickness, 7-mm Square PWAS.

Experimental A0 (cross), S0 (circle), and A1 (cross) data. b)

Theoretical values (solid lines) with PWAS length=6.4 mm ................. 277

Figure 9.7 Wave propagation from the Oscilloscope at 450 kHz and 570 kHz for

the 1200x1200-mm, 3-mm thick plate.................................................... 279

Figure 9.8 Waves propagation for 1200x1200-mm and 500x500-mm plate, 3-

mm thick. (a) 270 kHz. (b) 570 kHz....................................................... 279

Figure 9.9 Aluminum plate 2024-T3 1200x1200-mm, 1-mm thick with

rectangular PWAS .................................................................................. 280

Figure 9.10 Plate 2024-T3, 1200x1200-mm, 1-mm thick. Rectangular PWAS (P1

transmitter, P2 receiver). (a) Group velocity: experimental and

theoretical values; (b) Experimental data for the tuning......................... 281

Figure 9.11 Tuning on plate 2024-T3, 1200x1200-mm, 1-mm thick; rectangular

PWAS (P1 transmitter, P2 receiver); experimental data for A0

(crosses) and S0 (circles); Solid lines, theoretical values with PWAS

length=22 mm ......................................................................................... 282

Figure 9.12 Tuning on plate 2024-T3, 1200x1200-mm, 1-mm thickness;

rectangular PWAS (P1 transmitter, P3 receiver); experimental A0

(crosses) and S0 (circles) data; Solid lines, theoretical values with

PWAS length~=4.5 mm.......................................................................... 283

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Figure 9.13 Tuning curves for an Aluminum plate 1-mm thick and a 7-mm square

PWAS. Blue circles: Experimental S0 mode data; Red crosses:

Experimental A0 mode data; Solid line: theoretical A0 and S0 tuning

values under ideal bond assumption. Equation (9.33); Dash line:

theoretical A0 and S0 values for shear lag assumption, Equation

(9.34); Dash dot line: theoretical A0 and S0 values for N generic

mode derivation, Equation (9.35). a) Bond thickness 1 μm; b) bond

thickness 30 μm. ..................................................................................... 286

Figure 9.14 Tuning curves for an Aluminum plate 1-mm thick and bond layer 30-

μm thick. Circles: Experimental S0 mode data; Crosses: Experimental

A0 mode data; Solid line: theoretical A0 and S0 tuning values under

ideal bond assumption, Equation (9.33); Dash line: theoretical A0 and

S0 values for shear lag assumption, Equation (9.34); Dash dot line:

theoretical A0 and S0 values for N generic mode derivation, Equation

(9.35). a) Real PWAS length 7-mm, effective PWAS length 6.4-mm.

b) Real PWAS length 5-mm, effective PWAS length ~4.5-mm. ........... 287

Figure 10.1 Example, using three-layer plate with semi-infinite half spaces.

(Lowe, 1995)........................................................................................... 291

Figure 10.2 Plate of an arbitrary number of layers with a plane wave propagating

in the x1-x3 plane at an angle η with respect to x3 axis. .......................... 293

Figure 10.3 Composite plate and the kth layer made of unidirectional fibers............. 295

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Figure 10.4 Comparison of dispersion curves predicted by layered model (transfer

matrix method) vs. isotropic classic theory (Rayleigh – Lamb

equation). a) One-layer aluminum plate 2024-T3, 1-mm thick. b)

Two-layer aluminum plate 2024-T3, 1-mm total thickness. Dash

lines: values derived from the Rayleigh – Lamb equation; Solid lines:

values derived from the transfer matrix method. .................................... 305

Figure 10.5 Dispersion curves for plate made of one unidirectional layer of 65%

graphite 35% epoxy (material properties from Nayfeh 1995) as

derived by our code. a) θ = 0°; b) θ = 18°; c) θ = 36°; d) θ = 90°.......... 306

Figure 10.6 Transfer matrix instability for high frequency-thickness products......... 307

Figure 10.7 Dispersion curves for first antisymmetric wave mode (A0)

propagating at different angles with respect to the fiber direction.

Plate material: 65% graphite 35% epoxy (material properties from

Nayfeh 1995). ......................................................................................... 308

Figure 10.8 Slowness curve and notation................................................................... 309

Figure 10.9 Slowness curve for unidirectional 65% graphite 35% epoxy plate.

Solid line: frequency thickness product of 400 kHz-mm; Dash line:

frequency thickness product of 1700 kHz-mm. Values are 41 10c ⋅ ........ 310

Figure 10.10 Wave front surface for unidirectional 65% graphite 35% epoxy plate.

Solid blue line: frequency thickness product of 400 kHz-mm; Dash

red line: frequency thickness product of 1700 kHz-mm......................... 310

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Figure 10.11 Dispersion curves for a quasi-isotropic plate [(0/45/90/-45)2s]. (a)

output from the program; (b) elaboration of S0, SH, A0 modes. ........... 312

Figure 10.12 Phase velocities for a quasi isotropic plate. Theoretical values for

0θ = ° , 90θ = ° , 45θ = ° , and 135θ = ° ....................................................... 312

Figure 10.13 Group velocities for a quasi isotropic plate. Experimental and

theoretical values for 0θ = ° , 90θ = ° , 45θ = ° , and 135θ = ° .................... 313

Figure 10.14 Plate subject to surface tractions............................................................. 315

Figure 10.15 Experiment layout for [(0/45/90/-45)2]S, of T300/5208 Uni Tape with

2.25-mm thickness and size 1240x1240-mm.......................................... 323

Figure 10.16 Tuning Experimental data for a round PWAS for different

propagation directions. a) quasi-A0 mode; b) Quasi-S0 mode and

quasi-SH0 mode...................................................................................... 324

Figure 10.17 Experimental and theoretical tuning values. a) Experimental data for

square PWAS. Triangles: quasi-A0 mode; Circles: quasi-S0 mode;

Squares: quasi-SH0 mode. b) Experimental vs. theoretical values for

first antisymmetric mode. ....................................................................... 325

Figure 11.1 Typical probability of detection (POD) curves for increasing damage.

(Grills, 2001)........................................................................................... 330

Figure 11.2 Transducer lay out and specimen dimensions (all dimension in mm).... 334

Figure 11.3 Seeded flaw location (A, B, C, D, E) in the composite specimens......... 336

xxiii

Figure 11.4 Statistical criteria. a) 95% confidence interval of probability of

detection for increasing values of n. b) Acceptance criteria. .................. 338

Figure 12.1 Survivability and performance of PWAS under thermal fatigue. a)

Indication of survivability through resumption of resonant properties

after submersion in liquid nitrogen (PWAS, AE-15, room

temperature). b) Wave propagation in composite for various thermal

environments. Comparison of a wave packet before, during, and after

submersion in liquid nitrogen. ................................................................ 342

Figure 12.2 Unidirectional composite strips with PWAS installed. a.) Hole in the

pitch-catch path; b.) Hole off-set from the pitch-catch path. .................. 343

Figure 12.3 Experimental setup for quasi-isotropic plate experiments. The damage

sites are marked as: (i) “Hole” for a through hole of increasing

diameter; and (ii) I1, I2 for two impacts of various energy levels.......... 344

Figure 12.4 Lap joint; Teflon patches location (crosses) and PWAS location

(circles). .................................................................................................. 345

Figure 12.5 Schematic of thick composite specimen and location of Teflon inserts

(crosses). ................................................................................................. 345

Figure 12.6 DI analysis of the damaged unidirectional composite strip. a.) Hole in

the pitch-catch path; b.) Hole off-set from the pitch-catch path. ............ 348

Figure 12.7. DI values at different sizes of the hole and PWAS pairs. a) Excitation

frequency of 54 kHz. b) Excitation frequency of 255 kHz. Circle:

PWAS pair 0-13; Triangle: PWAS pair 5-8. .......................................... 351

xxiv

Figure 12.8. DI values at different hole size, Frequency 54 kHz. Pulse – echo.......... 352

Figure 12.9. Impactor. a) Base impactor with hemispherical tip; b) barrel; c)

impactor assembled................................................................................. 353

Figure 12.10 Pitch-catch DI values as a function of the damage level for two

PWAS pairs. a) Impact at site A, excitation frequency of 54 kHz.

Circle: PWAS pair 12-11; Square: PWAS pair 9-10; b) Impact at site

A, excitation frequency of 225 kHz. Circle: PWAS pair 12-11;

Square: PWAS pair 9-10; c) Impact at site B, excitation frequency of

54 kHz. Circle: PWAS pair 10-3; Square: PWAS pair 5-8; d) Impact

at site B, excitation frequency of 225 kHz. Circle: PWAS pair 10-3;

Square: PWAS pair 5-8........................................................................... 356

Figure 12.11 Pulse-echo DI values as a function of the damage level for two PWAS

pairs at damage site A. Circle: excitation frequency of 54 kHz;

Square: excitation frequency of 225 kHz................................................ 357

Figure 12.12 DI values for different damage level (PWAS pair 02 – 00) on the

composite lap-joint specimen. a) Room temperature; b) Cryogenic

temperature. Square: Excitation frequency of 60 kHz; Circle

excitation frequency of 318 kHz............................................................. 358

xxv

Figure 12.13 Composite tank interface specimen, room temperature. a) Experiment

at room temperature; square: excitation frequency of 60 kHz; circle:

excitation frequency of 318 kHz. b) Experiment at cryogenic

temperature; square: excitation frequency of 75 kHz; circle: excitation

frequency of 318 kHz.............................................................................. 361

Figure 13.1 Installation strategy. a) Sensors layout on specimen (projection view).

b) Particular of sensors 16, 17, and ground on tube................................ 363

Figure 13.2 Impedance readings before the test......................................................... 365

Figure 13.3 Visual inspection of PWAS after reading #29. a) PWAS 1 broken; b)

PWAS 12 disconnected; c) PWAS 18 broken, PWAS 19

disconnected............................................................................................ 366

Figure 13.4 Impedance readings for PWAS 0............................................................ 367

Figure 13.5 Impedance readings for PWAS 2 and 10................................................ 367

Figure 13.6 Post-processing analysis. Gray PWAS: transmitters; Black PWAS:

bad wiring; Arrows: pitch-catch direction, from the transmitter to the

receiver.................................................................................................... 369

Figure 13.7 Pitch-catch data at cryogenic temperature and strain level about 7000

μin/in. Transmitter PWAS 4, receiver PWAS 11. Frequency 45 kHz.... 369

Figure 13.8 Pith-catch at ambient temperature and no load for PWAS 02

transmitter and PWAS 04 receiver (vertical wave propagation) at

different history times. Frequency 45 kHz. ............................................ 370

xxvi


transmitter and PWAS 10 receiver (horizontal wave propagation) at



transmitter and PWAS 12 receiver (oblique wave propagation) at



transmitter and PWAS 10 receiver (horizontal wave propagation) at

different history times. Frequency 165 kHz. .......................................... 372

Figure 14.1 Durability and survivability test. a) Specimen for durability and

survivability test. b) Durability setup...................................................... 376

Figure 14.2 Impedance readings of PWAS at cryogenic temperature under

uniaxial load............................................................................................ 376

Figure 14.3 The dome-barrel specimen on the drop table. a) Transverse shock; b)

In plane shock. ........................................................................................ 377

Figure 14.4 A typical accelerometer signal................................................................ 378

Figure 14.5 The Re Z vs. Frequency before and after the test ................................... 378

Figure 14.6 The required random vibration spectrum................................................ 379

Figure 14.7 Noise spectra. a) Required noise spectrum, b) Noise spectrum

collected by a microphone during the test .............................................. 380

Figure 15.1 Specimen with PWAS installed (A – F: PWAS location) ...................... 383

xxvii

Figure 15.2 Installation procedure for configuration b .............................................. 387

Figure 15.3 PWAS bonded to specimen #3. PWAS in configuration b has a less

amount of glue than that of PWAS in configuration a. The

capacitance is 2.80 nF and 2.69 nF respectively..................................... 388

Figure 15.4 QQ-plot and residual plot........................................................................ 389

Figure 15.5 Interaction plots. a) Interaction between PWAS location and bond

type. b) Interaction between specimen and bond type............................ 390

Figure 16.1 Plate with a crack depth 1d . .................................................................... 397

Figure 16.2 Particle displacement of the incident, reflected, and transmitted SH0

wave at f=1000 kHz. a) Distance from the crack x=2 mm; b) Distance

from the crack x=5 mm. ......................................................................... 401

Figure 16.3 Two semi-infinite layers with different thickness and material

properties. (After Ditri, 1996)................................................................. 402

Figure 17.1 Generalized rays from a source O to a receiver in location (r,z). (Pao

et al., 1979) ............................................................................................. 410

Figure 17.2 Generalized theory: model of a plate excited by a force concentrated

on the lower surface and normal to it, receiver on the top surface at a

longitudinal distance 4h from the force (h = plate thickness). a) Mode

configuration; b) First two paths (1+, 2-); c) First three paths (1+, 2-,

3+); d) First four paths (1+, 2-, 3+, 4-). (Pao et al., 1979)........................ 410

Figure A.1 Equilibrium of a small element of a plate............................................... 447

xxviii

Figure C.2 Energy density amplitude variation with distance from source. Solid

lines: Bessel functions; Dash line: functions proportional to ( )2 rπ ...... 469

Figure G.3 Interaction between two PWAS and the structure through the bonding

layer: model with interfacial shear stress, ( )xτ ...................................... 514

Figure H.4 Residual plot. a) Data not transformed; b) Data transformed, DI=DI2... 520

1

1 INTRODUCTION

This dissertation addresses the Lamb wave interaction between piezoelectric wafer active

sensor (PWAS) and host structure during structural health monitoring (SHM). The scope

of my research was to develop a better understanding of the engineering variables that

influence the interaction of PWAS transducers with the structure during transducer

operation. This is a key issue needed to develop more power/energy efficient SHM

systems. PWAS-structure interaction modeling is also fundamental in achieving single

mode excitation with multi-modal guided waves. This tuning is a requirement for most of

the SHM algorithms (time reversal, phase-array, and imaging).

1.1 MOTIVATION

SHM is an emerging research area with multiple applications in civil, mechanical, and

aerospace engineering. SHM systems are able to asses the state/integrity of a structure to

facilitate life-cycle management decisions (Hall 1999). SHM systems can inform the user

of the status of structure in real time and provide an estimate of the remaining useful life

of the structure. Benefits in the application of SHM include the possibility to extend the

damage-tolerance design philosophy used for aeronautical structures to other engineering

fields and to change the maintenance procedure for aircraft from schedule driven to

condition based. This will cut down the costs of maintenance and decrease the time

required for the structure to be off-line. In the particular case of SHM applied on

composite structures, the knowledge in real time of the growth of damage can be used in

2

combination with finite element models to predict the residual life of a composite

structure in service. Until now, there are no predictive models to determine the damage

growth in composite structures; the only available methods to determine the extent of

damage is by non destructive evaluation (NDE) methods. Studies of SHM on composites

are increasing and are developed by extending the theory derived for isotropic materials

and by direct application of SHM systems to complex anisotropic structures. However,

the extension of SHM methods to these structures is not always straight forward and

profound theoretical knowledge of the physics principles is required.

SHM sets out to determine the health of a structure by reading a network of sensors that

are permanently attached onto the structure and monitored over time. The SHM system

first performs a diagnosis of the structural safety and health, followed by prognosis of the

remaining life. SHM can be performed in a passive or active way. Passive SHM are a

network of sensors that “listen” to the structure to monitor whether the component is

signaling changes. Active SHM uses a network of active sensors that interrogate the

structural health through active sensors and thus determine the presence or absence of

damage.

The ultrasonic-based active SHM method uses PWAS to transmit and receive guided

waves in a thin-wall structure. PWAS are small, light transducers and they are less

expensive than conventional ultrasonic transducers. PWAS are permanently bonded or

embedded in a structure to perform on-demand structural interrogation. To perform

SHM, it is envisaged PWAS transducers are deployed over wide areas. Design of energy-

efficient autonomous PWAS networks requires an understanding and predictive modeling

of the power and energy transduction between the PWAS and the multi-mode guided

3

waves present in the structure during SHM. Up to date work on transducer SHM

technology (Dugnani, 2009; Liu et al., 2008; Lu et al., 2008; Park et al., 2009; Giurgiutiu,

2008) has not yet systematically addressed the modeling of power and energy

transduction. This topic has been addressed to a certain extent in classical NDE

(Viktorov, 1967; Auld, 1990; Rose, 1999). Classical NDE analysis has not studied in

detail the power flow between transducer and structure because the coupling-gel interface

did not have clearly predictable behavior; power was not generally an issue, since NDE

devices are not meant to operate autonomously on harvested power.

To address these issues, our research focus on developing a transducer-structure

interaction model that it is valid for any configuration (frequency of the wave

propagation, material of the structure, transducer geometry, and material). The interaction

between transducer and structure is determined through the derivation of the tuning

curves and the function of the load transferred from the transducer to the substrate

through the connecting media. The load transfer theory is developed through the use of

the guided-wave normal modes expansion (NME) theory presented by Auld (1990). The

knowledge of the behavior of the coupling-interface between PWAS and structure will

help us to determine the power flow in the SHM system.

1.2 RESEARCH GOAL, SCOPE, AND OBJECTIVES

The goal of this research is to understand, model, and predict the interaction between

piezoelectric wafer active sensors (PWAS), host structure, and ultrasonic guided waves

propagating in the structure.

4

The scope of this research is to develop predictive models of PWAS excitation of

ultrasonic guided waves, to apply them to the SHM of complex structures under extreme

environments, and to determine the reliability and survivability of SHM systems.

The objectives of this research are defined as follows:

1. To present the detail modeling of guided waves propagating in a generic medium.

Specify the derivation for both straight-crested and circular-crested guided waves.

2. To understand the power flow and energy conservation for guided waves

propagating in a media, the reciprocity relation of ultrasonic acoustic fields, and

the orthogonality relation for of both straight-crested and circular-crested guided

waves and develop appropriate mathematical formulations.

3. To derive, through normal modes expansion, the shear layer coupling between

PWAS and structure valid at any frequency and number of modes present.

4. To extend the theory for tuning PWAS transducers with ultrasonic guided waves

in composite structures.

5. To develop methods for using PWAS transducers to detect damage in composite

structures under different environment conditions and damage types.

6. To demonstrate that PWAS-based SHM systems can withstands qualification tests

for space applications.

7. To asses acoustic emission (AE) detection with PWAS transducers and explore

methods to determine the guided waves scattering from damage in the structure.

5

1.3 DISSERTATION LAYOUT

To accomplish the objectives set forth in the preceding section, the dissertation is

organized in sixteen chapters divided in three parts.

In Part I, we develop a generic formulation for ultrasonic guided waves propagating

in thin wall structures.

In Chapter 2, we present the generic derivation of the acoustic field equations. The

derivation is generic because we do specify neither the structure nor the coordinate

system.

In Chapter 3, the acoustic field equations are specified for the case of thin wall

structures and the solutions are derived for both straight-crested and circular-crested

guided waves propagating in an isotropic medium.

In Chapter 4, we derive the generic acoustic Poynting theorem valid for any wave

systems. The power flow formulation is then explicitly derived for both straight-crested

and circular-crested guided waves. The latter derivation has not been yet presented in

literature for structural guided waves: in our derivation, the waves propagate in the radial

direction and the wave front energy amplitude decreases with the increasing wave front

length.

In Chapter 5, we set the basis for modal analysis of wave fields excited by external

sources. We derive the real and complex reciprocity relations. As in the previous

chapters, the discussion is first held at the general level (no specific structure and

coordinate system); then, we derive the relations for guided waves propagating in

isotropic structures.

6

In Chapter 6, the theory developed in Chapter 5 is used to verify that wave guided

modes are a set of orthogonal functions. Orthogonality derivations without reference to

the particular solution are given.

In Part II, we address some unresolved issues of the PWAS SHM predictive

modeling.

In Chapter 7, we develop the theory of guided wave excitation with PWAS

transducers. First we present the main characteristics of PWAS transducers, then we

express PWAS excitation through the normal mode expansion theory.

In Chapter 8, we address the shear layer coupling between PWAS and structure. We

first define the problem, i.e., a PWAS transducer permanently bonded to a thin plate

exciting Lamb waves in the structure. We recall the shear-lag solution for the case of low

frequency approximation when only axial and flexural waves are present (Crawley et al.,

1987). We show the limitations of this solution and we extend the theory to the case of

two Lamb wave modes present, one symmetric and one antisymmetric. To overcome the

challenges of this new model, we derive a new method based on NME that is valid at any

frequency and in the presence of N generic modes. An extensive discussion on the

derived methods is provided along with a discussion on the stress distribution in the

bonding layer.

In Chapter 9, we address the theoretical prediction of tuning between PWAS and

structure. We derive three models for the prediction of tuning: the first is the simple pin-

force model; the second is the shear-lag model with low frequency approximation; the

third is the N generic model based on the shear-lag model derived in the previous chapter.

After discussing the main advantages of each model, we compared the prediction of

7

tuning between PWAS and guided waves with the experimental data. The results of our

improved model were in excellent agreement with the experimental data.

In Chapter 10, the acoustic field equations derived in Chapter 1 are extended to the

case of composite structures. We derive dispersion curves for various different types of

composite structures and compare the curves with literature results and experimental

data. We extend the NME theory to the case of composite structures and derived the

theoretical predictions of tuning between PWAS and composites. We show good

agreement between theoretical predictions and experimental data.

In Part III, we describe SHM applications and related issues.

In Chapter 11, we addressed the issue of reliability of SHM system. We discus the

lack of quality specifications for SHM inspections. Since the best practice so far is to

adapt the specifications derived for NDE, we consider an illustrative example of how

probability of detection (POD) curves can be derived for SHM of composite structures.

In Chapeter12, we explore the ability of PWAS-based SHM to perform during space

missions. We tested the PWAS performance for damage detection on different type of

composite structures and under different environmental conditions.

In Chapter 13, we tested the survivability of SHM systems in extreme environments

(up to -300 F) and under high stress conditions (up to 7000 μin/in) on subcomponent

specimen.

In Chapter 14, we assessed the durability of the entire PWAS-based SHM system

through the fatigue thermal loads, shock test, random vibrations, thermal stress, and

acoustic stress.

8

In Chapter 15, we discuss in situ reliability of PWAS transducers. We study the effect

of partial bonding between PWAS and structure on the electric capacitance and derive

capacitance ranges for both good and partial bond layer.

In Chapter 16, we present the mode decomposition method for shear horizontal waves

scattering from a crack and Lamb waves scattering from a change in material properties.

In Chapter 17, we develop models for acoustic emission in an infinite plate. The

models are developed through integral displacement theory and NME method.

9

PART I ELASTIC WAVES FOR STRUCTURAL HEALTH MONITORING

10

Elastic waves in solids have been extensively studied since the late eighteen century.

However, only in the last few decades, ultrasonic wave propagation for structural health

monitoring (SHM) have been started to be studied. Structural health monitoring is a non

destructive method to asses the integrity of a structure. The main difference between

SHM and the conventional non destructive evaluation (NDE) methods is that SHM can

be performed while the structure is in service. For this reason, the transducers used for

SHM are permanently connected to the structure itself. Aim of SHM system is to asses

the presence of damage, determine the geometry of the damage, and the quality, i.e.,

corrosion, crack, delamination. Different methods are available to determine the extent

and quality of damage, however, to obtain an efficient method (i.e., few transducer and

low power consumption with optimum detection capabilities), the knowledge of the wave

propagation mechanism must be known deeply.

The first part of my research is based on the fundamental knowledge of wave

propagation in elastic structures. We start from the acoustic fields equations derivation in

tensor notation. These equations are written in a generic form so that they can be

specified for the particular problem of interest. A specific derivation for the case of

guided waves propagating in isotropic materials is at the same time derived. The

derivation is made for both straight crested and circular crested guided waves since we

are interested to further extend the theory presented from 1D model to a 2D model.

Once the basis of wave propagation has been presented, we discuss in detail the

power flow and energy conservation of the wave field. In literature power flow and

11

energy conservation are a quite established research topics. However, we found a gap in

the power flow for circular crested wave propagation. Most of the derivations made in

cylindrical coordinates deal with the problem of wave propagation in cylinders or tubes.

In these cases, the treatment and solution of power flow is quite similar to that of straight

crested wave propagation since the wave front of the wave remains constant in length and

harmonic in the direction of the wave propagation (hence, along the tube or cylinder). In

the case of circular crested wave propagation, the derivation of power flow and, in

particular, the energy conservation theory are more complicated because the wave filed is

not harmonic in the direction of the wave propagation and the wave front length is not

constant.

We also derive the fundamental theorem of reciprocity relation for the guided wave

fields and we prove the orthogonality of the wave fields. The results are derived in a

generic formulation and for both straight and circular crested waves. The theory

developed in Part I is used in Part II to develop the normal mode expansion method for

1D and 2D model and for both isotropic and anisotropic materials. The normal mode

expansion model is used to derive tuning curves for any transducer-structure

configuration.

12

2 ACOUSTIC FIELD EQUATIONS

In this section we derive the acoustic wave field equations for a generic solid media

under the excitation of body forces F . To derive these equations, we need the expression

of the equation of motion, strain-displacement equation, and the Hooks’s law. Here, we

present the generic derivation of acoustic field equations that is valid for both isotropic

and anisotropic materials in any coordinate system.

2.1 EQUATION OF MOTION

Consider an elastic solid subjected to a volume force, the equation of motion in tensor

notation is given by

2

2tδρδ

∇ ⋅ = −uT F (2.1)

where ρ is the volume density; T is the stress tensor, ijT=T for , 1, 2,3i j = ; u is the

displacement vector, iu=u for 1,2,3i = ; and F is the body force (force per volume),

iF=F for 1,2,3i = . The operator ∇ is the gradient operator. This operator depends on

the particular coordinate system we are considering. Consider the rectangular coordinate

system xyz, where the coordinates x, y, and z coincide respectively with the coordinates 1,

2, and 3, the gradient is then expressed as

T

x y z∂ ∂ ∂⎧ ⎫∇ = ⎨ ⎬∂ ∂ ∂⎩ ⎭

(2.2)

13

and

xyxx xz

xx xy xzyx yy yz

xy yy yz

xz yz zzzyzx zz

TT Tx y zxT T T

T T TT T T

y x y zT T T

TT Tz x y z

∂⎧ ⎫∂ ∂⎧ ⎫∂ + +⎪ ⎪⎪ ⎪ ∂ ∂ ∂∂ ⎪ ⎪⎡ ⎤ ⎪ ⎪ ⎪ ⎪∂ ∂ ∂∂⎢ ⎥ ⎪ ⎪ ⎪ ⎪∇ ⋅ = ⋅∇ = = + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ∂ ∂ ∂ ∂⎪ ⎪ ⎪ ⎪⎢ ⎥⎣ ⎦ ⎪ ⎪ ⎪ ⎪∂ ∂∂ ∂+ +⎪ ⎪ ⎪ ⎪

∂⎩ ⎭ ∂ ∂ ∂⎪ ⎪⎩ ⎭

T T (2.3)

Note that Equation (2.3) took advantage of the property ∇ ⋅ = ⋅∇T T which is true since

the T matrix is symmetric. We can represent the stress tensor as a column using Voigt

notations, i.e.,

1

2

3

4

5

6

xx

yy

zz

yz

xz

xy

T TT TT TT TT TT T

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪

⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭

T (2.4)

and define the del dot operator for Voigt notations as

0 0 0

0 0 0

0 0 0

x z y

y z x

z y x

⎡ ⎤∂ ∂ ∂⎢ ⎥∂ ∂ ∂⎢ ⎥⎢ ⎥∂ ∂ ∂

∇⋅ = ⎢ ⎥∂ ∂ ∂⎢ ⎥⎢ ⎥∂ ∂ ∂⎢ ⎥∂ ∂ ∂⎣ ⎦

(2.5)

Premultiplication of Equation (2.4) by Equation (2.5) leads to the same expression as in

Equation (2.3), i.e.,

14

0 0 0

0 0 0

0 0 0

xx xyxx xz

yy

zz yx yy yz

yz

xz zyzx zz

xy

T TT TT x y zx z yT T T TTy z x x y zT TT T

z y x T x y z

∂⎧ ⎫ ⎧ ⎫∂ ∂⎡ ⎤∂ ∂ ∂ + +⎪ ⎪ ⎪ ⎪⎢ ⎥ ∂ ∂ ∂∂ ∂ ∂ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪∂ ∂ ∂⎢ ⎥∂ ∂ ∂ ⎪ ⎪ ⎪ ⎪∇ ⋅ = = + +⎨ ⎬ ⎨ ⎬⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥∂ ∂ ∂ ∂∂ ∂

+ +⎪ ⎪ ⎪ ⎪⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂⎣ ⎦ ⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭

T (2.6)

Otherwise, consider a cylindrical coordinate system r zθ , where the coordinates r, θ, and

z coincide respectively with the coordinates 1, 2, and 3. We define the del dot operator

directly in Voigt notation, i.e.,

( )

( )

( )

1 1 10 0

1 1 10 0 0

1 10 0 0

rr r r z r

rr z r r r

rz r r r

θ

θ

θ

⎡ ⎤∂ ∂ ∂−⎢ ⎥∂ ∂ ∂⎢ ⎥

∂ ∂ ∂⎢ ⎥∇⋅ = +⎢ ⎥∂ ∂ ∂⎢ ⎥∂ ∂ ∂⎢ ⎥

⎢ ⎥∂ ∂ ∂⎣ ⎦

(2.7)

and the stress tensor in Vogit notation, i.e.,

1

2

3

4

5

6

rr

zz

z

rz

r

T TT TT TT TT TT T

θθ

θ

θ

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

T (2.8)

Multiplication of Equation (2.8) by Equation (2.7) leads to the left hand side term of


15

rrr rz

r r z

zrz zz

T TrT Tr r r r zrT T T Tr r r r z

TrT Tr r r z

θθ θ

θ θ θθ θ

θ

θ

θ

θ

∂∂ ∂⎧ ⎫− + +⎪ ⎪∂ ∂ ∂⎪ ⎪∂ ∂ ∂⎪ ⎪∇ ⋅ = + + +⎨ ⎬∂ ∂ ∂⎪ ⎪

∂∂ ∂⎪ ⎪+ +⎪ ⎪∂ ∂ ∂⎩ ⎭

T (2.9)

For more details on the derivation of the equation of motion in cylindrical coordinates see

Appendix A. Note that the equation of motion (2.1) is valid in both coordinate systems

only in the Vogit notation.

2.2 STRAIN-DISPLACEMENT EQUATION

The strain and the displacement components are linked through the strain-displacement

relation. Call S the strain tensor, such that in Vogit notation

s= ∇S u (2.10)

where s∇ is the symmetric del operator defined in rectangular coordinates as

0 0 0

0 0 0

0 0 0

T

s

x z y

y z x

z y x

∂ ∂ ∂⎡ ⎤⎢ ⎥∂ ∂ ∂⎢ ⎥

∂ ∂ ∂⎢ ⎥∇ = ⎢ ⎥∂ ∂ ∂⎢ ⎥∂ ∂ ∂⎢ ⎥

⎢ ⎥∂ ∂ ∂⎣ ⎦

(2.11)

and in cylindrical coordinates as

1 20 0

1 20 0 0 2

10 0 0

T

s

r r z r

r z r r

z r r

θ

θ

θ

∂ ∂ ∂⎡ ⎤⎢ ⎥∂ ∂ ∂⎢ ⎥

∂ ∂ ∂⎢ ⎥∇ = −⎢ ⎥∂ ∂ ∂⎢ ⎥∂ ∂ ∂⎢ ⎥⎢ ⎥∂ ∂ ∂⎣ ⎦

(2.12)

16

2.3 HOOKE’S LAW

The generic Hooke’s law can be written as

ij ijkl kl ijkl klkl

c cσ ε ε= =∑ (2.13)

In tensor notation, Equation (2.13) can be written as

:=T c S (2.14)

where S and T are the 2nd rank strain and stress tensors, whereas c is the 4th rank

stiffness tensor. The double dot product designated by the period symbol : indicates that

the double index summation is applied. Conversely, we can write

:=S s T (2.15)

where 1−=s c is the 4th rank compliance tensor.

Through the use of Voigt matrix notation, the 4th rank stiffness tensor is reduced to a 2nd

rank tensor (i.e., a matrix) and the 2nd rank stress and strain tensors are reduced to 1st rank

tensors (i.e., rows or columns, as appropriate). It can be shown from fundamental

principles that the stiffness matrix should be symmetric. For isotropic materials (e.g., a

metallic plate), the stiffness matrix further simplifies to

11 12 13

12 22 23

13 23 33

44

55

66

0 0 00 0 00 0 0

0 0 0 0 00 0 0 0 00 0 0 0 0

c c cc c cc c c

cc

cc

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(2.16)

17

where 11 22 33 2c c c λ μ= = = + , 12 23 13c c c λ= = = , and 44 55 66c c c μ= = = , with λ and μ

being the Lame constants of the material.

2.4 ACOUSTIC FIELD EQUATIONS SUMMARY

The equation of motion (2.1) and the strain-displacement relation (2.10) form the acoustic

field equations, i.e.,

2

2

s

tρ

⎧ ∂∇ ⋅ = −⎪

∂⎨⎪∇ =⎩

uT F

u S (2.17)

Through use of the constitutive Equation (2.15) (Hooke’s law), we eliminate the

unknown strain S between Equations in (2.17) and we obtain the acoustic field equations

only in the unknown stress T and particle displacement u , i.e.,

2

2

:s

tρ

⎧ ∂∇ ⋅ = −⎪

∂⎨⎪∇ =⎩

uT F

u s T (2.18)

It is to note that Equation (2.18) is independent of the material under consideration and

the coordinate system used.

18

3 GUIDED WAVES IN PLATES

In this section, we apply the acoustic field equations derived in Section 2 to the case of

guided waves in isotropic material. Guided waves are waves that travel in a media

bounded by two surfaces at a given distance; hence the waves are guided between the top

and bottom surfaces. First, we consider the case of straight-crested guided waves in

rectangular coordinates, and then that of circular-crested guided waves in cylindrical

coordinates.

3.1 STRAIGHT-CRESTED GUIDED WAVES IN RECTANGULAR COORDINATES

In this part, we derive the acoustic field equations for plane waves propagating in an

isotropic infinite plate. The solution to the acoustic field equations is reported without

derivation since this can be found in several textbooks (Giurgiutiu, 2008; Graff, 1991).

Waves propagating in a plate of a finite thickness 2d are called guided waves because

they are guided between the top and bottom surfaces. Consider a rectangular coordinate

system such as that the x coordinate is along the wave propagation and the y coordinate is

parallel to the thickness of the plate. Figure 3.1 shows the coordinate system.

Figure 3.1 Plane wave notations

y d= +

y d= −

19

3.1.1 Equation of motion

Assume z-invariance such as 0z

∂=

∂, equation of motion (2.1) with the use of relation

(2.3) becomes

2

2

2

2

2

2

xyxx xx

xy yy yy

yzxz zz

TT u Fx y t

T T uF

x y tTT u F

x y t

ρ

ρ

ρ

∂⎧∂ ∂+ = −⎪ ∂ ∂ ∂⎪

⎪∂ ∂ ∂⎪ + = −⎨ ∂ ∂ ∂⎪⎪ ∂∂ ∂

+ = −⎪∂ ∂ ∂⎪⎩

(3.1)

System (3.1) has three equations: the first two are coupled through the term xyT , while the

third is uncoupled from the others. The first two equations are the equations of motion for

straight crested Lamb waves. These waves propagate along the x coordinate and they

have particle displacement in both the x and y directions (denoted by xu and yu

respectively). The third equation in system (3.1) represents the equation of motion for

straight crested shear horizontal (SH) waves. The SH waves propagate along the x

direction with particle displacement along the z direction (denoted by zu ).

3.1.2 Strain-displacement equation

Consider the strain-velocity relation (2.10), with the use of Equation (2.11) and the z-

invariant condition, this can be expanded as

20

0

xxx

yyy

zz

uSxu

Sy

S

∂⎧ =⎪ ∂⎪∂⎪ =⎨ ∂⎪

⎪ =⎪⎩

2

2

2

zyz

zxz

yxxy

uSyuSx

uuSy x

∂⎧ =⎪ ∂⎪∂⎪ =⎨ ∂⎪

∂⎪ ∂= +⎪ ∂ ∂⎩

(3.2)

We noticed that in Equation (3.1) the terms in xx , xy , and yy are not coupled with the

terms in xz , and yz , hence the fourth and fifth equation in (3.2) are decoupled from the

other three and they represent the strain-displacement relation for SH waves. Likewise,

the first, second and sixth equations represent the Lamb wave strain-displacement

relation.

3.1.3 Hooke’s law

Consider z-invariant plain strain conditions such as 3 0zzz

uS Sz

∂= = =

∂. With the help of

Equations (2.14) and (2.16), Hook’s law can be expressed as:

( )( )

11 12

12 22

13 23

44

55

66

2

2

2 2

2 22 2

xx xx yy xx yy

yy xx yy xx yy

zz xx yy xx yy

yz yz yz

xz xz xz

xy xy xy

T c S c S S S

T c S c S S S

T c S c S S S

T c S S

T c S ST c S S

λ μ λ

λ λ μ

λ λ

μ

μμ

⎧ = + = + +⎪

= + = + +⎪⎪ = + = +⎪⎨

= =⎪⎪ = =⎪⎪ = =⎩

(3.3)

3.1.4 Acoustic field equations

The acoustic filed equations are derived by substituting Hook’s law Equation (3.3) into

the Equation of motion (3.1), i.e.,

21

( )

( )

2

2

2

2

2

2

2 2

2 2

2 2

yy xyxx xx

xy yy yxxy

yzxz zz

S SS u Fx x y t

S S uS Fx y y t

SS u Fx y t

λ μ λ μ ρ

μ λ λ μ ρ

μ μ ρ

∂ ∂⎧ ∂ ∂+ + + = −⎪ ∂ ∂ ∂ ∂⎪

⎪ ∂ ∂ ∂∂⎪ + + + = −⎨ ∂ ∂ ∂ ∂⎪⎪ ∂∂ ∂

+ = −⎪∂ ∂ ∂⎪⎩

(3.4)

Substitute the strain-displacement Equation (3.2) into (3.4) and rearrange the terms to

obtain

( )

( )

2 22 2 2

2 2 2

2 2 22 2

2 2 2

2 2 2

2 2 2

2

2

y yx x xx

y y yx xy

z z zz

u uu u u Fx y x y x y t

u u uu u Fx y x y x y tu u u Fx y t

λ μ μ λ μ ρ

μ λ μ λ μ ρ

μ μ ρ

⎧ ∂ ∂∂ ∂ ∂+ + + + = −⎪

∂ ∂ ∂ ∂ ∂ ∂ ∂⎪⎪ ∂ ∂ ∂∂ ∂⎪ + + + + = −⎨ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎪⎪ ∂ ∂ ∂

+ = −⎪∂ ∂ ∂⎪⎩

(3.5)

The uncoupling between the acoustic wave equations is even more evident in Equation

(3.5). The first two equations depend on both x and y coordinates and they represent the

Lamb waves equations. The third Equation (3.5) depends only on z coordinate and

represents the SH waves acoustic field equations. The Lamb waves equations of motion

and the SH waves equation of motion can be solved separately.

To derive the particle displacement, we consider a plate not subject to body forces,

hence 0=F . Moreover, we assume that the top and bottom surfaces of the plate are free

surfaces, hence the boundary conditions are

0

0

0

yy y d

xy y d

yz y d

T

T

T

=±

=±

=±

⎧ =⎪⎪ =⎨⎪

=⎪⎩

(3.6)

22

3.1.5 Shear horizontal waves solutions

Solution for the SH waves are found from the third equation in (3.5), i.e.,

2 2 2

2 2 2 2

1z z z

s

u u ux y c t

∂ ∂ ∂+ =

∂ ∂ ∂ (3.7)

where sc is the phase velocity defined as

sc μρ

= , (3.8)

The displacement is assumed to be harmonic both in time and in the x coordinate, i.e.,

( )( , , ) ( ) i x tz zu x y t U y e ξ ω−= (3.9)

where ξ is the wavenumber and ω is the radial frequency. Solution of Equation (3.7)

with the assumption in Equation (3.9) is equal to

( , , ) sin cosA Sn ni x i xA S i t

z n n n nn

u x y t A ye B ye eξ ξ ωη η −⎡ ⎤= +⎣ ⎦∑ (3.10)

where superscript A and S stand for antisymmetric and symmetric mode respectively, and

2 2 2 2scη ω ξ= − . From the boundary condition (3.6) we found

( )

22

2 12

S

A

nd

nd

πη

πη

=

= + 0,1,n = K , (3.11)

Constants nA and nB can be determined through the normalization factor as discussed in

Appendix E. With the use of (3.11) and the definition of η , we find the expression of the

wave velocity with respect to the frequency to be

23

( )

2 2 2 2 2

22 2 2 2

24

4

16 2 1

Ss

s

As

s

fdc cf d n c

fdc cf d n c

ππ

π

π

=−

=− +

0,1,n = K , (3.12)

Figure 3.2 shows the dispersion curves for H waves propagating in aluminum plate. The

first symmetric mode is constant while all the other modes vary with the frequency and

the all present a cut-off frequency. Below the cut-off frequency, the mode becomes

evanescent, i.e., the wavenumber is imaginary.

Figure 3.2 Dispersion curves of SH waves propagating in an aluminum plate. Dash lines:

antisymmetric modes; solid lines: symmetric modes.

From solution of the SH waves particle displacement (3.10), we can derive the

stresses associated with the waves, i.e., xzT and yzT . Substitute Equation (3.10) into the

strain-displacement Equation (3.2) to get

fd (kHz-mm)

c/c s

2000 4000 60000.6

0.85

1.1

1.35

1.6

24

2 cos sin

2 sin cos

A Sn n

A Sn n

i x i xA A S S i tyz n n n n n n

n

i x i xA A S S i txz n n n n n n

n

S A ye B ye e

S i A ye B ye e

ξ ξ ω

ξ ξ ω

η η η η

ξ η ξ η

−

−

⎧ ⎡ ⎤= −⎣ ⎦⎪⎨

⎡ ⎤= +⎪ ⎣ ⎦⎩

∑

∑ (3.13)

Substitute (3.13) into the expression of the strain derived in Hook’s law Equation (3.3),

after rearrangement we obtain

cos sin

sin cos

A Sn n

A Sn n

i x i xA A S S i tyz n n n n n n

n

i x i xA A S S i txz n n n n n n

n

T A ye B ye e

T i A ye B ye e

ξ ξ ω

ξ ξ ω

μ η η η η

μ ξ η ξ η

−

−

⎧ ⎡ ⎤= −⎣ ⎦⎪⎨

⎡ ⎤= +⎪ ⎣ ⎦⎩

∑

∑ (3.14)

3.1.6 Lamb waves solutions

Solution for the Lamb waves are found from the first and second equations in (3.5), i.e.,

( )

( )

2 22 2 2

2 2 2

2 2 22 2

2 2 2

2

2

y yx x x

y y yx x

u uu u ux y x y x y t

u u uu ux y x y x y t

λ μ μ λ μ ρ

μ λ μ λ μ ρ

⎧ ∂ ∂∂ ∂ ∂+ + + + =⎪

∂ ∂ ∂ ∂ ∂ ∂ ∂⎪⎨

∂ ∂ ∂∂ ∂⎪ + + + + =⎪ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎩

(3.15)

The solution to Lamb waves equations of motion is found by assuming that the

displacement is time-harmonic and by expressing the displacement in the x and y

directions through two scalar potentials. The derivation of the particle displacement can

be found in many textbooks on wave propagation. We remand the reader to Giurgiutiu

(2008) for details. Hereunder we report the particle displacement solution for symmetric

and antisymmetric waves propagation respectively, i.e.,

( ) ( )

( ) ( )

( , , ) cos cos

( , , ) sin sin

Sn

Sn

i x tx n Sn Sn Sn Sn Sn

n

i x ty n Sn Sn Sn Sn Sn

n

u x y t B y R y e

u x y t i B y R y e

ξ ω

ξ ω

ξ α β β

α α ξ β

−

−

⎧ = − −⎪⎨

= − +⎪⎩

∑

∑ (3.16)

25

( ) ( )

( ) ( )

( , , ) sin sin

( , , ) cos cos

An

An

i x tx n An An An An An

n

i x ty n An An An An An

n

u x y t A y R y e

u x y t i A y R y e

ξ ω

ξ ω

ξ α β β

α α ξ β

−

−

⎧ = − −⎪⎨

= +⎪⎩

∑

∑ (3.17)

where

2

2 22pc

ωα ξ= − and 2

2 22sc

ωβ ξ= − , (3.18)

pc is the axial phase velocity defined as

2pc λ μ

ρ+

= (3.19)

The symmetric and antisymmetric eigenvalues (wavenumbers Snξ and Anξ ) are derived

from the solution of the Rayleigh-Lamb equation for symmetric and antisymmetric

modes respectively, i.e.,

( )22 2

2

tantan 4

dd

ξ βαβ ξ αβ

−= − , (3.20)

( )

2

22 2

tan 4tan

dd

α ξ αββ ξ β

= −−

, (3.21)

.The Rayleigh-Lamb equation is obtained by imposing the stress-free top and bottom

surfaces boundary conditions, Equation (3.6). Substitution of the eigenvalues into

Equation (3.6) yields the eigen-coefficients SnR and A

nR , i.e.,

( )2 2 cos

2 cosS S S

SS S S

dR

dξ β α

ξ β β

−= (3.22)

26

( )2 2 sin

2 sinA A A

AA A A

dR

dξ β α

ξ β β

−= (3.23)

Constants nA and nB can be determined through the normalization factor as discussed in

Appendix E.

The wavenumbers and phase velocities of the Lamb waves modes are derived from

the Rayleigh-Lamb Equations (3.20) and (3.21). The Rayleigh-Lamb roots can be real,

imaginary, or complex. Real roots are the wavenumbers of propagating waves; pure

imaginary roots are the wavenumbers of evanescent waves; and complex roots are the

wavenumbers of attenuated waves.

Figure 3.3 shows the phase velocity versus the frequency-thickness product of

propagating wave modes in an aluminum plate. Except for the first symmetric and

antisymmetric modes and the second symmetric mode (S1), all other modes present a

threshold value, cut-off frequency, below which the mode becomes evanescent (i.e.

imaginary wavenumber). The phase velocity of wave mode S1 approaches the cut-off

frequency limit from the left and has a minimum value of frequency at which double real

root is present. From this point, a complex branch of the S1 mode is seen originating (see

Graff 1975). Below the first cut-off frequency, only S0 and A0 propagating modes exist.

At low frequencies, the A0 mode can be approximated by the flexural plate waves where

the velocity changes with the square root of the frequency, while the S0 can be

approximated by the axial mode where the velocity is constant. When the dependence of

the S0 mode velocity with frequency is almost constant, the S0 mode can be considered

almost non-dispersive, while the A0 mode remains dispersive.

27

Figure 3.3 Dispersion curves of Lamb waves propagating in an aluminum plate. Dash lines:

antisymmetric modes; solid lines: symmetric modes. a) Dispersion curves for the

frequency range 0-4000 kHz-mm b) Dispersion curves below the first cut-off

frequency (fd<780 kHz-mm).

Figure 3.4 shows how the S0 and A0 waves change with time. The S0 wave amplitude

and number of peaks remain constant while the A0 wave amplitude decreases with time

and the A0 wave number of peaks increases with time..

Figure 3.4 S0 and A0 wave propagation at the low frequencies. a) Wave propagation of non-

dispersive S0 mode. b) Wave propagation of dispersive A0 mode

a) b) μsec μsec

c/c s

c/c s

a) b) (kHz-mm)fd (kHz-mm)fd

A0

A0 S0

S0

S1

S2 S3 A1

A2

A3

28

From solution of the Lamb waves particle displacement (3.16) and (3.17) we can derive

the stresses associated with the waves, i.e., xxT , xyT ,and yyT . Substitute Equation (3.16)

into the strain-displacement Equation (3.2) to get the symmetric strains, i.e.,

( ) ( )

( ) ( )

( )( ) ( )

2

2 2

cos cos

cos cos

2 2 sin sin

Sn

Sn

Sn

i x txx n Sn Sn Sn Sn Sn Sn

n

i x tyy n Sn Sn Sn Sn Sn Sn

n

i x txy n Sn Sn Sn Sn Sn Sn Sn

n

S i B y R y e

S i B y R y e

S B y R y e

ξ ω

ξ ω

ξ ω

ξ ξ α β β

α α ξ β β

ξ α α ξ β β

−

−

−

⎧= − −⎪

⎪⎪ = − +⎨⎪⎪ = + −⎪⎩

∑

∑

∑

(3.24)

Substitute the expression of the strain derived in Equation (3.24) into the Hook’s law

Equation (3.3), after rearrangement we obtain

( )( ) ( )

( )( ) ( )

( ) ( )

( )

2 2

2 2

2 2

2 2

2 cos 2 cos

2 cos 2 cos

cos

2 sin sin

Sn

Sn

Sn

i x txx n Sn Sn Sn Sn Sn Sn Sn

n

i x tyy n Sn Sn Sn Sn Sn Sn Sn

n

i x tzz n Sn Sn Sn

n

xy n Sn Sn Sn Sn Sn Sn

T i B y R y e

T i B y R y e

T i B ye

T B y R

ξ ω

ξ ω

ξ ω

λ μ ξ λα α μ ξ β β

λξ λ μ α α μ ξ β β

λ ξ α α

μ ξ α α ξ β β

−

−

−

⎡ ⎤= − + + −⎣ ⎦

⎡ ⎤= − + + +⎣ ⎦

= − +

= + −

∑

∑

∑

( ) ( )Sni x tSn

ny e ξ ω−

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∑

(3.25)

Note that

( ) ( )2 2 2 2 22 2λ μ ξ λα μ ξ β α+ + = + − (3.26)

( ) ( )2 2 2 22λξ λ μ α μ ξ β+ + = − − (3.27)

hence Equations (3.25) becomes

29

( ) ( )

( ) ( )

( ) ( )

( )( )

2 2 2

2 2

2 2

2 2

2 cos 2 cos

cos 2 cos

cos

2 sin sin

Sn

Sn

Sn

i x txx n Sn Sn Sn Sn Sn Sn Sn Sn

n

i x tyy n Sn Sn Sn Sn Sn Sn Sn

n

i x tzz n Sn Sn Sn

n

ixy n Sn Sn Sn Sn Sn Sn Sn

T i B y R y e

T i B y R y e

T i B ye

T B y R y e

ξ ω

ξ ω

ξ ω

μ ξ β α α ξ β β

μ ξ β α ξ β β

λ ξ α α


−

−

−

⎡ ⎤= − + − −⎣ ⎦

⎡ ⎤= − −⎣ ⎦

= − +

= + −

∑

∑

∑( )Sn x t

n

ξ ω−

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∑

(3.28)

Likewise, to obtain the antisymmetric stresses substitute Equation (3.17) into the strain-

displacement Equation (3.2) to get the antisymmetric strains, i.e.,

( ) ( )

( ) ( )

( )( ) ( )

2

2

2 2

sin sin

sin sin

2 2 cos cos

An

An

An

i x txx n An An An An An An

n

i x tyy n An An An An An An

n

i x txy n An An An An An An An

n

S i A y R y e

S i A y R y e

S A y R y e

ξ ω

ξ ω

ξ ω

ξ α ξ β β

α α ξ β β

ξ α α ξ β β

−

−

−

⎧= − −⎪

⎪⎪ = − +⎨⎪⎪ = − + −⎪⎩

∑

∑

∑

(3.29)



( )( ) ( )

( )( ) ( )

( )( )( )

2 2

2 2

2 2

2 2

2 sin 2 sin

2 sin 2 sin

sin

2 cos cos

An

An

An

i x txx n An An An An An An An

n

i x tyy n An An An An An An An

n

zz n An An Ann

i xxy n An An An An An An An

T i A y R y e

T A y R y e

T i A y

T A y R y e

ξ ω

ξ ω

ξ

λ μ ξ λα α μ ξ β β

λξ λ μ α α μ ξ β β

λ ξ α α


−

−

⎡ ⎤= − + + −⎣ ⎦

⎡ ⎤= − + + +⎣ ⎦

= − +

= − + −

∑

∑

∑( )t

n

ω−

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∑

(3.30)

Substitute Equations (3.26) and (3.27) into (3.30) to get

30

( ) ( )

( ) ( )

( )( )( ) ( )

2 2 2

2 2

2 2

2 2

2 sin 2 sin

sin 2 sin

sin

2 cos cos

An

An

An

i x txx n An An An An An An An An

n

i x tyy n An An An An An An An

n

zz n An An Ann

i x txy n An An An An An An An

n

T i A y R y e

T A y R y e

T i A y

T A y R y e

ξ ω

ξ ω

ξ ω


μ ξ β α ξ β β

λ ξ α α


−

−

−

⎡ ⎤= − + − −⎣ ⎦

⎡ ⎤= − −⎣ ⎦

= − +

= − + −

∑

∑

∑

∑

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

(3.31)

3.1.7 Approximate simplified solution

In this section, we present the simplified solution for axial and flexural waves, i.e. low

frequency range of S0 and A0 modes. For the case of the flexural wave, we will derive

the solution with and without the Euler-Bernoulli assumption.

3.1.7.1 Straight-crested wave plates: axial and shear waves

Consider a differential element of plate thickness t=2d subjected to normal and shear

forces as shown in Figure 3.5.

Figure 3.5 An element of plate subjected to forces (after Giurgiutiu, 2008)

The equations of motion are given by the equilibrium of forces of the plate element, i.e.,

2

2

2

2

x xzx x xz xz

z xzz z xz xz

N N uN dz N dx dz N dx N dz dx tdxdzx z t

N N wN dx N dz dx N dz N dx dz tdxdzz x t

ρ

ρ

⎧ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞− + + − + + =⎜ ⎟ ⎜ ⎟⎪ ∂ ∂ ∂⎪ ⎝ ⎠ ⎝ ⎠⎨

∂ ∂ ∂⎛ ⎞ ⎛ ⎞⎪− + + − + + =⎜ ⎟ ⎜ ⎟⎪ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎩

(3.32)

xx

NN dxx

∂+

∂xN

xzN

xzN

xzxz

NN dxx

∂+

∂

xzxz

NN dzz

∂+

∂

zN

zz

NN dz

z∂

+∂

dxdz x

z

y

dxdz

31

where u is the displacement in the x direction and w is the displacement in the z direction.

Cancelling the terms, the equation of motion reduces to

2

2

2

2

x xz

z xz

N N utx z t

N N wtz x t

ρ

ρ

⎧∂ ∂ ∂+ =⎪⎪ ∂ ∂ ∂

⎨∂ ∂ ∂⎪ + =⎪ ∂ ∂ ∂⎩

(3.33)

We assume that displacements u and w are uniform across the thickness. The force

resultants per unit width are given by

( , )

( , )

( , )

d

x xxdd

xz xzdd

z zzd

N T x z dy

N T x z dy

N T x z dy

−

−

−

⎧ =⎪⎪

=⎨⎪⎪ =⎩

∫∫∫

(3.34)

Recall the definition of the stresses, Equation (2.14), and consider normal stresses and the

shear stress xzT , i.e.,

( )( )

( )

2

2

2

2

xx xx yy zz

yy xx yy zz

zz xx yy zz

xz xz

T S S S

T S S S

T S S S

T S

λ μ λ λ

λ λ μ λ

λ λ λ μ

μ

⎧ = + + +⎪

= + + +⎪⎨

= + + +⎪⎪ =⎩

(3.35)

For free top and bottom surfaces, the stress in the y direction is assumed to be zero

( 0yyT = ) hence, solving the second equation in Equation (3.35), we obtain after

rearrangement

32

( )

242 2

22 4

2 22

xx xx zz

yy xx zz

zz xx zz

xz xz

T S S

S S S

T S S

T S

λ μ λμμλ μ λ μ

λλ μλμ λ μμ

λ μ λ μμ

+⎧ = +⎪ + +⎪⎪ = − +⎪ +⎨⎪ +

= +⎪+ +⎪

⎪ =⎩

(3.36)

Recalling the Lame relations defined as

( )( )

( )

1 1 2

2 1

E

EG

νλν ν

μν

⎧ =⎪ + −⎪⎨⎪ = =⎪ +⎩

(3.37)

the stress-strain relation, Equation (3.36), can be written as

( )

( )

( )

( )

2

2

1

1

1

22 1

xx xx zz

yy xx zz

zz xx zz

xz xz

ET S S

S S S

ET S S

ET S

ννν

ν

νν

ν

⎧ = +⎪ −⎪⎪ = − +⎪ −⎨⎪ = +⎪ −⎪⎪ =

+⎩

(3.38)

Substitute the stress-strain relation defined in Equation (3.38) and the strain-displacement

relation, Equation (3.3), into the force resultants, Equation (3.34), to get

( ) ( )

2 2

2 2

1 1

2 1 2 1

1 1

d

x d

d

xz d

d

z d

E u w Et u wN dyx z x z

E u w Et u wN dyz x z x

E u w Et u wN dyx z x z

ν νν ν

ν ν

ν νν ν

−

−

−

⎧ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟⎪ − ∂ ∂ − ∂ ∂⎝ ⎠ ⎝ ⎠⎪∂ ∂ ∂ ∂⎪ ⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟⎨ + ∂ ∂ + ∂ ∂⎝ ⎠ ⎝ ⎠⎪

⎪ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟⎪ − ∂ ∂ − ∂ ∂⎝ ⎠ ⎝ ⎠⎩

∫

∫

∫

(3.39)

33

Substitute Equation (3.39) into (3.33) to get the equilibrium equation in terms of the

particle displacements only, i.e.,

( )

( )

2 2 2 2 2

2 2 2 2

2 2 2 2 2

2 2 2 2

1 11 2 1

1 11 2 1

u w u w ux x z z x z E t

u w u w wx z z x z x E t

ρνν ν

ρνν ν

⎧ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂+ + + =⎜ ⎟ ⎜ ⎟⎪ − ∂ ∂ ∂ + ∂ ∂ ∂ ∂⎪ ⎝ ⎠ ⎝ ⎠

⎨⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂⎪ + + + =⎜ ⎟ ⎜ ⎟⎪ − ∂ ∂ ∂ + ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎩

(3.40)

If we consider the problem to be z-invariant, Equation (3.40) simplifies to

2 2

2 2 2

2 2

2 2 2

1

1L

S

u ux c t

w wx c t

⎧∂ ∂=⎪∂ ∂⎪

⎨∂ ∂⎪ =⎪ ∂ ∂⎩

(3.41)

where ( )2 21Lc E ρ ν⎡ ⎤= −⎣ ⎦ is the axial (or longitudinal) wave velocity and

( )[ ]2 2 1Sc E Gρ ν ρ= + = is the shear wave velocity. The first equation in (3.41) is

characterized by the axial displacement u only and it represents the equation of motion

for axial waves in plates; the second equation in (3.41) is characterized by the transverse

displacement w only and it represents the equation of motion for shear waves in plates.

3.1.7.2 Flexural waves

To derive the flexural waves, consider a differential element of plate thickness 2d. Figure

3.6 shows the moments and forces acting on the element.

34

Figure 3.6 An element of plate subjected to forces and moments (after Graff, 1991)

where v is the displacement in the y direction. Cancelling the terms, the equations of

motion reduce to

2

2

0

0

x z

x zxx

xzzz

Q Q vtx z t

M M Qx x

MM Qz z

ρ⎧∂ ∂ ∂

+ =⎪ ∂ ∂ ∂⎪⎪∂ ∂

+ − =⎨∂ ∂⎪

⎪ ∂∂− − =⎪ ∂ ∂⎩

(3.42)

Differentiate the second and third equations in Equation (3.42) and substitute the result in

the first to obtain the equilibrium relation in terms of moments, i.e.,

2 2 22 2

2 2 2x zx xzzM M MM vt

x z x zx z tρ∂ ∂ ∂∂ ∂

+ + − =∂ ∂ ∂ ∂∂ ∂ ∂

(3.43)

To solve the equation of motion, we must establish the relationship between moments

and deflections. We introduce the kinematics of the deformation, in particular we assume

that, when the element is subjected to pure bending, it deforms such as plane sections

remain plane and perpendicular to the mid plane (Euler-Bernoulli assumption). The

35

displacements in the x and z directions are due to the rotation of the section plane, and are

defined as

( , , )

( , , )

( , , )

x

z

vu y x z t yx

v v x z tvw y x z t yz

ψ

ψ

∂⎧ = − = −⎪ ∂⎪=⎨

⎪ ∂⎪ = − = −∂⎩

(3.44)

hence, the shear forces are

d d

x xyd d

u v v vQ T dz dyy x x x

μ μ μ− −

∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= = + = − +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠∫ ∫ 0

d

dd d

z yzd d

dy

v w v vQ T dz dyz y z z

μ μ μ

−

− −

=

∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= = + = − +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠

∫

∫ ∫ 0d

d

dy−

⎧⎪⎪⎨⎪ =⎪⎩

∫ (3.45)

This proves that in the Euler-Bernoulli assumption, the shear forces are equal to zero.

Remove the Euler-Bernoulli assumption made in Equation (3.44) and define the

displacements as a series expansion of the vertical displacement and its derivative ( ( )n is

the nth derivative), i.e.,

( )

0

( )

0

( )

0

( , , )

( , , )

( , , )

n nx

n

n n

n

n nz

n

u y x z t

v y v x z t

w y x z t

ψ

ψ

∞

=

∞

=

∞

=

⎧=⎪

⎪⎪⎪ =⎨⎪⎪⎪ =⎪⎩

∑

∑

∑

(3.46)

The shear forces this time are not equal to zero anymore and are given by

36

( )1 ( )

0 0

( )1 ( )

0 0

d d d nn n n

x xy xn nd d d

d d d nn n n

z yz zn nd d d

u v vQ T dz dz ny y dyy x x

w v vQ T dz dz ny y dyy z x

μ μ ψ

μ μ ψ

∞ ∞−

= =− − −

∞ ∞−

= =− − −

⎧ ⎛ ⎞∂ ∂ ∂⎛ ⎞= = + = +⎪ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎪⎨

⎛ ⎞∂ ∂ ∂⎛ ⎞⎪ = = + = +⎜ ⎟⎜ ⎟⎪ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎩

∑ ∑∫ ∫ ∫

∑ ∑∫ ∫ ∫ (3.47)

The form of the shear stress depends on the truncation of the power series. This is the

formulation for the Timoshenko beam theory that we will not pursue further. Under the

Euler-Bernoulli assumption, the moments can be defined as

d

x xxd

d

z zzd

d

xz zx xzd

M T ydy

M T ydy

M M T ydy

−

−

−

⎧=⎪

⎪⎪⎪ =⎨⎪⎪⎪ = − = −⎪⎩

∫

∫

∫

(3.48)

Substitute the strain-displacement relation (3.3) and the displacement definition, Equation

(3.44), into the stress equation in Equation (3.38) to obtain

( )

2 2

2 2 2

2 2

2 2

2 2

2 2 2

2

1

1

12

1

xx

yy

zz

xz

E v vT yx z

v vS yx z

E v vT yx z

E vT yx z

νν

νν

νν

ν

⎧ ⎛ ⎞∂ ∂= − +⎜ ⎟⎪ − ∂ ∂⎝ ⎠⎪

⎪ ⎛ ⎞∂ ∂= +⎪ ⎜ ⎟⎪ − ∂ ∂⎝ ⎠

⎨⎛ ⎞∂ ∂⎪ = − +⎜ ⎟⎪ − ∂ ∂⎝ ⎠

⎪∂⎪ = −⎪ + ∂ ∂⎩

(3.49)

With the use of Equation (3.49), Equation (3.48) yields

37

( )( )

2 2 2 22

2 2 2 2 2

2 2 2 22

2 2 2 2 2

2 22

1

1

2 11

d

xd

d

zd

d

xz zxd

E v v v vM y dy Dx z x z

E v v v vM y dy Dx z x z

E v vM M y dy Dx z x z

ν νν

ν νν

νν

−

−

−

⎧ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂= − + = − +⎪ ⎜ ⎟ ⎜ ⎟− ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎪

⎪ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎪ = − + = − +⎨ ⎜ ⎟ ⎜ ⎟− ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎪⎪ ∂ ∂⎪ = − = = −⎪ + ∂ ∂ ∂ ∂⎩

∫

∫

∫

(3.50)

where D is the flexural plate stiffness defined as ( )3 22 3 1D Ed ν⎡ ⎤= −⎣ ⎦ . With the use of

Equation (3.50), the equation of motion (3.43) becomes

4 4 4 2

4 2 2 4 4 212F

v v v vx x z z c t

∂ ∂ ∂ ∂+ + = −

∂ ∂ ∂ ∂ ∂ (3.51)

where cF is the flexural wave speed defined as ( )4Fc D tρ= . For straight crested flexural

wave, the problem is z-invariant, hence Equation (3.51) becomes

4 2

4 4 21

F

v vx c t

∂ ∂= −

∂ ∂ (3.52)

3.2 CIRCULAR-CRESTED GUIDED WAVES IN CYLINDRICAL COORDINATES

In this section, we perform the derivation of circular-crested guided waves in cylindrical

coordinate system. A detailed derivation of the solution of the acoustic field equations is

provided. For the case of Lamb waves solution, the derivation of the solution follows that

shown in Goodman (1951) and Giurgiutiu (2008) with the exception of assuming that the

solution follows sine and cosine functions instead of hyperbolic functions. This

difference allows us to show in a simpler way the similarities between the straight crested

solution and the circular crested one.

38

Consider a cylindrical coordinate system such as that the r coordinate is along the

wave propagation and the z coordinate is parallel to the thickness of the plate. Figure 3.7

shows the coordinate system.

Figure 3.7 Cylindrical wave notations

3.2.1 Equation of motion

Recall the equation of motion expressed in Equation (2.1), i.e.,

2

2tδρδ

∇ ⋅ = −uT F (3.53)

Assume θ-invariance such as 0θ∂

=∂

, use relation in Equation (2.9) and express the

equation of motion in extended form, i.e.,

2

2

2

2

2

2

1

1

1

rr rz rr

r z r

rz zz zz

TrT T u Fr r z r t

rT T T u Fr r z r t

rT T u Fr r z t

θθ

θ θ θ θθ

ρ

ρ

ρ

⎧ ∂ ∂ ∂+ − = −⎪ ∂ ∂ ∂⎪

⎪ ∂ ∂ ∂+ + = −⎨

∂ ∂ ∂⎪⎪ ∂ ∂ ∂

+ = −⎪∂ ∂ ∂⎩

(3.54)

or, by performing the derivate with respect to r,

r

z

Wave front

z d= +

z d= −

θ

39

2

2

2

2

2

2

2

rrrr rz rr

r z r

rz zz rz zz

T TT T u Fr z r t

T T T u Fr z r t

T T T u Fr z r t

θθ

θ θ θ θθ

ρ

ρ

ρ

⎧ −∂ ∂ ∂+ + = −⎪ ∂ ∂ ∂⎪

⎪∂ ∂ ∂+ + = −⎨

∂ ∂ ∂⎪⎪∂ ∂ ∂

+ + = −⎪∂ ∂ ∂⎩

(3.55)

System (3.55) has three equations: the first and third equations are coupled through the

term rzT , while the third is uncoupled from the others. As in rectangular coordinates, the

coupled equations are the equations of motion for circular crested Lamb waves. These

waves propagate along the r coordinate and they have particle displacement in both the r

and z direction (denoted by ru and zu respectively). The second equation in system

(3.55) represents the equation of motion for circular crested SH waves. The SH waves

propagate along the r direction with particle displacement along the θ direction (denoted

by uθ ). The Lamb waves equation of motion and the SH wave equation of motion can be

solved separately.

3.2.2 Strain-displacement equation

Consider the strain-velocity relation (2.10), with the use of Equation (2.12) and the θ-

invariant condition, this can be expanded as

rrr

r

zzz

uSr

uSruSz

θθ

∂⎧ =⎪ ∂⎪⎪ =⎨⎪

∂⎪ =⎪ ∂⎩

2

2

12

z

r zrz

r

uSz

u uSz ruS ur r

θθ

θθ θ

∂⎧ =⎪ ∂⎪∂ ∂⎪ = +⎨ ∂ ∂⎪∂⎪

= −⎪ ∂⎩

(3.56)

In Equation (3.55) we have seen that the terms in rr , rz , and zz are not coupled with

the terms in rθ , and zθ , hence, the third and fifth equation in (3.56) are decoupled from

40

the other three and they represent the strain-displacement relations for SH waves.

Likewise, the first, second and fourth equations represent the Lamb wave strain-

displacement relations.

Note that in cylindrical coordinates the strain along the invariant coordinate (θ ) is not

equal to zero.

3.2.3 Hooke’s law

Assume θ-invariant condition such as 0uθ

θ∂

=∂

. Substitute in the equation of Hook’s law

(2.14) the stiffness matrix (2.16) to obtain the stress-strain relation in cylindrical

coordinates, i.e.,

( )( )( )

11 12 13

12 22 23

13 12 33

44

55

66

2

2

22 22 22 2

rr rr zz rr zz

rr zz rr zz

zz rr zz rr zz

z z z

rz rz rz

r r r

T c S c S c S S S S

T c S c S c S S S S

T c S c S c S S S ST c S ST c S ST c S S

θθ θθ

θθ θθ θθ

θθ θθ

θ θ θ

θ θ θ

λ μ λ λ

λ λ μ λ

λ λ μ λμ

μμ

⎧ = + + = + + +⎪

= + + = + + +⎪⎪ = + + = + + +⎪⎨

= =⎪⎪ = =⎪

= =⎪⎩

(3.57)

3.2.4 Acoustic field equations

The acoustic filed equations are derived by substituting Hook’s law Equation (3.57) into

the Equation of motion (3.55), after rearrangement of the terms, we obtain

( )

( )

2

2

2

2

2

2

2 2 2

2 2 4

2 2 2

rrrr zz rz rr

r z r

rz rr zz rz zz

S S SS S S u Fr r r z r t

S S S u Fr z r t

SS S S S u Fr z z z r t

θθ θθ

θ θ θ θθ

θθ

λ μ λ λ μ μ ρ

μ μ μ ρ

μ λ λ μ λ μ ρ

⎧ ∂ −∂ ∂ ∂ ∂+ + + + + = −⎪ ∂ ∂ ∂ ∂ ∂⎪

⎪ ∂ ∂ ∂+ + = −⎨

∂ ∂ ∂⎪⎪ ∂∂ ∂ ∂ ∂

+ + + + + = −⎪∂ ∂ ∂ ∂ ∂⎩

(3.58)

41

Substitute the strain-displacement Equation (3.56) into (3.58) and rearrange the terms to

obtain the acoustic field equations in cylindrical coordinates

( ) ( ) ( ) ( )

( ) ( ) ( )

2 2 2 2

2 2 2 2

2 2 2

2 2 2 2

2 2 2 2

2 2 2

12 2 2

1

1 12

r r r z r rr

r r z z z zz

u u u u u u Fr r r r r z z t

u u u u u Fr z r r r t

u u u u u u Fr z r z r z r r t

θ θ θ θ θθ

λ μ λ μ λ μ λ μ μ ρ

μ μ μ μ ρ

λ μ λ μ μ λ μ μ ρ

⎧ ∂ ∂ ∂ ∂ ∂+ − + + + + + + = −⎪ ∂ ∂ ∂ ∂ ∂ ∂⎪

⎪ ∂ ∂ ∂ ∂+ + − = −⎨

∂ ∂ ∂ ∂⎪⎪ ∂ ∂ ∂ ∂ ∂ ∂

+ + + + + + + = −⎪∂ ∂ ∂ ∂ ∂ ∂ ∂⎩

(3.59)

The uncoupling between the acoustic wave equations is even more evident in Equation

(3.59). The first and third equation depend on both r and z coordinates and they represent

the Lamb wave equations. The second equation in (3.59) depends only on θ coordinate

and it represents SH waves equation. The Lamb waves equations of motion and the SH

waves equation of motion can be solved separately.

To derive the particle displacement we consider a plate not subject to body forces,

hence 0=F . Moreover, we assume that the top and bottom surfaces of the plate are free

surfaces, hence the boundary conditions are

0

0

0

zz z d

rz z d

z z d

T

T

Tθ

=±

=±

=±

⎧ =⎪⎪ =⎨⎪ =⎪⎩

(3.60)

3.2.5 Shear horizontal waves solutions

Solution to the SH waves equation is found from the second equation in (3.59), i.e.,

2 2 2

2 2 2 2 2

1 1

s

u u u u ur z r r r c t

θ θ θ θ θ∂ ∂ ∂ ∂+ + − =

∂ ∂ ∂ ∂ (3.61)

The displacement is assumed to be harmonic both in time and in the z coordinate, i.e.

42

( )( , , ) ( ) i z tu r z t g r e β ωθ ξ −= (3.62)

where 2 2 2 2scη ω ξ= − . Substitute solution (3.62) into Equation (3.61), divide by ( )i z te η ω− ,

and rearrange the terms to obtain

2

22 2

( ) 1 ( ) 1 ( ) 0g r g r g rr r r rξ ξ ξ ξ∂ ∂ ⎛ ⎞+ + − =⎜ ⎟

∂ ∂ ⎝ ⎠ (3.63)

Perform the following coordinate substitution

rζ ξ= (3.64)

Equation (3.63) becomes

2

22 2

( ) 1 ( ) 11 ( ) 0g g gζ ζξ ζζ ζ ζ ζ

⎡ ⎤∂ ∂ ⎛ ⎞+ + − =⎢ ⎥⎜ ⎟∂ ∂ ⎝ ⎠⎣ ⎦ (3.65)

Solution of Equation (3.65) is the Bessel function of order 1p = , i.e.,

( )1( )g r AJ rξ ξ= (3.66)

Hence, solution to Equation (3.61) takes the form

( )1( , , ) ( ) sin cos i tu r z t J r A z B z e ωθ ξ β β −= + (3.67)

Equation (3.67) must satisfy the stress-free top and bottom surfaces boundary conditions,

i.e.,

( ) 0zd

udzθ

θτ±

∂± = =

∂ (3.68)

Substitute solution (3.67) into the boundary conditions (3.68) to obtain

43

sin cos 0

sin cos 0A d B d

A d B dη η

η η+ =⎧

⎨− + =⎩ (3.69)

This system has nontrivial solution if the determinant equals zero, hence

sin cos 0d dη η = (3.70)

Solution to Equation (3.70) is given by

( ) ( )

22

2 1 2

S

A

nd

n d

πη

η π

⎧ =⎪⎨⎪ = +⎩

1, 2,n = L (3.71)

where, as usual, subscript S is for symmetric modes and subscript A is for antisymmetric

modes. Constant nA and nB can be found through the normalization as discussed in

Appendix E. Particle displacement (3.67) can now be written in extended form as

( )1 1( , , ) ( )sin ( )cosS A A A i tn n n n n n

nu r z t A J r z B J r z e ω

θ ξ η ξ η −= +∑ (3.72)

Particle displacement (3.72) behaves as 1( )J rξ in the radial direction. As r increases the

amplitude of the displacement decreases as

( )5 2 rπ (3.73)

The amplitude of the unit energy per circumferential length released by the source

decreases as the circular crested wave travels outward since the length of the wave front

increases as 2 rπ . However, if we consider the total circumferential energy this is given

by 2ruθπ , with the use of Equation (3.73), we see that the total energy is constant. Refer

to Appendix C for more details.

44

From the solution of SH waves particle displacement (3.72), we can derive the

stresses associated with the SH waves, i.e., rT θ and zTθ . Substitute Equation (3.72) into

the strain-displacement Equation (3.56) to get, after rearrangement,

( )1 1

0 1

0 1

2 cos ( ) cos ( )

2sin ( ) ( )2

2cos ( ) ( )

A A S A A A i tz n n n n n n n n

n

A S S Sn n n n n

i tr

n A A A An n n n n

S A zJ r B zJ r e

A z J r J rrS e

B z J r J rr

ωθ

ωθ

η η ξ η η ξ

η ξ ξ ξ

η ξ ξ ξ

−

−

⎧ = −⎪⎪ ⎛ ⎞⎡ ⎤⎪ −⎜ ⎟⎨ ⎢ ⎥⎣ ⎦⎜ ⎟⎪ =

⎜ ⎟⎪ ⎡ ⎤+ −⎜ ⎟⎪ ⎢ ⎥⎣ ⎦⎝ ⎠⎩

∑

∑ (3.74)

Substitute the expression of the strain in the stress-strain relation (3.57) to get

( )1 1

0 1

0 1

cos ( ) cos ( )

2sin ( ) ( )

2cos ( ) ( )

A A S A A A i tz n n n n n n n n

n

A S S Sn n n n n

i tr

n A A A An n n n n

T A zJ r B zJ r e

A z J r J rrT e

B z J r J rr

ωθ

ωθ

μ η η ξ η η ξ

η ξ ξ ξμ

η ξ ξ ξ

−

−

⎧ = −⎪⎪ ⎛ ⎞⎡ ⎤⎪ −⎜ ⎟⎨ ⎢ ⎥⎣ ⎦⎜ ⎟⎪ =

⎜ ⎟⎪ ⎡ ⎤+ −⎜ ⎟⎪ ⎢ ⎥⎣ ⎦⎝ ⎠⎩

∑

∑ (3.75)

Note that the behavior in the thickness direction of the particle displacement and that of

the stresses are the same in both rectangular and cylindrical coordinates. This was to be

expected since the derivation in rectangular coordinates is a particular case of that in

cylindrical coordinates: as the radial distance goes to infinity, the Bessel functions

become periodic and with constant amplitude and the term 1 0r → .

3.2.6 Lamb waves solutions

Solution for the circular crested Lamb waves are found from the first and third equations

in (3.59) assuming volume forces equal to zero, i.e.,

45

( ) ( ) ( ) ( )

( ) ( ) ( )

2 2 2 2

2 2 2 2

2 2 2 2

2 2 2

12 2 2

1 12

r r r z r r

r r z z z z

u u u u u ur r r r r z z t

u u u u u ur z r z r z r r t

λ μ λ μ λ μ λ μ μ ρ

λ μ λ μ μ λ μ μ ρ

⎧ ∂ ∂ ∂ ∂ ∂+ − + + + + + + =⎪⎪ ∂ ∂ ∂ ∂ ∂ ∂

⎨∂ ∂ ∂ ∂ ∂ ∂⎪ + + + + + + + =⎪ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎩

(3.76)

The wave equations can be expressed in terms of the dilatation Δ and the displacement

particles in radial and vertical directions. Recall that the dilatation of a material is defined

by

rr zzS S SθθΔ = + + (3.77)

With the use of (3.77) and the equation of motion defined in terms of strain, see Equation

(3.58), we obtain

( )

( )

2

2

2

2

2 2

2 2

rrrr zz rz r

rr zz rz rz z

S S SS S S ur r r r z r t

SS S S S uz z z z r r t

θθ θθ

θθ

λ μ μ μ μ μ μ ρ

λ μ μ μ μ μ μ ρ

⎧ ∂ −∂ ∂ ∂ ∂∂Δ+ + − − + + =⎪⎪ ∂ ∂ ∂ ∂ ∂ ∂

⎨∂∂ ∂ ∂ ∂∂Δ⎪ + − + − + + =⎪ ∂ ∂ ∂ ∂ ∂ ∂⎩

(3.78)

Equation (3.78) can be rearranged in this form

( )

( )

2

2

2

2

2 2 2 2 2

2 2 2 2 2

rrzz rz r

rr rz rz z

S S SS S ur r r z r t

S S S S uz z z r r t

θθ θθ

θθ

λ μ μ μ μ μ ρ

λ μ μ μ μ μ ρ

⎧ ∂ −∂ ∂ ∂∂Δ+ − − + + =⎪⎪ ∂ ∂ ∂ ∂ ∂

⎨∂ ∂ ∂ ∂∂Δ⎪ + − − + + =⎪ ∂ ∂ ∂ ∂ ∂⎩

(3.79)

The terms in strain can be proved to be equal to the curl of the rotation vector ω , i.e.,

( )

( )2 2

2

1 1 ˆ2 2

1 1 1 1 ˆ2 2

rrzz rz r z

rr rz rz r r z z

S S SS S u ur r z r z z r

S S S S u u u uz z r r r z r z r r r

θθ θθ

θθ

⎧ ∂ −∂ ∂ ∂ ∂∂ ⎛ ⎞− − + + = − = − ∇× ⋅⎜ ⎟⎪ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎪⎨

⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂⎪− − + + = − − + + = − ∇× ⋅⎜ ⎟⎪ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎩

ω r

ω z(3.80)

where the curl of the rotation vector is defined as

46

1 1ˆˆ ˆ

1ˆˆ ˆ

z r z r

r z

u ruu u u ur z z r r r

u ruu uz z r r r

θ θ

θ θ

θ θ∂ ∂∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞⎛ ⎞= ∇× = − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∂ ∂∂ ∂⎛ ⎞= − + − +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

ω u r θ z

r θ z (3.81)

Hence, Equation (3.78) becomes

( ) ( )

( ) ( )

2

2

2

2

ˆ2

ˆ2

r

z

ur t

uz t

λ μ μ ρ

λ μ μ ρ

⎧ ∂∂Δ+ − ∇× ⋅ =⎪⎪ ∂ ∂

⎨∂∂Δ⎪ + − ∇× ⋅ =⎪ ∂ ∂⎩

ω r

ω z (3.82)

or more generically

( ) ( )2

22t

λ μ μ ρ ∂+ ∇Δ − ∇× =

∂uω (3.83)

If the vector operation divergence is performed on the above, we obtain

( ) ( )2

222

tλ μ μ ρ ∂ ∇ ⋅

+ ∇ Δ − ∇ ⋅ ∇× =∂

uω (3.84)

Note that Equation (3.84) becomes

( )2

222

tλ μ ρ ∂ Δ

+ ∇ Δ =∂

(3.85)

since

( )1 z r r zr

u u u urur r z r r z

∂ ∂ ∂∂∇ ⋅ = + = + + = Δ

∂ ∂ ∂ ∂u (3.86)

and ( ) 0∇ ⋅ ∇× = where

2

22

1 rr r r z

∂ ∂ ∂⎛ ⎞∇ = +⎜ ⎟∂ ∂ ∂⎝ ⎠ (3.87)

47

Hence, the dilatation terms must satisfy the wave Equation (3.85) (Helmholtz equation).

We assume the dilatation to take two forms

( )0

cossin

i t zA J r e

zω αμ ξ

αλ μ− ⎧

Δ = ⎨+ ⎩ (3.88)

where 0 ( )J rξ is the Bessel function of order zero. The first form of the dilatation

corresponds to extensional wave propagation; the second form corresponds to the flexural

wave propagation. Substitute Equation (3.88) into the Helmholtz Equation (3.85) to get

( )( ) ( )

202

00 2 2 2

coscos cossinsin sin

i ti t i t

p

zJ rr z zz J rA er e J r ez zr r z c t

ωω ω

αξα αα ξμ ξα αλ μ

−− −

⎡ ⎤⎧⎛ ⎞∂ ∂ ⎨⎢ ⎥∂ ⎜ ⎟ ⎧ ⎧ ∂∂⎝ ⎠ ⎩⎢ ⎥+ =⎨ ⎨⎢ ⎥+ ∂ ∂ ∂⎩ ⎩⎣ ⎦(3.89)

Perform the derivate, use the first relation of Equation (3.18), and rearrange the terms to

obtain

( ) ( ) ( )2

20 002

cos1 0sin

i t zJ r J rA e J r

zr r rω αξ ξμ ξ ξ

αλ μ− ⎡ ⎤∂ ∂ ⎧

+ + = ⎨⎢ ⎥+ ∂ ∂⎣ ⎦ ⎩ (3.90)

The Bessel function of order 0p = , i.e. ( )0J rξ , is a solution of the equation in brackets,

hence Equation (3.90) is satisfied and the assumed dilatation forms (3.88) is a solution of

the Helmholtz equation (3.85).

We assume that the radial and thickness displacement solutions are of the form

( )( )

1 1

2 0

( )

( )

i tr

i tz

u Z z J r e

u Z z J r e

ω

ω

ξ

ξ

−

−

⎧ =⎪⎨

=⎪⎩ (3.91)

Recall the dilatation definition (3.77) and substitute the strain-displacement relations

(3.56) to get

48

r r zu u ur r z

∂ ∂Δ = + +

∂ ∂ (3.92)

Substitute the displacements forms (3.91) into the dilatation term (3.92) and perform the

derivate with respect to r and rearrange the terms, i.e.,

( ) ( )21 0 0

( )( ) i tZ zZ z J r J r ez

ωξ ξ ξ −∂⎡ ⎤Δ = +⎢ ⎥∂⎣ ⎦ (3.93)

Substitute Equation (3.93) into Equation (3.88) to get the following relation

21

cos( )( )sin

zZ zZ z Azz

αμξαλ μ

⎧∂+ = ⎨∂ + ⎩

(3.94)

This is the condition the two functions in z must satisfy in order to be true assumption in

Equation (3.88).

Substitute the strain-displacement relations, Equation (3.56), into (3.78) and by

rearranging the terms we get

( )

( )

22

2 2

22

2

r rr

zz

u uur r t

uuz t

λ μ μ μ ρ

λ μ μ ρ

⎧ ∂∂Δ+ + ∇ − =⎪⎪ ∂ ∂

⎨∂∂Δ⎪ + + ∇ =⎪ ∂ ∂⎩

(3.95)

Substitute Equations (3.91), (3.87) and (3.88) into the equation of motion (3.95), perform

the derivates with respect to r and t and rearrange the terms to get

( )

( )

221

1 12

222

0 22

cos( ) ( ) sin

sin( ) ( ) cos

i t

i t

zZ zJ r e Z z Azz

zZ zJ r e Z z Azz

ω

ω

αμ ξ β ξ

α

αμ ξ β α

α

−

−

⎧ ⎡ ⎤⎧∂+ =⎪ ⎨⎢ ⎥∂ ⎩⎪ ⎣ ⎦

⎨⎡ ⎤−⎧∂⎪ + = − ⎨⎢ ⎥⎪ ∂ ⎩⎣ ⎦⎩

(3.96)

Consider first the symmetric form of Equation (3.96) and simplify it to obtain

49

221

12

222

22

( ) ( ) cos

( ) ( ) sin

Z z Z z A zz

Z z Z z A zz

β ξ α

β α α

⎧∂+ =⎪⎪ ∂

⎨∂⎪ + =⎪ ∂⎩

(3.97)

Equation (3.97) is solved by first finding solution to the homogenous equation and then

by identifying a particular solution to the non-homogeneous equation. i.e.,

221

12

222

22

( ) ( ) 0

( ) ( ) 0

Z z Z zz

Z z Z zz

β

β

⎧∂+ =⎪⎪ ∂

⎨∂⎪ + =⎪ ∂⎩

(3.98)

The general solution of Equation (3.98) is of the form

( ) ( )( ) ( )

1 1 2

2 1 2

( ) sin cos( ) sin cos

Z z C z C zZ z E z E z

β β

β β

⎧ = +⎪⎨

= +⎪⎩ (3.99)

A particular solution of Equation (3.97) is

( )

( )

1 2 2

2 2 2

cos

sin

p

p

Z A z

Z A z

ξ αα β

α αα β

⎧ = −⎪ −⎪⎨⎪ = −⎪ −⎩

(3.100)

The total solution is equal to the sum of the general solution (3.99) and particular solution

(3.100). Note that, since the radial displacement should be symmetric and the thickness

displacement antisymmetric, constants C1 and E2 should be equal to zero. The total

solution is given by

( )

( )

1 2 2 2

2 1 2 2

( ) cos cos

( ) sin sin

Z z C z A z

Z z E z A z

ξβ αα β

αβ αα β

⎧ = −⎪ −⎪⎨⎪ = −⎪ −⎩

(3.101)

50

To obtain constants C2 and E1, substitute solution (3.101) into condition in Equation

(3.94), i.e.,

( ) ( )2 2

2 1 2 2cos cos cosC z E z A zμ ξ αξ β β β αλ μ α β

⎛ ⎞++ = +⎜ ⎟+ −⎝ ⎠

(3.102)

Note that

22 2

2 22 2

2 22 2 2 22 2

2 2

p s

s p

p s

c cc c

c c

ωξ ξξ α μ

ω ωα β λ μξ ξ

+ −+

= = = −− − +− − +

(3.103)

Hence, Equation (3.102) becomes after rearrangement

2 1C E βξ

= − (3.104)

The displacement solution for the symmetric case is given by Equations (3.91), (3.101),

and (3.104), i.e.,

( ) ( )

( ) ( )

1 12 2

1 02 2

cos cos

sin sin

i tr

i tz

u E z A z J r e

u E z A z J r e

ω

ω

β ξβ α ξξ α β

αβ α ξα β

−

−

⎧ ⎡ ⎤= − +⎪ ⎢ ⎥−⎪ ⎣ ⎦⎨

⎡ ⎤⎪ = −⎢ ⎥⎪ −⎣ ⎦⎩

(3.105)

Now consider the antisymmetric form of Equation (3.96), i.e.,

221

12

222

22

( ) ( ) sin

( ) ( ) cos

Z z Z z A zz

Z z Z z A zz

β ξ α

β α α

⎧∂+ =⎪⎪ ∂

⎨∂⎪ + = −⎪ ∂⎩

(3.106)

51

Solution to Equation (3.106) is found by first deriving the solution to the homogenous

equations; hence the general solutions is the same as Equation (3.99) while the particular

solution is

1 2 2

2 2 2

sin

cos

p

p

Z A z

Z A z

ξ αα β

α αα β

⎧ = −⎪ −⎪⎨⎪ =⎪ −⎩

(3.107)

The total solution is equal to the sum of the general and particular solution. Note that

since the radial displacement should be antisymmetric and the thickness displacement

symmetric, constants C2 and E1 should be equal to zero. The total solution is given by

( )

( )

1 1 2 2

2 2 2 2

( ) sin sin

( ) cos cos

Z z C z A z

Z z E z A z

ξβ αα β

αβ αα β

⎧ = −⎪ −⎪⎨⎪ = +⎪ −⎩

(3.108)

To obtain constants C1 and E2, substitute solution (3.108) into condition in Equation

(3.94)

( ) ( )2 2

1 2 2 2sin sin sinC z E z A zξ α μξ β β β αα β λ μ

⎛ ⎞+− = +⎜ ⎟− +⎝ ⎠

(3.109)

Use Equation (3.103) and simplify to obtain

1 2C E βξ

= (3.110)

Hence, the displacement solution in the antisymmetric case are given by

52

( ) ( )

( ) ( )

2 12 2

2 02 2

sin sin

cos cos

i tr

i tz

u E z A z J r e

u E z A z J r e

ω

ω

β ξβ α ξξ α β

αβ α ξα β

−

−

⎧ ⎛ ⎞= −⎪ ⎜ ⎟−⎪ ⎝ ⎠⎨

⎛ ⎞⎪ = +⎜ ⎟⎪ −⎝ ⎠⎩

(3.111)

From the application of the boundary conditions (3.60) we derive the eigenvalues

(wavenumbers ξ ) of the wave modes.

3.2.6.1 Eigenvalues for symmetric modes

Recall that the boundary conditions for free top and bottom surfaces, i.e.,

0

0zz z d

rz z d

T

T=±

=±

⎧ =⎪⎨

=⎪⎩, (3.112)

Use the stress-strain relation, and the strain-displacement equations to get zzT and rzT in

terms of particle displacement, i.e.,

( )2r z r

zz

r zrz

u u uTr z ru uTz r

λ λ μ λ

μ

∂ ∂⎧ = + + +⎪ ∂ ∂⎪⎨ ∂ ∂⎛ ⎞⎪ = +⎜ ⎟⎪ ∂ ∂⎝ ⎠⎩

, (3.113)

Substitute the symmetric particle displacement Equation (3.105) and rearrange the terms

to get

( ) ( ) ( )

( ) ( )

2 2

1 02 2

2 2

1 12 2

2cos 2 cos

2 sin sin

i tzz

i trz

T A z E z J r e

T A z E z J r e

ω

ω

λ μ α λξα μ β β ξ

α β

ξα ξ βμ α β ξα β ξ

−

−

⎧ ⎡ ⎤+ += − −⎪ ⎢ ⎥−⎪ ⎣ ⎦

⎨⎡ ⎤−⎪ = −⎢ ⎥⎪ −⎣ ⎦⎩

(3.114)

Use relation (3.27) into Equation (3.114) to get

53

( ) ( )

( ) ( )

2 2

1 02 2

2 2

1 12 2

cos 2 cos

2 sin sin

i tzz

i trz

T A z E z J r e

T A z E z J r e

ω

ω

β ξμ α β β ξα β


−

−

⎧ ⎡ ⎤−= − −⎪ ⎢ ⎥−⎪ ⎣ ⎦

⎨⎡ ⎤−⎪ = −⎢ ⎥⎪ −⎣ ⎦⎩

(3.115)

Substitute Equation (3.115) into the boundary condition (3.112), i.e.,

( )

( )

2 2

12 2

2 2

12 2

cos 2 cos 0

2 sin sin 0

A d E d

A d E d

ξ β α β βα β

ξα ξ βα βα β ξ

⎧ −− − =⎪ −⎪

⎨−⎪ − =⎪ −⎩

(3.116)

System in Equation (3.116) has no banal solution if the determinant of the coefficients A

and E1 is equal to zero, hence

( )

( )

2 2

2 2

2 2

2 2

cos 2 cos0

2 sin sin

d d

d d

ξ β α β βα β


−− −

−=

−−

−

(3.117)

This yields the characteristic equation

( )22 2

2

tantan 4

dd

ξ βαβ ξ αβ

−= − (3.118)

This is the Rayleigh-Lamb equation for symmetric mode. This equation is the same as for

rectangular coordinates, see Equation (3.20). Substitution of the eigenvalues in Equation

(3.116) yields the eigen coefficient ratio in the form

( )( )

2 21*

cos2 cos

dEA d

ξ β αβ β−

= − (3.119)

where

54

*2 2

AAα β

=−

, (3.120)

Define the coefficient ratio RS as

( )

( )2 2

*1

cos2 cosS

dR E A

dξ β α

ξβξ β−

= − = , (3.121)

Use Equations (3.119), (3.120), and (3.121) to obtain the symmetric displacement

equation in the form

( ) ( )

( ) ( )

*1

*0

cos cos

sin sin

i tr Sn Sn Sn Sn Sn Sn Sn

n

i tz Sn Sn Sn Sn Sn Sn Sn

n

u A z R z J r e

u A z R z J r e

ω

ω

ξ α β β ξ

α α ξ β ξ

−

−

⎧ ⎡ ⎤= − −⎣ ⎦⎪⎨

⎡ ⎤= − +⎪ ⎣ ⎦⎩

∑

∑ (3.122)

Note that the behavior in the thickness direction of the particle displacements in Equation

(3.122) are identical to those derived for straight crested waves, Equations (3.16). The

factor i that appears in the straight crested formulation does not need to be here because

the Bessel functions 0J and 1J are in quadrature.

Constants *A can be determined through the normalization factor as discussed in

Appendix E.

3.2.6.2 Eigenvalues for antisymmetric modes

To derive the eigenvalues for the antisymmetric modes, substitute the antisymmetric

particle displacement Equation (3.111) into the stresses in Equation (3.113) and rearrange

the terms to get

55

( ) ( )

( )

2 2

2 02 2

2 2

2 12 2

2sin 2 sin

2 cos cos

i tzz

i trz

T A z E z J r e

T A z E z J r e

ω

ω

λξ λ μ αα μ β β ξ

α β


−

−

⎧ ⎡ ⎤+ += − +⎪ ⎢ ⎥−⎪ ⎣ ⎦

⎨⎛ ⎞−⎪ = − +⎜ ⎟⎪ −⎝ ⎠⎩

(3.123)

Use relation (3.27) to rearrange Equation (3.123), i.e.,

( )

( )

2 2

2 02 2

2 2

2 12 2

sin 2 sin

2 cos cos

i tzz

i trz

T A z E z J r e

T A z E z J r e

ω

ω

ξ βμ α β β ξα β


−

−

⎧ ⎡ ⎤−= − − +⎪ ⎢ ⎥−⎪ ⎣ ⎦

⎨⎛ ⎞−⎪ = − +⎜ ⎟⎪ −⎝ ⎠⎩

(3.124)

Substitute Equation (3.124) into the boundary condition (3.112)

2 2

22 2

2 2

22 2

sin 2 sin 0

2 cos cos 0

A d E d

A d E d

ξ β α β βα β


⎧ −− + =⎪ −⎪

⎨−⎪ + =⎪ −⎩

(3.125)

System in Equation (3.125) has no banal solution if the determinant of the coefficients A

and E2 is equal to zero, hence

2 2

2 2

2 2

2 2

sin 2 sin0

2 cos cos

d d

d d

ξ β α β βα β


−−

−=

−−

(3.126)

This yields the characteristic equation

( )

2

22 2

tan 4tan

dd

α ξ αββ ξ β

= −−

(3.127)

56

This is the Rayleigh-Lamb equation for antisymmetric modes. This equation is the same

as for rectangular coordinates, see Equation (3.21). Substitution of the eigenvalues in

Equation (3.125) yields the eigen coefficient ratio in the form

( )2 2

2*

sin2 sin

dEA d

ξ β αβ β

−= (3.128)

Define the coefficient ratio RA as

( )2 2

*2

sin2 sinA

dR E A

dξ β α

ξβξ β−

= = (3.129)

Use Equations (3.128), (3.120), and (3.129) to obtain the antisymmetric displacement

equation in the form

( ) ( )

( ) ( )

*1

*0

sin sin

cos cos

i tr Sn Sn Sn Sn Sn Sn Sn

n

i tz Sn Sn Sn Sn Sn Sn Sn

n

u A z R z J r e

u A z R z J r e

ω

ω

ξ α β β ξ

α α ξ β ξ

−

−

⎧ ⎡ ⎤= − −⎣ ⎦⎪⎨

⎡ ⎤= +⎪ ⎣ ⎦⎩

∑

∑ (3.130)

Note that the behavior of the stresses in the thickness direction in Equation (3.122) are

identical to those derived for straight crested waves, Equation (3.17). The factor i that

appears in the straight crested formulation does not need to be here because the Bessel

functions 0J and 1J are in quadrature.

3.2.6.3 Lamb wave stresses for symmetric modes

From solution of the Lamb waves particle displacement (3.105) and (3.111) we derive the

stresses associated with the waves, i.e., rrT , rzT ,and zzT . First derive the stress

components for the symmetric modes. In Section 3.2.6.1 we found the expressions for zzT

and rzT ; with the use of Equation (3.122) zzT and rzT become

57

( ) ( )

( ) ( )

* 2 20

* 2 21

cos 2 cos

2 sin sin

i tzz Sn Sn Sn Sn Sn Sn Sn Sn Sn

n

i trz Sn Sn Sn Sn Sn Sn Sn Sn Sn

n

T A z R z J r e

T A z R z J r e

ω

ω

μ ξ β α ξ β β ξ

μ ξ α α ξ β β ξ

−

−

⎧ ⎡ ⎤= − −⎣ ⎦⎪⎨

⎡ ⎤= + −⎪ ⎣ ⎦⎩

∑

∑ (3.131)

To derive rrT , substitute the displacement solutions in Equation (3.122) into the strain-

displacement Equation (3.56), i.e.,

( ) ( ) ( )

( ) ( )

( ) ( )

1*0

1*

* 20

sin sin

sin sin

sin sin

Sn i trr Sn Sn Sn Sn Sn Sn Sn Sn

n

Sn i tSn Sn Sn Sn Sn Sn

n

i tzz Sn Sn Sn Sn Sn Sn Sn Sn

n

J rS A z R z J r e

rJ r

S A z R z er

S A z R z J r e

ω

ωθθ

ω

ξξ α β β ξ ξ

ξξ α β β

α α ξ β β ξ

−

−

−

⎧ ⎡ ⎤⎡ ⎤= − − −⎪ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦⎪

⎪ ⎡ ⎤⎪ = − −⎨ ⎢ ⎥⎣ ⎦⎪⎪ ⎡ ⎤= − +⎣ ⎦⎪⎪⎩

∑

∑

∑

(3.132)



( ) ( )

( ) ( )

2 2 20

*

1

2 sin 2 sin

2 sin sin

Sn Sn Sn Sn Sn Sn Sn Sn Sni t

rr SnSnn

Sn Sn Sn Sn Sn

z R z J rT A eJ r

z R zr

ω

ξ β α α ξ β β ξμ ξ

ξ α β β

−

⎧ ⎫⎡ ⎤− + − −⎣ ⎦⎪ ⎪= ⎨ ⎬

⎪ ⎪+ −⎩ ⎭

∑ (3.133)

Note that while the normal stress zzT in the z direction and the shear component rzT

depend on the radial coordinate through respectively the Bessel function of order 0 and 1,

the normal stress in the radial direction rrT depends on both Bessel functions. As r

increases the magnitude of the first term of rrT decreases with as 4 5 r while the

second term decreases as 31 r , hence the contribution of the second term is soon

negligible with respect to the contribution from the first term.

58

The straight crested solution derived in rectangular coordinate is the approximation of

the circular crested solution as the radial distance goes to infinity. This is confirmed by

the fact that the term function of z in the displacement Equations (3.122) and (3.130) are

the same as those derived in the particle displacement in rectangular coordinates,

Equation (3.16). The stresses for symmetric circular-crested guided waves are

summarized hereunder

( ) ( )

( ) ( )

( ) ( )

2 2 20

*

1

* 2 20

*

2 sin 2 sin

2 sin sin

cos 2 cos

2 sin


rr SnSnn

Sn Sn Sn Sn Sn


n

rz Sn Sn Sn Sn Sn Sn

z R z J rT A eJ r

z R zr

T A z R z J r e

T A z R

ω

ω


ξ α β β


μ ξ α α ξ

−

−

⎧ ⎫⎡ ⎤− + − −⎣ ⎦⎪ ⎪= ⎨ ⎬

⎪ ⎪+ −⎩ ⎭⎡ ⎤= − −⎣ ⎦

= +

∑

∑

( ) ( )2 21sin i t

Sn Sn Snn

z J r e ωβ β ξ −

⎧⎪⎪⎪⎪⎪⎨⎪⎪

⎡ ⎤−⎪ ⎣ ⎦⎪⎪⎩

∑

(3.134)

3.2.6.4 Lamb wave stresses for antisymmetric modes

In Section 3.2.6.2 we found the expression for zzT and rzT for antisymmetric modes rzT ;

with the use of Equation (3.130) zzT and rzT become

( ) ( )

( )( ) ( )

* 2 20

* 2 21

sin 2 sin

2 cos cos


n

i trz Sn Sn Sn Sn Sn Sn Sn Sn Sn

n

T A z R z J r e

T A z R z J r e

ω

ω


μ ξ α α ξ β β ξ

−

−

⎧ ⎡ ⎤= − −⎣ ⎦⎪⎨

= − + −⎪⎩

∑

∑ (3.135)

To derive rrT , substitute the displacement solution in Equation (3.130) into the strain-

displacement Equation (3.56), i.e.,

59

( ) ( )

( ) ( )

( ) ( )

( ) ( )

* 20

1*

1*

* 20

sin sin

sin sin

sin sin

sin sin

Sn Sn Sn Sn Sn Sn Sn Sni t

rr Snn Sn Sn Sn Sn Sn Sn

Sn i tSn Sn Sn Sn Sn Sn

n

izz Sn Sn Sn Sn Sn Sn Sn Sn

A z R z J rS eJ r

A z R zr

J rS A z R z e

r

S A z R z J r e

ω

ωθθ

ω

ξ α ξ β β ξ

ξξ α β β

ξξ α β β

α α ξ β β ξ

−

−

−

⎡ ⎤−⎢ ⎥= − ⎢ ⎥− −⎢ ⎥⎣ ⎦

⎡ ⎤= − −⎢ ⎥⎣ ⎦

⎡ ⎤= − +⎣ ⎦

∑

∑t

n

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

∑

(3.136)



( ) ( )

( ) ( )

2 2 20

*

1

2 sin 2 sin

2 sin sin


rr SnSnn

Sn Sn Sn Sn Sn

z R z J rT A eJ r

z R zr

ωξ β α α ξ β β ξ

μ ξξ α β β

−

⎧ ⎫⎡ ⎤+ − −⎣ ⎦⎪ ⎪= − ⎨ ⎬

⎪ ⎪− −⎩ ⎭

∑ (3.137)

The stresses for antisymmetric circular-crested guided waves are summarized hereunder

( ) ( )

( ) ( )

( ) ( )

2 2 20

*

1

* 2 20

*

2 sin 2 sin

2 sin sin

sin 2 sin

2 cos


rr SnSnn

Sn Sn Sn Sn Sn


n

rz Sn Sn Sn Sn Sn S

z R z J rT A eJ r

z R zr

T A z R z J r e

T A z R

ω

ω


ξ α β β


μ ξ α α ξ

−

−

⎧ ⎫⎡ ⎤− + − +⎣ ⎦⎪ ⎪= ⎨ ⎬

⎪ ⎪+ −⎩ ⎭⎡ ⎤= − −⎣ ⎦

= − +

∑

∑

( )( ) ( )2 21cos i t

n Sn Sn Snn

z J r e ωβ β ξ −

⎧⎪⎪⎪⎪⎪⎨⎪⎪

−⎪⎪⎪⎩

∑

(3.138)

Note that dependence on the thickness of the plate in the normal and shear stress is the

same in both rectangular and cylindrical coordinates (compare zzT with yyT and rzT with

xyT ). The first term in the radial normal stress is the same as that of the normal stress

along x in rectangular coordinates. However, in the radial normal stress, there is an extra

term given by

60

( )( ) ( ) ( )* 1 12 sin sin 2 ( )A rJ r J r

A R z z u zr rξ ξ

μ β β ξ α μ− − = (3.139)

This contribution is due to the angular strain component that in cylindrical coordinates is

not equal to zero, see Equation (3.56).

61

4 POWER FLOW AND ENERGY CONSERVATION – THE ACOUSTIC

POYNTING THEOREM

During wave propagation, the power emanated from the source of the waves must equal

the sum of the rate of change of energy stored in the wave field, the power flow carried

by the wave, and the power dissipated through loss mechanisms. This simple concept can

be expressed in the form of the Acoustic Poynting Theorem, which we present next

following Auld (1973)

Recall the equation of motion (2.1) and the strain-displacement relation (2.10).

2

2

s

tρ

⎧ ∂∇ ⋅ = −⎪

∂⎨⎪∇ =⎩

uT F

u S (4.1)

Recall that the particle velocity is given by

t

∂=

∂uv (4.2)

Use relation (4.2) into Equation (4.1) and perform the derivate with respect to t to the

second equation of (4.1), i.e.,

s

t

t

ρ ∂⎧∇ ⋅ = −⎪⎪ ∂⎨ ∂⎪∇ =⎪ ∂⎩

vT F

Sv (4.3)

62

Take the single dot product of the first equation of (4.3) with the velocity vector v , and

the double dot product of the second equation with the stress tensor T to get

( )

( ): :s

t

t

ρ ∂⎧ ⋅ ∇ ⋅ = ⋅ − ⋅⎪⎪ ∂⎨ ∂⎪ ∇ =⎪ ∂⎩

vv T v v F

ST v T (4.4)

Add the two equations in (4.4) to obtain

( ) ( ): :s t tρ ∂ ∂

⋅ ∇ ⋅ + ∇ = ⋅ + − ⋅∂ ∂v Sv T T v v T v F (4.5)

Recall the distributive property of the del operator (see Appendix B.2 for details)

( ) ( ) : s∇ ⋅ ⋅ = ⋅ ∇ ⋅ + ∇v T v T T v (4.6)

Note that identity (4.6) requires that T be a symmetric 2nd rank tensor. Substitution of

Equation (4.6) into Equation (4.5) yields

( ) :t t

ρ ∂ ∂∇ ⋅ ⋅ = ⋅ + − ⋅

∂ ∂v Sv T v T v F (4.7)

Integrate Equation (4.7) over a control volume V to get

( ) :V V

dV dVt t

ρ ∂ ∂⎛ ⎞∇ ⋅ = ⋅ + − ⋅⎜ ⎟∂ ∂⎝ ⎠∫ ∫v Sv T v T v F (4.8)

Recall the divergence theorem

ˆV S

dV dS∇ ⋅ = ⋅∫ ∫a a n (4.9)

where a is a vector and n is the unit vector normal outwards to the surface. Applying the

divergence theorem (4.9) to the LHS of Equation (4.8) we get

63

( ) ˆ :S V V V

dS dV dV dVt t

ρ ∂ ∂⋅ ⋅ = ⋅ + + − ⋅

∂ ∂∫ ∫ ∫ ∫v Sv T n v T v F (4.10)

The terms in Equation (4.10) can be identified with the following power and energy

definitions:

inV

P dV= ⋅∫ v F (power supplied to the volume V by the source) (4.11)

v

V V

u dVt t

ρ∂ ∂= ⋅

∂ ∂∫ ∫vv (rate of change of kinetic energy density of the wave) (4.12)

:S

V

u dVt t

∂ ∂=

∂ ∂∫ST (rate of change of strain energy density stored in the wave) (4.13)

The outward power flow through the control surface S of normal unit vector n is

calculated as the product between the traction vector nT and the velocity vector v , i.e.,

out SP dS= − ⋅∫ nv T (outward power flow through the closed surface S) (4.14)

The negative sign is related to the sign convention on surface tractions. Referring to

Figure 4.1, we observe that outward tractions represent the force exerted by medium 2

onto medium 1, whereas the outward power flow should involve the force exerted by

medium 1 onto medium 2, i.e., the negative of the outward tractions.

Note that the power flow outP of Equation (4.14) represents the instantaneous power

flow. Using complex notations, we define the complex power flow as (see Auld (1973)

vol. II page 155)

12complex S

P dS= − ⋅∫ nv T% (complex power) (4.15)

64

Note that the complex power flow is valid for time harmonic variations (i.e., i te ω ) while

Equation (4.14) is valid for arbitrary time dependence.

The real part of the complex power represents the time-averaged power a.k.a. average

power (see Appendix C.2), i.e.,

( )1 Re2av S

P dS= − ⋅∫ nv T% (average power) (4.16)

The imaginary part of the complex power represents the peak value of the reactive power.

Figure 4.1 Coordinate notation. Power flow from 1 to 2 is ˆ ˆdS dS− ⋅ ⋅ = ⋅v T n P n (after Auld 1990)

Recall the definition of tractions nT in terms of stresses T , i.e.,

ˆ= ⋅nT T n (4.17)

Substitution of Equation (4.17) into Equation (4.14) yields

( )ôut S

P dS= − ⋅ ⋅∫ v T n (4.18)

Equation (4.18) can be rearranged as

( ) ôut SP dS= − ⋅ ⋅∫ v T n (4.19)

dS

1

2

n vˆnT = ⋅T n

65

Equation (4.19) can be written in more compact form if we introduce the power flow

density function P (acoustic Poynting vector) defined as

= − ⋅P v T (power flow density a.k.a. acoustic Poynting vector) (4.20)

Substitution of Equation (4.20) into Equation (4.19) yields the expression of outward

power flow in terms of the acoustic Poynting vector, i.e.,

ôut SP dS= ⋅∫ P n (4.21)

Note:

1 : :2S

V

u dV= ∫S c S (strain energy stored in the system) (4.22)

: : :S

V V

u dV dVt t t

∂ ∂ ∂= =

∂ ∂ ∂∫ ∫S SS c T (time derivative of strain energy) (4.23)

12v

V

u dVρ= ⋅∫ v v (rate of change of kinetic energy density of the wave) (4.24)

In view of the above, we can express Equation (4.10) in terms of power flows and energy

change rates. Recall Equation (4.10), i.e.,

( ) ˆ :S V V V

dS dV dV dVt t

ρ ∂ ∂⋅ ⋅ = ⋅ + + − ⋅

∂ ∂∫ ∫ ∫ ∫v Sv T n v T v F (4.25)

Substituting Equations (4.11), (4.13), (4.12), (4.19) into Equation (4.25) yields

v Sout in

u uP Pt t

∂ ∂− = + −

∂ ∂ (4.26)

Upon rearrangement, Equation (4.26) becomes the Poynting theorem, i.e.,

66

v Sin out

u u UP Pt t t

∂ ∂ ∂− = + =

∂ ∂ ∂ (Poynting Theorem) (4.27)

where inP , outP , /vu t∂ ∂ , /Su t∂ ∂ are given by Equations (4.11)-(4.19) and U is the total

energy density defined as the sum between kinetic and elastic energy densities, i.e.,

v SU u u= + (4.28)

Equation (4.27) represents the principle of energy conservation for a non-dissipative

medium. If dissipation is also present, like in most natural phenomena, then a power

dissipation term dP needs to be added and Equation (4.27) becomes

v Sin out d

u u UP P Pt t t

∂ ∂ ∂− − = + =

∂ ∂ ∂ (Poynting Theorem with dissipation) (4.29)

The complex Poynting Theorem is expressed using complex power formulation, i.e., the

complex Poynting vector

2

− ⋅=

v TP%

(complex Poynting vector) (4.30)

Then, following Auld (1973), we write

( ) ( ){ } ( )ˆ2 S v d inpeak peak av

S

dS i u u P Pω⋅⎛ ⎞ ⋅ − − + =⎜ ⎟⎝ ⎠∫

v T n%

(complex Poynting theorem) (4.31)

or

( ) ( ){ } ( )ˆ S v d inpeak peak avS

dS i u u P Pω⋅ − − + =∫ P n (complex Poynting theorem) (4.32)

67

4.1 POWER FLOW IN RECTANGULAR COORDINATES

Consider a section of area dxdydz of a rectangular plate, as in Figure 4.2. The are six

surfaces that determines the rectangular sections, these surfaces are denoted by normals

ˆxn± , ˆyn± , and ˆzn± .

Figure 4.2 Rectangular section dxdydz of a plate of thickness 2d. a) Section notations; b)

Power flow through surface with normal nx.

The power flows thorough the six surfaces; however, no power flows through the top and

bottom ( y d= ± ) ˆyn± free surfaces.

4.1.1 Shear horizontal waves

Consider straight crested shear horizontal (SH) waves propagating in a rectangular plate.

The SH velocity vector can be expressed as

{ }0 0 zv=v (4.33)

The stress matrix is defined as

0 00 0

0

xz

yz

xz yz

TT

T T

⎡ ⎤⎢ ⎥Τ = ⎢ ⎥⎢ ⎥⎣ ⎦

(4.34)

y

x

dz

dy

dx ˆyn+

ˆyn−

ˆxn+

ˆyn−

ˆyn+ˆxn−

z a) b)

68

The scalar product between velocity and stress matrix is

{ }0z xz z yzv T v T⋅ =v T (4.35)

4.1.1.1 Power flow along ˆxn

Consider the power flow through the surface of area dydz and with normal ˆxn due to

propagating shear horizontal waves (Figure 4.2b); from Equation (4.14) we get

( ) ˆz d

out xz d

P dydz− −

= − ⋅ ⋅∫ ∫ v T n (4.36)

Note that the normal ˆxn is defined as

{ }ˆ 1 0 0x =n (4.37)

Multiply the normal in the x direction by the velocity-stress product to get

( ) ˆ x z xzv T⋅ ⋅ =v T n (4.38)

Substitute Equation (4.38) into (4.36), i.e.,

z d

out z xzz d

P v T dydz− −

= − ∫ ∫ (4.39)

Since the problem is z-invariant (straight crested wave with the wave front parallel to z

axis), both velocity and stress are not dependent on z. Hence, we can consider the power

flow per unit wave front length, i.e.,

d

xout z xz

d

P v T dy−

= − ∫ (4.40)

69

Equation (4.40) is the power flow per unit length of shear horizontal waves propagating

in the x direction.

Recall the solution through separation of variables to the wave equation

2 2 2

2 2 2

u u ux y t

μ μ ρ∂ ∂ ∂+ =

∂ ∂ ∂ (4.41)

i.e.,

( ) ( )

( ) ( )

sin cos( , , )

sin cos

A Sn n

A Sn n

i x t i x tA Sn n n n

zi x t i x tA Sn

n n n n

A ye B yeu x y t

C ye D ye

ξ ω ξ ω

ξ ω ξ ω

η η

η η

− −

+ +

⎧ ⎫⎡ ⎤+⎢ ⎥⎪ ⎪⎣ ⎦= ⎨ ⎬⎡ ⎤⎪ ⎪+ +⎢ ⎥⎣ ⎦⎩ ⎭

∑ (4.42)

where ( )ni x te ξ ω− is for forward propagating modes and ( )ni x te ξ ω+ is for backward

propagating modes. Without loss of generality, consider only one generic symmetric

wave mode; for notation simplification we will drop the superscript S, hence solution

(4.42) can be written as

( ) ( )[ ]( , , ) ( ) i x t i x tzu x y t Y y Be Deξ ω ξ ω− += + (4.43)

where ( )Y y is ( )( ) cosY y yη= . Solution to Equation (4.41) can be derived directly in

terms of sine and cosine as (for one symmetric mode propagating forward and one

symmetric mode propagating backward)

[ ][ ]1 2 3 4( , , ) ( ) cos( ) sin( ) cos( ) sin( )zu x y t Y y A x A x A t A tξ ξ ω ω= + + (4.44)

Solution in Equation (4.44) and that in Equation (4.43) are equivalent. Perform the

multiplication of the terms dependent on x and t in Equation (4.44), i.e.,

70

1 3 1 4

2 3 2 4

cos( )cos( ) cos( )sin( )( , , ) ( )

sin( )cos( ) sin( )sin( )z

A A x t A A x tu x y t Y y

A A x t A A x tξ ω ξ ω

ξ ω ξ ω+⎡ ⎤

= ⎢ ⎥+ +⎣ ⎦ (4.45)

Substitute the expression of the particle displacement (4.45) into the particle velocity

Equation (4.2), i.e.

1 3 1 4

2 3 2 4

cos( )sin( ) cos( )cos( )( , , ) ( )

sin( )sin( ) sin( ) cos( )z

z

A A x t A A x tuv x y t Y yA A x t A A x tt

ξ ω ξ ωω

ξ ω ξ ω−⎡ ⎤∂

= = − ⎢ ⎥+ −∂ ⎣ ⎦ (4.46)

and recall that

1 3 1 4

2 3 2 4

sin( ) cos( ) sin( )sin( )( , , ) ( )

cos( )cos( ) cos( )sin( )z

xz

A A x t A A x tuT x y t Y yA A x t A A x tx

ξ ω ξ ωξ

ξ ω ξ ω+⎡ ⎤∂

= = − ⎢ ⎥− −∂ ⎣ ⎦ (4.47)

4.1.1.1.1 Average power flow

Equations (4.46) and (4.47) represent respectively the particle velocity and stress due to

both backward and forward propagating wave. We want to derive the average power flow

due to the presence of both backward and forward propagating waves. Consider the

expression of the average power flow, i.e.,

0

1( ) ( , , ) ( , , )d

av z xzd

P x v x y t T x y t dydtΤ

−

= − ⋅Τ ∫ ∫ (4.48)

The product of velocity by stress becomes

1 3 1 42

2 3 2 4

1 3 1 4

2 3 2 4

cos( )sin( ) cos( )cos( )( , , ) ( , , ) ( )

sin( )sin( ) sin( )cos( )

sin( )cos( ) sin( )sin( )cos( )cos( ) cos( )sin( )

z xz

A A x t A A x tv x y t T x y t Y y

A A x t A A x t

A A x t A A x tA A x t A A x t

ξ ω ξ ωωξ

ξ ω ξ ω

ξ ω ξ ωξ ω ξ ω

−⎡ ⎤= ⎢ ⎥+ −⎣ ⎦

+⎡ ⎤⎢ ⎥− −⎣ ⎦

(4.49)

or, by rearranging the terms,

71

( )

( )( )

( )

2 22 23 41 22

2 2 2 21 2 3 4

sin( ) cos( )sin( ) cos( )( )

sin ( ) cos ( ) sin ( ) cos ( )

A A t tA A x xvT Y y

A A x x A A t t

ω ωξ ξξω

ξ ξ ω ω

⎡ ⎤⎡ ⎤ −−= ⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥+ − + −⎣ ⎦ ⎣ ⎦ (4.50)

Substitute Equation (4.50) into (4.48) to obtain the average power flow, i.e.,

( )( )

( )( )

2 21 22

2 21 2

2 23 4

2 20 3 4

sin( ) cos( )( ) ( )

2 sin ( ) cos ( )

sin( ) cos( )2sin ( ) cos ( )

d

avd

T

A A x xP x Y y dy

A A x x

A A t tdt

T A A t t

ξ ξωξ

ξ ξ

ω ω

ω ω

−

⎡ ⎤−⎛ ⎞= − ⋅⎢ ⎥⎜ ⎟

⎢ ⎥+ −⎝ ⎠ ⎣ ⎦⎡ ⎤−

⋅ ⎢ ⎥⎢ ⎥+ −⎣ ⎦

∫

∫ (4.51)

or, by solving the integral,

( )

( )

2 21 22

3 42 21 2

sin( )cos( ) 1 1( ) ( ) 2 02 2 2sin ( ) cos ( )

d

avd

A A x xP x Y y dy A A

A A x x

ξ ξωξξ ξ−

⎡ ⎤−⎛ ⎞ ⎛ ⎞= − − =⎢ ⎥ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎢ ⎥+ −⎝ ⎠ ⎣ ⎦∫ (4.52)

Equation (4.52) indicates that the average power flow due to the presence of both

forward and backward propagating wave is equal to zero.

4.1.1.1.2 Average power flow of the forward and backward propagating waves

We have seen in the previous section that the average power flow is equal to zero. Now

we want to explicit the contributions form both backward and forward propagating modes

to the average power flow. For this motive, the terms in the particle displacement,

Equation (4.44), are transformed such that the contributions due to forward and backward

propagating waves are explicit. The terms in Equation (4.45) are transformed such that

( ) ( )1 3 1 31 3 cos( ) cos( ) cos cos

2 2A A A AA A x t x t x tξ ω ξ ω ξ ω= − + + (4.53)

The particle displacement can be rewritten as

72

( ) ( )

( ) ( )

1 3 2 4 1 3 2 4

1 4 2 3 2 3 1 4

cos cos2 2( , , ) ( )

sin sin2 2

A A A A A A A Ax t x tu x y t Y y

A A A A A A A Ax t x t

ξ ω ξ ω

ξ ω ξ ω

+ −⎡ ⎤− + +⎢ ⎥= ⎢ ⎥

+ −⎢ ⎥+ + −⎢ ⎥⎣ ⎦

(4.54)

Rename the constant as

1 3 2 41 2

A A A AB += , 2 3 1 4

2 2A A A AB +

= , 1 4 2 33 2

A A A AB += , 1 3 2 4

4 2A A A AB −

= (4.55)

From Equation (4.54) and (4.55) we obtain the expression of velocity and stress as

( ) ( )

( ) ( )1 2

3 4

sin cos( , , ) ( )

cos sin

B x t B x tv x y t Y y

B x t B x t

ξ ω ξ ωω

ξ ω ξ ω

⎡ ⎤− − −= ⎢ ⎥

+ + − +⎢ ⎥⎣ ⎦ (4.56)

( ) ( )( ) ( )

1 2

3 4

sin cos( , , ) ( )

cos sin

B x t B x tT x y t Y y

B x t B x t

ξ ω ξ ωξ

ξ ω ξ ω

⎡ ⎤− − + −= ⎢ ⎥

+ + − +⎢ ⎥⎣ ⎦ (4.57)

Note that the constants 1B and 2B multiply the terms of the forward propagating wave

while 3B and 4B multiply the terms of the backward propagating wave.

Substitute Equation (4.56) and (4.57) into (4.48) and rearrange the terms to obtain the

average power flow, i.e.,

( ) ( )( ) ( )

( ) ( )( ) ( )

2 2 2 21 2

1 222 2 2 2

0 3 4

3 4

sin cos2 sin cos2( ) ( )

2 cos sin2 sin cos

d

avd

B x t B x tB B x t x t

P x Y y dy dtB x t B x tB B x t x t

ξ ω ξ ω

ξ ω ξ ωωξξ ω ξ ω

ξ ω ξ ω

Τ

−

⎡ ⎤− + −⎢ ⎥− − −⎛ ⎞ ⎢ ⎥= ⎜ ⎟ ⎢ ⎥Τ − + − +⎝ ⎠ ⎢ ⎥

⎢ ⎥+ + +⎣ ⎦

∫ ∫ (4.58)

Recall the following integrals

( ) ( ) ( )( )( ) ( ) ( )( )

2

2

sin sin cos 2

cos sin cos 2

x t dx t x t x t

x t dx t x t x t

ξ ω ω ξ ω ξ ω ω

ξ ω ω ξ ω ξ ω ω

± = ± ±

± = ± ± ±

∫∫

m (4.59)

73

Perform the integral in Equation (4.58) with the use of (4.59) to get

( ) ( )

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

21

22

22

23 3 4

22

4 1 20

sin cos2

sin cos2 2( ) ( )

2 sin cos sin22

sin cos sin22

T

d

avd

t x t x tB

t x t x tBP x Y y dy

t x t x t x tB B B

t x t x t x tB B B

ω ξ ω ξ ωω

ω ξ ω ξ ωωξ ω

ω ξ ω ξ ω ξ ωω ω

ω ξ ω ξ ω ξ ωω ω

−

⎡ ⎤+ − −⎢ ⎥⎢ ⎥

− − −⎢ ⎥+⎢ ⎥⎛ ⎞⎢ ⎥= ⎜ ⎟

Τ + + + +⎢ ⎥⎝ ⎠ − +⎢ ⎥⎢ ⎥

− + + −⎢ ⎥− +⎢ ⎥⎣ ⎦

∫

(4.60)

after substitution and rearrangement, we obtain

( )2 2 2 2 21 2 3 4( ) ( )

2

d

avd

P x Y y dy B B B Bωξ

−

⎛ ⎞= + − −⎜ ⎟

⎝ ⎠∫ (4.61)

As stated before, the contribution from the forward propagating mode is given by 1B and

2B , while the contribution due to the backward propagating mode is given by 3B and 4B .

Note that from Equation (4.55) we have

( )( )2 2 2 2

1 2 3 42 2 2 21 2 3 4 4

A A A AB B B B

+ ++ = + = (4.62)

Hence Equation (4.61) becomes

( ) ( )( ) ( )( )2 2 2 2 2 2 2 21 2 3 4 1 2 4 3 2, ( ) 0

4 4 2

d

avd

A A A A A A A AP x y Y y dyωξ

−

⎡ ⎤⎛ ⎞+ + + += − =⎢ ⎥⎜ ⎟

⎣ ⎦⎝ ⎠∫ (4.63)

As expected, the average power flow is equal to zero. Call the average power flow due to

forward propagating mode as

74

( )( )2 2 2 2

1 2 3 4 2 ( )4 2

davforward

d

A A A AP Y y dyωξ

−

⎛ ⎞+ += ⎜ ⎟

⎝ ⎠∫ (4.64)

and the average power flow due to backward propagating mode as

( )( )2 2 2 2

1 2 3 4 2 ( )4 2

dav

backwardd

A A A AP Y y dyωξ

−

⎛ ⎞+ += − ⎜ ⎟

⎝ ⎠∫ (4.65)


0av avav forward backwardP P P= + = (4.66)

and hence

av avbackward forwardP P= − (4.67)

It is to note that, the average power flow is given by the sum of the forward and

backward propagating wave average power flow. These two average power flows are

constant in space and are equal in modulus but with opposite sign. The average power

flow due to a forward or a backward propagating wave is equal to a constant multiplied

by a term that depends on the wave mode under consideration, i.e.,

2 ( )2

d

d

Y y dyωξ

−∫ (4.68)

This term is present in the expression of the average power flow of a propagating wave

and in that of a standing wave.

4.1.1.1.3 Generic formulation

The particle displacement can be expressed by not explicitly determining the terms

dependent on x, i.e.,

75

[ ]( , , ) ( ) ( ) cos( ) sin( )u x y t X x Y y t tω ω= + (4.69)

where

( ) cos( ) sin( )X x x xξ ξ= + (4.70)

Hence, velocity and stress are respectively

[ ]( , , ) ( ) ( ) sin( ) cos( )zv x y t X x Y y t tω ω ω= − + (4.71)

[ ]( , , ) ( ) ( ) cos( ) sin( )xzT x y t X x Y y t tω ω′= + (4.72)

The product of velocity by stress becomes

2 2 2( , , ) ( , , ) ( ) ( ) sin ( ) cos ( )z xzv x y t T x y t X x X x Y t tω ω ω′ ⎡ ⎤= − −⎣ ⎦ (4.73)

Substitute this expression in the average power flow Equation (4.48) to get

2 2 2

0

( ) ( )( ) sin ( ) cos ( )d T

avd

X x X xP x Y dy t t dtT

ω ω ω−

⎛ ⎞ ′⎡ ⎤= − −⎜ ⎟ ⎣ ⎦

⎝ ⎠∫ ∫ (4.74)

solve the integral to get

( )2 ( ) ( )( ) 1 1 02

d

avd

X x X xP x Y dyω−

⎛ ⎞ ′= − − =⎜ ⎟

⎝ ⎠∫ (4.75)

Note that in this case

[ ][ ] ( )( ) ( ) cos( ) sin( ) sin( ) cos( ) cos 2X x X x x x x x xξ ξ ξ ξ ξ ξ ξ′ = + − + = (4.76)

Let’s consider expression (4.74) and substitute Equation (4.76) into it (note that

2 2sin ( ) cos ( ) cos(2 )t t tω ω ω− = − ), i.e.,

76

( ) ( )2

0

2( ) cos 2 cos 22

d T

avd

P x Y dy x t dtT

ωξ ξ ω−

⎛ ⎞= ⎜ ⎟

⎝ ⎠∫ ∫ (4.77)

Note that for forward propagating mode the average power flow was equal to a constant

multiplied by the term in Equation (4.68).

The average total power flow derived in Equation (4.77) is a constant (in this case

zero) multiplied by the term in Equation (4.68). We will expect than that the average

power flow due to one of the propagating wave derived in circular coordinate is equal to

a constant multiplied by a term equal to half the product of the radial frequency, the

wavenumber, and the through the thickness particle displacement contribution.

The term dependent on x in Equation (4.77) is plotted in Figure 4.3. However, it is to

point out that the average power flow does not depend on x because Equation (4.77) is

equal to zero. If we had considered only the forward propagating mode, the dependence

on x in the average power flow would not have been existent.

Figure 4.3 Average power flow apparent variation with x. The abscissa is equal to

cos(2 )XX xξ′ = .

0 5 10 15 20

2−

1−

1

2

( )fx kHz m⋅

XX ′

77

The term in x is explicitly present in Equation (4.77) because the solution was not

transformed in the D’Alambert solution (see also the result in Equation (4.52)

4.1.1.2 Power flow along ˆxn (complex)

Let consider the complex power flow through the surface with normal ˆxn due to


( )1 ˆ2

y d

out xy d

P dydz− −

= − ⋅ ⋅∫ ∫ v T n% (4.78)

where the ~ sign indicate the conjugate.

Multiply the normal in the x direction, Equation (4.37), by the velocity-stress product to

get

( ) ˆ x z xzv T⋅ ⋅ =v T n% % (4.79)


12

z d

out z xzz d

P v T dydz− −

= − ∫ ∫ % (4.80)

Since both the velocity and the stress do not dependent on z, we can consider the power

flow per unit length, i.e.,

1( , , ) ( , , ) ( , , )2

dx

out z xzd

P x y t v x y t T x y t dy−

= − ∫ % (4.81)

Equation (4.81) is the complex power flow per unit length of shear horizontal waves

propagating in the x direction.

The time-average power flow is

78

0 0

1 1 1( , , ) ( , , ) ( , , )2

dx

av out z xzd

P x y t P dt v x y t T x y t dydtΤ Τ

−

= = −Τ Τ∫ ∫ ∫ % (4.82)

Let write the expression of velocity and stress as

( )

( )

( , , ) ( )

( , , ) ( )

i x tz z

i x txz xz

v x y t v y e

T x y t T y e

ξ ω

ξ ω

−

−

=

= (4.83)

and substitute this expression in Equation (4.82) to get

( )1( , , ) ( ) ( )2

di x

av z xzd

P x y t e v y T y dyξ ξ−

−

= − ∫% % (4.84)

The acoustic complex Poynting vector is defined as

= − ⋅P v T% (4.85)

If the fields are harmonic, the time-average value of the Poynting vector is

( )0

1, ( , ) ( , )av x y x y x y dtΤ

− = ⋅Τ ∫P v T% (4.86)

Substitute expression (4.83) into Equation (4.86), i.e.

( ) ( ) ( )

0

( ) ( ), i x t i x tx z xzav

v y T yP x y e e dtξ ω ξ ωΤ

− − −− =Τ ∫

%% (4.87)

Solve the integral and get

( ) ( ) ( ) ˆ, ( ) ( ) Rei xxav z xz xP x y v y T y e ξ ξ−− = = ⋅ ⋅v T n%% % (4.88)


power while the imaginary part of the complex power represents the peak value of the

reactive power.

79

4.1.1.3 The acoustic Poynting theorem

In the case of SH waves let’s consider the equation of motion and the strain-displacement

equation derived for this case (see Equations (3.1) and (3.2)).

2

2

zxz

zyz

yzxz zz

uSxuSyTT u F

x y tρ

⎧ ∂⎪ =∂⎪

⎪ ∂⎪ =⎨ ∂⎪⎪ ∂∂ ∂

+ = −⎪∂ ∂ ∂⎪⎩

(4.89)

Perform the time derivative of the first to equations and use relation (4.2) to get

xz z

yz z

yzxz zz

S vt x

S vt y

TT v Fx y t

ρ

⎧∂ ∂=⎪

∂ ∂⎪⎪∂ ∂⎪ =⎨ ∂ ∂⎪⎪ ∂∂ ∂

+ = −⎪∂ ∂ ∂⎪⎩

(4.90)

Follow procedure in Section 4, multiply the first equation by xzT , the second by yzT , and

the third by zv , i.e.,

xzzxz xz

yzzyz yz

yzxz zz z z z z

SvT Tx t

SvT Ty t

TT vv v v v Fx y t

ρ

⎧ ∂∂=⎪

∂ ∂⎪⎪ ∂∂⎪ =⎨ ∂ ∂⎪⎪ ∂∂ ∂

+ = −⎪∂ ∂ ∂⎪⎩

(4.91)

Sum the three equations to get

80

yz yzxz xzz z zz xz z yz z xz yz z z

T ST Sv v vv T v T v T T v Fx x y y t t t

ρ∂ ∂∂ ∂∂ ∂ ∂

+ + + = + + −∂ ∂ ∂ ∂ ∂ ∂ ∂

(4.92)

Note that the first four terms on the right-hand side can be grouped two by two. Finally

obtain:

z yz yzz xz xzzz xz yz z z

v T Sv T Svv T T v Fx y t t t

ρ∂ ∂∂ ∂∂

+ = + + −∂ ∂ ∂ ∂ ∂

(4.93)

It is easy to see that Equation (4.93) is the expression of Equation (4.7) for SH waves.

Subsequently, the acoustic Poynting vector of Equation (4.20) is written for straight-

crested Lamb waves in the form

( ) ( )ˆ ˆz xz z yzv T v T⎡ ⎤= − ⋅ = − +⎣ ⎦P v T x z (4.94)

4.1.2 Lamb waves

Consider Lamb waves propagating in a rectangular plate. The Lamb wave velocity vector

is

{ }0x yv v=v (4.95)


00

0 0 0

xx xy

xy yy

T TT T

⎡ ⎤⎢ ⎥Τ = ⎢ ⎥⎢ ⎥⎣ ⎦

(4.96)

The scalar product between velocity and stress matrix is

{ }0x xx y xy x xy y yyv T v T v T v T⋅ = + +v T (4.97)

81

4.1.2.1 Power flow along ˆxn

The power flow through the surface of area dydz and with normal ˆxn due to propagating

Lamb waves is given by Equation (4.36) where the normal ˆxn is defined in Equation

(4.37) (see Figure 4.2b). Use the normal ˆxn definition (4.37) and product (4.97) into

power flow Equation (4.36) to get

( )z d

out x xx y xyz d

P v T v T dydz− −

= − +∫ ∫ (4.98)

As for SH waves, the problem is z-invariant and hence both the velocity and the stresses

do not depend on z. Consider the power flow per unit wave front length defined as

( )d

xout x xx y xy

d

P v T v T dy−

= − +∫ (4.99)

Equation (4.99) is the power flow per unit length of Lamb waves propagating in the x

direction.

Recall the solution through separation of variables to the Lamb wave equations and for

simplicity consider only symmetric modes (we will omit subscript S), i.e.,

( ) ( )

( ) ( )

( , , ) ( ) ( )

( , , ) ( ) ( )

n n

n n

i x t i x tx n nx n nx

n n

i x t i x ty n ny n ny

n n

u x y t A Y y e B Y y e

u x y t i C Y y e i D Y y e

ξ ω ξ ω

ξ ω ξ ω

− +

− +

⎧ = − −⎪⎨

= − −⎪⎩

∑ ∑

∑ ∑ (4.100)

where ( )nxY y and ( )nyY y are respectively the displacement behavior in the y coordinate

for the axial and thickness particle displacement. For notation simplicity, we will

consider only one mode and we will omit the subscript n. The dependence on the x and t

82

variables can be expressed in terms of sine and cosine functions, hence Equation (4.100)

becomes

[ ][ ][ ][ ]

1 2 3 4

1 2 3 4

( , , ) ( ) cos( ) sin( ) cos( ) sin( )

( , , ) ( ) cos( ) sin( ) cos( ) sin( )x x

y y

u x y t Y y A x A x A t A t

u x y t Y y A x A x A t A t

ξ ξ ω ω

ξ ξ ω ω

⎧ = + +⎪⎨

= + +⎪⎩ (4.101)

Perform the multiplication of the terms dependent on x and t in Equation (4.101), i.e.,

1 3 1 4

2 3 2 4

1 3 1 4

2 3 2 4

cos( ) cos( ) cos( )sin( )( , , ) ( )

sin( )cos( ) sin( )sin( )

cos( )cos( ) cos( )sin( )( , , ) ( )

sin( ) cos( ) sin( )sin( )

x x

y y


A A x t A A x t


A A x t A A x t



⎧ +⎡ ⎤=⎪ ⎢ ⎥+ +⎪ ⎣ ⎦

⎨+⎡ ⎤⎪ = ⎢ ⎥⎪ + +⎣ ⎦⎩

(4.102)

Substitute the expression of the particle displacement (4.102) into the particle velocity

Equation (4.2), i.e.

1 3 1 4

2 3 2 4

1 3 1 4

2 3 2 4

cos( )sin( ) cos( ) cos( )( , , ) ( )

sin( )sin( ) sin( )cos( )

cos( )sin( ) cos( ) cos( )( , , ) ( )

sin( )sin( ) sin( ) cos(

xx x

yy y

A A x t A A x tuv x y t Y yA A x t A A x tt

A A x t A A x tuv x y t Y y

A A x t A A xt

ξ ω ξ ωω

ξ ω ξ ω

ξ ω ξ ωω

ξ ω ξ ω

−⎡ ⎤∂= = − ⎢ ⎥+ −∂ ⎣ ⎦

−∂= = −

+ −∂ )t⎡ ⎤⎢ ⎥⎣ ⎦

(4.103)

and recall from Equations (3.2) and (3.3) that

1 3 1 4

2 3 2 4

1 3 1 4

2 3 2 4

cos( )cos( ) cos( )sin( )( )sin( ) cos( ) sin( )sin( )

sin( ) cos( ) sin( )sin( )( )

cos( ) cos( ) cos( )sin( )

x

yxxy

y

A A x t A A x tY yA A x t A A x tyuuT

y x A A x t A A x tY y

A A x t A A x t


μ μξ ω ξ ω

ξξ ω ξ ω

+⎡ ⎤∂⎢ ⎥+ +∂∂⎛ ⎞∂ ⎣ ⎦= + =⎜ ⎟∂ ∂ − −⎡ ⎤⎝ ⎠ + ⎢+ +⎣

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎥⎜ ⎟⎦⎝ ⎠

(4.104)

4.1.2.1.1 Average power flow Equations (4.103) and (4.104) represent respectively the particle velocity and stress due

to both backward and forward propagating wave. We want to derive the average power

83

flow due to the presence of both backward and forward propagating waves. Consider the

expression of the average power flow, i.e.,

( )0

1( )d

av x xx y xyd

P x v T v T dydtΤ

−

= − +Τ ∫ ∫ (4.105)

The product of velocity by stress becomes, after rearrangement of the terms,

( )1

1 2 22 3 41

2 221 3 4

22

sin( )cos( ) sin( ) cos( )cos( )

sin( )cos( ) cos ( ) sin ( )sin( )

x xx y xy

A xY

A x A A t tA xv T v T

A xA x A A t tYA x

ξξ ω ωξ

ωξξ ω ω

ξ

⎧ ⎫−⎡ ⎤⎪ ⎪⎢ ⎥ ⎡ ⎤+ −⎡ ⎤⎪ ⎣ ⎦ ⎪+ = − ⎢ ⎥⎨ ⎬ ⎢ ⎥+⎡ ⎤ ⎡ ⎤⎢ ⎥− −⎣ ⎦⎪ ⎪ ⎣ ⎦⎣ ⎦+ ⎢ ⎥⎪ ⎪+⎣ ⎦⎩ ⎭

(4.106)

Substitute Equation (4.106) into (4.99) to obtain the average power flow, i.e.,

23 1

12241

22 0 1

23 4 22

sin( )sin( )cos( )cos( )cos( )

( )sin( ) cos( )cos ( )

sin( )sin ( )

d

avd

A A xt t YA xAA x

P x dydtA x A xt YA A A xt

ξω ω

ξξωξ ξω

ξω

Τ

−

⎡ ⎤⎛ ⎞ ⎧ − ⎫⎡ ⎤⎢ ⎥⎜ ⎟ ⎪ ⎪⎢ ⎥⎜ ⎟ +−⎡ ⎤ ⎢ ⎥ ⎪ ⎣ ⎦ ⎪⎝ ⎠= − ⎨ ⎬⎢ ⎥ ⎢ ⎥+Τ ⎡ ⎤⎡ ⎤⎣ ⎦ ⎪ ⎪⎢ ⎥ +− ⎢ ⎥⎢ ⎥ ⎪ ⎪⎢ ⎥ +⎣ ⎦− ⎩ ⎭⎣ ⎦⎣ ⎦

∫ ∫ (4.107)

or, by solving the integral,

[ ]

11

23 4 1 2

12

2

sin( )cos( )1 1( ) cos( ) sin( ) 0

2 2 cos( )sin( )

d

avd

A xY

A xP x A A A x A x dy

A xY

A x

ξξ

ω ξ ξξ

ξ−

⎧ − ⎫⎡ ⎤⎪ ⎪⎢ ⎥+⎪ ⎣ ⎦ ⎪⎡ ⎤= + − =⎨ ⎬⎢ ⎥⎣ ⎦ ⎡ ⎤⎪ ⎪+ ⎢ ⎥⎪ ⎪+⎣ ⎦⎩ ⎭

∫ (4.108)

Equation (4.108) indicates that the average power flow due to the presence of both

forward and backward propagating wave is equal to zero.

4.1.2.1.2 Average power flow of the forward and backward propagating waves

We have seen in the previous section that the average power flow is equal to zero.

Now we want to explicit the contributions to the average power flow form both backward

84

and forward propagating modes. For this motive, the terms in the particle displacement

Equation (4.102) can be transformed so that the contributions due to forward and

backward propagating waves are more explicit. Note that we can transform the terms in

Equation (4.102) as in Equation (4.53). With the use of Equations (4.53), (4.55) into

Equation (4.102) we can write the particle displacement as

( ) ( )( ) ( )( ) ( )( ) ( )

1 4

3 2

1 4

3 2

cos cos( , , ) ( )

sin sin

cos cos( , , ) ( )

sin sin

x x

y y

B x t B x tu x y t Y y

B x t B x t

B x t B x tu x y t Y y

B x t B x t

ξ ω ξ ω

ξ ω ξ ω

ξ ω ξ ω

ξ ω ξ ω

⎧ ⎡ ⎤− + +=⎪ ⎢ ⎥

+ + −⎢ ⎥⎪ ⎣ ⎦⎨

⎡ ⎤− + +⎪ = ⎢ ⎥⎪ + + −⎢ ⎥⎣ ⎦⎩

(4.109)

From Equation (4.109) we obtain the expression of velocity and stress as

( ) ( )( ) ( )

( ) ( )( ) ( )

1 4

3 2

1 4

3 2

sin sin( , , ) ( )

cos cos

sin sin( , , ) ( )

cos cos

x x

y y


B x t B x t


B x t B x t

ξ ω ξ ωω

ξ ω ξ ω

ξ ω ξ ωω

ξ ω ξ ω

⎧ ⎡ ⎤− − +=⎪ ⎢ ⎥

+ + − −⎢ ⎥⎪ ⎣ ⎦⎨

⎡ ⎤− − +⎪ = ⎢ ⎥⎪ + + − −⎢ ⎥⎣ ⎦⎩

(4.110)

( ) ( )( ) ( )

( ) ( )( ) ( )

1 4

3 2

1 4

3 2

cos cos( )sin sin

( , , )sin sin

( )cos cos

x

xy

y

B x t B x tY yy B x t B x t

T x y tB x t B x t

Y yB x t B x t

ξ ω ξ ω

ξ ω ξ ωμ

ξ ω ξ ωξ

ξ ω ξ ω

⎛ ⎞⎡ ⎤− + +∂⎜ ⎟⎢ ⎥

∂ + + −⎢ ⎥⎜ ⎟⎣ ⎦= ⎜ ⎟⎡ ⎤− − − +⎜ ⎟+ ⎢ ⎥⎜ ⎟+ + −⎢ ⎥⎣ ⎦⎝ ⎠

(4.111)

Substitute Equation (4.110) and (4.111) into (4.105) to obtain the average power flow,

i.e.,

85

( ) ( )

( ) ( )

( ) ( ) ( )( ) ( )

( )( ) ( )

( ) ( )

23 4

120 1 2

2 21 2

2 21 2

1 3 4 2

23

cos sin

sin cos

sin cos

sin cos( )cos cos

sin sin

d

d

av

B x t B x tdt Y dy

B x t B x t

B B x t x t

B B x t x tP xx t x t

B B B Bx t x t

B B

ξ ω ξ ωξ

ξ ω ξ ω

ξ ω ξ ω

ω ξ ω ξ ω

ξ ω ξ ω

ξ ω ξ ω

Τ

−

⎡ ⎤⎡ ⎤+ − +⎣ ⎦⎢ ⎥⎢ ⎥⎡ ⎤− − − −⎣ ⎦⎣ ⎦

⎡ ⎤− − −⎢ ⎥

⎡ ⎤⎢ ⎥+ − − −⎣ ⎦⎣ ⎦= −Τ ⎡ ⎤+ −

+ + − ⎢ ⎥+ − +⎣ ⎦

−+

∫ ∫

( ) ( ) ( )( ) ( )

20

24

2 23 4

sin cos

cos sin

d

d

dt Y dy

x t x t

B B x t x t

ξ ω ξ ω

ξ ω ξ ω

Τ

−

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎡ ⎤⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎡ ⎤+ +⎪ ⎪⎢ ⎥⎢ ⎥⎪ ⎪⎢ ⎥⎡ ⎤⎢ ⎥+ + − +⎪ ⎪⎢ ⎥⎣ ⎦⎣ ⎦⎣ ⎦⎩ ⎭

∫ ∫

(4.112)

where

( ) 2 2

1

2

( ) 2 ( ) ( )

( ) ( )( ) ( ) ( )

x y

y xx y

Y y Y y Y y

Y y Y yY y Y y Y yy y

λ μ μ

λ μ

⎧ = + +⎪

∂⎨ ∂= +⎪ ∂ ∂⎩

(4.113)

The second time integral in Equation (4.112) is equal to zero, hence the first integral

yields

( )2 2 2 21 1 2 3 4( )

2

d

avd

P x Y dy B B B Bωξ

−

⎛ ⎞= + − −⎜ ⎟

⎝ ⎠∫ (4.114)

As stated before, the contribution from the forward propagating mode is given by 1B and

2B , while the contribution due to the backward propagating mode is given by 3B and 4B .

Note that from Equation (4.55) we have

( )( )2 2 2 2

1 2 3 42 2 2 21 2 3 4 4

A A A AB B B B

+ ++ = + = (4.115)


86

( ) ( )( ) ( )( )2 2 2 2 2 2 2 21 2 3 4 1 2 4 3

1, ( ) 04 4 2

d

avd

A A A A A A A AP x y Y y dyωξ

−

⎡ ⎤ ⎛ ⎞+ + + += − =⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎝ ⎠

∫ (4.116)

As expected, the average power flow is equal to zero. Call the average power flow due to

forward propagating mode as

( )( )2 2 2 2

1 2 3 41( )

4 2

davforward

d

A A A AP Y y dyωξ

−

⎛ ⎞+ += ⎜ ⎟

⎝ ⎠∫ (4.117)

and the average power flow due to backward propagating mode as

( )( )2 2 2 2

1 2 3 41( )

4 2

dav

backwardd

A A A AP Y y dyωξ

−

⎛ ⎞+ += − ⎜ ⎟

⎝ ⎠∫ (4.118)


0av avav forward backwardP P P= + = (4.119)

and hence

av avbackward forwardP P= − (4.120)

4.1.2.2 Power flow along ˆxn (complex)

Let consider the complex power flow through the surface with normal ˆxn due to

propagating Lamb waves (Figure 4.2b); from Equation (4.14) we get

( )1 ˆ2

y d

out xy d

P dydz− −

= − ⋅ ⋅∫ ∫ v T n% (4.121)

Following the procedure for real power flow in the x direction, we define the power flow

per unit length as

87

( )1( , , )2

dx

out x xx y xyd

P x y t v T v T dy−

= − +∫ % % (4.122)

Write the expression of velocity and stress as

( )

( )

( , , ) ( )

( , , ) ( )

i x t

i x t

x y t y e

x y t y e

ξ ω

ξ ω

−

−

=

=

v v

T T (4.123)

By substituting the expressions of the velocity and stress as expressed in (4.123), we

obtain

( )( , , ) ( ) ( ) ( ) ( )d

i xxout x xx y xy

d

P x y t e v y T y v y T y dyξ ξ−

−

⎡ ⎤= − +⎣ ⎦∫% % % (4.124)

The average power flow is written as

0

1( , , )T

x xav out outP x y t P dt P

T= =∫ (4.125)

The acoustic complex Poynting vector is defined as

= − ⋅P v T% (4.126)

If the fields are harmonic, the time-average value of the Poynting vector is

( )0

1, ( , ) ( , )av x y x y x y dtΤ

− = ⋅Τ ∫P v T% (4.127)

Substitute expression (4.123) into Equation (4.127), i.e.

( ) ( ) ( )

0

( ) ( ) ( ) ( ), x xx y xy i x t i x tx

av

v y T y v y T yP x y e e dtξ ω ξ ω

Τ− − −

⎡ ⎤+⎣ ⎦− =Τ ∫

%% %

(4.128)

Solve the integral and get

88

( ) ( ) ( ) ˆ, ( ) ( ) ( ) ( ) Rei xxav x xx y xy xx y v y T y v y T y e ξ ξ−⎡ ⎤− = + = ⋅ ⋅⎣ ⎦P v T n%% % (4.129)


power while the imaginary part of the complex power represents the peak value of the

reactive power.


In the case of Lamb waves consider the derivative of the strain-velocity Equation (3.2)

with respect to t and the equation of motion (3.1) (through use of relation (4.2)), i.e.,

xx x

yy y

xy yx

xyxx xx

xy yy yy

S vt x

S vt y

S vvt y x

TT v Fx y t

T T vF

x y t

ρ

ρ

⎧∂ ∂=⎪

∂ ∂⎪⎪∂ ∂

=⎪∂ ∂⎪

⎪∂ ∂∂⎪ = +⎨ ∂ ∂ ∂⎪⎪ ∂∂ ∂

+ = −⎪∂ ∂ ∂⎪

⎪∂ ∂ ∂⎪ + = −

∂ ∂ ∂⎪⎩

(4.130)

Following the general procedure in Section 4, we multiply the first line by xxT , the

second by yyT , the third by xyT , the fourth by xv , and the fifth by yv , i.e.,

89

x xxxx xx

y yyyy yy

y xyxxy xy

xyxx xx x x x x

xy yy yy y y y y

v ST Tx tv S

T Ty t

v SvT Ty x t

TT vv v v v Fx y t

T T vv v v v F

x y t

ρ

ρ

⎧ ∂ ∂=⎪

∂ ∂⎪⎪ ∂ ∂

=⎪∂ ∂⎪

⎪ ∂ ∂⎛ ⎞∂⎪ + =⎨ ⎜ ⎟∂ ∂ ∂⎝ ⎠⎪⎪ ∂∂ ∂⎪ + = −

∂ ∂ ∂⎪⎪ ∂ ∂ ∂⎪ + = −⎪ ∂ ∂ ∂⎩

(4.131)

We sum the five equations in system (4.131) to get

xy xy yy y yxx x xx x y y xy xx yy

y xy yyx xxx x x y y y xy xx yy

T T T v vT v vv v v v T T Tx y x y y x x y

v S Sv Sv v F v v F T T Tt t t t t

ρ ρ

∂ ∂ ∂ ∂ ∂⎛ ⎞∂ ∂ ∂+ + + + + + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠

∂ ∂ ∂∂ ∂= − + − + + +

∂ ∂ ∂ ∂ ∂

(4.132)

We notice that the first seven terms on the left-hand side can be grouped two by two, i.e.,

( ) ( )x xx y xy x xy y yy

y yy xyx xxx y xx yy xy x x y y

v T v T v T v Tx y

v S Sv Sv v T T T v F v Ft t t t t

ρ ρ

∂ + ∂ ++

∂ ∂∂ ∂ ∂∂ ∂

= + + + + − −∂ ∂ ∂ ∂ ∂

(4.133)

It is easy to see that Equation (4.133) is the expression for straight-crested Lamb waves

of the general Equation (4.7). Subsequently, the acoustic Poynting vector of Equation

(4.20) is written for straight-crested Lamb waves in the form

( ) ( )ˆ ˆx xx y xy x xy y yyv T v T v T v T⎡ ⎤= − ⋅ = − + + +⎣ ⎦P v T x y (4.134)

90

4.2 POWER FLOW IN CYLINDRICAL COORDINATES

While power flow derivation in rectangular coordinates is treated in many textbooks

(Giurgiutiu, 2008; Auld, 1990), power flow in cylindrical coordinates is considered only

for the particular case in which the wave propagates in a tube or a cylinder, i.e., the waver

propagates in the z direction. In this section, we will consider a circular crested wave

propagating in the radial direction. The wave front length of the wave increases with the

radial distance, while the energy and, hence, the power of the wave remain constant.

Consider a section of area rdzdθ of a circular plate, as in Figure 4.4. The are five

surfaces that determines the circular sections, these surfaces are denoted by normal ˆrn ,

nθ± , and ˆzn± .

Figure 4.4 Circular section rdzdθ of a plate of thickness 2d. a) Section notations; b) Power flow

through surface with normal nr.

The power flows thorough the five surfaces; however, no power flows through the top

and bottom ˆzn± free surfaces.


Consider circular crested SH waves propagating in a circular plate. The SH velocity

vector is

rb)

ˆrn

rdθˆzn− dz

nθ+ˆzn+

nθ−

rz

rθa)

91

{ }0 0vθ=v (4.135)


0 0

00 0

r

r z

z

TT T

T

θ

θ θ

θ

⎡ ⎤⎢ ⎥Τ = ⎢ ⎥⎢ ⎥⎣ ⎦

(4.136)

Derive the scalar product between velocity and stress matrix to get

{ }0r zv T v Tθ θ θ θ⋅ =v T (4.137)

4.2.1.1 Power flow along ˆrn

Consider the circumferential power flow through the surface with normal ˆrn due to


( )2

0

ˆd

out rd

P rd dzπ

θ−

= − ⋅ ⋅∫ ∫ v T n (4.138)

Note that the normal ˆrn is defined as

{ }ˆ 1 0 0r =n (4.139)

Multiply the normal in the r direction by the velocity-stress product to get

( ) ˆ r rv Tθ θ⋅ ⋅ =v T n (4.140)


2

0

d

out rd

P rv T d dzπ

θ θ θ−

= − ∫ ∫ (4.141)

92

Since the problem is θ - invariant, both velocity and stress do not depend on θ, we can

perform the integration with respect to θ, i.e.,

2d

out rd

P r v T dzθ θπ−

= − ∫ (4.142)

Equation (4.142) is the circumferential power flow of shear horizontal waves propagating

in the r direction. Equation (4.142) can be expresses as

2 rout outP rPπ= (4.143)

where routP is the radial power flow per unit length (line density) in the r direction, i.e.,

d

rout r

d

P v T dzθ θ−

= − ∫ (4.144)

The term 2 rπ is the length/circumference of the circular wave front.

Recall the general solution of circular crested SH waves is given by Equation (3.67), i.e.,

( )1( , , ) ( ) sin cos i tu r z t J r A z B z e ωθ ξ η η −= + (4.145)

We can write Equation (4.145) in a generic form, i.e.,

[ ]3 4( , , ) ( ) ( ) cos( ) sin( )u r z t R r Z z A t A tθ ω ω= + (4.146)

where R is a function of r and Z is a function of z and they are

( )1

( ) sin cos( )

Z z A z B zR r J r

η ηξ

= +

= (4.147)

Accordingly, the velocity and stress expressions become

93

[ ]3 4( , , ) sin( ) cos( )uv r z t RZ A t A ttθ

θ ω ω ω∂= = − +

∂ (4.148)

[ ]3 4( , , ) cos( ) sin( )ruT r z t R Z A t A trθ

θ ω ω∂ ′= = +∂

(4.149)

where RRr

∂′ =∂

. The product of velocity and stress is

( ) ( )2 2 2 2 23 4 4 3cos ( ) sin ( ) sin( ) cos( )xv T RR Z A A t t A A t tθ θ ω ω ω ω ω⎡ ⎤′= − − + −⎣ ⎦ (4.150)

Substitute Equation (4.150) into Equation (4.144), to get

( ) ( )2 2 2 2 23 4 4 3cos ( ) sin ( ) sin( )cos( )

dr

outd

P RR A A t t A A t t Z dzω ω ω ω ω−

⎡ ⎤′= − + −⎣ ⎦ ∫ (4.151)

Let consider the circumferential power flow, i.e.

( ) ( )2 2 2 2 23 4 4 32 cos ( ) sin ( ) sin( )cos( )

d

outd

P rRR A A t t A A t t Z dzπω ω ω ω ω−

⎡ ⎤′= − + −⎣ ⎦ ∫ (4.152)

The radial power flow per unit length varies with r as RR′ while the circumferential

power flow varies with r as rRR′ . Figure 4.5 shows how both power flows change with

the radial distance. As the distance from the origin increases, the radial power flow per

unit length behaves as 1 r . Hence, the circumferential power flow in the r direction has a

spatial almost harmonic behavior. This is due to the fact that the circular crested front

length increases as 2 rπ as the wave travels outward while the energy of the wave

remains constant.

94

routP

1 r

outP

r (m) 0 0.05 0.1 0.15 0.2 0.25

0

1

2

3

4

Figure 4.5 Power flow in the r direction as a function of the radius (Symmetric SH0 mode for an

Aluminum with wave propagating at 100 kHz).

Consider the average power flow given by

0

1 Tr

av outP P dtT

= ∫ (4.153)

with the use of (4.150) the average power flow becomes

( ) ( )2 2 2 2 23 4 4 3

0

cos ( ) sin ( ) sin( )cos( )d T

avd

RRP Z dz A A t t A A t t dtT

ω ω ω ω ω−

′ ⎡ ⎤= − − + −⎣ ⎦∫ ∫ (4.154)

From the derivation of the average power flow in rectangular coordinate, we will expect

that the average power flow in circular coordinate is zero if both backward and forward

propagating waves are considered or a constant not equal to zero if only one of the two is

considered. If we solve the integral in Equation (4.154) we obtain

( )2 1 1 02

d

avd

RRP Z dzω

−

′= − − =∫ (4.155)

95

The average power flow is zero because the wave represented in Equation (4.146) is a

standing wave.

Figure 4.6 shows how Equations (4.74) and (4.155) change with the frequency-radius

product. As f r⋅ increases RR′ approaches XX ′ , or we can say that XX ′ is an

approximation of RR′ at high f r⋅ product. Moreover, as the frequency increases, the

two curves become indistinguishable at lower values of the radius.

Figure 4.6 Variation of RR′ (solid red line) and XX ′ (dashed blue line) with respect to

frequency-radius product

4.2.1.1.1 Large radius value approximation Note that for large values of r the Bessel functions can be approximated as

( ) ( ) ( )12 3 1cos cos sin

4J r r r r

r rξ ξ π ξ ξ

πξ πξ⎛ ⎞ ⎡ ⎤− = − −⎜ ⎟ ⎣ ⎦⎝ ⎠

(4.156)

From the second of Equation (4.147) we find

f r⋅ (kHz m)

0 5 10 15 20

2−

1−

1

2

96

( ) ( )

( ) ( )

1 2

1 2

1 cos sin

sin cos

R A r A rr

R A r A rr

ξ ξπξ

ξ ξ ξπ

⎡ ⎤= − +⎣ ⎦

′ ⎡ ⎤= −⎣ ⎦

(4.157)

where A1 and A2 are arbitrary constants introduced for convenience; the negative sign is

incorporated into constant A2. Substitute expressions in (4.157) into the expression of the

velocity (4.148) and stress (4.149) ,i.e.,

( ) ( ) [ ]

( ) ( ) [ ]

1 2 3 4

1 2 3 4

1( , , ) cos sin sin( ) cos( )

( , , ) sin cos cos( ) sin( )r

uv r z t Z A r A r A t A tt r

uT r z t Z A r A r A t A tr r

θ

θ

ω ξ ξ ω ωπξ

ξ ξ ξ ω ωπ

∂⎡ ⎤= = − + − +⎣ ⎦∂

∂⎡ ⎤= = − +⎣ ⎦∂

(4.158)

Multiply the terms in t with those in x, after we rearranged the terms and we retained only

the forward propagating terms, we get

( ) ( )

( ) ( )

1 3 2 4 1 4 2 3

1 3 2 4 1 4 2 3

( , , ) sin cos2 2

( , , ) sin cos2 2r

A A A A A A A AZv r z t r t r tr

A A A A A A A AZT r z t r t r tr

θ

θ

ω ξ ω ξ ωπξ

ξ ξ ω ξ ωπξ

⎡ ⎤+ −⎛ ⎞ ⎛ ⎞= − − + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤+ −⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

(4.159)


( )

( )

( )

( )

1 3 2 4 1 3 2 4

2

1 4 2 3 1 4 2 3

sin sin2 2

cos cos2 2

dr

outd

A A A A A A A Ar t r tP Z dz

r A A A A A A A Ar t r t

ξ ω ξ ωωξπ

ξ ω ξ ω −

⎡ + ⎤ ⎡ + ⎤⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥=− −⎢ ⎥ ⎢ ⎥⎛ ⎞ ⎛ ⎞+ − + −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

∫ (4.160)

Hence the circumferential power flow is

97

( )

( )

21 3 2 4

2

1 4 2 3

sin22 2

cos2

dr

out outd

A A A A r tP rP Z dz

A A A A r t

ξ ωπ ωξ

ξ ω −

+⎡ ⎤−⎢ ⎥= = ⎢ ⎥

−⎢ ⎥+ −⎢ ⎥⎣ ⎦

∫ (4.161)

Multiply the terms in brackets and rearrange the terms to obtain

( ) ( )

( ) ( )

2 22 21 3 2 4 1 4 2 3

2

1 4 2 3 1 3 2 4

sin cos2 22

2 sin cos2 2

d

outd

A A A A A A A Ar t r tP Z dz

A A A A A A A A r t r t

ξ ω ξ ωωξ

ξ ω ξ ω −

⎡ ⎤+ −⎛ ⎞ ⎛ ⎞− + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥=

⎢ ⎥− +⎛ ⎞⎛ ⎞+ − −⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

∫ (4.162)

The time average power flow is (after integration and rearrangement)

2 2

2 1 3 2 4 1 4 2 322 2 2

d

avd

A A A A A A A AP Z dzωξ

−

⎡ ⎤⎛ ⎞ + −⎛ ⎞ ⎛ ⎞= +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎝ ⎠∫ (4.163)

The average power flow of the circumferential power flow of the forward propagating

wave for large r approximation is a constant not equal to zero and it is equal to that

derived for straight-crested waves, see Equation (4.61). The two that multiply the

circular-crested power flow is due to have considered the circumferential power flow.

4.2.1.1.2 Outward propagating wave through complex form

From the derivation of the average power flow we noticed that only when it is possible to

decompose the wave in forward and backward propagating waves, it is possible to

determine their average power flow contributions.

From Section 4.2.1.1.1, we have seen that asymptotically it is possible to decompose

the Bessel function in forward and backward propagating waves. An alternative way to

write solution to the cylindrical wave equation is by using the complex form (Hildebrand

,1964), i.e.,

98

( ) ( )1 21 1 2 1( ) ( ) ( )R r B H r B H rξ ξ= + (4.164)

where the Hankel functions are defined as

( )

( )

11 1 1

21 1 1

( ) ( ) ( ) Hankel function of first kind

( ) ( ) ( ) Hankel function of second kind

H r J r iY r

H r J r iY r

ξ ξ ξ

ξ ξ ξ

= +

= − (4.165)

and 1( )Y rξ is the Bessel function of second kind. Solution in Equation (4.146) can be

written as

( ) ( )1 21 1 2 1( , , ) ( ) ( ) ( )i t i tu r z t Z z B e H r B e H rω ω

θ ξ ξ⎡ ⎤= +⎣ ⎦ (4.166)

If we consider the asymptotic expressions of the Hankel function of first and second kind

given by

( ) ( )

( ) ( )

31 4

1

32 4

1

2

2

i r

i r

H r er

H r er

ξ π

ξ π

ξπξ

ξπξ

⎛ ⎞−⎜ ⎟⎝ ⎠

⎛ ⎞− −⎜ ⎟⎝ ⎠

(4.167)

the asymptotic expression of the wave equation solution Equation (4.166) becomes

( ) ( )3

41 2

2( , , ) ( )i r t i r ti e eu r z t Z z e B B

r r

ξ ω ξ ωπ

θ πξ

+ − −− ⎡ ⎤+⎢ ⎥

⎣ ⎦ (4.168)

Hence, for large values of r the Hankel function of first kind represents the inward

propagating wave, ( )11H , while the Hankel function of the second kind represents the

outward propagating wave, ( )21H ,. The amplitude of the oscillations is inversely

proportional to r .

99

To continue our analysis we will write the wave propagation solution with the explicit

real and imaginary parts of the complex form (Hildebrand, 1964), i.e.,

[ ][ ]

[ ][ ]

1 1 1 1

1 1 1 1

2 1 2 1

2 1 2 1

( ) cos ( )sin

( )sin ( ) cos( , , )

( ) cos ( )sin

( )sin ( )cos

A J r t B Y r t

i C J r t D Y r tu r z t Z

A J r t B Y r t

i C J r t D Y r t

θ

ξ ω ξ ω

ξ ω ξ ω

ξ ω ξ ω

ξ ω ξ ω

⎧ ⎫−⎪ ⎪

+ +⎪ ⎪= ⎨ ⎬

+ +⎪ ⎪⎪ ⎪+ −⎩ ⎭

(4.169)

where the first two terms in brackets are the inward propagating wave and the second two

terms the outward propagating wave.

We will derive the average power flow for the circular crested outward propagating

wave, i.e.,

[ ]

[ ]2 1 2 1

2 1 2 1

( ) cos ( )sin( , , )

( )sin ( )cos

A J r t B Y r tu r z t Z

i C J r t D Y r tθ

ξ ω ξ ω

ξ ω ξ ω

⎧ ⎫+⎪ ⎪= ⎨ ⎬+ −⎪ ⎪⎩ ⎭

(4.170)

The velocity and stress expressions become

[ ]

[ ]2 1 2 1

2 1 2 1

( )sin ( )cos( , , )

( )cos ( )sin

A J r t B Y r tuv r z t Zt i C J r t D Y r tθ

θ

ξ ω ξ ωω

ξ ω ξ ω

⎧ ⎫− +∂ ⎪ ⎪= = ⎨ ⎬∂ + +⎪ ⎪⎩ ⎭ (4.171)

1 12 0 2 0

1 12 0 2 0

( ) ( )( ) cos ( ) sin

( ) ( )( ) sin ( ) cosr

J r Y rA J r t B Y r tr ruT Z

r J r Y ri C J r t D Y r tr r

θθ

ξ ξξ ξ ω ξ ξ ω

ξ ξξ ξ ω ξ ξ ω

⎧ ⎫⎡ ⎤⎡ ⎤ ⎡ ⎤− + −⎪ ⎪⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∂ ⎪⎣ ⎦ ⎪= = ⎨ ⎬∂ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎪ ⎪+ − − −⎢ ⎥⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦⎣ ⎦⎩ ⎭

(4.172)

The product of velocity and stress is, after rearrangement,

{ }2 2 2( , , ) ( , , ) ( )sin cos ( )sin ( )cosxv r z t T r z t Z K r t t L r t F r tθ θ ω ω ω ω ω= + + (4.173)

where

100

( )

( )

( )

2 2 12 2 1 0

2 2 12 2 1 0

1 12 2 2 2 1 0 0 1

12 2 1 0 2 2 1

( )( ) ( )

( )( ) ( ) ( )

2 ( ) ( )( ) ( ) ( ) ( )

( )( ) ( ) ( )( )

Y rB D Y r Y rrJ rK r A C J r J r

rJ r Y ri A D B C J r Y r J r Y r

rY rA B J r Y r iA C J r

rL r

ξξ ξ ξ

ξξ ξ ξ

ξ ξξ ξ ξ ξ ξ ξ

ξξ ξ ξ ξ

⎡ ⎤⎡ ⎤+ −⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎡ ⎤= − + −⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎡ ⎤+ + + −⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦

⎡ ⎤− − −⎢ ⎥⎣ ⎦=

10

1 12 2 1 0 2 2 1 0

1 12 2 1 0 2 2 1 0

12 2 1 0

( )( )

( ) ( )( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( )( )

( )( ) ( )

J rJ rr

Y r J riB D Y r Y r C D Y r J rr r

J r Y rA B Y r J r iD B Y r Y rr rF rJ riA C J r J r

r

ξξ ξ



ξξ ξ ξ

⎡ ⎤⎡ ⎤−⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎡ ⎤ ⎡ ⎤+ − − −⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

⎡ ⎤ ⎡ ⎤− − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦=⎡ ⎤+ −⎢ ⎥⎣ ⎦

12 2 1 0

( )( ) ( ) Y rC D J r Y rrξξ ξ ξ

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎡ ⎤+ −⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦

(4.174)

Substitute Equation (4.173) into Equation (4.144), to get

{ }2 2 2sin cos sin cosd

rout

d

P K t t L t F t Z dzω ω ω ω ω−

= − + + ∫ (4.175)

The circumferential power flow becomes

{ }2 2 22 2 sin cos sin cosd

rout out

d

P rP r K t t L t F t Z dzπ π ω ω ω ω ω−

= = − + + ∫ (4.176)

We are interested in deriving the average power flow, i.e.,

{ }2 2 2

0 0

1 2 sin cos sin cosT d T

av outd

rP P dt Z dz K t t L t F t dtT T

π ω ω ω ω ω−

= = − + +∫ ∫ ∫ (4.177)

Solution of the time integral gives

( ) 2d

avd

P r L F Z dzπ ω−

= − + ∫ (4.178)

or, with the use of Equation (4.174)

101

( )[ ] 22 2 2 2 0 1 1 0( ) ( ) ( ) ( )

d

avd

P r C D A B J r Y r J r Y r Z dzπ ωξ ξ ξ ξ ξ−

= − − ∫ (4.179)

It could be mathematically proven that

[ ]0 1 1 0( ) ( ) ( ) ( ) 2r J r Y r J r Y rπ ξ ξ ξ ξ− = (4.180)

Hence, the average power flow for an outward propagating wave is given by

( ) 22 2 2 22

d

avd

P C D A B Z dzωξ−

= − ∫ (4.181)

4.2.1.1.3 Outward propagating wave through Bessel function decomposition

An different way to derive the average power flow of the outward circular-crested wave

is through decomposition of the Bessel function. Write the particle displacement (4.146)

as a function of sine functions, i.e.,

( )( , , ) , ( )u r z t f r t Z z= (4.182)

where

( ) ( ) ( ) 1, sin sin for i i ir r i r i i if r t F k r j t k r j t r r rξ π ω ξ π ω +⎡ ⎤= + − + + + ≤ <⎣ ⎦ (4.183)

where ir is the ith zero of ( )1J rξ , ir

F is a scale factor such as

( ) [ ]1max , max ( )cos( )f r t J r tξ ω⎡ ⎤ =⎣ ⎦ ,

1

iri i

kr r

π

+

=−

, and 1

1

1

if is even

2 if is odd

i

i ii

i i

i i

r ir r

jr r ir r

+

+

+

⎧−⎪ −⎪= ⎨ −⎪⎪ −⎩

(4.184)

102

Hence, ir

k determines the period of the sine function and j the phase. Figure 4.7 shows

the Bessel function approximation between the first two zeros.

Figure 4.7 Bessel function ( )1J rξ approximated with the sum of two sine functions (forward

and backward propagating waves)

Note that Equation (4.183) is composed of both backward and forward propagating

waves. Consider only the forward propagating wave, Equation (4.182) becomes

( ) 1( , , ) ( )sin for i ir r i i iu r z t F Z z k r j t r r rξ π ω += + − ≤ < (4.186)

The corresponding velocity and stress are

( ) 1( , , ) ( )cos for i ir r i i i

uv r z t F Z z k r j t r r rtθ ω ξ π ω +

∂= = − + − ≤ <

∂ (4.187)

( ) 1( , , ) ( )cos for i ir r r i i i

uT r z t k F Z z k r j t r r rrθ ξ ξ π ω +

∂= = + − ≤ <

∂ (4.188)

The expression of the average power flow is

2

2 1

2( ) ( , , ) ( , , )i

T d

av z xzT d

rP r v r z t T r z t dydtT

π

−

= −Τ − ∫ ∫ (4.189)

where 1 iT j π ω= and ( )2 2iT j π π ω= − T2 are

0 2 4

0.5

Eq Eq Eq Eq

103

In order to incorporate the term r in Equation (4.189), we rewrite it as

[ ]2

2 1

1( ) 2 ( , , ) ( , , )i

T d

av z xzT d

P r rv r z t T r z t dydtT

π−

= −Τ − ∫ ∫ (4.190)

The term in brackets can be expressed as

( )( , , )2 ( , , ) 2 2 , ( )ru r z trv r z t g r t Z zt t

π π π ξ ω∂ ∂ ⎡ ⎤= = ⎣ ⎦∂ ∂ (4.191)

where

( ) ( ) ( ) 1, sin sin for i i ir r i r i i ig r t G k r j t k r j t r r rξ π ω ξ π ω +⎡ ⎤= + − + + + ≤ <⎣ ⎦ (4.192)

and ir

G is a scale factor such as ( ) [ ]1max , max ( )cos( )g r t rJ r tξ ω⎡ ⎤ =⎣ ⎦ . The velocity can

be expressed as

( ) 12 ( , , ) 2 ( )cos for i iz r r i irv r z t G Z z k r j t r r rπ ω π ξ π ω += − + − ≤ < (4.193)


( ) ( )2

1

2 21

2 1

4( ) cos for

2i i i

dr r r

av i i id T

k F GP r Z z dy k r j t dt r r r

T Tπωξ ξ π ω

Τ

+−

⎛ ⎞= + − ≤ <⎜ ⎟

−⎝ ⎠∫ ∫ (4.194)

solve the time integral to obtain

21( ) ( ) 2 for

2 i i i

d

av r r r i id

P r Z z dy G k F r r rωξ π +−

⎛ ⎞= ≤ <⎜ ⎟

⎝ ⎠∫ (4.195)

Note that the terms ir

G and ir

F vary respectively as

( )1i

ir i iG rα= − and ( )1

i

i ir

i

Frα

= − (4.196)

104

Hence, the multiplication term i i ir r rG k F in Equation (4.195) can be written as

( ) ( )1 1i i i i i

i i ir r r r i r ik G F k r k

rαα α= − − = (4.197)

where for the first 4 is we have

0

1

2

3

0

1

2

3

1.067

0.989

0.995

0.997

r r

r r

r r

r r

k k

k k

k k

k k

α α

α α

α α

α α

∞

∞

∞

∞

∞

∞

∞

∞

=

=

=

=

(4.198)

The difference between the infinite value and the value at 3i = is less than 0.3%. We can

consider the product ir ik constα . Moreover, it is important to note that the average

power flow for the forward propagating wave is equal to a constant term multiplied by

2 ( )2

d

d

Z z dyωξ

−∫ (4.199)

as predicted in section 4.1.1.1.3.

If we had not considered term r in expression (4.190), the resulting average power flow

would be

( ) 2 2, , ( ) 22 i i

d

av r rd

P r z t Z z dy k Fωξ π−

⎛ ⎞= ⎜ ⎟

⎝ ⎠∫ (4.200)

Figure 4.8 shows the variation of the average power flow as function of radius per

wavenumber. It is to note that the average power flow derived in Equation (4.195) is

almost constant (dashed line). If we had not considered the term 2 rπ in Equation (4.190)

, the average power flow obtained would have had decreased with the radius, solid line

105

Equation (4.200). This is because, in this case, the derived average power flow is average

power flow per unit circumferential length (solid line) and circumferential length

increases linearly with increasing r. Hence, the power flow in (4.200) is inversely

proportional to r.

Figure 4.8 Average power flow as a function of rξ .

4.2.1.2 Power flow along ˆzn

Let consider the power flow through the upper surface with normal ˆzn due to propagating

shear horizontal waves, from Equation (4.14) we get

( )2

0 0

ˆr

out zP rd drπ

θ= − ⋅ ⋅∫ ∫ v T n (4.201)

Note that the normal ˆzn is defined as

{ }ˆ 0 0 1z =n (4.202)

Multiply the normal in the z direction by the velocity-stress product to get

( )

2

, ,

( )i i i

avr r rd

d

P r z t k F G

Z z dyπωξ−

=

∫

( ) 2

2

, ,

( )i i

avr rd

d

P r z t k F

Z z dyπωξ−

=

∫

rξ0 5 10 15 20 25 30 35 40 45 50

0.025

0.05

0.075

0.1

0.125

0.15

0.175

0.2

12 zπ

106

( ) ˆ z zv Tθ θ⋅ ⋅ =v T n (4.203)


2

0 0

r

out zP rv T d drπ

θ θ θ= − ∫ ∫ (4.204)

Since both the velocity and the stress do not dependent on θ, we can perform the

integration with respect to θ, i.e.,

0

2r

out zP rv T drθ θπ= − ∫ (4.205)

Equation (4.142) is the power flow in the z direction of shear horizontal waves. The

power flow given by Equation (4.205) depends on z because both ( , )zT r zθ and ( , )v r zθ

depend on z. To calculate power flow through the top and bottom surfaces, make z d= ±

in Equation (4.205). However, recall that we assumed stress free surfaces at z d= ± ,

hence

( , ) 0zT r dθ ± = (4.206)

Assumption (4.206) implies that Equation (4.205) become null at z d= ± . i.e. there is no

power flow in the z direction through the top and bottom surfaces.


In the case of shear horizontal waves, consider the equation of motion (3.54) and the

strain-velocity Equation (3.56), i.e.,

107

2

2

2

2

r

z

r zr

u uSr ruSz

T T uT Fr z r t

θ θθ

θθ

θ θ θθ θρ

∂⎧ = −⎪ ∂⎪∂⎪ =⎨ ∂⎪

⎪∂ ∂ ∂+ + = −⎪ ∂ ∂ ∂⎩

(4.207)


2

2

r

z

r zr

S v vt r rS vt z

T T vT Fr z r t

θ θ θ

θ θ

θ θ θθ θρ

∂ ∂⎧ = −⎪ ∂ ∂⎪∂ ∂⎪

=⎨∂ ∂⎪

⎪∂ ∂ ∂+ + = −⎪ ∂ ∂ ∂⎩

(4.208)

We multiply the first line by rT θ , the second by zTθ , and the third by vθ , i.e.,

2

2

rr r r

zz z

r zr

v v ST T Tr r tv ST Tz t

T T vv v v T v v Fr z r t

θ θ θθ θ θ

θ θθ θ

θ θ θθ θ θ θ θ θ θρ

∂ ∂⎧ − =⎪ ∂ ∂⎪∂ ∂⎪

=⎨∂ ∂⎪

⎪ ∂ ∂ ∂+ + = −⎪ ∂ ∂ ∂⎩

(4.209)

We sum the three equations

2

2

r zr r z r

r zr z

v v v T TT T T v v v Tr r z r z r

S S vT T v v Ft t t

θ θ θ θ θθ θ θ θ θ θ θ

θ θ θθ θ θ θ θρ

∂ ∂ ∂ ∂− + + + + =

∂ ∂ ∂ ∂∂ ∂ ∂

= + + −∂ ∂ ∂

(4.210)

We notice that the first four terms on the right-hand side can be grouped two by two.

Finally we obtain

( )1 2z r zr r z

v T S S vrv T T T v v Fr r z t t t

θ θ θ θ θθ θ θ θ θ θ θρ∂ ∂ ∂ ∂∂

+ = + + −∂ ∂ ∂ ∂ ∂

(4.211)

108

The acoustic Poynting vector (4.20) for circular crested SH waves can be written in the

form

( ) ( ) ˆˆr zrv T v Tθ θ θ θ⎡ ⎤= − ⋅ = − +⎣ ⎦P v T r θ (4.212)

4.2.2 Lamb waves

Consider Lamb waves propagating in a circular plate. The velocity vector is

{ }0r zv v=v (4.213)


0

0 00

rr rz

rz zz

T TT

T Tθθ

⎡ ⎤⎢ ⎥Τ = ⎢ ⎥⎢ ⎥⎣ ⎦

(4.214)

Derive the scalar product between velocity and stress matrix to get

{ }0r rr z rz r rz z zzv T v T v T v T⋅ = + +v T (4.215)

4.2.2.1 Power flow along ˆrn

Following the same procedure as for the shear horizontal waves, we derive the power

flow in the r direction. Multiply the normal in the r direction, Equation (4.139), by the

velocity-stress product (4.215) to get

( ) ˆ r r rr z rzv T v T⋅ ⋅ = +v T n (4.216)


( )2

0

d

out r rr z rzd

P r v T v T d dzπ

θ−

= − +∫ ∫ (4.217)

109



( )2d

out r rr z rzd

P r v T v T dzπ−

= − +∫ (4.218)

Equation (4.218) is the power flow of Lamb waves propagating in the r direction. The

term 2 rπ is the length/circumference of the circular wave front; hence

2 rout outP rPπ= − (4.219)

where routP is the unit power flow in the r direction.

The same considerations derived for SH waves hold for Lamb waves. However, due to

the complexity of the circular-crested Lamb wave equations, we will not derive the

average power flow.

4.2.2.2 Power flow along ˆzn

Following the same procedure as for the shear horizontal waves, we derive the power

flow in the z direction. Multiply the normal in the z direction, Equation (4.202), by the

velocity-stress product (4.215) to get

( ) ˆ z r rz z zzv T v T⋅ ⋅ = +v T n (4.220)


( )2

0 0

r

out r rz z zzP r v T v T d drπ

θ= − +∫ ∫ (4.221)

110



( )0

2r

out r rz z zzP r v T v T drπ= − +∫ (4.222)

Equation (4.222) is the power flow of Lamb waves propagating in the z direction. The

power flow given by Equation (4.222) is dependant on z because ( , )rzT r z , ( , )zzT r z ,

( , )rv r z and ( , )zv r z depend on z. To calculate power flow through the top and bottom

surfaces, make z d= ± in Equation (4.222). However, recall that we assumed stress free

surfaces at z d= ± , hence

( , ) ( , ) 0rz zzT r d T r d± = ± = (4.223)

Assumption (4.223) implies that Equation (4.222) become null at z d= ± . i.e. there is no

power flow in the z direction through the top and bottom surfaces.


In the case of Lamb waves let’s consider the equation of motion (3.54) and the strain-

velocity Equation (3.56), i.e.,

111

2

2

2

2

2

rrr

r

zzz

r zzr

rrrr rz rr

zr zz rz zz

uSr

uSruSzu uSz r

T TT T u Fr z r t

T T T u Fr z r t

θθ

θθ ρ

ρ

∂⎧ =⎪ ∂⎪⎪ =⎪⎪ ∂⎪ =⎪ ∂⎨ ∂ ∂⎪ = +

∂ ∂⎪⎪ −∂ ∂ ∂⎪ + + = −

∂ ∂ ∂⎪⎪∂ ∂ ∂⎪ + + = −

∂ ∂ ∂⎩

(4.224)


2

rr r

r

zz z

zr r z

rrrr rz rr

zr zz rz zz

S vt r

S vt r

S vt zS v vt z r

T TT T v Fr z r t

T T T v Fr z r t

θθ

θθ ρ

ρ

∂ ∂⎧ =⎪ ∂ ∂⎪∂⎪ =⎪ ∂

⎪∂ ∂⎪ =⎪ ∂ ∂⎨ ∂ ∂ ∂⎪ = +⎪ ∂ ∂ ∂⎪ −∂ ∂ ∂⎪ + + = −

∂ ∂ ∂⎪⎪∂ ∂ ∂

+ + = −⎪∂ ∂ ∂⎩

(4.225)

We multiply the first line by rrT , the second by Tθθ , the third by zzT , the fourth by rzT ,

the fifth by xv and the fifth by zv , i.e.,

112

2

r rrrr rr

r

z zzzz zz

r z zrrz rz

rrrr rz rr r r r r r

zr zz rz zz z z z z z

v ST Tr t

SvT Tr tv ST Tz tv v ST Tz r t

T TT T vv v v v v Fr z r t

T T T vv v v v v Fr z r t

θθθθ θθ

θθ ρ

ρ

∂ ∂⎧ =⎪ ∂ ∂⎪∂⎪ =⎪ ∂

⎪ ∂ ∂⎪ =⎪ ∂ ∂⎨ ∂ ∂ ∂⎛ ⎞⎪ + =⎜ ⎟⎪ ∂ ∂ ∂⎝ ⎠⎪ −∂ ∂ ∂⎪ + + = −⎪ ∂ ∂ ∂⎪ ∂ ∂ ∂⎪ + + = −

∂ ∂ ∂⎩

(4.226)

We sum the five equations in system (4.131) and rearrange the terms to get

( ) ( )

2

r rr z zr r rz z zz r rr z rz

rr zz zr r zrr zz rz r r r z z z

v T v T v T v T v T v Tr z r

SS S S v vT T T T v v F v v Ft t t t t t

θθθθ ρ ρ

⎧∂ + ∂ + ++ +⎪⎪ ∂ ∂⎨∂∂ ∂ ∂ ∂ ∂⎪= + + + + − + −⎪ ∂ ∂ ∂ ∂ ∂ ∂⎩

(4.227)

Note that

( )1 A ArAr r r r

∂ ∂= +

∂ ∂ (4.228)

hence we obtain

( ) ( )1

2

r rr z zr r rz z zz rrrr

zz zr r zzz rz r z r r z z

r v T v T v T v T SST Tr r z t t

S S v vT T v v v F v Ft t t t

θθθθ

ρ ρ

⎧ ∂ + ∂ + ∂∂+ = +⎪⎪ ∂ ∂ ∂ ∂⎨

∂ ∂ ∂ ∂⎪+ + + + − −⎪ ∂ ∂ ∂ ∂⎩

(4.229)

The acoustic Poynting vector (4.20) for circular crested Lamb waves can be written in the

form

( ) ( )ˆ ˆr rr z zr r rz z zzr v T v T v T v T⎡ ⎤= − ⋅ = − + + +⎣ ⎦P v T r z (4.230)

113

5 RECIPROCITY RELATION

Modal analysis is a powerful tool to study the wave fields excited by an external source

or the scattering from a defect. To develop this theory, the wave fields must be expressed

in terms of the superposition of wave guide modes; hence the wave fields are expressed

as a Fourier expansion of normal modes.

To proceed, two conditions must be verified; first, that the set of modal distribution

functions is complete, and second, that the wave guide modes are a set of orthogonal

functions. We can assume that the set of modal distribution functions is complete. To

prove this, we should show that arbitrary fields distributions can be expanded in this way,

we remand to Courant and Hilbert (1953) for a discussion on this subject.

To prove that the wave guides modes are a set of orthogonal functions, we will first

derive the real and complex reciprocity relations.

To explain what the term reciprocity means in acoustic field, consider a generic body

Ω , and two forces 1F and 2F applied at points 1P and 2P , respectively.

Figure 5.1 Reciprocity relation

F1 F2 u12

u21

Ω 1

2

114

Call 12u the displacement of point 1P due to force 2F , and 21u the displacement of

point 2P due to force 1F . In its most elementary form, the mechanics reciprocity principle

states that the work done at point 1P by force 1F upon the displacement induced by force

2F is the same as the work done at point 2P by force 2F upon the displacement induced

by force 1F , i.e.,

1 12 2 21⋅ = ⋅F u F u (5.1)

In acoustics, the acoustic reciprocity principle (http://www.ta.tudelft.nl

PrivatePages/C.P.A.Wapenaar/Reciprocity2/index2.htm) states that an acoustic response

remains the same when the source and receiver are interchanged. However, this is just a

special case of a more general reciprocity theorem, formulated in 1873 by Lord Rayleigh.

In its most general form, acoustic reciprocity principle establishes a relation between two

acoustic states that could occur in one and the same spatial domain (De Hoop, A. T.

1988). The sources, the medium parameters, and the wave fields may be different in each

of the states.

The real and complex reciprocity relations are used for different applications. The

real reciprocity relation can be applied to the case of lossy and lossless media with

symmetric constitutive matrices with the exception of lossless media with rotary activity,

while the complex reciprocity relation can be applied only to lossless media, hence it

requires real constitutive matrices. Moreover, the real reciprocity relation is mostly used

in scattering analysis while the complex reciprocity relation is more suitable for

waveguide and resonator mode analysis and for velocity and frequency perturbation

(Auld, 1990).

115

5.1 REAL RECIPROCITY RELATION

Consider two acoustic sources 1F and 2F which produce two acoustic fields 1 1,v T and

2 2,v T . Recall the equation of motion (2.1) and strain-displacement relation (2.10), and

write them for sources 1F and 2F , i.e.,

11 1

11s

t

t

ρ ∂⎧∇ ⋅ = −⎪⎪ ∂⎨∂⎪ = ∇⎪ ∂⎩

vT F

S v (acoustic field equations for source 1F ) (5.2)

22 2

22s

t

t

ρ ∂⎧∇ ⋅ = −⎪⎪ ∂⎨∂⎪ = ∇⎪ ∂⎩

vT F

S v (acoustic field equations for source 2F ) (5.3)

Following the formalism of Equation (4.4), multiply Equation (5.2) by the field 2 2,v T

and Equation (5.3) by the field 1 1,v T , i.e.,

12 1 1

112

: s

t

t

ρ ∂⎧⋅ ∇ ⋅ = −⎪⎪ ∂⎨∂⎪ = ∇⎪ ∂⎩

vv T F

S vT and

21 2 2

221

: s

t

t

ρ ∂⎧⋅ ∇ ⋅ = −⎪⎪ ∂⎨∂⎪ = ∇⎪ ∂⎩

vv T F

S vT (5.4)

In other words, we cross-multiply the equations due to one acoustic source by the field

produced by the other acoustic source. We obtain

( ) 1

2 1 2 2 1

12 2 1: : s

t

t

ρ ∂⎧ ⋅ ∇ ⋅ = ⋅ − ⋅⎪⎪ ∂⎨ ∂⎪ = ∇⎪ ∂⎩

vv T v v F

ST T v and

( ) 21 2 1 1 2

21 1 2: : s

t

t

ρ ∂⎧ ⋅ ∇ ⋅ = ⋅ − ⋅⎪⎪ ∂⎨ ∂⎪ = ∇⎪ ∂⎩

vv T v v F

ST T v (5.5)

Subtract the first system from the second to get

116

( ) ( ) 1 2

2 1 1 2 2 1 2 1 1 2

1 22 1 2 1 1 2: : : :s s

t t

t t

ρ ρ

δδ

∂ ∂⎧ ⋅ ∇ ⋅ − ⋅ ∇ ⋅ = ⋅ − ⋅ − ⋅ + ⋅⎪⎪ ∂ ∂⎨ ∂⎪ − = ∇ − ∇⎪ ∂⎩

v vv T v T v v v F v F

S ST T T v T v (5.6)

Add the two lines in Equation (5.6); upon rearrangement, obtain

( ) ( )2 1 1 2 1 2 2 1

1 2 1 22 1 2 1 2 1 1 2

: :

: :

s s

t t t tρ ρ

⋅ ∇ ⋅ + ∇ − ⋅ ∇ ⋅ − ∇

∂ ∂ ∂ ∂= ⋅ − ⋅ − + − ⋅ + ⋅

∂ ∂ ∂ ∂

v T T v v T T vv v S Sv v T T v F v F

(5.7)

Recall the distributive property (4.6) and apply it to the mixed products, i.e.,

( ) ( )( ) ( )

2 1 2 1 1 2

1 2 1 2 2 1

:

:s

s

∇ ⋅ ⋅ = ⋅ ∇ ⋅ + ∇

∇ ⋅ ⋅ = ⋅ ∇ ⋅ + ∇

v T v T T v

v T v T T v (5.8)

Apply rules (5.8) to Equation (5.7) and get

( ) 1 2 2 12 1 1 2 2 1 1 2 2 1 1 2: :

t t t tρ ρ∂ ∂ ∂ ∂

∇ ⋅ ⋅ − ⋅ = ⋅ − ⋅ + − − ⋅ + ⋅∂ ∂ ∂ ∂v v S Sv T v T v v T T v F v F (5.9)

Assume that the sources are time-harmonic, i.e., ( , , ) i tx y z e ω=F F , the resulting field is

also harmonic, i.e.,

( , , ) i tx y z e ω=v v , ( , , ) i tx y z e ω=T T (5.10)

For time-harmonic functions of the form ( , , ) i tf f x y z e ω= , the differentiation with

respect to t is obtained by multiplication by iω , i.e.,

( , , ) ( , , ) ( , , )i t i t i tf f x y z e f x y z e i f x y z e i ft t t

ω ω ωω ω∂ ∂ ∂= = = =

∂ ∂ ∂ (5.11)

Using Equation (5.11) into Equation (5.9) yields

( ) ( )2 1 1 2 2 1 1 2iωρ∇ ⋅ − ⋅ = ⋅ − ⋅v T v T v v v v ( )1 2 2 1 2 1 1 2: :iω+ − − ⋅ + ⋅T S T S v F v F (5.12)

117

i.e.,

( ) ( )2 1 1 2 1 2 2 1 2 1 1 2: :iω∇ ⋅ − ⋅ = − − ⋅ + ⋅v T v T T S T S v F v F (5.13)

Recall Equation (2.15), i.e., :=S s T , and write

1 2 1 2

2 1 2 1

: : :: : :

==

T S T s TT S T s T

(5.14)

We assume that the compliance matrix s is symmetric, we have

1 2 2 1: : : :=T s T T s T (5.15)


1 2 2 1: :=T S T S (5.16)

Equation (5.16) indicates that the iω term in Equation (5.13) cancels out; Equation (5.13)

becomes

( )2 1 1 2 1 2 2 1∇ ⋅ − ⋅ = ⋅ − ⋅v T v T v F v F (5.17)

where in cylindrical coordinates

( )1 r zrA Ar r z

∂ ∂∇ = +

∂ ∂A (5.18)

Equation (5.17) is the real reciprocity relation.

5.2 COMPLEX RECIPROCITY RELATION

To obtain the reciprocity relation, we follow the same procedure as in Section 5.1, but do

the cross-multiplication of the two fields using complex conjugates (see Appendix A).

118

We take the first field to be due to source 1F and the second field to be due to source 2F% ,

which is the conjugate of 2F , i.e.,

11 1

11s

t

t

ρ ∂⎧∇ ⋅ = −⎪⎪ ∂⎨ ∂⎪∇ =⎪ ∂⎩

vT F

Sv (acoustic field equations for source 1F ) (5.19)

22 2

22s

t

t

ρ ∂⎧∇ ⋅ = −⎪ ∂⎪⎨

∂⎪∇ =⎪ ∂⎩

vT F

Sv

%% %

%%

(acoustic field equations for source 2F% ) (5.20)

Following the formalism of Equation (4.4), multiply Equation (5.19) by the field 2 2,v T%%

and Equation (5.20) by the field 1 1,v T , i.e.,

12 1 1

112

: s

t

t

ρ ∂⎧⋅ ∇ ⋅ = −⎪⎪ ∂⎨ ∂⎪∇ =⎪ ∂⎩

vv T F

SvT

%

% and

21 2 2

221

: s

t

t

ρ ∂⎧⋅ ∇ ⋅ = −⎪ ∂⎪⎨

∂⎪∇ =⎪ ∂⎩

vv T F

SvT

%% %

%%

(5.21)

In other words, we cross-multiply the equations due to one acoustic source by the field

produced by the other acoustic source. We obtain:

( ) 1

2 1 2 2 1

12 1 2: :s

t

t

ρ ∂⎧ ⋅ ∇ ⋅ = ⋅ − ⋅⎪⎪ ∂⎨ ∂⎪ ∇ =⎪ ∂⎩

vv T v v F

ST v T

% % %

% % (5.22)

( ) 2

1 2 1 1 2

21 2 1: :s

t

t

ρ ∂⎧ ⋅ ∇ ⋅ = ⋅ − ⋅⎪ ∂⎪⎨

∂⎪ ∇ =⎪ ∂⎩

vv T v v F

ST v T

%% %

%%

(5.23)

119

Add Equations (5.22) and (5.23) to get

( ) ( ) 1 2

2 1 1 2 2 1 2 1 1 2

1 22 1 1 2 2 1: : : :s s

t t

t t

ρ ρ∂ ∂⎧ ⋅ ∇ ⋅ + ⋅ ∇ ⋅ = ⋅ + ⋅ − ⋅ − ⋅⎪ ∂ ∂⎪⎨

∂ ∂⎪ ∇ + ∇ = +⎪ ∂ ∂⎩

v vv T v T v v v F v F

S ST v T v T T

%% %% % %

%% %%

(5.24)

Add the two lines in Equation (5.24); upon rearrangement, obtain

( ) ( )2 1 1 2 1 2 2 1

1 2 1 22 1 2 1 2 1 1 2

: :

: :

s s

t t t tρ ρ

⋅ ∇ ⋅ + ∇ + ⋅ ∇ ⋅ + ∇

∂ ∂ ∂ ∂= ⋅ + ⋅ + + − ⋅ − ⋅

∂ ∂ ∂ ∂

v T T v v T T v

v v S Sv v T T v F v F

% %% %

%% % %% % (5.25)

Recall the distributive property (4.6) and apply it to the mixed products, i.e.,

( ) ( )( ) ( )

2 1 2 1 1 2

1 2 1 2 2 1

:

:s

s

∇ ⋅ ⋅ = ⋅ ∇ ⋅ + ∇

∇ ⋅ ⋅ = ⋅ ∇ ⋅ + ∇

v T v T T v

v T v T T v

% % %

% % % (5.26)

Apply rules (5.26) to Equation (5.25) and get

( ) ( ) 1 2 2 12 1 1 2 2 1 1 2 2 1 1 2: :

t t t tρ ρ∂ ∂ ∂ ∂

∇ ⋅ ⋅ + ∇ ⋅ ⋅ = ⋅ + ⋅ + + − ⋅ − ⋅∂ ∂ ∂ ∂v v S Sv T v T v v T T v F v F

%%% % %% % (5.27)

Recall Equation (2.15), i.e., :=S s T , and write

1 2 1 2

2 1 2 1

: : :

: : :

=

=

T S T s T

T S T s T

% %

% % (5.28)

Apply differentiation with respect to t as appropriate to the terms in Equation (5.28)

2 2

1 1

1 12 2

: : :

: : :

t t

t t

∂ ∂=

∂ ∂∂ ∂

=∂ ∂

S TT T s

S TT T s

% %

% % (5.29)

Upon addition, Equation (5.29) yields

120

2 1 2 11 2 1 2: : : : : :

t t t t∂ ∂ ∂ ∂

+ = +∂ ∂ ∂ ∂S S T TT T T s T s% %

% % (5.30)

In this case, we assume that the compliance matrix s is not only symmetric as in the real

reciprocity relation, but also real. We have

2 21 1: : : :

t t∂ ∂

=∂ ∂T TT s s T% %

(5.31)

Using Equation (5.31) into Equation (5.30) yields

( )2 1 2 1 2 11 2 1 2 1 2 2 1: : : : : : : : :

t t t t t t t∂ ∂ ∂ ∂ ∂ ∂ ∂

+ = + = + =∂ ∂ ∂ ∂ ∂ ∂ ∂S S T T T ST T s T T s S T T S% % %

% % % % (5.32)

On the other hand, we notice that

( ) ( ) ( )2 1 1 2 2 1 1 2∇ ⋅ ⋅ + ∇ ⋅ ⋅ = ∇ ⋅ ⋅ + ⋅v T v T v T v T% %% % (5.33)

Substitution of Equations (5.32) and (5.33) into Equation (5.27) yields the complex

reciprocity relation in the form

( ) ( )2 1 1 2 2 1 2 1 2 1 1 2:t

ρ∂∇ ⋅ + ⋅ = ⋅ + − ⋅ − ⋅

∂v T v T v v T S v F v F% % %% % % (5.34)

Equivalent formulations of Equation (5.34) exist. One equivalent formulation would be

obtained by having the subscripts 1 and 2 interchanged, i.e.,

( ) ( )1 2 2 1 1 2 1 2 1 2 2 1:t

ρ∂∇ ⋅ + ⋅ = ⋅ + − ⋅ − ⋅

∂v T v T v v T S v F v F% % %% % % (5.35)

Another equivalent formulation of Equation (5.34) would be obtained by interchanging

the conjugate operation, i.e.,

( ) ( )2 1 1 2 2 1 2 1 2 1 1 2:t

ρ∂∇ ⋅ + ⋅ = ⋅ + − ⋅ − ⋅

∂v T v T v v T S v F v F%% %% % % (5.36)

121

A third option would be to interchange both the subscripts and the conjugate operation,

i.e.,

( ) ( )1 2 2 1 1 2 1 2 1 2 2 1:t

ρ∂∇ ⋅ + ⋅ = ⋅ + − ⋅ − ⋅

∂v T v T v v T S v F v F%% %% % % (5.37)

Adding Equations (5.34) and (5.37) we get the symmetric form

( ) ( )2 1 1 2 2 1 1 2 2 1 1 2 2 1 1 21 : :2 t

ρ ρ∂∇ ⋅ + ⋅ = ⋅ + ⋅ + + − ⋅ − ⋅

∂v T v T v v v v T S T S v F v F%% % %% % % % (5.38)

Other symmetric forms are also possible, e.g., by adding Equations (5.34) with (5.35) and

using the complex-numbers property given by Appendix B.3, i.e., a b a b⋅ = ⋅ %% ; in this

case, we get

( ) ( )2 1 1 2 2 1 1 2 2 1 1 2 2 1 1 21 : :2 t

ρ ρ∂∇ ⋅ + ⋅ = ⋅ + ⋅ + + − ⋅ − ⋅

∂v T v T v v v v T S T S v F v F% % % %% % % % (5.39)

For time harmonic fields, Equation (5.37) simplifies as follows. Recall Equation (5.10),

i.e.,

1 1( , , ) i tx y z e ω=F F , 1 1( , , ) i tx y z e ω=v v , 1 1( , , ) i tx y z e ω=T T , 1 1( , , ) i tx y z e ω=S S (5.40)

2 2 ( , , ) i tx y z e ω=F F , 2 2 ( , , ) i tx y z e ω=v v , 2 2 ( , , ) i tx y z e ω=T T , 2 2 ( , , ) i tx y z e ω=S S (5.41)

Take the conjugate of Equation (5.41) to get

2 2 ( , , ) i tx y z e ω−=F F% % , 2 2 ( , , ) i tx y z e ω−=v v% % , 2 2 ( , , ) i tx y z e ω−=T T% % , 2 2 ( , , ) i tx y z e ω−=S S% % (5.42)

Substitute Equations (5.40), (5.42) into Equation (5.34) to get

( )

( )2 1 1 2

2 1 2 1 2 1 1 2:

i t i t i t i t

i t i t i t i t i t i t i t i t

e e e e

e e e e e e e et

ω ω ω ω

ω ω ω ω ω ω ω ωρ

− −

− − − −

∇ ⋅ + ⋅

∂= ⋅ + − ⋅ − ⋅

∂

v T v T

v v T S v F v F

%%

% %% % (5.43)

122

It is apparent that the time dependent terms i te ω and i te ω− in Equation (5.43) cancel out,

and Equation (5.43) becomes

( ) ( ) ( )2 1 1 2 2 1 2 1 2 1 1 2:t

ρ∂∇ ⋅ + ⋅ = ⋅ + − ⋅ + ⋅

∂v T v T v v T S v F v F% % %% % % (5.44)

Since the terms in the first parentheses on the right-hand side do not depend on t , then

their derivates with respect to t vanish. Hence, the complex reciprocity relation for time

varying harmonic functions takes the simpler form

( ) ( )2 1 1 2 2 1 1 2∇ ⋅ + ⋅ = − ⋅ + ⋅v T v T v F v F% %% % (5.45)

The complex reciprocity relation of Equation (5.45) resembles the real reciprocity

relation of Equation (5.17), only that some signs are different, as appropriate for using

complex conjugate expressions.

5.3 REAL RECIPROCITY RELATION IN RECTANGULAR COORDINATES


Recall Equation (4.90), i.e.,

xz z

yz z

yzxz zz

S vt x

S vt y

TT v Fx y t

ρ

⎧∂ ∂=⎪

∂ ∂⎪⎪∂ ∂⎪ =⎨ ∂ ∂⎪⎪ ∂∂ ∂

+ = −⎪∂ ∂ ∂⎪⎩

(5.46)

Cross-multiply the two fields as in Equation (5.4) to obtain

123

1 12 2

1 12 2

11 12 2 2 2 1

xz zxz xz

yz zyz yz

yzxz zz z z z z

S vT Tt x

S vT Tt y

TT vv v v v Fx y t

ρ

⎧ ∂ ∂=⎪

∂ ∂⎪⎪ ∂ ∂⎪ =⎨ ∂ ∂⎪⎪ ∂∂ ∂⎪ + = −

∂ ∂ ∂⎪⎩

and

2 21 1

2 21 1

22 21 1 1 1 2

xz zxz xz

yz zyz yz

yzxz zz z z z z

S vT Tt x

S vT Tt y

TT vv v v v Fx y t

ρ

⎧ ∂ ∂=⎪

∂ ∂⎪⎪ ∂ ∂⎪ =⎨ ∂ ∂⎪⎪ ∂∂ ∂⎪ + = −

∂ ∂ ∂⎪⎩

(5.47)

Subtract the two system of Equation (5.47) to get

1 2 1 22 1 2 15 5

1 2 1 22 1 2 14 4

1 21 2 1 22 2 1 1 2 2 1 1 1 2

z zxz xz xz xz

z zyz yz yz yz

yz yzxz xz z zz z z z z z z z z z

S S v vT T T Tt t x x

S S v vT T T Tt t y y

T TT T v vv v v v v v F v v Fx y x y t t

ρ ρ

⎧ ∂ ∂ ∂ ∂⎪ − = −∂ ∂ ∂ ∂⎪

⎪ ∂ ∂ ∂ ∂⎪ − = −⎨ ∂ ∂ ∂ ∂⎪⎪ ∂ ∂∂ ∂ ∂ ∂⎪ + − − = − − +

∂ ∂ ∂ ∂ ∂ ∂⎪⎩

(5.48)

Add up the equations in the system (5.48) and rearrange the terms to obtain

( ) ( )2 1 1 22 1 1 2

1 2 1 21 22 1 2 1 2 1 1 2 2 1

z yz z yzz xz z xz

yz yz xz xzz zz z yz yz xz xz z z z z

v T v Tv T v Tx y

S S S Sv vv v T T T T v F v Ft t t t t t

ρ ρ

∂ −∂ −+ =

∂ ∂

∂ ∂ ∂ ∂∂ ∂− − + − + + −

∂ ∂ ∂ ∂ ∂ ∂

(5.49)

Assume time-harmonic fields, hence the derivatives with respect to t become

( ) ( )2 1 1 22 1 1 2

2 1 1 2

z yz z yzz xz z xz

z z z z

v T v Tv T v Tx y

i v v v vωρ

∂ −∂ −+ =

∂ ∂

− 1 2 2 1yz yz yz yzT S T S+ − 1 2 2 1

xz xz xz xzT S T S+ −( ) 1 2 2 1z z z zv F v F+ −

(5.50)

Equation (5.50) yields the real reciprocity relation for shear waves, i.e.,

( ) ( )2 1 1 22 1 1 2

1 2 2 1z yz z yzz xz z xzz z z z

v T v Tv T v Tv F v F

x y∂ −∂ −

+ = −∂ ∂

(5.51)

124

5.3.2 Lamb waves

Recall the system of Equations (4.131), i.e.,

x xxxx xx

y yyyy yy

y xyxxy xy

xyxx xx x x x x

xy yy yy y y y y

v ST Tx tv S

T Ty t

v SvT Ty x t

TT vv v v v Fx y t

T T vv v v v F

x y t

ρ

ρ

⎧ ∂ ∂=⎪

∂ ∂⎪⎪ ∂ ∂

=⎪∂ ∂⎪

⎪ ∂ ∂⎛ ⎞∂⎪ + =⎨ ⎜ ⎟∂ ∂ ∂⎝ ⎠⎪⎪ ∂∂ ∂⎪ + = −

∂ ∂ ∂⎪⎪ ∂ ∂ ∂⎪ + = −⎪ ∂ ∂ ∂⎩

(5.52)

Using Equation (5.52), assign superscripts 1 and 2 to achieve the cross-multiplication of

the two fields as in Equation (5.4), i.e.,

1 12 2

1 12 2

1 112 2

11 12 2 2 2 1

1 1 12 2 2 2 1

xx xxx xx

yy yyy yy

xy yxxy xy

xyxx xx x x x x

xy yy yy y y y y

S vT Tt x

S vT T

t y

S vvT Tt y x

TT vv v v v Fx y t

T T vv v v v F

x y t

ρ

ρ

⎧ ∂ ∂⎪ =∂ ∂⎪

⎪ ∂ ∂⎪ =

∂ ∂⎪⎪

⎛ ⎞∂ ∂∂⎪ = +⎜ ⎟⎨ ⎜ ⎟∂ ∂ ∂⎪ ⎝ ⎠⎪ ∂∂ ∂⎪ + = −⎪ ∂ ∂ ∂⎪

∂ ∂ ∂⎪ + = −⎪ ∂ ∂ ∂⎩

and

2 21 1

2 21 1

2 221 1

22 21 1 1 1 2

2 2 21 1 1 1 2

xx xxx xx

yy yyy yy

xy yxxy xy

xyxx xx x x x x

xy yy yy y y y y

S vT Tt x

S vT T

t y

S vvT Tt y x

TT vv v v v Fx y t

T T vv v v v F

x y t

ρ

ρ

⎧ ∂ ∂⎪ =∂ ∂⎪

⎪ ∂ ∂⎪ =

∂ ∂⎪⎪

⎛ ⎞∂ ∂∂⎪ = +⎜ ⎟⎨ ⎜ ⎟∂ ∂ ∂⎪ ⎝ ⎠⎪ ∂∂ ∂⎪ + = −⎪ ∂ ∂ ∂⎪

∂ ∂ ∂⎪ + = −⎪ ∂ ∂ ∂⎩

(5.53)

Subtract the two systems in Equation (5.53) to get

125

1 2 1 22 1 2 1

1 2 1 22 1 2 1

1 2 1 21 22 1 2 1

11 22 2 1

xx xx x xxx xx xx xx

yy yy y yyy yy yy yy

xy xy y yx xxy xy xy xy

xyxx xxx x x x

S S v vT T T Tt t x x

S S v vT T T T

t t y y

S S v vv vT T T Tt t y x y x

TT Tv v v vx y x

∂ ∂ ∂ ∂− = −

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

− = −∂ ∂ ∂ ∂

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂∂ ∂− = + − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

∂∂ ∂+ − −

∂ ∂ ∂

2 1 21 2 1 2 1 1 2

1 1 2 2 1 22 2 1 1 2 1 2 1 1 2

xy x xx x x x x x

xy yy xy yy y yy y y y y y y y y y

T v vv v v F v Fy t t

T T T T v vv v v v v v v F v F

x y x y t t

ρ ρ

ρ ρ

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪ ∂ ∂ ∂⎪ = − − +⎪ ∂ ∂ ∂⎪

∂ ∂ ∂ ∂ ∂ ∂⎪ + − − = − − +⎪ ∂ ∂ ∂ ∂ ∂ ∂⎩

(5.54)

Add up the equations in the system (5.54) and rearrange the terms to get

( ) ( ) ( ) ( )2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2

2 1 1 2 2 1 1 2 1 2 1 2

x xx y xy x xx y xy x xy y yy x xy y yy

x x x x y y y y x x x x

v T v T v T v T v T v T v T v Tx x y y

v F v F v F v F i v v v vωρ

∂ + ∂ + ∂ + ∂ +− + −

∂ ∂ ∂ ∂

= − + − + + − 1 2 1 2y y y yv v v v+ −( )

2 1 1 2xx xx xx xxi T S T Sω− − 2 1 1 2

yy yy yy yyT S T S+ − 2 1 1 2xy xy xy xyT S T S+ −( )

(5.55)


( ) ( ) ( ) ( )2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2

2 1 1 2 2 1 1 2 1 2 1 2


x x x x y y y y x x x x

v T v T v T v T v T v T v T v Tx x y y

v F v F v F v F i v v v vωρ

∂ + ∂ + ∂ + ∂ +− + −

∂ ∂ ∂ ∂

= − + − + + − 1 2 1 2y y y yv v v v+ −( )

2 1 1 2xx xx xx xxi T S T Sω− − 2 1 1 2

yy yy yy yyT S T S+ − 2 1 1 2xy xy xy xyT S T S+ −( )

(5.56)

Equation (5.56) yields the real reciprocity relation for Lamb waves, i.e.,

( ) ( ) ( ) ( )

( ) ( )

2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2

2 1 2 1 1 2 1 2


x x y y x x y y

v T v T v T v T v T v T v T v Tx y

v F v F v F v F

∂ ∂⎡ ⎤ ⎡ ⎤+ − + + + − +⎣ ⎦ ⎣ ⎦∂ ∂

= − + + + (5.57)

126

5.4 COMPLEX RECIPROCITY RELATION IN RECTANGULAR COORDINATES


Consider system (4.90). Assign superscripts 1 and 2 to achieve the cross-multiplication of


112 2

112 2

11 12 2 2 2 1

xzzxz xz

yzzyz yz

yzxz zz z z z z

SvT Tx t

SvT Ty t

TT vv v v v Fx y t

ρ

⎧ ∂∂=⎪

∂ ∂⎪⎪ ∂∂⎪ =⎨ ∂ ∂⎪⎪ ∂∂ ∂⎪ + = −

∂ ∂ ∂⎪⎩

% %

% %

% % % %

and

221 1

221 1

22 21 1 1 1 2

xzzxz xz

yzzyz yz

yzxz zz z z z z

SvT Tx t

SvT Ty t

TT vv v v v Fx y t

ρ

⎧ ∂∂=⎪

∂ ∂⎪⎪ ∂∂⎪ =⎨ ∂ ∂⎪⎪ ∂∂ ∂⎪ + = −

∂ ∂ ∂⎪⎩

%%

%%

%% % %

(5.58)

Add up the two systems in Equation (5.58)

1 21 22 1 2 15 5

1 21 22 1 2 1

1 21 2 1 22 2 1 1 2 2 1 1 1 2

z zxz xz xz xz

yz yzz zyz yz yz yz

yz yzxz xz z zz z z z z z z z z z

S Sv vT T T Tx x t t

S Sv vT T T Ty y t t

T TT T v vv v v v v v F v v Fx y x y t t

ρ ρ

⎧ ∂ ∂∂ ∂+ = +⎪

∂ ∂ ∂ ∂⎪⎪ ∂ ∂∂ ∂⎪ + = +⎨ ∂ ∂ ∂ ∂⎪⎪ ∂ ∂∂ ∂ ∂ ∂⎪ + + + = − + −

∂ ∂ ∂ ∂ ∂ ∂⎪⎩

%%% %

%%% %

%% % %% % % %

(5.59)

Sum up the equation in system above and rearrange the terms

1 21 2 1 2 1 22 2 1 1 2 1 2 1

1 21 21 22 2 1 1 1 2 2 1 2 15 5

yz yzxz xz z z z zz z z z xz xz yz yz

yz yzz zz z z z z z xz xz yz yz

T TT T v v v vv v v v T T T Tx y x y x x y y

S SS Sv vv v F v v F T T T Tt t t t t t

ρ ρ

∂ ∂∂ ∂ ∂ ∂ ∂ ∂+ + + + + + + =

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂∂ ∂∂ ∂− + − + + + +

∂ ∂ ∂ ∂ ∂ ∂

%% % %% %% %

%%% % % %% %

(5.60)

Upon rearrangement, Equation (5.60) becomes the complex reciprocity relation for shear

waves, i.e.,

( ) ( ) ( )2 1 1 2 2 1 1 2 2 1 2 1 2 1 2 1 1 2z xz z xz z yz z yz z z yz yz xz xz z z z zv T v T v T v T v v T S T S v F v F

x y tρ∂ ∂ ∂

+ + + = + + − −∂ ∂ ∂

% % % % %% % % % (5.61)

127

An alternative symmetric formulation of Equation (5.61) is

( ) ( )

( )

2 1 1 2 2 1 1 2

2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 212

z xz z xz z yz z yz

z z z z yz yz xz xz yz yz xz xz z z z z

v T v T v T v Tx y

v v v v T S T S S T S T v F v Ft

ρ ρ

∂ ∂+ + +

∂ ∂∂ ⎡ ⎤= + + + + + − −⎣ ⎦∂

% %% %

% %% % %% % %

(5.62)

For time harmonic fields, Equation (5.62) simplifies to the form

( ) ( )2 1 1 2 2 1 1 2 2 1 1 2z xz z xz z yz z yz z z z zv T v T v T v T v F v F

x y∂ ∂

+ + + = − −∂ ∂

% % %% % % (5.63)

5.4.2 Lamb waves

Recall Equation (4.131) and follow the general procedure described at the beginning of

Section 5.2.

1 12 2

1 12 2

1 112 2

11 12 2 2 2 1

1 1 12 2 2 2 1

x xxxx xx

y yyyy yy

y xyxxy xy

xyxx xx x x x x

xy yy yy y y y y

v ST Tx tv S

T Ty t

v SvT Ty x t

TT vv v v v Fx y t

T T vv v v v F

x y t

ρ

ρ

⎧ ∂ ∂⎪ =∂ ∂⎪

⎪ ∂ ∂⎪ =

∂ ∂⎪⎪

⎛ ⎞∂ ∂∂⎪ + =⎜ ⎟⎨ ⎜ ⎟∂ ∂ ∂⎪ ⎝ ⎠⎪ ∂∂ ∂⎪ + = −⎪ ∂ ∂ ∂⎪

∂ ∂ ∂⎪ + = −⎪ ∂ ∂ ∂⎩

% %

% %

% %

% % % %

% % % %

and

2 21 1

2 21 1

2 221 1

22 21 1 1 1 2

2 2 21 1 1 1 2

x xxxx xx

y yyyy yy

y xyxxy xy

xyxx xx x x x x

xy yy yy y y y y

v ST Tx tv S

T Ty t

v SvT Ty x t

TT vv v v v Fx y t

T T vv v v v F

x y t

ρ

ρ

⎧ ∂ ∂⎪ =∂ ∂⎪

⎪ ∂ ∂⎪ =

∂ ∂⎪⎪

⎛ ⎞∂ ∂∂⎪ + =⎜ ⎟⎨ ⎜ ⎟∂ ∂ ∂⎪ ⎝ ⎠⎪ ∂∂ ∂⎪ + = −⎪ ∂ ∂ ∂⎪

∂ ∂ ∂⎪ + = −⎪ ∂ ∂ ∂⎩

% %

% %

% %

% % % %

% % % %

(5.64)

Add the two systems in Equation (5.64), i.e.,

128

2 1 2 11 2 1 2

2 1 2 11 2 1 2

2 1 2 12 11 2 1 2

221 1

x x xx xxxx xx xx xx

y y yy yyyy yy yy yy

y y xy xyx xxy xy xy xy

xyxxx x

v v S ST T T Tx x t tv v S S

T T T Ty y t t

v v S Sv vT T T Ty x y x t t

TTv vx

∂ ∂ ∂ ∂+ = +

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

+ = +∂ ∂ ∂ ∂

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂∂ ∂+ + + = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

∂∂+

∂ ∂

%% % %

%%% %

%%% % %

%% 11 2 12 2 1 2 1 2 1 2

2 2 1 1 2 11 1 2 2 1 2 1 2 1 2

xyxx x xx x x x x x x x

xy yy xy yy y yy y y y y y y y y y

TT v vv v v v F v v Fy x y t t

T T T T v vv v v v v v F v v F

x y x y t t

ρ ρ

ρ ρ

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪ ∂∂ ∂ ∂⎪ + + = − + −⎪ ∂ ∂ ∂ ∂⎪

∂ ∂ ∂ ∂ ∂ ∂⎪+ + + = − + −⎪ ∂ ∂ ∂ ∂ ∂ ∂⎩

% %% % % %

% % %%% % % %

(5.65)

Sum up the equations in system above and rearrange the terms

( ) ( )

1 2 2 1 2 1 2 1 1 2 1 22 1 1 2

2 1 1 2 2 1 1 2 1 2 1 2 1 2 1 2 1 2

y xy y xy x xy y yy x xy y yyx xx x xx

x x x x y y y y x x y y xx xx yy yy xy xy

v T v T v T v T v T v Tv T v Tx x x x y y y y

v F v F v F v F v v v v T S T S T St t

ρ

∂ ∂ ∂ ∂ ∂ ∂∂ ∂+ + + + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂

= − − − − + + + + +∂ ∂

% % %% % % %%

% % %% %% % % %

(5.66)

Upon rearrangement, Equation (5.66) yields the complex reciprocity relation for Lamb

waves, i.e.,

( ) ( )

( ) ( )

2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2

1 2 1 2 1 2 1 2 1 2 2 1 1 2 2 1 1 2


x x y y xx xx yy yy xy xy x x x x y y y y


v v v v T S T S T S v F v F v F v Ft t

ρ

∂ ∂+ + + + + + +

∂ ∂∂ ∂

= + + + + − − − −∂ ∂

% % % %% % % %

% % % % %% % % %

(5.67)

For time harmonic fields, Equation (5.67) simplifies to the form

( ) ( )2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2

2 1 1 2 2 1 1 2


x x x x y y y y


v F v F v F v F

∂ ∂+ + + + + + +

∂ ∂

= − − − −

% % % %% % % %

% %% %

(5.68)

The proof of this simplification is similar to that used to arrive at the simplified general

expression (5.45) of the complex reciprocity relation (5.34).

129

5.5 REAL RECIPROCITY RELATION IN CYLINDRICAL COORDINATES


Recall the system of Equations (4.208), i.e.,

2

2

r

z

r zr

S v vt r rS vt z

T T vT Fr z r t

θ θ θ

θ θ

θ θ θθ θρ

∂ ∂⎧ = −⎪ ∂ ∂⎪∂ ∂⎪

=⎨∂ ∂⎪

⎪∂ ∂ ∂+ + = −⎪ ∂ ∂ ∂⎩

(5.69)



1 1 12 2 2

1 12 2

1 12 2 2 1

12 2 1

2

2

rr r r

zz z

r zr

S v vT T Tt r rS vT Tt z

T Tv v v Tr z r

vv v Ft

θ θ θθ θ θ

θ θθ θ

θ θθ θ θ θ

θθ θ θρ

⎧ ∂ ∂= −⎪ ∂ ∂⎪

⎪ ∂ ∂=⎪⎪ ∂ ∂

⎨∂ ∂⎪ + + =⎪ ∂ ∂

⎪∂⎪= −⎪ ∂⎩

and

2 2 21 1 1

2 21 1

2 21 1 1 2

21 1 2

2

2

rr r r

zz z

r zr


T Tv v v Tr z r

vv v Ft

θ θ θθ θ θ

θ θθ θ

θ θθ θ θ θ

θθ θ θρ

⎧ ∂ ∂= −⎪ ∂ ∂⎪

⎪ ∂ ∂=⎪⎪ ∂ ∂

⎨∂ ∂⎪ + + =⎪ ∂ ∂

⎪∂⎪= −⎪ ∂⎩

(5.70)

Subtract the two system of Equation (5.70) to get

1 2 1 1 2 22 1 2 2 1 1

1 2 1 22 1 2 1

1 1 2 22 2 2 1 1 1 1 2

12 2 1 1

2 2

2 2

r rr r r r r r

z zz z z z

r z r zr r

S S v v v vT T T T T Tt t r r r rS S v vT T T Tt t z z

T T T Tv v v T v v v Tr z r r z r

v vv v F vt

θ θ θ θ θ θθ θ θ θ θ θ

θ θ θ θθ θ θ θ

θ θ θ θθ θ θ θ θ θ θ θ

θ θθ θ θ θρ ρ

∂ ∂ ∂ ∂− = − − +

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

− = −∂ ∂ ∂ ∂

∂ ∂ ∂ ∂+ + − − − =

∂ ∂ ∂ ∂∂ ∂

= − −∂

21 2v F

t θ θ

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪ +⎪ ∂⎩

(5.71)

Add up the equations in the system (5.71) and rearrange the terms to obtain

130

2 1 1 2 2 1 1 2

1 2 1 2 1 22 1 2 1 2 1 2 1 1 2

1 1

2 2

r r z z

r r z zr r z z

rv T rv T v T v Tr r r r z z

S S S S v vT T T T v v v F v Ft t t t t t

θ θ θ θ θ θ θ θ

θ θ θ θ θ θθ θ θ θ θ θ θ θ θ θρ ρ

∂ ∂ ∂ ∂− + − =

∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂

− + − + + − − +∂ ∂ ∂ ∂ ∂ ∂

(5.72)


2 1 1 2 2 1 1 2

1 2 2 1

1 1r r z z

r r r r

rv T rv T v T v Tr r r r z z

i T S T S


θ θ θ θω

∂ ∂ ∂ ∂− + − =

∂ ∂ ∂ ∂

+ − 1 2 2 12 2z z z zT S T Sθ θ θ θ+ − 1 2 1 2v v v vθ θ θ θρ ρ+ −( ) 2 1 1 2v F v Fθ θ θ θ− + (5.73)

Equation (5.73) yields the real reciprocity relation for shear waves, i.e.,

( ) ( )2 1 1 2 2 1 1 2

1 2 2 11 r r r rr v T v T v T v Tv F v F

r r zθ θ θ θ θ θ θ θ

θ θ θ θ

∂ − ∂ −+ = −

∂ ∂ (5.74)

5.5.2 Lamb waves

Recall the system of Equations (4.131), assign superscripts 1 and 2 to achieve the cross-

multiplication of the two fields as in Equation (5.4), i.e.,

1 12 2

1 12 2

1 12 2

1 1 12 2

1 11 1 12 2 2 2 2 1

1 1 1 12 2 2 2 2

2

rr rrr rr

r

zz zzz zz

rz r zrz rz


rz zz rz zz z z z z z

S vT Tt r

S vT Tt r

S vT Tt zS v vT Tt z r



θθθθ θθ

θθ ρ

ρ

∂ ∂=

∂ ∂∂

=∂

∂ ∂=

∂ ∂⎛ ⎞∂ ∂ ∂

= +⎜ ⎟∂ ∂ ∂⎝ ⎠−∂ ∂ ∂

+ + = −∂ ∂ ∂

∂ ∂ ∂+ + = −

∂ ∂ ∂1

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

(5.75)

131

2 21 1

2 21 1

2 21 1

2 2 21 1

2 22 2 21 1 1 1 1 2

2 2 2 21 1 1 1 1

2

rr rrr rr

r

zz zzz zz

rz r zrz rz


rz zz rz zz z z z z z

S vT Tt r

S vT Tt r

S vT Tt zS v vT Tt z r



θθθθ θθ

θθ ρ

ρ

∂ ∂=

∂ ∂∂

=∂

∂ ∂=

∂ ∂⎛ ⎞∂ ∂ ∂

= +⎜ ⎟∂ ∂ ∂⎝ ⎠−∂ ∂ ∂

+ + = −∂ ∂ ∂

∂ ∂ ∂+ + = −

∂ ∂ ∂2

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

(5.76)

Subtract the two systems in Equation (5.53) to get

1 2 1 22 1 2 1

1 2 1 22 1 2 1

1 2 1 22 1 2 1

1 2 1 1 2 22 1 2 12 2

rr rr r rrr rr rr rr

r r

zz zz z zzz zz zz zz

rz rz r z r zrz rz rz rz

S S v vT T T Tt t r r

S S v vT T T Tt t r r

S S v vT T T Tt t z zS S v v v vT T T Tt t z r z r

θθ θθθθ θθ θθ θθ

∂ ∂ ∂ ∂− = −

∂ ∂ ∂ ∂∂ ∂

− = −∂ ∂

∂ ∂ ∂ ∂− = −

∂ ∂ ∂ ∂⎛ ⎞ ⎛∂ ∂ ∂ ∂ ∂ ∂

− = + − +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝1 1 2 21 1 2 2

2 2 2 1 1 1

1 22 2 1 1 1 2

1 1 1 2 2 22 2 2 1 1 1

12 2 1 1

rr rrrr rz rr rzr r r r r r

r rr r r r r r

rz zz rz rz zz rzz z z z z z

z zz z z z

T T T TT T T Tv v v v v vr z r r z r

v vv v F v v Ft t

T T T T T Tv v v v v vr z r r z r

v vv v F vt

θθ θθ

ρ ρ

ρ ρ

⎞⎜ ⎟

⎠− −∂ ∂ ∂ ∂

+ + − − − =∂ ∂ ∂ ∂

∂ ∂= − − +

∂ ∂∂ ∂ ∂ ∂

+ + − − − =∂ ∂ ∂ ∂

∂ ∂= − −

∂

21 2z zv F

t

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪ +⎪ ∂⎩ (5.77)

Add up the equations in the system (5.54) and rearrange the terms to get

132

2 1 1 2 1 2 2 1 1 2 1 2 2 1 2 1

2 1 2 1 1 2 1 22 1 1 2 2 1 1 2

2 12 11 2 1 2

r rr r rr z zz z zz r rz z rz r rz z rz

r rr z rz r rr z rzr r r r z z z z

rr rrrr rr

v T v T v T v T v T v T v T v Tr r z z z r z r

v T v T v T v T v F v F v F v Fr r r r

S SS ST T T Tt t t t

θθ θθθθ θθ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂− − + − − + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + − − = − + − +

∂ ∂∂ ∂+ − + −

∂ ∂ ∂ ∂

1 22 1

1 2 1 2 1 22 1 2 1 2 12 2

zz zzzz zz

rz rz r r z zrz rz r r z z

S ST Tt t

S S v v v vT T v v v vt t t t t t

ρ ρ ρ ρ

∂ ∂− +

∂ ∂∂ ∂ ∂ ∂ ∂ ∂

− + + − + −∂ ∂ ∂ ∂ ∂ ∂

(5.78)


2 1 1 2 1 2 2 1 1 2 1 2 2 1 2 1

2 1 2 1 1 2 1 22 1 1 2 2 1 1 2

1 2 2 1 2 1 1 2 2 1

r rr r rr z zz z zz r rz z rz r rz z rz


rr rr rr rr zz zz

v T v T v T v T v T v T v T v Tr r z z z r z r

v T v T v T v T v F v F v F v Fr r r r

i T S T S T S T S T Sθθ θθ θθ θθω

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂− − + − − + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + − − = − + − +

+ − − + − +( )1 2 2 1 1 22 2zz zz rz rz rz rzT S T S T S− +

( )1 2 1 2 1 2 1 2r r r r z z z zi v v v v v v v vωρ+ − + −

(5.79)

Equation (5.56) yields the real reciprocity relation for circular crested Lamb waves, i.e.,

( ) ( ) ( ) ( )

( ) ( )

2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2

2 1 2 1 1 2 1 22 1 1 2 2 1 1 2

r rr z rz r rr z rz r rz z zz r rz z zz


v T v T v T v T v T v T v T v Tr z

v T v T v T v Tv F v F v F v F

r

∂ ∂⎡ ⎤ ⎡ ⎤+ − + + + − +⎣ ⎦ ⎣ ⎦∂ ∂+ − +

+ = − + − +

(5.80)

or

( ) ( ) ( )2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2

2 1 1 2 2 1 1 2

1r rr z rz r rr z rz r rz z zz r rz z zz

r r r r z z z z

r v T v T v T v T v T v T v T v Tr r z

v F v F v F v F

∂ ∂⎡ ⎤ ⎡ ⎤+ − − + + − +⎣ ⎦ ⎣ ⎦∂ ∂= − + − +

(5.81)

133

5.6 COMPLEX RECIPROCITY RELATION IN CYLINDRICAL COORDINATES




1 1 12 2 2

1 12 2

1 12 2 2 1

12 2 1

2

2

rr r r

zz z

r zr


T Tv v v Tr z r

vv v Ft

θ θ θθ θ θ

θ θθ θ

θ θθ θ θ θ

θθ θ θρ

⎧ ∂ ∂= −⎪ ∂ ∂⎪

⎪ ∂ ∂=⎪⎪ ∂ ∂

⎨∂ ∂⎪ + + =⎪ ∂ ∂

⎪∂⎪= −⎪ ∂⎩

% % %

% %

% % %

% %

and

2 2 21 1 1

2 21 1

2 21 1 1 2

21 1 2

2

2

rr r r

zz z

r zr


T Tv v v Tr z r

vv v Ft

θ θ θθ θ θ

θ θθ θ

θ θθ θ θ θ

θθ θ θρ

⎧ ∂ ∂= −⎪ ∂ ∂⎪

⎪ ∂ ∂=⎪⎪ ∂ ∂

⎨∂ ∂⎪ + + =⎪ ∂ ∂

⎪∂⎪= −⎪ ∂⎩

% % %

% %

% %%

% %

(5.82)

Add the two systems in Equation (5.82)

1 2 1 1 2 22 1 2 2 1 1

1 2 1 22 1 2 1

1 1 2 22 2 2 1 1 1 1 2

12

2 2

2 2

r rr r r r r r

z zz z z z

r z r zr r

S S v v v vT T T T T Tt t r r r rS S v vT T T Tt t z z

T T T Tv v v T v v v Tr z r r z r

vv

θ θ θ θ θ θθ θ θ θ θ θ

θ θ θ θθ θ θ θ

θ θ θ θθ θ θ θ θ θ θ θ

θθρ

∂ ∂ ∂ ∂+ = − + −

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

+ = +∂ ∂ ∂ ∂

∂ ∂ ∂ ∂+ + + + + =

∂ ∂ ∂ ∂∂

=

% % %% % %

% %% %

% %%% % %

%2

2 1 1 1 2vv F v v Ft t

θθ θ θ θ θρ

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪

∂⎪ − + −⎪ ∂ ∂⎩% %%

(5.83)

Sum up the equation in system (5.83) and rearrange the terms

( ) ( )1 2 2 1 2 1 1 21 2 2 1

1 2 1 2 1 22 1 2 1 2 2 1 1 1 22 2

r r z zr r

z z r rz z r r

v T v T v T v Tv T v Tr r z

S S S S v vT T T T v v F v v Ft t t t t t

θ θ θ θ θ θ θ θθ θ θ θ

θ θ θ θ θ θθ θ θ θ θ θ θ θ θ θρ ρ

∂ + ∂ +++ + =

∂ ∂∂ ∂ ∂ ∂ ∂ ∂

+ + + + − + −∂ ∂ ∂ ∂ ∂ ∂

% %%% %%

% % %% % %% %

(5.84)

Upon rearrangement, Equation (5.84) becomes the complex reciprocity relation for shear

waves, i.e.,

134

( ) ( )

( )

1 2 2 1 1 2 2 1

1 2 2 1 1 2 2 1 1 2

1

2

r r z z

z z r r

r v T v T v T v Tr r z

T S T S v v v F v Ft


θ θ θ θ θ θ θ θ θ θρ

∂ ∂⎡ ⎤+ + + =⎣ ⎦∂ ∂∂

= + + − −∂

% %% %

% % %% %

(5.85)

For time harmonic fields, Equation (5.67) simplifies to the form (multiply everything by r

( ) ( ) ( )1 2 2 1 1 2 2 1 2 1 1 21r r z zr v T v T v T v T v F v F

r r zθ θ θ θ θ θ θ θ θ θ θ θ∂ ∂⎡ ⎤+ + + = − +⎣ ⎦∂ ∂

% % %% % % (5.86)

5.6.2 Lamb waves

Recall Equation (4.131), assign superscripts 1 and 2 to achieve the cross-multiplication of


1 12 2

112 2

1 12 2

1 1 12 2

1 11 12 2 2

12 2 1

1 1 12 2 2

2

r rrrr rr

r

z zzzz zz

r z rzrz rz

rrrr rzr r r

rr r r

rz zz rzz z z

v ST Tr t


T TT Tv v vr z r

vv v Ft

T T Tv v vr z r

θθθθ θθ

θθ

ρ

∂ ∂=

∂ ∂∂

=∂

∂ ∂=

∂ ∂⎛ ⎞∂ ∂ ∂

+ =⎜ ⎟∂ ∂ ∂⎝ ⎠−∂ ∂

+ + =∂ ∂

∂= −

∂∂ ∂

+ +∂ ∂

% %

% %

% %

% %

% % %

% %

% % %

12 2 1zz z z

vv v Ft

ρ

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪ =⎪⎪

∂⎪= −⎪ ∂⎩

% % and

2 21 1

221 1

2 21 1

2 2 21 1

2 22 21 1 1

21 1 2

2 21 1 1

2

r rrrr rr

r

z zzzz zz

r z rzrz rz

rrrr rzr r r

rr r r

rz zz rzz z z

v ST Tr t


T TT Tv v vr z r

vv v Ft

T T Tv v vr z

θθθθ θθ

θθ

ρ

∂ ∂=

∂ ∂∂

=∂

∂ ∂=

∂ ∂⎛ ⎞∂ ∂ ∂

+ =⎜ ⎟∂ ∂ ∂⎝ ⎠−∂ ∂

+ + =∂ ∂

∂= −

∂∂ ∂

+ +∂ ∂

%%

%%

%%

%% %

% %% %

% %

% % % 2

21 1 2zz z z

rvv v Ft

ρ

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪ =⎪⎪

∂⎪= −⎪ ∂⎩% %

(5.87)

Add the two systems in Equation (5.64), i.e.,

135

1 2 1 22 1 2 1

1 21 22 1 2 1

1 2 1 22 1 2 1

1 1 2 22 1 22

r r rr rrrr rr rr rr

r r

z z zz zzzz zz zz zz

r z r zrz rz rz

v v S ST T T Tr r t t

S Sv vT T T Tr r t tv v S ST T T Tz z t tv v v vT T Tz r z r

θθ θθθθ θθ θθ θθ

∂ ∂ ∂ ∂+ = +

∂ ∂ ∂ ∂∂ ∂

+ = +∂ ∂

∂ ∂ ∂ ∂+ = +

∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂

+ + + =⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

%%% %

%%% %

%%% %

% %% %1 2

1

1 1 2 21 1 2 22 2 2 1 1 1

1 22 2 1 1 1 2

1 1 1 2 22 2 2 1 1

2rz rzrz

rr rrrr rz rr rzr r r r r r

r rr r r r r r

rz zz rz rz zzz z z z z z

S STt t

T T T TT T T Tv v v v v vr z r r z r

v vv v F v v Ft t

T T T T Tv v v v v vr z r r z

θθ θθ

ρ ρ

∂ ∂+

∂ ∂− −∂ ∂ ∂ ∂

+ + + + + =∂ ∂ ∂ ∂

∂ ∂= − + −

∂ ∂∂ ∂ ∂ ∂

+ + + + +∂ ∂ ∂ ∂

%

% %% %% % %

% %% %

% %% % %

21

1 22 2 1 1 1 2

rz

z zz z z z z z

Tr

v vv v F v v Ft t

ρ ρ

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪ =⎪⎪

∂ ∂⎪= − + −⎪ ∂ ∂⎩

%

% %% % (5.88)

Sum up the equations in system (5.88) and rearrange the terms

( )

( ) ( )

( )

2 1 1 2 2 1 1 22 1 1 2 2 1 1 2

1 2 2 1 2 1 1 2 1 2 1 2 1 2 1 2

1 2 1 2 2 1 1 2 2

2

r rr r rr z rz z rzr rr r rr z rz z rz

z zz z zz r rz r rz rr rr zz zz rz rz

r r z z r r r r z

v T v T v T v Tv T v T v T v Tr r

v T v T v T v T T S T S T S T Sz t

v v v v v F v F v Ft

θθ θθ

ρ

+ + +∂+ + + +

∂∂ ∂

+ + + + = + + +∂ ∂

∂+ + − − −

∂

% %% %% %% %

% % % %% %% %

%% % % % 1 1 2z z zv F− %

(5.89)

Upon rearrangement, Equation (5.89) yields the complex reciprocity relation for circular

crested Lamb waves, i.e.,

( ) ( )

( ) ( )

2 1 1 2 2 1 1 2 1 2 2 1 2 1 1 2

1 2 1 2 1 2 1 2 1 2 1 2 2 1 1 2

2 1 1 2

1

2

r rr r rr z rz z rz z zz z zz r rz r rz

rr rr zz zz rz rz r r z z r r r r

z z z z


T S T S T S T S v v v v v F v Ft t

v F v F

θθ θθ ρ

∂ ∂⎡ ⎤+ + + + + + +⎣ ⎦∂ ∂∂ ∂

= + + + + + − −∂ ∂

− −

% % % %% % % %

% % % % %% % %

%%

(5.90)

136

For time-harmonic fields, the terms differentiated with respect to time are no longer

dependent of time, hence

( ) ( )

( )

2 1 1 2 2 1 1 2 1 2 2 1 2 1 1 2

2 1 1 2 2 1 1 2

1r rr r rr z rz z rz z zz z zz r rz r rz

r r r r z z z z


v F v F v F v F

∂ ∂⎡ ⎤+ + + + + + +⎣ ⎦∂ ∂= − + + +

% % % %% % % %

% % % %

(5.91)

137

6 ORTHOGONALITY RELATION

The proof of orthogonality is usually done without the need to derive the particular

solution but it is demonstrated through a generic solution that satisfies both the equation

of motion and the boundary conditions. In Appendix D.1, we show the derivation of

orthogonality proof for various vibration problems. In this Section we show first the

orthogonality derivation for both straight crested and circular crested shear horizontal

(SH) waves under the assumption of separation of variables. From the orthogonality poof

for vibration we can easily see that the mathematics becomes quite complicated when the

problem considered is no longer in rectangular coordinates but in cylindrical coordinates.

For this reason, the case of Lamb waves is only derived through the reciprocity relation.

In the second and third part of this Section, we show the proof of orthogonality

relation through the use of the reciprocity relations and by assuming to know the solution

variation in the direction of the wave propagation. Note that here we use the complex

reciprocity relation to derive the orthogonality relation. A different orthogonality relation

it would have been found by using the real reciprocity relation (see Appendix D.3).

6.1 ORTHOGONALITY RELATION WITHOUT ASSUMPTIONS ON THE SOLUTION

In this Section, the orthogonality relation for shear horizontal waves is shown first for

straight-crested waves and then for circular-crested waves. The only assumption retained

on the solution is that the solution is found through separation of variables. We show only

138

the derivation for SH waves since the derivation for Lamb waves is mathematically

difficult.

6.1.1 Straight-crested waves

6.1.1.1 Orthogonality proof

The equation of motion for SH waves in rectangular coordinate is

2

2yzxz z

TT ux y t

ρ∂∂ ∂

+ =∂ ∂ ∂

(6.1)

for simplicity of notation, we define later on zu u= . Assume time-harmonic wave

proportional to i te ω− ; hence Equation (6.1) becomes

2yzxz TT ux y

ρω∂∂

+ = −∂ ∂

(6.2)

Recall the relation between stress and displacement as

xz

yz

uTxuTy

μ

μ

∂⎧ =⎪ ∂⎪⎨ ∂⎪ =

∂⎪⎩

(6.3)

Substitute Equation (6.3) into Equation (6.2) to obtain

2 2

22 2

u u ux y

μ μ ρω∂ ∂+ = −

∂ ∂ (6.4)

Divide Equation (6.4) by μ and use 2sc ρ μ= ; hence,

2 2 2

2 2 2s

u u ux y c

ω∂ ∂+ = −

∂ ∂ (6.5)

139

Stress-free boundary conditions are imposed at the top and bottom surfaces, i.e.,

( , ) 0yzT x d± = (6.6)

Considering Equation (6.3), we get

0d

uy ±

∂=

∂ (6.7)

Assume the problem accepts natural modes of wave propagation at certain natural

frequencies, i.e.,

( , ); ; 1,2,3j jU x y jξ = K (6.8)

At this stage, we choose not to detail the exact form of the general expression of ( , )jU x y

and of the characteristic equation of jξ , although they can be easily deduced (e.g.,

Giurgiutiu V. “Structural Health Monitoring” pages 191-192, Equations (6.42), (6.50).)

Our aim is to develop a generic orthogonality proof for the natural modes ( , )jU x y that

does not depend on their particular form.

Consider two separate mode shapes, ( , )pU x y and ( , )qU x y , such as they satisfy

Equation (6.5) and boundary condition (6.7), i.e.,

2 2 2

2 2 2

2 2 2

2 2 2

p pp

s

q qq

s

U UU

x y c

U UU

x y c

ω

ω

⎧∂ ∂+ = −⎪

∂ ∂⎪⎨

∂ ∂⎪ + = −⎪ ∂ ∂⎩

and 0

0

p

d

q

d

Uy

Uy

±

±

⎧∂=⎪ ∂⎪

⎨∂⎪ =⎪ ∂⎩

(6.9)

Multiply the first equation of the first system in Equation (6.9) by ( , )qU x y and the

second by ( , )pU x y , hence

140

2 2 2

2 2 2

2 2 2

2 2 2

p pq q p q

s

q qp p q p

s

U UU U U U

x y c

U UU U U U

x y c

ω

ω

⎧∂ ∂+ = −⎪

∂ ∂⎪⎨

∂ ∂⎪ + = −⎪ ∂ ∂⎩

(6.10)

Integrate over the thickness, i.e.

2 2 2

2 2 2

2 2 2

2 2 2

d dp p

q q p qsd d

d dq q

p p q psd d

U UU U dy U U dy

x y c

U UU U dy U U dy

x y c

ω

ω

− −

− −

⎧ ⎛ ⎞∂ ∂+ = −⎪ ⎜ ⎟⎜ ⎟∂ ∂⎪ ⎝ ⎠

⎨⎛ ⎞∂ ∂⎪

+ = −⎜ ⎟⎪ ⎜ ⎟∂ ∂⎝ ⎠⎩

∫ ∫

∫ ∫ (6.11)

Integrate by parts

dp

qd

UU

y −

∂∂

2 2

2 2

d dp p q

q p qsd d

dq

pd

U U UU dy U U dy

y yx c

UU

y

ω

− −

−

⎛ ⎞∂ ∂ ∂+ − = −⎜ ⎟

∂ ∂∂⎝ ⎠

∂∂

∫ ∫2 2

2 2

d dq q p

p q psd d

U U UU dy U U dy

y yx cω

− −

⎧⎪⎪⎪⎨⎪ ⎛ ⎞∂ ∂ ∂

+ − = −⎜ ⎟⎪∂ ∂∂⎝ ⎠⎪⎩

∫ ∫

(6.12)

Subtract the first line from the second to get

2 2 2 2

2 2 2 2

dp q

q ps sd

U UU U dy

x x c cω ω

−

⎛ ⎞∂ ∂ ⎛ ⎞− = − −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∫d

p qd

U U dy−∫ (6.13)

Assume separation of variable such that

( , ) ( ) ( )U x y X x Y y= (6.14)

Use Equation (6.14) into Equation (6.13) to get

( ) 0d

p p q q p p q qd

X Y X Y X Y X Y dy−

′′ ′′− =∫ (6.15)

141

where 2

2

d XXdx

′′ = , 2

2

d YYdy

′′ = . Since pX and qX do not depend on y , we can factor

them out, i.e.

0d

p qp q p q

p q d

X XX X Y Y dy

X X −

′′ ′′⎛ ⎞− =⎜ ⎟

⎝ ⎠∫ (6.16)

Assume , 0p qX X ≠ and divide Equation (6.16) by the product p qX X to get

0d

p qp q

p q d

X XY Y dy

X X −

′′ ′′⎛ ⎞− =⎜ ⎟

⎝ ⎠∫ (6.17)

Apply the separation of variable assumption (6.14) to system (6.10) to get

2

2

p p p p p ps

q q q q q qs

X Y X Y X Yc

X Y X Y X Yc

ω

ω

⎧ ′′ ′′+ = −⎪⎪⎨⎪ ′′ ′′+ = −⎪⎩

(6.18)

Without loss of generality, we can assume 0XY ≠ and divide Equation (6.18) by XY ,

i.e.,

2

2

p p

p p s

q q

q q s

X YX Y c

X YX Y c

ω

ω

′′ ′′⎧+ = −⎪

⎪⎨ ′′ ′′⎪ + = −⎪⎩

(6.19)

subtract the two equations and rearrange the terms, i.e.,

p q p q

p q p q

X X Y YX X Y Y

′′ ′′ ′′ ′′⎛ ⎞− = − −⎜ ⎟

⎝ ⎠ (6.20)

Substitute expression (6.20) into Equation (6.17)

142

0d

p qp q

p q d

Y YY Y dy

Y Y −

′′ ′′⎛ ⎞− − =⎜ ⎟

⎝ ⎠∫ (6.21)

The following two situations exist:

i. If p q≠ , then 0p q

p q

Y YY Y

′′ ′′⎛ ⎞− ≠⎜ ⎟

⎝ ⎠; hence

0d

p qd

Y Y dy−

=∫ (6.22)

ii. If p q= , then 0p q

p q

Y YY Y

′′ ′′⎛ ⎞− =⎜ ⎟

⎝ ⎠; hence

( )2 0d

pd

Y dy−

≠∫ (6.23)

This is the normalization factor.

6.1.1.2 Real power flow

We can obtain the same result through the real reciprocity relation. Consider the real

reciprocity relation for shear waves in the absence of body forces, i.e.,

( ) ( ) 0p q q p p q q pz xz z xz z yz z yzv T v T v T v T

x y∂ ∂

− + − =∂ ∂

(6.24)

Integrate with respect to y

( ) ( ) 0d dp q q p p q q p

z xz z xz z yz z yz dd

v T v T dy v T v Tx −

−

∂− + − =

∂ ∫ (6.25)

Apply the boundary condition (6.6), hence

143

( ) 0d

p q q pz xz z xz

d

v T v T dyx −

∂− =

∂ ∫ (6.26)

The integral in Equation (6.26) is the power flow in the x direction. It can easily seen that

for p q= , the equality is satisfied; we want to prove that this is true also for p q≠ .

Assume separation of variables, hence, with the help of Equation (6.14) we write

( , ) ( ) ( )

( , )( , ) ( ) ( )

z

zxz

v x y i X x Y yu x yT x y X x Y y

x

ω

μ μ

= −∂ ′= =

∂

(6.27)

Substitute expressions in Equation (6.27) into Equation (6.26) to get

( ) 0d

p p q q q q p pd

i X Y X Y X Y X Y dyx

μω−

∂ ′ ′− − =∂ ∫ (6.28)

Factor out the term p qY Y , i.e.:

( ) 0d

p q q p p qd

i X X X X Y Y dyx

μω−

∂ ′ ′− − =∂ ∫ (6.29)

Bring out of the y integral the term dependent on x, perform the derivative w.r.t. x to get

p qi X Xμω ′ ′− p q q pX X X X′′ ′ ′+ −( ) 0d

q p p qd

X X Y Y dy−

′′− =∫ (6.30)

Upon rearrangement, we obtain

0d

p q q p p qd

i X X X X Y Y dyμω−

′′ ′′− − =⎡ ⎤⎣ ⎦ ∫ (6.31)

The following two cases apply:

144

i. For p q≠ , 0p q q pX X X X′′ ′′− ≠⎡ ⎤⎣ ⎦ , hence 0d

p qd

Y Y dy−

=∫

ii. For p q= , 0p q q pX X X X′′ ′′− =⎡ ⎤⎣ ⎦ and

( )2 0d

pd

Y dy−

≠∫ (6.32)


We will now prove that the normalization factor (6.32) is related to the power flow ppP in

the x direction and defined as

1 ( ) ( )2

d

pp p pd

P V y T y dy−

= ∫ (6.33)

Note that Equation (6.27) can be written as

( , ) ( ) ( ) ( ) ( )( , ) ( ) ( ) ( ) ( )

z

xz

v x y i X x Y y X x V yT x y X x Y y X x T y

ωμ

= − =′ ′= =

(6.34)

where

( ) ( )V y i Y yω= − and ( ) ( )T y Y yμ= (6.35)

Substitute Equation (6.34) into (6.33) to obtain

( )2( )2

d

pp pd

iP Y y dyωμ

−

−= ∫ (6.36)

Q.E.D.

145

6.1.1.3 Complex power flow

We can obtain the same result through the complex reciprocity relation. Consider the

complex reciprocity relation for shear waves in the absence of body forces, i.e.,

( ) ( ) 0p q q p p q q pz xz z xz z yz z yzv T v T v T v T

x y∂ ∂

+ + + =∂ ∂

% %% % (6.37)

Integrate with respect to y to get

( ) ( ) 0d d

p q q p p q q pz xz z xz z yz z yz d

d

v T v T dy v T v Tx −

−

∂+ + + =

∂ ∫ % %% % (6.38)


( ) 0d

p q q pz xz z xz

d

v T v T dyx −

∂+ =

∂ ∫ %% (6.39)

The integral in Equation (6.39) is the power flow in the x direction. We want to prove

that the equality is always true.

Assume separation of variables as in Equation (6.14). Substitute the Equation (6.27)

expressions into Equation (6.39) to get after rearrangement

( ) 0d

p p q q q q p pd

i X Y X Y X Y X Y dyx

μω−

∂ ′ ′− =∂ ∫ % % % % (6.40)

Factor out the term p qY Y% , i.e.:

( ) 0d

p q q p p qd

i X X X X Y Y dyx

μω−

∂ ′ ′− =∂ ∫ % % % (6.41)

Bring out of the y integral the term dependent on x, perform the derivative w.r.t. x to get

146

p qi X Xμω ′ ′%p q q pX X X X′′ ′ ′+ −% % 0

d

q p p qd

X X Y Y dy−

⎡ ⎤′′− =⎣ ⎦ ∫% % (6.42)

Upon rearrangement, we obtain

0d

p q q p p qd

i X X X X Y Y dyμω−

⎡ ⎤′′ ′′− =⎣ ⎦ ∫% % % (6.43)


i. For p q≠ , 0p q q pX X X X⎡ ⎤′′ ′′− ≠⎣ ⎦% % , hence 0

d

p qd

Y Y dy−

=∫ % .

ii. For p q= , 0p q q pX X X X⎡ ⎤′′ ′′− =⎣ ⎦% % and

0d

p qd

Y Y dy−

≠∫ % (6.44)


Similarly as in the real reciprocity relation, we can prove that the normalization factor

(6.44) is related to the power flow ppP in the x direction and defined as

1 ( ) ( )2

d

pp p pd

P V y T y dy−

= ∫ % (6.45)

With the use of relation in Equation (6.34), Equation (6.45) becomes

( ) ( )2

d

pp p pd

iP Y y Y y dyωμ

−

= ∫ % (6.46)

Q.E.D.

147

6.1.2 Circular-crested waves

6.1.2.1 Orthogonality proof

The equation of motion for circular-crested shear horizontal wave in absence of external

force and under the assumption of time harmonic waves proportional to i te ω− is of the

form

2

2

2yrr

TT uTr y r t

θθ θθ ρ

∂∂ ∂+ + =

∂ ∂ ∂ (6.47)

for simplicity of notation, we define later on u uθ= . Assume time-harmonic wave

proportional to i te ω− ; hence Equation (6.47) becomes

22r zr

T T T ur z r

θ θθ ρω∂ ∂

+ + = −∂ ∂

(6.48)

Recall the relation between stress and displacement as

r

z

u uTr r

uTz

θ

θ

μ

μ

⎧ ∂⎛ ⎞= −⎜ ⎟⎪⎪ ∂⎝ ⎠⎨

∂⎪ =⎪ ∂⎩

(6.49)

Substitute Equation (6.49) into Equation (6.48) to obtain

2 2

22 2 2

1u u u u ur r r z r

μ ρω⎛ ⎞∂ ∂ ∂

+ + − = −⎜ ⎟∂ ∂ ∂⎝ ⎠ (6.50)

Divide Equation (6.50) by μ and use 2sc ρ μ= ; hence,

2 2 2

2 2 2 2

1

s

u u u u ur r r z r c

ω∂ ∂ ∂+ + − = −

∂ ∂ ∂ (6.51)

148

Stress-free boundary conditions are imposed at the top and bottom surfaces, i.e.

( , , ) 0zT r d tθ ± = ; considering Equation(6.49), we get

( , ) 0rd

uT x dzθ

±

∂± = =

∂ (6.52)

Assume the problem accepts natural modes of wave propagation at certain natural

frequencies, i.e.,

( , ); ; 1,2,3j jU r z jξ = K (6.53)

At this stage, we choose not to detail the exact form of the general expression of ( , )jU r z

and of the characteristic equation of jξ . Our aim is to develop a generic orthogonality

proof for the natural modes ( , )jU r z that does not depend on their particular form.

Consider two separate mode shapes, ( , )pU r z and ( , )qU r z , such as they satisfy Equation

(6.51) and boundary condition (6.52), i.e.,

( , ) 0

( , ) 0

pr

d

qr

d

UT x d

zU

T x dz

θ

θ

±

±

∂± = =

∂∂

± = =∂

(6.54)

Substitute the mode shapes in the wave Equation (6.51), i.e.,

2 2 2

2 2 2 2

2 2 2

2 2 2 2

1

1

p p p pp

s

q q q qq

s

U U U UU

r z r r r c

U U U UU

r z r r r c

ω

ω

∂ ∂ ∂+ + − = −

∂ ∂ ∂

∂ ∂ ∂+ + − = −

∂ ∂ ∂

(6.55)

149

Multiply the first equation by ( )qU r and the second by ( )pU r ; multiply by r, and

perform integration with respect to z to get

2 2 2

2 2 2 2

2 2 2

2 2 2 2

1

1

d dp p p p q

q q q p qsd d

d dq q q q p

p p p q psd d

U U U U UU U U dz U U dz

r z r r r c

U U U U UU U U dz U U dz

r z r r r c

ω

ω− −

− −

⎛ ⎞∂ ∂ ∂+ + − = −⎜ ⎟

∂ ∂ ∂⎝ ⎠

⎛ ⎞∂ ∂ ∂+ + − = −⎜ ⎟

∂ ∂ ∂⎝ ⎠

∫ ∫

∫ ∫ (6.56)

Integrate by parts

dp

qd

UU

z −

∂∂

2 2

2 2 21d d

p p q p p qq q p q

sd d

dq

pd

U U U U U UU U dz U U dz

r z z r r r c

UU

z

ω

− −

−

⎛ ⎞∂ ∂ ∂ ∂+ − + − = −⎜ ⎟

∂ ∂ ∂ ∂⎝ ⎠

∂∂

∫ ∫2 2

2 2 21d d

q q p q q pp p q p

sd d

U U U U U UU U dz U U dz

r z z r r r cω

− −

⎛ ⎞∂ ∂ ∂ ∂+ − + − = −⎜ ⎟

∂ ∂ ∂ ∂⎝ ⎠∫ ∫

(6.57)

Subtract the first line from the second to get

2 2

2 21 1 0

dp p q q

q q p pd

U U U UU U U U dz

r r r r r r−

⎡ ⎤⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂+ − + =⎢ ⎥⎜ ⎟ ⎜ ⎟

∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦∫ (6.58)

Assume separation of variables such that

( , ) ( ) ( )U r z R r Z z= (6.59)

Substitute expression (6.59) into Equation (6.58)

2

2

1

01

p p q qd p p q q p p q q

p p q qdq q p p q q p p

R Z R ZR Z R Z R Z R Z

r r dzR Z R Z

R Z R Z R Z R Zr r

−

⎛ ⎞′′ ′+ −⎜ ⎟⎜ ⎟ =⎜ ⎟′′ ′− − +⎜ ⎟⎝ ⎠

∫ (6.60)

Assume 0p qR R ≠ and divide Equation (6.60) by the product p qR R to get

150

2 21 1 0

dp p p q q q p q

p q p q q p q pp p q qd

R R Z Z R R Z ZZ Z Z Z Z Z Z Z dz

R r R r R r R r−

⎛ ⎞′′ ′ ′′ ′⎜ ⎟+ − − − + =⎜ ⎟⎝ ⎠

∫ (6.61)

Rearrange the terms and bring out of the y integral the terms not dependent on z to obtain

2 21 1 1 1 0

dp p q q

p qp p q q d

R R R RZ Z dz

R r R r R r R r −

⎡ ⎤⎛ ⎞′′ ′ ′′ ′⎛ ⎞⎢ ⎥⎜ ⎟+ − − + − =⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∫ (6.62)

Apply separation of variables assumption (6.59) to system (6.55) to get

2 2

2 2

1

1

p pp p p p p p p p

s

q qq q q q q q q q

s

R ZR Z R Z R Z R Z

r r cR Z

R Z R Z R Z R Zr r c

ω

ω

⎧ ′′ ′ ′′+ + − = −⎪⎪⎨⎪ ′′ ′ ′′+ + − = −⎪⎩

(6.63)

where 2

2

d RRdr

′′ = , 2

2

d ZZdz

′′ = . Without loss of generality, we can assume 0RZ ≠ and

divide Equation (6.63) by RZ , i.e.,

2 2

2 2

1 1

1 1

p p p

p p p s

q q q

q q q s

R R ZR r R Z r c

R R ZR r R Z r c

ω

ω

′′ ′ ′′⎧+ + − = −⎪

⎪⎨ ′′ ′ ′′⎪ + + − = −⎪⎩

(6.64)

Subtract the two equations and rearrange the terms, i.e.,

2 21 1 1 1p p q q p q

p p q q p q

R R R R Z ZR r R r R r R r Z Z

′′ ′ ′′ ′ ′′ ′′⎛ ⎞ ⎛ ⎞+ − − + − = − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (6.65)

Substitute expressions (6.65) into Equation (6.62)

0d

q pp q

q p d

Z ZZ Z dz

Z Z −

′′ ′′⎛ ⎞− =⎜ ⎟

⎝ ⎠∫ (6.66)

151


i. If p q≠ , then 0q p

q p

Z ZZ Z

′′ ′′⎛ ⎞− ≠⎜ ⎟

⎝ ⎠; hence

0d

p qd

Z Z dz−

=∫ (6.67)

ii. If p q= , then 0q p

q p

Z ZZ Z

′′ ′′⎛ ⎞− =⎜ ⎟

⎝ ⎠ and

0d

p qd

Z Z dz−

≠∫ (6.68)


6.1.2.2 Real reciprocity relation

We can obtain the same result through the real reciprocity relation. Consider the real

reciprocity relation for shear waves in the absence of body forces, i.e.:

( ) ( )1 0p q q p p q q pr r z zr v T v T v T v T

r r zθ θ θ θ θ θ θ θ∂ ∂⎡ ⎤− + − =⎣ ⎦∂ ∂

(6.69)


( ) ( )1 0d dp q q p p q q p

r r z z dd

r v T v T dz v T v Tr r θ θ θ θ θ θ θ θ −

−

∂ ⎡ ⎤− + − =⎣ ⎦∂ ∫ (6.70)


( ) 0d

p q q pr r

d

r v T v T dzr θ θ θ θ

−

∂ ⎡ ⎤− =⎣ ⎦∂ ∫ (6.71)

Note that the power flow of shear waves propagating in the r direction is equal to

152

2 0d

rd

P r v T dzθ θπ−

= =∫ (6.72)

Hence the integral in Equation (6.71) is related to the power flow in the r direction. It can

easily seen that for p q= , the equality in Equation (6.71) is satisfied; we want to prove

that this is true also for p q≠ .

Assume separation of variable; hence, with the help of Equation (6.59), we write

( , ) ( ) ( )

( , ) ( , ) ( )( , ) ( ) ( )

z

z zr

v r z i R r Z zu r z u r z R rT r z R r Z z

r r rθ

ω

μ μ

= −

∂⎛ ⎞ ⎛ ⎞′= − = −⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

(6.73)

Substitute Equation (6.73) expressions into Equation (6.71) to get

( ) ( )

( ) ( ) 0d

q pp p q q q q p p

d

R r R rr i R Z R r Z i R Z R r Z dz

r r rωμ ωμ

−

⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞∂ ′ ′− − + − =⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦∫ (6.74)

Factor out the term p qZ Z , i.e.:

( ) ( )

( ) ( ) 0d

q pp q q p p q

d

R r R ri r R R r R R r Z Z dz

r r rμω

−

⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞∂ ′ ′− − − − =⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦∫ (6.75)

Bring out the z integral the terms dependent on r, perform the derivative w.r.t. r, and

rearrange to get

( ) 0d

p q p q p q p q p qd

i R R R R r R R R R Z Z dzμω−

′ ′ ′′ ′′⎡ ⎤− − + − =⎣ ⎦ ∫ (6.76)


153

i. For p q≠ , recall Equation (6.67), i.e. 0d

q pd

Z Z dz−

=∫ ; hence Equation (6.76) is always

satisfied and therefore Equation (6.71).

ii. For p q= , ( ) 0p q p q p q p qR R R R r R R R R′ ′ ′′ ′′⎡ ⎤− + − =⎣ ⎦ and ( )2 0d

pd

Z dz−

≠∫ .

Note that notation in Equation (6.73) can be written as

( , ) ( ) ( ) ( ) ( )

( ) ( )( , ) ( ) ( ) ( ) ( )

z z

r r

v r z i R r Z z R r V zR r R rT r z R r Z z R r T z

r rθ θ

ω

μ

= − =

⎛ ⎞ ⎛ ⎞′ ′= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(6.77)

Equation (6.76) can be rewritten as

1 0d

q p q pp q

q p q p d

R R R Rr i Z Z dz

r R R R Rω μ

−

′ ′ ′′ ′′⎡ ⎤⎛ ⎞− + − − =⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦∫ (6.78)

or multiplying and dividing by 2π

1 2 ( ) ( ) 02

dq p q p p q

r rq p q p d

R R R R r v z T z dzr R R R R θ

ππ −

′ ′ ′′ ′′⎡ ⎤⎛ ⎞− + − =⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦∫ (6.79)

and define the real power flow as:

( ) ( )d

p ppp r r

d

P r v z T z dzθπ−

= ∫ (6.80)

6.1.2.3 Complex reciprocity relation

We can obtain the same result through the complex reciprocity relation. Consider the

complex reciprocity relation for shear waves in the absence of body forces, i.e.,

154

( ) ( )1 0p q q p p q q pr r z zr v T v T v T v T

r r zθ θ θ θ θ θ θ θ∂ ∂⎡ ⎤+ + + =⎣ ⎦∂ ∂

% %% % (6.81)


( ) ( )1 0d d

p q q p p q q pr r z z d

d

r v T v T dz v T v Tr r θ θ θ θ θ θ θ θ −

−

∂ ⎡ ⎤+ + + =⎣ ⎦∂ ∫ % %% % (6.82)


( ) 0d

p q q pr r

d


−

∂ ⎡ ⎤+ =⎣ ⎦∂ ∫ %% (6.83)

Equation (6.83) is the power flow in the z direction. We want to prove that equality is

always satisfied. Assume separation of variable as in Equation (6.59) and substitute these

expressions into Equation (6.83) to get

0d

q pp p q q q q p p

d

R Rr i R Z R Z i R Z R Z dz

r r rω μ ω μ

−

⎡ ⎤⎛ ⎞⎛ ⎞⎛ ⎞∂ ′ ′⎢ ⎥− − − =⎜ ⎟⎜ ⎟⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦∫

%% % % % (6.84)

Factor out the term p qZ Z% , i.e.:

( ) 0d

p q q p q pd

i r R R R R Z Z dzr

μω−

∂ ⎡ ⎤′ ′− =⎣ ⎦∂ ∫ % % % (6.85)

Bring out from the z integral the terms dependent on r, perform the derivative w.r.t. r, and

rearrange to get

0d

p q q p p q q p q pd

i R R R R rR R rR R Z Z dzμω−

⎡ ⎤′ ′ ′′ ′′− + − =⎣ ⎦ ∫% % % % % (6.86)


155

i. For p q≠ , 0p q q p p q q pR R R R rR R r R′ ′ ′′ ′′− + − ≠% % % % , hence 0d

q pd

Z Z dz−

=∫ % ; Equation (6.86) is

always satisfied.

ii. For p q= , 0p q q p p q q pR R R R rR R rR R′ ′ ′′ ′′− + − =% % % % and 0d

q pd

Z Z dz−

≠∫ % .

Equation (6.86) can be written as

0d

p q q pp q q p p q

d

R R R RR R R R r i Z Z dz

r rω μ

−

⎡ ⎤′ ′′′ ′′− − + − − =⎢ ⎥

⎣ ⎦ ∫% %

% % % (6.87)

Through the use of notation in Equation (6.77), Equation (6.87) becomes

0d

p q q p p q q p p qd

R R R R rR R rR R r v T dz−

⎡ ⎤′ ′ ′′ ′′− − + − =⎣ ⎦ ∫% % % % % (6.88)

Note that the term d

p qd

r v T dz−∫ % is proportional to the complex power flow.

6.1.3 Reduction to the Sturm-Liouville problem

In this section we will show that the SH wave equation, either straight-crested or circular-

crested, can be reduced to the Liouville equation (see Appendix D.2). For the case of

Lamb waves, the resuction to a Sturm-Liouville problem is achieved through the use of

potentials (method possible only for isotropic materials).

6.1.3.1 Straight-crested SH waves

Consider the equation of motion of the straight-crested SH waves as defined in Equation

(6.5), i.e.,

156

2 2 2

2 2 2s

u u ux y c

ω∂ ∂+ = −

∂ ∂ (6.89)

and the solution through separation of variables defined as

( , ) ( ) ( )u x y X x Y y= (6.90)

Substitute solution in Equation (6.90) into the equation of motion to obtain

2

2s

X Y XY XYcω′′ ′′+ = − (6.91)

Divide Equation (6.91) by Y and regroup the terms to get

2

2 0s

YX XY c

ω′′⎛ ⎞′′ + + =⎜ ⎟

⎝ ⎠ (6.92)

This is the Liouville equation defined in (D.77) where functions p, q, and r are defined as

( ) 1( )( ) 1

p xq x Y Yr x

=′′=

= (6.93)

and the separation constant λ is equal to 2 2scω .

6.1.3.2 Circular-crested SH waves

Consider the equation of motion of the circular-crested SH waves as defined in Equation

(6.51), i.e.,

2 2 2

2 2 2 2

1

s

u u u u ur r r z r c

ω∂ ∂ ∂+ + − = −

∂ ∂ ∂ (6.94)

and the solution through separation of variables defined as

( , ) ( ) ( )u r z R z Z z= (6.95)

157

Substitute solution in Equation (6.95) into the equation of motion to obtain

2

2 2

1 1

s

R Z R Z RZ RZ RZr r c

ω′′ ′ ′′+ + − = − (6.96)

Divide Equation (6.91) by Z and regroup the terms to get

2

2 2

1 1 0s

ZR R Rr Z r c

ω′′⎛ ⎞′′ ′+ + − + =⎜ ⎟

⎝ ⎠ (6.97)

Note that the first two terms can be grouped together to get

( )2

2

1 0s

d ZrR r r Rdr Z r c

ω′′⎛ ⎞′ + − + =⎜ ⎟

⎝ ⎠ (6.98)

This is the Liouville equation defined in (D.77) where functions p, q, and r are defined as

( )( )( ) 1( )

p r rq r r Z Z rr r r

=′′= −

=

(6.99)

and the separation constant λ is equal to 2 2scω .

6.2 ORTHOGONALITY RELATION IN RECTANGULAR COORDINATES

In this Section, we derive the orthogonality relations for straight-crested waves through

the assumption that solutions 1 and 2 are generic time-harmonic and space-harmonic

guided-wave modes (e.g., plate guided waves), i.e.,

1

2

( , , , ) ( )

( , , , ) ( )

n

m

i x i tn

i x i tm

x y z t y e e

x y z t y e e

ξ ω

ξ ω

−

−

=

=

v v

v v (6.100)

158

In the generic case, the wavenumbers and the amplitudes are assumed to be complex

( ,m nξ ξ ∈ , ,n m ∈v v ). The strains and stress will also be harmonic, i.e.,

1

2

( , , , ) ( )

( , , , ) ( )

n

m

i x i tn

i x i tm

x y z t y e e

x y z t y e e

ξ ω

ξ ω

−

−

=

=

T T

T T (6.101)

Consider now the complex conjugates of field 2 and write

2 ( , , , ) ( ) mi x i tmx y z t y e eξ ω+ −=v v %

% % (6.102)

2 ( , , , ) ( ) mi x i tmx y z t y e eξ ω+ −=T T %% % (6.103)

Recall the complex reciprocity relation for time-harmonic functions as given by Equation

(5.45) and set the source terms equal to zero ( 1 2= =F F 0 ), hence

( )2 1 1 2 0∇ ⋅ + ⋅ =v T v T%% (6.104)

Write in extended form the del operator and perform the derivative with respect to x

taking into consideration Equations (6.100) through (6.103), i.e.,

( ) ( ) ( )

( )

2 1 1 2 2 1 1 2 2 1 1 2

2 1 2 1 1 2 1 2 2 1 1 2

ˆ ˆ

ˆ ˆ( ) ( )m n n m

x y

i i i iy

ξ ξ ξ ξ

∂ ∂∇ ⋅ + ⋅ = ⋅ + ⋅ ⋅ + ⋅ + ⋅ ⋅

∂ ∂∂⎡ ⎤= ⋅ + ⋅ − + − ⋅ + ⋅ ⋅ + ⋅ + ⋅ ⋅ =⎣ ⎦ ∂

v T v T v T v T x v T v T y

v T v T v T v T x v T v T y 0

% % %% % %

% %% % %% % %

(6.105)

where x and y are the unit vectors in the x and y directions. Simplification of Equation

(6.105) yields

( )( ) ( )2 1 1 2 2 1 1 2ˆ ˆ 0m niy

ξ ξ ∂− ⋅ + ⋅ ⋅ + ⋅ + ⋅ ⋅ =

∂v T v T x v T v T y% % %% % (6.106)

Substitution of Equations (6.100) through (6.103) into Equation (6.106) yields

159

( ) ( ) ( )

( ) ( )

ˆ( ) ( ) ( ) ( )

ˆ( ) ( ) ( ) ( ) 0

m n

m n

i xm n m n n m

i xm n n m

i e y y y y

e y y y yy

ξ ξ

ξ ξ

ξ ξ + −

+ −

− ⋅ + ⋅ ⋅

∂+ ⋅ + ⋅ ⋅ =

∂

v T v T x

v T v T y

%

%

% %%

%% (6.107)

Since the exponential function ( )m ni xe ξ ξ+ −%

is non zero, we can divide Equation (6.107) by

( )m ni xe ξ ξ+ −%

and get

( )( )

( )

ˆ( ) ( ) ( ) ( )

ˆ( ) ( ) ( ) ( ) 0

n m m n n m

m n n m

i y y y y

y y y yy

ξ ξ− − ⋅ + ⋅ ⋅

∂+ ⋅ + ⋅ ⋅ =

∂

v T v T x

v T v T y

% %%

%% (6.108)

Integrate Equation (6.108) with respect to y to get

( ) ( )

( )

ˆ( ) ( ) ( ) ( )

ˆ( ) ( ) ( ) ( )

d

n m m n n md

d

m n n m d

i y y y y dy

y y y y

ξ ξ−

−

− − ⋅ + ⋅ ⋅ =

= − ⋅ + ⋅ ⋅

∫ v T v T x

v T v T y

% %%

%%

(6.109)

Denote the integral in Equation (6.109) by mnP , i.e.,

( )1 ˆ( ) ( ) ( ) ( )4

d

mn m n n md

P y y y y dy−

= − ⋅ + ⋅ ⋅∫ v T v T x%% (6.110)


( ) ( ) ˆ4 ( ) ( ) ( ) ( )d

n m mn m n n m di P y y y yξ ξ

−− = − ⋅ + ⋅ ⋅v T v T y% %% (6.111)

If boundary conditions in Equation (6.111) at y d= ± are either traction free ( ˆ⋅ =T y 0 )

or rigid ( =v 0 ), then the RHS of Equation (6.111) vanishes and we get

( ) 4 0n m mni Pξ ξ− =% (6.112)

160

If m nξ ξ≠% , then one can divide by ( )n mξ ξ− % and Equation (6.112) becomes the

orthogonality relation, i.e.,

0mnP = for m nξ ξ≠% (6.113)

The wavenumbers of guided waves always occur in pairs having equal value and

opposite signs. By convention, the modes that propagate or decay in the x+ direction are

numbered with positive integers (and negative integers for those in x− direction).

For undamped propagating modes, the wavenumbers are real ( ,m nξ ξ ∈ ), and the

conjugate is just the same as the original number ( m mξ ξ=% ). In this case, Equation (6.112)

becomes

( ) 4 0n m mni Pξ ξ− = (6.114)

If the two modes are distinct ( m n≠ ), then 0n mξ ξ− ≠ ; one may divide Equation (6.114)

by ( )n mξ ξ− and get

0mnP = for m n≠ (6.115)

This is the orthogonality relation for undamped propagating modes.

If m n= , then 0n mξ ξ− = and relation (6.115) no longer applies. In this case nnP is

nonzero. In fact, the real part of nnP represent the average power flow carried by the

guided wave, since

ˆ ˆ ˆRe Re Re4 2

d d dn n n n n n

nn avd d d

P dy dy dy P− − −

⎛ ⎞⋅ + ⋅ − ⋅= − ⋅ = ⋅ = ⋅ =⎜ ⎟

⎝ ⎠∫ ∫ ∫

v T v T v Tx x P x%% %

(6.116)

161

where P is the complex Poynting vector defined by Equation (4.30). The average power

definition of Equation (4.16) was used. In writing Equation (6.116), we use the complex-

number property of Appendix B.3, i.e., a b a b⋅ = ⋅ %% .

For evanescent modes, the wavenumbers ,m nξ ξ are imaginary; hence we can write that

m m m

n n n

ξ ξ ξ

ξ ξ ξ−

−

= − = −

= − = −

%

% (6.117)

from (6.112) the orthogonality relation becomes

0mnP = for m n≠ − (6.118)

A non propagating mode can not transport energy along a waveguide. The average power

flow for imaginary wavenumbers is carried by the cross-product terms between the field

of the positive evanescent mode M and the reflected mode –M.

The orthogonality relation in Equation (6.116) can be used to derive the

normalization factor for Lamb waves. A detail derivation of the normalization factor is

made in Appendix E.

Note that the derivation of the orthogonality relation is valid for Lamb wave

propagating in layered waveguide structures, where the material has arbitrary anisotropy,

but the properties of the media are z-invariant.


Recall the complex reciprocity relation of Equation and set the source terms equal to zero

( 1 2= =F F 0 ), hence

162

( ) ( )2 1 1 2 2 1 1 2 0z xz z xz z yz z yzv T v T v T v Tx y

∂ ∂+ + + =

∂ ∂% %% % (6.119)

Assume that solutions 1 and 2 are free modes such that:

1

2

ˆ( , , , ) ( )

ˆ( , , , ) ( )

n

m

i xn i tz

i xm i tz

x y z t v y e e

x y z t v y e e

ξ ω

ξ ω

−

−

=

=

v z

v z %% %

(6.120)

1

0 0 ( )( , , , ) 0 0 ( )

( ) ( ) 0

n

nxz

i xn i tyz

n nxz yz

T yx y z t T y e e

T y T y

ξ ω−

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

T (6.121)

2

0 0 ( )( , , , ) 0 0 ( )

( ) ( ) 0

n

mxz

i xm i tyz

m mxz yz

T yx y z t T y e e

T y T y

ξ ω−

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

T %

%

% %

% % (6.122)

Substituting in (6.119) we obtain after rearrangement

( ) ( )

( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 0

n m

n m

i xm n n mz xz z xz

i xm n n mz yz z yz

v y T y v y T y ex

v y T y v y T y ey

ξ ξ

ξ ξ

− −

− −

∂ ⎡ ⎤+⎢ ⎥⎣ ⎦∂∂ ⎡ ⎤+ + =⎣ ⎦∂

%

%

%%

%% (6.123)

i.e.,

( )( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 0

m n n mz yz z yzm n n m

n m z xz z xz

v y T y v y T yi v y T y v y T y

yξ ξ

∂ +− − + + =

∂

%%% %% (6.124)


( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )d d

m n n m m n n mn m z xz z xz z yz z yz d

d

i v y T y v y T y dy v y T y v y T yξ ξ−

−

− − + = − +∫% % %% % (6.125)

163

Since the shear wave modes satisfy the stress free boundary conditions ( 0yzT = ), the

RHS of Equation (6.139) vanishes and we get

( ) ( )( ) ( ) ( ) ( ) 0d

m n n mn m z xz z xz

d

i v y T y v y T y dyξ ξ−

− − + =∫% %% (6.126)

Define

( )1 ( ) ( ) ( ) ( )4

dm n n m

nm z xz z xzd

P v y T y v y T y dy−

= − +∫ %% (6.127)


( )4 0n m mni Pξ ξ− =% (6.128)

For propagating modes, the wavenumbers ,m nξ ξ are real; hence, Equation (6.128)

becomes

( ) 4 0n m mni Pξ ξ− = (6.129)

For different modes, i.e. m n≠ , the wavenumbers are also different, m nξ ξ≠ , and we can

divide Equation (6.129) by ( )m nξ ξ− ; hence, we get the orthogonality condition

0mnP = for m n≠ (6.130)

For the same mode, i.e. m n= , the wavenumbers are the same, m nξ ξ= and

( ) 0m nξ ξ− = ; in this case, Equation (6.129) implies that 0mnP ≠ . In fact, the real part of

mnP represents the time-averaged power carried by the wave, i.e.,

164

( )1Re ( ) ( ) ( ) ( )4

dm n n m

nn z xz z xz avd

P v y T y v y T y dy P−

⎡ ⎤= − + =⎢ ⎥

⎣ ⎦∫ %% (6.131)

Since m n= and since a b a b⋅ = ⋅ %% (see Appendix B.3), Equation (6.145) simplifies to

1Re ( ) ( )2

dn n

av z xzd

P v y T y dy−

⎡ ⎤= −⎢ ⎥

⎣ ⎦∫ % (6.132)

From the average power flow defined in Equation (6.132) it is possible to derive the wave

modes normalization factor. (Appendix E gives a full derivation of the normalization

factor).

6.2.2 Lamb waves

Recall the complex reciprocity relation of Equation (5.68)and set the source terms equal

to zero ( 1 2= =F F 0 ), hence

( ) ( )1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 0y xy y xy x xx x xx y yy y yy x xy x xyv T v T v T v T v T v T v T v Tx y

∂ ∂+ + + + + + + =

∂ ∂% % % %% % % % (6.133)

Assume that solutions 1 and 2 are free modes of non-dissipative Lamb waves such that

( )( )

1

2

ˆ ˆ( , , , ) ( ) ( )

ˆ ˆ( , , , ) ( ) ( )

n

m

i xn n i tx y

i xm m i tx y

x y z t v y v y e e

x y z t v y v y e e

ξ ω

ξ ω

−

−

= +

= +

v x y

v x y %% % %

(6.134)

1

( ) ( ) 0( , , , ) ( ) ( ) 0

0 0 ( )

n

n nxx yx

i xn n i tyx yy

nzz

T y T yx y z t T y T y e e

T y

ξ ω−

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

T (6.135)

2

( ) ( ) 0( , , , ) ( ) ( ) 0

0 0 ( )

m

m mxx yx

i xm m i tyx yy

mzz

T y T yx y z t T y T y e e

T y

ξ ω−

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

T %

% %

% % %

% (6.136)

165

Substituting in (6.133), we obtain after rearrangement

( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0

n m

n m

i xn m m n n m m ny xy y xy x xx x xx

i xn m m n n m m ny yy y yy x xy x xy

v y T y v y T y v y T y v y T y ex

v y T y v y T y v y T y v y T y ey

ξ ξ

ξ ξ

− −

− −

∂ ⎡ ⎤+ + +⎢ ⎥⎣ ⎦∂⎡ ⎤∂

+ + + + =⎢ ⎥∂⎣ ⎦

%

%

% %% %

% %% %

(6.137)

i.e.,

( )( )

( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

n m m n n m m nn m y xy y xy x xx x xx

n m m n n m m ny yy y yy x xy x xy

i v y T y v y T y v y T y v y T y

v y T y v y T y v y T y v y T yy

ξ ξ− − + + + =

∂− + + +

∂

% % %% %

% %% % (6.138)


( ) ( )

( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

dn m m n n m m n

n m y xy y xy x xx x xxd

dn m m n n m m ny yy y yy x xy x xy d

i v y T y v y T y v y T y v y T y dy

v y T y v y T y v y T y v y T y

ξ ξ−

−

− − + + + =

− + + +

∫% % %% %

% %% %

(6.139)

Since the Lamb wave modes satisfy the stress free boundary conditions ( 0xy yyT T= = ),

the right-hand side of Equation (6.139) vanishes and we get

( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0d

n m m n n m m nn m y xy y xy x xx x xx

d

i v y T y v y T y v y T y v y T y dyξ ξ−

− − + + + =∫% % %% % (6.140)

Define

( )1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4

dn m m n n m m n

nm y xy y xy x xx x xxd

P v y T y v y T y v y T y v y T y dy−

= − + + +∫ % %% % (6.141)


( )4 0n m mni Pξ ξ− =% (6.142)

166

For undamped propagating modes, the wavenumbers ,m nξ ξ are real;. hence, Equation

(6.142) becomes

( ) 4 0n m mni Pξ ξ− = (6.143)

For different modes, i.e. m n≠ , the wavenumbers are also different, m nξ ξ≠ , and we can

divide Equation (6.143) by ( )m nξ ξ− ; hence, we get the orthogonality condition

0mnP = for m n≠ (6.144)

For the same mode, i.e. m n= , the wavenumbers are the same, m nξ ξ= and

( ) 0m nξ ξ− = ; in this case, Equation (6.143) implies that 0mnP ≠ . In fact, the real part of

mnP represents the time-averaged power carried by the wave, i.e.,

( )1Re ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4

dn m m n n m m n

nm y xy y xy x xx x xx avd

P v y T y v y T y v y T y v y T y dy P−

⎡ ⎤= − + + + =⎢ ⎥

⎣ ⎦∫ % %% % (6.145)

Since m n= and since ( ) ( )Re Rea b a b⋅ = ⋅ %% (see Appendix B.3), Equation (6.145)

simplifies to

( )1Re ( ) ( ) ( ) ( )2

dn n n n

av x xx y xyd

P v y T y v y T y dy−

⎡ ⎤= − +⎢ ⎥

⎣ ⎦∫ % % (6.146)

From the average power flow defined in Equation (6.146) it is possible to derive the

normalization factor. (Appendix E gives a full derivation of the normalization factor.)

167

6.3 ORTHOGONALITY RELATION IN CYLINDRICAL COORDINATES

In this Section, we derive the orthogonality relations for circular-crested waves through

the assumption that solutions 1 and 2 are generic time-harmonic guided-wave modes and

varying radially as the Bessel function.

For circular crested wave it is not possible to derive a generic formulation of the

orthogonality relation as in Equation (6.114). We directly derive the complex orthogonal

relation for the case of SH waves and Lamb waves.


Recall the complex reciprocity relation of Equation (5.86) and set the source terms equal


( ) ( )1 2 2 1 1 2 2 1 0r r z zr v T v T r v T v Tr zθ θ θ θ θ θ θ θ

∂ ∂⎡ ⎤+ + + =⎣ ⎦∂ ∂% %% % (6.147)

Integrate Equation (6.147) with respect to z to get

( ) ( )1 2 2 1 1 2 2 1 0d

d

r r z z dd

r v T v T dz r v T v Tr θ θ θ θ θ θ θ θ −

−

∂ ⎡ ⎤+ + + =⎣ ⎦∂ ∫ % %% % (6.148)

Since the shear wave modes satisfy the stress free boundary conditions ( 0zTθ = ), the

right-hand side of Equation (6.148) vanishes and we get

( )1 2 2 1 0d

r rd


−

∂ ⎡ ⎤+ =⎣ ⎦∂ ∫ % % (6.149)

1

1 12

2 1

( , ) ( , ) ( ) ( )

( , ) ( , ) ( ) ( )n n

m m

r z v r z i Z z J r

r z v r z i Z z J rθ

θ

ω ξ

ω ξ

= =

= =

v

v (6.150)

where

168

( ) ( ) sin cosrZ z v z T A z B zθ θ β β= = = + (6.151)

and, from Equation (3.75),

1 11 0

2 12 0

( )( , ) ( ) ( ) 2

( )( , ) ( ) ( ) 2

nr n n n

mr m m m

J rr z T Z z J rr

J rr z T Z z J rr

θ

θ

ξμ ξ ξ

ξμ ξ ξ

⎛ ⎞= = −⎜ ⎟⎝ ⎠⎛ ⎞= = −⎜ ⎟⎝ ⎠

T

T (6.152)


11 0

11 0

( )( ) ( ) 20

( )( ) ( ) 2

md n n m m m

d nm m n n n

J ri Z J r Z J rrr dz

r J ri Z J r Z J rr

ξωμ ξ ξ ξ

ξωμ ξ ξ ξ−

⎡ ⎤⎛ ⎞⎛ ⎞− −⎢ ⎥⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎢ ⎥⎜ ⎟ =

⎢ ⎥⎜ ⎟∂ ⎛ ⎞+ −⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

∫

%% %%

%%

(6.153)

Rearrange the terms, factor out the product m nZ Z and divide by the term iωμ , i.e.,

( )0 1 1 0( ) ( ) ( ) ( ) 0d

m m n n m n m nd

r J r J r J r J r Z Z dzr

ξ ξ ξ ξ ξ ξ−

∂ ⎡ ⎤− =⎣ ⎦∂ ∫ % % % % (6.154)

Bring out of the z integral the terms dependent on r, and perform the derivative w.r.t. r, to

get

0 1 1 0

2 0 11 1 0 0

2 1 01 1 0 0

( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( ) 0

( ) ( )( ) ( ) ( ) ( )

m m n n m n

dm nm m n m n m n m

m nd

m nn m n m n m n n

J r J r J r J r

J r J rJ r J r J r J r Z Z dzrrJ r J rJ r J r J r J r

r

ξ ξ ξ ξ ξ ξ

ξ ξξ ξ ξ ξ ξ ξ ξ ξ

ξ ξξ ξ ξ ξ ξ ξ ξ ξ−

⎡ ⎤−⎢ ⎥

⎛ ⎞⎢ ⎥− + −⎜ ⎟⎢ ⎥ =⎜ ⎟+⎢ ⎥⎜ ⎟⎢ ⎥+ − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∫

% % %

%% % % % % %

%% % %

(6.155)

Rearrange the terms to obtain

( )2 21 1( ) ( ) 0

d

n m m n m nd

r J r J r Z Z dzξ ξ ξ ξ−

− =∫% % % (6.156)

169

With the use of (6.151), the integral in Equation (6.156) can be written as

( )1 1 Re4 2

d dm n m n

m n r r nmd d

Z Z dz v T v T dz Pθ θ θ θ− −

− = − + =∫ ∫% %% (6.157)

This is the Poynting vector in the radial direction for circular-crested waves as defined in

(4.212).

i. If n m≠ , then ( )2 2 0n mξ ξ− ≠% ; hence 0d

m n nmd

Z Z dz P−

= =∫ .

ii. If n m= , then ( )2 2 0n mξ ξ− ≠% and ( )2 0d

m mmd

Z dz P−

= ≠∫ . This is the

normalization factor.

6.3.2 Lamb waves

Recall the complex reciprocity relation of Equation (5.91) and set the source terms equal


( ) ( )2 1 1 2 2 1 1 2 1 2 2 1 2 1 1 21 0r rr r rr z rz z rz z zz z zz r rz r rzr v T v T v T v T v T v T v T v Tr r z

∂ ∂⎡ ⎤+ + + + + + + =⎣ ⎦∂ ∂% % % %% % % % (6.158)


( ) ( )2 1 1 2 2 1 1 2 1 2 2 1 2 1 1 21 0d

d

r rr r rr z rz z rz z zz z zz r rz r rz dd

r v T v T v T v T dz v T v T v T v Tr r −

−

∂ ⎡ ⎤+ + + + + + + =⎣ ⎦∂ ∫ % % % %% % % % (6.159)

Since the Lamb wave modes satisfy the stress free boundary conditions ( 0rz zzT T= = ),

the right-hand side of Equation (6.159) vanishes and we get

( )2 1 1 2 2 1 1 2 0d

r rr r rr z rz z rzd

r v T v T v T v T dzr −

∂ ⎡ ⎤+ + + =⎣ ⎦∂ ∫ % %% % (6.160)

170

Assume that solutions 1 and 2 are free modes of non-dissipative Lamb waves such that

( ) ( )( ) ( )

1 11 1 0

2 22 1 0

ˆ ˆˆ ˆ( , ) ( ) ( ) ( ) ( )

ˆ ˆˆ ˆ( , ) ( ) ( ) ( ) ( )

r zr z n n n n

r zr z m m m m

r z v v v z J r v z J r


ξ ξ

ξ ξ

= + = +

= + = +

v r z r z

v r z r z (6.161)

where from Equations (3.122) and (3.130) we have

( )( )

*

*

( ) cos cos

( ) sin sin

Srn Sn Sn Sn Sn Sn Sn

Szn Sn Sn Sn Sn Sn Sn

v z i A z R z

v z i A z R z

ω ξ α β β

ω α α ξ β

⎧ = − −⎪⎨

= − +⎪⎩ (6.162)

for symmetric modes and

( )

( )

*

*

( ) sin sin

( ) cos cos

Arn Sn Sn Sn Sn Sn Sn

Azn Sn Sn Sn Sn Sn Sn

v z A z R z

v z A z R z

ξ α β β

α α ξ β

⎧ = −⎪⎨

= − +⎪⎩ (6.163)

for antisymmetric modes. In Equation (6.161) we do not use the subscript S or A since it

is written in a generic form. From the derivation of stresses in Section 3.2.6, we can write

the stresses as

11 0

12 0

1 1

2 1

( )2( , ) ( ) ( ) ( )

( )2( , ) ( ) ( ) ( )

( , ) ( ) ( )

( , ) ( ) ( )

rr rr r nn n n

rr rr r mm m m

rz rzn n

rz rzm m

J rT r z T z J r v zi r


T r z T z J r

T r z T z J r

ξμμ ξω

ξμμ ξω

μ ξ

μ ξ

= −

= −

=

=

(6.164)

where

( )

( )

* 2 2 2

* 2 2

( ) 2 cos 2 cos

( ) 2 sin sin

rrn Sn Sn Sn Sn Sn Sn Sn Sn Sn

rzn Sn Sn Sn Sn Sn Sn Sn Sn

T z A z R z

T z A z R z

β ξ α α ξ β β

α ξ α ξ β β

⎡ ⎤= − + − −⎣ ⎦⎡ ⎤= + −⎣ ⎦

(6.165)


171

( )

( )

* 2 2 2

* 2 2

( ) 2 sin 2 sin

( ) 2 cos cos

rrn An An An An An An An An An

rzn An An An An An An An An

T z A z R z

T z A z R z

ξ β α α ξ β β

α ξ α ξ β β

⎡ ⎤= − + − −⎣ ⎦⎡ ⎤= − + −⎣ ⎦

(6.166)

for antisymmetric modes.

Substituting expression of velocity and stress in the expression of the average power flow

Equation (6.160), we obtain

11 0

11 0

0 1 0 1

( )2( ) ( ) ( ) ( ) ( )

( )2( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

r rr r nm m n n n

r rr r mn n m m m

z rz z rzm m n n n n m m

J rv z J r T z J r v zi r

J rr v z J r T z J r v zr i r

v z J r T z J r v z J r T z J r

ξμξ μ ξω

ξμξ μ ξω

ξ μ ξ ξ μ ξ

⎡ ⎛ ⎞⎡ ⎤−⎢ ⎜ ⎟⎢ ⎥⎣ ⎦⎢ ⎜ ⎟⎢ ⎜ ⎟⎡ ⎤∂

+ +⎜ ⎟⎢ ⎥∂ ⎣ ⎦⎜ ⎟⎜ ⎟+ +⎝ ⎠⎣

%%

%%% %

% %%%

0d

d

dz−

⎤⎥⎥⎥

=⎢ ⎥⎢ ⎥⎢ ⎥

⎦

∫ (6.167)

Rearrange the terms in Equation (6.167) to get

( )

( )0 1

1 0

( ) ( ) ( ) ( ) ( ) ( )0

( ) ( ) ( ) ( ) ( ) ( )

r rr z rzdm n n m n m

r rr z rzd n m m n n m

v z T z v z T z J r J rr dz

r v z T z v z T z J r J r

ξ ξ

ξ ξ−

⎡ ⎤⎛ ⎞+∂ ⎢ ⎥⎜ ⎟ =⎜ ⎟∂ ⎢ ⎥+ +⎝ ⎠⎣ ⎦

∫%%%

%% % (6.168)

Differentiate with respect to r

( )( )

( )( )0 0 1 1

0 0 1 1

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

r rr z rzdm n n m m n m n n m

r rr z rzd n m m n n n m m n m

v z T z v z T z J r J r J r J rdz

v z T z v z T z J r J r J r J r

ξ ξ ξ ξ ξ ξ


⎡ ⎤+ −⎢ ⎥ =⎢ ⎥+ + −⎣ ⎦

∫% % %%%

% % %% %(6.169)

For propagating modes the wavenumbers are real and the Bessel functions are always

real, moreover the stresses are in quadrature with the velocities. Without loss of

generality, assume that the stresses are real, and call v i uω= − , hence Equation (6.169)

becomes

( )( )

( )( )0 0 1 1

0 0 1 1

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

r rr z rzdm n n m m n m n n m

r rr z rzd n m m n n n m m n m

u z T z u z T z J r J r J r J ri dz

u z T z u z T z J r J r J r J r

ξ ξ ξ ξ ξ ξω


⎡ ⎤− + −⎢ ⎥ =⎢ ⎥+ − −⎣ ⎦

∫ (6.170)

172

i. If m n= , Equation (6.170) becomes

( )0 0 1 1

0 0 1 1

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 0

( ) ( ) ( ) ( )

dn n n n n n r rr z rz

n n n nn n n n n n d

J r J r J r J ri u z T z u z T z dz

J r J r J r J rξ ξ ξ ξ ξ ξ

ωξ ξ ξ ξ ξ ξ −

−⎛ ⎞ ⎡ ⎤− =⎜ ⎟ ⎣ ⎦− +⎝ ⎠∫ (6.171)

where 0 0 1 1

0 0 1 1

( ) ( ) ( ) ( )0

( ) ( ) ( ) ( )n n n n n n

n n n n n n

J r J r J r J rJ r J r J r J r

ξ ξ ξ ξ ξ ξξ ξ ξ ξ ξ ξ

−⎛ ⎞=⎜ ⎟− +⎝ ⎠

and

( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0d d

r rr z rz r rr z rzn n n n n n n n nn

d d

i u z T z u z T z dz v z T z v z T z dz Pω− −

⎡ ⎤⎡ ⎤− = + = ≠⎣ ⎦ ⎣ ⎦∫ ∫ % % .

Hence, the normalization factor is.

( )( ) ( ) ( ) ( )d

r rr z rzn n n n

d

v z T z v z T z dz−

⎡ ⎤+⎣ ⎦∫ % (6.172)

ii. If m n≠ , then 0 0 1 1

0 0 1 1

( ) ( ) ( ) ( ) 0( ) ( ) ( ) ( ) 0

m n m n n m

n n m m n m



− ≠− ≠

hence

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0d d

r rr z rz r rr z rzn m m n m n n m

d d

v z T z v z T z dz v z T z v z T z dz− −

⎡ ⎤ ⎡ ⎤+ = + =⎣ ⎦ ⎣ ⎦∫ ∫% % .

In section 3.2.6 we have found that the particle displacement (hence the velocity) and the

stresses dependence on the variable z in cylindrical coordinates was exactly the same as

that derived in rectangular coordinates; hence

( ) ( )( ) ( )

rr xx

rz xy

T TT T

≡≡

and ( ) ( )( ) ( )

r x

z y

v vv v

≡≡

(6.173)

This means that normalization factor obtained does not depend on the selected coordinate

system, i.e.,

( ) ( )d d

r rr z rz x xx y xyn n n n n n n n

d d

v T v T dz v T v T dy− −

+ = +∫ ∫% %% % (6.174)

173

The time-averaged power flow does not depend on the coordinates system.

174

PART II PWAS-BASED STRUCTURAL HEALTH MONITORING

175

The main topic of my research concerns structural health monitoring using piezoelectric

wafer active sensors (PWAS). PWAS are piezoelectric transducers that can be

permanently attached to the structure through an adhesive layer. In Part II of the

dissertation, we discuss the interaction between the PWAS and the structure and how

transducers excite guided waves through a bonding layer. These concepts are at the basis

for the development of an efficient SHM system in terms of energy used, number of

PWAS needed for the detection, PWAS geometry determination, and SHM method

configuration.

First, we derive the normal mode expansion model for the case of waves excited in a

structure by either surface or volume forces. It is to note that the derivation of NME

presented here in the generic form is valid for both isotropic and anisotropic materials.

The NME theory is based on the reciprocity relation theorem and on the assumption

of knowing how the wave varies along the propagation direction (harmonically or not).

Hence, the derivation is different whether we consider rectangular or circular crested

waves and both methods are reported in the following sections.

Through the use of NME theory, we extend the theory of shear lag coupling between

PWAS and structure form the simple case of axial and flexural waves excited in the

structure, to the more comprehensive case of Lamb waves.

The shear lag parameter is a fundamental element in the derivation of the tuning

between the structure and the PWAS. Tuning is the result of the coupling between PWAS

176

and Lamb waves. Maximum coupling between PWAS and Lamb waves occurs when the

PWAS length is an odd multiple of the half wavelength. Since different Lamb wave

modes have different wavelengths, which vary with frequency, it is possible to selectively

excite various Lamb wave modes at various frequency, i.e., to tune the PWAS into one or

another Lamb wave mode. The shear lag parameter depends on the geometry of the

transducer, the structure, and the bonding layer, and also on the excitation frequency;

hence its effect on the tuning curves reflects the effects of the SHM configuration

considered on the excitation and can be a useful tool for determining the best SHM

configuration.

So far, theoretical tuning curves have been provided for isotropic materials. The

derivation of this curves on composite plates have been postulated by Xi (2002), but the

solution provided is mathematically difficult. Here, we present a novel method to obtain

tuning curves on composites plates. The method proposed is based on the NME theory

and the knowledge of the dispersion curves of guided waves in composite plates. An

explicit derivation of NME for composite plates with PWAS was not available before and

it is derived in detail here. The derivation of dispersion curves of guided waves in

composite plates is a well known problem and several methods are available for its

solution. In this dissertation we will make use of the method described by Nayfeh (1995)/

177

7 PWAS EXCITATION OF GUIDED WAVES

In this section, we focus on the excitation of guided waves through the use of

piezoelectric transducers. The excitation is expressed through the normal mode expansion

method in which the acoustic field is represented as the superposition of all the acoustic

field modes. To obtain the expression of the acoustic field, we must specify the type of

transducer we take under consideration. In our case, we derive the normal mode

expansion method for the case of a piezoelectric wafer active sensors that can be

permanently attached on the surface of the structure or can be embedded in the volume of

the structure.

7.1 PIEZOELECTRIC WAFER ACTIVE SENSORS CHARACTERISTICS

Guided waves in a structure can be excited with different kind of transducers, i.e., wedge

and comb transducers, interdigital electrode array transducers, piezoelectrically driven

surface-wave transducers, etc. The focus of my research is based primarily on

piezoelectric transducers and, in particular, on the lead zirconate titanate (PZT)

piezoelectric wafer active sensor (PWAS) and the gallium orthophosphate (GaPO4)

PWAS. The material characteristics of the transducer determine the type of excitation and

hence of wave that is transmitted to the structure. Piezoelectric materials are able to

generate an electric field when subject to a mechanical stress (direct effect) or they can

generate a mechanical strain in response to an applied electric field (converse effect).

178

Through the direct and converse effect, piezoelectric transducers can be used to either

generate or sense a propagating wave in a structure.

PZT PWAS are perovskite ceramics with the lead at the corner, the oxygen on the

faces, and the zirconate/titanium ion in the center (see Figure 7.1a). Below Curie

temperature (TC) the zirconate/titanium ion shifts from the center to an offset position so

that the center of the positive and negative charge no longer coincides, yielding a dipole

moment in the c direction (dashed line in Figure 7.1a).

Figure 7.1 Lead Zirconite titanate PWAS. a) atomic structure of PZT for temperature below

Curie temperature. (www.piezo-kinetics.com). b) PWAS transducer notations

This form of the perovskite structure is called ferroelectric phase. At temperature

above Curie temperature, the perovskite is in the paraelectric phase and no electrical

polarity is present.

Material characteristics of the APC-850 PZT PWAS used in our laboratory are

reported in Table 7.1 where the coordinate system used is depicted in Figure 7.1b. The

electrodes on the PWAS are on the faces with normal 3.

1

3

2

a) b)

1

3 Electrode

179

Table 7.1 Material properties

Property APC-850 GaPO4

ρ (Kg/m3) 7600 3570

d33 (10-12m/V) 290 4.5

d31 (10-12m/V) -125 -4.5

d32 (10-12m/V) -125

d14 (10-12m/V) -1.9

d15 (10-12m/V) 480 -3.8

d36 (10-12m/V) 1.9

g33 (10-12m/V) 26.5

g31 (10-12m/V) -11

g15 (10-12m/V) 38

ν 0.3

TC (C°) 325

The relation between electrical and mechanical variables is defined, in Vogit notation, as

1 31

2 321

3 332

43

5 15

6

0 00 00 00 0 0

0 00 0 0

S dS d

ES d

ES

ES dS

⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎧ ⎫⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥

⎩ ⎭⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎣ ⎦⎩ ⎭

(7.1)

where ijd are the coupling coefficient between electrical and mechanical variables.

Recall that the relation between Vogit and matrix notation is

1 11 4 232 22 5 133 33 6 12

→ →→ →→ →

(7.2)

180

With electrodes on the PWAS on the top and bottom surface with normal 3, the

possible obtainable strains are normal strain in all directions, i.e.,

1 31 3

2 32 3

3 33 3

S d ES d ES d E

=⎧⎪ =⎨⎪ =⎩

(7.3)

As an electric field is applied in direction 3, the PWAS undergoes dilatation and

contraction in directions 1, 2, and 3. From this consideration, we see that excitation

through PWAS can only produce Lamb wave in the structures but no shear horizontal

waves.

Gallium orthophosphate PWAS is considered the “high temperature brother of

quartz”. The crystal structure can be derived from that of quartz by substituting the Si

atoms with Ga and P. This results in a total length in the c direction double of that of the

quartz (see Figure 7.2a). GaPO4 transducers have high piezoelectric sensitivity and high

thermal stability up to temperatures above 970°C. This PWAS does not have

pyroelectricity, hence it will not cause interference to the sensor signal. However, the

material characteristics of the ijd coefficients are an order of magnitude smaller than

those of PZT PWAS, hence the transmitted signal is significant smaller.

The GaPO4 transducers used in our laboratory had a triple layer structure: electrode,

GaPO4 thin film crystal, and electrode, where the electrodes were sputtered Pt-layers.

181

Figure 7.2 GaPO4 PWAS. a) GaPO4 crystal structure. (www.roditi.com), b) Transducer

deformation for an electric field in direction 3.

Material characteristics of gallium orthophosphate are reported in Table 7.1. The

relation between electrical and mechanical variables is defined as

1 31

21

3 332

4 143

5 15

6 36

0 00 0 00 0

0 00 0

0 0

S dS

ES d

ES d

ES dS d

⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎧ ⎫⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪= ⎢ ⎥⎨ ⎬ ⎨ ⎬

⎢ ⎥⎪ ⎪ ⎪ ⎪⎩ ⎭⎢ ⎥⎪ ⎪

⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎩ ⎭ ⎣ ⎦

(7.4)

If the electrodes on GaPO4 PWAS are on the top and bottom surface with normal 3,

the possible obtainable strains are normal strain along directions 1 and 3, and shear 12,

i.e.,

1 31 3

3 33 3

6 36 3

S d ES d ES d E

=⎧⎪ =⎨⎪ =⎩

(7.5)

1

2

13

a) b)

E3 +

–

182

The GaPO4 PWAS undergoes dilatation along direction 3 and contraction along

direction 1; it withstand shear on the plane normal to 3 (see Figure 7.2b). From this

consideration, we see that excitation through GaPO4 PWAS can produce either Lamb

waves and shear horizontal waves in the structures.

7.2 PWAS EXCITATION OF STRAIGHT-CRESTED GUIDED WAVES

In this section, we show the normal mode expansion (NME) method that can be used to

derive the wave guided fields excited by an arbitrary distribution of mechanical and

electrical sources. We consider straight crested guided waves. After the generic

derivation of NME and its specific derivation for the case of SH waves and Lamb waves,

we will apply the NME method to the case of guide waves excited by piezoelectric wafer

active sensors (PWAS). The NME method will be used both for PWAS installed on the

surface of the structure and for PWAS embedded in the structure.

In this section, the NME method is derived from the complex reciprocity relation for

time harmonic guided waves as derived in Section 5.2 Equation (5.45), i.e.,

( )2 1 1 2 2 1 1 2∇ ⋅ + ⋅ = − ⋅ − ⋅v T v T v F v F% %% % (7.6)

Multiplication by 1− both sides of Equation (7.6) and expansion of the del operator

yields

( ) ( )2 1 1 2 2 1 1 2 2 1 1 2ˆ ˆy x

∂ ∂− ⋅ − ⋅ ⋅ + − ⋅ − ⋅ ⋅ = +

∂ ∂v T v T y v T v T x v F v F% % %% % % (7.7)

Consider guided waves that propagate in a plate along direction x . The guided waves can

be excited at the acoustic boundaries by:

• Volume sources, F

183

• Traction forces, ˆ⋅T y

• Velocity sources, v

It is assumed that solution “1” is the field excited by the sources. Let assume that the

excited field can be represented by the modal expansion

1 1( , ) ( ) ( )m mm

x y a x y= = ∑v v v (7.8)

1 1( , ) ( ) ( )m mm

x y a x y= = ∑T T T (7.9)

where ( )ma x are the x-dependent modal participation factors that depend on the mode

under consideration and the excitation used to generate the field. The modal participation

factors are the same for the all the acoustic fields. The y-dependent terms, ( )m yv , are

assumed to be known and depend only on the mode considered. It is assumed that

solution “2” is homogeneous ( 2 0=F ) and is represented by just a single mode, the nth

forward mode of the free plate, i.e.,

2 ( , ) ( ) ni xnx y y e ξ−=v v (7.10)

2 ( , ) ( ) ni xnx y y e ξ−=T T (7.11)

Integrating Equation (7.7) with respect to y, we obtain

( ) ( )2 1 1 2 2 1 1 2 2 1ˆ ˆd d d

d d d

dy dy dyy x− − −

∂ ∂− ⋅ − ⋅ ⋅ + − ⋅ − ⋅ ⋅ =

∂ ∂∫ ∫ ∫v T v T y v T v T x v F% %% % % (7.12)

Note that the first integral in Equation (7.12) can be evaluated exactly, hence Equation

(7.12) becomes

184

( ) ( )2 1 1 2 2 1 1 2 2 1ˆ ˆd dd

dd d

dy dyx−

− −

∂− ⋅ − ⋅ ⋅ + − ⋅ − ⋅ ⋅ =

∂∫ ∫v T v T y v T v T x v F% %% % % (7.13)

Substitution of Equations (7.10)-(7.11) into Equation (7.13) and rearrangement yields

( )

( )

1 1

1 1 1

ˆ( ) ( )

ˆ( ) ( ) ( )

n

n n

d i xn n d

d di x i x

n n nd d

y y e

e y y dy y e dyx

ξ

ξ ξ

−

− −

− ⋅ − ⋅ ⋅

∂ ⎡ ⎤+ − ⋅ − ⋅ ⋅ =⎣ ⎦∂∫ ∫

v T v T y

v T v T x v F

%

% %

%%

%% % (7.14)

Substitution of Equations (7.8)-(7.9) into the integral in Equation (7.14) yields

( )1 1

1

ˆ( ) ( )

ˆ( ) ( ) ( ) ( ) ( ) ( )

( )

n

n n

n

d i xn n d

di x i x

n m m m m nm md

di x

nd

y y e

y e a x y a x y y e dyx

y e dy

ξ

ξ ξ

ξ

−

−

−

− ⋅ − ⋅ ⋅

∂ ⎛ ⎞+ − ⋅ − ⋅ ⋅ =⎜ ⎟∂ ⎝ ⎠

=

∑ ∑∫

∫

v T v T y

v T v T x

v F

%

% %

%

%%

%%

%

(7.15)

or, by factoring out the first integral the term ( )ni xme a xξ% and rearranging the terms

( )

( )

1 1

1

ˆ( ) ( )

ˆ( ) ( ) ( ) ( ) ( ) ( )

n

n n

d i xn n d

d di x i x

m n m m n nm d d

y y e

e a x y y y y dy e y dyx

ξ

ξ ξ

−

− −

− ⋅ − ⋅ ⋅

⎡ ⎤∂+ − ⋅ − ⋅ ⋅ =⎢ ⎥∂ ⎣ ⎦

∑ ∫ ∫

v T v T y

v T v T x v F

%

% %

%%

%% % (7.16)

Substitute into Equation (7.16) the orthogonality relation in Equation (6.110) to get

( )1 1 1ˆ 4 ( )n n n

dd i x i x i xn n m mn nd m d

e e a x P e dyx

ξ ξ ξ− − −

−−

∂− ⋅ − ⋅ ⋅ + = ⋅

∂ ∑ ∫v T v T y v F% % %%% % (7.17)

The summation in (7.17) has only one nonzero term that is for the propagating mode n

( nξ real) for which 0nnP ≠ . Hence, Equation (7.17) becomes

( )1 1 1ˆ 4 ( )n n n

dd i x i x i xn n nn n nd

d

e P e a x e dyx

ξ ξ ξ

−−

∂ ⎡ ⎤− ⋅ − ⋅ ⋅ + = ⋅⎣ ⎦∂ ∫v T v T y v F% % %%% % (7.18)

185

Dividing Equation (7.18) by i xe ξ and rearranging we get

( )1 1 1ˆ4 ( )dd

nn n n n n ndd

P i a x dyx

ξ−

−

∂⎛ ⎞+ = ⋅ + ⋅ ⋅ + ⋅⎜ ⎟∂⎝ ⎠ ∫v T v T y v F% %% % (7.19)

The first term on the right hand side of Equation (7.19) is the forcing function due to the

surfaces forces; the second term is the forcing function due to the volume sources.

Equation (7.19) is a first order ODE; solution of the ODE expressed by Equation (7.19) is

obtained using the integrating factor method described in Appendix B.4. Comparison of

Equation (7.19) with the standard ODE form of Equation (B.30) of Appendix B.4 yields

the integrating factor ( ) n n

P x dx i dx i xIF e e eξ ξ∫ ∫= = = . Substitution of Equation (7.19) into

Equation (B.55) of Appendix B.4 yields

( )( ) ( )n ni x i xna x e f x e dx Cξ ξ−= +∫

% % (7.20)

where

( )1 1 11 1ˆ( )

4 4

dd

n n ndnn nn d

f x dyP P−

−

= ⋅ + ⋅ ⋅ + ⋅∫v T v T y v F%% % (7.21)

Substituting Equation (7.21) into Equation (7.20) and absorbing the constant C into the

integral sign as the undefined lower limit c allows us to write the general solution of

Equation (7.19) in the form

( )1 1 1ˆ( )4

nn

x di x d i xn n n nd

nn c d

ea x dy e dxP

ξξ

−

−−

⎡ ⎤= ⋅ + ⋅ ⋅ + ⋅⎢ ⎥

⎣ ⎦∫ ∫v T v T y v F

%%%% % forward wave solution(7.22)

Note that the solution expressed by Equation (7.22) is a forward propagating wave since

it contains the factor ni xe ξ− . The above argument can be equally applied to backward

186

propagating waves. Assume that solution 2 was taken to be a backward wave: instead of

Equation (7.28) we would choose

2

2

( , ) ( )

( , ) ( )

n

n

i xnx x

i xny y

v x y v y e

v x y v y e

ξ

ξ

⎧ =⎪⎨

=⎪⎩ (backward propagating wave) (7.23)

It is apparent that Equation (7.23) can be obtained from Equation (7.28) by changing nξ

into nξ− . Performing this change in Equation (7.22) yields the backward wave solution,

i.e.,

( )1 1 1ˆ( )4

nn

x di x d i xn n n nd

nn c d

ea x dy e dxP

ξξ−

−−

⎡ ⎤= ⋅ + ⋅ ⋅ + ⋅⎢ ⎥

⎣ ⎦∫ ∫v T v T y v F

%%%% % backward wave solution (7.24)

Equations (7.22) and (7.24) are the field amplitude for an arbitrary wave guide

transducer. Once the particular transducer have been selected, the amplitude constant c

can be determined.


Recall the complex reciprocity relation of Equation (5.63), i.e.,

( ) ( )2 1 1 2 2 1 1 2 2 1 1 2z xz z xz z yz z yz z z z zv T v T v T v T v F v F

x y∂ ∂

+ + + = − −∂ ∂

% % %% % % (7.25)

Assume that solution “1” is generated by the force 1F , whereas solution “2” is a free

solution, i.e., 2 =F 0 . Integrating Equation (7.25) with respect to y, we obtain:

( ) ( )2 1 1 2 2 1 1 2 2 1d dd

z xz z xz z yz z yz z zdd d

v T v T dy v T v T v F dyx −

− −

∂+ + + = −

∂ ∫ ∫% %% % % (7.26)

Assume that solution “1” , can be expressed as an expansion of SH wave modes, i.e.,

187

1

1

( , ) ( ) ( )

( , ) ( ) ( )

mz z m z

m

mxz xz m xz

m

v v x y a x v y

T T x y a x T y

⎧ = =⎪⎨

= =⎪⎩

∑

∑ (7.27)

where the superscript m designates the thm mode. Also assume that solution “2” is the

thn Lamb wave mode with wavenumber nξ , i.e.,

2

2

2

( , ) ( )

( )

( )

n

n

n

i xnz z

i xnxz xz

i xnyz yz

v x y v y e

T T y e

T T y e

ξ

ξ

ξ

−

−

−

⎧ =⎪

=⎨⎪ =⎩

(forward propagating wave) (7.28)

where the superscript n designates the thn mode. Substitution of Equations (7.28) into

Equation (7.26) yields after rearranging the terms

( ) ( )1 1 1 1

1

( ) ( ) ( ) ( )

( )

n n

n

d di x i xn n n n

z xz z xz z yz z yz dd

di x n

z zd

e v y T v T y dy e v y T v T yx

e v y F dy

ξ ξ

ξ

− −

−−

−

−

⎡ ⎤∂+ + + =⎢ ⎥∂ ⎣ ⎦

= −

∫

∫

% %

%

% %% %

%

(7.29)

Substitution of Equations (7.27) into Equation (7.29) gives

( )

( )1 1 1

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

n

n n

di x n m m n

m z xz z xzm d

ddi x i xn n n

z yz z yz z zdd

e a x v y T y v y T y dyx

e v y T v T y e v y F dy

ξ

ξ ξ

−

−

− −

−−

⎡ ⎤∂+⎢ ⎥∂ ⎣ ⎦

+ + = −

∑ ∫

∫

%

% %

%%

%% %

(7.30)

The integral in Equation (7.30) is the integral that appears in the expression of mnP in

Equation (6.131) from the Section 6.2.1 dealing with the shear horizontal wave

orthogonality. Substitution of Equation (6.131) into Equation (7.30) yields

188

( )1 1 14 ( ) ( ) ( ) ( )n n n

ddi x i x i xn n n

m nm z yz z yz z zdm d

e a x P e v y T v T y e v y F dyx

ξ ξ ξ− − −

−−

∂ ⎡ ⎤− + + = −∂ ⎢ ⎥⎣ ⎦

∑ ∫% % %%% % (7.31)

The summation in (7.31) has only one nonzero term corresponding to the propagating

mode n ( nξ real); hence, Equation (7.31) becomes, after rearrangement and division by

ni xe ξ− % ,

( )1 1 14 ( ) ( ) ( ) ( )dd

n n nnn n n z yz z yz z zd

d

P i a x v y T v T y v y F dyx

ξ−

−

∂⎛ ⎞− = + +⎜ ⎟∂⎝ ⎠ ∫% %% % (7.32)

In Equation (7.32), the first term of the right hand side is the forcing function due to the

boundary conditions at the upper and lower surfaces; the second term is the forcing

function due to the volume sources.

The solution of the ODE has been presented in the generic formulation in Section 7.2.

We will derive the solution of the specific case in Section 7.4.

7.2.2 Lamb waves


( ) ( )2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2

2 1 1 2 2 1 1 2


x x x x y y y y


v F v F v F v F

∂ ∂+ + + + + + +

∂ ∂

= − − − −

% % % %% % % %

% %% %

(7.33)


solution, i.e., 2 =F 0 . Integrating Equation (7.25) with respect to y, we obtain:

( ) ( )

( )

1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1

2 1 2 1

d d

y xy y xy x xx x xx y yy y yy x xy x xy dd

d

y y x xd

v T v T v T v T dy v T v T v T v Tx

v F v F dy

−−

−

⎡ ⎤∂+ + + + + + +⎢ ⎥∂⎣ ⎦

= − −

∫

∫

% % % %% % % %

% %

(7.34)

189

Assume that solution “1” , can be expressed as an expansion of Lamb wave modes, i.e.,

1

1

( , ) ( ) ( )

( , ) ( ) ( )

mx x m x

m

my y m y

m

v v x y a x v y

v v x y a x v y

⎧ = =⎪⎨

= =⎪⎩

∑

∑ (7.35)

1

1

1

( , ) ( ) ( )

( , ) ( ) ( )

( , ) ( ) ( )

mxx xx m xx

m

myy yy m yy

m

mxy xy m xy

m

T T x y a x T y

T T x y a x T y

T T x y a x T y

⎧ = =⎪⎪⎪ = =⎨⎪⎪ = =⎪⎩

∑

∑

∑

(7.36)

where the superscript m designates the thm mode.

Also assume that solution “2” is the thn Lamb wave mode with wavenumber nξ , i.e.,

2

2

( , ) ( )

( , ) ( )

n

n

i xnx x

i xny y

v x y v y e

v x y v y e

ξ

ξ

−

−

⎧ =⎪⎨

=⎪⎩ (forward propagating wave) (7.37)

2

2

2

( )

( )

( )

n

n

n

i xnxx xx

i xnyy yy

i xnxy xy

T T y e

T T y e

T T y e

ξ

ξ

ξ

−

−

−

⎧ =⎪⎪ =⎨⎪

=⎪⎩

(7.38)

where the superscript n designates the thn mode. Substitution of Equations (7.37) and

(7.38) into Equation (7.34) yields after rearrangement of the terms

( )

( )

( )

1 1 1 1

1 1 1 1

1 1

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

n

n

n

di x n n n n

y xy y xy x xx x xxd

di x n n n n

y yy y yy x xy x xy d

di x n n

y y x xd

e v T y v y T v T y v y T dyx

e v T y v y T v T y v y T

e v y F v y F dy

ξ

ξ

ξ

−

−

−

⎡ ⎤∂+ + +⎢ ⎥∂ ⎣ ⎦

+ + + +

= − −

∫

∫

%

%

%

% %% %

% %% %

% %

(7.39)

Substitution of Equations (7.35) and (7.36) into Equation (7.39) gives

190

( )

( )

( )

1 1 1 1

1 1

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

n

n

n

di x m n n m m n n m

m y xy y xy x xx x xxm d

di x n n n n


di x n n

y y x xd

e a x v y T y v y T y v y T y v y T y dyx

e v y T y v y T y v y T y v y T y

e v y F y v y F y dy

ξ

ξ

ξ

−

−

−

⎡ ⎤∂+ + +⎢ ⎥∂ ⎣ ⎦

+ + + +

= − −

∑ ∫

∫

%

%

%

% %% %

% %% %

% %

(7.40)

The first integral in Equation (7.40) is the integral that appears in the expression of mnP in

Equation (6.145) from the Section 6.2.2 dealing with the Lamb wave orthogonality.


( )

( )

1 1 1 1

1 1

4 ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

n

n

n

i xm nm

m

di x n n n ny yy y yy x xy x xy d

di x n n

y y x xd

e a x Px

e v y T y v y T y v y T y v y T y

e v y F y v y F y dy

ξ

ξ

ξ

−

−

∂ ⎡ ⎤− ⎢ ⎥∂ ⎣ ⎦

+ + + + =

− −

∑

∫

%

%

%

% %% %

% %

(7.41)


mode n ( nξ real); hence, Equation (7.31) becomes, after rearrangement,

( )

1 1

1 1

1 1

( ) ( ) ( ) ( )4 ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

dn ny yy y yy

nn n n n nx xy x xy d

dn ny y x x

d

v y T y v y T yP i a x

x v y T y v y T y

v y F y v y F y dy

ξ−

−

⎛ ⎞+∂⎛ ⎞ ⎜ ⎟+ = =⎜ ⎟ ⎜ ⎟∂⎝ ⎠ + +⎝ ⎠

+ +∫

% %%

% %

% %

(7.42)




The solution of the ODE has been presented in the generic formulation in Section 7.2.

We will derive the solution of the specific case in Section 7.4.

191

7.3 PWAS EXCITATION OF CIRCULAR-CRESTED GUIDED WAVES

In this section, we derive the normal mode expansion for circular crested waves. This

provides an extension to the case of the straight crested waves. For circular crested waves

the derivation can not be general since their dependence to the radial component depends

on the type of wave considered. We first present the case for SH waves and then that for

Lamb waves.



( ) ( ) ( )1 2 2 1 1 2 2 1 2 1 1 2r r z zr v T v T r v T v T r v F v F

r zθ θ θ θ θ θ θ θ θ θ θ θ∂ ∂⎡ ⎤+ + + = − +⎣ ⎦∂ ∂

% % %% % % (7.43)


solution, i.e., 2 =F 0 . Integrating Equation (7.43) with respect to z, we obtain:

( ) ( )1 2 2 1 1 2 2 1 2 1d dd

r r z z dd d

r v T v T dz r v T v T r v F dzr θ θ θ θ θ θ θ θ θ θ−

− −

∂ ⎡ ⎤+ + + = −⎣ ⎦∂ ∫ ∫% %% % % (7.44)

Assume that solution “1” can be expressed as an expansion of Lamb wave modes, i.e.,

1

1

( , ) ( ) ( )

( , ) ( ) ( )

mm

m

mr r m r

m

v v r z a r v z

T T r z a r T z

θ θ θ

θ θ θ

⎧ = =⎪⎨

= =⎪⎩

∑

∑ (7.45)

where the superscript m designates the thm mode. Also assume that solution “2” is the

thn Lamb wave mode with wavenumber nξ , i.e.,

192

( )( )( )

20

20

20

( , ) ( )

( , ) ( )

( , ) ( )

nn

nr r n

nz z n

v r z v z J r

T r z T z J r

T r z T z J r

θ θ

θ θ

θ θ

ξ

ξ

ξ

⎧ =⎪⎪ =⎨⎪ =⎪⎩

(7.46)

where the superscript n designates the thn mode. Equation (7.46) is the expression of a

standing wave. To consider only outward propagating modes, we recall the results of

Section 4.2.1.1.2, Equation (4.166) and we write (7.46) through the Hankel functions of

the second kind, i.e.,

( )( )( )

2 (2)0

2 (2)0

2 (2)0

( , ) ( )

( , ) ( )

( , ) ( )

nn

nr r n

nz z n

v r z v z H r

T r z T z H r

T r z T z H r

θ θ

θ θ

θ θ

ξ

ξ

ξ

⎧ =⎪⎪ =⎨⎪ =⎪⎩

(forward propagating wave) (7.47)

where

( ) ( ) ( )(2)0 0 0n n nH r J r iY rξ ξ ξ= − (7.48)

Substitution of Equations (7.47) into Equation (7.44) and rearranging of terms yields

( ) ( )

( )( ) ( )

(2) 1 10

(2) 1 1 (2) 10 0

( ) ( )

( ) ( ) ( )

dn n

n r rd

ddn n n

n z z ndd

rH r v T z v z T dzr

rH r v T z v z T rH r v z F dz

θ θ θ θ

θ θ θ θ θ θ

ξ

ξ ξ

−

−−

⎡ ⎤∂+⎢ ⎥∂ ⎣ ⎦

+ + = −

∫

∫

% % %

% % %% %

(7.49)

Substitution of Equations (7.45) into Equation (7.49) gives

( ) ( )

( )( ) ( )

(2)0

(2) 1 1 (2) 10 0

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

dm n n m

n m r rm d

ddn n n

n z z ndd

rH r a r v z T z v z T z dzr

rH r v T z v z T rH r v z F dz

θ θ θ θ

θ θ θ θ θ θ

ξ

ξ ξ

−

−−

⎡ ⎤∂+⎢ ⎥∂ ⎣ ⎦

+ + = −

∑ ∫

∫

% % %

% % %% %

(7.50)

193

The integral in Equation (7.50) is the integral that appears in the expression of mnP in

Equation (6.157) Section 6.3.1 dealing with the shear horizontal wave orthogonality.


( )

( )( ) ( )

(2)0

(2) 1 1 (2) 10 0

2 ( )

( ) ( ) ( )

n m nmm

ddn n n

n z z ndd

rH r a r Pr

rH r v T z v z T rH r v z F dzθ θ θ θ θ θ

ξ

ξ ξ−

−

∂ ⎡ ⎤−∂ ⎢ ⎥⎣ ⎦

+ + = −

∑

∫

%

% % %% %

(7.51)


mode n ( nξ real); hence, Equation (7.51) becomes, after rearrangement and division by

( )(2)0 nH rξ% ,

( )( )

( )(2)1 1 1 1(2)0

12 ( ) ( ) ( ) ( )ddn n n n

nn n n z z dn d

H rP a r v T z v z T v z F dz

r H r r θ θ θ θ θ θ

ξξ

ξ −−

⎡ ⎤∂− + = + −⎢ ⎥

∂⎢ ⎥⎣ ⎦∫

%% % %

% (7.52)

Note that for large values of r the terms in square brackets in Equation (7.52) becomes

( )( )

(2)1(2)0

1nn i r

n nrn

H re

r H rξξ

ξ ξξ →∞− ⎯⎯⎯→

%

% (7.53)

and Equation (7.52) becomes equal to the expression for straight crested waves given by

Equation (7.32).




A solution of the ODE is found by solving the homogeneous ODE. A solution of the

homogeneous equation is given by

194

( )0 (2)

0

1

n

yrH rξ

=%

(7.54)

The complete solution is given by

( ) ( ) ( )(2) 1 1 1

0(2)0

1( ) ( ) ( ) ( )x dd

n n nn z z d

n c d

a r rH r v T z v z T v z F dz drrH r θ θ θ θ θ θξ

ξ −−

⎡ ⎤= + −⎢ ⎥

⎣ ⎦∫ ∫% % % %

% (7.55)

7.3.2 Lamb waves

Consider now circular-crested Lamb waves propagating in a plate. Recall the complex

reciprocity relation of Equation (5.91), i.e.,

( ) ( )

( )

2 1 1 2 2 1 1 2 1 2 2 1 2 1 1 2

2 1 1 2 2 1 1 2

1r rr r rr z rz z rz z zz z zz r rz r rz

r r r r z z z z


v F v F v F v F

∂ ∂⎡ ⎤+ + + + + + +⎣ ⎦∂ ∂= − + + +

% % % %% % % %

% % % %

(7.56)


solution, i.e., 2 =F 0 . Integrating Equation (7.56) with respect to y, we obtain

( ) ( )

( )

2 1 1 2 2 1 1 2 1 2 2 1 2 1 1 2

2 1 2 1

1 d d

r rr r rr z rz z rz z zz z zz r rz r rz dd

d

r r z zd

r v T v T v T v T dz v T v T v T v Tr r

v F v F dz

−−

−

⎡ ⎤∂+ + + + + + +⎢ ⎥∂ ⎣ ⎦

= − +

∫

∫

% % % %% % % %

% %

(7.57)

Assume that solution “1”, can be expressed as an expansion of Lamb wave modes, i.e.,

1

1

( , ) ( ) ( )

( , ) ( ) ( )

mr r m r

m

mz z m z

m

v v r z a r v z

v v r z a r v z

⎧ = =⎪⎨

= =⎪⎩

∑

∑ (7.58)

195

1

1

1

( )( , ) ( ) ( ) ( )

( , ) ( ) ( )

( , ) ( ) ( )

m mmrr rr m rr r

m m

mzz zz m zz

m

mrz rz m rz

m

a rT T r z a r T z v zi r

T T r z a r T z

T T r z a r T z

ω⎧ = = +⎪⎪⎪ = =⎨⎪⎪ = =⎪⎩

∑ ∑

∑

∑

(7.59)

where the superscript m designates the thm mode.

Also assume that solution “2” is the thn outward Lamb wave mode with wavenumber nξ ,

i.e.,

2 (2)

1

2 (2)0

( , ) ( ) ( )

( , ) ( ) ( )

nr r n

nz z n

v r z v z H r

v r z v z H r

ξ

ξ

⎧ =⎪⎨

=⎪⎩ (7.60)

(2)2 (2) 1

0

2 (2)0

2 (2)1

( )( ) ( ) ( )

( ) ( )

( ) ( )

n n nrr rr n r

nzz zz n

nrz rz n

H rT T z H r v zi r

T T z H r

T T z H r

ξξω

ξ

ξ

⎧= +⎪

⎪⎪ =⎨⎪ =⎪⎪⎩

(7.61)

where the superscript n designates the thn mode. Substitution of Equations (7.60), (7.61)

into Equation (7.57) yields

( ) ( )

(2)(2) 1 1 (2) 11 0

(2) 1 1 (2)0 1

1 1 (2) 1 1 (2)0 1

( )( ) ( ) ( ) ( )1

( ) ( ) ( ) ( )

( ) (

n n nd nr n rr r rr n r

d n nz n rz z rz n

n n n nz zz z zz n r rz r rz

H rv H r T v T z H r T zr dzr

r rv z H r T v T z H r

v T v T H r v T v T H

ξξ ξ

ξ ξ

ξ

−

⎡ ⎤⎛ ⎞⎡ ⎤+ +∂ ⎢ ⎥⎜ ⎟⎢ ⎥⎣ ⎦⎢ ⎥⎜ ⎟∂ ⎜ ⎟⎢ ⎥+ +⎝ ⎠⎣ ⎦

+ + + +

∫%

% % % %%

% % %%

% % % %% %( )

( )(2) 1 (2) 11 0

)

( ) ( ) ( )

d

n dd

n nr n r z n z

d

r

v H r F v z H r F dz

ξ

ξ ξ

−

−

= − +∫ % %% %

(7.62)

Substitution of Equations (7.58), (7.59) into Equation (7.62) gives

196

( ) ( )( )

(2)1

(2)0

1 1 (2) 1 1 (2)0 1

(2)1

( ) ( ) ( ) ( ) ( )1 ( )( ) ( ) ( ) ( ) ( )

( ) ( )

(

n m m ndr rr z rz n

m m n n mm d r rr z rz n

dn n n n

z zz z zz n r rz r rz n d

nr

v z T z v z T z H rr a r dz

r r v z T z v z T z H r

v T v T H r v T v T H r

v H

ξ

ξ

ξ ξ

ξ

−

−

⎡ ⎤⎛ ⎞⎡ ⎤+ +∂ ⎣ ⎦⎢ ⎥⎜ ⎟⎜ ⎟∂ ⎢ ⎥⎡ ⎤+ +⎣ ⎦⎝ ⎠⎣ ⎦

+ + + +

= −

∑ ∫% %%

% %%

% % % %% %

%%( )1 (2) 10) ( ) ( )

dn

n r z n zd

r F v z H r F dzξ−

+∫ %%

(7.63)

The z integral in Equation (7.63) is the integral that appears in Equation (6.169) dealing

with the Lamb wave orthogonality. Hence, the summation in (7.63) has only one nonzero

term corresponding to the propagating mode n ( nξ real); Equation (7.63) becomes, after

rearrangement

( ) ( )( )

( )

(2) (2)1 0

1 1 (2) 1 1 (2)0 1

(2) 1 (2) 11 0

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

nnn n n

dn n n n

z zz z zz n r rz r rz n dd

n nr n r z n z

d

P r H r H r a rr r

v T v T H r v T v T H r

v H r F v z H r F dz

ξ ξ

ξ ξ

ξ ξ

−

−

∂ ⎡ ⎤⎡ ⎤+⎣ ⎦⎣ ⎦∂

+ + + +

= − +∫

% %

% % % %% %

% %% %

(7.64)

To solve Equation (7.64), we consider the particular case of force on top surface and

parallel to the plate only (Figure 7.3), hence 1 0=F and 1 0zzT = , and the traction free

condition; hence, ˆn z d=±⋅ =T z 0 , i.e., 0n n

zz rzT T= = .

Figure 7.3 Lamb waves wave front and external load Trz applied on the surface of the structure

r z

Trz

Trz

wave front

197

Equation (7.64) simplifies to

(2) (2) 1 (2)1 0 1

( )1 ( ) ( ) ( ) ( )nr

n n n rz nnn

v dr H r H r a r T H rr r P

ξ ξ ξ∂ ⎡ ⎤⎡ ⎤+ = −⎣ ⎦⎣ ⎦∂%% % % (7.65)

Perform the derivative with respect to r and divide by (2)1 ( )nH rξ% to obtain

(2) (2) (2)

10 0 0(2) (2) (2)1 1 1

( ) ( ) ( ) ( )1 1 ( )( ) ( ) ( )

nn n n r

n n rzn n n nn

H r H r H r v da r TH r rH r H r r P

ξ ξ ξξξ ξ ξ

⎡ ⎤⎡ ⎤ ⎡ ⎤ ∂− + + + = −⎢ ⎥⎢ ⎥ ⎢ ⎥ ∂⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

% % % %% % %

(7.66)

Note that as r becomes large the term in square brackets becomes

(2) (2) (2)0 0 0(2) (2) (2)1 1 1

( ) ( ) ( ) 21 1 2( ) ( ) ( ) 2

ni rn n nn n

n n n

H r H r H r eH r rH r H r r r

ξξ ξ ξξ ξξ ξ ξ

⎡ ⎤ ∂ ∂⎛ ⎞− + + + = +⎜ ⎟⎢ ⎥ ∂ ∂⎝ ⎠⎣ ⎦

% % %

% % % (7.67)

Solution of the inhomogeneous ODE Equation (7.66) is found by solving the

homogeneous ODE. A solution of the homogeneous ODE is given by

0 (2) (2)1 0

1( ) ( )n n

yr H r H rξ ξ

=⎡ ⎤+⎣ ⎦% %

(7.68)

The complete solution is given by

(2) (2) 11 0(2) (2)

1 0

( )( ) ( ) ( )( ) ( )

rnr

n n n rzcn n nn

v da r r H r H r T drr H r H r P

ξ ξξ ξ

⎡ ⎤= +⎣ ⎦⎡ ⎤+⎣ ⎦∫

% % %% %

(7.69)

7.4 NORMAL MODE EXPANSION MODEL WITH SURFACES FORCES

In this section, we derive the normal mode expansion model when only surfaces forces

are present. For simplicity, we consider straight-crested guided waves as derived in

Section 7.2. However, through the use of the results in Section 7.3, these results can be

extended to the case of circular-crested guided waves.

198

We assume that the volume source are zero, i.e., 1( , )x y =F 0 ; Equation (7.19) becomes

( )1 1 ˆ4 ( )d

nn n n n n dP i a x

xξ

−

∂⎛ ⎞+ = + ⋅⎜ ⎟∂⎝ ⎠v T v T y% %% (7.70)

Where y is the unit vector in the y direction.

Recall that the orthogonality relation (6.112) is obtaining by requiring that the normal

modes of the plate (layer) satisfy the traction free condition; hence,

ˆn y d=±⋅ =T y 0 (7.71)

Using Equation (7.71), we can express the right-hand side of Equation (7.70) as

1 1ˆ ˆn n⋅ ⋅ + ⋅ ⋅v T y v T y%%( ) ( )1 ˆd d

n dd −−= ⋅ ⋅v T y% (7.72)

The boundary conditions (7.72) depend on the number of transducers used and on their

location on the surface of the structure.

7.4.1 One PWAS on the top surface

Consider a finite PWAS of length 2al a= applied at the upper surface and centered at the

origin of the x axis as shown in Figure 7.4

Figure 7.4 Surface forces due to a PWAS bonded on the top surface of the structure.

The transducer is bonded to the structure through an adhesive layer that is able to

transmit only shear stress. The surface tractions for this problem take the form

x

y 2a

τ

199

0

0 1 0 00

xx xy xz xy x

xy yz

xz yz zz yz z

T T T T tT TT T T T t

⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥= = = ⎨ ⎬⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭⎩ ⎭⎣ ⎦ ⎩ ⎭

t (7.73)

where

( ) ( ) ( )

0 z x

x x at x t x

otherwiseτ⎧ ≤⎪= = ⎨

⎪⎩ (7.74)

Equation (7.74) implies that

( )x =t 0 for a x≤ (7.75)

In view of Equation (7.74), the traction force at the upper and lower surfaces can be

expresses as

1 ˆ( , ) ( )x d x⋅ =T y t (upper surface) (7.76)

1 ˆ( , ) 0x d− ⋅ =T y (lower surface) (7.77)

where ( )xt is an externally applied surface traction given by Equation (7.74). Hence, the

right-hand side of Equation (7.70) becomes

( )1 ˆ ( ) ( )d

n ndd x

−⋅ ⋅ = ⋅v T y v t% % (7.78)


4 ( ) ( ) ( )nn n n nP i a x d xx

ξ∂⎛ ⎞+ = ⋅⎜ ⎟∂⎝ ⎠v t% % (7.79)

This is a first order ODE; upon rearranging, we obtain

( ) 1( ) ( ) ( )4

nn n n

nn

a x i a x d xx P

ξ∂+ = ⋅

∂v t% % (7.80)

200

The solution of the ODE expressed by Equation (7.80) is obtained using the integrating

factor method described in Appendix B.4 and used in Section 7.2. Comparison of

Equation (7.80) with the ODE solution Equation (7.22) yields the general solution for one

PWAS on the top surface, i.e.,

( )( ) ( )4

n n

xi x i xn

nnn c

da x e e x dxP

ξ ξ−= ⋅ ∫v t% %% forward wave solution (7.81)


it contains the factor ni xe ξ− . Since the PWAS placed at the x -axis origin is the only

acoustic source, it is apparent that waves will have to emanate outwards from the PWAS.

Inside the PWAS region ( a x a− < < ), the amplitude of the waves will vary with x ;

however, outside the PWAS region, the wave amplitude stays constant. Hence, the

amplitude ( )na x has to satisfy the following boundary condition

( ) 0 for na x x a= ≤ − b.c. on forward wave solution (7.82)

Applying the boundary condition (7.82) to solution (7.81) yields

( )( ) ( ) 04

n n

ai a i xn

nnn c

da a e e x dxP

ξ ξ−

−− = ⋅ =∫v t% (7.83)

Denote by ( )nF x the integrant of the integral in Equation (7.83), i.e.,

( ) ( )ni xnF x e x dxξ−= ∫ t (7.84)

Hence, after rearrangement, Equation (7.83) becomes

( ) ( )F c F a= − (7.85)

201

Equation (7.85) implies that Equation (7.81) can be written with the lower limit c equal

to a− , i.e.,

( )( ) ( )4

n n

xi x i xn

nnn a

da x e e x dxP

ξ ξ−

−

= ⋅ ∫v t% (7.86)

Note that Equation (7.75) implies that the upper limit on the integral in Equation (7.86)

cannot exceed a since the excitation ( )xt vanishes for a x≤ . This means that the

function ( )na x has constant magnitude outside the PWAS excitation region

( ( ) constna x = for a x< ). However, inside the excitation region, the function ( )na x

varies. To cover all possibilities, we write

( ) ( ) for 4

( )( ) ( ) for 4

0 for

n n

n n

ai x i xn

nn ax

i x i xnn

nn a

d e e x dx a xP

da x e e x dx a x aP

x a

ξ ξ

ξ ξ

−

−

−+

−

⎧⋅ <⎪

⎪⎪⎪= ⋅ − ≤ ≤⎨⎪⎪⎪ < −⎪⎩

∫

∫

v t

v t

%

% (forward wave solution) (7.87)

where superscript + signifies waves propagating in the positive x direction. It is apparent

that the forward wave is zero in the rear of the PWAS, since there are no acoustic sources

to generate it before the PWAS.

The above argument can be equally applied to backward propagating waves. Assume that

solution 2 was taken to be a backward wave: instead of Equation (7.28) we would choose

2

2

( , ) ( )

( , ) ( )

n

n

i xnx x

i xny y

v x y v y e

v x y v y e

ξ

ξ

⎧ =⎪⎨

=⎪⎩ (backward propagating wave) (7.88)

202

It is apparent that Equation (7.88) can be obtained from Equation (7.28) by changing nξ

into nξ− . Performing this change in Equation (7.81) yields the backward wave solution,

i.e.,

( )( ) ( )4

n n

xi x i xn

nnn c

da x e e x dxP

ξ ξ−= ⋅ ∫v t% backward wave solution (7.89)

with the boundary condition

( ) 0 for na x a x= ≤ b.c. on backward wave solution (7.90)

Applying the boundary conditions of Equation (7.90) to the backward wave solution of

Equation (7.89) yields the lower limit c equal to a , i.e.,

( ) ( )( ) ( ) ( )4 4

n n n n

x ai x i x i x i xn n

nnn nna x

d da x e e x dx e e x dxP P

ξ ξ ξ ξ− −= ⋅ = − ⋅∫ ∫v vt t% % (7.91)

Hence, the backward wave solution can be written as

0 for

( )( ) ( ) for 4

( ) ( ) for 4

n n

n n

axi x i xn

nnn x

axi x i xn

nn a

a x

v da x e e x dx a x aP

v d e e x dx x aP

ξ ξ

ξ ξ

τ

τ

−−

−

−

⎧ <⎪⎪⎪⎪= − ⋅ − ≤ ≤⎨⎪⎪⎪− ⋅ < −⎪⎩

∫

∫

%

%

(backward wave solution)(7.92)

where superscript − signifies waves propagating in the negative x direction. It is

apparent that the backward wave is zero in front of the PWAS, since there are no acoustic

sources to generate it before the PWAS.

If we are only interested in the waves outside the PWAS excitation region, then the

modal participation factors take the constant magnitude forms

203

( )( ) ( )4

n n

ai x i xn

nnn a

da x e x dx eP

ξ ξ−+

−

⎡ ⎤= ⋅⎢ ⎥

⎣ ⎦∫

v t% forward wave for x a≥ (7.93)

( )( ) ( )4

n n

ai x i xn

nnn a

da x e x dx eP

ξ ξ−−

−

⎡ ⎤= − ⋅⎢ ⎥

⎣ ⎦∫

v t%

backward wave for x a≤ − (7.94)

7.4.2 Two PWAS (upper and lower surface)

Consider the case of two finite PWAS of length 2al a= applied at the upper and lower

surface and centered at the origin of the x axis as shown in Figure 7.5.

Figure 7.5 Surface forces due to a PWAS bonded on the top surface and a second on the

bottom surface of the structure.

The surface tractions for this problem take the form

0

0 1 0 00

xx xy xz xy x

xy yz

xz yz zz yz z

T T T T tT TT T T T t

⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥= = = ⎨ ⎬⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭⎩ ⎭⎣ ⎦ ⎩ ⎭

t (7.95)

where

( )

( , ) ( , ) ( , )0 z x x

x x at x d t x d t x d

otherwiseτ⎧ ≤⎪= = − − = ⎨

⎪⎩ in-phase (7.96)

( )

( , ) ( , )0 z x

x x at x d t x d

otherwiseτ⎧ ≤⎪= ± = ⎨

⎪⎩ out of phase (7.97)

Equations (7.96) and (7.97) implies that

x

y 2a

τ

τ

204

( )x =t 0 for a x≤ (7.98)

In view of Equations (7.96) and (7.97), the traction force at the upper and lower surfaces

can be expresses as:

1 ˆ( , ) ( , )x d x d⋅ =T y t (upper surface) (7.99)

1 ˆ( , ) ( , )x d x d− ⋅ = −T y t (lower surface) (7.100)

where ( , )x d±t is an externally applied surface traction given by Equation (7.95). Hence,

the right-hand side of Equation (7.70) becomes

( )1 ˆ ( ) ( , ) ( ) ( , )d

n n ndd x d d x d

−⋅ ⋅ = ⋅ − − ⋅ −v T y v t v t% % % (7.101)


4 ( ) ( ) ( , ) ( ) ( , )nn n n n nP i a x d x d d x dx

ξ∂⎛ ⎞+ = ⋅ − − ⋅ −⎜ ⎟∂⎝ ⎠v t v t% % (7.102)


[ ]( ) 1( ) ( ) ( , ) ( ) ( , )4

nn n n n

nn

a x i a x d x d d x dx P

ξ∂+ = ⋅ − − ⋅ −

∂v t v t% % (7.103)

where for the case of the two PWAS excitation in phase, Equation (7.103) specializes in

( ) ( ) ( )( ) ( , )4

n n nn n

nn

a x d di a x x dx P

ξ∂ + −+ = ⋅

∂v v t% % in-phase (7.104)

and for the case of the two PWAS excitation out of phase, Equation (7.103) specializes in

( ) ( ) ( )( ) ( , )4

n n nn n

nn

a x d di a x x dx P

ξ∂ − −+ = ⋅

∂v v t% % out of phase (7.105)

205


factor method described in Appendix B.4. It yields

( ) ( ) ( ) for 4

( ) ( )( ) ( ) for 4

0 for

n n

n n

ai x i xn n

nn ax

i x i xn nn

nn a

d d e e x dx a xP

d da x e e x dx a x aP

x a

ξ ξ

ξ ξ

−

−

−+

−

⎧ ± −⋅ <⎪

⎪⎪ ± −⎪= ⋅ − ≤ ≤⎨⎪⎪⎪ < −⎪⎩

∫

∫

v v t

v v t

% %

% % (forward wave solution)(7.106)

where superscript + signifies waves propagating in the positive x direction.

0 for

( ) ( )( ) ( ) for 4

( ) ( ) ( ) for 4

n n

n n

ai x i xn n

nnn x

ai x i xn n

nn a

a x

d da x e e x dx a x aP

d d e e x dx x aP

ξ ξ

ξ ξ

τ

τ

−−

−

−

⎧ <⎪⎪⎪ ± −⎪= − − ≤ ≤⎨⎪⎪ ± −⎪− < −⎪⎩

∫

∫

v v

v v

% %

% %


where superscript − signifies waves propagating in the negative x direction and the ± is

for in-phase and out of phase excitation respectively.

Note that since for in-phase excitation ( ) ( )n nd d= −v v% % and for out of phase excitation

( ) ( )n nd d= − −v v% % , Equations (7.106) and (7.107) become

( ) ( ) for 2

( )( ) ( ) for 2

0 for

n n

n n

ai x i xn

nn ax

i x i xnn

nn a

d e e x dx a xP


x a

ξ ξ

ξ ξ

τ

τ

−

−

−+

−

⎧<⎪

⎪⎪⎪= − ≤ ≤⎨⎪⎪⎪ < −⎪⎩

∫

∫

v

v

%


206

0 for

( )( ) ( ) for 2

( ) ( ) for 2

n n

n n

ai x i xn

nnn x

ai x i xn

nn a

a x


d e e x dx x aP

ξ ξ

ξ ξ

τ

τ

−−

−

−

⎧ <⎪⎪⎪⎪= − − ≤ ≤⎨⎪⎪⎪− < −⎪⎩

∫

∫

v

v

%

%


7.4.3 Shear horizontal waves: normal mode expansion model with surfaces forces

In this section we will refine the model derived in Section 7.4.1 for the case when only

SH waves are excited in the structure by the transducer.

In the case of surface PWAS excitation as described in Equation (7.74), the volume

source is zero 1( , )x y =F 0 . Hence, Equation (7.32) becomes

( )1 14 ( ) ( ) ( )d

n nnn n n z yz z yz d

P i a x v y T v T yx

ξ−

∂⎛ ⎞− = +⎜ ⎟∂⎝ ⎠% %% (7.110)


modes of the plate (layer) satisfy the traction free condition, i.e., ˆ( ) 0n d ⋅ =T y . Hence,

0nyz d

T±

= and the corresponding term in Equation (7.110) vanishes to yield

( )14 ( ) ( )dn

nn n n z yz dP i a x v y T

xξ

−

∂⎛ ⎞− =⎜ ⎟∂⎝ ⎠% % (7.111)

For the surface PWAS that excites shear yzT at the upper surface, Equation (7.74)

indicates that the shear component is zero on the lower surface, 0yz dT

−= , and non-zero

at the upper surface, ( )yz zdT t x= . Hence, Equation (7.111) becomes

4 ( ) ( ) ( , ) ( ) ( )n nnn n n z yz z zP i a x v d T x d v d t x

xξ∂⎛ ⎞− = =⎜ ⎟∂⎝ ⎠% % % (7.112)

207

where ( )zt x is given by Equation (7.74). This is a first-order ODE of the form

( ) ( )( ) ( )4

nn z

n n znn

a x v di a x t xx P

ξ∂+ =

∂%

(7.113)

Integration of Equation (7.113) is done with the integrating factor method described in

Appendix B.4. In fact, the whole process resembles closely the process described in

Equations (7.80) through (7.94). Hence,

( ) ( ) for 4

( )( ) ( ) for 4

0 for

n n

n n

ani x i xz

znn a

xni x i xz

n znn a

v d e e t x dx a xP

v da x e e t x dx a x aP

x a

ξ ξ

ξ ξ

−

−

−+

−

⎧<⎪

⎪⎪⎪= − ≤ ≤⎨⎪⎪⎪ < −⎪⎩

∫

∫

%

% (forward wave solution)(7.114)

0 for

( )( ) ( ) for 4

( ) ( ) for 4

n n

n n

ani x i xz

n znn x

ani x i xz

znn a

a x


v d e e t x dx x aP

ξ ξ

ξ ξ

−−

−

−

⎧ <⎪⎪⎪⎪= − − ≤ ≤⎨⎪⎪⎪− < −⎪⎩

∫

∫

%

%


If we are only interested in the forward solution outside the excitation region, then

( )( ) ( )4

n n

n ai x i xz

n znn a

v da x e t x dx e

Pξ ξ−+

−

⎡ ⎤= ⎢ ⎥

⎣ ⎦∫

% (7.116)

Recalling Equation (7.27), we write the total particle velocity as

( )( , ) ( ) ( )4

n n

n ai x i xz

z nn nn a

v dx y e t x dx y e

Pξ ξ−+

−

⎡ ⎤⎛ ⎞= ⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦∑ ∫v v

% (7.117)

208

where ( )n yv is the velocity modeshape of the thn mode, i.e.,

( ) ( )nn zy v y=v (7.118)

7.4.4 Lamb waves: normal mode expansion model with surfaces forces

In this section we will refine the model derived in Section 7.4.1 for the case when only

Lamb waves are excited in the structure by the transducer.

In the case of surface PWAS excitation as described in Equation (7.74), the volume

source is zero 1( , )x y =F 0 . Hence, Equation (7.42) becomes

( )

4 ( )

( , ) ( ) ( ) ( , ) ( , ) ( ) ( ) ( , )

nn n n

dn n n n


P i a xx

v x y T y v y T x y v x y T y v y T x y

ξ

−

∂⎛ ⎞+⎜ ⎟∂⎝ ⎠

= + + +% %% %

(7.119)


modes of the plate (layer) satisfy the traction free condition, i.e., ˆ( ) 0n d ⋅ =T y . Hence,

0nyy d

T±

= , 0nxy d

T±

= and the corresponding terms in Equation (7.119) vanish to yield

( )4 ( ) ( ) ( , ) ( ) ( , )dn n

nn n n y yy x xy dP i a x v y T x y v y T x y

xξ

−

∂⎛ ⎞+ = +⎜ ⎟∂⎝ ⎠% % (7.120)

For the surface PWAS at the upper surface, Equation (7.74) indicates that the normal

component of the traction force is zero, i.e. 0yy dT

±= , whereas the shear component is

zero on the lower surface, 0xy dT

−= , and non-zero at the upper surface, ( )xy xd

T t x= .

Hence, Equation (7.120) becomes

4 ( ) ( ) ( , ) ( ) ( )n nnn n n x xy x xP i a x v d T x d v d t x

xξ ∂⎛ ⎞+ = =⎜ ⎟∂⎝ ⎠

% % (7.121)

209

where ( )xt x is given by Equation (7.74). This is a first-order ODE of the form

( ) ( )( ) ( )4

nn x

n n xnn

a x v di a x t xx P

ξ∂+ =

∂%

(7.122)

Integration of Equation (7.122) is done with the integrating factor method described in

Appendix B.4. In fact, the whole process resembles closely the process describe in

Equations (7.80) through (7.94). Hence,

( ) ( ) for 4

( )( ) ( ) for 4

0 for

n n

n n

ani x i xx

xnn a

xni x i xx

n xnn a

v d e e t x dx a xP


x a

ξ ξ

ξ ξ

−

−

−+

−

⎧<⎪

⎪⎪⎪= − ≤ ≤⎨⎪⎪⎪ < −⎪⎩

∫

∫

%


0 for

( )( ) ( ) for 4

( ) ( ) for 4

n n

n n

ani x i xx

n xnn x

ani x i xx

xnn a

a x


v d e e t x dx x aP

ξ ξ

ξ ξ

−−

−

−

⎧ <⎪⎪⎪⎪= − − ≤ ≤⎨⎪⎪⎪− < −⎪⎩

∫

∫

%

%

(backward wave solution) (7.124)

If we are only interested in the forward solution outside the excitation region, then

( )( ) ( )4

n n

n ai x i xx

n xnn a

v da x e t x dx e

Pξ ξ−+

−

⎡ ⎤= ⎢ ⎥

⎣ ⎦∫

% (7.125)

Recalling Equation (7.27), we write the total particle velocity as

( )( , ) ( ) ( )4

n n

n ai x i xx

x nn nn a

v dx y e t x dx y e

Pξ ξ−+

−

⎡ ⎤⎛ ⎞= ⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦∑ ∫v v

% (7.126)

210

where ( )n yv is the velocity modeshape of the thn mode, i.e.,

( )

( )( )

nx

n ny

v yy

v y⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

v (7.127)

7.5 NORMAL MODE EXPANSION MODEL WITH VOLUME FORCES

In this section, we consider straight-crested guided waves excited in the structure by

volume forces only. In this case, we assume that the traction source are zero, i.e.,

1n ⋅ =v T 0% ; recall that the orthogonality relation (6.112) was obtained by requiring that

the normal modes of the plate (layer) satisfy the traction free condition; hence,

ˆn y d=±⋅ =T y 0 (7.128)

hence, Equation (7.19) becomes

14 ( )d

nn n n nd

P i a x dyx

ξ−

∂⎛ ⎞+ = ⋅⎜ ⎟∂⎝ ⎠ ∫ v F% (7.129)

Let consider a medium with embedded PWAS at depth py from the x axis as shown in

Figure 7.6. We assume that the thickness of the PWAS is much smaller than the thickness

of the plate; hence we assume that the shear forces are applied at the same y location.

Figure 7.6 Volume forces due to a PWAS embedded in the structure.

x

y

yp

2a

x

y

yp τ

τ

211

For a finite PWAS of length 2al a= embedded in the structure at py and centered at the

origin of the x axis, the volume force takes the form

1

( ) and ( , ) ( , )

0 p

x

x x a y yx y F x y

otherwise

τ⎧ ≤ =⎪= = ⎨⎪⎩

F (7.130)

The product on the right hand side of Equation (7.129) becomes

1

( ) and ( ) ( ) ( )

0

d dpx x

n n p n pd d

x x a y ydy v y dy v y x

otherwise

ττ

− −

⎧ ≤ =⎪⋅ = =⎨⎪⎩

∫ ∫v F% % % (7.131)


4 ( ) 2 ( ) ( )xnn n n n pP i a x v y x

xξ τ∂⎛ ⎞+ =⎜ ⎟∂⎝ ⎠

% (7.132)

In Equation (7.132), we inserted two times the value of the surface forces because the

PWAS contributes on top and bottom surface. Equation (7.132) is a first order ODE;

upon rearranging, we obtain

( )( ) ( ) ( )

2

xn pn

n nnn

v ya x i a x xx P

ξ τ∂+ =

∂

% (7.133)

Solution of the ODE expressed by Equation (7.133) is similar to the one obtained for one

PWAS on top surface (Section 7.4), i.e.,

( )( ) for

2( )

( ) ( ) for 2

0 for

n n

n n

an p i x i x

nn ax

n p i x i xn

nn a

v ye e x dx a x

Pv y

a x e e x dx a x aP

x a

ξ ξ

ξ ξ

−

−

−+

−

⎧<⎪

⎪⎪⎪= − ≤ ≤⎨⎪⎪⎪ < −⎪⎩

∫

∫

t

t

% %

% %

%


212

( )( ) for

2( )

( ) ( ) for 2

0 for

n n

n n

an p i x i x

nn aa

n p i x i xn

nn x

v ye e x dx a x

Pv y

a x e e x dx a x aP

x a

ξ ξ

ξ ξ

−

−

−−

⎧− <⎪

⎪⎪⎪= − − ≤ ≤⎨⎪⎪⎪ < −⎪⎩

∫

∫

t

t

% %

% %

%

% (backward wave solution) (7.135)

213

8 SHEAR LAYER COUPLING BETWEEN PWAS AND STRUCTURE

Scope of this section is to understand how ultrasonic excitation is transmitted from a

PWAS into a thin-wall structure through the adhesive layer (Figure 8.1), and how it is

distributed into the Lamb wave modes excited in the structure.

Crawley and De Luis (1987) developed an analytical model of the coupling between

wafer piezoelectric actuators and thin-wall structural members. The configuration studied

was of two piezoelectric elements bonded on both sides of an elastic structure. They

assumed that the strain distribution in the piezoelectric actuator was a linear distribution

across the thickness (Euler-Bernoulli linear flexural or uniform extension) and developed

a shear lag solution for the interfacial stress τ between the PWAS and the structure. The

shear lag parameter Γ was found to depend on modal repartition number α which took

the value 1α = for symmetric (i.e., axial) excitation and 3α = for antisymmetric (i.e.,

flexural) excitation. This initial analysis was further detailed by Crawley and Anderson

(1990). Giurgiutiu (2005) extended Crawley and de Luis (1987) and Crawley and

Anderson (1990) theory to the case of only one piezoelectric element bonded to the thin-

wall structure by calculating the total effect as a superposition of symmetric and

antisymmetric contributions and found the value of α for a single-sided PWAS

excitation to be 4α = .

Refinements of Crawley and deLuis (1987) and Crawley and Anderson (1990)

approach have been reported in Luo and Tong (2002), Tong and Luo (2003), and Ryu

214

and Wang (2004). Luo and Tong (2002) and Tong and Luo (2003) studied both static and

dynamic solution of a piezoelectric smart beam and introduced the peel stress effect but

still within the limitations of the Euler-Bernoulli theory of bending. Ryu and Wang

(2004) analyzed the interfacial stress induced by a surface-bonded piezoelectric actuator

on a curved beam. They used the variational principle to derive the governing equations

and the boundary conditions, but did not seem to go beyond axial-flexural combination.

Crawley and de Luis (1987) analyzed this situation under the assumption of axial and

flexural waves which correspond to constant and linear displacement distributions across

the thickness, respectively. Such axial and flexural waves are the low-frequency

approximations to lowest-order symmetric and antisymmetric Lamb waves, S0 and A0.

Assume the PWAS has thickness at half-length a , and elastic modulus aE ; the structure

has thickness 2t d= , and elastic modulus E ; the adhesive bonding layer has thickness bt

and shear modulus bG (Figure 8.1), Crawley and de Luis (1987) derived a shear lag

expression that depend on modal repartition number α , stiffness ratio a a

EtE t

ψ = and

shear lag parameter 2 b

b a a

Gt t E

α ψψ+

Γ = , i.e.,

sinh( )cosh

aa ISA

t xx E aa a

ψτ εα ψ

Γ= Γ

+ Γ (8.1)

where the induced-strain ISAε is

31ISA

a

d Vt

ε = (8.2)

and V is the applied voltage and d31 is the piezoelectric constant.

215

Although the solution of Equation (8.1) is only valid at low values of the frequency-

thickness product (i.e., where the axial and flexural wave approximation holds), this

solution has been subsequently used by other authors for describing the shear-lag transfer

at ultrasonic frequencies where the axial and flexural approximation to the S0 and A0

modes no longer holds and where more than these two fundamental modes may be

present (e.g., Giurgiutiu, 2005; Raghavan and Cesnik, 2005). The justification for using

this solution was that simply no better solution exists. Here we overcome the limitations

of the current shear-lag model and derive a generic solution for the ultrasonic excitation

transmitted between a PWAS and a thin-wall structure through an adhesive layer in the

presence of multiple guided Lamb-wave modes.

PWAS

-a +a

x

τ(x)eiωt ta

t=2d

tby=+d

y=-d

Figure 8.1 Interaction between the PWAS and the structure through the bonding layer

8.1 PROBLEM DEFINITION

Assume a PWAS attached to the upper surface of a thin-wall structure as illustrated in

Figure 8.1. The PWAS is subjected to harmonic electric excitation of angular frequency

ω , i.e., i te ω . The PWAS induces a time-harmonic shear stress boundary condition

( )( )

x i t

y

t xe

t xω⎧ ⎫⎪ ⎪= ⎨ ⎬

⎪ ⎪⎩ ⎭t (8.3)

where xt and yt are the surface tractions on the upper surface of the structure,

216

( )

( )0 x

x x at x

otherwiseτ⎧ ≤⎪= ⎨

⎪⎩ (8.4)

( ) 0yt x = (8.5)

The function ( )xτ is the shear stress induced by the PWAS through the bonding layer.

The PWAS excitation induced induces ultrasonic guided Lamb waves in the structure of

time harmonic variation i te ω . We seek solution in terms of an expansion in Lamb wave

modes, i.e.,

1

1

( , ) ( ) ( )

( , ) ( ) ( )

Nn

x n xn

Nn

y n yn

v x y a x v y

v x y a x v y

=

=

⎧ =⎪⎪⎨⎪ =⎪⎩

∑

∑ (velocity) (8.6)

1

1

1

( , ) ( ) ( )

( , ) ( ) ( )

( , ) ( ) ( )

Nn

xx n xxn

Nn

yy n yyn

Nn

xy n xyn

T x y a x T y

T x y a x T y

T x y a x T y

=

=

=

⎧ =⎪⎪⎪

=⎨⎪⎪

=⎪⎩

∑

∑

∑

(stress) (8.7)

where the superscript n designates the thn mode. It is assumed that, at angular frequency

ω , only N Lamb wave modes are present in the thin-wall structure. The modal

participation factors of Equations (8.6), (8.7) are found as

( ) ( ) for 4

( )( ) ( ) for 4

0 for

n n

n n

ani x i xx

nn axn

i x i xxn

nn a

v d e e x dx a xP


x a

ξ ξ

ξ ξ

τ

τ

−

−

−+

−

⎧<⎪

⎪⎪⎪= ⋅ − ≤ ≤⎨⎪⎪⎪ < −⎪⎩

∫

∫

%


217

0 for

( )( ) ( ) for 4

( ) ( ) for 4

n n

n n

axi x i xn

nnn x

axi x i xn

nn a

a x


v d e e x dx x aP

ξ ξ

ξ ξ

τ

τ

−−

−

−

⎧ <⎪⎪⎪⎪= − ⋅ − ≤ ≤⎨⎪⎪⎪− ⋅ < −⎪⎩

∫

∫

%

%

(backward wave solution) (8.9)

where

( )1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4

dn n n n n n n n

nn y xy y xy x xx x xxd

P v y T y v y T y v y T y v y T y dy−

= − + + +∫ % %% % (8.10)

The dimensions of nnP are [ ]nnP =velocity × stress × length = W/m . The modal

participation factors ( )na x calculated with Equations (8.8) and (8.9) should be

nondimensional; upon verification, we get

[ ] velocity( ) velocity stress lengthna x =

× × stress × length = 1 (8.11)

The problem in hand is to establish an expression for ( )xτ . If we know ( )xτ , then

Equation (8.8), (8.9) can be used to find the modal participation factors to be used in the

general solution of Equations (8.6), (8.7). To achieve this, we will first use general

elasticity principles to establish a set of differential equations and then solve these

equations subject to the boundary conditions.

8.2 SHEAR-LAG SOLUTION FOR AXIAL AND FLEXURAL MODES

Assume that in a plate of thickness 2d have been excited two modes, the axial mode and

the flexural mode. Assume also that the flexural stress distribution follows the Bernoulli-

Euler assumption, i.e., plane sections remain plane and perpendicular to the mid-plane.

218

As shown in Figure 8.2a, the axial stress has constant amplitude axσ across the thickness

and the flexural amplitude flexσ and linear distribution, flexyd

σ ; the total stress is given by

superposition, i.e.,

( , ) ( ) ( )ax flexyx y x xd

σ σ σ= + (8.12)

The stress resultants are calculated as

- - -

( ) ( , ) ( ) ( )d d d

x ax flexd d d

yN x x y dy x dy x dyd

σ σ σ+ + +

= = +∫ ∫ ∫ ( )axt xσ= (8.13)

- -

-

( ) ( , ) ( ) ( )

( )

d dz ax flexd d

daxd

yM x x y ydy x x ydyd

x ydy

σ σ σ

σ

+ +

+

⎛ ⎞= = +⎜ ⎟⎝ ⎠

=

∫ ∫

∫2

-

2-

( )

1( ) ( )3

dflexd

dflex flexd

y x dydtdx y dy x

d

σ

σ σ

+

+

+

= =

∫

∫

(8.14)

Figure 8.2 Forces and moments acting in the plate. a) Stress distribution of the axial and

flexural modes. a) Equilibrium of an infinitesimal element.

Equilibrium of the infinitesimal element of Figure 8.2b yields

0.6− 0.4− 0.2− 0

0.4−

0.2−

0.2

0.4

Force

σx(S0)σx(A0)

Moment z zM dM+zM

xN x xN dN+

dx

τ

Plat

e th

ickn

ess t

=1m

m

a) b)

dx

219

0

0x

z

NM d

ττ

′ + =⎧⎨ ′ + =⎩

(8.15)

For the case of a PWAS embedded in the structure, the equilibrium of the infinitesimal

element would be

1

2 02 0

x

z

NM d

ττ

′ + =⎧⎨ ′ + =⎩

(8.16)

where ( ) ( )x

∂ ⋅′⋅ =∂

, and 1d is the distance from the PWAS to the neutral axis of the

infinitesimal element. However, since the derivations are quite similar, we will proceed

considering only the case of a PWAS bonded on the top surface of the plate. Substitution

of Equations (8.13) and (8.14) into Equation (8.15) yields

0

3 0ax

flex

ttσ τσ τ

′ + =⎧⎪⎨ ′ + =⎪⎩

(8.17)

Addition of Equations in (8.17) gives the equation of equilibrium of the infinitesimal

element, i.e.,

( ) 4 0ax flext σ σ τ′ ′+ + = (8.18)

Evaluation of the total stress, Equation (8.12), at y d= gives the total stress in the

structure at the upper surface, i.e.,

( ) ax flex ax flexy dy d

yyd

σ σ σ σ σ=

=

= + = + (8.19)

Substitution of Equation (8.19) into Equation (8.18) gives

0tσ ατ′ + = (8.20)

220

where

1 3 4α = + = (8.21)

In the PWAS, simple equilibrium considerations yield

0a at σ τ′ − = (8.22)

Stress-strain relations in the structure and PWAS are

( )a a a ISA

EE

σ εσ ε ε

=⎧⎪⎨ = −⎪⎩

(8.23)

Substitution of Equation (8.23) into Equations (8.20) and (8.22) yields

00a a a

tEt E

ε ατε τ′ + =⎧

⎨ ′ − =⎩ (8.24)

The shear stress-strain relation in the bonding layer is given by

( )bb a

b

GG u ut

τ γ= = − (8.25)

Differentiating Equation (8.25) with respect to x yields

( ) ( )b ba a

b b

G Gu ut t

τ ε ε′ ′ ′= − = − (8.26)

Equation (8.26) can be solved for aε , i.e.,

ba

b

tG

ε τ ε′= + (8.27)

Substituting Equation (8.27) into Equation (8.24) yields

221

0b

a a a ab

tEtt E t EG

αε τ

τ ε τ

⎧ ′ = −⎪⎪⎨⎪ ′′ ′+ − =⎪⎩

(8.28)

Substitution of the first equation of Equation (8.28) into the second yields the ODE in

terms of the shear transferred from the PWAS to the structure, i.e.,

0ba a a a

b

tt E t EG tE

ατ τ τ′′ − − = (8.29)

Recall that the stiffness ratio is defined as

a a

EtE t

ψ = (8.30)

and the shear lag parameter as

2 b

b a a

Gt t E

α ψψ+

Γ = (8.31)

Hence, Equation (8.29) becomes after rearrangement

2( ) ( ) 0x xτ τ′′ − Γ = (8.32)

This is an ordinary differential equation of the shear transferred from the PWAS to the

structure.

8.2.1 Shear stress differential equation solution

The solution of Equation (8.32) is of the form

1 2( ) sinh coshx c x c xτ = Γ + Γ (8.33)

222

The constants 1c and 2c are determined from the boundary conditions. In our case, the

boundary conditions at x a= ± are

( )

( )0

0a a

a

σ

σ

⎧ ± =⎪⎨

± =⎪⎩ (8.34)

Substitution of Equation (8.33) into Equation (8.34) yields after rearrangement

( )

( ) 0a ISAa

a

ε ε

ε

⎧ ± =⎪⎨

± =⎪⎩ (8.35)

Recalling the definition uε ′= ; hence, Equation (8.35) becomes

( )( ) 0

a ISAu a

u a

ε′⎧ ± =⎪⎨

′ ± =⎪⎩ (8.36)

Equation (8.36) can be used to establish the boundary conditions in terms of τ ; recall the

definition of the shear strain inside the adhesive layer, i.e.,

a

b

u ut

γ −= (8.37)

Upon differentiation, Equation (8.37) becomes

( )1a

b

u ut

γ ′ ′ ′= − (8.38)

Adding up the equations in Equations (8.36) yields

( ) ( )a ISAu a u a ε′ ′± − ± = (8.39)


223

( ) ( ) ( )1 ISAa

b b

a u a u at t

εγ ′ ′ ′⎡ ⎤± = ± − ± =⎣ ⎦ (8.40)

Recalling that bGτ γ= , Equation (8.40) gives

( ) ( ) ISAb b

b

a G a Gt

ετ γ′ ′± = ± = (8.41)

Substitution of Equation (8.33) in to Equation (8.41) yields a system of two algebraic

equations in the unknown constants 1c and 2c , i.e.,

1 2

1 2

cosh sinh

cosh sinh

bISA

b

bISA

b

Gc a c atGc a c at

ε

ε

⎧ Γ Γ + Γ Γ =⎪⎪⎨⎪ Γ Γ − Γ Γ =⎪⎩

(8.42)

Upon solution, we see that 2 0c = and

1 coshb

ISAb

Gct a

ε=Γ Γ

(8.43)

Equation (8.43) gives the constants 1c and 2c needed in Equation (8.33). Substitution of

Equation (8.43) into Equation (8.33) recovers the solution of Crawley and deLuis (1987),

i.e.,

sinh( ) sinhcosh cosh

b ISA a a ISA

b

G a E tx ax xt a a a aε εψτ

α ψΓ Γ

= = ΓΓ Γ + Γ

(8.44)

QED.

8.2.2 Distribution of the shear stress transferred through the bonding layer

Consider the shear stress transmitted by the PWAS to the structure as described through

the shear lag solution with low frequency approximation. From Equation (8.44) we note

224

that the shear stress depend on: bond layer material properties (Gb) and bond layer

thickness (tb); PWAS material properties (Ea) and geometry properties (ta, a); structure

material properties (E) and thickness (t); and the applied voltage. For a typical PWAS

application in our experiment the values of the parameters above are reported in Table

8.1.

Table 8.1 Shear stress parameters

Cyanoacrylate adhesive APC-850 PWAS Aluminum structure

Gb tb Ea ta a E t

1 μm 2 GPa

100 μm 63 GPa 0.2 mm 3.5 mm 70 GPa 1 mm

Figure 8.3 shows how the shear stress distribution varies with PWAS length and

thickness of the bond layer. The stress is rapidly transmitted from the PWAS to the

structure at the end tips of the actuator. How fast and how much shear stress is transferred

depends on the type of PWAS used, of bond layer, and the structure under study.

Figure 8.3 Normalized shear strain as a function of the normalized PWAS position and bond

layer thickness (All other parameters are defined in Table 8.1).

Nor

mal

ized

stra

in τ

(x)

Normalized position

1− 0.5− 0 0.5 1

1−

0.5−

0.5

1

tb = 100 μm tb = 1 μm

225

Hereunder we analyze the effect of the major parameters on the shear stress

transmission from the PWAS to the plate. Figure 8.4 shows how the shear stress

transferred by the PWAS to the structure varies with the different parameters. The desired

transmission is achieved when the transmission is at the tips of the PWAS and maximum

excitation transferred happens. The parameters that influence the relative stiffness

parameter ψ (i.e. E, Ea, t, and ta) influence also the amount of excitation transferred from

the PWAS to the structures. Greater stiffness ratios (structure more rigid than the PWAS)

result in greater transfer of the excitation, however, where the excitation is transmitted

along the PWAS length (closer or not to the tips) is not effected.

The length of the PWAS and the bond layer parameters (Gb and tb) change the

percentage of the PWAS length where the shear stress is transmitted, but they do not

effect the amount of shear transmitted. In particular, an increase of a or of Gb shifts the

transmission of the stress closer to the tips; for low values of the thickness of bond layer,

tb, the shear stress transfer is concentrated more to the tips of the transducer.

As the bond thickness decreases, aΓ increases (see Equation (8.31)). In the limit,

aΓ → ∞ , we can assume that the load transfer takes place at the end of the actuator. This

situation is the ideal bond limit or pin-force model, i.e.,

( ) ( )[ ]0( )x x a x aτ τ δ δ= − − + (8.45)

where δ is the Dirac impulse function and 0τ is obtained from Equation (8.44), i.e.,

0a a ISAE t

aεψτ

α ψ=

+ (8.46)

226

We report a concise summary of the major effect of each parameter on the shear stress

transfer.

Figure 8.4 Effect of the different parameters on the shear stress transmission. The abscissa is

the normalized position of the PWAS length (in the graph is shown only the portion

close to the actuator tip:0.8 to 1.)

0.8 0.85 0.9 0.95

1−

0.8−

0.6−

0.4−

0.2−

0.8 0.85 0.9 0.95

1−

0.8−

0.6−

0.4−

0.2−

0.8 0.85 0.9 0.95

1−

0.8−

0.6−

0.4−

0.2−

0.8 0.85 0.9 0.95

1−

0.8−

0.6−

0.4−

0.2−

0 0.2 0.4 0.6 0.8

1−

0.8−

0.6−

0.4−

0.2−

0.8 0.85 0.9 0.95

1−

0.8−

0.6−

0.4−

0.2−

tb=1μm

Gb=2GPa

a=3.5mm

ta=0.2mm

t=1mm

tb=100μm

0.6% tb=10μm

Gb=200GPa

Gb=20GPa

ta=0.4mm

ta=1mm

Ea=43mm

Ea=63mm

Ea=103mm

a=7.5mm

a=10mm

t=30mm

t=10mm

Nor

mal

ized

stra

in τ

(x)

Nor

mal

ized

stra

in τ

(x)

Nor

mal

ized

stra

in τ

(x)

Nor

mal

ized

stra

in τ

(x)

Nor

mal

ized

stra

in τ

(x)

Nor

mal

ized

stra

in τ

(x)

Normalized position Normalized position



227

Effect of the bond layer stiffness Gb: as the stiffness of the bond layer increases the

transmission of the shear stress is more rapid and closer to the tip of the PWAS; hence

the percentage of the length of the PWAS in which the shear is transmitted decreases

with the stiffness of the bond layer. The total shear stress transmitted does not change

with the bond thickness.

Effect of the bond layer thickness tb: as the thickness of the bond layer increases the

transmission of the shear stress is slower and more distant to the tip of the PWAS; hence

the percentage of the length of the PWAS in which the shear is transmitted increases with

the thickness of the bond layer. However, the total shear stress transmitted does not

change with the bond thickness.

Effect of the PWAS length a: as the length of the PWAS increases, the percentage of the

length of the PWAS in which the shear is transmitted decreases. There is no change in the

shear transferred.

Effect of the PWAS thickness ta: as the thickness of the PWAS increases, both the

stiffness ratio ψ and the induced-strain ISAε decrease as the inverse of the thickness. As

the PWAS thickness increases, the percentage of the length of the PWAS in which the

shear is transmitted does not change significantly, however, the capacity to transmit the

excitation to the structure decreases (An increase of 50% in thickness gives a decrease of

the shear transferred of 24%.)

Effect of the PWAS stiffness Ea: as stiffness of the PWAS increases, the percentage of

the length of the PWAS in which the shear is transmitted does not change significantly,

228

however the excitation transmitted by the sensor increases (For an increase of 40% in

stiffness gives an increase of the shear transferred of 39%).

Effect of the structure thickness t: as the thickness of the structure increases, the

stiffness ratio ψ increases linearly with the thickness. The effect of the structure

thickness does not change percentage of the length of the PWAS in which the shear is

transmitted. The stress transfer increases with the structure thickness; for an increment of

thickness of the 90% (from 1mm to 1 cm) the shear transferred increases of 40%.)

8.2.3 Alternative formulation of shear stress analysis

In this section we derive the shear stress transferred from the PWAS to the structure

through a more generic analysis. We start with the equilibrium of the PWAS; recall


( ) ( ) 0a at x xσ τ′ − = (8.47)

where ( )a xσ is the stress in the PWAS. Stress-strain relation in the actuating PWAS is

( )a a a ISAEσ ε ε= − (8.48)

where ISAε is the actuating induced strain in the PWAS. Substitution of Equation (8.48)

into Equations (8.47) yields

0a a at E ε τ′ − = (8.49)


( )b

b ab

GG u ut

τ γ= = − (8.50)

229

Differentiating Equation (8.50) with respect to x yields

( ) ( )b b

a ab b

G Gu ut t

τ ε ε′ ′ ′= − = − (8.51)

Equation (8.51) can be solved for aε , i.e.,

b

ab

tG

ε τ ε′= + (8.52)


0ba a a a

b

tt E t EG

τ ε τ′′ ′+ − = (8.53)

Stress-strain relation in the structure is

Eσ ε= (8.54)

We assume that the following relation exists at the upper surface ( y d= ) where the

PWAS and the structure interact

( , ) ( ) 0t x d xσ ατ′ + = (8.55)

where ( , )x dσ is the direct stress in the structure, ( )xτ is the shear stress in the shear

layer, and α is a numerical constant to be determined later.

Substitution of Equation (8.54) into Equations (8.55) yields

0tEε ατ′ + = (8.56)

Rearrangement of Equation (8.56) gives

tEαε τ′ = −

(8.57)

230

Substitution of Equations (8.57) into Equation (8.53) yields the differential equation for

τ , i.e.,

0ba a a a

b

tt E t EG tE

ατ τ τ′′ − − = (8.58)

Note that Equation (8.58) is the same as Equation (8.29) derived in Section 8.2, hence we

will obtain the same solution, i.e.,

0sinh( ) sinh

coshb ISA

b

G a xx xt a aετ τΓ

= = ΓΓ Γ

(8.59)

where

01

coshb ISA

b

G at a aετ =

Γ Γ (8.60)

Substitution in Equation (8.31) into Equation (8.59) recovers the solution of Crawley and

de Luis (1987), i.e.,

sinh( )cosh

aa ISA

t xx E aa a

ψτ εα ψ

Γ= Γ

+ Γ (8.61)

Solution of Equation (8.61) is less generic because it depends on α , and assumes

Equation (8.31) to calculate Γ .

8.2.4 Limits of the classic solution

The classic solution derived before was developed by taking into consideration the axial

mode and the flexural mode. At low frequency the first symmetric Lamb wave mode (S0)

approximates the axial mode and its stress distribution is the same as the one assumed,

i.e. constant across the thickness (see Figure 8.2 a). Likewise, at low frequency the first

antisymmetric Lamb wave mode (A0) approximates the flexural mode and its stress

231

distribution is the same as the one assumed, i.e. linear varying across the thickness (see

Figure 8.2 a). As the frequency increases, the stress distributions of the two modes are no

longer linear across the thickness of the plate. Figure 8.5 shows that stress distributions of

the two modes, S0 and A0, at the frequency-thickness product of 780 kHz-mm.

Figure 8.5 Stress distribution of the first symmetric and antisymmetric modes at frequency-

thickness product of 780 kHz-mm.

The approximation of the first two Lamb wave modes with the axial and flexural

wave modes no longer holds. A new shear lag derivation is needed in order to verify the

magnitude of the error when the classic solution is used at high frequencies. In the

formulation of the shear transfer it is the modal repartition number α that defines the

contribution of each mode present in the transfer of the shear from PWAS to structure.

For this motive we are interested to derive a formulation for α for the case of two Lamb

wave modes present at frequencies below the first cut-off frequency. Since α is present

only in the shear lag parameter Γ , we will also seek a direct solution to define this

parameter.

Plat

e th

ickn

ess t

=1m

m

106× 1− 106× 0 1 106×

0.4−

0.2−

0.2

0.4

Force

σx(S0)

σx(A0)

Moment

232

8.3 SHEAR-LAG SOLUTION FOR TWO MODES, ONE SYMMETRIC AND THE OTHER

ANTISYMMETRIC

We want to derive the modal repartition number for the case of only the first symmetric

and first antisymmetric modes present and for any frequency below the A1 cut-off

frequency. To calculate α assume there are two stress modes present, one symmetric,

( )S yσ , and another antisymmetric, ( )A yσ . At this stage we remove the assumption that

the stress distributions are linear, but we retain the Bernoulli-Euler assumption. The total

stress is given by superposition, i.e.,

( , ) ( ) ( ) ( ) ( )S S A Ax y a x y a x yσ σ σ= + (8.62)

The stress resultants are calculated as follows.

8.3.1 Axial force analysis

- - -

( ) ( , ) ( ) ( ) ( ) ( )d d d

x S S A Ad d dN x x y dy a x y dy a x y dyσ σ σ

+ + += = +∫ ∫ ∫ ( )S St a x= Λ (8.63)

where SΛ is a modeshape constant given by

-

1 ( )d

S Sdy dy

tσ

+Λ = ∫ (8.64)

and it represents the average stress distribution. Substitution of Equations (8.63) into

Equation (8.15) yields after rearrangement

1( ) ( ) 0SS

t a x xτ′ + =Λ

(8.65)

233

8.3.2 Bending moment analysis

- -( ) ( , ) ( ) ( )

d dz S Sd d

M x x y ydy a x y ydyσ σ+ +

= =∫ ∫ -

-

( ) ( )

( ) ( ) ( )

dA Ad

dA A A Ad

a x y ydy

a x y ydy td a x

σ

σ

+

+

+

= = Λ

∫

∫ (8.66)

where AΛ is a modeshape constant given by

-

1 ( )d

A Ady ydy

tdσ

+Λ = ∫ (8.67)

and it represents the equivalent stress distribution that gives the same resultant moment.

Substitution of Equation (8.66) into Equation (8.15) yields after rearrangement

1( ) ( ) 0AA

t a x xτ′ + =Λ

(8.68)

8.3.3 Derivation of α

Evaluation of Equation (8.62) at y d= gives the total stress in the structure at the upper

surface, i.e.,

( , ) ( ) ( ) ( ) ( )S S A Ax d a x d a x dσ σ σ= + (8.69)

Differentiation of Equation (8.69) with respect to x yields

( , ) ( ) ( ) ( ) ( )S S A Ax d a x d a x dσ σ σ′ ′ ′= + (8.70)

Multiplication of Equation (8.65) by ( )S dσ and of Equation (8.68) by ( )A dσ followed

by addition yields

[ ] ( ) ( )( ) ( ) ( ) ( ) ( ) 0S AS S A A

S A

d dt a x d a x d xσ σσ σ τ⎡ ⎤

′ ′+ + + =⎢ ⎥Λ Λ⎣ ⎦ (8.71)

234

Substitution of Equation (8.70) into (8.71) and rearrangement yields

( ), ( ) 0t x d xσ ατ′ + = (8.72)

where

- -

( ) ( )( ) ( )1 1( ) ( )

S SA Ad d

S AS Ad d

d dd d

y dy y ydyt td

σ σσ σασ σ

+ += + = +

Λ Λ ∫ ∫ (8.73)

Equation (8.73) gives a generic expression for α that depends only on the stress

distributions ( )S yσ and ( )A yσ . The value of α determined through Equation (8.73) can

be substituted into Equation (8.31) to determine the shear lag parameter Γ .

For low frequency approximation Equation (8.73) reduces to Equation (8.21), i.e., 4. To

prove it, recall that at low frequencies the symmetric mode and the antisymmetric mode

can be approximated with the axial and flexural modes respectively. The expression of

the stress distributions across the thickness is

( )

( ) ( )

S ax

A flex flex

y constyy yd

σ σ

σ σ σ

= =⎧⎪⎨

= =⎪⎩

(8.74)


2

2 --

ax Addax A

ddy dydy

t td

σ σα σ σ ++= +

∫∫ (8.75)

Integrating the integrals in Equation (8.75) we obtain

2

32 23

t tdd d

α = + (8.76)

235

By rearranging the terms, Equation (8.76) reduces to Equation (8.21), i.e.,

1 3 4α = + = (8.77)

8.3.4 Derivation of Γ without using α ; two modes, one symmetric and the other

antisymmetric

We seek an expression for Γ needed in the differential equation for τ without appealing

to Equation (8.55), i.e., without using α . Substitution of Equations (8.70) and (8.54) into

Equation (8.53) gives

[ ]( ) ( ) ( ) ( ) 0b aa a a S S A A

b

t Et E t a x d a x dG E

τ σ σ τ′′ ′ ′+ + − = (8.78)

Substitution of Equations (8.65), (8.68) into Equation (8.78) yields after rearrangement

( ) ( )1 1 0b a a S A

b a a S A

G t E d dt t E t E

σ στ τ⎧ ⎫⎡ ⎤⎪ ⎪′′ − + + =⎨ ⎬⎢ ⎥Λ Λ⎪ ⎪⎣ ⎦⎩ ⎭

(8.79)

Denote

2 ( ) ( )1 1b a a S A

b a a S A

G t E d dt t E t E

σ σ⎧ ⎫⎡ ⎤⎪ ⎪Γ = + +⎨ ⎬⎢ ⎥Λ Λ⎪ ⎪⎣ ⎦⎩ ⎭ (8.80)

Substitution of Equation (8.80) into Equation (8.79) yields the differential equation for

τ , i.e.,

2( ) ( ) 0x xτ τ′′ − Γ = (8.81)

Note that Equation (8.80) is the equivalent of Equation (8.31), and that Equation (8.81) is

the same as Equation (8.32). Of course, Equation (8.80) can be processed to look like

Equation (8.31) by introducing the coefficient α as defined in Equation (8.73).

236

The crux of this approach has been the ability to express ( )Sa x′ and ( )Aa x′ in terms of

( )xτ which, upon substitution in the expression for ( , )x dσ ′ permitted the resolution of ε ′

in the differential equation for τ .

8.3.5 Limits of the solution with two modes present

The solution derived in Section 8.3 was developed for frequency below the cut-off

frequency of the second antisymmetric mode (A1). At frequencies above that value, the

number of modes present in the structure is greater than two. Figure 8.6 shows the stress

distribution of the first three Lamb wave modes for a frequency-thickness product values

below the first symmetric cut-off frequency.

Figure 8.6 Stress distribution of the first three Lamb wave modes (A0, S0, and A1) at

frequency-thickness product of 1600 kHzmm.

The approach used so far for deriving the shear lag solution is no longer valid for the case

of N modes. To prove it, assume that the general solution consists of the superposition of

N generic modes, i.e.,

Plat

e th

ickn

ess t

=1m

m

2 107× 1− 107× 0 1 107× 2 1×

0.4−

0.2−

0.2

0.4

Force

σx(S0)

σx(A0)

Moment

σx(A1)

237

1

( , ) ( ) ( )N

n nn

x y a x yσ σ=

= ∑ (8.82)

where Equation (8.82) represents the xxT component of Equation (8.7). The stress

derivative evaluated at the upper surface y d= is

1

( , ) ( ) ( )N

n nn

x d a x dσ σ=

′ ′= ∑ (8.83)

The stress resultants are calculated as follows

- -

- -

( ) ( , ) ( ) ( ) ( )

( ) ( , ) ( ) ( ) ( )

d d Sx n n n nd d

d d Az n n n nd d

N x x y dy a x y dy t a x

M x x y ydy a x y ydy td a x

σ σ

σ σ

+ +

+ +

⎧ = = = Λ⎪⎨⎪ = = = Λ⎩

∑ ∑∫ ∫∑ ∑∫ ∫

(8.84)

where SnΛ and A

nΛ are modeshape constants given respectively by

-

-

1 ( )

1 ( )

dSn nd

dAn nd

y dyt

y ydytd

σ

σ

+

+

⎧Λ =⎪⎪⎨⎪Λ =⎪⎩

∫

∫, 1,...,n N= (8.85)

Substitution of Equations (8.84) into Equation (8.15) yields after rearrangement

( ) ( ) 0

( ) ( ) 0

Sn n

An n

t a x x

t a x x

τ

τ

⎧ ′Λ + =⎪⎨

′Λ + =⎪⎩

∑∑

(8.86)

Note that the system in Equation (8.86) has N unknowns, i.e., it is 2N − indeterminate.

Hence, it is no possible to solve the system for ( )na x′ .

To obtain the shear lag for the case of N generic modes we will express the stresses as

expansions of Lamb wave modes (Equation (8.7)) where the modal participation factors

expressions are known.

238

8.4 SHEAR-LAG SOLUTION FOR N GENERIC MODES

Recall Equations (8.8) and (8.9) over the interval a x a− ≤ ≤ , i.e.,

( )( ) ( )4

n n

xni x i xx

nnn a

v da x e e x dxP

ξ ξ τ−+

−

= ⋅ ∫% (8.87)

( )( ) ( )4

n n

ani x i xx

nnn x

v da x e e x dxP

ξ ξ τ−− = − ⋅ ∫% (8.88)

Note that the field amplitudes above are for one PWAS bonded on the surface of the

plate. If we are interested in the shear-lag solution when the PWAS is embedded in the

structure we should use instead of Equations (8.8) and (8.9), Equations (7.134) and

(7.135). Substitution of Equation (8.87) and (8.88) into Equation (8.82) gives

1 1

( , ) ( ) ( ) ( ) ( )N N

n n n nn n

x y a x y a x yσ σ σ+ −

= =

= +∑ ∑ (8.89)

Recall the stress-strain relation for the structure as defined in Equation (8.54) and apply it

to Equation (8.89) to get the strain in the structure at the upper surface, ε , i.e.,

1 1

1( ) ( ) ( ) ( ) ( )N N

n n n nn n

x a x y a x yE

ε σ σ+ −

= =

⎡ ⎤= +⎢ ⎥⎣ ⎦∑ ∑ (8.90)

Differentiation of Equation (8.90) with respect to x gives the strain rate ε ′ , i.e.,

1 1

1( ) ( ) ( ) ( ) ( )N N

n n n nn n

x a x d a x dE

ε σ σ+ −

= =

⎡ ⎤′ ′ ′= +⎢ ⎥⎣ ⎦∑ ∑ (8.91)

Taking the derivative of Equation (8.87) and (8.88) with respect to x gives (see

Appendix B.5)

239

( ) ( )( ) ( ) ( )4 4

n n

xn ni x i xx x

n nnn nna

v d v da x i e e x dx xP P

ξ ξξ τ τ−+

−

′ = − +∫% % (8.92)

( ) ( )( ) ( ) ( )4 4

n n

an ni x i xx x

n nnn nnx


ξ ξξ τ τ−− = − +∫% % (8.93)

Substitution of Equation (8.92) and (8.93) into Equation (8.91) yields

1

( )1( ) ( ) ( ) 2 ( ) ( )4

n n n n

x anNi x i x i x i xx

n n nn nn a x

v dx i e e x dx i e e x dx x dE P

ξ ξ ξ ξε ξ τ ξ τ τ σ− −

= −

⎧ ⎫⎡ ⎤⎪ ⎪′ = − − +⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

∑ ∫ ∫%

(8.94)

Recall the shear relation as defined in Equation (8.53), i.e.,

0ba a a a

b

tt E t EG

τ ε τ′′ ′+ − = (8.95)

Substitution of Equation (8.94) into Equation (8.95) gives

1

( )( ) ( ) 0

4( ) 2

n n

n n

xi x i x

nnNab a a x

a a nanb nn i x i x

nx

i e e x dxt t E v dt E dG E P

i e e x dx

ξ ξ

ξ ξ

ξ ττ σ τ

ξ τ τ

−

−

= −

⎡ ⎤−⎢ ⎥

⎢ ⎥′′ + − =⎢ ⎥⎢ ⎥− +⎢ ⎥⎣ ⎦

∫∑

∫

% (8.96)

Factoring ( )xτ out of the last two terms of Equation (8.96) and dividing by ba a

b

tt EG

leads

to

1 1

( ) ( )( ) ( ) ( ) ( ) 1 ( ) 02

nN Nb b a a x n

n n n nn nb a b a nn

G G t E v d dx i a x a x d xt E t t E E P

στ ξ σ τ+ −

= =

⎡ ⎤′′ ⎡ ⎤− − + − =⎢ ⎥⎣ ⎦

⎣ ⎦∑ ∑ %

(8.97)

Denote the modal repartition number as

1

( ) ( )2

nNa a x

nn nn

t E v d dE P

α σ=

= − ∑ % (8.98)

240

hence the shear lag parameter is defined in Equation (8.31). Moreover, define

( ) ( )4

nb n x n

nb nn

G v d dt E P

ξ ση =%

(8.99)

Note that α calculated with Equation (8.98) is adimensional since

[ ] stress velocity stresslength =1stress velocity stress length

α ×= × ×

× × (8.100)

Substitution of Equations (8.98), (8.31), and (8.99) into Equation (8.97) yields the

differential equation for τ , i.e.,

2

1( ) ( ) ( ) ( ) 0n n n n

x aNi x i x i x i x

nn a x

x x i e e x dx e e x dxξ ξ ξ ξτ τ η τ τ− −

= −

⎡ ⎤′′ − Γ − + =⎢ ⎥

⎣ ⎦∑ ∫ ∫ (8.101)

8.4.1 Low frequencies approximation

For low frequencies, long wavelength motion, the wavenumber approaches zero

( 0s Aξ ξ= → ). Hence

( ) ( ) 04

nb x n

n nb nn

G v d dt E P

ση ξ= =%

(8.102)

The third term in Equation (8.101) can be ignored. The equation of the stress equilibrium

(8.101) becomes

2( ) ( ) 0x xτ τ′′ − Γ = (8.103)

Equation (8.103) is equal to Equation (8.32) if we can prove that the modal repartition

number Equation (8.103) is equal to that of Equation (8.21). Hence, we should prove that

1

( ) ( ) 42

nNx

nn nn

v dt dP

σ=

− =∑ % (8.104)

241

At low frequency the stress across the thickness can be expressed as

( )

( ) ( )

S ax

A flex flex

y constyy yd

σ σ

σ σ σ

= =⎧⎪⎨

= =⎪⎩

(8.105)

We also assume that at the low frequencies

( ) 0

( ) 0

axy

flexxy

v y

yσ

⎧ =⎪⎨

=⎪⎩ (8.106)

For propagating modes, the average power flow for the nth mode is expresses as

1Re ( ) ( ) ( ) ( )2

dn n n n

nn x x y xyd

P v y y v y y dyσ σ−

⎡ ⎤= − +⎣ ⎦∫ % % (8.107)

Substituting the expressions of the stresses at low frequencies in Equation (8.107), we

obtain

2

2

Re symmetric mode2

Re antisymmetric mode2

dax ax

SSd

dflex flex

AAd

vP dy

v yP dyd

σ

σ−

−

⎧= −⎪

⎪⎨⎪ = −⎪⎩

∫

∫

%

% (8.108)

Integrate Equations (8.108) to get

Re symmetric mode

2

Re antisymmetric mode6

ax axSS

flex flexAA

tvP

tvP

σ

σ

⎧ = −⎪⎪⎨⎪ = −⎪⎩

%

% (8.109)

Substitute Equations (8.105), (8.106), and (8.109) into Equation (8.104) to get

( ) ( )

3 4Re Re

flex flexax ax

ax ax flex flex

vvv v

σσσ σ

+ =%%

% % (8.110)

242

Note that

( )

( ) ( )( )

( )( )

Re Im Im1

Re Re Rea a aa

a a a−

= = −%

% (8.111)

If Rea ∈ Equation (8.111) reduces to

( )

1Re

aa

=%

% (8.112)

If Ima ∈ Equation (8.111) reduces to

( )Rea

a= ∞

%

% (8.113)

For the case of velocity and stresses situation of Equation (8.112) is always verified.


4 4= (8.114)

If we remove assumptions in Equation (8.106) we notice that at low frequencies

( ) ( ) ( ) ( )flex flex flex flex flexx x y xy y flex

yv y y v y y vd

σ σ σ= =% % % (8.115)

Hence the second equation in Equation (8.108) becomes

2

2Re Re3

dflex flex

AA flex flexd

tvyP v dyd

σσ

−

⎡ ⎤= − = −⎢ ⎥

⎣ ⎦∫

%% (8.116)

Substitute the first of Equation (8.109) and Equation (8.116) into Equation (8.104) to

obtain

( ) ( )

3 1 1.5 2.5 4Re 2 Re

flex flexax ax

ax ax flex flex

vvv v

σσσ σ

+ = + = ≠%%

% % (8.117)

243

The discrepancy between the exact solution and the approximate solution can be

explained by recalling that the approximate solution had as assumption that the flexural

mode had a linearly varying Bernoulli-Euler strain distribution across the thickness, i.e.,

plane sections remain plane and perpendicular to the mid plane. No Jourawski strain

distribution is considered.

8.4.2 Modal repartition number

In the previous sections we have derived the modal repartition number through three

different methods: classic or α at low frequencies approximation; α for frequencies

below first cut-off frequency (A0 and S0 only); and α for any frequency (N modes

present). Each derivation gave a new formulation for the repartition number. We have

already shown the behavior of the three formulations for the frequency that is close to

zero. In this section, we will present how the modal repartition number behaves as the

frequency increases.

Figure 8.7a shows how the modal repartition number (α ) varies with frequency. Line

I is the modal repartition number derived from the classic solution; its value is a constant

independent of the frequency. Line II represents the repartition number when only the

first symmetric and antisymmetric modes are present (as derived in Equation (8.73)). The

repartition number is equal to 4 at low frequencies and then it increases with the

increasing of frequency. It is interesting to note that, while the symmetric mode

contribution to the repartition number does not change with frequency (line II in Figure

8.7b), the variation of α is entirely due to the contribution of the antisymmetric mode

(line II in Figure 8.7b). For frequency values beyond ~800 kHz-mm this solution is no

longer valid because a third mode, A1, is present.

244

Figure 8.7 Repartition mode number as a function of frequency. (I): classic solution 4α = ; (II):

2 modes solution Equation (8.73) for S0 and A0; (III): N generic modes solution

Equation (8.98) for S0, A0, and A1; (IV): N generic modes solution Equation (8.98)

for S0, A0, and A1 and contribution form shear stress equal to zero in the power

flow. a) Total repartition number. b) Repartition number divided between α for the

antisymmetric modes and α for the symmetric modes.

Line III in figure is the value of the repartition number derived with the N generic

modes formulation. From 0 to ~800 kHz only S0 and A0 are present, at the cut-off

frequency, the repartition number value has a discontinuity that is due to the presence of

A1 (only the reparation number contribution from the antisymmetric modes in Figure

8.7b has this discontinuity). The discontinuity is an artifact of the model that will be

explained in more details later. The starting value of the repartition number is 2.5 and its

value decreases as the frequency increases. The presence of A1 increases the total value

of the repartition number but there is almost no change in curvature. If the product

( ) ( )n ny xyv y yσ% in the antisymmetric power flow derivation of Equation (8.107) is set to

zero, the repartition number curve changes significantly (line IV). The starting value is 4

as for the classic derivation and the repartition number increases with the frequency

0 200 400 600 800

1

2

3

4

α

( )fd kHzmm ( )fd kHzmm

α

0 200 400 600 800

2

4

6

8

a) b)

(I)(II)

(III)

(IV) Antisymmetric modes

Symmetric modes

245

increase. However, curves IV and II are still different; the explanation for this

discrepancy is still to be found.

In the case of N generic modes present, the contribution due to the symmetric mode is

no longer constant with frequency, but decreases with it, on the other end, the

contribution due to the antisymmetric mode gives an almost constant repartition number.

The effect is the opposite of what as been found with the simplified derivation for two

modes; the symmetric mode effects the variation of the total repartition number.

We showed that in the N generic modes derivation (curve III in Figure 8.7) there is a

discontinuity of the modal repartition number at the cut-off frequency. To explain this

behavior, we first observe the dispersion curves in the frequency-velocity plane, Figure

8.8a.

Figure 8.8 Dispersion curves for an aluminum plate. a) Frequency-phase velocity plane; b)

Wavenumber-radial frequency plane. Left plane imaginary wavenumbers, right plane

real wavenumbers.

Figure 8.8a shows the dispersion curves for an aluminum plate in the frequency-phase

velocity plane. At frequency below ~800 kHz-mm only A0 and S0 modes are present, at

c/c s

ωd

(kH

z-m

m)

fd (kHz-mm) 6 4 2 0 2 4 6

Im(ξd) Re(ξd)a) b)

A0

A1 A2 A3

S0

S1 S2 S3

S0

A0

S1

S2

S3

A1

A3

246

frequency above ~800 kHz-mm, the second antisymmetric mode A1 appears. The phase

velocity of A1 goes at infinity as the frequency approaches the cut-off frequency.

Figure 8.8b shows the dispersion curves in the wavenumber-radial frequency plane.

The right plane of the picture represents real wavenumbers and the lines shown are the

same of Figure 8.8a transformed in the new plane. On the left side of the picture are

shown the imaginary dispersion curves. The wave, for imaginary wavenumber, is an

evanescent wave, the exponential term i xe ξ becomes xe ξ− and the wave amplitude

decreases rapidly with increasing x. At low radial frequencies (hence low frequencies)

only one imaginary evanescent mode is present and it is antisymmetric. The evanescent

mode “connects” the A0 and A1 modes in the imaginary plane.

The modal repartition number could be derived taking into consideration the

contribution from the imaginary antisymmetric mode at low frequencies.

Figure 8.9a shows the total repartition number for the different formulations. The

presence of the imaginary A1 in the derivation (Line III) has removed from the graph the

discontinuity at the cut-off frequency. From Figure 8.9b we see that the A1

antisymmetric repartition number is uniform , the sum of the A0 and A1 repartitions

number across the frequency is a constant equal to the one derived in the classic

formulation ( 0 1 3A Aα α+ = ). The variation in the total repartition number in Figure 8.9a

(line III) is due only to the symmetric mode.

247

Figure 8.9 Repartition mode number as a function of frequency with imaginary A1. (I): classic

solution 4α = ; (II): two modes solution, Equation (8.73), for S0 and A0; (III): N

generic modes solution Equation (8.98) for S0, A0, and A1. a) Total repartition

number. b) Repartition number divided between α for the antisymmetric modes and

α for the symmetric modes. (IV): contribution from A1.

8.4.3 Approximate theory for plate flexural waves

In the derivation of the shear lag parameters for the case of only two modes present

(Section 8.3) it was retained the Bernoulli-Euler assumption. In this paragraph we will

show how the repartition number changes if this assumption is removed.

In Section 3.1.7.2 we have shown that under the Euler-Bernoulli assumption the shear

forces are equal to zero, but if the assumption is removed this is not true anymore. Let

consider a plate subjected to bending.

Let assume that the shear is not equal to zero (we discard the Euler-Bernoulli

assumption). For simplicity let consider a differential element of length dx and unit width

subjected to bending. Figure 8.10a shows the stress distribution across the thickness.

0 200 400 600 800

1

2

3

4

5

α

Antisymmetric modes

α

Symmetric modes 0 200 400 600 800

2

4

6

8

( )fd kHzmm

(I)

(II)

(III)(IV)

a) b) ( )fd kHzmm

248

Figure 8.10 Differential element of length dx. a) normal stress due to bending; b) Shear force

due to the presence of stress.

The stress distribution varies in the x direction; we can assume that after a length dx

the stress is incremented by a quantity xdσ . If we consider an element of length dx and

thickness d y− (Figure 8.10b), the net result of the stress across the thickness is a force F

given by

( )d d

x x x xy yF d dy d dyσ σ σ σ= + − =∫ ∫ (8.118)

Figure 8.11 Forces and moments on a plate. a) Shear stress sign convention. b) Moment

balance.

y

σx σx+d σx

x

dx

F F

τyx

τyx

x

τ(x)

Q

Q M

M+dMb)a)

dx τxy

τxyd-y

d+y

y

d σx σx+d σx

x y

σxσx+d σx

x

dx dx

F F

a) b)

d-y

249

On the face of the element of length dx and thickness d y+ , there is a force of same

magnitude F but opposite direction. The external forces F are balanced by the shear stress

distribution along the dx length (see Figure 8.11a), i.e.

yxdx Fτ = − (8.119)

Hence the shear stress can be derived by substituting Equation (8.118) into Equation

(8.119) (Note that xy yxτ τ= ), i.e.,

( , )d xxy y

x y dyx

στ ∂= −

∂∫ (8.120)

The normal stress distribution across the thickness can be expresses as

( )( , ) ,x xx y x d y dσ σ= (8.121)

By plugging Equation (8.121) into Equation (8.120) we obtain the expression of the

Jourawski shear stress

2

2( , ) ( , )( , ) 1

2dx x

xy y

x d x dy d yx y dyx d x d

σ στ⎛ ⎞∂ ∂

= − = − −⎜ ⎟∂ ∂ ⎝ ⎠∫ (8.122)

In Section 8.3 we made the Bernoulli-Euler assumption and we neglected the presence of

the Jourawski shear stress. If we consider the first antisymmetric Lamb wave mode at

low frequencies, we see that although small, the shear stress is present and its distribution

across the plate thickness is exactly the Jourawski shear stress derived in (8.122). Figure

8.12 shows the normal and shear stress distributions of the first antisymmetric mode.

Their slope and magnitude are exactly those derived by Equations (8.121) and (8.122).

The component of the force due to the presence of shear is quite small compared with the

couple from the normal stress.

250

Figure 8.12 Normal stress distribution and shear stress distribution across the plate thickness for

the first antisymmetric mode for three different frequencies. Solid black line: normal

stress ( , )x x yσ as in Equation (8.121); Dot line: shear stress ( , )xy x yτ as in

Equation (8.122); Dash-dot line: normal stress distribution of A0 as from Equation

(3.31); Dash line: shear stress distribution of A0 as from Equation (3.31).

As the frequency increases, the contribution from shear increases, but the exact shear

distribution still does not differ from the approximate Jourawski stress. At frequency-

thickness product equal 110 kHz-mm, the difference between the resultant force of the

exact shear and the resultant force of approximate one is 10%; at frequency-thickness

product equal 180 kHz-mm, the difference between the resultant momentum of the exact

normal stress and the resultant force of approximate one is 10%. At higher frequencies

not only the magnitude of the resultant force or momentum are different, but also the

slopes of the distributions. Consider the net resultant force at low frequency (around 15

kHz-mm) due to the A0 shear stress, this is equal to about 1.5 MPa (1.5% difference

between exact and approximate) while the net momentum due to the normal stress is

2− 106× 0 2 106×

0.5−

0.5

4 103× 2− 103× 0 2 103× 4 103×

0.5−

0.5

0.02− 0 0.02

0.5−

0.5

Plat

e th

ickn

ess (

mm

)

Stress (GPa) Stress (GPa) Stress (GPa)

1 kHz mmfd = − 110 kHz mmfd = − 1000 kHz mmfd = −

251

equal to about 2 MPa-mm (0.7% difference between exact and approximate), the two

values are close enough justify the need to remove the Bernoulli-Euler assumption.

To derive the modal repartition number when the Jourawski shear is present, assume

the displacement to be of the form

2

2

( , , )

( , , )

xvu y x z t yx

yv v x z td

ψ ∂⎧ = =⎪ ∂⎪⎨⎪ = −⎪⎩

(8.123)

With this assumption, the shear stress becomes

2

21xyu v v yy x x d

τ μ μ⎛ ⎞∂ ∂ ∂⎛ ⎞= + = −⎜ ⎟⎜ ⎟∂ ∂ ∂ ⎝ ⎠⎝ ⎠

(8.124)

The shear stress resultant is a force in the y direction. This is written as

2

2( , ) ( , )1

2 3

d dx x

x xyd d

x d x dd y tdQ dy dyx xd

σ στ− −

⎛ ⎞∂ ∂= = − − = −⎜ ⎟∂ ∂⎝ ⎠∫ ∫ (8.125)

The normal stress resultant is a net zero force in the x direction and a moment M, i.e.

( ) ( )2

( , ) , ,3

d d

x x x xd d

y tdM x y ydy x d dy x dd

σ σ σ− −

= = =∫ ∫ (8.126)

The moment due to the external force is

( )2tdM x dxτ τ= (8.127)

The balance of the shear force, the bending moment, and the external force due to the

PWAS gives (see Figure 8.11b)

0M dM M Qdx dMτ+ − − + = (8.128)

252

or

0dMdM Qdx dx

τ− + = (8.129)

Substituting Equations (8.125), (8.126), and (8.127) into the equilibrium expression we

obtain

( ), ( , ) ( ) 03 3 2

x xd x d x dtd td t xdx x

σ σ τ∂+ + =

∂ (8.130)

Simplify expression (8.130) to get the final expression

( , ) 3 ( ) 02

x x dt xx

σ τ∂+ =

∂ (8.131)

The shear lag coefficient α is now equal to half the shear lag coefficient without the

Jourawski shear stress.

Procedure above can be generalized for a generic antisymmetric stress distribution

( , , )x x y tσ . The shear force and the moment expressions are respectively:

-

( )( , )( ) ( , )

( )

d d

xd d yx

xydx

y dydyx dQ x x y dyx d

σσσ

σ−∂

= = −∂

∫ ∫∫ (8.132)

( )

( ) ( , ) ( , )( )

d

xdd

z x xxd

y ydyM x x y ydy x d

d

σσ σ

σ−

−

= =∫

∫ (8.133)

Substitution of Equations (8.132) and (8.133) into Equation (8.129), yields

253

2( , ) ( ) ( ) 0

2 ( ) ( )

x xd d

x xd y

x d t dt xx

y y y dy dy

σ σ τσ σ

−

∂+ =

∂ ⎡ ⎤+⎢ ⎥

⎢ ⎥⎣ ⎦∫ ∫

(8.134)

Denote the shear lag parameter as

2 ( )1

2( ) ( )

xd d d

x xd d y

t d

y ydy y dy dy

σασ σ

− −

=⎡ ⎤

+ ⎢ ⎥⎢ ⎥⎣ ⎦

∫ ∫ ∫ (8.135)

Consider the shear stress distribution due to the antisymmetric lamb wave mode, i.e.

( )2 2 2( ) 2 sin( ) 2 sin( )Ax y y D yσ μ ξ β α α ξβ β⎡ ⎤= − + − −⎣ ⎦ (8.136)

The first integral in Equation (8.135) is

2 2 22 sin( )cos( )( ) 2

sin( )2 cos( )

dAx

d

dd dy ydy

dD d d

ξ β α ααα ασ μ

βξ ββ

−

⎡ ⎤+ − ⎛ ⎞−⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥=⎢ ⎥⎛ ⎞− −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

∫ (8.137)

The shear stress component function of y is

( ) ( )2 2 2 cos( ) cos( )( ) 2 2 cos( ) cos( )d

Ax

y

d yy dy D d yα ασ μ ξ β α ξ β βα−⎡ ⎤= + − − −⎢ ⎥⎣ ⎦∫ (8.138)

With the use of (8.138), the second integral in Equation (8.135) becomes, after

rearrangement

2 2 22 sin( )cos( )( ) 2

sin( )2 cos( )

d dAx

d y

dd dy dydy

dD d d

ξ β α ααα ασ μ

βξ ββ

−

⎡ ⎤+ − ⎛ ⎞−⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥=⎢ ⎥⎛ ⎞− −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

∫ ∫ (8.139)

254

Note that the two integrals (Equation (8.137) and (8.139)) are equal, i.e.,

( ) ( )d d d

A Ax x

y d y

y ydy y dy dyσ σ−

⎡ ⎤= ⎢ ⎥

⎢ ⎥⎣ ⎦∫ ∫ ∫ (8.140)

Hence Equation (8.134) can be simplified as

( , ) ( ) 0x x dt xx

σ ατ∂+ =

∂ (8.141)

with modal repartition number

2 ( )

4 ( )

xd

xd

t d

y ydy

σασ

−

=

∫ (8.142)

Figure 8.13 shows how the alpha parameter changes with frequency and the effect of

removing the Bernoulli-Euler assumption from derivation in Section 8.3.

Figure 8.13 Repartition mode number due to the A0 mode as a function of frequency. (a): classic

solution with Bernoulli-Euler assumption, 3α = ; (b): 2 modes solution Equation

(8.73) for A0; (c): N generic mode solution Equation (8.98) for A0; (d): 2 modes

solution Equation (8.142) for A0 (normal + shear stress).

α

( )fd kHzmm0 200 400 600 800

1

2

3

4

(a)

(b)

(c)

(d)

255

The repartition mode number with the addition of the shear stress (dash-dot green line)

starts at the low frequencies from 1.5 as derived for the more generic solution (dash blue

line). The repartition number still increases with the frequency as it was found for the

derivation for two modes A0 and S0, but its increment is less than before (before the A1

cut-off frequency the difference in repartition number increment was about 36%). The

discrepancy between the new derivation and the exact for N generic mode lies in the fact

that we assumed that the shear stress distribution is exactly that derived by Jourawski,

beyond fd=110 kHz-mm (solid black vertical line) the force difference between exact and

approximate shear is above 10%, while the difference in repartition number (dash-dot

versus dash line) is of 6%.

8.4.4 Numerical solution of the integral-differential equation

We want to solve the shear stress equilibrium of the bonding layer in the presence of N

wavemodes. Recall the general expression as in Equation (8.101), i.e.

2

1( ) ( ) ( ) ( ) 0n n n n

x aNi x i t i x i t

nn a x

x x i e e t dt e e t dtξ ξ ξ ξτ τ η τ τ− −

= −

⎡ ⎤′′ − Γ − + =⎢ ⎥

⎢ ⎥⎣ ⎦∑ ∫ ∫ (8.143)

with boundary conditions

( )

( )0 0

ISAb

b

a Gt

ετ

τ

⎧ ′ ± =⎪⎨⎪ =⎩

(8.144)

To solve Equation (8.143), we apply the variational iteration method (VIM) as shown in

He (2007) and Wang and He (2007).

The correction functional is

256

2

10

1

( ) ( )

( ) ( ) ( )( ) ( )n n n n

i ixx aN

i i i x i t i x i tn

n a x

s s

x x si e e t dt e e t dtξ ξ ξ ξ

τ τ

τ τ λη τ τ+ − −

= −

⎡ ⎤′′ − Γ −⎢ ⎥

= + ⎛ ⎞⎢ ⎥− +⎜ ⎟⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∫ ∑ ∫ ∫ (8.145)

where iτ% is considered as restricted variations, which means 0iτ =% . λ is a general

Lagrangian multiplier which can be identified optimally via the iteration theory.

First, we find the Lagrangian multiplier by calculating variation with respect to iτ ,

noticing that 0iδτ = , yields

2

10

1

( ) ( )

( ) ( ) ( )( ) ( )n n n n

i ixx aN


n a x

s s

x x s dsi e e t dt e e t dtξ ξ ξ ξ

τ τ

δτ δτ δ λη τ τ+ − −

= −

⎡ ⎤′′ − Γ −⎢ ⎥

= + ⎡ ⎤⎢ ⎥− +⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦

∫ ∑ ∫ ∫% % (8.146)

or,

21

0

22

20

( ) ( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

x

i i i i

x

i i i is xs x

x x s s s ds

d s d sx s s s s s dsds ds

δτ δτ δ λ τ τ

λ λδτ λ δτ δτ λ δτ

+

==

⎡ ⎤′′= + − Γ⎣ ⎦

⎡ ⎤′= + − + − Γ⎢ ⎥

⎣ ⎦

∫

∫ (8.147)

We obtain the following stationary conditions:

22

2( , ) ( , ) 0

( , )1 0

( , ) 0s x

s x

d x s x sdsd x s

dsx s

λ λ

λ

λ=

=

⎧− Γ =⎪

⎪⎪

− =⎨⎪⎪ =⎪⎩

(8.148)

Solve the above differential equation to find the Lagrange multiplier. The general

solution is

257

( , ) sinh ( ) cosh ( )x s A s x B s xλ = Γ − + Γ − (8.149)

Applying the boundary conditions yields

( , ) 0

( , ) cosh ( ) 1

x x Bd x x A s x

ds

λλ

= =⎧⎪⎨

= Γ Γ − =⎪⎩

(8.150)

hence, rearranging the terms

1

0

A

B

⎧ =⎪Γ⎨

⎪ =⎩

(8.151)

Finally we obtain the Lagrangian multiplier

sinh ( )( , ) s xx sλ Γ −=

Γ (8.152)

Substitute Equation (8.152) into Equation (8.145), for simplicity let consider the presence

of one wavemode only, we obtain the following iteration formula

2

10

1

( ) ( )sinh ( )( ) ( )

( ) ( )n n n n

i ixx aN


n a x

s ss xx x ds

i e e t dt e e t dtξ ξ ξ ξ

τ τ

τ τη τ τ+ − −

= −

⎡ ⎤′′ − Γ⎢ ⎥Γ −

= + ⎡ ⎤⎢ ⎥Γ − +⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦

∫ ∑ ∫ ∫% %(8.153)

Assume that the approximate solution to be of the type

0 ( ) sinhx A xτ = Γ (8.154)

Substitute the approximate solution in the iteration formula and rearrange the terms to

obtain, i.e.,

258

11 0

sinh

( ) sinh sinh ( )

sinh

n n

n

si s i t

xNa

n an i s i t

s

e e tdtiAx A x s x ds

e e tdt

ξ ξ

ξ ξ

τ η

−

−

= −

⎡ ⎤Γ⎢ ⎥

⎢ ⎥= Γ − Γ − ⎢ ⎥Γ ⎢ ⎥+ Γ⎢ ⎥⎣ ⎦

∫∑ ∫

∫ (8.155)

Solve the integral in dt in Equation (8.155) and substitute the result in the original

equation to get

1

1 0

( ) sinh

sinh ( )2

s s a i a a i ai s

xN

n a i a a i a s sn i s

x A x

e e e eei i i iiA s x ds

e e e eei i i i

ξ ξξ

ξ ξξ

τ

ξ ξ ξ ξη

ξ ξ ξ ξ

Γ −Γ −Γ − Γ −−

Γ − −Γ − Γ −Γ=

= Γ

⎡ ⎤⎛ ⎞+ − +⎢ ⎥⎜ ⎟Γ + Γ − Γ + Γ −⎝ ⎠⎢ ⎥− Γ − ⎢ ⎥Γ ⎛ ⎞⎢ ⎥+ + − −⎜ ⎟⎢ ⎥Γ − Γ + Γ − Γ +⎝ ⎠⎣ ⎦

∑ ∫ (8.156)

Solve the integral in ds in Equation (8.156), after rearrangement we get

( ) ( ) ( )

( ) ( )

1

2 2 2 22 2

1

2 2 2 2 2 2

( ) sinh

2 sinhsinh

2 cosh cosh2 sin

a ai a

N

n a an i a

x A x

i e e ax ei iAi

x x a e ei ie xi i

ξ

ξ

τ

ξξξ ξ ξ

η

ξ ξξ ξ ξ ξ

Γ −Γ−

Γ −Γ= −

= Γ

⎧ ⎫⎡ ⎤⎛ ⎞ΓΓ − − − −⎪ ⎪⎢ ⎥⎜ ⎟⎜ ⎟Γ +Γ Γ + Γ − Γ +⎪ ⎪⎢ ⎥⎪ ⎝ ⎠ ⎪⎣ ⎦− ⎨ ⎬Γ ⎛ ⎞Γ Γ⎪ ⎪+ + + +⎜ ⎟⎪ ⎪⎜ ⎟Γ + Γ + Γ − Γ +⎪ ⎪⎝ ⎠⎩ ⎭

∑ (8.157)

This is the first approximate solution. Let’s apply the boundary conditions to derive

constant A. First derivative Equation (8.157) with respect to x, i.e.,

( ) ( ) ( )

( ) ( )

1

2 2 2 22 2

2 21

2 2 2 2

( ) cosh

2 sinhcosh

cosh sinh22

coshcos

a ai a

N

nn

a ai a

x A x

i e e ax ei i

Ai x x xi

a e ei e xi i

ξ

ξ

τ

ξξξ ξ ξ

η η ξξ

ξ ξξ ξ ξ

Γ −Γ−

=

Γ −Γ−

′ = Γ Γ

⎧ ⎫⎡ ⎤⎛ ⎞ΓΓ Γ − − − −⎪ ⎪⎢ ⎥⎜ ⎟⎜ ⎟Γ +Γ Γ + Γ − Γ +⎪ ⎪⎢ ⎥⎝ ⎠⎣ ⎦

⎪ ⎪Γ + Γ Γ⎪ ⎪− +⎨ ⎬

Γ Γ +⎪ ⎪⎪ ⎪⎛ ⎞Γ⎪ ⎪+ + +⎜ ⎟⎜ ⎟⎪ ⎪Γ + Γ − Γ +⎝ ⎠⎩ ⎭

∑ (8.158)

259

Apply the boundary conditions (8.144), i.e.,

( ) ( ) ( )

( ) ( )

1

2 2 2 22 2

2 21

2 2 2 2

( ) cosh

2 sinhcosh

cosh sinh22

coshcos

a ai a

N

nn

a ai a

b IS

a A a

i e e aa ei i

iA a a ai

a e ei e ai i

G

ξ

ξ

τ

ξξξ ξ ξ

η ξξ

ξ ξξ ξ ξ

ε

Γ −Γ−

=

Γ −Γ−

′ ± = Γ Γ

⎧ ⎫⎡ ⎤⎛ ⎞Γ−Γ Γ + − −⎪ ⎪⎢ ⎥⎜ ⎟⎜ ⎟Γ +Γ Γ + Γ − Γ +⎪ ⎪⎢ ⎥⎝ ⎠⎣ ⎦

⎪ ⎪Γ + Γ Γ⎪ ⎪− +⎨ ⎬

Γ Γ +⎪ ⎪⎪ ⎪⎛ ⎞Γ⎪ ⎪+ + +⎜ ⎟⎜ ⎟⎪ ⎪Γ + Γ − Γ +⎝ ⎠⎩ ⎭

=

∑

A

bt

(8.159)

Solution of Equation (8.159) yields

b ISA

b

GAt Qε

= (8.160)

where

( ) ( ) ( )

( ) ( )

2 2 2 22 2

2 21

2 2 2 2

cosh

2 sinhcosh

cosh sinh22

coshcos

a ai a

N

nn

a ai a

Q a

i e e aa ei i

i a a ai

a e ei e ai i

ξ

ξ

ξξξ ξ ξ

η ξξ

ξ ξξ ξ ξ

Γ −Γ−

=

Γ −Γ−

= Γ Γ

⎧ ⎫⎡ ⎤⎛ ⎞Γ−Γ Γ + − −⎪ ⎪⎢ ⎥⎜ ⎟⎜ ⎟Γ +Γ Γ + Γ − Γ +⎪ ⎪⎢ ⎥⎝ ⎠⎣ ⎦

⎪ ⎪Γ + Γ Γ⎪ ⎪− +⎨ ⎬

Γ Γ +⎪ ⎪⎪ ⎪⎛ ⎞Γ⎪ ⎪+ + +⎜ ⎟⎜ ⎟⎪ ⎪Γ + Γ − Γ +⎝ ⎠⎩ ⎭

∑ (8.161)

Hence the first iteration is written as

260

( ) ( ) ( )

( ) ( )

1

2 2 2 22 2

1

2 2 2 2 2 2

( ) sinh

2 sinhsinh

2 cosh cosh2 sin

b ISA

b

a ai a

Nb ISA

n a anb i a

Gx xt Q

i e e ax ei iGi

t Q x x a e ei ie xi i

ξ

ξ

ετ

ξξξ ξ ξε η

ξ ξξ ξ ξ ξ

Γ −Γ−

Γ −Γ= −

= Γ

⎧ ⎫⎡ ⎤⎛ ⎞ΓΓ − − − −⎪ ⎪⎢ ⎥⎜ ⎟⎜ ⎟Γ +Γ Γ + Γ − Γ +⎪ ⎪⎢ ⎥⎪ ⎝ ⎠ ⎪⎣ ⎦− ⎨ ⎬Γ ⎛ ⎞Γ Γ⎪ ⎪+ + + +⎜ ⎟⎪ ⎪⎜ ⎟Γ + Γ + Γ − Γ +⎪ ⎪⎝ ⎠⎩ ⎭

∑ (8.162)

Note that for 0bt → , i.e., aΓ → ∞ , Equation (8.162) becomes

( )

31

2 sinh sinhlim cosh 1 cosh2 cos2 2

i aN

n i aa n

i e a aQ a i ai e ai a

ξ

ξ

ξη

ξ ξξ

−

−Γ →∞ =

⎧ ⎫⎡ ⎤− − Γ − Γ⎪ ⎪⎢ ⎥≈ Γ Γ − ≈ Γ Γ⎨ ⎬⎢ ⎥Γ + +⎪ ⎪⎢ ⎥Γ⎣ ⎦⎩ ⎭

∑ (8.163)

and the interfacial shear stress becomes

( )

1

4

lim ( ) sinhcosh

sinh 2 sinh sinhcosh

2

b ISAa b

i a

b ISAn

b

Gx xt a

x i e a aG ait

ξ

ετ

ξε η

Γ →∞

−

≈ ΓΓ Γ

Γ− − Γ − Γ

Γ Γ−Γ 1 cosh2 2 sin

cosh

N

n i axi x ie xa

ξξ ξ= −

⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥⎪ ⎪⎣ ⎦⎨ ⎬

Γ⎪ ⎪+ +⎪ ⎪Γ⎩ ⎭

∑ (8.164)

With the use of the shear lag definition in Equation (8.31) and Equation (8.46), the

interfacial shear stress simplifies as

( ) ( )[ ]1 0sinhlim ( )cosh

aa ISAa

t xx E a x a x aa a

ψτ ε τ δ δα ψΓ →∞

Γ≈ Γ ≈ − − +

+ Γ (8.165)

This is the expression of the shear stress under ideal bond assumption.

8.5 STRESS DISTRIBUTION IN THE BONDING LAYER

The effect of the adhesive, actuator, and structure parameters on the shear stress

distribution have been extensively discussed in Section 8.2.1. In this Section, we focus on

261

the difference in shear stress transferred between the derivation for low frequencies

approximation, Equation (8.44), and that for N generic modes, Equation (8.162).

Figure 8.14a shows how the shear stress transferred by the PWAS to the structure

varies with frequency. Lines I and II are the shear stress distribution as derived from

Equation (8.44) and from Equation (8.162) for fd=1 kHz-mm. The difference between the

two curves is due to the different values in the modal repartition number. As fd increases,

the shear stress curves shifts down till it reaches the cut-off frequency (line III). After the

cut-off frequency, the shear stress is closer to the approximate solution (lines I and IV

respectively) and then it continues to shift downwards as the fd product increases.

Figure 8.14b shows the percentage difference between the two equations in load

transmitted for different frequency values (gray area in Figure 8.14a). At low frequencies,

the difference is about 14%, this difference is entirely due to the different modal

repartition number at low frequencies (4 vs. 2.5). If we use the value of 2.5 for α in the

low frequencies approximation, the two curves (I and II) become identical. As the

frequency increases, the percentage of PWAS length used to transfer the load does not

change but the amount of load transferred increases linearly till the first cut-off frequency

(fd-~800 kHz-mm) up to 20%. Figure 8.14b shows that beyond the first cut-off

frequency, (A1 is propagating), there is a sudden discontinuity that reflects the

discontinuity in the derivation of the modal repartition number. The percentage difference

drops to 3% and than it start to increase again with the same curvature.

262

Figure 8.14 Shear stress variation with frequency. a) Shear stress transmitted by the PWAS to

the structure through a bond layer. (I): shear stress derived for low frequency

approximation ( 4α = ), Equation (8.44); All other curves: shear stress derived in the

generic N mode formulation ( fα = (frequency,no. of modes)), Equation (8.162). (II)

fd=1 kHz-mm; (III) fd=783 kHz-mm before the cut-off frequency; (IV) fd=784 kHz-mm

after the cut-off frequency). b) Percentage difference in load transmitted at different

frequency-thickness products.

Figure 8.15a shows the change in shear stress transfer with frequency when the

imaginary branch of A1 is considered. There is no significant difference in the shear

curves for different derivation methods and different frequencies.

Figure 8.15b shows that there is no discontinuity at the cut-off frequency and the total

load transferred from the actuator to the structure increases till 8% in the frequency-

thickness range 0-1000 kHz-mm.

(I)

(II)

Perc

enta

ge

diff

eren

ce

oftra

nsm

itted

load

( )fd kHzmm

( )xτ

2·Area = transmitted load

a) b) 0 200 400 600 800

5

10

15

20

(III) (IV)

x a

263

Figure 8.15 Interfacial shear stress distribution and percentage of change a) Shear stress

transmitted by the PWAS to the structure through a bond layer with imaginary A1.

(I): shear stress derived for low frequency approximation ( 4α = ); (II): shear stress

derived in the generic N mode formulation ( fα = (frequency, no. of modes)) for

different frequencies (fd=1; 200; 781; 850 (solid line); 1000 kHz-mm (dash-dot ]

line)). b) Percentage difference in load transmitted at different frequency-thickness

products.

Perc

enta

ge

diff

eren

ce

oftra

nsm

itted

load

( )fd kHzmm

( )xτ

x a

2·Area = transmitted load

a) b)

(I)

(II) 0 200 400 600 800

10

20

264

9 TUNED GUIDED WAVES EXCITED BY PWAS

This section deals with the aspect of the interaction between PWAS and structure during

the active SHM process, i.e., tuning between the PWAS and the Lamb waves traveling in

the structure. The PWAS, under electric excitation, transfers the oscillatory contractions

and expansion to the bonded layer and the layer to the metal surface. In this process

several factors influences the wave behavior: thickness of the bonding layer, geometry of

the PWAS, thickness and material of the plate. The result of the influence of all these

factors is the tuning of the PWAS with the material. This phenomenon has been studied

recently by Giurgiutiu (2005) who developed the theory of the interaction of the PWAS

with the structure for a rectangular PWAS with an infinite dimension. Lately Raghavan

and Cesnik (2004) extended these results to the case of a circular transducer.

In the first part of this section, we present the different models to represent the load

transferred from the PWAS to the structure through a bond layer. First we consider the

simple case of ideal bonding solution which permits the use of the pin-force model. Then

we present the case in which the bond layer has a finite thickness. The load transfer

models are used to derive the theoretical tuning curves. We show that the same

theoretical curves can be obtained through the Fourier transformation method as derived

by Giurgituiu (2005) and through the NME method presented in the present dissertation.

Let’s consider the excitation provided by a PWAS bonded through an adhesive layer

to the top surface of a plate. The excitation can be modeled in different ways. Hereunder,

265

we report first the case of ideal bonding solution in which we assume the bonding layer to

be small; second the case in which the bonding layer can assume different thickness but

we retain the low frequency approximation of the shear lag; finally, we will show the

case of non ideal bonding conditions under the shear lag exact solution.

9.1 SHEAR TRANSFER THROUGH BOND LAYER

9.1.1 Ideal bond solution

Assume the PWAS to be attached to the structure ideally, hence assume ideal bond. The

shear stress transfer takes place into an infinitesimal regions at the PWAS tips, i.e.,

( ) ( )0 if

( , )0 if

x

x a x a x at x d

x a

τ δ δ⎧ − − + ≤⎡ ⎤⎪ ⎣ ⎦= ⎨>⎪⎩

(9.1)

where 0a

a ISAt Ea

ψτ εα ψ

=+

. Substituting (9.1) into the expression of the filed amplitude,

Equation (7.125), we obtain:

( ) ( )0( )( )

4n n

ani x i xx

nnn a

v da x e x a x a e dxP

ξ ξτ δ δ−+

−

= − − +⎡ ⎤⎣ ⎦∫%

(9.2)

In Equation (9.2) the term 0τ depends on the excitation, term ( )4

nx

nn

v dP

% is the excitability

function of mode n (depends on the mode excited and not on the source used for

excitation), and term

( ) ( ) n

ai x

na

F x a x a e dxξδ δ−

= − − +⎡ ⎤⎣ ⎦∫ . (9.3)

is the Fourier integral. Its solution is

266

( ) ( ) ( ) ( )

2 sin

n n n

n n

a a ai x i x i x

na a a

i a i an

F x a x a e dx x a e dx x a e dx

e e i a

ξ ξ ξ

ξ ξ

δ δ δ δ

ξ− − −

−

= − − + = − − +⎡ ⎤⎣ ⎦

= − =

∫ ∫ ∫ (9.4)

Substitute this value in Equation (9.2) to obtain

0( )( ) sin2

ni xnx

n nnn

v d ea x i aP

ξ

τ ξ−

+ = −%

(9.5)

Equation (9.5) represents the PWAS-Lamb wave tuning amplitude for ideal bonding

conditions.

9.1.2 Shear-lag stress excitation with low frequency approximation

Consider the excitation provided by a PWAS. If we describe this phenomenon through

the shear lag model, the shear stress in the bonding layer is assumed of the form, see

Equation (8.59),

1 sinh if ( , )

0 if x

x x at x d

x a

τ⎧ Γ ≤⎪= ⎨>⎪⎩

(9.6)

where 1 0 cosha

aτ τ Γ

=Γ

. Substituting Equation (9.6) into Equation (7.125) we obtain

( )1( )( ) sinh

4n n

ani x i xx

nnn a

v da x e x e dxP

ξ ξτ −

−

= Γ∫%

(9.7)

where the Fourier integral ( )n nF ξ is a tuning function that depends on the relation

between the PWAS size, 2a , and the modal wavenumber, nξ , i.e.,

( ) ( )sinh n

ai x

n na

F x e dxξξ−

= Γ∫ for a x< (9.8)

267

Integral (9.8) can be solved explicitly to get

( ) 2 2

sin cosh sinh cos2 n n nn n

n

a a a aF i ξ ξ ξξξ

Γ Γ − Γ=

Γ + (9.9)


1 2 2

( ) sin cosh sinh cos( )2

ni xnx n n n

nnn n

v d e a a a aa x iP

ξ ξ ξ ξτξ

−+ Γ Γ − Γ

=Γ +

% (9.10)

For the shear lag case we provide the complete solution, hence solution for x<a. as in

Equation (7.123). Note that, for the domain a x a− ≤ ≤ , i.e., under the PWAS, the

function nF takes the form

( ) ( ), sinh n

xi x

n na

F x x e dxξξ−

= Γ∫ for a x a− ≤ ≤ (9.11)

Upon integration, Equation (9.11) becomes

( ) ( )( ) ( )1, sinh

2

n nn

xx x i x ii x

n nn na a

e eF x x e dxi i

ξ ξξξ

ξ ξ

Γ+ − Γ−

− −

⎡ ⎤= Γ = +⎢ ⎥Γ + Γ −⎣ ⎦

∫ (9.12)

Processing of Equation (9.12) gives the closed-form solution

( )( ) ( ) ( ) ( )1,

2

n n n nx i a i x i a i

n nn n

e e e eF xi i

ξ ξ ξ ξ

ξξ ξ

Γ+ − Γ+ − Γ− Γ−⎡ ⎤− −= +⎢ ⎥Γ + Γ −⎣ ⎦

(9.13)

Substituting Equation (9.13) into Equation (7.123) for the case of wave propagation

inside the excitation region, gives

( ) ( ) ( ) ( )

1( )( ) for 8

n n n nn x i a i x i a ii xnx

nnn n n

v d e e e e ea x a x aP i i

ξ ξ ξ ξξ

τξ ξ

Γ+ − Γ+ − Γ− Γ−−+ ⎡ ⎤− −

= + − ≤ ≤⎢ ⎥Γ + Γ −⎣ ⎦

%(9.14)

268

Similarly, we can develop closed-form solutions for the backward wave. Substituting

Equation (9.6) into Equation (7.124) for x a< − , gives

( )1( )( ) sinh

4n n

ani x i xx

nnn a

v da x e x e dxP

ξ ξτ −−

−

= Γ∫%

(9.15)

The tuning function for the backward wave is defined as

( ) ( )sinh n

ai x

n na

F x e dxξξ −

−

= Γ∫ for a x> (9.16)

Integral (9.16) can be solved explicitly and becomes after rearrangement

( ) 2 2

cos sinh sin cosh2 n n nn n

n

a a a aF i ξ ξ ξξξ

Γ − Γ Γ=

Γ + (9.17)


1 2 2

( ) sinh cos cosh sin( )2

ni xnx n n n

nnn n

v d e a a a aa x iP

ξ ξ ξ ξτξ

− Γ − Γ Γ= −

Γ +%

(9.18)

For backward wave in the domain a x a− ≤ ≤ , i.e., under the PWAS, the function nF

takes the form

( ) ( ), sin n

ai x

n nx

F x x e dxξξ −= Γ∫ for a x a− ≤ ≤ (9.19)

Upon integration, Equation (9.11) becomes

( ) ( ) ( ) ( )( ) ( )1 1, sin

2 2

n nn nn

aa a x i x ix i x ii x

n nn nx x x

e eF x x e dx e e dxi i

ξ ξξ ξξξ

ξ ξ

Γ− − Γ+Γ− − Γ+− ⎡ ⎤

⎡ ⎤= Γ = − = +⎢ ⎥⎣ ⎦ Γ − Γ +⎣ ⎦∫ ∫ (9.20)

Processing of Equation (9.12) gives the closed-form solution

269

( )( ) ( ) ( ) ( )1,

2

n n n na i x i a i x i

n nn n

e e e eF xi i

ξ ξ ξ ξ

ξξ ξ

Γ− Γ− − Γ+ − Γ+⎡ ⎤− −= +⎢ ⎥Γ − Γ +⎣ ⎦

(9.21)

Note that the ideal bonding solution is the limit case of the shear lag solution as Γ goes

to infinity. To prove it, take the limit of Equation (9.10) for Γ → ∞ , i.e.,

( )2 2

( ) sin cosh sinh coslim ( ) lim2 cosh

ni xna a ISA x n n n

na ann n

t E v d e a a a aa x i aa P a

ξε ξ ξ ξψα ψ ξ

−+

Γ →∞ Γ →∞

Γ Γ − Γ= Γ

+ Γ + Γ%

(9.22)

since ( )( )2

sinhlim 0

cosha

aΓ→∞

Γ=

Γ Γ, hence Equation (9.22) becomes

0( ) ( )lim ( ) sin sin2 2

n ni x i xn na x x

n a ISA n nann nn

t v d e v d ea x i E a a ia aa P P

ξ ξψ ε ξ τ ξα ψ

− −+

Γ →∞= =

+% %

(9.23)

This is the same expression derived for forward propagating modes exited by a PWAS

ideally bonded on the top surface of a plate as in Equation (9.5).

9.1.3 Shear-lag stress excitation for N generic modes

Consider the shear stress in the bonding layer as derived for N generic modes, Equation

(8.162). Substitution of Equation (8.162) into Equation (7.74) yields

( )

( ) ( )

( )

2 22 2

12 2

2 2 2 2

2

2 sinh

1 sinh2

( ) cosh cosh2 sin

2 sin

n

n

n

n

i an

N nnna a

n i a

n n

b ISAx i a

b n nn nn

a ai a

nn

i aei x

e eei i

Gt x x x at Q e x

e ee xi

ξ

ξ

ξ

ξ

ξξξη

ξ ξε

ξ ξξ ξη

ξξ

−

Γ −Γ= −

−

Γ −Γ−

⎧ ⎫Γ⎡ ⎤− +⎪ ⎪⎢ ⎥Γ +Γ Γ +⎪ ⎪⎢ ⎥− Γ⎨ ⎬⎢ ⎥⎛ ⎞Γ⎪ ⎪⎢ ⎥− −⎜ ⎟⎜ ⎟⎪ ⎪Γ − Γ +⎢ ⎥⎝ ⎠⎣ ⎦⎩ ⎭= Γ Γ

+Γ + Γ +

+Γ + +

Γ −

∑

( )1

2

0

N

n

n

x a

i

otherwise

ξ=

⎧⎪⎪⎪⎪⎪⎪ ≤⎪⎨ ⎡ ⎤⎪ ⎢ ⎥⎪ ⎢ ⎥⎪ ⎢ ⎥⎛ ⎞⎪ ⎢ ⎥⎜ ⎟⎜ ⎟Γ +⎪ ⎢ ⎥⎝ ⎠⎣ ⎦⎪⎪⎩

∑

(9.24)

270

The amplitude derived through normal mode expansion is given by Equation (7.125).

Substitute the expression of the excitation (9.24) into (7.125) to obtain

( ) ( )2( )4

n

ni xx b ISA

n n nnn b

v d Ga x e FP t Q

ξε ξ−+ =Γ

% (9.25)

where the tuning function Fn is defined as

( )

( ) ( )

2 22 2

12 2

2 2 2 2

2 sinh

( ) 1 sinh2

2 coshcosh sin

2

m

n

m

n m n

i am

aM mm i xmn n a a

m i a a

m m

a ai x i a i xm

mm ma am

i aeiF e xdx

e eei i

ae x xdx e e xdx

ξ

ξ

ξ

ξ ξ ξ

ξξξηξ

ξ ξ

ξ ξξ ξη

−

−Γ −Γ

= − −

− −

− −

⎧ ⎫Γ⎡ ⎤− +⎪ ⎪⎢ ⎥Γ +Γ Γ +⎪ ⎪⎢ ⎥= − Γ⎨ ⎬⎢ ⎥⎛ ⎞Γ⎪ ⎪⎢ ⎥− −⎜ ⎟⎜ ⎟⎪ ⎪Γ − Γ +⎢ ⎥⎝ ⎠⎣ ⎦⎩ ⎭

ΓΓ +

Γ + Γ ++

Γ

∑ ∫

∫ ∫

( ) ( )1

2 2 sinm n

M

aa am i a i x

mam m

e ee e xdxi i

ξ ξ ξξ ξ

Γ −Γ= − −

−

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎛ ⎞⎢ ⎥+ +⎜ ⎟⎜ ⎟⎢ ⎥Γ − Γ +⎝ ⎠⎣ ⎦

∑∫

(9.26)

Note that the first integral in Equation (9.26) is the same as that in Equation (9.8).

Finally, the tuning function (9.26) can be written as

( )

( ) ( )

( )

2 22 2

2 21

2 2

( ) (

2 2

2 sinh cos sinhsin cosh( ) 2 1

2

2 1cos cosh2

2

m

m

n

i am n nM mmm n

n n a am i a n

m m

i a

nmn

n m

m

i a a aei a aF i

e eei i

e eii a a a

ξ

ξ

ξ

ξ ξ ξξξη ξξ

ξξ ξ

ξξ ξξ ξ

η

−

Γ −Γ= −

Γ− −

⎧ ⎫Γ⎡ ⎤ Γ− +⎪ ⎪⎢ ⎥Γ +Γ Γ + −Γ Γ⎪ ⎪⎢ ⎥= −⎨ ⎬⎢ ⎥⎛ ⎞Γ Γ +⎪ ⎪⎢ ⎥− −⎜ ⎟⎜ ⎟⎪ ⎪Γ − Γ +⎢ ⎥⎝ ⎠⎣ ⎦⎩ ⎭

−Γ −

Γ −Γ +

+Γ

∑

( ) ( )

( )

( )

)

( ) ( )

2 22 2

sin

coshsin

n

n n

m

i a

ni a i a

n n

m m n

a am ni a

m m m m n

m n

i

e ei i

a

a e eei i a

ξ

ξ ξ

ξ

ξ

ξ ξ

ξ ξξ ξ

ξ ξ ξ ξ ξξ ξ

Γ+

− Γ− Γ+

=

Γ −Γ−

⎡ ⎤⎛ ⎞⎛ ⎞⎢ ⎥⎜ ⎟⎜ ⎟Γ +⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎜ ⎟− +⎜ ⎟⎜ ⎟⎜ ⎟Γ − Γ +⎢ ⎥⎝ ⎠⎝ ⎠⎢ ⎥

⎛ ⎞−⎡ ⎤⎢ ⎥⎣ ⎦⎜ ⎟⎢ ⎥−⎛ ⎞Γ ⎜ ⎟⎢ ⎥+ + +⎜ ⎟⎜ ⎟⎢ ⎥⎜ ⎟Γ + Γ − Γ + +⎡ ⎤⎝ ⎠ ⎣ ⎦⎜ ⎟⎢ ⎥+⎜ ⎟+⎢ ⎥⎝ ⎠⎣ ⎦

1

M

∑

(9.27)

271

9.2 PWAS – LAMB WAVES TUNING

The tuning between structure and PWAS is the selective Lamb mode excitation with

PWAS transducer. Under electric excitation, the PWAS undergoes oscillatory

contractions and expansions which are transferred to the structure through the bonding

layer and thus excite Lamb waves into the structure. In this process, several factors

influence the behavior of the excited wave: the thickness of the bonding layer, the

geometry of the PWAS, the thickness and material of the structure. The result of the

influence of all these factors is the tuning of the PWAS with various Lamb wave modes

in the material. Figure 9.1 shows the coupling between PWAS and two Lamb wave

modes, S0 and A0. Maximum coupling between PWAS and the Lamb mode happens

when the PWAS length is an odd multiple of half the wavelength λ . Since different

wave modes have different wavelengths, which vary with frequency, the opportunity

arises for selectively exciting various Lamb modes at various frequencies.

This phenomenon has been studied by Giurgiutiu (2003) who developed the theory of

the interaction of a rectangular PWAS with straight-crested Lamb waves. Lately,

Raghavan and Cesnik (2004) extended these results to the case of a circular transducer

coupled with circular-crested Lamb waves. Both methods were developed through the

Fourier integral transformation of the wave fields. In this section, we will apply a novel

method to obtain the tuning curves, i.e., normal mode expansion model. We will show

numerically that there is no difference between the two formulations. However, a

theoretical proof is provided in the Appendix F.

272

Figure 9.1 S0 and A0 particle displacement and interaction of PWAS with Lamb waves. (Bao

2003)

To obtain the tuning curves we must derive the strain in the structure. The strain in the

structure due to the excitation is given by

( , , )xx x

u v x y t dtx x

ε ∂ ∂= =

∂ ∂ ∫ (9.28)

Recall the expression of velocity derived through normal mode expansion in Equation

(7.126), hence the expression of the strain becomes

1( ) ( ) ( ) ( )i t i tx n x n n x

n na x v y e dt a x v y e

xω ωε ξ

ω− −∂

= =∂ ∑ ∑∫ (9.29)

or in expanded form using Equation (9.5)

h

= 2

d

λ/2

V(t)PWAS ~

V(t)PWAS ~

λ/2

h =

2d

S0

A0

273

( ) ( )0

( ) ( ) sin2

ni x tx nxx n n

n nn

v y v dia a eP

ξ ωε τ ξ ξω

−= ∑%%

(9.30)

This represents the tuning expression of the strain in the structure excited by the PWAS

under the ideal bonding conditions. The tuning expression for the strain was derived by

Giurgiutiu (2003) through Fourier transformation method, i.e.,

( ) ( )( ) ( ) ( )0

', sin ni x tn nx n

n n n

Nax t a eD

ξ ωξτε ξμ ξ

−= − ∑%

(9.31)

Where nD′ is the derivate with respect to the wavenumber of the Rayleigh-Lamb

equation, ( )2 2 cos cosnN d dξβ ξ β α β= + for antisymmetric modes and

( )2 2 sin sinnN d dξβ ξ β α β= + for symmetric modes. In Appendix F we prove

analytically that the expression of the strain derived through normal mode expansion is

the same as that derived with the Fourier transformation.

0 1000 2000 3000 4000 5000 6000

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16Modal expansionFourier transformationModal expansionFourier transformation

Symmetric mode

f (kHz)

Stra

in

0 1000 2000 3000 4000 5000 6000

0.02

0.04

0.06

0.08

0.1

0.12

Modal expansionFourier transformationModal expansionFourier transformation

Antisymmetric mode

f (kHz)

Stra

in

Figure 9.2 Comparison of tuning curves for the strain excited by a PWAS derived through the

Fourier transformation model and the normal mode expansion method.

274

Figure 9.2 shows the numerical derivation of the strain for S0 and A0 modes. The solid

lines are the strain values calculated with Equation (9.30), the dotted lines are the strain

values calculated with Equation (9.31). The two derivations are identical

9.2.1 Experimental results and theoretical predictions

Pitch-catch experiments were performed in which one PWAS served as transmitter and

another PWAS served as receiver. The predicted values of the tuning curves were

compared with the experimental results. The signal used in the experiments was a

Hanning-windowed tone burst with 3 counts. The signal was generated with a function

generator (Hewlett Packard 33120A) and sent through an amplifier (Krohn-Hite model

7602) to the transmitter PWAS. A data acquisition instrument (Tektronix TDS5034B)

was used to measure the signal from the receiver PWAS. Several plates were used in the

experiments: (1) aluminum alloy 2024-T3 with 1-mm thickness and 1200x1060-mm size;

(2) aluminum alloy 6061-T8 with 3-mm thickness and 500×500-mm size; (3) aluminum

alloy 2024-T3 with 3-mm thickness and 1200×1200-mm size. In each experiment, we

used a pair of PWAS at a distance of 250 mm from one another. The frequency of the

signal was swept from 10 to 700 kHz in steps of 20 kHz. At each frequency, we collected

the wave amplitude and the time of flight for both the symmetric mode and the

antisymmetric modes.

9.2.1.1 Square PWAS

Square PWAS 7-mm long, 0.2-mm thick (American Piezo Ceramics APC-850) were

used on two aluminum 2024-T3 plates of different thickness (1 mm and 3 mm) and one

aluminum 6061-T8 plate of 3-mm thickness.

275

9.2.1.1.1 Experiments on 2024-T3 plate with 1-mm thickness and 1200×1060-mm size

Figure 9.3 shows the configuration for the square PWAS on the 2024-T3 aluminum alloy

plate 1-mm thick. The PWAS were located at the center of the plate in order to avoid

interference with the reflection from the boundaries.

Figure 9.3 Aluminum plate 2024-T3 1-mm thick with square, rectangular and round PWAS

The group velocities of the S0 mode were detected with no difficulties at each

frequency. The A0 mode was followed closely at each frequency, but, for frequencies

where the wave amplitude was closer to zero, the experimental values were more distant

from the predicted values. Figure 9.4 shows the experimental data (cresses and circles)

and predicted values (solid lines) of the wave amplitude for the S0 and A0 modes. For the

theoretical predictions, we used an effective PWAS length of 6.4 mm. For this effective

PWAS length value, we obtain the best agreement between experiments and predictions.

In the development of the theory, it was assumed that there was ideal bonding between

the PWAS and the plate. This assumption means that the stresses between the transducers

and the plate are fully transferred at the PWAS ends. In reality, the stresses are

transferred over a region adjacent to the PWAS ends (Figure 8.3).

P1

1200 mm

1200

mm

250mm 250mm

250mm

P

P2

276

Figure 9.4 Tuning for aluminum 2024-T3, 1-mm thickness, 7-mm square PWAS; experimental

A0 (cross) and S0 (circle) data; theoretical values (solid lines) for 6.4 mm PWAS.

The experimental and theoretical values of the tuning are in good agreement (Figure

9.4). The first minimum of the A0 mode, both in the experimental graph and in the

predicted graph, is found around 210 kHz. At this frequency, the S0 mode amplitude is

nonzero and increasing. The theory also predicts the S0 maximum should happen at the

same frequency as the second A0 maximum; this prediction was also verified by the

experiments.

9.2.1.1.2 Experiments on 2024-T3 plate with 3-mm thickness and 1200×1200-mm size

In this thicker plate, three Lamb wave modes (S0, A0, A1) exist in the testing frequency

range. Figure 9.5 shows the group velocities for the S0, A0 and A1 modes. The

experimental data are close to the predicted values for frequencies up to 550 kHz. Above

this frequency, the group velocities of the three Lamb wave modes come into a common

nexus. Hence, the three waves are too close and too dispersive to be measured accurately.

f (kHz)0 100 200 300 400 500 600 700

2

4

6

8

A0

S0

277

In particular, it was found difficult to determine which wave represents the A0 mode and

which represents the A1 mode.

0

1000

2000

3000

4000

5000

6000

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700

freq (KHz)

Cg

(m/s

)

Anti 0 Cg Sym 0 Cg Anti 1 Cg

Anti 0 Cg teoric Sym 0 Cg teoric Anti 1 Cg teoric

Figure 9.5 Group velocity: Aluminum 2024-T3, 3-mm thick, 7-mm square PWAS

Figure 9.6 Aluminum 2024-T3, 3-mm thickness, 7-mm Square PWAS. Experimental A0 (cross),

S0 (circle), and A1 (cross) data. b) Theoretical values (solid lines) with PWAS

length=6.4 mm

f (kHz) 0 100 200 300 400 500 600 700

1

2

A0S0

A1

278

Figure 9.6 shows the experimental and predicted data of the wave amplitude for A0,

S0 modes. The experimental and predicted values are in accordance up to 550 kHz. The

S0 maximum is close to the A0 minimum at around 360 kHz. The A1 mode has also been

detected.

9.2.1.1.3 Experiments on 6061-T8 plate with 3-mm thick and 500x500-mm size

The results on the plate 500×500-mm, 3-mm thick were similar to those on the plate

1200×1200-mm, 3-mm thick except for the presence of boundary reflections. The data

followed the predicted values quite closely. At frequencies between 500 kHz and 700

kHz, both plates showed the presence of three modes, S0, A0, and A1. Figure 9.5 shows

that at these high frequencies, their group velocities are close to each other and that both

the S0 mode and the A1 mode are dispersive. The three wave packets are close to each

other and a superposition effect starts to manifest, e.g., the tail of one wave packet

interferes with the head of the next one. This superposition forms apparent decreases and

increases of the actual packet amplitude. For example, Figure 9.7 shows the three wave

packets at two different frequencies, 450 kHz and 570 kHz for the 1200x1200-mm, 3-mm

thick plate. At 450 kHz, it is possible to determine the location and amplitude of the S0

mode while the superposition effect of the S0 tail with the A0 and A1 modes makes it

difficult to determine the location and amplitude of the A0 and A1 waves. At 550 kHz, it

is possible to determine the location and amplitude of S0 and that of a second wave,

which could be either the A0 or the A1 mode. The distinction between A0 and A1 modes

is difficult to determine, because it is difficult to follow their progression along the

dispersion curves during the change of frequency. The third wave location and amplitude

is approximate because the tails of the two other modes superpose with the third mode.

279

Figure 9.7 Wave propagation from the Oscilloscope at 450 kHz and 570 kHz for the

1200x1200-mm, 3-mm thick plate.

The effects described above were even more pronounced in the small plate of size

500 mm × 500 mm. Above 450 kHz, it was difficult to locate the three waves, and the

collected data seemed to be more distant from the predicted values. Moreover, the signal

was disturbed by the reflection from the boundaries. Figure 9.8 compares the wave

propagation of a 250 kHz tone burst in two plates of different size but both 3-mm thick.

The boundary effects were much more pronounced in the small plate, where the

reflection from the boundary was already affecting the slower A0 mode. At 570 kHz, the

superposition of the waves and the presence of the boundary reflection in the small plate

made it quite difficult to determine the location and amplitude of the three modes.

Figure 9.8 Waves propagation for 1200x1200-mm and 500x500-mm plate, 3-mm thick. (a) 270

kHz. (b) 570 kHz

a) b)

Big plate 1200x1200 mm

Small plate 500x500 mm

280

9.2.1.2 Round PWAS

Experiments with round PWAS diameter 7-mm, 0.2-mm thick (American Piezo Ceramics

APC-850) were performed on two aluminum 2024-T3 plates of different thickness (1 mm

and 3 mm). The results were found to be quite similar to those for square PWAS and, for

sake of brevity, will not be reported here.

9.2.1.3 Rectangular PWAS

Rectangular PWAS of high aspect ratio were tested to examine the directional tuning of

Lamb waves. Three rectangular PWAS of 25×5-mm size, and 0.15-mm thickness (Steiner

& Martin) were used. The experiment configuration is shown in Figure 9.9. PWAS P1

was the transmitter and PWAS P2 and P3 were the receivers.

Figure 9.9 Aluminum plate 2024-T3 1200x1200-mm, 1-mm thick with rectangular PWAS

9.2.1.3.1 Transmitter P1, receiver P2

Figure 9.10a shows the experimental and predicted group velocity values. The A0 mode

has been detected well for frequencies below 400 kHz. The S0 mode shows a dispersion

behavior in the experimental data at low frequency. The experimental data of the tuning

(Figure 9.10b) are quite different from the predicted values (see solid lines in Figure

9.11). The transmitted signal amplitude from P1 to P2 was small compared with that of

the other experiments.

P1

P2

P325 mm

250 mm

5 mm250 mm

281

Figure 9.10 Plate 2024-T3, 1200x1200-mm, 1-mm thick. Rectangular PWAS (P1 transmitter, P2

receiver). (a) Group velocity: experimental and theoretical values; (b) Experimental

data for the tuning.

A new experiment was conducted sweeping the frequency from 25 kHz to 250 kHz at

steps of 3 kHz. The intent of the new experiment was to visualize the three jumps of the

A0 mode as shown in Figure 9.11 (solid lines).

Figure 9.11 shows the experimental values of the wave amplitude for frequency up to

250 kHz taken with steps of 3 kHz. The small steps we used to collect the data let us

detect the three maximum in the A0 mode that where not visible in the first graph. The

first maximum is in accordance with the predicted values, whiles the second and third are

at higher frequency than that predicted. The S0 maximum is in accordance with the

predicted values. The value of the theoretical PWAS that best predicts the experimental

behavior is 22 mm. It is interesting to note that when the receiver is along the line of the

bigger dimension of the transmitter, the PWAS behaves as a square PWAS 25x25-mm

long.

0

1000

2000

3000

4000

5000

6000

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700freq (KHz)

Cg

(m/s

)

Anti Cg Sym CgAnti Cg teoric Sym Cg teoric

0

1

2

3

4

5

6

7

8

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700

Freq (KHz)

Volts

(mV)

A0 S0

b) a)

282

Figure 9.11 Tuning on plate 2024-T3, 1200x1200-mm, 1-mm thick; rectangular PWAS (P1

transmitter, P2 receiver); experimental data for A0 (crosses) and S0 (circles); Solid

lines, theoretical values with PWAS length=22 mm

9.2.1.3.2 Transmitter P1, receiver P3

Figure 9.12 shows the experimental and theoretical values of the wave amplitude for the

A0 and S0 modes when P1 is the transmitter and P3 is the receiver. Both the A0

minimum and the A0 maximum are in accordance with the predicted values. The S0

maximum amplitude location is different between experimental and predicted values.

Regarding the tuning, the predicted values were useful in detecting the frequency range to

be used. The value of the theoretical PWAS that best predicts the experimental behavior

seemed to be 4.5 mm.

0 100 200

1

2

A0

S0

f (kHz)

283

Figure 9.12 Tuning on plate 2024-T3, 1200x1200-mm, 1-mm thickness; rectangular PWAS (P1

transmitter, P3 receiver); experimental A0 (crosses) and S0 (circles) data; Solid

lines, theoretical values with PWAS length~=4.5 mm

9.2.2 Effect of ideal bond assumption on prediction

During the experiments it was noticed that the best concordance between the

experimental data and the predicted curves was achieved for theoretical PWAS length

smaller than that of the real transducer. In Table 9.1 are reported the values of the real

PWAS length, the theoretical values used, the percentages of the effective length of the

real PWAS, and the complimentary non effective lengths. The PWAS transmits the stress

to the plate on average at ~10% of its length before the borders. The adjustment of the

real PWAS length was necessary because in the development of the theory, it was

supposed that the stress induced by the PWAS was transferred to the structure at the end

of the PWAS itself. Next Section will present a more detailed explanation of this

phenomenon.

0 200 400 600

10

20

30

A0

S0

f (kHz)

284

Table 9.1 Actual and effective PWAS length

Real PWAS length

Effective PWAS length

% of effective PWAS

% of non effective PWAS

25 mm 22 mm 88% 12%

7 mm 6.4 mm 91.4% 8.6%

5 mm 4.5 mm 90% 10%

9.2.3 Tuning curves through different shear lag parameters

To show the effect of the bonding layer on the tuning curves, we will compare the tuning

curves as derived from the ideal bond solution with those derived with both the shear lag

solution and the N generic modes solution.

For the discussion of the tuning curves of the structure-PWAS system we limit our

focus on the case of only one PWAS bonded on the top surface of the structure (see

Figure 8.1). The case where two PWAS are attached to the structure on the opposite sides

of the plate is briefly discussed in Appendix F.

We recall that, the strain excited in the structure by the PWAS is equal to Equation

(9.29), i.e.,

( ) ( ) i txx n n

n

v y a x e ωε ξω

= ∑ (9.32)

Under the assumption of ideal bond through PWAS and structure, the strain excited by

the PWAS in the structure becomes, through the use of Equation (9.5),

( ) ( ) ( )0

( ) sin2

n

ni t xxx

x n nn nn

v dv yia a eP

ω ξε τ ξ ξω

−= ∑%

(9.33)

285

Substituting the expression of the field amplitude for shear lag varying with frequency,

Equation (9.10), into Equation (9.32) yields

( ) ( )1 2 2

( ) cosh sin sinh cos2

n

ni t xxx n n n

x nn nn n

v dv y a a a ai eP

ω ξξ ξ ξε τ ξω ξ

−Γ Γ − Γ=

Γ +∑%

(9.34)

This is the tuning curves for shear lag varying with frequency.

Finally, we consider the solution derived for the case of N modes presents, Equation

(9.25), the strain becomes

( ) ( )2

( )4

nn i xx i tb ISA x

x n n nnb nn

v dG v y e F et P Q

ξωεε ξ ξ

ω

−

=Γ∑

% (9.35)

Figure 9.13a shows the experimental tuning curves for the first antisymmetric (cross) and

symmetric (circle) mode. In figure are also reported the tuning prediction made through

Equation (9.33), (9.34), and (9.35). The theoretical amplitude of the curves have been

scaled such as the first antisymmetric peak amplitude was the same as the experimental

one (multiplication factor is 5.3 for Equation (9.33) and 2.6 for both Equation (9.34) and

Equation (9.35)). Both the maxima and the zeros of the antisymmetric mode prediction

curves are off the experimental values, while the symmetric prediction curves are more

close to the expected values. The prediction curves derived with Equations (9.34) and

(9.35) are almost coincident for any frequency and they are closer to the solution through

ideal bonding assumption at the low frequencies.

In Figure 9.13b, the predicted curves are plotted for a thicker bond thickness (tb=30

μm). The first antisymmetric maxima and minimum are now coincident with the

experimental values, while the symmetric maxima have not changed significantly its

286

location. The same result could be obtained with Equation (9.33) by changing the PWAS

effective length from 7-mm to 6.4-mm.

Figure 9.13 Tuning curves for an Aluminum plate 1-mm thick and a 7-mm square PWAS. Blue

circles: Experimental S0 mode data; Red crosses: Experimental A0 mode data;

Solid line: theoretical A0 and S0 tuning values under ideal bond assumption.

Equation (9.33); Dash line: theoretical A0 and S0 values for shear lag assumption,

Equation (9.34); Dash dot line: theoretical A0 and S0 values for N generic mode

derivation, Equation (9.35). a) Bond thickness 1 μm; b) bond thickness 30 μm.

Figure 9.13a shows the tuning curves under the assumption of ideal bond condition

with PWAS effective length of 6.4-mm. The maxima and minima locations are

coincident with those derived with the shear lag solution with bond thickness tb=30 μm.

A thickness of 30 μm of the bond layer it is not an ideal bond condition, hence the

assumption made in Equation (9.33) is no longer valid and an effective PWAS length is

needed. We obtain analog results for a 5-mm PWAS, the effective length is about 4.5-

mm for the ideal bond solution, while in the shear lag model the bond thickness is equal

to 30-μm (see Figure 9.13b). The prediction curves, derived through Equation (9.34) or

( )fd kHz mm ( )fd kHz mma) 0 100 200 300 400 500 600 700

2

4

6

8

0 100 200 300 400 500 600 700

2

4

6

8

A0

S0

A0

S0

b)

287

(9.35), show that the effective PWAS length used in the experiments in Section 9.2.1

where in fact needed because of the ideal bonding assumption. The shear lag model is an

effective tool to verify the thickness of the bonding between the PWAS and structure. In

all the figures, there is almost no difference between the predictions made through

Equation (9.34) and Equation (9.35).

Figure 9.14 Tuning curves for an Aluminum plate 1-mm thick and bond layer 30-μm thick.

Circles: Experimental S0 mode data; Crosses: Experimental A0 mode data; Solid

line: theoretical A0 and S0 tuning values under ideal bond assumption, Equation

(9.33); Dash line: theoretical A0 and S0 values for shear lag assumption, Equation

(9.34); Dash dot line: theoretical A0 and S0 values for N generic mode derivation,

Equation (9.35). a) Real PWAS length 7-mm, effective PWAS length 6.4-mm. b)

Real PWAS length 5-mm, effective PWAS length ~4.5-mm.

0 200 400 600

10

20

30

0 200 400 600

2

4

6

8

( )fd kHz mm

A0

S0

( )fd kHz mm

A0 S0

b) a)

288

10 TUNED GUIDED WAVES IN COMPOSITE PLATES

Since the use of composite materials in aeronautical and space structures has increased

considerably in the last decade, it is important that structural health monitoring

techniques are developed for non isotropic systems as well. As for the isotropic material,

in anisotropic materials both the dispersion curves of the wave guides and the interaction

between structure and actuator, i.e. tuning curves, are needed to understand how to

perform SHM.

In the first part of this section we discuss how the dispersion curves can be derived

for composite plates and we make some consideration on the behavior of guided waves in

composite plates.

In the second part of this section, we present how normal mode expansion can be

adapted to derive tuning curves for composite plates. So far, the derivation of theoretical

curves for composite was limited only to the formulation of the method. This method,

however, was mathematically difficult to solve and seemed not practical to use. The new

method proposed, on the contrary, is quite simple and requires the knowledge of the

dispersion curves and the contribution of the power flow of each mode present in the

structure. In this section we present a first preliminary derivation of theoretical curve for

a quasi isotropic plate and we compare the theoretical predictions with the experimental

data.

289

10.1 DISPERSION CURVES IN COMPOSITE PLATES

Modeling of the dispersion curves in multilayered media have been under consideration

since the late eighteen century. The multilayered in consideration had two layers of semi

infinite material. Layered plate media composed of more then two materials have been

studied only recently.

The development of inspection techniques in composites are based on the study of

complicated wave mechanisms and relies strongly on the use of predictive modeling

tools. The response method and the modal methods are the two most used inspection

techniques. In the first the reflection and transmission characteristics of the plate are

examined; in the second, the plate wave propagation properties of the system are

evaluated. Both models make use of the matrix formulation which describes elastic

waves in layered media with arbitrary number of layers.

When a wave is excited in a bulk isotropic material, three types of waves are present:

longitudinal, shear vertical and shear horizontal. In the case of an isotropic thin plate, the

longitudinal and the shear vertical waves are coupled; their interaction forms two

different families of mode propagation: the symmetric and antisymmetric Lamb wave

modes. The shear horizontal wave is decoupled from the other twos and can be treated

separately. In a generic anisotropic plate the three waves are coupled. The coupling

among the waves is present in orthotropic or higher than orthotropic material symmetry

unless the propagation is in the direction of the material symmetry. In this case, the shear

horizontal is decoupled from the longitudinal and shear horizontal wave, the solution of

the dispersion curves is simpler than the generic anisotropic case, but still a close form

solution of the dispersion curve equation is not available.

290

Lowe (1995) reviewed the available matrix techniques for modeling ultrasonic waves

in multilayered media when the layers in the plate were made of isotropic layers. The

methods he presented are those used also for generic anisotropic materials, even if in this

case the mathematical computation is more difficult.

Tang et al. (1989) presented an approximate method for the derivation of Lamb wave

phase velocities in composite laminates. The method was developed by using the

approximation of the elementary shear deformation plate theory and hence was used to

obtain the lowest symmetric S0 mode and antisymmetric A0 mode.

Wave propagation in multilayered media with an arbitrary number of flat layers was

derived by Thomson (1950) and corrected by Haskell (1953). They described the

displacements and the stresses at the bottom of the layer with those at the top of the layer

through a transfer matrix (TM). The coupling of the TM of each layers of the media with

one another gives the single matrix of the complete system. The solution obtained

showed however instability for large layer thickness and high frequencies. This problem

was caused by the poor conditioning of the TM due to the combination of both decaying

and growing coefficients in the presence of evanescent waves.

In the transfer matrix method, the field equations for the displacements and stresses in

a flat isotropic elastic solid layer are expressed as the superposition of the fields of four

bulk waves within the layer. The approach therefore is to derive the field equations for

bulk waves, which are solutions to the wave equation in an infinite medium, and then to

introduce the boundary conditions at an interface between two layers (Snell’s law), so

defining the rules for coupling between layers and for the superposition of the bulk

291

waves. The analysis of the layers is restricted to two dimensions, with the imposition of

plane strain and motion in the plane only.

At each interface are assumed eight waves: longitudinal and shear waves arriving

from “above” the interface and leaving “below” the interface and (L+, S+), similarly,

longitudinal and shear waves arriving from below the interface and leaving above the

interface (L-, S-). There are thus four waves in each layer of the multilayered plate

(Figure 10.1). Snell’s law requires that for interaction of the waves they must all share the

same frequency and spatial properties in the x1 direction at each interface. It follows that

all displacement and stress equations have the same ω and the same k1 component of

wavenumber, being k1 the projection of the wavenumber of the bulk wave onto the

interface. The angles of incidence, transmission, and reflection of homogeneous bulk

waves in the layers are than constrained.

Figure 10.1 Example, using three-layer plate with semi-infinite half spaces. (Lowe, 1995)

292

An alternative to the Thomson-Haskell formulation is the global matrix proposed by

Knopoff. In the global matrix all the equation of all the layers of the structure are

considered. The system matrix consist of 4(n-1) equations, where n is the total number of

layers. There is no a priori assumption on the interdependence between the sets of

equations for each interface. The solution is carried out on the full matrix, addressing all

the equations concurrently, i.e.,

[ ][ ] [ ]

[ ] [ ][ ]

11 2

22 3

33 4

4

4 55

0

b t

b t

b t

b b

AD DA

D DA

D DA

D D A

−

+

⎡ ⎤⎡ ⎤⎡ ⎤ −⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥ =⎢ ⎥− ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎡ ⎤− ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

(10.1)

where [D] is 4x4 field matrix that relates the amplitude of the partial waves to the

displacement and stress fields in a layer and Ai is the amplitude of the partial waves in

layer i. This technique is robust but slow to compute when there are many layers and the

matrix is large.

Nayfeh (1991; 1995) extended the Thomson-Hasakell formulation to the case of

anisotropic material and composites of anisotropic layers. In this formulation, each layer

of the plate can posses up to as low as monoclinic symmetry. The wave is allowed to

propagate along an arbitrary angle η (Figure 10.2) from the normal of the plate as well as

along any azimuthal angle θ (Figure 10.3).

293

Figure 10.2 Plate of an arbitrary number of layers with a plane wave propagating in the x1-x3

plane at an angle η with respect to x3 axis.

Both methods present the same characteristic that a solution of the function does not

strictly prove the existence of a modal solution, but only that the system matrix is

singular. Furthermore the calculation of the determinant for the modal solution needs the

use of a good algorithm because the aim of the problem is to find the zero of the

determinant while the matrix is frequently close to being singular.

Kausel (1986) and later Rokhlin et al. (2002) and Wang et al (2001) found a method

to resolve the numerical instability of the TM by introducing the layer stiffness matrix

(SM) and using an efficient recursive algorithm to calculate the global stiffness matrix for

an arbitrary anisotropic layered structure. In this method a layer SM is used to replace the

layer TM. The SM relates the stresses at the top and the bottom of the layer with the

displacements at the top and bottom layer; the terms in the matrix have only

exponentially decaying terms and the matrix have the same dimension and simplicity of

the TM. This method is unconditionally stable and is slightly more computationally

efficient than the TM method. For each layer, the local coordinate origin is settled at the

top of the jth layer for waves propagating along the –z direction and at the bottom of the

jth layer for waves propagating along the +z direction. This selection of coordinate system

η

294

is very important for eliminating the numerical overflow of the exponential terms when

the waves become non-homogeneous (large fd). In this way, the exponential terms are

normalized and the non-homogeneous exponentials are equal to one at the interface and

decay toward the opposite surface of the layer. The SM is defined as a matrix which

relates the stresses at the top and bottom of its layer to the displacements of the top and

bottom,

1 1j jj

j j

u

u

σ

σ− −⎡ ⎤ ⎡ ⎤

=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

K (10.2)

where K is the stiffness matrix, σj is the stress at the bottom surface of layer j, and σj-1 is

the stress at the top of the layer. The TM of each layer has the principal diagonal terms

depending on 1

z jik he+

that for large fd goes to zero and make the TM singular. The SM

principal diagonal has not those terms and it is independent of layer thickness, thus the

SM is always regular. Through a recursive algorithm it is possible to calculate the total

SM for a layered system.

10.1.1 Derivation of dispersion curves

In this section, we use the method of the transfer matrix applied to anisotropic materials

as reported in Nayfeh (1991) and Nayfeh (1995). For low fd products, the method is

stable. Other methods, as the stiffness matrix, have proved to be stable and more

efficient.

The main problem in the derivation of dispersion curves in composite plates is that

the wave velocity depends on the fiber direction in the layers and on the layers stack

sequence. For a plate made of one layer made of unidirectional fibers, the wave speed of

295

the wave propagating in the material depends on the angle θ between fibers direction and

wave propagation direction. Hence, for each angle θ , different dispersion curves will be

derived.

To obtain the dispersion curves in a plate made of more then one layer, for each layer

in the plate, we must define a relation between displacements and stresses at the bottom

surface and those at the top surface. Then, through the Snell low, the continuity of

displacements, and the relation derived for each layer between stresses and

displacements, we relate the stresses and the displacements at the bottom surface of the

plate to those at the top surface of the plate. By imposing the stress free boundary

surfaces, we obtain the dispersion curves for the plate for a given propagation direction.

Hereunder, we explain in more detail the procedure described.

Consider a composite plate made of N layers of unidirectional fibers like the one

depicted in Figure 10.3.

Figure 10.3 Composite plate and the kth layer made of unidirectional fibers.

Each layer of the composite plate is made of unidirectional fibers, hence they are

layers of orthotropic material. If we select a coordinate system 1 2 3, ,x x x′ ′ ′ such as 1x′ is

parallel to the fiber direction, the stiffness matrix of the kth layer can be written as

296

11 12 13

12 22 23

13 23 33

44

55

66

' ' ' 0 0 0' ' ' 0 0 0' ' ' 0 0 0

' :0 0 0 ' 0 00 0 0 0 ' 00 0 0 0 0 '

c c cc c cc c c

Cc

cc

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(10.3)

Consider a wave propagating at an angle θ with respect to the fiber direction (see Figure

10.3). Define a new coordinate system 1 2 3, ,x x x such as 1x form an angle θ with 1x′ . The

stiffness matrix in the new coordinate system (global) is found by doing a coordinate

transformation. Let T1 and T2 be the transformation matrices, the stiffness matrix

becomes

11 12 13 16

12 22 23 26

13 23 33 3611 2

44 45

45 44

16 26 36 66

0 00 00 0

'0 0 0 00 0 0 0

0 0

c c c cc c c cc c c c

C T C Tc cc c

c c c c

−

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(10.4)

Assume that the wave propagation direction is parallel to the plane 1 2x x , moreover

assume that the displacement solution can be written as

( ) ( ) ( )1 31 2 3 1 2 3, , , , i x x vtu u u U U U e ξ α+ −= (10.5)

where ξ is the wavenumber in direction 1x , v ω ξ= is the phase velocity, ω is the

circular frequency, α is an unknown parameter equal to the ratio between the

wavenumber in direction x3 and the wavenumber in direction x1, and Ui is the

displacement amplitude.

297

Recall the general formulation of the equation of motion (Equation (2.17)) and assume

that the external forces are equal to zero ( 0=F ), i.e.,

2

2

s

tρ

⎧ ∂∇ ⋅ =⎪

∂⎨⎪∇ =⎩

uT

u S (10.6)

To remove the stress matrix in (10.6) make use of the hook’s relation (2.14) to get

( )

2

2:

s

tρ

⎧ ∂∇ ⋅ =⎪

∂⎨⎪∇ =⎩

uc S

u S (10.7)

Substitute the second of Equation (10.7) into the first to obtain

( )2

2: s

s

tρ

⎧ ∂∇ ⋅ ∇ =⎪

∂⎨⎪∇ =⎩

uc u

u S (10.8)

The first equation in (10.8) is the equation of motion with the only unknown the particle

displacement u . Equation (10.8) can be written in explicit form making use of the

stiffness matrix in Equation (10.4); after rearrangement we get

298

( )

( ) ( )

( )

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

11 66 55 16 16 12 66 262 2 2 2 21 2 3 1 2 1 1 2 2

2 22 23 32 1

45 13 55 45 362 23 1 3 2 3

2 2 2 21 1 1 1

16 12 66 26 452 21 1 2 2

2u u u u u u uc c c c c c c cx x x x x x x x x

u uu uc c c c cx x x x x t

u u u uc c c c cx x x x x

ρ

∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂∂ ∂+ + + + + =

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂+ + + +

∂ ∂ ∂ ∂ ∂

( ) ( )

( ) ( ) ( ) ( )

2 2 22 2 2

66 22 442 2 2 23 1 2 3

2 22 23 32 2

26 23 44 45 36 21 2 2 3 1 3

22 2 2 231 1 2 2

13 55 36 45 36 45 23 44 55 21 3 2 3 1 3 2 3 1

2 23

44 3322

2

u u uc c cx x x

u uu uc c c c cx x x x x x t

uu u u uc c c c c c c c cx x x x x x x x x

u uc cx

ρ

∂ ∂ ∂+ + +

∂ ∂ ∂

∂ ∂∂ ∂+ + + + + =

∂ ∂ ∂ ∂ ∂ ∂ ∂

∂∂ ∂ ∂ ∂+ + + + + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂+ +

∂

2 23 3 3

452 23 1 2

2 u ucx x x t

ρ

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪ ∂ ∂

+ =⎪∂ ∂ ∂ ∂⎩

(10.9)

Substitute the wave solution (10.5) in the equation of motion (10.9) to obtain after

rearrangement

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

2 2 211 55 1 16 45 2 13 55 3

2 2 216 45 1 66 44 2 45 36 3

2 213 55 1 36 45 2 55 33 3

0

0

0

c c v U c c U c c U

c c U c c v U c c U

c c U c c U c c v U

α ρ α α

α α ρ α

α α α ρ

⎧ + − + + + + =⎪⎪

+ + + − + + =⎨⎪

+ + + + + − =⎪⎩

(10.10)

If the material coordinate and the global coordinate systems coincide, the stiffness

coefficients c16, c26, c36, and c45 are equal to zero, hence Equation (10.10) can be written

as

( ) ( )( )( ) ( )

2 211 55 1 13 55 3

2 266 44 2

2 213 55 1 55 33 3

0

0

0

c c v U c c U

c c v U

c c U c c v U

α ρ α

α ρ

α α ρ

⎧ + − + + =⎪⎪

+ − =⎨⎪

+ + + − =⎪⎩

(10.11)

The second equation in (10.11) is decoupled from the other twos. This means that the SH

wave is decoupled by the other two modes of propagation and the mathematical

formulation is simpler.

299

The system in Equation (10.10) accepts non trivial solution if the determinant is equal

to zero, i.e., after rearrangement,

6 4 21 2 3 0A A Aα α α+ + + = (10.12)

where

( ) ( ) ( )( )( ) ( ) ( ) ( )

( )( ) ( )( )( )( ) ( )( ) ( )

2 2 255 44 55 55 33 66 44 33 11 16 45 33

2 22 245 36 13 55 45 13 55 44 55 45 36 45 55

1

2 2 2 255 66 55 44 11 55

2 2 233 11 66 36 45 13 55 16 66 13 5

2

2

2

2

c c c v c c c v c c c v c c c

c c c c c c c c c v c c c cA

c c v c v c c v c v

c c v c v c c c c c c v c c

A

ρ ρ ρ

ρ

ρ ρ ρ ρ

ρ ρ ρ

− + − + − −

+ + + − + − − − +=

Δ− − + − −

+ − − + + + − − +

=

( )( ) ( )( )

( )( )( ) ( )

25

22 2 255 16 45 16 33 11 36 45

2 2 2 2 211 66 55 55 16

3

233 44 55 33 45

2 c v c c c c c v c c

c v c v c v c v cA

c c c c c

ρ ρ

ρ ρ ρ ρ

− − − − − +Δ

− − − − −=

ΔΔ = −

(10.13)

If we assume to know the velocity of the wave, the coefficients of Equation (10.12) are

known and their solution give three pairs of roots, i.e.,

2 1

4 3

6 5

α αα αα α

= −= −= −

(10.14)

Since the determinant of Equation (10.10) is equal to zero for any qα , q=1,2,…6, we find

the solution to the system to be

( )( ) ( )( )( )( ) ( )( )( )( ) ( )( )

( )( ) ( )( )

2 2 216 45 13 55 11 55 45 362

2 2 21 16 45 45 36 66 44 13 55

2 2 213 55 16 45 11 55 36 453

2 2 2 21 13 55 36 45 55 33 16 45

q qqq

q q q

q q q qqq

q q q q

c c c c c c v c cUV

U c c c c c c v c c

c c c c c c v c cUW

U c c c c c c v c c

α α ρ

α α ρ

α α α ρ α

α α ρ α

⎧ + + − + − +⎪ = =

+ + − + − +⎪⎪⎨

+ + − + − +⎪= =⎪ + + − + − +⎪⎩

(10.15)

300

Note that the following relations exist

2 1 4 3 6 5

2 1 4 3 6 5

V V V V V VW W W W W W

= = == − = − = −

(10.16)

With the use of Equation (10.15), solution (10.5) can be written as

( ) ( ) ( )1 36

1 2 3 11

, , 1, , qi x x vtq q q

qu u u V W U e ξ α+ −

=

= ∑ (10.17)

Through the use of the displacement strain relation (2.10) and the stress-strain relation

(2.14) we obtain the stress-displacement relation, i.e.,

31 2 1 211 12 13 16

1 2 3 2 1

31 2 1 212 23 23 26

1 2 3 2 11

31 2 1 2213 23 33 36

1 2 3 2 13

4 244

5 3

6

uu u u uc c c cx x x x x

uu u u uc c c cx x x x xT

uu u u uT c c c cx x x x xT

T ucT xT

∂∂ ∂ ∂ ∂⎛ ⎞+ + + +⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠∂∂ ∂ ∂ ∂⎛ ⎞+ + + +⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠⎧ ⎫

⎪ ⎪ ∂∂ ∂ ∂ ∂⎛ ⎞⎪ ⎪ + + + +⎜ ⎟∂ ∂ ∂ ∂ ∂⎪ ⎪ ⎝ ⎠⎪ ⎪ =⎨ ⎬∂⎪ ⎪

⎪ ⎪ ∂⎪ ⎪⎪ ⎪⎩ ⎭

3 3145

2 3 1

3 32 145 55

3 2 3 1

31 2 1 216 26 36 66

1 2 3 2 1

u uucx x x

u uu uc cx x x x

uu u u uc c c cx x x x x

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬

∂ ∂∂⎛ ⎞ ⎛ ⎞⎪ ⎪+ + +⎜ ⎟ ⎜ ⎟⎪ ⎪∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎪ ⎪∂ ∂∂ ∂⎪ ⎪⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟⎪ ⎪∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎪ ⎪

⎪ ⎪∂∂ ∂ ∂ ∂⎛ ⎞+ + + +⎪ ⎪⎜ ⎟∂ ∂ ∂ ∂ ∂⎪ ⎪⎝ ⎠⎩ ⎭

(10.18)

Substitute solution (10.17) into (10.18) to get after rearrangement

( )( )

( )1 3

11 13 161

12 23 262

13 33 3631

44 454

45 555

6 16 36 66

q

q q q

q q q

q q q i x x vtq

q q q qq

q q q q

q q q

c c W c VTc c W c VTc c W c VT

i U ec V c WTc V c WT

T c c W c V

ξ α

ααα

ξ α α

α α

α

+ −

+ +⎧ ⎫⎧ ⎫⎪ ⎪⎪ ⎪ + +⎪ ⎪⎪ ⎪⎪ ⎪+ +⎪ ⎪⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬+ +

⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪+ +⎪ ⎪ ⎪ ⎪⎪ ⎪ + +⎪ ⎪⎩ ⎭ ⎩ ⎭

∑ (10.19)

301

Since we want to relate the stresses on the top surface with those on the bottom, we

consider only the stresses on the plane normal to direction 3, i.e., T1, T4, and T5, and from

Equation (10.19)these are

( ) ( ) ( ) ( )1 36

* * * * * *33 13 23 3 5 4 1 2 3 1

1, , , , , , qi x x vt

q q q qq

T T T D D D U e ξ ασ σ σ + −

=

= = ∑ (10.20)

where * iσ σ ξ= and

( )( )

1 13 33 36

2 45 55

3 44 45

q q q q

q q q q q

q q q q q

D c c W c V

D c V c W

D c V c W

α

α α

α α

= + +

= + +

= + +

(10.21)

Combine Equations (10.17), (10.20), and (10.16) to write the displacements and stresses

expanded matrix form, i.e.,

1 3

1 3

3 3

3 3

1 11

1 1 3 3 5 52 11

1 1 3 3 5 53 13*

11 11 13 13 15 1533 13*

21 21 23 23 25 2513 15*

31 31 31 31 35 3523

1 1 1 1 1 1 i x

i x

i x

i x

i

u U eV V V V V Vu U eW W W W W Wu U eD D D D D D U eD D D D D D U eD D D D D D

ξα

ξα

ξα

ξα

ξ

σσσ

−

−

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −

=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥

− − −⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

( )1

5 3

5 315

i x vt

x

i x

e

U e

ξ

α

ξα

−

−

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(10.22)

For convenience, call the displacement and stress vector on the left hand side of Equation

(10.22) P, the 6x6 matrix X, the vector of the U1i elements U, and the diagonal matrix

whose elements are 3ii xe ξα D. Equation (10.22) can be written in a more compact form as

k k k kP X D U= (10.23)

Through this equation, it is possible to link the displacements and the stresses of the

bottom layer to those of the top layer. Call kD− the diagonal matrix for the case x3 = 0 (in

302

this case kD I− = ) and kD+ the diagonal matrix for the case x3 = dk . For the upper and

lower layer of the kth layer we have, respectively,

k k k k kP X D U X U− −= = (10.24)

k k kP X D U+ += (10.25)

Solving the first equation for the displacement vector and substituting it into Equation

(10.25) gives the relation between the displacements and stresses in the upper surface and

the displacements and stresses in the lower surface of the layer, i.e.,

k k kP A P+ −= (10.26)

where

1−= kkkk XDXA (10.27)

Appling the above procedure for each layer, it is possible to relate the displacements and

the stresses at the upper surface of the layered plate to those of the its lower surface via

the transfer matrix multiplication

1 1n nA A A A−= L (10.28)

The total transfer matrix expression is

{ }{ }

[ ] [ ][ ] [ ]

{ }{ }

uu u

u

u uA AA A

σ

σ σσσ σ

+ −

+ −

⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥⎣ ⎦⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

(10.29)

In order to obtain the dispersion curve, we must impose stress free upper and bottom

surface, this leads to the characteristic equation

303

0uA σ = (10.30)

In general, matrix uA σ is a complex matrix. Equation (10.30) means that the absolute

value of the determinant must be equal to zero. The determinant of uA σ is an implicit

relation between the wave number ξ and the phase velocity v. Equation (10.30) is quite

complicated and it is not possible to solve analytically in an explicit form. Numerical root

searching tools have to be used to search for the phase velocity for a given wavenumber.

We developed a computer code that derives the dispersion curves. The inputs are the

vector of the orientations of the layers and the material properties of each layer. The code

computes the transfer matrix for each layer in the global coordinate system and the total

transfer matrix (10.28) for different value of velocity and frequency. Through root search,

the final value of the velocity is found.

10.1.2 Dispersion curves for isotropic layers

Consider the case of a layer of an isotropic material, hence 11 22 33c c c= = , 12 13 23c c c= = ,

66 55 44c c c= = , and 66 11 122c c c= − , Equation (10.10) becomes

( ) ( )( )( ) ( )

2 211 66 1 11 66 3

2 266 66 2

2 211 66 1 66 11 3

0

0

0

c c v U c c U

c c v U

c c U c c v U

α ρ α

α ρ

α α ρ

⎧ + − + − =⎪⎪

+ − =⎨⎪

− + + − =⎪⎩

(10.31)

Note that, the second equation in the system is decoupled from the other two and it can be

solved separately. A non trivial solution to system (10.31) is found if the determinant of

the characteristic equation is equal to zero, i.e., if

4 21 2 0A Aα α+ + = (10.32)

304

where

( ) ( ) ( )

( )( )

22 211 66 66 66 11 66

1

2 211 66

2

c c v c c v c cA

c v c vA

ρ ρ

ρ ρ

− + − − −=

Δ− −

=Δ

(10.33)

Recall that 66c μ= , 11 2c λ μ= + , 2 2pc λ μ

ρ+

= , and 2sc μ ρ= , solutions of Equation

(10.33) are

2

1 2 2

2 2

3 4 211

5 6 1

1

1 1

s

p

vc

v vc c

α α

ρα α

α α α

= − = −

= − = − = −

= − =

(10.34)

and the solution to system (10.31) is

333 3

13

355

15 5

1

UWUUWU

α

α

⎧ = =⎪⎪⎨⎪ = = −⎪⎩

(10.35)

The stress coefficients (10.21) simplify to

( )( )

( )

213 13 3 33 3 5

23 55 3 3 3

15 13 5 33 52

525 55 5 5

5

1

22

1

D c c W

D c WD c c W

D c W

α μ α

α μα

α μ

αα μα

= + = −

= + =

= + = −

−= + =

(10.36)

The expanded matrix form for the isotropic plate becomes (we do not consider the SH

wave)

305

( )

3 3

3 31

5 3

5 3

1 11

3 3 3 5 5 13*33 13 13 15 15 13*13 23 23 25 25 15

1 1 1 1 i x

i xi x vt

i x

i x

u U eu W W W W U e

eD D D D U eD D D D U e

ξα

ξαξ

ξα

ξα

σσ

−−

−

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥− − ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− − ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(10.37)

Figure 10.4 shows that the dispersion curves derived through transfer matrix method are

exactly the same as those derived through solution of the Rayleigh – Lamb equation.

Same results are obtained either we consider a single layer aluminum plate of 1-mm

thickness or a two-layer aluminum plate of total thickness 1 mm. The derivation for

isotropic layer is useful if we want to study joints of metal layers.

Figure 10.4 Comparison of dispersion curves predicted by layered model (transfer matrix

method) vs. isotropic classic theory (Rayleigh – Lamb equation). a) One-layer

aluminum plate 2024-T3, 1-mm thick. b) Two-layer aluminum plate 2024-T3, 1-mm

total thickness. Dash lines: values derived from the Rayleigh – Lamb equation; Solid

lines: values derived from the transfer matrix method.

10.1.3 Dispersion curves for unidirectional composite plates

Let consider a plate made of one layer of unidirectional fibers. If we consider the wave

propagation in the direction of the fiber (θ = 0°) or perpendicular to the fiber (θ = 90°)

0 1000 2000 3000

2000

4000

6000

Phas

e ve

loci

ty (m

/s)

f (kHz) 0 1000 2000 3000

2000

4000

6000

f (kHz)

Phas

e ve

loci

ty (m

/s)

One-layer Two-layer

306

the shear horizontal wave is decoupled from the other two modes and they can be derived

separately (see Equation (10.11).) The code developed can compute the dispersion curves

for plate made of one unidirectional layer. We tested our computer code results versus

results found in Nayfeh (1995) for graphite-epoxy plate θ=0 and θ=45. The values

derived with the code where the same as shows in the book.

Figure 10.5 shows the dispersion curves derived for a unidirectional composite plate

made of one layer of 65% graphite 35% epoxy for different wave propagation direction.

Figure 10.5 Dispersion curves for plate made of one unidirectional layer of 65% graphite 35%

epoxy (material properties from Nayfeh 1995) as derived by our code. a) θ = 0°; b) θ

= 18°; c) θ = 36°; d) θ = 90°.

0 1 2 3 4 5 6 7 80

1000

2000

3000

4000

5000

6000

0 1 2 3 4 5 6 7 80

1000

2000

3000

4000

5000

6000

0 1 2 3 4 5 6 7 80

1000

2000

3000

4000

5000

6000

ξd ξd

ξd ξd

cs

cs

a) b)

c) d)

307

For the case of wave propagating along the fiber direction or transversely to the fiber

direction, the quasi-SH wave is decoupled from the other two waves and it is possible to

derive the quasi-antisymmetric and quasi symmetric mode separately. The value of the

phase velocity of the quasi S0 mode at low frequencies decreases as the wave

propagation angle increases. The quasi SH0 wave phase velocity is not constant trough

the frequency range for wave propagation angle close to 45 degrees.

As mentioned the transfer matrix method is not stable for high frequency-thickness

products, in this case we obtain a plot similar to Figure 10.6.

Figure 10.6 Transfer matrix instability for high frequency-thickness products.

10.1.4 Group velocity in composite plates

Figure 10.7 shows the dispersion curves of the first quasi antisymmetric wave mode (A0)

propagating in a composite plate made of graphite epoxy. The dispersion curves shown

are for different angles of propagation (θ ) with respect to the fiber direction.

ξd

c (m

m/s

ec)

308

Figure 10.7 Dispersion curves for first antisymmetric wave mode (A0) propagating at different

angles with respect to the fiber direction. Plate material: 65% graphite 35% epoxy

(material properties from Nayfeh 1995).

The phase velocity is higher when the wave propagates along the fiber direction. As

the angle of the wave propagation direction increases, the phase velocity decreases till

reaching a minimum in the direction perpendicular to the fiber. This is due to the fact that

along the fiber the material stiffness is greater than all the other directions and it

decreases while θ increases.

Assume we are interest in finding the group velocity of the wave propagating along

direction θ with respect to the fiber direction. From literature (Rose 1999) we know that

the group velocity vector is perpendicular to the phase slowness curve. The phase

slowness is the inverse of the phase velocity; hence, the phase slowness curve shows the

dependence of the relative arrival time of a plane wave on the direction of wave

propagation.

c (m

/s)

fd

θ = 90°

θ = 54° θ = 36°

θ = 18°θ = 0°

θ

309

From the slowness curve (Figure 10.8), we find for the point of interest P the

perpendicular Ec at the slowness curve point P. The angle of the perpendicular to the

slowness curve at point P is equal to ψ . Hence, the group velocity magnitude is given by

cosE

ccφ

= (10.38)

where φ ψ θ= − . Knowing the magnitude of Ec and the angle ψ for each point on the

slowness curve, it is possible to construct the wave surface.

Figure 10.8 Slowness curve and notation

Figure 10.9 shows the slowness curve for the 65% graphite 35% epoxy unidirectional

plate. The slowness curve is derived from the inverse of the phase velocity of the wave

for any angle of propagation at a given frequency thickness product. The slowness curve

is different for each frequency thickness product.

Figure 10.10 shows the wave surface for the 65% graphite 35% epoxy unidirectional

plate at frequency thickness product of 400 kHz-mm (solid line) and 1700 kHz-mm (dash

line) derived through Equation (10.38).

Ec

φ

θ ψ

1 ccP

310

Figure 10.9 Slowness curve for unidirectional 65% graphite 35% epoxy plate. Solid line:

frequency thickness product of 400 kHz-mm; Dash line: frequency thickness product

of 1700 kHz-mm. Values are 41 10c ⋅ .

Figure 10.10 Wave front surface for unidirectional 65% graphite 35% epoxy plate. Solid blue line:

frequency thickness product of 400 kHz-mm; Dash red line: frequency thickness

product of 1700 kHz-mm.

As shown in Figure 10.10, the wave surface is quite different from that of the slowness

surface. Hence, there is difference between the phase velocity propagation directions and

Fiber directionψ

Fiber direction θ

1c

311

amplitude and the group velocity propagation directions and amplitude. This phenomena

is quite important for directional transducers such as wedge transducers, comb

transducers, or rectangular PWAS transducers

10.1.5 Dispersion curves for quasi-isotropic composite plates

Experimental values and theoretical predictions of dispersion curves have been derived

for a quasi-isotropic plate [(0/45/90/-45)2s] Uni Tape T300/5208 with 2.25-mm thickness

and 1240×1240-mm size. The material properties are reported in Table 10.1.

Table 10.1 Ply material properties (Herakovich 1998)

T300/5208

Density (g/cm3) 1.54

Axial Modulus E1 (GPa) 132

Transverse Modulus E2 (GPa) 10.8

Poisson’s ratio ν12 0.24

Poisson’s ratio ν23 0.59

Shear Modulus G12 (GPa) 5.65

Shear Modulus G23 (GPa) 3.38

Modulus ratio E1/E2 12.3

Axial tensile strength XT (MPa) 1513

Transverse tensile strength YT (MPa) 43.4

Strength ratio XT/YT 35

The dispersion curves obtained from the code developed in our laboratory are

represented in Figure 10.11. From the output values of the program it is possible to

extract the first 3 modes of interest and convert the plot from wavenumber-thickness to

frequency-thickness representation. From the phase velocity it is possible to calculate the

group velocity curves as shown in Section 10.1.4.

312

Figure 10.11 Dispersion curves for a quasi-isotropic plate [(0/45/90/-45)2s]. (a) output from the

program; (b) elaboration of S0, SH, A0 modes.

Due to the properties of the quasi-isotropic materials, we expect that the phase

velocity magnitude to be almost constant for any propagation angle. The theoretical

values of the phase velocities are plotted in Figure 10.12.

Figure 10.12 Phase velocities for a quasi isotropic plate. Theoretical values for 0θ = ° , 90θ = ° ,

45θ = ° , and 135θ = °

S0

A0

SH

S0

A0

SH

ξd

(a)

(b)

0

1000

2000

3000

4000

5000

6000

0 100 200 300 400 500 600 700f (kHz-mm)

c (m

/s)

quasi S0

quasi SH0

quasi A0

0θ = °90θ = °

45θ = °135θ = °

313

From the dispersion curves it is possible to derive the wave surfaces. Note that since

the slowness curve is almost a circle, the group velocity directions will be the same as

those of the phase velocity.

Figure 10.13 Group velocities for a quasi isotropic plate. Experimental and theoretical values for

0θ = ° , 90θ = ° , 45θ = ° , and 135θ = °

Figure 10.13 shows the experimental and theoretical values of the group velocity for

the 16 layers composite plates. The group velocity is constant at the low frequencies. The

A0 group velocity is well predicted by the theoretical values. The S0 group velocities are

close but lower than the predicted ones. When the quasi-S0 velocity is close to that of the

quasi-SH velocity it becomes difficult to distinguish the two waves and determine their

velocities and amplitudes.

10.2 PWAS – GUIDED WAVES TUNING

In this section we present the novel formulation for deriving tuning curves in composite

plates. First we will briefly present the theory of excitation of guided waves in

0

1000

2000

3000

4000

5000

6000

0 100 200 300 400 500 600 700f (kHz)

c g (m

/s)

0θ = °

90θ = °45θ = °

135θ = °

314

composites plates as derived by Xi (2002) and we will show its limitations. Then, we will

extend the NME method to the case of composite plates.

A series of experiments where performed to determine the behavior of the tuning

between PWAS and structure in composite materials. A preliminary theoretical

development of tuning on composite has been developed and comparison with

experiments has been performed.

10.2.1 Excitation of guided waves

To derive PWAS tuning on isotropic plate it is possible to derive through Fourier

transformation method a close form solution as shown in Giurgiutiu (2008) and in

Section 7 of this dissertation. Xi (2002) extended the integral transform solution derived

for isotropic plates to the case of composite plates.

The analytical solution of the inverse transform of the Fourier integral is in this case

to be solved numerically. The integral transform solution follow the global matrix

procedure; first the displacement solutions are transformed in the wavenumber domain

(Fourier transform) and then the global system equation for the entire laminate is

determined as AC T= % , where A is the global matrix for the composite plate, C consists

of constant vectors for the layers to be determined from the boundary conditions on the

upper and lower surfaces of the plate, and T% is the Fourier transform of the external

force. Vector C is obtained from the solution of AC T= % . Knowing the vector of

constants, it is possible to determine the displacement functions in the wavenumber

domain for all the layers in the laminate. To obtain the displacement in the space domain,

the following inverse Fourier transform integration must be solved.

315

1( , ) ( , )2

ikxU z x U z k e dkπ

∞ −

−∞= ∫ % (10.39)

Solution of equation (10.39) can be done numerically because analytical solution is not

possible. Moreover, even if a numerical method is used, a proper treatment is needed, as

the integrand goes to infinity at the poles on the integral axis.

A different method, NME, can be used to determine the transducer frequencies for

any kind of plates. Hereunder we will apply the NME developed in Sections 7 and 0 to

the case of composite plates. First we will recall some basic concept of the NME theory.

Figure 10.14 Plate subject to surface tractions

10.2.1.1 Power flow

One of the parameter that appears in the tuning through normal mode expansion is the

power flow of the wave mode under consideration in the direction of propagation of the

wave, i.e., x. Hereunder, we reassume the derivation of the power flow through the proof

of orthogonality of the guided wave modes. The proof of orthogonality requires one

general acoustic field theorems, i.e., the complex reciprocity relation. Recall the

expression of the complex reciprocity relation as expressed in Equation (5.37), i.e.,

( ) ( )1 2 2 1 1 2 1 2 1 2 2 1:t

ρ∂∇ ⋅ + ⋅ = ⋅ + − ⋅ − ⋅

∂v T v T v v T S v F v F%% %% % % (10.40)

2d

y

x

ty tx

316

where v1, T1 are field solutions driven by source F1, and v2, T2 are field solutions driven

by sources F2. The tilde sign above a quantity signifies complex conjugate, i.e., if

c a ib= + , then c a ib= −% .

Assume that the source terms are equal to zero, i.e. 1 2 0= =F F , and that all field

quantities varies as i te ω , therefore it

ω∂→

∂. It is apparent that the time dependent terms

i te ω and i te ω− in Equation (10.40) cancel out, and the first term on the right hand side

becomes

( ) ( )1 2 1 2 1 2 1 2: : 0i t i t i t i te e e et t

ω ω ω ωρ ρ− −∂ ∂⋅ + = ⋅ + =

∂ ∂v v T S v v T S% %% % (10.41)

With these assumptions, Equation (10.40) becomes

( )1 2 2 1 0∇ ⋅ + ⋅ =v T v T% % (10.42)

Assume that solutions “1” and “2” are free modes with propagating factors ξm and ξn

respectively, and hence they can be written as

1

2

( , )

( , )

m

n

i xm

i xn

e y z

e y z

ξ

ξ

−

−

=

=

v v

v v and 1

2

( , )

( , )

m

n

i xm

i xn

e y z

e y z

ξ

ξ

−

−

=

=

T T

T T (10.43)

The structure under consideration is a layered waveguide structures with arbitrary

anisotropy and inhomogeneity. We can assume that the properties do not vary along z

direction. In the case of anisotropy, this can be achieved by choosing as reference axis the

principal coordinate system, hence Equation (10.43) becomes

1

2

( )

( )

m

n

i xm

i xn

e y

e y

ξ

ξ

−

−

=

=

v v

v v and 1

2

( )

( )

m

n

i xm

i xn

e y

e y

ξ

ξ

−

−

=

=

T T

T T (10.44)

317

Substitute the field expressions in Equation (10.44) into the complex reciprocity relation

(10.42) and explicit the del operator to obtain after rearrangement

( )( ) ( ) ( ) ( )m n m ni x i xm n n m m n n m m ni e e

yξ ξ ξ ξδξ ξ

δ− − − −− − ⋅ − ⋅ ⋅ = − ⋅ − ⋅ ⋅v T v T x v T v T y

% %% % %% % (10.45)

Integrate with respect to y Equation (10.45) to get

( ) ( )4d

m n mn n m m n di Pξ ξ

−− = − ⋅ − ⋅ ⋅v T v T y% %% (10.46)

where Pmn is power flow defined as

( )1 Re4

d

mn n m m nd

P dy−

= − ⋅ − ⋅ ⋅∫ v T v T x%% (10.47)

Assume either stress-free or rigid acoustic boundary conditions, i.e. 0⋅ =T y or 0=v at

y d= ± ; the right-hand side of Equation (10.46) is zero. Equation (10.46) becomes

( )4 0m n mni Pξ ξ− =% . (10.48)

This is the expression of the orthogonality relation for the waveguide modes. Equation

(10.48) is satisfied for

i. m nξ ξ= % if 0mnP ≠ ;

ii. 0mnP = if m nξ ξ≠ % .

The frequency spectrum shows that the waves modes occur in pairs with equal and

opposite wavenumber ξ. For propagating modes ξm is real, then Equation (10.47) can be

written as

318

( )2

2

1Re2

b

nn n nb

P dy−

= − ⋅ ⋅∫ v T x% (10.49)

nnP is nonzero and represents the average power flow of the nth mode in the x direction

per unit waveguide width (in the z direction) (see Figure 10.14).

10.2.1.2 Tuning of guided waves

For the study of excitation of waveguide modes, the starting point is the complex

reciprocity relation (10.40). In this case we retain terms F1 and F2. but we still assume

time harmonic waves. Relation (10.40) becomes

( ) ( )2 1 1 2 2 1 1 2 2 1 1 2y xδ δ

δ δ− ⋅ − ⋅ ⋅ + − ⋅ − ⋅ ⋅ = +v T v T y v T v T x v F v F% % %% % % (10.50)

Consider the case of a PWAS bonded on the top surface of a composite plate. In this

case, the wave guides can be excited at the acoustic boundaries by traction forces only,

⋅T y . We assume that the excited field (solution “1”) can be represented by mode

expansion as in Equation (7.8), i.e.,

1 1

1 1

( , ) ( ) ( )

( , ) ( ) ( )

n nn

n nn

x y a x y

x y a x y

= =

= =

∑

∑

v v v

T T T (10.51)

We assume also that solution “2” is of the type

2 2

2 2

( , ) ( )

( , ) ( )

n

n

i xn

i xn

x y y e

x y y e

ξ

ξ

−

−

= =

= =

v v v

T T T with 2 0=F (10.52)

Integrating Equation (10.50) with respect to y, we obtain

( ) ( )2 1 1 2 2 1 1 2 0d d

d d

dy dyy x

δ δδ δ− −

− ⋅ − ⋅ ⋅ + − ⋅ − ⋅ ⋅ =∫ ∫v T v T y v T v T x% %% % (10.53)

319

Substituting solutions (10.51) and (10.52) into Equation (10.53), we obtain

( )

( )

1 1

( ) 0

n

n

d i xn n d

di x

m n m m nm d

e

e a x dyx

ξ

ξδδ

−

−

− ⋅ − ⋅ ⋅

⎛ ⎞+ − − ⋅ =⎜ ⎟

⎝ ⎠⋅ ⋅∑ ∫

v T v T y

v T v T x

%

%

%%

%% (10.54)

Recall Equation (10.49), hence we get

( )1 1 ( ) 0n nd i x i x

n n m nmd m

e e a x Px

ξ ξδδ−

⎛ ⎞− ⋅ − ⋅ ⋅ + =⎜ ⎟⎝ ⎠

∑v T v T y % %%% (10.55)

According to the orthogonality relation (10.48), the summation in (10.55) has only one

nonzero term. Considering the propagating mode n (ξn real), Equation (10.55) can be

written as

( )4 ( )d

nn n n n n dP i a x

xδ ξδ −

⎛ ⎞+ = ⋅ + ⋅ ⋅⎜ ⎟⎝ ⎠

v T v T y%% (10.56)

where

( )1 1Re Re4 2

d d

nn n n n n n nd d

P dy dy− −

⎡ ⎤ ⎡ ⎤= − ⋅ − ⋅ ⋅ = − ⋅ ⋅⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦∫ ∫v T v T x v T x%% % (10.57)

Assume that the anisotropic plate is loaded over a finite portion in the y direction on the

upper surface by an infinite width traction force in the x direction

( ) ( ) ( )i t i ty xx e t x t x eω ω⎡ ⎤⋅ = = +⎣ ⎦T y t y x (10.58)

The right end side of Equation (10.56) becomes

( ) ( )( ) ( , ) ( , ) ( ) ( )d

n n ndy x y x y y d x

−⋅ ⋅ + ⋅ ⋅ = ⋅v T y v T y v t%% % (10.59)

320

The second term on the left-hand side is zero because we assumed traction free boundary.

Finally Equation (10.56) becomes

( )4 ( ) ( )nn n n nP i a x d xx

δ ξδ

⎛ ⎞+ = ⋅⎜ ⎟⎝ ⎠

v t% (10.60)

This is a first-order ODE that governs the amplitudes of the general modes. Its solutions

is

( )( ) ( )4

nn

xi xix

n nnn c

ea x d e dP

ξη η η

−

= ⋅ ∫v t% (10.61)

Where c is a constant used to satisfy the boundary conditions. Let the external tractions t

to be nonzero only in the interval a x a− ≤ ≤ , we can write the solution as (see Section

7.1)

( )( ) ( ) for 4

n n

ai x i xn

nnn a

da x e e x dx x aP

ξ ξ−+

−

= ⋅ >∫v t% (forward wave solution) (10.62)

The strain, hence the tuning curves, on the top surface of the plate is given by

1( ) ( ) ( ) ( )i t i tx n x n n x

n na x v y e dt a x v y e

xω ωε ξ

ω∂

= =∂ ∑ ∑∫ (10.63)

or in expanded form using Equation (10.62)

( ) ( ) ( ) for 4

n n

ai x i xi tn

x x nnnn a

d v y e e e x dx x aP

ξ ξωε ξω

−

−

= >∑ ∫v t%

(10.64)

This represents the tuning expression of the strain in the composite plate excited by the

PWAS. This derivation is formally equal to the case of an isotropic plate. The number of

modes present depends on the material properties of the composite plate. For the case of a

321

composite plate made of one layer of unidirectional fibers, the PWAS will excite only

Lamb modes (symmetric and antisymmetric) if we consider propagation along the fibers

or transverse to the fibers. In all other cases, three waves will be present.

The main difficult in solving Equation (10.64) lies in the derivation of the average

power flow. The average power flow is given by the integral over the plate thickness of

the velocity by the stress, and it must be performed numerically.

Recall the definition of the average power flow as given by Equation (10.57), i.e.,

( )1 ˆRe2

d

nn n ndP dy

−= − ⋅ ⋅∫ v T x% (10.65)

Consider the nth wave mode propagating in the kth layer of the composite plate (for

simplicity of notation we drop the subscript n), the integrand of Equation (10.65) is given

by

( ) 1 1 2 6 3 4ˆ v T v T v T⋅ ⋅ = + +v T x% % % % (10.66)

where the velocities are defined as

( ) ( ) ( )1 36

1 2 3 11

, , 1, , qi x x vtq q q

qv v v i v V W U e ξ αξ + −

=

= − ∑ (10.67)

and the stresses are defined as

( ) ( )1 3

1 11 13 16

4 44 45 1

6 16 36 66

q

q q qi x x vt

q q q q qq

q q q

T c c W c VT i c V c W U eT c c W c V

ξ αα

ξ α α

α

+ −

+ +⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= + +⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪+ +⎩ ⎭ ⎩ ⎭

∑ (10.68)

Once the dispersion curves are known, the stress and the velocity in each layer for each

mode are known and, hence, the average power flow can be computed. It is to be

322

emphasized that the derivation of the tuning curves presented here does not depend on the

particular method used to derive the dispersion curves.

The integral in Equation (10.64) depends on the assumption made on the bond layer

between PWAS and structure. For the composite plate, we assume that the thickness of

the bond layer approaching zero, i.e. we assume ideal bond conditions.

In the case of ideal bonding solution, the shear stress in the bonding layer is

concentrated at the ends of the PWAS tips. We can use the pin-force model to represent

the load transferred form the PWAS to the structure, i.e.,

( ) ( )0 if

( , )0 if

a x a x a x ax d

x a

τ δ δ⎧ − − + ≤⎡ ⎤⎪ ⎣ ⎦= ⎨>⎪⎩

xt (10.69)

Substituting (10.69) into Equation (10.64), we obtain:

( ) ( )0

00

( ) ( )4

n n n

ai x i x i xi tn

x x nnnn a

da v y e e x a e dx x a e dxP

ξ β ξωε τ ξ δ δω

− −

−

⎡ ⎤= − − +⎢ ⎥

⎣ ⎦∑ ∫ ∫

v% (10.70)

where, the term 0aτ is a constant depending on the excitation, the term ( ) 4nx nnv d P% is the

excitability function of mode n (depends on the mode excited and not on the source used

for excitation), and the term in square brackets is the Fourier integral of the excitation.

The integral solution is

( ) ( )0

0

2 sinn n

ai x i x

na

x a e dx x a e dx i aβ ξδ δ ξ−

−

− − + = ±∫ ∫ (10.71)

Substituting this result into Equation (10.64), we obtain:

( ) ( ) sin2

ni xi tnx x n n

nnn

di v y e e aP

ξωε ξ ξω

−= ∑v% (10.72)

323

10.2.2 Experimental results and theoretical predictions

A set of experiments where performed to verify the presence of the tuning between

PWAS and an anisotropic host structure. Pitch-catch experiments were performed in

which one PWAS served as transmitter and another PWAS served as receiver. The signal

used in the experiments was a Hanning-windowed tone burst with 3 counts. The signal

was generated with a function generator (Hewlett Packard 33120A) and sent through an

amplifier (Krohn-Hite model 7602) to the transmitter PWAS. A data acquisition

instrument (Tektronix TDS5034B) was used to measure the signal measured by the

receiver PWAS. The plate used in the experiments was a quasi-isotropic composite plate

[(0/45/90/-45)2]S, of T300/5208 Uni Tape with 2.25-mm thickness and size 1240×1240-

mm (material properties are reported in Table 10.1). Figure 10.15 shows the layout of the

experiments, the figure represents the central part of the composite plate.

Figure 10.15 Experiment layout for [(0/45/90/-45)2]S, of T300/5208 Uni Tape with 2.25-mm

thickness and size 1240x1240-mm.

0º

90º

T

S2

R1

S1

R3

R2

R4 R5

S3

324

The PWAS denoted with the letter T was the transmitter while the others were

receivers. The distance between the receivers and the transmitter was 250 mm. The angle

between the receivers was 22.5º. The frequency of the signal was swept from 15 to 600

kHz in steps of 15 kHz. At each frequency, we collected the wave amplitude and the time

of flight for the waves present in the plate. A problem, faced in the experiments, was the

efficacy of the ground. To obtain a strong signal the ground was provided by bonding a

sheet of copper on the composite surface. In this way the signal was strong and consistent

during the experiments.

10.2.2.1 Round PWAS

Experiments with round PWAS diameter 7-mm, 0.2-mm thick (American Piezo Ceramics

APC-850) were performed with the layout shown in Figure 10.15. Three waves were

detected: quasi-S0, quasi-A0, and quasi-SH0.

Figure 10.16 Tuning Experimental data for a round PWAS for different propagation directions. a)

quasi-A0 mode; b) Quasi-S0 mode and quasi-SH0 mode

Figure 10.16 shows the waves amplitudes as detected by PWAS R1, R2, R3, R4, and

R5, corresponding to the directions 0º, 22.5 º, 45º, 67.5 º, and 90º. The quasi-A0 mode

0

2

4

6

8

10

12

14

16

18

0 100 200 300 400 500 6000

2

4

6

8

10

12

14

16

18

0 100 200 300 400 500 600

f (kHz) f (kHz)

V (m

V)

a) b)

0°

22°

45° 67°

90°

0° 22°

45°

67°

90°

325

reached its maximum before the quasi S0 mode maximum. The quasi SH wave had

amplitude one forth of the S0 mode.

10.2.2.2 Square PWAS

The layout of the experiment is shown in Figure 10.15, transducer S1 was used as

transmitter while transducer S2 was used as receiver. Figure 10.17a shows the

experimental amplitudes of the three waves. The A0 mode extinguishes as soon as the

quasi-SH wave appears. The wasi-S0 mode has a maximum at 450 kHz and then

decreases. The quasi-A0 and quasi-S0 modes have a slope similar to that of a metallic

plate.

Figure 10.17 Experimental and theoretical tuning values. a) Experimental data for square PWAS.

Triangles: quasi-A0 mode; Circles: quasi-S0 mode; Squares: quasi-SH0 mode. b)

Experimental vs. theoretical values for first antisymmetric mode.

Figure 10.17b shows the comparison between experimental and theoretical values for

quasi-antisymmetric mode. The theoretical curve follows the behavior of the

experimental one. Tuning curves through NME method seems to be a promising tool to

predict the interaction between PWAS and structure.

0

1

2

3

4

5

6

7

8

9

10

11

12

0 100 200 300 400 500 600 700f (kHz)

V (m

V)

A0 S0

SH

f (kHz) 0 100 200 300

5

10

a) b)

326

PART III STRUCTURAL HEALTH MONITORING ISSUES AND

APPLICATIONS

327

In Part III we discuss issues and applications of SHM.

First we address the problem of reliability of structural health monitoring methods.

SHM technologies have been studied in research laboratories and it has been used in

some applications. In order for SHM methods to be used extensively in application fields,

the quality of SHM inspections must be ensured. Non destructive evaluation (NDE)

inspections quality has been ensured through specification requirements that control the

inspection process and results. However, so far, SHM methods have been left without

precise guidelines and the best practice has been to follow the specification provided for

NDE. One of the NDE quality control tools is the probability of detection curves (POD).

We show the procedure to develop POD curves for SHM through permanent attached

PWAS on composite panels.

In this Part, we show that the SHM inspection method using PWAS is not only

capable of detecting damage but also to function in any environment. In particular, our

research focuses on the applicability of SHM using PWAS for space applications. We

present the experiments we performed to determine the PWAS damage detection ability

under extreme environment conditions.

A set of experiments was performed to asses the survivability and durability of SHM

systems for real space applications. The set of experiments were determined from the

space mission guidelines for the determination of space-qualified NDE techniques. The

results show that SHM through PWAS is able to be a space-qualified SHM method.

328

In NDE methods, the bond layer between transducer and structure is not taken into

consideration because NDE sensors are not permanently attached to the structure. In

SHM using PWAS, the sensors are permanently attached to the structure and hence it is

important to be aware of the quality of the bond layer before and during the SHM

process. Here, we present a set of experiments developed to determine the effect of

partial bonding of the PWAS to the structure using capacitance measurements. From the

analysis of the results, we conclude that capacitance measurements can be an effectively

integrated in SHM methods to check whether a detection of damage is due to a real

damage or a change in the transducer bonding quality (true call or false call). We

determine confidence intervals that can be used as reference values to determine the

status of the bonding between transducer and structure.

In Part III, we also address the problem of SH waves and Lamb was scattering from

damage. In particular we show the solution for the case of a non-through the thickness

crack in a plate when a SH wave is incident. For the more difficult case of Lamb waves

scattering from discontinuities, we present the formulation of the problem, but no

solution is provided.

In most of our experiments, the PWAS have been used as active sensor. In this case,

the PWAS interrogates the structure on demand with an excitation frequency determined

by the user. PWAS transducers can be also used as passive sensors. Since a crack

propagating in a structure or an object impacting a structure releases a sudden strain

energy in the structure, the propagating strain wave can be detected by PWAS

transducers placed in the structure. We present two theoretical models to predict the wave

propagation when such events occur in a structure.

329

11 RELIABILITY OF STRUCTURAL HEALTH MONITORING

Lately SHM methods have received increase attention and many significant

improvements have been made by the scientific community in terms of capability to

detect, locate, and determine the size of damage in structures. SHM technologies are

slowly transitioning from the research labs to the application fields in civil, naval,

nuclear, aeronautical, and aerospace engineering. Although SHM methods are a mature

technique, still they lack of specification requirements to control the inspection process

and the quality of the inspection results.

Hereunder we highlight the state of the art of the specification for SHM methods and

we discuss the differences between NDE and SHM inspection requirements. We show

how a first set of specification for SHM could be derived and in particular how POD

curves for SHM with permanent PWAS can be derived.

11.1 SPECIFICATIONS FOR QUALITY STRUCTURAL HEALTH MONITORING INSPECTION

In most of the engineering fields, and especially in the aerospace and aeronautical fields,

the design philosophy is based on the damage tolerant design approach. This approach

ensures safe operation in the presence of flows (Gallagher et al. 1984). In the 70’s the

reliability procedures for non-destructive inspection (NDI) was based on the principle

that the structural component as-manufactured is considered to have a flaw of length a0.

The length of the flaw is determined by the inspection capability of the manufacturing

process. The flaw length will increase during the component service till a critical size

330

after a determined time t0 of service. An inspection through noninvasive methods is

expected to be performed after a time equal to half t0. The inspection should be able to

detect flaws of a length aNDE, where length aNDE is based on the knowledge of the

probability of detection curves for that particular NDE method used (ASM Handbook

Vol.17). Figure 11.1 shows a typical POD for increasing length damage. As the damage

size increases, the probability (hence the ability) that the NDE method will detect the

damage increases. In an ideal condition, we would like the curve to step to high

probability rates quickly and at small damage lengths.

Figure 11.1 Typical probability of detection (POD) curves for increasing damage. (Grills, 2001)

So far, no standards have been defined for SHM inspections; in the absence of a

precise guide lines, the general approach to determine the reliability of SHM methods is

to rely on the standards derived for NDE methods. However, SHM methods are quite

different from NDE methods and hence the inspection guidelines used so far are not a

good assessment of SHM capabilities. The lack of a standard in SHM is one of the

motives that make difficult the transition from pure research to application.

331

The main difference between NDE and SHM is that SHM can be performed on the

component while on service and no shut down is required, hence, theoretically, the

component could be inspected continuously. However, since a great quantity of data is

collected every time a scan of the structure is performed, in reality, the SHM can not run

continuously. To determine the time interval between two inspections, we must know the

probability of detecting the critical flaw size. For this motive POD curves for SHM

methods are still needed for field applications. Since SHM is performed while the

structure is on service, the POD curves are not only structure specific but also affected by

thermal and mechanical loads.

The performance of NDE methods is influenced mostly by the human factor, i.e., the

operator that performs the NDI. On the other side, SHM is influenced by both the method

used to install permanently the sensors and the transducers location on the structure.

These two aspects would lead to the desire to have POD curves for each SHM

manufacture (those who install the transducers on the structure) and for each structure

components with SHM transducer on it.

In NDE techniques the geometry of the structure under inspection does not influence

the outcome of the inspection; on the contrary, SHM is mostly geometry driven, meaning

that each structure geometry needs its particular SHM configuration.

Since the transducers are permanently attached to the structures, the sensors and the

bonding layer undergo degradation due to aging, corrosion, temperature cycling, and

vibrations. Reliability test on PWAS have been performed in order to asses the capability

of the transducer to survive different environments (Lin et al. 2009). However, a

systematic design of experiment to asses the reliability of the SHM structure (transducer-

332

bond-structure component) during service is still needed. The SHM system as well

should be under routine schedule accordingly with the survivability and reliability results.

A last aspect that should be proved is the algorithms robustness of the software used

for damage detection.

Table 11.1 shows the basic steps needed to make SHM method with permanent

attached transducers a reliable system for health monitoring inspection. (Chambers et al.

2006, Kessler S. 2005)

Table 11.1 Health monitoring reliability needs.

Factor under test Experiments result SHM

guideline output

Geometry beam, plate, joints, welds, rivet hole, elbows, etc

Material metallic, composite (unidirectional, layup, etc)

SHM method

Phase-array, sparse array, pitch-catch, pulse-echo,

electro mechanical impedance

SHM algorithm

Minimum flaw size

Inspection interval for structure

Transducer-bond-wiring system Aging, corrosion

SHM system – structural

component Load, vibrations, fatigue

Survivability

Sensor installation method

Reliability

Inspection interval for

SHM system

A POD curve should be determined for each given configuration of SHM method,

structure, and SHM algorithm. In this way the minimum detectable flaw size for each

333

particular configuration can be determined. Survivability and reliability tests will give the

maximum safe life span for the SHM method, or the durability of the system.

11.2 PROBABILITY OF DETECTION CURVES

To determine the POD curves for a well established NDE method, several specimens

with different flaw sizes are needed. Often the term POD is used just to refer to a limit set

of experiment where a specimen (either with one flaw or multiple flaws) for each

different geometry of interest is used to determine whether the SHM system is able or not

(one hit-miss data) to detect damage.

In a more comprehensive design of experiment aimed to derive a single POD curve,

we will need N identical specimens with different flaw sizes in the same location with

identical SHM system configuration. If we are also interested in other geometry

characteristics of the flaw (such as depth or delamination location) another set of

experiments is needed.

Consider, as an example, we want to determine a suit of experiments for determining

the probability of detection (POD) of damage in composite specimens using SHM

methodology.

We specify that the SHM methodology for damage detection under study relies on

permanently-attached unobtrusive minimally invasive transducers that are left in place on

the structure and are interrogated at will. In particular, these experiments will use

piezoelectric wafer active sensors. Other SHM transducers may be used as long as they

meet the description "permanently-attached unobtrusive minimally invasive that are left

in place on the structure and are interrogated at will".

334

We set that the damage detection methodology will rely on the following techniques

that can be used with a single PWAS transducers installation:

• pitch-catch

• pulse-echo

• phased array

• electromechanical (E/M) impedance

The data analysis will rely on algorithms that can identify the presence of damage, locate

the damage, and characterize it.

First we need to determine the SHM set up used in the experiments. In this case,

PWAS transducers will be used as transmitters and/or receivers. The setup will be:

• Eight (8) individual PWAS forming a sparse network (7-mm diameter, 0.2-mm

thick)

• Four (4) rectangular PWAS arrays (50-mm by 5 mm)

The transducers installation is shown in Figure 11.2.

Figure 11.2 Transducer lay out and specimen dimensions (all dimension in mm).

700

500

150

200

100

150

335

In the next step we select the specimens. In this case we want to derive a first basic

knowledge of the SHM system capabilities on damage detection on composites. We

select:

• Specimen material: conventional carbon/graphite fiber epoxy matrix prepeg, each

ply 0.127-mm.

• Specimen dimensions: 500x700-mm

• Extra length is left in the vertical direction for mounting in tensile testing

machine. Adhesively bonded wedge-in attached to the ends

• Four different composite plates:

1. Unidirectional (04)s, eight plies (thickness ~1mm)

2. Cross-ply (02/902)s, eight plies (thickness ~1mm)

3. Balanced #1: (0/45/-45/90)s, eight plies (thickness ~1mm)

4. Quasi-isotropic: (0/45/90/-45)s, eight plies (thickness ~1mm)

In the third step, we determine the damage type and location. The damage will be a

seeded damage consisting of 20-mm round Teflon inserts. The damage will be placed in

the specimens as shown in Figure 11.3. Five positions are considered, as indicated. The

seeded damage will be used to produce initial damage and to make it propagate through

the application of load.

336

Figure 11.3 Seeded flaw location (A, B, C, D, E) in the composite specimens

The seeded damage will be placed in two thickness positions: symmetric, i.e. at t/2

where t is the specimen thickness; asymmetric, i.e., at t/5 from the top.

Last step is the most important; we select the number n of specimens needed to

achieve statistical significance. For each configuration, it is possible to derive a POD

curve (ASM Handbook Vol.17). The SHM method is applied to a number of tests articles

with different flaws. The POD curves are generated from the results of the campaign of

tests. To generate the POD curves and the CI of the resulting curve, we need to perform

the same test on several specimens.

Table 11.2 Summary of the specimen configurations (Note: A, B, C, D, E: seed location)

Unidirectional Cross-ply Balanced Quasi-isotropic

Flaw depth (0)s (0/90)s (45/0/-45)s (0/45/90/-45)s

t/2 n·ABCDE n·ABCDE n·ABCDE n·ABCDE

t/5 n·ABCDE n·ABCDE n·ABCDE n·ABCDE

To construct the POD curves, we must determine the probability to determine a defect

with the chosen SHM method for different damage size and the CI interval. Table 11.3

A: Centered B: Edge side C: Corner

E: Edge off-setD: Edge up

PWAS

Seeded fault

337

reports the 95% CI for different values of the probability (P) of detection and different

sample sizes (n). A graphical representation of these results is given in Figure 11.4a.

Table 11.3 95% CI amplitude for different sample sizes (n) and probabilities

Probabilities, P

0.10 0.25 0.5 0.75 0.90

10 0 0.3 0 0.5 0.2 0.8 0.5 1 0.7 1

20 0 0.2 0.1 0.4 0.3 0.7 0.6 0.9 0.8 1

30 0.03 0.2 0.13 0.4 0.37 0.63 0.6 0.87 0.8 0.97

40 0.03 0.17 0.15 0.38 0.38 0.62 0.62 0.85 0.82 0.97

50 0.04 0.18 0.16 0.36 0.38 0.62 0.64 0.84 0.82 0.96

60 0.03 0.17 0.17 0.35 0.4 0.6 0.65 0.83 0.83 0.97

70 0.04 0.16 0.17 0.34 0.4 0.6 0.66 0.83 0.84 0.96

80 0.05 0.16 0.17 0.33 0.41 0.59 0.68 0.82 0.84 0.95

90 0.06 0.16 0.18 0.32 0.41 0.59 0.68 0.82 0.84 0.94

n

100 0.05 0.15 0.18 0.32 0.42 0.58 0.68 0.82 0.85 0.95

For example, if the SHM method is able to detect a flaw with a probability of 0.90

(10 out of 100 are not detected), with a sample size of 20, there is a 95% of probability

that the observed probability of detecting a defect is between (0.8, 1). As we can see this

interval is quite large; things become worst as the probability of detection decreases (for

n=20 and p=0.5 the interval is from 0.3 to 0.7) (see Figure 11.4a).

338

Figure 11.4 Statistical criteria. a) 95% confidence interval of probability of detection for

increasing values of n. b) Acceptance criteria.

In order to obtain smaller CI we must choose high sample size. Once the desired

sample size is chosen, we must determine a criterion to determine whether the SHM

method detects or not the defect.

As an example, consider the case in which we are using pitch-catch method to

perform SHM on a specimen. The control variable it is the amplitude of the transmitted

wave. As the damage increases the amplitude is expected to decrease. The wave

amplitude is subject to oscillation due to the noise on the signal and other factors that are

not correlated with the damage. In order to determine the critical value of the wave

amplitude below which we can say that the difference is due to noise, we must create a

baseline of readings. Through the baseline readings it is possible to determine the signal

distribution and the percentile of the distribution. We are interested in high percentile

because, we will say that a reading does not belong to the baseline distribution if its value

is above a critical value that is the value of the 0.99 percentile (see Figure 11.4b).

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

lower and upper limit

prob

abili

ty o

f det

ectio

n

Sample size n, 95% CI1020304050

60708090100

Lower limit valuesas n increases

Upper limit valuesas n increases

0 5 10

0.0

0.1

0.2

0.3

0.4

0.5

Den

sity

0 5 10

0.0

0.1

0.2

0.3

0.4

0.5

Den

sity

Discrimination threshold

Noise orbaseline Signal

Criteria

339

To obtain the noise distribution or baseline, we must perform on each specimen N

data collection in the undamaged condition. Then the specimens are damaged and the

value of the amplitude is compared with the critical value.

A “bootstrap” technique is used to determine the critical value and the best number of

baseline to record.

As we can see from the example the number of specimen proposed is considerable

high. A thoroughly determination of POD curves for SHM methods is not only time

consuming but also expensive. However, we think that in the future more strict

requirements for the implementation of SHM on real application will require rigorous

POD derivations to make SHM as reliable and well accepted in the industrial community

as NDE technologies are now days.

340

12 SPACE QUALIFIED NON-DESTRUCTIVE EVALUATION & STRUCTURAL

HEALTH MONITORING

One of the objectives of my research was to determine the technology readiness level

(TRL) of structural health monitoring through permanent attached PWAS for space

applications. The aim was to achieve the validation of system/subsystem/component in

relevant space environments (TRL 5). In this section, we show the experimental results of

the subsystem/component validation program.

12.1 INTRODUCTION

Previous research (Cuc et al., 2005; Kessler et al., 2001; Zhangqing and Ye (2005), Matt

et al., 2005) investigated the possibility of using embedded ultrasonic non-destructive

evaluation and the opportunity for developing embedded structural health monitoring for

damage detection. Wave propagation methods were used for detection of cracks,

corrosion and disbonds in stiffened metallic panels. The ability to detect cracks under

bolts and rivets was also investigated. It was found that successful damage detection can

be achieved using wave propagation methods as well as the electromechanical impedance

method. A comparison of the damage detection methods for various damage types is

given in Table 12.1.

341

Table 12.1 Summary of PWAS damage detection methods. (Cuc et al., 2005)

Wave Propagation Standing Wave Method Damage Pitch-Catch Pulse-echo Phased-Array EM Impedance

Disbond Fair Excellent -- Excellent Cracks -- -- Excellent Fair Corrosion -- -- -- Excellent Crack under bolt

Excellent Fair -- Fair

Delamination Excellent Excellent -- Excellent

As most of the space applications are moving towards composite material, a new set

of experiments was determined to validate the SHM system for space application on

composite structures. The space structural components are subjected to high loads and

extreme low temperatures, i.e., cryogenic temperature i.e., T=-300F (-185 C). We wanted

to prove: first, that the SHM system developed in our laboratory was able to perform

damage detection of different kinds on composite structures; second, that it was able to

withstand cryogenic temperatures (CT); and, third, that at the same time was able to

perform damage detection on composites at cryogenic temperatures.

12.2 SUBSYSTEM/COMPONENT SPECIFICATION

The basic element used for damage detection in these experiments was a round 7-mm

diameter, 0.2-mm thick PWAS. We proved that the piezoelectric material was able to

maintain actuation abilities in the cryogenic environments. To reproduce the cryogenic

temperature, we used containers immersed in liquid nitrogen since its liquid temperature

is approximately at the cryogenic temperature.

To check the ability to withstand CT and to operate in those conditions, a set of

experiments was performed in which a PWAS attached to a composite strip was

immerged ten times in a container with liquid nitrogen. Impedance data where taken

342

before each submersion. Figure 12.1a shows impedance signatures taken. It can be seen

that the material retained its peaks and their relative frequency location.

Figure 12.1b shows pitch-catch wave propagation before, during, and after

submersion of the specimen in liquid nitrogen. The data collected showed that the PWAS

was able to perform at cryogenic temperatures. It should be noted that the signal

propagated in the nitrogen had smaller peak to peak amplitude, but at the same time had

proportionately smaller background noise. The amplitude at room temperature (RT) after

submersion in liquid nitrogen did not return to original amplitudes. These experiments

were conducted with the transducers in direct contact with the liquid nitrogen. During

submersion some leakage of the wave in to the liquid can happen. In real application the

transducer will be not in contact with the liquid nitrogen.

Figure 12.1 Survivability and performance of PWAS under thermal fatigue. a) Indication of

survivability through resumption of resonant properties after submersion in liquid

nitrogen (PWAS, AE-15, room temperature). b) Wave propagation in composite for

various thermal environments. Comparison of a wave packet before, during, and

after submersion in liquid nitrogen.

0

500

1000

1500

2000

2500

3000

5 10 15 20 25 30 35 40Frequency (kHz)

Impe

danc

e (O

hms)

Baseline 1 Cycle2 Cyc les 3 Cycles4 Cyc les 5 Cycles6 Cyc les 7 Cycles8 Cyc les 9 Cycles10 Cycles

-25-20-15-10-505

10152025

0 50 100 150 200

Time (microseconds)

Ampl

itude

(mV

)

BaselineCryogenic1 Cycle

a) b)

343

The adhesive layer between the PWAS and the structure was a critical aspect.

Incorrect thickness, porosity, poor chemical preparation, etc. lead to poor transmission of

shear energy. The adhesive selected was Vishay M-Bond AE-15 (2-component) since it

was able to retain its properties even at the required temperatures. Su/Pb solder was used

for standard RT applications. Since at CT lead becomes brittle, indium was used in CT

applications because of the materials ability to retain mechanical properties even at CT.

In order to test the performance of the PWAS on various types of composite materials

and structures; four different test specimens were utilized. The first type of specimen was

a composite strip made of unidirectional fibers. The specimen was 400 x 51-mm (16”x2”)

and 1-mm thick. The fibers direction was parallel to the longest direction of the

specimen. A schematic of this specimen is shown in Figure 12.2. Two composite strips

were used with two PWAS installed on each at a distance of 150 mm. We used the

unidirectional strips to test damage detection of through holes at room temperature. In

one of the strip the hole was in line with two PWAS, in the second specimen the hole was

off-set from the pitch-catch path.

Figure 12.2 Unidirectional composite strips with PWAS installed. a.) Hole in the pitch-catch path;

b.) Hole off-set from the pitch-catch path.

Fiber direction

PWAS 0 PWAS 1

Pitch-catch path Damage

Fiber direction

PWAS 0 PWAS 1

a)

b)

344

The second type of specimen used was a quasi-isotropic composite plate [(0/45/90/-

45)2]S, of T300/5208 Uni Tape with 2.25-mm thickness and 1.200×1.200-mm size

(4’x4’). The composite panel specimen was used to investigate the PWAS damage

detection performance at room temperatures both for through-hole detection and impact

damage detection. A schematic of the test specimen and experimental setup used during

the experiments are shown in Figure 12.3.

Figure 12.3 Experimental setup for quasi-isotropic plate experiments. The damage sites are

marked as: (i) “Hole” for a through hole of increasing diameter; and (ii) I1, I2 for two

impacts of various energy levels.

The third type of specimen used during the experimental characterization of the

PWAS was a composite lap joint. For this specimen, material and layers lay-up was not

specified. The geometry of the specimen is shown in Figure 12.4. The specimen was built

with 16 seeded defects inside (Teflon patches), 8 on two rows distant respectively 76 mm

(3”) and 152 mm (6”) from the longitudinal edge. The thickness of the joint was about 13

mm (5/8”). The 8 patches in each row are approximately equidistant. We performed the

PWAS ComputerTektronix GPIB GPIB

8-pin ribbon

t

Parallel PortASCU2-PWAS

HP

33120

8-channel

345

following tests on this specimen: damage detection at room temperature with PWAS pair

1; damage detection at cryogenic temperature with PWAS pair 2; damage detection at

cryogenic temperature under uni-axial load with PWAS pair 3.

Figure 12.4 Lap joint; Teflon patches location (crosses) and PWAS location (circles).

The fourth type of specimen (Figure 12.5) utilized during the experimental testing

was a thick plate. As for the previous specimen, material and layers lay-up was not

specified. The composite tank interface specimen had plate of dimension 305 x 229-mm

(12”x9”) and thickness about 7 mm (1/4”). The specimen was fabricated with 16 patches

of different sizes located between various plies. Two experiments were performed on the

specimen: patch detection at room temperature; patch detection at cryogenic temperature.

Figure 12.5 Schematic of thick composite specimen and location of Teflon inserts (crosses).

P01 P00

346

Table 12.2 summarize the specimens used, the type of damages, the environmental

condition, and the methods used in the experiments. The data collected in each

experiment were analyzed using damage index (DI) software developed in our lab. The

DI software was used to assess the severity of the damage in each test run. The DI is a

scalar quantity that results from the comparative processing of the signal under

consideration. The damage metric should reveal the difference between readings

(impedance spectrum or wave packets) due to the presence of damage. Ideally, the DI

would be a metric, which captures only the spectral features that are directly modified by

the damage presence, while neglecting the variations due to normal operation conditions

(i.e., statistical difference within a population of specimens, and expected changes in

temperature, pressure, ambient vibrations, etc.). To date, several damage metrics have

been used to compare impedance spectra or wave packages and assess the presence of

damage. Among them, the most popular are the root mean square deviation (RMSD), the

power, the mean absolute percentage deviation (MAPD), and the correlation coefficient

deviation (CCD) (Giurgiutiu, 2008). In our experiments we have used the RMSD DI

shown in Equation (12.1).

( ) ( ){ }

( ){ }

20

20

Re Re

Re

i in

in

S SRMSD

S

−=

∑

∑ (12.1)

RMSD yields a scalar number, which represent the relationship between the compared

readings. The advantage of using this method is that the data do not need any

preprocessing, i.e., the data obtained from the measurement equipment can be directly

used to calculate the DI.

347

Table 12.2 Summary of experiments discussed in this paper.

Specimen Damage Environment Loading Methods

Unidirectional strips Hole RT Free P-C

Hole Composite panel

Impact damage RT Free P-C, P-E

RT Free Lap-joint Impact damage

CT Free P-C

RT Thick plate Delamination

CT Free P-C

12.3 DAMAGE DETECTION EXPERIMENTS ON TEST SPECIMENS

12.3.1 Unidirectional composite strips

The initial testing of the PWAS based damage detection began with unidirectional strips

shown in Figure 12.2. In both strips we installed two round PWAS 150 mm apart and we

used the pitch-catch method to detect the damage. In the first experiment, we determined

the smallest through-hole diameter that was detectable by the PWAS when the through-

hole was centered with the PWAS pair (see Figure 12.2a). In the second experiment, we

determined the smallest detectible hole diameter when the hole was offset 20 mm with

respect to the pitch-catch path (see Figure 12.2b).

Five baseline readings were taken when the strips were undamaged; then, a hole with

0.8-mm diameter was drilled on each specimen. The holes were enlarged in 11 steps until

they reached 6.4 mm in diameter. For each step, we took five pitch-catch readings. Table

12.3 reports the dimensions of the hole for each step and reading.

348

Table 12.3 Hole sizes for corresponding readings in the unidirectional composite strip

experiments.

Step # Readings Hole size (mm) Step # Readings Hole size (mm)

0 00 – 04 -- 6 31 – 35 3.2

1 05 – 09 0.8 7 36 – 40 3.6

2 10 – 14 1.5 8 41 – 45 4.0

3 15 – 19 1.6 9 46 – 50 4.8

4 20 – 25 2.0 10 51 – 55 5.5

5 26 – 30 2.4 11 56 – 60 6.4

On the unidirectional composite strips, the excitation signal used was a 3 count tone

burst at 480 kHz, which resulted in the strongest S0 wave packet. The signals were

analyzed using the RMSD DI and the results are shown in Figure 12.6 for both the

centered and off-center hole cases. Each dot in the graph represents a reading (there are 5

readings for each step to indicate reproducibility). The first 5 readings are the baseline

readings, i.e., the strip without damage.

Figure 12.6a shows that the DI increases monotonically with the increasing hole size

which will allow for easy interpretation of the DI in relation to the damage size.

0

0.05

0.1

0.15

0.2

0 5 10 15 20 25 30 35 40 45 50Reading #

DI v

alue

Figure 12.6 DI analysis of the damaged unidirectional composite strip. a.) Hole in the pitch-catch

path; b.) Hole off-set from the pitch-catch path.

349

Figure 12.6b shows that initially the DI value for off-axis hole increases

monotonically, but then the DI value plateaus at 1.5 mm and it jumps again when the hole

diameter reaches 3.2 mm. This indicates that, in unidirectional composites, off-axis

damage increase may be more difficult to detect and identify compared to centered

damage.

Since in every experiment conducted, the variance within each step is quite small,

from now on we will report only the mean value of the readings in each step.

12.3.2 Quasi-isotropic composite laminate

For the quasi-isotropic composite plate, two different kinds of damage, through-hole and

impact damage, were investigated at room temperature conditions. Twelve PWAS were

installed in a sparse array on the quasi-isotropic composite plate as shown in Figure 12.3;

the distance between the 6 PWAS pairs was 30 mm. In this case, the data was collected

automatically through ASCU2 system with an input voltage of the signal from the

function generator of 11 V (Figure 12.3). The 11-V limit was the maximum input voltage

that it is possible to send through ASCU2. The excitation signal used during the

interrogations was a 3 count tone burst at central frequency 54 kHz and 255 kHz. These

frequencies were selected through Lamb wave tuning experiments to maximize the A0,

and S0 wave modes. At 54 kHz, it was possible to obtain the maximum pseudo A0 mode;

at 255 kHz we obtained the maximum pseudo S0.

12.3.2.1 Detection of through-holes

In the through-hole detection case, data was collected from PWAS 0, 1, 5, 8, 12, and 13

(Figure 12.3). Each PWAS was in turn a transmitter and a receiver. Readings were taken

with the plate in an undamaged state and the plate in a damaged state. Four baseline

350

readings were taken in the undamaged configuration. A hole was drilled between PWAS

1 and 12. The location of the hole was halfway between these two PWAS. The diameter

of the hole was increased in 13 steps. At each step four readings were recorded. Table

12.4 reports for each step, the number of readings recorded, and the hole dimension. Each

reading was compared to the baseline reading 0 through DI analysis. The DI value was

computed through RMSD DI of Equation (12.1).

Table 12.4 Hole diameters corresponding to the quasi-isotropic plate damage detection

experiment.

Step Reading # Hole size in mil [mm] Step Reading # Hole size in mil [mm]

1 00 – 03 0 2 04 – 07 032 [0.8]

3 08 – 11 059 [1.5] 4 12 – 15 063 [1.6]

5 16 – 19 078 [2.9] 6 20 – 23 109 [2.8]

7 24 – 28 125 [3.2] 8 29 – 32 141 [3.5]

9 33 – 36 156 [4.0] 10 37 – 40 172 [4.4]

11 41 – 44 188 [4.8] 12 45 – 48 203 [5.2]

13 49 – 52 219 [5.5]

For the pitch-catch analysis we took in consideration only the data coming from the

following PWAS configuration: PWAS 0 transmitter, PWAS 13 receiver; PWAS 1

transmitter, PWAS 12 receiver; PWAS 5 transmitter, PWAS 8 receiver.

This experiment allowed us to determine the minimum hole diameter that the two

PWAS pairs (0 – 13, 1 – 12) were able to detect. We used PWAS pair 5 – 8 to check

whether there was any difference between the PWAS pairs close to the damage and those

351

far away. We wanted to prove that any change detected through PWAS pairs 0-13 and 1-

12 was due only to the increase of damage size and not to other factors.

At an excitation frequency of 54 kHz, only the A0 mode is present. The wave

velocity of the A0 mode in this material is 1580 m/sec; the wavelength is 29.3 mm.

Figure 12.7a shows the DI values for two PWAS pairs (1 – 12, 5 – 8). As the hole

diameter increases, the DI values for the PWAS pair close to the hole increase while the

DIs for the PWAS pair 05 – 08 remain almost the same. We analyzed the data with

statistical software (SAS) and we observed that with a significance of 99%, PWAS pair 1

– 12 can detect the presence of the hole when its diameter is 2.8 mm while PWAS pair 0

– 13 could detect the hole at diameter 3.2-mm with the same significance level. There

was no significant difference between the DI values of PWAS pair 5 – 8. For an

explanation of the results obtained with SAS see Appendix H.

Figure 12.7. DI values at different sizes of the hole and PWAS pairs. a) Excitation frequency of

54 kHz. b) Excitation frequency of 255 kHz. Circle: PWAS pair 0-13; Triangle: PWAS

pair 5-8.

0

0.05

0.1

0.15

0.2

0.25

0 0.8 1.6 2.9 2.8 3.2 4 4.4 4.8 5.2 5.50

0.05

0.1

0.15

0.2

0.25

0 0.8 1.6 2.9 2.8 3.2 3.5 4 4.4 4.8 5.2 5.5

Hole size (mm)

DI DI

Hole size (mm) b) a)

352

At an excitation frequency of 255 kHz, the S0 mode has maximum amplitude. The

wave velocity is about 6000 m/sec, the wavelength is 23.5 mm. Figure 12.7b shows the

DI values for two different PWAS pairs (0 – 13, 5 – 8). As the hole diameter increases,

the DI values for the two PWAS pairs close to the hole increases while the DI for the

PWAS pair 5 – 8 remain almost the same. With a significance of 99%, PWAS pair 0 – 13

and PWAS pair 1 – 12 (not shown in the graph) could detect the presence of the hole

when its diameter was 3.2 mm. There was no significant difference between the DI

values of PWAS pair 5 – 8.

Figure 12.8. DI values at different hole size, Frequency 54 kHz. Pulse – echo.

Pulse – echo analysis was performed for PWAS 0 – 1. PWAS 0 was the transmitter

while PWAS 1 was the receiver. Figure 12.8 shows the change of DI values with hole

size for an excitation frequency of 54 kHz. Analyzing the data we found that there was

significant difference between step 1 (baseline) and step 7. We could detect the hole

when its diameter was 3.2 mm with 99% confidence.

Hole size (mm)

DI

0

1

2

3

4

5

6

0 0.8 1.6 2.9 2.8 3.2 3.5 4 4.4 4.8 5.2 5.5

353

12.3.2.2 Detection of impact damage

Impact damage was the second form of damage that was studied on the quasi-isotropic

composite plate. The impact damaged was applied to the plate using the impactor shown

in Figure 12.9. The impactor had a hemispherical tip of 12.7 mm in diameter (0.5”) and

its weight was 391 g (13.79 oz). The impactor weight could be increased by adding

barrels (Figure 12.9b) to the base configuration of Figure 12.9a. Each barrel weighted

500 g. (1 lb 1.63 oz); a total of 3 barrels could be assembled on the impactor.

Figure 12.9. Impactor. a) Base impactor with hemispherical tip; b) barrel; c) impactor assembled.

Two impact damages on different locations were produced on the plate with two

different impactor configurations (respectively two barrels and one barrel). The impactor

used for damage site A had a total weight of 1391 g (3 lb 1.1 oz). Two different impact

damage states were created at this site by dropping the impactor from different heights.

The first impact damage state had an impact energy level of 6 ft-lb and hit the plate at

about 3.5 m/sec (11 ft/sec); the second impact damage state had an impact energy level of

12 ft-lb and hit the plate at about 5 m/sec (16 ft/sec).

a) b) c)

354

The impactor used for damage site B had a total weight of 890 g (1 lb 15.5 oz)). The

first impact damage state had an impact energy level of 6 ft-lb and hit the plate at about

4.3 m/sec (14 ft/sec); the second impact damage state had an impact energy level of 12 ft-

lb and hit the plate at about 6 m/sec (20 ft/sec). Table 12.5 shows the energy and velocity

levels for the damage states at both damage sites. For both damage site A and damage

site B, we recorded 11 baseline readings and 10 readings for each energy level. The

readings were again collected through the ASCU2 system. The input voltage of the signal

was limited to 11 V.

Table 12.5 Summary of impact test parameters on quasi-isotropic plate specimen.

Damage site Readings Energy m-Kg (ft-lb) Velocity m/sec (ft/sec) Step

00-10 1

11-20 0.83 (6) 3.5 (11) 2

A

21-30 1.66 (12) 5 (16) 3

00-10 1

11-20 0.83 (6) 4.3 (14) 2

B

21-30 1.66 (12) 6 (20) 3

The first impact site (Damage A) was produced between PWAS 12 and PWAS 11

(see Figure 12.3 for reference). No visual damage was produced at 6 ft-lb energy level.

After the second impact at energy level of 12 ft-lb, damage could be seen on the opposite

surface of the plate. We took the readings for PWAS pairs 11 – 12 and 9 – 10. Each

PWAS of each pair was used once as a transmitter and once as a receiver. The second

damage site (Damage B) was produced between PWAS 3 and PWAS 10 (see Figure 12.3

for reference). We collected readings from PWAS pairs 3 – 10 and 5 - 8. No visible

355

damage was produced after the two impacts. However, the presence of damage in the

plate structure was registered through standard ultrasonic methods.

Figure 12.10a shows the DI values for damage site A for an excitation frequency of

54 kHz. There is little difference between the DI values of the PWAS pair (9 – 10) far

from the impact damage. The PWAS pair with the damage in between (11 – 12) shows a

significant change in DI values after the second impact (energy level 12 ft-lb) indicating

that it is possible to detect the damage after the second impact. The DI values for damage

site B (Figure 12.10c ) at 54 kHz were qualitatively the same as for damage site A.

Figure 12.10b shows the DI values for the three different steps for damage site A or an

excitation frequency of 225 kHz. The PWAS pair that is far from the impact damage does

not show much difference between the DI values of the three steps. The PWAS pair with

the damage in between (11 – 12) shows a change in DI values after the second impact

(energy level 12 ft-lb); however, the change is not significant as in the case of frequency

54 kHz. The pseudo S0 mode is less sensitive to this kind of damage in the composite

panel. Again, similar results were obtained for damage site B (Figure 12.10d).

356

Figure 12.10 Pitch-catch DI values as a function of the damage level for two PWAS pairs. a)

Impact at site A, excitation frequency of 54 kHz. Circle: PWAS pair 12-11; Square:

PWAS pair 9-10; b) Impact at site A, excitation frequency of 225 kHz. Circle: PWAS

pair 12-11; Square: PWAS pair 9-10; c) Impact at site B, excitation frequency of 54

kHz. Circle: PWAS pair 10-3; Square: PWAS pair 5-8; d) Impact at site B, excitation

frequency of 225 kHz. Circle: PWAS pair 10-3; Square: PWAS pair 5-8

Pulse – echo analysis was performed for PWAS 11 – 10 on damage site A. PWAS 11

was used as transmitter while PWAS 10 was the receiver. The frequency used where 54

kHz and 225 kHz. The latter was used instead of 255 kHz because it gave better results

for the S0 mode. Figure 12.11 shows the DI values at different step and excitation

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 6 120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 6 12

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 6 12

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 6 12

DI

Energy (lb-ft) Energy (lb-ft) a) b)

Energy (lb-ft) Energy (lb-ft) c) d)

DI

357

frequencies for damage site A. There is a statistically significant difference between step

1 (baseline) and other two steps for the case of 54 kHz, while there is significant

difference between Step 1 and Step3 (impact at 12 ft-lb) for the case of 225 kHz. As in

the pitch-catch method, the S0 mode was less sensitive to impact damage.

Figure 12.11 Pulse-echo DI values as a function of the damage level for two PWAS pairs at

damage site A. Circle: excitation frequency of 54 kHz; Square: excitation frequency

of 225 kHz.

12.3.3 Composite lap-joint

Detection of impact damage was performed on a composite lap-joint specimen. The

impactor configuration used in this case was the same as that shown in Table 12.5. A

total of 11 readings were taken in the undamaged baseline configuration, 10 readings

were taken after the impact with energy level 6 ft-lb, and 10 readings were recorded after

the impact at 12 ft-lb. The first reading of the baseline configuration was used as the

reference reading for the DI analysis.

The location of the PWAS and the impact sites on the lap-joint are shown in Figure

12.4. Two columns of PWAS were installed; each column of PWAS was bonded close to

0

0.2

0.4

0.6

0.8

1

1.2

0 6 12

DI

Energy (lb-ft)a)

358

one of the edges of the joint. The distance between the columns was 203 mm (8”). Each

impact damages were located between PWAS pairs. The input voltage of the signal for

these experiments was increased to 18 V to obtain a better signal to noise ratio, hence

manual scan of the PWAS was performed. We used the Lamb wave tuning method to

select the frequencies at which there was only the presence of one mode. We found that

such conditions existed at 60 kHz and 318 kHz. The wave speed at 60 kHz was 1175

m/sec, the wavelength 21.8 mm. The wave speed at 318 kHz was 3065 m/sec, the

wavelength 10 mm. PWAS pair 1 in Figure 12.4 was used for damage detection at room

temperature.

Figure 12.12a shows the DI values for the two different frequencies at room

temperature. Both low and high frequencies were able to detect impacts at 6 ft-lb and 12

ft-lb with a significance level of 99%.

Figure 12.12 DI values for different damage level (PWAS pair 02 – 00) on the composite lap-joint

specimen. a) Room temperature; b) Cryogenic temperature. Square: Excitation

frequency of 60 kHz; Circle excitation frequency of 318 kHz.

0

0.1

0.2

0.3

0.4

0.5

0 6 120

0.2

0.4

0.6

0.8

1

1.2

0 6 12

Energy (lb-ft)

DI

Energy (lb-ft)

DI

a) b)

359

The absolute difference at low frequency between step 1 and step 2 is much higher

than the same difference at high frequency. This indicates that, similar to the quasi-

isotropic plate case, lower frequencies are more sensitive to impact damage at room

temperature than the high frequency excitations. Since impact damage is a complicated

form of damage involving multiple damage modes in the composite material, the DI

values are sensitive to the pitch-catch paths used

Similar tests were performed using the PWAS system for damage detection on the

lap-joint at cryogenic temperatures, below -150° C. In this case we used PWAS pair 2

indicated in Figure 12.4 and the damaged was located between the pair. Two frequencies

were selected through tuning: 60 kHz and 318 kHz. Figure 12.12b shows the DI values

for the two different frequencies. Both low and high frequencies were able to detect the

impacts produced at different energy levels (6 ft-lb and 12 ft-lb) with a significance level

of 99 %.

As shown in Figure 12.12, impact damage was detectable by the PWAS system. Both

high and low frequency excitations were sensitive to the impact damage; however,

excitation at 60 kHz showed higher sensitivity in the RT case compared to the 318 kHz

excitation.

12.3.4 Thick composite plate specimen

Nine PWAS were installed on the composite tank specimen (Figure 12.5). The

experiments were performed to detect the presence of Teflon patches which were

incorporated during manufacture of the specimen to simulate delaminations. Two

experiments were performed on the specimen: Delamination detection at room

temperature; delamination detection at cryogenic temperature. Readings taken with

360

PWAS pair 1 – 4 and 7 – 4 were used as a baseline for the material because there were no

patches in the wave path between these pairs.

Based on the sample configuration, the following notation will be used: Step 1

denotes the 4 DI values of the PWAS pair with no patch in between; step 2 denotes the 5

DI values of the PWAS pair with patch in between and closer to the free edge of the plate

(PWAS 0, 5, and 8); step 3 denotes the 5 DI values of the remaining pairs, patch location

deeper in the thickness (PWAS 2, 3, and 6). The composite tank interface specimen was

scanned at two different frequencies: 60 kHz and 318 kHz. The speed of the wave at 60

kHz was about 2680 m/sec and the wavelength of the wave about 45 mm. At high

frequency (318 kHz), it was not possible to determine the velocity. The experiments were

again performed with an input voltage of 18 V to improve the signal to noise ratio. Figure

12.13a shows the DI values for the composite tank specimen at room temperature. Step 2

refers to the data recorded with PWAS pair 5– 0. From the analysis of the DI values we

see that the low frequency is more sensitive to the patch depth, especially when the

patches are large. The high frequency was more sensitive to the patch presence, but it was

affected by their depth and dimension.

A similar experiment was conducted at cryogenic temperatures (Figure 12.13b).

Readings were taken with temperatures below -150° C. We used a different frequency for

the low frequency case (75 kHz) because we the cryogenic temperature caused a shift in

the frequency of the maximum amplitude of the A0 mode. Figure 12.13b shows how DI

index changed with the different steps. We found that there was significant difference

between the steps; the PWAS were able to detect the presence of delamination. From the

DI values we determined that both frequencies could detect the presence of delamination;

361

however at 318 kHz there was greater sensitivity. The depth or the dimension of the

patches did not affect the DI values.

Figure 12.13 Composite tank interface specimen, room temperature. a) Experiment at room

temperature; square: excitation frequency of 60 kHz; circle: excitation frequency of

318 kHz. b) Experiment at cryogenic temperature; square: excitation frequency of 75

kHz; circle: excitation frequency of 318 kHz.

Based on the damage detection results presented here, it is shown that PWAS based

sparse arrays are effective for detecting multiple types of damage (through-holes, impact

damage, and delaminations) in complex composite materials used in spacecraft

applications. In particular, results were shown for damage detection of through-holes in

unidirectional composite strips at room temperature, detection of through-holes and

impact damage in a quasi-isotropic plate at room temperature, detection of impact

damage on a composite lap-joint specimen at room and cryogenic temperatures, and

detection of simulated delaminations in a composite tank interface at room and cryogenic

temperatures. These results indicate that a PWAS based array would be effective and

reliable for structural health monitoring on composite space vehicles.

0

0.2

0.4

0.6

0.8

1

0 6 12

0

0.2

0.4

0.6

0.8

1

0 6 12

Energy (lb-ft)

DI

Energy (lb-ft)

DI

a) b)

362

13 SURVIVABILITY OF SHM SYSTEMS

An objective of my research was to determine the technology readiness level (TRL) of

structural health monitoring through permanent attached PWAS for space applications. In

Section 12 we showed the experimental results of the subsystem/component validation

program. In this section, we will present the experimental results of the system validation

program or the survivability capability of SHM through PWAS in space applications.

13.1 TEST SPECIFICATIONS

The PWAS health monitoring system was tested on a subcomponent test of a space fuel

tank. The specimen was thermal-mechanically cycled to cryogenic temperatures of about

-300F and peak strains around 7000 μin/in.

The sensors network was installed on the tank an year before the actual test. The

allowed area for sensor installation was limited to four columns located at 90-degree

increments around the tank. Based upon this constraint a simple strategy for sensor

installation was formulated and carried out on the tank as shown in Figure 13.1. Sixteen

pairs of PWAS were installed along four rows at 90 degree from one another. Three

ground locations were also installed: close to PWAS 16, to PWAS 7, and to PWAS 31.

The sensors were installed over the course of a few days. Each location required

approximately 1.5 hours of work to bond the sensors (total net time 24 hours). The

sensors were installed using a vacuum curing blanket. The adhesive used was Vishay AE-

363

15 because its operating temperature range is -452 F to 200F (-269 to 95 C) and its

elongation capability is 2% (20000 μstrain) at -320 F (-195C) far above the test

maximum strain (9000 μstrain). The wire used to connect the PWAS was 34 gage wire,

not cryogenic rated and the solder used was 97In3Ag, selected after several test at

cryogenic temperature.

Figure 13.1 Installation strategy. a) Sensors layout on specimen (projection view). b) Particular

of sensors 16, 17, and ground on tube.

13.2 TEST PROCEDURE

Different scan methods were predetermined for the test. The requirements were: fast scan

(of the order of 10 min); ability to scan most of the area of the tube. Here under we report

the principal scan method used during the test.

The test lasted 4 days; impedance readings for each PWAS were taken at the

beginning and end of each day. Table 13.2 reports the scans taken and the test

environment for both pitch-catch and impedance data collection.

a) b)

364

Table 13.1 Full-scan, 12 min (for 1000 sample at 200 Hz) (T=transmitter, R=receivers)

T R T R

1 0 2 9 10 25 26 17 6 17 9 10 16 18 25 26 1 22

2 1 3 4 9 10 12 25 26 28 18 9 10 12 17 19 20 25 26 28

4 3 5 6 11 12 14 27 28 30 20 11 12 14 19 21 22 27 28 30

6 5 7 13 14 29 30 22 01 22 13 14 21 23 29 30 6 17

9 1 2 8 10 17 18 25 14 25 1 2 17 18 24 26 1 30

10 1 2 4 9 11 12 17 18 20 26 1 2 4 17 18 20 25 27 28

12 3 4 6 11 13 14 19 20 22 28 3 4 6 19 20 22 27 29 30

14 5 6 13 15 21 22 30 9 30 5 6 21 22 29 31 25 14

Table 13.2 Test sequence for impedance.

Reading Strain (μin/in) Temperature (F) Reading Strain (μin/in) Temperature (F)

Imp 0 87.5 18 – 19 few ~-300

00 – 04 87.5 20 ~6-7000 ~-270

05 87.5 – -297 21 ~-272

06 – 07 >-289 Imp 3 ~-272

08 ~2400 ~-287 Imp 4 ~-175

09 few ~-287 22 ~-200

10 ~4-5000 ~-307 23 ~6-7000 ~-308

Imp 1 ~-245 24 Few ~-308

Imp 2 87.5 25 ~6000 ~-308

11 – 12 87.5 26 ~300 ~-306

13 few ~-310 27 ~6-7000 ~-307

14 ~5000 ~-300 28 ~-305

15 ~600 ~-300 Imp 5 87.5

16 – 17 ~6000 ~-300 29 87.5

365

There were a total of seven cycles with strain above 5000 μin/in and temperature

below 300 F. There were a total of three cycles with temperature below 300 F (Between

reading impedance 3 and 4 the tube was at room temperature).

13.3 RESULTS

Before the test started the status of each PWAS was checked. After one year from the

installation and after the specimen (hence the PWAS) had been in contact with water for

several days, all the PWAS were visually bonded to the tube. Capacitance and impedance

readings were taken to check the quality of the bonding between the transducers and the

structure. The capacitance readings were all within the range required.

The impedance readings (Figure 13.2) showed PWAS 16 had a problem in the solder

connection. Action was taken and it was made a new wire-PWAS connection.

Figure 13.2 Impedance readings before the test.

After the recording of reading 29 in Table 13.2, visual inspection was performed on

the SHM system. Of the 32 PWAS installed, five presented a wire disconnection due to

-1000

-900

-800

-700

-600

-500

-400

-300

-200

-100

00 50 100 150 200 250 300 350 400 450 500

F (kH )

P_00 P_01P_02 P_03

P_04 P_05P_06 P_07

P_08 P_09P_10 P_11P_12 P_13

P_14 P_15P_16 P_17P_18 P_19

P_20 P_21P_22 P_23

P_24 P_25P_26 P_27P_28 P_29

P_30 P_31

PWAS 16

Frequency (kHz)

Im (z

)

P_00 P_01P_02 P_03P_04 P_05P_06 P_07

P_08 P_09P_10 P_11P_12 P_13P_14 P_15P_16 P_17P_18 P_19P_20 P_21P_22 P_23

P_24 P_25P_26 P_27P_28 P_29P_30 P_31

366

the solder disconnection from the PWAS (Figure 13.3a); one was broken with the wire

attached to the detached part of the PWAS (Figure 13.3b); one was broken with the wire

still on the part of the sensor attached to the structure (see Figure 13.3c). The tube

exploded few pressure cycles after the images were taken.

Figure 13.3 Visual inspection of PWAS after reading #29. a) PWAS 1 broken; b) PWAS 12

disconnected; c) PWAS 18 broken, PWAS 19 disconnected.

For each PWAS, six impedance readings were taken in the frequency range 1 kHz –

500 kHz at different history times. During post-processing, plots of the real part of E/M

impedance were assembled. The real part of E/M impedance, Re(Z), measured at the

PWAS terminals reflects with fidelity the mechanical impedance of the structure at the

PWAS location (Giurgiutiu and Zagrai 2001).

Hereunder we report the graph of the six impedance readings for PWAS 0, 2, and 10

these PWAS were not visually broken, disconnected, or disbanded from the tube.

Impedance 0 corresponds to the impedance taken before the test is started. Impedance 1

was taken after two cycles at 5000 μin/in and one cycle at temperature below -300 F.

There is no much difference between these two readings. Impedance 2 was taken when

the specimen was at ambient temperature and without load. The reading was taken the

day after impedance 1 and there was no history change in between. Impedance 3 was

a) b) c)

367

taken at about -270 F and after 3 cycles at about 6000 μin/in. Impedance 4 was taken

while the tank was filling with liquid nitrogen and without load. The reading was taken

the day after impedance 3. Impedance 5 was taken at the end of the test at room

temperature and with no load. The SHM system has withstood other 3 cycles with strain

above 6000 μin/in.

Figure 13.4 Impedance readings for PWAS 0

Figure 13.5 Impedance readings for PWAS 2 and 10

0

50

100

150

200

250

300

0 50 100 150 200 250 300

Impedance 0Impedance 1Impedance 2Impedance 3Impedance 4Impedance 5

Frequency (kHz)

Re

(z)

0

100

200

0 50 100 150 200 250 300

Imp 0 PWAS 02 Imp 1 PWAS 02Imp 2 PWAS 02 Imp 3 PWAS 02Imp 4 PWAS 02 Imp 5 PWAS 02Imp 0 PWAS 10 Imp 1 PWAS 10Imp 2 PWAS 10 Imp 3 PWAS 10Imp 4 PWAS 10 Imp 5 PWAS 10

Frequency (kHz)

Re

(z)

368

From Figure 13.4 and Figure 13.5, we can see that after the first two cycles at high

micro-strains, the SHM system reveals a new resonance frequency at about 100 kHz.

This resonance frequency is evident only when the structure is not subjected to load and

critical temperatures (impedance 2, 4, and 5). The same behavior can be seen in all the

PWAS that were still electronically connected.

While impedance results for 1 and 3 are similar to those of impedance 0, the baseline;

impedance 2, 4, and 5 show a new resonance frequency. The resonance frequency has

low energy for impedance 2, but it increases considerably in impedance 4 and 5. These

three readings have in common the environment and loading conditions. The tube is at

rest and the temperature has not reached the cryogenic level, the walls of the structure are

no more under tension. In previous studies (Giurgiutiu et al., 2003) we have seen that a

structural change in the substrate corresponds to knew frequency resonance in the

impedance spectrum. We could then link the new resonance frequency to a structural

change in the tube. For completeness of the exposure we should had analyzed the

quality of the bonding of the PWAS to the structure through microscope. However, the

PWAS were destroyed during the explosion of the tube.

Pith-catch data have been collected at two frequencies. The frequencies were

determined through tuning of the PWAS and the structure. We found experimentally that

we had maximum of the pseudo A0 mode at 45 kHz. For the pseudo S0 mode, we

selected the frequency, 165 kHz, at which the pseudo S0 mode is maximum and at the

same time the pseudo A0 mode and SH mode are minimum. Figure 13.6 shows the

readings we have analyzed. We discarded the readings from and to the PWAS that were

visually disconnected from the data acquisition system or those who were broken.

369

Figure 13.6 Post-processing analysis. Gray PWAS: transmitters; Black PWAS: bad wiring;

Arrows: pitch-catch direction, from the transmitter to the receiver.

We have noticed that when the tube is under high load levels and cryogenic

temperature, the pitch-catch analysis can not be performed (Figure 13.7) because the

wave amplitude is too attenuated and not visible.

Figure 13.7 Pitch-catch data at cryogenic temperature and strain level about 7000 μin/in.

Transmitter PWAS 4, receiver PWAS 11. Frequency 45 kHz.

V (m

V)

Time (μsec)

Readings 25 and 27

0 1 10 4−× 2 10 4−× 3 10 4−× 4 10 4−×

0.1

0.2

0.3

Readings 20 and 23

Readings 16 and 17

370

At ambient temperature and with the tube at rest and the wave propagating vertically

(towards the top or bottom) the first wave is almost coincident with the initial burst

because the wave speed is high (see Figure 13.8).

Figure 13.8 Pith-catch at ambient temperature and no load for PWAS 02 transmitter and PWAS

04 receiver (vertical wave propagation) at different history times. Frequency 45 kHz.

When we consider a pitch-catch along the horizontal direction, the wave speed

decreases considerably with respect to the vertical direction (-39%). When the pitch catch

is oblique, the wave speed decreases even more, but not so dramatically (-46%).

All the data analyzed appear qualitatively similar. Here we present only the case of

pitch-catch when PWAS 2 is transmitter and PWAS 10 and 12 are receiver. Figure 13.10

shows the wave propagating from PWAS 2 to PWAS 10, along the circumference. The

A0 mode is present for each reading, but the wave amplitude decreases as the number of

fatigue and thermal cycle increases. While the wave amplitude in the baseline reading

(Reading 0) is about 0.14 mV, in the last reading (Reading 29) the amplitude is about

-0.1

-0.06

-0.02

0.02

0.06

0.1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Reading 0Reading 11Reading 12Reading 28Reading 29

V (m

V)

Time (μsec)

371

0.07 mV. All the PWAS present the same kind of attenuation of the wave amplitude. This

can be due to either a structural change of the tube material or a degradation of the

PWAS capability to transmit-receive the signal. However, both pitch-catch signal and

impedance readings seem to be those of a healthy PWAS.


10 receiver (horizontal wave propagation) at different history times. Frequency 45

kHz.

Similar results are obtained considering a wave propagating obliquely with respect to

the circumference (Figure 13.11).

-0.1

-0.06

-0.02

0.02

0.06

0.1

0.00015 0.0002 0.00025 0.0003 0.00035


V (m

V)

Time (μsec)

372


12 receiver (oblique wave propagation) at different history times. Frequency 45 kHz.

At 165 kHz the wave amplitude is small. The noise to signal ratio is still low, hence it

is possible to recognize the same amplitude attenuation with the cycles as for the case of

45 kHz (Figure 13.11).


10 receiver (horizontal wave propagation) at different history times. Frequency 165

kHz.

-0.06

-0.02

0.02

0.06

0.1

0.00015 0.0002 0.00025 0.0003 0.00035


V (m

V)

Time (μsec)

-0.1

-0.06

-0.02

0.02

0.06

0.1

8.00E-05 1.00E-04 1.20E-04 1.40E-04 1.60E-04 1.80E-04 2.00E-04


V (m

V)

Time (μsec)

373

13.4 CONCLUSION

The SHM system installed on the tank tube was subjected to seven high strain cycles and

three cryogenic temperature cycles. At the end of the test, visual inspection was

performed. Five of the thirty-two PWAS installed were electronically disconnected, two

PWAS were broken. The PWAS proved to be working till the end of the catastrophic

experiment. The impedance readings show a new resonance peak at about 100 kHz, after

the first two cycles at high micro strains. It is not possible to asses with certainty whether

the new resonance is due to a structural change in the tube or to a PWAS – bond

degradation. However, the pitch-catch readings were consistent after different high load-

low temperature cycles, the new resonance peak of the impedance readings were at the

same frequency for all the connected PWAS (even the broken one PWAS 18), all these

seems to suggest a structural change in the tube under test. If this is the case, the PWAS

system have detected a structural change at history 22, after four high strains cycles and

three low temperature cycles. It is to be noted that however the standard SHM (ultrasonic

method) used to check the tube integrity did not reveal any damage in the structure.

The experiment conducted shown good survivability of the PWAS-bond system

either under harsh environmental condition (extreme temperature, aging, and liquid

contact) and extreme loads.

374

14 DURABILITY OF SHM SYSTEMS

To obtain space qualification, any non destructive evaluation methods must demonstrate

the durability of the system in different conditions. In Giurgiutiu (2004) and Giurgiutiu

and Lin (2004) have been presented initial durability and survivability tests - thermal

cycle loading, exposure to the environment, exposure to operational fluids such as oils,

hydraulics, fuels, etc.

In this Section we will present the durability of SHM through PWAS for space

applications.

14.1 REQUIREMENTS

For space certification of the NDE and health monitoring system, the entire SHM system

(hence transducer, bond layer, and wiring) must be subjected to durability test. From the

International Space Station Program (SSP) a summary of documents has been used to

provide requirements for the operating environment (thermal, acoustic, impact), hardware

selection, and system level integration (voltage drops, EMI). A notional test plan is

provided in Table 14.1.

375

Table 14.1 Notional test plan for space certification of NDE system

14.2 MISSION PROFILE

In order to obtain space qualification, tensile experiments on aluminum dog-bone bars

were performed to measure both the adhesive and PWAS strength. Data such as

impedance have been taken from the sensors at each static strain increment at cryogenic

temperature. The increments were of 250 μ strain up to 5,000 μ strain.

The specimen used was an aluminum dog bone beam. A PWAS was installed at the

middle of the dog bone while, on the other side, a strain gage was installed (Figure

14.1a).

376

Figure 14.1 Durability and survivability test. a) Specimen for durability and survivability test. b)

Durability setup.

In order to perform the test at cryogenic temperature under static load, a container for

nitrogen liquid as been fabricated. Figure 14.1b shows the dog bone specimen under

uniaxial load and at cryogenic temperature.

Figure 14.2 Impedance readings of PWAS at cryogenic temperature under uniaxial load

10

100

1000

0 50000 100000 150000 200000 250000 300000 350000 400000 450000 500000

0 us 500 us1000 us 1500 us2100 us 2500 us3100 us 4000 us5000 us

a) b)

377

From Figure 14.2 we can see that the bond quality did not change when the level

reached 5000 us and that it was possible to excite the PWAS at that level of strain.

14.3 SHOCK TEST

The ground handling shock test required that the system underwent a 20-g shock pulse in

both directions of each of the three orthogonal axes. A MTS 846 shock test system was

used for this test. It consisted of a drop table and a base with a pressurized cylinder. The

specimen used was the tank-dome joint as described in Figure 12.5, the experimental

setup is shown in Figure 14.3.

Figure 14.3 The dome-barrel specimen on the drop table. a) Transverse shock; b) In plane

shock.

An accelerometer was used to record the magnitude of the shock. A typical

accelerometer signal is shown in Figure 14.4. The drop test achieved a shock pulse that

was higher than 20-g’s meeting the requirement for the test.

b)a)

378

Figure 14.4 A typical accelerometer signal

To determine whether the SHM system survived the shock, a scan of the real part of

the impedance versus frequency was taken before and after the shock. Figure 14.5 shows

the graph of the real impedance versus frequency. Before and after scan, the impedance

does not differ, hence, no change in the SHM system has happened. The sensor network

can survive a 20-g shock pulse without sustaining damage.

Figure 14.5 The Re Z vs. Frequency before and after the test

379

14.4 RANDOM VIBRATION TEST

The objective of this test is to demonstrate system survivability of the sensor network

while it experiences vibrations similar to those on a launch vehicle. The required

spectrum is shown in Figure 14.6.

Figure 14.6 The required random vibration spectrum

A vibration exciter was used to complete this test. An accelerometer was used to

determine if the exciter could produce the needed amplitudes. The results of that test

showed that the exciter did not have the capability of producing the amplitudes required,

hence, the test was not performed.

14.5 THERMAL TEST

The thermal environment test requires that the thermal operating conditions of a

cryogenic tank be replicated to test for durability and survivability. Unlike the previous

two tests, the system must be operating while in this environment. To demonstrate that

the system can operate in this environment, an impact test was done on two different

specimens. Results are reported in Sections 12.3.3 and 12.3.4. The experiments

Vibration Spe ctrum

0.01

0.110 100 1000 10000

Frequency (Hz)

ASD

(g^2

/Hz)

Frequency (Hz) ASD (g2/Hz)

20 0.0125

20-50 +6 dB/oct

50-600 0.075

600-2000 -4.5 dB/oct

2000 0.0125

Overall 9.1 Grms b) a)

380

demonstrated that the sensors can not only survive the thermal environment but also

operate in this extreme environment, fulfilling the requirements for this test.

14.6 ACOUSTIC ENVIRONMENT TEST

The acoustic environment test mimics the acoustic noise environment experienced by the

structure of a launch vehicle. Figure 14.7a shows the required spectrum. During the test,

all the quoted sound energy levels are produced at the same time for the duration of an

actual launch, approximately three minutes.

Figure 14.7 Noise spectra. a) Required noise spectrum, b) Noise spectrum collected by a

microphone during the test

In Figure 14.7b is shown the recorded spectrum of the actual test. The recorded spectrum

met all the required sounds levels. A scan of the real impedance before and after the

exposure to acoustic environment was conducted. The scans recorded before and after the

acoustic test did not differ on a level that would indicate that damage was incurred. It can

be concluded that SHM system can withstand an acoustic environment experienced on a

launch vehicle.

b) a)

381

15 EFFECT OF PARTIAL BONDING BETWEEN TRANSDUCER AND

STRUCTURE ON CAPACITANCE

Structural health monitoring (SHM) methods are used to determine the health of a

structure. Transducers, attached to the surface of a structure, are used in a passive way

(listen to the structure) or in an active way (interrogate the structure). In our researches,

we use a particular kind of transducer called piezoelectric wafer active sensor (PWAS). A

network of PWAS is bonded through an adhesive layer to the structure. The sensor uses

the piezoelectric principle to convert a difference in voltage into a strain change. The

PWAS can be excited at different frequencies so that different modes of excitation are

activated.

The methods of interrogation of a structure through PWAS are pitch-catch, pulse-

echo, and electromechanical impedance (E/M). The latter method can be used also to

determine the quality of the bonding between the transducer and the structure.

Most of the applications for SHM are on structural components subjected to big

variation in the environment conditions (from high temperatures to cryogenic

temperatures (~-198 C)) and different strain levels (from 0 to 3000 μstrain). The structure

component, the PWAS, and the bonding layer have different thermal expansion

coefficients and different young modulus. They will stretch of a different amount and

under certain conditions the PWAS will disbond partially or fully from the structure. The

signal received by the half detached PWAS or fully detached PWAS is different from the

382

signal received when the PWAS is well attached. From the pitch-catch method or the

pulse-echo method it is not possible to determine whether the change in the signal is due

to damage in the structure or to a change in the conditions of the PWAS – structure

bonding.

The state of the art for this technique is that if a transducer is well bonded to the

structure, the E/M reading shows suddenly change in the real impedance amplitude and a

discontinuity of the imaginary impedance at the resonance frequency of the structure

(Giurgiutiu, 2008). The impedance spectrum of a PWAS with free boundary conditions

presents a resonance at 296 kHz that it is not present when the PWAS is attached to a

structure (Giurgiutiu, 2008). However, we do not know how the impedance spectrum

changes when the PWAS bonding degrades and the PWAS becomes partially attached to

the structure or totally detached but without free boundary conditions. Moreover, it is

difficult to define a general procedure to ensure debonding detection of the transducers

through E/M.

A more easy way to determine the quality of the bonding is through capacitance. We

know that a well bonded PWAS has a capacitance value of about 2.67 nF. However, we

do not know how different it is the capacitance of a partially detached PWAS or of a

detached one.

In the present section we will try to establish how it is possible to determine if a

PWAS is well bonded or not through capacitance method.

Future SHM technique will integrate the use of capacitance check when variations in

pitch-catch, or pulse-echo, or E/M values are found during data collection. If the

capacitance of the transducers has changed to a degree that is evidence of bonding

383

degradation, the change in the SHM readings will be disregarded and a false call for

damage in the structure will be avoided.

The experiment performed had one factor with three levels (a: fully bonded, b: half

bonded, c: not bonded). The response we were interested in was the capacitance variation

and of the structure (PWAS + strip of metal).

The experimental unit is a PWAS attached to the structure. We will use identical

strips of Aluminum metal 160x50x1 mm. On each strip, 6 PWAS are installed (see

Figure 15.1).

Figure 15.1 Specimen with PWAS installed (A – F: PWAS location)

The position of the PWAS on the strip can affect the results recorded; we will

randomize the PWAS location on the strip.

15.1 POWER ANALYSIS AND SAMPLE SIZE

We performed a power analysis to check that the experiment setup was sufficient to

guarantee a power of 0.8 for an alpha level of 0.05. We used SAS to determine the power

as described in Gatti et al. (1998). The SAS program used was:

A B C A B C

384

PROC IML;

F = FINV(PR, DF1, DF2, 0);

POWER = 1 - PROBF(F, DF1, DF2, NCP);

PRINT 'F VALUE = ' F;

PRINT 'POWER = ' POWER;

Where DF1 is the numerator degrees of freedom (r-1), DF2 was the denominator

degrees of freedom (rn-r), n was the number of replication per cell, PR = 1-α, and NCP

= λ represented the non centrality parameter. We could find λ from the relation

2 2k

k nλτ σ =∑ , where k kτ μ μ= −

In order to determine the power, we should know the variance of the experiment and

the value of the response difference that determines our rejection region. In order to

proceed we need an estimation of the variance.

15.2 POPULATION VARIANCE

To obtain the variance of the test we performed a preliminary investigation of the

variance of the response on PWAS already installed on different specimens. We took in

consideration PWAS installed on specimens with same thickness and dimensions and

attached with the same adhesive. For this purpose, we used PWAS already installed on

specimens available in our laboratory. From this preliminary experiment it was possible

not only to estimate the variance but also to estimate 2k

kτ∑ .

We measured the capacitance of 24 PWAS installed on different plates. Results of the

preliminary experiment are reported in Table 15.1.

385

Table 15.1 Capacitance variance.

Capacitance (nF)

Mean Variance

2k

kτ∑ 2 2

kk

τ σ∑

2.679 0.0155 0.358 23

Table 15.2 reports the different values of power we obtain for different values of n.

Table 15.2 Power as a function of n for 3r = , 0.05α = .

n 2 2k

knλ τ σ= ∑ Power (from SAS)

2 46 0.917

3 69 0.999

4 92 1

We can obtain a high value of power (~0.92) with n = 2, r = 3, α = 0.05.

15.3 EXPERIMENTAL SETUP

The experiment setup is reported in Table 15.3, where a represents the PWAS full

bonded, b represents the PWAS partially bonded, and c represents the PWAS detached

(see Figure 15.1). Capital letters A, B, and C represent PWAS location on the specimen

as described in Figure 15.1. There are a total of 18 PWAS (6 for each factor level, 6 for

each strip and 3 for each location).

Table 15.3 Experiment setup. 1 – 3: specimen identification number; a – c: type of bonding

Location

A B C C B A

1 a b c c b a

2 c a b b a c

Spec

imen

3 b c a a c b

386

Because factor level c is a PWAS wired but detached from the structure, it can

happen that the PWAS is not in contact with the structure. In order to avoid this case, we

taped the PWAS to the structure. This reflected realistic environment conditions. Real

structures are often in contact with other foam layers for thermal insulation and or

protection. A piece of tape will be applied to all PWAS used in the experiment.

15.4 EXPERIMENT

The PWAS were installed by the same operator. The procedure to install PWAS in

configuration a was the standard procedure used to install strain gages:

1. sand

2. apply conditioner

3. apply neutralizer

4. apply catalyst

5. apply adhesive (M-bond 200)

6. hold the PWAS under pressure with the finger for 90 sec.

The procedure to install PWAS in configuration b was the same as that for PWAS in

configuration a. To obtain partial bonding, a piece of tape was applied on the bottom half

of the sanded surface (Figure 15.2 b). The part of surface under the tape was not cleaned

as required and we did not apply catalyst to it (Figure 15.2 c-e). A small amount of glue

was applied only on the top half of the surface (Figure 15.2 f). After we had applied the

glue we removed the tape and pressed the PWAS on the surface such that the excessive

amount of glue would follow the arrow direction in Figure 15.2 g.

387

Figure 15.2 Installation procedure for configuration b

We tested the installation procedure on an aluminum plate. We were able to obtain

partially bonded PWAS, but the amount of surface not bonded was not constant. Small

variation in adhesive would lead to big difference in bonding extension. However, this

was considered to be the most reliable and consistent method to obtain partially bonded

PWAS. At the end of the experiment, we will destroy the PWAS in configuration b to

check the extension of the bonding.

After the installation, we recorded the capacitance value of the PWAS. Table 15.4

reports the findings. As we can see, the PWAS not bonded present higher capacitance

while the PWAS full bonded present the lower values of capacitance. PWAS in

configuration b have capacitance that varies form the higher values of PWAS in

Sand

Tape

Conditioner

Neutralizer a) b) c)

d) e) f)

PWAS

g)

Catalyst

M bond 200

388

configuration a to the lower values of PWAS in configuration c. Configuration b presents

the highest standard deviation.

Table 15.4 Capacitance (nF) of the PWAS

Specimen

1 2 3 mean std

a 2.65 2.72 2.70 2.68 2.69 2.69 2.69 0.02

b 2.89 2.81 3.13 2.99 3.09 2.80 2.95 0.14

c 3.31 3.25 3.32 3.32 3.27 3.24 3.28 0.04

We inspect the PWAS in configuration b. We found that the PWAS with a smaller

amount of adhesive around the edge of the transducer had smaller capacitance, while a

less amount of adhesive around the edge corresponded to a higher capacitance. Figure

15.3 shows an example of this situation. The dark gray shadow around the PWAS is the

hardened glue. PWAS b on specimen #3 has a smaller shadow and it is all located in the

upper part of the PWAS itself. PWAS b on specimen #5 has a small shadow but it is

almost all around the PWAS. If we compare their capacitance we see that PWAS b on

specimen #3 has a higher capacitance (2.80 nF) than PWAS b on specimen #5 (2.67 nF).

Figure 15.3 PWAS bonded to specimen #3. PWAS in configuration b has a less amount of glue

than that of PWAS in configuration a. The capacitance is 2.80 nF and 2.69 nF

respectively.

389

15.5 ANALYSIS

In order to perform the analysis, we must check the assumptions. We found that the

normality assumption is not completely met.

-2 -1 0 1 2

-1. 500E-01

-1. 000E-01

-5. 000E-02

0

5. 0000E-02

1. 0000E-01

1. 5000E-01

resid

Normal Quant i l es

Figure 15.4 QQ-plot and residual plot

From the plot of the residual we checked that the variance of the residuals was

constant.

Figure 15.5 show the interaction plots between the bond type (a = full bonded, b =

partially bonded, c = detached) and respectively PWAS location and the specimen. From

the plots we deduce that there is no interaction between the factor bond and the blocks

specimen and location.

390

Figure 15.5 Interaction plots. a) Interaction between PWAS location and bond type. b)

Interaction between specimen and bond type.

We perform the analysis of the latin square with replications within the cell with the

PROC GLM function in SAS. The SAS program use was:

Proc glm;

class Spec Loc Bond;

model Cap=Spec Loc Bond;

run;

The output is:

Source DF Type III SS Mean Square F Value Pr > F

Specimen 2 0.0229 0.01145 1.89 0.1968

Blocks

Location 2 0.0190 0.00951 1.57 0.2509

Factor Bond 2 1.0729 0.53646 88.63 <0.0001

Cap1

2. 6

2. 7

2. 8

2. 9

3. 0

3. 1

3. 2

3. 3

3. 4

1 2 3

B

B

B

C

C

C

B C

Bond: a, b, c

Location

3.4 3.3 3.2 3.1 3.0 2.9 2.8 2.7 2.6

A

p1

6

7

8

9

0

1

2

3

4

1 2 3

B

B

B

C

C

C

2 3

Specimen

1

391

The contrasts were orthogonal because type I and Type III gave the same results. We

did not consider interaction factors because we were only interest to check that the blocks

did not affect the response. Moreover, the blocks and the treatments were fixed.

From the Type III table we can see that the blocks are not significant and do not

influence the response (p-values > 0.05). The factor bond type is significant. Different

bonds give different capacitance values.

We can perform a Tukey multiple comparison to check which pairs of bond types are

significantly different.

Output:

Least Squares Means for effect Bond

Pr > |t| for H0: LSMean(i)=LSMean(j)

Dependent Variable: Capacitance

i/j 1 2 3

1 0.0003 <.0001

2 0.0003 <.0001

3 <.0001 <.0001

Bond Cap LSMEAN 99% Confidence Limits

a 2.688333 2.589686 2.786981

b 2.951667 2.853019 3.050314

c 3.285000 3.186353 3.383647

As we can see, there was significant difference among the three different type of

bond. We could use the confidence interval to derive prediction intervals to be used as a

392

rule to check the goodness of the bonding. We can say that we are 99% confident that a

PWAS well bonded has a capacitance value between 2.58 nF and 2.78 nF.

From theses results, we concluded that capacitance methods can be an effective tool

to use along with SHM method to check whether a detection of damage was due to a real

damage or a change in the transducer bonding quality (true call or false call). From the

analysis above we derived 99% confidence interval that could be used to derive the

reference values to determine the status of the bonding between transducer and structure.

The analysis we performed had as assumption the normality of the data. This assumption

was not completely met, hence the CI interval we derived were not to be trusted

completely. For further analysis, a non parametric model should be taken into account to

verify the goodness of the CI derived in this section.

393

16 GUIDED WAVES SCATTERING FROM DAMAGE

Lamb waves scattering from a defect has been studied through Mindlin theory or through

boundary element model, finite difference model and finite element model. The problem

under investigation has complex wave interactions and usually complex geometries,

hence numerical methods represent the only viable approach for understanding the

multiple reflections and diffractions of the ultrasonic waves within the component. Most

of the work done to date is on circular holes or defects with a regular shape. Among the

few papers on wave scattering from crack, Messerery et al. (2005) focused on Rayleigh

wave propagation.

Lamb waves excited in a plate have more then one mode of propagation. When the waves

interact with a defect in the structure the waves are modified by mode conversion.

Therefore, the received signal generally contains more than one mode, and the

proportions of the different modes present is modified by mode conversion at defects and

other impedance changes. The wave modes are also generally dispersive, which means

that the shape of a propagating wave changes with distance along the propagation path.

When mode conversion took place, the geometry of the structure under consideration is

important. If the geometry of the structure is symmetric, only the same family modes as

an incident mode are capable of carrying each portion of scattered energy, so if the

incident wave is antisymmetric one, there is no chance that the reflection of S0 mode

could occur. The incident wave mode and the converted wave mode overall must have

394

the inverse trend to the other, satisfying the energy conservation. This is because each of

those scattered modes would share its own portion of scattered energy with the other

mode through mode conversion, based on a fixed amount of incident energy.

In order to perform health monitoring it is important to decide which mode(s) and

frequency of the excitation to use. The sensitivity of different modes in different

frequency-thickness regions is a parameter that enables to determine the best testing

regime for a particular type of defect. The S0 mode at low frequencies has the

characteristic of low dispersion and low leakage of energy if the plate is fluid loaded;

moreover, the stresses due to S0 mode are almost uniform through the thickness of the

plate so that its sensitivity to a defect is not dependent on the through the thickness

location of the defect. Lowe et al. (2002) and Alleyne et al. (1992) studied the Lamb

wave scattering from a notch through finite element simulations.

Lowe et al. (2002) studied through finite element model the reflection of Lamb mode S0

from a rectangular notch in a plate. The notch is assumed to be rectangular in section

(with zero width in the case of a crack), to be infinitely long and aligned normal to the

direction of wave propagation. Lowe also assumed plane strain in the plane of the particle

motion of the Lamb wave. The displacement in the in-plane direction was monitored at

the mid-thickness of the plate, thus ensuring that only the symmetric S0 propagating

mode was detected, since the antisymmetric A0 mode has zero in-plane displacement at

this depth.

An alternative to finite element model are the boundary element model. This method

seems to be a powerful tool for describing Lamb wave mode conversion phenomena from

an edge. Cho et al. (1996) used the boundary element method to study the mode

395

conversion phenomena of Lamb waves from a free edge. They applied their method to

the study of multimodes reflection from free edge in a semi-infinite steel plate. They

found that S1 is the mode which is less affected by the other modes’ appearance in mode

conversion. Cho (2000) also studied the mode conversion in a plate with thickness

variation through the hybrid boundary element method. The elastodynamic interior

boundary value problem is formulated as a hybrid boundary integral equation in

conjunction with the NME technique based on the Lamb wave propagation equation.

Vermula and Norris (2005) studied flexural wave scattering on thin plates using Mindlin

theory. Also McKeon et al. (1999) and Hinders (1996) used the Mindlin theory to study

Lamb waves scattering from a through hole and rivets, respectively. Mindlin theory

includes shear-deformation and rotary-inertia effects, as in the Timoshenko beam theory.

The equations of motion restrict the deformation to three degree of freedom, and they are

obtained by averaging the exact equations of elasticity across the plate thickness.

Grahn (2002) used a 3D approach to studied the scattering problem of an incident plane

S0 Lamb wave in a plate with a circular partly through-thickness. The wave fields in the

outer part outside the hole and in the inner part beneath the hole are expanded in the

possible Lamb modes and the horizontally polarized shear modes. Both propagating

modes and evanescent modes were included in the expansions. The expansion

coefficients were obtained by utilizing the boundary conditions at the hole boundary and

the continuity conditions below the hole. He derived a linear system of equations for the

expansion coefficients by projecting the different boundary conditions onto an orthogonal

set of projection functions.

396

Moreau et al. (2008) extended the 3D approach to more complex geometry through the

use of both orthogonality relations and finite element modes.

In the method presented the scattering coefficient of Lamb waves generated by the

interaction of incident wave with scatterers in a plate were directly determined by using a

FEM or a BEM in conjunction with normal mode expansion. Gunawan et al. (2004)

proposed a mode-exciting method where all Lamb modes are simultaneously excited by

appropriate boundary conditions given on both ends of a finite plate. The excited Lamb

wave modes constitute a system of equations which is solved to determine the scattering

coefficients for all Lamb modes.

Hereunder we report a preliminary modeling of Lamb waves scattering from a sudden

change in material properties in the plate and SH waves scattering from a crack on a

plate.

In this section, we present some preliminary modeling of Lamb waves and SH waves

scattering from inhomogeneity and damage following Ditri (1996). First, we extend the

derivation made by Ditri for SH waves to the case of a non-through the thickness crack in

the plate. Second, we present the generic derivation of Lamb waves scattering when a

difference in both thickness and material properties is encountered by a propagating

wave.

16.1 MODE DECOMPOSITION OF INCIDENT, REFLECTED, AND TRANSMITTED WAVES:

SH WAVES SCATTERING FROM A CRACK

Consider a plate of thickness d with a crack of depth 1d as depicted in Figure 16.1. For

convenience with consider the plate as formed of two plates 0Ω and 1Ω of same

397

thickness and material properties perfectly joined at the boundary region Γ (dash line in

Figure 16.1).

Figure 16.1 Plate with a crack depth 1d .

The particle displacement field is only in the z direction and it is denoted with 0zu for

region 0Ω and 1zu for region 1Ω . The particle displacement field must satisfy the equation

of motion in Equation (3.5) for SH waves under the assumption of external forces equal

to zero, i.e.,

2 2 2

2 2 2z z zu u u

x y tρμ

∂ ∂ ∂+ =

∂ ∂ ∂ (16.1)

The displacement fields can be expressed as a summation of all the positive and negative

travelling or decaying normal modes, i.e.,

( )( , , ) ( )n ni x i x i tz n n zn

nu x y t a e b e u y e

α αξ ξα α α α ω−= +∑ (16.2)

where { }0,1α ∈ ; the terms with amplitude nza represent the positive travelling or

decaying modes for the displacements in the z direction; the terms with amplitude nzb

represent the negative travelling or decaying modes for the displacements in the z

direction. The modes are traveling if nξ is real or decaying if nξ is imaginary. The sign

of the wavenumber is taken into account in the exponential term.

0Ω

1ddΓ

x

y

1Ω

Crack

398

Each term in Equation (16.2) must satisfy the third equation in (16.1) provided

( )

222

2

222

2

symmtric modes

2 1 antisymmtric modes2

s

s

nc d

nc d

ω πξ

ω πξ

⎛ ⎞= − ⎜ ⎟⎝ ⎠

⎛ ⎞+= − ⎜ ⎟

⎝ ⎠

(16.3)

The terms in Equation (16.2) must satisfy the traction-free boundary conditions at the

upper and lower surfaces of each semi-plate, i.e.,

0

0

0

0

zyz y

y

zyz y d

y d

uTy

uTyα

μ

μ

==

==

⎧ ∂= =⎪ ∂⎪

⎨∂⎪ = =⎪ ∂⎩

(16.4)

In addition, we must impose continuity of displacements and stresses at the common

boundary Γ , i.e., for the displacements using Equation (16.2)

( ) ( )0 0 0 1 1 1( ) ( )n n nz n n nzn n

a b u y a b u y+ = +∑ ∑ (16.5)

The non zero tractions at the boundary Γ is xzT given by

zxz

uTx

α

μ ∂=

∂ (16.6)

Requiring continuity at Γ leads to

0 10 0xz xzx x

T T= =

= (16.7)

or, by substituting Equation (16.2) into (16.7) and after rearrangement

( ) ( )0 0 0 0 1 1 1 1( ) ( )n n n zn n n n znn n

a b u y a b u yξ ξ− + = − +∑ ∑ (16.8)

399

Last condition to be satisfied is that the portion of the plate at the crack interface be

traction-free, hence

0

00

110

0 \

0 \xz x

xz x

T y

T y=

=

= ∈∂Ω Γ

= ∈∂Ω Γ (16.9)

where 0∂Ω is the right boundary of 0Ω and 1∂Ω is the left boundary of 1Ω . With the use

of Equation (16.2), Equation (16.9) becomes

( )

( )

0 0 0 00

1 1 1 11

( ) 0 \

( ) 0 \

n n n znn

n n n znn

a b u y y

a b u y y

ξ

ξ

− + = ∈∂Ω Γ

− + = ∈∂Ω Γ

∑

∑ (16.10)

Consider the case in which a single propagating Lamb waves mode m is generated in

plate 0Ω with unit amplitude and propagating in the positive x direction. Hence, we

impose the following conditions

0

1 0n nm

n

a

b

δ=

= (16.11)

where nmδ is the delta Kronecker. Substitution of Equation (16.11) into Equations (16.5),

(16.8), and (16.10) yields after rearrangement

0 0 1 1 0( ) ( ) ( )n nz n nz mzn n

b u y a u y u y− = −∑ ∑ (16.12)

0 0 0 1 1 1 0 0( ) ( ) ( )n n zn n n zn m zmn n

b u y a u y u yξ ξ ξ+ =∑ ∑ (16.13)

0 0 0 0 00

1 1 11

( ) ( ) \

( ) 0 \

n n zn m zmn

n n znn

b u y u y y

a u y y

ξ ξ

ξ

= ∈∂Ω Γ

= ∈∂Ω Γ

∑

∑ (16.14)

400

To solve this problem, assume that there are N modes reflected and N modes transmitted.

The total numbers of unknown are 2N, hence to solve this problem we need 2N number

of equations. For each y ∈Γ two equations are created; continuity of displacements and

stresses, Equations (16.12) and (16.13). For each 0y δ∈ Ω Γ one equation is created,

stress-free boundary, first equation of (16.14). For each 1y δ∈ Ω Γ one equation is

created, stress-free boundary, second equation of (16.14). We apply the interface

conditions (16.12) and (16.13) at k points ky ∈Γ and the stress-free conditions at k points

ky αδ∈ Ω Γ , giving a total of 4m equations. To be able to solve the equations we must

have 4m=2N.

The system of equations can be written as

[ ]{ } { }A x b= (16.15)

where { } { }0 1 1 0 1 1, , , , , , , Tk kx b b b a a a− −= K K is the column vector of amplitude reflection

and transmission factors; the square [ ]A matrix is given by

( )

( )

( )

( )

( )

( )

0, 1

1, 1

0 0, 1

1 1, 1

0 02 , 1

2 ,

3 ,

1 13 , 1

( )

( )

( )

( )

( )

0

0

( )

i j ij z

i j N ij z

i k j n ij z

i k j N n ij z

i k j n i kj z

i k j N

i k j

i k j N n i kj z

A u y

A u y

A u y

A u y

A u y

A

A

A u y

ξ

ξ

ξ

ξ

−

+ −

+ −

+ + −

+ +−

+ +

+

+ + +−

=

= −

=

=

=

=

=

=

(16.16)

vector { }b is given by

401

0

0 0

0 02

3

( )

( )

( )0

i mz i

i k m mz i

i k m mz i k

i k

b u y

b u y

b u yb

ξ

ξ+

+ +

+

= −

=

==

(16.17)

Consider a 10-mm plate with a crack of depth d1=3/5d. Figure 16.2 shows the particle

displacement of the incident, reflected, and transmitted wave.

Figure 16.2 Particle displacement of the incident, reflected, and transmitted SH0 wave at f=1000

kHz. a) Distance from the crack x=2 mm; b) Distance from the crack x=5 mm.

The plots were computed retaining 238 modes for each wave. The incident SH0 wave is

symmetric and so it is the transmitted wave. The reflected wave amplitude is greater in

the region of the thickness where the crack is present. For a thoroughly analysis, we

should derive the energy of the three waves and evaluate the error made in the

assumption.

200− 0 200 400

2

4

6

8

10

100− 0 100 200 300 400

2

4

6

8

10

y

uz/βm uz/βm a) b)

Transmitted

Reflected

Incident Crack

402

16.2 MODE DECOMPOSITION OF INCIDENT, REFLECTED, AND TRANSMITTED WAVES:

LAMB WAVES SCATTERING FROM CHANGE IN MATERIAL PROPERTIES

Consider two semi-infinite flat layers of isotropic material as depicted in Figure 16.1

perfectly joined at the common boundary Γ .

Figure 16.3 Two semi-infinite layers with different thickness and material properties. (After Ditri,

1996)

The two layers denoted by regions 0Ω and 1Ω can have different thickness and material

properties. The particle displacement fields are denoted as 0xu , 0

yu , and 0zu for region 0Ω

and 1xu , 1

yu , and 1zu for region 1Ω . The particle displacement fields must satisfy the

equations of motion in Equation (3.5) under the assumption of external forces equal to

zero, i.e., Equation Chapter 13 Section 1

( ) ( )

( ) ( )

22 2 2

2 2 2

2 2 22

2 2 2

2 2 2

2 2 2

2

2

yx x x

y y yx

z z z

uu u ux y x y t

u u uux y x y t

u u ux y t

αα α α

α α αα

α α α

λ μ μ λ μ ρ

λ μ μ λ μ ρ

μ μ ρ

⎧ ∂∂ ∂ ∂+ + + + =⎪

∂ ∂ ∂ ∂ ∂⎪⎪ ∂ ∂ ∂∂⎪ + + + + =⎨ ∂ ∂ ∂ ∂ ∂⎪⎪ ∂ ∂ ∂

+ =⎪∂ ∂ ∂⎪⎩

(16.18)

where { }0,1α ∈ .

0Ω 1Ω0d 1dΓ

x

y

403

Consider the equation of motions for Lamb waves propagating in an isotropic plate. The

displacement fields in each region can be expressed as a summation of all the positive

and negative travelling or decaying normal modes, i.e.,

( )

( )( , , ) ( )

( , , ) ( )

n n

n n

i x i xn n n i tx x x x

n

i x i xn n n i ty y y y

n

u x y t a e b e u y e

u x y t a e b e u y e

α α

α α

ξ ξα α α α ω

ξ ξα α α α ω

−

−

= +

= +

∑

∑ (16.19)

where the terms with amplitude nxa and n

ya represent the positive travelling or decaying

modes for the displacements in the x and y direction respectively; the terms with

amplitude nxb and n

yb represent the negative travelling or decaying modes for the

displacements in the x and y direction respectively. The mode will be traveling if nξ is

real or decaying if nξ is imaginary. The sign of the wavenumber is taken into account in

the exponential term.

Each term in Equation (16.19) must satisfy the first two equations in (16.1) provided that

the wavenumbers are solution of the Rayleigh Lamb waves Equations (3.20) and (3.21).

The terms in Equation (16.19) must satisfy the traction-free boundary conditions at the

upper and lower surfaces of each semi-plate, i.e.,

( )

( )

00

00

2 0

2 0

0

0

yxyy y

y

yxyy y d

y d

yxxy y

y

yxxy y d

y d

uuTx y

uuTx y

uuTy x

uuTy x

αα

αα

α

α

α

α

λ λ μ

λ λ μ

μ

μ

==

==

==

==

⎧ ∂⎡ ⎤∂= + + =⎪ ⎢ ⎥∂ ∂⎣ ⎦⎪

⎪ ∂⎡ ⎤∂⎪ = + + =⎢ ⎥⎪ ∂ ∂⎣ ⎦⎪⎨

∂⎛ ⎞∂⎪ = + =⎜ ⎟⎪ ∂ ∂⎝ ⎠⎪⎪ ∂⎛ ⎞∂⎪ = + =⎜ ⎟∂ ∂⎪ ⎝ ⎠⎩

(16.20)

404

We must impose continuity of displacements and stresses at the common boundary Γ ,

i.e., for the displacements using Equation (16.19)

( ) ( )

( ) ( )

0 0 0 1 1 1

0 0 1 1

( ) ( )

( ) ( )

n n n n n nx x x x x x

n n

n n n n n nx x y x x y

n n

a b u y a b u y

a b u y a b u yα α

+ = +

+ + +

∑ ∑

∑ ∑ (16.21)

The non zero tractions at the boundary Γ are xxT and xyT given by

( )2 yx

xx a

yxxy

uuTx y

uuTy x

αα

α γ

αα

α

λ μ λ

μ

⎧ ∂∂= + +⎪

∂ ∂⎪⎨

⎛ ⎞∂∂⎪ = +⎜ ⎟⎪ ∂ ∂⎝ ⎠⎩

(16.22)

Requiring continuity at Γ leads to

0 1

0 0

0 1

0 0

xx xxx x

xy xyx x

T T

T T= =

= =

⎧ =⎪⎨

=⎪⎩ (16.23)

or, by substituting Equation (16.19) into (16.23) and after rearrangement

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

00 0 0 0 0 0

0 0 0

11 1 1 1 1

1 1 1

00 0 0 0 0 0

11 1 1 1 1 1

( )2 ( )

( )2 ( )

( ) ( )

( ) ( )

nyn n n n n

n x x x y yn n

nyn n n n n

n x x x y yn n

nn n n n nxx x n y y y

n n


n n

u yi a b u y a b

yu y

i a b u y a by

u ya b i a b u yy

u ya b i a b u yy

α

ξ λ μ λ

ξ λ μ λ

ξ

ξ

∂+ − + + + =

∂

∂= + − + + +

∂

∂+ + − + =

∂

∂= + + − +

∂

∑ ∑

∑ ∑

∑ ∑

∑ ∑

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

(16.24)

Last condition to be satisfied is that the portion of the thicker plate that extends below the

interface region must be traction-free, hence

405

1

011

0

0 \

0xx x

xy x

Ty

T=

=

⎧ =⎪ ∈∂Ω Γ⎨=⎪⎩

(16.25)

where 1∂Ω is the left boundary of 1Ω . With the use of Equation (16.19), Equation

(16.25) becomes

( )1 1 1 1

1 1 1

111 1 1 1

2 0 \

0

nyn n n

n x x yn n

nn n nxx n y y

n n

ui a u a

yy

ua i a uy

α

ξ λ μ λ

ξ

⎧ ∂− + + =⎪

∂⎪ ∈∂Ω Γ⎨∂⎪ − =⎪ ∂⎩

∑ ∑

∑ ∑ (16.26)

Consider the case in which a single propagating Lamb waves mode m is generated in

one of the two layers 0Ω with unit amplitude and propagating in the positive x direction.

Hence, we impose the following conditions

0 0

1 1 0

n nx y nm

x y

a a

b b

δ= =

= = (16.27)

where nmδ is the delta Kronecker. Substitution of Equation (16.27) into Equations(16.21),

(16.24), and (16.26) yields

0 0 0 0 1 1

0 0 0 1

( ) ( ) ( )

( ) ( ) ( )

m m n n n nx x x x x x

n nm m n n n ny y x y x y

n n

a u y b u y a u y

a u y b u y a u yα α

+ =

+ +

∑ ∑

∑ ∑ (16.28)

( )

( )

0 00 0 0 0 0 0 0

0 0 0

11 1 1 1

1 1 1

0 0 10 0 0 0 0 0 0 0 1 1 1 1

2

2

m ny ym m n n m n

n x x x x y yn n

nyn n n

n x x yn n

m n nm n m m n n n n nx x xx x n y y n y y x n y y

n n n n

u ui a u b u a b

y y

ui a u a

yu u ua b i a u i b u a i a uy y y

ξ λ μ λ

ξ λ μ λ

ξ ξ ξ

⎧ ⎡ ⎤∂ ∂⎡ ⎤+ − + + + =⎪ ⎢ ⎥

∂ ∂⎢ ⎥⎣ ⎦ ⎣ ⎦⎪⎪ ∂⎪= − + +⎨ ∂⎪⎪ ∂ ∂ ∂

+ − + = −∂ ∂ ∂

⎩

∑ ∑

∑ ∑

∑ ∑ ∑ ∑⎪⎪

(16.29)

406

( ) ( )

( ) ( )

1 1 11 1 1

111 1 1 1 1 1

2 0 \

0

n n

nyi x i xn n n n

n x x y yn n


n n

ui a u a e b e

yy

ua b i a b uy

α αα

ξ ξα αξ λ μ λ

ξ

−⎧ ∂+ + + =⎪

∂⎪ ∈∂Ω Γ⎨∂⎪ + + + =⎪ ∂⎩

∑ ∑

∑ ∑ (16.30)

Solution for the Lamb waves mode is not straight forward as for the SH waves case even

if we consider the simple case of equal thickness but different material. The intent of this

section was to formulate the problem, but no solution is derived.

407

17 ACOUSTIC EMISSION IN INFINITE PLATE

Acoustic emission (AE) is a phenomenon arising from a rapid release of strain energy

within a material. This can due to a crack propagating or an object impacting the

structure. PWAS transducers can be used either as active sensors to interrogate the health

of the structure on demand or as passive sensors that listens for AE. In AE this energy

radiates from the source in the form of elastic waves. The response is principally affected

by the angular direction of the force and the distance between source and response.

Acoustic emission have been studied extensively, the angular direction of the force

has been studied experimentally by Prosser (1991) and Gorman et al. (1991). They

measured the out of plane displacement of the wave with force angle with respect to the

surface varying from 90 to 0 with steps of 30 degrees. To catch the response when the

force was at 0 degree, the transducer was located not on the surface of the pate but on the

edge of it. They noticed that vertical forces gave larger flexural waves while horizontal

forces gave larger longitudinal waves. This led the authors to assume that the relative

amplitude could give information about source orientation. However, this was not

proven.

It was also found that the distance between the receiver and the source affects the

response by changing the sign of the response. If the antisymmetric mode of the structure

has been excited, the response will be antisymmetric, and the sign of the displacement

will change as the distance between the source and the receiver changes.

408

Another interesting founding was made by Pao et al. (1979) (ray theory). The location

of the source and receiver with respect to the surfaces of the plate varies the response. In

particular, the authors found that it would be difficult to distinguish the response due to

the configuration with source in the middle of the plane or source on the surface and

receiver on the opposite surface. A third configuration, with both source and receiver on

the same surface, is instead significantly different from that of the buried force because

the Rayleigh wave is now present in the response.

Medick (1961) and Gorman et al. (1990) developed a method to simulate AE called

normal mode solution for classic plate theory. This method is based on the approximate

solution for flexural plate mode and low frequency assumption. The out-of-plane

displacement is modeled through the classic plate theory in which the in-plane

displacements are neglected and a shear coefficient is taken into account to introduce the

shear effect. This method is suited for thin homogeneous isotropic plates. The load is

assumed to be concentrated in one point and normal to the surface. If we further assume

that the load can be expressed as ( )Pf t , the model for the acoustic emission is

represented as

4 ( ) ( )( , ) ( , )2

Pf t rD w r t hw r trδρ

π∇ + =&& (17.1)

where w is the displacement in the out-of-plane direction, D is the flexural plate

stiffness, and ρ is the material density.

This method is quite simple to implement. A Matlab program has been written in our

lab, this program can compute the out-of-plane displacement recorded by the transducers

409

for a given plate dimensions, source and receiver location. Our transducer however can

sense the in-plane displacements, hence this model is not useful for our applications.

Two different methods have been developed so far that are able to model in-plane

displacements.

Pao et al. (1979) and Pao (1978) have studied transient wave in elastic plate through

the generalized ray solution. The solution is derived for any mode present and it can be

applied for thick or thin plates, however, the plate must be assumed to be infinite. The

model will derive the exact solution as long as the time duration it is modeled is

sufficiently small. The load configuration can be that of a concentrated force, a double

force on any surface of the plate, or a dilatation force.

The solution is derived by summing the contribution due to all possible reflections of

a wave from the source to the receiving point (see Figure 17.1). The model is based on

the generalized ray solution in which a generalized path is specified by the path along

which the wave propagates and by the modes of the wave motion. The modes considered

are that of a pressure or a shear wave. For a given source and receiver point in the plate,

an infinite number of paths can be drawn. The greater the number of paths is taken in

consideration the longer the time interval that the model can predict. Figure 17.2 shows

the theoretical derivation for a plate of thickness h with a force normal to the bottom

surface and a receiver in the top surface at distance 4h from the force in the r direction.

As the bold arrow in the simulation graphs denotes the time beyond which the results are

no more exact. Note that the complexity of the derivation increases rapidly with the

number of paths.

410

Figure 17.1 Generalized rays from a source O to a receiver in location (r,z). (Pao et al., 1979)

Figure 17.2 Generalized theory: model of a plate excited by a force concentrated on the lower

surface and normal to it, receiver on the top surface at a longitudinal distance 4h

from the force (h = plate thickness). a) Mode configuration; b) First two paths (1+, 2-);

c) First three paths (1+, 2-, 3+); d) First four paths (1+, 2-, 3+, 4-). (Pao et al., 1979)

h1+

2-

3+ 4-

a) b)

c) d)

r

411

A FORTRAN program for generalized ray solution was developed by Hsu (1985).

This is still the reference program used to derive the in-plane displacement for wave

excited by a lead break on the top surface.

Another method, derived by Pursey (1957) and Gakenheimer et al. (1969), is based

on the integral displacement of the displacement components. However, the numerical

solution of such equations is difficult to obtain and no derivation for Lamb wave is

available.

In the next Sections we report first the derivation of the integral displacement of the

displacement components for Lamb wave excited by a load modeled as a step function;

second the formulation of AE made through NME as derived in Section 7.

17.1 ACOUSTIC EMISSION THROUGH INTEGRAL DISPLACEMENT

The integral solution is based on Fourier and Laplace transformations. The Navier

equations, in terms of Lame’ potentials φ and zψ , are transformed in frequency and

wavenumber domain. We imposed a step load on the top surface as boundary condition

and we assumed the system to be at rest initially.

Consider Navier equation in the form

( ) 2u u uλ μ μ ρ+ ∇∇ + ∇ = && (17.2)

and consider time invariance in the z direction. Every solution of equation (17.2) admits

representation

u φ ψ= ∇ + ∇× (17.3)

where the Lame’ potentials satisfy the wave equations

412

22

2 2

22

2 2

1

1p

s

c t

c t

φφ

ψψ

⎧ ∂∇ =⎪ ∂⎪

⎨∂⎪∇ =⎪ ∂⎩

(17.4)

and the divergence condition

0ψ∇ ⋅ = (17.5)

Assume the following initial conditions

( , ,0) 0

( , ,0) 0

( , ,0) 0

( , ,0) 0

x y

x ytx y

x yt

φφ

ψψ

=⎧⎪∂⎪ =⎪ ∂⎨ =⎪⎪∂

=⎪∂⎩

(17.6)

Recall the Laplace transformation defined as

0

( ) ( ) ptLf p f t e dt

∞−= ∫ (17.7)

and the Fourier transformation defined as

( ) ( ) i xFf f x e dxξξ

∞−

−∞

= ∫ (17.8)

where p is the Laplace transform parameter and ξ is the real Fourier parameter. Apply

the transformation to Equation (17.4) to get after rearrangement of the terms

22

2

22

2

0

0

y

y

φ α φ

ψ β ψ

⎧∂− =⎪∂⎪

⎨∂⎪ − =⎪ ∂⎩

(17.9)

413

where we have defined 2

22p

pc

α ξ= + and 2

22s

pc

β ξ= + . Solution of Equation (17.9) is

of the type

sinh cosh

sinh coshj j j

A y B yC y D y

φ α αψ β β

= +⎧⎨ = +⎩

j=x,y,z (17.10)

where constants A, B, Cj, and Dj are to be determined through the boundary conditions.

We consider the external load due to a vertical force applied on the top surface of the

plate, i.e.

[ ]1( , , ) ( ) ( ) ( )

( , , ) 0yy

xy

x d t x H t H t t

x d t

τ δ

τ

⎧ = − − −⎪⎨

=⎪⎩ (17.11)

With the use of the Laplace and Fourier transformations, the boundary conditions in


11( , , )

( , , ) 0

t p

yyF

xyF

ed pp

d p

τ ξ

τ ξ

−⎧ −= −⎪

⎨⎪ =⎩

(17.12)

Recall that the stress relation in tensor notation can be defined as

( ), ,ij ij i j j iu u uτ λ δ μ= ∇ + + (17.13)

Substitute Equation (17.3) into (17.13) to get

( )

22

2

22

2

2 2

2

zyy

zxy z

iy y

iy y

ψφτ λξ φ λ μ ξ μ

ψφτ μ ξ ξ ψ

⎧ ∂∂= − + + +⎪ ∂ ∂⎪

⎨⎛ ⎞∂∂⎪ = − −⎜ ⎟⎪ ∂ ∂⎝ ⎠⎩

(17.14)

414

Through the use of the boundary conditions (17.12), solution (17.10), and Fourier

transform (17.8), the divergence condition (17.5) and stresses (17.14) become

( ) ( )

( )

( ) ( )( )

( )

1

2 2 2 2 2 2 2 2 2 2 2 2

2

2 2 2 2 2 2 2 2 2 2 2 2

2

2 2

2 sinh 2 cosh

12 cosh sinh

2 sinh 2 cosh

2 cosh sinh 0

2 cosh 2 sinh si

s p p s p p

t p

s z z

p s p p s p

s z z

z

A c c c d B c c c d

ei c C d D dp

A c c c d B c c c d

i c C d D d

i A d i B d C

ξ ξ α α ξ ξ α α

ξ β β βρ

ξ ξ α α ξ ξ α α

ξβ β β

ξ α α ξ α α β ξ

−

− + + − + +

−+ + = −

− − − − − +

+ − =

+ − +

( )( )

( )

2 2

2 2

2 2

nh

cosh 0

2 cosh 2 sinh sinh

cosh 0

z

z

z

d

D d

i A d i B d C d

D d

β

β ξ β

ξ α α ξ α α β ξ β

β ξ β

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪

−⎪⎪− + =⎪⎪ − + + −⎪⎪− + =⎩

(17.15)

Since ( )2 2 2 2 2 2 2 2 22p s p sc c c cξ ξ α ξ β− − = − + , Equation (17.15) can be written as

( ) ( )

( )

( )

( )

( )

( )

1

2 2 2 2

2

22 2 2

22 2 2

2 2

2 2

sinh cosh

12 cosh sinh

sinh cosh 4 cosh sinh2 cosh

cosh sinh 4 sinh cosh 02 sinh

cosh2 cosh

s

t p

z zs

zz

zz

z

z

A d B d

ei C d D dc p

d d D d dDi d

d d C d dCi d

dA Di d

B C

ξ β α ξ β α

ξβ β βρ

β ξ α β ξ αβ α βξα α

β ξ α β ξ αβ α βξα α

β ξ βξα α

β ξ

−

+ + + +

−+ + = −

+ −

+ −− =

+=

+=

inh2 sinh

di d

βξα α

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

(17.16)

Note that by applying the relations sinh sinx i ix= − , cosh cosx ix= , and by calling

c iα α= , the denominator of the second Equation in (17.16) can be written as

415

( )( )

22 2 21

22 2 22

sin cos 4 cos sin

cos sin 4 sin cos

c c c c c c c

c c c c c c c

R d d i d d

R d d i d d

β ξ α β ξ α β α β

β ξ α β ξ α β α β

⎧ = − +⎪⎨⎪ = − +⎩

(17.17)

These are the Rayleigh – Lamb equations and hence Equation (17.16) becomes

( )

( )

1

1

1

1

22

21

2 2

21

2 2

22

sinh 1

cosh 1

cosh 12

sinh 12

t p

zs

t p

zs

t p

s

t p

s

i d eCD c p

i d eDD c p

d eAD c p

d eBD c p

ξα αρ

ξα αρ

β ξ βρ

β ξ βρ

−

−

−

−

⎧ −= −⎪

⎪⎪ −

= −⎪⎪⎨

+ −⎪ = −⎪⎪⎪ + −

= −⎪⎩

(17.18)

Substituting equations (17.18) into (17.10), yields the solution

( ) 1

1

2 2

21 2

22 1

1 cosh sinhsinh cosh2

1 sinh coshsinh cosh

t p

s

t p

zs

e d dy yc p D D

e d di y yc p D D

β ξ β βφ α αρ

α αψ ξα β βρ

−

−

⎧ + − ⎡ ⎤= − +⎪ ⎢ ⎥⎪ ⎣ ⎦⎨

− ⎡ ⎤⎪ = − +⎢ ⎥⎪ ⎣ ⎦⎩

%

%

(17.19)

To find the solution in time and space domain we must perform the inverse Laplace

transformation and inverse Fourier transformation, i.e.,

( )1

1

12 22

0

2

22

0

1

cosh sinh1 1( , , )

sinh4 cosh

sinh sinh1 1( , , )

cosh2 cosh

t pi x tp

s

t pi x tp

zs

d yRex y t e d e dp

dic p yR

d yRex y t e d e dp

dc p yR

ξ

ξ

β αφ β ξ ξ

βπ ρ α

α βψ ξα ξ

απ ρ β

∞ ∞−

−∞

∞ ∞−

−∞

⎧ ⎡ ⎤⎪ ⎢ ⎥−⎪ ⎢ ⎥= − +⎪ ⎢ ⎥+⎪ ⎢ ⎥⎪ ⎣ ⎦⎨

⎡ ⎤⎪⎢ ⎥⎪ − ⎢ ⎥= −⎪⎢ ⎥⎪ +⎢ ⎥⎪ ⎣ ⎦⎩

∫ ∫

∫ ∫

(17.20)

416

The Fourier transformation is performed by using the residual theorem. The denominator

of the solutions is the Rayleigh-Lamb equation for the derivation of dispersion curves. By

taking the dispersion values solutions as poles, the inverse Fourier transformation is

given by the sum of all the contributions of the different modes excited in the structures,

i.e.

( ) ( )[ ]

( )[ ]

1

1

2 21

20

2

22

0

1

cosh sinh1( , , )

sinh4 cosh

sinh sinh1( , , )

cosh2 cosh

n

n

nt t ptp

ni x n n

n nsn

n

nn t t ptp

ni x nz n

n nsn

n

d yD e e

x y t e dpdic py

D

d yD e ex y t e dp

dc pyD

ξ

ξ

β αβ ξ

φβπ ρ α

α βαψ ξ

απ ρ β

−∞

∞ −

⎧ ⎡ ⎤⎪ ⎢ ⎥′ + −⎪ ⎢ ⎥= −⎪ ⎢ ⎥

+⎪ ⎢ ⎥′⎪ ⎣ ⎦⎨

⎡ ⎤⎪⎢ ⎥⎪ ′ −⎢ ⎥= −⎢ ⎥+⎢ ⎥′⎣ ⎦⎩

∑ ∫

∑ ∫⎪⎪⎪

(17.21)

The inverse Laplace transformation is not banal, and it can only be performed

numerically. Knowing the potentials it is possible to determine the particle displacement

on the surface of the plate. This method is however computationally heavy.

17.2 ACOUSTIC EMISSION THROUGH NORMAL MODE EXPANSION

Prosser et al. 1999 presented the plate theory modeling of acoustic emission waveforms.

This theory uses the Mindlin plate theory to derive the flexural response to a lead break

on the surface of the structure. The limit of their derivation is that only out-of-plane

displacements can be derived.

Here we propose an alternative model in which, through the use of NME method, we

derive the acoustic wave fields for the case of an acoustic emission.

Let the volume source be zero, i.e., 1( , )x y =F 0 ; hence, Equation (7.19) becomes

417

( )1 1 ˆ4 ( )d

nn n n n n dP i a x

xξ

−

∂⎛ ⎞+ = + ⋅⎜ ⎟∂⎝ ⎠v T v T y%% (17.22)

where y is the unit vector in the y direction. Recall that the orthogonality relation (6.112)

is obtaining by requiring that the normal modes of the plate satisfy the traction free

condition; hence,

ˆn y d=±⋅ =T y 0 (17.23)

Using Equation (17.23), we can express the right-hand side of Equation (17.22) as

1 1ˆ ˆn n⋅ ⋅ + ⋅ ⋅v T y v T y%%( ) ( )1 ˆd d

n dd −−= ⋅ ⋅v T y%

The tractions at the lower and upper surfaces, 1 ˆy d=±

= ⋅t T y , are prescribed from the

boundary conditions. For a lead break load applied at the upper surface and centered at

the origin of the x axis, the surface tractions take the form

x

y

tt

⎧ ⎫= ⎨ ⎬

⎩ ⎭t , where 0xt = and

( ) 0( )

0 y

t x xt x

otherwise=⎧

= ⎨⎩

(17.24)

In view of Equation (17.24), the traction force at the upper and lower surfaces can be

expresses as:

1 ˆ( , ) ( )x d x⋅ =T y t (upper surface) (17.25)

1 ˆ( , ) 0x d− ⋅ =T y (lower surface) (17.26)

where ( )xt is an externally applied surface traction given by Equation (17.24). Hence,

the right-hand side of Equation (17.22) becomes

418

( )1 ˆ ( ) ( )d

n ndd x

−⋅ ⋅ = ⋅v T y v t% % (17.27)


4 ( ) ( ) ( )nn n n nP i a x d xx

ξ∂⎛ ⎞+ = ⋅⎜ ⎟∂⎝ ⎠v t% (17.28)


( ) 1( ) ( ) ( )4

nn n n

nn

a x i a x d xx P

ξ∂+ = ⋅

∂v t% (17.29)


factor method. Comparison of Equation (17.29) with the standard ODE form yields

( )( ) ( )4

n n

xi x i xn

nnn c

da x e e x dxP

ξ ξ−= ⋅ ∫v t% forward wave solution (17.30)


it contains the factor ni xe ξ− . Since the lead break at the x -axis origin is the only acoustic

source, it is apparent that waves will have to emanate outwards from the load. The wave

amplitude stays constant. Hence, the amplitude ( )na x has to satisfy the following

boundary condition

( ) 0 for 0na x x= ≤ b.c. on forward wave solution (17.31)

Applying the boundary condition (17.31) to the solution (17.30) yields

0( )(0) ( ) 0

4n ni a i xn

nnn c

da e e x dxP

ξ ξ−= ⋅ =∫v t% (17.32)

Denote by ( )nF x the integrant of the integral in Equation (17.32), i.e.,

419

( ) ( )ni xnF x e x dxξ−= ∫ t (17.33)

Hence,

0

( ) (0) ( ) 0ni xn n

c

e x dx F F cξ = − =∫ t (17.34)

i.e.,

( ) (0)F c F= (17.35)

Equation (17.35) implies that Equation (17.30) can be written with the lower limit c

equal to 0 , i.e.,

0

( )( ) ( )4

n ni x i xnn

nn

da x e e x dxP

ξ ξ∞

−+ = ⋅ ∫v t% (forward wave solution) (17.36)

where superscript + signifies waves propagating in the positive x direction. The above

argument can be equally applied to backward propagating waves.

0( )( ) ( )

4n n

xi x i xn

nnn

v da x e e t x dxP

ξ ξ−−

−∞

= − ⋅ ∫% (backward wave solution) (17.37)

where superscript − signifies waves propagating in the negative x direction.

In the case of lead break solution the load can be written as

( ) ˆ if 0

(0, )0 P x x

dotherwise

δ =⎧⎪= ⎨⎪⎩

yt (17.38)

Substituting (17.38) into (17.36) we obtain:

( )0

( )( )

4n n

ny i x i x

nnn

Pv da x e x e dx

Pξ ξδ

∞−

+ = ∫%

(17.39)

420

The integral in the wave field amplitude (17.39) is the Fourier integral of the Dirac delta

function; it can be solved and it is equal to one. Hence, Equation (17.39) becomes

( )

( )4

ni xny

nnn

v d ea x P

P

ξ−+ =

% (17.40)

From Equation (17.40) we derive the expression of the particle displacement, i.e.,

( )( )

( , , ) ( ) ( ) ( )4

ni x tny

nn nn

v d ex y t a x y P y

P

ξ ω− −+= =∑u u u

% (17.41)

Theoretical predictions and experimental results should be compared to validate the

theory developed.

421

18 CONCLUSIONS AND FUTURE WORK

This dissertation has addressed the fundamental studies in the Lamb-wave interaction

between piezoelectric wafer active sensor (PWAS) and host structure during structural

health monitoring (SHM.). The application of PWAS-based SHM for the detection of

damages in structures guided waves methods has experienced an ascending trend.

However, to achieve the large deployment of these SHM techniques and to be able to

perform in-situ and online SHM, the physics of wave excitation in the structure must be

fully understood.

The discussion in this dissertation began with the generic formulation for ultrasonic

guided waves in thin wall structures. The formulation was generic because, unlike many

authors, in many parts of our derivation (power flow, reciprocity theorem, orthogonality,

etc.) we stayed away from specifying the actual mathematical expressions of the guided

wave modes and maintained a generic formulation throughout. We also studied the power

flow and energy conservation mechanisms associated with the guided waves with

particular interest on power flow of circular crested guided waves.

The dissertation continues with the extension of the normal modes expansion (NME)

model to the case of PWAS bonded to or embedded in the structure. We developed a

theoretical prediction model of the shear transfer from PWAS to the structure through a

bond layer without limitations on the frequency and the number of modes present. We

solved the resulting integro-differential equation for shear lag transfer; we applied these

422

results to predicting the tuning between guided waves and PWAS and obtained excellent

agreement with experimental results. Another novel aspect covered in this dissertation is

that of guided waves in composite materials. The NME theory is extended to the case of

composites and we developed a generic formulation for the tuning curves that was not

directly dependent on the composite layup and could be easily extended to various

composite formulations. We conducted carefully-planned experiments on composites

with different orientations. The comparison between our predictions and experiments was

quite good.

In the last part of the dissertation, SHM issues and applications are presented. We

discussed the reliability of SHM systems and the lack of quality specifications for SHM

inspections with particular focus on the case of composites SHM. We determined

experimentally the ability of PWAS to detect damage in various composite specimens.

We tested the performance of PWAS for damage detection on composite plates, on

unidirectional composite strips, on quasi-isotropic plates, on lap-joints junctions, and

composite tank sections. We also tested the ability of PWAS transducers to operate under

extreme environments and high stress conditions, i.e. the survivability of PWAS-based

SHM. We proved the durability of the entire PWAS-based SHM system under various

different load conditions. We also tested the influence of bond degradation on PWAS

electrical capacitance as installed on the structure, which gives a measure of the quality

of the PWAS installation, a key feature in PWAS-based SHM. We developed theoretical

models for shear horizontal waves scattering from a crack and Lamb waves scattering

from change in material properties. We studied the acoustic emission (AE) in infinite

plate and we used NME to model AE phenomena.

423

A review of the main results of this dissertation is given next.

18.1 RESEARCH CONCLUSIONS

18.1.1 Elastic waves for structural health monitoring

The aim of the discussion was to provide first a generic derivation valid for guided waves

propagation in a layered media assuming an orthogonal coordinate system. First, the

theory of guided waves propagation was thoroughly studied; then, a more detail

derivation for both shear horizontal waves and Lamb waves was provided for the case of

isotropic materials considering both straight-crested and circular-crested guided waves.

The major findings are listed as follows:

(1.) Acoustic field equations independent of material of the media and orthogonal

coordinate system can be derived only in Vogit notations.

(2.) Solution of the circular-crested Lamb waves can be obtained by assuming that the

dilatation varies with the thickness as a sine and cosine function. There is no need

to assume hyperbolic functions.

(3.) From the solution of the circular shear horizontal waves, we notice that the

particle displacements and stresses distributions across the thickness are equal to

those derived for straight crested shear horizontal waves.

(4.) The particle displacements variations across the thickness are the same for both

straight-crested and circular-crested Lamb waves.

(5.) The stresses distributions across the thickness are the same for both straight-

crested and circular-crested Lamb waves, except for the normal stress in the

direction of the propagation of the wave. The circular-crested Lamb waves radial

424

stress has an additional term that becomes negligible as the distance from the

source increases.

18.1.2 Power flow and orthogonality relations

The major findings of the research on power flow and orthogonality relations are listed as

follows:

(1.) For time-harmonic guided waves, half of the real part of the complex power flow

represents the time-averaged power over a time period T.

(2.) The average power flow of a guided wave through a surface is equal to zero if

both forward and backward propagating waves are considering.

(3.) To obtain the average power flow the contribution from forward and backward

waves must be separated.

(4.) For straight-crested waves, the D’Alambert formulation of the solution provides a

straightforward way to obtain the wave separation between forward and backward

waves.

(5.) Power flow for circular crested wave is present in literature of structural guided

waves only for wave propagating in cylinders or tubes. No derivation is present in

literature for wave propagating radially.

(6.) No D’Alambert solution is possible for circular crested waves. However, through

the use of the complex form of the solution of circular-crested waves, we were

able to separate the inward propagating wave from the outward propagating wave

and obtain the average power flow.

425

(7.) Real and complex reciprocity relations can be derived without the need to specify

the material of the medium or the orthogonal coordinate system.

(8.) Guided waves form a complete set of orthogonal functions. If the wave is a

propagating wave (i.e., real wavenumber), the power flow is carried by the cross-

product of the propagating mode with itself. If the wave is an evanescent wave

(i.e., imaginary wavenumber), the power flow is carried by the cross product of

the positive evanescent mode and its reflected mode (negative wavenumber).

(9.) For straight-crested waves is possible to prove the orthogonality of the guided

wave functions without specifying the particular solution. This is not possible for

circular-crested waves.

(10.) The normalization factor is the same in any coordinate system.

18.1.3 Excitation of guided waves and shear layer coupling between PWAS and

structure

The major findings of the research on excitation of guided waves and shear layer

coupling between PWAS and structure are listed as follows:

(1.) PWAS bonded on an isotropic plate can only excite Lamb waves.

(2.) The NME theory has been extended to circular crested waves.

(3.) NME is a useful method to study guided wave excited by one or two PWAS

installed on the surface of a structure or embedded in a structure.

(4.) The shear-lag model is provided for any frequency and number of wave modes

present through the use of NME theory.

426

(5.) There is difference at low frequency between the shear-lag derivation without

assumptions and the shear-lag derived with the low frequency approximation. The

difference is due to the Euler-Bernoulli assumption made in the latter.

(6.) The influence of the different parameters of the PWAS-bond-system on the shear

transfer distribution is discussed in detail.

(7.) The importance of the evanescent modes in the derivation of the shear-lag is

reported in detail.

18.1.4 Tuned guided waves in structures

The major findings of the research on tuned guided waves in structures are listed as

follows:

(1.) Forcing functions for ideal bond conditions, shear-lag parameter with low

frequencies approximation, and shear-lag for N generic modes are derived and

compared.

(2.) Tuning curves are derived through the NME method. The formulation gives the

same results as the integral Fourier derivation. This is shown numerically and

analytically.

(3.) Experimental data and theoretical predictions are compared for various PWAS

geometry and plate thickness. There is agreement between predictions and

experiments.

(4.) Tuning curves predictions through ideal bond assumptions need a correction

factor in the PWAS length. However, the correction factor is not needed when the

shear-lag model is used.

427

(5.) Theoretical tuning curves are developed for composite plates.

(6.) The tuning curve formulation does not depend on the particular method used to

extract the dispersion curves of the guided waves propagating in the composite

plate.

(7.) A program to extract dispersion and group velocities curves in composite plates of

layered materials is available.

(8.) Predictions and experimental data for quasi-antisymmetric wave propagating in a

quasi-isotropic plate made of 16 layers are compared. The data are in good

agreement.

18.1.5 Reliability of SHM

The major findings of the research on reliability of SHM are listed as follows:

(1.) Specifications for the reliability of quality SHM inspections are not available.

(2.) SHM relies on the specifications derived for NDE systems; however, these are not

always valid for SHM systems.

(3.) Probability of detection (POD) curves have not been derived for any SHM

system. The major problem is that POD curves for SHM are influenced by many

factors. To obtain POD curves, several experiments for each factor should be

considered. Cost and time limitations have been the major drawbacks.

18.1.6 Space-qualified SHM

(1.) PWAS are able to perform damage detection at cryogenic temperatures.

(2.) Electro mechanical impedance does not change with thermal fatigue.

428

(3.) PWAS are able to detect damage due to holes, impact, and delamination in

composite structures of different thickness, layer lay-up, and geometry.

(4.) SHM using PWAS is a space-qualified system for damage detection.

18.1.7 Survivability and durability of SHM systems

(1.) SHM systems with PWAS can survive extreme low temperatures and stress

conditions and still be operative.

(2.) The weak points in the PWAS-based SHM system are the solder between PWAS

and electric wire and the bond layer between PWAS and structure.

(3.) Through the combined use of E/M impedance and wave propagation it is possible

to asses the functionality o f the SHM system and the health of the structure.

(4.) SHM using PWAS can withstand space qualification tests and pass the lunch

requirement for the system.

(5.) Ranges of capacitance of the PWAS bonded on an aluminum structure are given

for good bond, partial bond, and non-bonded PWAS. These ranges can be used as

a reference for the operator.

18.1.8 Scattering from a crack and acoustic emission detection

(1.) Theoretical formulation of Lamb wave scattering is not possible without the use

of finite element models.

(2.) A simple derivation for scattering from a partial crack is derived for SH waves.

(3.) The generic problem for Lamb waves scattering from material inhomogeneity and

thickness change is provided.

429

(4.) A review of the available AE modeling shows that few models are applicable for

PWAS AE detection.

(5.) An integral Fourier formulation is derived for modeling the in-plane particle

displacement due to an AE.

(6.) An alternative formulation is derived through NME theory.

18.2 MAJOR CONTRIBUTIONS

This dissertation has brought major contributions to the fundamental studies in the Lamb-

wave interaction between PWAS and host structure during SHM. Some of these are novel

and had not been found in the specialized literature. Hereunder, we report a list of the five

most important contributions.

(1.) Development of the shear-lag solution for N generic guided wave modes present.

(2.) Extension of the derivation of tuning curves to anisotropic materials through

NME. (For the first time)

(3.) Derivation of orthogonality relations without specific mathematical expressions.

(For the first time)

(4.) Application of statistical models to damage identification algorithms

(5.) Demonstration through experiments that PWAS-based SHM can be applied to

composites in extreme environments towards space qualification.

430

18.3 RECOMMENDATION FOR FUTURE WORK

The research presented in this dissertation can be used in the development of an efficient

PWAS-based SHM system. The following tasks are recommended to be undertaken to

continue the research:

(1.) Extend the shear-lag solution to the circular crested guided waves.

(2.) Extend the shear-lag solution to anisotropic materials.

(3.) Explore the contribution of complex wavenumbers and higher order imaginary

wave modes to the shear-lag solution.

(4.) Experimentally verify the shear-lag derived with the N generic model.

(5.) Extend the derivation of tuning curve through NME to circular crested waves and

compare with the integral Fourier derivation in 2-D.

(6.) Investigate which specifications are needed to obtain a reliable system, in

particular determine whether POD curves are needed for SHM or if alternative

methods can be found.

(7.) A more reliable connection between PWAS and electric wire and bond between

PWAS and structure should be explored.

(8.) Capacitance interval of the good bond of PWAS on different materials should be

provided and used to determine the quality of the bond.

(9.) Scattering models with finite elements methods should be implemented. These

can be useful to determine the best PWAS lay out for damage detection in a

structure.

431

(10.) The AE model through integral Fourier method and NME method should be

verified against experimental data.

432

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APPENDIX

442

A EQUATION OF MOTION IN CYLINDRICAL COORDINATES In the first section of the appendix we will show the general formulation for the equation

of motion as described in continuum mechanics. In the second part of the appendix we

will derive the equation of motion in cylindrical coordinates from equilibrium

considerations.

A.1 EQUATIONS OF MOTION IN CONTINUUM MECHANICS The equations of motion in continuum mechanics are derived from the conservation of

linear momentum. Consider a region R in space where exists a material volume of density

ρ having surface tractions and body forces acting upon it. Denote with iu the

displacement of the material volume. The second law of Newton states that the rate of

change of linear momentum equals the resultant force on the volume. The change in

linear momentum can be expressed as

2

2 iRu d

tρ τ∂

∂ ∫∫∫ (A.1)

where dτ is an element of volume and ρ is the material density per unit volume. The

force applied to the volume is equal to

jij iS Rn dS f dσ ρ τ+∫∫ ∫∫∫ (A.2)

where ijσ are the contravartiant components of the stress tensor, jn is the normal to the

surface S, dS is an element of the surface area, jf are the body forces per unit mass.

443

Note that ( )j nij in tσ = are the component of the surface traction forces ( )nt

r associated with

the plane with normal n. Combining Equations (A.1) and (A.2), Newton’s second law

becomes

2

2j

i ij iR S Ru d n dS f d

tρ τ σ ρ τ∂

= +∂ ∫∫∫ ∫∫ ∫∫∫ (A.3)

Recall the Gauss divergence theorem, i.e.,

,j

i j iV SF d F n dSτ =∫∫∫ ∫∫ (A.4)

Use Equation (A.4) in Equation (A.3) to obtain

2

,2 i ij j iR R Ru d d f d

tρ τ σ τ ρ τ∂

= +∂ ∫∫∫ ∫∫∫ ∫∫∫ (A.5)

Rearrange the terms and note that the time derivative can be brought inside the volume

integral, hence we get

2

,2 0iij j iR

u f dt

ρ σ ρ τ⎛ ⎞∂

− − =⎜ ⎟∂⎝ ⎠∫∫∫ (A.6)

Since the integral must be true for any given region R of the volume, Equation (A.6)

implies

2

,2 0iij j i

u ft

ρ σ ρ∂− − =

∂ (A.7)

and in absence of external forces, Equation (A.7) simplifies as

2

, 2 0iij j

ut

σ ρ ∂− =

∂ (A.8)

444

The derivate if the contravariant components of the stress tensor can be expressed as (see

Heibockel)

[ ], 2 2 2 2

1 ,ij mjij j iij

i j m jii

g g ij mx h h h hg g

σ σσ

⎛ ⎞∂= −⎜ ⎟∂ ⎝ ⎠

(A.9)

where ijg are the metric components of the coordinate system considered, i iih g= , and

the second term in Equation (A.9) can be expressed as

[ ]2

2 2 2 2 2 2

1 1,2 2

mj mj jm ij jj jimj i m i

m j m j j j

g g hgij mh h h h x x x h h xσ σ σ∂ ∂ ∂⎛ ⎞∂

= + + =⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ (A.10)

Hence, the stress tensor derivate is equal to

2

, 2 2 2 2

1 12

ij jj jij j iij i

i j j jii

hg g

x h h h h xg gσ σ

σ∂⎛ ⎞∂

= −⎜ ⎟∂ ∂⎝ ⎠ (A.11)

In order to express Equation (A.11) in terms of physical components in orthogonal

systems, we must perform the following transformation

( )

( )ij i j

ii

ii

ij h h

b i hbg

σ σ=

= (A.12)

Using (A.12), Equation (A.11) becomes

23 3

, 21 1

1 1 ( ) 1 ( )2

jij j iij i

j jii i j j

hij jjg gg x h h h xg

σ σσ= =

∂∂ ⎛ ⎞= −⎜ ⎟∂ ∂⎝ ⎠∑ ∑ (A.13)

Substitute Equation (A.13) into the equation of motion (A.7) to obtain

2 23 3

2 21 1

1 1 ( ) 1 ( ) ( ) 02

j iiij i

j jii i j j ii

h hij jj u ig gg x h h h x g tg

σ σ ρ= =

∂∂ ∂⎛ ⎞ − − =⎜ ⎟∂ ∂ ∂⎝ ⎠∑ ∑ (A.14)

445

Equation (A.14) is the equation of motion in any orthogonal coordinate system. Recall

that for any orthogonal system it is 0ijg = for i j≠ . Expand the equation of motion to

get a system of three equations, i.e.,

2 23 31

11 2 1 21 111 1 11

2 23 32

22 2 2 21 122 2 22

23

33 21 33 3

1 1 ( ) 1 ( ) ( ) 02

1 1 ( ) 1 ( ) ( ) 02

1 1 ( ) 1 ( )2

jj

j jj j

jj

j jj j

jj

j j j



hij jjg gg x h h hg

σ σ ρ

σ σ ρ

σ σ

= =

= =

=

∂∂ ∂⎛ ⎞ − − =⎜ ⎟∂ ∂ ∂⎝ ⎠∂∂ ∂⎛ ⎞ − − =⎜ ⎟∂ ∂ ∂⎝ ⎠∂∂ ⎛ ⎞ −⎜ ⎟∂ ∂⎝ ⎠

∑ ∑

∑ ∑

∑23

33 2

1 33

(3) 0j

h ux g t

ρ=

⎧⎪⎪⎪⎪⎨⎪⎪

∂⎪ − =⎪ ∂⎩

∑

(A.15)

In Cartesian coordinates we have that

11 22 33

1 2 3

11 22 33

11

1

g g gh h hg g g g

= = =⎧⎪ = = =⎨⎪ = =⎩

(A.16)

With the use of relations in Equation (A.16), the equations of motion (A.15) become

2

2

2

21 2 3

2

21 2 3

0

0

0

xyxx xz x

xy yy yz y

yzxz zz z

ux y z t

ux x x t

ux x x t

σσ σ ρ

σ σ σρ

σσ σ ρ

∂⎧∂ ∂ ∂+ + − =⎪ ∂ ∂ ∂ ∂⎪

⎪∂ ∂ ∂ ∂⎪ + + − =⎨∂ ∂ ∂ ∂⎪

⎪ ∂∂ ∂ ∂⎪ + + − =∂ ∂ ∂ ∂⎪⎩

(A.17)

In cylindrical coordinates we have that

11 33 1 3

222 2

211 22 33

1g g h h

g h r

g g g g r

= = = =⎧⎪

= =⎨⎪ = =⎩

(A.18)


446

2

2

2

2

2

2

1 0

1 2 0

1 0

r rr irr rz

r z r

zrz zz rz z

ur r r r t

ur r z r t

ur r z r t

θ θθ

θ θθ θ θ θ

θ

σ σ σσ σ ρθ

σ σ σ σ ρθ

σσ σ σ ρθ

∂ − ∂∂ ∂+ + + − =

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

+ + + − =∂ ∂ ∂ ∂

∂∂ ∂ ∂+ + + − =

∂ ∂ ∂ ∂

(A.19)

In spherical coordinates we have that

211 1

2 222 2

2 2 233 3

4 211 22 33

1

sin

sin

g h

g h

g h

g g g g

ρ

ρ θ

ρ θ

⎧ = =⎪

= =⎪⎨

= =⎪⎪ = =⎩

(A.20)


( )

2

3 2

2

3 2

2

1 1 1 cos 2 0sin sin

1 1 1 cos3 0sin sin

1 1 1 cos3 2sin sin

rux t

ux t

ρρ ρθ ρφρθ ρρ θθ φφ

ρθ θφθθ θρθ θθ φφ

ρφ θφ φφρφ θφ

σ σ σ θ σ σ σ σ ηρ ρ θ ρ θ ρ θ

σ σσ θσ σ σ ηρ ρ θ ρ θ ρ θ

σ σ σ θσ σ ρρ ρ θ ρ θ φ ρ θ

∂ ∂ ∂ ∂⎛ ⎞+ + + + − − − =⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠∂ ∂∂ ∂⎡ ⎤+ + + + − − =⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦∂ ∂ ∂ ∂⎛ ⎞+ + + + −⎜ ⎟∂ ∂ ∂ ⎝ ⎠ 2

(3) 0ut

⎧⎪⎪⎪⎪⎨⎪⎪

=⎪∂⎪⎩

(A.21)

A.2 EQUATIONS OF MOTION FROM EQUILIBRIUM CONSIDERATION Consider a plate in a cylindrical coordinate system. The distance r of a point P of the

plate from the origin O is defined by the angle θ between r and an axis Ox fixed in the

plate (Figure A.1). The distance OP is equal to r.

Consider the equilibrium of the volume element in the radial direction. The normal

stress ( )1rσ acts on the surface 1r d dzθ , where 1r is the radium of the side 1. Hence the

force on surface 1 along the radium is equal to

( ) 11r r d dzσ θ (A.22)

447

Figure A.1 Equilibrium of a small element of a plate.

Similarly, the radial force on surface 3 is given by

( ) 33r r d dzσ θ− (A.23)

The component of the normal force ( )2θσ on surface 2 along the radii gives a negative

contribute to the net force, i.e.,

( )2sin

2ddrdzθθσ− (A.24)

The component of the normal force ( )4θσ on surface 4 along the radii gives a negative

contribute to the net force, i.e.,

( )4sin

2ddrdzθθσ− (A.25)

The shear stress ( )2rθτ on surface 2 and the shear stress ( )4rθτ on surface 4 give a net

contribution given by

( ) ( )2 4r r drdzθ θτ τ⎡ ⎤−⎣ ⎦ (A.26)

x

z

O

P 1

2

3 4

5

6

dθ θ

y

x

y

O

P

2

3 4

dθθ

1

(σr)1

(σr)3

(σθ)4

(σθ)2

(τrθ)4

(τrθ)2(τrθ)1

(τrθ)3

(τθz)5(τθz)2(τrz)5

(τrz)1

(σz)5

(τθz)4

(τrz)3

5

448

The shear stress ( )5rzτ on surfaces 5 and the shear stress ( )6rzτ on surface 6 give a net

contribution given by

( ) ( )5 6rz rz rd drτ τ θ⎡ ⎤−⎣ ⎦ (A.27)

Sum all the forces along the radial direction and add the body forces contribution R to

obtain

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

1 31 3 2 4 2 4

5 6

sin2 2

0

r r r r

rz rz

d dr r d dz drdz drdz

drd Rrd drdz

θ θ θ θθ θσ σ θ σ σ τ τ

τ τ θ θ

⎡ ⎤⎡ ⎤ ⎡ ⎤− − − + −⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦⎡ ⎤+ − + =⎣ ⎦

(A.28)

Divide Equation (A.28) by the volume drd dzθ to get

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 31 3 2 4

5 62 4

sin sin2 2

0

r r

r r rz rz

r r d ddr d d

r Rrd dz

θ θ

θ θ

σ σ σ σθ θθ θ

τ τ τ τθ

−− − +

− −+ + + =

(A.29)

Note that Equation (A.29) can be alternatively written as

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1 3 5 62 4 2 4 0

2r r r r rz rzr r

r Rrdr d dz

θ θ θ θσ σ σ σ τ τ τ τθ

− + − −− + + + = (A.30)

As the dimensions of the element get smaller, the terms in Equation (A.30) can be

approximated as after rearrangement

1 0r rr rz Rr r z r

θ θτ σ σσ τθ

∂ −∂ ∂+ + + + =

∂ ∂ ∂ (A.31)

Consider now the equilibrium of the volume element in the angular direction. We have

the following contributions: Normal force on surface 2

449

( )2drdzθσ (A.32)

Normal force on surface 4

( )4drdzθσ− (A.33)

Contribution from the shear forces on surfaces 1 and 3

( ) ( )1 31 3r rr r d dzθ θτ τ θ⎡ ⎤−⎣ ⎦ (A.34)


( ) ( )2 4sin

2r rddrdzθ θθτ τ⎡ ⎤+⎣ ⎦ (A.35)


( ) ( )5 6z z rd drθ θτ τ θ⎡ ⎤−⎣ ⎦ (A.36)

The total force in the angular direction plus the component of the body force S per unit

volume divided by the element volume gives

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1 31 3 5 62 4 2 4 0

2r r z zr rr r

r Srd dr dz

θ θ θ θθ θ θ θτ τ τ τσ σ τ τθ

− −− ++ + + + = (A.37)


approximated as after rearrangement

1 2 0r r z Sr r r z

θ θ θ θτ σ τ τθ

∂ ∂ ∂+ + + + =

∂ ∂ ∂ (A.38)

The equilibrium of the volume element in the z direction is given by the following

contributions: Normal force on surface 5

450

( )5z drrdσ θ (A.39)

Normal force on surface 6

( )6z drrdσ θ− (A.40)


( ) ( )1 31 3rz rzr r d dzτ τ θ⎡ ⎤−⎣ ⎦ (A.41)


( ) ( )2 4z z drdzθ θτ τ⎡ ⎤−⎣ ⎦ (A.42)

The total force in the z direction plus the component of the body force Z per unit volume

divided by the element volume gives

( ) ( ) ( ) ( ) ( ) ( )1 35 6 1 3 2 4 0z z rz rz z zr r

r Zrdz dr d

θ θσ σ τ τ τ τθ

− − −+ + + = (A.43)


approximated as

0zz rzrr Zrz r

θτσ τθ

∂∂ ∂+ + + =

∂ ∂ ∂ (A.44)

or

1 0zrz z rz Zr r z r

θττ σ τθ

∂∂ ∂+ + + + =

∂ ∂ ∂ (A.45)

Equations (A.31), (A.38), and (A.45) are the equation of motion of mass subjected to

body forces expressed in cylindrical coordinates.

451

B MATHEMATIC CONCEPTS

B.1 DEL OPERATOR NOTATION FOR VOGIT MATRIX FORMALISM The Vogit notation is a transformation of the vector notation. The order of the tensor is

reduced by one in the transformation, such as a first order tensor becomes a vector and a

second order tensor becomes a first order tensor. The transformation is done by following

the rule

1 42 53 6

xx yzyy xzzz xy

→ →→ →→ →

(B.1)

Consider the first order tensor of the stresses. In vector notation the stress tensor is

defined as

xx xy xz

xy yy yz

xz yz zz

T T TT T TT T T

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

T (B.2)

In Vogit notation Equation (B.2) becomes

1

2

3

4

5

6

xx

yy

zz

yz

xz

xy

T TT TT TT TT TT T

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪

⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭

T (B.3)

452

Consider the second order tensor of the stiffness coefficient, i.e., ijklc=c . For

convenience we consider the case in which the stiffness tensor is symmetric. In Vogit

notation the second order tensor becomes

11 12 13 14 15 16

22 23 24 25 26

33 34 35 36

44 45 46

55 56

66

c c c c c cc c c c c

c c c cc c c

c cc

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

c (B.4)

B.1.1 Rectangular coordinates Divergence of a tensor A in vector notation

xyxx xz

xx yx xzyx yy yz

yx yy yz

xz yz zzzyzx zz

AA Ax y z

A A AA A A

A A Ax y z x y z

A A AAA A

x y z

∂⎧ ⎫∂ ∂+ +⎪ ⎪∂ ∂ ∂⎪ ⎪⎡ ⎤ ⎪ ⎪∂ ∂ ∂∂ ∂ ∂ ⎪ ⎪⎧ ⎫ ⎢ ⎥∇ ⋅ = = + +⎨ ⎬⎨ ⎬ ⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂⎩ ⎭ ⎪ ⎪⎢ ⎥⎣ ⎦ ⎪ ⎪∂∂ ∂+ +⎪ ⎪

∂ ∂ ∂⎪ ⎪⎩ ⎭

A (B.5)

Divergence of a tensor A in Vogit notation

1 6 51

2

3 6 2 4

4

5 5 34

6

0 0 0

0 0 0

0 0 0

A A AAA x y zx z yA A A AAy z x x y zA A AA

z y x x y zA

⎧ ⎫ ⎧ ⎫⎡ ⎤ ∂ ∂∂∂ ∂ ∂+ +⎪ ⎪ ⎪ ⎪⎢ ⎥ ∂ ∂ ∂∂ ∂ ∂ ⎪ ⎪ ⎪ ⎪⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥ ∂ ∂ ∂∂ ∂ ∂ ⎪ ⎪∇ ⋅ = = + +⎨ ⎬ ⎨ ⎬⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥∂ ∂ ∂ ∂ ∂∂

+ +⎪ ⎪ ⎪ ⎪⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂⎪ ⎪⎣ ⎦ ⎩ ⎭⎩ ⎭

A (B.6)

Curl of a vector A in vector notation

453

yz

xx z

y

zy xx y z

AAi j k y zx A

A AAy x y z z x

A A AA A Ax yz

∂⎧ ⎫∂∂⎧ ⎫ −⎪ ⎪⎪ ⎪ ∂ ∂∂ ⎪ ⎪⎧ ⎫⎪ ⎪⎪ ⎪∂ ∂∂ ∂ ∂ ∂⎪ ⎪⎪ ⎪∇× = × = = −⎨ ⎬⎨ ⎬⎨ ⎬∂ ∂ ∂ ∂ ∂ ∂⎪ ⎪⎪ ⎪⎪ ⎪ ⎩ ⎭ ⎪ ⎪∂⎪ ⎪ ∂∂ −⎪ ⎪⎪ ⎪ ∂ ∂∂⎩ ⎭ ⎩ ⎭

A (B.7)

Curl of a vector A in Vogit notation

0

0

0

yz

xx z

y

zy x

AAy zz y AA AA

z x z xA A A

y x x y

∂⎧ ⎫∂∂ ∂⎡ ⎤ −− ⎪ ⎪⎢ ⎥ ∂ ∂∂ ∂ ⎪ ⎪⎢ ⎥ ⎧ ⎫⎪ ⎪∂ ∂∂ ∂ ⎪ ⎪⎢ ⎥∇× = − = −⎨ ⎬⎨ ⎬⎢ ⎥∂ ∂ ∂ ∂⎪ ⎪⎪ ⎪⎢ ⎥ ⎩ ⎭ ⎪ ⎪∂ ∂ ∂ ∂⎢ ⎥− −⎪ ⎪⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦ ⎩ ⎭

A (B.8)

Symmetric Del operator in Vogit notation

1

2

3

3 2

3 1

2 1

0 0

0 0

0 0

0

0

0

x

s y

z

AxxAyyAAzz A

A AAz y y z

A Az x x z

A Ay x x y

∂∂ ⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥ ∂∂ ⎪ ⎪⎢ ⎥

∂∂ ⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥ ∂∂⎪ ⎪⎢ ⎥

∂∂ ⎪ ⎪⎢ ⎥ ⎧ ⎫ ⎪ ⎪⎢ ⎥ ∂∂ ⎪ ⎪ ⎪ ⎪∇ = =⎢ ⎥ ⎨ ⎬ ⎨ ⎬∂∂ ∂ ∂⎢ ⎥ ⎪ ⎪ ⎪ ⎪+⎩ ⎭⎢ ⎥∂ ∂ ∂ ∂⎪ ⎪⎢ ⎥ ⎪ ⎪∂ ∂ ∂ ∂⎢ ⎥ ⎪ ⎪+⎢ ⎥ ⎪ ⎪∂ ∂ ∂ ∂⎢ ⎥ ⎪ ⎪∂ ∂ ∂ ∂⎢ ⎥ ⎪ ⎪+∂ ∂ ∂ ∂⎢ ⎥ ⎪ ⎪⎣ ⎦ ⎩ ⎭

A (B.9)

The symmetric Del operator is used in the strain-displacement relation, s= ∇S u , i.e.,

454

1

2

1

2 31

32

3 243

5

3 16

2 1

0 0

0 0

0 0

2 022 0

0

uxxuyyuuzz u

u uuz y y z

u uz x x z

u uy x x y

εεεεεε

∂∂ ⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥ ∂∂ ⎪ ⎪⎢ ⎥

∂∂ ⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎧ ⎫ ∂∂⎪⎢ ⎥⎪ ⎪

∂∂ ⎪⎢ ⎥⎪ ⎪ ⎧ ⎫ ⎪⎢ ⎥⎪ ⎪ ∂∂⎪ ⎪ ⎪ ⎪ ⎪= =⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬∂∂ ∂ ∂⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ +⎩ ⎭⎢ ⎥∂ ∂ ∂ ∂⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪∂ ∂ ∂ ∂⎪ ⎪ ⎢ ⎥ ⎪⎩ ⎭ +⎢ ⎥ ⎪∂ ∂ ∂ ∂⎢ ⎥ ⎪∂ ∂ ∂ ∂⎢ ⎥ ⎪ +∂ ∂ ∂ ∂⎢ ⎥ ⎪⎣ ⎦ ⎩

⎪⎪⎪⎪

⎪⎪⎪⎪⎪⎪⎪⎪⎭

(B.10)

B.1.2 Cylindrical coordinates For the cylindrical coordinates, the vector notation is not always appropriate. For this

motive we will only present the Vogit notation.

Divergence of a tensor A in Vogit notation

( )

( )

( )

1

2

3

4

5

6

1 0 0

10 0 0

0 0 0

1 1

1 1

1 1

rrr rz

r r z

zrz zz

ArAr r r z rArAr z r r r

r Az r r r A

A ArA Ar r r r z

rA A A Ar r r r z

ArA Ar r r z

θθ θ

θ θ θθ θ

θ

θ

θ

θ

θ

θ

θ

⎧ ⎫⎡ ⎤∂ ⋅ ∂ ∂− ⎪ ⎪⎢ ⎥∂ ∂ ∂ ⎪ ⎪⎢ ⎥

⎪ ⎪∂ ∂ ∂ ⋅ ⎪ ⎪⎢ ⎥∇ ⋅ = + =⎨ ⎬⎢ ⎥∂ ∂ ∂ ⎪ ⎪⎢ ⎥⎪ ⎪∂ ∂ ∂ ⋅⎢ ⎥⎪ ⎪⎢ ⎥∂ ∂ ∂⎣ ⎦ ⎪ ⎪⎩ ⎭

∂∂ ∂⎧ − + +⎪ ∂ ∂ ∂⎪∂ ∂ ∂

= + + +⎨ ∂ ∂ ∂∂∂ ∂

+ +∂ ∂ ∂

A

⎫⎪⎪

⎪ ⎪⎬

⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

(B.11)

Curl of a vector A in Vogit notation

455

( ) ( )

110

0

1 1 1 10

yz

rr z

zr

AAr zz r A

A AAz r z r

Ar rA Ar r r r r r

θ

θ

θθ

θ θ

∂⎧ ⎫∂∂ ∂⎡ ⎤ −− ⎪ ⎪⎢ ⎥ ∂ ∂∂ ∂ ⎪ ⎪⎧ ⎫⎢ ⎥∂ ∂∂ ∂ ⎪ ⎪⎪ ⎪⎢ ⎥∇× = − = −⎨ ⎬⎨ ⎬⎢ ⎥∂ ∂ ∂ ∂⎪ ⎪⎪ ⎪⎢ ⎥ ⎩ ⎭ ⎪ ⎪∂ ∂ ⋅ ∂ ∂⎢ ⎥− −⎪ ⎪⎢ ⎥∂ ∂⎣ ⎦ ∂ ∂⎩ ⎭

A (B.12)

Symmetric Del operator in Vogit notation

( )

( ) ( )

21

31

23 2

3

3 1

2 2 1

0 0

1 11 1 0

0 0

1 12 0

0

2 2 20 2

r

T

s

Arr

AAr rr r

AAzz A

A AAz r r z

A Az r r z

r A r A Ar r r r r

θθ

θ θ

θ θ

∂∂ ⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥ ∂∂ ⎪ ⎪⎢ ⎥

∂∂ ⎪ ⎪⎢ ⎥ +⎪⎢ ⎥ ∂∂⎪⎢ ⎥ ∂∂ ⎪⎢ ⎥ ⎧ ⎫⎡ ⎤ ⎪∇ + ∇ ⎢ ⎥ ∂∂ ⎪ ⎪⎣ ⎦∇ = = =⎨ ⎬ ⎨ ⎬⎢ ⎥ ∂∂ ∂ ∂⎪ ⎪ ⎪⎢ ⎥ +⎩ ⎭∂ ∂ ⎪ ∂ ∂⎢ ⎥⎪⎢ ⎥∂ ∂ ∂ ∂⎪ +⎢ ⎥∂ ∂ ∂ ∂⎪⎢ ⎥⎪⎢ ⎥∂ ⋅ ∂ ⋅∂ ∂⎪ +⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦ ⎩

A AA

⎪⎪⎪⎪

⎪⎪⎪⎪⎪⎪⎪⎭

(B.13)

The symmetric Del operator is used in the strain-displacement relation, s= ∇S u , i.e.,

( ) ( )

211

2 31

32

3 243

5

3 16

2 2 1

0 0

1 11 1 0

0 0

1 12 022 0

2 2 20 2

rurr

uur rr r

uuzz u

u uuz r r z

u uz r r z

r A r u ur r r r r

ε θθεεε

θ θεε

θ θ

∂∂⎡ ⎤⎢ ⎥ ∂∂⎢ ⎥

∂∂⎢ ⎥ +⎧ ⎫ ⎢ ⎥ ∂∂⎪ ⎪ ⎢ ⎥ ∂∂⎪ ⎪ ⎢ ⎥ ⎧ ⎫⎪ ⎪ ⎢ ⎥ ∂∂⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎢ ⎥ ∂∂ ∂ ∂⎪ ⎪ ⎪ ⎪⎢ ⎥ +⎩ ⎭∂ ∂⎪ ⎪ ∂ ∂⎢ ⎥⎪ ⎪ ⎢ ⎥∂ ∂ ∂ ∂⎪ ⎪ +⎩ ⎭ ⎢ ⎥∂ ∂ ∂ ∂⎢ ⎥

⎢ ⎥∂ ⋅ ∂ ⋅∂ ∂+⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

(B.14)

456

B.2 DISTRIBUTIVE PROPERTY OF THE DEL OPERATOR In this section we demonstrate the distributive property of the del operator when applied

to the velocity-stress product, i.e.,

( ) ( ) : s∇ ⋅ ⋅ = ⋅ ∇ ⋅ + ∇v T v T T v (B.15)

Recall the expressions of the vector velocity v , the symmetric 2nd rank stress tensor T in

Cartesian and Voigt notations, the del operator ∇ , and the symmetric del operator s∇ ,

i.e.,

{ }x y zv v v=v (B.16)

xx xy xz

xy yy yz

xz yz zz

T T TT T TT T T

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

T (B.17)

{ }xx yy zz yz xz xyT T T T T T=T (B.18)

T

x y z∂ ∂ ∂⎧ ⎫∇ = ⎨ ⎬∂ ∂ ∂⎩ ⎭

(B.19)

0 0 0

0 0 0

0 0 0

T

s

x z y

y z x

z y x

∂ ∂ ∂⎡ ⎤⎢ ⎥∂ ∂ ∂⎢ ⎥

∂ ∂ ∂⎢ ⎥∇ = ⎢ ⎥∂ ∂ ∂⎢ ⎥∂ ∂ ∂⎢ ⎥

⎢ ⎥∂ ∂ ∂⎣ ⎦

(B.20)

Consider the first term on the right hand side of Equation (B.15), this becomes

457

( ) ( )xx xy xz x

xy yy yz y

xz yz zz z

xxy xy yy yz yzxx xz xx zz

y

z

T T T vT T T v

x y zT T T v

vT T T T TT T T T v

x y z x y z x y zv

⎡ ⎤ ⎧ ⎫⎢ ⎥ ⎪ ⎪⎧ ⎫∂ ∂ ∂

⋅ ∇ ⋅ = ∇ ⋅ ⋅ = ⎢ ⎥⎨ ⎬ ⎨ ⎬∂ ∂ ∂⎩ ⎭ ⎢ ⎥ ⎪ ⎪⎩ ⎭⎢ ⎥⎣ ⎦

⎧ ⎫∂ ∂ ∂ ∂ ∂⎧ ⎫⎪ ⎪∂ ∂ ∂ ∂

= + + + + + +⎨ ⎬⎨ ⎬∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎩ ⎭⎪ ⎪⎩ ⎭

v T T v

(B.21)

or

( ) xy xy yy yz yzxx xz xz zzx y z

T T T T TT T T Tv v v

x y z x y z x y z∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂

⋅ ∇ ⋅ = + + + + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠v T (B.22)

Consider the second term on the right-hand side of Equation (B.15)

0 0

0 0

0 0: :

0

0

0

x

yT Txx xx

yy yy zx

zz zzs y

yyz yzz

xz xz

xy xy

vxx

vT T yyT T vvT T zz v vT T

vz y zT TT T

z x

y x

∂∂⎡ ⎤⎢ ⎥ ∂∂⎢ ⎥ ∂∂⎢ ⎥

⎧ ⎫ ⎧ ⎫⎢ ⎥ ∂∂⎪ ⎪ ⎪ ⎪⎢ ⎥∂⎪ ⎪ ⎪ ⎪∂⎢ ⎥ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ ∂⎪ ⎪∂⎪ ⎪ ⎪ ⎪∇ = =⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ∂∂ ∂ ∂⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ +⎩ ⎭⎢ ⎥∂ ∂⎪ ⎪ ⎪ ⎪ ∂

⎢ ⎥⎪ ⎪ ⎪ ⎪∂ ∂⎢ ⎥⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎢ ⎥∂ ∂⎢ ⎥∂ ∂⎢ ⎥∂ ∂⎢ ⎥⎣ ⎦

T vz

x z

yx

vy

v vz x

vvy x

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪∂⎪ ⎪∂ ∂⎪ ⎪+⎪ ⎪∂ ∂⎪ ⎪∂∂⎪ ⎪+⎪ ⎪∂ ∂⎩ ⎭

(B.23)

Expansion of Equation (B.23) yields

: yx z

s xx yy zz

y yx xz zyz xz xy

vv vT T Tx y z

v vv vv vT T Tz y z x y x

∂∂ ∂∇ = + +

∂ ∂ ∂

∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂∂ ∂⎛ ⎞+ + + + + +⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

T v (B.24)

Add Equations (B.24) and (B.22) and rearrange the terms to get

458

( ) ( ) ( )

( )

: s x xx y xy z xz x xy y yy z yz

x xz y yz z zz

v T v T v T v T v T v Tx y

v T v T v Tz

∂ ∂⋅ ∇ ⋅ + ∇ = + + + + +

∂ ∂∂

+ + +∂

v T T v (B.25)

Derive the explicit formulation of the left-hand side of Equation (B.15), i.e.,

( ) ( )

T

xx xy xz x

xy yy yz y

xz yz zz z

x T T T vT T T v

yT T T v

z

⎧ ⎫∂⎪ ⎪∂ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎪ ⎪∂⎪ ⎪∇ ⋅ ⋅ = ∇ ⋅ ⋅ = ⎢ ⎥⎨ ⎬ ⎨ ⎬∂ ⎢ ⎥⎪ ⎪ ⎪ ⎪

⎩ ⎭⎢ ⎥⎪ ⎪ ⎣ ⎦∂⎪ ⎪

∂⎩ ⎭

v T T v (B.26)

Perform the multiplications to get

( ) ( ) ( )

( )

x xx y xy z xz x xy y yy z yz

x xz y yz z zz

v T v T v T v T v T v Tx y

v T v T v Tz

∂ ∂∇ ⋅ ⋅ = + + + + +

∂ ∂∂

+ + +∂

v T (B.27)

Equation (B.27) is equal to Equation (B.25).

B.3 COMPLEX NUMBERS PROPERTY

( ) ( )( )Re Re Rer i r i r r i ra b a ia b ib a b ia b⋅ = − + = −⎡ ⎤⎣ ⎦% r iia b+( )( ) ( )( )Re Re Re

i i r r i i

r i r i r r i r

a b a b a b

a b a ia b ib a b ia b

+ = +

⋅ = + − = +⎡ ⎤⎣ ⎦%

r iia b−( )i i r r i ia b a b a b+ = +(B.28)

i.e.,

a b a b⋅ = ⋅ %% (B.29)

B.4 ODE SOLUTION We recall the solution of a ordinary differential equation (ODE) according to Zill and

Cullen (2001). Consider the ODE

459

( ) ( )y P x y f x′ + = (B.30)

See the complete solution in the form of the sum between the complementary solution cy

and the particular solution py , i.e.,

c py y y= + (B.31)

The complementary solution cy is the solution of the homogenous equation, i.e.,

( ) 0y P x y′ + = (B.32)

upon rearrangement we get

( )y P xy′

= − (B.33)

Integration of Equation (B.33) yields

1ln ( )y P x dx C= − +∫ (B.34)

where 1C in and arbitrary constant. Manipulation of Equation (B.34) yields

1( ) ( )P x dx C P x dxy e Ce

− + −∫ ∫= = where 1CC e= (B.35)

Denote

( )

1P x dx

y e−∫= (B.36)

Substitution of Equation (B.36) into Equation (B.35) yields the complementary solution

1( ) ( )cy x C y x= (B.37)

To obtain the particular solution, assume the following form

460

1( ) ( ) ( )py x u x y x= (B.38)

where ( )u x is to be determined. We will substitute Equation (B.38) into Equation (B.30)

and solve for ( )u x . First, calculate the derivative, i.e.,

1 1py u y u y′ ′ ′= + (B.39)

Note that Equation (B.32) implies

y P y′ = − (B.40)

Since 1( )y x satisfies Equation (B.32), then it also satisfies Equation (B.40), i.e.,

1 1y P y′ = − (B.41)

Substitution of Equation (B.41) into Equation (B.39) yields

1 1py u y u Py′ ′= + (B.42)

Substitution of Equations (B.38) and (B.42) into Equation (B.30) gives

1 1u y u P y′ − 1Puy+ f= (B.43)

or 1( ) ( ) ( )u x y x f x′ = , upon rearrangement we get

1

( )( )( )

f xu xy x

′ = (B.44)

Upon integration, Equation (B.44) yields

1

( )( )( )

f xu x dxy x

= ∫ (B.45)

Substitution of Equation (B.45) into Equation (B.38) gives

461

11

( )( ) ( )( )p

f xy x y x dxy x

= ∫ (B.46)

Combining Equations (B.35) and (B.46) into Equation (B.31) gives

1 1 11 1

( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( )c p

f x f xy x y x y x Cy x y x dx y x C dxy x y x

⎛ ⎞= + = + = +⎜ ⎟

⎝ ⎠∫ ∫ (B.47)

Substituting Equation (B.36) into Equation (B.47) yields the complete solution of

Equation (B.30), i.e.,

( )( ) ( )( ) ( )P x dx P x dxy x e f x e dx C−= +∫ (B.48)

Equation (B.48) expresses the more general concept of integrating factor (IF),

( )P x dx

IF e∫= (B.49)

The integrating factor concept works as follows. Multiply the original ODE (B.30) by the

integrating factor IF given by Equation (B.49), i.e.,

( ) ( ) ( )( ) ( )

P x dx P x dxy P x y e f x e∫ ∫′ + = (B.50)

Upon expansion, Equation (B.50) becomes

( ) ( ) ( )

( ) ( )P x dx P x dx P x dx

y e y P x e f x e∫ ∫ ∫′ + = (B.51)

Note that the LHS of Equation (B.51) is an exact differential; hence, Equation (B.51) can

be written as

( ) ( )( )

P x dx P x dxd y e f x edx

⎛ ⎞∫ ∫=⎜ ⎟⎝ ⎠

(B.52)

Upon integration, Equation (B.52) yields

462

( ) ( )

( )P x dx P x dx

y e f x e dx C∫ ∫= +∫ (B.53)

Solving for y , Equation (B.53) gives

( ) ( )

( ) ( )P x dx P x dx

y x e f x e dx C− ⎛ ⎞∫ ∫= +⎜ ⎟

⎝ ⎠∫ (B.54)

which is the same as Equation (B.48). Hence the solution of Equation (B.30) with the use

of the integrating factor IF is expressed in the following general form

( )1( ) ( )y x IF f x IF dx C−= +∫ (B.55)

Recall the expression of Equation (17.29).

( ) ( )( ) ( )4

n nn n

nn

a x di a x xx P

ξ∂− = ⋅

∂v t% (B.56)

This is a first order non homogeneous ordinary differential equation. To solve this

equation, first solve the homogeneous part of the ODE

( ) ( ) 0nn n

a x i a xx

ξ∂− =

∂ (B.57)

A generic solution is

( ) ni xna x Ae ξ= (B.58)

where A is a constant to be determined. Call

( )4n

nnn

dP

=vF% (B.59)

Substitute Equations (B.58) and (B.59) into Equation (17.29)

463

( ) ( )n

n

i xi xn

nAe a x Ae x

x

ξξ

−−∂

= ⋅∂

F t (B.60)

Integrate both sides w.r.t. x

( ) ( )n n

xi x i x

n nc

Ae a x A e x dxξ ξ− −= ⋅ ∫F t (B.61)

Rearrange equation (B.61) and obtain solution to Equation (17.29)

( )( ) ( )4

n n

xi x i xn

nnn c

da x e e x dxP

ξ ξ−= ⋅ ∫v t% (B.62)

B.5 DERIVATIVE OF AN INTEGRAL

( ) ( )x

af x g x dx= ∫ (B.63)

( ) ( ) ( ) ( ) ( ) ( )x x x x x

a a xf f x x f x g x dx g x dx g x dx g x x

+Δ +ΔΔ = + Δ − = − = ≈ Δ∫ ∫ ∫ (B.64)

0 0

1( ) lim lim ( ) ( )x x

xx x

ff x g x dx g xx x

+Δ

Δ → Δ →

Δ′ = = =Δ Δ ∫ (B.65)

464

C POWER AND ENERGY

C.1 POWER AND ENERGY WAVE FRONTS

C.1.1 Plane waves Consider a plate in a rectangular coordinate system xyz and a plane wave propagating in

it. Assume y and z invariance, hence 0y

∂=

∂, 0

z∂

=∂

. The wave equation is given by

2 2

2 2 2

1u ux c t

∂ ∂=

∂ ∂ (C.1)

where c is the speed of the wave. This equation accepts the harmonic solution

( ) ( )1 2( , ) i x t i x tu x t C e C eξ ω ξ ω− += + (C.2)

where C1 and C2 are constant to be determined through the boundary conditions.

Consider only a forward propagating mode, Equation (C.2) can be written as

( )1( , ) cosu x t C x tξ ω= − (C.3)

The energy density of a wave is given by the sum of kinetic energy and potential energy

that, for harmonic waves, are equal; hence the energy density is given by two times the

kinetic energy, hence,

( )2 2 2 21

1( ) 2 sin2

e x u C x tρ ρ ω ξ ω⎛ ⎞= = −⎜ ⎟⎝ ⎠

& (C.4)

Assume that at t=t0 the distance traveled is given by 0 0x k tω= + where k is a constant

equal to k x tξ ω= − ; after a period 2T π= the distance traveled is

465

1 0 0x k t T x Tω= + + = + , after n periods, the distance traveled is 0nx x nT= + , the

energy density becomes

( )2 2 21 0( ) sine x C k eρ ω= = (C.5)

where ( )2 2 20 1 sine C kρ ω= . The energy density amplitude is constant and does not

depend on the distance. The total energy is constant as well and it is given by the energy

density per area of the wave front. The wave front area is equal to the area of the plane

source, i.e., a constant equal to A; hence the total energy is given by

0( ) ( )E x e x A e A= = (C.6)

The power flow through a surface of area A, is given by the product of the particle

velocity, the stress, and the surface area, i.e.,

( )2 21

( , ) ( , )( , ) sinu x t u x tP x t EA EAC x tx t

ξω ξ ω∂ ∂= − = −

∂ ∂ (C.7)

Note that the power flow can be written as ( , ) ( , ) ( , )EP x t E x t cE x tξρω

= = .

C.1.2 Spherical waves Consider a spherical wave propagating in a medium from a point source, consider also a

spherical coordinate system rφθ . We assume spherical symmetry, hence, the wave is φ

and θ invariant, i.e., 0φ∂

=∂

, 0θ∂

=∂

. The wave equation is given by

( ) ( )2 2

2 2 2

1ru rur c t

∂ ∂=

∂ ∂ (C.8)

where c is the speed of the wave. This equation accepts the harmonic solution

466

( ) ( )1 2( , ) i r t i r tC Cu r t e er r

ξ ω ξ ω− += + (C.9)

where A1 and A2 are constant to be determined through the boundary conditions. The

wave amplitude in Equation (C.9) decreases as 1 r or with distance from the source.

Consider only an outward propagating mode, Equation (C.9) can be written as

( )1( , ) cosCu r t r tr

ξ ω= − (C.10)

The energy density of a spherical wave is given by the sum of kinetic energy and

potential energy that, for harmonic waves, are equal; hence the energy density is given by

two times the kinetic energy, hence,

( )2 2

2 212

1( ) 2 sin2

Ce r u r trωρ ρ ξ ω⎛ ⎞= = −⎜ ⎟

⎝ ⎠& (C.11)

Assume that t=t0 the radius is 0 0r k tω= + where k is a constant equal to k r tξ ω= − ;

after a period 2T π= the radius becomes 1 0 0r k t T r Tω= + + = + , after n periods, the

radius becomes 0nr r nT= + and the energy density becomes

( )2 2 2

21 0 02 2( ) sin

n n

C r ee r kr rωρ= = (C.12)

where ( ) ( )2 2 2 20 1 0 sine C r kρ ω= . The energy density amplitude decreases with the square

of the distance from the wave source. However, the total energy is constant and is given

by the energy density per area of the wave front. The wave front area is equal to the area

of the sphere of radius r, i.e., 24A rπ= , hence the total energy is given by

2 20 0( ) ( )4 4E r e r r r eπ π= = (C.13)

467

The energy density decreases with r2 because the area of the spherical wave front

increases with the square of the distance, hence the constant value of the total energy

must be spread on a larger surface as the radius increases.



( )

( )( )2

1 2

1 cos( , ) ( , )( , ) sinsin

r tu r t u r tP r t EA EAC r trr t r r t

ξ ωω ξ ωξ ξ ω

⎡ ⎤−∂ ∂ ⎢ ⎥= − = −⎢ ⎥∂ ∂ + −⎢ ⎥⎣ ⎦

(C.14)

Recall that the surface area is equal to 24 rπ , the power flow becomes

( ) ( ) ( )2 21

cos sin( , ) 4 sinr t r tP r t E C r tr

ξ ω ξ ωπ ω ξ ξ ω

⎡ ⎤− −= + −⎢ ⎥⎣ ⎦

(C.15)

For small values r, the power flow goes to infinity, as the radius increases, the first term

in Equation (C.15) goes to zero and the power flow can be written as

( , ) ( , ) ( , )EP r t E r t cE r tξρω

= = as for the plane waves.

C.1.3 Cylindrical waves Consider a cylindrical wave propagating in a medium from a point source, consider also a

cylindrical coordinate system r zθ . Assume cylindrical symmetry, hence the wave is θ

and z invariant, i.e., 0θ∂

=∂

, 0z

∂=

∂. The wave equation is given by

2

2 2

1 1u urr r r c t

∂ ∂ ∂⎛ ⎞ =⎜ ⎟∂ ∂ ∂⎝ ⎠

(C.16)

where c is the speed of the wave. This equation accepts as solutions the Bessel functions

468

( ) ( )1 0 2 0( , ) i t i tu r t C J r e C Y r eω ωξ ξ− −= + (C.17)

where C1 and C2 are constant to be determined through the boundary conditions, 0J is the

Bessel function of the first kind and order zero, and 0Y is the Bessel function of the

second kind and order zero. Equation (C.17) can be written in the complex form as

( ) ( ) ( ) ( )1 21 0 2 0( , ) i t i tu r t C H r e C H r eω ωξ ξ− −= + (C.18)

where the Henkel functions are given by

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

10 0 0

20 0 0

H r J r iY r

H r J r iY r

ξ ξ ξ

ξ ξ ξ

= +

= − (C.19)

the first function of Equation (C.19) is for inward propagating modes, while the second is

for outward propagating modes. Consider only an outward propagating mode, Equation

(C.18) can be written as

( ) ( ) ( ) ( )1 0 2 0( , ) cos sinu r t C J r t C Y r tξ ω ξ ω= + (C.20)

For cylindrical coordinates it is not possible to express the wave solution in terms of

D’Alambert solution. The energy density of a cylindrical wave is given by the sum of

kinetic energy and potential energy and being the cylindrical wave only harmonic in time,

potential and kinetic energy are not equal. Potential energy density is given by

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 22 2 2 22 2 1 1 2 1

1 2 1 1

cos sin1 1( , )2 2 2 sin cos

C J r t C Y r tv r t u

C C J r Y r t t

ξ ω ξ ωρ ξ ρ

ξ ξ ω ω

⎡ ⎤⎡ ⎤ ⎡ ⎤+⎣ ⎦ ⎣ ⎦′ ⎢ ⎥= =⎢ ⎥+⎣ ⎦

(C.21)

Kinetic energy density is given by

469

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 22 2 2 22 2 1 0 2 0

1 2 0 0

sin cos1 12 2 2 sin cos

C J r t C Y r tk u

C C J r Y r t t

ξ ω ξ ωρ ω ρ

ξ ξ ω ω

⎡ ⎤⎡ ⎤ ⎡ ⎤+⎣ ⎦ ⎣ ⎦⎢ ⎥= =⎢ ⎥−⎣ ⎦

& (C.22)

The total energy density is given by the sum of potential end kinetic energy density, i.e.,

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2 22 2 2 2 21 1 2 0

2 22 2 2 2 22 1 1 0

2 21 2 1 1 0 0

cos1( , ) sin2

2 sin cos

C J r C Y r t

e r t C Y r C J r t

C C J r Y r J r Y r t t

ξ ξ ω ξ ω

ρ ξ ξ ω ξ ω

ξ ξ ξ ω ξ ξ ω ω

⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤+⎣ ⎦ ⎣ ⎦⎣ ⎦⎢ ⎥⎢ ⎥⎡ ⎤⎡ ⎤ ⎡ ⎤= + +⎣ ⎦ ⎣ ⎦⎣ ⎦⎢ ⎥⎢ ⎥⎡ ⎤+ −⎣ ⎦⎣ ⎦

(C.23)

The terms in Equation (C.23) that depend on the radial direction varies with r as shown

( )2 rπ , see Figure C.2. The energy density amplitude decreases with the radial distance

from the sources as function proportional to ( )1 rπ .

Figure C.2 Energy density amplitude variation with distance from source. Solid lines: Bessel

functions; Dash line: functions proportional to ( )2 rπ

However, the total energy is constant; this is given by the energy density per area of

the wave front. The wave front area per unit width is equal to the area of the circle of

radius r, i.e., 2 rπ , hence the total energy is given by

0 10 20 30 40 50

0.05

0.1

0 10 20 30 40 50

0.05

0.1

rξ rξ

( ) 20a J rξ⎡ ⎤⎣ ⎦

( ) ( )0 02cJ r Y rξ ξ

( ) 20b Y rξ⎡ ⎤⎣ ⎦

( ) 21a J rξ⎡ ⎤⎣ ⎦

( ) ( )1 12cJ r Y rξ ξ

( ) 21b Y rξ⎡ ⎤⎣ ⎦

2arπ 2a

rπ

2brπ

2brπ

2crπ

2crπ

470

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

2 22 2 2 2 21 1 2 0

2 22 2 2 2 22 1 1 0

2 21 2 1 1 0 0

cos

( , ) 2 sin

sin 2

C J r C Y r t

E r t re r C Y r C J r t

C C J r Y r J r Y r t

ξ ξ ω ξ ω

π π ρ ξ ξ ω ξ ω

ξ ξ ξ ω ξ ξ ω

⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤+⎣ ⎦ ⎣ ⎦⎣ ⎦⎢ ⎥⎢ ⎥⎡ ⎤⎡ ⎤ ⎡ ⎤= = + +⎣ ⎦ ⎣ ⎦⎣ ⎦⎢ ⎥⎢ ⎥⎡ ⎤+ −⎣ ⎦⎣ ⎦

(C.24)

With the use of consideration in Figure C.2, the total energy can be written as

( ) ( ) ( ) ( )

( ) ( )

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

2 21 2

cos sin( , ) 2

sin 2

C C t C C tE r t

C C t

ξ ω ω ξ ω ωρ

ξ ω ω

⎡ ⎤+ + +≈ ⎢ ⎥

+ −⎢ ⎥⎣ ⎦ (C.25)

The energy density decreases with r because the area of the cylinder wave front increases

linearly with the distance from the source, hence the constant value of the total energy

must be spread on a larger surface as the radius increases.



( ) ( )

( ) ( )( ) ( )

( ) ( )1 1 1 0

2 1 2 0

cos sin( , ) ( , )( , )sin cos

C J r t C J r tu r t u r tP r t EA EAr t C Y r t C Y r t

ξ ω ξ ωξω

ξ ω ξ ω

⎡ ⎤ ⎡ ⎤−∂ ∂= − = ⎢ ⎥ ⎢ ⎥

∂ ∂ + +⎣ ⎦ ⎣ ⎦ (C.26)

or after rearrangement

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2 22 0 1 1 0 1

2 21 2 1 0 0 1

sin cos( , )

cos sin

C Y r Y r C J r J r t tP r t EA

C C J r Y r t J r Y r t

ξ ξ ξ ξ ω ωξω

ξ ξ ω ξ ξ ω

⎡ ⎤⎡ ⎤−⎣ ⎦⎢ ⎥=⎡ ⎤⎢ ⎥+ −⎣ ⎦⎣ ⎦

(C.27)

Note that ( ) ( )0 1Y r Y rξ ξ and ( ) ( )0 1J r J rξ ξ varies as ( )1 rπ , while ( ) ( )1 0J r Y rξ ξ and

( ) ( )0 1J r Y rξ ξ− as ( )2 rπ . Recall that the surface area is equal to 2 rπ , hence the power

flow becomes

( ) ( ) ( )2 22 1 1 2( ) 2 sin cos 2P t E C C t t C Cξω ω ω⎡ ⎤≈ − +⎣ ⎦ (C.28)

471

C.2 AVERAGE POWER FLOW

C.2.1 Basic definitions Let consider two media separated by a surface S. The force exerted by medium 1 on

medium 2 is dS− ⋅T n and the power delivered through dS from medium 1 to medium 2

is dS− ⋅ ⋅v T n where v is the velocity field, T is the stress tensor of 2nd rank, and n is the

normal direction to the surface S (see Figure 4.1). The power flow density in the direction

of n is

⋅ = − ⋅ ⋅P n v T n (C.29)

where the acoustic Poynting vector P is defined as

= − ⋅P v T (C.30)

Let v and T be time harmonic, hence their expression is

( )( )

0

0

( ) cos

( ) cosv

v

t t

t t

ω φ

ω φ

= +

= +

v v

T T (C.31)

where 30 ∈v 3 3

0 ∈ ×T are real amplitudes. Note that ( )tv and ( )tT are not

necessarily in phase, i.e., v Tφ φ≠ . Substitution of Equation (C.31) into Equation (C.30)

yields

( ) ( )0 0( ) ( ) ( ) cos cosv Tt t t t tω φ ω φ− = ⋅ = ⋅ + +P v T v T (C.32)

Recall the trigonometric identities

cos( ) cos cos sin sin2cos cos cos( ) cos( )

a b a b a ba b a b a b± =

= − + +m

(C.33)

In view of Equation (C.33), we can write Equation (C.32) as

472

( ) ( )

( ) ( )

10 02

1 10 0 0 02 2

( ) cos cos 2

cos cos 2v T v T

v T v T

t t

t

φ φ ω φ φ

φ φ ω φ φ

⎡ ⎤− = ⋅ − + + +⎣ ⎦= ⋅ − + ⋅ + +

P v T

v T v T (C.34)

The function ( )tP defined by Equation (C.34) is a harmonic function that contains a

constant part, ( )10 02 cosactive v Tφ φ− = ⋅ −P v T , and a harmonic part

( )10 02 cos 2reactive v Ttω φ φ− = ⋅ + +P v T that oscillates with twice the frequency of the

original variables given by Equation (C.31). To calculate the time-average value of ( )tP ,

we use the formula

( ) ( )1 1 1

0 020( ) cos cos 2

T

av v T v TT Tt dt tφ φ ω φ φ− = − = − + + +∫P P v T

( )0

10 02 cos

T

v T

dt

φ φ

⎡ ⎤⎣ ⎦

= −

∫v T

(C.35)

where T is the period of oscillation, 2 /T π ω= . Note that the time-averaged value equals

the active part, i.e., av active=P P .

C.2.2 Complex notations To take advantage of complex notations, we will utilize the implied convention that only

the real part of the complex exponential function is used, i.e.,

( ) ( )( )( ) ( )( )

0 0

0 0

( ) cos Re

( ) cos Re

v

T

i tv

i tT

t t e

t t e

ω φ

ω φ

ω φ

ω φ

+

+

= + =

= + =

v v v

T T T (C.36)

where we have used the Euler formulae, i.e.,

cos

2

ix ixe ex−+

=,

sin2

ix ixe exi

−−=

(C.37)

Define the complex variables

473

( )

( )0

0

( )

( )

v

T

i t

i t

t e

t e

ω φ

ω φ

+

+

=

=

v v

T T (C.38)

and the complex amplitudes

0 0

0 0

v

T

i

i

e

e

φ

φ

=

=

v v

T T (C.39)

We can rewrite Equation (C.36) with the use of Equations (C.38), (C.39), i.e.,

( )( ) ( ) ( )( )( ) ( ) ( )

0 0 0

0 0 0

( ) Re Re Re

( ) Re Re Re

v v

T T

i t i i t i t

i t i i t i t

t e e e e

t e e e e

ω φ φ ω ω

ω φ φ ω ω

+

+

= = =

= = =

v v v v

T T T T (C.40)

In view of the relation between Equations (C.31) and (C.40), we will use interchangeably

the cosinusoidal formulation of Equation (C.31) and the complex formulation of Equation

(C.40); when referring to the complex formulation of Equation (C.40), the Re sign will be

implied, but not explicitly written.

Equation (C.35) can be written in complex notations as

[ ] ( ) ( )( )

( )( ) ( )1 1

0 0 0 02 2

1 10 0 0 02 2

( ) ( ) cos Re

Re Re

v T

v T

iav v Tav

i i

t t e

e e

φ φ

φ φ

φ φ −

−

− = ⋅ = ⋅ − = ⋅

⎡ ⎤= = ⋅⎣ ⎦

P v T v T v T

v T v T% (C.41)

where T% is the complex conjugate of T . Equation (C.41) can be expressed in terms of

the complex variables of Equation (C.38) using the products ( ) ( )t tv T% and ( ) ( )t tv T%% .

Since the harmonic exponential i te ω is common in both variables, the products ( ) ( )t tv T%

and ( ) ( )t tv T%% do not contain it because i te ω and i te ω− cancel out, e.g.,

0 0 0 0( ) ( ) i t i tt t e eω ω−= ⋅ = ⋅v T v T v T% % % (C.42)

474

Using Equation (C.42), we can write Equation (C.41) as

[ ] ( )1 10 02 2( ) ( ) Re Re ( ) ( )av av

t t t t⎡ ⎤− = ⋅ = ⋅ = ⋅⎣ ⎦P v T v T v T% % (C.43)

Equation (C.43) indicates the general result that the time–averaged product of two

complex harmonic variables is the real part of half the product between one variable and

the conjugate of the other variable, i.e.,

[ ] 1 12 2( ) ( ) Re ( ) ( ) Re ( ) ( )

avt t t t t t⎡ ⎤ ⎡ ⎤⋅ = ⋅ = ⋅⎣ ⎦⎣ ⎦v T v T v T% % (C.44)

In view of the above, we define the complex Poynting vector, i.e.,

( )1 1 1

0 0 0 0 0 02 2 21 1

0 0 0 02 2cos( ) sin( )

v v TTi ii

v T v T

e e ei

φ φ φφ

φ φ φ φ

−−− = ⋅ = = ⋅

= ⋅ − + ⋅ −

P v T v T v Tv T v T

% % (C.45)

The real part of the complex Poynting vector equals the time-averaged Poynting vector,

Reav =P P (active Poynting factor). The magnitude of the complex Poynting vector

equals the peak value of the oscillating part of the reactive Poynting vector,

Impeakreactive =P P .

475

D ORTHOGONALITY FOR VIBRATION AND WAVE PROBLEMS

D.1 ORTHOGONALITY PROOF FOR SOME VIBRATION PROBLEMS

D.1.1 Straight-crested axial vibration of rectangular plates

The problem is y-invariant and depends only on x, i.e.,

( , , ) ( , )( , , ) 0

u x y t u x tv x y t

→≡

(D.1)

The wave equation for this particular problem is

2Lc u u′′ = && (D.2)

where ( )2

21LEc

ρ ν=

−. Assume that the solution is time harmonic, such as,

ˆ( , ) ( ) i tu x t u x e ω= (D.3)

Equation (D.2) becomes

2ˆ ˆ 0Lu uγ′′ + = (D.4)

where the wavenumber is defined as 2

22LLc

ωγ = . Solution to Equation (D.4) is given by

ˆ( ) sin cosL Lu x A x B xγ γ= + (D.5)

476

where A and B are constant to be determined through the boundary conditions. Assume

that the boundary conditions for the rectangular plate are stress free conditions at the

edges. Recall that

2 ( , )1

d

x xxd

EhN dy u x tσν−

′= =−∫ (D.6)

Hence the boundary conditions are

(0, ) 0( , ) 0

u tu l t

′ =⎧⎨ ′ =⎩

(D.7)

Substitute Equation (D.5) into the boundary conditions and solve to get

0sin 0L L

AB lγ γ

=⎧⎨ =⎩

(D.8)

System (D.8) is verified for

nLnlπγ = (D.9)

With the use of (D.8) and (D.9), Equation (D.5) can be written as

( )( , ) cos i tn nL

nu x t B x e ωγ= ∑ (D.10)

Consider two separate mode shapes, ( )pU x and ( )qU x , such as they satisfy Equation

(D.2) and (D.8) i.e.,

2 2

2 2

L p p p

L q q q

c U U

c U U

ω

ω

′′ = −

′′ = − (D.11)

Multiply the fist of Equation (D.11) by ( )qU x and the second by ( )pU x , hence

477

2 2

2 2

L p q p p q

L q p q q p

c U U U U

c U U U U

ω

ω

′′ = −

′′ = − (D.12)

Integrate both equations over the area. Since the problem is y-invariant, we only need to

integrate over x, i.e.

2 2

0 0

2 2

0 0

l l

L p q p p q

l l

L q p q q p

c U U dx U U dx

c U U dx U U dx

ω

ω

′′ = −

′′ = −

∫ ∫

∫ ∫ (D.13)

Integrate by part

2 2 20

0 0

2 2 20

0 0

l llL p q L p q p p q

l llL q p L q p q q p

c U U c U U dx U U dx

c U U c U U dx U U dx

ω

ω

′ ′ ′− = −

′ ′ ′− = −

∫ ∫

∫ ∫ (D.14)

Subtract the first line from the second

( ) ( )2 2 2

00

llL p q q p p q p qc U U U U U U dxω ω′ ′− = − − ∫ (D.15)

From the boundary conditions (D.8) we obtain

( )2 2

0

0l

p q p qU U dxω ω− =∫ (D.16)

For distinct mode numbers, p q≠ , the frequencies are also distinct, 2 2p qω ω≠ , and hence

Equation (D.16) implies

0

0l

p qU U dx =∫ , p q≠ (D.17)

478

this is the orthogonality condition. To derive the normalization factor, consider two

distinct solutions given by Equation (D.10), i.e.,

( ) cos

( ) cosp p pL

p q qL

U x B x

U x B x

γ

γ

=

= (D.18)

Substitute Equation (D.18) into the orthogonality relation (D.17) and solve the integral to

get

( ) ( )0

sin sincos cos

2

lmL nL mL nLm n

m n mL nLmL nL mL nL

l lB BB B x xdxγ γ γ γ

γ γγ γ γ γ

⎡ ⎤+ −= +⎢ ⎥+ −⎣ ⎦

∫ (D.19)

Note that

( )

( )

mL nL

mL nL

m nl

m nl

πγ γ

πγ γ

⎧ + = +⎪⎪⎨⎪ − = −⎪⎩

(D.20)

Substitution of Equation (D.20) into Equation (D.19) yields

( ) ( )

0

sin sincos cos2

lm n

m n mL nLB B l m n m nB B x xdx

m n m nπ πγ γ

π⎡ ⎤+ −

= +⎢ ⎥+ −⎣ ⎦∫ (D.21)

i. For m n≠ 0

cos cos 0l

m n mL nLB B x xdxγ γ =∫

i. For m n= 0

sin 2cos cos2 2

lm n

m n mL nLB B l nB B x xdx

nπγ γ

π=∫

( )sin m nm n

π−+

−

2

2nB l

π⎡ ⎤⎢ ⎥ =⎢ ⎥⎣ ⎦

.

The normalization factor is 2nB l= .

479

D.1.2 Axial vibration of circular plates

Assume that the problem is axial symmetric, hence θ -invariant and it depends only on r,

i.e., ( , , ) ( , )u r t u r tθ → and ( , , ) 0v r tθ ≡ . The wave equation for this particular problem is

22

1L

uc u u ur r

⎛ ⎞′′ ′+ − =⎜ ⎟⎝ ⎠

&& (D.22)

where ( )2

21LEc

ρ ν=

−. Assume that the solution is time-harmonic, such as,

ˆ( , ) ( ) i tu r t u r e ω= . Equation (D.22) becomes

( )2 2 2ˆ ˆ ˆ1 0r u ru r uγ′′ ′+ + − = (D.23)

where the wavenumber is defined as 2

22Lc

ωγ = . Perform the following change of variable

x rγ= (D.24)

and recall that

2 2 2

dx dr

dx dr

γ

γ

=

= (D.25)

Equation (D.23) becomes

( )2 2ˆ ˆ ˆ1 0x u xu x u′′ ′+ + − = (D.26)

Solution to Equation (D.26) is given by

( ) ( )1 1ˆ( )u x AJ x BY x= + (D.27)

However the Bessel function of the second kind has infinite value at 0x rγ= = and has

to be discarded. Equation (D.27) becomes

480

( )1ˆ( )u r AJ rγ= (D.28)

where A is a constant to be determined from the boundary conditions. We assume that the

boundary conditions are

2( , )( ) ( , ) 0

1

d

r rrd

Eh u a tN a dy u a ta

σ νν−

⎛ ⎞′= = + =⎜ ⎟⎝ ⎠−∫ (D.29)

Substitute Equation (D.28) into (D.29) and obtain

( )( )

( )2 11 0A J a J a

aνγ γ γ⎛ ⎞+

− + =⎜ ⎟⎝ ⎠

(D.30)

Equation (D.30) is verified for

( ) ( ) ( )2 11 0aJ a J aγ γ ν γ− + = (D.31)

Considering the boundary conditions (D.31), solution (D.28) can be written as

( )1ˆ( ) n nn

u r A J rγ= ∑ (D.32)

where nγ satisfies Equation (D.31).

Consider two separate mode shapes, ( )pU r and ( )qU r , such as they satisfy Equation

(D.22) and (D.29) i.e.,

2 2

2 2

pL p p p p

qL q q q q

Uc rU U rU

rU

c rU U rUr

ω

ω

⎛ ⎞′′ ′+ − = −⎜ ⎟

⎝ ⎠⎛ ⎞

′′ ′+ − = −⎜ ⎟⎝ ⎠

(D.33)

Multiply the fist of Equation (D.33) by ( )qU r and the second by ( )pU r , hence

481

2 2

2 2

p qL p q p q p p q

q pL q p q p q q p

U Uc rU U U U rU U

rU U

c rU U U U rU Ur

ω

ω

⎛ ⎞′′ ′+ − = −⎜ ⎟

⎝ ⎠⎛ ⎞

′′ ′+ − = −⎜ ⎟⎝ ⎠

(D.34)

Integrate over the area. Since the problem is axisymmetric, we only need to integrate

radially, i.e.

2 2

0 0

2 2

0 0

a ap q

L p q p q p p q

a aq p

L q p q p q q p

U Uc rU U U U dr rU U dr

r

U Uc rU U U U dr rU U dr

r

ω

ω

⎛ ⎞′′ ′+ − = −⎜ ⎟

⎝ ⎠

⎛ ⎞′′ ′+ − = −⎜ ⎟

⎝ ⎠

∫ ∫

∫ ∫ (D.35)

Integrate by part and rearrange the terms to get

2 2 20

0 0

2 2 20

0 0

a aa p qL p q L p q p p q

a aa q pL q p L q p q q p

U Uc rU U c rU U dr rU U dr

r

U Uc rU U c rU U dr rU U dr

r

ω

ω

⎛ ⎞′ ′ ′+ − − = −⎜ ⎟

⎝ ⎠

⎛ ⎞′ ′ ′+ − − = −⎜ ⎟

⎝ ⎠

∫ ∫

∫ ∫ (D.36)

Subtract the first line from the second, i.e.,

( ) ( )2 2 2

00

aaL p q q p p q p qc a U U U U rU U drω ω′ ′− = − − ∫ (D.37)

Note that from (D.29)

( , )( , ) 0

( , )( , ) 0

pp q

qq p

u a tu a t u

au a t

u a t ua

ν

ν

′ + = ×

′ + = × −

(D.38)

Sum the equations in (D.38) to obtain

482

( )p q q p p q q pau u u u u u u u

aν′ ′− + −( ) 0

a= (D.39)

hence

( ) 0p q q p au u u u′ ′− = (D.40)

With the use of Equation (D.40), Equation (D.37) becomes

( )2 2

0

0a

p q p qrU U drω ω− =∫ (D.41)

For same mode shapes (p=q) the equation is verified, for different mode shapes ( p q≠ )

we should have

0

0a

q prU U dr =∫ (D.42)

Note that for the specific solution (D.32) Equation (D.41) becomes

( ) ( ) ( )2 21 1

0

0a

q p p q p qA A rJ r J r drω ω γ γ− =∫ (D.43)

For p q≠ the integral is equal to 0, for p q=

( ) ( ) ( )

( ) ( ) ( )

222 2 2 2 21 1 12 2

0 02

2 2 2 2 2 2 21 12

1 12

2

aa

q q q q qq

qq q q q

q

A r J r dr A r J r J rr

Aa J a a J a

νγ γ γγ

γ ν γ γ γγ

⎡ ⎤⎧ ⎫⎛ ⎞⎪ ⎪′⎡ ⎤ = − + =⎢ ⎥⎜ ⎟⎨ ⎬⎣ ⎦ ⎜ ⎟⎢ ⎥⎪ ⎪⎝ ⎠⎩ ⎭⎣ ⎦

⎡ ⎤′− +⎣ ⎦

∫ (D.44)

with nγ is a real zero of Equation (D.31).

483

D.1.3 Flexural vibration of rectangular plates

D.1.3.1 Infinite aspect ratio (straight-crested flexural vibration)

Consider a plate with infinite aspect ratio. The problem is assumed to be y-invariant and

the wave equation is

4Fc w w′′′′ = − && (D.45)

where 4F

Dchρ

= . Solution to Equation (D.45) is

( )1 2 3 4( , ) F F F Fi x i x x x i tw x t A e A e A e A e eγ γ γ γ ω− −= + + + (D.46)

where FFc

ωγ = . Assume the following boundary conditions for a supported plate, i.e.,

(0, ) 0( , ) 0

(0, ) (0, ) 0( , ) ( , ) 0

x

x

w tw a tM t Dw tM a t Dw a t

=⎧⎪ =⎪⎨ ′′= − =⎪⎪ ′′= − =⎩

(D.47)

Substitute Equation (D.46) into (D.47) and rearrange the terms.

1 2 3 4

1 2 3 42 2 2 2

1 2 3 42 2 2 2

1 2 3 4

0

0

0

0

F F F F

F F F F

i a i a a a

F F F Fi a i a a a

F F F F

A A A A

A e A e A e A e

A A A A

A e A e A e A e

γ γ γ γ

γ γ γ γ

γ γ γ γ

γ γ γ γ

− −

− −

+ + + =⎧⎪

+ + + =⎪⎨

− − + + =⎪⎪− − + + =⎩

(D.48)

A non trivial solution to system (D.48) is found if the determinant of the characteristic

equation is equal to zero, i.e.,

484

2 2 2 2

2 2 2 2

1 1 1 1

0F F F F

F F F F

i a i a a a

F F F Fi a i a a a

F F F F

e e e e

e e e e

γ γ γ γ

γ γ γ γ

γ γ γ γ

γ γ γ γ

− −

− −

=− −

− −

(D.49)

Solution of the determinant yields ( )( )44 0F F F Fa a i a i aF e e e eγ γ γ γγ − −− − = , hence

sinh sin 0F Fa aγ γ = or

Fnnaπγ = (D.50)

Consider two separate mode shapes, ( )pW x and ( )qW x , such as they satisfy Equation

(D.45) and boundary conditions (D.47). Multiply the fist of equation by ( )qW x and the

second by ( )pW x , i.e.,

4 2

4 2

ivF p q p p q

ivF q p q q p

c W W W W

c W W W W

ω

ω

⎧ =⎪⎨

=⎪⎩ (D.51)

Integrate through the length of the plate to get

4 2

0 0

4 2

0 0

a aiv

F p q p p q

a aiv

F q p q q p

c W W dx W W dx

c W W dx W W dx

ω

ω

⎧=⎪

⎪⎨⎪ =⎪⎩

∫ ∫

∫ ∫ (D.52)

Perform integration by parts and obtain

4 2

0 0

4 2

0 0

a a

F p q p p q

a a

F q p q q p

c W W dx W W dx

c W W dx W W dx

ω

ω

⎧′′ ′′ =⎪

⎪⎨⎪ ′′ ′′ =⎪⎩

∫ ∫

∫ ∫ (D.53)

485

Subtract the two equations to get

( )2 2

0

0a

p q p qW W dxω ω− =∫ (D.54)

This is the orthogonality relation for a simply supported plate with infinite aspect ratio.

D.1.3.2 General flexural vibration of simply supported plate

Consider a rectangular plate with infinite aspect ratio, the wave equation is

( )4 4 0wγ∇ − = (D.55)

Boundary conditions for a free plate are given by

(0, ) 0( , ) 0( ,0) 0( , ) 0

w yw a yw xw x b

=⎧⎪ =⎪⎨ =⎪⎪ =⎩

( )( )( )( )

0, 0

, 0

,0 0

, 0

x

x

y

y

M y

M a y

M x

M x b

⎧ =⎪

=⎪⎨ =⎪⎪ =⎩

(D.56)

Solution to Equation (D.55) is written as

( , , ) sin sini tmn

m x n yw x y t A ea b

ω π π= (D.57)

Rewrite Equation (D.55) in a more extended form, i.e.,

4 4 4

4 24 2 2 42Fw w wc w

x x y yω

⎛ ⎞∂ ∂ ∂+ + =⎜ ⎟

∂ ∂ ∂ ∂⎝ ⎠ (D.58)

Consider two separate mode shapes, ( )pW x and ( )qW x , such as they satisfy Equation

(D.58) and boundary conditions (D.56). Multiply the fist of equation (D.58) by ( )qW x

and the second by ( )pW x , i.e.,

486

4 4 44 2

4 2 2 4

4 4 44 2

4 2 2 4

2

2

p p pF q q q p p q

q q qF p p p q q p

W W Wc W W W W W

x x y y

W W Wc W W W W W

x x y y

ω

ω

⎧ ⎛ ⎞∂ ∂ ∂⎪ + + =⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎪ ⎝ ⎠⎨

⎛ ⎞∂ ∂ ∂⎪+ + =⎜ ⎟⎪ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎩

(D.59)


4 4 44 2

4 2 2 40 0 0 0

4 4 44 2

4 2 2 40 0 0 0

2

2

a b a bp p p

F q q q p p q

a b a bq q q

F p p p q q p

W W Wc W W W dydx W W dydx

x x y y

W W Wc W W W dydx W W dydx

x x y y

ω

ω

⎧ ⎛ ⎞∂ ∂ ∂⎪ + + =⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎪ ⎝ ⎠⎨

⎛ ⎞∂ ∂ ∂⎪+ + =⎜ ⎟⎪ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎩

∫ ∫ ∫ ∫

∫ ∫ ∫ ∫ (D.60)

Perform integration by parts and obtain

2 2 2 2 2 24 2

2 2 2 20 0 0 0

2 2 2 2 2 24 2

2 2 2 20 0 0 0

2

2

a b a bp q p q p q

F p p q

a b a bq p q p q p

F q q p

W W W W W Wc dydx W W dydx

x y x yx x y y

W W W W W Wc dydx W W dydx

x y x yx x y y

ω

ω

⎧ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎪ + + =⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂∂ ∂ ∂ ∂⎪ ⎝ ⎠⎨

⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎪− + =⎜ ⎟⎪ ⎜ ⎟∂ ∂ ∂ ∂∂ ∂ ∂ ∂⎝ ⎠⎩

∫ ∫ ∫ ∫

∫ ∫ ∫ ∫ (D.61)

Subtract the two equations to obtain the orthogonality relation for a rectangular plate, i.e.,

( )2 2

0 0

0a b

p q p qW W dydxω ω− =∫ ∫ (D.62)

D.1.4 Flexural vibration of circular plates

Consider a circular plate in a cylindrical coordinate system. The equation of motion for a

circular plate is given by

4 0D w hwρ∇ + =&& (D.63)

Assume time-harmonic vibrations and expand the bi-harmonic operator assuming θ -

invariance

487

4 3 2

4 24 3 2 2 3

2 1 1Fc w w

r rr r r r rω

⎛ ⎞∂ ∂ ∂ ∂+ − + =⎜ ⎟∂∂ ∂ ∂⎝ ⎠

(D.64)

Assume that the circular plate is a free plate such as the boundary conditions are

( )3 2

23 2 2

1 1( ) 0ra a

w w wV a wr r rr r r

⎛ ⎞∂ ∂ ∂ ∂= ∇ = + − =⎜ ⎟∂ ∂∂ ∂⎝ ⎠

(D.65)

and

2

21( ) 0r

a

w wM a Dr rr

ν⎡ ⎤∂ ∂

= − + =⎢ ⎥∂∂⎣ ⎦ (D.66)

Consider two separate mode shapes, ( , )pW r θ and ( , )qW r θ , such as they satisfy Equation

(D.64) and boundary conditions (D.65) (D.66) and substitute them into Equation (D.64),

i.e.,

4 3 24 2

4 3 2 2 3

4 3 24 2

4 3 2 2 3

2 1 1

2 1 1

F p p p

F q q q

c W Wr rr r r r r

c W Wr rr r r r r

ω

ω

⎛ ⎞∂ ∂ ∂ ∂+ − + =⎜ ⎟∂∂ ∂ ∂⎝ ⎠

⎛ ⎞∂ ∂ ∂ ∂+ − + =⎜ ⎟∂∂ ∂ ∂⎝ ⎠

(D.67)

Multiply the fist of equation (D.67) by ( , )qW r θ and the second by ( , )pW r θ , i.e.,

4 3 24 2

4 3 2 2

4 3 24 2

4 3 2 2

1 12

1 12

p p p pF q p p q

q q q qF p q q p

W W W Wc r W rW W

r rr r r rW W W W

c r W rW Wr rr r r r

ω

ω

⎧ ⎛ ⎞∂ ∂ ∂ ∂⎪ + − + =⎜ ⎟

∂∂ ∂ ∂⎪ ⎝ ⎠⎨

⎛ ⎞∂ ∂ ∂ ∂⎪+ − + =⎜ ⎟⎪ ∂∂ ∂ ∂⎝ ⎠⎩

(D.68)


488

4 3 2 2

4 3 2 2 40 0

4 3 2 2

4 3 2 2 40 0

1 12

1 12

a ap p p p p

q p qF

a aq q q q p

p q pF

W W W Wr W dr rW W dr

r rr r r r c

W W W Wr W dr rW W dr

r rr r r r c

ω

ω

⎧ ⎛ ⎞∂ ∂ ∂ ∂⎪ + − + =⎜ ⎟

∂∂ ∂ ∂⎝ ⎠⎪⎨

⎛ ⎞∂ ∂ ∂ ∂⎪+ − + =⎜ ⎟⎪ ∂∂ ∂ ∂⎝ ⎠⎩

∫ ∫

∫ ∫ (D.69)

Perform integration by parts and obtain after rearrangement

3 2 2 2

3 2 2 2 20

2 2 2

2 2 40 0

3 2 2 2

3 2 2 2 2

1 1 1

1

1 1

p p p p q p pq q q q q

a

a ap q p q p

p qF

q q q q p qp p p p

a

W W W W W W Wa W W W W W

a r r r rr r a r r

W W W Wr dr rW W dr

r r rr r c

W W W W W Wa W W W W

a r rr r a r r

ω

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂+ − − + +⎜ ⎟ ⎜ ⎟

∂ ∂ ∂∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

⎛ ⎞∂ ∂ ∂ ∂+ − =⎜ ⎟

∂ ∂∂ ∂⎝ ⎠

⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂+ − − +⎜ ⎟

∂ ∂∂ ∂ ∂ ∂⎝ ⎠

∫ ∫

0

2 2 2

2 2 40 0

1

1

qp

a aq p q p q

q pF

WW

r r

W W W Wr dr rW W dr

r r rr r cω

⎧⎪⎪⎪⎪⎪⎪⎨

⎛ ⎞∂⎪+⎜ ⎟⎪ ∂⎝ ⎠⎪

⎪ ⎛ ⎞∂ ∂ ∂ ∂⎪+ − =⎜ ⎟⎪ ∂ ∂∂ ∂⎝ ⎠⎩

∫ ∫

(D.70)

From the boundary conditions (D.65) and the relation (0) (0) 0W AJ= = , Equation

(D.70) becomes

2 2 2 2

2 2 2 40 0

2 2 2 2

2 2 2 40 0

1

1

a ap q p q p q p

p qa F

a aq p q p q p q

q pa F

W W W W W Wa r dr rW W dr

r r r rr r r c

W W W W W Wa r dr rW W dr

r r r rr r r c

ω

ω

⎧ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎪− + − =⎜ ⎟

∂ ∂ ∂∂ ∂ ∂⎝ ⎠⎪⎨

⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎪− + − =⎜ ⎟⎪ ∂ ∂ ∂∂ ∂ ∂⎝ ⎠⎩

∫ ∫

∫ ∫ (D.71)

Subtract the second of (D.71) from the first to get

( )2 22 2

2 2 40

ap qp q q p

p qa F

W W W Wa rW W dr

r rr r c

ω ω⎛ ⎞ −∂ ∂ ∂ ∂− − =⎜ ⎟

∂ ∂∂ ∂⎝ ⎠ ∫ (D.72)

489

Recall the boundary condition (D.66), substitute the two wave modes ( , )pW r θ and

( , )qW r θ into it and multiply the first by the radial derivative of ( , )pW r θ and the second

by the radial derivative of ( , )qW r θ , i.e.,

2

2

2

2

1 0

1 0

p p q

q q p

W W Wr r rr

W W Wr r rr

ν

ν

∂ ∂ ∂+ = ×

∂ ∂∂∂ ∂ ∂

+ = ×∂ ∂∂

(D.73)

Sum up the equations to obtain

2 2

2 2 0p q q pW W W Wr rr r

∂ ∂ ∂ ∂− =

∂ ∂∂ ∂ (D.74)

Hence Equation (D.72) becomes

( )2 2

0

0a

p q p qrW W drω ω− =∫ (D.75)

D.2 STURM-LIOUVILLE PROBLEM

Consider the homogeneous, second-order, linear equation for the scalar field ψ of the

form

2 2 0kψ ψ∇ + = (D.76)

where k can be either 0, a real constant, or a function of coordinates.

Through separation of variables, Equation (D.76) can be written in the form

[ ]( ) ( ) ( ) 0d dp z q z r zdz dz

ψ λ ψ⎡ ⎤ + + =⎢ ⎥⎣ ⎦ (D.77)

490

This equation is the Liouville equation. Functions p, q, and r are characteristic of the

coordinate used in the separation; p and r are always defined positive. The parameter λ

depends on the boundary conditions. With the use of Equation (D.77) and the procedure

described in Morse et al (1953), we find the orthogonality relation

0b

n ma

rdzψ ψ =∫ (D.78)

If nψ and mψ belong to different eigenvalues ( n mλ λ≠ ) they are orthogonal.

In general, if a problem can be written as in Equation (D.77), the mode solutions form a

set of mutually orthogonal functions.

D.3 ORTHOGONALITY RELATION FROM THE REAL RECIPROCITY RELATION

D.3.1 Orthogonality relation in rectangular coordinates

Assume that solutions 1 and 2 are generic time-harmonic guided-wave modes (e.g., plate

guided waves), i.e.,

1

2

( , , , ) ( )

( , , , ) ( )

n

m

i x i tn

i x i tm

x y z t y e e

x y z t y e e

ξ ω

ξ ω

−

−

=

=

v v

v v (D.79)

In the generic case, the wavenumbers and the amplitudes are assumed to be complex

( ,m nξ ξ ∈ , ,n m ∈v v ). The strains and stress will also be harmonic, i.e.,

1

2

( , , , ) ( )

( , , , ) ( )

n

m

i x i tn

i x i tm

x y z t y e e

x y z t y e e

ξ ω

ξ ω

−

−

=

=

T T

T T (D.80)

Recall the real reciprocity relation for time-harmonic functions as given by Equation

(5.17) and set the source terms equal to zero ( 1 2= =F F 0 ), i.e.,

491

( )2 1 1 2 0∇ ⋅ − ⋅ =v T v T (D.81)

Substitution of Equations (D.79) and (D.80) into Equation (6.104) yields

( ) ( ) ( )

( ) ( )

2 1 1 2 2 1 1 2 2 1 1 2

2 1 2 1 1 2 1 2 2 1 1 2

ˆ ˆ

ˆ ˆ 0n m n m

x y

i i i iy

ξ ξ ξ ξ

∂ ∂∇ ⋅ − ⋅ = ⋅ − ⋅ ⋅ + ⋅ − ⋅ ⋅

∂ ∂∂

= − ⋅ − ⋅ + ⋅ + ⋅ ⋅ + ⋅ − ⋅ ⋅ =∂

v T v T v T v T x v T v T y

v T v T v T v T x v T v T y(D.82)

where x and y are the unit vectors in the x and y directions. Simplification of Equation

(D.82) yields

( )( ) ( )2 1 1 2 2 1 1 2ˆ ˆ 0n miy

ξ ξ ∂− + ⋅ − ⋅ ⋅ + ⋅ − ⋅ ⋅ =

∂v T v T x v T v T y (D.83)

Substitution of Equations (D.79) and (D.80) into Equation (D.83) yields

( ) ( ) ( )

( ) ( )

ˆ( ) ( ) ( ) ( )

ˆ( ) ( ) ( ) ( ) 0

n m

n m

i x i tn m m n n m

i x i tm n n m

i e e v y y y y

e e v y y y yy

ξ ξ ω

ξ ξ ω

ξ ξ − +

− +

− + − ⋅

∂+ − ⋅ =

∂

T v T x

T v T y (D.84)

Since the exponential function ( )n mi x i te eξ ξ ω− + is non zero, we can divide Equation (D.84)

by ( )n mi x i te eξ ξ ω− + and get

( )( )

( )

ˆ( ) ( ) ( ) ( )

ˆ( ) ( ) ( ) ( ) 0

n m m n n m

m n n m

i y y y y

y y y yy

ξ ξ− + − ⋅

∂+ − ⋅ =

∂

v T v T x

v T v T y (D.85)

Integrate Equation (D.85) with respect to y to get

( ) ( ) ( )ˆ ˆ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )d

d

n m m n n m m n n m dd

i y y y y dy y y y yξ ξ−

−

− + − ⋅ = − − ⋅∫ v T v T x v T v T y (D.86)

492

If boundary conditions in Equation (D.86) at y d= ± are either traction free ( ˆ⋅ =T y 0 ) or

rigid ( =v 0 ), then the right-hand side of Equation (D.86) vanishes and we get

( ) ( ) ˆ( ) ( ) ( ) ( ) 0d

n m m n n md

i y y y y dyξ ξ−

− + − ⋅ =∫ v T v T x (D.87)

If m n nξ ξ ξ−≠ − = , then one can divide by ( )n mξ ξ+ and Equation (D.87) becomes the

orthogonality relation, i.e.,

( ) ˆ( ) ( ) ( ) ( ) 0d

m n n md

y y y y dy−

− ⋅ =∫ v T v T x for m nξ ξ≠ − (D.88)

The wavenumbers of guided waves always occur in pairs having equal value and

opposite signs. By convention, the modes that propagate or decay in the x+ direction are

numbered with positive integers (and negative integers for those in x− direction).

If the m n≠ − , then 0n mξ ξ+ ≠ ; one may divide Equation (D.87) by ( )n mξ ξ− and get

( ) ˆ( ) ( ) ( ) ( ) 0d

m n n md

y y y y dy−

− ⋅ =∫ v T v T x for m n≠ − (D.89)

This is the orthogonality relation for undamped propagating modes.

If m n= − , then 0n m n n n nξ ξ ξ ξ ξ ξ−+ = + = − = and relation (D.88) no longer applies. In

this case ( ) ˆ( ) ( ) ( ) ( )d

m n n md

y y y y dy−

− ⋅∫ v T v T x is nonzero. The orthogonality relation from

the real reciprocity relation relates the wave propagating in the forward direction with

that propagating in the backward direction.

493

Note that the derivation of the orthogonality relation is valid for Lamb wave propagating

in layered waveguide structures, where the material has arbitrary anisotropy, but the

properties of the media are z-invariant.

D.3.2 Orthogonality relation in cylindrical coordinates

For circular crested wave it is not possible to derive a generic formulation of the

orthogonality relation. We directly derive the relation for the case of SH waves and Lamb

waves.

D.3.2.1 Shear horizontal waves

Recall the real reciprocity relation of Equation (5.74) and set the source terms equal to

zero ( 1 2= =F F 0 ), hence

( ) ( )1 2 2 1 1 2 2 1 0r r z zr v T v T r v T v Tr zθ θ θ θ θ θ θ θ

∂ ∂⎡ ⎤− + − =⎣ ⎦∂ ∂ (D.90)

Integrate Equation (D.90) with respect to z to get

( ) ( )1 2 2 1 1 2 2 1 0d

d

r r z z dd

r v T v T dz r v T v Tr θ θ θ θ θ θ θ θ −

−

∂ ⎡ ⎤− + − =⎣ ⎦∂ ∫ (D.91)

Since the shear wave modes satisfy the stress free boundary conditions ( 0zTθ = ), the

right-hand side of Equation (D.91) vanishes and we get

( )1 2 2 1 0d

r rd


−

∂ ⎡ ⎤− =⎣ ⎦∂ ∫ (D.92)

Let assume that solutions 1 and 2 are free modes such that (from Equation (3.67)):

1

1 12

2 1

( , ) ( , ) ( ) ( )

( , ) ( , ) ( ) ( )n n

m m

r z v r z i Z z J r

r z v r z i Z z J rθ

θ

ω ξ

ω ξ

= =

= =

v

v (D.93)

494

where ( ) sin cosrZ z v T A z B zθ θ η η= = = + and, from Equation (3.75),

1 11 0

2 12 0

( )( , ) ( ) ( ) 2

( )( , ) ( ) ( ) 2

nr n n n

mr m m m

J rr z T Z z J rr

J rr z T Z z J rr

θ

θ

ξμ ξ ξ

ξμ ξ ξ

⎛ ⎞= = −⎜ ⎟⎝ ⎠⎛ ⎞= = −⎜ ⎟⎝ ⎠

T

T (D.94)

Substitute Equations (D.93) and (D.94) into (D.92) to get

11 0

11 0

( )( ) ( ) 20

( )( ) ( ) 2

md n n m m m

d nm m n n n

J ri Z J r Z J rrr dz

r J ri Z J r Z J rr

ξωμ ξ ξ ξ

ξωμ ξ ξ ξ−

⎡ ⎤⎛ ⎞⎛ ⎞−⎜ ⎟⎢ ⎥⎜ ⎟∂ ⎝ ⎠⎢ ⎥⎜ ⎟ =⎢ ⎥∂ ⎜ ⎟⎛ ⎞− −⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

∫ (D.95)

Rearrange the terms and factor out the product m nZ Z , i.e.,

( )0 1 1 0( ) ( ) ( ) ( ) 0d

m m n n m n m nd

i r J r J r J r J r Z Z dzr

ωμ ξ ξ ξ ξ ξ ξ−

∂⎡ ⎤− =⎣ ⎦∂ ∫ (D.96)

Bring out of the z integral the terms dependent on r, divide by the term iωμ , and perform

the derivative with respect to r, to get

( )2 21 1( ) ( ) 0

d

n m m n m nd

r J r J r Z Z dzξ ξ ξ ξ−

− =∫ (D.97)

iii. If n m≠ , then ( )2 2 0n mξ ξ− ≠ ; hence 0d

m nd

Z Z dz−

=∫ .

iv. If n m= , then ( )2 2 0n mξ ξ− ≠ and ( )2 0d

md

Z dz−

≠∫ . This is the normalization factor.

495

D.3.2.2 Lamb waves

Recall the real reciprocity relation of Equation (5.81) and set the source terms equal to

zero ( 1 2= =F F 0 ), hence

( ) ( ) ( )2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2 0r rr z rz r rr z rz r rz z zz r rz z zzr v T v T v T v T r v T v T v T v Tr z

∂ ∂⎡ ⎤ ⎡ ⎤+ − − + + − + =⎣ ⎦ ⎣ ⎦∂ ∂ (D.98)


( ) ( ) ( )2 1 2 1 1 2 1 2 2 1 2 1 1 2 1 2 0d d

r rr z rz r rr z rz r rz z zz r rz z zz dd

r v T v T v T v T dz r v T v T v T v Tr −

−

∂ ⎡ ⎤ ⎡ ⎤+ − − + + − + =⎣ ⎦ ⎣ ⎦∂ ∫ (D.99)

Since the Lamb wave modes satisfy the stress free boundary conditions ( 0rz zzT T= = ),

the right-hand side of Equation (D.99) vanishes and we get

( )2 1 2 1 1 2 1 2 0d

r rr z rz r rr z rzd

r v T v T v T v T dzr −

∂ ⎡ ⎤+ − − =⎣ ⎦∂ ∫ (D.100)

Let assume that solutions 1 and 2 are free modes of non-dissipative Lamb waves such

that

( ) ( )( ) ( )

1 11 1 0

2 22 1 0

ˆ ˆˆ ˆ( , ) ( ) ( ) ( ) ( )

ˆ ˆˆ ˆ( , ) ( ) ( ) ( ) ( )

r zr z n n n n

r zr z m m m m



ξ ξ

ξ ξ

= + = +

= + = +

v r z r z

v r z r z (D.101)

where from Equations (3.122) and (3.130) we have

( )( )

*

*

( ) cos cos

( ) sin sin

Srn Sn Sn Sn Sn Sn Sn

Szn Sn Sn Sn Sn Sn Sn

v z i A z R z

v z i A z R z

ω ξ α β β

ω α α ξ β

⎧ = −⎪⎨

= +⎪⎩ (D.102)


496

( )

( )

*

*

( ) sin sin

( ) cos cos

Arn Sn Sn Sn Sn Sn Sn

Azn Sn Sn Sn Sn Sn Sn

v z i A z R z

v z i A z R z

ω ξ α β β

ω α α ξ β

⎧ = −⎪⎨

= − +⎪⎩ (D.103)

for antisymmetric modes. In Equation (D.101) we do not use the subscript S or A since

the Equation in (D.102) and (D.103) are formally equal. From stress derivation in Section

3.2.6, we can write the stresses as

11 0

12 0

1 1

2 1

( )( , ) ( ) ( ) 2 ( )

( )( , ) ( ) ( ) 2 ( )

( , ) ( ) ( )

( , ) ( ) ( )

rr rr r nn n n

rr rr r mm m m

rz rzn n

rz rzm m



T r z T z J r

T r z T z J r

ξμ ξ μωξμ ξ μω

μ ξ

μ ξ

= −

= −

=

=

(D.104)

where

( )

( )

* 2 2 2

* 2 2

( ) 2 cos 2 cos

( ) 2 sin sin

rrn Sn Sn Sn Sn Sn Sn Sn Sn Sn

rzn Sn Sn Sn Sn Sn Sn Sn Sn

T z A z R z

T z A z R z

β ξ α α ξ β β

α ξ α ξ β β

⎡ ⎤= − + − −⎣ ⎦⎡ ⎤= + −⎣ ⎦

(D.105)


( )

( )

* 2 2 2

* 2 2

( ) 2 sin 2 sin

( ) 2 cos cos

rrn An An An An An An An An An

rzn An An An An An An An An

T z A z R z

T z A z R z

ξ β α α ξ β β

α ξ α ξ β β

⎡ ⎤= − + − −⎣ ⎦⎡ ⎤= − + −⎣ ⎦

(D.106)

for antisymmetric modes. Substituting in (D.100), we obtain

11 0 0 1

11 0 0 1

( )( ) ( ) 2 ( ) ( )0

( )( ) ( ) 2 ( ) ( )

r rr r z rznd m m n n n m m n n

r rr r z rzd mn n m m m n n m m

J rv J r T J r v v J r T J ri rr dz

r J rv J r T J r v v J r T J ri r

ξξ ξ ξ ξωμ

ξξ ξ ξ ξω

−

⎡ ⎤⎛ ⎞⎛ ⎞+ +⎜ ⎟⎢ ⎥⎜ ⎟∂ ⎝ ⎠⎢ ⎥⎜ ⎟ =⎢ ⎥∂ ⎜ ⎟⎛ ⎞− + −⎜ ⎟⎢ ⎥⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

∫ (D.107)

Rearrange the terms in Equation (D.107) to get

497

( ) ( )1 0 0 1( ) ( ) ( ) ( ) 0d

r rr z rz z rz r rrm n n m m n m n n m m n

d

r v T v T J r J r v T v T J r J r dzr

ξ ξ ξ ξ−

∂ ⎡ ⎤⎡ ⎤− + − =⎣ ⎦⎣ ⎦∂ ∫ (D.108)

Differentiate with respect to r:

( )( )

( )( )0 0 1 1

1 1 0 0

( ) ( ) ( ) ( )0

( ) ( ) ( ) ( )

r rr z rzdm n n m m m n n m n

z rz r rrd m n n m m m n n m n

v T v T J r J r J r J rdz

v T v T J r J r J r J r

ξ ξ ξ ξ ξ ξ


⎡ ⎤− −⎢ ⎥ =⎢ ⎥− − −⎣ ⎦

∫ (D.109)

i. If m n≠ , then ( )0 0 1 1( ) ( ) ( ) ( ) 0m m n n m nJ r J r J r J rξ ξ ξ ξ ξ ξ− ≠ and

( )1 1 0 0( ) ( ) ( ) ( ) 0m m n n m nJ r J r J r J rξ ξ ξ ξ ξ ξ− ≠ ; hence

( ) ( ) 0d d

r rr z rz z rz r rrm n n m m n n m

d d

v T v T dz v T v T dz− −

− = − =∫ ∫

ii. If m n= , Equation (D.108) becomes

( )0 0 1 1

1 1 0 0

( ) ( ) ( ) ( )0

( ) ( ) ( ) ( )

dn n n n n n r rr z rz

n n n nn n n n n n d

J r J r J r J rv T v T dz

J r J r J r J rξ ξ ξ ξ ξ ξ

ξ ξ ξ ξ ξ ξ −

−⎛ ⎞ ⎡ ⎤− =⎜ ⎟ ⎣ ⎦+ −⎝ ⎠∫ (D.110)

since

0 0 1 1

1 1 0 0

( ) ( ) ( ) ( )0

( ) ( ) ( ) ( )n n n n n n

n n n n n n



−⎛ ⎞=⎜ ⎟+ −⎝ ⎠

(D.111)

hence ( ) 0d

r rr z rzn n n n

d

v T v T dz−

− ≠∫ . This is the normalization factor.

498

E NORMALIZATION FACTOR In this appendix we will derive the normalization factor for shear horizontal waves and

Lamb waves. Since the average power flow is the same in rectangular and cylindrical

coordinates, we will perform the derivation only for rectangular coordinates.

E.1 SHEAR HORIZONTAL WAVES

In this section we verify that the Poynting vector is zero for n m≠ and none zero for

m n= . Consider a mode shapes, i.e.,

( , , ) sin cos SA i xi xj A S i tz j j j jU x y t A ye B ye eξξ ωη η⎡ ⎤= +⎣ ⎦ (E.1)

Where Aη and Sη are defined in Equations (3.11). We will derive the power flow and

the normalization factor for both symmetric and antisymmetric modes.

E.1.1 Symmetric modes The velocity and stress are expressed as:

( , , ) ( , )

( , , ) ( , )

j j jz zS zS

j j j i txz zS S zS

v U x y t i U x yt

T U x y t i U x y ex

ω

ω

μ μξ

∂= =

∂∂

= =∂

, 0,1, 2j = L (E.2)

where

( , ) cos Si xj SzS j jU x y B ye ξη= (E.3)

Consider two mode shapes Um and Un, i.e.,

499

( , ) cos

( , ) cos

S

S

i xm SzS m m

i xn SzS n n

U x y B ye

U x y B ye

ξ

ξ

η

η

=

= (E.4)

hence the time-averaged power defined in Equation (6.131) becomes after rearrangement

( )

( ) ( )2

n m dS S m n

nm zS zSd

P U y U y dyωμ ξ ξ

−

+= − ∫

%% (E.5)

Substitute the expression of Um and Un defined in Equation (E.4) into (E.5) to get

( )

cos cos2

n m dS S S S

nm m m n nd

P B yB ydyωμ ξ ξ

η η−

+= − ∫

%% % (E.6)

Recall the definition of the symmetric wavenumber, i.e.,

22

S nd

πη = 0,1,n = K , (E.7)

From Equation (E.7) we notice that Sη is real and hence Sξ is real because we assume

propagating modes. In lieu of this consideration, Equation (E.7) becomes

( )

cos cos2

n m dS S S S

nm m n m nd

P B B y ydyωμ ξ ξ

η η−

+= − ∫ (E.8)

Solution of the above integral is:

( ) ( ) ( )sin sin

2

n m S S S SS S m n m n

nm m n S S S Sm n m n

d dP B B

ωμ ξ ξ η η η η

η η η η

⎡ ⎤+ − +⎢ ⎥= − +

− +⎢ ⎥⎣ ⎦ (E.9)

Notice that from Equation (E.7) we get

500

( ) ( )

( ) ( )

2 2 22 2

2 2 22 2

S Sm n

S Sm n

m n m nd d

m n m nd d

π πη η

π πη η

⎧ − = − = −⎪⎪⎨⎪ + = + = +⎪⎩

(E.10)

hence

( ) ( )

( )( )

( )sin sin

2

n mS S

nm m n

d m n m nP B B

m n m nωμ ξ ξ π π

π π

+ ⎡ ⎤− += − +⎢ ⎥

− +⎢ ⎥⎣ ⎦ (E.11)

For m ≠ n the value in parenthesis is equal to zero. For m = n we obtain:

( )( )

( )( )

2

0

sin sinlimn

nn S n m n

m n m nP dB

m n m nπ π

ωμξπ π− →

− += − +

− +2n

S ndBωμξ⎡ ⎤⎢ ⎥ = −⎢ ⎥⎣ ⎦

(E.12)

This is the normalization factor. If we set 1nnP = , we can derive the value of Bn.

1n n

S

Bdωμξ

= (E.13)

E.1.2 Antisymmetric modes Similarly, for the antisymmetric mode, the velocity and stress are expressed as

( , , ) ( , )

( , , ) ( , )

j j jz zA zA

j j j i txz zA A zA

v U x y t i U x yt

T U x y t i U x y ex

ω

ω

μ μξ

∂= =

∂∂

= =∂

(E.14)

where

( , ) sin Ai xj AzA j jU x y A ye ξη= (E.15)

Consider two mode shapes Um and Un, i.e.,

501

( , ) sin

( , ) sin

A

A

i xm AzA m m

i xn AzA n n

U x y A ye

U x y A ye

ξ

ξ

η

η

=

= (E.16)

hence the time-averaged power defined in Equation (6.131) becomes

( )

( ) ( )2

n m dA A m n

nm zA zAd

P U y U y dyωμ ξ ξ

−

+= − ∫

%% (E.17)

Substitute the expression Um and Un defined in Equation (E.16) into (E.17). Recall the

definition of the antisymmetric wavenumber, i.e.,

( )2 12

A nd

πη = + 0,1,n = K , (E.18)

notice that Aη is real and Aξ is real because we assume propagating modes, i.e.,

( )

sin sin2

n m dA A A A

nm m n m nd

P A A y ydyωμ ξ ξ

η η−

+= − ∫ (E.19)

Solution of the above integral is:

( ) ( ) ( )sin sin

2

n m A A A AA A m n m n

nm m n A A A Am n m n

d dP A A

ωμ ξ ξ η η η η

η η η η

⎡ ⎤+ − +⎢ ⎥= − −

− +⎢ ⎥⎣ ⎦ (E.20)

Notice that

( ) ( )

( ) ( )

2 1 2 1 22 2

2 1 2 1 2 12 2

A Am n

A Am n

m n m nd d

m n m nd d

π πη η

π πη η

⎧ − = + − − = −⎪⎪⎨⎪ + = + + + = + +⎪⎩

(E.21)

Hence

502

( ) ( )

( )( )

( )sin sin 1

2 1

n mA A

nm m n

d m n m nP A A

m n m nωμ ξ ξ π π

π π

+ ⎡ ⎤− + += − +⎢ ⎥

− + +⎢ ⎥⎣ ⎦ (E.22)

For m ≠ n the value in parenthesis is equal to zero. For m = n we obtain:

( )( )

2 sin 100 1

nnn A n

m nP dA

m nπ

ωμξπ

+ += − +

+ +2n

A ndAωμξ⎡ ⎤⎢ ⎥ =⎢ ⎥⎣ ⎦

(E.23)

This is the normalization factor for antisymmetric modes. We can normalize to 1 to find

the value of An.

1n n

A

Adωμξ

= (E.24)

E.2 LAMB WAVES In this section, we will derive the time-averaged power for Lamb wave modes. Then, we

will do an analytical verification of the orthogonality condition for some particular Lamb

wave mode solutions. However, we will not be able to verify the orthogonality relation

analytically for any arbitrary Lamb wave modes like we did in Sections E.1 for SH

waves; Lamb wave modes have more intricate expressions; thus, in the general case, one

has to rely on numerical verification of the orthogonality condition..

E.2.1 Symmetric modes Consider the expression of the particle displacement for symmetric modes, i.e.,

( ) ( )

( ) ( )

( ) cos cos

( ) sin sin

S

S

i x tSx S S S S S

i x tSy S S S S S

u y B y R y e

u y iB y R y e

ξ ω

ξ ω

ξ α β β

α α ξ β

−

−

⎧ = − −⎪⎨

= − +⎪⎩ (E.25)

Where SR is defined in Equation (3.22) and B is a constant to be determined. The

corresponding particle velocities are

503

( ) ( )

( ) ( )

cos cos

sin sin

S

S

i x tSxSx S S S S S

i x tSySy S S S S S

uv i B y R y et

uv B y R y e

t

ξ ω

ξ ω

ω ξ α β β

ω α α ξ β

−

−

∂⎧ = = −⎪⎪ ∂⎨ ∂⎪ = = − +⎪ ∂⎩

(E.26)

The corresponding stresses are

( ) ( )

( )( ) ( )

2 2 2

2 2

2 cos 2 cos

2 sin sin

s

s

i x tSxx S S S S S S S S

i x tSxy S S S S S S S

T i B y R y e

T B y R y e

ξ ω

ξ ω



−

−

⎧ ⎡ ⎤= − + − −⎣ ⎦⎪⎨

= + −⎪⎩

(E.27)

Let substitute the expression of the velocities and of the stresses in Equation (6.146).

( ) ( )( )

( ) ( )

2

2 2

2 2 2

2sin sin 2 sin sin

Re 2 coscos cos

2 cos

nn

S S S S S S S S S S S Sd

S S S Sd S S S S S

S S S S

BP

y R y y R y

dyyy R y

R y

ωμ

α α ξ β ξ α α ξ β β

ξ β α αξ α β β

ξ β β−

=

⎛ ⎞+ + −⎜ ⎟⎜ ⎟× ⎡ ⎤+ −⎜ ⎟+ − ⎢ ⎥⎜ ⎟−⎢ ⎥⎣ ⎦⎝ ⎠

∫

%%% %

% %%%

(E.28)

Assume that every component is real, hence rearranging the terms we obtain

( )( ) ( )

( )

2 2 2 2 2

2 2 2 2 2 2 2 2 2

02 2 2 2 2

2 sin 3 2 cos cos

sin 2 cos

2 cos 3 sin sin

S S s S S S S S S Sd

nn S S S S S S S S S S

S S S S s S S S S S

y R y y

P B R y y dy

R y R y y

ξ α α β ξ β α α β

ωμ ξ ξ β β ξ ξ β α α

ξ β β ξ β α α β

⎡ ⎤⎛ ⎞− + −⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟= + − + + −⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟+ + −⎝ ⎠⎣ ⎦

∫ (E.29)

Perform the integral and rearrange the terms to get

2

2nn SBP Vωμ

= (E.30)

where

504

( )( ) ( )

( )

( )

2 2 2 2 2 2

2 2 2

2

sin cos1 3

sin cos4

4 sin cos 2 3 cos sin

S SS S S S S S S S

S

S SS S S S S

S

S S S S S S S S S S

d dd R R

d dV

R d d R d d

β βξ ξ β ξ ξ ββ

α αξ ξ β αα

α β α β ξ β α β

⎡ ⎤+ + − −⎢ ⎥⎢ ⎥⎢ ⎥

= + + −⎢ ⎥⎢ ⎥⎢ ⎥+ − +⎢ ⎥⎣ ⎦

(E.31)

Note that although we have derived the solution for a particular case, this is valid also for

Re or Imα ∈ , Im or Reβ ∈ (hence Im or ReR ∈ ). This is the normalization factor, if

we normalize to 1, we derive the unknown constant, i.e.,

2n

Sn

BVωμ

= (E.32)

E.2.2 Antisymmetric modes Consider the expression of the particle displacement for antisymmetric modes, i.e.,

( ) ( )

( ) ( )

( , , ) sin sin

( , , ) cos cos

A

A

i x tAx A A A A A

i x tAy A A A A A

u x y t A y R y e

u x y t iA y R y e

ξ ω

ξ ω

ξ α β β

α α ξ β

−

−

⎧ = −⎪⎨

= +⎪⎩ (E.33)

Where AR is define in Equation (3.23) and A is a constant to be determined. The

corresponding particle velocities are:

( ) ( )

( ) ( )

( , , ) sin sin

( , , ) cos cos

A

A

i x tA Axx A A A A A

Ay i x tAy A A A A A

uv x y t i A y R y et

uv x y t A y R y e

t

ξ ω

ξ ω

ω ξ α β β

ω α α ξ β

−

−

∂⎧ = = − −⎪⎪ ∂⎨ ∂⎪ = = +⎪ ∂⎩

(E.34)

The corresponding stresses are

( ) ( )

( )( ) ( )

2 2 2

2 2

2 sin 2 sin

2 cos cos

A

A

i x txxA A A A A A A A A

i x txyA A A A A A A A

T i A y R y e

T A y R y e

ξ ω

ξ ω



−

−

⎧ ⎡ ⎤= − + − +⎣ ⎦⎪⎨

= + −⎪⎩

(E.35)

505

Substitute the expression of the velocities and of the stresses in Equation (6.146).

( ) ( )( )

( ) ( )

2

2 2

2 2 2

2cos cos 2 cos cos

Re 2 sinsin sin

2 sin

nn

A A A A A A A A A A A Ad

A A A Ad A A A A A

A A A A

AP

y R y y R y

dyyy R y

R y

μω


ξ β α αξ α β β

ξ β β−

= −

⎛ ⎞+ + −⎜ ⎟⎜ ⎟× ⎡ ⎤− + −⎜ ⎟− − ⎢ ⎥⎜ ⎟+⎢ ⎥⎣ ⎦⎝ ⎠

∫

%%% %

% %%%

(E.36)

Assume: , Reα β ∈ , hence ReR ∈ , Equation (E.36) becomes after rearranging the terms

( )( )

( )( )

2 2 2 2 2 2

2 22

2 2 2 2 2 2 2

2 2 2

2 cos cos

3 cos cos

2 2 sin 2 sin

3 2 sin sin

A A A A A A A A

dA A A A A A

nnd A A A A A A A A A

A A A A A A A

y R y

R y yAP dyy R y

R y y

ξ α α ξ ξ β β

ξ β α α βμωξ ξ β α α ξ β β

ξ β α β α β

−

⎛ ⎞+ −⎜ ⎟⎜ ⎟+ −⎜ ⎟= −⎜ ⎟+ + − +⎜ ⎟⎜ ⎟− + −⎝ ⎠

∫ (E.37)

Solve the integral and after rearrangement obtain

2

2nn AAP Vμω

= (E.38)

where

( )( ) ( )

( )( )

2 2 2 2 2 2

2 2 2

2 2

sin cos1 3

sin cos4

4 cos sin 2 3 sin cos

A AA A A A A A A A

A

A AA A A A A

A

A A A A A A A A A A

d dd R R

d dV

R d d R d d

β βξ ξ β ξ ξ ββ

α αξ ξ β αα

α β α β ξ β α β

⎡ ⎤− + + − −⎢ ⎥⎢ ⎥⎢ ⎥

= − + + −⎢ ⎥⎢ ⎥⎢ ⎥+ − +⎢ ⎥⎣ ⎦

(E.39)

Note that

If Reα ∈ then ( , )nnP f cω= (E.40)

If Imα ∈ then ( , )nnP f cω= − (E.41)

506

Equation (E.38) is the normalization factor. If we normalize to 1, we derive the unknown

constant, i.e.,

2

A

AVμω

= (E.42)

E.2.3 Derivation for some particular Lamb wave modes We prove explicitly the orthogonality relation for some special solution of the Rayleigh-

Lamb equations. For convenience we treat only the symmetric case (we omit the lower

case S).

Consider two Lamb wave modes Up and Uq, i.e.,

( ) ( )

( ) ( )

( ) cos cos

( ) sin sin

p

q

i x tp p p p p p

i x tq q q q q q

U y B y R y e

U y iB y R y e

ξ ω

ξ ω

ξ α β β

α α ξ β

−

−

⎧ = − −⎪⎨⎪ = − +⎩

(E.43)

Substitute Up and Uq, in the averaged time power flow Equation (6.146) to get

( ) ( )( )

( ) ( )( ) ( )( )

( ) ( )

2

2 2

2 2 2

2 2

2 2 2

2sin sin 2 sin sin

cos cos 2 cos 2 cosRe

sin sin 2 sin sin

cos cos 2 cos 2

qp

q q q q q p p p p p p p

q q q q q p p p p p p p p

p p p p p q q q q q q q

p p p p p q q q q

BP

y R y y R y

y R y y R y

y R y y R y

y R y y

ωμ


ξ α β β ξ β α α ξ β β


ξ α β β ξ β α α

= ×

+ + −

⎡ ⎤+ − + − −⎣ ⎦

+ + −

+ − + − − cos

d

d

q q q q

dy

R yξ β β

−

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

⎡ ⎤⎜ ⎟⎣ ⎦⎝ ⎠

∫ (E.44)

After solving the integral and rearrangement of the terms, we obtain

507

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )( ) ( )

2 2

2 2

2 2 2 2

2

sin sin2

sin sin2

sin sin2

sin

2

q p q pp q p q

q p q p

q p q pq p q q q p

q p q p

q p q pp q p p q p

q p q p

q pp q p p q p q q

qp

d d

d dR

d dR

dR R

BP

α α α αα α ξ ξ

α α α α

β α β αα ξ β ξ ξ

β α β α

α β α βα ξ β ξ ξ

α β α β

β βξ β ξ ξ ξ β

β

ωμ

⎛ ⎞− +⎜ ⎟+ −⎜ ⎟− +⎝ ⎠

⎛ ⎞− +⎜ ⎟+ − + −⎜ ⎟− +⎝ ⎠⎛ ⎞− +⎜ ⎟+ − + −⎜ ⎟− +⎝ ⎠

−+ − + −

=

( )

( ) ( )

( )

( )

( ) ( ) ( )

( ) ( )

2 2 2 2 2 2

2 2 2

2 2 2

sin

sin

2 2sin

sin sin2 2

sin2 2

q p

q p q p

q p

q pq p p p p q q q

q p

q p

q p q pp p q q q p q

q p q p

q pq q p p p q p

q

d

d

d

d dR

dR

β ββ β β

α α

α αξ ξ β α ξ ξ β α

α αα α

α β α ββ ξ β α ξ ξ

α β α β

β αβ ξ β α ξ ξ

β α

⎛ ⎞+⎜ ⎟−⎜ ⎟− +⎝ ⎠

⎛ ⎞−⎜ ⎟

−⎜ ⎟⎡ ⎤+ + − + + − ⎜ ⎟⎣ ⎦ +⎜ ⎟

+⎜ ⎟+⎝ ⎠⎛ ⎞− +⎜ ⎟− + − + +⎜ ⎟− +⎝ ⎠

−− + − +

−( )

( ) ( ) ( )

sin

sin sin2

q p

p q p

q p q pp q q p p q

q p q p

d

d dR R

β αβ α

β β β ββ β ξ ξ

β β β β

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎛ ⎞+⎢ ⎥⎜ ⎟+

⎜ ⎟+⎢ ⎥⎝ ⎠⎢ ⎥⎛ ⎞⎢ ⎥− +⎜ ⎟+ + +⎢ ⎥⎜ ⎟− +⎢ ⎥⎝ ⎠⎣ ⎦

(E.45)

Recall the Rayleigh-Lamb equation for symmetric modes, i.e.,

( )22 2 2cos sin 4 sin cos 0d d d dξ β α β ξ αβ α β− + = (E.46)

As stated before, we do not consider the first symmetric mode but only those modes that

are purely equivolume and composed of SV waves reflecting at 45 angles of incidence.

These are the solution when the branches cross the ray OL in figure 8.9 in Graff (1991).

For these modes, solutions of the Rayleigh-Lamb equation are

508

ndmd

πα

πβ

⎧ =⎪⎪⎨⎪ =⎪⎩

and ( )

( )

2 12

2 12

rd

sd

πα

πβ

⎧ = +⎪⎪⎨⎪ = +⎪⎩

where m, n, r, and s = 1,2,3… (E.47)

For ppdπα = , q

qdπα = , p

pdπβ = , and q

qdπβ = we obtain

( )

( )

p q

p q

p qd

p qd

πα α

πα α

⎧ + = +⎪⎪⎨⎪ − = −⎪⎩

, ( )

( )

p q

p q

p qd

p qd

πβ β

πβ β

⎧ + = +⎪⎪⎨⎪ − = −⎪⎩

(E.48)

If p ≠ q, the sine terms are equal to zero and hence Ppq is zero. If q = p relation (E.48)

becomes

2

0

p q

p q

pdπα α

α α

⎧ + =⎪⎨⎪ − =⎩

, 2

0

p q

p q

pd

πβ β

β β

⎧ + =⎪⎨⎪ − =⎩

, 2

0

p q

p q

pdπα β

α β

⎧ + =⎪⎨⎪ − =⎩

(E.49)

Recall that

0 0

sin coslim lim1x x

xd d xd dx→ →

= = (E.50)

Hence Equation (E.45) becomes after rearranging the terms

( )( )2 2 2 21pp p p p pP B d Rωμ ξ ξ β= + + (E.51)

Recall from Equation (3.22) that

2 2

2p p

pp p

Rξ β

ξ β−

= (E.52)

Substitution of Equation (E.52) into Equation (E.51) yields

509

( )( )2 2 4 4

224

p p p ppp

p p

P B dξ β ξ β

ωμξ β

+ += (E.53)

E.2.4 Orthogonality relation for one antisymmetric mode and one symmetric mode The orthogonality proof is easily demonstrated if we consider a symmetric mode and an

antisymmetric mode.

Substitute the expressions of the symmetric and antisymmetric velocities (Equations

(E.26) and (E.34)) and stresses (Equations (E.27) and (E.35)) in the power flow

expression (6.146) and rearrange the terms to get.

( )

( )( )

( )

( )

2 2

2 2 2

2 2

2 sin cos

2 sin cos

2 cos sin

2cos sin

2

2 sin cosRe4

S A

A s A S A

SA A s A A S A

A s A A S A

S S A A AS A

s A

i xs s S A S A S A

nm

A

y y

R y y

R y y

Ry y

ABe R y yPξ ξ

ξ ξ α α αα

ξ β ξ ξ α β

ξ ξ β β β

β ξ β αβ α

ξ ξ

μω ξ β ξ ξ α β α

ξ

−

⎛ ⎞+⎜ ⎟−⎜ ⎟+ − +⎝ ⎠

⎡ ⎤+⎢ ⎥

+ ⎛ ⎞⎢ ⎥+ −−⎜ ⎟⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦

− − +=

−

% %

% %%

% %%

% %%

% %

( )( )

( )( )

( )

2 2

2 2

2 2 2

2 2 2

2 2 2

sin cos

2cos sin

2

2 2 cos sin

d

dA S

S A S A

s s A

A A A S

S A

s s s A

s s s S A A A S A

dy

R R y y

y y

R y y

β ξβ β

ξ β ξ

ξ β α ξα α

ξ β α ξ

ξ β α ξ ξ β α β

−

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

⎡ ⎤−⎜ ⎟⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟+ −⎣ ⎦⎜ ⎟

⎜ ⎟⎡ ⎤+ −⎜ ⎟⎢ ⎥+⎜ ⎟⎢ ⎥+ + −⎣ ⎦⎜ ⎟⎜ ⎟− + − +⎝ ⎠

∫%

%%

% %%

% %%

(E.54)

Note that each terms in the integrand is made of an odd function, hence

0nmP = (E.55)

510

F STRAIN DERIVATION THROUGH NME AND FOURIER TRANSFORMATION

In this appendix we prove analytically that the expression of the strain derived through

normal mode expansion is the same as that derived with the Fourier transformation.

Consider for simplicity only one symmetric mode, we want to prove that

( ) ( )( ) ( )0

0 '

( )( ) sin sin2

s ss sxsx s s

ss S s

Nv d aia v y a aP D

ξξ ττ ξ ξω μ ξ

≡ −%

(F.1)

where

( )2 2 cos cosSN d dξβ ξ β α β= + (F.2)

and SD′ is the derivate with respect to the wavenumber of the symmetric Rayleigh-Lamb

equation. SD′ is hence given by

( )( ) ( )

2 2 3

3 2 2

2 22 2 2 2

4 2 sin cos 4 sin sin

4 cos cos 8 cos sin

sin sin cos cos

SD d d d d d

d d d d d

d d d d d d

β αξ ξ ξ βα α β ξ α α βα β

ξ β α β ξ β ξ α β

ξ ξξ β α β ξ β α βα β

⎛ ⎞′ = + − −⎜ ⎟⎝ ⎠

+ − −

− − + −

(F.3)

Recall the expression of velocity Equation (E.26) and that of the average power flow

Equation (E.30) as derived in appendix Section E.1.1 and substitute their expressions in

The expression in the right hand side of (F.1) (or Equation (9.30) in Section 9.2). After

rearrangement we get

511

( )( )( ) ( ) ( )

( ) ( ) ( )

( ) ( )

0

2 2 2 2 2 2

2 2 2 2 2

cos cos cos cos sinsin cos1 3

4 sin cossin cos4 2 3 cos sin

ni z t

nz

d R d d R d a ead id dd R R

R d dd d R d d

ξ ωξ ξ α β β ξ α β β ξτεβ βμ ξ ξ β ξ ξ β

βαβ α β

α αξ ξ β α ξ β α βα

−− −=

⎡ ⎤+ + − −⎢ ⎥⎢ ⎥+⎢ ⎥

⎢ ⎥⎢ ⎥+ + − − +⎣ ⎦

%% (F.4)

Perform the multiplications at the denominator, substitute the expression of R as derived

in Equation (3.22) to obtain

( )( ) ( ) ( )

( ) ( )

( )( ) ( )

( )

( )

22 2 20

22 2 22 2

2 2 2

2 2 2

22 2 22 2

2 2 2

2 22 2

2 2

cos sin4

cos1

4 co s

4 sin cos

cos3 sin cos

4 co s

cos4 sin cos

2 cos

cos3

cos

ni z t

nz

ai d a ed

dd

d

d d

dd d

d

dd d

d

dd

ξ ωτ αβ ξ β α ξξμε

ξ β αξαβ ξ β

ξ β β

ξβ ξ β α α α

ξ β αξα ξ β β β

ξ β β

ξ β αα β α β

ξβ β

ξ β ααβ

ξβ β

−+=

⎛ ⎞−⎜ ⎟+ +⎜ ⎟⎝ ⎠

+ + −

−− −

−+

−− ( )2 2 cos sind dξ β α β

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥+⎢ ⎥⎣ ⎦

(F.5)

Multiply the denominator by the term coscos

dd

ξ βξ β

and make use of Equation (F.2) to get

512

( )

( ) ( ) ( )

( ) ( )

( ) ( )

( )( )

( )

2 2

20

22 2 22 2

2 2 2

22 2 22 2

2 2 2

2 2 2

2 22 2

2 2

cossin

4 cos

cos1

4 co s

cos3 sin cos

4 co s

4 sin cos

cos4 sin cos

2 cos

cos2

2 c

ns i z t

nz

N da e

a dd id

dd

dd d

d

d d

dd d

d

d

ξ ωα ξ β αξ

τ ξ βεμ ξ β α

ξαβ ξ βξ β β

ξ β αξα ξ β β β

ξ β β

ξβ ξ β α α α

ξ β αα β α β

ξβ β

ξ β αξβ

−+

=⎛ ⎞−⎜ ⎟+ +⎜ ⎟⎝ ⎠

−− −

+ + −

−+

−− ( )2 23 cos sin

osd d

dαβ ξ β α β

β

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥+⎢ ⎥⎣ ⎦

(F.6)

Recall the expression of the Raylegh-Lamb equation, i.e. equality

( )22 2 2cos sin 4 sin cos 0d d d dξ β α β ξ αβ α β− + = (F.7)

Transform Equation (F.6) such as

( ) ( ) ( )

( ) ( ) ( )

( )

022 2

2 2 2 22

32 2

sin

cos7 cos sin

cos4 cos 4 2 sin cos

cos

ni z ts

nz

N a ead id d

d dd

d d d dd

ξ ωξτεμ ξ ξ β α ξ β ξ ξ β α β

β β βξ β β ξβ ξ α α β

α α

−

=−

− + −

+ + −

(F.8)

We have to prove that the denominator of equation above is equal to Ds’. From (F.3) we

have

( ) ( )

( )

( )

22 232 2

32 2

22 2 2

cos4 4 cos2 sin coscos cos

4 sin cos 8 cos sin

cos sin 4 sin cos sincos

S

d dd dD d dd d

d d d d

d d d d d dd

ξ ξ β αξβ ξ β βξ α α βα α β β

ξ α α β ξ β ξ α ββ

ξ ξ β α β ξ αβ α β ββ β

−′ = − + +

+ − −

⎡ ⎤− − +⎢ ⎥⎣ ⎦

(F.9)

513

Or after rearranging and using the equality in (F.7)

( ) ( )

( )( )

22 232 2

2 2 22

cos4 4 cos2 sin coscos cos

7 cos sin

S

d dd dD d dd d

d d

ξ ξ β αξβ ξ β βξ α α βα α β β

ξ ξ β ξ β α ββ

−′ = − + +

− + −

(F.10)

This is the same expression of the denominator in (F.8), hence the expressions of the

strain derived through different methods are the same.

514

G STRUCTURE EXCITED BY TWO PWAS Consider the case in which two PWAS are attached to the structure on the opposite sides

of the plate.

PWAS

-a +a

x

τ(x)eiωt ta

t=2d

tb y=+d

y=-d

PWAS τ(x)eiωt

Figure G.3 Interaction between two PWAS and the structure through the bonding layer: model

with interfacial shear stress, ( )xτ

If the PWAS are excited in phase only symmetric modes propagate in the structure, if

they are excited with opposite phase, only antisymmetric modes are present in the

structure.

G.1 SYMMETRIC MODE Consider the case in which the PWAS are excited in phase. In the PWAS, simple

equilibrium considerations yield

0a at σ τ′ − = (G.1)

Stress-strain relations in the structure and PWAS are

( )a a a ISA

EE

σ εσ ε ε

=⎧⎪⎨ = −⎪⎩

(G.2)

515

The equilibrium of the structure is given by

2 0xN τ′ + = (G.3)

where

-

( ) ( , ) ( )d

x sdN x x y dy t a xσ

+= = Λ∫ (G.4)

and the modeshape constant is given by

-

1 ( )d

S Sdy dy

tσ

+Λ = ∫ (G.5)

hence Equation (G.3) becomes

( , ) 2 0( )S

St x d

dσ τ

σΛ ′ + = (G.6)

Substitution of Equation (G.6) and (G.1) into (G.2) yields

2 ( ) 0

0

S

S

a a a

dtE

t E

σε τ

ε τ

⎧ ′ + =⎪ Λ⎨⎪ ′ − =⎩

(G.7)


( )bb a

b

GG u ut

τ γ= = − (G.8)

Differentiating Equation (G.8) with respect to x yields

( ) ( )b ba a

b b

G Gu ut t

τ ε ε′ ′ ′= − = − (G.9)

Equation (G.9) can be solved for aε , i.e.,

516

ba

b

tG

ε τ ε′= + (G.10)

Substituting Equation (G.10) into Equation (G.7) yields

2 ( )

0

S

S

ba a a a

b

dtE

tt E t EG

σε τ

τ ε τ

⎧ ′ = −⎪ Λ⎪⎨⎪ ′′ ′+ − =⎪⎩

(G.11)

Solving for ( )xτ the system in Equation (G.11)

2 ( )1 1 0b Sa a

b S

t dt EG

στ τψ

⎛ ⎞′′ − + =⎜ ⎟Λ⎝ ⎠ (G.12)

where a a

EtE t

ψ = . Denote

2 2 ( )b S S

b a a

G dt t E

σ ψψ

Λ +Γ = (G.13)

Substitution of Equation (G.13) into Equation (G.12) yields a differential equation for τ ,

i.e.,

2( ) ( ) 0x xτ τ′′ − Γ = (G.14)

In the case in which two modes are present (1 and 2), the problem is solved with the use

of the NME method. Recall that over the interval a x a− ≤ ≤

( )( ) ( )2

n n

xni x i xx

nnn a

v da x e e x dxP

ξ ξ τ−+

−

= ∫% (G.15)

( )( ) ( )2

n n

ani x i xx

nnn x

v da x e e x dxP

ξ ξ τ−− = − ∫% (G.16)

517

The strain equation in the structure can be written as:

1 1 1 1 2 1 2 11( , ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )x d d a x d a x d a x d a xE

ε σ σ σ σ+ − + −⎡ ⎤= + + +⎣ ⎦ (G.17)

Recall that

( ) ( )( ) ( ) ( )2 2

n n

xn ni x i xx x

n nnn nna


ξ ξξ τ τ−+

−

′ = − +∫% % (G.18)

( ) ( )( ) ( ) ( )2 2

n n

an ni x i xx x

n nnn nnx


ξ ξξ τ τ−− = − +∫% % (G.19)

Substitution of Equation (G.17) into Equation (G.11) yields after rearrangement

1 1

2 2

111

111

22 2

222

( ) ( )( ) ( ) ( )2

1 0( ) ( ) ( ) ( ) ( )

2

xni x i xxx

nn aba a n x

b i x i xx x

nn a

v d dv d d e e x dxPPt t itt E

G v d d v d d e e x dxP P

ξ ξ

ξ ξ

σσ ξ ττ τ

ψ ψσ σ ξ τ

±

−

±

−

⎡ ⎤⎡ ⎤⎛ ⎞+⎢ ⎥⎢ ⎥⎜ ⎟

⎢ ⎥⎢ ⎥⎜ ⎟′′ − − − =⎢ ⎥⎢ ⎥⎜ ⎟⎢ ⎥⎢ ⎥+⎜ ⎟ +⎜ ⎟ ⎢ ⎥⎢ ⎥⎝ ⎠⎣ ⎦ ⎣ ⎦

∫

∫

m

m

%%

% %(G.20)

G.2 ANTISYMMETRIC MODE Consider the case in which the PWAS are excited out of phase. The derivation for the

case of one antisymmetric mode is similar to the one derived in section G.1. In this case

the equilibrium of the structure is

2 0zM dτ′ + = (G.21)

where

-

( ) ( , ) ( )d

z A AdM x x y ydy td a xσ

+= = Λ∫ (G.22)

and AΛ is mode-shape constant given by

518

-

1 ( )d

A Ady ydy

tdσ

+Λ = ∫ , 1,...,n N= (G.23)

Equation (G.21) can be written as

( , ) 2 0( )A

AA

t x dd

σ τσ

Λ ′ + = (G.24)

As we can see Equation (G.24) is formally equal to Equation (G.6), the same results are

derived by substituting the index S with index A.

519

H STATISTICAL DATA ANALYSIS Hereunder we report the SAS code used to determine when there was significance

difference between the DI’s of the baseline readings and other DI values with a

significance level of α . We report the case of damage detection of a hole in the quasi-

isotropic plate. First, the data are loaded in the program. Although we do not need the

PWAS factor, we retain it here for convenience. Factor is Step that represents the hole

size. Variable DI is the damage index value that compare reading ## to the baseline

reading 0. PWAS 1 refers to the pair P0_P13where the first PWAS is the transmitter and

the second is the receiver. Input of the data:

DATA H02_018P1;

INPUT Step DI PWAS;cards;

1 0.057 1

1 0.032 1

1 0.036 1

2 0.062 1

2 0.028 1

2 0.033 1

2 0.049 1

4 0.061 1

4 0.059 1

4 0.055 1

4 0.075 1

5 0.077 1

5 0.067 1

5 0.055 1

5 0.058 1

6 0.057 1

6 0.056 1

6 0.052 1

7 0.095 1

7 0.094 1

7 0.096 1

7 0.098 1

7 0.102 1

8 0.121 1

8 0.1 1

8 0.09 1

8 0.081 1

9 0.156 1

9 0.134 1

9 0.134 1

9 0.138 1

10 0.152 1

10 0.15 1

10 0.168 1

10 0.152 1

11 0.186 1

11 0.187 1

11 0.187 1

12 0.2 1

12 0.21 1

12 0.207 1

12 0.2 1

12 0.198 1

13 0.223 1

13 0.222 1

520

; run;

A generalized linear model is used to fit the data where the factor is the size of the hole

(Step) and the variable of interest is the DI value. The output of the generalized linear

model is saved in variable diagnostic.

PROC GLM data = H02_018P1;

CLASS Step;

MODEL DI = Step;

LSMEANS Step;

MEANS Step / HOVTEST=BF;

OUTPUT OUT=diagnost p=ybar r=resid; run;

We assume that the data are normally distributed. We have few data for each step hence

we can not verify that our assumption is correct. We plot the residuals to verify the

assumption of equal variance (Figure H.4a).

symbol1 v=circle l=32 c = black;

PROC GPLOT data=diagnost;

PLOT resid*ybar/vref=0; run;

Figure H.4 Residual plot. a) Data not transformed; b) Data transformed, DI=DI2.

a) b)

521

The variance of the DI values is not constant hence this assumption is not verified. We

perform a transformation of variables to verify the equal variance assumption. We

consider the squared DI values; in this case the variance assumption is more close to the

assumed (Figure H.4b). We continue our analysis through the Tuckey confidence interval

multiple comparison. Since there is only one factor we do not need to check the

interaction in Tuckey’s test of additivity.

PROC GLM DATA = H02_018P1;

CLASS Step;

MODEL DI = Step PWAS;

LSMEANS Step;

MEANS Step / ALPHA=0.01 TUKEY CLDIFF;

run;

Output

Step compari

son

Difference Between Means

Simultaneous 99% Confidence Limits

Step comparis

on

Difference

Between Means


7 - 12 -0.03114 -0.037331 -0.024941 *** 4 - 12 -0.03613 -0.043058 -0.029205 ***

7 - 11 -0.02542 -0.034182 -0.01666 *** 4 - 11 -0.03042 -0.039708 -0.021124 ***

7 - 10 -0.01325 -0.02018 -0.006327 *** 4 - 10 -0.01825 -0.025836 -0.010662 ***

7 - 9 -0.00895 -0.015878 -0.002025 *** 4 - 9 -0.01395 -0.021534 -0.00636 ***

7 - 8 0.001268 -0.005658 0.008194 4 - 8 -0.00373 -0.011315 0.00386

7 - 6 0.006412 -0.002349 0.015173 4 - 7 -0.005 -0.011922 0.001931

7 - 5 0.005622 -0.001304 0.012548 4 - 6 0.001417 -0.007875 0.010709

7 - 4 0.004995 -0.001931 0.011922 4 - 5 0.000627 -0.006961 0.008214

7 - 2 0.007956 0.00103 0.014882 *** 4 - 2 0.002961 -0.004627 0.010548

7 - 1 0.007276 0.00035 0.014202 *** 4 - 1 0.002281 -0.005307 0.009868

6 - 12 -0.03755 -0.046309 -0.028787 *** 2 - 12 -0.03909 -0.046018 -0.032166 ***

6 - 11 -0.03183 -0.042563 -0.021103 *** 2 - 11 -0.03338 -0.042669 -0.024084 ***

522

Step compari

son

Difference Between Means


Step comparis

on

Difference

Between Means


6 - 10 -0.01967 -0.028958 -0.010374 *** 2 - 10 -0.02121 -0.028797 -0.013622 ***

6 - 9 -0.01536 -0.024656 -0.006072 *** 2 - 9 -0.01691 -0.024495 -0.00932 ***

6 - 8 -0.00515 -0.014437 0.004148 2 - 8 -0.00669 -0.014275 0.000899

6 - 7 -0.00641 -0.015173 0.002349 2 - 7 -0.00796 -0.014882 -0.00103 ***

6 - 5 -0.00079 -0.010083 0.008502 2 - 6 -0.00154 -0.010836 0.007749

6 - 4 -0.00142 -0.010709 0.007875 2 - 5 -0.00233 -0.009921 0.005253

6 - 2 0.001544 -0.007749 0.010836 2 - 4 -0.00296 -0.010548 0.004627

6 - 1 0.000863 -0.008429 0.010156 2 - 1 -0.00068 -0.008267 0.006907

5 - 12 -0.03676 -0.043684 -0.029832 *** 1 - 12 -0.03841 -0.045338 -0.031486 ***

5 - 11 -0.03104 -0.040335 -0.02175 *** 1 - 11 -0.0327 -0.041989 -0.023404 ***

5 - 10 -0.01888 -0.026463 -0.011288 *** 1 - 10 -0.02053 -0.028117 -0.012942 ***

5 - 9 -0.01457 -0.022161 -0.006986 *** 1 - 9 -0.01623 -0.023815 -0.00864 ***

5 - 8 -0.00435 -0.011941 0.003233 1 - 8 -0.00601 -0.013595 0.001579

5 - 7 -0.00562 -0.012548 0.001304 1 - 7 -0.00728 -0.014202 -0.00035 ***

5 - 6 0.00079 -0.008502 0.010083 1 - 6 -0.00086 -0.010156 0.008429

5 - 4 -0.00063 -0.008214 0.006961 1 - 5 -0.00165 -0.009241 0.005933

5 - 2 0.002334 -0.005253 0.009921 1 - 4 -0.00228 -0.009868 0.005307

5 - 1 0.001654 -0.005933 0.009241 1 - 2 0.00068 -0.006907 0.008267

The three asterisks on the right are for those comparisons that are significantly different.

Form the Tuckey multiple comparison we see that there is significant difference between

step 1 and step 7 with a significance level of 99%.

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