ONLY FOR INFORMATION PURPOSE

INTRODUCTION TO FINITE ELEMENT METHODS

CONSTANT STRAIN TRIANGLE

Introduction to Finite Elements

Constant Strain Triangle (CST)

Summary:

• Computation of shape functions for constant strain triangle• Properties of the shape functions • Computation of strain-displacement matrix• Computation of element stiffness matrix• Computation of nodal loads due to body forces• Computation of nodal loads due to traction• Recommendations for use• Example problems

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

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 =

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

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,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

4444444444444444444 34444444444444444444 21xyxyxyxy

yyyy

xxxx

y

x

γεε

ε

dBε =

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=σ

dBε =

∫=eV

k dVBDBT

443442143421

S

eT

b

e

f

S ST

f

V

T dSTdVXf ∫∫ += NN

Constant Strain Triangle (CST) : Simplest 2D finite element

• 3 nodes per element• 2 dofs per node (each node can move in x- and y- directions)• Hence 6 dofs per element

x

y

u3

v3

v1

u1

u2

v2

2

3

1

(x,y)

vu

(x1,y1)

(x2,y2)

(x3,y3)

166212 dNu ××× =

=

=

3

3

2

2

1

1

321

321

v

u

v

u

v

u

N0N0N0

0N0N0N

y)(x,v

y)(x,uu

The displacement approximation in terms of shape functions is

=

321

321

N0N0N0

0N0N0NN

1 1 2 2 3 3u (x,y) u u uN N N≈ + +

1 1 2 2 3 3v(x,y) v v vN N N≈ + +

Formula for the shape functions are

A

ycxbaN

A

ycxbaN

A

ycxbaN

2

2

2

3333

2222

1111

++=

++=

++=

12321312213

31213231132

23132123321

33

22

11

x1

x1

x1

det2

1

xxcyybyxyxa

xxcyybyxyxa

xxcyybyxyxa

y

y

y

triangleofareaA

−=−=−=−=−=−=−=−=−=

==

where

x

y

u3

v3

v1

u1

u2

v2

2

3

1

(x,y)

vu

(x1,y1)

(x2,y2)

(x3,y3)

Properties of the shape functions:

1. The shape functions N1, N2 and N3 are linear functions of x and y

x

y

2

3

1

1

N1

2

3

1

N2

12

3

1

1

N3

=nodesotherat

inodeat

0

''1N i

2. At every point in the domain

yy

xx

i

i

=

=

=

∑

∑

∑

=

=

=

3

1ii

3

1ii

3

1ii

N

N

1N

3. Geometric interpretation of the shape functionsAt any point P(x,y) that the shape functions are evaluated,

x

y

2

3

1P (x,y)

A1A3

A2

A

AA

AA

A

33

22

11

N

N

N

=

=

=

Approximation of the strains

xy

u

v

u v

x

y

x

B dy

y x

εεεγ

∂ ∂

∂ = = ≈ ∂ ∂ ∂ + ∂ ∂

=

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

332211

321

321

332211

321

321

000

000

2

1

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

y)(x,N0

y)(x,N0

y)(x,N0

0y)(x,N

0 y)(x,N

0y)(x,N

bcbcbc

ccc

bbb

A

xyxyxy

yyy

xxx

B

Element stresses (constant inside each element)

dBD=σ

Inside each element, all components of strain are constant: hence the name Constant Strain Triangle

IMPORTANT NOTE:1. The displacement field is continuous across element boundaries2. The strains and stresses are NOT continuous across element boundaries

Element stiffness matrix

∫=eV

k dVBDBT

AtkeV

BDBdVBDB TT == ∫ t=thickness of the elementA=surface area of the element

Since B is constant

t

A

443442143421

S

eT

b

e

f

S ST

f

V

T dSTdVXf ∫∫ += NN

Element nodal load vector

Element nodal load vector due to body forces

∫∫ ==ee A

T

V

T

bdAXtdVXf NN

=

=

∫∫∫∫∫∫

e

e

e

e

e

e

A b

A a

A b

A a

A b

A a

yb

xb

yb

xb

yb

xb

b

dAXNt

dAXNt

dAXNt

dAXNt

dAXNt

dAXNt

f

f

f

f

f

f

f

3

3

2

2

1

1

3

3

2

2

1

1

x

y

fb3x

fb3y

fb1y

fb1x

fb2x

fb2y

2

3

1

(x,y)

XbXa

EXAMPLE:

If X a=1 and Xb=0

=

=

=

=

∫

∫

∫

∫∫∫∫∫∫

03

03

03

0

0

0

3

2

1

3

3

2

2

1

1

3

3

2

2

1

1

tA

tA

tA

dANt

dANt

dANt

dAXNt

dAXNt

dAXNt

dAXNt

dAXNt

dAXNt

f

f

f

f

f

f

f

e

e

e

e

e

e

e

e

e

A

A

A

A b

A a

A b

A a

A b

A a

yb

xb

yb

xb

yb

xb

b

Element nodal load vector due to traction

∫=e

TS ST

SdSTf N

EXAMPLE:

x

y

fS3x

fS3y

fS1y

fS1x

2

3

1 ∫− −

=el S

along

T

SdSTtf

31 31N

Element nodal load vector due to traction

EXAMPLE:

x

y

fS3x

2

31

∫− −

=el S

along

T

SdSTtf

32 32N

fS3y

fS2x

fS2y

(2,0)

(2,2)

(0,0)

=0

1ST

tt

dyNtfex l alongS

=××

=

= ∫−

−

122

1

)1(32

2 322

0

0

3

3

2

=

=

=

y

x

y

S

S

S

f

tf

f

Similarly, compute

1

2

Recommendations for use of CST

1. Use in areas where strain gradients are small

2. Use in mesh transition areas (fine mesh to coarse mesh)

3. Avoid CST in critical areas of structures (e.g., stress concentrations, edges of holes, corners)

4. In general CSTs are not recommended for general analysis purposes as a very large number of these elements are required for reasonable accuracy.

Example

x

y

El 1

El 2

1

23

4

300 psi1000 lb

3 in

2 inThickness (t) = 0.5 inE= 30×106 psiν=0.25

(a) Compute the unknown nodal displacements.(b) Compute the stresses in the two elements.

Realize that this is a plane stress problem and therefore we need to use

psiE

D 72

10

2.100

02.38.0

08.02.3

2

100

01

01

1×

=

−−=

νν

ν

ν

Step 1: Node-element connectivity chart

ELEMENT Node 1 Node 2 Node 3 Area (sqin)

1 1 2 4 3

2 3 4 2 3

Node x y

1 3 0

2 3 2

3 0 2

4 0 0

Nodal coordinates

Step 2: Compute strain-displacement matrices for the elements

=

332211

321

321

000

000

2

1

bcbcbc

ccc

bbb

AB

Recall

123312231

213132321

xxcxxcxxc

yybyybyyb

−=−=−=−=−=−=

with

For Element #1:

1(1)

2(2)

4(3)(local numbers within brackets)

0;3;3

0;2;0

321

321

======

xxx

yyy

Hence

033

202

321

321

==−=−===

ccc

bbb

−−−

−=

200323

003030

020002

6

1)1(BTherefore

For Element #2:

−−−

−=

200323

003030

020002

6

1)2(B

Step 3: Compute element stiffness matrices

7

)1(T)1()1(T)1()1(

10

2.0

05333.0

02.02.1

3.00045.0

2.02.02.13.04.1

3.05333.02.045.05.09833.0

BDB)5.0)(3(BDB

×

−−−−

−−−

=

== Atk

u1 u2 u4 v4v2v1

7

)2(T)2()2(T)2()2(

10

2.0

05333.0

02.02.1

3.00045.0

2.02.02.13.04.1

3.05333.02.045.05.09833.0

BDB)5.0)(3(BDB

×

−−−−

−−−

=

== Atk

u3 u4 u2 v2v4v3

Step 4: Assemble the global stiffness matrix corresponding to the nonzero degrees of freedom

014433 ===== vvuvuNotice that

Hence we need to calculate only a small (3x3) stiffness matrix

710

4.102.0

0983.045.0

2.045.0983.0

×

−−

=K

u1 u2v2

u1

u2

v2

Step 5: Compute consistent nodal loads

=

y

x

x

f

f

f

f

2

2

1

=

yf2

0

0

ySy ff2

10002 +−=

The consistent nodal load due to traction on the edge 3-2

lbx

dxx

dxN

tdxNf

x

x

xS y

2252

950

250

3150

)5.0)(300(

)300(

3

0

2

3

0

3

0 233

3

0 2332

−=

−=

−=

−=

−=

−=

∫

∫

∫

=

= −

= −

3 2

3232

xN =

−

lb

ffySy

1225

100022

−=

+−=Hence

Step 6: Solve the system equations to obtain the unknown nodal loads

fdK =

−=

−−

×1225

0

0

4.102.0

0983.045.0

2.045.0983.0

10

2

2

17

v

u

u

Solve to get

×−××

=

−

−

−

in

in

in

v

u

u

4

4

4

2

2

1

109084.0

101069.0

102337.0

Step 7: Compute the stresses in the elements

)1()1()1( dBD=σ

With

[ ][ ]00109084.0101069.00102337.0

d444

442211)1(

−−− ×−××=

= vuvuvuT

Calculate

psi

−−−

=1.76

1.1391

1.114)1(σ

In Element #1

)2()2()2( dBD=σ

With

[ ][ ]44

224433)2(

109084.0101069.00000

d−− ×−×=

= vuvuvuT

Calculate

psi

−=

35.363

52.28

1.114)2(σ

In Element #2

Notice that the stresses are constant in each element