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!