CHAP 5
Finite Element Analysis of Contact Problem
Nam-Ho Kim
1
Introduction
• Contact is boundary nonlinearity
– The graph of contact force versus displacement becomes vertical
– Both displacement and contact force are unknown in the interface
• Objective of contact analysis
1.
Whether two or more bodies are in contact
2. Where the location or region of contact is
3. How much contact force or pressure occurs in the interface
4. If there is a relative motion after contact in the interface
•
Finite element analysis procedure for contact problem
1.
Find whether a material point in the boundary of a body is in
contact with the other body
2. If it is in contact, the corresponding contact force must be
calculated
2
Introduction
• Equilibrium of elastic system:
– finding a displacement field that minimizes the potential energy
• Contact condition (constrained minimization)
– the potential is minimized while satisfying the contact constraint
• Convert to unconstrained optimization
– Can be solved using either the penalty method or Lagrange
multiplier method
• Slave-master concept for contact implementation
– the nodes on the slave boundary cannot penetrate the surface
elements on the master boundary
3
Goals
• Learn the computational difficulty in boundary nonlinearity
• Understand the concept of variational inequality and its
relation with the constrained optimization
• Learn how to impose contact constraint and friction
constraints using penalty method
• Understand difference between Lagrange multiplier
method and penalty method
• Learn how to integrate contact constraint with the
structural variational equation
• Learn how to implement the contact constraints in finite
element analysis
• Understand collocational integration
4
5.2
1D CONTACT EXAMPLES
5
Contact Problem – Boundary Nonlinearity
• Contact problem is categorized as boundary nonlinearity
• Body 1 cannot penetrate Body 2 (impenetrability)
Body 2
Body 1
Contact stress (compressive)
• Why nonlinear?
Contact boundary
– Both contact boundary and contact
stress are unknown!!!
Contact force
– Abrupt change in contact force
(difficult in NR iteration)
Penetration
6
Why are contact problems difficult?
• Unknown boundary
– Contact boundary is unknown a priori
– It is a part of solution
– Candidate boundary is often given
Body 1
Body 2
• Abrupt change in force
– Extremely discontinuous force profile
– When contact occurs, contact force
cannot be determined from displacement
– Similar to incompressibility (Lagrange
multiplier or penalty method)
• Discrete boundary
Contact force
Penetration
– In continuum, contact boundary varies smoothly
– In numerical model, contact boundary varies node to node
– Very sensitive to boundary discretization
7
Contact of a Cantilever Beam with a Rigid Block
• q = 1 kN/m, L = 1 m, EI = 105 N∙m2, initial gap  = 1 mm
• Trial-and-error solution
– First assume that the deflection is smaller than the gap
qx2
vN (x) 
(x2  6L2  4Lx),
24EI
qL4
vN (L) 
 0.00125m
8EI
– Since vN(L) > , the assumption is wrong, the beam will be in
contact
8
Cantilever Beam Contact with a Rigid Block cont.
• Trial-and-error solution cont.
– Now contact occurs. Contact in one-point (tip).
– Contact force, l, to prevent penetration
lx2
l
vc (x) 
(3L  x), vc (L) 
6EI
3  105
– Determine the contact force from tip displacement = gap
vtip  vN (L)  vc (L)  0.00125 
l
5
3  10
 0.001  
l  75N
v(x)  vN (x)  vc (x)
9
Cantilever Beam Contact with a Rigid Block cont.
• Solution using contact constraint
– Treat both contact force and gap as unknown and add constraint
2
qx2
l
x
v(x) 
(x2  6L2  4Lx) 
(3L  x)
24EI
6EI
– When l = 0, no contact. Contact occurs when l > 0. l < 0 impossible
– Gap condition:
g  vtip    0
10
Cantilever Beam Contact with a Rigid Block cont.
• Solution using contact constraint cont.
– Contact condition
No penetration:
g≤0
Positive contact force: l ≥ 0
Consistency condition: lg = 0
– Lagrange multiplier method
l 

lg  l  0.00025 
0
5 
3  10 

– When l = 0N  g = 0.00025 > 0  violate contact condition
– When l = 75N  g = 0  satisfy contact condition
Lagrange multiplier, l, is the contact force
11
Cantilever Beam Contact with a Rigid Block cont.
• Penalty method
– Small penetration is allowed, and contact force is proportional to it
– Penetration function
1
N   g  g 
2
N = 0 when g ≤ 0
N = g when g > 0
– Contact force
l  KN N
– From
KN: penalty parameter
g  vtip    0
KN
1
g  0.00025 
 g  g
5 2
3  10
– Gap depends on penalty parameter
12
Cantilever Beam Contact with a Rigid Block cont.
• Penalty method cont.
– Large penalty allows small penetration
Penalty parameter
Penetration (m)
Contact force (N)
3×105
1.25×10−4
37.50
3×106
2.27×10−5
68.18
3×107
2.48×10−6
74.26
3×108
2.50×10−7
74.92
3×109
2.50×10−8
75.00
13
Beam Contact with Friction
• Sequence: q is applied first, followed by the axial load P
• Assume no friction
PL
no-friction
utip

 1.0mm
EA
• Frictional constraint
Stick condition: t  l  0, utip  0
Slip condition:
t  l  0, utip  0
Consistency condition: u
tip (t
P=100N, =0.5
t: tangential friction force
 l)  0
14
Beam Contact with Friction cont.
1. Trial-and-error solution
– First assume stick condition
utip
PL tL


 0,
EA EA
Violate t

t  P  100N
 l  0
– Next, try slip condition t  l  37.5N
utip
PL tL


 0.625mm
EA EA
Satisfies utip > 0, therefore, valid
15
Beam Contact with Friction cont.
2. Solution using frictional constraint (Lagrange multiplier)
– Use consistency condition utip (t  l)  0
– Choose utip as a Lagrange multiplier and t-l as a constraint
EA


utip  P 
utip  l   0
L


– When utip = 0, t = P, and t – l = 62.5 > 0, violate the stick condition
– When
utip
(P  l)L

 0.625mm  0
EA
t = l and the slip condition is satisfied, valid solution
16
Beam Contact with Friction cont.
3. Penalty method
– Penalize when t – l > 0
1
T   t  l  t  l 
2
– Slip displacement and frictional force
utip  KT T
KT: penalty parameter for tangential slip
– When t – l < 0 (stick), utip = 0 (no penalization)
– When t – l > 0 (slip), penalize to stay close t = l
– Friction force
EA
t P
utip
L
17
Beam Contact with Friction cont.
• Penalty method cont.
– Tip displacement
– For large KT,
utip
utip
KT L(P  l)

L  KT EA
(P  l)L

EA
t P
EA
utip  l
L
Penalty parameter
Tip displacement (m)
Frictional force (N)
1×10−4
5.68×10−4
43.18
1×10−3
6.19×10−4
38.12
1×10−2
6.24×10−4
37.56
1×10−1
6.25×10−4
37.50
1×100
6.25×10−4
37.50
18
Observations
• Due to unknown contact boundary, contact point should be
found using either direct search (trial-and-error) or
nonlinear constraint equation
– Both methods requires iterative process to find contact boundary
and contact force
– Contact function replace the abrupt change in contact condition
with a smooth but highly nonlinear function
• Friction force calculation depends on the sequence of load
application (Path-dependent)
• Friction function regularizes the discontinuous friction
behavior to a smooth one
19
5.3
GENERAL FORMULATION OF
CONTACT PROBLEMS
20
General Contact Formulation
• Contact between a flexible body and a rigid body
• Point x
W contacts to xc
rigid surface (param x)
xc  xc (x)
• How to find xc(x)
xc,x
t
et 

t
xc,x
(xc )  ( x  xc (xc ))T et (xc )  0
Unit tangent vector
– Closest projection onto the rigid surface
W
Gc
x
en
xc0
x
gn e
t
xc
gt
Rigid Body
21
Contact Formulation cont.
• Gap function
gn  ( x  xc (xc ))T en (xc )
• Impenetrability condition
gn  0,
x  Gc
• Tangential slip function
gt  t 0 (xc  xc0 )
boundary that has a
possibility of contact
Parameter at the
previous contact point
W
Gc
x
en
xc0
x
gn e
t
xc
Rigid Body
gt
22
Ex) Project to a Parabola
• Projection of x={3, 1} to y =
• Let xc = {x,
et 
xc,x
xc,x
x2
4
x2}T

y = x2
3
1 
 
2 2x
1  4x  
1
2
xc
en
gn
x
1
 2x 
en  et  k 
 
0
0
2 1
1  4x  
• Projection point
3
3

x

2
x
(x)  ( x  xc (x))T et (x) 
0
1  4x2
1
• Gap
gn  ( x  xc )T en 
x2c  6xc  1
1  4x2c
1
2
3
4
xc = 1.29
xc = {1.29, 1.66}
 1.83
23
Variational Inequality
• Governing equation
     fb  0

u  0
 n  f s

xW
x  Gg
x  Gs
• Contact conditions (small deformation)
uT en  gn  0


n  0
 x  Gc
n ( uT en  gn )  0 
• Contact set

(convex)
K  w  [H1 (W)]N w
G
g
 0 and wT en  gn  0 on Gc

satisfies all kinematic constraints (displacement conditions)
24
Variational Inequality cont.
• Variational equation with ū = w – u (i.e., w is arbitrary)
W ijij (w  u) dW  W ij,j (wi  ui ) dW  G G
s
c
ijnj (wi  ui ) dG
ij,j  fi b ( x  W),
W ijij (w  u) dW 
G
c
G

G

G
ijnj (wi  ui ) dG 
( w  u) 
c
c
c
G
c
ijnj  fi s ( x  Gs )
ijnj (wi  ui ) dG
n (wn  un ) dG
n (wn  gn  un  gn ) dG
Since it is arbitrary, we
don’t know the value, but
it is non-negative
n (wn  gn ) dG  0
W ijij (w  u) dW 
( w  u),
w  K
Variational inequality for contact problem
25
Illustration of Projection
a( u, w  u)  ( w  u),
w  K
• If the solution u’ is out of set
it is projected to u on
,
• Beam deflection example
(rigid block with initial gap of 1mm)
• v’(x): beam deflection without rigid block v 
– Contact condition is violated (penetration to the block)
• v(x): projection of v’(x) onto convex set
– by applying the contact force
x 10-3
0
v(x)
v'(x)
0.4
0.8
1.2
0
Rigid block
0.2
0.4
0.6
0.8
1
26
Variational Inequality cont.
• For large deformation problem

K  w  [H1 (W)]N
w
G
g

 0 and ( x  xc (xc ))T en  0 on Gc .
• Variational inequality is not easy to solve directly
• We will show that V.I. is equivalent to constrained
optimization of total potential energy
• The constraint will be imposed using either penalty method
or Lagrange multiplier method
27
Potential Energy and Directional Derivative
• Potential energy
( u)  21 a( u, u)  ( u)
• Directional derivative
d
d
 ( u  v ) 
 0
 dd  21 a( u  v, u  v )  ( u  v ) 
 21 a( u, v )  21 a( v, u)  ( u)
 0
 a( u, v )  ( u)
• Directional derivative of potential energy
D( u), v  a( u, v )  ( v )
• For variational inequality
a( u, w  u)  ( w  u)  D( u), w  u  0,
w  K
28
Equivalence
• V.I. is equivalent to constrained optimization
– For arbitrary w
( w)  ( u)  21 a( w, w)  ( w)  21 a( u, u)  ( u)  a( u, w  u)  a( u, w  u)
 a( u, w  u)  ( w  u)  21 a( w, w)  a( u, w)  21 a( u, u)
 D( u), w  u  21 a( w  u, w  u)

( w)  ( u)  D( u), w  u
non-negative
w  K
• Thus, (u) is the smallest potential energy in
( u)  min ( w)  min  21 a( w, w)  ( w) 
wK
wK
– Unique solution if and only if (w) is a convex function and set
closed convex
is
29
Constrained Optimization
• PMPE minimizes the potential energy in the kinematically
admissible space
• Contact problem minimizes the same potential energy in
the contact constraint set

K  w  [H1 (W)]N w
G
g
 0 and ( x  xc )T en  0 on Gc

gn  0 on Gc
• The constrained optimization problem can be converted
into unconstrained optimization problem using the penalty
method or Lagrange multiplier method
– If gn < 0, penalize (u) using
1
1
2
P  n  gn dG  t  gt2 dG
GC
GC
2
2
penalty parameter
30
Constrained Optimization cont.
• Penalized unconstrained optimization problem
( u)  min ( w)  min  ( w)  P( w) 
wK
w
unconstrained
constrained
Solution space w/o contact
Contact
force
u
uNoContact
31
Ex) Beam Deflection with Rigid Block
• q = 1 kN/m, L = 1 m, EI = 105 N∙m2, initial gap  = 1 mm
• Assumed deflection: v(x)  a2x2  a3x3  a4 x 4
• Penalty function:
P  21 n gn2
gn    vtip    a2  a3  a4
• Penalized potential energy
L
1 L
1
2
 a    P   EI(v,xx ) dx   qv dx  n gn2
0
2 0
2
32
Ex) Beam Deflection with Rigid Block
• Stationary condition
 a
0
ai
 4EI  n 6EI  n

 6EI  n 12EI  n
 8EI   18EI  
n
n

1
8EI  n   a2   3 q  n  

  1
18EI  n   a3    4 q  n  
144
 a   1 q   
EI


n  4 
n 
5
5
• Penetration: a2 + a3 + a4 − 
• Contact force: −ngn
Penalty
parameter
a1
a2
a3
Penetration (m)
Contact force
(N)
3×105
2.31×10−3
-1.60×10−3
4.17×10−4
1.25×10−4
37.50
3×106
2.16×10−3
-1.55×10−3
4.17×10−4
2.27×10−5
68.18
3×107
2.13×10−3
-1.54×10−3
4.17×10−4
2.48×10−6
74.26
3×108
2.13×10−3
-1.54×10−3
4.17×10−4
2.50×10−7
74.92
3×109
2.13×10−3
-1.54×10−3
4.17×10−4
2.50×10−8
75.00
True value
2.13×10−3
-1.54×10−3
4.17×10−4
0.0
75.00
33
Variational Equation
• Structural Equilibrium
a( u, u )  ( u )  P( u; u )  0
( u; u )  P( u; u )  0
P( u; u )  n  gn gn dG  t  gt gt dG  b( u, u )
GC
bN ( u, u )
GC
bT ( u, u ),
need to express in terms
of u and ū
• Variational equation
a( u, u )  b( u, u )  ( u ),
u 
34
Frictionless Contact Formulation
• Variation of the normal gap
gn  ( x  xc (x))T en
gn  u T en
• Normal contact form
bN ( u, u )   gn u T en dG
Gc
35
Linearization
• It is clear that bN is independent of energy form
– The same bN can be used for elastic or plastic problem
• It is nonlinear with respect to u (need linearization)
• Increment of gap function
Dgn  D  ( x  xc (x))T en   DuT en  ( x  xc (x))T Den
Dgn ( u; Du)  DuT en
Den
et
x  xc
en
– We assume that the contact boundary is straight line Den = 0
– This is true for linear finite elements
Rigid surface
36
Linearization cont.
• Linearization of contact form
bN ( u, u )  
Gc
g
T
u
en dG

bN* (u; Du, u )   u T en enT Du dG
Gc
• N-R iteration
a* (u; Du, u )  bN* ( u; Du, u )  ( u )  a( u, u )  bN ( u, u ),
u 
37
Ex) Frictionless Contact of a Block
• Calculate displacement, penetration and contact force at
y
q
the contact interface.
• EA = 105N, n = 0, q = 1.0kN/m, plane strain
• Plane strain: u  {ux , uy }T u  {ux , uy }T
• Contact boundary: en  {0, 1}T
x  x,
xc  { xc , 0}T ,
Elastic
body
0
x  { xc ,uy }T
Rigid body
Contact
boundary
1
x
• Gap function: gn  ( x  xc )T en  uy
• Contact form: b ( u, u )   1 u u
N
n 0 y y
• Penalized potential energy
1
1
T
  P    D dA   ( q)uy
0
2 A
y 0
dx
1 2
1
dx  n  gn
y 1
0
2
y 1
dx
38
Ex) Frictionless Contact of a Block
• From n = 0, D becomes a diagonal matrix, decoupled x & y
• Since load is only y-direction, xx = gxy = 0
• Linear displacement in y-direction
uy  a0  a1 y
uy  a0  a1 y
• Penalized potential
P 
1
1
A Ea1a1 dA  0 (q)(a0  a1 ) dx  n 0 a0a0 dx  0
– Need to satisfy for arbitrary a0 , a1
q
q
a0   , a1  
n
EA
q
q
uy  

y, 0  y  1
n EA
n gn  nuy
y 0
q
As n increases,
penetration decreases but
contact force remains constant
39
Frictional Contact Formulation
• frictional contact depends on load history
• Frictional interface law – regularization of Coulomb law
• Friction form
bT ( u, u )  t  gt gt dG
  enT xc,xx
GC
gt  t 0 xc  nu T et
bT ( u, u )  t  ngt u T et dG
Gc
  etT xc,xx
g  enT xc,xxx
c t
2
 gn 
n  t t0
c
-ngn
t
gt
40
Friction Force
• During stick condition, fT = tgt (t: regularization param.)
t gt  n gn
• When slip occurs, t gt  n gn
• Modified friction form
 t  ngt u T et dG,
if t gt  n gn
Gc

bT ( u, u )  
T

sgn(g
)
n
g
u
et dG,
otherwise .

n
t G
n
c
41
Linearization of Stick Form
• Increment of slip
Dgt ( u; Du)  t 0 Dxc  n etT Du
• Increment of tangential vector

Det   e3  Den  en ( etT Du)
c
• Incremental slip form for the stick condition
bT* ( u; Du, u )  t  n2 u T et etT Du dG
Gc
ngt T
t 
u ( en etT  et enT )Du dG
Gc c
ngt
2
T
T
t 
(
g
t

2

)g


t
u
e
e
n
t t Du d G
Gc c2


42
Linearization of Slip Form
• Parameter for slip condition: t  n sgn(gt ).
• Friction form: bT ( u,u )  t  n gn u T et dG
Gc
• Linearized slip form (not symmetric!!)
bT* ( u; Du, u )  t  nu T et enT Du dG
Gc
ngn T
 t 
u ( en etT  et enT )Du dG
Gc c
ngn
2
T
T
 t 
(
g
t

2

)g


t
u
e
e
n
t t Du d G
Gc c2


43
Ex) Frictional Slip of a Cantilever Beam
• Distributed load q  axial load P, t = 106,  = 0.5
• Contact force: Fc = –ngn = 75N
• Penalized potential energy (axial alone)
1
 a   EA(u,x ) dx  Pu(L)  t gt2
0
2
L
a 
L
2
0 EAu,xu,x dx  Pu(L)  tgtgt x L
x L
 0,
u 
P=100N
44
Ex) Frictional Slip of a Cantilever Beam
• Linear axial displacement field: u(x) = a0 + a1x
u(0)  0  u(x)  a1x u,x  a1 u,x  a1
• Tangential slip in terms of displacement
– Parametric coordinate x has an origin at x = L, and it has the same
length as the x-coordinate
xc,x  t  t 0  1
xc0  0
gt  xc  u(L)  a1
• Assume the stick condition: t gt  n gn
a1 (EAa1  t a1  P)  0,
a1 
Px
u(x) 
 9.09  10 5 x
EA  t
t gt  90.9  37.5  n gn
The assumption is violated!!
45
Ex) Frictional Slip of a Cantilever Beam
• Assume the slip condition:
a 
L
0 EAu,xu,x dx  Pu(L)  n sgn(gt )gngt x L
 0, u 
– With contact force = 70N, n gn  37.5N
a1 (EAa1  P  37.5)  0,
a1 
a1  62.5  10 5
utip  0.625mm
46
5.4
FINITE ELEMENT
FORMULATION OF
CONTACT PROBLEMS
47
Finite Element Formulation
• Slave-Master contact
– The rigid body has fixed or prescribed displacement
– Point x is projected onto the piecewise linear segments of the
rigid body with xc (x = xc) as the projected point
– Unit normal and tangent vectors
et 
x2  x1
,
L
en  e3  et
Flexible body
x
g
x0
x1
xc0
x
en
xc
L
x2
et Rigid body
48
Finite Element Formulation cont.
• Parameter at contact point
xc 
1
( x  x1 )T et
L
• Gap function
gn  ( x  x1 )T en  0
Impenetrability condition
• Penalty function
NC

1
P   g
2 I 1
2

 I
– Note: we don’t actually integrate the penalty function. We simply
added the integrand at all contact node
– This is called collocation (a kind of integration)
– In collocation, the integrand is function value × weight
49
Finite Element Formulation cont.
• Contact form (normal)
bN ( u, u ) 
NC

I 1
gn
T
d
en

I  {d }T { fc }
– (gn): contact force, proportional to the violation
– Contact form is a virtual work done by contact force through
normal virtual displacement
• Linearization
D gn

Heaviside step function
H(x) = 1 if x > 0
= 0 otherwise
 H( gn )enT Du  H( gn )enT Dd
bN* ( u; Du, u )

NC
  H(gn ) d
I 1
 { d }T [Kc ]{ Dd}
T

en enT Dd
I
 {d }
T
NC
  H(gn ) en enT I { Dd}
I 1
Contact stiffness
50
Finite Element Formulation cont.
• Global finite element matrix equation for increment
{d }T [KT  Kc ]{ Dd}  { d }T { fext  fint  fc }
– Since the contact forms are independent of constitutive relation,
the above equation can be applied for different materials
• A similar approach can be used for flexible-flexible body
contact
– One body being selected as a slave and the other as a master
• Computational challenge in finding contact points
(for example, out of 10,000 possible master segments, how
can we find the one that will actually in contact?)
51
Finite Element Formulation cont.
• Frictional slip
gt  l0 ( xc  xc0 )
• Friction force and tangent stiffness (stick condition)
ftc  t gt et
ktc  t et etT
• Friction force and tangent stiffness (slip condition)
ftc  n sgn(gt )gn et ,
if t gt  n gn
ktc  n sgn(gt )et enT
52
5.6
CONTACT ANALYSIS
PROCEDURE AND MODELING
ISSUES
53
Types of Contact Interface
• Weld contact
–
–
–
–
A slave node is bonded to the master segment (no relative motion)
Conceptually same with rigid-link or MPC
For contact purpose, it allows a slight elastic deformation
Decompose forces in normal and tangential directions
• Rough contact
– Similar to weld, but the contact can be separated
• Stick contact
– The relative motion is within an elastic deformation
– Tangent stiffness is symmetric,
• Slip contact
– The relative motion is governed by Coulomb friction model
– Tangent stiffness become unsymmetric
54
Contact Search
• Easiest case
– User can specify which slave node will contact with which master
segment
– This is only possible when deformation is small and no relative
motion exists in the contact surface
– Slave and master nodes are often located at the same position and
connected by a compression-only spring (node-to-node contact)
– Works for very limited cases
• General case
– User does not know which slave node will contact with which
master segment
– But, user can specify candidates
– Then, the contact algorithm searches for contacting master
segment for each slave node
– Time consuming process, because this needs to be done at every
iteration
55
Contact Search cont.
Node-to-surface
contact search
Surface-to-surface
contact search
56
Slave-Master Contact
• Theoretically, there is no need to distinguish Body 1 from Body 2
• However, the distinction is often made for numerical convenience
• One body is called a slave body, while the other body is called a
master body
• Contact condition: the slave body cannot penetrate into the master
body
• The master body can penetrate into the slave body (physically not
possible, but numerically it’s not checked)
Master
Slave
57
Slave-Master Contact cont.
• Contact condition between a slave node and a master segment
• In 2D, contact pair is often given in terms of {x, x1, x2}
• Slave node x is projected onto the piecewise linear segments of the
master segment with xc (x = xc) as the projected point
• Gap: g  ( x  x1 )  en  0
• g > 0: no contact
• g < 0: contact
Slave
x
x0
x1
x
xc
g
en
0
xc
x2
et
L
Master
58
Contact Formulation (Two-Step Procedure)
1. Search nodes/segments that violate contact constraint
Body 1
Violated nodes
Contact candidates
Body 2
2. Apply contact force for the violated nodes/segments
(contact force)
Contact force
59
Contact Tolerance and Load Increment
• Contact tolerance
– Minimum distance to search for contact (1% of element length)
Out of
contact
Within
tolerance
Out of
contact
• Load increment and contact detection
– Too large load increment may miss contact detection
60
Contact Force
• For those contacting pairs, penetration needs to be
corrected by applying a force (contact force)
• More penetration needs more force
• Penalty-based contact force (compression-only spring)
FC  Kn g
• Penalty parameter (Kn): Contact stiffness
– It allows a small penetration (g < 0)
– It depends on material stiffness
– The bigger Kn, the less allowed penetration
x1
g<0
xc
x
x2
FC
61
Contact Stiffness
• Contact stiffness depends on the material stiffness of
contacting two bodies
• Large contact stiffness reduces penetration, but can
cause problem in convergence
• Proper contact stiffness can be determined from allowed
penetration (need experience)
• Normally expressed as a scalar multiple of material’s
elastic modulus
Kn  SF  E
SF  1.0
• Start with small initial SF and increase it gradually until
reasonable penetration
62
Lagrange Multiplier Method
• In penalty method, the contact force is calculated from penetration
– Contact force is a function of deformation
• Lagrange multiplier method can impose contact condition exactly
– Contact force is a Lagrange multiplier to impose impenetrability condition
– Contact force is an independent variable
• Complimentary condition
FC g  0
g  0
FC  0
contact
g  0
FC  0
no contact
• Stiffness matrix is positive semi-definite
K
 T
A
A  d  F 
    
0   FC   0 
• Contact force is applied in the normal direction to the master segment
63
Observations
• Contact force is an internal force at the interface
– Newton’s 3rd law: equal and opposite forces act on interface
F
pC1
qC1
F
pC2
qC2
Np
Nq
i=1
i=1
 pci   qci
qC2
F
• Due to discretization, force distribution can be different,
but the resultants should be the same
64
Contact Formulation
• Add contact force as an external force
• Ex) Linear elastic materials
F
[K]{d}  {F}  {FC (d)}
Internal
force
External
force
pC1
Contact
force
qC1
• Contact force depends on
displacement (nonlinear, unknown)
[K  KC ]{ Dd}  {F}  {FC (d)}  [K]{d}
Tangent stiffness
pC2
qC2
qC2
F
Residual
 F 
[KC ]   C  Contact stiffness
 d 
65
Friction Force
• So far, contact force is applied to the normal direction
– It is independent of load history (potential problem)
• Friction force is produced by a relative motion in the
interface
– Friction force is applied to the parallel direction
– It depends on load history (path dependent)
• Coulomb friction model
Ff  FC
 FC
Friction
force
Body 1
Stick
Slip
Friction force
Relative
motion
Contact force
66
Friction Force cont.
• Coulomb friction force is indeterminate when two bodies
are stick (no unique determination of friction force)
• In reality, there is a small elastic deformation before slip
• Regularized friction model
– Similar to elasto-perfectly-plastic model
Ff  Kt s
 FC
Stick
Kt: tangential stiffness
Slip
Friction
force
FC
Kt
Relative
motion
67
Tangential Stiffness
• Tangential stiffness determines the stick case
Ff  Kt s
Stick
 FC Slip
• It is related to shear strength of the material
Kt  SF  E
SF  0.5
• Contact surface with a large Kt behaves like a rigid body
• Small Kt elongate elastic stick condition too much (inaccurate)
Friction
force
FC
Kt
Relative
motion
68
Selection of Master and Slave
• Contact constraint
– A slave node CANNOT penetrate the master segment
– A master node CAN penetrate the slave segment
– There is not much difference in a fine mesh, but the results can
be quite different in a coarse mesh
• How to choose master and slave
– Rigid surface to a master
– Convex surface to a slave
– Fine mesh to a slave
Master
Slave
Slave
Master
69
Selection of Master and Slave
• How to prevent penetration?
– Can define master-slave pair twice by changing the role
– Some surface-to-surface contact algorithms use this
– Careful in defining master-slave pairs
Master-slave pair 2
Master-slave pair 1
70
Flexible or Rigid Bodies?
• Flexible-Flexible Contact
– Body1 and Body2 have a similar stiffness and both can deform
• Flexible-Rigid Contact
– Stiffness of Body2 is significantly larger than that of Body1
– Body2 can be assumed to be a rigid body
(no deformation but can move)
 107

– Rubber contacting to steel


7
10


• Why not flexible-flexible?


1
– When two bodies have a large difference in


1 

stiffness, the matrix becomes ill-conditioned
– Enough to model contacting surface only for Body2
• Friction in the interface
– Friction makes the contact analysis path-dependent
– Careful in the order of load application
71
Effect of discretization
• Contact stress (pressure) is
high on the edge
Uniform pressure
Slave body
• Contact stress is sensitive
to discretization
Non-uniform
contact stress
72
Rigid-Body Motion
• Contact constraint is different from displacement BC
– Contact force is proportional to the penetration amount
– A slave body between two rigid-bodies can either fly out or
oscillate between two surfaces
– Always better to remove rigid-body motion without contact
Rigid master
Rigid master
Rigid master
Rigid master
73
Rigid-Body Motion
• When a body has rigid-body motion, an initial gap can
cause singular matrix (infinite/very large displacements)
• Same is true for initial overlap
Rigid master
Rigid master
74
Rigid-Body Motion
• Removing rigid-body motion
– A small, artificial bar elements can be used to remove rigid-body
motion without affecting analysis results much
Contact stress
at bushing due to
shaft bending
75
Convergence Difficulty at a Corner
• Convergence iteration is stable when a variable (force)
varies smoothly
• The slope of finite elements are discontinuous along the
curved surface
• This can cause oscillation in residual force (not converging)
• Need to make the corner smooth using either higherorder elements or many linear elements
– About 10 elements in 90 degrees, or use higher-order elements
76
Curved Contact Surface
• Curved surface
– Contact pressure is VERY sensitive to the curvature
– Linear elements often yield unsmooth contact pressure
distribution
– Less quadratic elements is better than many linear elements
Linear elements
Quadratic elements
77
Summary
• Contact condition is a rough boundary nonlinearity due to
discontinuous contact force and unknown contact region
– Both force and displacement on the contact boundary are unknown
– Contact search is necessary at each iteration
• Penalty method or Lagrange multiplier method can be used to
represent the contact constraint
– Penalty method allows a small penetration, but easy to implement
– Lagrange multiplier method can impose contact condition accurately, but
requires additional variables and the matrix become positive semi-definite
• Numerically, slave-master concept is used along with collocation
integration (at slave nodes)
• Friction makes the contact problem path-dependent
• Discrete boundary and rigid-body motion makes the contact problem
difficult to solve
78