Wprowadzenie do MES Introduction to Finite Element Concepts

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Introduction to Finite Element Concepts

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Overview of the basic mechanical /stress terms

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Review : Stress valuesReview : Stress valuesStresses

Normal Stress : normal to cutting plane

Shear Stress : tangential to cutting plane

Normal and Shear Stress values at Point Pdepend on :

- Location of Point P on the cutting plane- Cutting direction

=> Stress behavior at Point P can be described by the Stress tensor S

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Review : Stress valuesReview : Stress values

x

y

z

zy

xy

xz

zx

yx

yz

xx

zz

yy

Normal Stress values:xx = x yy = y zz = z

Shear Stress values: xy = yx

xz = zx

yz = zy

The normal of a face and the normalstress vector have the same direction.

Shear Stresses in two perpendicular cutting directions at a point have the same value.

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Stress Tensor S :

6 independent stress values from 3 perpendicularcutting planes at a point describe the stress behavior completely.

X xy xz

S = xy y yz

xz yz z

The stress tensor is symmetrical .

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Example : Plane Stress

= stress behavior in a plane membrane

A plane membrane is under loading onlyin its plane.

The thickness of the membrane is smallcompared to its two other dimensions.

No forces in z-direction : xz = yz = z = 0

x xy

S = xy y

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Coordinate Transformation (Plane Stress):

= ½ (x+ y) + ½ (x- y)cos2 + xysin2

= ½ (x+ y) - ½ (x- y)cos2 - xysin2

= - ½ (x- y)sin2 + xycos2

- Angle between x- and -Axis

x

y

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Review : Stress valuesReview : Stress values Invariant of Transformation (Plane Stress) :

x+ y = + = 1+ 2 Sum of normal stress values

x2

+ 2xy2 + y

2 = 2 + 22 + 2 = 1

2+ 22

Special Case : Hydrostatic Stress behavior

x= y = = Same normal stress values in allcutting directions

xy = = 0 Shear stress values are equal Zero

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Principal Stresses (Plane Stress) :

= Maximum values of Normal Stresses under the condition :

d/d = 0 and d/d = 0

this leads to : - (x- y)sin2 + 2xycos2 = 0

2xy => tan 2* = tan 2(* + /2) = x- y

* and (* + /2) are two perpendicular cutting directions, called Principal Directions .

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Review : Stress valuesReview : Stress valuesTrigonometric transformation :

1 x- y cos2* = = [ 1 + tan22* ]½

[ (x- y )2 + 4xy2 ]½

tan2* 2xy sin2* = = [ 1 + tan22* ]½

[ (x- y )2 + 4xy2 ]½

Principal Stresses 1 and 2 : 1 > 2 1,2 = ½ (x+ y) ½ [ (x- y )2 + 4xy

2 ]½

(*) = (* + /2) = 0 !! No Shear Stresses

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Principal Shear Stress (Plane Stress) :

= Maximum values of Shear Stresses under the condition :

d/d = 0

this leads to : - (x- y)cos2 - 2xysin2 = 0

x- y => tan2** = tan2(** + /2) = - 2xy

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Review : Stress valuesReview : Stress values Because : tan2** = - 1 / tan2*

2** and 2* are oriented perpendicular to each other,this results in an angle of 45o between the direction of maximum shear stress ** and the direction of maximum normal stress *.

Principal shear stress max :

max = ½ [ (x- y )2 + 4xy2 ]½

or max = ½ (1- 2)

Using ** in and results in a Non-zero value ofnormal stress value M :

M(**) = ½ (x + y) = ½ (1 + 2)

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Equivalent stresses V :

There is a 3–dimensional stress behaviour in an arbitrary part.

Experimental Data have been received from a tension specimen (1-dimensional stress behaviour).

The equivalent stress is used to compare 3-dimensional stress behavior with the 1-dimensional stress behaviour of the tension test.

Different strength hypotheses have been developed.

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Normal stress hypothesis :

Assumption : The maximum Normal stress value isresponsible for the material load

V = 1

Used for : brittle materials

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Shear stress hypothesis : (1864 described by H.Tresca)

Assumption : The the material load is characterizedby the maximum stress value.

Plane Stress: Max = ½ (1 - 2 ) => V = 1 - 2

V = [( x - y)2 + 4 xy

2 ]½

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von Mises stress hypothesis :(named after Huber (1872-1950), v. Mises (1883-1953)and Hencky (1885-1951))

Assumption : The material load is characterized bythe energy which is used for the changeof the shape without a change in thevolume of the part

Plane Stress: V = [ 1

2 + 22

- 12 ]½ = [ x

2 + y2

- xy + 3 xy2

]½

Used for : ductile materials

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Review : Strain valuesReview : Strain valuesDeformation in space :

Kinematical relationStrains :

x = u/x , y = v/y , z = w/z

Angular distortions :

xy = u/y + v/x , xz = u/z + w/x , yz = v/z + w/y

Strain tensor V : symmetric

x ½xy ½xz

V = ½ xy y ½ yz

½ xz ½yz z

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Review : Strain valuesReview : Strain values Plane Strain :

dx

dy

u u+(u/x)dx

v

v+(v/y)dy

v+(v/x)dx

(v/x)dx

u+(u/y)dx(u/y)dy

x

y

P

P’Q

Q’

R

R’

S

S’

/2-xy

Assumption : Small Deformation

Strain :

x = u/x , y = v/y

Gliding or Shearing(Angular distortion)

= u/y , = v/x

xy = + = u/y + v/x

Strain tensor

x ½xy

V = ½ xy y

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Review : Strain valuesReview : Strain values

Coordinate transformation (Plane Strain) :

= ½(x+ y) + ½(x- y)cos2 + ½xysin2

= ½(x+ y) - ½(x- y)cos2 - ½xysin2

½ = - ½(x- y)sin2 + ½xycos2

- Angle between x- and - axis

x

y

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Principal strains (Plane Strain) : = Maximum strain values under the condition :

d/d = 0 and d/d = 0

this leads to : - (x- y)sin2* + xycos2* = 0

xy => tan 2* = tan 2(* + /2) = x - y

* and ( * + /2 ) are two perpendicular oriented cutting directions.

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Principal strains (Plane Strain) 1 and 2

1,2 = ½ (x+ y) ½ [ (x- y )2 + xy2 ]½

(*) = (* + /2) = 0 !! No angular distortion

Maximum angular distortion : d/d = 0

- (x- y)cos2** - xysin2** = 0 x- y => tan 2** = tan 2(** + /2) = - xy

=> M(**) = ½ (x + y) = ½ (1 + 2) ( Strains 0 )

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Introduction to FEMIntroduction to FEM

FEM = Finite Element Method

FEM is a numerical approximation method,which is used for the calculation and optimizationof the structural behavior of mechanical parts !

Finite Element = Discrete structural description of continua withhelp of mathematical formulations

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Historical OverviewHistorical Overview

1950´s: Application of matrix methods for structural analysis

1950´s: Introduction of the displacement- and stiffness method for complex aerospace structures

1960: Birth date of the name “Finite Elements“ 1970´s: FEM applications mainly in the aerospace and the

automotive industry (NASA -> NASTRAN) 1980´s: Introduction of powerful computer graphics 1997: FEM is the standard tool for structural analysis

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Seminar ContentsSeminar Contents

Basic Theory of FEM Element Types and their Use Geometric model Finite Element model Material Data Element Properties Loads Boundary Conditions Postprocessing

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Finite Elements have simple geometric shapes, like triangle, rectangle, cube.

The points at the corners are called Node or Grid point.Nodes connect the elements.

The description of the structural behaviour inside an element is done by the calculation of the nodal displacements (= discretization) in combination withspecial shape function (= mathematical functions usedfor integration across the element region)


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Analytical preparation

Principle of virtual work=Minimal change of energy due to external load

A static equilibrium between internal and external forces is reached when the Elastic Potential becomes minimal; this means the first derivation of is Zero.

/ui = 0 - System of equations with n equations for n unknown displacements

ui - Displacement of the i-th degree of freedom


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Elastic Potential (Potential Energy) = ½ T dV - uTfK dV - uTfV dV - uTfA dA

Strain vector : x , y , z , xy , yz , zx

Stress vector : x , y , z , xy , yz , zx u Vector of displacements : ux , uy , uz , rx , ry, rz

fK Vector of external forces and moments on nodesfV Vector of inertial forces (Acceleration, Rotation)fA Vector of forces on faces (Pressure)

(Thermal load is not considered here.)


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Continuous – Discrete displacement relations

The continuous displacement field in an element u(x,y,z)will be described by Shape functions N(x,y,z) and the discrete nodal displacements of the element ui.

u(x,y,z) = N(x,y,z) * ui


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Strain – Displacement relations

Strains can be derived from the displacementaccording to the theory of linear elasticity by theuse of the Cauchy-Matrix D (Differential operator).

(x,y,z) = D * u(x,y,z)

Using the Continuous – Discrete displacement relation

(x,y,z) = DN(x,y,z) * ui


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Stress – Strain relation

For a linear material the Hooke’s Matrix E statesthe relation between stress and strain.

(x,y,z) = E * (x,y,z)

Using the Strain – Displacement relation

(x,y,z) = ED * u(x,y,z)

Using the Continuous – Discrete displacement relation (x,y,z) = EDN(x,y,z) * ui


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Getting system of linear equations

All preceding equations will be put into the equation of theelastic potential . Creating the derivates /ui = 0 weget the basic equation of the FE-method in the linear static.

K ui = F

Global stiffness matrix : Sum of elemental stiffness matrixes

K = (DN)TE(DN) dV (Maxwell´s Law : Matrix is symmetric) External nodal forces (inertial loads and loads on faces are been represented by equivalent nodal forces

F = NTfK dV + NTfV dV + NTfA dA


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Example : Triangular element (Plane stress)

Y

X

ux2

ux3

ux1

uy2

uy3

uy1


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Assumption : Linear shape function

ux(x,y) = a0 + a1x + a2y

uy(x,y) = b0 + b1x + b2y

re-written in matrix form : a0 b0

a1 b1

ux(x,y) = [ 1 x y ] a2 uy(x,y) = [ 1 x y ] b2


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Using linear shape function for all 3 nodes :

ux1 = a0 + a1x1 + a2y1 uy1 = b0 + b1x1 + b2y1

ux2 = a0 + a1x2 + a2y2 uy2 = b0 + b1x2 + b2y2 ux3 = a0 + a1x3 + a2y3 uy3 = b0 + b1x3 + b2y3 re-written in matrix form : ux1 1 x1 y1 a0 uy1 1 x1 y1 b0 ux2 = 1 x2 y2 a1 uy1 = 1 x2 y2 b1 ux3 1 x3 y3 a2 uy1 1 x3 y3 b2

uxi = G * a uyi = G * b


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a and b as function of uxi and uyi :

a = G-1 * uxi b = G-1 * uyi

Getting the Continuous – Discrete displacement relations : ux1 ux2 ux(x,y) = [ 1 x y ] G-1 uxi = N(x,y) uxi = [ N1 N2 N3 ] ux3

uy1 uy2 uy(x,y) = [ 1 x y ] G-1 uyi = N(x,y) uyi = [ N1 N2 N3 ] uy3


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Strains (Plane stress) :

x(x,y) = ux(x,y)/xy(x,y) = uy(x,y)/yxy(x,y) = ux(x,y)/y + uy(x,y)/x

re-written in matrix form :

x(x,y) /x 0 ux(x,y) y(x,y) = 0 /y uy(x,y) xy(x,y) /y /x

= D * u


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Using discrete nodal displacements

ux1 ux2 ux3 x (x,y) N1,x N2,x N3,x 0 0 0 uy1 y (x,y) = 0 0 0 N1,y N2,y N3,y uy2 xy(x,y) N1,y N2,y N3,y N1,x N2,x N3,x uy3

Linear Shape function Ni = Ni(x,y)

Derivatives : Ni,x = const. , Ni,y = const.

=> constant strain value over the element region


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Stresses , isotropic Material

x (x,y) E 1 0 x (x,y)

y (x,y) = 1 0 y (x,y) xy(x,y) 1-2 0 0 (1- )/2 xy(x,y)

= H *

Shear Modulus : G = E/ 2(1+ )

Poisson Ratio : 0 < < 0.5

Because all strains (x,y) are constant values, all stress value are constant over the element region as well !!


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How does FEM work?How does FEM work?

Solve a number of simple problems, add them all up and get the answer of a complex problem

Divide a complex problem into simple ones Divide complex geometry into simple objects which we

can understand (Lines, Squares, Cubes)Use the computer to do millions (and millions, …) of

numerical operationsUse modern hardware equipment to present the results

graphically

Calculate the area of a circle

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Mathematical ModelMathematical ModelFormulation of simple

elements

K = Spring stiffness (Ea/L).

U = Spring elongation

F = Spring force

K * U = F

F

u

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Mathematical ModelMathematical ModelDividing of the geometry into simple

elements and assembling all elements

[K] = Stiffness matrix of the part

(Sum of all elements)

{U} = Components of the displacements of the single nodes of the part

{F} = Components of the loads of the single nodes of the part

[K] *{U} = {F}

Solving the matrix equation withthousands of unknowns

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Theoretical BackgroundTheoretical Background A given problem is discretized by dividing the

original domain into simply shaped subdomains, the so called elements.

Each element it quite simple, and the program can figure out its mechanical properties quite easily

By summation of all the element contributions one gets the whole model behavior

ux

uy

X

Y

[ k ]e { u }e = { f }e

element level

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TerminologyTerminology

Each element is connected to its neighbour only at a number of points, called nodes

Each node has 6 independent possibilities to move: 3 translational and 3 rotational

These independent possibilities to move are called degrees of freedom (DOF’s)

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TerminologyTerminology

TranslationsTx, Ty, and Tz (1, 2, and 3)

Rotations Rx, Ry, and Rz (4, 5, and 6)

Forces

Fx, Fy, and Fz

Moments

Mx, My, and Mz

Tx ,Fx

Ry ,My

Ty ,Fy

Rx ,Mx

Tz ,Fz

Rz ,Mz

Y

X

Z Cartesian Coordinate System

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Steps in an FE AnalysisSteps in an FE Analysis

Geometry

Elements

Loads

Constraints

Materials

AnalysisModel

Solver Analysis Results

DisplacementsStressesForcesStrains

Contour PlotsX-Y-PlotsListings

Properties

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FEM Program algorithmFEM Program algorithm Represent continuous model as a collection of

elements and connections Formulate element stiffness matrices [k] Assemble all element stiffness matrices to a global

stiffness matrix [K] Generate load vector [F] Solve matrix [F]=[K][x] with respect to [x] Calculate element stresses and strains

We shall illustrate this using a simple example...

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Continuous ModelContinuous Model

L = 50 mm

L = 50 mm

A = 20mm2

E = 210000 N/mm2

A = 50mm2

E = 210000 N/mm2

F = 1000 N

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Discrete modelDiscrete model

We make one-dimensional elements with two

degrees of freedom

Element 2

Element 1

Node 2

Node 1

Node 3A = 20mm2

E = 210000 N/mm2

L = 50 mm

A = 50mm2

E = 210000 N/mm2

L = 50 mm

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Element stiffness matricesElement stiffness matrices

L

AE

L

AEL

AE

L

AE

k

21 /1000

8484

8484mmNk

2

2 /1000210210

210210mmNk

For one element, the stiffness relation is

Each of the element matrices then becomes:

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Global stiffness matrixGlobal stiffness matrix

mmNmmNK /1000

2102100

2102100

000

/1000

000

08484

08484

mmNK /1000

2102100

21029484

08484

The Global stiffness matrix is assembled by combining the matrices at the appropriate degrees of freedom for each node

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Apply boundary conditionsApply boundary conditions

Node 1’s DOF is constrained, and is therefore a nulled, giving the effective stiffness matrix

3

2

1

x

x

x

x

3

2

0

x

xx

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Load VectorLoad Vector

NF

1000

0

0

The only force is acting at Node 3, giving the Load Vector

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Solve the matrix equationSolve the matrix equation

xKF

mm

x

x

x

017.0

012.0

0

3

2

1

The simple equation is now solved, giving the deformations

N

x

x 1000

0

2102100

21029484

08484

1000

0

0

3

2

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Calculate element forces and stressesCalculate element forces and stresses

222

211

2

1

/0.21

/4.50

00010.050

)012.0(017.0

00024.050

0012.0

mmNE

mmNE

L

LL

L

Using well known formula, the strains and stresses are calculated for each element

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What an FEA user has to specify :

Choose the element typeCreation of the Finite Element MeshSpecification of the material propertiesSpecification of the element propertiesVerification of the Mesh qualityApplication of the loads and boundary conditionsSpecification of the options of the desired analysis type Request of the desired results Interpretation of the results


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The Software Returns:The Software Returns:

Static or time-dependent: Deformations Stresses Strains

Others: Eigenfrequencies Stability/Buckling Load Factors

Other Types of Analysis will Return other Results

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Element TypesElement Types

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Standard ElementsStandard Elements

First order

Second order

Line Triangle/Quadrilateral Hexahedral Tetrahedral ( tri/quad) ( hex ) ( tet )

Line Triangle/Quadrilateral Hexahedral 10-noded Tetrahedral

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Element selection criteriaElement selection criteria

Which element type should be used?

Beams, plates or solids?Quadrilateral or triangle?Hexahedral, pentahedral or tetrahedral ?Higher or lower order?

The choice of the element type is strongly dependent

on the structure to be analyzed

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Beam ElementsBeam Elements 1 - Dimensional elements 6 DOFS First and second order Cross section constant or varying along the element

length Transfers axial forces, torsional and bending moments Neutral line can be moved away from the node (“Offset”) Degrees of freedom may be released at the end of the

elements Assumes non-deforming cross section when deforming

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Typical Beam ModelsTypical Beam Models

Trusses

Stiffeners

Frames

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Examples of Beam ModelsExamples of Beam Models

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Shell ElementsShell Elements

2 - Dimensional Elements5 DOFS (No “drilling” DOF)Thickness constant or varying across elementTriangle and Quadrilateral basis shapeFirst and second orderNeutral line can be moved away from the nodes (“Offset”)Assumes constant thickness when deforming

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Typical Shell ModelsTypical Shell Models

Automotive parts

Thin shells, thinwalled pressure vessels

Aircraft components

Thin plane or curvedmetal sheets

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Examples of shell modelsExamples of shell models

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Examples of shell models, 2Examples of shell models, 2

Courtesy of AdtranzCourtesy of Adtranz

Courtesy of Adtranz

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Solid ElementsSolid Elements

3 - Dimensional elements3 DOFS (no rotational DOFs) Tet and cube basic shapesFirst and second order

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Typical Solid ModelsTypical Solid Models

Thick walledpressure vessels

Thick platesand consoles

Cast iron parts and fittings

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Examples of Solid ModelsExamples of Solid Models

Courtesy of Volvo Car

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Special ElementsSpecial Elements

Spring Damper

Concentrated Mass

Gap(Point-Point-Contact)

Rigid / Interpolation

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Element Selection GuidelinesElement Selection Guidelines

Choose the simplest element type if you can make a shell model - do it If you can make a beam model - do it

For structural analysis, don’t use first order tetra-elements.

(But they are ok for thermal analysis)

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Example:Example:Deciding between 1D, 2D and 3D Deciding between 1D, 2D and 3D element modelselement models

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The ModelThe Model We have the following model:

It consist of rectangular beams thatare 100 mm wide and 200 mm highwith thickness 5 mm. The longestbeams are 2000 mm and the total width of the model is 1000 mm. The centerline of the crossing bars are located 550 mm from each end.

The beams are welded together, but the weld fillets are to be neglected due to their small size

The structure carries a load of 2000 kg evenly distributed over the two cross bars

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The problemThe problem

We want to investigate the maximum overall deflection and maximum stress

As a guideline for meshing, we do not want element aspect ratios greater than 5

Now lets do 1D, 2D and 3D calculations and see what

effort they take and what answers they yield!

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Beam ModelBeam Model

Element size 100 mm

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Shell ModelShell Model

Element size 33 mm

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Solid ModelSolid Model

Element size 25 mm

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Comparing the three analysesComparing the three analyses1D 2D 3D

Number of nodes 60 9072 66015

Number of elements 60 3024 33003

DOF 336 44640 196317

Disk usage for solving 1 MB 195 MB 124 MB

Memory usage for solving 16 MB 21 MB 1530 MB

Modelling time 20 min 30 min 60 min

Solving time 2 sec 70 sec 2000 sec

Postprosessing time 15 min 2 min 1 min

Total time 35 min 33 min 94 min

Maximum deflection 0.0984 mm 0.143 mm 0.135 mm

The calculations were conducted using MSC.PATRAN V8.5, MSC.NASTRAN V70.5. Hardware were HP Kayak 450 MHz, 256 MB RAM, MS/NT 4.0.

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Comparing the three analysisComparing the three analysisBeam Model

Minmal modelling effort, accurate calculation results

Does not calculate beam cross section deformation (which happens in this case)

Postprocessing somewhat awkward and time consuming, could present errors

Plate ModelHigher modelling effort, good calculation results

Must have control of what is “top“ and “bottom“ of elements

Solid ModelHighest modelling effort, good calculation results

Does not give any significant additional results compared to plate model

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Geometric ModelGeometric Model

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GeometryGeometry

The geometry defining the model in question can be:

Imported into the FE program Created using the FE program

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GeometryGeometry

It is of greatest importance that small features not important for the calculation of results the user is looking for should be neglected

A CAD MODEL is not the same as a CAE MODEL!!!

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Coordinate SystemsCoordinate Systems

Different systems available Default is the global Cartesian system

Y

Z

X y

x

z

Y

Z

X

z

r

Y

Z

X

r

Cartesian Cylindrical Spherical

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Discretization of the geometric Discretization of the geometric model (Meshing)model (Meshing)

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Element TypeElement Type

Before meshing the user must chose between differentelement types: Beam, Shell or Solid First or second order Triangle or quadrilateral basis shape

When this is done, meshing of the structure can begin

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Element SizeElement Size

Before the mesh routine is activated, the user specifiesthe element size. This can be done in a number of ways:

Global element size Preferred number of elements along an edge Varying element length along an edge Allowed deviation from geometry

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Automatic meshing routinesAutomatic meshing routinesUnmeshed… and meshed geometry

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Using meshing refinementUsing meshing refinement

Examples of: Curvature based

mesh refinement Desired number

of elements along edges - uniform and varying

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Generation of Volume Structures Generation of Volume Structures from Meshed 2D-Geometryfrom Meshed 2D-Geometry

Revolving shell mesh...

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Generation of Volume Structures Generation of Volume Structures from Meshed 2D-Geometryfrom Meshed 2D-Geometry

Extruding shell mesh...

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Guidelines for Element SizeGuidelines for Element Size In areas with high stress variations a finer mesh is

needed To represent a circle properly the maximum angle

between nodes should not be more than 15 degrees

When doing dynamic analysis, generate enough nodes to represent deformation results (5 nodes per half period)

<15o

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Conclusions for Element SizeConclusions for Element Size

There are two basic facts: A higher number of smaller elements gives a higher

accuracy (within reasonable numbers) A higher number elements needs more CPU time to

compute

Neither of the two can be compromised, certainly not theanswer - and projects tends to have deadlines also…The user needs to be competent using all the meshingcapabilities of the FE program to get a mesh that is goodenough but not too good!

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Element qualityElement quality

Each element has a predefined behaviour valid forundistorted element shapes.

During meshing, elements are usually stretched or twistedto some degree.

Is our assumption really valid ??We must see to this ourselves, during each calculation!

Let´s take a look at a few examples...

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Element DistortionElement DistortionUsing too coarse a mesh will result in the mesher makingdistorted elements in order to try to stay within parentgeometry

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Element DistortionElement Distortion

Transitioning from large to small elements usually leads todistorted elements.Examples below shows an ambiguous and an acceptabletransition.

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Types of Element DistortionsTypes of Element Distortions

Warping

ah

d b

Taper

Skew and Internal Angles

a

b

Aspect Ratio

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Guidelines for Element DistortionsGuidelines for Element Distortions

Skew and Internal AnglesAvoid deviations from the „ideal“ angle > 30°

Aspect RatioAvoid ratio of edge length > 5:1

a

b

Tet CollapseAvoid ratio of longest edge to the shortest height > 10Otherwise the results may be very bad or the solver will stop

longest edge

shortest height

60°

60° 60°

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Example:Example:

Influence of Mesh on calculationInfluence of Mesh on calculationof resultsof results

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The ModelThe Model

Investigate the spring stiffness of this spring

Use several element sizes, both first and second order

Observe that when element size increases, they become both distorted and do not follow the geometry

2 mm

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First order elementsFirst order elements3 mm 0.5 mm

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Second order elementsSecond order elements

5 mm 0.75 mm

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Convergence ResultsConvergence Results

0

10

20

30

40

50

60

5 4,5 4 3,5 3 2,5 2 1,5 1 0,75 0,5

Element size [mm]

Spr

ing

stiff

ness

[N/m

m]

1. Order

2. Order

We need many first order elements to converge in structural analysis

Second order elements seem powerful

First order elementsare also less capableof following geometric curves

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ConclusionsConclusionsWe have learnt that:The calculation of results is dependent on both shape

and number of elements in modelToo coarse a mesh, too distorted a mesh, or even a

fine but distorted mesh - will yield wrong answer!

The user must, for each calculation, ensure:Adequate element shapesAdequate element numbers

There are no universal rules of thumb for the above, it is based on user experience

Introduction to Finite Element Method

Defining Model LoadsDefining Model Loads

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Definition of LoadsDefinition of Loads

The loads are the reasons why we do calculations on the models. We want to calculate the response on the model due to the loads

The loads can be in terms of:

Force Displacement Pressure Temperature Heat flux etc etc

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Definition of LoadsDefinition of Loads

Loads can be applied to points , surfaces or bodies They can be constant, or time dependent

p

F

G

t

F

t

F

t

F


Defining Model ConstraintsDefining Model Constraints

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Constraints Constraints

By constraining our model we model the interface of the model to the rest of the world

Constraints can be given to both displacements and rotations

If the model is not constrained, it will yield numerical problems (“singularities”) during solving, as shown below

Unstable Stable

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ConstraintsConstraints

Fixed Connection

X

Y

Transfers vertical andhorizontal forces andmoments

No displacementor rotation possible

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X

Y

Pinned connection

Rotation possible,no displacements

Horizontal and verticalreaction forces,no reaction moments

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Sliding and pinned connection

X

Y

Rotation and horizontal displacement possible,no vertical displacement Vertical reaction force,

no reaction moments orhorizontal forces

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Constraints, ExampleConstraints, ExamplePinned constraints can be done in two ways:

A B

Inner edge given Rigid Elements

displacement R=0 made from inner

referenced to a edge to center

cylindrical point. This point

coordinate is given displacement

system in the Ux, Uy, Uz,

center Rx, Ry = 0


Defining MaterialsDefining Materials

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Material Model SelectionMaterial Model Selection

Isotropic Orthotropic (also "Composite"- plates/shells) Anisotropic Hyperelastic Manual data import User-modifiable material libraries

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Linear Material AssumptionsLinear Material AssumptionsConditions in the range of validityof linear static calculations:

Linear stress-strain-curve (Hooke’s Law) (Valid as long as the resulting stresses are below yield)

Displacements are very small compared to the size of the part or its thickness

The principal of superposition is valid !

E

Theory of Elasticity(Hooke’s Law)

F

u

F = k u

x = E x

+ =

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Linear Material PropertiesLinear Material Properties

Young‘s modulus, E

Shear modulus, G

Poisson ratio, Relationship between axial and transverse

strain: =

y

x

y

xP P

ν)(12E

G

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Non-linear MaterialsNon-linear MaterialsOutside the range of validity of linear

static calculations: Nonlinear geometric effects (stability,

large displacements)

Nonlinear material behaviour (elasticity and/or plasticity)

Time- and temperature-dependent material behaviour (creep)

Contact with/without friction

The principle of superposition is no longer valid !

Eo

K

+

Stress,

Strain,

Tension

Limit

YieldLlimit

0.2% Offset

LinearRange = E


Defining Element PropertiesDefining Element Properties

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PropertiesProperties

In addition to the element shapes extra information is needed in order to describe the model completely.

The extra information needed is dependent on element type, but almost always material is an example of a property that needs to be input

The extra element information is input as element properties

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Beam Element PropertiesBeam Element Properties

Beam elements needs to input:

MaterialCross sectional Area (A) - transfers axial forces Moments of Inertia (I) - transfers bending and shearTorsional Moment of Inertia (J) - transfers torsional

forces

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Shell Element PropertiesShell Element Properties

Shell Element needs to input:

Material Thickness

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Solid Element PropertiesSolid Element Properties

Solid elements needs to input:

Material


Conducting the analysisConducting the analysis

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Checking the modelChecking the model

Before the analysis is run, the model should be

checked. A few items might be good to verify:

Check for coincident nodes and/or elements Check the element quality Check the mass or volume of the model ???

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Types of AnalysisTypes of Analysis

Linear Static Analysis

Calculation of stresses Calculation of deformations

”Linear” means that the calculated results are linear proportional to the applied loads.

”Static” means that the loading is not time dependent

or the time dependence is negligible.

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Types of AnalysisTypes of Analysis Normal Mode Analysis

Calculation of the Eigenvalues Eigenvalues are frequencies at which the structure would start to oscillate if a periodic loading would be applied.

Calculation of the Eigenmodes Eigenmodes are the deformed shape of the structure at

a certain eigenfrequency.

Eigenfrequency analysis is used to determine the behavior of parts due to dynamic loading.

If there is an eigenfrequency of the part close to the frequencyof the loading resonance effects may result and the loadings may be substantially enlarged.

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Types of AnalysisTypes of Analysis

Stability and Buckling Analysis

Calculation of Critical Stability and Buckling Loads Load factor at which the structure fails by instability(buckling).

Calculation of Stability and Buckling Shapes Shape of the buckled structure at the point of critical loading.

In a linear static analysis it is assumed that the structure is in stable equilibrium and will return to its original shape.

A structure will be unstable (buckling) if for a certainload combination the deformation increases withoutincreasing the loading.

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Finding the answer..Finding the answer..

After the model is created in the preprocessor, the problem is sent to the solver

The solver calculates the answer to the problem, and this can be viewed and interpreted in the postprocessor


Postprocessing Calculation ResultsPostprocessing Calculation Results

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Postprocessing of ResultsPostprocessing of Results

Postprocessing can be done as:

Plot of Deformed Structure Contour Plots Animated Display of Deformations and Contours X-Y Plots Result Listings

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zelem yelem

xelem

Results for Beam ElementsResults for Beam Elements

Beam elements has no geometrical extensions other than lengthWhen postprocessing stresses the user must select at which point the stresses shall be shown, such as top left, bottom right and so on.

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Results for Shell ElementsResults for Shell ElementsWhen shell elements are used in plots, the results are displayes for ”top”, ”bottom” or ”middle” of the elements.Looking at the shells from the other side by rotating the model will yield the same plot.The user must selcet which position results shall be presented at, and verify what is the element top by checking the element normal vector. Shown are the stresses in the Y-direction. They should be tensile on the positive z-side and compressive on the negative z-side of the model.

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Results for Solid ElementsResults for Solid Elements

Solid elements require no special postprocessing attention from the user.

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Different presentations of same calculationDifferent presentations of same calculation

1. Continuous spectrum

2. Discrete spectrum

3. Element fill

4. Real results

1. 2.

3. 4.

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Checking the answerChecking the answer

Check the stress gradient. If high stress gradients are present, it might be wise to refine the mesh locally and re-run the analysis

A re-run analysis with a finer mesh that yields the same answer is indicative of a converged answer


Special Modelling TechniquesSpecial Modelling Techniques

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Special Modeling TechniquesSpecial Modeling Techniques

Symmetry

Cases for which the geometry and the constraints of a structure are identical with respect to one or more axes or planes

Symmetry constraints are a function of the orientation of the coordinate axis

Use of Symmetry may be restricted for some types of analysis. (Normally not applied for stability or dynamic analysis)

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Exploiting half symmetryExploiting half symmetry

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Exploiting Axis-symmetryExploiting Axis-symmetry

This CAD model… could have this CAE model!

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Guidelines for deciding the Guidelines for deciding the analysis modelanalysis model

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Analysis models Analysis models

Analysis assumptions should be based on knowledge and experience

Always verify the plausibility of the results before you make design decisions depending on them

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Definition of the ProblemDefinition of the Problem

2-D simulation

- Plane Stress

- Plane Strain

- Axis-symmetric

3-D simulation and modeling

- Simulation with elastic beams

- Use of symmetry

- Plate or solid models

What is the most efficient analysis method to get the desired results?

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How to get a better understanding How to get a better understanding of the modelof the model

Try small realistic examples where analytical solutions exist to test the behaviour of unknown element types and unknown solution types.

Divide the complete structure into substructures to better understand the behavior of the model.

Apply load combinations separately to understand their individual influence and combine them again later for sensitivity studies

Vary the constraints/boundary conditions and do separate analyses when the situation is not well understood and unclear.

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Model VerificationModel Verification

Simple test calculations done by hand for comparison

Pre-estimate the behaviour of the part and try to interpret possible deviations from the expected results

Always use your engineering common sense !

(Are the orders of magnitude of the results correct? Does the deformation look plausible?, etc.)

Compare with prototypes

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Example:Example:Choosing calculation modelChoosing calculation model

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How to decide the analysis modelHow to decide the analysis modelThis is thenew InternationalSpace Station,modeled inthe CADsystemUnigraphics

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Beam ModelBeam ModelExample of possible Beam Model:

Analysis of Solar Array Truss

Structure

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Shell ModelShell ModelExample of Shell model:

Analysis of crew compartment

module

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Solid ModelSolid ModelExample of Solid model:

Analysis of lifting

lug

Documents

Wprowadzenie do MES Introduction to Finite Element Concepts