Large deformations and contact problems in Mentat & MARCpiet/inf/mrc/pdf/MaMeLdco2011BW.pdf · 2011. 5. 9. · 1 Large deformations 1.1 Background : Incremental load When the deformation

Large deformations and contact problems

in

Mentat & MARC

Tutorial with Background and Exercises

Eindhoven University of TechnologyDepartment of Mechanical EngineeringPiet Schreurs May 5, 2011

Contents

1 Large deformations 31.1 Background : Incremental load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Plate with central hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Modeling, analysis and results . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Background : Iterative analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Contact problems 112.1 Background : The CONTACT-option in MARC . . . . . . . . . . . . . . . . . . . . 112.2 Compressing a thick-walled tube with a rigid die . . . . . . . . . . . . . . . . . . . 12

2.2.1 Modeling, analysis and results . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3.1 Compressing a thick-walled tube with a deformable die . . . . . . . . . . . 17

3 Elasto-plastic material behavior 183.1 Background : Yield criterion and hardening . . . . . . . . . . . . . . . . . . . . . . 183.2 Plate with central hole; elasto-plastic material behavior . . . . . . . . . . . . . . . 19

3.2.1 Modeling, analysis and results . . . . . . . . . . . . . . . . . . . . . . . . . . 203.3 Forming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.3.1 Modeling, analysis and results . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2

1 Large deformations

1.1 Background : Incremental load

When the deformation of a structure is so small that its initial geometry hardly changes, thedeformation is referred to as ”geometrically linear”.

In this tutorial we will analyze problems, where deformations are large. This is referred to as”geometrically nonlinear” deformation. For the analysis to yield accurate results, it will be neededto prescribe the final load in a number of incremental steps.

3

1.2 Plate with central hole

The figure shows a square plate with a central circular hole. Dimensions are indicated in thefigure.

Poisson’s ratio = 0.3 [-]Young’s modulus = 210 [GPa]

u [mm]

16 [cm]

x

y

thickness = 1 [cm]

10 [cm]

The plate is loaded in its plane. We will prescribe the horizontal displacement of the left- andright edge (x = ±8 [cm]) to be u = 10 [mm] in 10 steps of 1 [mm]. Upper and lower edge are freeto move, so the y-displacement is unconstrained.

The material behavior is assumed to be isotropic, linearly elastic, according to Hooke’s law,with Young’s modulus E = 210 [GPa] and Poisson’s ratio ν = 0.3.

Only the upper right quarter of the plate is modeled, which necessitates the introduction ofsymmetry conditions on the edges x = 0 and y = 0.

4

1.2.1 Modeling, analysis and results

Model the upper right quarter of the plate. Make proper BOUNDARY CONDI-

TIONS for the symmetry conditions.

The displacement of the right edge is discribed to be 10 [mm] after 10 equalincremental steps.

The displacement is prescribed as a function of the (fictitious) time t. To thispurpose a TABLE of type ’time’ is made, leading to a figure with the horizontaltime-axis and the vertical function-axis f(t). The total prescribed displace-ment on time t is then : u(t) = u0f(t), where u0 is the displacement whichis prescribed in FIXED DISPLACEMENT. We want to prescribe a linear functionf(t) = t and do this by using a (FORMULA).

TABLES

NEW

1 INDEPENDENT VARIABLE

TYPE

time

(INDEPENDENT VARIABLE V1)

MAX

Enter maximum value for v1 : 10(FUNCTION VALUE F)

MAX

Enter maximum value for F : 10FORMULA

ENTER

Enter formula : v1 This leads to a linear f(t)SHOW MODEL

RETURN

NEW

FIXED DISPLACEMENT This is apply3 !DISPLACEMENT X

Enter value for ’x’ : 1 This is u0 in [mm]TABLE

table1

OK

(NODES) ADD

Select the right edge.

5

The prescribed displacement apply4 is defined as a function of the (fictitious)time, which is defined in a TABLE of type ”time”. The total analysis time ischosen to be 10 ’seconds’, so the time axis has a maximum length of 10. Thenumber of steps is 10. Load case time and number of incremental steps isspecified in (ANALYSIS) LOADCASE. Each increment takes 1 ”second” and theincremental displacement is 1 [mm].

(MAIN MENU) (ANALYSIS)

LOADCASES

NEW

MECHANICAL

STATIC

TOTAL LOADCASE TIME : 10(FIXED) # STEPS : 10

The loadcase lcase1 can be made with earlier defined BOUNDARY CONDITIONS.The convergence criterion is set on a lower value than the default one, to invokesome interations in the solution process.

LOADS

Select apply1, apply2, apply3

OK

CONVERGENCE TESTING

RELATIVE FORCE TOLERANCE

0.0001OK

OK

In (ANALYSIS) JOBS, the analysis is further specified and carried out. The de-fined loadcase lcase1 is selected. Total loadcase time is 10 ”seconds”, which issubdivided in 10 incremental steps.There should be no loads in the INITIAL LOADS section, because these loadswill be applied at time t = 0 and remain on the structure during the totalanalysis period. INITIAL LOADS defines the so-called ”null-increment”, whereonly zero-boundary conditions may be applied.


JOBS

NEW

MECHANICAL

(LOADCASES) Select lcase1

INITIAL LOADS

CLEAR Deselect apply1, apply2, apply3

OK

PLANE STRESS

OK

MAIN

6

Now we are ready and SAVE the model as platelargedef. Run MARC and lookat the results with Mentat. Be sure that AUTO MAGNIFY is OFF : the deformationis large enough to be clearly visible without any enlargement.The results of the subsequent incremental steps can be seen by repeatedlyclicking on NEXT or by pushing the MONITOR button.

A HISTORY PLOT can provide a lot of insight in the deformation process. Thevalue of a nodal variable can be plotted as a function of increment number ortime.

HISTORY PLOT

SET NODES

Select (lm) node(s) (eg. the upper left node).

Close the selection with END LIST.

REWIND

COLLECT DATA

Enter first history increment : 0Enter last history increment : 10Enter increment step size : 1NODES/VARIABLES

ADD 1-NODE CURVE

Enter History-Plot node : select (lm) node in (NODES)

Enter X-axis variable : Increment

Enter Y-axis variable : Displacement y

FIT

Make other plots. Remove plots with CLEAR CURVES.

7

1.3 Background : Iterative analysis

Looking at the results reveals that the deformation of the plate is large, so geometrically nonlinear.Because the element stiffness matrix depends on the element geometry, it changes during theanalysis. In the .log file it can be seen that only in the first increment the line start of assembly isprinted. This indicates that the matrix is calculated only once and not updated afterwards. Thismeans that the results cannot be accurate. We can change this by using some options in Mentat.

Close the results file and open the initial model file. We go to the JOBS section to make somechanges.

FILES

RESTORE or OPEN : platelargedefMAIN


JOBS

MECHANICAL

ANALYSIS OPTIONS

(NONLINEAR PROCEDURE)

LARGE STRAIN

OK

Ready ! Save the model and run the job. When it is finished, look at the .logfile.

We see that the stiffness matrix is made at the beginning of each increment (start of assembly). Itis also obvious that each incremental analysis takes more time. In each increment lines are printedconcerning convergence.

All this results from the application of the option LARGE STRAIN. Use of this option implies thatlarge rotations (max 10o) and fairly large strains (max 10%, incremental 2%) can be handeledquite well. In each increment the iterative solution technique is used, where in a number of stepssubsequent solutions for the equilibrium equations are determined. When these subsequent so-lutions are gradually better, the iteration process is said to converge; The iterative procedure isnever used in the first or null- increment, so that deformations in this increment must always bevery small. The influence of stresses in the stiffness matrix are taken into account and a so-calledresidual load correction is applied to compensate for the inaccuracy of the iterative solution.

The option UPDATED LAGRANGE is set by default. It implies that the Updated Lagrange formulationis used, which means that the begin-increment state is taken as a reference for strains. The optionis not available for all elements. In MARC manual Volume B it can be found which elements usethis option.

When it is deselected, and only LARGE STRAIN is used, the initial undeformed state is used as areference for the strains. In the material model, the 2nd-Piola-Kirchhoff stress is related to theGreen-Lagrange strain and also the input material parameters relate these variables. In ”jobresults” strain and stress are respectively Green-Lagrange strain and 2de-Piola-Kirchhoff stress.Cauchy stresses are sometimes calculated (see Volume B of the MARC manual, which describesthe elements).

A short explanation.

In the deformed state the nodal displacements u˜

have to be determined such that nodal equilibrium

is satisfied. This implies that the internal nodal forces f˜

imust equal the external nodal forces f

˜e.

8

For a linear problem, the internal forces are a linear function of the nodal displacements, whichcan then be solved directly from :

f˜

i= f

˜i(u˜) = Ku

˜= f

˜e

where K is the constant stiffness matrix.For a non-linear problem, with large deformations and/or non-linear material behavior, the

internal forces are no longer a linear function of the nodal displacements. The nodal equilibriumis represented by a set of nonlinear equations :

f˜

i(u˜) = f

˜e

To determine the nodal displacements, such that this set of equations is satisfied, an iterativeprocedure is followed, which is illustrated in the figure below, where u

˜, f

˜i

and f˜

ehave to be

considered to be ”measures” for the columns proper. The origin of the plot represents the knownstate at the beginning of the increment. The nodal displacements w.r.t. the undeformed state areu˜

b. The external load is increased with the incremental load ∆f˜.

f˜

f˜

e

f˜

i

u˜

o u˜

u˜

b

∆f˜

The solution u˜

= u˜

o at the intersection of f˜

iand f

˜e

cannot be determined directly, starting fromu˜

b. However, it is possible to determine, in a number of iterative steps, a sequence of increasinglybetter solutions, as is indicated in the figure below.

f˜

e

f˜ f

˜i

u˜∗

δu˜

u˜

b u˜

u˜

o

9

The tangent at f˜

i(u˜) represents the stiffness matrix K(u

˜), which has to be calculated at the

beginning of each iteration. In each iteration, a linear set of equations has to be solved to determinethe change of the last approximation u

˜∗, yielding the iterative nodal displacements

K(u˜∗)δu

˜= f

˜e− f

˜i(u˜∗) = r

˜∗

When the calculated solution u∗ is ”good enough” acording to a certain criterion, the iterationprocess is said to be converged. The convergence limit has to be specified in the input file. Alsoother parameters of the iterative procedure can be changed, but default values are mostly good.

The figure above shows a converging iterative solution process : the subsequent approximationsapproach the exact solution u

˜o. The difference between f

˜i

and f˜

e, the residual or residual load

r˜∗, is decreasing. It may happen, and it indeed often does, that the subsequent solutions do not

improve, but are becoming less good approximations. When this happens, the program MARCwill stop after a (specified) maximum number of iteration steps, printing the error message exit

number 3002. We can try to reach a solution with more iteration steps or a smaller incrementalload.

When the convergence criterion is satisfied, the approximate solution u˜

c will not exactly concidewith the unknown exact solution u

˜o. This means that the nodal equilibrium is not satisfied exactly.

The difference between f˜

eand f

˜iis taken into account in the next increment, which is referred to

as residual load correction.

10

2 Contact problems

2.1 Background : The CONTACT-option in MARC

The option CONTACT can be used to detect and take into account contact between various bodiesor between parts of one and the same body, during an analysis. Contact conditions like frictionand adhesion can be taken care of. Contact bodies can be rigid or deformable. The option isespecially useful, when it is not known on beforehand where or when contact will occur. Usingthe option CONTACT is associated with some rules :

- At least one deformable body is needed in contact.- When the model is made, the deformable bodies must be defined before the rigid bodies.- One node or element can be a part of only one contact body.- No coordinate axis transformation is allowed in a contact node.- No tying or servolink can be defined in a node which may come into contact with a rigid

body.- Contact bodies are defined internally in the program MARC during the first or null-incrrment.

When CONTACT is used, it is not allowed to apply INITIAL LOADS which are not zero.

For accurate results and good convergence the next recommandations are given in the MARCmanual :

- Deformable contact bodies with smaller elements in the (potential) contact region must bedefined before contact bodies with larger elements.

- Define smaller deformable contact bodies before larger ones.- Define deformable contact bodies with lower stiffness before those with higher stiffness.- Define first the convex contact bodies.- Use CONTACT TABLES to decrease analyses times.

11

2.2 Compressing a thick-walled tube with a rigid die

A thick-walled tube with a uniform wall thickness of 2 [cm] is shown in the figure below. Relevantdimensions are indicated in the figure.

6 [cm]

10 [cm]

contact body III : rigid

contact body II : rigid

Poisson’s ratio = 0.45 [-]

contact body I : deformable

Young’s modulus = 5 [GPa]

The tube is made of a compliant elastic material with given Young’s modulus E and Poisson’sratio ν : E = 5 [GPa] and ν = 0.45 [-].

The tube, which sits on a rigid floor, is compressed by a die, which moves vertically downward,until the height is reduced to 70% of the undeformed outer tube diameter. It can be assumed thatthe material behavior remains elastic during the deformation.

During deformation, the contact area between tube and floor and tube and die will increasegradually. To analyze this contact problem, some contact bodies (CONTACT BODIES), indicated inthe figure, have to be defined in the option CONTACT. The floor and the die are modeled as rigid,while the tube is obviously deformable.

12


(MAIN MENU) (PREPROCESSING)

MESH GENERATION

We define the coordinate system with grid points interspaced at 0.01 [m].


(COORDINATE SYSTEM)

SET

U DOMAIN : -0.1, 0.1U SPACING : 0.01V DOMAIN : -0.1, 0.1V SPACING : 0.01GRID

FILL

RETURN

The necessary CURVES are straight lines (LINE) and circles (CENTER/POINT).

Define the two lines.

Define the two circles.

The circles are drawn with 10 straight line sections. Although this is not a verynice circle to look at, it is really a circle in the analysis. However, it is possibleto make a smoother circle.

PLOT

CURVES SETTINGS

(PREDEFINED SETTINGS)

HIGH

REGENERATE

MAIN

To generate an element mesh, we use a SURFACE and the CONVERT option. Thechosen DIVISION is not very fine, but suits our needs.


MESH GENERATION

SURFACE TYPE

RULED

RETURN

(SRFS) ADD : select CURVES

ELEMENT CLASS

QUAD(4)

RETURN

CONVERT

DIVISIONS

Enter the number of convert divisions in U and V : 24, 5(GEOMETRY?MESH) SURFACES TO ELEMENTS

select CURVES

RETURN

13

Never forget to SWEEP coinciding nodes and other things. Use CHECK to seewhether elements are UPSIDE DOWN and flip them if necessary.

SWEEP etc.CHECK etc.

The thickness of the plane strain tube is prescribed next.


GEOMETRIC PROPERTIES

NEW

PLANAR

PLANE STRAIN

THICKNESS 0.5OK

ADD

(ALL) EXIST.


BOUNDARY CONDITIONS

MECHANICAL

NEW

FIXED DISPLACEMENT

Fix the point of the tube, which touches the floor (apply1).

To prevent rigid rotation (rolling), we fix the points at 0 ≤ y ≤ 2 [cm] above the

bottom point in x-direction (apply2).


MATERIAL PROPERTIES (2x)NEW

ISOTROPIC

YOUNG’S MODULUS : 5e9POISSON’S RATIO : 0.45OK

(ELEMENTS) ADD

(ALL) EXIST.

The CONTACT BODIES are defined.


CONTACT

CONTACT BODIES

NEW

DEFORMABLE

OK

(ELEMENTS) ADD

(ALL) EXIST.

NEW

RIGID

OK

14

(CURVES) ADD : select floorNEW

RIGID

OK

(CURVES) ADD : select die

In a rigid contact body, the material is found at one side of the contact lineor surface. To visualize the side where the material is located, we use theoption ID CONTACT. The resulting ”hairy lines” indicate the material, and itshould obviously be inside the body. When the material (so the lines) are atthe wrong side, the line CURVE) or surface has to be ”flipped” in FLIP CURVES.Set ID CONTACT OFF (toggle) after checking.

The movement of the rigid die is applied by prescribing its velocity as a functionof time, in this case only in negative y-direction. This is done with a TABLE oftype time having a function value f(t) This function f is multiplied with theconstant velocity v0 to give the velocity as a function of time : v(t) = v0f(t).We assume the velocity to be constant, so we make f(t) = 1 and end up withv(t) = v0. The displacement of the die is : u(t) = v0t.

TABLES

NEW


TABLE TYPE time OK

XMAX : 50 This is the maximum time !YMAX : 10FORMULA

ENTER : 1SHOW MODEL

RETURN

The velocity v0 is prescribed. It is possible to move the die towards thetube in the null-increment, by defining an (APPROACH VELOCITY) in negativey-direction.

(RIGID BODY / BODY CONTROL)

RIGID

VELOCITY

PARAMETERS

(VELOCITY) Y : -0.0025 This is the velocity v0 in [m/s]

(VELOCITY) TABLE table1 OK

(APPROACH VELOCITY) Y : -1OK

OK

MAIN

By prescribing v0 during 12 ”seconds”, the total die displacement becomesu(t = 12) = 12∗ v0 = 0.03 [m], which reduces the height of the tube to 70 % ofits undeformed outer diameter.

15

LOADCASES

NEW

MECHANICAL

STATIC

LOADS

Select apply1, apply2 OK

TOTAL LOADCASE TIME : 12(FIXED) # STEPS : 12OK

OK

MAIN

JOBS

ELEMENT TYPES

MECHANICAL

PLANE STRAIN SOLID

(PLANE STRAIN FULL INTEGRATION) 11

OK

(ALL) EXIST.

RETURN

NEW

MECHANICAL

(LOADCASES) Select lcase1

ANALYSIS OPTIONS

LARGE STRAIN

OK

JOB RESULTS

Select job results

INITIAL LOADS

CLEAR

(ANALYSIS DIMENSION)

PLAIN STRAIN

OK

MAIN

SAVE the model as tube1. After running MARC the results can be observed.

16

2.3 Exercise

2.3.1 Compressing a thick-walled tube with a deformable die

The tube is now compressed by a deformable die. Dimensions are indicated in the figure below.

Young’s modulus = 5 [GPa]Poisson’s ratio = 0.45 [-]

10 [cm]

contact body III : rigid

contact body I : deformable

contact body II : deformable

6 [cm]

1 [cm]

The die is assumed to behave elastically with Young’s modulus E = 20 [GPa] and Poisson’s ratioν = 0.3 [-].

The downward motion of the die is realized by prescribing nodal displacements in FIXED DIS-

PLACEMENT as a function of (fictitious) time, using a ”time” table. Rigid motion of the die has tobe prevented.

Using default options for convergence testing will lead to problems, due to the default conver-gence criterion (RESIDUAL LOADS). Can you explain why these problems occur? Convergence canbe reached by choosing RESIDUAL DISPLACEMENTS as convergence criterion (in LOAD CASE).

17

3 Elasto-plastic material behavior

3.1 Background : Yield criterion and hardening

When deformation is large, the material will deform plastically resulting in permanent deforma-tion after load release. A tensile bar will yield, i.e. start to deform plastically, when the axialstress σ equals the initial yield stress σy0. For two- and three-dimensional stress states the stresscomponents are combined into a scalar effective or equivalent stress, σ̄, which is compared to theinitial yield stress in a yield criterion. When compression and tension results in yielding for thesame absolute value of σ̄, the yield criterion can be written as :

σ̄2 = σ2y0

The much used Von Mises equivalent stress is formulated as a function of the principal stressesσ1, σ2 and σ3. The yield criterion is :

12{(σ1 − σ2)

2 + (σ2 − σ3)2 + (σ3 − σ1)

2} = σ2y0

When the stresses increase after reaching the initial yield stress, the yield criterion remains sat-isfied. The increased equivalent stress equals the actual yield stress σy , which is higher than theinitial value. The yield stress increases due to plastic deformation. This is described by a hard-ening law, which relates σy to the effective plastic strain ε̄p. For linear isotropic hardening therelation is :

σy = σy0 + Hε̄p

where H is a constant hardening coefficient.

18

3.2 Plate with central hole; elasto-plastic material behavior

The figure shows the square plate with the central circular hole.

Poisson’s ratio = 0.3 [-]Young’s modulus = 210 [GPa]

u [mm]

16 [cm]

x

y

thickness = 1 [cm]

10 [cm]

We will prescribe the horizontal displacement of the left- and right edge (x = ±8 [cm]) to beu = 10 [mm] in 10 steps of 1 [mm]. Upper and lower edge are free to move, so the y-displacementis unconstrained.

The material is isotropic. Young’s modulus E, Poisson’s ratio ν and the initial yield stress σy0

are : E = 210 [GPa], ν = 0.3 [-] and σy0 = 500 [MPa] . Upon reaching the initial yield stress, thematerial hardens according to the linear isotropic hardening rule :

σy = σy0 + Hε̄p

where H is the hardening constant : H = E/20 = 10.5 [GPa].

19


Make the model of the upper right quarter of the plate or load it from an already savedmodel.

Prescribe BOUNDARY CONDITIONS for a total displacement of the right edge of 10

[mm] in 10 incremental steps.

Material properties are specified for all elements.


MATERIAL PROPERTIES

ISOTROPIC

YOUNG’S MODULUS : 210000POISSON’S RATIO : 0.3

(PLASTICITY) ELASTIC-PLASTIC

(YIELD SURFACE) VON MISES

(HARDENING RULE) ISOTROPIC

INITIAL YIELD STRESS

Enter value for ’yield-stress’ : 500OK

OK

We make a TABLE, where the vertical axis is f = σy/σy0 and the horizontalaxis is the equivalent plastic strain ε̄p. During the analysis, the function valuef is multiplied with the initial yield stress σy0, resulting in the current valueof the yield stress σy.For linear hardening we get :

f =σy

σy0= 1 +

H

σy0ε̄p

TABLES

NEW


TYPE

eq plastic strain This is the equivalent plastic strain ε̄p

(INDEPENDENT VARIABLE V1)

MAX

Enter maximum value of v1 : 2FORMULA ENTER

Enter formula : 1 + (210000/20)/(500) * v1FIT

SHOW MODEL

RETURN

ISOTROPIC


TABLE Select eq.pl.strain table OK

OK

20

OK

MAIN

Define the loadcase.

Define the JOB.

Select in JOB RESULTS also Plastic Strain (epl-strain).

The deformation of the plate is large. Even when we do not activate theoption LARGE STRAIN, the iterative procedure is used in each increment. It isused automatically when plastic deformation occurs.

Selecting the ADDITIVE plasticity procedure (default) implies that the materialmodel is a relation between the rate of the Cauch stress (σ̇) and the strain rate(ε̇). The stress increment is determined by integration of this relation. Becausethis integration is done numerically, the incremental strains cannot be to large.


JOBS

MECHANICAL

ANALYSIS OPTIONS

LARGE STRAIN

OK

SAVE the model (eg. as platelargedefelpl), run MARC and look at the results.Use HISTORY PLOT to visualize the time-dependent values of nodal variables.

Try to use exponential hardening according to :

σy

σy0=

(

1 +ε̄p

ε0

)n

with ε0 = 0.01 and n = 0.23.

Do the analysis with a prescribed load on the right edge. Choose EDGE LOAD

with pressure = −250 [MPa]. Reduce the load to zero affter reaching themaximum value and consider the permanent deformation of the plate.

21

3.3 Forming

The figure below shows a circular plate with a central hole, which is modeled with axisymmetricelements. The plate is fixed between rigid plate holders. The plate is deformed by a rigid punch,which moves from left to right through the hole. Dimensions are indicated in the figure.

4

10

2.5

0.5plate

plate holder plate holder

punch

3

0.5

Material parameters for the plate material are :

Young’s modulus : E = 210000 [N/mm2]Poisson’s ratio : ν = 0.3 [-]initial yield stress : σy0 = 300 [N/mm2]

During elastoplastic deformation the material hardening is isotropic, with constant hardening pa-rameter H = E/20 [N/mm2].

22


MESH GENERATION

(COORDINATE SYSTEM)

SET

U DOMAIN : -10, 10U SPACING : 0.5V DOMAIN : -10, 10V SPACING : 0.5GRID

FILL

RETURN

Define the curves for the plate holder.

Define the element mesh.

Define the curve (POLYLINE) for the punch.

Use DIVISIONS : 20 8 .

MATERIAL PROPERTIES (2x)NEW

ISOTROPIC

YOUNG’S MODULLUS : 210000POISSON’S RATIO : 0.3(PLASTICITY) ELASTIC-PLASTIC

(YIELD SURFACE) VON MISES

(HARDENING RULE) ISOTROPIC

INITIAL YIELD STRESS : 300OK

OK

TABLES

NEW


TABLE TYPE eq plastic strain OK

(LIMITS) XMAX : 3(FORMULA)

ENTER

Enter formula : 1 + 210000/(20 * 300) * v1FIT

SHOW MODEL

RETURN

ISOTROPIC


TABLE (PLASTIC STRAIN) Select eq.pl.strain tablle OK OK

(ELEMENTS) ADD

(ALL) EXIST

MAIN

GEOMETRIC PROPERTIES

AXISYMMETRIC

NEW

SOLID

23

CONSTANT DILATATION

OK

(ELEMENTS) ADD

(ALL) EXIST

MAIN

CONTACT

CONTACT BODIES

Define the deformable contact body (the element mesh).

Define the two contact bodies for the plate holders.

Define the contact body for the punch.

Define the TABLE for a constant velocity of the punch.

Check the orientation of the rigid contact bodies and use FLIP CURVES if necessary.

(RIGID BODY / BODY CONTROL)

VELOCITY

PARAMETERS

(VELOCITY) X : 0.5(VELOCITY) TABLE time

(INITIAL VELOCITY) X : 1OK

OK

MAIN

Define the LOADCASE.

Define the JOB.

Run the analysis and observe the results.

Run the same analysis but now with retraction of the punch into its initialposition.

24

Documents

Large deformations and contact problems in Mentat & MARCpiet/inf/mrc/pdf/MaMeLdco2011BW.pdf · 2011. 5. 9. · 1 Large deformations 1.1 Background : Incremental load When the deformation