875.317 computational mechanics - baunat.boku.ac.at · transcend the classical boundaries of solid...

University of Natural resources and Applied Sciences, Institut of Structural Engineering Alfred Strauss

875.317 Computational Mechanics

STRAUSS


Time Schedule

Mo 20. Feb. 2006 9:00 -14:00 Mo 27. Feb. 2006 8:30 -12:00 Do 02. Mrz. 2006 13:30 -17:00

Evaluation VO5-7 Exercises + Oral Exam about the Main components of the Course

UE 17. 19. AprilMATLAB Excercises - to be solved in ABAQUS


1Overview


Computational Mechanics

Nanomechanicsmolecular and atomic levelsParticle Physics and chemistryCrystallographic and granular levelDesign and fabrication of materials and micro devices

Continuum mechanicsBodies at the macro levelMicrostructure is homogenized by phenomenological averagesSolid and fluid mechanicsapplied sciences approaches

Computational Structural Mechanics Emphasizes Technological Applications

Computational Fluid Mechanics Emphasizes Equilibrium and Motion of Fluids


Computational Mechanics

Multiphysics

Transcend the classicalboundaries of solid and fluid mechanics

Interaction of fluid and structures

Systems

Identifies mechanicalobjects

Natural or artificialBuildingsBridgesCarsairplanes


FEM Direct Stiffness MethodComputational Solid and Structural Mechanics

A convenient subdivision of problems in ComputationalSolid and Structural Mechanics (CSM) is

ComputationalSolid and StructuralMechanics (CSM)

Static(quasi Static)

Dynamic(inertial forces are timedependent)


FEM Direct Stiffness MethodComputational Solid and Structural Mechanics

A further subdivision of problems in ComputationalSolid and Structural Mechanics (CSM) is

ComputationalSolid and StructuralMechanics (CSM)

Linearstatic problems in which the response islinear in the cause-and effect

NonlinearProblems outside this domain


FEM Direct Stiffness MethodCSM Linear Statics

For the numerical simulation on the computer we must now chose aspatial discretization method

CSM Linear Statics

Finite Element MethodFinite Difference MethodBoundary Element MethodFinite Volume MethodSpectral MethodMesh-Free Method


FEM Direct Stiffness Method

DisplacementFEM Formulation Equilibrium

MixedHybrid

CSM Linear Statics by FEMHaving selected the FEM for discretization, we must next pick a formulation and a solution method

StiffnessFEM Solution Flexibility

Mixed


Summarizing

Computational Structural MechanicsLinear static problemsSpatially discretized by displacement-formulated FEMSolved by the stiffness method


What is a Finite Element?

Archimedes' problem (circa 250 B.C.): rectification of thecircle as limit of inscribed regular polygons


IDEALIZATION & DISCRETIZATION


FEM Direct Stiffness Method

Two Interpretations of FEM for Teaching

Mathematical

Numerical approximation of aBoundary Value Problem byRitz-Galerkin discretizationwith functions of local support

Physical

Breakdown of structural systeminto components (elements) andreconstruction by the assemblyprocess


The Physical FEM. The physical system (left) is thesource of the simulation process.

The ideal mathematical model(should one go to thetrouble of constructingit) is inessential.


Model updating process in the Physical FEM.


The Mathematical FEM. The mathematical model(top) is the source of the simulation process.

Discrete model and solution follow fromit. The ideal physicalsystem (shouldone go to thetrouble of exhibiting it) isinessential.


Synergy Between Mathematicaland Physical FEM

Combining physical and mathematicalmodeling through multilevelFEM.


EXERCISES

EXERCISE 1.1 Work out Archimedes problem

EXERCISE 1.2 Selected components of somesubstructures.


2The Direct

Stiffness Method:Breakdown


The direct Stiffness Method (DSM)

Importance: DSM is used by all major commercial FEM codesA democratic method, works the same no matter what theelement

Obvious decision: use the truss to teach the DSM

Bar (truss member) element,2 nodes, 4 DOFs

Tricubic brick element,64 nodes, 192 DOFs


Model Based Simulation(a simplification of diagrams of Chapter 1)


The Direct Stiffness Method (DSM) StepsStarting with: Idealization

DisconnectionLocalizationMember (Element) Formation

GlobalizationMergeApplication of BCsSolutionRecovery of Derived Quantities

Breakdown

Assembly &Solution


A Physical Plane Truss

Too complicated to do by hand. We willUse a simpler one to illustrate DSM steps


FEM Direct Stiffness MethodThe Example Truss: Physical Model

(Loads not shown)


FEM Direct Stiffness MethodThe Example Truss: FEM ModelNodes, Elements and DOFs

1 2

3

fx1 , ux1 fx2 , ux2

fx3 , ux3fy3 , uy3

fy1 , uy1 fy2 , uy2

(3) (2)

(1)

L(3) = 10 * 2E (3) A (3) = 200*2 L(2) = 10

E (2) A (2) = 50

L(1) = 10 E (1) A (1) = 100

x

y


FEM Direct Stiffness MethodThe Example Truss: FEM Model BCsApplied Loads and Supports

1 2

3 fx3 = 2fy3 = 1

x

y


Master (Global) Stiffness Equations

=

y3

x3

y2

x2

y1

x1

ff

ff

ff

f

=

y3

x3

y2

x2

y1

x1

uu

uu

uu

u

=

y3

x3

y2

x2

y1

x1

y3y3y3x3y3y2y3x2y3y1y3x1

x3y3x3x3x3y2x3x2x3y1x3x1





y3

x3

y2

x2

y1

x1

uu

uu

uu

KKKKKK

KKKKKK

KKKKKK

KKKKKK

KKKKKK

KKKKKK

ff

ff

ff

Linear structure:

Nodalforces

Master stiffnessmatrix

Nodaldisplacements

or f = K u


Member (Element) Stiffness Equations

uKf =

=

jy

jx

iy

ix

jyjyjxjyiyjyixjy

jyjxjxjxiyjxjxjx

jyiyjxiyiyiyixiy

jyixjxixiyixixix

jy

jx

iy

ix

u

u

u

u

KKKK

KKKK

KKKK

KKKK

f

f

f

f


First Two Breakdown Steps:Disconnection and Localization

These Steps are conceptual


The 2-Node Truss (Bar) Element

k = EA / L Equivalentspring stiffness


Truss (Bar) Element Formulationby Mechanics of Materials (MoM)

ixjxixjxs uud,ffFd,LAEdkF =====

Element stiffnessequations in localcoordinates

Element stiffnessmatrix in localcoordinates

=

jy

jx

iy

ix

jy

jx

iy

ix

u

u

u

u

00000101

00000101

LAE

f

f

f

f

=

00000101

00000101

LAEK


EXERCISE 2.1 Explain why arbitrarily ..

EXERCISE 2.2 Show that the sum of the entries of

EXERCISE 2.3 Using matrix algebra derive .

EXERCISE 2.4 By direct matrix multiplication .

EXERCISE 2.5 The transformation equations ..

EXERCISES


3The Direct

Stiffness Method:Assembly and Solution


The Direct Stiffness Method (DSM) Steps(recalled for convenience)

DisconnectionLocalizationMember (Element) Formation

GlobalizationMergeApplication of BCsSolutionRecovery of Derived Quantities

Breakdown

Assembly &Solution

conceptualsteps

processingsteps

post-processingsteps


Globalization Step:Displacement Transformation

xi = uxi c + uyi s, yi = uxi s + uyi cxj = uxj c + uyj s, yj = uxj s + uyj c

in which c = cos s = sin

Node displacements transform as


Globalization Step: DisplacementTransformation (contd)

c s 0 0s c 0 00 0 c s0 0 s c

uxiuyiuxjuyj

xiyixjyj

=

Note:global on RHS,local on LHS

In matrix form

or e = Teue


Globalization Step: Force Transformation

c -s 0 0s c 0 00 0 c -s0 0 s c

fxifyifxjfyj

=

Node forces transform as

or fe = (Te )T f-e

f-xi

f-yi

f-xj

f-yj Note:

global on LHS,local on RHS


Globalization: Congruential Transformation of Element Stiffness Matrices

fe = (Te )T f-ee = Teue

K-e e = f

-e

Ke = (Te)T K-e Te

c2 sc -c2 -scsc s2 -sc -s2-c2 -sc c2 sc

-sc -s2 sc s2=Ke E

e Ae

Le


The Example Truss FEM ModelRecall from Chapter 2

1 2

3

fx1 , ux1 fx2 , ux2

fx3 , ux3fy3 , uy3

fy1 , uy1 fy2 , uy2

(3) (2)

(1)

L(3) = 10 * 2E (3) A (3) = 200*2 L(2) = 10

E (2) A (2) = 50

L(1) = 10 E (1) A (1) = 100

x

y


Globalized Element Stiffness Equations1 0 1 00 0 0 01 0 1 00 0 0 0

fx1 (1)fy1 (1)fx2 (1)fy2 (1)

= 10

ux1 (1)uy1 (1)ux2 (1)uy2 (1)

0 0 0 00 1 0 -11 0 1 00 0 0 0

fx2 (2)fy2 (2)fx3 (2)fy3 (2)

= 5

ux2 (2)uy2 (2)ux3 (2)uy3 (2)

for Example Truss

0.5 0.5 -0.5 -0.50.5 0.5 -0.5 -0.5-0.5 -0.5 0.5 0.5-0.5 -0.5 0.5 0.5

fx1 (3)fy1 (3)fx3 (3)fy3 (3)

= 20

ux1 (3)uy1 (3)ux3 (3)uy3 (3)


Assembly Rules

1. Compatibility: The joint displacements of all membersmeeting at a joint must be the same

2. Equilibrium: The sum of forces exerted by all membersthat meet at a joint must balance the external force appliedto that joint.

To apply these rules in assembly by hand, it is convenient to augment theelement stiffness equations as shown for the example truss in the nextslide.


Expanded Element Stiffness Equations10 0 10 0 0 0

0 0 0 0 0 010 0 10 0 0 0

0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

fx1 (1)fy1 (1)fx2 (1)fy2 (1)fx3 (1)fy3 (1)

=

ux1 (1)uy1 (1)ux2 (1)uy2 (1)ux3 (1)uy3 (1)

0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 5 0 -50 0 0 0 0 00 0 0 -5 0 5

fx1 (2)fy1 (2)fx2 (2)fy2 (2)fx3 (2)fy3 (2)

=

ux1 (2)uy1 (2)ux2 (2)uy2 (2)ux3 (2)uy3 (2)

10 10 0 0 -10 -1010 10 0 0 -10 -10

0 0 0 0 0 00 0 0 0 0 0

-10 -10 0 0 10 10-10 -10 0 0 10 10

=

fx1 (3)fy1 (3)fx2 (3)fy2 (3)fx3 (3)fy3 (3)

ux1 (3)uy1 (3)ux2 (3)uy2 (3)ux3 (3)uy3 (3)


Expanded Element Stiffness Equations10 0 10 0 0 0

0 0 0 0 0 010 0 10 0 0 0

0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

fx1 (1)fy1 (1)fx2 (1)fy2 (1)fx3 (1)fy3 (1)

=

ux1 uy1 ux2uy2 ux3uy3

0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 5 0 -50 0 0 0 0 00 0 0 -5 0 5

fx1 (2)fy1 (2)fx2 (2)fy2 (2)fx3 (2)fy3 (2)

=

ux1 uy1ux2uy2ux3uy3

10 10 0 0 -10 -1010 10 0 0 -10 -10

0 0 0 0 0 00 0 0 0 0 0

-10 -10 0 0 10 10-10 -10 0 0 10 10

=

fx1 (3)fy1 (3)fx2 (3)fy2 (3)fx3 (3)fy3 (3)

ux1 uy1ux2 uy2ux3uy3

To apply compatibility, dropthe member index from thenodal displacements

f (1) = K(1) u

f (2) = K(2) u

f (3) = K(3) u


Next, Apply Equilibrium Rule

Applying this to all joints (see Notes):

Be carful with + directionsof internal forces!

f = f (1) + f (2) + f ( 3 )


Forming the Master Stiffness Equationsthrough Equilibrium Rule

f = f (1)+f (2)+f ( 3 ) = (K (1)+K (2)+K ( 3 )) u = K u

20 10 -10 0 -10 -1010 10 0 0 -10 -10-10 0 10 0 0 0

0 0 0 5 0 -5-10 -10 0 0 10 10-10 -10 0 -5 10 15

=

fx1 fy1 fx2 fy2 fx3fy3



Applying Support and LoadingBoundary Conditions to Example Truss

1 2

3 fx3 = 2fy3 = 1

x

y

ux1 = uy1 = uy2 = 0Displacement BCs:

Force BCs:fx2 = 0, fx3 = 2, fy3 = 1

Recall:


Where Do Boundary Conditions go?ux1 = uy1 = uy2 = 0

fx2 = 0, fx3 = 2, fy3 = 1

20 10 -10 0 -10 -1010 10 0 0 -10 -10-10 0 10 0 0 0

0 0 0 5 0 -5-10 -10 0 0 10 10-10 -10 0 -5 10 15

=




Reduced Master Stiffness Equationsfor Hand Computation

Strike out rows and columns pertaining to known displacements:

Solve by Gauss elimination for unknown node displacements

10 0 00 10 100 10 15

=fx2 fx3fy3

ux2 ux3uy3

=021

K u = f Reducedstiffnessequations


Solve for Unknown Node Displacementsand Complete the Displacement Vector

ux2 ux3uy3

=0

0.4-0.2

=

00000.4

-0.2


Expand with knowndisplacement BCs

=u


Recovery of Node Forces Including Reactions20 10 -10 0 -10 -1010 10 0 0 -10 -10-10 0 10 0 0 0

0 0 0 5 0 -5-10 -10 0 0 10 10-10 -10 0 -5 10 15

=

-2-20121

0000

0.4-0.2

1 2

3 fx3 = 2fy3 = 1

xy

Recall:


Recovery of Internal Forces(Axial Forces in Truss Members)

For each member (element) e = (1), (2), (3)

1. extract ue from u

2. transform to local(element) displacements

e = Teue

de= exj exi

3. compute elongation

4. compute axial force pe


Computer Oriented Assembly andSolution in Actual FEM Codes

K stored in special sparse format

(for example skyline format studied in Part III)

Assembly done by freedom pointers (Sec 3.5.1)

Equations for supports are not physically deleted (Sec 3.5.2)

Next slide explains this for the example truss


Computer Oriented Modificationof Master Stiffness Equations

20 10 -10 0 -10 -1010 10 0 0 -10 -10-10 0 10 0 0 0

0 0 0 5 0 -5-10 -10 0 0 10 10-10 -10 0 -5 10 15

=

fx1 fy1 0fy2 21

00ux2 0ux3uy3

ux1 = uy1 = uy2 = 0

fx2 = 0, fx3 = 2, fy3 = 1

(freedoms 1, 2, 4)

zero out rows and columns 1, 2 and 4store 1's on diagonal


Computer Oriented Modificationof Master Stiffness Equations (contd)

1 0 0 0 0 00 1 0 0 0 00 0 10 0 0 00 0 0 1 0 00 0 0 0 10 100 0 0 0 10 15

=

000021

Modified masterstiffness equations

same u as inoriginal equationsK u = f

00ux2 0ux3uy3


EXERCISES

EXERCISE 3.6

EXERCISE 3.7 With MATLAB


4The Direct

Stiffness Method:Additional Topics


Prescribed Nonzero Displacementsin Example Truss


Prescribed Nonzero Displacements

Recall the master stiffness equations

The displacement B.Cs are now

20 10 -10 0 -10 -1010 10 0 0 -10 -10-10 0 10 0 0 0

0 0 0 5 0 -5-10 -10 0 0 10 10-10 -10 0 -5 10 15

=



ux1 = 0, uy1 = -0.5, uy2 = 0.4


Prescribed NZ Displacements (contd)

Remove rows 1,2,4 but (for now) keep columns

20 10 -10 0 -10 -1010 10 0 0 -10 -10-10 0 10 0 0 0

0 0 0 5 0 -5-10 -10 0 0 10 10-10 -10 0 -5 10 15

=

fx1 fy1 0fy2 21

0-0.5ux2 0.4ux3uy3

-10 0 10 0 0 0-10 -10 0 0 10 10-10 -10 0 -5 10 15

0-0.5ux2 0.4ux3uy3

=021


Prescribed Nonzero Displacements

Transfer effect of known displacements to RHS, and delete columns

Solving gives

10 0 0 0 10 100 10 15

=021

ux2 ux3uy3

-(-10)*0+0*(-0.5)+0*0.4(-10)*0+(-10)*(-0.5)+0*0.4(-10)*0+(-10)*(-0.5)+(-5)*0.4

=0-3-2

ux2 ux3uy3

=0

-0.50.2



Complete the displacement vector with known values

ux2 ux3uy3

=0

-0.50.2

==u

0-0.500.4

-0.50.2




Recovery of reaction forces and internal

Member forces proceeds are before

In summary, the only changes to the DSM

is in the application of displacement

boundary conditions before solve


Initial Force Effects (also called Initial Strain& Initial Stress Effects by FEM authors)

Thermomechanical effects

Moisture effects

Prestress effects

Lack of fit

Residual Stresses

Internal actuators (smart structure)

For specificity we will study in detail only the first one:

thermomechanical effects


Thermomechanical Effects on Bar Element


Thermomechanical Effects on Bar Element (contd)

Axial strain is sum of mechanical and thermal:


Thermomechanical Effects on Bar Element (contd)


Incorporating Thermomechanical Effectsinto the Element Stiffness Equations


Physical Interpretation ofThermal Force Vector

Think of a fully restrained bar when heated


Matrix Form of Element Stiffness Equationsin Local System


Assembly Rules withThermomechanical Effects

1. Compatibility : The joint displacements of all membersmeeting at a joint must be the same

2. Equilibrium : The sum of effective forces exerted by all members that meet at a joint must balance the external force applied to that joint.

No changes in application of 1. To account for 2,

the thermal forces are globalized and added to the

Mechanical forces during the merge process


Master Stiffness Equations withThermomechanical Effects

Solve for node displacements u. If you want to retrieve

the mechanical node forces including reactions subtract

the thermal force:

For computation of internal forces and stresses seeworked out Examples


Worked out Examples #1 in Notes


Worked out Examples #1 in Notes (contd)


Generalization: Initial Force Effects in the DSM


Summary: Treating Initial ForceEffects in the DSM


Chapter 6


Constructing MoM Members

What are MoM Members?

Skeletal structural members whose stiffness equations can beconstructed by Mechanics of Materials (MoM) methods

Can be locally modeled as 1D elements


MoM Members Tend to Look Alike

One dimension (longitudinal) much larger than the other two(transverse)

y

z

x

z

longitudina

l direction

y

x


But Receive Different Names according to Structural FunctionBars: transmit axial forces

Beams: transmit bending

Shafts: transmit torque

Spars (aka Webs): transmit shear

Beam - Columns: transmit bending and axial force


Common Features of MoM Finite Element Models

End quantities aredefined at the joints

y

z

x

(e)

i jz

y

x

Internal quantities aredefined in the member


Governing Matrix Equations for simple MoMElement

From node displacement to internal deformations (strains)v = Bu Kinematic

From deformations to internal forcesp = S v Constitutive

From internal forces to node forcesf = A p Equilibrium

If f andu are PVW (virtual Work) conjugate, B = A


Tonti Diagram of Governing Matrix Equations forsimplex MoM Element

f = AT S B u =Ku

p = S v

v = Bu f = AT p

u f

v p

Stiffness

Constitutive

Kinematic Equilibrium


Elimination of the Internal Quantities

v and p gives the Element Stiffness Equations through simple MatrixMultiplications

f = AT S B u =Ku

K = AT S Bif B = A

K = BT S B


The Bar Element Revisited

EA

i j

y

x

x

y

z

F

-F

fxj uxjfxi uxi

- L -

(a)

(b)

Axial rigidity EA, length L


The Bar Element Revisited

[ ]

FFf

f

SddL

EAF

u

ud

T

xj

xi

xj

xi

Af

uB

=

=

=

==

=

=

11

11

===

1111

LEASSK TT BBBA

Can be expanded to the 4 x 4 of Chapter 2 by adding two zero rows andcolumns to accomodate uyi and uyj


The Spar (a.k.a. Shear Web) Element

GA

i j

y

x

fyj uyjfyi uyi

- L -

(b)

x

y

z-V

Shear rigidity GA, length L

V

(a)


The Spar Element

[ ]

=

==

=

=

=

=

==

=

=

uKuBAf

Af

uB

yj

yisT

yj

yi

T

yj

yi

s

yj

yi

u

uL

GASf

f

VVf

f

SAGV

u

uL

1111

11

111

=

1111

LGAsK


The Shaft Element

GJ

i j

y

xmxi xi

- L -

(b)

mxj xj

xz

T Torsional rigidity GJ, length L

T

(a)m

xi xi

mxj

xj

y


Matrix Equation for Non-Simplex MoM Element

From node displacement to internal deformations (at each section)v = Bu Kinematic

From deformations to internal forces (at each section)p = R v Constitutive

From internal forces to node forcesdf = AT dp Equilibrium


Tonti Diagram of Matrix Equations for Non-simplex MoM Element (with A=B)

f =0L

BT R B dxu

p = R v

v = Bu df = BT dp

u f

v p

Stiffness

Constitutive(at each section)

Kinematic(at each section)

Equilibrium


Chapter 7


FEM Terminology

degrees of freedom (abbrv: DOF)

state (primary) variables: displacements in mechanics

conjungate variables: forces in mechanics

stiffness matrixK u = f

master stiffness equations K u = fM+ fI


Physical significance of vectors u and f in miscellaneous FEM applications

Application State (DOF) vector Forcing vector fProblem u represents reptesents

Structures and solid mechanics Displacement Mechanical forceHeat conduction Temperature Heat fluxAcoustic fluid Displacement potential Particle velocityPotential flows Pressure Particle velocityGeneral flows Velocity FluxesElectrostatics Electric potential Charge densityMagnetistatics Magnetical potential Magnetic intensity


Where FEM fits (from Chapter 1)

Physicalsystem

DISCRETIZATION SOLUTIONIDEALIZATION

Mathem.system

Discretemodel

DiscretesolutionFEM

RealizationIdentification

Continuifica-tion Solution error

Discretization + Solution error

Modelling +Discretization + Solution error


Idealization Process (from Chapter 2)

Physical system

Mathematical system

IDEALIZATION


Mathematical Model Definition

Traditional definition:

Scaled fabrication version of a physical system(think of a car or train model)

Simulation oriented definition:

A model is a symbolic device built to simulate and predict aspects of behavior of a system


Implicit Modeling

Physicalsystem

System discretemodel

Completesolution

Mathem.model

Componentdiscretemodel

Componentequation

Component

Level

System

Level

FEM Library of Black Box FEM Code


Recall the Breakdown DSM Steps

DisconnectionLocalizationMember (Element) Formation generic elements

Let Stop Here andStudy Generic Elements next

Breakdown


Because most of the Remaining DSM Steps

GlobalizationMergeApplication of BCsSolutionRecovery of Node Forces

are Element Independence


Attributes of Mechanical Finite Elements

DimensionalityNodes serve two purposes:

gemetric definitionhome for DOFs (connectors)

Degrees of freedom (DOFs) of freedomsConjugate node forces

Material proportiesFabrication proporties


Element geometry is defined by node locations

1D

2D

2D

3D


Classification of Mechanical Finite Elements

Primitive StructuralContinuumSpecialMacroelementsSubstructures

Superelements


Primitive Structural Elements(often built from MoM models)

Physical structuralComponent

MathematicalModel Name

Finite ElementDiscretization

bar

beam

tube, pipe

spar (web)

shear panel


Continuum Elements

Physical Finite Elementidealization

Physical Finite Elementidealization

Plates 3D solids


Special Elements

Crack element Infinite element Honeycomb panel

double node Infinity


Macroelements


Substructures


Boundary Conditions (BCs)

The most difficult topic for FEM programm users(the devil hides in the boundary)

EssentialTwo types:

Natural


Chapter 8


FEM Modeling: Mesh, Loads and BCs

General Modeling Rules

Finite Element Mesh Layouts

Distributed LoadsNode by Node (NbN) lumpingElement by Element (EbE) lumping

Displacement BCssuppressing rigid body motionstaking advantage of symmetry and antisymmetry


General FEM Modeling Rules

Use the simplest elements that will do the job

Never, never use complicated of special elements unless you areabsolutely sure of what you are doing

Use the coarsest mesh that will capture the dominant behavior of thephysical model, particularly in design situations

Three word summary: Keep it simple


Another Justification for Simplicity

In product design situations several FEM models of increasingrefinement will be set up as design evolves

Ergo, do not overkill at the beginning


Where finer meshes should be used

Cutouts Cracks Vicinity of concentrated loads entrant corners

Load transfer (bondedjoints, welds, anchors, etc.) Abrubt thickness changes Material interfaces

weld


Avoid 2D/3D Elements of bad aspect ratio

Good Bad


Elements must not cross interfaces

No OK

Pysical interface


Element geometry preferences

Other things being equal, prefer

in 2D: quadrilaterals over triangles

in 3D: bricks over wedgeswedges over tetrahedra

(Elements do not file discrimination suits)


Node by Node distributed load lumping

21 43 65 87Boundary

Finite element mesh

Distributed loadintensity (loadacts downwardson boundary)

Nodal force f5 at 5 is set to P, themagnitude of thecrosshachedarea under theload curve. Thisarea goeshalfway overadjacentelement sides

f5=P


Element by Element distributed load lumping

21 43 65 87Boundary

Finite element mesh

Distributed loadintensity (loadacts downwardson boundary)

Force P has magnitude of thecrosshachedarea under theload curve and acts at itscentroid. f5

(e)f4

(e)

centroid C of crosshatchedareaP

a b

L=a+b

f4(e)=(b/L)P f5

(e)=(a/L)P

4 5P

C

Details of element force computations


Boundary Conditions (BCs)

The most difficult topic for FEM programm users(the devil hides in the boundary)

EssentialTwo types:

Natural


Boundary Conditions

Essential vs. Natural

Recipe:

If a BC involves one or more DOF in a direct way, it is essential and goes to the Left Hand Side (LHS) ofK u = f

Otherwise it is natural and goes to the Right Hand Side (RHS) of K u = f


Minimum support conditions to suppress rigidbody motions in 2D

(b)(a) (c)y

y

x x


Minimum support conditions to suppress rigidbody motions in 3D

x

z

y


Visualizing Symmetry and Antisymmetryconditions in 2D

A A A A A Adisplacement

vectors

symmetryline

antisymmetryline

loads


Example of application of symmetry BCs

y

x

A B

C D

A B

C D


Example of application of antisymmetry BCs

x

A B

C D

A B

C DVertical (y) motion of one nodesuch as C of D may beconstrained to suppress y-RBM

y

x


Breaking up Point loads at symmetry BCs

x

A B

C D

y

PP

A B

C D P/2P/2P/2

P/2P/2


Breaking up Point loads at antisymmetry BCs

x

A B

C D

y

x

A B

C D

A B

C D

? ?

PP

2P


Boundary Conditions

Essential vs. Natural

Recipe:

1. If a BC involves one or more DOF in a direct way, it is essentialand goes to the Left Hand Side (LHS) of K u = f

2. Otherwise it is natural and goes to the Right Hand Side (RHS) of K u = f


Chapter 9


MultiFreedom Constrains I

Single freedom constraint examplesux4 = 0 linear, homogeneousuy9 = 0.6 linear, non homogeneous

Multifreedom constaint examplesux2 = 0.5 uy2 linear, homogeneousux2 - ux4 + ux6 = 0.25 linear, non homogeneous(x5+ux5-x3-ux3) + (y5+uy5-y3-uy3) = 0 non linear homogeneous


Sources of Multifreedom

Skew displacementCoupling nonmatched FEM meshesGlobal-local and multiscale analysisIncompressibilityModel reduction


MFC Application Methods

Master Slave Elimination Chapter 9

Penalty Function AugmentationChapter 10

Lagrange Multiplier Adjustment


Example 1D Structure to illustrate MFCs

Multifreedom constraint:

(2)(1) (3) (4) (5) (6)u1,f1 u2,f2 u3,f3 u4,f4 u5,f5 u6,f6 u7,f7

1 2 3 4 5 6 7x

u2 = u6 or u2 - u6 = 0Linear homogeneous MFC


Example 1D Structure to illustrate MFCs

Unconstrained master stiffness equation: K u = f

(2)(1) (3) (4) (5) (6)u1,f1 u2,f2 u3,f3 u4,f4 u5,f5 u6,f6 u7,f7

1 2 3 4 5 6 7x

K11 K12 0 0 0 0 0 u1 f1K12 K22 K23 0 u2 f20 K23 K33 K34 0 u3 f30 0 K34 K44 K45 0 u4 f40 0 0 K45 K55 K56 0 u5 f50 0 0 0 K56 K66 K67 u6 f60 0 0 0 0 K67 K77 u7 f7

K11 K12 0 0 0 0 0 u1 f1K12 K22 K23 0 u2 f20 K23 K33 K34 0 u3 f30 0 K34 K44 K45 0 u4 f40 0 0 K45 K55 K56 0 u5 f50 0 0 0 K56 K66 K67 u6 f60 0 0 0 0 K67 K77 u7 f7

K11 K12 0 0 0 0 0 u1 f1K12 K22 K23 0 0 0 0 u2 f20 K23 K33 K34 0 0 0 u3 f30 0 K34 K44 K45 0 0 u4 f40 0 0 K45 K55 K56 0 u5 f50 0 0 0 K56 K66 K67 u6 f60 0 0 0 0 K67 K77 u7 f7

=


Master Slave Method for Exampel Structure

Taking u2 as master:

u1 1 0 0 0 0 0 u1u2 0 1 0 0 0 0 u2u3 0 0 1 0 0 0 u3u4 = 0 0 0 1 0 0 u4u5 0 0 0 0 1 0 u5u6 0 1 0 0 0 0 u6u7 0 0 0 0 0 1 u7

u = T


Forming the Modified Stiffness Equations

Unconstrainted master stiffness equation: K u = f

Master slave transformation: u = T

Congruential transformation: K = TT K T

f = TT f

Modified stiffness equations: K = f

^

^ ^

^


Modified Stiffness Equations for Example Struct.

K u = f

K11 K12 0 0 0 0 u1 f1K12 K22+K66 K23 0 K56 K67 u2 f2+f60 K23 K33 K34 0 0 u3 f30 0 K34 K44 K45 0 u4 f40 K56 0 K45 K55 K56 u5 f50 K67 0 0 0 K66 u7 f7

=

^ ^ ^

in full

Solve for u, then recover u = T u^ ^


Multiple MFCsSupposeu2 - u6 = 0 , u1 + 4u4 = 0, 2u3 + u4 + u5 = 0Pick 3, 4 and 6 as slaves:u6 = u2, u4 = -1/4u1, u3 = -1/2(u4 + u5) = 1/8u1 1/2u5Put in matrix form:

u1 1 0 0 0u2 0 1 0 0u3 1/8 0 -1/2 0u4 = -1/4 0 0 0u5 0 0 1 0u6 0 1 0 0u7 0 0 0 1

u1u2u5u7

This isu = T then proceedas before


Non homogeneous MFCs

u2 u6 = 0.2

u1 1 0 0 0 0 0 u1 0u2 0 1 0 0 0 0 u2 0u3 0 0 1 0 0 0 u3 0u4 = 0 0 0 1 0 0 u4 0u5 0 0 0 0 1 0 u5 0u6 0 1 0 0 0 0 u6 -0.2u7 0 0 0 0 0 1 u7 0

Pick again u6 as slave, put into matrix form:

+



u = T + g g = gap vector

Premultiply both sides by TT K, replace K u = f and pass data to RHS.This gives:

K u = f

with K = TT K T and f = TT (g K g)a modified force vector

^ ^ ^

^ ^



For the example

K11 K12 0 0 0 0K12 K22+K66 K23 0 K56 K670 K23 K33 K34 0 00 0 K34 K44 K45 00 K56 0 K45 K55 K560 K67 0 0 0 K66

=

u1u2u3u4u5u7

f1f2+f6 - 0.2K66

f3f4

f5 - 0.2K56f7 - 0.2K67

Solve for , then recover u = T +g


Model Reduction Example

2 master DOFs to be retained

(2)(1) (3) (4) (5) (6)u1,f1 u2,f2 u3,f3 u4,f4 u5,f5 u6,f6 u7,f7

1 2 3 4 5 6 7x

u1,f1 u7,f7

1x

u1,f1 u7,f7

1 7x

5 slave DOFs to be eliminated

Reduced Model



Lots of slaves, few masters. Only masters are left. Example of perviousslide:

=

u1u2u3u4u5u6u7

1 05/6 1/64/6 2/63/6 3/62/6 4/61/6 5/60 1

u1u75 slaves 2 masters



Applying the congruential transformation we get the reduced stiffnessequation:

u1

u7

K11 K17

K17 K77

^ ^

^ ^

f1

f7

^

^=


Assessment of Master Slave Method

ADVANTAGESexact if precautions takeneasy to understandretains positive definitenessimportant application to model reduction

DISADVANTAGESrequires user decisionsmessy implementation for general MFCshinders sparsity of master stiffness equationssensitive to constraint dependencerestricted to linear constraints


Chapter 10


MultiFreedom Constrains II

under the homogeneous MFC:

(2)(1) (3) (4) (5) (6)u1,f1 u2,f2 u3,f3 u4,f4 u5,f5 u6,f6 u7,f7

1 2 3 4 5 6 7x

u2 = u6 or u2 - u6 = 0

Recall the example structure:


Penalty Function Method

w = the penalty weight assigned to the constraint

u2

u6

1

-1

f2(7)

f6(7)

=

(2)(1) (3) (4) (5) (6)u1,f1 u2,f2 u3,f3 u4,f4 u5,f5 u6,f6 u7,f7

1 2 3 4 5 6 7x

-1

1w

penalty element of axial rigidity w

(7)


Penalty Function Method

Upon merging the penalty element the modified stiffness equation are:

K11 K12 0 0 0 0 0K12 K22+w K23 0 0 -w 00 K23 K33 K34 0 0 00 0 K34 K44 K45 0 00 0 0 K45 K55 K56 00 -w 0 0 K56 K66+w K670 0 0 0 0 K67 K77

=

u1u2u3u4u5u6u7

f1f2f3f4f5 f6f7

This modified system is submitted to the equation solver.Note that u retains the same arrangement of DOFs


But which penalty weight to use?


Penalty Function Method General MFCs

u3u5u6

31-4

=9 3 -123 1 -4-12 -4 16

[ 3 1 -4 ]TPremultiply both sides by Penalty element

stiffness equations

u3u5u6

3 u3 + u5 4 u6 = 1

[ 3 1 -4 ] = 1


Penalty Function Method General MFCs

Scale by w and merge:

K11 K12 0 0 0 0 0K12 K22 K23 0 0 0 00 K23 K33+9w K34 3w -12w 00 0 K34 K44 K45 0 00 0 3w K45 K55+w K56-4w 00 0 -12w 0 K56-4w K66+16w K670 0 0 0 0 K67 K77

=

u1u2u3u4u5u6u7

f1f2

f3+3wf4

f5+wf6-4w

f7


Assessment of Penalty Function Method

ADVANTAGESgeneral applicationeasy to impelment using FE library and standard assemblerno change in vector of unknownsretains positive definitenessinsensitive to constraint dependence

DISADVANTAGESselection of weights left to user big burdenaccuracy limited by ill - conditioning


Langrange Multiplier Method

(2)(1) (3) (4) (5) (6)u1,f1 u2,f2 u3,f3 u4,f4 u5,f5 u6,f6 u7,f7

1 2 3 4 5 6 7x

force pair that enforces MFC

K11 K12 0 0 0 0 0K12 K22 K23 0 0 0 00 K23 K33 K34 0 0 00 0 K34 K44 K45 0 00 0 0 K45 K55 K56 00 0 0 0 K56 K66 K670 0 0 0 0 K67 K77

=

u1u2u3u4u5u6u7

f1f2-f3f4f5

f6+f7



This is now a system of 7 equations and 8 unknowns.Need an extra equation: the MFC

K11 K12 0 0 0 0 0K12 K22 K23 0 0 0 00 K23 K33 K34 0 0 00 0 K34 K44 K45 0 00 0 0 K45 K55 K56 00 0 0 0 K56 K66 K670 0 0 0 0 K67 K77

=

u1u2u3u4u5u6u7

f1f2f3f4f5 f6f7

Because is unknown, it is passed to the LHS and appended to thenode displacement vector:

01000-10



This is the multipier augmentade system. The new coefficient matrix iscalled the bordered stiffness.

Append MCF as additional equation:

=

000100010000000

1000000000000000000010000000000

7

6

5

4

3

2

1

7

6

5

4

3

2

1

7767

676656

565545

454434

343323

232212

1211

fffffff

uuuuuuu

KKKKK

KKKKKK

KKKKKK

KK



Recipe step #1:append the 3 constrains

=

130

04103008000050

010001000000

0000000000000000000000000

7

6

5

4

3

2

1

7

6

5

4

3

2

1

7767

676656

565545

454434

343323

232212

1211

fffffff

uuuuuuu

KKKKK

KKKKKK

KKKKKK

KK

Three MFCs: u2 u6 = 0, 5 u2 8 u7 = 3, 3 u3 +u5 4 u6 = 1


Langrange Multiplier MethodRecipe step #2:append the multipliers, symmetrize and fill

=

130

000041030000080000500000100010080000000010000000000000000003000000051000000000000

7

6

5

4

3

2

1

3

2

1

7

6

5

4

3

2

1

7767

676656

565545

454434

343323

232212

1211

fffffff

uuuuuuu

KKKKK

KKKKKK

KKKKKK

KK


Assessment of Langrange Multiplier Method

ADVANTAGESgeneral applicationexactno user decisions (black box)

DISADVANTAGESdifficult implementationsadditional unknownsloses positive definitenesssensitive to constraint dependence


MFC Application Methods, Assessment Summary

excellentfair

small to noneexcellent

high

noyes

excellentgoodhigh

mediocrenone

yesno

fairpoor to fair

highvariable

high

yesyes

GeneralityEase of implementationSensitivity to user decisionsAccurancySensitivity as regardsConstraint dependenceRetains positive definitenessModifies unknown vector

Langrange Multiplier

Penalty FunctionMaster SlaveElimination


Chapter 11


Superelements and Global-Local Analysis

Two extremes:

Macroelements bottom up

Substructures top down


Substructures

Substructures was invented in the aerospace industry, in the early1960s.


Substructures


Among other things, to take advantage of repetition


Multilevel FEM substructuring was invented in theNorwegian Offshore Industry in the mid/late 60s


Static condensation

A universal way to eliminate internal DOFs

b b b b b

i

i

b

b

b

b

Substructure Macroelement


Multistate Rockets naturally decompose intosubstructures


Static Condensation by Matrix Algebra

=

i

b

i

b

iiib

bibb

ff

uu

KKKK

)(1 bibiiii uKfKu =

~~

bbbbb fuK =

iiibibb

ibiibibbbb

fKKff

KKKKK1

~

1~

=

=

Solve for interior displacement from 2nd matrix equation

Partition

replace into first matrix equation

where

Condensed stiffness equation


Static Condensation by symmetric GaussElimination

=

0463

8413472112523126

4

3

2

1

uuuu

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

( )

( )

( )

=

8404

8106

8303

8447

8412

8431

8142

8115

8132

8341

8312

8336

3

2

1

uuu

=

463

525

25

25

839

819

25

819

839

3

2

1

uuu

Task: eliminate u4

Condensedequation

8 is called thepivot


Static Condensation by symmetric GaussElimination

=

463

525

25

25

839

819

25

819

839

3

2

1

uuu

Task: eliminate u3

( )( ) ( )( )

( )( ) ( )( )

( )

( )

=

52/546

52/543

52/52/5

839

52/52/5

819

52/52/5

819

52/52/5

839

2

1

uu

=

85

829

829

829

829

2

1

uu Condensed

equation


Static Condensation Module Results on Notes Example

=

8413472112523126

K

=

0463

f

=

525

25

25

839

819

25

819

839

K

=

463

f

=

829

829

829

829

K

=

85

f

Upon condensation DOF 4:

Before condensation:

Upon condensation DOF 3:


Global Local analysis (an instance of multiscaleanalysis)

Example structure: panel with small holes


Standard (one stage) FEM analysis

coarse mesh finer mesh


Global Local (two stage) FEM analysis

Global analysis with a coarse mesh, ignoring holes, followed by localanalysis of the vicinity of the holes with finer meshes (next slide)


Local analysis

BCs of displacement or (better) of force type using results extractedfrom the global analysis


Chapter 12


Variational formulation of Bar Element

Bar Member Variational Derivation

y

zCross section

xP

Longitudinal axis


Bar Member

q(x)

x

Laxial rigidity EA

P

cross section

u(x)


The Bar revisited - NotationQuantity Meaningx Longitudinal bar axis *(.) d(.)/dxu(x) Axial displacementq(x) Distributed axial force, given per unit of bar lengthL Total length of bar memberE Elastic modulusA Cross section area, may vary with xEA Axial rigidity = du/dx = u Infinitesimal axial strain = E = E u Axial stressp = A = EA = EA u Internal axial forceP Prescribed end load

*) x is used in this Chapterinstead ofx to simplifythe notation


Tonti diagram of governing equations

Prescribedend

displacement

Axial strains(x)

Axial displacement

u(x)

Axial force p(x)

Destributed

Axial Load

Prescribedend loads

DisplacementBCs


Constitutive

= u p + q = 0

unknown given (problem data)

Force BCs


Potential Energy of the Bar Member

Internal energy (= strain energy)

External Work

Total potential energy

( ) dxuEAudxuEAudxpULL L

'

0

''

0 0

'

21

21

21

===

=L

qudxW0

WU =


Concept of kinematically admissible variation

u(x) is kinematically admissible if u(x) and u(x) + u(x)(i) are continuous over bar length, i.e. u(x) C0 in x [0, L](ii) satisfy exactly displacement BC; in the figure, u(0) = 0

u

xLu(0) = 0

u(x) +u(x)u(x)

u(x) u(L)


The Minimum Potential Energy (MPE) principle

The MPE principle states that the actual displacement solution u*(x)that satisfies the governing equations is that which renders the TPE (Total Potential Energy) functional [u] stationary:

with respect to admissible variations u = u* + u of the exactdisplacement solution u*(x)

0== WU *uu =if


FEM discretization of Bar Member

(2)(1) (3) (4)u1,f1 u2,f2 u3,f3 u4,f4 u5,f5

1 2 3 4 5


FEM displacement trial function

Axial displacement plotted normal to x for visualization convenience

(2)(1) (3) (4) (5)u1,f1 u2,f2 u3,f3 u4,f4 u5,f5

1 2 3 4 5

u

xu1 = 0

u1 u2 u3 u2

End node 1 assumed fixed u(x)


Element shape functions

l = Le

(e)i j

1 0

0 1

N1e

N2e


Total potential energy principle and decomposition over elements

0== WU *uu =if

0...

...)()2()1(

)()2()1(

=+++=

+++=e

e

Ne

N

but

and

from fundamental lemma of variational calculus, eachelement variation must vanish, giving

0== eee WU


Element shape functions


Chapter 13


A Beam is a structural member designed to resistprimarily transverse loads


Transverse load are transported to supports byflexural action

Compressivestress

Tensilestress

Neutral surface


Beam configuration and models

Configuration

Spatial (general beams)Plane (this Chapter)

Models

Bernoulli EulerTimoshenko (advanced topic: described in Chapter but not covered

in course)


Plane Beam Terminology

L

y,v

x,u z

y,vq(x)

Neutral surface

Beam cross section

Symmetryplane

Neutral surface


Common support conditions

Simply supported

Cantilever


Basic relations for Bernoulli Euler model of plane beam

Plus equilibrium equation M = q (not used specifically in FEM)

=

=

=

)()()(

)(

),(),( '

xvy

xvyv

xvxxvy

yxvyxu

EIM

Eydx

vdEyE

ydx

vdyxvy

xu

=

===

==

=

=

2

2

2

2

2

2


Kinematics of Bernoulli Euler beam

y,v

x,u

x

y

Cross section

=dv/dx=v

v(x,y)=v(x,0)

P(x,y)

P(x+u,y+v)

L


Tonti diagram for Bernoulli Euler model of plane beam (strong form)

Prescribedend

displacement

Curature(x)

Transversedisplacement

v(x)

Bending Moment M(x)

Distributedtransverseload q(x)

Prescribedend loads

DisplacementBCs


Constitutive

= v M = q

Force BCsM = EI


Total potential energy of beam member

=

=

===

L

LV

L

xxxx

qvdxW

dxEI

dxxvEIdxMdVU

0

0

2

2

02

2

21

21

21

21

WU =Internal External

Internal energy due to bending

External energy due to transverse load q


Degrees of freedom of plane beam element

2

1

1

2

v1 v2

=

2

2

1

1

v

v

ue


Bernoulli Euler kinematics of plane beamelement

y,v

x,u

x

y

P(x,y)

P(x+u,y+v)

L

1

1

2

v1 v2

2

EI


Shape functions in terms of natural coordinate

[ ]

122

2

1

1

2211

=

=

=

Lx

Nuv

v

NNNNv eeevee

ve

Introduce the natural(isoparametric) coordinate

( ) ( )

( ) ( )

+=

+=

1181)(

2141)(

22

22

LN

N

e

ev( ) ( )

( ) ( )

+=

+=

1181)(

2141)(

21

21

LN

N

e

ev


Shape element function plots

( ) ( )

( ) ( )

( ) ( )

( ) ( )

+=

+=

+=

+=

1181)(

2141)(

1181)(

2141)(

22

22

21

21

LN

N

LN

N

e

ev

e

ev

)(1 evN

)(1 eN

)(2 evN

)(2 eN

=-1 =1

v1=1

v2=1

1=1

2=1


Getting Curvatures from displacementinterpolation

edefee

vee

v

ee

e

Buv

v

dxNd

dxNd

dxNd

dxNd

udx

Nddx

udNudx

Nddx

xvd

=

=

=+==

2

2

1

1

22

2

22

2

21

2

21

2

2

2

2

2

2

2

2

2 )(

dfd

Lddf

dxd

Lddf

dxLd

dxxfd

ddf

Ldxd

ddf

dxxdf

)(4)(2)()/2()(

)(2)()(

2

22

2

=

+=

==

+= 1361361

LLLB

Applying the chain rule

to differentiate the shape function we get

1 x 4 curvature displacementmatrix


Element stiffness and consistent node force

eTeeeTee fuuKu =21

Varying the element TPE

we get

LdqNqdxNf

LdBEIBBdxEIBK

LTTe

LTTe

21

21

0

1

1

0

1

1

==

==


Analytical computation of prismatic beam elementstiffness

( ) ( )( ) ( ) ( )

( )( )

=

++

+

=

2

22

3

1

122

2

2222

22

3

4612

264612612

13.13636

19136131363613636

2

LL

LLLLL

LEI

d

LsymmL

LLLL

LEIK e

(prismatic means constant EI)


Mathematica Script for symbolic computation of prismatic plane beam element stiffness

ClearAll [EI,l,];

B={{6*, (3*-1)*l, -6*, (3*+1)*l}}/l^2;

Ke=(EI*l/2)*Integrate [Transpose [B].B,{,-1,1}];

Ke=Simplify[Ke];

Print[Ke for prismatic beam:];

Print[Ke//MatrixForm];

Ke for prismatic beam:

Corroborates the results from hand integration

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

4626

612612

2646

612612

22

2323

22

2323


Analytical computation of consistent node force vector for uniform load q

( ) ( )

( ) ( )

( ) ( )

( ) ( )

=

+

+

+

+

==

L

LqL

d

L

LqLNdqLf e

12121

12121

1

1

1

1

2

2

2

2

1181

2141

1181

2141

21

21

fixed end moments


Mathematica Script for symbolic computation of consistent node force vector for uniform load q

ClearAll [q,l,];

Ne={{2*(1-)^2*(2+), (1-)^2*(1+)*l,

2*(q-)^2*(2-),-(1+)^2*(1-)*l}}/8;

fe=(q*l/2)*Integrate [Ne,{,-1,1}]; fe=Simplify [fe];

Print[fe^T for uniform load q:];

Print[fe//MatrixForm];

fe^T for uniform load:

Force vector printed as row vector to save space

122122

22 qqqq llll


14The Plane Stress

Problem


Plate in Plane Stress

Inplane dimensions: in x,y plane

Thi

ckne

ssdi

men

sion

ortr

ansv

erse

dim

ensi

on


Mathematical Idealization asa Two Dimensional Problem


Plane Stress Physical Assumptions

Plate is flat and has a symmetry plane (the midplane)

All loads and support conditions are midplane symmetric

Thickness dimension is much smaller than inplane dimensions

Inplane displacements, strains and stresses uniform throughthickness

Transverse stresses and negligible, set to 0

Unessential but used in this course:

Plate fabricated of homogeneous material through thickness

xzzz , yz


Notation for stresses, strains, forces, displacements


Inplane Forces are Obtained byStress Integration Through Thickness


Plane Stress Boundary Conditions


The Plane Stress ProblemGiven:

geometry

material properties

wall fabrication (thickness only for homogenous plates)

applied body forces

boundary conditions:

prescribed boundary forces or tractions

prescribed displacements

Find

inplane displacements

inplane strains

inplane stresses and / or internal forces


Matrix Notation for Internal Fields


Governing Plane Stress ElasticityEquations in Matrix Form


Strong-Form Tonti Diagram ofPlane Stress Governing Equations


TPE-Based Weak Form Diagram ofPlane Stress Governing Equations


Total Potential Energy of Platein Plane Stress


Discretization into Plane StressFinite Elements


Plane Stress Element Geometriesand Node Configurations


Total Potential Energy of Plane Stress Element


Constructing a Displacement Assumed Element


Element Construction (contd)

Differentiate the displacement interpolation wrt x,y

to get the strain-displacement relation


Element Construction (contd)Element total potential energy

Element stiffness matrix

Consistent node force vector


Requirements on Finite Element Shape Functions

Interpolation ConditionNi takes an value 1 at node i, 0 at all other nodes

Continuity (intra- and inter-element)and Completeness Conditions

are covered later in the course (Chs. 18-19)


MATLAB - BASICS


Matrizen und Magische Quadrate

Geben Drers Matrix A als eine Liste von Elementen. Sie mssen nur wenige Grundregeln beachten:

Die Elemente werden durch Leerzeichen oder Kommas getrennt Ein Strichpunkt markiert das Ende einer Zeile Die gesamte Matrix wird von eckigen Klammern [ ] eingeschlossen

Um die Matrix von Drer einzugeben, tippen Sie einfach:

A=[16 3 2 13;5 10 11 8;9 6 7 12;4 15 14 1]

MATLAB zeigt sofort die eingegebene Matrix an:


Summe, Transposition und Spur

Vielleicht haben Sie schon bemerkt, dass die speziellen Eigenschaften des Magischen Quadrats mit der Summierung der Elemente in verschiedenen Richtungen zusammenhngt. Bilden Sie die Summe jeder Zeile oder Spalte. Sie werden immer die selbe Zahl erhalten. berprfen Sie dies mit MATLAB. Der erste Befehl ist:

erzeugt Spaltensumme

sum(A)

erzeugt Zeilensumme da Transponierte von A verwendet

sum(A)


Summe der Diagonalen einer Matrix

Die Summe der Elemente der Hauptdiagonalen, die Spur der Matrix A, wird einfach mit Hilfe der diag-Funktion ermittelt

sum(diag(A))

Die Gegendiagonale ist mathematisch nicht so wichtig, deshalb gibt es keine MATLAB-Funktion zu ihrer Summierung. Die Funktion fliplr, die eigentlich fr Graphik-Anwendungen gedacht war, hilft hier jedoch weiter. Sie spiegelt die Matrix an ihrer Mittelachse. Dadurch wird die Gegendia-gonale zur Hauptdiagonalen und ihre Spur ist

sum(diag(fliplr(A)))

Matrix im Holzschnitt von Drer ist wirklich ein magisches Quadrat ist


Indizes

Das Element in Zeile i und der Spalte j der Matrix A wird mit A(i, j) bezeichnet.

Die Summe einer Spalte der Matrix A kann somit auch wie folgt modelliert werden:

A(1,4) + A(2,4) +A (3,4) + A(4,4)

Es ist auerdem mglich, Elemente einer Matrix mit einem einfachen Index anzusprechen, etwa A(k).

In diesem Fall wird das Zahlenfeld A als ein langer Spaltenvektor betrachtet

A(8) welcher Eintrag etspricht dies in der Matrix?


INDEX

Falls Sie einen Wert auerhalb der Matrix aufrufen wollen, gibt MATLAB einen Fehler aus:

t=A(4,5)

Man kann jedoch einen Wert auerhalb der Feldgrenzen speichern. Die Gre der Matrix wird fr das neue Element passend erweitert:


Der Doppelpunkt-Operator

Der Ausdruck: 1:10ist ein Zeilenvektor, der die Zahlen eins bis zehn der Reihe nach enthlt:

Um andere Abstnde zu erhalten, geben Sie eine beliebige, evtl. auch negative Schrittweite an. Z.B ergibt

100:-7:50

0:pi/4:pi


Der Doppelpunkt-Operator

Indexausdrcke, die den Doppelpunktoperator enthalten, beziehen sich auf Ausschnitte von Mat-rizen, sog. Teilmatrizen.

A(1:3,2)

? sum(A(1:4,4))

Der alleinstehende Doppelpunkt bezieht sich auf alle Matrixelemente in einer Zeile oder Spalte.

Das Schlsselwort end bezieht sich auf die letzte Zeile oder Spalte.

Magische Matrizesum(A(1:16))/4


Die magicFunktion

Magische Quadrate nahezu beliebiger Gre

B=magic(4)

Drers Matrix - Spalten vertauschenA=B(:,[1,3,2,4])

Diese Anweisung hat bewirkt, dass in jeder Zeile von Matrix B die Elemente in der Reihenfolge 1, 3, 2, 4 angeordnet wurden.


bungsaufgaben

1. Erzeugen Sie ein magisches Quadrat der Gre 6x6 und weisen Sie alle Eigenschaften von magischen Quadraten nach.

2. Erzeugen Sie mit der Funktion rand eine 5x5-Zufallsmatrix A. Welches sind die Werte der folgenden Ausdrcke?

a) A(2,:) b) A(:,5) c) A(:,1) d) A([1,5]) e) A(1,1:2:5) f) A(4:-1:1,5:-1:1)


Operatoren

In Ausdrcken gelten die gelufigen arithmetischen Operatoren und Priorittsregeln (Punkt- vor Strichrechnung usw.), die durch Klammern abgendert werden knnen. Arithmetische Operatoren sind

+ Addition - Subtraktion * Multiplikation / Division \ Linke Division ^ Potenzierung ' Transposition (mit Konjugation) () Klammern zur Festlegung der Auswertungsreihenfolge


Funktionen

Um sich eine Liste aller elementaren mathematischen Funktionen ausgeben zu lassen, gibt man

help elfunein. Fr eine Liste der speziellen Funktionen gibt man help specfunein. Das Repertoire an elementaren Matrizen zeigt help elmatan.


Arbeiten mit Matrizen

zeros Alle Elemente sind Nullen (Nullmatrix) ones Alle Elemente sind Einsen rand Gleichverteilte Zufallszahlen aus [0,1[ randn normalverteilte Zufallszahlen (Mittelwert 0 und Varianz 1)

Z=zeros(4,2)


Zusammensetzen von Matrizen

Der Verkettungsoperator ist das eckige Klammerpaar [ ]. Fangen Sie mit dem magischen Quadrat A aus Kapitel 1 an und erstellen Sie

B=[A A+32; A+48 A+16]

Das Resultat ist eine 8x8 Matrix, die Sie erhalten, wenn Sie die 4 Untermatrizen aneinanderhn-gen.


Lschen von Zeilen und Spalten

Sie knnen Zeilen und Spalten einer Matrix lschen, indem Sie diesen einfach ein leeres Paar eckiger Klammern zuweisen. Fangen Sie anmit:

X=A;

Lschen Sie anschlieend die 2. Spalte von X in dem Sie

X(:,2)=[ ]

schreiben.


Der format - Befehl

Der format-Befehl kontrolliert das numerische Format der durch MATLAB dargestellten Zahlen.

format short 1.33330format short e 1.3333e+000 1.2345e-006format short g 1.33331.2345e-006format long 1.333333333333330.00000123450000format long e 1.33333333333333e+0001.23450000000000e-006format long g 1.333333333333331.2345e-006format bank 1.330.00format rat 4/31/810045format hex 3ff55555555555553eb4b6231abfd271format compactDieser Befehl unterdrckt alle Leerzeilen, die normalerweise zur bersichtlichkeit dienen sollen.

Auf diese Weise ist es mglich, mehr Informationen im Fenster zu sehen. Mit dem Befehl format +


Eine Matrix ist ein zweidimensionales Feld, das eine lineare Transfor

875.317 computational mechanics - baunat.boku.ac.at · transcend the classical boundaries of solid...

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