Finite Element Modelling - HZG...WMSReport0003.doc, Brocks, 27.08.02 - - 6 Kinematics Body A body B...

Technical Note GKSS/WMS/00/03interner Bericht

Finite Element Modelling

Lecture NotesFaculty of Engineering Kiel

W. Brocks

Februar 2000

Institut für Werkstofforschung

GKSS-Forschungszentrum Geesthacht

WMSReport0003.doc, Brocks, 27.08.02 - - 2

Inhalte Wiederholung von Grundbegriffen der Festigkeitslehre und Einführung in dieKontinuumsmechanik fester Körper: Spannungs- und Verzerrungstensor,Gleichgewichtsbedingungen, elastisches Stoffgesetz,Formulierung des Festigkeitsproblems als Randwertaufgabe und alsVariationsaufgabe (Prinzip der virtuellen Arbeiten)Einführung in die Methode der finiten Elemente (FEM): Diskretisierung desKontinuums, Verschiebungsansätze in den Elementen, (elastisches) Stoffgesetz,Steifigkeitsmatrix und Lastvektoren, Lösung des GleichungssystemsAufbau eines FE-Programmes: erforderliche Eingaben, Programmablauf,ErgebnisdatenEinführung in die Anwendung des FE-Programmsystems ANSYS mit Übungenam Rechner: Erstellung von FE-Netzen und sonstigen Eingabedaten,Durchführung von FE-Rechnungen, Ergebnisdarstellung undErgebnisauswertung.Selbständige Bearbeitung einfacher Probleme: Stab unter Zugbelastung,Biegebelastung einer eingespannten Scheibe mit Rechteckquerschnitt, gelochteScheibe unter zweiachsialem Zug.

empfohlene Literatur (deutsch):

J. ALTENBACH und H. ALTENBACH: "Einführung in die Kontinuumsmechanik",Teubner Studienbücher Mechanik, Stuttgart 1994.K. KNOTHE, H. WESSELS: "Finite Elemente, Einführung für Ingenieure", Springer-Verlag, Berlin, 1991.K.J. BATHE: "Finite-Elemente-Methoden", Springer-Verlag, Berlin, 1986.O.C. ZIENKIEWICZ: "Methode der Finiten Elemente", Carl-Hanser-Verlag,München, 1984.

========================================================Contents repetition of fundamentals of strength of materials and introduction into

continuum mechanics of solids: stress and strain tensors, balance equantions,constitutive equations of linear elasticity,formulation as boundary value problem and variational problem (principle ofvirtual work)introduction into the finite element method (FEM): discretization of thecontinuum, shape and displacement functions, isoparametric elements, (elastic)constitutive equations, local and global stiffnes matrices and load vectors, solutionof equationsgeneral structure of a FE programme: input data, flow diagram of problemsolution, resultsapplication of the FE programme ANSYS with computer exercises: meshes andother input data, FE simulations, presentation, evaluation and interpretation ofresults,individual working on case and parameter studies: tensile bars, bending of aclamped rectangular panel, panel with hole under biaxial tension

recommended literature (english):

W.M. LAI, D. RUBIN, and E. KREMPL: "Introduction to Continuum Mechanics",Pergamon Press, 1993.M.E. GURTIN: "An introduction to Continuum Mechanics", Academic Press, NewYork, 1981.K.J. BATHE, "Finite Element Procedures", Prentice Hall, New Jersey 07632, 1996.O.C. ZIENKIEWICZ: "The Finite Elemente Method", McGraw-Hill Book Comp.,Maidenhead (UK), 1977.


Contents

Introduction to continuum mechanics

• kinematics

• forces and stresses

• balance equations

• constitutive equations, HOOKE's law of elasticity

Variational principles

• variational calculus

• principle of virtual work

FE modelling and computational procedure

Shape functions:

• orthogonal elements

• general isoparametric elements

Nonlinear FE Analyses

• geometrical nonlinearity

• material nonlinearity

Examples

• plane panel under bending load

• plane panel with hole under biaxial tension

Appendix

• Notation

• Physical units (SI units)


Continuum Mechanics

Continuum mechanics is a

phenomenological field theory.

Basing on observed phenomena, mathematical modelsfor the mechanical behaviour of matter are formulated.

As everybody knows, the behaviour of matter isdetermined by interactions of atoms and molecules.However, an engineering modeling cannot be done onthis level.

The discretely structured matter is hence represented by

a phenomenological model, i.e. the continuum;

this is done by averaging its properties in space.

For describing the mechanical behaviour ofheterogeneous materials with strong local gradients ofthe microstructure a phenomenological theory is notsufficient, in general. For this purpose,

micromechanical models on a "meso level"between a microstructural and a phenomenologicalmodeling are developed and applied.


Equations of Continuum Mechanics

1. material independent principles

kinematics• body• configuration• motion• deformation

kinetics• external actions: forces• internal reactions: stresses

balance equations (conservation laws)• matter• momentum• moment of momentum• energy / work / power• entropy (2nd principle of thermodynamics)

2. material dependent equations

constitutive equations: solids, fluids, gasesreversible, irreversible processes

The combination of all principles leads to theformulation of initial boundary value problems.


KinematicsBody

A body B is a three-dimensional differentiablemanifold, the elements of which are called particlesx ∈ B.

A body B is endowed with a non-negative scalarmeasure, m, which is called the mass of the body.

ConfigurationA configuration χχχχ of a body B is a smooth

homeomorphism of B onto a region, BB ⊂ EE3, of

three-dimensional EUKLID ean space EE3, x = χχχχ(x),

called the region occupied by the body B in theconfiguration χχχχ.

MotionA motion of a body B is a one-parameter family tχχχχof configurations, x = tχχχχ(x) = χχχχ(x,t). The realparameter t is the time.

Deformation

The "local" deformation results from observing aninfinitesimal vicinity of a particle in its presentconfiguration with respect to a referenceconfiguration.


Deformation

Any general motion of a body includes a deformationwhich can be described as follows. A particle, x, which isidentified by its place

0 0x = ( )χ x in a referenceconfiguration at time t=0, occupies the place

t t t tt tx x x x x u= ( ) = ( ) = ( ) = ( ) = +−χ χ χ φ φx 0 1 0 0 0 0

0( ), ,

at the time t.The mapping tφφφφ is called deformation,

and 0t u is the displacement of the particle.

Observing an infinitesimal vicinity of the particle in thecourse of motion,

t t t tt

d d dx x x x xx

x+ = +( ) ≈ ( ) + ⋅φ φ φ0 0 00

0∂∂

leads to the definition of the deformation gradient,

0 00

0 0t

tt tF

xu I H I= = ∇ + = +∂

∂φ

Various tensor-valued measures of the local deformation,commonly addressed as strains, can be derived from

0t F.


Strain Tensors polar decomposition

0 0 0 0 0t t t t tF R U V R= ⋅ = ⋅

with 0 0t t TR R I⋅ =

right CAUCHY-GREEN Tensor

0 0 0 02t t T t t T TC F F U I H H H H= ⋅ = = + + + ⋅

• GREEN's quadratic strain tensor

012 0

2

12

00

00

00

00

012

00

00

t t

t t T t t T

t t t T

E U I

u u u u

E u u

( )G = −( )= ∇ + ∇( ) + ∇( ) ⋅ ∇( )[ ]= + ∇( ) ⋅ ∇( )0t E strain tensor for small deformations

• HENCKY's logarithmic ("true") strain tensor

0 0t H tE U( ) ln= ( )

tensor of deformation ratest t t t t TD v v= ∇ + ∇( )[ ]1

2

t t T t tD F E F= ⋅ ⋅− −0 0

1˙ ( )G


Forces and Stresses

The concept describes the action of the outside worldon the body B in motion and the interaction betweenthe different parts P of the body.

(a) external body force

F b xb t dV( ) ( , ) ,P P B

P

= ⊂∫ ρ

over the part P of B in the configuration χχχχ.

(b) contact force

F t xc dS( ) ( ; )P P

P

= ∫∂

extended over the boundary ∂P of P .

(c) total resultant force F F F( ) ( ) ( )P P P= +b c

(d) stress principle t x t x n( ; ) ( , )P =

where n is the exteriour unit normal vector at thepoint x on the boundary ∂P of PThis implies the existence of a stress-tensor fieldS(x) such that the stress vector t(x,n) may beexpressed by

t x n n S x( , ) ( )= ⋅


Principal Stresses and Invariants

"eigenvalue" problem S n n⋅ = σcondition det S I−( ) =σ 0

characteristic equation σ σ σ31

22 3 0− − − =J J J

invariants of stress tensor

J

J

J

ii

ij ij ii jj

ij jk ki

1

212

2 2 12

313

( ) tr( )

( ) tr( ) tr ( )

( ) det( )

S S

S S S

S S

= =

= −[ ] = −( )= =

σ

σ σ σ σ

σ σ σ

three real solutions ("eigenvalues") = principal stresses

σ σ σI II III≥ ≥with corresponding principal directions n n nI II III, ,

S n n⋅ = =i i i i I II IIIσ ( , , )

S n n=

σσ

σ

I

II

III

i j

0 0

0 0

0 0

invariants in terms of principal stresses

J

J

J

I II III

I II II III I III

I II III

1

2

3

( )

( )

( )

S

S

S

= + +

= − + +( )=

σ σ σ

σ σ σ σ σ σσ σ σ


stress deviator ′ = −S S I13 σ ii

hydrostatic stress σ σh ii= 13

invariants of stress deviator

J

J

J

ij ij

ij jk ki ij jk ki

1

212

16 11 22

222 33

233 11

2

122

232

132

313

13

0′( ) =

′( ) = ′ ′

= −( ) + −( ) + −( ) +[+ + + ]

′( ) = ′ ′ ′ =

S

S

S

σ σ

σ σ σ σ σ σ

σ σ σ

σ σ σ σ σ σ

VON M ISES equivalent (effective) stress

σ e J= ′3 2 ( )S

VON M ISES yield condition

σ e R≤ 0

upper limit of purely elastic deformation

beginning plastic deformation

R0 = yield strength


plane stress

S e e=

σ σσ σ

xx xy

xy yy i j

0

0

0 0 0

principal stresses

σσ

σ σ σ σ σI

IIxx yy xx yy xy

= +( ) ± −( ) +12

214

2 2

direction of first (maximum) principal stress

tan 22

0ϕσ

σ σ=

−xy

yy xx

maximum shear stress

τ σ σ σ σ σmax = −( ) + = −( )14

2 2 12xx yy xy I II

direction of maximum shear stress

ϕ ϕ π1 0 4

= ±

VON M ISES equivalent (effective) stress

σ σ σ σ σ σ

σ σ σ σ

e xx xx xx yy xy

I II I II

= + − +

= + −

2 2 2

2 2

3


Balance Equations

conservation of mass

global: any finite part P of B

˙ ( )m

ddt

dVPP

= =∫ ρ 0

local (equation of continuity)˙ div ˙ρ ρ+ =x 0

balance of momentum

global

F x( ) ˙P

P

= ∫ddt

dVρ

local (CAUCHY's law of motion)div ˙S b x+ =ρ ρ

balance of moment of momentumfor non-polar media

global

x b x t x x× + × = ×∫∫ ∫ρ ρ

∂

dV dSddt

dVPP P

˙

local (symmetry of stress tensor)S S= T


General Principlesgoverning the mechanical behaviour of materials

1. material frame-indifference

Constitutive equations must be invariant underchanges of frame reference. If a constitutiveequation is satisfied for a process with a motion anda symmetric stress tensor given by

x S S= χ( , ), = ( , )xt txthen it must be satisfied also for the motion andstress tensor given by

˜ ˜ ˜ ( ) ( )

˜ ˜ ˜ ) )

˜

x c Q

S S Q S Q

= + ⋅

⋅ ⋅= −

χ( , ) = χ( , ) ,

= ( , ) = ( ( , ) (

x x

x x

t t t t

t t t t

t t

T

τ

2. determinism

The stress in a body is determined by the history ofthe motion of that body.

3. local action

In determining the stress at a given particle x, the

motion outside an arbitrary neighbourhood of x maybe disregarded.


Constitutive Equations

principle of determinism:

S( , ) ( , ),x xyt

t

= =−∞f

ττχ x, y ∈ B

transformation of time: τ = −t s

S( , ) ( , ),x xyt t s

s

= − =

∞f

0

χ

principle of material frame-indifference:

Q Q Q⋅ ⋅ = ⋅ f fT χ

S( , ) ( , ) ( , )x xyt t s t s

s

= − − − =

∞f

0

χ χ

principle of local action:

Ω : ; ( (), )y x x y− ≤ = =δ χ χx y

S F F( , ) ( , ( ,...) )x x xt

s

t s t s t s= ∇( ) =

∞

−∞− −

−∞−f

0

"simple materials" S F( , ) ( )x xt

s

t s= =

∞

−∞−f

0

principle of fading memory:

the memory of a simple material fades in time


Hooke's Law of Elasticity

general: S C E= ⋅⋅< >4

, σ εij ijkl klC=

E C S=

⋅⋅

< > −4 1

, ε σij ijkl klC= −1

C< >4

stiffness tensor (4th order)

C< > −

4 1

compliance tensor (4th order)

isotropic: C G K Gijkl ik jl ij kl= + −( )2 23δ δ δ δ

or

S E E I= +−

21 2

Gν

ν(tr )

σ ε νν

ε δij ij kk ijG= +−

21 2

( )

E S S I= −+

12 1G

νν

(tr )

ε σ νν

σ δij ij kk ijG= −

+

12 1

( )

G = shear modulusK = bulk modulusν = POISSON's ratio


Material Parameters for Linear ElasticityElastizitätskonstanten

λ = µ = E = ν = K = G =

λ, µ λ µ µ λ µλ µ3 2+( )

+λ

λ µ2 +( )λ µ+ 2

3µ

G, K K G− 23

G 33

K G

K G

⋅+

3 26 2

K G

K G

−+

K G

E, ν Eνν υ( )( )1 1 2+ −

E

2 1( )+ νE ν E

3 1 2( )− νE

2 1( )+ ν

λ, µ LAMÉ's coefficients LAMÉsche KonstantenG shear modulus SchubmodulK bulk modulus KompressionsmodulE YOUNG's modulus Elastizitätsmodulν POISSON's ratio Querkontraktionszahl


Variational Principles

variational principles in mechanics

Ø replace the (differential) equations of motion

or equilibrium, as there are

• balance of momentum,

• balance of angular momentum,

• CAUCHY's field equations;

Ø are extremum principles for energy type quantities,

like

• work,

• kinetic energy,

• potential energy.

Differential equations of motion can be established by

methods of variational calculus.


Variational Calculusproblem: find a set of functions xi(t), i = 1, ..., n

for which the integral

I F t x x x x dtn n

t

t

= ( )∫ , ,..., , ˙ ,..., ˙1 1

0

1

becomes an extremum under given "boundary" conditionsx t x x t xi i i i( ) ; ( )0

01

1= = (b.c.)

definition of "varied" functions

x t x t t t ti i i i i( ) ( ) ( ) ( ) ( )= + = =εξ ξ ξwith 0 1 0

ξi t( ) are arbitrary, differentiable functionsmeeting the b.c.

⇒ I F t x x dtt

t

( ) , ,..., ˙ ˙ ,...ε εξ εξ= + +( )∫ 1 1 1 1

0

1

the condition for I becoming an extremum

is ∂∂ε ε

I

=

=0

0

δ ξ δ ξx t x ti i i i= =( ) ; ˙ ˙ ( ) are variations of xi ; xi

δ ∂∂ε ε

II=

=0

is the (first) variation of I

δI = 0 is the variational problem


the variational problem leads to

δ ∂∂ε

∂∂

ξ ∂∂

ξε

II F

xFx

dti

ii

i

t

t

=

= +

==

∫0

0

1

0˙

˙

partial integration yields

˙˙ ˙ ˙

ξ ∂∂

ξ ∂∂

ξ ∂∂i

it

t

ii t

t

iit

tFx

dtFx

ddt

Fx

dt

=

−

∫ ∫

0

1

0

1

0

1

where [..] = 0 due to b.c.

δ ξ ∂∂

∂∂

IFx

ddt

Fx

dtii it

t

= −

=∫ ˙0

1

0

and as ξi t( ) are arbitrary ("test functions")

⇒ EULER's differential equationof the variational problem

∂∂

∂∂

Fx

ddt

Fxi i

− =˙

0


Variational Problem in Mechanics

u x x= −( ) ( )t t0 displacement vektorw arbitrary "virtual" displacement,

• independent of t• w( )t0 0= in the reference configuration

Z energy functional

δ εε

∂ ε∂εε ε

ZZ Z Z= + −

= +

→ =lim

( ) ( ) ( )0

0

u w u u w

variation of Z at u in w direction(GÂTEAUX derivative)

δ δ ∂ ε∂ε ε

u xu w

w= = + ==

( )

0

virtual displacement

(1) δ α αδZ Z( , ) ( , )u w u w=(2) δ δ δZ Z Z( , ) ( , ) ( , )u w w u w u w1 2 1 2+ = +

example: Z = ⋅ − ⋅12 u u b u

δε

ε ε εε

Z = + ⋅ + − ⋅ + − ⋅ + ⋅[ ]→

lim ( ) ( ) ( )0

12

12

1u w u w b u w u u b u

= ⋅ + ⋅ − ⋅[ ]→

limε

ε0

12u w w w b w

= − ⋅( )u b uδ δZ is linear in δu


Principle of Virtual WorkCAUCHY's field equations of motion

∂σ∂

ρ ρij

ij jx

b u+ = ˙ , j = 1, 2, 3

multiply by virtual displacement δuj and integrate over

the volume V:

∂σ∂

δ ρ δ ρ δij

ij

V

j j

V

j j

Vx

u dV b u dV u u dV∫ ∫ ∫+ = ˙

∂σ∂

δ ∂∂

σ δ σ∂ δ

∂ij

ij

V iij j

V

ijj

iVx

u dVx

u dVu

xdV∫ ∫ ∫= ( ) −

( )

GAUß' theorem

∂∂

σ δ σ δ δ∂ ∂x

u dV n u dA t u dAi

ij j

V

i ij j

V

j j

V

( ) = =∫ ∫ ∫

σ∂ δ

∂σ δ

∂∂

σ δεijj

iij

j

iVV

ij ij

V

u

xdV

u

xdV dV

( )= =∫∫ ∫

ρ δ ρ δ δ ρ˙ ˙ ˙ ˙u u dVddt

u u dV u u dVj j

V

j j

V

j j

V∫ ∫ ∫= − 1

2

δ δ δ δA W P E− = −


virtual work of external forces

δ δ ρ δ∂

A t u dA b u dVj j

V

j j

V

= +∫ ∫virtual work of stresses (virtual strain energy)

δ σ δεW dVij ij

V

= ∫virtual power of momentum

δ ρ δPddt

u u dVj j

V

= ∫ ˙

virtual kinetic energy

δ δ ρE u u dVj j

V

= ∫ 12 ˙ ˙

virtual work of mass acceleration

δ δ δ ρ δB P E u u dVj j

V

= − = ∫ ˙


δ A W P E− − +( ) = 0

The variation of the energy functional

(A - W - P + E) vanishes.

or

The energy functional (A - W - P + E) becomes an

extremum (minimum) among all admissible

states (virtual displacements).

special cases:

• rigid body δW = 0

• elastic body δ ε δεW C dVijkl kl ij

V

= ∫

W C dVijkl ij kl

V

= ∫12 ε ε elastic strain energy

• static problem (equilibrium)

δ δ δB P E= − = 0


Example: plane panel

LH

p

thickness = 1

( , )

( , ),

tboundary

=

=−

≤ ≤ =

t x y

t x y px

y x L y H

0

0

displacement field ( , )

( , )u =

u x y

u x yx

y

boundary conditions (b.c.) u y u yx y( , ) ( , )0 0 0= =

dimensionless coordinates ξ η= =xL

yH

;

global shape functions, ϕ ξ ηi ( , ), fulfilling b. c.

ϕ ξ ϕ ξ ϕ ξηϕ ξ ϕ ξ η ϕ ξη

1 22

3

43

52

62

= = == = =

, ,

, ,

Φ =

ϕ ϕ ϕ ϕ ϕ ϕϕ ϕ ϕ ϕ ϕ ϕ

1 2 3 4 5 6

1 2 3 4 5 6

0 0 0 0 0 0

0 0 0 0 0 0


series expansion of displacement field

u =

==

+=

∑

∑

α ϕ

α ϕα

i ii

i ii

1

6

61

6 Φ with .

.α

α

α

=

1

12

α is (12×1) matrix of unknowns

strain matrix

εεεγ

α α

∂∂∂∂

∂∂

∂∂

=

=+

= = =

xx

yy

xy

uxuy

uy

ux

x

y

x y

D u D BΦdifferential operator

D =

=

∂∂

∂∂

∂∂

∂∂

∂∂ξ

∂∂η

∂∂η

∂∂ξ

x

y

y x

L

H

H L

0

0

0

0

B =

1 2 3 2

2

2 1 2 3 2

2 2

2

2 2 2

0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0

L L L L L L

H H H

H H H L L L L L L

ξ η ξ ξη η

ξ ξ ξη

ξ ξ ξη ξ η ξ ξη η


stress matrix and HOOKE's law

σσσσ

ε=

=xx

yy

xy

C

C stiffness matrix

for plane stress:σ zz = 0

C C=−

−( )

=E T

1

1 0

1 0

0 0 12

12

ν

νν

ν

for plane strain: ε zz = 0

( )( )

C C=+ −

−−

−( )

=E T

1 1 2

1 0

1 0

0 0 1 212

ν ν

ν νν ν

ν



δ δ ξ δ αηξ

A LdT T= [ ] ===∫ t u f1

0

1

virtual work of stresses

δ σ δ ε ξ η α δ αηξ

W Ld HdT T= ===∫∫

0

1

0

1

K

"generalized" forces

f t( ) ( ) t ( )T T TLd L d

pL

= =

= − ( )

==

==

∫ ∫ξ ξ ξ ξ ξηξ

ηξ

Φ Φ1

0

1

1

0

1

12

13

12

14

13

120 0 0 0 0 0

global stiffness matrix

K B C B K= ===∫∫LH d dT T

ηξ

ξ η0

1

0

1

principle of virtual work

δ δ α δ αA W T T− = −( ) f K

as δ α is arbitrary

K fα =

linear system of equations for unknowns αi


FE Modelling and Computational Procedure

Finite element methods (FEM) base on variational principles for minimizing some potential like,

e.g., the potential energy of a mechanical systems.

Variational methods replace the solution of the corresponding boundary value problem.

For the so-called "deformation methods", FEM is based upon the

Principle of Virtual Work

δ δ ρ δ δ

∂

Π = − ⋅⋅ ( ) + ⋅ + ⋅ =∫ ∫ ∫S u b u t ugradB B B

dV dV dA 0

• stress-tensor field T(x),

• external body forces, b(x), defined for any x ∈ B,

• contact forces or tractions t(x), defined for any x on the boundary ∂B.

• The continuous body B is separated by imaginary lines or surfaces into a number, K, of

(finite) elements Bk.

The union

B = B

=k

k

K

1U

is the finite model of the body.

• The elements are assumed to be interconnected at a discrete number, N, of nodal points, xn,

situated on their boundaries.

The displacements of these nodal points

u u xn n n N= =( ), ,...1

are the basic unknown parameters of the problem.

• A set of functions, ϕ ik( )( )ξ , i = 1, ...Nk, ξξξξ being local coordinates, is chosen to


define uniquely the state of displacement, ˜ ( )( )u k ξ , within each element, k, in terms of its nodal

displacements, uik( ) ,

˜ ( ) ( )( ) ( ) ( )u uki

kik

i

Nk

ξ ξ==∑ϕ

1

with u uik k

ik

ik

jk

ij( ) ( ) ( ) ( ) ( )˜ ,= ( ) ( ) =ξ ξϕ δ .

• The displacement functions uniquely define the state of deformation, i.e. some strain tensor,˜ ( )( )E k ξ , within each element in terms of the nodal displacements, e.g., small strain

˜ ( )( )( )

( )( )

( )Ex

ux

uk ik

ik i

k

ik

T

i

Nk

ξξ

ξξ

ξ= ⋅ ⋅ + ⋅ ⋅

=

∑12 1

∂ϕ∂

∂∂

∂ϕ∂

∂∂

,

with ∂∂

ξx

= JACOBIAN matrix.

• These strains, together with the constitutive properties of the material, determine the state of

stress, ( )( )S k ξ , throughout the element and also on its boundaries.

• A system of forces, tik( ) concentrated at the nodes equilibrating the boundary stresses,

˜ ˜( ) ( ) ( )t n Sk k k= ⋅ , is determined, resulting in a force-displacement or "stiffness" relationship for

each element.

t C uik

ijk

i

N

jk

k( ) ( ) ( )=

=∑

1

.

• Nodal displacements, uik( ) , nodal forces, ti

k( ) , and element stiffnesses, Cijk( ) , are

assembled according to the conditions of connectivity, u uik

ink

nn

N

A( ) ( )==

∑1

,

for all elements to compose the system of equations ensuring the conditions of compatibility

and equilibrium throughout.

˜ ( ) ˜ ( ) ( )( )u x u x x u= == =

∑ ∑k

k

K

n nn

N

1 1

ψ

where

˜ ( )( )( ) ( )

u x u xk ik

ik

i

N

k

k

= ∈

=∑ϕ

1

0

B

else, ψ ϕn i

kin

k

i

N

k

K

Ak

( ) ( ) ( )x ===∑∑

11

• Any system of nodal displacements listed for the whole structure in which all the elements

participate, automatically satisfies the condition of compatibility.


As the equilibrium condition has already been satisfied within each element all that is

necessary is to establish equilibrium at the nodes of the structure. This is done by the

principle of virtual work.

• The resulting equations governing the mechanical behaviour of the entire structure contain the

nodal displacements as unknowns.

s f p 0n n n n N+ + = =, ,...,1

s S

xf b p tn

nn n n ndV dV dA= − ⋅ = =∫ ∫ ∫∂ψ

∂ρ ψ ψ

∂B B B

; ;

A solution of this system of equations provides an approximate solution of the fields of

displacements, strains and stresses throughout the domain of the body.


Shape Functionsfor Orthogonal Elements

x

yu

x

y

a/2 ξ

b/2 η

1 2

34

bk

ak

xk

yk

element (k):

global coordinatesx a x x a

y b y y bk k k k

k k k k

− ≤ ≤ +− ≤ ≤ +

12

12

12

12

local coordinates: − ≤ ≤ +1 1ξ η,

ξ η= −( ) = −( )2 2a

x xb

y yk

kk

k,


4-Node Elements

"linear" plane elements: Nk = 4

shape functionsϕ ξ η δi j j ij( , ) = i, j = 1, ..., 4

ϕ ξ η ξ η

ϕ ξ η ξ η

ϕ ξ η ξ η

ϕ ξ η ξ η

114

214

314

414

1 1

1 1

1 1

1 1

( , )

( , )

( , )

( , )

= −( ) −( )= +( ) −( )= +( ) +( )= −( ) +( )

displacement field: ˆu u= Φ k

u x y

u x y

u

u

u

u

x

y

xk

yk

xk

yk

( , )

( , )

..

..

( )

( )

( )

( )

=

ϕ ϕ ϕ ϕϕ ϕ ϕ ϕ

1 2 3 4

1 2 3 4

1

1

4

4

0 0 0 0

0 0 0 0

strain matrix

ˆ ˆε Φ=

= = =εεγ

xx

yy

xy

k kD u D u B u

u k nodal displacement matrix of element (k)


differential operator D =

=

∂∂

∂∂

∂∂

∂∂

∂∂ξ

∂∂η

∂∂η

∂∂ξ

x

y

y x

a

b

b a

k

k

k k

0

0

0

0

2

2

2 2

ˆ ˆ..

..u A u A = =

k k k

x

y

Nx

Ny

u

u

u

u

1

1

u global nodal displacement matrixA k (8 × 2N) incidence or connectivity matrix

N = total number of nodes2 N = number of degrees of freedom

ˆε = B A uk

stress matrix and HOOKE 's law

ˆσ ε=

= = σσσ

xx

yy

xy

k k kC C B A u


δ δA W− = 0



δ δ δ δ∂ ∂

A ds dsT

S

T

S

kk

KT

k

= = =∫ ∫∑=

ˆ ˆ ˆt u t u f uΦ1

nodal forces ˆ f A t( )= ∫∑=

kT T

Sk

K

s dsk

Φ∂1

virtual work of stresses

δ δ δW dx dyT

y b

b

x a

a

k

KT

k

k

k

k

= ==−

+

=−

+

=∫∫∑ ˆ ˆσ ε

12

12

12

12

1

u K u

element stiffness matrix

K B C B Kkk k T

k kTa b

d d= ==−

+

=−

+

∫∫41

1

1

1

ξ ηηξ


K A K A K= ==

∑ kT

kk

K

kT

1

δ δ δA W T T− = −( ) =ˆ ˆ ˆf u K u 0

as δˆu is arbitrary

ˆ ˆK u f=

linear system of equations for 2N unknowns u


1 2

1 2

34

1 2

34

2

1

4 6

3 5

x

y

global nodes (1), ..., (6)

elements [1], [2]

nodal coordinates (1): 0 , H

(2): 0 , 0

(3): L/2 , H

(4): L/2 , 0

(5): L , H

(6): L , 0

elements [1]: (2) (4) (3) (1)

[2]: (4) (6) (5) (3)

a L b H kk k= = =2 1 2, , ,

displacement b.c.: u u u ux y x y1 1 2 2 0= = = =


nodal displacements

ˆ

( )

( )

( )

( )

( )

( )

( )

( )

u k

xk

yk

xk

yk

xk

yk

xk

yk

u

u

u

u

u

u

u

u

=

1

1

2

2

3

3

4

4

, ˆu =

0

0

0

0

3

3

4

4

5

5

6

6

u

u

u

u

u

u

u

u

x

y

x

y

x

y

x

y

ˆ ˆu A u = k k

A =

2

0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0 0 0 0 1

0 0 0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 0 0 1 0 0

0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0 0


element stiffness matrix (k = 1, 2)

K B C B KkT

k kTLH

d d= ==−

+

=−

+

∫∫81

1

1

1

ξ ηηξ

global stiffness matrix K A K A K= ==

∑ kT

k kk

T

1

2

element nodal forces

ˆ f t( )kT

ks dsk

= ∫ Φ∂B

pressure load ( )

t k p=

−

0

ξ

ˆ ˆf ( ) t( ) rkT

k k

Ld= +

==−

+

∫4 11

1

Φ ξ ξ ξη

ξ

or interpolation for arbitrary function )t(ξ

) ˆt(ξ η η= ==1 1Φ t k , ˆt k = (8×1) matrix

ˆ ˆ ˆf rkT

k k

Ld= ( )

+ =

=−

+

∫4 11

1

Φ Φη

ξ

ξ t

reaction forces (k = 1, only)

ˆr 1 1 1 4 40 0 0 0Tx y x yr r r r= ( )

global nodal forces ˆ ˆf A f= =

∑ kT

kk 1

2


system of equations

K K K K K K K K

K K K K K K K K

K K K K K K K K

K K K

1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8

2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8

3 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8

4 1 4 2

0 0 0 0

0 0 0 0

0 0 0 0

, , , , , , , ,

, , , , , , , ,

, , , , , , , ,

, , 44 3 4 4 4 5 4 6 4 7 4 8

5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 5 10 5 11 5 12

6 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8 6 9 6 10

0 0 0 0, , , , , ,

, , , , , , , , , , , ,

, , , , , , , , , ,

K K K K K

K K K K K K K K K K K K

K K K K K K K K K K K66 11 6 12

7 1 7 2 7 3 7 4 7 5 7 6 7 7 7 8 7 9 7 10 7 11 7 12

8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 8 9 8 10 8 11 8 12

9 5 9 60 0 0 0

, ,

, , , , , , , , , , , ,

, , , , , , , , , , , ,

, ,

K



K K K99 7 9 8 9 9 9 10 9 11 9 12

10 5 10 6 10 7 10 8 10 9 10 10 10 11 10 12

11 5 11 6 12 7 11 8 11 9 11 10 11 11 11 12

12 5 12 6 12 7 12

0 0 0 0

0 0 0 0

0 0 0 0

, , , , , ,

, , , , , , , ,

, , , , , , , ,

, , , ,

K K K K K

K K K K K K K K

K K K K K K K K

K K K K 88 12 9 12 10 12 11 12 12

3

3

4

4

5

5

6

6

0

0

0

0

K K K K

u

u

u

u

u

u

u

u

x

y

x

y

x

y

x

y, , , ,

=

+

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

1

3

5

1

1

2

2

f

f

f

r

r

r

r

y

y

y

x

y

x

y


Lagrangian Interpolation Polynomials

gin j

j ijj i

n( )

,

( )ξξ ξξ ξ

=−−=

≠

+

∏1

1

n = order, i = 1, ..., n+1

1D: "truss" elementn = 1 (linear):

1 2ξ2 nodes ξ ξ1 21 1= − = +;

i = 1: ϕ ξ ξ ξ1 11 1

2 1( ) ( )( )= = −( )g

i = 2: ϕ ξ ξ ξ2 21 1

2 1( ) ( )( )= = +( )g

n = 2 (quadratic):

1 23

ξ

3 nodes ξ ξ ξ1 2 31 1 0= − = + =; ; 1

i=1: ϕ ξ ξ ξ ξ ξ ξ1 12 1

212

12

21 1 1( ) ( )( )= = − −( ) = −( ) − −( )g

i=2: ϕ ξ ξ ξ ξ ξ ξ2 22 1

212

12

21 1 1( ) ( )( )= = + +( ) = +( ) − −( )g

i=3: ϕ ξ ξ ξ ξ ξ3 32 21 1 1( ) ( )( )= = +( ) −( ) = −g

Φ = ( )ϕ ϕ ϕ1 2 3


2D: quadrilateral elementn = 1 (linear) 4 nodes

ξ ηξ ηξ ηξ η

1 1

2 2

3 3

4 4

1 1

1 1

1 1

1 1

= − = −= + = −= + = += − = +

ξ

η

1 2

34

ϕ ξ η ϕ ξ ηϕ ξ η ϕ ξ η

ξ η ξ ηξ η ξ η

1 3

2 4

11

11

11

21

21

11

21

21

( , ) ( , )

( , ) ( , )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

=

g g g g

g g g g

=−( ) −( ) −( ) +( )+( ) −( ) +( ) +( )

14

14

14

14

1 1 1 1

1 1 1 1

ξ η ξ ηξ η ξ η

ˆu u= Φ k

Φ =

ϕ ϕ ϕ ϕϕ ϕ ϕ ϕ

1 2 3 4

1 2 3 4

0 0 0 0

0 0 0 0


n = 2 (quadratic): 9 nodes

ξ ηξ ηξ ηξ ηξ η

5 5

6 6

7 7

8 8

9 9

0 1

1 0

0 1

1 0

0 0

= = −= + == = +

= − == =

ξ

η

1 2

34

5

6

7

8 9

ϕ ϕ ϕϕ ϕ ϕϕ ϕ ϕ

ξ η ξ η ξ ηξ η ξ η ξ

1 3 8

2 4 6

5 7 9

12

12

12

22

12

32

22

12

22

32

22

3

=

g g g g g g

g g g g g g

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

( ) ( ) ( ) ( ) ( )

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

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

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

( )

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

2

32

12

32

22

32

32

ηξ η ξ η ξ ηg g g g g g

... ... ...

... ... ...Φ =

ϕ ϕ ϕϕ ϕ

1 2 9

1 9

0 0

0 0 0


3D: "brick" element

ϕ ξ η ζ ξ η ζ



ϕ ξ η ζ

1 1 1 1

2 2 1 1

3 2 2 1

4 1

( , , ) ( ) ( ) ( )

( , , ) ( ) ( ) ( )

( , , ) ( ) ( ) ( )

( , , )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( )

=

=

=

=

g g g

g g g

g g g

g

n n n

n n n

n n n

n (( ) ( ) ( )

( , , ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )

ξ η ζ


g g

g g g

n n

n n n

2 1

5 1 1 2=etc.

n = 1 (linear) 8 nodes

ξ

η

5 6

78

34

1 2

ζ

... ... ...

... ... ...

... ... ...

Φ =

ϕ ϕϕ ϕ

ϕ ϕ

1 8

1 8

1 8

0 0 0 0

0 0 0 0

0 0 0 0

n = 2 (quadratic) 27 nodes


Note: for quadratic (and higher order) 2D and 3D

elements, internal nodes may be omitted to

reduce the number of degrees of freedom by

means of which some higher polynomial terms

vanish

"boundary node" elements

⇒ 2D quadratic: 8 nodes term ξ η2 2 vanishes

1 ξ ξ 2

η ξη ξ η2

η2 ξη2 ξ η2 2

⇒ 3D quadratic: 20 nodes


include only if node i is defined

ϕ1 =14 1 1−( ) −( )ξ η − 1

2 5ϕ − 12 8ϕ

ϕ2 =14 1 1+( ) −( )ξ η − 1

2 5ϕ − 12 6ϕ

ϕ3 =14 1 1+( ) +( )ξ η − 1

2 6ϕ − 12 7ϕ

ϕ4 =14 1 1−( ) +( )ξ η − 1

2 7ϕ − 12 8ϕ

ϕ5 =14

21 1−( ) −( )ξ η

ϕ6 =14

21 1+( ) −( )ξ η

ϕ7 =14

21 1−( ) +( )ξ η

ϕ8 =14

21 1−( ) −( )ξ η

Interpolation functions of four to eight variable-number-nodes for 2D element


General Isoparametric Elementscurvilinear distorted element

ξ

η

1 2

34

x

y

1

2

3

4

x f y f z fx y z= = =( , , ) ; ( , , ) ; ( , , )ξ η ζ ξ η ζ ξ η ζ

interpolation by shape functions: ˆx x= Ψ k

isoparametric elements: Ψ = Φ , ψ ϕi i=

2D: x =

x

yˆ ...

( )

( )

( )

( )

x k

k

k

Nk

Nk

x

y

x

y

=

1

1

3D: x =

x

y

z

ˆ

...

...

( )

( )

( )x k

k

k

k

x

y

z=

1

1

1


differential operator (2D) D =

∂∂

∂∂

∂∂

∂∂

x

y

y x

0

0

∂∂

∂∂ξ

∂ξ∂

∂∂η

∂η∂

∂∂ζ

∂ζ∂x x x x

= + +

∂∂ξ∂

∂η∂

∂ζ

∂∂ξ

∂∂ξ

∂∂ξ

∂∂η

∂∂η

∂∂η

∂∂ζ

∂∂ζ

∂∂ζ

∂∂∂∂∂∂

=

f f f

f f f

f f f

x

y

z

x y z

x y z

x y z

∂∂ξ

∂∂

=

Jx

⇒ ∂∂

∂∂ξx

J

=

− 1

J = JACOBI matrix, det J J= ≠ 0

2D: JJ

− =−

−

1 1∂∂η

∂∂ξ

∂∂η

∂∂ξ

f f

f f

y y

x x

J = −∂∂ξ

∂∂η

∂∂ξ

∂∂η

f f f fx y y x

∂∂ξ

∂ϕ∂ξ

fxx i

i

N

ik

k

==∑

1

( ), ... etc.


calculation of

strain matrix ˆ ˆε Φ= =D u B uk k

stress matrix ˆσ ε= =C C B uk k k


K B C BkT

k dVk

= ∫∫∫B

transformation of the volume differential:

2D: dV BdA Bdx dy B d d= = = J ξ η(B = thickness)

3D: dV dx dy dz d d d= = J ξ η ζ

condition: J > 0 , J positive definite


K B C B JkT

k d d d==−

+

=−

+

=−

+

∫∫∫ ξ η ζζηξ 1

1

1

1

1

1

integration numerically by GAUSS quadrature


K A K A==

∑ kT

kk

K

k1


GAUSS Quadrature

a function f(ξ) is integrated approximately by a

weighted sum of its values at sampling points ξj

f d w fj jj

n

( ) ( )ξ ξ ξ−

+

=∫ ∑=1

1

1

the formula integrates a polynomial of order (2n-1)

exactly

sampling points (GAUSS points):

number

n

coordinates

ξj

weight

wj

1 0.000 000 2.000 000

2 ±0.577 350 1.000 000

3 ±0.774 597

0.000 000

0.555 556

0.888 889

4 ±0.861 136

±0.339 981

0.347 855

0.652 145

2D: f d d w w fj k j kk

n

j

n

( ) ( , )ξ ξ η ξ ηξη =−

+

=−

+

==∫∫ ∑∑=

1

1

1

1

11


Nonlinear FE Analyses

nonlinear structural behavioura) geometrical (large displacements and/or large strain)

b) constitutive nonlinearity (material)

incremental formulation

quasi-static process (equilibrium),

"time" t ≥ 0 monotonously increasing parameter of

load history


time

time

t A t W t

t t A t t W t t

: ( ) ( )

: ( ) ( )

δ δδ δ

− =+ + − + =

0

0∆ ∆ ∆

f t t f t f tt

( ) ( ) ˙+ ≈ +∆ ∆

t t t→ + ∆ : δ δ∆ ∆A W− = 0

FE formulation:

) K( u u f∆ ∆=


(a) geometrical non-linearity

0rtr

tu0

t =0 t

O

t+∆t

t+∆tr

∆u

0ut+∆t

configuration t t t t t+ += + = +∆ ∆ ∆r r u r u00

total displacement 0 0t t t+ = +∆ ∆u u u

total strain (GREEN's quadratic strain)

000

00

00

00

0 0

12

t tt t t t t t

t t t t

++ +

+ +

=

+

+

⋅

= +

∆∆ ∆

∆ ∆

Eu

r

u

rur

ur

E G

G

T T

( ) ∂∂

∂∂

∂∂

∂∂

left subscript: reference state

left superscript: actual state

reference state 0: Total LAGRANGEan Formulation

reference state t: Updated LAGRANGEan Formulation


Updated LAGRANGEan Formulation

reference state t

tt t

tt

tt

t+ = + = +∆ ∆ ∆S S S T S

S = 2nd PIOLA-KIRCHHOFF stresses

T = CAUCHY stresses

tt t

t t+ = ≈∆ ∆ ∆E E EG G( ) ( )

virtual work during t t t→ + ∆ (surface tractions only):

δ δ δ∂ ∂

∆ ∆ ∆∆ ∆A dA dAtt t

tt t

V Vt t

= ⋅ = ⋅+ +∫ ∫t u t u

δ δ

δ δ

∆ ∆

∆ ∆ ∆

∆ ∆W dV

dV dV

tt t

tt t

V

tt

t

V

t t

V

t

t t

= ⋅⋅

= ⋅⋅ + ⋅⋅

+ +∫

∫ ∫

S E

S E S E

G( )

∆ ∆t u S E⋅ − ⋅⋅ = − =∫ ∫δ δ δ δ∂

t

V

tt

t

V

dA dV A t W tt t

( ) ( ) 0

∆ ∆ ∆ ∆t t

V

t

V

dV dAt t

S E t u u⋅⋅ − ⋅ +( ) =∫ ∫δ δ∂

0

constitutive relation: ∆ ∆t t tS C E= ⋅⋅(incrementally linear)

t ) K( u u f∆ ∆ ∆=

nonlinear system of equations


⇒ iterative solution (e.g. NEWTON)

∆ ∆ u u K r(j) (j)

= ==

−

=∑ ∑j

i

tj

i

1

1

1

residual force (out-of balance force)

( )r f f u u(j) (j)

= − ++t t t∆ ∆convergence criterion

r RTOL and / or u DTOL(i) (i)

≤ ≤∆

f

u

∆u(1)

∆u(2)

∆u

∆ft f

t+ ∆tf

tu t+ ∆tu

r(1)


(b) constitutive non-linearity

elastic- plastic material behaviour(VON MISES, PRANDTL, REUSS)

1. uniaxial tensile test

RF

σ

εεF

εe

εp

loading

unloading

yield condtion: σ ε≤ =R R Rp( ) , ( )0 F

HOOKE's law: σε σ

ε ε σ=≤

−( ) >

E R

E R

for

forF

pF

loading condtion:˙ , ˙

˙ , ˙

σ εσ ε

> >< =

0 0

0 0

p

p

loading

unloading


2. general (multiaxial) stress state

yield condtion: σ ε≤ R p( ) ; σ σ σ= ′ ′32 ij ij

VON MISES equivalent (effective) stress

′ = −σ σ σij ij kk13 deviatoric stresses

HOOKE's law:

˙˙ ( )

˙ ˙ ( )σ

ε νν

ε δ σ

ε ε νν

ε δ σij

ij kk ij

ij ijp

kk ij

G R

G R=

+−

≤

−( ) +−

>

21 2

21 2

for

for

F

F

loading condtion:′ > >′ < =

σ σ εσ σ ε

ij ij ijp

ij ij ijp

˙ , ˙

˙ , ˙

0 0

0 0

loading

unloading

flow rule ˙˙ ˙

ε εσ

σ σσ

σijp

p

ij p ijE= ′ = ′3

232

EE E

E EEp =

−=t ,

d

tt d

σε

σ ε σ ε ε ε εij ijp p p

ijp

ijp˙ ˙ ˙ ˙ ˙= ⇒ = 2

3

ε ijp is deviatoric, i.e. εkk

p = 0

"true" stress-strain curve required for UL formulation:

true stresses σ = FA

vs logarithmic strain ε = lnLL0


LH

p

A plane panel of dimensions

length L = 200 mm, height H = 100 mm, thickness B = 1mm

is clamped at x = 0 and loaded by a constant pressure p = 100 MPa at y = H

The material is isotropic, elastic with

YOUNG's modulus E = 218 400 MPa and POISSON's ratio ν = 0.3.

Establish the system of equations of the respective finite element model and calculate the

displacement, stress and strain field by applying the FE code ANSYS.

1. analytical solution for a model of two linear elements

1 2

1 2

34

1 2

34

2

1

4 6

3 5

x

y


1.1 Calculate the elastic stiffness matrix

C =

C C C

C C C

C C C

11 12 13

12 22 23

13 23 33

for plane stress conditions.

1.2 Calculate the (symmetric) element stiffness matrix for one element k

K B C Bkk k T

k

a bd d=

=−

+

=−

+

∫∫41

1

1

1

ξ ηηξ

1.3 Calculate the global stiffness matrix (see table on page 3)

K A K A==

∑ kT

kk

K

k1

1.4 Calculate the column matrix of nodal forces (see table on page 3)

f A t= ∫∑=

k

T T

Sk

K

dsk

Φ∂1

.

2. FE solution by the ANSYS code

Solve the same problem by the FE code ANSYS and compare the results with analytical values

(see lecture notes), especially the displacement uy of node (6) and the stresses at the nodes (1)

and (2) in the table on page 4.

2.1 Use the model of two linear plane stress elements as on page 1.

2.2 Use two quadratic (8-node) elements instead of linear elements.

2.3 Use two other refined meshes and quadratic elements; explain and display the meshes.


analytical solution (theory of bending)

bending stress

σ xxz

z

x yM x

Iy

H( , )

( )= − −

2

M xpBL x

LxLz ( ) = − −

+

2 2

21 2

σ σ σxx xx xy H xx x

y

z

z

M

Wmax ( )

= = = ===

==

0 00

01200 MPa

section modulus W IH

BHz z= = =2

610

6

2 4

mm3

deflection1. simple bending (no shear)

u xpBL

EIxL

xL

xLy

b

z

( ) = −

−

+

4 2 3 4

246 4

u u LpBL

EIyb

yz

,max ( ) .= = =4

81 10 mm

2. shear

u xpLGH

xL

xLy

s ( ) = −

−

2 2

22

u u LpLGHy

sys,max ( ) .= = =

2

20 24 mm

total u u uy yb

ysmax ,max ,max .= + = 1 34 mm


0.0 0.2 0.4 0.6 0.8 1.0-15

-10

-5

0

x / L

u y /

L *

(E /

p)

bendingsheartotal

rectangular panel

exercise #1:


exercise #2:

biaxially loaded panel with circular hole

x

y

px

py

px

py

2 H

2 H

R

H = 50 mm, R = 10 mm

plane stress, B = 1 mm

py = 100 MPa, px = β py

E = 200000MPa, ν = 0.3


analytical solution for "infinite" plate: R << H,

uniaxial tension (β = 0)

σ θ θ

σ θ θ

σ θ

θθ

θ

rr y

y

r y

r pRr

Rr

Rr

r pRr

Rr

r p

( , ) cos

( , ) cos

( , )

= −

− −

+

= +

+ +

=

12

2 2 4

12

2 4

12

1 1 4 3 2

1 1 3 2

11 3 3 22 4

+

−

Rr

Rr

sin θ

x

y

r

θ

1 2 3 4-1

0

1

2

3

r / R

σ ij /

py

σrr (θ=0)

σθθ (θ=0)

σrr (θ=π/2)

σθθ (θ=π/2)


FE solution: symmetry conditions ⇒1/4 model

x

y

py

px

boundary conditions:

u uy y x x= == =0 0

0 0;


Notation30.10.99

tensors, vektors, skalars - general

scalar latin or greek, small or capital (italics) a , H, α ,

vector

symbolic latin or greek, small,

bold or underlined x , x , σσσσn , σ n

indexed latin or greek, small (italics) xi , α i

x = xi ei , x = xi ei

2nd order tensor

symbolic latin or greek, capital,

bold or underlined T , T

indexed latin or greek, capital or small (italics) Tij , σij

T = Tij ei ej , T = Tij ei ej

4th order tensor

symbolic latin or greek, capital,

bold with overhead <4> C< >4

or double underlined C

indexed latin or greek, capital Cijkl

C< >4

= Cijkl ei ej ek el , C = Cijkl ei ej ek el


vector and tensor algebra and analysis

scalar product

α = u⋅v = ui vi , t = n⋅S = niσij ej,

A = B⋅C = Bij Cjk ei ek = Aikei ek

double scalar product

α = B⋅⋅C = Bij Cji or α = B : C ...

T = C< >4

⋅⋅E = Cijkl Ekl ei ej or S = C : E ...

tensorial product

T = a b = ai bj ei ej

C< >4

= A B = Aij Bkl ei ej ek el

a b⋅c = (a b)⋅c = a (b⋅c)

spatial derivatives

NABLA operator ∇ = eiix

∂∂

v = grad ϕ = ∇∇∇∇ ϕ = ∂ϕ∂xi

ei = ϕ,i ei

F = grad v = ∇∇∇∇ v = ∂∂

v

xi

j

ei ej = vi,j ei ej

a = div v = ∇∇∇∇⋅v = = vi,i = ∂∂

v

xi

i

w = div T = ∇∇∇∇⋅T = ∂∂T

xij

i

ej = Tij,i ej

convention of summation is used in all cases

A B A Bij jk ij jkj

==

∑1

3

, C Ckk kkk

==

∑1

3

, A B A Bij ji ij jiji

===

∑∑1

3

1

3


special tensors and invariants

2nd order unit tensor I = δij ei ej = ei ei

T⋅I = I ⋅T = T

deviator of a tensor T: T' = ′Tij ei ej = (Tij - Tkk δij ) ei ej = T - T⋅⋅I

transposed tensor TT = Tji ei ej

inverse tensor T-1 , T-1

T⋅T-1 = T-1⋅T = I

1st invariant (trace) of a tensor J1(T) = tr (T)

2nd invariant of a tensor J212

2( ) tr trT T T= −( )2

3rd invariant (determinant) of a tensor J3(T) = det (T)


tensors used in continuum mechanics

deformation gradient F = I + grad u = (δij + ui,j ) ei ej

GREEN's strain tensor E F F I e e( ) T, , , ,

Gi j j i i k k j i ju u u u= ⋅ −( ) = + +( )1

212

linear strain tensor E u u e e e e E= −( ) = +( ) = =12

12grad grad , ,

T u ui j j i i j ij i jTε

strain rate tensor D e e v v= = −( )dij i j12 grad gradT

for small strains D E e e= =˙ ε ij i j

elastic and plastic part ˙ ˙ ˙ ˙ ˙E E E e e= + = +( )e p e pε εij ij i j

accumulated effective ("equivalent") plastic strain

ε ε ε ε τ τep p p p= =∫ ∫p

ij ij

t t

d J d23

0

43 2

0

˙ ˙ ( ˙ )E

CAUCHY ("true") stress tensor S = σij ei ej = ST = σji ei ej

hydrostatic stress σ σh = = =13

13

13 1kk Jtr ( )S S

VON MISES effective ("equivalent") stress

σ σ σ σe = = ′ = ′ ′3 232J ij ij( )S

2nd PIOLA KIRCHHOFF stress tensor T F F S F T= ⋅ ⋅ =− −det( ) 1 T T

for small strains T S=


plastic yielding

yield curve (uniaxial tensile test) R(εp)

with ε ε ε σp e

p= = −E

yield strength (at initial plastic flow) R0 = R(0)

especially

lower yield strength ReL

0.2% proof stress Rp0.2

plastic potential, flow potential (VON MISES)

isotropic hardening Φ = − = ′ ′ − = ′ ⋅ ⋅ ′ −σ ε σ σ ε εe p p p2 2 3

22 2R R Rij ij( ) ( ) ( )S S

kinematic hardening Φ = − = ′ − ′( ) ⋅ ⋅ ′ − ′( ) −σ e2

02 3

2 02R RS X S X

with X = "back stress" tensor

associated flow rule

˙ ˙ε λ ∂∂σij

ij

p =′

Φ , ˙ ˙ES

p =′

λ ∂∂Φ


matrix notation

general: (n×m) matrix: A or A

n = number of rows

m = number of columns

. . .

. . . . .

. . . . .

. . .

A =

A A

A A

m

n nm

11 1

1

. .

. . . .

. . . .

. . . .

. .

A T

n

m nm

A A

A A

=

11 1

1

special: (n×1) matrix u or u

.u =

u

un

1

.u Tnu u= ( )1

the elements of a matrix do not form the components of a tensor or vector, in general

products

u vTi i

n

u v= = ∑α1

u and v: (n×1) ; α scalar

. . .

. . . . .

. . . . .

. . .

u v AT

m

n n m

u v u v

u v u v

= =

1 1 1

1

u: ( n×1) , v: (m×1) ; A: ( n×m)

....v A u= =

=

=

∑

∑

A v

A v

i ii

m

ni ii

m

11

1

A: ( n×m) ; u: ( m×1) , v: (n×1)

v u AT T T= A T : (m×n) ; u T : (1×m) , v T : (1×n)

. .

.... . . ....

. .

C A B= =

= =

= =

∑ ∑

∑ ∑

A B A B

A B A B

i ii

m

i imi

m

ni ii

m

ni imi

m

1 11

11

11 1

A: ( n×m) ; B: ( m×p) ; C: ( n×p)


Physical Units (SI-Units)ISO 1000 (1973)

SI = Système International d'Unités

SI units are

the seven basic units and coherently derived units, i.e.

by a factor of 1

Basic Quantities and Units

basic quantity SI basic unit

name symbol

length meter m

mass kilogramm kg

time second s

thermodyn. temperature Kelvin K

amperage, intensity ofelectric current

Ampere A

amount of substance Mol mol

luminous intensity Candela cd


Decimal Parts and Multiples of SI Units

Parts and multiples of SI units which are generated bymultiplication with the factors

10±1, 10±2, 10±3n (n = 1, 2, ...)have special names and symbols.They are composed by prefixes.

faktor prefix symbol

10-15 femto f10-12 pico p10-9 nano n10-6 micro µ10-3 milli m

10-2 centi c10-1 deci d101 deca da102 hecto h

103 kilo k106 mega M109 giga G1012 tera T

Multiples by the factors10±3n (n = 1, 2, ...)

are to be preferred!


Derived Quantities and Units

Derived units are formed by products or ratios of basic units.The same holds for the unit symbols.

quantity SI unit relation

name symbol

frequency Hertz Hz 1 Hz = 1 s-1

force Newton N 1 N = 1 kg m s-2

pressure, stress Pascal Pa 1 Pa = 1 N / m2

energy, work, amount of heat Joule J 1 J = 1 N m

power, heat flow Watt W 1 W = 1 J / s

electric charge, quantity of electricity Coulomb C 1 C = 1 A s

electric potential, voltage Volt V 1 V = 1 J / C

electric capacity Farad F 1 F = 1 C / V

electric resistance Ohm Ω 1 Ω = 1 V / A

magnetic flux Weber Wb 1 Wb = 1 V s

magnetic flux density Tesla T 1 T = 1 Wb / m2

inductivity Henry H 1 H = 1 Wb / A


Examples for FE meshes

3D FE mesh at the notch of a side-notched flat tensile panel

3D FE mesh of a centre-notched tensile panel


2D FE mesh of a C(T) specimen, half model accounting for symmetry

Detail of the FE mesh of the C(T) specimen at the crack tip with collapsed elements

1

2

3 1

2

3


2D FE mesh of a biaxially loaded cruciform specimen with two-fold

symmetry

Detail of the FE mesh of the cruciform specimen at the crack tip with a

regular arrangement of elements


3D FE mesh of a tubular joint under 3-point bending, having one

symmetry plane

32896HEXAEDERELEMENTE

Detail of the FE mesh of the semi-elliptical surface flaw in the weldment of

the tubular joint

ANSICHTSYMMETRIE-EBENE

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