1 Some Ideas Behind Finite Element Analysis. 2 The Engineering Approach Consider a 2-D example in linear elasticity: A flat plate subjected to forces

1

Some Ideas Behind Finite Element Analysis

2

The Engineering Approach Consider a 2-D example in linear elasticity:

A flat plate subjected to forces applied to its edge

The forces are in the plane of the plate

The plate is divided up into triangular elements

3

The Engineering Approach

Consider the element e with nodes i, j and m.

4

The Engineering Approach

The element communicates with its neighbours by forces

At node i, the force { } has two components, and

parallel to the x and y axes.

All six components of force are represented by a vector

eiF iU iV

5

The Force Vector

The vector of nodal forces can be expressed in a number of ways

where

em

ej

ei

e

F

F

F

F

i

iei V

UF

6

The Force Vector

The force vector for the complete element is written as,

m

m

j

j

i

i

e

V

U

V

U

V

U

F

7

The Displacement Vector

Each node has two displacement components, u and v (2D

analysis)

The displacement of node i parallel to the x-axis is denoted by

and that parallel to the y-axis by

The vector of nodal displacements, is

where

m

j

ie

i

ii v

u

iU

iV

e

8


The complete displacement vector is

m

m

j

j

i

i

e

v

u

v

u

v

u

9


If the nodes are displaced by the same vector, the element is

translated without strain

This is known as rigid body displacement

If the nodes rotate by the same amount, the element rotates without

strain

This is known as a rigid body rotation

The nodal forces are zero for a rigid translation or rotation

The mechanical properties of the material are needed to make the

link between nodal displacements and nodal forces

10

The stiffness matrix

The nodal forces are related to the nodal displacements by the

element stiffness matrix, [K]e

m

i

m

i

v

u

a

aa

aa

V

U

.

.

.

.

||

||

||

||

||

||

.

.

.

.

61

2221

1211

11

The Finite Element Equation

The 6x6 element stiffness matrix is patitioned into nine 2x2

submatrices, so that

where the partitions of the stiffness matrix is,

m

j

i

emm

emj

emi

ejm

ejj

eji

eim

eij

eii

e

em

ej

ei

kkk

kkk

kkk

F

F

F

2221

1211

aa

aak eii

12

The Finite Element Process

Once the element stiffness matrices are calculated then they are

assembled into a global stiffness matrix for the entire model

The externally applied force {F1} produces nodal forces {Fl}, {Fm} and

{Fn} acting on the elements l, m and n

13


For a simple mesh consisting of three elements the global stiffness

matrix is found as follows

14


The element stiffness matrices are;

133

123

122

113

112

111

1

ksym

kk

kkk

k

Element 1 Element 2 Element 3

244

234

233

224

223

222

2

ksym

kk

kkk

k

355

345

244

335

334

333

3

ksym

kk

kkk

k

15


The global stiffness matrix, K, is assembled using the principle of

superposition:

355

354

353

345

344

244

343

243

242

335

234

232

333

233

133

232

132

131

224

223

123

222

122

121

113

112

111

00

0

0

00

kkk

kkkkkk

kkkkkkkkk

kkkkkk

kkk

K

16


In this way, if the vector of external forces applied to the various

nodes is [F1 F2 F3 F4 F5 ]T ,then

5

4

3

2

1

35

34

24

33

23

13

21

11

11

5

4

3

2

1

K

F

FF

FFF

FF

F

F

F

F

F

F

17


It is possible to solve for the nodal forces given a complete set of

nodal displacements

The more common problem is to solve for the nodal displacements

given a set of externally applied forces

18


The stiffness matrix has a strong leading diagonal

The matrix is square, symmetric and is positive definite

At this point, the assembled stiffness matrix is singular

No solution can be obtained

Physically the model is not fixed in space

All rigid body motions must be restrained before a solution is

possible

19


Adequate restraints have to be inserted

Only one set of nodal displacements is possible for any set of

imposed nodal forces.

The nodal displacements are found by solving a set of simultaneous

equations

From the displacements the strains are calculated

From the strains the corresponding stresses are found

20

The Principles of Virtual Displacement

Consider a particle acted on by a number of forces,

A virtual displacement of the particle is imagined to be so small that the

forces applied to the particle remain unchanged in magnitude and

direction.

21


In equilibrium the resultant force acting on the particle is zero

The work done by a force is equal to the work done by its

components

Forces acting on particle are considered part of net force

The net force is zero (for equilibrium)

The virtual work done by the virtual displacement is zero

22


Consider an elastic body represented by a system of two

particles joined by an elastic spring

A possible virtual displacement is indicated by the vector ds

involving displacement of the particle 2.

23


The force exerted by the spring is equal to the spring constant, k,

multiplied by the extension of the spring , Δ

Applying the principle of virtual work

0... 21 sksFsF

24


The individual contributions have been given in terms of the

appropriate scalar product

(kΔ).δs represents the change in the stored energy in the spring

The change in the elastic stored energy caused by the virtual

displacement has to be taken into account

0... 21 sksFsF

25


Consider an elastic solid in equilibrium before and after the

imposition of a virtual displacement to a system of forces applied to

it.

26


The virtual displacement gives rise to virtual strains

The virtual strains are compatible with the virtual displacements of

the forces acting on the solid

The stresses remain unchanged by the virtual displacement

The change in strain energy per unit volume caused by the virtual

displacement is}{ { } .{ }T

27


The total change in the stored strain energy in the body is

If there are n external forces, each with components in the

reference x, y, and z direction, such as

and each undergoes its own virtual displacement

Vol

T dVolU

Tziyixii PPPP ,,,

Tziyixii ssss ,,,

28


The work done by the external forces equals the increase in the

stored energy, by the Principle of Virtual Work,

The left-hand side of this equation can be replaced by an integral

over an area in the case of distributed loads

The Principle of Virtual Displacements is used to find nodal forces

in formulated element stiffness matrices

dVoldsPn

i Vol

Ti

Ti

1

29

The Theory of Minimum Total Potential Energy

It was found using the Principle of Virtual Work that

The left-hand side of this equation is the work done by the external forces

as a result of the set of virtual displacements

This is the loss in the potential energy of the system of external forces

The right-hand side is the increase in the strain energy of the elastic body

dVoldsPn

i Vol

Ti

Ti

1

30

The Theory of Minimum Total Potential Energy

If, is the potential energy of the system, then

In the equilibrium condition the potential energy of the system has

an extreme value of a maximum or a minimum.

01

dVolspdTn

i Vol

iT

i

31

Shape Function

The displacement of any point within an element is a function of the

nodal displacements and of the position of the point. This is

expressed by means of shape functions.

32

Shape Function

For any point P with coordinates (x, y) the displacement can be

expressed as

where

and

Each shape function is a function of position, Ni=Ni (x,y)

uNuNuNuNu Tmmjjii

mjiT NNNN

mjiT uuuu

33

Conditions the shape functions must fulfil

At node i, the displacement is given in terms of the nodal

displacements

For this to be true for any value of nodal displacement, it can be

seen that Ni must equal 1 and the shape functions at all other

nodes zero. The same is true for all the other shape functions, i.e.

mmjjiii uNuNuNu

pq

pqforyxN qqp

0

1

34


The displacement must be continuous between elements

The shape function of a node is zero along a boundary not

containing that node.

The shape functions should be able to represent rigid translation,

i.e.

This condition requires that Ni + Nj + Nm=1 for all points in the

element

mji

mjimmjjii NNNNNN

35


The shape functions must be able to represent rigid rotation

36


A point P, with coordinates (x, y) is subjected to a small rotation

about the origin, causing it to move through a distance rө. The

component of the displacement in the x direction, u, is given by

and since

This can be written as;

sinru

r

ysin

yu

37


Similarly for the nodal displacement

By substituting these values into the nodal displacement equation,

hence

Similarly

mm

jj

ii

yu

yu

yu

yuNuNuNu mmjjii

mmjjii yNyNyNy

mmjjii xNxNxNx

38


For triangular elements, suitable shape functions can be found from the

area and sub-areas of the triangle, these are known as area shape

functions

( )

( )imp

jimj

areaN

area

39

Strain-Displacement Relationship

• The strains at any point in a 2-D state of strain are given by

• In matrix form

x

v

y

u

y

v

x

uxyyx

,,

v

u

xy

y

x

xy

y

x

0

0

40


For a three node triangular element the strain-displacement

relationship can be expressed as,

m

m

j

j

i

i

mmjjii

mji

mji

v

u

v

u

v

u

x

N

y

N

x

N

y

N

x

N

y

Ny

N

y

N

y

Nx

N

x

N

x

N

000

000

41


The strain-displacement relationship is generally expressed in

matrix form as,

where [B] is called the strain matrix

Similarly the virtual displacements and strains are related by,

eB

TeTTB**

42


The strains are related to the stresses by Hooke’s law. The

elasticity matrix [D] for plane stress is,

where E is Young’s modulus of elasticity and is Poisson ratio

The [D] matrix is always symmetrical even when the material

exhibits anisotropic (but linear) elasticity.

2

1 0

1 01

10 0

2

x

y

xy

vE

v D

v

43


The stress can be expressed in terms of the displacement by

combining the strain-displacement and stress-strain relationships;

The stresses and strains throughout the element can be calculated

from the nodal displacements.

eBD

44

Use of the principle of virtual displacements

Imagine a set of virtual nodal displacements is applied to the element.

Since it is considered that the deformation of the element is caused by

the nodal forces, we equate the virtual work done by these forces when

the nodes undergo their virtual displacements to increase the elastic

potential energy of the element.

{e*} is the virtual strains caused by the virtual nodal

displacements {d*}e

dVolvVuUFT

Vol

mmiieeT ** .......

45


In general {e*} will be a function of position within the element,

although in this case, stress and strain are constant over the

element.

The stress-displacement relationship is;

The virtual strain-displacement relationship is;

eBD

TeTTB**

46


Thus

The nodal displacements are not functions of position;

Since this is true for any arbitrary set of we have

dVolBDBdVol eTeT

Vol

T

Vol

**

eVol

TeTT

Vol

dVolBDBdVol **

eVol

Te dVolBDBF

* e

47


This is the finite element equation for the element

From the definition of the element stiffness matrix,

by comparing previous equations we have the stiffness matrix

eVol

Te dVolBDBF

eee KF

Vol

Te dVolBDBK

48


u and v are linear functions of x and y

The strains and stresses are therefore constant

[B]T[D][B] is thus constant

The components of [K]e are therefore the components of [B]T[D][B]

multiplied by the volume of the element

In other higher order elements this is not strictly true and numerical

methods of integration are frequently used

[K]e must be symmetrical ([K]eT=[K]e)

49

The Use of the Principle of Minimum Potential Energy

The elastic energy stored per unit volume of a deformed body is

In the system consisting of the triangular element and the nodal

forces, taking the zero of potential energy to be when the points of

application of the nodal forces are at the respective nodes in the

un-deformed element. Then the potential energy of the system is,

TU2

1

dVolFT

Vol

eeT 2

1

50


The potential energy can be written as,

and are constant over the element and can be taken

outside the integral

Vol

eTeTeeT dVolBDBF 2

1

eVol

TeTeeT dVolBDBF 2

1

* e * eT

51


Let us denote the square 6x6 matrix in the integral by [A]

In equilibrium is a minimum with respect to any of the nodal

displacements

Vol

T dVolBDBA

0.....

mii vvu

52


Taking the first of these as an example and noting that

then

sr

sr

u

u

s

r

0

1

TeT

ei

i

A

AUu

0000012

1

0000012

10

53


This can be written explicitly as

It has been noted that [A] is symmetrical and the last two terms of

this equation are therefore equal

Aofcolumnfirst

AofrowfirstUu

eT

ei

i

2

1

2

1

ei AofrowfirstU 2

1

54


In general

By comparing this with the equation

we see that

ei AF

eVol

Te dVolBDBF

e T

V

K A B D B dV

55


In this particular case of the uniform-strain triangular element, since

the integral is a constant in this equation

It can be written as

is the area of the triangular element and t is its thickness

ei AF

tBDBK Te

Documents

1 Some Ideas Behind Finite Element Analysis. 2 The Engineering Approach Consider a 2-D example in linear elasticity: A flat plate subjected to forces