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Modeling of contact interaction with friction
Author: Margarita KovtunSaint-Petersburg State Polytechnic University
Objective
• Contact problems are highly nonlinear, their solution is always expensive in terms of time and resources
• Development of the algorithm based on the minimization of deformation energy of mechanical system
• Two main spheres of application: simulation of riveting process, modeling of sliding between soil layers
Contents
• Problem statement
• Rigidity matrix concept
• Computational procedure
• Applications
- One-dimensional problem
- Sliding
- Rode bending
• Future improvements
initial gap
contact zonecontact zone
Problem statement
Unknown variables: U – vector of displacements in the contact zone
WU minarg= , where W –
deformation energy of the mechanical system
Two parts are initially separated, distance between them – initial gap
Bodies may slide relative to each other according to the dry friction law
F
– applied loads
FixationsElastic body
Area of possible contact Computational nodes
Rigidity matrix (1)
Thus, we substitute the body by its rigidity matrix, having the equality:
UKF ⋅=
matrixrigidity - , :energy nDeformatio KUKUW T ⋅⋅=21
FixationsElastic body
Area of possible contact Computational nodes
Rigidity matrix (2)
( )nznynxzyxzyxT fffffffffF ,,,,,,,,, 222111 L=
Loads can be applied in computational nodes in any direction
FixationsElastic body
Area of possible contact Computational nodes
Rigidity matrix (2)
Loads can be applied in computational nodes in any direction
( )nznynxzyxzyxT fffffffffF ,,,,,,,,, 222111 L=
FixationsElastic body
Area of possible contact Computational nodes
Rigidity matrix (2)
Displacements of nodes are calculated in all direction
( )nznynxzyxzyxT uuuuuuuuuU ,,,,,,,,, 222111 L=
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
z
y
x
uuu
1
1
1
Main idea
FUK =⋅
Energy of deformation: min21
→⋅⋅= UKUW T
Forces acting on the system are balanced:
Displacements in normal direction are restricted: Δ≤NU
Forces in contact nodes should obey Coulomb law: nFF μτ ≤
initial gap
F
contact zone
F
трF
x1 x2 x3
One-dimensional problem
elasticF
∑∑=
−
−
=
−+=⋅=N
iii
N
iiN
TN uucucucKUUcxxxW
11
1
1
2221 22
),,( K
where c
– spring rate, m
– mass
1,,1,)()( 11 −=≤−−−≤− −+ Nimguucuucmg iiii Kμμ
Energy of deformation:
mguucFmg NN μμ ≤−−≤− − )( 1
elasticF
Elastic forceFriction force
Results
№F
(H)
(м) (м) (м) Scheme of bodies displacements
1 3 0 0 0
2 6 0 0 2
3 9 0 1 6
4 14 2 8 18
1x 2x 3x
F
трF
x1 x2 x3
F
трFтрF
x1 x2 x3
F
трF трF
x1 x2 x3
F
трF трF трF
x1 x2 x3
трF
2D sliding
Rigid foundation
F
• 2D deformable body• Made of isotropic material• Moves on the rigid foundation• Gap between bodies initially is closed
• Compute rigidity matrix by mentioned procedure
• All the nodes are contact ones:• Gap is zero:
• Couloumb-Mohr law is to be satisfied in each node:
2D sliding
min21
→⋅⋅= UKUW T
0=yU
nFF μτ ≤
Rigid foundation
F
Results comparison
friction coefficient = const
0
0.1
0.2
0.3
0.4
0.5
0.6
0 500 1000 1500 2000 2500 3000 3500
Force
Error, %
ANSYS
ni
ANSYScalc
ni
U
UU
,1
,1
max
max
=
=−
=ε :error Relative
Results comparison
Force = const
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60
friction coefficient
Error,%
ANSYS
ni
ANSYScalc
ni
U
UU
,1
,1
max
max
=
=−
=ε :error Relative
Results comparison
0.00E+00
5.00E-04
1.00E-03
1.50E-03
2.00E-03
2.50E-03
3.00E-03
3.50E-03
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Node number
Ux
Force = 50H
Force = 200H
Force = 800H
Force = 3200H
Dependence of displacements in X-direction on the force value
Results comparisonDependence of displacements in X-direction on
the value of friction coefficient
0.00E+00
1.00E-05
2.00E-05
3.00E-05
4.00E-05
5.00E-05
6.00E-05
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20Node number
Ux
mu=0.05mu=0.2mu=0.5mu=1
2D sliding
Body deformation for mu = 1 Body deformation for mu = 0.1
Rode bending
• Contact problem
• Zone of possible contact depends on applied loads
• It is necessary to derive iterative procedure
Fixations
x
y
x1
x2
xNApplied load
1 mm
5 mm
Gap elimination
Calculation of sliding distances
Iterative procedure
FUK =⋅Equilibrium equations:
Unknown forces appear:
contact
contact
ffFUK ,+=⋅
- reaction force or friction force
Coulomb-Mohr law: nFF μτ ≤
contact nodes nodes in equilibrium
Iterative procedure
Compute force so that one node comes in contact
minF
FUK =⋅Parts are separated:
Substitute the equality by unequality in new contact node
FUK =⋅nFF μτ ≤
If F
> minF
true
false
Done
Results
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0.0016
0.0018
0 0.0025 0.005 0.0075 0.01 0.0125 0.015 0.0175 0.02 0.0225 0.025
Friction coefficient
Ux
sliding
sticking
Dependence of displacement in X-direction on the friction coefficient value in the last node
Results
Dependence of displacement in X-direction on the force value in the last node
0
0.002
0.004
0.006
0.008
0.01
0.012
1000 1500 2000 2500 3000 3500 4000 4500 5000
Force
Dis
plac
emen
ts
mu=0.15mu = 0
Results
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0 10000 20000 30000 40000 50000 60000 70000
Force
Error
Dependence of relative error on the force value
ANSYS
ni
ANSYScalc
ni
U
UU
,1
,1
max
max
=
=−
=ε :error Relative
Deformation shapes
One contact nodeTwo contact nodes
Future improvements and investigations
• Sliding: - modeling two flexible bodies interaction- modeling systems with refined meshes- 3D geometry simulation
• Riveting process simulation:- modeling 2D and 3D objects - optimizing the procedure of definingcontact zone
Thank you!