Engineering Analysis with Boundary Elementssharif.ir/~ataie/Printed-Articles/EABE_2012.pdf · Seepage Boundary element method Anisotropy Heterogeneous Multi-domain Multi-region Zoned

Engineering Analysis with Boundary Elements 36 (2012) 812–824

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Engineering Analysis with Boundary Elements

0955-79

doi:10.1

n Corr

E-m

journal homepage: www.elsevier.com/locate/enganabound

Three dimensional flow in anisotropic zoned porous media using boundaryelement method

K. Rafiezadeh a, B. Ataie-Ashtiani a,b,n

a Department of Civil Engineering, Sharif University of Technology, P.O. Box 11155-9313, Tehran, Iranb National Centre for Groundwater Research & Training, Flinders University, GPO Box 2100, Adelaide SA 5001, Australia

a r t i c l e i n f o

Article history:

Received 25 July 2011

Accepted 10 December 2011Available online 11 January 2012

Keywords:

Seepage

Boundary element method

Anisotropy

Heterogeneous

Multi-domain

Multi-region

Zoned

Dam

River

Sheet-pile

97/$ - see front matter & 2011 Elsevier Ltd. A

016/j.enganabound.2011.12.002

esponding author.

ail address: [email protected] (B. Ataie-Ashtian

a b s t r a c t

Coupling the adjacent zones for seepage analysis in porous media needs compatibility and equilibrium

equations (equality of potential on coinciding nodes and conservation of flowing mass between zones,

respectively). When stretched coordinate transformation is applied to the anisotropic zones, the

Dirichlet boundary conditions remain unchanged, but the Neumann boundary condition should also be

transformed. Similarly in a zoned problem, for the interface between zones, compatibility equations

remain unchanged during the transformation while the equilibrium equations should be transformed.

In this paper, transformed Neumann boundary conditions and equilibrium equations for the interface of

neighbor anisotropic zones for seepage problems have been developed in three dimensions. A computer

program for seepage analysis of zoned anisotropic media based on the Boundary Element Method is

developed. The code is used to solve several examples with isotropic and anisotropic zones. Some

examples are also solved by finite element method for verification. Illustrated results show the ability

and accuracy of the mathematical and the numerical model for solving different types of applied three-

dimensional seepage problems that arise in engineering practice.

& 2011 Elsevier Ltd. All rights reserved.

1. Introduction

Boundary element method (BEM) is a powerful tool for solvingseepage problems. The main advantage of boundary element methodis the form of equations, which are written on the boundary. As aresult, discretizing is also done only on the boundary and conse-quently much less effort is needed for preprocessing. As the numberof nodes and element decreases, the computational cost alsodecreases. In seepage problems, when the medium is assumed tobe a rigid homogeneous isotropic saturated and the Darcy’s law isemployed, the governing equation is Laplace equation [1] andboundary element method for solving Laplace equation is vastlyprovided in the literature, for example see [2]. In practice, especiallywhen we are dealing with seepage problems, lots of anisotropicproblems arise. To stress on the importance of seepage analysis inanisotropic media, we may state that both soil deposits in the natureand man-made soil structures are often anisotropic from the seepagepoint of view. Sedimentary deposits are often anisotropic with flowoccurring more readily along the plane of deposition than acrossit [3]. In most of man-made earth structures it is also the case,because the soil is deposited and compacted in layers. In compacted

ll rights reserved.

i).

earth fills- for example- the ratio Kh/Kv (horizontal to verticalhydraulic conductivity ratio) may exceed 20 [4].

Governing equation for an anisotropic medium is Kx@2j/

@x2þKy@

2j/@y2þKz@

2j/@z2¼0, which is not the Laplace equation [5].

Two strategies have been used to solve this equation by boundaryelement method. The first one-step strategy is to find the funda-mental solution of the governing equation and then derive theboundary element formulations as well, for example see [5]and [6]. While the fundamental solution of the Laplace equation is1/4pr, [5] used 1/{4p(kxkykz)

1/2(x2/kxþy2/kyþz2/kz)1/2} as the funda-

mental solution to solve Kx@2j/@x2

þKy@2j/@y2

þKz@2j/@z2

¼0directly by BEM. The author of [6] tried to achieve weakly singularintegral equations. The governing equations were established basedon a pair of weakly singular weak-form integral equations for fluidpressure and fluid flux. More complicated set of fundamentalsolutions for the problem were used [6]. In the first step of thesecond strategy, the geometry of the problem is transformed. Aspecial transform is usually used that the governing equation in thetransformed geometry would be the Laplace equation. It means thatsolving the Laplace equation in the transformed geometry is equiva-lent to solving the above-mentioned anisotropic governing equationin the original geometry. The geometrical transformation is normallydone using coordinates transformation. Such a coordinate transfor-mation has been extensively used in classical soil mechanics forgraphical solutions of seepage problems by flow nets [7]. While thefirst strategy is completed in one step, the second strategy needs

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Nomenclature

j potential headn outward normal to boundary vector@j/@n normal to boundary gradient of potential headq fluxK hydraulic conductivityKx, Ky, Kz hydraulic conductivity in principal axes directions

Kh, Kv hydraulic conductivity in horizontal and verticaldirections

r Nabla operator: (@/@x, @/@y, @/@z)r2 Laplacian operator: @2/@x2

þ@2/@y2þ@2/@z2

G Green’s function{N} vector of shape functionsx, Z elemental local coordinatesr distance from source point to integration point

K. Rafiezadeh, B. Ataie-Ashtiani / Engineering Analysis with Boundary Elements 36 (2012) 812–824 813

preprocessing for coordinates transformations. Although in the firststrategy, the formulations are straight forward and do not need anygeometrical transformations, as authors know, this strategy has onlybeen used with simple confined steady-state problems. For furtherdevelopments in more complicated problems such as free surfaceproblems or dealing with unsaturated media, it is much morepractical to use the standard Laplace equation when we are goingto upgrade to three-dimensions. As there are many successfuldevelopments for two-dimensional isotropic cases, such previousdevelopments can then be implemented for more complicatedanisotropic problems when the second strategy of coordinate trans-formation is used. In addition, since lots of reliable codes and librarieshave been previously developed in BEM for the Laplace equation, as amatter of a practical code development, the more standard Laplaceequation is mostly preferred. The method of coordinate transforma-tion has been successfully used in boundary element method forLaplace’s equation in the literature for single-domain media, forexample see [8] and [9]. Using this strategy reduces the extensivework of developing a brand new code based on more complicatedfundamental solutions to providing a simple preprocessing code forgeometrical domain transformation and a simple post processor fordoing the reverse transformation. This strategy has been usedsuccessfully for seepage applications in [10] for three-dimensionalaquifers. In [10], although the three-dimensional aquifers were dealt,the effective dimension of an aquifer system was reduced to two byuse of the Dupuit assumptions. The authors of [10] discussed that inmany practical cases sufficiently accurate results could be obtainedwith less than a full three-dimensional analysis. Based on the fact thatthe horizontal dimensions of the aquifer are often much larger thanthe vertical dimensions, they assumed a ‘nearly horizontal’ flow as agood approximation. After assuming a nearly horizontal flow, theirproblem was reduced to two dimensions in the horizontal plane.They used the two-dimensional stretched coordinate transformationto deal with anisotropy. After the transformation, the new two-dimensional mapped region was used for BEM calculations and thegoverning equation on the mapped domain was the two-dimensionalLaplace equation. As the domain transforms, the inter-zonal equili-brium equations should also be transformed. They developed thetransformed inter-zonal equilibrium equations in two dimensions.They consequently used the two-dimensional BEM formulations forthe three-dimensional problem. Although the nearly horizontalassumption is a good approximation for problems like aquifers, inmany other Civil Engineering structures like dams, sheet-piles, etc. itis not the case because the horizontal dimensions are not much largerthan the vertical dimensions. Thus many problems can only be solvedby a full three-dimensional analysis. In this paper we present theinter-zonal equilibrium equations for three-dimensions and provide afull-three dimensional seepage analysis.

Similar to anisotropy, the heterogeneity is also a commonproblem in applications of seepage analysis. In engineering practicemost of the times, the media is constructed from multiple regions ofdifferent characteristics while each individual region is itself homo-geneous. Examples of these media are earth dams including a claycore and random fill embankments, concrete dams or sheet piles onlayers of different hydraulic conductivities or aquifers containing

different layers of different ages and characteristics. When dealingwith such problems, the media can be divided into zones [11].Within each zone we can write Laplace equation with a coordinatetransformation for anisotropic zones or without a coordinate trans-formation for isotropic zones. While boundary conditions are avail-able on exterior boundaries no boundary conditions are on theinterface boundaries of each zone. Additional equations are used forcoupling adjacent zones with each others. These equations areneeded to ensure the equality of potential head in nodes occupyingthe same position and conservation of flowing mass between thetwo zones [12]. The boundary element equations of all zones thenare assembled in a global system of equations for solution.

When there are two isotropic neighbor zones, the compatibilityequation (equating the head) is j(1)

¼j(2) and the equilibriumequation (conserving the mass) is K(1)(@j/@n)(1)

¼�K(2)(@j/@n)(2) inwhich the superscripts show the zone number. When the zones areanisotropic, the compatibility equation remains unchanged becauseit is independent of hydraulic conductivities, but enhancements inthe equilibrium equation should be considered while the hydraulicconductivities in different directions are not equal. Two differentscenarios have been used in two-dimensional problems: [13–15]

used the two-dimensional equivalent hydraulic conductivityffiffiffiffiffiffiffiffiffiffiffiKxKy

pfor anisotropic media leading to the condition

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiK ð1Þx K ð1Þy

qð@j=@nÞð1Þ ¼

�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiK ð2Þx K ð2Þy

qð@j=@nÞð2Þ as the equilibrium condition. [10] and [16]

also dealt the problem in two dimensions but used a differentscenario. They first calculated the equivalent normal to boundarygradients in stretched coordinates based on the normal to boundarygradients in un-stretched coordinates and then used aKx as theequivalent hydraulic conductivity in which a depends on the inter-face edge angle to the x-direction.

One of the main objectives of this paper is to develop thethree-dimensional transformed Neumann boundary conditionsand inter-zonal equilibrium equations when stretched coordinatetransform has been used for anisotropic media. (See Sections 4and 5) Similar set of equations for two-dimensional problemswere presented before in [10] and [16]. Another objective of thiswork is to develop a computer code for solving general seepageproblems for zoned anisotropic media in three dimensions usingboundary element method (see Section 7). The code has beendeveloped by the authors and three applied examples have beensolved by the developed code and the illustrated numericalresults have been provided. Verifications with fine mesh FEMhave been performed for some examples to show the accuracy ofthe formulations and reliability of the code (see Section 8).

This paper is the first result of an ongoing research project fordeveloping a boundary element code to be able to do the unconfinedseepage analysis for zoned anisotropic media incorporating unsatu-rated effects in applied civil engineering scale.

2. Governing equation

When focusing on applications of flow in porous media in Civiland Environmental engineering, we assume the porous media as a

Fig. 1. Spatial angle on a corner.

Fig. 2. Local coordinate system.

K. Rafiezadeh, B. Ataie-Ashtiani / Engineering Analysis with Boundary Elements 36 (2012) 812–824814

continuum media and study the macroscopic behavior of flowrather the microscopic one. Assuming a saturated homogeneousanisotropic media (which the principal axes of the anisotropy areparallel to coordinates axes) and neglecting the water compres-sibility and soil skeleton deformations, the combination of massconservation and Darcy’s Law yields to:

Kx@2j@x2

!þKy

@2j@y2

!þKz

@2j@z2

!¼ 0 ð1Þ

in which j is the potential and Kx, Ky, Kz are the hydraulicconductivity of porous media in x, y and z directions, respectively(See [5] for derivation).

If we use the stretched coordinates transformation as follows:

xn ¼Kz

Kx

� �1=2

x, yn ¼Kz

Ky

� �1=2

y, zn ¼ z ð2Þ

then we will have the familiar Laplace equation, which governsthe seepage in saturated porous media in stretched coordinates

r2nj¼ 0 ð3Þ

where rn2 is the Laplacian operator in stretched coordinates.

When the coordinate axes and the principal axes of the aniso-tropy are not parallel to each other, we may first apply a rotationto the coordinate axes to make them parallel to the principal axesof the anisotropy. The same methodology has been used in [10]and [16]. Authors in [8] and [9] used another coordinate trans-formation that the rotation is included in the transformation.They used their transformation for single domain problems. Theirone-step more complicated transformation is equivalent to thepresented two-step (a) rotation of axes and (b) using the simpletransformation in Eq. (2). In the special case that the principalaxes of anisotropy are parallel to the coordinate axes, theirtransformation will be simplified to that is given in Eq. (2). Sincecompaction in engineering earth structures usually happens invertical direction, the principal axes of anisotropy are parallel tothe coordinate axes in most cases and one occasionally needs arotation. So because the two transformations are mathematicallyequivalent, it is more economical to use the transformation inEq. (2).

3. BEM formulations

Let having a governing equation in the form F(j)¼0, which isdefined on a region O with boundary G. If any function G exists,which satisfies F(G)¼d in which d denotes the Dirac deltafunction, then we have

cijiþ

ZG

G@j@n

dG¼ZGj @G

@ndG ð4Þ

here G is named as the Green’s function of F and is equal to 1/4pr

when the region is three dimensional and the governing equationis the Laplace equation and r is the distance from any point i to theintegration point on the boundary. ci is between 0 and 1 and isequal to 1 if the point i is inside the boundary, 1/2 if point i is on asmooth boundary and is equal to zero when point i is outside theboundary. If point i is on a non-smooth part of the boundary(a boundary corner), a sphere is assumed with the center point i

and radius e (Fig. 1). The coefficient ci is then equal to limit of theratio of the area of the part of the sphere which is inside theboundary to the total area of the sphere when e approaches to zero.

It is clear that the potential in any point in the region caneasily be found if j and @j/@n are both known on the boundary.Of course in well posed Cauchy potential problems always onlyone is available by the boundary conditions. (In Dirichlet bound-ary conditions j is prescribed on the boundary while @j/@n is

unknown, and in Neumann Boundary conditions only @j/@n isprescribed while j is unknown, and in mixed boundary condi-tions, in some parts of the boundary only j is prescribed and inremaining parts only @j/@n is given). By the following approachwe complete the data on the boundary in a three dimensionalproblem.

In the first step we can mesh the boundary with triangularelements, and then we will have n nodes and m elements. If wewrite Eq. (4) for each of nodes of the boundary then we will haven equations and 2n unknowns (j and @j/@n for each node). Fromthe 2n unknowns, n of them are prescribed by the boundaryconditions, and so there remain n unknowns with n equations,which can be solved for the unknowns.

Defining a x�Z local coordinates for each triangular element(Fig. 2), the global coordinates, head and normal gradient of headof any point on the element can be approximated by theappropriate nodal values as following:

x¼ fNgTx1

x2

x3

8><>:

9>=>;, y¼ fNgT

y1

y2

y3

8><>:

9>=>;, z¼ fNgT

z1

z2

z3

8><>:

9>=>;

j¼ fNgTj1

j2

j3

8><>:

9>=>;,

@j@n¼ fNgT

ð@j=@nÞ1ð@j=@nÞ2ð@j=@nÞ3

8><>:

9>=>; ð5Þ

in which indices 1, 2 and 3 show the element node number and{N} is the vector of shape functions that is given by

fNg ¼

N1

N2

N3

8><>:

9>=>;¼

1�x�ZxZ

8><>:

9>=>; ð6Þ

Eq. (4) then can be evaluated in the following form:

ciFiþXme ¼ 1

fagTe fFge ¼Xme ¼ 1

fbgTe f@j=@nge ð7Þ


in which

fage ¼

a1

a2

a3

8><>:

9>=>;¼ 2Ae

ZZAm

e

N1

N2

N3

8><>:

9>=>;

@

@n

1

r

� �dxdZ ð8Þ

fbge ¼

b1

b2

b3

8><>:

9>=>;¼ 2Ae

ZZAm

e

N1

N2

N3

8><>:

9>=>;

1

rdxdZ ð9Þ

Integrating element by element, while the integration pointdoes not coincide on the source point, integrals in (8) and (9) canbe easily calculated by numerical integration methods, such asGauss quadrature method. But when the integrating point coin-cides on the source point, r vanishes and integral in (9) containsthe term 1/r and is singular. Integral in (8) contains @(1/r)/@n thatis equal tor(1/r).n and since r(1/r) is in the plane of element andn is normal to the plane of element, the dot product vanishes andthis integral is not singular and can be evaluated by Gaussquadrature method. But for integral in (9), when the integrationpoint coincides the source point, from the three shape functions,two of them will be zero and thus the integrand is equal to zeroand no singularities exist but one of the shape functions is equalto 1 and thus the integral will be singular. Assume that Nk (k¼1 or2 or 3) is the shape function that is equal to one. Then theintegrant will be Nk/r and we will have

bk ¼ 2Ae

ZZAm

e

Nk

rdxdZ¼ 2Ae

ZZAm

e

Nk�1

rdxdZþ2Ae

ZZAm

e

1

rdxdZ

ð10Þ

The first integral is not singular and can be evaluated numeri-cally by Gauss quadrature method but the second is singular andshould be evaluated analytically by transforming to the polarcoordinates.

4. Anisotropy

When stretched coordinates transformation is incorporated todeal with the anisotropic problems, the definition of Neumannboundary conditions needs more care. In fact, while the Dirichletboundary conditions are applied to the equation with no change,the Neumann boundary conditions need to be transformed.

While transformation (2) is used, considering a¼(Kz/Kx)1/2 andb¼(Kz/Ky)1/2 we have

@j@x¼ a @j

@xn

,@j@y¼ b

@j@yn

,@j@z¼@j@zn

ð11Þ

We have

@j@xn

¼@j@n

nxnþ@j@s

sxnþ@j@t

txn

@j@yn

¼@j@n

nynþ@j@s

synþ@j@t

tyn

@j@zn¼@j@n

nznþ@j@s

sznþ@j@t

tzn ð12Þ

where

nxn¼ nn:i, nyn

¼ nn:j, nzn ¼ nn:k

sxn¼ sn:i, syn

¼ sn:j, szn ¼ sn:k

txn¼ tn:i, syn

¼ tn:j, szn ¼ tn:k ð13Þ

and n is the unit normal to boundary vector and s and t are thetwo tangential unit vectors of the boundary. We also have

@j@n¼@j@x

nxþ@j@y

nyþ@j@z

nz ð14Þ

Using (12) and (13) in (14) we will have

@j@n¼@j@nn

ðanxnxnþbnynyn

þnznzn Þþ@j@snðanxsxn

þbnysynþnzszn Þ

þ@j@tnðanxtxn

þbnytynþnztzn Þ ð15Þ

From definition of transformation we have

sx

sz¼ a sxn

szn

,tx

tz¼ a txn

tzn

,nz

nx¼ a nzn

nxn

sy

sz¼ b

syn

szn

,ty

tz¼ b

tyn

tzn

,nz

ny¼ b

nzn

nyn

ð16Þ

Using (16) and the normality of n and s, we have

anxsxnþbnysyn

þnzszn ¼ szn ðanxsxn=sznþbnysyn

=sznþnzÞ

¼ szn ðnxsx=szþnysy=szþnzÞ ¼ ðnxsxþnysyþnzszÞ ¼szn

szn:s¼ 0

Similarly from the normality of n to t we haveanxtxn

þbnytynþnztzn ¼ 0:

Thus Eq. (15) will be simplified to

@j@n¼ C

@j@nn

ð17Þ

where

C ¼ ðanxnxnþbnynyn

þnznzn Þ ð18Þ

Substituting values of nx and ny from (16) into (18) we willhave

C ¼ nz=nzn ð19Þ

Since n and nn are unit vectors we have

nz ¼ 1þnx

nz

� �2

þny

nz

� �2" #�1=2

nzn ¼ 1þnxn

nzn

� �2

þnyn

nzn

� �2" #�1=2

ð20Þ

Thus we may finally reach to

C ¼nxn

a

� �2

þnyn

b

� �2

þn2yn

" #�1=2

ð21Þ

@j@n¼

nxn

a

� �2

þnyn

b

� �2

þn2yn

" #�1=2@j@nn

ð22Þ

It means that the Neumann boundary conditions should becorrected by Eq. (22) when a stretched coordinate transform (2) isgoing to be used to transform the non-Laplace governing equation offlow in anisotropic media (1) to the well known Laplace equation.

5. Treating the interface

While the media for seepage analysis consists of more thanone zone and the zones have interface with each other, for thenodes on the interface there is no boundary conditions. But therestill exist two consistency equations: the compatibility andequilibrium equations. The compatibility equation states thatfor any node i in zone k, which is placed on the interface andcoincides on node j in zone l, the potential should be the same, i.e.

jðkÞi ¼jðlÞj ð23Þ

in which the superscripts show the zone index and the subscriptsshow the node index. The equilibrium equation states the con-servation of mass. To conserve the mass transport we also should


satisfy the equality of normal to interface fluxes, i.e.

qðkÞi ¼ K ðkÞe

@ji

@n

� �ðkÞ¼ �K ðlÞe

@jj

@n

� �ðlÞ¼ �qðlÞj ð24Þ

where Ke means the equivalent hydraulic conductivity in thenormal to interface direction and the minus sign means that themass exiting from one zone is entering the other one. In isotropiccase Ke is equal to the hydraulic conductivity of the media whilein anisotropic media, when we are using the stretched coordi-nates transformation in (2), Ke¼Kz [16]. Using the definition of C

given in Eq. (21) we will have:

CðkÞK ðkÞz

@ji

@nn

� �ðkÞ¼ �CðlÞK ðlÞz

@jj

@nn

� �ðlÞð25Þ

Eqs. (23) and (25) are the two consistency equations for anytwo coinciding nodes on the interface of the two neighboringzones which hereafter we name them as inter-zonal conditions.

In case when multiple zones with distinct hydraulic conductiv-ities are joined together and share one point as intersection point(for example a sphere divided into 8 sub-regions with a commoncorner point at the center), a singularity happens at the intersectionpoint. From a practical point of view, in most calculations theintersection point is simply avoided, but additional nodes are takenon each of the inter-zonal boundaries close to the intersection. Theresults may not be entirely accurate very close to the intersectionbut are usually more than accurate enough for practical purposes[16]. The same strategy (additional nodes close to the corner point)

Table 1Main objects of the program.

Properties/methods Description

NodeID Holds user defined ID of the no

Zone index Holds the index of the zone it

Original coordinates Holds the user defined coordin

Coordinates Holds the calculated stretched

Head Holds the potential head of the

Gradient Holds the normal gradient of t

Boundary condition Holds the boundary condition

Belongs to elem index Hold index of elements, which

ElementID Holds user defined ID of the el

Zone index Holds the index of the zone it

Nodes index Holds indexes of the three nod

Inward guide point Holds the coordinates of a poin

element is facing to inside of t

Unit normal vector Calculates the unit normal vect

Area Calculates the area of the elem

Integration points global coors Calculates 13-point global coor

Calc integrals Calculate {a}e and {b}e of the e

Calculates normal and singular

ZoneID Holds user defined ID of the zo

Hydraulic conductivity Holds hydraulic conductivities

Node collection Defines a collection of all node

Element collection Defines a collection of all elem

Interior point collection Holds all interior points in whi

HC Holds [H]e matrix.

GC Holds [G]e matrix.

Derive equations Calculates [H]e and [G]e matric

Interior solution Calculates interior solutions for

ZonesAssemble equations Assembles equations and form

Solve Solves Eq. (29) for {u}.

Extract Finds if each unknown in {u} is

zone and assigns the calculated

Export log file Prints out all input and calcula

Print interior solution Prints out the interior heads an

Draw geometry Exports a file that can be used

can be used for corner points on the boundary when differentgradients are prescribed as boundary conditions.

6. Assembling equations

For each node in the zone we can write Eq. (7). A system ofequation then will be formed. We may arrange this system ofequation in the matrix form to reach

½H�fFg ¼ ½G�f@j=@ng ð26Þ

Investigating elements in [H] and [G], it is clear that they arefunctions of geometry only and not functions of boundary condi-tions. Now assume the zone has a no flux boundary condition forwhole of its boundary. The solution to this problem will be aconstant head on the zone. Applying these conditions (ji¼cte,

qi¼0) to Eq. (26) we may see that the sum elements of each rowin [H] will be zero. As [H] is not a function of boundary elements, thesame result is true for any configuration of boundary elements. Thisfact can help us to avoid calculating ci as it is difficult in many cases(especially in three dimensions). Calculation of ci only happens onthe main diagonal of the matrix [H]. Thus we first calculate allelements of [H] by integrals in Eq. (8) except the main diagonalelements and then calculate the main diagonal elements from theabove-mentioned relation on each row of [H].

de (any string variable).

belongs to.

ates of the node.

coordinates of the node.

node.

he potential head.

of the node (Head_Prescribed/Gradient_Prescribed or Interzone).

the node belongs to.

ement (any string variable).

belongs to.

es of the element.

t defined by the user to define, which side of the

he zone.

or of the element.

ent.

dinates for Gauss integration.

lement based on the source point coordinates.

integrals.

ne (any string variable).

in three dimensions.

objects in the zone.

ent objects in the zone.

ch the user wants solution

es.

user-defined interioir points.

s the [K], [U] and {k} matrices.

related to head or gradient of which node in which

value to it.

ted boundary conditions in a file.

d gradients in a file.

in AutoCADs for drawing the geometry of the problem.

Fig. 3. Flowchart of the program.


After forming Eq. (26) for s zone in the media we can assemblethem in the following global system of equations:

½H�1 0 0 0

0 ½H�2 0 0

0 0 & 0

0 0 0 ½H�s

266664

377775

n�n

fjg1

fjg2

^

fjgk

8>>>><>>>>:

9>>>>=>>>>;

n�1

¼

½G�1 0 0 0

0 ½G�2 0 0

0 0 & 0

0 0 0 ½G�s

266664

377775

n�n

f@j=@ng1

f@j=@ng2

^

f@j=@ngs

8>>>><>>>>:

9>>>>=>>>>;

n�1

ð27Þ

Now assume that kth element and lth element in the globalmatrix in Eq. (26) are two neighbor nodes on the boundaries oftwo neighbor zones. We should apply the compatibility Eq. (23)and the equilibrium Eq. (25) in (27). This happens by removingthe lth column of the global [H] and adding it to the kth column,removing the lth row of {j}, removing the lth column of the

global [G], multiplying it by �CðkÞK ðkÞz =CðlÞK ðlÞz and adding it to the

kth column and removing the lth row of {q}. After repeating thisprocess for ni times in which ni is number of inter-zones, we thenhave an equation in the following form:

½H�n�ðn�niÞfFgðn�niÞ�1 ¼ ½G�n�ðn�niÞf@j=@ngðn�niÞ�1 ð28Þ

There are n equations and n unknowns in Eq. (28). Accordingto the prescribed boundary conditions, some unknowns are inright-hand side of the equations and some in the left-hand side ofit. So we should transfer all unknown variables to the left-handside and all known variables to the right-hand side. Then we willhave an equation in the following form:

½U�n�nfugn�1 ¼ ½K�ðnÞ�ðn�2niÞf@j=@ngðn�2niÞ�1 ¼ fkgðn�2niÞ�1 ð29Þ

which is a set of n equations with n unknowns that can be solvedfor unknowns.

7. Model development

A computer program was developed by the authors to do theanalyses on practical seepage problems, which deal with zonedanisotropic media using the BEM formulation and inter-zonal rela-tions described in the previous sections. Heterogeneity was consid-ered by employing different zones. Different zones can have differenthydraulic conductivities and can be isotropic or anisotropic, but eachzone should be homogeneous itself.

An ASCII data file should be prepared by the user to define theproblem for the program. The data file is consisted from datablocks for defining the physical model. The user is required toprovide an ID and hydraulic conductivities for each zone; ID,Coordinates and boundary conditions (head prescribed/gradientprescribed/inter-zone) for each node; ID, three node IDs for eachelement; IDs of inter-zone nodes on neighbor zones, which arepinned together; and coordinates of interior points, which theuser wants the solution for them. Some optional data blocks canalso be added for automatic mesh generation. 13-point Gaussianquadrature method was used for numerical integration and Gausselimination method was used for solving the system of equations.

Object-oriented concepts have been used in the programming.Different objects with a set of properties and methods have beendefined to effectively handle the computations. The quantity ofobjects and their properties and methods are too large to beaddressed in this paper but the most important objects with moreimportant properties and methods are listed in Table 1. Aflowchart of the program, which shows the general algorithm ofthe program is given also in Fig. 3.

8. Solved examples

Three numerical examples are analyzed to demonstrate thevalidity of the formulations that are presented in the paper. Allthree examples are analyzed in two isotropic and anisotropic cases.

8.1. Sheet-pile

In the first example we consider a sheet-pile problem. It isassumed that a sheet-pile of depth 4 m is driven into a canal of8 m water depth. The canal is placed on a layer of soil withhydraulic conductivity of Kx¼Ky¼Kz¼3�10�5 m/s for isotropiccase and Kx¼Ky¼3�10�5 m/s, Kz¼3�10�6 m/s for anisotropiccase. The thickness of this soil layer is 8 m and is placed on thebedrock, which is assumed to be impervious. Fig. 4 shows a 3Dsketch of the problem while a 2D section is provided in Fig. 5.

To model the problem of zero thickness sheet-pile, since thewall is infinitely thin, the left-hand side nodes occupy the samepositions as the right-hand side nodes. The result is that thecoefficient matrix has identical rows, and is therefore singular.The matrix is ‘correct’ in the sense that the correct solution wouldsatisfy the BIEM equations; however, the singular matrix leads toan infinite number of solutions, and the correct solution is onlyone of them. Displacing the left-hand side and the right-hand sidepoints vertically so that they do not occupy the same point butare collinear does not improve the situation [16]. Although suchproblems can be solved by zero-thickness elements in FEM (forexample see [17]) or nonlinear special elements in BEM [16], thisproblem can also be solved by incorporating two separate zonesin the left-hand side and right-hand side of the sheet-pile, i.e.downstream and upstream zones, respectively. Nodes on thesezones, which are on the sheet-pile have zero-flux Neumann

Fig. 4. Three-dimensional sketch of the sheet-pile problem.

Fig. 5. Section view of the sheet-pile example.

Fig. 6. Sheet-pile problem boundary mesh.


boundary conditions whilst the nodes on the interface betweenzones have the inter-zonal boundary conditions. The boundaryelement mesh of the model is shown in Fig. 6. Two zones (upstreamand downstream), 2292 nodes and 4576 triangular linear elementswere used in the model in which 1146 nodes and 2288 elementswere in each upstream and downstream zones and 100 nodes (50for each zone) were on the interface between the two zones. Datumhas been taken at z¼0 and the boundary conditions for these twozones are as in Table 2.

Executing the model, results are shown graphically in Fig. 7.Fig. 7(a) shows the total head on the bedrock (z¼0) for the isotropiccase while Fig. 7(e) shows it for the anisotropic case. The nature ofthis problem is two-dimensional and these results confirm it. As it isseen in Fig. 7(a) and (e), the total head on the bedrock is constant ona specific x-coordinate for any y-coordinate.

Total head for isotropic and anisotropic cases on a verticalplane on the axis of the river (y¼0 plane) are given inFig. 7(b) and (d), respectively. Flux vectors on this plane are also


given in Fig. 7(c) and (f). Comparing the results for isotropic andanisotropic cases, it is shown that there is a large difference in thepattern of the potential heads. In the anisotropic case, because thehydraulic conductivity in the vertical direction is taken one-tenth ofthe hydraulic conductivity in horizontal directions, water can moreeasily seep through horizontal directions versus through verticaldirection.

A sensitivity analysis has been performed for this example to seethe effect of longitudinal discretization on the accuracy of results.The results provided above are for the case that the x-axis is dividedinto 60 segments. The discretization on the y-axis and the z-axis hasnot changed in this analysis and the analysis is performed for theanisotropic case. Different discretizations for the x-axis have beenconsidered in which the x-axis is divided into 4, 6, 10, 20, 40, 60 and120 segments. To see the effect of mesh size along the x-axis, thepotential on an interior point is investigated. As the potential for aninterior point in BEM is calculated as a function of all boundarypotentials and normal fluxes, comparing the potential on an interiorpoint can reflect the global accuracy of the modeling. The interiorpoint of coordinates (�10, 0, 2) is considered for comparison. Totalhead values of the interior point for different meshes have beenreported in Table 3. Errors in the table have been calculated withrespect to the case that number of segments in the x-dir is 120. As thenumber of the segments along the x-axis increases it means that we

Table 2Boundary conditions of example 1.

Face Boundary condition

Upstream zone (�25 moxo0)

z¼0 @j/@n¼0

z¼8 m j¼10 m

x¼�25 m j¼10 m

x¼0 m, 0 mozo4 m Inter-zone conditions

x¼0 m, 4 mozo8 m @j/@n¼0

y¼�4 m @j/@n¼0

y¼4 m @j/@n¼0

Downstream zone (0oxo25 m)

z¼0 @j/@n¼0

z¼8 m j¼0

x¼25 m j¼0

x¼0 m, 0 mozo4 m Inter-zone conditions

x¼0 m, 4 mozo8 m @j/@n¼0

y¼�4 m @j/@n¼0

y¼4 m @j/@n¼0

Fig. 7. (a) Isotropic – potential head contour at z¼0, (b) isotropic – potential head co

contour at z¼0, (e) anisotropic – potential head contour at y¼0 and (f) anisotropic –

have a finer mesh, a more number of nodes and elements in themodel and hence more cost of calculations. Elapsed time ofcomputing has also been reported in Table 3 for comparison. Theelapsed time includes reading the user input file and writing all theoutput files.

8.2. Gravity dam

Although in the first example we had to use zoned media to beable to model the sheet-pile with BEM, the hydraulic conductiv-ities for both upstream and downstream zones were the same. Inour second example we are going to show how the presentedformulations in Section 5 work when two zones with differentproperties are coupled to each other. To do this we would want tomodel seepage beneath a gravity dam on top of a layeredfoundation. The dam is built on a river, which the reservoir depthis 10 m and the water depth in downstream side is zero. Founda-tion which the dam is built on consists of two layer of soil each4 m thick and over the bedrock, which is assumed to be imper-vious. In the isotropic case hydraulic conductivities of the toplayer of foundation are Kx¼Ky¼Kz¼3�10�5 m/s while hydraulicconductivities of the bottom layer of foundation are Kx¼Ky¼Kz¼

3�10�6 m/s. In the anisotropic case hydraulic conductivities of thetop layer of foundation are Kx¼Ky¼3�10�5 m/s, Kz¼3�10�6 m/swhile hydraulic conductivities of the bottom layer of foundation areKx¼Ky¼3�10�6 m/s, Kz¼3�10�7 m/s.

A three-dimensional sketch of the problem is provided in Fig. 8and sections of the geometry are given in Fig. 9. Three-dimen-sional mesh of the boundary is shown in Fig. 10. Two zones (topand bottom), 2371 nodes (1186 nodes for each zone) and 4736

ntour at y¼0, (c) isotropic – flux vectors at y¼0, (d) anisotropic – potential head

flux vectors at y¼0.

Table 3Effect for mesh fineness in the x-direction on accuracy.

Number of segments

in the x-direction

Elapsed time (s) Total head (m) Error (%)

4 2 2.73518 7.12

6 3 2.62083 2.65

10 6 2.57953 1.03

20 14 2.55972 0.25

40 48 2.55477 0.06

60 106 2.55381 0.02

120 522 2.55327 –

Fig. 8. Three-dimensional sketch for gravity dam example.

Fig. 9. Section view of the gravity dam example.

Fig. 10. BEM mesh for Gravity dam example.



Bottom layer

x¼�50 m j¼10 m

x¼50 m j¼0

y¼�10 m @j/@n¼0

y¼10 m @j/@n¼0

z¼4 m Inter-zone conditions

z¼0 @j/@n¼0

Top layer

x¼�50 m j¼10 m

x¼50 m j¼0

y¼�10 m @j/@n¼0

y¼10 m @j/@n¼0

z¼4 m Inter-zone conditions

z¼8 m, �50 moxo�10 m j¼10 m

z¼8 m, �10 moxo10 m @j/@n¼0

z¼8 m, 10 moxo50 m j¼0


linear triangular elements (2368 elements for each zone) and 726nodes on the interface of the two zones (363 nodes for each zone)were used in the model.

Datum has been taken z¼0 and the boundary conditions forthese two zones are as in Table 4.

The model was run and the graphical results are shown inFig. 11. The potential head on the bedrock (z¼0) and on theinterface between the two layers of the foundation (z¼4 m) forthe isotropic case are shown in Fig. 11(a) and (b), respectively.The potential head on the bedrock (z¼0) and on the interfacebetween the two layers of the foundation (z¼4 m) for theanisotropic case are also shown in Fig. 11(d) and (e), respectively.Distribution of the potential head on a vertical plane passingthrough the river axis is also shown in Fig. 11(c) and (f) for theisotropic and anisotropic media, respectively. Again and similar toexample 8.1, as there is no variation in the hydraulic conductiv-ities, geometry and boundary conditions along the y-axis, theproblem has a two-dimensional nature. The results provided inFig. 11 (a), (b), (d) and (f) also agree with this fact and it can beseen that the potential head is constant along the y-axis. Compar-ing the results for isotropic and anisotropic cases shows a greatdifference on the distribution of potential head. Because thehydraulic conductivity in the vertical direction is taken one-tenth

of the hydraulic conductivity in horizontal directions, water canmore easily seep through horizontal directions versus throughvertical direction.

Fig. 11. (a) Isotropic – potential head contour at bedrock (plane: z¼0), (b) isotropic – potential head contour at interface (plane: z¼4 m), (c) isotropic – potential head

contour on river mid-plane (plane: y¼0), (d) anisotropic – potential head contour on bedrock (plane: z¼0), (e) anisotropic – potential head contour at interface (plane:

z¼4 m) and (f) anisotropic – potential head contour on river mid-plane (plane: y¼0).

Fig. 12. Total head comparison (a) anisotropic and isotropic cases at y¼0, z¼0

(b) anisotropic and isotropic cases at y¼0, z¼4 m.


To verify the results, a two-dimensional fine mesh FEM analysiswith Seep/W application was performed. We have checked the FEMresults by a FEM mesh, which the element sizes were halved and gotapproximately the same output. This shows that the FEM mesh isfine enough. The FEM mesh includes 909 nodes and 800 quad-rilateral two dimensional elements. Calculated values of BEM andFEM values of total head are compared along a horizontal line alongthe river axis on the bedrock (y¼0, z¼0) and on the interface of thetwo soil layers (y¼0, z¼4 m) in both isotropic and anisotropic cases.The values of the potential head in then plotted in Fig. 12. Thecomparison of the results of our BEM model with the FEM modelshows the accuracy of the model and inter-zonal formulations.

8.3. River with banks

Although the computer program used and the undertakenmodels in the previous two examples were all three-dimensional,both examples could be modeled by a two-dimensional computerprogram and formulations. Here we are giving an example that isthree-dimensional in nature and cannot be analyzed with two-dimensional models. Similar to the gravity dam problem, againwe have a river with a barrier (dam), which does not allow thewater to flow to the downstream of the river through the canal.The water seeps through the foundation and banks to the down-stream. Presence of the banks in this example makes it a purethree-dimensional problem that cannot be modeled in twodimensions. It has been shown for the same problem that thetwo-dimensional simplification of the problem brings errorswhich are not negligible [18]. The river is of 10 m width and thewater depth in the upstream is 2 m. Water depth at the down-stream is assumed to be zero. The river is assumed to have twobanks of 10 m width and the depth of bottom pervious layer ofthe river is 6 m. Again we assume two isotropic and anisotropiccases. In isotropic case, the hydraulic conductivity of the soil isKx¼Ky¼Kz¼3�10�5 m/s while in the anisotropic case the hydrau-lic conductivities are taken Kx¼Ky¼3�10�5 m/s, Kz¼3�10�6 m/s.Three-dimensional sketch of the problem is shown in Figs. 13and 14. The BEM mesh is shown in Fig. 15. 2778 nodes and 5552linear triangular elements were used in the model.

Datum has been taken z¼0 and the boundary conditions forthe problem are given in Table 5.

After the model has been executed for the problem, thegraphical results are provided in Fig. 16. The potential headcontours are shown in several planes. The head contour at thebedrock (z¼0) is shown in Fig. 16(a) for isotropic case and inFig. 16(d) for anisotropic case. The potential head contour on a

Fig. 13. Three-dimensional sketch for river with banks example.

Fig. 14. Geometrical dimensions.

Fig. 15. BEM mesh for river banks example.


vertical plane passing through the river axis is shown in Fig. 16(b)and (e) for isotropic case and anisotropic case, respectively. Thepotential head in the top view of the region is also shown inFig. 16(c) and (f) for isotropic case and anisotropic case, respectively.The top view shows the potential head contours of the river bed andthe top surfaces of the two banks. The results in this example show adifference with the two previous examples. The previous twoexamples had a two-dimensional nature, i.e. although they werethree-dimensional problems, they also could be solved by two-dimensional analyses. In this example the geometry and boundary

conditions are not constant in different y-coordinates. This fact isreflected in the results also. The head contours are not constantalong a specific y-coordinate. This shows that the program cannot beanalyzed with a two-dimensional analysis. Anisotropy also plays alarge role in results. Comparing potential head in the bedrock, i.e.Fig. 16(a) and (d), shows a large difference. The hydraulic conduc-tivity in the anisotropic case is one-tenth of the hydraulic conduc-tivity in the isotropic case in vertical direction while the hydraulicconductivities are the same in horizontal directions (both the x- andy-directions). The different pattern of the potential head in


anisotropic case is because the water can seep more easily inhorizontal directions than the vertical directions. There is a largedifference between the two cases on the bedrock potentials. Suchdifference is more in the centerline of the river in comparison withthe farther areas in beneath of the banks. The reason is that the areabelow the river is vertically under the river bed which has abruptchanges in boundary conditions and so the potential in this area ismore dominated by the vertical hydraulic conductivity. As the



x¼�30 m j¼8 m

x¼30 m j¼6 m

y¼�15 m @j/@n¼0

y¼15 m @j/@n¼0

y¼�5 m, 6 mozo8 m, �30 moxo�6 m j¼2 m

y¼�5 m, 6 mozo8 m, �6 moxo6 m @j/@n¼0

y¼�5 m, 6 mozo8 m, 6 moxo30 m j¼0

y¼5 m, 6 mozo8 m, �30 moxo�6 m j¼2 m

y¼5 m, 6 mozo8 m, �6 moxo6 m @j/@n¼0

y¼5 m, 6 mozo8 m, 6 moxo30 m j¼0

z¼0 @j/@n¼0

z¼8 m @j/@n¼0

y¼6 m (�30 moxo6 m) j¼2 m

y¼6 m (�6 moxoþ6 m) @j/@n¼0

y¼6 m (þ6 moxoþ30 m) j¼0

Fig. 16. (a) Isotropic – potential head contour at bedrock (plane: z¼0), (b) isotropic – po

contour in the top-view, (d) anisotropic – potential head contour at bedrock (plane: z¼

(f) anisotropic – potential head contour in the top-view.

vertical hydraulic conductivities are different in isotropic andanisotropic cases, the difference on the bed is normally great. Thedifference in potential head patterns in areas below the banks on thebedrock is also great but the difference is less than the area belowthe river. This is because that the pattern of the potential head inthese areas is under effect of both horizontal and vertical distancesbetween these areas and the river bed area. The equal hydraulicconductivities in horizontal directions can slightly balance thedifference, which is implied by the vertical hydraulic conductivitydifference. On the contrary in top view on banks, the differencebetween the potential head patterns in isotropic and anisotropiccases is little. This is because that these areas have more horizontaldistance than vertical distance with the river bed and thus thepattern of the potential head is mainly ruled by the horizontalhydraulic conductivities which are equal on isotropic and aniso-tropic cases.

To verify the results of our BEM analysis, two-dimensionalFEM models cannot be used as the nature of the geometry of theproblem is three-dimensional. Thus a three-dimensional finiteelement analyses has been conducted for isotropic and anisotro-pic cases incorporating a fine mesh. We have used the Seep3DFEM application for verification and checked the results by amesh, which the element sizes were halved and got the sameoutput. This shows that the FEM mesh is fine enough. 15,921nodes and 13,200 hexahedron elements were used in the FEMmesh. Calculated values of BEM and FEM values of total head hasbeen compared on the river axis on the bedrock (y¼0, z¼0). Thesecalculated values have been plotted in Fig. 17 for comparison.

tential head contour on rivermid-plane (plane: y¼0), (c) isotropic – potential head

0), (e) anisotropic – potential head contour on river mid-plane (plane: y¼0) and

Fig. 17. Comparison of results of a FEM and BEM on river axis on the bedrock.


Results show that the calculated values of FEM and BEM analysesare almost equal.

9. Discussions and conclusions

The goal of this research was to develop a computer program,which is capable of solving three-dimensional confined problems,which are anisotropic and contain multiple domains with differ-ent hydraulic conductivities. The code will then be upgraded forunconfined problems in multi-domain anisotropic zones in thefuture. As the theory is well-developed for two dimensionalunconfined problems in isotropic media, we used the stretched-coordinate transformation to deal with Laplace equation for BEM.The interface equilibrium equations in stretched coordinates andtransformed Neumann boundary condition are presented in thepaper. These equations then are used in the traditional BEMframework for Laplace equation to develop a computer programfor analyses. Equilibrium equations for three-dimensional multi-domain problems have been used by various examples andverified with FEM analyses. Object-oriented concepts have beenused in the programming to be more easily expandable for morecomplicated problems in the future. The advantages of thepresented model are:

–
Using BEM makes it possible to discritise only the boundaries. – Using a stretched coordinate transform makes it possible to
reduce the governing equation of the seepage problem inanisotropic media to Laplace equation that has been success-fully solved by two and three dimensional BEM previously.

–
Using Laplace equation makes it easier for further expansionsto the model for unconfined seepage as it has been usedsuccessfully for two-dimensional geometries.
–
Developing the inter-zonal equilibrium equations in three-dimensions lets the true coupling of the neighbor anisotropicdomains in multi-domain problems.
However there are some limitations in the model:

–
Darcy flow has been considered. Non-Darcy flows or unsatu-rated seepage cannot be modeled.
–
The model can only be used with confined problems. The free-surface and moving boundary problems cannot be solved withour model in the present format.
–
Multi-domain strategy is used to deal with heterogeneity. Eachzone should be homogeneous inside but can be anisotropic.
–
The program can only receive Dirichlet, Neumann or inter-zonal boundary conditions. The boundary conditions cannot bea function of time and thus only steady-state problems can besolved.
We have solved several examples of multi-domain seepageproblems that are frequently arisen in the engineering practice.The results show that the model is capable of modeling suchproblems with thousands of nodes and elements. We haveverified our results with 2D and 3D FEM models and showedthat the results are accurate.

In the future we wish to improve the program to be able tomodel transient unconfined multi-domain anisotropic problems.

Acknowledgments

We are grateful for constructive comments of Professor A.Cheng and two anonymous reviewers, which helped improvingthe final manuscript.

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[17] Luo Guan-Yang, Hong Cao. Using zero-thickness elements to simulatesuspended cut-off walls in a regional seepage field. Comput Geotechnics2006;33(6–7):305–15.

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Documents

Engineering Analysis with Boundary Elementssharif.ir/~ataie/Printed-Articles/EABE_2012.pdf · Seepage Boundary element method Anisotropy Heterogeneous Multi-domain Multi-region Zoned