FEM Finite Element Methods Introduction - vsb.cz · 1 VŠ – Technical University of Ostrava...

1

VŠB – Technical University of Ostrava Faculty of Mechanical Engineering

FEM Finite Element Methods

Introduction

2

Content:

1. Static 2-D Problem 2. A Static Solution of 2-D Beam with ANSYS 3. Modal Analysis 4. Solution of Natural Frequencies of a Cantilever Beam by FEM 5. Contacts in FEM 6. FEM - Tool for Buckling Solution

1. STATIC 2-D PROBLEM

FEM Computation INTRODUCTION

We cannot solve this plane problem with an analytical

(exact) solution as we would with any other partial differential equations

There is stress due to the object’s complex shape… !!

WHY ??

4 FEM Computation INTRODUCTION

BACKGROUND OF FEM WE HAVE A FINITE NUMBER OF ELEMENTS

THEREFOR…

FINITE

ELEMENTS

METHOD

A MESH OF

ELEMENTS

NOW IT CAN BE SOLVED THROUGH

A NUMERICAL WAY = A SYSTEM OF

LINEAR

EQUATIONS


JOINING ELEMENTS TOGETHER


ELEMENTS

ARE

JOINED

TOGETHER

ONLY IN

THE

NODES!!!

6

ELEMENTS ARE CONNECTED ONLY

IN THE NODES


7

REALY PROBLEM CAN’T BE SOLVED WITHOUT

BOUNDARY CONDITIONS


WHAT DO

WE MEAN

BY…

PRESSURE

LOADING DISPLACEMENT

8

WE ARE DISCUSSING THE “BOUNDARY PROBLEM”

AND

SOLVING IT !!


9

RESULTS: DISPLACEMENT


10

EQUIVALENT STRESS


11

Static & Dynamic Problems

FK

1 11 2 12 3 13 1

1 21 2 22 3 23 2

1 1

k k k f

k k k f

k k fn n n n n n n

....................

....................

, ,

. . . . .

. . . . .

. . . . .

.......................+

12

Static & Dynamic Problems

tFKM

ntnnnnnnnnnnn

t

t

fkkmm

fkkmm

fkkmm

,1,1,1,1-n

2222221222211

1222111122111

...

. . . . . .

. . . . . .

. . . . . .

............

............

13

Eigenvalue Problem

02 ORMK

Buckling Problems

FKK G

14

EXAMPLES:

CAR BRAKE STRUCTURE

15

EXAMPLES:

CAR FOR CRASH SIMULATION

16

EXAMPLES: MATERIAL

NONLINEAR PROBLEM

17

TYPES OF FINITE ELEMENTS

2

1

LINK (TRUSS, ROD, SPAR) PIPE

18


BEAM 3D

19


SHELL

SOLID

20

Easy Example Leads to Stiffness

Matrix of the Beam Element

A

q

ℓ

EJ

B

EJy qx

IV

x( ) ( )

0 ,0 yy o

0" ,0" yy o 0 ,0 ooo MM

BOUNDARY CONDITIONS

o o

x

xo

d dxyEJdxEJ

M."

2

1

2

1 2

)(

2

)(POTENCIAL ENERGY OF

INTERNAL FORCES

POTENCIAL ENERGY OF

EXTERNAL FORCES

o

xxz dxyq ,)()(

21

EXACT SOLUTION: qyEJ IV

x )(.

1)(. cqxyEJ III

x

21

2)(

2

1. cxcqxyEJ x

II

32

2

1

3

)(2

1

6

1. cxcxcqxyEJ I

x

EJ y qx c x c x c x cx. .( ) 1

24

1

6

1

2

4

1

3

2

2

3 4

.6

1

240 a 0 3

3

1

4

4

ccq

c THE FIRST BC y=0 .2

11 qc

.24

1

12

1

24

1 334 xqxqqxEJy

AFTER THE SECOND BC 0" ,0" yy o

EXACT SOLUTION !

22

y ax i x ii

n

( ) ( )

1

RITZ SOLUTION – VARIATION CALCULUS

naaaF ,...,, 21POTENCIAL ENERGY OF STRUCTURE

a = UNKNOWN !

.,...,2,1 ,0 niai

.2

21)( xxaxxay x WE SUPPOSE :

o oo

dxxaaEJdxyqdxyEJaa2

21

2

21 342

1"

2

1,

l

o

aaq

aaaaEJldxxxaxxaq .212

2 4

2

3

1

22

221

2

1

2

21

23

RITZ SOLUTION – VARIATION CALCULUS

CONTINUED

06

1)2(2 3

21

1

qaaEJa

.012

1)2(2 42

21

2

qaaEJa

0 ,24

2

2

1 aEJ

qa

xxEJ

qy x )(

24

2

)(

ERROR TO EXACT SOLUTION IS 16%

24

A BEAM ELEMENT

Δ1 Δ2 Δ3

Δ4

w b bx b x b xx( ) . 0 1 2

2

3

3BETTER WAY

x

Ld

dx

L a .

,3

3

2

210)( aaaaw APROXIMATION EQUATION

323232321 ,23 ,2 ,231.

LLSN

TT

SHAPE FUNCTION !

25

STIFFNESS MATRIX OF BEAM ELEMENT

.

4.

6,12

2,64

6,12,6,12

2

22

3

Lsym

L

LLL

LL

L

EJk

LOAD VECTOR:

LTT

xx fdNqLLdwqdxwqz0

1

0

1

0

)()()()()( ,..

1

0

)( ,. dNqLfTT .

12,

2,

12,

2

22

1

qLqLqLqL

fT

L

q

f2=qL2/12

f1=qL/2

f4= -qL2/12

f3=qL/2

26

STATIC MATRIX EQUATION

.2

1

TT

zdp fk

n

i

n

i

n

i

i

T

iii

T

ipi fk1 1 1

.2

1n = NUMBER OF ELEMENTS

,2

1G

T

G

T

G FK

,0 FK G

G

FK G

27

SOLUTION OF OUR EASY EXAMPLE

1 2

1

2

3

4

1

2

3

4

Numbers of the

elements:

Local DP:

Global DP

by element

Global DP:

a) without

the influence

of peripheral

conditions

1

2

3

4

5

6

b) with consideration

of peripheral

conditions

1

2

3

4

3

4

5

6

0

1

2

0

0

3

28

SOLUTION OF OUR EASY EXAMPLE/cont’d


00020100

20222120

10121110

00020100

0

2

1

0

33

312312

2/33

312312

8

22

22

31

lEJk

0 2 1 0

Numbers of DOF parameters for

the second element:

33303032

03000002

03000002

23202022

3

0

0

2

33

312312

2/33

312312

8

22

22

32

EJk

Numbers of DOF parameters for

the first element:

2 0 0 3



2

2

3

30

3243

038

EJK

GLOBAL STIFFNESS MATRIX

,48

,2

,48

22

qqq

FT

GLOBAL LOAD VECTOR

48

2

48

.

30

3243

038

2

2

3

2

1

2

2

3

q

q

q

EJ

FEM STATIC EQUATION:

SOLUTION FOR UNKNOWN DOF

1 3 ,.

24

1 ,

384

5 ,

24

1 3

3

4

2

3

1EJ

q

EJ

q

EJ

q



INTERNAL FORCES

,...

22

2

22

2

2

2

2

2

TT

o gL

EJ

d

Nd

L

EJ

d

wd

L

EJEJ

dx

d

d

wdM

)62( ,126 ,)64( ,126

2

2

LLgd

Nd TT

.......

33

3

3

3

3

3

3

3

TT

o pL

EJ

d

Nd

L

EJ

dx

d

d

wdEJ

dx

wdEJ

dx

dMT

.6 ,12 ,6 ,12

3

3

LLd

qdp

d

NdT

TT

432122646)( LL

L

EJpM o

.612612)()( 43213 LL

L

EJkTpT



Global DOF

FEM - 2 ELEMENTS

FEM - 2 ELEMENTS

EXACT

32

Numbers of theElements

Global DOF

with Bounadary Conditions


___________________________________________________________________________________

A Static Solution of 2-D Beam with ANSYS 1

A STATIC SOLUTION OF 2-D BEAM WITH ANSYS

TOPIC:

The topic is to learn basic commands in the ANSYS programme package for an easy static example of a plane

beam.

For our beam (see figure) solve internal forces – axial force N, shearing force Q and bending moment

Mb - normal stress and maximum deflection y. Find out the place of maximum deflection and the

place of maximum normal stress. The cross-section is a steel rectangle with a height of H = 100 mm

and width B = 20 mm. Young's modulus is E = 210 GPa and Poisson's ratio ν = 0,3.

1. Enter title for display

Utility Menu > File > Change Title …

/title, EASY 2-D BEAM

2. Open preprocessor

ANSYS Main Menu > Preprocessor

3. Enter graphic points "Keypoints"

Preprocessor > (-Modeling-) Create > Keypoints > In Active CS…

Graphic points in 5 important places on our beam, see next figure

Coordinates of keypoints (solution in mm a MPa)

No. of KP X coord. mm Y coord. mm Z coord. mm

1 0 0 0

2 2000 0 0

3 7000 0 0

4 9000 0 0

5 11000 0 0

Notice: You can put values in shape: 2E3

___________________________________________________________________________________


4. Enter lines "Lines"

Preprocessor > (-Modeling-) Create > (-Lines-) Lines > Straight Line > Apply…

After entering the command Straight Line the very important Selection menu appears,, which we

will use very often. You can choose entities individually "by mouse", or by number, by "Box",

"Circle" or "Polygon". After opening this small window, the cursor will change shape with or

unselect .

1

1 22 33 44 5XYZ

JAN 27 201208:29:11

LINES

TYPE NUM

Notice: Numbering of objects

Plot_Ctrls > Numbers … >

Keypoint No. 1 is

hidden under

coordinate system

___________________________________________________________________________________


For numbering, we have to switch "Off" to "On", and in the window with "[/NUM] Numbering

shown with", change to "Colors & Numbers". You can see it in the next figure for Keypoints.

Next command Plot > Keypoints > Keypoints appear Keypoints with numbers.

5. Enter "Type of Elements"

Preprocessor > Element Type > Add/Edit/Delete > Add > Beam > (2D elastic 3)

For our example we are choosing plane beam “BEAM3”. It has three DOF in the starting node as

well as in the ending nodes (displacement in directions X, Y and rotation around Z). Now we open

"Options", (see next figure) and we change the K6 key to "Include output" (we want to get the

internal forces in the results file)

6. Enter section characteristic of element "Real Constants"

Preprocessor > Real Constants … > Add …

In the window "Real constants for BEAM3" you have to enter the next section characteristics: cross-

section area, second moment of area, height of the section. The ratio for the shearing force influence

is "SHEARZ" (the shearing force is neglected when the value is zero). The SHEARZ parametr for

the rectangular section has a value of 1.2 . Values can be entered into the system through the

"ANSYS Input window", (see next figure), type H=100 and then B=20.

___________________________________________________________________________________


Notice: You can check typing parameters :Parameters > Scalar Parameters …

7. Enter "Material Properties"

Preprocessor > Material Props > (-Constant-) Isotropic > Specify material number 1 > : Young’s modulus EX 2.1E5 and Poisson’s ratio NUXY 0.3

8. Enter size of elements for meshing ("Define Mesh Size")

Preprocessor > (-Meshing-) Size Cntrls > (-Lines-) All lines … We chose an element edge length of 250 mm

After this command, lines are divided into a number of elements, but no elements exist yet!

___________________________________________________________________________________


1

X

Y

Z

JEDNODUCHY ROVINNY NOSNIK

NOV 18 2001

09:42:29

ELEMENTS

ELEM NUM

1

1 22 33 44 5XYZ

JAN 27 201208:30:23

LINES

TYPE NUM

9. Creation of " Mesh "

Preprocessor > (-Meshing-) Mesh > Lines > choose "Pick All"

We can choose a method for drawing and numbering elements by Plot Cntrls > Numbering … see next figure

Y

1Z2X3 4 5 6 7 8 9 1011121314151617181920212223242526272829303132333435363738394041424344

The numbering is unnoticed, so we change it from "Colors & numbers" to "Colors only". To draw the

true height of elements:

Plot Cntrls > Style > Size and Shape … "ESHAPE" put ON and SCALE on 1.

___________________________________________________________________________________


ENTER BOUNDARY CONDITIONS

1. Analysis type

Solution > New Analysis > Static

2. Loading

(-Loads-) Apply > (-Structural-) Force/Moment > On Keypoints …We must draw

Keypoints (Keypoints) and enter force and moment (see following tables):

We can draw force and moment through the following commands:

Plot Ctrnls > Symbols > For Individual: Applied BC's

___________________________________________________________________________________


Force and moment:

3. Enter join (support, constraint, coupling, binding)

(-Loads-) Apply > (-Structural-) Displacement > On Keypoints …After drawing

Keypoints with numbers, we can remove DOF after the next figure:

Remove two DOF for the joint in place of KP2: UX and UY, in the dark blue figure. Rotation about

axis Z is free. Only one DOF is removed in KP5, namely UY. Constrained displacements are colored

light-blue.

1

1 2 3 4 5XYZ

JAN 27 201208:32:24

POINTS

POIN NUM

U

4. Enter pressure (continuous) load

(-Loads-) Apply > (-Structural-) Pressure > On Beams …

___________________________________________________________________________________


For easy choice of loading elements, we draw lines by means of the "LPLO" command. After that we

can choose elements through the selection box. We only enter the pressure value 2 N/mm (=2 kN/m)

into the window. „LKEY“ = 1 do not change.

5. SOLUTION

(-Solve-) Current LS …

The window with the specification of our example is open. We needn’t change anything. The

solution will start by clicking on "OK" in the next window.

The solution will finish with the opening of the small "Solution is done" window.

ANALYSIS OF RESULTS

1. Open postprocessor

ANSYS Main Menu > General Postprocessor…

2. Drawing deformed shape

Plot Results > Deformed Shape… > (Def + undef edge )

TIME=1 DMX =259.122 Y

You can find the maximum deflection on the upper left, but not its location. We will get a better

course of deflection by means of the "Vector Plot"

___________________________________________________________________________________


Plot Results > (-Vector Plot-) Predefined … > ( DOF solution –Translation U )

2. Drawing the course of internal forces and stresses

Element Table > Define Table … > ( Add )

To obtain the drawing results we have to create an "ETABLE". In bottom window we have to enter

special commands ( e.g. SMISC,2 ). This is found in the ANSYS manual for BEAM3.

___________________________________________________________________________________


Axial force in the starting node: SMISC,1 (title NI ), Axial force in the ending node: SMISC,7 (title

NJ ). Shearing force in the starting node: SMISC,2 (title TI ), Shearing force in the ending node:

SMISC,8 (title TJ ). Bending moment in the starting node: SMISC,6 (title MZI ), Bending moment in

the starting node: SMISC,12 (title MZJ ). Max. normal stress (tension+bending) in the starting node

NMISC,1 (title SMAXI), Max. normal stress (tension+bending) in the ending node: NMISC,3 (title

SMAXJ). Min. normal stress (compression+bending) in the starting node: NMISC,2 (title SMINI),

Min. normal stress (compression+bending) in the ending node: NMISC,4 (title SMINJ).

We can see the course of the shearing forces after the command PLLS,TI,TJ,1

___________________________________________________________________________________


We can see the course of bending moment after the command PLLS,MZI,MZJ,1

We can see the course of the maximum normal stress after the command PLLS,SMAXI,SMAXJ,1

___________________________________________________________________________________


3. PROGRAMME (macro) for drawing internal forces and stresses BEAM3

! Macros have been created in the internal ANSYS Programme Design Language (APDL)

! by means of using "ETABLE" commands

/POST1 ! open postprocessor

SET,1,1 ! open results of the first loading step

ETABLE,NI,SMISC,1 ! definition of axial force in starting node I

ETABLE,NJ,SMISC,7 ! definition of axial force in ending node J

ETABLE,TI,SMISC,2 ! definition of shearing force in starting node I

ETABLE,TJ,SMISC,8 ! definition of shearing force in ending node J

ETABLE,MZI,SMISC,6 ! definition of bending moment in starting node I

ETABLE,MZJ,SMISC,12 ! definition of bending moment in ending node J

ETABLE,SMAXI,NMISC,1 ! definition of max. normal stress in starting node I

ETABLE,SMAXJ,NMISC,3 ! definition of max. normal stress in ending node J

ETABLE,SMINI,NMISC,2 ! definition of min. normal stress in starting node I

ETABLE,SMINJ,NMISC,4 ! definition of min. normal stress in ending node J

PLLS,NI,NJ ! drawing course of axial forces N on display

/WAIT,8 ! wait 8 sec and looking on display

PLLS,TI,TJ ! drawing course of shearing forces Q on display


PLLS,MZI,MZJ ! drawing course of bending moment Mb on display


PLLS,SMAXI,SMAXJ ! drawing course of maximum normal stress on display


PLLS,SMINI,SMINJ ! drawing course of minimum normal stress on display

/EOF ! macro end

Macros can be created in any text editor using ASCII code with the *.mac extension. It has been saved

in the working directory. Open a macro by typing the title without the extension.

Modal Analysis 1

3. MODAL ANALYSIS

• Modal analysis and its purpose.

• Learn how to do a modal analysis in ANSYS.

• Work modal analysis exercises.

Modal Analysis 2

Description & Purpose

A modal analysis is a technique used to determine the vibration characteristics of structures:

1. natural frequencies

2. mode shapes

Modal Analysis 3

Benefits of modal analysis • Allows the design to avoid resonant vibrations or

to vibrate at a specified frequency (speaker box, for example).

• Gives engineers an idea of how the design will respond to different types of dynamic loads.

• Helps in calculating solution controls (time steps, etc.) for other dynamic analyses.

Recommendation: Because a structure’s vibration characteristics determine how it responds to any type of dynamic load, it is generally recommended to perform a modal analysis first before trying any other dynamic analysis.

Modal Analysis 4

Terminology • A “mode” refers to the pair of one natural

frequency and corresponding mode shape.

• A structure can have any number of modes, up to the number of DOF in the model.

Assumptions & Restrictions • The structure is linear (i.e. constant stiffness and mass). • There is no damping. • The structure has no time varying forces, displacements, pressures, or temperatures applied (free vibration).

Modal Analysis 5

Example: Cantilever beam

mode 1, f1 = 16.53 Hz, {0}1 ---->

mode 2, f2 = 103.4 Hz, {0}2 ---->

mode 3, f3 = 288.8 Hz, {0}3 ---->

Modal Analysis 6

Theory development • Start with the linear general equation of

motion:

• Assume free vibrations, and ignore damping:

• Assume harmonic motion:

tFKDM

0 KM

)sin(-

)cos(

)sin(

0

2

0

0

iiii

iiii

iii

t

t

t

Modal Analysis 7

• Substitute and simplify

0)(

0)sin()sin(

0

0

2

00

2

ii

iiiiiii

KM

tKtM

KM

Theory development

Last equality is satisfied if {0}i = 0 (trivial, implies no vibration) or if

0det 2 MK i

This is an eigenvalue problem which may be solved for up to N eigenvalues i

2 and N eigenvectors {0}i where N is the number of DOF.

Modal Analysis 8

Theory development

Mode shapes can be normalized either to the mass matrix or to unity, where the largest component of the vector {0}i is set to 1.

100 i

T

iM

Modal Analysis 9

Eigenvalues & Eigenvectors

• The square roots of the eigenvalues are i , the structure’s natural circular frequencies (rad/s).

• Natural frequencies fi can then calculated as

f = /2π (cycles/s)

• The eigenvectors {0}i represent the mode shapes, i.e. the shape assumed by the structure when vibrating at frequency fi.

Natural Frequencies of a cantilever Beam 1

4. Solution of the Natural Frequencies of a Cantilever Beam Using FEM

A cantilever as shown has a length L, cross-sectional area S, cross-sectional moment of

inertia J, is the material density and Yang's modulus E. Let us consider the so-called

Bernoulli beam vibration (ie, neglecting the effect of shear forces on the deformation and

ignoring the rotary inertia effects). The exact value of its own circular frequencies are

obtained by solving the frequency equation

S

JE

L

ii

.

..frequency circular own itsapply to where,0cosh*cos1

2

2

For our beam based 1 = 0.5968. and 2 = 1.4942. , the angular frequency is

S

JE

LS

JE

L .

..

03521,22,

.

..

51526,32221

Figure 1 Discretization of the cantilever beam

Now we will proceed to the discretization of the cantilever beam with the aid of the

deformation variant of FEM. To simplify the solution we choose the coarsest possible

discretization for the cantilever and to create one finite element. We will take the known

definition of the stiffness matrix [k] and mass matrix [m] of the bending element,

provided the use of a cubic polynomial approximation for the deflection of the beam. The

matrices have the following form:

.

4.

612

264

612612

2

22

3

Lsym

L

LLL

LL

L

EJk .

4.

22156

3134

135422156

4202

22

Lsym

L

LLL

LL

LSm

Symbols r1 (deflection) and r2 (rotation) at the end of the bracket indicate the global

deformation parameters (DOF). Since the boundary condition ("enshrine") in the initial

node of the element prevents both deflection and rotation, the numbers of the

corresponding deformation parameters are zero and the rows and columns of the

element's matrices are "not reflected" in the matrices of the resulting beam. The end of

the beam that is our ending node of our finite element is free, therefore the DOF numbers

are 1 and 2 The resulting matrices of our beam will have after the following form after

discretization and the application of boundary conditions:

L

S, J,

r1

r2

Natural Frequencies of a cantilever Beam 2

23 46

612

LL

L

L

EJK and

2422

22156

420 LL

LLSM

We get the frequency equation for the calculation of its own circular frequencies

F)a 2(FEM) by developing the determinant: . 0det 2

)( MK FEM

This equation has the following form

015120.1224 2232

)(

264

)( JEEJmLmL pFEMpFEM , where mp = .L.S is the

mass of the element. The result of the quadratic equation for unknown 1,2(FEM) are the

following:

S

JE

LFEM

.

.53273,32)(1

and

S

JE

LFEM

.

.80689,342)(2

.

By comparing the exact values on the circular frequency Ωi with the values of its own

circular frequency of the beam discretized by FEM F), we come to these

conclusions. The first frequency deviation from an exact solution is Δ1 ≈ 0.5% and for the

second it is Δ2 ≈ 58% (the error was calculated using the formula Δi = (F / Ωi ) .

100). In view of using the "coarsest possible" discretization cantilever by one finite

element, the value of the first eigenfrequency is a "surprisingly good" match with the

theoretical result. The second natural eigenfrequency is already useless. We would have

to use a finer discretization of the beam (at least two, preferably more, elements). We

obtain the first eigenshape from the equation

0

0

2

12

)(1r

rMK FEM , where r1 and r2 are amplitudes of harmonics vibration

of the first eigenshape. They are linearly dependent and therefore can only obtain their

ratio at a specified value of deformation parameter, for example, r1 = 1 r2 = released

1378/L. The shape of the oscillation is shown in Figure 2

Figure 2 The first eigenshape

Note: When cutting the beam into two finite elements, we get the value of its own

circular frequencies with the following error to the exact solution: Serial Number of

eigenfrequencies 1. 2. 3. 4.

Error to the

exact solution % 0,07 0,85 21,8 80,4

r

1

r2

5. CONTACTS in FEM

EASY EXAMPLE

EJ

EJ

F

2

1

STARTING EXAMPLE 1 2 3

{P0}

u1 u2 u3

Detail

1 2 3

1

2

3 P1

P2 P3

P0

=

(1)

0

0

0

0

0

0

0

0

0

000000

010000

000000

000100

000000

000001

0

0

0

0

3

3

2

2

1

1

0

3

2

1

P

P

P

u

u

u

P

PuQ

P

P

P

P

{P} ….. Vector of Final Gap [Q] ..... Identification Matrix (only 0 and 1) {u} ….. Deformation Parameters in Bending {P0} …… Vector of Starting Gap

2 Contacts in FEM

STARTING EXAMPLE

If the last equation (1) is valid, then a gaps exists between

beams. Then it holds generally to the

classic static matrix equation of FME.

FuK

vectorloading .... F

ntsdisplacemeunknown of vector .... u

matrix stiffness global .... K

3 Contacts in FEM

• PENALTY METHOD ANSYS

• LAGRANGE MULTIPLIER METHOD ANSYS

• AUGMENTED LAGRANGE METHOD ANSYS

• PARTITIONING (SEMIANALYTICAL) METHOD

MATSOL

METHODS FOR CONTACT SOLUTION

Contacts in FEM 4

EJ

EJ

F 1

2

PENALTY METHOD

Contacts in FEM 5

For two beams and artificial spring we get

bb CCC

111

1

bb CCCC

1111

2

For conditions c1 c2 must be

Penalty parameter which has been chosen too large can lead to ill-conditioning of the equation system.

Stiffness of one beam is cb = F3/3EJ. Stiffness of two beams if they have contact is

Disadvantage of easy Penalty method

Contacts in FEM 6

Energy of the system: p = ½{u}T[K]{u} - {u}T{F} + ½{P}T[]{P}

PENALTY METHOD - EQUATIONS

= ½{u}T[K]{u} - {u}T{F} + ½{u}T[Q]T[][Q]{u} + {u}T[Q]T[]{P0} + ½{P0}T[K]{P0}

We institute from equation (1) for {P} = [Q]{u} + {P0}

We use the minimum principle of potencial energy:

P / {u} = 0

We get: P / {u} = [K]{u} – {F} + [Q]T[][Q]{u} + [Q]T[]{P0} = 0

[Kp]{u} = {Fp}

[Kp] = [K] + [Q]T[][Q] and {Fp} = {F} + [Q]T[]{P0}

7 Contacts in FEM

R12=

R21= EJ

F 1

EJ

2

LAGRANGE MULTIPLIER METHOD

are Lagrange multipliers !

Contacts in FEM 8

LAGRANGE MULTIPLIER METHOD

Energy of the system: = ½{u}T[K]{u} - {u}T{F} + {}T{P}

= ½{u}T[K]{u} - {u}T{F} + {}T[Q]{u} + {}T{P0}

Again we institute from equation (1) for {P} = [Q]{u} + {P0}

Minimum principle of potencial energy: / {u} = 0

We get: / {u} = [K]{u} – {F} + [Q]T{} = 0

{u} = [K]-1{{F} - [Q]T{} } Very important equation (2)

The second variation step: / {} = [Q]{u} + {P0} = 0

We get:

0

0 P

Fu

Q

QKT

[K]{u} = {F}

9

AUGMENTED LAGRANGE METHOD

Energy of the system: A = ½{u}T[K]{u} - {u}T{F} + {}T{P} + ½{}T[1/]{}


Minimum principle of potencial energy: A / {u} = 0 and A / {} = 0

We get:

01

P

Fu

Q

QKT

[K A]{u} = {F}

1/ -> compliance [mm/N]

10

PARTITIONING (SEMIANALYTICAL) METHOD

Energy of the system: S = ½{u}T[K]{u} - {u}T{F} + {}T{P}


and equation (2)

Minimum principle of potencial energy: S / {} = 0

We get only one unknown vector : {}

01PFKQQKQ

TT

[K S]{} = {FS}

{u} = [K]-1{{F} - [Q]T{} }

S / {} = - [Q]T[K]-1[Q]T{} + {P0} + [Q][K]-T{F} = {0}

Big advantages: the number of << u

11 Contacts in FEM

8

7

2

1

8

7

6

5

4

3

2

1

2

4,4

2

3,4

2

2,4

2

1,4

2

3,4

2

3,3

2

2,3

2

1,3

2

4,2

2

3,2

2

2,2

2

1,2

2

4,1

2

3,1

2

2,1

2

1,1

1

4,4

1

3,4

1

2,4

1

1,4

1

3,4

1

3,3

1

2,3

1

1,3

1

4,2

1

3,2

1

2,2

1

1,2

1

4,1

1

3,1

1

2,1

1

1,1

M

R

0

0

0

F

M

R

0

0w

w

w

0

0w

kkkk0000

kkkk0000

kkkk0000

kkkk0000

0000kkkk

0000kkkk

0000kkkk

0000kkkk

8

7

6

5

4

3

2

1

REAL EXAMPLE – creation of global matrixes [K], {u} and {F}

φ4

w5 q

φ8

φ2

w1 w3 F

l l

φ6

w7

F

Beam stiffness k =1.75 10 6 Nmm-1

12 Contacts in FEM

REAL EXAMPLE -> open gap

46-

2

m10 8.333J

m01.0S

m1l

m1.0d

φ4

w5 q

φ8

φ2

w1 w3

F

l l

φ6

w7

F


00wEJ2

Fl

EJ3

Flw 65

2

4

3

3

0

0

0

4600

61200

0046

00612

6

5

4

3

6

5

4

3

2

23

2

23 F

w

w

lEJ

l

EJ

l

EJ

l

EJ

lEJ

l

EJ

l

EJ

l

EJ

Only for q+w3 > 0 -> open gap

13 Contacts in FEM

REAL EXAMPLE -> PENALTY METHOD

6543

0

0

53

0101

wwu

QuQPP

qPwwqP

T

0

0

4600

6120

0046

0612

6

5

4

3

6

5

4

3

2

23

2

23

q

qF

w

w

lEJ

l

EJ

l

EJ

l

EJ

lEJ

l

EJ

l

EJ

l

EJ

= spring stiffness

14 Contacts in FEM

REAL EXAMPLE -> PENALTY METHOD -> RESULTS

Spring stiffness

(Nmm-1)

error (%) for

F=2.5F0

105 41.71

106 31.43

107 8.91

108 1.10

Small stiffness = 105 Big stiffness = 107

Small stiffness 105

Big stiffness 107

q

F

-w3


F0 2.5F0

2q

10 5

10 6

10 7

Exact solution

Spring stiffness

15 Contacts in FEM

REAL EXAMPLE -> PENALTY METHOD -> RESULTS

16 Contacts in FEM

EXAMPLE -> AUGMENTED LAGRANGE METHOD -> RESULTS

01

P

Fu

Q

QKT

[Q] => {1 0 -1 0}

{} =>

[1/] => 1/

q

F

w

w

lEJ

l

EJ

l

EJ

l

EJ

lEJ

l

EJ

l

EJ

l

EJ

0

0

0

/10101

04600

16200

00046

100612

6

5

4

3

2

23

2

23

17 Contacts in FEM


23

3

3

3

3

l

EJEJ

Flq

F

w

26)(

3

3

q

EJ

Flw

18 Contacts in FEM


19 Contacts in FEM

EASY EXAMPLE -> PARTITIONED METHOD -> RESULTS

first step: solution of contact force [K S]{} = {FS}

[KS ] = [Q][K]-1 [Q]T = 2. 3 / 3EJ

{FS } = [Q][K]-1 {F} + { P0} = -F. 3 / 3EJ + q

= - F /2 + 3/2 .q . EJ/3

second step : solution of displacement {u} {u} = [K]-1{{F} - {} [Q]}

0

1

0

1

0

0

0

6300

3200

0063

0032

6 323

21

2

2

6

5

4

3

l

EJqF

F

l

ll

l

ll

EJ

lw

w

20 Contacts in FEM

introduction of linear buckling

geometric stiffness matrix for straight-lined elements

practical examples

nonlinear buckling

practical examples

6. FEM - Tool for Buckling Solution

1.0 BASIC FEM MATRIX EQUATIONS IN STATICS

approximation of displacement

rNaxu T

(1)

[x] …matrix of approximation function

{a} …constant values of approximation polynomial

[N] …shape function

{r} … vector of DOF for element

strain-displacement relations

rBuε (2)

{ε}… vector of strain components

[δ] … operation matrix

[B] … matrix

elastic stress-strain relations

σCε (3)

[C] …value of the material parameters

{σ} … vector of stress components

2 FEM - Tool for Buckling Solution

1.0 BASIC FEM MATRIX EQUATIONS IN STATICS

strain energy

rkrrBCBrσεTTT

V

T

d dV2

1)(

2

1dV

2

1 . (4)

potential of external loading

V

TTT

V

T

z frzNrzu -dV)(-dV (5)

{z} … loading components

{f} … equivalent nodal loading

basic static equation of FEM

variation of the total potential functional

frkr

r

0 , (6)

and for the solving system

FRK . (7)


• Intuition Y1>Y2

• Stiffness of tensioned bar is less than compressed bar i.e. influence of

geometric stiffness matrix [[K] + [KG]]

Matrix [ KG ] depends on loading vector { F }. The increasing of loading

vector we can describe as { Fo } i.e. {F} = λ . { Fo } and then real values of [

KG ] is now λ . [ KG ].

1.1 „LINEAR“ BUCKLING

V V F F

Y1 Y2

oGKK Fr oG FKKr 1

0det GKK oKR FF

0 rGKK 4

1.2 GEOMETRIC STIFFNESS MATRIX [kG] OF BAR ELEMENT

The bar element has very easy basic equations. The approximation function can be

a linear i.e. u = b0 + b1ξ, where ξ is a new variable ξ = x/L ( L is lenght of element ).

Equation (1) has a form

rNT

r

ru

2

11 , (9)

Fig. 2

v

r3 r1

r2

r4

u


rNuT

r

r

r

r

v

u

4

3

2

1

0

0

10

01

. (10)

2

2

1

x

v

x

ux , xx E .

dxx

v

x

v

x

u

x

uAEdx

x

v

x

uAELL

d

4

0

22

0

22

4

1

22

1

2

rkrrkr

rrrr

GT

TT

TT

dL

N

L

AE

2

1

2

1

1010

0000

1010

0000

2

1

0000

0101

0000

0101

2

1

(13)


(11) (12)

6

12

L , AE

v2*L, AE

r1

Fo

000

022

22

1

2211

11

22det 0

F

LL

AE (14)

[K] [KG]

From equation (14) we obtain for λ :

00

2612,0221

1

F

AE

F

AE

(15)


simple example



simple example


FRKKK GNG .

1.4 NONLINEAR GEOMETRIC STIFFNESS MATRIX [kGN]

OF BAR ELEMENT

(16)

(17)

4

4

1

x

v,

1010

0000

1010

0000

2

2

423rr

L

AEGNk .


1.5 GEOMETRIC STIFFNESS MATRIX [kGo] OF PLANE FRAME

ELEMENT (bending + tension)

Z,w

r4

r1

r2

r5

X,u

r3

r6

r

T

LLw

u

rN

r

r

r

r

r

r

w

6

5

4

3

2

1

)(23

00

0)2(

0

2310

0132323232

2

2

2

2

1

x

wz

x

w

x

ux (24)

(19) 10 FEM - Tool for Buckling Solution

rGorrorrTr rkrrkrrkrTTT

LL L L

L

A

L

A

d

x

wEAdx

x

w

x

uEAdx

x

wEJdx

x

uEA

dxdAx

w

x

u

x

w

x

wz

x

w

x

u

x

wz

x

w

x

uE

dxdAx

wz

x

w

x

uE

2

1

2

1

2

1

4

1

2222

.24

1

2

.2

1

2

0

4

0 0 0

22

2

22

0

22

2

2

2

24

2

2

2

22

0

22

2

2

stiffness matrix stiffness matrix geometric stiffness matrix

of bar element of bending element of bending element




15

2

10

10

3010

10

10

1

5

60

10

1

5

60

0000003010

10

15

2

10

10

10

1

5

60

10

1

5

60

000000

LLLL

LLLL

NoGk . (27)




L

E, J

Fo r1

r2

0

15

2

10

105

6

1

46

612

det 2

2

23

LL

L

L

L

EJ

L

EJL

EJ

L

EJ

, (27)



SIMPLE EXAMPLE


FKRE = π2EJ/(4L2) ≈ 2,4674 EJ/L2

Only 1 element, 2 degrees of freedom

FKR = 2,486 EJ/L2

difference from exact solution cca 0,8 %



SIMPLE EXAMPLE


1.6 LINEAR BUCKLING - PRACTICAL EXAMPLES

a) buckling of steel arch tunnel support - influence of filling (stowing)

good stowing λ1 = 31

Support without stowing λ1 = 19,5


BULK MATERIAL DENSITY

= 1200 kg/m3 , f = 1,3

ANGLE OF INTERNAL

FRICTION = 25O

STEEL DENSITY = 7850

kg/m3 , f = 1,1

ELEMENT TYPES: SHELL,

BEAMS, CONTACT

MASS

TOTAL No. of ELEMENTS:

4362 ( SHELLS=4036)

DOF 22248

THE HIGHEST LEVEL OF

ASH -2m BELOW

THE ROOF

ASH VOLUME ~508 m3

MASSASH 792,5 t

GRAV.FORCEASH 7,93MN


a) buckling of the ash silo structure


• Collapse shape

• = 1,32


a) buckling of an ash silo structure - collapse shape


2.0 NONLINEAR BUCKLING ANALYSIS

2.1 algorithm

FRK R .

in

nr

in

a

ni

T

n,i ,, ZFFRK (30)

Three types of nonlinearities:

geometric (large strain, ….)

material properties ( creep, hyperelasticity, bilinearn or multilinear models, … )

structural (contact, …).

n … n-th step of loadingi … i- th iteration step into one loading step

[KT

n,i] … tangential stiffness matrix

{ΔRi} … addition of DOF vector in i- th iteration step

{Fan} … final value of loading vector into n-th step

{Fnr

n,i} … value of loading vector into n-th step and i-th iteration step

{Zn,i} … rest value of loading vector into n-th step and i-th iteration step


Ri

R Ri+1 Ri+2

Zn,i+1

Zn,i+2

iii RRR 1 . (31)

refZin Z., Z and/or refRi R.R , kde (32) a (33)


2.1 algorithm



2.2 course of solution

The critical vector of the load is obtained when a "very small" loading

vector {Fan} creates an extreme increase of the DOF vector {ΔRi}.

The solution is non-convergent, the solution is divergent. The last value

of the loading vector when the solution is convergent – this is the value

of the real critical vector {Fkr}.


Shape of arch support

Detail of clamping connection

2.0 NONLINEAR BUCKLING - practical examples

2.3 steel arch tunnel support


Course of loading

enlarged 3 times

Course of displacement for two

nodes in contact in clamping

connection


2.3 steel arch tunnel support


R1 = 1724 mm

90o

R2= 2494 mm

74o

545

545

3900

3448


2.4 grid tunnel support in shotcret tunnels


Shape of collapse ( enlarged 10 x )




Cumulative plastic strain

( displacement enlarged 3 x )



25

3.0 Conclusion Linear buckling analysis should be the first step of the

solution. Then the second step, namely the nonlinear

solution, should be made. We get two very important results

from linear buckling – namely:

• values of the first critical loading vector are “top

values” for a nonlinear solution

• shape of collapse from linear buckling should be a

starting imperfection for a nonlinear solution


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FEM Finite Element Methods Introduction - vsb.cz · 1 VŠ – Technical University of Ostrava...

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