a ﬁnite element approach to solve contact problems in geotechnical engineering

8/2/2019 A finite element approach to solve contact problems in geotechnical engineering

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INTERNATIONAL JOURNAL FOR NUMERICAL AND ANALYTICAL METHODS IN GEOMECHANICSInt. J. Numer. Anal. Meth. Geomech., 2005; 29:525550Published online in Wiley InterScience (www.interscience.wiley.com). DOI: 10.1002/nag.424

A finite element approach to solve contact problems in

geotechnical engineering

Jianqiang Maon,y

Department of Geotechnical Engineering, Southwest Jiaotong University, Chengdu, Peoples Republic of China

SUMMARY

This paper presents a finite element approach to solve geotechnical problems with interfaces. Thebehaviours of interfaces obey the MohrCoulomb law. The FEM formulae are constructed by means ofthe principle of virtual displacement with contact boundary. To meet displacement compatibilityconditions on contact boundary, independent degrees of freedom are taken as unknowns in FEM

equations, instead of conventional nodal displacements. Examples on pressure distribution beneath a rigidstrip footing, lateral earth pressure on retaining walls, behaviours of axially loaded bored piles, a shield-driven metro tunnel, and interaction of a sliding slope with the tunnels going through it are solved withthis method. The results show good agreement with analytical solutions or with in situ test results.Copyright# 2005 John Wiley & Sons, Ltd.

KEY WORDS: interface; contact problem; principle of virtual displacement; finite element; geotechnical

1. INTRODUCTION

The interfaces, such as interface between the structure and geotechnical media, the joints in rock

mass, failure surfaces in soil mass, etc., effect the mechanical behaviours of the structures and

the surrounding geotechnical media significantly. Various numerical interface models have been

developed by Goodman et al. [1], Zienkiewicz et al. [2], Clough and Duncan [3], Ghaboussi et al.

[4] and Desai et al. [5]. These models or their derivative forms have been widely applied in

geotechnical engineering, as they can be readily incorporated into an FEM process. However, a

common and distinct disadvantage of these models is that additional constitutive equations and

mechanical parameters have to be employed for the interfaces.

As we know, in spite of the variety of interfaces, the following characters are essential for

an interface model, as shown in Figure 1: (1) if the tangential force at a point on interface

reaches the limiting value of resistance, relative slip will occur, else will keep in sticking status

(2) whether relative slip occurs or not, the contact bodies cannot penetrate each other in

any way.

Received 21 January 2004Revised 10 November 2004Copyright# 2005 John Wiley & Sons, Ltd.

yE-mail: [email protected]

nCorrespondence to: J. Mao, Department of Geotechnical Engineering, Southwest Jiaotong University, Chengdu,Peoples Republic of China.


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In fact, almost all the problems containing interfaces can be uniformly described as a contact

problem through the following mathematical form: assuming the two contact bodies are O and

O0; respectively, and on the contact boundary (i.e. the interface) Gc:

(1) n is the normal direction 2 O and s is the slipping direction. Accordingly, es and es0 are

unit vectors corresponding to s and s0 (opposite to s: Also we note that for two-

dimensional problems, the possible slipping direction is uniquely along the tangential

direction, while for three-dimensional problems, the slip will occur in the direction of

maximum tangential force, which will depend on the contact forces at the point.

(2) The contact forces on O0 are pn and ps; and on O are p0n and p0s:(3) The displacements in incremental form (for contact problems are nonlinear) are du0n; du

0s

and dun; dus: Moreover, d0n is the initial gap between contact bodies (for most of

geotechnical problems, d0n 0).

(4) The slip on the interface is govern by the MohrCoulomb law

jpsj cB mjpnj 1

where cB and m are cohesion and coefficients of friction of the interface, respectively.

When slip occurs, the cohesion at the slipping point will be lost, i.e. cB 0: The

conditions to be satisfied on interface are summarized in Table I.

There are many contact problems in mechanical and civil engineering, hence numerous efforts

has been made in last few decades to develop corresponding numerical algorithms. A number oftechniques such as iteration method [68], mathematics programming [912], penalty method

[13], Lagrange multiplier [1416], have been developed and used in mechanical engineering

problems. However, these methods are rarely applied in geotechnical engineering [17, 18] due to

their complex techniques in computation. A practical approach applicable to geotechnical

problems is presented in this paper.

np

np

np

np

sps

p

npn p

relative slip along tangential direction

contact status when slip does not occur

ns

cannot penetrate each other in normal direction

contact status when slip occurs

a pair of contact points

Figure 1. Illustration of the contact forces and displacements on interface.

Copyright # 2005 John Wiley & Sons, Ltd. Int. J. Numer. Anal. Meth. Geomech. 2005; 29:525550

J. MAO526


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2. FEM FORMULATION BASED ON THE PRINCIPLE OF VIRTUAL WORK

WITH CONTACT BOUNDARY

2.1. Principle of virtual work for contact problems

The principle of virtual work is one of the important principles in analytical mechanics

concerning the kinematics of particles or rigid bodies. For a deformable body, if the body is

divided into a series of infinitesimal elements and is taken as particles in sense of kinematics,

then the whole body will lead to an assembly of particles.

Figure 2 illustrates three different types of infinitesimal elements: (1) in bodies O and O0; (2) on

stress boundaries Gs and G0s; and (3) on contact boundary Gc: Gc consists of sticking part Gc1 and

slipping part Gc2 ; which Gc Gc1 [ Gc2 : The infinitesimal elements are same as those used in

continuous mechanics to construct the equations of equilibrium. The expressions of resultant

forces acting on different infinitesimal elements are summarized in Table II.

According to the principle of virtual displacement, the virtual work of whole system should be

equal to zero. So we obtainZO

r r %FF du dO

ZO

0

r0 r %FF0 du0 dO

ZGs

%pp p du dG

ZG

0s

%pp0 p0 du0 dG

ZGc1

pn p0n dun ps p

0s dus

p0n pn du0n p

0s ps du

0s dG

ZGc2

pn p0n dun p

0n pn du

0n ps mjp

0njes0 dus

p0s mjpnjes du0s dG 0 2

where du; du0 are virtual displacements which should satisfy the displacement conditions on

displacement boundary Gu; G0u and contact boundary Gc in prior, i.e.

du 0 on Gu; du0 0 on G0u 3a

Table I. Conditions to be satisfied on interface.

Traction DisplacementContact status

p p0 s n s

Sticking or ps5cB mjpnjes dun du0n

dus du0s

pn p0n

p0s5cB mjp0njes0

Slipping and ps mjpnjes jun u0nj d

0n dus=du

0s

ps p0s p

0s mjp

0njes0


A FINITE ELEMENT APPROACH TO SOLVE CONTACT PROBLEMS 527


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dun du0n; dus du

0s on Gc1 3b

dun du0n; dus=du

0s on Gc2 3c

Figure 2. Stresses and resultant forces on infinitesimal elements.

Table II. Resultant forces acting on infinitesimal elements.

Infinitesimal Direction of the Resultant of the forceselement resultant force on infinitesimal element Virtual work

In O x1; x2; x3 r r %FF dO r r %FF du dO

In O0 x1; x2; x3 r0 r %FF0 dO r0 r %FF0 du0 dO

On Gs x1; x2; x3 %pp p dG %pp p du dG

On G0s x1; x2; x3 %pp0 p0 dG %pp0 p0 du0 dG

On Gc1 (to O) n pn p0n dG pn p

0n dun dG

s ps p0s dG ps p

0s dus dG

On Gc1 (to O0) n0 p0n pn dG p

0n pn du

0n dG

s0 p0s ps dG p0s ps du

0s dG

On Gc2 (to O) n pn p0n dG pn p

0n dun dG

s ps mjp0njes0 dG ps mjp

0njes0 dus dG

On Gc2 (to O0) n0 p0n pn dG p

0n pn du

0n dG

s0 p0s mjpnjes dG p0s mjpnjes du

0s dG


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Substituting Equations (3b), (3c) and

ZO

r r du dO

ZG

p du dG

ZO

r : de dO

Z

G

pn dun ps dus dG Z

O

r : de dO 4a

ZO

r0 r du0 dO

ZG

p0 du0 dG

ZO

r0: de0 dO

ZG

p0n du0n p

0s du

0s dG

ZO

r0: de0 dO 4b

into Equation (2), finally we get

ZO

r : de dO Z

O0r0 : de0 dO

ZO

%FF du dO Z

O0 %FF0 du0 dO

ZGs

%pp du dG

ZG

0s

%pp0 du0 dG

ZGc1

pn p0n dun ps p

0s dus dG

ZGc2

pn p0n dun

mjp0njes0 dus mjpnjes du0s dG 5

The above equation form the basis to construct the FEM formulae in the next section.

In the case that one of the bodies, such as O0; is stiff enough to be taken as a rigid body, the

virtual displacement of a generic point in O0 may be expressed as

du0 dV0 R0 dH0 A0dV0

dH0

( ) A0 dU0 6

where V0 is the translation of the rigid body, and H0 is the rotation. R0 is not a tensor and its

components are

R0

ij

X3

k1

eijkx0

k

where eijk is Ricci Symbol, x0k k 1; 2; 3 is co-ordinate of the point (the origin is designated at

the centre of rotation).

The equation corresponding to rigid body-deformable body contact can be obtained by

substituting Equation (6) into (5). The expression is omitted here on.




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2.2. Finite element equilibrium equations

After the discretization, Equation (5) takes the following form:

XZO

e

O0e

BTr dO XZO

e

O0e

NT %FF dO XZG

e

sG0e

s

NT %pp dG

XZ

Gec1

NTnpn N0Tn p

0n dG

XZ

Gec1

NTs ps N0Ts p

0s dG

XZG

ec2

NTnpn N0Tn p

0n

XZ

Gec2

NTs mjpnj N0Ts mjp

0nj dG 7

where B is straindisplacement matrix, N (and N0 is displacement interpolation matrix, and

similarly Ns and N0sNn and N

0n are tangential (normal) displacement interpolation matrices.

Direct solution of equilibrium equation (7) is a very complicated procedure, hence somesimplifications are made.

In Equation (7), tractions pn;ps and p0n;p

0s on contact boundary G

ec1

are computed with the

stresses of different elements Oe and O0e; respectively, and are not equal. To simplify the

computation, takingPR

Gc1NTnpn;

PRGc1

NTs ps andPR

Gc1N0Tn p

0n;PR

Gc1N0Ts p

0s as nodal contact

forces corresponding to pn; ps and p0n; p

0s on Gc1 ; as shown in Figure 3, we can assume

approximately XZG

ec1

NTnpn N0Tn p

0n dG

XG

ec1

Pni P0ni 0 on Gc1

XZG

ec1

NT

sp

s N0T

sp0

s dG X

Gec1

Psi

P0

si 0 on G

c1

niPniP

siP

siP

n

s

e

e

e

e e

e

( )ni sP

( )ni sPniP

iniP

2

ec1

ec

(a) (b) (c)

Figure 3. Nodal contact forces on contact boundary: (a) Definition ofn and s at contact node i; (b) contactforces on Gec1 ; and (c) contact forces on G

ec2:


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Similarly, we obtainXZG

ec2

NTnpn N0Tn p

0n dG

XG

ec2

Pni P0ni 0 on Gc2

and ZG

ec2

NTs mjpnj dG XG

ec2

mjPnijs

ZGc2

N0Ts mjp0nj dG

XG

ec2

mjPnijs0

in which mjPnijs and mjPnijs0 are tangential nodal contact forces on Gc2 :

Finally, we obtain XZO

eO0eBTr dO F Rc 8

in which

F XZOeO0e

NT %FF dO XZGesG0es

NT %pp dG

is nodal force corresponding to the applied load, and

Rc XG

ec2

mjPnjs XG

ec2

mjPnjs0

is the nodal contact force on the Gc2 : If all contact nodes are in sticking state, Rc will be equal to

zero and the equation becomes same as conventional FEM. For other contact statues, the

expressions ofRc will be given in next section.

2.3. Computational procedure for elasto-plastic contact problems

By applying the initial stress method usually used for elasto-plastic materials to elasto-plastic

contact problems, Equation (8) can be expressed in the following incremental form:

KmDUm DFm DFmep DRmc 9

where superscript m denotes the application of mth load increment, DU is incremental nodal

displacement vector, DF is load increment vector, DRc is the increments ofRc; K is the stiffness

matrix which depends on the contact status (see Appendix A for details), and

DFmep XZ

OeO0e

BTDrme Drmep dO 10

is the residual force vector in which subscripts e and ep denote elastic and elasto-plastic,

respectively.

By uncoupling Equation (9) to contact iteration and elasto-plastic iteration , we obtain:

Kcmn dU

mn dRc

mn 11a

Kmn dUmn dFep

mn 11b

in which subscript n denotes iterative step and prefix d implies the increments of variables after

an iteration. The corresponding iterative procedure in a load step is illustrated in Figure 4.




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Theoretically, the iteration may begin with either contact iteration or elasto-plastic iteration

and go on till the contact status of all contact nodes does not change and there is no increase in

new failure points. In practices, it seems work better that the procedure begins with contact

iteration.

The iteration of Equation (11b) is same as normal elasto-plastic iteration procedure. But some

details concerning iteration of Equation (11a) will be elaborated further.

2.3.1. dU and Kc. Besides sticking status and slipping, there can exist separating status and

penetrating status. The former implies that a pair of formerly sticking nodes are separated as the

deformation develops. The latter implies that formerly apart nodes may penetrate each other in

the process of contact, since we cannot estimate the magnitude of the displacements in prior.

Obviously, penetrating status is not physically allowable, so the overlapping part should beforcibly pushed back (see the following text). The four types of contact statuses are illustrated

in Figure 5 and corresponding criterions are shown in Table III, in which P and U denote nodal

contact force and displacement, respectively, and CB is equivalent nodal cohesion corresponding

to cohesion cB:

To satisfy the displacement conditions on contact boundary, firstly, nodal displacement

increments are defined under local co-ordinate system, as shown in Figure 6. Secondly, as shown

in Table IV, independent displacement increments are taken as unknowns, instead of assigning

displacement increments for every displacements of contact nodal pairs. Hence, the order of

Equation (11a) is the sum of independent displacement increments and is variational with the

contact status.

Noted that dRcmn in Equation (11a) could not be straightaway evaluated with

dRcmn Rc

mn1 Rc

mn 12

for the order of Rcmn1 may not equal the order of Rc

mn due to the change of contact status

from n 1th to nth iteration.

In order to illustrate how to make Rcmn cooperated with Rc

mn1; here we assume that a

nodal pair i change from sticking status into slipping status after n 1th iteration.

Elasto-plastic iteration

with equations (11b)

Convergent?

contact iterationwith equation (11a)

Convergent?

T T

TT

F

F F

F

Iteration end

Contact status changedafter iteration with (11b) ?

Failure points increasedafter iteration with (11a) ?

Figure 4. Iterative procedure.


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The equation of n 1th iteration is

Kcmn1dU

mn1 dRc

mn1 13

d nU

d nU

dsU

d sU

e

e

Figure 6. Nodal displacement increments on contact boundary.

nP

sP

nP

nP

0

0

sticking slipping

separating penetrating

nS

Figure 5. Contact statuses in computation.

Table III. Contact status and corresponding criterion.

Status Criterion

Sticking Pn50 and jPsj5CB mjPnj

Slipping Pn50 and Ps5CB mjPnj

Separating Pn50

Penetrating Un U0n d

0n > 0




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in which

dUmn1 dUsi T

before nth iteration, we have:

Rcmn 0

T

Since the slipping nodal pair has two independent displacement increments in tangential

direction, thus:

dUmn dUsi dU0si

T

Accordingly, dRcmn will be evaluated with

dRcmn Rc

mn1 R

0c

mn 14

in which

Rcmn1 mjPnij mjPnij

T

and Rcmn in Equation (12) has been equivalently replaced with

R0cmn Psi Psi

T

as shown in Figure 7.

Table IV. Displacement increments of a contact nodal pair (for two-dimensional problem).

Defined displacement Number of independent The displacement conditionStatus increments increments should be satisfied

Sticking dUndU0n; dUsdU

0s 2 Un U

0n d0 0

Slipping dUndU0n; dUs; dU0s 3 Un U0n d0 0Separating dUn; dU

0n; dUs; dU

0s 4 Un U

0n d050

Penetrating dUn; dU0n; dUsdU

0s 3 Un U

0n d0 0

d siU

iP

s

d niU

d siU

i

i

d niU

d siU siP

e

c

d niU

e e

e

e

Figure 7. Equivalent transformation of displacements and forces at a sticking nodal pair.


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2.3.2. Rc. Among the above-mentioned four different status, penetrating status is

a special case. As shown in Figure 8, the overlapping part in normal direction at nodal pair

j is

DSnj Unj U0nj d0j 15

the normal incremental displacements which have to be pushed back are dSnj2 O and

dS0nj2 O0; respectively,thus

dSnj dS0nj DSnj 16a

or

dSnj ZDSnj dS0nj Z0DSnj Z Z0 1 16b

In computation, it is difficult to determine the exact values of Z and Z0 beforehand, but that has

little influence on the final result as long as the load increments are small enough. Usually, if the

stiffness of the two contact objects are close, we may take Z Z0 12: While if one (such as O0) is

much stiffer than the other O; we may take Z 1 and Z0 0; so that dSnj DSnj and dS0nj 0:

Adjusted contactboundary

njS

i

j d njS

n

s

d njS

cd siR

cd niR

e

e

Figure 8. Diagram for evaluating dRc at penetrating status.

Table V. Nodal tractions Rc at nodal pair i:

O O0

Status Normal Tangential Normal Tangential

Sticking Pni Psi Pni Psi

Slipping Pni mjPnij Pni mjPnij

Separating 0 0 0 0

Penetrating dRcin PNP

j

kinjn dSnj k0injn

dS0nj

dRcis PNP

j

kis jn dSnj k0is jn

dS0nj




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Sequentially, dRc is evaluated with

dRcin

XNP

j

kin jn dSnj k0in jn

dS0nj 17a

dRcis XNP

j

kis jn dSnj k0is jn

dS0nj 17b

where subscript n and s denote normal and tangential directions, respectively, and i is all nodes

(including j itself) of the element to which j belongs. k denotes stiffness component. NP is the

sum of penetrating node pairs. Finally, nodal contact forces corresponding to different contact

status are summarized in Table V.

2.4. Demonstration of the behaviours of interface when taken as a contact problem

To demonstrcate the effects to treat interface as a contact problem, a simple example similar to

direct shear test is given in Figure 9. Initially the vertical load s is applied. As the horizontal

load t increases, relative slip between the contact blocks will occur and this will lead to fast

increase of the horizontal displacement of upper block (represented with point A on the

interface). When load t reaches to tmax 2ms 0:6s; the displacement of upper block becomes

0.5

0.5interface)( 3.0= Square 1 1

10000

0.2

E

=

=A

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.50.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8 No interface

3.0=

2.0=

1.0=

Load

Horizontal displacement (10-4)

(a)

(b) (c)

Figure 9. A simple example to demonstrate the behaviours of interface by taking as contact problem:(a) general description; (b) finite element mesh; and (c) relation between horizontal displacement

and load under different vertical loads.


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infinite, which implies that all nodes on interface have slipped. It is noted that no special stress

strain relation is introduced to simulate the mechanical behaviours of interface as in most

interface models at present and the properties of interface fully depends on cB; m of the interface

and properties of the deformable bodies.

3. EXAMPLES

In this section, several representative examples are presented to assess the validity of the

method developed in this paper. In most examples, the geotechnical materials are treated as

2.4 m

Interface

Infinite

P

M

3

o

26MPa

0 3

18kN/m

30

0

E

.

c

=

=

=

=

=

(a) (b)

Figure 10. Rigid strip footing: (a) general description; and (b) finite element mesh (a half).

-1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.20246

8101214161820222426283032343638

Analytic solution (P=40kN,e=0)FEM solution (P=40kN,e=0)Analytic solution (P=40kN,e=0.4m)FEM solution (P=40 kN,e=0.4m)

Pressure(kPa)

Horizontal coordinate (m)

Figure 11. Pressure distributions beneath the footing cB 0; m 0:0:




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ideal elasto-plastic materials and obey MohrCoulomb yield criterion. Otherwise, four-node

isoparametric element is employed in computation. For material parameters, elastic modulus,

Poisson ratio, unit weight, internal friction angle and cohesion are denoted as E; n; g; f; c;

respectively.

3.1. Contact pressure beneath strip rigid footing

As shown in Figure 10, a rigid strip footing is applied with line distributed force P or moment

M: This is a rigid-deformable contact problem. The curves shown in Figure 11 are the contact

pressure distributions under load P and M; respectively, when the interface is frictionless m

0:0 and the foundation is elastic. The FEM and analytical solutions show good agreement.

Figure 12 shows the pressure distributions corresponding to the increase of P when the interface

is frictional m 0:2 and the foundation is elasto-plastic.

3.2. Lateral earth pressure on rigid retaining wall

At first, we compute a rigid retaining wall with frictionless interface m 0; as shown in

Figure 13. The computational lateral pressuredisplacement curve is shown in Figure 14. Also,

we can get the lateral pressure distributions on wall back corresponding to different

displacements of the wall. As an illustration, Figure 15 gives the pressure distributions when

the displacements are 0.0 and 0.004. We can see that all the computational results show very

good agreement with analytical solutions [19]. Especially, it is worthwhile to note that if the

interfaces are ignored (take m 1), the computational results (plotted with the dash line) will

deviate far away from the correct solution. Therefore, it is critical to simulate interfaces with

reasonable model in computation.

The second example is an in situ test for lateral pressure on retaining wall shown in Figures

16(a) and (b). Figure 17 depicts the computational result of the displacement of the wall and

-1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.20

50

100150

200

250

300

350

400

450

500

550

600

650

700

750

P=80kN/mP=240kN/mP=400kN/mP=560kN/mP=720kN/m

Pressure(kPa)

Horizontal coordinate (m)

Figure 12. Pressure distributions beneath the footing under different loads cB 0; m 0:2:


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soil, in which relative slip along the interfaces is obvious. Figure 18 shows the lateral

pressure distributions obtained with in situ test and computation, respectively. We may observe

the similarity between the test and computation results. Moreover, the maximum pressures

gained by computation and test are 44.1 and 44:4 kPa; respectively, though the locations

are different.

3.3. Behaviours of axially loaded bored piles

The two piles computed have come from a in situ test studying the behaviours of axially loaded

bored piles in the Xigeda mudstone , which is a special mudstone distributed in south of Sichuan

16.7mRigid Retaining Wall

Interface

backfill

Interface

Interface3

o

26MPa

0 3

18kN/m

300

E

.

c

=

=

=

==

(a)

(b)

Figure 13. Rigid retaining wall: (a) general description; and (b) finite element mesh.

-0.010 -0.005 0 .0000

20

40

60

80

100

120

-0.05 0.00 0.05 0.10 0.15 0.20 0.25 0.30

0

100

200

300

400

500

600

700

Towards the backfillaway from the backfill

Analytic solution

FEM solution

Earthpressure(kN)

Displacement of the wall (m)

Figure 14. Lateral earth pressure against displacement of the wall.




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province, Peoples Republic of China. The Pile 1 is impact boring pile and the Pile 2 is hand

digging pile with open end. To study the mudstone exclusively, the overburden around piles was

replaced with loose fill during the test. The piles were also wrapped with sheet iron to reduce

the friction between pile and fill. The characterization of the piles and strata are shown in

Figure 19(a). The elastic modulus E was obtained from an in situ plate loading test, and c; f

of the mudstone were measured in the lab. The other strength parameters of the interface

are shown in Table VI, in which m and mR are the coefficient of friction before and afterslip occurs.

Obviously, this can be computed as an axisymmetric problem. Figure 20 depicts the

development of relative slip range along the skin of Pile 1 with the increase of load, while

Figure 21 shows failure zone near the bottom of the pile. In Figure 22, we can see that

the computational loadsettlement curve of Pile 1 shows good agreement with the test curve on

the whole. However, when the load exceeds 6000 kN; the development of computational curve is

smoother than the test curve. In fact, as the load increases, failure surface occurs and grows in

the mudstone near the base of the pile, which cannot be completely simulated with the ideal

elasto-plastic model. Also, the result ignoring the interface is given in Figure 22, which is nearly

linear and has little similarity with the test curve.

Pile 2 was cast as an open-end pile designed to investigate the skin friction exclusively.

Unfortunately, the test load did not achieve ultimate load due to the under estimation of thepiles bearing capacity. As shown in Figure 23, the loadsettlement curve of test and

computation are almost linear before ultimate load. In computation, however, when the load

approaches 14 300 kN; the iteration becomes divergent, which implies that all nodes on interface

have slipped and the pile lost support so as to subside in whole.

5

4

3

2

1

0

-50 -40 -30 -20 -10 0 10 20 30 40

Analytical solutionEarth pressure at rest

(FEM solution, ignoring interfaces)

Earth pressure at rest

(FEM solution)

Earth pressure when the displacement

of the wall is 0.004m towards backfill

(FEM solution)

Depth(m)

Earth pressure (kPa)

Figure 15. Lateral earth pressure distributions.


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3.4. A shield-driven metro tunnel

As an example for contact problem with initial gap between the two contact bodies, a shield-

driven metro tunnel is computed. Usually, the gap between lining and surrounding, which

mainly arose due to the difference between the external diameter of shield machine and

that of the lining, can hardly be fully filled. The characterization of the tunnel, strata and

computational mesh are shown in Figure 24. In this computation, we take the maximum

gap d0max 25 mm: The loads from buildings, vehicles, etc. are simplified to a 20 kPadistributed load.

The settlement of ground surface, the internal forces in the lining, etc. can be obtained from

the computation results. However, we focus our interest on the contact process between the

lining and the surrounding strata here. In Figure 25, 20, 40, 100% are the percentages that the

excavation loads were released. The results showed that the contact begins at the top and

Excavation boundary

1:0.

16

3070mm

5000mm

5200mm

3200mm

1200mm

1:

0.65

Backfill

InterfaceInterface

)3.0( =

)5.0( =

Interface (=3.0)

3

o

15MPa

0 3

20kN/m

25

50kPa

E

.

c

=

=

=

=

=

3

o

5MPa

0 3

18kN/m

15

20kPa

E

.

c

=

=

=

=

=

3

o

2 5MPa

0 35

18 9kN/m

5

17kPa

E .

.

.

c

=

=

=

=

=

(a) (b)

(c)

Figure 16. A retaining wall in engineering: (a) profile; (b) parameters of the soils and interfaces;and (c) finite element mesh.




18/26

bottom of the tunnel, then develops towards the middle, and the lining and the surrounding

strata become fully contacted in the end.

3.5. Tunnels in landslide

The two tunnels are on the ChengduKunming Railway in southwest of Peoples Republic of

China. The first one was constructed in 1971. Since a heavy rains in 1991, the tunnel began to

0 10 20 30 40 500

1

2

3

4

5

6

7

8

9FEM solution

Result of in-situ test

Dist

ancetothebottom

ofthewall(m)

Horizontal pressure (kPa)

Figure 18. Horizontal earth pressure on the wall.

Original configuration

Deformed configuration of the

undisturbed soil or displacement

of the wall

Deformed configuration of the backfill

Figure 17. Displacement of the wall and soil.


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deform increasingly that the lining spilt and the track was severely bent and moved 0.51 :5 m

horizontally. Another tunnel was constructed in 1993, at a distance of 60 m from the old one.

The similar split occurred after the tunnel was completed in 1994. The subsequent geological

survey revealed that the tunnels were situated in a landslide with three sliding surfaces, as shownin Figure 26(a).

The three sliding surfaces and the interfaces between tunnel and stratum are taken as contact

interfaces in the computation. The final deformation of the landslide and the tunnels is depicted

in Figure 27, in which the slide of sliding bodies, the movement of old tunnel, and the distortion

of new tunnel are all salient.

Earth fill (Formerly boulder or cobble with earth)

Interface

Lightly weathered Xigeda mudstone

kPa7035kN/m5.183.0MPa150 o3 ===== cE

21m

Reinforced concrete

4m

6m

2m

(For pile 1)

(For pile2)

Intensively weathered Xigeda mudstone

Pile

(a) (b)

Figure 19. Two test piles: (a) general characterization of the piles and the strata; and (b) finite elementmesh (axisymmetric).

Table VI. The strength parameters of interface.

Pile 1 Pile 2

cB (kPa) m mR c (kPa) m mR

Earth fill 0 0.2 0.2 0 0.2 0.2Intensively weathered mudstone 100 0.8 0.7 100 0.7 0.45Lightly weathered mudstone 80 1.00 0.8 80 0.9 0.50




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2000kN 4000kN 6000kN 8000kN 10000kN

Figure 20. Slip range along Pile 1.

4000kN 6000kN 8000kN 10000kN 12000kN

Figure 21. Failure zone near the bottom of Pile 1.

35

30

25

20

15

10

5

00 2000 4000 6000 8000 10000 12000

Load (kN)

In-situ test

FEM

FEM(Ignoring interface)Settlement(mm)

Figure 22. The loadsettlement curve of Pile 1.


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4. CONCLUSION

Simulation of interface plays an important role in computation of most geotechnical problems. Theinvestigations showed that no additional constitutive relation for interface needs to be introduced

when taken as a contact problem. The mechanical behaviours of the interfaces completely depend on

the shear strength criterion (such as MohrCoulomb law) of interface and the properties of contact

bodies. Therefore, the required parameters are primarily cB and m (or other strength parameters)

which are relatively easy to obtain through tests comparing to the interface parameters defined by

Tunnel

0.0m

1.9m

5.2m

6.8m

27.85m

Weathered Silty Mudstone

Residual Soil

Earth Fill

Silty Mudstone

Lining

Excavation Boundary

Initial Gap

6250

mm

6000mm

5400mm

Grout0 maxAlluvial/ Diluvial Soil

(a) (b) (c)

Figure 24. Shield-driven metro tunnel: (a) geological section; (b) initial gap between the lining (filling) andthe surrounding; and (c) finite element mesh.

16

14

12

10

8

6

4

2

00 2000 4000 6000 8000 10000 12000 14000 16000

Load (kN)

FEM

In-situ testSettlement(mm)

Figure 23. The loadsettlement curve of Pile 2.




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The excavation boundary before deformation

The excavation boundary after deformation

The lining after deformation







(a) (b) (c)

Figure 25. The contact process between the lining and the surrounding: (a) 20%; (b) 40%; and (c) 100%.

Figure 26. Tunnels in a landslide: (a) description of the landslide and the tunnels;and (b) finite element mesh.


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other interface constitutive models. Otherwise, the technique developed in the paper make it easier to

hold the compatibility conditions on interface in the process of computation. Hence we conclude

that, the method presented in this paper is very practical for the application in engineering problems.

APPENDIX A: FORMULATION OF STIFFNESS MATRIX FOR

RIGID-DEFORMABLE CONTACT PROBLEM

The stiffness matrix for rigid-deformable contact problem consists of elements corresponding to

deformable body and rigid body and depends on the contact status. We can assume

Kc KRR KRD

KDR KDD

" #A1

in which subscript R and D denote rigid and deformable, respectively.

Plane problem were taken as an example to illustrate the procedure of formulating the

stiffness matrix. The displacement vector of the rigid body is

V V1 V2 Y T

in which V1 and V2 are rigid translations in the directions of x1 and x2; respectively, and

Y is rigid rotation. The displacement of a node on contact boundary Gc is (as shown in

Figure A1)

u un

us

( )

sin a cos a r cosa b

cos a sin a r sina b

" # V1V2

Y

8>>>:

9>>=>>;

An

As

" #V AV A2

0 20 40 60 80 100 120 140 160 180 200

0

20

40

60

80

100

120

Old tunnelNew tunnel

Verticaldistance(m)

Horizontal distance (m)

After excavation

of the new tunnelAfter excavation

of the old tunnel

After excavation

of the new tunnel0.12m

0.61m

0.51m

Figure 27. Deformations of the landslide and the tunnels.




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The contact force P Pn PsT at a contact node will apply force Q Q1 Q2 M

T on rigid

body. According to the definition ofA; we can obtain

Q Qn Qs ATn Pn A

Ts Ps A

TP A3

1. KDDAccording to the definition, the force at node i arising due to displacement at node j is

Pi Pn

Ps

( )i

knn kns

ksn kss

" #ijun

us

( )j kijn k

ijs

un

us

( )j

kijn T

kijs T

" # unus

( )j

k

ij

u

j

A4

in which k is the same as the stiffness component in conventional FEM, so we can obtain the

components ofKDD

kijDD k

ij A5

2. KDR and KRDNotice that the normal displacement un and tangential displacement us on Gc1 ; un on Gc2

depend on the rigid displacement V; us on Gc2 is independent. Therefore, the nodal force P at

node i arising due to V is

Pi XjGc1

kijuj XjGc2

kijnujn X

jGc1

kijAjV XjGc2

kijnAjnV

X

jGc1

kijAj X

jGc2

kijnAjn

0@

1AV A6

in which S means summing the nodes on Gc1 or Gc2 :

1Q

2Q

M

1x

2x

n

s

O

nPs

P

c

r

V

u = A V

s su = A V

sP

n

n n

P

T T

n n s sP P+ =A A Q

(a) (b)

Figure A1. Relations between the displacements (or forces) of rigid body and the displacements (or forces)of a node on the boundary.


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Finally, we obtain the components in KDR

kiDR X

jGc1

kijAj X

jGc2

kijnAjn A7

By symmetric condition, we have KDR KRD:3. KRRKRR represents the forces acting on rigid body arising from the unit displacement itself. By the

definition, we can obtain

KRR X

iGc1

XjGc1

AiTkijAj XiGc1

XjGc2

AiTkijnAn

X

iGc2

XjGc1

AinTkijAj

XiGc2

XjGc2

AinTkijnnAn A8

In fact, with Table A1, different types of stiffness elements can be easily obtained. However,

direct application of above formulae in programming is very inconvenient. Therefore,

equivalently we rewrite, for example

kijAj knn kns

ksn kss

" #ijAjn

Ajs

" #

kijnnAjn k

ijnsA

js

kijsnAjn k

ijssA

js

" #A9

AiTkij kijnnAjn

T kijsnAjs

T kijnsAjn

T kijssAjs

T A10

AiTkijAj kijnnAin

TAjn kijnsA

in

TAjs kijsnA

is

TAjn kijssA

is

TAjs A11

Take the evaluation of Equation (A9) as an example. Let subscripts l; m 1; 2 denote normal

direction n and tangential direction s; respectively, and p 1; 2; 3 denote rigid translations indirections of x1; x2 and rigid rotation, then Equation (A9) may be written as

keilp X2m1

kijlmA

jmp l; m 1; 2; p 1; 2; 3 A12

Table A1. The components to constitute elements of stiffness matrix.

D R

n s On Gc1 On Gc2

D n kijnn

kijns

kijn

TAj kijnnAj

n

s kijsn kijss k

ijs

TAj kijsnAjn

R On Gc1 AiTkijn A

iTkijs AiTkijAj AiTkijnA

jn

On Gc2 kijnnA

in

T kijnsAin

T AinTkijn

TAj AinTkijnnA

jn




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Hence the procedure is

(1) Compute stiffness component kijlm; which is same as the component in conventional FEM.

(2) Use Equation (A12) to evaluate keilp; which is the contribution of kijlm to Kc:

(3) Let subscripts L and P; which are corresponding to il and p; respectively, denote the

location of keilp in global stiffness matrix Kc; then we can assemble keilp to Kc with

KcLP KcLP keilp A13

in which implies evaluate, not equal to. The above procedure is corresponding to

the evaluation ofKDR in Kc: For KDR and KRR; the procedures are similar.

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a ﬁnite element approach to solve contact problems in geotechnical engineering

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