Plane, axisymmetric and 3D models

in

Mentat & MARC

Tutorial with Background and Exercises

Eindhoven University of TechnologyDepartment of Mechanical EngineeringPiet Schreurs January 28, 2010

Contents

1 Finite element method 3

2 Plane stress and plane strain 62.1 Background : Theory and element formulation . . . . . . . . . . . . . . . . . . . . 62.2 Plate with a central hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.1 Modelling and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2.3 Dynamic boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.3 Axle bearing with radial load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3.1 Modelling and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3.3 Quadratic elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.4 Orthotropic plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.5.1 Large thin plate with a cirular hole . . . . . . . . . . . . . . . . . . . . . . . 252.5.2 Plate with a square hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.5.3 Thick-walled pressurized cylinder . . . . . . . . . . . . . . . . . . . . . . . . 292.5.4 Rotating engineering parts . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 Axisymmetry 343.1 Background : Axisymmetric modelling . . . . . . . . . . . . . . . . . . . . . . . . . 343.2 Axle bearing with axial load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.2.1 Modelling, analysis and results . . . . . . . . . . . . . . . . . . . . . . . . . 383.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.3.1 Circular plate with a circular hole . . . . . . . . . . . . . . . . . . . . . . . 403.3.2 Rotating engineering parts . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4 Three-dimensional models 434.1 Background : Three-dimensional elements . . . . . . . . . . . . . . . . . . . . . . . 434.2 Three-dimensional model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.3.1 Inhomogeneous plate with a central circular hole . . . . . . . . . . . . . . . 46

2

1 Finite element method

In this section it is described how an approximate solution for the equilibrium equation can bedetermined with the finite element method. This description is very concise and detailed deriva-tions are not included.

Weighted residual formulation

The equilibrium equation reads :

~∇ · σ + ρ~q = ~0 ∀ ~x ∈ V

Analytical solutions for this partial differential equation only can be found for very simple prob-lems. For the more realistic cases only approximate solutions can be determined, using a numericaltechnique. The finite element method is widely used for this purpose. For the application of thismethod the differential equation is transformed into an integral equation. This transformation isformulated as follows :

The requirement that in each material point ~x ∈ V of the body the equi-librium equation

~∇ ·σ + ρ~q = ~0

must be satisfied, is equivalent to the requirement that for every allowedfunction ~w(~x) the integral over the volume V

∫

V

~w ·

[

~∇ ·σ + ρ~q]

dV = 0

is satisfied.

The function ~w(~x) is the weighting function and the integral the weighted residual formulation ofthe equilibrium equation. These names become obvious, when it is realized that for an approximatesolution, the equilibrium equation is not satisfied exactly and that an error ~∇ ·σ + ρ~q 6= ~0 willremain. This local error is smeared out over the whole volume V of the body by the integral,employing a weighting function ~w(~x).The weighted residual integral is the starting point for application of the finite element methodand will now be further elaborated.

Weak formulation

To relax the continuity requirements concerning the approximate stress field σ(~x), the derivativeof the stress tensor σ is removed from the integral. This can be done by simple partial integrationof the divergence term.

~∇ · (σ · ~w) = (~∇~w) : σT + ~w · (~∇ ·σ) → substitution →

∫

V

[

~∇ · (σ · ~w) − (~∇~w) : σT + ~w · ρ~q

]

dV = 0 ∀ ~w →

∫

V

(~∇~w)T : σ dV =

∫

V

~w · ρ~q dV +

∫

S

~n · σ · ~w dS ∀ ~w

Applying Gauss’ theorem transformed the first term into an integral over the boundary S of V ,where ~n is the unity vector in a boundary point.

3

The stress tensor can be related to the strain tensor ε, which can be expressed in the gradient ofthe displacement ~u of the material points. For linear elastic material behavior, we have

σ = 4C : ε = 4

C : (~∇~u)

where 4C is the fourth-order material tensor. In a boundary point we have ~n ·σ = ~p, where ~p is

the stress vector in that point. The weighted residual integral becomes :

∫

V

(~∇~w)T : 4C : (~∇~u) dV =

∫

V

~w · ρ~q dV +

∫

S

~w · ~p dS ∀ ~w

Discretisation

The integrals over V and S are written as a sum of integrals over a number of sub-volumesand -boundaries, in which V and S are subdivided. These volumes are the elements and theirboundaries – as far as these are located on the external boundary S of V –, with volume V e andarea Se, respectively. The discretised weighted residual integral becomes :

∑

e

∫

V e

(~∇~w)T : 4C : (~∇~u) dV e =

∑

e

∫

V e

~w · ρ~q dV e +∑

eS

∫

Se

~w · ~p dSe ∀ ~w

Interpolation and integration

The next step in deriving the finite element formulation is the interpolation of the displacement andthe weigthing function. This is done in each element, where a local element coordinate systemwith coordinates ξ

˜is used. For three-dimensional elements there are three local coordinates :

ξ˜

T = {ξ1, ξ2, ξ3}. Each of these so-called isoparametric coordinates has values between -1 and +1.The value of a variable – ~u and ~w – in an element point with coordinates ξ

˜is expressed in its

values in a number of discrete points of the element, the element nodes. The interpolation functionassociated with node i is N i(ξ

˜). For ~u and ~w and for their gradients, we can now write :

~u = N1(ξ˜)~u1 + N2(ξ

˜)~u2 + N3(ξ

˜)~u3 + · · · = N

˜T (ξ

˜)~u˜

e → ~∇~u =(

~∇N˜

)T

~u˜

e = ~B˜

T~u˜

e

~w = N1(ξ˜)~w1 + N2(ξ

˜)~w2 + N3(ξ

˜)~w3 + · · · = N

˜T (ξ

˜)~w˜

e → ~∇~w =(

~∇N˜

)T

~w˜

e = ~B˜

T~w˜

e

The columns ~u˜

e and ~w˜

e contain the nodal values of ~u and ~w. The displacement and the weightingfunction are interpolated in the same way, which is according to Galerkin’s method. Substitution ofthe interpolations in the contribution of one single element to the total weighted residual integral,gives :

∫

V e

( ~B˜

T~w˜

e)T : 4C : ( ~B

˜

T~u˜

e) dV e =

∫

V e

~w˜

eT

N˜

· ρ~q dV e +

∫

Se

~w˜

eT

N˜

· ~p dSe ∀ ~w˜

e →

~w˜

eT

·

∫

V e

~B˜

·4C ·

~B˜

TdV e

· ~u˜

e = ~w˜

eT

·

∫

V e

N˜

ρ~q dV e

+ ~w˜

eT

·

∫

Se

N˜

~p dSe

∀ ~w˜

e →

~w˜

eT

·Ke· ~u˜

e = ~w˜

eT

·~f˜

e

e∀ ~w

˜e

where Ke is the element stiffness matrix and ~f

˜

e

ethe column with the external nodal forces. Cal-

culating the element stiffness matrix implies integration over the element volume. The integral,whose integrand is a function of the local coordinates ξ

˜, can not be calculated analytically. It is

done numerically by evaluating the integrand in a number of discrete so-called integration points

4

and adding the results after multiplication by a weighting factor. For a three-dimensional elementwe can write this as

1∫

ξ1=−1

1∫

ξ2=−1

1∫

ξ3=−1

f(ξ1, ξ2, ξ3) dξ1dξ2dξ3 =∑

i

ci f(ξi1, ξ

i2, ξ

i3)

Assembling

During the assembling procedure the contribution of each element is added to the total weightedresidual integral. The components of the element columns ~w

˜

e and ~u˜

e are inserted at the properplaces in the system columns ~w

˜and ~u

˜, containing the values of ~w and ~u in all nodes. The element

nodal force ~f˜

e

eand the element stiffness matrix K

e are than added to the system column ~f˜

eand

the system stiffness matrix K, which results in :

~w˜

T·K · ~u

˜= ~w

˜T

·~f˜

e∀ ~w

˜

Set of equations

Because the above equation has to be satisfied for all possible weighting functions ~w˜, the unknown

components of ~u˜

and ~f˜

emust be determined such that the next set of algebraic equations is

satisfied :

K · ~u˜

= ~f˜

e

After taking into account the boundary conditions – prescribed components of ~u˜

and ~f˜

e–, the

unknown displacement components can be solved. It is obvious that all tensors and vectors firsthave to be written in components w.r.t. a suitable coordinate system.

Boundary conditions

No external forces are needed for a rigid translation of the structure. Writing the set of equationsin components, this means

k11 k12 k13 · · ·k21 k22 k23 · · ·k31 k32 k33 · · ·...

...... .

a

a

a...

=

000...

or Ka˜

= 0˜

The above relation indicates that K is a singular matrix, who’s determinant is zero. The set ofequations Ku

˜= f

˜e

can never be solved, because the inverse of the singular matrix does not exist.By prescribing enough boundary conditions to prevent rigid body motion, the resulting set ofequations can be solved, because the resulting stiffness matrix is no longer singular but regular.

5

2 Plane stress and plane strain

2.1 Background : Theory and element formulation

In the tutorial ”Truss and beam structures” trusses and beams are described and used to modelsimple structures with MSC.Marc/Mentat. Although the nodes of the truss and beam elementscan have a displacement (and rotation) in three directions, only one stress is relevant : the axialstress. For linear elastic material behavior, the stress in a truss is related to the axial strainε = ∆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 thedimensions of the plate in its plane. The figure below shows such a plate, with its plane in thexy-coordinate plane 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.In that case 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 constant over the thickness. The stress components working on a stresscube dx× dy × dz are the normal stresses σxx and σyy and the shear stresses σxy and σyx. It canbe shown that σxy = σyx, so only three stress components are relevant.

6

P

σ11

σ22

σ12

σ21

12

3

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

For linear 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

εxx

εyy

εxy

Due to the stresses σxx and σyy the thickness of the plate will change. The strain in thicknessdirection, 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,is suppressed. The plate is for instance glued to two parallel rigid bodies, which stay at the samedistance. 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 6= 0. The linear elastic materialbehavior is described by :

σxx

σyy

σxy

=E

(1 + ν)(1 − 2ν)

1 − ν ν 0ν 1 − ν 00 0 1 − 2ν

εxx

εyy

εxy

; σzz = ν(σxx + σyy)

It is noted that problems will occur when ν→0.5. Stresses will become infinite. This is notstrange, 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 thexy-plane. Also the deformation is in the xy-plane and is described by the x- and y-displacement

7

of a number of nodal points, situated on the edges of the elements. The displacement of aninternal point of the element is interpolated between the displacement of the nodes. For thisinterpolation, a local coordinate system is used with isoparametric coordinates ξ and η, whichhave 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(ξ, η) = N1(ξ, η)u1 + N2(ξ, η)u2 + N3(ξ, η)u3 + · · · =

n∑

i=1

N i(ξ, η)ui

v(ξ, η) = N1(ξ, η)v1 + N2(ξ, η)v2 + N3(ξ, η)v3 + · · · =

n∑

i=1

N i(ξ, η)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 implies integration of a function over the elementvolume. This integral can only be evaluated numerically : the integrand is calculated in a discretenumber of internal integration points and these values are added after multiplication with a cer-tain coefficient. The location of the integration points and the coefficients are fixed for a certainelement 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 withthe MSC.Marc program. The figure also shows the element in the so-called isoparametric space,

where points are identified with the local isoparametric coordinates ξ˜

=[

ξ η]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 onthe side can only vary linearly between the two nodal values. A 4-node quadrilateral is thereforecalled a linear element. The four interpolation functions, associated with the nodes are :

N1 = 14 (ξ − 1)(η − 1) ; N2 = − 1

4 (ξ + 1)(η − 1)

N3 = 14 (ξ + 1)(η + 1) ; N4 = − 1

4 (ξ − 1)(η + 1)

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

13

√3

21

43

1 2

4

3

ξ

η

13

√3

8

In a quadratic element, the displacement of an arbitrary point is interpolated using interpolationfunctions, which are quadratic in the local coordinates ξ and η. This element has 8 nodes and istherefore also referred to as an 8-node element. Four of these nodes are situated in the cornersof the element, the other four are in the middle of the element sides in the undeformed situation.The numbering of the local nodes is indicated in the figure below and is in accordance with theMSC.Marc program. The interpolation functions are :

N1 = 14 (ξ − 1)(η − 1)(−ξ − η − 1) ; N2 = 1

4 (ξ + 1)(η − 1)(−ξ + η + 1)

N3 = 14 (ξ + 1)(η + 1)(ξ + η − 1) ; N4 = 1

4 (ξ − 1)(η + 1)(ξ − η + 1)

N5 = 12 (ξ2 − 1)(η − 1) ; N6 = 1

2 (−ξ − 1)(η2 − 1)

N7 = 12 (ξ2 − 1)(−η − 1) ; N8 = 1

2 (ξ − 1)(η2 − 1)

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

1

η

ξ

2 1

η

ξ

43

2

21

98

3

15

√15

15

√15

5

64

7

86

4 37

5

6

7

8

5

9

2.2 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’smodulus and Poisson’s ratio are known : E = 210 [GPa] and ν = 0.3 [-].

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

10

2.2.1 Modelling and analysis

We start in the MAIN MENU and go to MESH GENERATION. We will use theCartesian coordinate system. This is shown on the computer screen as a two-dimensional grid in the xy-plane. In the grid points we can define (nodal) pointsand other things by clicking with the left mouse button. In the grid settingsthe grid dimensions and the spacing between the grid points can be specified.The program does not know of dimensions and units, so we have to choosethem and stick to them during the modelling and analysis. Here, dimensionswill be expressed in unit of meters. We choose the grid point spacing to be0.01 meter.

(MAIN MENU)(PREPROCESSING)

MESH GENERATION

(COORDINATE SYSTEM)

SET

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

FILL

RETURN

In the tutorial for ”Truss and beam structures” we used only CURVES, but thisis not enough any more; we have to do more. We want to define a SURFACE

and subdivide this into elements. A simple method is used here, which canbe applied in many cases. First two CURVES are defined, one POLY LINE and aquarter of a circle, an ARC. The POINTS which define the CURVES can be locatedin grid points by clicking the left mouse button.After defining the CURVES, a surface of the type RULED is made. The idea isthat a ”stick” is placed with its begin and end point on two separate CURVES

and is subsequently rolled over them, thus describing a SURFACE in space.

CURVE TYPE

POLYLINE

RETURN

(CRVS) ADD

Define three points of the POLYLINE.

Close the window with END LIST.

CURVE TYPE

(ARCS) CENTER/POINT/POINT

RETURN

(CRVS) ADD

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

When the ARC is drawn in the wrong direction, click UNDO and define the lattertwo ARC points in swapped order.

11

SURFACE TYPE

RULED

RETURN

(SRFS) 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 menu with FLIP CURVE. Thesurface must then be redefined. When this is done successfully, the surface isnow going to be converted into elements.

ELEMENT CLASS

QUAD(4)

RETURN

CONVERT

DIVISIONS : 8, 4(GEOMETRY/MESH) SURFACES TO ELEMENTS

Select SURFACE.

Close with END LIST.

After defining the element mesh, SWEEP has to be used to remove coincidingnodes and elements. It is recommended to go to the CHECK menu and checkwhether there are elements INSIDE OUT or UPSIDE DOWN. When this is the casethese elements must be ”fliped”.It is recommended to prevent future drawing of CURVES and SURFACES in thePLOT-menu.

(MAIN MENU)

BOUNDARY CONDITIONS

Prescribe the MECHANICAL 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 thiscase, because the nodes are located on a straight line. Do some experimentswith making boxes, by pressing the Control key during box drawing. Close theselection with END LIST.

(MAIN MENU)

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 defaultvalue, which is very high (σv0 = 1020 [Pa]).

12

(MAIN MENU)

GEOMETRIC PROPERTIES

PLANAR

NEW

PLANE STRESS

THICKNESS : 0.01OK

(ELEMENTS) ADD

(ALL) EXIST

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

(MAIM MENU) (ANALYSIS)

JOBS

ELEMENT TYPES

MECHANICAL

PLANE STRESS

(FULL INTEGRATION) 3

OK

(ALL) EXIST.

RETURN

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

NEW

MECHANICAL

- INITIAL LOADS : all ”applies”

- JOB RESULTS : stress, strain, von-mises

- (ANALYSIS DIMENSION) : PLAIN STRESS

MAIN

SAVE the model as plaatgat1.

13

The model can now be analysed. This is done by the finite element programMARC, by submitting the model in (MAIN MENU)(ANALYSIS) JOBS.

(ANALYSIS)

JOBS

RUN

SUBMIT 1

MONITOR

The program MARC is started and the model is analysed. In the RUN

menu we see some information about the analysis. In the status screen theword ”Running” is seen. When the status indicates ”Ready” the analysis isfinished. When everything has worked well, we see in the RUN menu the exitnumber 3004. After completion of the analysis, three fils are written by MARC :

plaatgat1 job1.logplaatgat1 job1.outplaatgat1 job1.t16

The results of the analysis can be visualised with Mentat.

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

14

2.2.2 Results

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

(MAIN MENU) (POSTPROCESSING)

RESULTS

OPEN DEFAULT

FILL

DEF & ORG

Because the deformation is very small – as it should be for a linear elasticanalysis –, we have to enlarge it in SETTINGS and when AUTOMATIC is used thescaling is slected so that a proper deformation is visible.Values of the variables selected in JOB RESULTS can be visualised. Selection isdone with 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 afunction of the x-distance along that edge. Try other possibilities.

PATH PLOT

NODE PATH

Enter first node in Path-Plot node path : select (lm) node1

Enter next node in Path-Plot node path (1) : select (lm) node2

etc. etc. close with # (END LIST)VARIABLES

ADD CURVE

Enter X-axis variable : Arc Length

Enter Y-axis variable : Displacement y

FIT

RETURN

MAIN

Close the .t16 file with CLOSE. It is very important to do this because problemsmay occur when loading a (new) model into Mentat. After closing the .t16file, the model file is restored automatically.

15

2.2.3 Dynamic boundary conditions

Instead of prescribing the displacement of the right edge of the plate, we can also apply a dis-tributed load.

If necessary, we load the model plaatgat1 in FILES with RESTORE or OPEN. Theprescribed displacement is replaced with a prescribed edge load, which is aforce per unit of area. In BOUNDARY CONDITIONS we make a new apply EDGE

LOAD. To use it we have to select it in INITIAL LOADS. In that case we removethe prescribed displacement apply3.

FILES

RESTORE or OPEN plaatgat1MAIN

(MAIN MENU) (PREPROCESSING)

BOUNDARY CONDITIONS

MECHANICAL

NEW

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 : selecteer EDGES

Remove appl3 in INITIAL LOADS and select apply4.

SAVE the model as plaatgat2, run MARC and look at results.

16

2.3 Axle bearing with radial load

An axle is fixed into a rubber ring in a rigid block as is shown in the figure below. Relevantdimensions are 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 isnot taken into account.

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

Which half part are you going to model?

17

2.3.1 Modelling and analysis

(MAIN MENU)(PREPROCESSING)

MESH GENERATION

The distance between the grid points is choosen in accordance with the dimen-sions of the model. In the grid points we can locate model POINTS.

(COORDINATE SYSTEM)

SET

U DOMAIN : -0.1, 0.1U SPACING : 0.005V DOMAIN : -0.1, 0.1V SPACING : 0.005GRID

FILL

RETURN

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

CURVE TYPE

(ARCS) CENTER/POINT/POINT

RETURN

(CRVS) 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 sur-face has some elements near the center point which are triangles. This is noproblem : a quad4 element can have two points coincideing.

ELEMENT CLASS

QUAD(4)

RETURN

CONVERT

DIVISIONS : 8, 8(GEOMETRY/MESH) SURFACES TO ELEMENTS

Select ring surface

Select the axle surface

RETURN

After generating the element mesh we have to use SWEEP and CHECK.

18

SWEEP

(SWEEP) ALL

(REMOVE UNUSED) NODES

(REMOVE UNUSED) POINTS

RETURN

CHECK

(CHECK ELEMENTS) UPSIDE DOWN

If necessary FLIP ELEMENTS.

RETURN

MAIN

Boundary conditions (note the symmetry), material properties and geometryparameters must be prescribed.

(MAIN MENU)(PREPROCESSING)

BOUNDARY CONDITIONS

MECHANICAL

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?

(MAIN MENU)(PREPROCESSING)

MATERIAL PROPERTIES

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

strangely named material4 by Mentat).

(MAIN MENU)(PREPROCESSING)

GEOMETRIC PROPERTIES

PLANAR

NEW

PLAIN STRAIN

THICKNESS : 0.1OK

(ELEMENTS) ADD

(ALL) EXIST

MAIN

In (ANALYSIS) JOBS we select element 11, a 4-node linear plane strain element.Then we select the variables, which we want as output in JOB RESULTS andspecify the load in INITIAL LOADS. Finally we indicate that the deformationstate is plane strain.

(MAIN MENU) (ANALYSIS)

JOBS

19

ELEMENT TYPES

MECHANICAL

PLAIN STRAIN SOLID

(PLANE STRAIN FULL INTEGRATION) 11

OK

(ALL) EXIST

OK

RETURN

NEW

MECHANICAL

- INITIAL LOADS : apply1, apply2, apply3

- JOB RESULTS : stress, strain, von-mises

- (ANALYSIS DIMENSION) : PLAIN STRAIN

MAIN

SAVE the model as asblok1 and run MARC. The results, available in the .t16files can then be loaded and the results can be visualised.

20

2.3.2 Results

Looking at Von Mises contour plots it is immedeately clear that the stresses inthe rubber are much lower than those in the axle. We can visualize the stressstate in the rubber in more detail by making the axle invissible in the SELECT

menu.

SELECT

ELEMENTS

Select the elements of the axle with a ”box”

Close with END LIST

MAKE INVISIBLE

Push MAKE INVISIBLE again to make everything visible.

21

2.3.3 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, usingquadratic elements with eight element nodes. We can adapt the model as-blok1. First CLOSE the .t16 file.

FILES

RESTORE or OPEN : asblok1MAIN

(MAIN MENU) (PREPROCESSING)

MESH GENERATION

CHANGE CLASS

QUAD(8)

ELEMENTS

(ALL) EXIST

RETURN

SWEEP

(SWEEP) ALL

(REMOVE UNUSED) NODES

(REMOVE UNUSED) POINTS

MAIN

(MAIN MENU) (PREPROCESSING)

BOUNDARY CONDITIONS

Some boundary conditions must be adapted, as there are no boundary condi-tions defined in the new nodes.

Adapt the boundary conditions.

(MAIN MEU) (ANALYSIS)

JOBS

ELEMENT TYPES

MECHANICAL

PLANE STRAIN SOLID

(PLANE STRAIN FULL INTEGRATION) 27

OK

(ALL) EXIST

RETURN

MAIN

SAVE the model and run MARC. Results can be visualised in Mentat.

22

2.4 Orthotropic plate

The figure below shows a square plate, which will be loaded in its plane. It can be assumed thata plane 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 fibres are embedded, which all have the sameorientation along the direction indicated as 1 in the ”material” 1, 2-coordinate system. Bothmatrix and fibres are linearly elastic with Young’s modulus and Poisson’s ratio Em, Ef , νm andνf , respectively. The volume fraction of the fibres is V . The angle between the 1-direction andthe x-axis is α = 10 degrees. In the 1, 2-coordinate system the material behaviour for plane stressis given by the next relation between stress and strain components :

σ11

σ22

σ12

=1

1 − ν12ν21

E1 ν21E1 0ν12E2 E2 0

0 0 (1 − ν12ν21)G12

ε11

ε22

γ12

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 behaviour.

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 usedto calculate 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

23

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 ismodelled in 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 thefollowing commands.

(MAIN MENU)

MATERIAL PROPERTIES

NEW

ORTHOTROPIC

E11 E1

E22 E2

E33 E3

NU12 ν12

NU23 ν23

NU31 ν31

G12 G12

G23 G23

G31 G31

ADD

(ALL) EXIST.

The orientation of the material coordinate system w.r.t. the global coordinate system must alsobe given. This is done per element in the submenu ORIENTATION. In this submenu we see thatthere are several (TYPE)s of orientations. The EDGE-types say that the 1-axis of the local coordinatesystem is defined by rotation over an ANGLE relative to a particular side of an element. Side 12 canbe seen in the plot of the element mesh as a half-arrow on the element edges. It is the side fromlocal node 1 to local node 2. We can also define the rotation of the local 1-axis relative to one ofthe global coordinate planes.

ORIENTATION

NEW

EDGE12

ANGLE 10ADD

(ALL) EXIST.

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

For all these loadcases, calculate the stress and strain components in the global and the materialcoordinate system.

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

24

2.5 Exercises

2.5.1 Large thin plate with a cirular hole

A large rectangular plate is loaded with a uniform stress σxx = σ as is shown in the figure below.In the center of the plate is a hole with radius a, much smaller than the dimensions of the plate.

x

y

r

θ

a

σσ

The stresses around the hole can be determined, using an Airy stress function approach. Therelevant stresses are expressed as components in a cylindrical coordinate system, with coordinatesr, measured from the center of the hole, and θ in the circumferential direction.

σrr =σ

2

[(

1 − a2

r2

)

+

(

1 + 3a4

r4− 4

a2

r2

)

cos(2θ)

]

σθθ =σ

2

[(

1 − a2

r2

)

−(

1 + 3a4

r4

)

cos(2θ)

]

σrθ = − σ

2

[

1 − 3a4

r4+ 2

a2

r2

]

sin(2θ)

The highest stress values are found in points at the hole edge r = a.

σθθ(r = a, θ = π2 ) = 3σ ; σθθ(r = a, θ = 0) = −σ

The Stress Concentration Factor (SCF) is :

Kt =σmax

σ= 3 [-]

where it is noted that Kt is independent of the hole diameter.

The plate thickness is 0.01 m, and its dimensions in x- and y-direction are 1 m and 0.5 m,respectively. Numerical values of other relevant parameters are listed in the table below.

a = 0.05 m σ = 1 kPa

E = 250 GPa ν = 0.3

Stress components are plotted for θ = 0 and for θ = π2 as a function of the radial distance r.

25

0 0.1 0.2 0.3 0.4 0.5−1000

−500

0

500

1000

r [m]

σ [P

a]

σrr

σtt

σrt

0 0.1 0.2 0.3 0.4 0.5−500

0

500

1000

1500

2000

2500

3000

r [m]

σ [P

a]

σrr

σtt

σrt

• Model the plate in Mentat. Model only a quarter of the plate and apply the proper symmetryconditions. Choose 4-node quadrilateral plane stress elements, ± 20 in x-direction and ± 10 iny-direction. Use a bias factor to refine the element mesh near the hole edge.(To get a ’nice’ element mesh, it might be wise to generate the mesh in two parts.)

Make a PATH PLOT of the stress components σrr and σθθ as a function of the radial coordinate,both for θ = 0 and for θ = π

2 . Compare the results with the analytical solution.

• Increase the number of elements using the same bias factor and quantify the relation betweenthe number of elements and the error in the numerical solution.

• Decrease the hole diameter to a = 0.01 m and repeat the analysis.

• Use 8-node quadrilateral elements and answer the former questions again.

26

2.5.2 Plate with a square hole

The figure below shows a plate and its dimensions in [mm]. The plate is clamped at the left edge(x = 0) and loaded with an edge load of 100 N/mm2 at the right edge. The plate is thin, so wecan assume a plane stress state in the material.Model the upper part of the plate and use the number of elements as indicated in the figure.

8

2040

40 80

8

dikte = 1

2000 N

20

D

FEBA

C20

5

10

8

104 8

8

The Young’s modulus and Poisson’s ratio of the material are E = 210000 MPa and ν = 0.3.

• Make TENSOR PLOTS of the maximum, the minimum and the major principal stresses.

(MAIN > RESULTS > MORE > TENSOR PLOT).

• Make a PATH PLOT of the Major Stress and the Von Mises Stress along the line ABCDEF .

Model the plate again and replace the square hole by a circular one, as indicated in the figure.Use approximately the same element mesh as before.

8

10100 N/mm2

27

• Make TENSOR PLOTS of the maximum, the minimum and the major principal stresses.

• Make a PATH PLOT of the Major Stress and the Von Mises Stress along the line ABCDEF .

The figure below shows the required PATH PLOTs, where x indicated the distance along the path.

0 20 40 60 80 100 120 140−150

−100

−50

0

50

100

150

200

x [mm]

σ [M

Pa]

σmj

σvm

0 20 40 60 80 100 120 140−150

−100

−50

0

50

100

150

200

x [mm]

σ [M

Pa]

σmj

σvm

28

2.5.3 Thick-walled pressurized cylinder

The figure shows a thick-walled cylinder with inner radius a and outer radius b. The inner pressureis pi and the external pressure is pe.

pe

a

b

rpi

The radial and tangential stresses are :

σrr =pia

2 − peb2

b2 − a2− (pi − pe)a

2b2

(b2 − a2)r2; σθθ =

pia2 − peb

2

b2 − a2+

(pi − pe)a2b2

(b2 − a2)r2

For an open ended cylinder, the axial stress is zero : σzz = 0. For a closed cylinder the axial stressis :

σzz =pia

2 − peb2

b2 − a2

The radial and tangential stresses are the same for both cases. The radial displacement for anopen ended cylinder is :

u =1 − ν

E

pia2 − peb

2

b2 − a2r +

1 + ν

E

(pi − pe)a2b2

(b2 − a2)r

The next table lists values for the relevant geometrical and material parameters, where E and ν

are Young’s modulus and Poisson’s ratio of the cylinder material. The height of the cylinder is0.5 m.

a = 0.25 m b = 0.50 m

E = 250 GPa ν = 0.33

pi = 100 MPa pe = 0 MPa

For these values, stresses and radial displacement are plotted as a function of the radius r.

0.25 0.3 0.35 0.4 0.45 0.5−1

−0.5

0

0.5

1

1.5

2x 10

8

r [m]

σ [P

a]

σrr

σtt

σzz

0.25 0.3 0.35 0.4 0.45 0.51.3

1.4

1.5

1.6

1.7

1.8

1.9

2x 10

−4

r [m]

u r [m

]

29

• Model a quarter of the cylinder in Mentat using 4-node quadrilateral plane stress elements andapply the proper boundary conditions for symmetry and edge load. Use 16 elements in tangentialdirection and 8 in radial direction.

Calculate the stresses, plot them in a path plot and compare the values with the analyticalsolution.

• Make all elements twice as small and compare again the stresses and displacement with thenumerical solution.

• Start again with the initial element mesh, however, using now 8-point quadrilateral elements.

30

2.5.4 Rotating engineering parts

The next figure shows a disc with a central hole.

r z

When the hickness t is uniform – not depending on the radius r –, the stress components have tosatisfy the next equilibrium equation is :

∂σrr

∂r+

σrr − σθθ

r+ ρω2r = 0

where ρ is the density and ω the constant angular velocity. Using Hooke’s law for the linearelastic material behaviour and the strain-displacement relations, this can be transformed into adifferential equation for the radial displacement u :

d2u

dr2+

1

r

du

dr− u

r2+

(

1 − ν2

E

)

ρω2r = 0

The general sulution is :

u = a1r +a2

r−

(

1 − ν2

E

)

ρω2r3

8

where a1 and a2 are integration constants.

σrr = c1 −c2

r2−

(

3 + ν

8

)

ρω2r2 ; σθθ = c1 +c2

r2−

(

1 + 3ν

8

)

ρω2r2 ; σzz = 0

where c1 and c2 are again integration constants, depending on a1 and a2.

Solid disc with constant thickness

For a disc with outer radius b, the displacement at r = 0 must remain finite, while the stressboundary condition is σrr(r = b) = 0. Integration constants can now be determined :

c1 =

(

3 + ν

8

)

ρω2b2 ; c2 = 0

The stresses are :

σrr =

(

3 + ν

8

)

ρω2(b2 − r2) ; σθθ =[

(3 + ν)b2 − (1 + 3ν)r2] ρω2

8; σzz = 0

The disc has an outer radius of 0.5 m and a thickness of 0.05 m. It rotates with an angular velocityof 6 cycles per second. Young’s modulus is E = 200 GPa and Poisson’s ratio is ν = 0.3. The massdensity is ρ = 7500 kg/m3.

31

The rotation is modelled as a CENTRIFUGAL LOAD in BOUNDARY CONDITONS.

(MAIN) BOUNDARY CONDITIONS

MECHANICAL

CENTRIFUGAL LOAD

ANGULAR VELOCITY (CYCLES/TIME) 6(AXIS OF ROTATION)

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

(ELEMENTS) ADD

(ALL) EXIST.

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

• Model the disc with plane stress elements and calculate the stresses. Present the radial and thetangential stress components in a path plot.

Constant thickness disc with a central hole

For a disc with outer radius b and a central hole with radius a, the stress boundary conditions areσrr(r = a) = σrr(r = b) = 0. Solving the integration constants leads to :

c1 =3 + ν

8ρω2(a2 + b2) ; c2 =

3 + ν

8ρω2a2b2

32

• Model the disc from the former exercise with a central hole having a radius of 0.25 m and calculateagain the radial and tangential stress components. Use plane stress elements and represent thestress values in a path plot.

Discs with variable thickness

For a disc with variable thickness t(r) the next equilibrium equation in radial direction can bederived :

∂(t(r)rσrr)

∂r− t(r)σθθ + ρω2t(r)r2 = 0

The disc has inner and outer radius a and b, respectively, and a thickness distribution t(r) = ta

2ar.

The general solution for the stresses is :

σrr =2c1

atard1 +

2c2

atard2 − 3 + ν

5 + νρω2r2 ; σθθ =

2c1

atad1r

d1 +2c2

atad2r

d2 − 1 + 3ν

5 + νρω2r2

with

d1 = − 12 +

√

54 + ν ; d2 = − 1

2 −√

54 + ν

The integration constants can be determined from the boundary conditions σrr(r = a) = σrr(r =b) = 0.

33

3 Axisymmetry

3.1 Background : Axisymmetric modelling

Many devices and components have a geometry which is symmetric w.r.t. an axis and are thuscalled axisymmetric. It is obvious that the position of material points of such an object can bestbe described in a cylindrical coordinate system. Coordinates are the distance z measured alongthe axis of symmetry, the distance r from and perpendicular to that axis and an angle φ, whichindicates the position in circumferential direction, with respect to an arbitrary starting point(φ = 0o : 0 ≤ φ ≤ 360o(= 2π[rad]) (see figure below).

r

z

r

φ

P

r

z

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

Modelling such axisymmetric problems in 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 modelledand it 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 charaterised 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.

34

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

σrr

σtt

σzz

σrz

=E

(1 + ν)(1 − 2ν)

1 − ν ν 0 ν

ν 1 − ν 0 ν

0 0 1 − ν 0ν ν 0 1 − 2ν

εrr

εtt

εzz

εrz

x = z-as

y = r-asσrr

P

σtt σzz

σrr

σrz

σrz

σzr

σzr

σtt σzz

Analysing axisymmetric problems with the finite element method implies that axisymmetric ele-ments must be used, which are defined in the cross-section. The deformation of such an element isdefined by the displacement in r- and z-direction of the element nodal points. We must be awarethat a nodal point 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 interpolatedbetwen the displacements of the element nodes. As with the plane elements, linear or quadraticinterpolation can be used. Again we have 4-node and 8-node elements with 4 and 9 integrationpoints respectively. In MARC these elements are indicated as element type 10 and 28. Thecross-section of a 4-node element is shown in the figure below.

35

1 2

34

4

ξ

η

1 2

3

13

√3

21

43

1 2

43

ξ

η

13

√3

In MARC/Mentat the strains in the integration points are defined as follows :

strain 1 = global zz-strain = εzz

strain 2 = global rr-strain = εrr

strain 3 = global tt-strain = εtt

strain 4 = global rz-strain = γrz

Stress components are defined in the same way.

36

3.2 Axle bearing 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 modelling and analysis, as indicated in thefigure below.

as

manchet

x

y

FA

NB. : The x-axis is the axial axis and the positive y-axis is the radial axis. This is theusual definition in Mentat/MARC.

37

3.2.1 Modelling, analysis and results

(MAIN MENU)(PREPROCESSING)

MESH GENERATION

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

CURVE TYPE

LINE

RETURN

(CRVS) 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.

(MAIN MENU) (PREPROCESSING)

BOUNDARY CONDITIONS

MECHANICAL

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 presribed in the usual way. Geo-metric parameters are not relevant for this axisymmetric analysis.

In (ANALYSIS) JOBS we select element 10, a 4-node quadrilateral axisymmetricelement. We indicate that the analysis is axisymmetric.

(MAIN MENU) (ANALYSIS)

JOBS

ELEMENT TYPES

MECHANICAL

AXISYMMETRIC SOLID

(FULL INTEGRATION) 10

OK

(ALL) EXIST

RETURN

RETURN

NEW

MECHANICAL

- INITIAL LOADS : apply1, apply2, apply3

- JOB RESULTS : stress, strain, von-mises

- (ANALYSIS DIMENSION) : AXISYMMETRIC

MAIN

38

SAVE the model as asblok2 and run MARC. The results, available in the .t16file can be loaded into Mentat and visualised.

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

39

3.3 Exercises

3.3.1 Circular plate with a circular hole

The next figure shows a plate with a central hole.

ba

The radial displacement and the radial and tangential stresses are given as a function of the radiusr :

ur = c1r +c2

r

σrr =E

1 − ν2

[

(1 + ν)c1 − (1 − ν)c2

r2

]

σtt =E

1 − ν2

[

(1 + ν)c1 + (1 − ν)c2

r2

]

where E is Young’s modulus and ν is Poisson’s ratio. The integration constants c1 and c2 aredetermined 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 = 0.3ub = 0.001 [m]

• Make an axisymetric model and calculate the deformation and stress state in the plate.

Compare the results to the exact solution.

The above model was axisymmetric and axisymmetric elements were used in the analysis (seefigure below).

40

y = radial

z = tangential

x = axial

It is also possible to use a plane stress or plane strain model to calculate the deformation and thestresses. In this case we model and analyse a quarter of the plate as shown in the figure below.

x

y

z

yx

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

(MAIN MENU) (PREPROCESSING)

BOUNDARY CONDITIONS

MECHANICAL

TRANSFORMS

NEW

CYLINDRICAL

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

[0 0 1].

(NODES) ADD select nodes at outer edge

MAIN

• Make a plane stress model and calculate the deformation and stress state in the plate.

Compare the results to the axisymmetric solution.

41

In the following two figures the stresses from a plane stress analysis are plotted for the planar andthe axisymmetric model.

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

0 0.02 0.04 0.06 0.08 0.1−1

0

1

2

3

4

5

6x 10

8

r [m]σ

[Pa]

σrr

σtt

σzz

3.3.2 Rotating engineering parts

• Model the rotating discs from the plane stress exercises with axisymmetric elements and calculatethe stresses.

Present the radial and the tangential stress components in a path plot.

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4 Three-dimensional models

4.1 Background : Three-dimensional elements

Each component (or structure) in real life has obviously dimensions in three coordinate directionsand is thus a three-dimensional solid object. In previous sections special cases were consideredwhere dimensions and load were such that a two-dimensional model could be used : plane stress,plane strain or axisymmetry.When such a simplification is not allowed, a complete three-dimensional model must be madeand analyzed. In that case the object is subdivided in three-dimensional elements : hexahedrals(bricks) and/or tetrahedrons (pyramids). Here, we will consider only brick elements, either with8 nodes in the corners or with 20 nodes in corners and midpoints of the edges (see figure below).

1

2

3

4

5

6

7

8

1

2

3

4

5

6

7

8

9 10

11

12

13 14

1516

17

18

19

20

In MARC these element have type numbers 7 and 21, respectively. A side of a brick element iscalled a face and face loads (forces per unit of area) can be applied on it.Displacement components of an internal point are interpolated between their values in the nodesand this interpolation is linear for the 8-node brick and quadratic for the 20-node element. Forintegration over the element volume, 8 and 27 integration points are used for the 8- and 20-nodebrick, respectively.

Subdividing a three-dimensional solid object in hexagonal elements can be done automatically inMentat (or sometimes preferably in Unigraphics). For simple objects and meshes another methodis easily used, which is based on expansion of a two-dimensional mesh in the third dimension.The Mentat command to do this is EXPAND in the MESH GENERATION menu. We can expand bytranslation and/or by rotation as will be shown in the examples.

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4.2 Three-dimensional model

We will model a rectangular plate, load it with a tensile load and calculate its deformation andstress state. First we will analyse it under the assumption of plane stress and plane strain. Thenwe will expand its geometry to three dimensions and analyse it with boundary conditions, similarto plane stress and plane strain.

(MAIN MENU)(PREPROCESSING)

MESH GENERATION

Define a coordinate grid with −0.1 ≤ u ≤ 0.1, −0.1 ≤ v ≤ 0.1 and u- and v-spacing

0.01. All distances being in meters.

ELEMENT CLASS

QUAD 4

OK

(ELEMS) ADD

Define one element of class QUAD 4 with corner nodes in (0 0 0), (0.1 0 0), (0.1 0.05

0), (0 0.05 0). Subdivide this element in 10 × 5 elements. SWEEP and CHECK the

mesh.

Boundary conditions are prescribed. The points on the edge y = 0 are not allowed to move iny-direction. The points on the edge x = 0 are not allowed to move in x-direction. The point(0, 0, 0) is completely fixed.The loading is done by a tensile edge load of -1e6 Pa at edge x = 0.1.

Define these boundary conditions.

Material properties for isotropic linear elastic material behaviour are specified : Young’s modulusE = 210 GPa and Poisson’s ratio ν = 0.3.

Because we first want to do a plane anlysis, we have to specify in GEOMETRIC PROPERTIES thethickness : 0.01 m. First we choose PLANAR PLANE STRESS THICKNESS. For the plane strain analysiswe have to change this to PLANAR PLANE STRAIN THICKNESS.

In the JOBS menu we select the ELEMENT TYPE. First we choose number 3 for plane stress andafterwards we select number 11 for plane strain.

In JOB RESULTS we choose ”stress” and ”total strain” as output parameters.

FILES SAVE AS ”plate2D” and run the model in MARC. Look up the relevant stresses and strainsin Mentat and repeat this for a plane strain analysis.

Results :

σ11 ε11 ε22

plane stress 1.000e6 4.762e-6 -1.429e-6plane strain 1.000e6 4.333e-6 -1.857e-6

Now the model ”plate2D” is opened again and adapted to three dimensions. We will save it lateras ”plate3D”.

We go to (MAIN MENU)(PREPROCESSING) and MESH GENERATION and then in the submenu EXPAND,where we expand the element mesh in the third (z-) direction. The number of element layers is

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chosen to be 5, with each layer being 0.002 m thick. The total thickness is thus 0.01 m.

(MAIN MENU)(PREPROCESSING)

MESH GENERATION

EXPAND

TRANSLATIONS : 0 0 0.002REPETITIONS : 5ELEMENTS

(ALL) EXISTING

RETURN

Now SWEEP the mesh and CHECK for INSIDE OUT elements.

When we rotate the model we will see that we have made a three-dimensional element mesh. Inthe BOUNDARY CONDITIONS menu we can visualize all boundary conditions and it becomes clearthat prescribed nodal point displacements have been copied with the nodes. It remains to be seenif this is what we want. The edge load, however, has disappeared and must be defined again butnow as a face load on the element faces on the right side of the model.

MATERIAL PROPERTIES are also copied. GEOMETRIC PROPERTIES must be removed as they are not rel-evant for our three-dimensional model.

The only thing which is left to do is selecting the correct element type in JOBS ELEMENT TYPES

3-D SOLID. We choose element 7, which is a 8-node brick element.

Save the model as ”plate3D” and run it in MARC.

When we look at the results, we see that the deformation and stress state differs from those inthe two-dimensional analyses. Obviously this is due to the boundary conditions.

Go back to the ”plate3D” model and make boundary conditions, which are analoguous to planestress and plane strain respectively. Compare the stresses and strains with the values from thetwo-dimensional analyses.

The same expansion which is described above, can also be applied to make a three-dimensionalmodel out of an axisymmetric one. In that case instead of TRANSLATIONS we have to use ROTATIONS

around the x-axis, which is the axis of axisymmetry in Mentat/MARC.

Take an axisymmetric model of a disc from he exercises and make a three-dimensional

disc out of it.

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4.3 Exercise

4.3.1 Inhomogeneous plate with a central circular hole

The figure below shows a square plate (dimensions 0.2 × 0.2 m) with a central circular hole (radius= 0.03 m). The plate is made from two different materials A and B. Material A is located betweenthe hole and a circle with radius 0.07 m. In the undeformed state the plate thickness is 0.04 meverywhere.

P

Q

A

B

0.2

0.07

x

y

σσ

0.03

The plate is axially loaded with an edge load σ = 50 MPa. A plane stress state is assumed toresult in the plate after loading.

Model a quarter of the plate [ 0 ≤ x ≤ 0.1 ; 0 ≤ y ≤ 0.1 ] and apply the correct

boundary conditions.

Make an element mesh with 10 quad4 elements in tangential direction and 20 elements

in radial direction.

The following table lists relevant material parameters.

Young’s modulus material A EA 400 GPaPoisson’s ratio material A νA 0.25 -Young’s modulus material B EA 200 GPaPoisson’s ratio material B νA 0.3 -

• What are the x- and y-displacements in points P and Q ?

• What is the maximum value of the stress in the material and in which point M(x, y) is this valuefound ?

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Besides the applied boundary load a new edge load in y-direction is applied on the edges [−0.1 <

x < 0.1 ; y = ±0.1].

• For which value of the new edge load is the stress in point M zero.

Convert the two-dimensional model into a three-dimensional model, using the EXPAND command.Take 10 elements over the thickness of the plate.

Calculate strains and stresses in the plate material.

• Make a PATH PLOT of the strain εzz over the thickness in the point with coordinates x = 0.03and y = 0.

• Make a PATH PLOT of the stress σzz over the thickness in the point with coordinates x = 0.03and y = 0.

• What can you conclude concerning the assumption of plane stress ?

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