nation taiwan ocean university department of harbor and river june 11, 2015 pp.1 null-field integral...

Nation Taiwan Ocean University

Department of Harbor and River

April 18, 2023 pp.1

Null-field Integral Equation Approach for Solving Stress Concentration Problems with Circular Boundaries

研究生 : 陳柏源指導教授 : 陳正宗博士日期 : 2005/06/29 15:00-16:20

國立台灣海洋大學河海工程學系結構組碩士班論文口試

Mechanics Sound Vibration Laboratory HRE. NTOUhttp://ind.ntou.edu.tw/~msvlab/ April 18, 2023 pp.2

Outlines

Review the researches of three theses Present method

Expansions of fundamental solution and boundary densityAdaptive observer systemVector decomposition techniqueLinear algebraic equationImage technique for solving scattering problems of half-plane

Numerical examplesLaplace problemHelmholtz problem

Green’s function for annular Laplace problemsConclusions


Outlines






Review the researches of three theses

MSVLABMSVLAB

Kao Jeng-HongKao Jeng-Hong Wu An-ChienWu An-Chien Chen Po-YuanChen Po-Yuan

Regularized meshleRegularized meshless methodss method

Null-field integral Null-field integral approachapproach

Laplace equation1. Anti-plane shear

problems2. Anti-plane

piezoelectricity problemsHelmholtz equation1. Acoustic eigenproblem --interior problem

Laplace equation1. Anti-plane

piezoelectricity and in-plane electrostatic problems

2. Anti-plane elasticity problems

Laplace equation1. Torsion problems2. Bending problemsHelmholtz equation1. Stress concentration factor

of cavity problems2. Surface amplitude of cavity

or inclusion problems --exterior problem


Comparison with Chen and Wu

Similar part Different part

Chen 1. Formulation

2. Cavity and/or inclusion problem

1. Laplace and Helmholtz problem

2. Half-plane problem

3. Application of problem

Wu 1. Only Laplace problem

2. Full-plane problem3. Application of problem


Organization of the thesis

ThesisThesis

Engineering problems

Laplace problems

Helmholtz problems

Green’s function

Chapter 5Derivation of the Green’sfunction for annularLaplace problems

Chapter 41. Half-plane problems with a

cavity subject to the incident SH-wave

2. Half-plane problems with inclusions subject to the incident SH-wave

Chapter 31. Torsion problem

with circular holes2. Bending problem

with circular holes

Semi-analytical approachSemi-analytical approach Analytical approachAnalytical approach

Null-field approachNull-field approach


Goal

To develop a systematic approach approach in conjunction with Fourier series, degenerate kernels and adaptive observer system for solving Laplace and Helmholtz problems with multiple circular boundaries.

Advantages ：Mesh free.Well-posed model Free of CPV and HPV.Elimination of boundary-layer effectExponential convergence


Outlines






Present approach

(x) (s, x) (s) (s)B

K dBj f=ò

Degenerate kernel Fundamental solution

CPV and HPV

No principal value?

0

(x) (s)(x) (s) (s)jB

jja dBbj f

¥

=

= åò

0

0

(s,x) (s) (x), x s

(s,x)

(s,x) (x) (s), x s

ij j

j

ej j

j

K a b

K

K a b

¥

=

¥

=

ìïï = <ïïïï=íïï = >ïïïïî

å

å


Present approach

W

x BÎ WÈ

x c BÎ W È

cW

cW

W

Exterior problem

Interior problem

2 (x) (s,x) (s) (s) (s,x) (s) (s)i i

B Bu T u dB U t dBp = -ò ò

0 (s,x) (s) (s) (s,x) (s) (s)B

e

B

eT u dB U t dB= -ò ò

x BÎ WÈ

x c BÎ W È

2 (x) (s,x) (s) (s) (s,x) (s) (s)e e


0 (s,x) (s) (s) (s,x) (s) (s)B

i

B

iT u dB U t dB= -ò ò

Advantages of degenerate kernelAdvantages of degenerate kernel

1.1. No principal valueNo principal value

2.2. Elimination of boundary-layer effectElimination of boundary-layer effect

3.3. Well-posedWell-posed


Outlines






Expansions of fundamental solution (2D)

Laplace problem--

Helmholtz problem--

(s,x) ln x s lnU r= - =

1

1

1( , ; , ) ln ( ) cos ( ),

(s, x)1

( , ; , ) ln ( ) cos ( ),

i m

m

e m

m

U R R m Rm R

UR

U R m Rm

rq r f q f r

q r f r q f rr

¥

=

¥

=

ìïï = - - ³ïïïï=íïï = - - >ïïïïî

å

åR

eUO

iU

q f

s

r

xx

(1)

0

(1)

0

( , ; , ) ( ) ( )cos( ( )), 2

(s,x)

( , ; , ) ( ) ( )cos( ( )), > 2

im m

m

em m

m

iU R J k H kR m R

Ui

U R H k J kR m R

(1)0( , ) ( ) 2U s x i H krp=-


Limiting process of the Helmholtz problem

Fundamental solution

Degenerate kernel

[ ]

(1)0( , ) ( ) 2

(2 / ) ln 2

ln ln

U s x i H kr

kr

k r

p

p p

=-

=-

= +

0(1)0 ( ) (2 / ) lnziH z zp®- ¾¾¾®

1

1

1( , ; , ) ln ln ( ) cos ( ),

(s,x)1

( , ; , ) ln ln ( ) cos ( ),

i m

m

e m

m

U R k R m Rm R

UR

U R k m Rm

rq r f q f r

q r f r q f rr

¥

=

¥

=

ìïï = + - - ³ïïïï=íïï = + - - >ïïïïî

å

å

( 0)k ®

Rigid body term


1

1

(s,x) (x) (s), s x

(s,x)

(s,x) (s) (x), x s

ii i

i

ei i

i

U a b

U

U a b

¥

=

¥

=

ìïï = ³ïïïï=íïï = >ïïïïî

å

å

1

1

(s,x) (x) , s x

(s,x)

(s,x) (x), x

(s)

( ) ss

ii

i

ei i

i

iT a

T

T b

b

a

¥

=

¥

=

ìïï = ³ïïïï¢

¢=íïï = >ïïïïî

å

å

( )1

x s

(s(s,x) (s,x) (x) (x)

(s), ( )

) (s)

s

i ei i

i

i i

i iT T a b

W

b

b

a

a

¥

=

Þ

¢ ¢-

¾ ¾®

-=

¾

å

sn

U(s,x)U(s,x) T(s,x)T(s,x)

0

0

( , ) ( ) cos( ( )),( ) ( )

(

> 2

( , )

( , ) ( ) ( ) cos( ( )), ) > 2

im m

m

em m

m

m

m

mm

kiT s x J k m RY kR iJ kR

J kR

T s xki

T s x Y k iJ k m R


Jump behavior between domain with complementary domain

2

0

2 2

(s,x) (s,x) cos

cos cos

cos

2 cos , x

i e

n n n n n n

m m m m

T T n Rd

kR J kR Y kR iJ kR n kR J kR Y kR iJ kR n

kR Y kR J kR Y kR J kR n

n B

2,m m m m m mW J kR Y kR Y kR J kR Y kR J kR

kR

x BÎ WÈ

x c BÎ W È

cW

W

2 (x) (s,x) (s) (s) (s,x) (s) (s)i i


0 (s,x) (s) (s) (s,x) (s) (s)B

e

B

eT u dB U t dB= -ò ò

2

0

2 2

(s,x) (s,x) cos

cos cos

0, x

i e

n n n n n n

U U n Rd

R J kR Y kR iJ kR n R J kR Y kR iJ kR n

B


Outlines






Image technique for solving half-plane problem

Alluvial

h a

SH-Wave

Matrix

-4 -2 2 4

-1

-0.5

0.5

1

-4 -2 2 4

-1

-0.5

0.5

1

Matrix

SH-Wave

SH-Wave

Free surface


Take free body

,M I M M I It tw w t t

I

I

w

t

Inclusion

Matrix

SH-Wave

SH-Wave

i r

i r

w w

t t

Matrix

SH-Wave

SH-Wave

Matrix

M M i rt

M M i rt

w w w w

t t t t


Linear algebraic system

Matrix field

Inclusion field

Two constrains

{ } { }M M i r M M i rt tU t t = T u u+ +é ù é ù- -ê ú ê úë û ë û

{ } { }I I I IU t = T ué ù é ùê ú ê úë û ë û

{ } { }M Itu = u

{ } { }M M I Itμ t = - μ té ù é ùê ú ê úë û ë û

M M M i rt

I I MtI

M I I

T -U 0 0 u u(x)

0 0 T -U t 0=

I 0 -I 0 u 0

0 μ 0 μ t 0

+é ùì ü ì üï ï ï ïï ï ï ïê úï ï ï ïê úï ï ï ïï ï ï ïï ï ï ïê úí ý í ýê úï ï ï ïï ï ï ïê úï ï ï ïê úï ï ï ïï ï ï ïê úï ï ï ïë ûî þ î þ

{ }i r

i r M M

i r

uu(x) T -U

t

++

+

ì üï ïï ï= í ýï ïï ïî þ


Flowchart of present method

0 [ (s, x) (s) (s, x) (s)] (s)B

T u U t dB= -òDegenerate kernel Fourier series

Collocation point and matching B.C.

Adaptive observer system

Linear algebraic equation

Fourier coefficientsPotential of

domain pointSurface

amplitude

Stress field

Vector decomposition

Numerical

Analytical


Outlines

Review the researches of three theses Mathematical formulation



Green’s function for annular Laplace problems Conclusions


Numerical examples

Laplace problem

Torsion problem for a bar

Bending problem for a cantilever beam Helmholtz problem

Half-plane problems with a cavity subject to the incident SH-wave

Half-plane problems with inclusions subject to

the incident SH-wave


Torsion problem for a bar--CMES, Vol. 12(2)

R

a

b2 k

N

p

qa

a


Torsional rigidity (one hole)

Exact solution [68] 0.97872 0.95137 0.90312 0.82473 0.76168 0.74454 0.72446 0.69968 0.66555

Present method

L=20 0.97872 0.95137 0.90312 0.82473 0.76168 0.74455 0.72451 0.69991 0.66705

L=10 0.97872 0.95137 0.90312 0.82476 0.76244 0.74603 0.72748 0.70616 0.68111

Caulk’s method(BIE) [1

4]

40 divisions 0.97872 0.95137 0.90316 0.82497 0.76252 0.74569 0.72605 0.70178 0.66732

20 divisions 0.97873 0.95140 0.90328 0.82574 0.76583 0.75057 0.73367 0.71473 0.69321

( )2 2

1 nk

N

kA B

k

Gx y dA dB

jj

m =

¶= + -

¶åò òTorsional rigidity:Torsional rigidity:


Torsional rigidity (equal angle)

Caulk (First-order Approximate) [14]

0.8661 0.8224 0.7934

Caulk (BIE formulation) [14]

0.8657 0.8214 0.7893

Present method (L=10)

0.8657 0.8214 0.7893


Torsional rigidity (Ling’s problem)

Caulk (First-order Approximate) [14]

0.8739 0.8741 0.7261

Caulk (BIE formulation) [14]

0.8713 0.8732 0.7261

Ling’s results 0.8809 0.8093 0.7305

Present method (L=10)

0.8712 0.8732 0.7244

Because there is no apparent reason for the unusually large difference in the second example, Ling’s rather lengthy calculations are probably in error here. --ASME JAM

10%


Convergence

0 10 20 30 40

Fourier term s

0.724

0.725

0.726

0.727

0.728

0.729

tors

iona

l rig

idity


Bending problem for a cantilever beam

b

a

R

2Y 1Y

BC


Stress concentration

0 4 8 12 160.6

1.2

1.8

2.4

Sc

0 4 8 12 160.6

1.2

1.8

2.4S

c

0 4 8 12 160

0.4

0.8

1.2

1.6

Sc

Point B Point C

zx ASc

Q

2 21 11

2 1 2 2zxy

Qy x y

x I x

Stress concentrationStress concentration


Stress concentration at point B

0.4 0.5 0.6 0.7

b

1.5

2

2.5

3

3.5

Sc

Theta=3*pi/8

Theta=pi/8

Theta=pi/4

Present

Present

method

method

Naghdi’s result

Naghdi’s result

ss

0.4 0.5 0.6 0.7

a

1.5

2

2.5

3

3.5

Sc

0.4 0.5 0.6 0.7

a

1.5

2

2.5

3

3.5

Sc

Steele &

Steele &

Bird

Bird

The two approaches disagree by as much 11%. The grounds for this discrepancy have not yet been identified.

--ASME Applied Mechanics Review


Advantages of the present method

- 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1 0

Log|Y 1 |

0.8

1.2

1.6

2

2.4

Sc

P o in t B

0 4 8 12 16 20

Fourier term s(L)

2.42

2.422

2.424

2.426

2.428

2.43

Sc

Elimination of boundary-layer effectElimination of boundary-layer effect Convergence test of Fourier seriesConvergence test of Fourier series


Two holes problem

D

0 1 2 3 4 5

D /2a

2

3

4

5

6

7

8

9

Sc

Tw o ho les

O ne hole

Present methodPresent method

Steele & Bird’s result [6]Steele & Bird’s result [6]

Point P


Contour of stress concentration Steele & Bird’s result [6]Steele & Bird’s result [6] Present methodPresent method


Numerical examples

Laplace problem

Torsion problem for a bar

Bending problem for a cantilever beam Helmholtz problem

Stress concentration factor of cavities problem subject to the incident SH-wave

Half-plane problems with inclusions subject to the incident SH-wave


A full-plane problem with two cavities subject to the incident SH-wave.

SH-wave

2a

1a

SH-wave

1a

2a

D

Case 1Case 1 Case 2Case 2

yw

tm

¥¥ =

i ikxyw e

tm

¥

=

BC of the Honein’s problem

BC of the present problem


Shear stress ( ) around the smaller cavity

0 1 2 3 4 5 6

- 2

0

2

4

6

8

stre

ss

ang le=90(k=0.001)

D =0.01

D =0.1

D =2

0 1 2 3 4 5 6

- 4

- 2

0

2

4

6

8

stre

ss

ang le=45 (k=0.001)

D =0.01

D =0.1

D =2

Case 1Case 1 Case 2Case 2

k=0.001k=0.001 k=0.001k=0.001

Present methodPresent method Present methodPresent methodHonein’s results Honein’s results

z

tz

wqs m

¶=

¶Shear stressShear stress


A half-plane problem with a circular cavity subject to incident SH-wave.

h

SH-wave

a

x

y


Nondimensional stress ( ) around the cavity

*z

0

90

180

270

0 2 4

0

90

180

270

0 1 2 3 4

0

90

180

270

0 1 2 3 4

0180

0 1 2 3

Lin and Liu’s Lin and Liu’s results [89] results [89]

Present Present method method

0 , / 1.5, 0.1h a ka 0 , / 12, 0.1h a ka 45 , / 1.5, 0.1h a ka 45 , / 12, 0.1h a ka

*

0

zz

qq

ss

s=nondimensional stress nondimensional stress


Limiting case of the half-plane problem

0

30

60

90

120

150

180

210

240

270

300

330

0 1 2 3 4 5

Pao and Mow’s result [70] (only half).

Limiting case of and

90

/ 100h a

A full-plane problem A full-plane problem with a cavity subject to with a cavity subject to horizontally SH-wavehorizontally SH-wave.


A half-plane problem with a semi-circular alluvial valley subject to the SH-wave

AlluvialMatrix

h a

SH-Wave

x

y

kac

a

2 2( ) ( ) 0,k w x x

Wave number

Dimensionless frequency

Governing equation

Velocity of shear wave

c


Surface amplitudes of the alluvial valley problem

- 4 - 3 - 2 - 1 0 1 2 3 4

x/a

0

5

10

15

20

Am

plit

ud

e

0

5

10

15

20-4 -3 -2 -1 0 1 2 3 4

- 4 - 3 - 2 - 1 0 1 2 3 4

x/a

0

5

10

15

20

Am

plit

ud

e

0

5

10

15

20-4 -3 -2 -1 0 1 2 3 4

- 4 - 3 - 2 - 1 0 1 2 3 4

x/a

0

5

10

15

20

Am

plit

ud

e

0

5

10

15

20-4 -3 -2 -1 0 1 2 3 4

- 4 - 3 - 2 - 1 0 1 2 3 4

x/a

0

5

10

15

20

Am

plit

ud

e

0

5

10

15

20-4 -3 -2 -1 0 1 2 3 4


Manoogian’s results [60]Manoogian’s results [60]

2, / 1/ 6, / 2 /3I M I M

0 30 60 90

2 2Re ImAmplitude w w

vertical horizontal


Limiting case of a canyon


Manoogian’s results [60]Manoogian’s results [60]

8/ 102, , / 2 /3IM MI

0 30 60 90

- 4 - 3 - 2 - 1 0 1 2 3 4

x/a

0

1

2

3

4

5

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4

0

1

2

3

4

5

- 4 - 3 - 2 - 1 0 1 2 3 4

x/a

0

1

2

3

4

5A

mp

litu

de

- 4 - 3 - 2 - 1 0 1 2 3 4

0

1

2

3

4

5

- 4 - 3 - 2 - 1 0 1 2 3 4

x/a

0

1

2

3

4

5

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4

0

1

2

3

4

5

- 4 - 3 - 2 - 1 0 1 2 3 4

x/a

0

1

2

3

4

5

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4

0

1

2

3

4

5

vertical horizontal


Limiting case of a rigid alluvial valley

- 4 - 3 - 2 - 1 0 1 2 3 4x / a

0

2

4

6

8

1 0

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4x / a

0

2

4

6

8

1 0

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4x / a

0

2

4

6

8

1 0

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4x / a

0

2

4

6

8

1 0

Am

plit

ud

e

Present methodPresent method4/ 102, , / 2 /3I MI M

0 30

60 90



Soft-basin effect

/ 2 /3, 2I M

/ 3/ 6I Mc c / 2 / 6I Mc c / 12 / 6I Mc c

/x a /x a /x a

14 18

3


A half-plane problem with two alluvial valleys subject to the incident SH-wave

Canyon

Matrix

3aSH-Wave

房 [93] 將正弦和餘弦函數的正交特性使用錯誤，以至於推導出錯誤的聯立方程，求得錯誤的結果。

-- 亞太學報曹 2004


Limiting case of two canyons


Tsaur et al.’s results [103]Tsaur et al.’s results [103]

8/ 102, , / 2 /3IM MI

0 30 60 90

- 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7x / a

0

2

4

6

8

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7x / a

0

2

4

6

8

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7x / a

0

2

4

6

8

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7x / a

0

2

4

6

8

Am

plit

ud

e

vertical horizontal


Surface displacements of two alluvial valleys

Present methodPresent method 2, / 1/ 6, / 2 /3I M I M

0 30

60 90

- 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7x / a

0

4

8

1 2

1 6

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7x / a

0

4

8

1 2

1 6

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7x / a

0

4

8

1 2

1 6

Am

plit

ud

e

- 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7x / a

0

4

8

1 2

1 6

Am

plit

ud

e


Inclusion

Matrixh

SH-Wave

a

x

y

A half-plane problem with a circular inclusion subject to the incident SH-wave


Surface displacements of a inclusion problem under the ground surface

- 3 - 2 - 1 0 1 2 3

x/a

0

1

2

3

4

5

6

Am

plit

ud

e

- 3 - 2 - 1 0 1 2 3

0

1

2

3

4

5

6

- 3 - 2 - 1 0 1 2 3

x/a

0

1

2

3

4

5

6

Am

plit

ud

e

- 3 - 2 - 1 0 1 2 3

0

1

2

3

4

5

6

- 3 - 2 - 1 0 1 2 3

x/a

0

1

2

3

4

5

6

Am

plit

ud

e

- 3 - 2 - 1 0 1 2 3

0

1

2

3

4

5

6

- 3 - 2 - 1 0 1 2 3

x/a

0

1

2

3

4

5

6

Am

plit

ud

e

0

1

2

3

4

5

6-3 -2 -1 0 1 2 3

Present methodPresent method 2, / 1/ 6, / 2 /3I M I M

Tsaur et al.’s results [102]Tsaur et al.’s results [102]

Manoogian and Lee’s results [62]Manoogian and Lee’s results [62]

0 30 60 90

When I solved this problem I could find no published results for comparison. I also verified my results using the limiting cases. I did not have the benefit of published results for comparing the intermediate cases. I would note that due to precision limits in the Fortran compiler that I was using at the time.

--Private communication


Limiting case of a cavity problem


Lee and Manoogian’s [53] for the cavity case.Lee and Manoogian’s [53] for the cavity case.

8/ 102, , / 2 /3IM MI

0 30 60 90

- 3 - 2 - 1 0 1 2 3x/a

0

1

2

3

4

5

6

Am

plit

ud

e

0

1

2

3

4

5

6-3 -2 -1 0 1 2 3

- 3 - 2 - 1 0 1 2 3x/a

0

1

2

3

4

5

6A

mp

litu

de

0

1

2

3

4

5

6-3 -2 -1 0 1 2 3

- 3 - 2 - 1 0 1 2 3x/a

0

1

2

3

4

5

6

Am

plit

ud

e

0

1

2

3

4

5

6-3 -2 -1 0 1 2 3

- 3 - 2 - 1 0 1 2 3x/a

0

1

2

3

4

5

6

Am

plit

ud

e

0

1

2

3

4

5

6-3 -2 -1 0 1 2 3

vertical horizontal


Limiting case of a rigid alluvial valley

Present methodPresent method4/ 102, , / 2 /3I MI M

0 30

60 90

- 3 - 2 - 1 0 1 2 3

x/a

0

1

2

3

4

5

6

Am

plit

ud

e

0

1

2

3

4

5

6-3 -2 -1 0 1 2 3

- 3 - 2 - 1 0 1 2 3

x/a

0

1

2

3

4

5

6

Am

plit

ud

e

0

1

2

3

4

5

6-3 -2 -1 0 1 2 3

- 3 - 2 - 1 0 1 2 3

x/a

0

1

2

3

4

5

6

Am

plit

ud

e

0

1

2

3

4

5

6-3 -2 -1 0 1 2 3

- 3 - 2 - 1 0 1 2 3

x/a

0

1

2

3

4

5

6

Am

plitu

de

0

1

2

3

4

5

6-3 -2 -1 0 1 2 3


A half-plane problem with two circular inclusions subject to the SH-wave

Matrix

Inclusionh

SH-Wave

a a

D

y

x


Surface amplitudes of two-inclusions problem

- 6 - 4 - 2 0 2 4 6

x/a

1

2

3

4

5

Am

plit

ud

e

h /a =1 .5 , D /a =2 .5

h

h

h

h

- 6 - 4 - 2 0 2 4 6

x/a

0

1

2

3

4

5

Am

plit

ud

e

h /a =1 .5 , D /a =2 .5

h

h

h

h

- 6 - 4 - 2 0 2 4 6

x/a

0

1

2

3

4

5

Am

plit

ud

e

h /a=1 .5 , D /a =2 .5

h

h

h

h

- 6 - 4 - 2 0 2 4 6

x/a

0

1

2

3

4

5

Am

plit

ud

e

h /a=1 .5 , D /a =2 .5

h

h

h

h

Present methodPresent method / 1/ 6, / 2 /3, / 2.5I M I M D a

0.1 0.25

0.75 1.25


Limiting case of two-cavities problem


Jiang et al. result [95]Jiang et al. result [95]

8 ,/ / 2 /3,10 / 2.5IM MI D a

0.1 0.25 0.75 1.25

- 6 - 4 - 2 0 2 4 6

x/a

0

1

2

3

4

Am

plit

ud

e

h /a =1 .5 , D /a =2 .5

h 0 o

h 30 o

h 60 o

h 90 o

- 6 - 4 - 2 0 2 4 6

x/a

0

1

2

3

4

5

Am

plit

ud

e

h /a =1 .5 , D /a =2 .5

h 0 o

h 30 o

h 60 o

h 90 o

- 6 - 4 - 2 0 2 4 6

x/a

0

1

2

3

4

5

6

Am

plit

ud

e

h /a =1 .5 , D /a =2 .5

h 0 o

h 30 o

h 60 o

h 90 o

- 6 - 4 - 2 0 2 4 6

x/a

0

1

2

3

4

5

Am

plit

ud

e

h/a=1.5, D /a=2.5

h 0 o

h 30 o

h 60 o

h 90 o


Outlines






Derivation of the Green’s function for annular Laplace problems

2 (x, ) (x )G

0G 0G

1B2B

s

(s, )0 (s,x) (s, ) (s) (s,x) (s) ( ,x)

B B

GT G dB U dB U

n

xx x

¶= - +

¶ò ò

Degenerate kernelDegenerate kernel Fourier seriesFourier series

2 (x,s) 2 (x s)U

1B2B


Null-field integral equation

0 0

1

0 1 2 2 ln

1cos cos sin sin

m mm m

m m m mm

bp ap b

R Ra ab p a p m m b q a q m m

m b b b b

0 0

1

0 ln 2 ln 2 ln

1cos cos sin sin

m mm m

m m m mm

R b bp a a p

a a a ab p a p m m b q a q m m

m b R b R

( , )x b

( , )x a


Contour plots for the annular Green’s function

-10 -8 -6 -4 -2 0 2 4 6 8 10-10

-8

-6

-4

-2

0

2

4

6

8

10

-10 -8 -6 -4 -2 0 2 4 6 8 10-10

-8

-6

-4

-2

0

2

4

6

8

10

Analytical solutionAnalytical solution ( )50L Semi-analytical solutionSemi-analytical solution ( )50L

4, 10, (0,7.5)a b


Two limiting cases of the annular Green’s function

0a

b

- 2 - 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5 2 2 . 5 3

- 2

- 1 . 5

- 1

- 0 . 5

0

0 . 5

1

1 . 5

2

- 1 - 0 . 8 - 0 . 6 - 0 . 4 - 0 . 2 0 0 . 2 0 . 4 0 . 6 0 . 8 1- 1

- 0 . 8

- 0 . 6

- 0 . 4

- 0 . 2

0

0 . 2

0 . 4

0 . 6

0 . 8

1

Chen & Wu’s[31] resultsChen & Wu’s[31] results

Exterior case

Interior case


Two limiting cases of the annular Green’s function

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Limiting case of the annular Limiting case of the annular Green’s function Green’s function

71, 10 , 20, (1.25,0 )a b L

Limiting case of the annular Limiting case of the annular Green’s function Green’s function

0.001, 1, 20, (0.8,0 )a b L


Outlines






Conclusions

A systematic way to solve the Laplace and Helmholtz problems with circular boundaries was proposed successfully in this thesis by using the null-field integral equation in conjunction with degenerate kernels and Fourier series.

The present method is more general for calculating the torsion and bending problems with arbitrary number of holes and various radii and positions than other approach.


Conclusions

When the wave number “k” approaches zero, the Helmholtz problem can be reduced to the Laplace problem. Laplace problem can be treated as a special case of the Helmholtz problem.

Our approach can deal with the cavity problem as a limiting of inclusion problem with zero shear modulus.


Some findingsLaplace Helmholtz

Ling1947

Analytical solution

Bird & Steele1992

房營光 1995

Analytical solution

Lee & Lee &

ManoogianManoogian19921992

Caulk1983Naghdi1991

Analytical solution

Tsaur et al.2004

Analytical solution

Present method

Present method (semi-analytical) Tsaur et al.

?


Further research

Following the successful experience of this thesis, extending to the problem of torsional rigidity of a bar with inclusions can be considered as a forum in the future.

The extension to hill scattering problem can be studied by using the present approach.

The Green’s function of eccentric case, mixed BC and multi-medium can be easily solved by using our approach.


Further research

The bi-observer expansion technique for the two point function of source and field systems may be suitable for the eccentric case in a more straightforward way free of adaptive observer system.

Our method can also be applied for problems with different boundaries. How to keep the orthogonal property is the main challenge.

Nation Taiwan Ocean University

Department of Harbor and River

April 18, 2023 pp.67

Thanks for your kind attentions.

You can get more information from our website.

http://msvlab.hre.ntou.edu.tw/


Separable form of fundamental solution (1D)

-10 10 20

2

4

6

8

10

Us,x

2

1

2

1

(x) (s), s x

(s,x)

(s) (x), x s

i ii

i ii

a b

U

a b

=

=

ìïï ³ïïïï=íïï >ïïïïî

å

å

1(s x), s x

1 2(s, x)12

(x s), x s2

U r

ìïï - ³ïïï= =íïï - >ïïïî

-10 10 20

-0.4

-0.2

0.2

0.4

Ts,x

s

Separable property

continuouscontinuous jumpjump

1, s x

2(s, x)1

, x s2

T

ìïï >ïïï=íï -ï >ïïïî


Derivation of degenerate kernel

Graf’s addition theoremComplex variable

s xs ( , ) , x ( , )R z zq r f= = = =

x sln x s ln z z- = - Real partReal part

x x xs x s s s

1s s s

1ln( ) ln[( )(1 )] ln( ) ln(1 ) ln( ) ( )m

m

z z zz z z z z

z z m z

¥

=

- = - = + - = - å

( )x

1 1 1 1s

1 1 1 1( ) ( ) ( ) [ ] ( ) cos ( )

im m m i m m

im m m m

z ee m

m z m R e m R m R

ff q

q

r r rq f

¥ ¥ ¥ ¥-

= = = =

= = = -å å å å

Real Real partpart

IfIf s xz z-

1

1

1( , ; , ) ln ( ) cos ( ),

(s, x)1

( , ; , ) ln ( ) cos ( ),

i m

m

e m

m

U R R m Rm R

UR

U R m Rm

rq r f q f r

q r f r q f rr

¥

=

¥

=

ìïï = - - ³ïïïï=íïï = - - >ïïïïî

å

å

ln R

2

2 3

1

1ln(1 ) (1 )

11 1

( )2 3

1 m

m

x dx x x dxx

x x x

xm

¥

=

- =- =- + + +-

=- + + +

=-

ò ò

å

L

L

0k ®

Bessel’s Bessel’s functionfunction


Fictitious frequencies

0 2 4 6 8

-8

-4

0

4

8Present m ethod (M =20)

BEM (N =60)


Derivation of the Poisson integral formula

G. E.: xxu ,0)(2

B. C. :

)(fu

a

Traditional method

R 'R

Image source

Null-field integral equation method

Reciprocal radii method

Poisson integral formula

Image concept

Methods

Free of image concept

Searching the image point

Degenerate kernel

2

0 22

22

)()cos(22

1),( df

aa

au

nation taiwan ocean university department of harbor and river june 11, 2015 pp.1 null-field integral...

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