MANE 4240 & CIVL 4240Introduction to Finite Elements

FEM Discretization of 2D Elasticity

Prof. Suvranu De

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

Recap: 2D Elasticity

x

y

Su

ST

Volume (V)u

v

x

px

py

Xa dV

Xb 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

Solution

Case 1: Stretch

2

yv

xu

Check that the new coordinates (in the deformed configuration)

2

2

'

'

yvyy

xuxx

Case 2: Pure shear/ 4

0

u y

v

Check that the new coordinates (in the deformed configuration)

'

'

/ 4x x u x y

y y v y

y)(x,v

y)(x,uu

xy

y

x

xy

y

x

0

0

xy

y

x

uDD

u

yxuu

LawStrain -Stress

Relationnt Displaceme-Strain

),(fieldntDisplaceme

Recap: 2D Elasticity

2

100

01

01

1 2

E

D

For plane stress(3 nonzero stress components)

2

2100

01

01

211

E

D

For plane strain(3 nonzero strain components)

VinXT 0 Equilibrium equations

Boundary conditions

1. Displacement boundary conditions: Displacements are specified on portion Su of the boundary

uspecified Sonuu

2. Traction (force) boundary conditions: Tractions are specified on portion ST of the boundaryNow, 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,Xb]T and surface tractions TS=[px,py]T, causing displacements u=[u,v]T and strains and stresses , the potential energy is defined as the strain energy minus the potential energy of the loads (X and TS)

U-W

TSS

T

V

T

V

T

dSTudVXu

dV

W

2

1U

x

y

Su

ST

Volume (V)u

v

x

px

py

Xa dV

Xb dVVolume element dV

Strain energy of the elastic body

V

T

V

T dVDdV 2

1

2

1U

DUsing the stress-strain law

In 2D plane stress/plane strain

V xyxyyyxx

V

xy

y

x

T

xy

y

x

V

T

dV

dV

dV

2

1

2

1

2

1U

Principle of minimum potential energy: Among all admissible displacement fields the one that satisfies the equilibrium equations also render the potential energy a minimum.

“admissible displacement field”: 1. first derivative of the displacement components exist2. 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

y

Su

STu

v

x

px

py

Element ‘e’

3

21

4

y

xvu

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

Total potential energy

Potential energy of element ‘e’:

TS

ST

V

T

V

T dSTudVXudV2

1

eT

ee S ST

V

T

V

Te dSTudVXudV

2

1

Total potential energy = sum of potential energies of the elements

e

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)

whereu1=u(x1,y1)v1=v(x1,y1)etc

uDD

u

LawStrain -Stress

Relationnt Displaceme-Strain

xy

y

x

xy

y

x

0

0

xy

y

x

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) and then obtain the approximations for and (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 WITHIN EACH 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

dNu

dBD

dBε Strain approximation

Stress approximation

Displacement approximation in terms of shape functions

u

v3

y

xv1

2

3

4

u1

u2

u3

u4

v4v2

v1

TASK 1: APPROXIMATE THE DISPLACEMENTS WITHIN EACH ELEMENT

44332211

44332211

vy)(x,N vy)(x,N vy)(x,N vy)(x,Ny)(x,v

u y)(x,Nu y)(x,Nu y)(x,Nu y)(x,Ny)(x,u

Displacement approximation within element ‘e’

44332211

44332211

vy)(x,N vy)(x,N vy)(x,N vy)(x,Ny)(x,v

u y)(x,Nu y)(x,Nu y)(x,Nu y)(x,Ny)(x,u

4

4

3

3

2

2

1

1

4321

4321

v

u

v

u

v

u

v

u

N0N0N0N0

0N0N0N0N

y)(x,v

y)(x,uu

dNu

We’ll derive specific expressions of the shape functions for different finite elements later

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

v y)(x,N

vy)(x,Ny)(x,v

u y)(x,N

u y)(x,N

u y)(x,N

u y)(x,Ny)(x,u

11

11

xy

44

33

22

11

y

44

33

22

11

x

xyxy

yyyyy

xxxxx

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

y)(x,N0

y)(x,N0

y)(x,N0

y)(x,N0

0y)(x,N

0y)(x,N

0 y)(x,N

0y)(x,N

xyxyxyxy

yyyy

xxxx

y

x

dBε

Compact approach to derive the B matrix:

NB

dBdNRelationnt Displaceme-Strain

dNufieldntDisplaceme

u

Stress approximation within the element ‘e’

DLawStrain -Stress

BD

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

eT

ee S ST

V

T

V

Te dSTudVXudV

2

1

dNu dBDdBε

eT

ee S ST

V

T

V

Te dSTdVXdV dNdNdBdBD

2

1)d(

fk

dSTdVXdV

dSTdVXdV

TTe

f

S ST

V

TT

k

V

TT

S STT

V

TT

V

TTe

eT

ee

eT

ee

ddd2

1)d(

NNddBDBd2

1

NdNddBDBd2

1)d(

Rearranging

From the Principle of Minimum Potential Energy

0dd

)d(

fke

Discrete equilibrium equation for element ‘e’ fk d

eV

k dVBDBT

Element stiffness matrix for element ‘e’

Element nodal load vector

S

eT

b

e

f

S ST

f

V

T dSTdVXf NN

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’

eA

k dABDBt T

Element nodal load vector

S

eT

b

e

f

l ST

f

A

T dlTdAXf NtNt

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

dA dV=tdA

The properties of the element stiffness matrix

1. The element stiffness matrix is singular and is therefore non-invertible2. The stiffness matrix is symmetric3. 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

34v4

v3

v2v1

u1u2

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 v1u2 v2

u3 u4v3 v4

1 1

1

1

11

N (x,y)0

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

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

u v

xB B

y

yx

 

eV

k dVBDBT

11 12 13 14 15 16 17 18

21 22 23 24 25 26 27 28

31 32 33 34 35 36 37 38

41 42 43 44 45 46 47 48

51 52 53 54 55 56 57 58

61 62 63 64 65 66 67 68

71 72 73 74 75 76 77 78

81 82 83 84 85 86 87 88

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 kk

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 v1u2 v2 u3

u4v3v4

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 T11 12 13

T T21 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

ELEMENT Node 1 Node 2 Node 3

1 1 2 3

2 2 3 4

u

v3

y

xv

1

2

4

3

u2

u4

u3

u1

v1v4

v2

Element #1

Element #2

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

v2

u3

v3

u1 v1 u2 v2 u3v3

u2

u3

v2

v3

u4

v4

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

v2

u3

v3

u4

v4

u1 v1 u2 v2 u3 u4v4v3

Finally, solve the system equations taking care of the Finally, solve the system equations taking care of the displacement boundary conditionsdisplacement boundary conditions..

Physical interpretation of the stiffness matrix

3y

3x

2y

2x

1y

1x

3

3

2

2

1

1

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

f

f

f

f

f

f

v

u

v

u

v

u

d

kkkkkk

kkkkkk

kkkkkk

kkkkkk

kkkkkk

kkkkkk

fk

Consider a single triangular element. The six corresponding equilibrium equations ( 2 equilibrium equations in the x- and y-directions at each node times the number of nodes) can be written symbolically as

x

y

u3

v3

v1

u1

u2

v2

2

3

1

Choose u1 = 1 and rest of the nodal displacements = 0

3y

3x

2y

2x

1y

1x

61

51

41

31

21

11

f

f

f

f

f

f

k

k

k

k

k

k

Hence, the first column of the stiffness matrix represents the nodal loads when u1=1 and all other dofs are fixed. This is the physical interpretation of the first column of the stiffness matrix. Similar interpretations exist for the other columns

x

y

u1=1

2

3

1

ijk = “Force” at d.o.f ‘i’ due to unit displacement at d.o.f ‘j’ keeping all the other d.o.fs fixed

Now consider the ith row of the matrix equation fk d

ix654321 fd iiiiii kkkkkk

This is the equation of equilibrium at the ith dof

Consistent and Lumped nodal loads

Recall that the nodal loads due to body forces and surface tractions

eT

e S ST

SV

T

bdSTfdVXf N;N

These are known as “consistent nodal loads”1. They are derived in a consistent manner using the Principle of Minimum Potential Energy2. The same shape functions used in the computation of the stiffness matrix are employed to compute these vectors

Example

1

2

3

x

yp per unit area

Traction distribution on the 1-2-3 edgepx= ppy= 0

We’ll see later that

232

22

221 2

)(;;

2

)(

b

ybyN

b

ybN

b

ybyN

b

b

N1 N2 N3

The consistent nodal loads are

32

)(3

4

32

)(

233

2

22

22

211

pbdy

b

ybypdyNpF

pbdy

b

ybpdyNpF

pbdy

b

ybypdyNpF

b

b

b

bx

b

b

b

bx

b

b

b

bx

1

2

3

x

y

b

b

pb/3

pb/3

4pb/3

The lumped nodal loads are

2

2

3

2

1

pbF

pbF

pbF

x

x

x

1

2

3

x

y

b

b

pb/2

pb/2

pb

Lumping produces poor results and will not be pursued further

Displacement approximation in terms of shape functions

Strain approximation in terms of strain-displacement matrix

Stress approximation

Summary: For each element

Element stiffness matrix

Element nodal load vector

dNu

dBD

dBε

eV

k dVBDBT

S

eT

b

e

f

S ST

f

V

T dSTdVXf NN