plane and axisymmetric models in mentat & marcpiet/inf/mrc/pdf/mamepa2014cl.pdf ·...

Plane and axisymmetric modelsin

Mentat & MARC

Tutorial with some Background

Eindhoven University of TechnologyDepartment of Mechanical EngineeringPiet J.G. SchreursLambert C.A. van Breemen March 6, 2014

Contents

1 Plane stress and plane strain 31.1 Background : Theory and element formulation . . . . . . . . . . . . . . . . . . . . . . 31.2 Plate with a central hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.1 Modeling and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.3 Edge load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3 Axle support with radial load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.3.1 Modeling and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.3.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.3.3 Quadratic elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.4 Orthotropic plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.5 Circular disc with prescribed radial edge displacement . . . . . . . . . . . . . . . . . . 21

1.5.1 Modeling and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.5.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.6 Rotating solid disc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.6.1 Modeling, analysis and results . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2 Axisymmetry 282.1 Background : Axisymmetric modelling . . . . . . . . . . . . . . . . . . . . . . . . . . 282.2 Axle support with axial load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.2.1 Modeling, analysis and results . . . . . . . . . . . . . . . . . . . . . . . . . . 322.3 Circular discs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2

1 Plane stress and plane strain

1.1 Background : Theory and element formulation

In the tutorial ”Truss and beam structures” trusses and beams are described and used to model simplestructures with MSC.Marc/Mentat. Although the nodes of the truss and beam elements can havea displacement (and rotation) in three directions, only one stress is relevant : the axial stress. Forlinear elastic material behavior, the stress in a truss is related to the axial strain ε = ∆l

l0, according to

σ = Eε, where E is the Young’s modulus, which can be measured in a tensile experiment.

Plane stress

Many structures are build from flat plates, having a thickness, which is much smaller than the dimen-sions of the plate in its plane. The figure below shows such a plate, with its plane in the xy-coordinateplane and with a uniform thickness h0.

z

yx

z

yx

h0

In many cases it is allowed to assume that the plate is loaded in its plane, as shown in the figure. Whenthe plate is isotropic, there will be no bending of the plate and it will stay flat after deformation.

xy

z

P

In a point P of the plate a ”column” is cut out, with its sides parallel to the xz- and yz-coordinateplanes, respectively, and with dimensions dx × dy × h0. This small part of the plate is loaded withstresses, so forces per area. Because the thickness of the plate is very small, it can be assumedthat these stresses are uniform over the thickness. The stress components working on a stress cubedx×dy×dz are the normal stresses σxx and σyy and the shear stresses σxy and σyx. It can be shownthat σxy = σyx, so only three stress components are relevant.

3

P

σxx

σyy

σxy

σyx

xy

z

Because there are no stresses working in z-direction, this stress state is referred to as plane stress. Thedeformation in point P is described by the normal strain components εxx and εyy and the shear strainεxy.

For linear isotropic elastic material behavior the next relation holds between stresses and strains : σxx

σyy

σxy

=E

1− ν2

1 ν 0ν 1 00 0 1− ν

εxxεyyεxy

Due to the stresses σxx and σyy the thickness of the plate will change. The strain in thickness direction,so in z-coordinate direction in our case, is :

εzz =∆h

h0= − ν

E(σxx + σyy)

Plane strain

The displacement in the z-direction, which is for our plate the direction perpendicular to its plane, issuppressed. The plate is for instance glued to two parallel rigid bodies, which stay at the same distance.In that case we have εzz = 0, and call the deformation a state of plane strain.

In a state of plane strain, there is generally a stress σzz = 0. The linear elastic material behavioris described by : σxx

σyy

σxy

=E

(1 + ν)(1− 2ν)

1− ν ν 0ν 1− ν 00 0 1− 2ν

εxxεyyεxy

; σzz = ν(σxx + σyy)

It is noted that problems will occur when ν approaches 0.5. Stresses will become infinite. This isnot strange, since in a linear elastic material the volume change will be zero for ν = 0.5. To preventnumerical problems in MSC.Marc, we use the option CONSTANT DILATATION in the menu GEOMETRY.

Interpolation and integration : the element stiffness matrix

When analyzing plane stress and plane strain problems with the finite element method we have touse plane stress or plane strain elements. These elements are in MSC.Marc always located in the

4

xy-plane. Also the deformation is in the xy-plane and is described by the x- and y-displacement ofa number of nodal points, situated on the edges of the elements. The displacement of an internalpoint of the element is interpolated between the displacement of the nodes. For this interpolation, alocal coordinate system is used with isoparametric coordinates ξ1 and ξ2, which have values between-1 and +1. In mathematical terms the interpolation of the displacement u and v in x- and y-direction,respectively, can be written as follows :

u(ξ1, ξ2) = N1(ξ1, ξ2)u1 +N2(ξ1, ξ2)u

2 +N3(ξ1, ξ2)u3 + · · · =

n∑i=1

N i(ξ1, ξ2)ui

v(ξ1, ξ2) = N1(ξ1, ξ2)v1 +N2(ξ1, ξ2)v

2 +N3(ξ1, ξ2)v3 + · · · =

n∑i=1

N i(ξ1, ξ2)vi

where N i are the interpolation functions, associated to node i, and ui and vi the nodal displacementcomponents. The number of element nodes is n.

Calculation of the element stiffness matrix, which relates nodal displacements to nodal forces,implies integration of a function over the element volume. This integral can only be evaluated numer-ically : the integrand is calculated in a discrete number of internal integration points and these valuesare added after multiplication with a certain coefficient. The location of the integration points and thecoefficients are fixed for a certain element type.

Linear and quadratic elements

The figure below shows a 4-node element with all nodes in the four corners of a quadrilateral withstraight sides. Nodal points have a local counterclockwise numbering, which is in accordance with theMSC.Marc program. The figure also shows the element in the so-called isoparametric space, where

points are identified with the local isoparametric coordinates ξ˜=

[ξ1 ξ2

]T.

1 2

34

4

ξ

η

1 2

3

Because there are only two nodes on one element side, the displacement of an arbitrary point on theside can only vary linearly between the two nodal values. A 4-node quadrilateral is therefore called alinear element. The four interpolation functions, associated with the nodes are :

N1 = 14 (ξ1 − 1)(ξ2 − 1) ; N2 = −1

4 (ξ1 + 1)(ξ2 − 1)

N3 = 14 (ξ1 + 1)(ξ2 + 1) ; N4 = −1

4 (ξ1 − 1)(ξ2 + 1)

The element has 4 integration points, which are shown in the figure below. Their local coordinates arefixed and determined such that the numerical integration is as accurate as possible.

13

√3

21

43

1 2

43

ξ

η

13

√3

5

In a quadratic element, the displacement of an arbitrary point is interpolated using interpolationfunctions, which are quadratic in the local coordinates ξ1 and ξ2. This element has 8 nodes and istherefore also referred to as an 8-node element. Four of these nodes are situated in the corners ofthe element, the other four are in the middle of the element sides in the undeformed situation. Thenumbering of the local nodes is indicated in the figure below and is in accordance with the MSC.Marcprogram. The interpolation functions are :

N1 = 14 (ξ1 − 1)(ξ2 − 1)(−ξ1 − ξ2 − 1) ; N2 = 1

4 (ξ1 + 1)(ξ2 − 1)(−ξ1 + ξ2 + 1)

N3 = 14 (ξ1 + 1)(ξ2 + 1)(ξ1 + ξ2 − 1) ; N4 = 1

4 (ξ1 − 1)(ξ2 + 1)(ξ1 − ξ2 + 1)

N5 = 12 (ξ

21 − 1)(ξ2 − 1) ; N6 = 1

2 (−ξ1 − 1)(ξ22 − 1)

N7 = 12 (ξ

21 − 1)(−ξ2 − 1) ; N8 = 1

2 (ξ1 − 1)(ξ22 − 1)

The 8-node element has 9 integration points, the location of which is indicated in the figure.

ξ1

ξ2

1 2

34

ξ1

ξ2

1 2

34

68

7

5

8

7

6

5

7 8 9

1 2 3

46

5

15

√15

15

√15

6

1.2 Plate with a central hole

The figure below shows a square plate with a central hole. Relevant dimensions are indicated.

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. Displacements of left and right edges (x = ±8 [cm]) are prescribed :u = ±0.1 [mm]. In y-direction the displacement of the edge points is free. Top and bottom edge(y = ±8 [cm]) are also free. This load leads to a plane stress state in the plate.

The material of the plate is isotropic and can be assumed to be linearly elastic. Young’s modulusand Poisson’s ratio are known : E = 210 [GPa] and ν = 0.3 [-].

Deformation of and stresses in the plate will be determined using MSC.Marc and therefore a modelmust be made with MSC.Mentat. Considering symmetry and load learns that only a quarter of theplate has to be modeled. We choose the part with 0 ≤ (x, y) ≤ 8 [cm]. The correct symmetryconditions must of course be prescribed as boundary conditions.

7

1.2.1 Modeling and analysis

GEOMETRY & MESH

We start in the GEOMETRY & MESH. We will use the Cartesian coordinate system.This is shown on the computer screen as a two-dimensional grid in the xy-plane. Inthe grid points we can define (nodal) points and other things by clicking with theleft mouse button. In the grid settings the grid dimensions and the spacing betweenthe grid points can be specified. The program does not know of dimensions andunits, so we have to choose them and stick to them during the modeling andanalysis. Here, dimensions will be expressed in unit of meters. We choose the gridpoint spacing to be 0.01 meter.

SET (Coordinate System)U DOMAIN : -0.1, 0.1U SPACING : 0.01V DOMAIN : -0.1, 0.1V SPACING : 0.012� GRID (Coordinate Systen)(Fill view)

In the tutorial for ”Truss and beam structures” we used only (CURVES), but this isnot enough any more; we have to do more. We want to define a (SURFACE) andsubdivide this into elements. A simple method is used here, which can be appliedin many cases. First two (CURVES) are defined, one POLYLINE and a quarter of acircle, an (ARC). The (POINTS) which define the (CURVES) can be located in gridpoints by clicking the left mouse button.After defining the (CURVES), a surface of the type RULED is made. The idea is thata ”stick” is placed with its begin and end point on two separate (CURVES) and issubsequently rolled over them, thus describing a (SURFACE) in space.

GEOMETRY & MESH (Basic Manipulation)(CURVES)

POLYLINE

(CURVES) ADD

Define three points of the POLYLINE.

Close the window with (End list).

(CURVES)

ARC CEN/PNT/PNT

(CURVES) ADD

Define the required (see Command-screen) points of the (ARC).

When the (ARC) is drawn in the wrong direction, click (Undo) and define thelatter two (ARC) points in swapped order.

(SURFACES)

RULED

8

(SURFACES) ADD

Click on the two (CURVES).

A ”strange” surface may appear because of different ”orientation” of the twocurves. In that case we have to (Undo) the last action (definition of the surface)and ”flip” one of the two curves in the CHECK (Operations) menu with FLIP CURVES.The surface must then be redefined. When this is done successfully, the surface isnow going to be converted into elements.

(ELEMENTS)

QUAD(4)

CONVERT (Operations)DIVISIONS : 8, 4(Convert)

SURFACES

(To)

ELEMENTS

CONVERT

Select (SURFACE).

Close with (End list).

After defining the element mesh, SWEEP (Operations) has to be used to removecoinciding nodes and elements. It is recommended to go to the CHECK (Operations)menu and check whether there are elements INSIDE OUT or UPSIDE DOWN. When thisis the case these elements must be ”flipped”.It is recommended to prevent future drawing of (CURVES) and (SURFACES) in theview → plot control-menu.

GEOMETRIC PROPERTIES

NEW (STRUCTURAL)

PLANAR

PLANE STRESS

PROPERTIES

Thickness : 0.01OK

(ELEMENTS) ADD

Enter geometry add element list : (All existing)

MATERIAL PROPERTIES

Prescribe the material parameters : E = 2.1e11 [Nm−2] and ν = 0.3 [-].

The initial yield stress does not have to be prescribed. Mentat uses a default value,which is very high (σv0 = 1020 [Pa]).

BOUNDARY CONDITIONS

9

Prescribe the NEW (STRUCTURAL) boundary conditions.

Besides the prescribed boundary conditions symmetry conditions have to be prescribed.

Selecting nodes after NODES ADD is done with the mouse. Hold the left mousebutton and draw a box around the nodes to be selected. This is easy in this case,because the nodes are located on a straight line. Do some experiments with makingboxes, by pressing the Control key during box drawing. Close the selection with(End list).

Model definition is now completed. We are going to specify the analysis.First we choose the element type. Element 3 is a plane stress element with 4 nodes.((QUAD(4))). As can be seen from the list, there are more elements with straightedges and four nodes.

JOBS

ELEMENT TYPES (Element Types)(ANALISYS DIMENSION)

PLANAR

SOLID

(PLANE STRESS FULL INTEGRATION) 3

OK

Enter element list : (All existing)

After choosing the element type, the mechanical ((MECHANICAL)) analysis is spec-ified further.

JOBS

NEW

STRUCTURAL

PROPERTIES

- INITIAL LOADS : all ”applies”

- JOB RESULTS : stress, strain, von-mises

- ANALYSIS DIMENSION : PLAIN STRESS

OK

file → save as → File name: plate1 → SAVE

10

The model can now be analyzed. This is done by the finite element programMSC.Marc, by submitting the model in JOBS.

JOBS

RUN

SUBMIT 1

MONITOR

The program MSC.Marc is started and the model is analyzed. In the (RUN)

menu we see some information about the analysis. In the status screen the word”Running” is seen. When the status indicates ”Ready” the analysis is finished.When everything has worked well, we see in the (RUN) menu the exit number3004. After completion of the analysis, three files are written by MSC.Marc :

plate1 job1.logplate1 job1.outplate1 job1.t16

The results of the analysis can be visualized with Mentat.

The file with extension .out contains the results in alpha-numerical format (ASCII). It is generally arather long file and is mostly only opened when an error has occurred during the analysis. The filewith extension .t16 contains the results which can be visualized and post-processed in Mentat. Thefile with the extension .log contains information about the analysis.

11

1.2.2 Results

We take a look at the analysis results in Mentat. The .t16 file must be opened.This can be done directly from via file → open default. It can also be openedfrom within the (MAIN)-menu in the submenu RESULTS. When MSC.Marc has beenstarted by Mentat (via RUN), the Post file can be opened with OPEN DEFAULT. WhenMentat has been closed and we want to open a new Post file, we have to use theOPEN-button.

RESULTS

file → open default(Fill view)

(DEFORMED SHAPE)

DEFORMED & ORIGINAL

Because the deformation is very small – as it should be for a linear elastic analysis–, we have to enlarge it in SETTINGS and when AUTOMATIC is used the scaling isselected so that a proper deformation is visible.Values of the variables selected in JOB RESULTS can be visualized. Selection is donewith SCALAR, TENSOR or VECTOR.

SCALAR

Equivalent Von Mises Stress

OK

CONTOUR BANDS

It is possible to make a plot of a variable along a path in the model, a PATH PLOT.Here we will plot the y-displacement of the top edge of the plate as a function ofthe x-distance along that edge. Try other possibilities.

PATH PLOT

NODE PATH

Enter first node in Path-Plot node path : select (lm) node1Enter next node in Path-Plot node path (1) : select (lm) node2etc. etc. close with # ( (End list))

ADD CURVE

Enter X-axis variable : Arc Length

Enter Y-axis variable : Displacement y

FIT

Close the .t16 file with file → close. It is very important to do this becauseproblems may occur when loading a (new) model into Mentat. After closing the.t16 file, the model file is restored automatically.

12

1.2.3 Edge load

Instead of prescribing the displacement of the right edge of the plate, we can also apply a distributedload.

If necessary, we load the model file → open → plate1. The prescribed displacementis replaced with a prescribed edge load, which is a force per unit of area. InBOUNDARY CONDITIONS we make a new EDGE LOAD. To use it we have toselect it in INITIAL LOADS. In that case we remove the prescribed displacement.

file → open → plate1BOUNDARY CONDITIONS

NEW (STRUCTURAL)

EDGE LOAD

PRESSURE

Enter value for ’p’ : -1e8OK

The given value must be negative for a tensile load.

(EDGES) ADD

Enter add apply element edge list : select (EDGES)

Remove apply3 in INITIAL LOADS and select apply4.

file → save as → File name: plate2 → SAVE

Run MSC.Marc and look at results.

13

1.3 Axle support with radial load

An axle is fixed into a rubber ring in a rigid block as is shown in the figure below. Relevant dimensionsare indicated.

y

z

x

0.15 [m]

FR

FR

0.1 [m]

0.05 [m]

Material properties of the axle material are known :

Young’s modulus : E = 2.1 · 1011 [Nm−2]Poisson’s ratio : ν = 0.3 [-]

The rubber material is assumed to be linearly elastic with the next material parameters given :

Young’s modulus : E = 1.0 · 107 [Nm−2]Poisson’s ratio : ν = 0.49 [-]

The axle is loaded by two radial forces, which are equal : FR = 15000 [N]. Bending of the axle is nottaken into account.

It is assumed that a plane strain deformation state exists in the rubber material. Due to symmetryin the yz-plane w.r.t. the y-axis, only one half of the axle and rubber ring has to be modeled.

Which half part are you going to model?

14


GEOMETRY & MESH

The distance between the grid points is chosen in accordance with the dimensionsof the model. In the grid points we can locate model (POINTS).

SET (Coordinate System)U DOMAIN : -0.1, 0.1U SPACING : 0.005V DOMAIN : -0.1, 0.1V SPACING : 0.0052� GRID (Coordinate Systen)(Fill view)

We define three (ARCS). Because we want to model a solid axle, we define the thirdarc of the type CENTER/POINT/POINT with radius zero by clicking three times on thecentral grid point (0, 0, 0). This ’virtual’ curve can be used to define a surface.

(CURVES)

ARC CEN/PNT/PNT

(CURVES) ADD

Define the half circle of the axle.

Define the half circle of the rubber ring.

Define the central arc with radius zero.

Define the surface between the two half-circles.

Define the surface between the axle-circle and the arc in the center point.

The two defined (SURFACES) are converted to 4-node elements. The axle surfacehas some elements near the center point which are triangles. This is no problem :a quad4 element can have two points coinciding.

(ELEMENTS)

QUAD(4)


SURFACES

(To)

ELEMENTS

CONVERT

Select ring surface

Select the axle surface

After generating the element mesh we have to use SWEEP (Operations) and CHECK

(Operations).

15

Geometry parameters, material properties, and boundary conditions (note the sym-metry) must be prescribed.


NEW (STRUCTURAL)

PLANAR

PLANE STRAIN

PROPERTIES

Thickness : 0.1OK

(ELEMENTS) ADD


MATERIAL PROPERTIES

Prescribe the elastic properties of the axle (material1) and the rubber ring (material2

strangely named material4 by Mentat).

BOUNDARY CONDITIONS

NEW (STRUCTURAL)

Suppress all displacements of the edge of the rubber ring (apply1).

Nodes on the symmetry axis can only have a displacement in y-direction (apply2).

Prescribe the force FR (apply3).

The question is where this POINT LOAD must be applied, i.e. in which node(s).Must we prescribe a total force of 15000 N or 30000 N in negative y-direction?

In JOBS we select element 11, a 4-node linear plane strain element. Then we selectthe variables, which we want as output in JOB RESULTS and specify the load inINITIAL LOADS. Finally we indicate that the deformation state is plane strain.

JOBS


PLANAR

SOLID

(PLANE STRAIN FULL INTEGRATION) 11

OK


NEW

STRUCTURAL

PROPERTIES

- INITIAL LOADS : apply1, apply2, apply3


- ANALYSIS DIMENSION : PLAIN STRAIN

file → save as → File name: support1 → SAVE and runMSC.Marc. The results, available in the .t16 files can then be loaded andthe results can be visualized.

16

1.3.2 Results

Looking at Von Mises contour plots it is immediately clear that the stresses in therubber are much lower than those in the axle. We can visualize the stress state inthe rubber in more detail by making the axle invisible in the SELECT menu.

view → visibilityELEMENTS (Make Invisible)

Select the elements of the axle with a ”box”

Close with (End list)

Push ELEMENTS (Make Visible) and (All existing) again to make everything visible.

17

1.3.3 Quadratic elements

The analysis is done using linear elements with four element nodes (QUAD(4)).More accurate results, mostly with fewer elements, can be reached, using quadraticelements with eight element nodes. We can adapt the model support1. First file→ CLOSE the .t16 file.

file → open → support1GEOMETRY & MESH

CHANGE CLASS (Operations)QUAD(8)

ELEMENTS

(All existing)

SWEEP (Operations)

BOUNDARY CONDITIONS

Some boundary conditions must be adapted, as there are no boundary conditionsdefined in the new nodes.

Adapt the boundary conditions.


PLANAR

SOLID

(PLANE STRAIN FULL INTEGRATION) 27

OK


Save the model and run MSC.Marc. Results can be visualized in Mentat.

18

1.4 Orthotropic plate

The figure below shows a square plate, which will be loaded in its plane. It can be assumed that aplane stress state exists in the plate : σzz = σxz = σyz = 0.

2 y

x

1

α

The plate’s material is a ”matrix” in which long fibers are embedded, which all have the same orientationalong the direction indicated as 1 in the ”material” 1, 2-coordinate system. Both matrix and fibers arelinearly elastic with Young’s modulus and Poisson’s ratio Em, Ef , νm and νf , respectively. The volumefraction of the fibers is V . The angle between the 1-direction and the x-axis is α = 10 degrees. Inthe 1, 2-coordinate system the material behavior for plane stress is given by the next relation betweenstress and strain components : σ11

σ22

σ12

=1

1− ν12ν21

E1 ν21E1 0ν12E2 E2 0

0 0 (1− ν12ν21)G12

ε11ε22γ12

The Young’s moduli, Poisson’s ratios and shear modulus are defined as :

E1 =σ11

ε11; E2 =

σ22

ε22; ν12 = −ε22

ε11; ν21 = −ε11

ε22; G12 =

σ12

γ12

The material stiffness matrix is symmetric :

ν21E1 = ν12E2

which leaves us with four independent material parameters describing this orthotropic behavior.

The material parameters can be calculated with the next formulas, which are based on the rule-of-mixtures :

E1 = V Ef + (1− V )Em ;1

E2=

V

Ef+

1− V

Em→ E2 =

EfEm

V Em + (1− V )Ef

ν21 = V νf + (1− V )νm ;1

G12=

V

Gf+

1− V

Gm→ G12 =

GfGm

V Gm + (1− V )Gf

The next table lists numerical values for material parameters where the above formulas are used tocalculate E1, E2, ν12 and G12.

Em = 70 GPa Ef = 500 GPa E1 = 242 GPa E2 = 106.7 GPa

νm = 0.4 νf = 0.25 ν21 = 0.34 G12 = 38.5 GPa

V = 0.4 ν12 = 0.77

19

In Mentat we also have to give values for E3, G13, G23, ν31 and ν23. These values are taken to bethe same as the ones for the matrix material.

E3 = E2 = 106.7 GPa ; G13 = G12 = 38.5 GPa ; G23 = Gm = 25 GPa

ν31 = ν21 = 0.34 ; ν23 = νm = 0.4

The sides of the plate have a length 1 m, and the thickness of the plate is 1 cm. The plate is modeledin plane stress with 10 × 10 4-node elements of type 3.

Input of orthotropic material parameters in the material coordinate system can be done with the fol-lowing commands.

MATERIAL PROPERTIES

NEW

STANDARD

STRUCTURAL

Type : ELASTIC-PLASTIC ORTHOTROPIC

E1 E1

E2 E2

E3 E3

NU12 ν12NU23 ν23NU31 ν31G12 G12

G23 G23

G31 G31

(ELEMENTS) ADD

Enter add material element list : (All existing)

The orientation of the material coordinate system w.r.t. the global coordinate system must also begiven. This is done per element in the submenu ORIENTATION. In this submenu we see that there areseveral TYPEs of orientations. The EDGE-types say that the 1-axis of the local coordinate system isdefined by rotation over an ANGLE relative to a particular side of an element. Side 12 can be seen in theplot of the element mesh as a half-arrow on the element edges. It is the side from local node 1 to localnode 2. We can also define the rotation of the local 1-axis relative to one of the global coordinate planes.

NEW (Orientations)EDGE12

ANGLE 10(ELEMENTS) ADD

Enter add material element list : (All existing)

Boundary conditions are defined for tensile loading in x- and y-direction and for simple shear loadingin x-direction. In the latter case the y-displacement of the upper boundary is suppressed. Tensile loadsand shear load are edge loads of 1 GPa.

For all these loadcases, calculate the stress and strain components in the global and the material co-ordinate system.

The calculated stress and strain components can be given in the global coordinate system (default) andin the ”preferred system”. This ”preferred system” coincides with the ”material coordinate system”which is defined above in the NEW (Orientations) submenu.

20

1.5 Circular disc with prescribed radial edge displacement

The next figure shows a circular disc with outer radius b having a central circular hole with radius a.

ba

For linear isotropic material behavior the analytical solution for radial displacement and radial andtangential stresses are given as a function of the radius r :

ur = c1r +c2r

; σrr =E

1− ν2

[(1 + ν)c1 − (1− ν)

c2r2

]; σtt =

E

1− ν2

[(1 + ν)c1 + (1− ν)

c2r2

]where E is Young’s modulus and ν is Poisson’s ratio. The integration constants c1 and c2 are deter-mined by the boundary conditions. For ur(r = b) = ub ; σrr(r = a) = 0 we have :

c1 =(1− ν)b

(1− ν)b2 + (1 + ν)a2ub ; c2 =

(1 + ν)a2b

(1− ν)b2 + (1 + ν)a2ub

The next parameter values are given :

a = 0.01 [m] | b = 0.1 [m] | d = 0.01 [m] | E = 2 · 1011 [Pa] | ν = 0.3 [-] | ub = 0.001 [m]

For both a plane stress or a plane strain situation deformation and the stresses can be calculated withplanar elements. Due to the axial symmetry of geometry and loading, a representative part of the disccan be modeled and analyzed. In this case we model and analyze a quarter of the plate as shown inthe figure below.

x

y

z

yx

21


We start in the GEOMETRY & MESH. We will use the Cartesian coordinate system.Proceed as in the example ’Plate with a central hole’ and make the geometry usingtwo arcs and a ruled surface between them.Use CONVERT (Operations) to make the mesh with QUAD(4) elements:

GEOMETRY & MESH

GEOMETRY & MESH

(ELEMENTS)

QUAD(4)


SURFACES

(To)

ELEMENTS

Select (SURFACE).

Close with (End list).

After defining the element mesh, SWEEP (Operations) has to be used to removecoinciding nodes and elements. It is recommended to go to the CHECK (Operations)menu and check whether there are elements UPSIDE DOWN. When this is the casethese elements must be ”flipped”.It is recommended to prevent future drawing of (CURVES) and (SURFACES) in theview → Plot Control-menu.

In this example we want to analyze a plane stress situation. This has to be selectedwhen we specify the thickness.


NEW (STRUCTURAL)

PLANAR

PLANE STRESS

PROPERTIES

Thickness : 0.01OK

(ELEMENTS) ADD


In BOUNDARY CONDITIONS we have to suppress the X-displacement on thevertical symmetry section and the Y -displacement on the horizontal symmetrysection.

To prescribe the radial displacement on the outer edge, a local coordinate systemhas to be used, which can be defined in TRANSFORMS.

22

TABLES & COORD. SYST.

(New) (Coordinate System)CYLINDRICAL (R,Phi,Z)

Give three points to define the axis of the cylindrical coordinate system, eg. [0 0 0], [0 0 1]

and [1 0 0]. Now we need to transform the nodes on the outer edge to the cylindrical

system.

TOOLBOX

TRANSFORMATIONS (General)NEW2� Coordinate System

COORDINATE SYSTEM

crdsyst1

NODES ADD select nodes at outer edge

Insert the material properties as usual.Although the material is isotropic and there is no real need for a material coordinatesystem, we will define one. The reason is that we want MSC.Marc to calculatethe radial and tangential stresses, which are the stresses in the ”preferred system”.This coordinate system can be defined in all elements analogously, due to the factthat the elements are all oriented in the same way.In the submenu ORIENTATION we use EDGE12 for all elements with ANGLE zero. The”preferred stress”component-11 is then the tangential stress and the component-22the radial stress.

MATERIAL PROPERTIES

(New) (Orientations)EDGE12

ANGLE 0ADD

(All existing)

We are going to specify the analysis.First we choose the element type. Element 3 is a plane stress element with 4 nodes.((QUAD(4))). As can be seen from the list, there are more elements with straightedges and four nodes.

JOBS


PLANAR

SOLID

(PLANE STRESS FULL INTEGRATION) 3

OK


After choosing the element type, the mechanical ((MECHANICAL)) analysis is spec-ified further.

23

NEW

STRUCTURAL

PROPERTIES

- INITIAL LOADS : all ”applies”

- JOB RESULTS : stress, preferred stress, strain, von-mises

- ANALYSIS DIMENSION : PLANE STRESS

file → save as → File name: Disc1 → SAVE

24

1.5.2 Results

After analysis the result can be visualized.The deformation can be shown as a deformed mesh. CONTOUR BANDS can be set tovisualize displacements, strains, stresses and preferred stresses. Numerical valuescan also be viewed in nodal points.

In a PATH PLOT the stresses can be plotted against the radius. The result is shownin the figure below.

0 0.02 0.04 0.06 0.08 0.1−1

0

1

2

3

4

5

6x 10

8

r [m]

σ [P

a]

σrr

σtt

σzz

25

1.6 Rotating solid disc

A solid circular disc with radius b and uniform thickness, is rotated with a constant radial velocity ω[rad/s] about its central axis perpendicular to its plane. The material is isotropic and linearly elasticwith Young’s modulus E and Poisson’s ratio ν. The density is ρ.

z

ω

b

r

r

The general solution is :

u = c1r +c2r

−(1− ν2

E

)ρω2r3

8

σrr =E

1− νc1 −

E

1 + ν

c2r2

− 3 + ν

8ρω2r2 ; σtt =

E

1− νc1 +

E

1 + ν

c2r2

− 1 + 3ν

8ρω2r2

where c1 and c2 are integration constants. For a solid disc the displacement at r = 0 must remain finite,while the stress boundary condition is σrr(r = b) = 0. Integration constants can now be determined :

c1 =3 + ν

1 + ν

1− ν2

E

1

8ρω2b2 ; c2 = 0

The next parameter values are given for this example :

b = 0.05 [m] | f = 6 [c/s] | E = 200 [GPa] | ν = 0.3 [-] | ρ = 7500 [kg/m3] |

26

1.6.1 Modeling, analysis and results

In the menu MATERIAL PROPERTIES the density of the material has to be specified.Again the axial symmetry allows the analysis of the quarter plate, using proper boundary conditions onthe section lines.The rotation is modeled as a CENTRIFUGAL LOAD in BOUNDARY CONDITONS. The rotation rate isgiven in rotations (= cycles) per second. The rotation axis has to be defined by two points in thethree-dimensional space.

BOUNDARY CONDITIONS

NEW (STRUCTURAL)

CENTRIFUGAL LOAD2� Angular Frequency (Cycles/Time) 6Axis Of Rotation

X1 0Y1 0Z1 0X2 0Y2 0Z2 1OK

ELEMENTS ADD


Stress components for the analytical solution are plotted against the radius in the figure below.

0 0.1 0.2 0.3 0.4 0.50

2

4

6

8

10

12x 10

5

r [m]

σ [P

a]

σrr

σtt

27

2 Axisymmetry

2.1 Background : Axisymmetric modelling

Many devices and components have a geometry which is symmetric w.r.t. an axis and are thus calledaxisymmetric. It is obvious that the position of material points of such an object can best be described ina cylindrical coordinate system. Coordinates are the distance z measured along the axis of symmetry,the distance r from and perpendicular to that axis and an angle θ, which indicates the position incircumferential direction, with respect to an arbitrary starting point (θ = 0o : 0 ≤ θ ≤ 360o(= 2π[rad])(see figure below).

z

r

θ

P

r

z

r

When the load is independent of the angle θ, it is also called axisymmetric. With the additionalassumption that there is no rotation around the z-axis, we only have to consider and model the cross-section of the object, when we want to analyze its mechanical behavior (see figure).

Modeling such axisymmetric problems in MSC.Marc, the z-axis is oriented in the x-direction andthe r-axis in the y-direction. Only one half of the geometry of the cross-section must be modeled andit must be located in the half-space y > 0, as is shown in the figure below.

r r

z

x = z-as

y = r-as

The stress state in a material point is characterized by four stress components, which are indicatedon the faces of a stress cube in the figure below. This stress cube is shown two times, once in thecross-section and once three-dimensional.

28

The deformation is described by four strain components, which are defined in accordance with thestress components. For isotropic linear elastic material behavior we have :

σrr

σtt

σzz

σrz

=E

(1 + ν)(1− 2ν)

1− ν ν 0 νν 1− ν 0 ν0 0 1− ν 0ν ν 0 1− 2ν

εrrεttεzzεrz

x = z-as

y = r-asσrr

P

σtt σzz

σrr

σrz

σrz

σzr

σzr

σtt σzz

Analyzing axisymmetric problems with the finite element method implies that axisymmetric elementsmust be used, which are defined in the cross-section. The deformation of such an element is definedby the displacement in r- and z-direction of the element nodal points. We must be aware that a nodalpoint is in fact a nodal ring as is indicated in the figure below.

x = z-as

y = r-as

In the cross-section of the element the axial and radial displacement of a point is interpolated betweenthe displacements of the element nodes. As with the planar elements, linear or quadratic interpolationcan be used. Again we have 4-node and 8-node elements with 4 and 9 integration points respectively.In MSC.Marc these elements are indicated as element type 10 and 28. The cross-section of a 4-nodeelement is shown in the figure below.

29

1 2

34

4

ξ

η

1 2

3

13

√3

21

43

1 2

43

ξ

η

13

√3

In MSC.Marc/Mentat the strains in the integration points are defined as follows :

strain 1 = global zz-strain = εzzstrain 2 = global rr-strain = εrrstrain 3 = global tt-strain = εttstrain 4 = global rz-strain = γrz

Stress components are defined in the same way.

30

2.2 Axle support with axial load

The load on the axle is assumed to be an axial force : FA = 20000 [N], as is indicated in the figurebelow.

y

z

x

FA

The geometry and the load allow for an axisymmetric Modeling and analysis, as indicated in the figurebelow.

as

manchet

x

yFA

NB. : The x-axis is the axial axis and the positive y-axis is the radial axis. This is the usualdefinition in MSC.Marc/Mentat.

31

2.2.1 Modeling, analysis and results

GEOMETRY & MESH

Define the Cartesian coordinate system with grid point spacing 0.005 [m].

(CURVES)

LINE

(CURVES) ADD

Define three (LINES) parallel to the x-axis between −0.05 < x < 0.05 for y = 0,

y = 0.025 and y = 0.075.

Define (SURFACES) between these (CURVES).

Convert the surfaces in elements of type QUAD(4) with DIVISIONS [8, 4].

SWEEP and CHECK the mesh.

BOUNDARY CONDITIONS

NEW (STRUCTURAL)

Fix the top edge of the rubber part (apply1).

Nodes on the axial axis (y = 0) can only have a displacement in the x-direction (= axial)

(apply2).

Apply the axial force in one point of the axle (apply3).

Material properties of axle and rubber are prescribed in the usual way. Geometricparameters are not relevant for this axisymmetric analysis.

In JOBS we select element 10, a 4-node quadrilateral axisymmetric element. Weindicate that the analysis is axisymmetric.

JOBS

ELEMENT TYPES (Element Types)(ANALASYS DIMENSION)

AXISYMMETRIC

SOLID

(FULL INTEGRATION) 10

OK

Enter element list : (All exisiting)

NEW

STRUCTURAL

PROPERTIES

- INITIAL LOADS : apply1, apply2, apply3


- ANALYSIS DIMENSION : AXISYMMETRIC

Save the model as Support2 and run MSC.Marc. The results, available in the .t16file can be loaded into Mentat and visualized.

Change the model to use quadratic 8-node elements. For axisymmetric analyses in MARC this iselement 28.

32

2.3 Circular discs

A circular disc can be modeled with axi-symmetric elements. In fact these elements are ’ring’-elements.What we model on the screen is the two-dimensional cross-section of the model.

In MSC.Mentat the axial-direction always coincides with the horizontal x-axis. Only the part of thecross-section above this axis is modeled and converted in a finite element mesh.

y = radial

z = tangential

x = axial

For a disc with uniform thickness, this half-cross-section is just a rectangle, with the thickness as thedimension in the x-direction (= axial direction) and the inner ad outer radius as the dimensions in they-direction (= radial direction). It would of course be no problem to model a disc with a nonuniformthickness.

Boundary conditions have to be applied to prevent rigid body movement. It is obvious that therecan not be a rigid body movement in radial direction, but in axial direction we have to prevent therigid body displacement. When the disc is solid, it is advised to prescribe the radial displacements onthe axis to be zero.

Edge displacements and edge loads can be prescribed straightforwardly in BOUNDARY CONDI-

TONS. Rotation leads to radial accelerations, which are modeled in MSC.Mentat as body forces inCENTRIFUGAL LOAD. We have to provide the rotational velocity as number of rotations (= cycles)per second. We also have to define the axis of rotation. This is a straight line and thus defined withtwo points in space. Below we have the x-axis as the rotation axis and the number of cycles/second is 6.

BOUNDARY CONDITIONS

NEW (STRUCTURAL)

CENTRIFUGAL LOAD2� Angular Frequency (Cycles/Time) 6Axis Of Rotation

X1 0Y1 0Z1 0X2 1Y2 0Z2 0OK

ELEMENTS ADD


33

plane and axisymmetric models in mentat & marcpiet/inf/mrc/pdf/mamepa2014cl.pdf ·...

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