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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.
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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
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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.
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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
The excavation boundary before deformation
The excavation boundary after deformation
The lining after deformation
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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