numerical aspects of a poroelastic time domain boundary element formulation

8/13/2019 Numerical Aspects of a Poroelastic Time Domain Boundary Element Formulation

1/35

Institute of Applied Mechanics, TU Braunschweighttp://www.infam.tu-braunschweig.de

Numerical Aspects of a Poroelastic Time Domain

Boundary Element Formulation

Martin Schanz, Dobromil Pryl, Lars Kielhorn

Adaptive Fast Boundary Element Methods in Industrial Applications

Sollerhaus, 29.9.-2.10.2004

amin Institut fr Angewandte Mechanik

Technische Universitt Braunschweig
http://www.infam.tu-braunschweig.de/


2/35

amin

2/25Content

Governing equations

Biots theory Differential equation

Poroelastic Boundary Element Method

Boundary integral equation

Spatial shape functions

Convolution Quadrature Method

Numerical results

Dimensionless variables

Mixed shape functions

Rock foundation in a soil half-space


3/35

amin

3/25Biots theory of poroelastic continua

constitutive equationi j=Si j+

Fi j

i j= G (ui,j+ uj,i) +

K2

3G

uk,kpi j=uk,k+

2

Rp

equilibrium=s(1) + f

i j,j+ Fi=2

t2ui+ f

t

wi

Darcys law

qi=

p,i+ f

2

t2

ui+a+ f

t

wi

continuity equation

t + qi,i=a

pressure gradient

Flux

Nomenclature

i j total stress qi specific flux ui solid displacement fluid strain

Biots stress coefficient p pore pressure wi seepage velocity Fi bulk body force

G,K shear, bulk modulus porosity a apparent mass density bulk density


4/35

amin

4/25Governing equations

representation in Laplace domain (L{f(t)}= f)

Gui,j j+

K+

1

3G

uj,i j () p,i s2

(f) ui=

Fi

fsp,ii

2s

Rp () sui,i=a

Bui

p

=Fia


5/35

amin



Gui,j j+

K+

1

3G

uj,i j () p,i s2

(f) ui=

Fi

fsp,ii

2s

Rp () sui,i=a

Bui

p

=Fia

weaksingular fundamental solutions

USi j= 1

+8E(1)

r,ir,j+ i j(34) 1

r+O

r0

PF = fs4

1

r+O

r0

TFi =fs

2

8

121

() r,ir,n+ ni

+

1

12

1

r+O

r0

QSj= 1+

8

E(1

) (1

2) (r,nr,j

nj)

2 (1

) (r,nr,j+ nj)

1

r

+Or0


6/35

amin



Gui,j j+

K+

1

3G

uj,i j () p,i s

2

(f)ui=

Fi

fsp,ii

2s

Rp () sui,i=a

Bui

p

=Fia

weaksingular fundamental solutions

USi j= 1

+8E(1)

r,ir,j+ i j(34) 1r +Or0 PF =

f

s

4

1

r +O

r0

TFi =fs

2

8

121

() r,ir,n+ ni

+

1

12

1

r+O

r0

QSj= 1+

8E(1

) (12) (r,nr,jnj)2 (1) (r,nr,j+ nj)1

r+Or

0

strongsingular fundamental solutions

TSi j= [(12) i j+3r,ir,j] r,n+ (12) (r,jni r,inj)

8 (1) r2 +O

r0

QF = 14

r,n

r2+O

r0


7/35

amin

5/25Content

Governing equations

Biots theory

Differential equation





Numerical results





8/35

amin

6/25Boundary integral equation

x1x2

x3

Fd t

n

t

u=u+ t

weighted residuals

GTBui(x,s)

p (x,s)

d= 0

with G =

USi j(x,y,s) U

Fi (x,y,s)

PSj(x,y,s) PF (x,y,s)


9/35

amin

6/25Boundary integral equation

x1x2

x3

Fd t

n

t

u=u+ t

weighted residuals

GTBui(x,s)

p (x,s)

d= 0

with G =

USi j(x,y,s) U

Fi (x,y,s)

PSj(x,y,s) PF (x,y,s)

two partial integrations singular behavior transformation to time domain

t

0

USi j(t

,y,x)

PSj(t

,y,x)

UFi (t ,y,x) PF (t ,y,x)ti(,x)

q (,x)d

d

=

t

0

C

TSi j(t,y,x) QSj(t ,y,x)

TFi (t ,y,x) QF (t ,y,x)

ui(,x)

p (,x)

dd +

ci j(y) 0

0 c (y)

ui(t,y)

p (t,y)


10/35

amin

7/25Spatial discretization

spatial discretization

ui(x,t) =E

e=1

F

f=1

Nfe (x) ue fi (t)

ti(x,t) =E

e=1

F

f=1

Nfe (x) te fi (t)

p (x,t) =E

e=1

F

f=1

Nfe (x)pe f(t)

q (x,t) =E

e=1

F

f=1

Nfe (x) qe f(t)

e.g. linear ansatz function

N1

e(,) =

1

4 (1) (1)N2e(,) =

1

4(1) (1+ )

N3e(,) =1

4(1+ ) (1+ )

N4

e(,) =1

4 (1+ ) (1)

S i l di i i


11/35

amin

7/25Spatial discretization

spatial discretization

ui(x,t) =E

e=1

F

f=1

Nfe (x) ue fi (t)

ti(x,t) =E

e=1

F

f=1

Nfe (x) te fi (t)

p (x,t) =E

e=1

F

f=1

Nfe (x)pe f(t)

q (x,t) =E

e=1

F

f=1

Nfe (x) qe f(t)

e.g. linear ansatz function

N1

e(,) =

1

4 (1) (1)N2e(,) =

1

4(1) (1+ )

N3e(,) =1

4(1+ ) (1+ )

N4

e(,) =1

4 (1+ ) (1)

discretized integral equation

ci j(y) ui(y,t)c (y)p (y,t)

=E

e=1

F

f=1

t0

USi j(x,y, t) PS

j(x,y, t

)UFi (x,y, t) PF (x,y,t )

Nfe (x)dte f

i (

)qe f ()

d

t

0

C

TSi j(x,y,t ) QSj(x,y,t)

TFi (x,y,t) QF (x,y,t )

Nfe (x)d

ue fi ()

pe f ()

d

Mi d h f i


12/35

amin

8/25Mixed shape functions

Isoparametric elements - shape functions identical for all quantities and geometry

Mixed elements using different shape functions for different quantities

(common for finite elements), e.g. Nfe (x)linear foru,tand constant for p,q

1 0 1

1

ko-2D1 0 1

1

li-2D1 0 1

1

lk-2D

lk-dr

Shape functions in 2-d and 3-d

Element uNfe,

tNfe

pNfe,

qNfe

ko-2D, ko-dr constant constant

li-2D, li-dr linear linear

lk-2D, lk-dr linear constant

amC l ti Q d t M th d


13/35

amin

9/25Convolution Quadrature Method

quadrature rule forn= 0,1, . . . ,Ntime steps:

y (t) = f(t)g (t) =t

0

f(t ) g ()d y (nt) =n

k=0nk f,tg (kt)

Lubich, C. (1988): Convolution Quadrature and Discretized Operational Calculus. I., Numerische Mathematik, Vol.

52, 129145

amC l ti Q d t M th d


14/35

amin

9/25Convolution Quadrature Method

quadrature rule forn= 0,1, . . . ,Ntime steps:

y (t) = f(t)g (t) =t

0

f(t ) g ()d y (nt) =n

k=0nk f,tg (kt)

integration weight:

n

f,t

=

1

2i

|z|=R

f(z)

t

zn

1d

zR

n

L

L1

=0

fRe

i 2L

t

e

in 2L

(z)A-stable multi step method, e.g. BDF 2: (z) = 322z + 1

2z2

ttime step size of equal duration

L=Neffective choice for determiningn (FFT) RN =with1010

Lubich, C. (1988): Convolution Quadrature and Discretized Operational Calculus. I., Numerische Mathematik, Vol.

52, 129145

amTemporal discretization


15/35

amin

10/25Temporal discretization

temporal discretizationwith Convolution Quadrature Method yields for n= 0,1, . . . ,N

ci j(y) ui(nt)c (y)p (nt)

=

E

e=1

F

f=1

n

k=0

e fn

kUSi j,y,t

e fn

kPSj,y,t

e fnk

UFi ,y,t e fnkPF,y,t

te fi (kt)

qe f (kt)

e fnk

TSi j ,y,t

e fnkQSj ,y,t

e fnk

TFi ,y,t

e fnkQF,y,t

ue fi (kt)

pe f (kt)

with integration weights, e.g.

e fnk

Ui j,y,t

=Rkn

L

L1=0

Ui j

x,y,

Rei

2L

t

Nfe (x)d ei(nk) 2L

amTemporal discretization


16/35

amin

10/25Temporal discretization

temporal discretizationwith Convolution Quadrature Method yields for n= 0,1, . . . ,N

ci j(y) ui(nt)c (y)p (nt)

=

E

e=1

F

f=1

n

k=0

e fn

kUSi j,y,t

e fn

kPSj,y,t

e fnk

UFi ,y,t e fnkPF,y,t

te fi (kt)

qe f (kt)

e fnk

TSi j ,y,t

e fnkQSj ,y,t

e fnk

TFi ,y,t

e fnkQF,y,t

ue fi (kt)

pe f (kt)

with integration weights, e.g.

e fnk

Ui j,y,t

=Rkn

L

L1=0

Ui j

x,y,

Rei

2L

t

Nfe (x)d ei(nk) 2L

quadrature formula regular integrals: Gauss formula weak singular integrals: Regularization with polar coordinate transformation strong singular integrals: Formula by G UIGGIANIand GIGANTE

amTime stepping procedure


17/35

amin

11/25Time stepping procedure

solution with point collocation, i.e. movingy in every node and solving the system in each time

step (U,T,u, tare generalized variables here)

0(T)u(t) =0(U) t(t)

1(T)u(t) + 0(T)u(2t) =1(U) t(t) + 0(U) t(2t)

2(T)u(t) + 1(T)u(2t) + 0(T)u(3t) =2(U) t(t) + 1(U) t(2t) + 0(U) t(t)

...

amTime stepping procedure


18/35

ain

11/25Time stepping procedure

solution with point collocation, i.e. movingy in every node and solving the system in each time

step (U,T,u, tare generalized variables here)

0(T)u(t) =0(U) t(t)

1(T)u(t) + 0(T)u(2t) =1(U) t(t) + 0(U) t(2t)

2(T)u(t) + 1(T)u(2t) + 0(T)u(3t) =2(U) t(t) + 1(U) t(2t) + 0(U) t(t)

...

finalrecursion formula

0(C)dn =0(D) d

n +n

m=1

m(U) t

nm m(T)unm

n= 1,2, . . . ,N

with the vector of unknown boundary data dn

and the known boundary data dn

in each time step

amContent


19/35

in12/25

Content

Governing equations

Biots theory

Differential equation





Numerical results




amiColumn: Problem description


20/35

in13/25

Column: Problem description

ty=1 Nm2H(t),permeable

free,impermeable

fixed, impermeable

3m

x

y

K[ Nm2 ] G[ Nm2 ] [kgm3 ] R[ Nm2 ] f [kgm3 ] [ m4

Ns ]

rock 8 109 6 109 2458 0.19 4.7108 1000 0.867 1.91010soil 2.1 108 9.8107 1884 0.48 1.2109 1000 0.981 3.55109

324 elements on 188 nodes

amiDimensionless variables


21/35

in14/25



x= x

A

t= t

B

K=K

C

G=G

C

= A2

B2

C

=BC

A2

Fall 1, 2, 3all material data O()

A=2

C B=

2 C=

1

K+

4

3G +

2

2R

= 1,103,103

aminDimensionless variables


22/35

in14/25



x= x

A

t= t

B

K=K

C

G=G

C

= A2

B2

C

=BC

A2


A=2

C B=

2 C=

1

K+

4

3G +

2

2R

= 1,103,103

Fall 4, 5

only normalization of modules

A= 1 B= 1 C=E= 9KG

6K+ G= 1,10



23/35

in14/25



x= x

A

t= t

B

K=K

C

G=G

C

= A2

B2

C

=BC

A2


A=2

C B=

2 C=

1

K+

4

3G +

2

2R

= 1,103,103

Fall 4, 5


A= 1 B= 1 C=E= 9KG

6K+ G= 1,10

Fall 6scaling of Youngs modules to the permeability

A= 1 B= 1 C=

E



24/35

in14/25



x= x

A

t= t

B

K=K

C

G=G

C

= A2

B2

C

=BC

A2


A=2

C B=

2 C=

1

K+

4

3G +

2

2R

= 1,103,103

Fall 4, 5


A= 1 B= 1 C=E= 9KG

6K+ G= 1,10

Fall 6scaling of Youngs modules to the permeability

A= 1 B= 1 C=

E

Fall 7simple normalizationA=rmaxmaximum radius B= temaximum time C=E

aminCondition number in frequency domain


25/35

in15/25

Condition number in frequency domain

0 1000 2000 3000 4000 5000 6000

10E6

10E11

10E16

10E21

10E26

10E6

10E11

10E16

10E21

10E26

Fall 7

Fall 6

Fall 5

Fall 4

Fall 3

Fall 2

Fall 1

Frequency [1/s]

Cond

itionnumber

aminCondition number in time domain


26/35

in16/25

0 0.005 0.01 0.015 0.02 0.025

10E8

10E14

10E20

10E26

10E32

10E8

10E14

10E20

10E26

10E32

Fall 7

Fall 6

Fall 5

Fall 4

Fall 3

Fall 2

Fall 1

Time t [s]

Conditionnumber

amin

1 /2Condition number: Different parameters


27/35

17/25p

0 0.005 0.01 0.015 0.02 0.025

10E9

10E15

10E21

10E27

10E33

10E9

10E15

10E21

10E27

10E33

Fall 4

Fall 3

Fall 2

Fall 1

Time t [s]

Conditionnumber

Fall 1= A=rmax B= 1 C= 1 Fall 2

= A= 1 B= tmax C= 1

Fall 3= A=rmax B= tmax C= 1 Fall 4

= A= 1 B= 1 C=E

amin

18/25Mixed Elements in 2-d


28/35

18/25

0.0 0.005 0.01 0.015 0.02

time t [s]

-1.6*10-9

-1.4*10-9

-1.2*10-9

-1.0*10-9

-8.0*10-10

-6.0*10-10

-4.0*10-10

-2.0*10-10

0.0*100

displacementuy

[m]

l i-2D, t=0.00012s li-2D 128, t=0.00005s

lk-2D, t=0.00012s lk-2D 128, t=0.00005s

ko-2D, t=0.0004s ko-2D 128, t=0.00005s

analytic 1-d

0.0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.0

time t [s]

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

porepressurep[N/m2]

l i-2D, t=0.00012s li-2D 128, t=0.00005slk-2D, t=0.00012s lk-2D 128, t=0.00005s

ko-2D, t=0.0004s ko-2D 128, t=0.00005s

analytic 1-d

amin

19/25Mixed Elements in 3-d: Fixed surfaces


29/35

19/25

0.0 0.005 0.01 0.015 0.02

time t [s]

-1.5*10-9

-1.0*10-9

-5.0*10-10

0.0*100

displacementu

y[m

]

li-dr, t=0.0002s

lk-dr, t=0.0002s

ko-dr, t=0.0002s

analytic 1-d

00.25

0.50.75

1

0

1

2

3

0

0.25

0.5

0.75

1

00.25

0.5

1

2

700 elementson 392 nodes

0.0 0.005 0.01 0.015 0.

time t [s]

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

por

epressurep[N/m2]

li-dr, t=0.0002slk-dr, t=0.0002s

ko-dr, t=0.0002s

analytic 1-d

amin

20/25Mixed Elements in 3-d: Free surfaces


30/35

20/25

0.0 0.02 0.04 0.06 0.08 0.1 0.12 0.14

time t [s]

0.0*100

5.0*10-9

1.0*10-8

1.5*10-8

displacementu

y[m

]

li-dr, t=0.0005s

lk-dr, t=0.0005s

ko-dr, t=0.0005s

0.0 0.02 0.04 0.06 0.08 0.1 0.12 0.14

time t [s]

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

por

epressurep[N/m2]

li-dr, t=0.0005s

lk-dr, t=0.0005s

ko-dr, t=0.0005s

amin

21/25Half space: Problem description


31/35

21/25

2m

3m

2m

1m

2m

5m

p(t)

p(t)

p(t)

p(t)

0,02s

p(t)

t

1,25MN

Rock foundation embedded in a soil half-space

Foundation: 124 elements on 68 nodesHalf space: 334 elements on 192 nodes

load

amin

22/25Half space: Condition number


32/35

22/25

0 0.02 0.04 0.06 0.08 0.1 0.12

10E18

10E20

10E22

10E24

10E26

10E18

10E20

10E22

10E24

10E26

Fall 2

Fall 1

Time t [s]

Cond

itionnumber

Fall 1=old dimensionless variables Fall 2

=new more simpler suggestion

amin

23/25Half space: Numerical results


33/35

23/25

amin

24/25Summary


34/35

/

Poroelastic BEM

Biots theory

Based on Convolution Quadrature Method

Only Laplace transformed fundamental solutions are required


Normalization w.r.t. time, space, and Youngs modulus

Largest influence due to the normalization to Youngs modulus


Only sometimes improvement of stability and accuracy

Very CPU-time consuming

No justification for numerical effort

I tit t f A li d M h i TU B h i


35/35

Institute of Applied Mechanics, TU Braunschweighttp://www.infam.tu-braunschweig.de

Numerical Aspects of a Poroelastic Time Domain

Boundary Element Formulation

Martin Schanz, Dobromil Pryl, Lars Kielhorn

Adaptive Fast Boundary Element Methods in Industrial Applications

Sollerhaus, 29.9.-2.10.2004

amin Institut fr Angewandte Mechanik

Technische Universitt Braunschweig
http://www.infam.tu-braunschweig.de/

numerical aspects of a poroelastic time domain boundary element formulation

Documents