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POTENTIAL ENERGY APPLIED IN FINITE ELEMENT METHOD

bySuhaimi Abu Bakar, PhD

INTRODUCTION

Finite difference method is suitable to used if the structure to be analysed has the specific governing differential equation.

For structures which are so complex that is difficult or impossible to determine the governing differential equation, a powerful method for analysing such complex structures is the finite element method.

The way finite element analysis obtains the temperatures, stresses, flows, or other desired unknown parameters in the finite element model are by minimizing an energy functional. An energy functional consists of all the energies associated with the particular finite element model. Based on the law of conservation of energy, the finite element energy functional must equal zero.

The finite element method obtains the correct solution for any finite element model by minimizing the energy functional. The minimum of the functional is found by setting the derivative of the functional with respect to the unknown grid point potential for zero. Thus, the basic equation for finite element analysis is:

0Fp

∂=

∂

where F is the energy functional and p is the unknown grid point potential (In mechanics, the potential is displacement) to be calculated. This is based on the principle of virtual work, which states that if a particle is under equilibrium, under a set of a system of forces, then for any displacement, the virtual work is zero. Each finite element will have its own unique energy functional.

PRINCIPLE OF MINIMUM POTENTIAL ENERGY

For conservative systems, of all the kinematically admissible displacement fields, those corresponding to equilibrium extremize the total potential energy. If the extremum condition is a minimum, the equilibrium state is stable.

Satisfy the single-valued nature of displacements (compatibility) and the boundary conditions.

SPRING EXAMPLE

δW

spring stiffness, k

212

0 :

0

k W

ddd k Wd

δ δ

δ

δδ

Π = −

Π=

Π= − =

We get equation of equilibrium satisfy equilibrium condition

internal force in the spring

external force

minimum potential energy

work done

strain energy

potential energy

δ is an equilibrium displacement, if displacement < δ, the spring still moving (inequilibrium) has kinetic energy

inequilibrium

equilibrium

A

STRAIN ENERGY

When loads are applied to a body, they will deform the materials. Provided no energy is lost in the form of heat, the external work done by the loads will be converted into internal work called strain energy. This energy, which is always positive, is stored in the body and is caused by the action of either normal or shear stress.

STRAIN ENERGY DUE TO NORMAL STRESS

dy

dz

dxThe force σxdydz does work on an extension εxdxσx

Work done during deformation, dW= Area of triangle OAB

Aσxdydz12 x xdW dxdydzσ ε=

or12 x xdW dVσ ε=

εxdxStrain energy is an amount of energy stored in the material due to work done

O B

In general, if the body is subjected only to a uniaxial normal stress σ, acting in a specified direction, the strain energy in the body is then

2VW dVσε

= ∫

Also, if the material behaves in a linear-elastic manner, Hooke’s law applies, σ=Eε, and therefore:

2

2VW dV

Eσ

= ∫

STRAIN ENERGY DUE TO SHEAR STRESS

τγ

γdzdxdy

The force τdxdy does work on a shear slip γdzdz

Work done during deformation, dW= Area of triangle OAB

( )12

dW dxdy dzτ γ=

or12

dW dVτγ=

Aτdxdy

γdz Strain energy is an amount of energy stored in the material due to work done

O B

In general, if the body is subjected only to a shear stress τ, the strain energy due to shear stress in the body is then

2VW dVτγ

= ∫

Also, if the material behaves in a linear-elastic manner, Hooke’s law applies, τ=Gγ, therefore:

2

2VW dV

Gτ

= ∫

STRAIN ENERGY DUE TO MULTIPLE STRESSES

σx

σy

τxy

τxz

τzxτzy

τyxτyz

σz

( )12 x x y y z z xy xy yz yz xz xzdW σ ε σ ε σ ε τ γ τ γ τ γ= + + + + +

or

12

TdW dV= ∫σ εΩ

z

y

wherex

x

y

z

xy

yz

xz

σσστττ

=

σ

x

y

z

xy

yz

xz

εεεγγγ

=

ε

STRAIN ENERGY – PURE BENDING

22

2

2

2

2

1212

12

12

1212

x xV

x

A

A

A

dU dV

My dA dxI E

My My dAdxI EI

M y dAdxEI

M IdxEIM dxEI

σ ε

σ

=

= =

=

=

=

∫

∫

∫

∫

dxx

le

q

dAN.A

x

Strain energy stored due to bending of a very small length of beam element, dx

STRAIN ENERGY IN BEAM DUE TO TRANSVERSE SHEAR

dAN.Ay

z

xτ

VQIb

τ =

22 12 2V V

VQW dV dAdxG G Ibτ = =

∫ ∫x

V 2 2

2 20 2L

A

V QW dA dxGI b

=

∫ ∫

2

2 2s A

A Qf dAI b

= ∫Define the form factor for shear as

Substituting into the above equation, we get2

0 2L sf VW dx

GA= ∫

EXTERNAL WORK – BAR ELEMENT

Consider the work done by an axial force applied to the end of the bar. As the magnitude of force, F is gradually increased from zero to some limiting value F=P, the bar displaced from x=0 to x=∆.

External Work,W F

( )

( )( )

0

0

0

2 2

1212

dx

kx dx

k xdx

k

k

P

∆

∆

∆

=

=

=

= ∆

= ∆ ∆

= ∆

∫∫∫

F

O ∆

P

x ∆

P

x

EXTERNAL WORK DUE TO BODY FORCES

x

bx

bz

by displacementdx dy

dz

original position

final position

z

y

bx, by, bz = Body forces per unit volume with respect to x, y and z axes, respectively

x

y

z

ddd

=

dx

y

z

bbb

=

b Work due to body forces =T dV

Ω∫d b

EXTERNAL WORK DUE TO SURFACE TRACTIONS

x

qx

qz

qy

dS qx, qy, qz = Surface tractions with respect to x, y and z axes, respectively

y

z

x

y

z

qqq

=

q Work due to surface tractions =T dS

Γ∫d q

CONSERVATION OF ENERGY

Most energy methods used in mechanics are based on a balance of energy, often referred to as the conservation of energy. The energy developed by heat effects will be neglected. As a result, if a loading is applied slowly to a body, so that kinetic energy can also be neglected, then physically the external loads tend to deform the body so that the loads do external work We as they are displaced. This external work caused by the loads is transformed into internal workor strain energy Wi, which is stored in the body. Furthermore, when the loads are removed, the strain energy restores the body back to its original undeformed position, provided the material’s elastic limit is not exceeded. The conservation of energy for the body can therefore be stated mathematically as

e iW W=

CONSERVATION OF ENERGY - EXAMPLEP

∆

Beam

2

0

12 2

L MP dxEI

∆ = ∫

P∆

We assumed that the load is gradually increased from 0 to P

Truss

212 2

N LPAE

∆ =∑

DIFFERENT CONCEPT IN ENERGY METHODExternal Work = Internal Work

21 12 2

Wx kx=Wxk

=

strain energyGradually increase the load up to W (let say 0 kN to W kN)

Equilibrium Principle

Internal load in spring

External load

Initial position

Kinetic energy zeros at this pointEquilibrium position

velocity, v=0

W kx= Wxk

=

Potential Energy Concept21

2

0

kx Wx

d kx Wdx

Π = −

Π= − =

gives Wxk

=

static

dynamic

gives

Same answergives

THE FOUR-NODE QUADRILATERAL (2D)

u111 2

34

u21

u12

u22

u14 u1

3

u1

u2ξ

η

1 2 3 41 1 1 2 1 3 1 4 1u N u N u N u N u= + + +

1 2 3 42 1 2 2 2 3 2 4 2u N u N u N u N u= + + +

1

2

uu

=

d

1112212231324142

uuuuuuuu

=

u

1 2 3 4

1 2 3 4

0 0 0 00 0 0 0N N N N

N N N N

=

N

( )( )11 1 14

N ξ η= − −

( )( )21 1 14

N ξ η= + −

( )( )31 1 14

N ξ η= + +

( )( )41 1 14

N ξ η= − +

u24 u2

3

d=Nu

where

N=Shape function

1

2

0

0x

y

xy

xuuy

y x

εεγ

∂ ∂

∂ = ∂ ∂ ∂ ∂ ∂ 0

0

x

y

y x

∂ ∂

∂ = ∂

∂ ∂ ∂ ∂

Lwhere= Ldε

= LNuε

where= Buε =B LN

STRAIN-STRESS RELATIONSHIP

1 0 0 0

1 0 0 0

1 0 0 0

10 0 0 0 0

10 0 0 0 0

10 0 0 0 0

x x

y y

z z

xy xy

yz yz

zx zx

E E E

E E E

E E E

G

G

G

ν ν

ν νε σε σν νε σγ τγ τγ τ

− − − − − − =

STRESS-STRAIN RELATIONSHIP

1 0

1 0

10 0

x x

y y

xy xy

E E

E E

G

ν

ε σνε σ

γ τ

− = −

For plane stress:

σ = Dε

2

1 01 0

110 0

2

x x

y y

xy xy

Eσ ν εσ ν ε

ντ ν γ

= − −

POTENTIAL ENERGY

( )

( )

( ) ( ) ( )

( )

( ) ( )

( )

12121212121212

T T T

T T T

T T T

T T T

TT T T T T

TT T T T T

TT

dV dV dS

dV dV dS

dV dV dS

dV dV dS

dV dV dS

dV dV dS

dV

Ω Ω Γ

Ω Ω Γ

Ω Ω Γ

Ω Ω Γ

Ω Ω Γ

Ω Ω Γ

Ω

Π = − −

= − −

= − −

= − −

= − −

= − −

= −

∫ ∫ ∫

∫ ∫ ∫

∫ ∫ ∫

∫ ∫ ∫

∫ ∫ ∫

∫ ∫ ∫

∫ Τ

σ ε

ε ε

Β

TT

d b d q

D d b d q

DBu Bu d b d q

DBu Bu Nu b Nu q

u DB Bu u N b u N q

u DB B u u N b u N q

u DB u T T T TdV dSΩ Γ

−∫ ∫u N b u N q

( ): T T TNote PQ Q P=

ee

Π = Π∑

Thus,

( )

( )

( )

12

12

12

12

e e

e e

e e

TT T T T T Te

TT T T T T

TT T T T

T T

dV dV dS

dV dV dS

dV dV dS

Ω Ω Γ

Ω Ω Γ

Ω Ω Γ

Π = − −

= − −

= − +

= −

∫ ∫ ∫

∫ ∫ ∫

∫ ∫ ∫

Τ

Τ

Β

Β

u B DB u u N b u N q

u DB u u N b u N q

u DB u u N b N q

u ku u f

k=element stiffness

12

T Te −u ku u f :

( )

1 1 1 1 1 41 11 1 1 12 2 1 1 21 1 1 1 1 4

1 1 42 21 1 2 22 2 2 2 21 1 2 2 2

4 1 4 1 4 42 1 1 2 2 2 2 2

12

n

ne n

n n nn

u k u u k u u k u

u k u u k u u k uu f u f u f

u k u u k u u k u

+ + + +

+ + + + Π = − + + + + + + + Note: n=2x4=8

Consider 4 nodes/elementΠ =from

11

0 :e

u∂Π

=∂

Taking the derivative

( )1 1 4 1 411 1 12 2 1 2 2 21 2 1 11

1

1 1 1 1 02 2 2 2

en nk u k u k u u k u k f

u∂Π = + + + + + + − = ∂

or

( ) ( )1 1 411 1 12 2 1 2 1 0 ( 1)nk u k u k u f A+ + + − =

Taking the derivative 12

0 :e

u∂Π

=∂

( )

1 1 1 1 1 41 11 1 1 12 2 1 1 21 1 1 1 1 4

1 1 42 21 1 2 22 2 2 2 21 1 2 2 2

4 1 4 1 4 42 1 1 2 2 2 2 2

12

n

ne n

n n nn

u k u u k u u k u

u k u u k u u k uu f u f u f

u k u u k u u k u

+ + + +

+ + + + Π = − + + + + + + +

original equation

( )1 1 1 4 41 12 21 1 22 2 2 2 2 2 21

2

1 1 1 1 02 2 2 2

en nu k k u k u k u u k f

u∂Π = + + + + + + − = ∂

( ) ( )1 1 421 1 22 2 2 2 2 0 ( 2)nk u k u k u f A+ + + − =or

42

0 :e

u∂Π

=∂

Similarly, taking the derivative

( ) ( )1 1 41 1 2 2 2 0 ( ' ')n n nn nk u k u k u f A n+ + + − =

Collecting equations A1, A2, …, A’n’ one gets the element equilibrium equation:

ku-f = 0

vector

( )e

TdV

Ω

= ∫ ΤΒk DB

where

e

T TdV dSΩ Γ

= +∫ ∫f N b N q

( )

( ) ( )

( ) ( ) ( )

( ) ( )

e

e

e

e

e

T

TT T

TT T T

T T

T

dV

dV

dV

dV

dV

Ω

Ω

Ω

Ω

Ω

=

=

=

=

=

∫

∫

∫

∫

∫

ΤΒk DB

DB B

B D B

B D B

B DBT =D D

0e

u∂Π

=∂

by minimizing the potential energy,

Load vector

stiffness matrix

PROCEDURE OF FINITE ELEMENT ANALYSIS

• Assemble the element equilibrium equation (ku=f) for all finite elements in the structure and form global equilibrium equation, KU=F

• Consider boundary condition and modify the K matrix, Uand F vectors. Form the following equation: AX=B where X = unknown vectors, B = known vectors.

• Solve the equation AX=B• Calculate other parameters – displacements, strains, stresses

etc