1

Full Derivation of the Helmholtz Potential Approach to the Analysis

of Guided Wave Generation during Acoustic Emission Events

Mohammad Faisal Haider1, Victor Giurgiutiu2

Department of Mechanical Engineering, University of South Carolina, 300 Main Street,

Columbia, SC 29208

Report# USC-ME-LAMSS-2001-101, September 2017.

1 Graduate Research Assistant, Department of Mechanical Engineering, University of South Carolina, 300 Main Street,

Columbia, SC 29208

2 Professor, Department of Mechanical Engineering, University of South Carolina, 300 Main Street, Columbia, SC

29208

2

ABSTRACT

This research addresses the predictive simulation of acoustic emission (AE) guided waves that

appear due to sudden energy release during incremental crack propagation. The Helmholtz

decomposition approach is applied to the inhomogeneous elastodynamic Navier-Lame equation

for both the displacement field and body forces. For the displacement field, we use the usual

decomposition in terms of unknown scalar and vector potentials, and H . For the body forces,

we hypothesize that they can be also expressed in terms of excitation scalar and vector potentials, *A and *B . It is shown that these excitation potentials can be traced to the energy released during

an incremental crack propagation. Thus, the inhomogeneous Navier-Lame equation has been

transformed into a system of inhomogeneous wave equations in terms of known excitation

potentials *A , *B and unknown potentials , H . The solution is readily obtained through direct

and inverse Fourier transforms and application of the residue theorem. A numerical study of the

AE guided wave propagation in a 6 mm thick 304-steel plate is conducted. A Gaussian pulse is

used to model the growth of the excitation potentials during the AE event; as a result, the actual

excitation potential follows the error function variation in the time domain. The numerical studies

show that peak amplitude of A0 signal is higher than the peak amplitude of S0 signal and the peak

amplitude of bulk wave is not significant compared to S0 and A0 peak amplitudes. In addition, the

effect of the source depth and propagating distance on guided waves is also investigated.

KEYWORDS: Acoustic emission, Crack propagation, Lamb wave, Helmholtz Potentials, Wave

propagation, SHM, NDT, Navier-Lame equations.

TABLE OF CONTENTS

1 Introduction ..............................................................................................................................4

1.1 State of the art ..................................................................................................................4

1.2 Scope of this research ......................................................................................................6

1.3 Description of excitation potentials as strain energy released from a crack ....................7

2 Formation of pressure and shear potentials ..........................................................................8

3 Solution in terms of displacement ........................................................................................11

3.1 Symmetric Lamb wave solution ....................................................................................14

3.2 Anti-symmetric Lamb wave solution.............................................................................14

3.3 Displacement solution in the wavenumber domain .......................................................15

3.4 Displacement solution in the physical domain ..............................................................16

3.5 Effect of source depth ....................................................................................................18

3.6 AE guided wave propagation .........................................................................................20

4 Numerical studies ...................................................................................................................22

4.1 Time dependent excitation potentials ............................................................................22

4.2 Phase velocity and group velocity dispersion curves ....................................................24

4.3 Example of AE guided wave propagation in a 6mm plate ............................................24

4.4 Effect of source depth ....................................................................................................26

4.5 Effect of higher order Lamb wave mode .......................................................................29

4.6 Effect of propagation distance .......................................................................................29

5 Summary, Conclusions, and Future Work ..........................................................................31

3

5.1 Summary ........................................................................................................................31

5.2 Conclusions ....................................................................................................................31

5.3 Future work ....................................................................................................................32

Acknowledgement ........................................................................................................................32

References .....................................................................................................................................33

Appendix A ...................................................................................................................................36

Appendix B ...................................................................................................................................42

Appendix C ...................................................................................................................................46

4

1 INTRODUCTION

Acoustic emission (AE) has been used widely in structural health monitoring (SHM) and

nondestructive testing (NDT) for the detection of crack propagation and to prevent ultimate failure

[1]-[5]. Acoustic emission (AE) signals are therefore needed to study extensively in order to

characterize the crack in an aging structure. This paper presents a theoretical formulation of AE

guided wave due to AE events such as crack extension in a structure. Lamb (1917) [6] derived

Rayleigh-Lamb wave equations from Navier-Lame elastodynamic equations in an elastic plate.

The presence of body force makes the homogeneous Navier-Lame elastodynamic equations into

inhomogeneous equations. Helmholtz decomposition principle was used to decompose

displacement to unknown scalar and vector potentials, and force vectors to known excitation scaler

and vector potentials [7]. There are two types of potentials acting in a plate for straight crested

Lamb waves: pressure potential and shear potential. The assumption of the straight-crested guided

wave makes the problem z-invariant (plane strain assumption). Excitation potentials for body

forces can be generalized as localized AE event. Excitation potentials can be traced to energy

released from the tip of the crack during crack propagation. These potentials satisfy the

inhomogeneous wave equations. In this research, inhomogeneous wave equations for unknown

potentials were solved due to known generalized excitation potentials in a form, suitable for

numerical calculation. The theoretical formulation shows that elastic waves generated in a plate

using excitation potentials followed the Rayleigh-Lamb equations.

In numerical studies, the one-dimensional (1D) (straight crested) AE guided wave propagation was

modeled in order to simulate the out-of-plane displacement that would be recorded by an AE

sensor placed on the plate surface at some distance away from the source. A real AE source releases

energy at certain time rate as a pulse over a finite time period. If the time rate of released energy

is known, then this time rate of released energy can be decomposed to time rate of pressure

excitation potential and shear excitation potential. The time profile of excitation potential is

calculated from the time rate of excitation potential. Out-of-plane displacement was calculated

numerically on the top surface of the plate in a 6mm thick 304-stainless steel plate. A very short

peak time (3 s ) of time rate of excitation potential was considered during the case study.

Parameter studies were performed to evaluate: (a) the effect of the pressure and shear potentials;

(b) the effect of the thickness-wise location of the excitation potential sources varying from mid-

plane to the top surface (source depth effect); (c) the effect of higher propagating modes and (d)

the effect of propagating distance away from the source.

1.1 STATE OF THE ART

The basic concepts and equations of elastodynamics have been explained by several authors [6],

[8]-[14]. Elastic waves emission from a damage process, a common phenomenon in earthquake

seismology. As a result, major references can be found from earthquake seismology. Aki and

Richards [14] have presented a comprehensive study on elastic wave fields due to seismic sources

in “Quantitative Seismology” book. A generalized theory of solution for the elastodynamic Green

function due to point dislocation source in a homogeneous isotropic bounded medium is discussed

in that book. But the solution of green function was not obtained in an elastic plate. Vvendensyaya

[15] developed a system of forces, which is equivalent to the rupture accompanied by slipping in

the theory of dislocation of Nabarro [16]. Another notable contribution is the development of the

force equivalent of dynamic elastic dislocation to the earthquake mechanism by Maruyama [17].

5

Lamb [18] presented propagation of vibrations over the surface of semi-infinite elastic solids. The

vibrations were assumed due to an arbitrary application of force at a point. Rice [19] discussed a

theory of elastic wave emission from damage (e.g., slip and micro cracking). A general

representation of the displacement field of an AE event was presented in terms of damage process

in the source region. The solution is obtained in an unbounded medium.

Miklowitz [20], Weaver and Pao [21] presented the response of transient loads of an infinite elastic

plate, using double integral transforms. Based on the generalized theory of AE, the elastodynamic

solution due to internal crack or fault can be analyzed by suitable Green function solution [22].

This Green function solution was obtained for either infinite media or half space media. Later,

Ohtsu et al. [23] characterized the source of AE on the basis of that generalized theory. AE

waveforms due to instantaneous formation of a dislocation were presented in that paper. In an

inverse study, deconvolution analysis was done to characterize the AE source. Johnson [24]

provided a complete solution to the three-dimensional Lamb’s problem, the problem of

determining the elastic disturbance resulting from a point source in a half space. Frank Roth [25]

extended the theory of dislocations to model deformations on the surface of a layered half-space

to calculate deformation inside the medium. Lamé constants for each layer were chosen

independently of each other. However, a theoretical representation of wave propagation in a plate

due to body force is not presented.

Wave propagation in an elastic plate is well known from Lamb [6] classical work. Achenbach [8],

Giurgiutiu [9], Graff [10] and Viktorov [11] considered Lamb waves, which exists in an elastic

plate with traction free boundaries. Achenbach [8] presented Lamb wave in an isotropic elastic

layer generated by a time harmonic internal/surface point load or line load. Displacements are

obtained directly as summations over symmetric and antisymmetric modes of wave propagation.

Elastodynamic reciprocity is used in order to obtain the coefficients of the wave mode expansion.

Bai et al. [26] presented three-dimensional steady state Green functions for Lamb wave in a layered

isotropic plate. The elastodynamic response of a layered isotropic plate to a source point load

having an arbitrary direction was studied in that paper. A semi-analytical finite element (SAFE)

method was used to formulate the governing equations. Using this method, in-plane displacements

were accommodated by means of an analytical double integral Fourier transform, while, the anti-

plane displacement approximated by using finite elements. They have used same modal

summation technique of eigenvectors as described by Liu and Achenbach [27], Achenbach [8].

Jacobs et al. [28] presented an analytical methodology by incorporating a time dependent acoustic

emission signal as a source model to represent an actual crack propagation and arrest event. A

review paper on the signal analysis used in acoustic emission (AE) for materials research field

published by Ono [29]. This paper reviewed recent progress in methods of signal analysis used in

acoustic emission such as deformation, fracture, phase transformation, coating, film, friction, wear,

corrosion and stress corrosion for materials research. A novel experimental mechanics technique

using scanning electron microscopy (SEM) in conjunction with AE monitoring is discussed by

Wisner et al. [30]. The objective is to investigate microstructure sensitive mechanical behavior and

damage of metals, in order to validate AE related information. An acoustic emission-based

technique maybe used to predict the residual fatigue life ceramic-matrix-composite at high

temperature [31]. Transducer characterization may be necessary for high tempratute AE

applications [32]. Cuadra et al. [33] proposed a 3D computational model to quantify the energy

associated with AE source by energy balance and energy flux approach. The time profile of

available energy as AE source was obtained for the first increment of crack. Khalifa et al. [34]

6

proposed a formulation for modeling the AE sources and for the propagation of guided or Rayleigh

waves. They used an exact analytical solution from a fracture-mechanics based model to obtain

the crack opening displacement. Green’s functions for Rayleigh wave are calculated using

reciprocity considerations without the use of integral transform techniques.

There are numerous publications based on finite element analysis for detecting AE signals. For

example, Hamsted et al. [35] reported wave propagation due to buried monopole and dipole

sources with finite element technique. Hill et al. [36] compared waveforms captured by AE

transducer for a step force on the plate surface by finite element modeling. Hamstad [37] presented

frequencies and amplitudes of AE signals on a plate as a function of source rise time. In that paper,

an exponential increase in peak amplitude with source rise time was reported. Sause et al. [38]

presented a finite element approach for modeling of acoustic emission sources and signal

propagation in hybrid multi-layered plates. They presented various parameter studies using

different layup configurations of multi-layered hybrid plates and validated the results to

analytically calculated dispersion curve results.

1.2 SCOPE OF THIS RESEARCH

There are numerous publications to characterize the source from a crack growth or damage.

Various deformation sources in solids, such as single forces, step force, point force, dipoles and

moments can be represented as AE sources [23], [39]-[41]. A moment tensor analysis to acoustic

emission (AE) has been studied to elucidate crack types and orientations of AE sources [42]. AE

source characteristics are unknown and the detected AE signals depend on the types of AE source,

propagation media, and the sensor response. Therefore extracting AE source feature from a

recorded AE waveform is always challenging. In this paper, we proposed a new technique to

predict the AE waveform using excitation potentials. Excitation potentials can be traced to the

energy released during an incremental crack propagation. The main contribution of this paper is to

simulate AE elastic waves due to energy released during crack propagation by Helmholtz potential

approach, for the first time to the authors’ best knowledge. The time profile of available energy as

AE source from a crack can be obtained analytically or from a 3D computational model [33].

Our analytical model is developed based on integral transform using Helmholtz potentials. The

solution is obtained through direct and inverse Fourier transforms and application of the residue

theorem. The resulting solution is a series expansion containing the superposition of all the Lamb

waves modes and bulk waves existing for the particular frequency-thickness combination under

consideration. It is worth mentioning that there are other methods that can be used to solve wave

equation of a structure, for example, Green’s functions for forced loading, semi analytical method;

normal mode expansion method etc. Green’s functions for forced loading can be solved through

an integral formulation relying on the elastodynamic reciprocity principle or integral transform.

However, the Green function for forced loading problem requires the knowledge of the point load

source that generated the AE event. Extracting information about the point load source that

generated an actual AE event recorded in practice is quite challenging. Our approach bypassed this

difficulty because it only needs and estimation of the energy released from the tip of the crack

during a crack propagation increment. Our proposed method novelty lies on calculating AE

waveform using energy released from the tip of the crack. Semi analytical method requires

extensive computational effort and has convergence issue at high frequency. Integral transform

provides an exact solution, and easy to solve numerically.

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1.3 DESCRIPTION OF EXCITATION POTENTIALS AS STRAIN ENERGY RELEASED FROM A CRACK

When a crack grows into a solid, a region of material adjacent to the free surfaces is unloaded, and

it releases strain energy. The strain energy is the energy that must be supplied to a crack tip for it

to grow and it must be balanced by the amount of energy dissipated due to the formation of new

surfaces and other dissipative processes such as plasticity. Fracture mechanics allows calculating

strain energy released during propagation of a crack. Figure 1a shows a plate with an existing crack

length of 2a and Figure 1b shows the typical energy release rate with crack half-length [43]. The

amount of energy released from a crack, is to be calculated as follow [43]

2 2( )

f

S

a tU

E

(1)

Figure 1 (a)-(d) plate with existing crack length 2a, energy release rate with crack length, time rate of energy from a crack, and total released energy from a crack.

Therefore, the amount of energy release from crack depends on: applied stress level ( f ), initial

crack half length ( a ), incremental crack length ( da ) and modulus of elasticity ( E ). For small

8

increment of crack length da (Figure 1a) the incremental strain energy is to be calculated by

following formula

22 ( )f

S

a tdU da

E

(2)

A particular example could be 1 mm thick plate under 200 MPa stress with existing 2 mm crack

length, if crack grows by 1μm from both ends, then the total amount of energy released from the

crack tip is 1.4 J . This incremental strain energy can be released at different time rate at different

time depending on crack growth and types of material. A real AE source releases energy during a

finite time period. If time rate of energy released is known (Figure 1c), and integration of that with

respect to time gives the time profile of energy (Figure 1d) released from a crack. Total released

energy from a crack can be decomposed to shear excitation potential and pressure excitation

potential. The proceeding section presents a theoretical formulation for Lamb wave solution using

time profile of excitation potentials: a Helmholtz potential technique.

2 FORMATION OF PRESSURE AND SHEAR POTENTIALS

Navier-Lame equations in vector form for Cartesian coordinates is given as

2( ) u u u (3)

where, ˆˆ ˆx y zu u i u j u k , with i , j , k being unit vectors in the x , y , z directions respectively,

, are Lamé constants, is the density.

If the body force is present, then the Navier-Lame equations are to be written as follow

2( ) u u f u (4)

Helmholtz decomposition states that [7], any vector can be resolved into the sum of an irrotational

(curl-free) vector field and a solenoidal vector field, where an irrotational vector field has a scalar

potential and a solenoidal vector field has a vector potential. Potentials are useful and convenient

in several wave theory derivations [9], [10].

Assume that the displacement u can be expressed in terms of two potential functions, a scalar

potential and a vector potential H ,

Φ+ Φu grad curl H H (5)

Where

x y zH H i H j H k (6)

Equation (5) is known as the Helmholtz equation, and is complemented by the uniqueness

condition, i.e.,

0H (7)

Introducing additional scalar and vector potentials Aand B for body force f [10], [14]

+ Af grad A curl B B (8)

9

Here,

x y zB B i B j B k (9)

The uniqueness condition is

0B (10)

Using the Equations (5) and (8) into Equation (4), wave equations for potentials are to be found

[Appendix A]

2 2

Pc A (11)

2 2

Sc H B H (12)

Here, 2 2

;Pc

2

Sc

The expanded form of Equations (11) and (12) are given in Appendix A. Equations (11) and (12)

are the wave equations for the scalar potential and the vector potential respectively. Equation (11)

indicates that the scalar potential, , propagates with the pressure wave speed, pc due to

excitation potential A , whereas Equation (12) indicates that the vector potential, H , propagates

with the shear wave speed, sc due to excitation potential B .

The excitation potentials are indeed a true description of energy released from a crack. The unit of

the excitation potentials are J/kg . Inhomogeneous wave equations for potentials are well known

in electrodynamics and electromagnetic fields. Wolski [44], Jackson [45] and Uman et al. [46]

treated inhomogeneous wave equation for scalar and vector potential with the presence of current

and charge density. The source can be treated as localized event and Green function can be used

as the solution of inhomogeneous wave equations. In this paper, wave equations for scalar

potential and vector potential H need to be solved with the presence of known source potentials

Aand B .

The derivation Lamb wave equation will be provided by solving wave Equations (11) and (12) for

the potentials. In this derivation we will consider generation of straight crested lamb wave due to

time harmonic excitation potentials. The assumption of straight-crested waves makes the problem

z-invariant. For P+SV waves (Lamb wave) [Appendix A: Equations (A8)-(A19)], the relevant

potentials are , , ,z zH A B . Equations (11) and (12) are condensed into two equations, i.e.,

2 2

Pc A (13)

2 2

S z z zc H B H (14)

Upon rearranging

2

2 2

1 1

P P

Ac c

(15)

10

2

2 2

1 1z z z

S S

H B Hc c

(16)

The P waves and SV waves give rise to the Lamb waves, which consist of a pattern of standing

waves in the thickness direction (Lamb wave modes) behaving like traveling waves in the x

direction in a plate of thickness h=2d (Figure 2). Excitation potentials are considered as z invariant

at 0; 0x y . For time harmonic source, equations for excitation potentials are

* 2 ( ) ( ) i t

PA c A x y e (17)

* 2 ( ) ( ) i tz s zB c B x y e (18)

Here A and zB are the source amplitude.

Figure 2 Plate of thickness 2d in which straight crested Lamb waves (P+SV) propagate in the x direction due to concentrated potentials at the origin.

Since the potentials are the time harmonic excitation, the response potentials , zH , have the

harmonic response, i.e.,

( , , ) ( , )

( , , ) ( , )z z

i t

i t

x y t x y

H x y t H x y

e

e

(19)

Then, Equations (15) and (16) become

2 2 2

2 2 2( ) ( )

P

A x yx y c

(20)

2 2 2

2 2 2( ) ( )z z

z z

S

H HH B x y

x y c

(21)

Equations (20) and (21) must be solved subject to zero-stress boundary conditions at the free top

and bottom surfaces of the plate, i.e.,

0, 0yy xyy d y d

(22)

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3 SOLUTION IN TERMS OF DISPLACEMENT

Taking Fourier transform of Equations (20) and (21), in x direction, ( , ) ( , )x y y and

( , ) ( , )H x y H y

2

2

2( , ) ( , ) ( , ) ( )

p

y y y A yc

(23)

2

2

2( , ) ( , ) ( , ) ( )z z z z

S

H y H y H y B yc

(24)

Here, ( ) ( )[ ( ) ]i xA y A y x e dx

and ( ) ( )[ ( ) ]i x

z zB y B y x e dx

Upon rearranging

2( , ) ( , ) ( )py y A y (25)

2( , ) ( , ) ( )z s z zH y H y B y (26)

Here,

22 2

2

22 2

2

p

p

s

s

c

c

(27)

Equations (25) and (26) are the second order ordinary differential equation (ODE) in y direction.

The total solution of Equations (25) and (26) consists of representation of two solutions [47]:

(a) The complementary solution for the homogeneous equation

(b) A particular solution of that satisfies source effect

Solution can be assumed as

0 1( , ) ( , ) ( , )y y y (28)

0 1( , ) ( , ) ( , )z z zH y i H y H y (29)

The functions 0 and 0zH satisfy the corresponding homogeneous differential equations

2

0 0( , ) ( , ) 0py y (30)

2

0 0( , ) ( , ) 0z s zH y H y (31)

Equations (30) and (31) give the complementary solutions, i.e.,

12

0 1 2( , ) C sin C cosp py y y (32)

0 1 2( , ) D sin D cosz s sH y y (33)

The functions 1( , )y and 1( , )zH y from Equation (29) satisfy the corresponding

inhomogeneous equations

2

11 1( , ) ( , ) ( ) ( )py y A y y (34)

2

1 1( , ) ( , ) ( ) ( )z s z zH y H y B y y (35)

Here, 1( , )y and 1( , )zH y are the Fourier transform of the free field depth dependent function.

The solution of Equation (34) is obtained using Fourier transform in y direction

2 2

1 1( , ) ( , )p A (36)

Here, ( ) i yA A y e dy

2 2

( , )p

A

(37)

The solution of Equation (35) is obtained in a similar way

2 2

1 1( , ) ( , )z s z zH H B

1 2 2( , ) z

z

s

BH

(38)

In order to get the solution in y domain, taking inverse Fourier transform of Equations (37) and

(38), i.e.,

1 2 2

1( , )

2

i y

p

Ay e d

(39)

1 2 2

1( , )

2

i yzz

s

BH y e d

(40)

The evaluation of the integral in Equations (39) and (40) can be done by the residue theorem using

a close contour [48]-[52] [Appendix B]. Integration formula for Equations (39) and (40) are

1 2 2

1( , ) sin

2 2

i y

p

p p

A Ay e d y

(41)

1 2 2

1( , ) sin

2 2

i yz zz s

s s

B BH e d y

(42)

13

Equations (41), (42) represent the particular solution. The total solution is superposition of

complementary solution Equations (32), (33) and particular solution Equations (41), (42), i.e.,

1 2( , ) C sin C cos sin2

p p p

p

Ay y y y

(43)

1 2( , ) D sin D cos sin

2

zz s s s

s

BH y i y y y

(44)

The coefficients 1 2 1 2C , C , D , D of the complimentary solutions Equations (32), (33) are to be

determined by using the Fourier transformed boundary conditions i.e.,

0, 0yy xyy d y d

(45)

After applying the boundary conditions [Appendix A: Equations (A20)-(A29)] we get

2 2

1 2

1 2

( ) C sin C cos sin2

2 D cos D sin cos2

0

s p p p

p

zs s s s s s

s

Ad d d

Bdd d

(46)

2 2

1 2

1 2

( ) C sin C cos sin2

2 D cos D sin cos2

0

s p p p

p

z

s s s s s s

s

Ad d d

Bdd d

(47)

1 2

2 2

1 2

2 C cos C sin cos2

( ) D sin D cos sin 02

p p p p p p

p

s s s

zs

s

Ad d d

d dB

d

(48)

1 2

2 2

1 2

2 C cos C sin cos2

( ) D sin D cos sin 02

p p p p p p

p

s s s

zs

s

Ad d d

d dB

d

(49)

Equations (46)-(49) are a set of four equations with four unknowns. The equations can be separated

into a couple of two equations with two unknowns, one for symmetric motion and one for anti-

symmetric motion.

14

3.1 SYMMETRIC LAMB WAVE SOLUTION

Addition of the Equations (46) and (47), and subtraction of the Equations (48) from(49), yields

2 2 2 2

2 2

2

1

( ) 2 ( )

2 ( )

cos cos C sin cos2

sin sin D0

s s

s

p s s p z s

pp p sd

Ad d d B d

d

(50)

Equation (50) represents an algebraic system that can be solved for 2C , 1D provided the system

determinant does not vanish

2 2

2 2

( )cos 2 cos0

2 sin ( )sin

s p s s

p p s s

s

d d

dD

Let,

2 2

( ) sin cos2

sS p z s

p

PA

d B d

(51)

SP is the source term for the symmetric solution which contains source potentials A and ZB .

Then,

2 2

2 2

2

1

( ) 2

2 ( )

sin cos

sin cos 0C

D

s

s

s

s s s S

p p pd

D

d d P

d

(52)

2 2

2

1

( )sin1

2 sin

C

D

S s s

S p ps

P d

P dD

(53)

Here,

2 2

2 2 2 2

2 2

( )cos 2 cos( ) cos sin 4 cos sin

2 sin ( )sin

s p s s

s p s s p s p

p p s s

s

d dd d d d

dD

(54)

By equating to zero the ( )sD term of Equation (54), one gets the symmetric Rayleigh -Lamb

equation.

3.2 ANTI-SYMMETRIC LAMB WAVE SOLUTION

Subtraction of the Equations (46) from(47), and addition of the Equations (48) and (49), yields

2 2

2 2 2 2

1

2

( ) 2

2 ( ) ( )

0sin sin C

cos cos D cos sin2

s

s s

p s s

zp p s p s

s

d dB

d A d d

(55)

15

Equation (55) represents an algebraic system that can be solved for 1C , 2D provided the system

determinant does not vanish

2 2

2 2

( )sin 2 sin0

2 cos ( )cos

s p s s

p p s s

A

d d

dD

Let,

2 2cos ( ) sin

2

zA p s s

s

BP A d d

(56)

AP is the source term for the anti-symmetric solution which contains source potentials A and ZB

.

Then,

2 2

2 2

1

2

( ) 2

2 ( )

cos sin 0

cos sinC

D

s

s

A

s s s

p p p Ad

D

d d

d P

(57)

2 2

1

2

2 sin1

( )sin

C

D

A s s

A s pA

P d

P dD

(58)

Here,

2 2

2 2 2 2

2 2

( )sin 2 sin( ) sin cos 4 sin cos

2 cos ( )cos

s p s s

s p s s p s p

p p s s

A

d dd d d d

dD

(59)

By equating to zero the ( )AD term of Equation (59), one gets the anti-symmetric Rayleigh -

Lamb equation.

3.3 DISPLACEMENT SOLUTION IN THE WAVENUMBER DOMAIN

Equation (A14) [Appendix A] gives the out-of-plane displacement yu , in terms of potentials, i.e.,

z

y

Hu

y x

(60)

In ‘x’ direction wavenumber domain the displacement yu becomes

y zu i Hy

(61)

By substituting of Equations (43), (44) into Equation (61) yields the expressions for out-of-plane

displacement in the wavenumber domain in terms of coefficients 1 2 1 2C , C , D , D which are

functions of the wavenumber ξ, i.e.,

16

1 2 1 2C cos C sin D sin D cos cos sin2 2

zy p p p p s s p s

s

BAu y y y y d d

(62)

Evaluation of Equation (62) at the plate top surface, y y d

u

:

1 2 1 2C cos C sin D sin D cos cos sin2 2

zy p p p p s s p s

s

BAu d d d d d d

(63)

The complete solution of displacement is the superposition of the symmetric, antisymmetric and

bulk wave solution. By using Equations (53) and (58) into Equation (63) and obtain

cos sin2 2

S A zy S A p s

S A s

N N BAu P P d d

D D

(64)

Where,

2 2 2

2 2

2 sin sin ( ) sin sin

2 sin cos ( )sin cos

s p p s s p p s

A s p s p s p s

N d d d d

N d d d d

(65)

3.4 DISPLACEMENT SOLUTION IN THE PHYSICAL DOMAIN

Displacement solution in the physical domain is obtained by taking inverse Fourier transform of

Equation (64), i.e.,

1

cos sin2 2 2

i xS A zy S A p s

S A s

N N BAu P P d d e d

D D

(66)

Let,

1 2 3yu I I I (67)

Where,

1

1

2

i xS AS A

S A

N NI P P e d

D D

(68)

2

1cos

2 2

i x

p

AI d e d

(69)

3

1sin

2 2

i xzs

s

BI d e d

(70)

The integration of Equations (68), (69), (70) are presented in Appendix C. After integrating

Equations (68), (69), (70), the displacement yu becomes

17

1 11 1

2 2 2 2 2 2 2 22 22 21 0

0 0

Y ( ) J ( )4 2

(2

( )( )( ) ( )

( ) ( )

As AS j jj

j

y

z

P p S S

z

AS jji xAi xjSS A

j jS AS Aj js j A

u

BAd x d x d x x d x d

c c c c

iB

NNi P e P e

D D

1 1

2 2 2 22 21) ( ) Y ( )( )

S S

x x d x dc c

(71)

After rearranging Equation (71) and reinstating explicitly the harmonic time dependency, we get,

1 11 1

2 2 2 2 2 2 2 22 22 21 0

( )( )

0 0

Y ( ) J ( )4 2

( )( )( ) ( )

( ) ( )

As AS j jj

j

y

i t i tz

P P S S

AS jji x tAi x tjSS A

j jS AS Aj js j A

u

BAd x d x d e x x d x d e

c c c c

i

NNi P e P e

D D

1 1

2 2 2 22 21( ) ( ) Y ( )( )

2

i tz

S S

Bx x d x d e

c c

(72)

Assume that,

y y L y A y Bu u u u

Where,

( )( )

0 0

( )( )( ) ( )

( ) ( )

As AS j jj

j

y L

AS jji x tAi x tjSS A

j jS AS Aj js j A

uNN

i P e P eD D

(73)

1

2

1 12 2 2 22 2

1J4

i t

y A

p p

Au x d x d d x e

c c

(74)

1 1

2 2 2 22 20

1 1

2 2 2 22 21

( ) J ( )2

( ) ( ) Y ( )( )2

i tzy B

S S

i tz

S S

Bu x x d x d e

c c

iBx x d x d e

c c

(75)

y Lu is the out-of-plane displacement containing Lamb wave mode and ,y A y Bu u are the out-of-plane

displacements of bulk wave for excitation potentials A and zB respectively.

Theoretical derivation reveals that, bulk wave is present in addition to Lamb wave. The Lamb

waves, which consist of a pattern of standing waves in the thickness direction (Lamb wave modes)

18

behaving like traveling waves in the x direction in a plate. Guided waves can propagate long

distances, yields an easy inspection of a wide variety of structures. Whereas, bulk waves generate

from a source and directly reach to the AE sensor without interfering the boundaries. Bulk waves

are categorized into longitudinal (pressure) and transverse (shear) waves. The displacements y Au

and y Bu are the longitudinal and transverse bulk waves respectively.

3.5 EFFECT OF SOURCE DEPTH

If the source is present at different location other than mid-plane i.e., 0y y (Figure 3) then

equations for excitation potentials become

* 2

0( ) ( ) i tpA c A x y y e (76)

* 2

0( ) ( ) i tz s zB c B x y y e (77)

Recall Equations (25) and (26) for new location of excitation potentials

2

0( , ) ( , ) ( )py y A y y (78)

2

0( , ) ( , ) ( )z s z zH y H y B y y (79)

Complementary solution of Equation (78) and (79) will be the same as before. Only particular

solution will change due to the source in right hand side.

Figure 3 Plate of thickness 2d in which straight crested Lamb waves (P+SV) propagate

in the x direction due to concentrated potentials at 0y y

Fourier shift transform states that,

0

0{ ( )} { ( )}i y

F f y y e F f y

(80)

By using Fourier shift transform theorem Equation (80), the particular solutions of Equations (78)

, (79) become

0

1 2 2

1( , )

2

i yi y

p

Aey e d

(81)

19

0

1 2 2

1( , )

2

i yi yz

z

s

B eH y e d

(82)

Upon rearranging,

0( )

1 2 2

1( , )

2

i y y

p

Ay e d

(83)

0( )

1 2 2

1( , )

2

i y yzz

s

BH y e d

(84)

Integration of Equations (83) and (84) can be done by using residue theorem. The detailed solution

is already presented before from Equations (39) to (42). For brevity, only the final results of

Equations (83), (84) are presented here. From residue theorem the integrands become,

0( )

1 02 2

1( , ) sin

2 2

i y y

p

p p

A Ay e d y y

(85)

Similarly,

0( )

1 02 2

1( , ) sin

2 2

i y yz zz s

s s

B BH e d y y

(86)

The corresponding source terms, Equations (51) and (56) for symmetric and antisymmetric

solutions will be

2 2

1 1( ) sin cos2

S s p z s

p

AP d B d

(87)

2 2

1 1cos ( ) sin2

zA p s s

s

BP A d d

(88)

Here, 1 0d d y

The corresponding displacement Equations (73), (74), (75) become

( )( )

0 0

( )( )( ) ( )

( ) ( )

As AS j jj

j

y L

AS jji x tAi x tjSS A

j jS AS Aj js j A

uNN

i P e P eD D

(89)

1 1

2 2 2 22 21 1 1Y

4 p p

i tc cy A

Au x d x d e

(90)

1 1

2 2 2 22 21 0 1

1 1

2 2 2 22 21 1 1

( ) J ( )2

( ) ( ) Y ( )( )2

s s

s s

i tzc cy B

i tzc c

Bu x x d x d e

iBx x d x d e

(91)

20

3.6 AE GUIDED WAVE PROPAGATION

AE guided waves will be generated by the AE event. AE guided waves will propagate through the

structure according to the structural transfer function. The out-of-plane displacement of the guided

waves can be captured by conventional AE transducer installed on the surface of the structure as

shown in Figure 4.

Figure 4 Acoustic emission propagation and detection by a sensor installed on a structure

For numerical analysis, a 304-stainless steel plate with 6 mm thickness was chosen as a case study.

The signal was received at 500 mm distance from the source. The excitation source was located at

the mid-plane of the plate. A very short peak time of 3 s was used for this case study. Later, the

effects of source depth and propagation distance were included into the study. A detailed

methodology of the numerical analysis is described in Figure 5. In numerical analysis, direct

Fourier transforms of the time domain excitation signal was performed to get the frequency

contents of the signal and an inverse Fourier transform was done to get the time domain output

signal from frequency response of out-of-plane displacement.

21

Figure 5 Methodology for guided wave analysis

22

4 NUMERICAL STUDIES

This section includes guided wave simulation in a plate due to excitation potentials.

4.1 TIME DEPENDENT EXCITATION POTENTIALS

The time-dependent excitation potentials depend on time-dependent energy released from a crack.

A real AE source releases energy during a finite time period. At the beginning, the rate of energy

released from a crack increases sharply with time and reaches a maximum peak value within very

short time; then decreases asymptotically toward the steady state value, usually zero. Time rate of

energy released from a crack can be modeled as a Gaussian pulse. Figure 6a and Figure 7a show

the time rate of pressure potential and shear potential. The corresponding equations are,

2

2

2

2*

2

0

*2

0

t

zZ

tA

A t et

BB t e

t

(92)

Here 0A and 0ZB are the scaling factors. Time profile of the potentials are to be evaluated by

integrating Equation (92), i.e.,

2

2

2

2

0

0

( ) 2

( ) 2

t

t

z z

tA A erf te

tB B erf te

(93)

Figure 6 (a) Time rate of pressure potential *A t and (b) Time profile of pressure

potential ( *A )

23

Figure 7 (a) Time rate of shear potential *

zB t and (b) Time profile of shear potential

*( )zB

Time profile of potential is shown in Figure 6b and Figure 7b. Time profile of AE source may

follow cosine bell function [37], [38] or error function. In this research, a Gaussian pulse is used

to model the growth of the excitation potentials during the AE event; as a result, the actual

excitation potential follows the error function variation in the time domain. Cuadra et al. [33]

obtained similar time profile of AE energy from a crack by using 3D computation method.

The key characteristics of the excitation potentials are:

peak time: time required to reach the time rate of potential to maximum value

rise time: time required to reach the potential to 98% of maximum value or steady state

value

peak value: maximum value of time rate of potential

maximum potential: maximum value of time profile of potential

Amplitude of the source A and zB for unit width is,

2

2

2

2

02

02

( ) 2

( ) 2

t

p

t

z z

s

h tA A erf tec

h tB B erf tec

(94)

Here, plate thickness 2h d .

24

4.2 PHASE VELOCITY AND GROUP VELOCITY DISPERSION CURVES

Solution of the Rayleigh-Lamb equation for symmetric and antisymmetric modes (Equations (54)

and (59)) yields the wave number and hence the phase velocity for each given frequency. Multiple

solutions exist, hence multiple Lamb modes also exist. Differentiation of phase velocity with

respect to frequency yields the group velocity. The phase velocity dispersion curves and group

velocity dispersion curves are shown in Figure 8a and Figure 8b respectively. The existence of

certain Lamb mode depends on the plate thickness and frequency. The fundamental S0 and A0

modes will always exist. The phase velocity (Figure 8a) is associated with the phase difference

between the vibrations observed at two different points during the passage of the wave. The phase

velocity is used to calculate the wave length of each mode.

(a) (b)

Figure 8 Lamb wave dispersion curves: (a) phase velocity and (b) group velocity of symmetric and anti-symmetric Lamb wave modes.

The fundamental wave mode of symmetric (S0) and anti-symmetric (A0) is considered in this

analysis. The signal is received at some distance away from the point of excitation potentials, only

the modes with real-valued wavenumbers are included in the simulation. The corresponding modes

for imaginary and complex-valued wavenumbers are ignored because they propagate with

decaying amplitude; at sufficiently far from the excitation point their amplitudes are negligible.

Since different Lamb wave modes traveling with different wave speeds exist simultaneously, the

excitation potentials will generate S0 and A0 wave packets. The group velocity (Figure 8b) of

Lamb waves is important when examining the traveling of Lamb wave packets. These wave

packets will travel independently through the plate and will arrive at different times. Due to the

multi-mode character of guided Lamb wave propagation, the received signal has at least two

separate wave packets, S0 and A0 All wave modes propagate independently in the structure. The

final waveform will be the superposition of all the propagating waves and will have the

contribution from each Lamb wave mode.

4.3 EXAMPLE OF AE GUIDED WAVE PROPAGATION IN A 6MM PLATE

A test case example is presented in this section to show how AE guided waves propagate over a

certain distance using excitation potentials. A 304-stainless steel plate with 6 mm thickness was

chosen for this purpose. The signal was received at a distance 500 mm away from the source.

Excitation sources were located at mid-plane of the plate ( 1 3 mmd ). Time profile excitation

potentials (Figure 6b and Figure 7b) were used to simulate AE waves in the plate. Excitation

25

potential can be calculated from the time rate of potential released during crack propagation. Peak

time, rise time, peak value and maximum potential of excitation potentials are 3 s , 6.6 s , 0.28

W/kg and 1 J/kg . Based on this information a numerical study on AE Lamb wave propagation

was conducted. Peak time or rise time is one of the major characteristics of AE source. Several

researchers investigate potential effect of rise time on AE signal by varying the source rise time

in-between 0.1 s to 15 s [37]- [39], [55].

(a) Pressure potential excitation (b) Shear potential excitation

Figure 9 Lamb waves (S0 and A0 mode) and bulk waves propagation at 500 mm distance in 6 mm 304-steel plate for (a) pressure potential excitation (b) shear potential excitation (peak time = 3 s ) located at mid plane.

In this research, each excitation potential was considered separately to simulate the Lamb waves.

Total released energy from a crack can be decomposed to pressure and shear excitation potentials.

This research presents the individual effect of the unit amplitude of potentials (1 J/kg ) on AE

signals. However, distributing the energy between the pressure and shear potentials for real AE

26

source is very important and we recommend it for future study for specific damage assumptions

(e.g., a crack propagating under Mode I fracture may have a different mix of shear/pressure

potentials than a crack propagating under Mode II or Mode III fracture modes; similarly for mixed-

mode fracture).

Figure 9a shows the Lamb wave (S0 and A0 mode) and bulk wave propagation using pressure

excitation potential. Signals are normalized by their individual peak amplitudes, i.e., amplitude/

peak amplitude. It can be inferred from the figures that both A0 and S0 are dispersive. Bulk wave

shows nondispersive behavior, but the peak amplitude is not significant compared to peak S0 and

A0 amplitude. Figure 9b shows the Lamb wave (S0 and A0 mode) and bulk wave propagation

using shear potential only. Peak amplitude of A0 is higher than peak S0 amplitude while using

shear potential only (Figure 9b). By comparing Figure 9a and Figure 9b the important observation

can be made as, pressure excitation potential has more contribution to the peak S0 amplitude over

shear excitation potentials whereas, shear excitation potential has more contribution to the peak

A0 amplitude. The notable characteristic is that, peak amplitude of A0 is higher than the peak S0

amplitude for both excitation potentials. It should be noted that, the peak amplitude of the bulk

wave is much smaller than the peak A0 and S0 amplitudes for both pressure and shear excitation

potentials located at the mid-plane. But their contribution to the S0 and A0 peak amplitude might

change due to change in source depth and propagating distance. The following sections discuss

the effect of source depth, higher propagating mode and propagating distance in AE guided wave

propagation using excitation potentials.

4.4 EFFECT OF SOURCE DEPTH

To study the effect of source depth, a 6 mm thick plate was chosen. The signal was received at 500

mm distance from the source. The excitation sources were located at the top surface, 1.5 mm from

the top surface, and 3 mm from the top surface (mid plane). Figure 9, Figure 10, Figure 11 show

the out-of-plane displacement (S0 and A0 mode, bulk wave) vs time at 500 mm propagation

distance in 6mm thick plate for mid-plane, top surface and 1.5 mm depth from top surface source

location. The summary of the results are listed in Table 1. Signals are normalized by their

individual peak amplitudes, i.e., amplitude/ peak amplitude. From Figure 10 it is observed that,

the A0 mode appears only while using pressure potential on the top surface whereas, S0 mode

appears only while using shear potential. Therefore, pressure potential contributes to the peak A0

amplitude and shear potential contributes to the peak S0 amplitude in the final waveform. Another

notable characteristic is that, the pressure potential contributes to the high amplitude of the trailing

edge (low frequency component) of A0 wave packet for top-surface source. With increasing source

depth, the effect of high amplitude of low frequency decreases (Figure 9a and Figure 11a) in A0

signal. By comparing the amplitude scales of Figure 9 and Figure 11, the increase in peak A0

amplitude and the decrease in peak S0 amplitude with increasing source depth while using pressure

excitation potential only.

Figure 9b, Figure 10b and Figure 11b show some qualitative and quantitative changes in spectrum

while using shear excitation potentials only. The amplitude of S0 signal decreases and A0 signal

increases with increasing source depth while using shear excitation potential only (Figure 11b and

Figure 9b). Qualitative change in S0 signal with source depth refers to the change in frequency

content of the signal. It should be noted that, peak amplitude of S0 and A0 signal using shear

excitation potential located at 1.5 mm depth from the top surface is more significant compared to

pressure potential located at same location (Figure 11). However, with increasing source depth, S0

27

signals amplitude become more significant due to pressure excitation potential over shear

excitation potential (Figure 9 and Figure 11). For all AE source location, the shear potential part

of the AE source has more contribution to the peak A0 amplitude than pressure potential. Peak

amplitude of bulk waves increases with increasing source depth (Figure 9 and Figure 11).

However, the amplitude of bulk wave is much smaller than the peak A0 and S0 amplitudes.

Therefore, peak amplitude of bulk wave may not appear in real AE signal.

(a) Pressure potential excitation (b) Shear potential excitation

Figure 10 Lamb waves (S0 and A0 mode) and bulk waves propagation at 500 mm distance in 6 mm 304-steel plate for (a) pressure potential excitation (b) shear potential excitation (peak time = 3 s ) located on top surface.

28

(a) Pressure potential excitation (b) Shear potential excitation

Figure 11 Lamb waves (S0 and A0 mode) and bulk waves propagation at 500 mm distance in 6 mm 304-steel plate for (a) pressure potential excitation (b) shear potential excitation (peak time = 3 s ) located at 1.5 mm depth from top

surface.

Table 1 Comparison of peak amplitudes of Lamb waves (S0 and A0 mode) and bulk waves for different source depth

Source depth

Peak S0 amplitude

(m)

Peak A0 amplitude

(m)

Peak bulk wave amplitude

(m)

Pressure Shear Pressure Shear Pressure Shear

Top surface

source - 5.0e-15 7.3e-15 - - -

1.5 mm from

top surface 7.5e-16 3.8e-15 7.4e-15 1.07e-14 7.14e-19 2.37e-17

Mid-plane

source 2.5e-15 1.9e-15 1.6e-14 2.5e-14 1.43e-18 2.4e-17

29

4.5 EFFECT OF HIGHER ORDER LAMB WAVE MODE

Since AE signals have a wideband response; higher order modes should appear in addition to the

fundamental S0 and A0 Lamb wave modes. This section discusses the effect of higher order modes

on AE signal due to shear potential excitation and pressure potential excitation. A real AE sensor

does not have ultimate high frequency response; therefore, the signals were filtered at 10 kHz and

700 kHz frequency. Figure 12 shows the higher order Lamb wave (S1 and A1) modes at 500 mm

distance in 6 mm 304-steel plate. The A1 signal is more dispersive for the shear excitation potential

than the pressure excitation potential. The peak amplitude of the A1 mode is higher than the peak

amplitude of the S1 mode for both shear and pressure potentials. The peak S1 and A1 amplitudes

for pressure potential are higher than the shear potential. By comparing Figure 9 and Figure 12 it

can be inferred that, peak amplitude of fundamental Lamb wave mode (S0 and A0) is higher than

peak S1 and A1 amplitude. Therefore, peak amplitude of higher order mode may not be significant

in real AE signal.

(a) Pressure potential excitation (b) Shear potential excitation

Figure 12 Higher order Lamb waves (S1 and A1) modes at 500 mm distance in 6 mm 304-steel plate for (a) pressure potential excitation (b) shear potential excitation (peak time = 3 s ) located at mid-plane.

4.6 EFFECT OF PROPAGATION DISTANCE

The attenuation of the peak amplitude of the signal as a function of propagating distance was also

determined. Figure 13 and Figure 14 show the normalized peak amplitude of out-of-displacement

(S0, A0, bulk wave) against propagation distance from 100 mm to 500 mm for pressure potential,

shear potential respectively. The amplitudes are normalized by their individual peak amplitudes.

30

A 6 mm thick 304-steel plate with both excitation potentials are considered. Potentials are located

at mid-plane with peak time of 3 s .

Figure 13 Effect of pressure excitation potential: variation of out-of-plane displacement (S0, A0 and bulk wave) with propagation distance in 6 mm 304- steel plate for source (peak time = 3 s ) located at the mid-plane.

Figure 14 Effect of shear excitation potential: variation of out-of-plane displacement (S0, A0 and bulk wave) with propagation distance in a 6-mm 304-steel plate for source (peak time = 3 s ) located at the mid-plane.

0

0.2

0.4

0.6

0.8

1

100 200 300 400 500

Ou

t o

f p

lan

e d

isp

lace

men

t

(Norm

aliz

ed p

eak a

mpli

tude)

Propagation distance (mm)

S0

A0

Bulk wave

0

0.2

0.4

0.6

0.8

1

100 200 300 400 500

Out

of

pla

ne

dis

pla

cem

ent

(Norm

aliz

ed p

eak a

mpli

tude)

Propagation distance (mm)

S0

A0

Bulk wave

31

Figure 13 and Figure 14 show significant attenuation of S0, A0, and bulk wave signal with

propagating distance 100 mm to 500 mm. However, larger attenuation of peak bulk wave

amplitude is observed compared to peak S0 and A0 amplitudes. Therefore, far from the source

bulk wave becomes less significant and may not be captured by the AE transducer. Another

important observation is that, peak A0 amplitude attenuates more than peak S0 amplitude while

using pressure potential only whereas, peak S0 amplitude attenuates more than peak A0 amplitude

while using shear potential only. The attenuation of the peak amplitude is expected due to

dispersion of the signal.

5 SUMMARY, CONCLUSIONS, AND FUTURE WORK

5.1 SUMMARY

The guided waves generated by an acoustic emission (AE) event were analyzed through a

Helmholtz potential approach. The inhomogeneous elastodynamic Navier-Lame equation was

expressed as a system of wave equations in terms of unknown scalar and vector solution potentials,

, H , and known scalar and vector excitation potentials, *A , *B . The excitation potentials *A , *B were traced to the energy released during an incremental crack propagation.

The solution was readily obtained through direct and inverse Fourier transforms and application

of the residue theorem. The resulting solution took the form of a series expansion containing the

superposition of all the Lamb waves modes and bulk waves existing for the particular frequency-

thickness combination under consideration.

A numerical study of the AE guided wave propagation in a 6 mm thick 304-steel plate was

conducted in order to predict the out-of-plane displacement that would be recorded by an AE

sensor placed on the plate surface at some distance away from the source. Parameter studies were

performed to evaluate: (a) the effect of the pressure and shear potentials; (b) the effect of the

thickness-wise location of the excitation potential sources varying from mid-plane to the top

surface (source depth effect); (c) the effect of higher propagating mode; (d) the effect of

propagating distance away from the source.

5.2 CONCLUSIONS

The present work has shown that pressure and shear source potentials can be used to model the

guided wave generation and propagation due to an AE event associated with incremental crack

growth. The advantage of this approach is to decouple the inhomogeneous Navier-Lame equations.

The numerical studies performed over a range of parameters have shown that:

i. Both A0 and S0 modes generated by an AE excitation in a 6-mm steel plate are

dispersive. The high frequency components of the AE excitation are responsible for the

dispersive part of the S0 mode, whereas the low frequency components of the AE

excitation are responsible for the dispersive part of the A0 mode.

ii. The peak amplitude of A0 mode is higher than the peak amplitude of the S0 mode for

all cases.

iii. The amplitude of bulk wave is much smaller than peak A0 and S0 amplitude.

Therefore, peak amplitude of bulk waves may not be significant in real AE signal.

iv. For mid-plane AE source location, the shear potential part of the AE source has more

contribution to the peak A0 amplitude than pressure potential whereas the pressure

32

potential part of the AE source has more contribution to the peak S0 amplitude than

shear potential.

v. With an increase in the source depth, the peak A0 and S0 amplitude increases by

considering pressure potential only, whereas, the peak A0 increases and S0 decreases

by using shear potential only.

vi. If the AE source is located at top surface, the effect of the excitation pressure and shear

potentials on the S0 and A0 modes seems to be decoupled: pressure potential does not

contribute to S0 and bulk wave amplitude whereas shear potential does not contribute

to the A0 and bulk wave amplitude.

vii. For top-surface AE source, pressure excitation potential has contribution to the high

amplitude low-frequency component of the A0 wave packet. This contribution

decreases as the source depth increases.

5.3 FUTURE WORK

Substantial future work is still needed to verify the hypotheses and substantiate the calculation of

the AE source potentials that produce the guided wave excitation. The extensive experimental AE

monitoring data existing in the literature should be explored to find actual physical signals that

could be compared with numerical predictions in order to extract factual data about the amplitude

and time-evolution of the AE source potentials. A frequency analysis of time domain signal should

be done to analyze the frequency content of the captured AE signals. Frequency content may help

to distinguish different source types and source location. If necessary, additional experiments with

wider band AE sensors should be conducted. An inverse algorithm could then be developed to

characterize the AE source during crack propagation. The source characterization can provide

information about amount of energy released from the crack. Therefore, it may help to generate a

qualitative as well quantitative description of the crack propagation phenomenon. A further

extensive study on the effect of plate thickness, AE rise time, and AE source depth would be

recommended.

Straight crested Lamb wave (1-D wave guided propagation) was considered in this paper.

However, realistic AE guided wave propagation usually takes place events in 2D geometries

(pressure vessels, containment vessels, etc.). Hence, the present theory should be extended to

circular crested Lamb waves (2-D wave guided propagation). The contribution of pressure and

shear excitation potentials to the total energy release from a crack needs to be studied through

types of materials and crack types.

ACKNOWLEDGEMENT

The authors are grateful for the financial support from US Department of Energy (DOE), Office

of Nuclear Energy, under grant numbers DE-NE 0000726 and DE-NE 0008400.

33

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36

APPENDIX A

WAVE EQUATIONS FOR POTENTIALS

Recall the inhomogeneous Navier-Lame equations in vector form

2( ) u u f u (A1)

Upon substitution of Equations (5), (8) into Equation (A1), Equation (A1) becomes,

2 2 2( ) ( ) ( ) ( ) ( )H A B H (A2)

Upon rearrangement of Equation (A2)

2 2[( 2 )( ) - ] ( ) 0A H B H (A3)

For Equation (A3) to hold at any place and any time, the components in parentheses must be

independently zero, i.e.,

2( 2 )( ) - 0A (A4)

2 0H B H (A5)

Upon division by and rearrangement,

2 2

Pc A (A6)

2 2

Sc H B H (A7)

Here, 2 2

;Pc

2

Sc

Equation (A6) and (A7) are the wave equations for the scalar potential and the vector potential

respectively.

The derivation Lamb wave equation will be provided by solving wave Equations (A6) and (A7)

for the potentials. In this derivation, we will consider generation of straight crested Lamb wave

due to time harmonic excitation potentials. The assumption of straight-crested waves makes the

problem z-invariant.

Upon expansion of Equation (4), i.e.,

37

22 2 2 2 22

2 2 2 2 2

2 2 2 2 22 2

2 2 2 2 2

222

2

( )

( )

( )

yx x x x xzx

y y y y yx zy

yxz

uu u u u uuf

x y x zt x x y z

u u u u uu uf

x y y zt y x y z

uuu

x z z yt

2 2 2 2

2 2 2 2z z z z

z

u u u uf

z x y z

(A8)

Under z invariant condition 0z

, Equation (A8) becomes

22 2 2 2

2 2 2 2

2 2 2 22

2 2 2 2

2 2 2

2 2 2

( )

( )

yx x x xx

y y y yxy

z z zz

uu u u uf

x yt x x y

u u u uuf

x yt y x y

u u uf

t x y

(A9)

Corresponding displacement and force equations from Equations (5) and (8), under z invariant

condition can be written as

zx

zy

y xz

Hu

x y

Hu

y x

H Hu

x y

(A10)

zx

zy

y xz

BAf

x y

BAf

y x

B Bf

x y

(A11)

Corresponding potentials Equations (11) and (12) become

2 2

pc A (A12)

38

2 2

2 2

2 2

S x x x

S y y y

S z z z

c H B H

c H B H

c H B H

(A13)

Equation (A10) implies that, it is possible to separate the solution into two parts:

(1) Solution for xu and yu , with excitation force xf and yf , which depend on the potentials

, , ,z zH BA . The solution will be the combination of pressure wave P represented by

potential and a shear vertical wave SV represented by potential zH due to excitation

potentials , zBA respectively.

(2) Solution for zu due to excitation force zf which depend on the potentials , , ,x y x yH H B B .

The solution for zu displacement will give shear horizontal wave SH represented by

potential ,x yH H due to excitation potentials ,x yB B respectively.

For P+SV waves in a plate,

zx

zy

Hu

x y

Hu

y x

(A14)

**

**

zx

zy

BAf

x y

BAf

y x

(A15)

, ( , )x y zu u f H (A16)

, ( , )x y zf f f A B (A17)

For P+SV waves (Lamb wave), the relevant potentials are , , ,z zH A B . Equations (A12) and

(A13) are condensed into two equations, i.e.,

2 2

Pc A (A18)

2 2

S z z zc H B H (A19)

39

APPLYING BOUNDARY CONDITIONS

Stress equations in terms of displacements for z-invariant motions are

2 ;yx

yy

yxxy

uu

x y

uu

y x

(A20)

Using Equation (A14) into Equation (A20) gives the stress expression in terms of potentials, i.e.,

22 2

2 2

2 22

2 2

2 2 ;

2

zyy

z zxy

H

x yx y

H H

x y x y

(A21)

Apply the x-domain Fourier transform to Equation (A21) and obtain

2

2

2

22

2

2 2 ;

2

zyy

zxy z

Hi

yy

Hi H

y y

(A22)

The differentiation properties of Equations (43), (44) are

1 2

22 2 2 2

1 22

( , )C cos C sin cos

2

( , )C sin C cos ( ) sin ( , )

2

p p p p p p

p

p p p p p p p

p

y Ay y y

y

y Ay y y y

y

(A23)

1 2

22 2 2 2

1 22

( , ) D cos D sin cos2

( , )D sin D cos ( ) sin ( , )

2

z zs s s s s s

s

z zs s s s s s s z

s

H By i y y y

y

H y Bi y y y H y

y

(A24)

Using Equations (43), (44) and differentiation property Equations (A23), (A24) into Equation

(A22)i.e.,

40

2 2

1 2

1 2

1 2

2 2

1 2

( ) C sin C cos sin2

2 D cos D sin cos2

2 C cos C sin cos2

( ) D sin D cos sin2

yy

s p p p

p

z

s s s s s s

s

xy

p p p p p p

p

s s s

zs

s

Ay y y

By y y

Ay y y

i

y yB

y

(A25)

Using Equation (A25) into the boundary conditions, Equation (A25) yields

0yy y d

2 2

1 2

1 2

( ) C sin C cos sin2

2 D cos D sin cos2

0

s p p p

p

z

s s s s s s

s

Ad d d

Bdd d

(A26)

0yy y d

2 2

1 2

1 2

( ) C sin C cos sin2

2 D cos D sin cos2

0

s p p p

p

z

s s s s s s

s

Ad d d

Bdd d

(A27)

0xy y d

1 2

2 2

1 2

2 C cos C sin cos2

( ) D sin D cos sin 02

p p p p p p

p

s s s

zs

s

Ad d d

d dB

d

(A28)

41

0xy y d

1 2

2 2

1 2

2 C cos C sin cos2

( ) D sin D cos sin 02

p p p p p p

p

s s s

zs

s

Ad d d

d dB

d

(A29)

42

APPENDIX B

INVERSE FOURIER TRANSFORM IN Y- DIRECTION

The evaluation of the integral in Equation (39) is to be done by the residue theorem using a close

contour. Since the poles are on the real axis, the integral is to be evaluated as

2 2

real axis

ARes ( ( ); )i y

j

p

e d i f

(B1)

Res ( ( )) ( ) ( ))j jf f (B2)

Here j are the poles of on the real axis. Poles of for Equation (B1) are

p (B3)

To evaluate the integration of Equation (39) two cases are considered as follow,

Case 1: 0y

1 2 2

1( , )

2

i y

p

Ay e d

(B4)

In order to evaluate the contour integration, an arc at the infinity on the upper half plane (Figure

B1) is added because the function converges in upper half circle.

According to residue theorem the integrant is

2 2

real axis

ARes ( ( ); )i y

j

p

e d i f

(B5)

Here j are the poles of 2 2

A( ) i y

p

f e

43

Figure B1 Close contour for evaluating the inverse Fourier transform by residue theorem for 0y

Using the poles, residue theorem gives,

2 2

A( ) ( )

( )( ) ( )( )

i y i y i y

p p

p p p p pp p

A Ae d i e i e

(B6)

Upon rearrangement

2 2

A

2 2

p pi y i y

i y

p p

A e ee d

i i

(B7)

Using Euler formula in Equation(B7), i.e.,

2 2

Asini y

p

p p

Ae d y

(B8)

Case 2: 0y

For 0y ; y y

1 2 2

1( , )

2

i y

p

Ay e d

(B9)

This time an arc at the infinity on the upper half plane cannot be added because the function does

not converge in upper half circle. To evaluate the contour integration of above function an arc at

the infinity on the lower half plane (Figure B2) can be added because the function converges in

lower half circle.

44

Figure B2 Close contour for evaluating the inverse Fourier transform by residue theorem for 0y

2 2

real axis

ARes ( ( ); )

i y

j

p

e d i f

(B10)

Using the poles, residue theorem gives

2 2

A( )

( )( )

( )( )( )

i y i y

p

p p p

i y

p

p p

p

p

Ae d i e

Ai e

(B11)

Upon rearrangement

2 2 2 2

p py y

y

i ii

p p

A A e ee d

i i

(B12)

Using Euler formula in Equation (B12), i.e.,

2 2

sini y

p

p p

A Ae d y

(B13)

By observing both cases the following summary can be made:

For 0y

2 2

sini y

p

p p

A Ae d y

(B14)

For 0y

45

2 2

sini y

p

p p

A Ae d y

(B15)

Therefore, whether, y is positive or negative, the contour integral result is same. Integration

formula for Equation (39) becomes

1 2 2

1( , ) sin

2 2

i y

p

p p

A Ay e d y

(B16)

Similarly from Equation (40), the evaluation of the integral is as follow

1 2 2

1( , ) sin

2 2

i yz zz s

s s

B BH e d y

(B17)

46

APPENDIX C

INVERSE FOURIER TRANSFORM IN X- DIRECTION

The integrant in Equation (68) is singular at the roots of SD and AD , i.e.,

0SD (C1)

0AD (C2)

Equations (C1) and (C2) are the Rayleigh-Lamb equations for symmetric and anti-symmetric

modes.

The evaluation of integral in Equation (68) can be done by residue theorem using a close contour

as shown in Figure C1. The positive roots correspond to forward propagating waves. To satisfy

the radiation boundary condition at x , that is no incoming waves from infinity, the negative

real poles are avoided in the contour ( 0)x .

The integration of Equation (66) can be written as the sum of the residues,

1

( ) ( )1( ) ( )

2 ( ) ( )

12 Res( )

2k

i x i xS AS A

S A

k

N NI P e d P e d

D D

i

(C3)

For the poles k , Equation (C3) becomes

1 Res( )k

kI i

(C4)

Figure C1 Close contour for evaluating the inverse Fourier transform by residue theorem for (a) 0x (b) 0x

47

From residue theorem, if ( )

( )( )

N zf z

D z and z a is a simple pole then

( )Res( ) , where ( )

( )z a

N a dDa D a

D a dz

. Applying residue theorem to obtain the integration of

Equation (C3) we get

1

0 0

( )( )( ) ( )

( ) ( )

As AS j jj

j

AS jji xAi xjSS A

j jS AS Aj js j A

INN

i P e P eD D

(C5)

The summation of Equation (C5) is taken over the symmetric and anti-symmetric positive real

wavenumbers j

S and j

A . At given frequency , there are 0,1,2,...., Sj j symmetric Lamb

wave modes and 0,1,2,...., Aj j anti-symmetric Lamb wave modes.

Using Euler formula in Equation (69) and obtain

2

1cos cos sin

2 2p

AI d x i x d

(C6)

By using even and odd function property of the integrant, Equation (C6) becomes

2

0

cos cos2

p

AI d xd

(C7)

2

2

2

0

cos cos2

p

AI d c xd

(C8)

Apply integration by parts to Equation (C8)

2 2cos ; cos ; p

duu d c v x uvd u vd vd d

d

(C9)

After integrating by parts using Equation (C9), Equation (C8) becomes

22

22

20

sin

sin2

p

p

d cA d

I xdx

c

(C10)

The integration of Equation (C10) is to be evaluated with the aid of ref [53], pp 35 (23) for

12 and using 1

2

2( )J sin( )x x i.e.,

1 1

2 2 2 22 22 1Y

4p p

AI c d x d c x d

(C11)

Here 1Y is the Bessel function of second kind of order 1.

Upon rearrangement of Equation (70)

48

2 2

32 2

1sin

2 2

i xzs

s

BI c d e d

c

(C12)

The integration of Equation (C12) is to be evaluated by contour integration with brunch cut. The

poles are sc . The closed path for contour integral is shown in Figure C2 [51]. The

corresponding arguments are listed in Table 1.

Table 2 Arguments for inverse Fourier transform

Line 2 2

sc

DE

+ i

2 2

sc

EF

+

2 2

sc

GH

+ 2 2

sc

HI

+ i

2 2

sc

Figure C2 Close contour for evaluating the inverse Fourier transform for 0x

After applying the arguments as shown in Table 1 the Equation (C12) becomes

49

2 2

32 2

2 22 2/

2 22 20 0

2 2/

2 20

1sin

2 2

sin sin1 1cos

sinsin

s

s

i xzs

s

c

z s z sx

s s

c

z s

s

BI c d e d

c

B d c B d ce d x d

c c

B d cix d

c

(C13)

Equation (C13) turns into three integrals. Let assume that

3 31 32 33I I I I (C14)

Where,

2 2

312 2

0

sin s xz

s

d cBI e d

c

(C15)

2 2/

322 2

0

sincos

sc

sz

s

d cBI x d

c

(C16)

2 2/

332 2

0

sinsin

sc

sz

s

d ciBI x d

c

(C17)

Equation (C15) accepts imaginary wave number. The imaginary wavenumber correspond to non-

propagating evanescent waves and will not be considered here.

The integrant in Equation (C16) can be evaluated by with the aid of ref [53], [54], i.e.,

1 1

2 2 2 22 232 0( ) J ( )

2

z

S S

BI x x d x d

c c

(C18)

Here 0J is the Bessel function of first kind of order 0.

Equation (C17) is to be determined from ref [53], pp. 35 (23), by putting 12 and using the

identity of Bessel function, 12

J ( ) 2 sinx x x i.e.,

1 1

2 2 2 22 233 1( ) ( ) Y ( )

2

z

S S

iBI x x d x d

c c

(C19)

Using Equations (67), (C5), (C11), (C18), (C19) the total displacement becomes

50

1 12 2 2 22 2

1

1 1

2 2 2 22 20

0 0

Y4

( ) J ( )2

( )( )( ) ( )

( ) ( )

As AS j jj

j

y

P P

z

S S

AS jji xAi xjSS A

j jS AS Aj js j A

u

Ad x d x d

c c

Bx x d x d

c c

i

NNi P e P e

D D

1 1

2 2 2 22 21( ) ( ) Y ( )( )

2

z

S S

Bx x d x d

c c

(C20)