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Introduction to Finite Elements

FEM Discretization of

2D Elasticity

Reading assignment:

Lecture notes

Summary:

• FEM Formulation of 2D elasticity (plane stress/strain)

•Displacement approximation

•Strain and stress approximation

•Derivation of element stiffness matrix and nodal load vector

•Assembling the global stiffness matrix

• Application of boundary conditions

• Physical interpretation of the stiffness matrix

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Recap: 2D Elasticity

x

y

Su

ST

Volume (V)u

v

x

px

py

Xa dV

X b dVVolume

element dV Su: Portion of the

boundary on which

displacements are

prescribed (zero or

nonzero)

ST: Portion of the

boundary on which

tractions are prescribed

(zero or nonzero)

Examples: concept of displacement field

x

y

3

2 1

4

2

2

Example

For the square block shown above, determine u and v for the

following displacements

x

y

1

4

Case 1: StretchCase 2: Pure sheary

2

21/2

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Solution

Case 1: Stretch

2

yv

xu

−=

=

Check that the new coordinates (in the deformed configuration)

2

2

'

'

yv y y

xu x x

=+=

=+=

Case 2: Pure shear

/ 4

0

u y

v

=

=

Check that the new coordinates (in the deformed configuration)

'

'

/ 4 x x u x y

y y v y

= + = +

= + =

=y)(x,v

y)(x,uu

=

xy

y

x

γ

ε

ε

ε

∂

∂

∂

∂∂

∂∂

∂

=∂

x y

y

x

0

0

=

xy

y

x

τ

σ

σ

σ

u D D

u

y xuu

∂==

∂=

=

ε σ

ε

LawStrain-Stress

RelationntDisplaceme-Strain

),(fieldntDisplaceme

Recap: 2D Elasticity

−−=

2

100

01

01

1 2 ν ν

ν

ν

E D

For plane stress

(3 nonzero stress components)

( )( )

−−

−

−+=

2

2100

01

01

211 ν ν ν

ν ν

ν ν

E D

For plane strain

(3 nonzero strain components)

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V in X T

0=+∂ σ Equilibrium equations

Boundary conditions

1. Displacement boundary conditions: Displacements are specified on

portion Su of the boundary

u

specified S onuu =

2. Traction (force) boundary conditions: Tractions are specified on

portion ST of the boundary

Now, how do I express this mathematically?

Strong formulation

But in finite element analysis we DO NOT work with the strong

formulation (why?), instead we use an equivalent Principle of

Minimum Potential Energy

Principle of Minimum Potential Energy (2D)

Definition: For a linear elastic body subjected to body forces

X=[Xa,X b]T and surface tractions TS=[px,py]

T, causing

displacements u=[u,v]T and strains ε and stresses σ, the potentialenergy Π is defined as the strain energy minus the potential energyof the loads (X and TS)

Π=U-W

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∫∫

∫

+=

=

T S S

T

V

T

V

T

dS T udV X u

dV

W2

1U ε σ

x

y

Su

ST

Volume (V)u

v

x

px

py

Xa dV

X b dVVolume

element dV

Strain energy of the elastic body

∫∫ == V T

V

T dV DdV ε ε ε σ

2

1

2

1U

ε σ D=Using the stress-strain law

In 2D plane stress/plane strain

( )∫

∫

∫

++=

=

=

V xy xy y y x x

V

xy

y

x

T

xy

y

x

V

T

dV

dV

dV

γ τ ε σ ε σ

γ

ε

ε

τ

σ

σ

ε σ

2

1

21

2

1U

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Principle of minimum potential energy: Among all admissible

displacement fields the one that satisfies the equilibrium equations

also render the potential energy Π a minimum.

“admissible displacement field”:

1. first derivative of the displacement components exist

2. satisfies the boundary conditions on Su

Finite element formulation for 2D:

Step 1: Divide the body into finite elements connected to each

other through special points (“nodes”)

x

ySu

STu

v

x

px

py

Element ‘e’

3

2

1

4

y

xv

u

1

2

3

4

u1

u2

u3

u4

v4

v3

v2

v1

=

4

4

3

3

2

2

1

1

v

u

v

u

v

u

v

u

d

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Total potential energy

Potential energy of element ‘e’:

∫∫∫ −−=Π T S S T

V

T

V

T dS T udV X udV ε σ 21

∫∫∫ −−=Π eT

ee S S

T

V

T

V

T

e dS T udV X udV ε σ 2

1

Total potential energy = sum of potential energies of the elements

∑ Π=Πe

e

This term may or may not be present

depending on whether the element is

actually on ST

Step 2: Describe the behavior of each element (i.e., derive the

stiffness matrix of each element and the nodal load vector).

Inside the element ‘e’

y

x

v

u

1

2

3

4

u1

u2

u3

u4

v4

v3

v2

v1

=y)(x,v

y)(x,uu

Displacement at any point x=(x,y)

=

4

4

3

3

2

2

1

1

v

u

v

u

v

u

v

u

d

Nodal displacement vector

(x1,y1)

(x2,y2)

(x4,y4)

(x3,y3)

where

u1=u(x1,y1)

v1=v(x1,y1)

etc

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u D D

u

∂==

∂=

ε σ

ε

LawStrain-Stress

RelationntDisplaceme-Strain

=

xy

y

x

γ

ε

ε

ε

∂

∂

∂

∂∂

∂∂

∂

=∂

x y

y

x

0

0

=

xy

y

x

τ

σ

σ

σ

If we knew u then we could compute the strains and stresses within the

element. But I DO NOT KNOW u!!

Hence we need to approximate u first (using shape functions) andthen obtain the approximations for ε and σ (recall the case of a 1D bar)

This is accomplished in the following 3 Tasks in the next slide

Recall

TASK 1: APPROXIMATE THE DISPLACEMENTS WITHINEACH ELEMENT

TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN

EACH ELEMENT

TASK 3: DERIVE THE STIFFNESS MATRIX OF EACH

ELEMENT USING THE PRINCIPLE OF MIN. POT ENERGY

We’ll see these for a generic element in 2D today and then derive

expressions for specific finite elements in the next few classes

Displacement approximation in terms of shape functions

d Nu =

dBD=σ

dBε =Strain approximation

Stress approximation

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u

v3

y

xv

1

2

3

4

u1

u2

u3

u4

v4v2

v1

TASK 1: APPROXIMATE THE DISPLACEMENTS

WITHIN EACH ELEMENT

44332211

44332211

vy)(x, Nvy)(x, Nvy)(x, Nvy)(x, Ny)(x,v

uy)(x, Nuy)(x, Nuy)(x, Nuy)(x, Ny)(x,u

+++≈

+++≈

Displacement approximation within element ‘e’

44332211

44332211

vy)(x, Nvy)(x, Nvy)(x, Nvy)(x, Ny)(x,v

uy)(x, Nuy)(x, Nuy)(x, Nuy)(x, Ny)(x,u

+++≈+++≈

=

=

4

4

3

3

2

2

1

1

4321

4321

vu

v

u

v

u

v

u

N0 N0 N0 N0

0 N0 N0 N0 N

y)(x,v

y)(x,uu

d Nu =

We’ll derive specific expressions of the shape functions for

different finite elements later

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TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN

EACH ELEMENT

......vy)(x, N

uy)(x, Ny)(x,vy)(x,u

vy)(x, N

vy)(x, N

vy)(x, N

vy)(x, Ny)(x,v

uy)(x, N

uy)(x, N

uy)(x, N

uy)(x, Ny)(x,u

11

11

xy

44

33

22

11

y

44

33

22

11

x

+∂

∂+

∂

∂≈

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

∂

∂≈

∂

∂=

∂

∂+

∂

∂+

∂

∂+

∂

∂≈

∂

∂=

x y x y

y y y y y

x x x x x

γ

ε

ε

Approximation of the strain in element ‘e’

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂∂

∂

∂

∂

∂

∂

∂

∂∂

∂

∂

∂

∂

∂

∂

∂

=

=

4

4

3

3

2

2

1

1

B

44332211

4321

4321

xy

v

u

v

u

v

u

v

u

y)(x, Ny)(x, Ny)(x, Ny)(x, Ny)(x, Ny)(x, Ny)(x, Ny)(x, N

y)(x, N0

y)(x, N0

y)(x, N0

y)(x, N0

0y)(x, N

0y)(x, N

0y)(x, N

0y)(x, N

x y x y x y x y

y y y y

x x x x

y

x

γ

ε

ε

ε

dBε =

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Compact approach to derive the B matrix:

( )

NB

dBd NRelationntDisplaceme-Strain

d NufieldntDisplaceme

∂=

=∂=∂=

=

uε

Stress approximation within the element ‘e’

ε σ D=LawStrain-Stress

ε σ B D=⇒

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Potential energy of element ‘e’:

TASK 3: DERIVE THE STIFFNESS MATRIX OF EACH

ELEMENT USING THE PRINCIPLE OF MININUM

POTENTIAL ENERGY

Lets plug in the approximations

∫∫∫ −−=Π eT

ee S S

T

V

T

V

T

e dS T udV X udV ε σ 2

1

d Nu = dBD=σ dBε =

( ) ( ) ( ) ( )∫∫∫ −−≈Π eT ee S S T

V

T

V

T

e dS T dV X dV d Nd NdBdBD21)d(

f k

dS T dV X dV

dS T dV X dV

T T

e

f

S S

T

V

T T

k

V

T T

S S

T T

V

T T

V

T T

e

eT

ee

eT

ee

ddd2

1)d(

N NddBDBd2

1

Nd NddBDBd2

1)d(

−=Π⇒

+−

=

−

−

≈Π

∫∫∫

∫∫∫

Rearranging

From the Principle of Minimum Potential Energy

0d

d

)d(=−=

∂

Π∂ f k e

Discrete equilibrium equation for element ‘e’ f k =d

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∫= eV k dVBDBT

Element stiffness matrix for element ‘e’

Element nodal load vector

S

eT

b

e

f

S S

T

f

V

T dS T dV X f ∫∫ += N N

Due to body force Due to surface traction

STe

e

For a 2D element, the size of the k matrix is

2 x number of nodes of the element

Question: If there are ‘n’ nodes per element, then what is the size of

the stiffness matrix of that element?

If the element is of thickness ‘t’

∫= e Ak dABDBtT


S

eT

b

e

f

l S

T

f

A

T dl T dA X f ∫∫ += Nt Nt

Due to body force Due to surface traction

For a 2D element, the size of the k matrix is

2 x number of nodes of the element

t

dAdV=tdA

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The properties of the element stiffness matrix

1. The element stiffness matrix is singular and is therefore non-

invertible

2. The stiffness matrix is symmetric

3. Sum of any row (or column) of the stiffness matrix is zero!

(why?)

∫= eV k dVBDBT

The B-matrix (strain-displacement) corresponding to this element is

We will denote the columns of the B-matrix as

Computation of the terms in the stiffness matrix of 2D elements

x

y

(x,y)

v

u

1 2

34v

4v

3

v2

v1

u1

u2

u3

u4

1 2 3 4

1 2 3 4

1 1 2 2 3 3 4 4

N (x,y) N (x,y) N (x,y) N (x,y)0 0 0 0

N (x,y) N (x,y) N (x,y) N (x,y)0 0 0 0

N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y)

x x x x

y y y y

y x y x y x y x

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

u1 v1 u2 v2

u3

u4

v3 v4

1 1

1

1

1

1

N (x ,y)0

N (x,y)0 ; ; and so on...

N (x ,y) N (x,y)

u v

x

B B y

y x

∂

∂ ∂ = = ∂ ∂ ∂ ∂ ∂

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∫= eV k dVBDBT

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 4 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

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

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

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

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 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 k k k

k k k k k k k k

=

u1

v1

u2

v2

u3

u4

v3

v4

u1 v1

u2

v2 u3 u4v3 v4

The stiffness matrix corresponding to this element is

which has the following form

1 1 1 1 1 2

1 1 1 1

T T T

11 12 13

T T

21 21

B D B dV; B D B dV; B D B dV,...

B D B dV; B D B dV;.....

e e e

e e

u u u v u uV V V

v u v vV V

k k k

k k

= = =

= =

∫ ∫ ∫

∫ ∫

The individual entries of the stiffness matrix may be computed as follows

Step 3: Assemble the element stiffness matrices into the global

stiffness matrix of the entire structure

For this create a node-element connectivity chart exactly as in 1D

3

2

Node 2

422

3

Node 3

11

Node 1ELEMENT

u

v3

y

xv

1

2

4

3

u2

u4

u3

u1

v1v4

v2

Element #1

Element #2

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Stiffness matrix of element 1 Stiffness matrix of element 2

There are 6 degrees of freedom (dof ) per element (2 per node)

=)2(

k

=)1(k

u1

u2

v1

v2u3v3

u1 v1 u2 v2 u3 v3

u2

u3

v2

v3u4v4

v2 u3 v3 u4 v4u2

88

K

×

=

Global stiffness matrix

How do you incorporate boundary conditions?

Exactly as in 1D

)2(k

)1(k

u1

v1

u2

v2u3

v3u4

v4

u1 v1 u2 v2 u3 u4v4v3

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Finally, solve the system equations taking care of theFinally, solve the system equations taking care of the

displacement boundary conditionsdisplacement boundary conditions..


Strain approximation in terms of strain-displacement matrix

Stress approximation

Summary: For each element

Element stiffness matrix


d Nu =

dBD=σ

dBε =

∫= eV k dVBDBT

S

eT

b

e

f

S S

T

f

V

T dS T dV X f ∫∫ += N N

10 FEM2&3D

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