EMA 405Introduction

SyllabusTextbook: nonePrerequisites: EMA 214; 303, 304,

or 306; EMA 202 or 221Room: 2261 Engineering HallTime: TR 11-12:15Course Materials:

ecow2.engr.wisc.edu

InstructorsJake Blanchard, Room 143 ERB, phone: 263-0391e-mail: blanchard@engr.wisc.eduoffice hours: TBD

GradingHomeworks – 40%Quiz – 20%Design Problem – 20%Final Project – 20%

ScheduleTopicsIntroductionFEA TheoryIntro to ANSYSTrussesPlane Stress/StrainAxisymmetric3-D ProblemsBeamsPlatesHeat TransferMultiple Load StepsPlasticity

The finite element methodBegan in 1940’s to help solve

problems in elasticity and structures

It has evolved to solve nonlinear, thermal, structural, and electromagnetic problems

Key commercial codes are ANSYS, ABAQUS, Nastran, etc.

We’ll use ANSYS, but other codes are as good or better (…a “religious” question)

The Process

Build a model◦Geometry◦Material Properties◦Discretization/mesh◦Boundary conditions◦Load

SolvePostprocessing

Structural ElementsTrussBea

msPlana

r3-DPlate

ElementsTrussBea

mPlan

arShellBrick

Finite Element Fundamentals

The building block of FEM is the element stiffness matrix

1

3

2

a

a

3

3

2

2

1

1

666261

262221

161211

3

3

2

2

1

1

v

u

v

u

v

u

kkk

kkk

kkk

f

f

f

f

f

f

y

x

y

x

y

x

Now Put Several Together

45

1

23

6

987

UKF

Global Stiffness

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

f

f

f

f

f

f

f

f

f

f

f

f

f

f

f

f

f

f

F

9

9

8

8

7

7

6

6

5

5

4

4

3

3

2

2

1

1

9

9

8

8

7

7

6

6

5

5

4

4

3

3

2

2

1

1

v

u

v

u

v

u

v

u

v

u

v

u

v

u

v

u

v

u

U

[K] is a composite of the element stiffness elements

Once K is known, we can choose forces and calculate displacements, or choose displacements and calculate forces

Boundary conditions are needed to allow solution

Element Stiffness

1

3

2

a

a

u3

u2u1

v2v1

v3

f3x

f1x f2x

f2y

f1y

f3y

y

x

How Do We Get Element Stiffness?

accvaccu

accvaccu

cvcu

ayx

yax

yx

ycxccyxv

ycxccyxu

assume

643313

542212

4111

33

22

11

654

321

;

;

;

;0

0;

0;0

),(

),(

3

2

1

3

2

1

1

1

3

2

1

3

2

1

101

011

001

101

011

001

01

01

001

][

u

u

ua

ac

c

c

a

aA

uAc

c

c

c

a

a

u

u

u

cAu

Coordinates of element corners

Substitute coordinates into assumed functions

Rewrite as matrix

equation

Continued…

3

2

1

3

2

1

3

2

1

3

2

1

101

011

00

11

1

101

011

001

u

u

ua

yxa

u

c

c

c

yxu

u

u

ua

ac

c

c

yuuxuuaua

yxu

uu

uu

au

yxa

u

31211

31

21

1

1),(

11

Rewrite assumed functions

Substitute

Multiply

Continued

332211

3

2

1

332211

321

),(

1

),(

1),(

vNvNvNyxv

Similarlya

yN

a

xN

a

y

a

xN

uNuNuNyxu

yuxuuyxaa

yxu

Collect terms

Stress-Strain

332211

332211

3

2

1

),(

),(

1

vNvNvNyxv

uNuNuNyxua

yN

a

xN

a

y

a

xN

a

v

a

v

a

u

a

u

x

v

y

u

a

v

a

v

y

va

u

a

u

x

u

ux

Nu

x

Nu

x

N

x

u

xy

y

x

x

2131

31

21

33

22

11

Stress-Strain

BDBtAdVBDBk

aB

v

u

v

u

v

u

B

TT

V

xy

y

x

011011

100010

0001011

3

3

2

2

1

1

Comes from minimizing total potential energy

(variational principles)

Material Properties

[D] comes from the stress-strain equations

For a linear, elastic, isotropic material

dVDU

ED

D

T

V

xy

y

x

xy

y

x

2

12

100

01

01

1][

][

2

Strain Energy

Final Result for Our Case

200222

011011

011011

200222

211231

211213

14][

2

1

200222

011011

011011

200222

211231

211213

12][

2

2

22

Etk

aA

a

AEtk

or

3

3

2

2

1

1

2

3

3

2

2

1

1

200222

011011

011011

200222

211231

211213

14

v

u

v

u

v

u

Et

f

f

f

f

f

f

y

x

y

x

y

x

Examples

12

3

3

2

2

1

1

2

1

1

2

1

3

14u

Et

f

f

f

f

f

f

y

x

y

x

y

x

u1

Examples

v1

12

3

3

2

2

1

1

2

1

1

2

3

1

14v

Et

f

f

f

f

f

f

y

x

y

x

y

x

Prescribe forces

F

ProcessWhat do we know? – v1=v2=0; f3y=F; all

horizontal forces are 0Remove rigid body motion – arbitrarily set

u1=0 to remove horizontal translation; hence, f1x is a reaction

Reduce matrix to essential elements for calculating unknown displacements – cross out rows with unknown reactions and columns with displacements that are 0

Solve for displacementsBack-solve for reaction forces

Equations

3

3

2

2

3

3

2

22

1

1

202

010

202

140

0

0

0

0

200222

011011

011011

200222

211231

211213

14

0

0

v

u

uEt

F

or

v

u

uEt

F

f

f

f

y

y

x

Solution

1

02

3

3

2

Et

F

v

u

u

Putting 2 Together

a

a

1

2

1 2

3 4

Element 2 Stiffness Matrix

1

3

2

1 (4)

3 (2)

2(3)Rotate

180o

c s 0 0 0 0-s c 0 0 0 00 0 c s 0 00 0 -s c 0 00 0 0 0 c s0 0 0 0 -s c

T =

K’ = TTKTFor 180o rotationK’=KJust rearrange the rows and columns top correspond to global numbering scheme (in red).

4

4

3

3

2

2

1

1

2

4

4

3

3

2

2

1

1

00000000

00000000

00200222

00011011

00011011

00200222

00211231

00211213

14

v

u

vu

v

u

v

u

Et

f

f

ff

f

f

f

f

y

x

y

x

y

x

y

x

4

4

3

3

2

2

1

1

2

4

4

3

3

2

2

1

1

31122100

13122100

11100100

22022000

22022000

11100100

00000000

00000000

14

v

u

vu

v

u

v

u

Et

f

f

ff

f

f

f

f

y

x

y

x

y

x

y

x

Ele

men

t 1

Ele

men

t 2

Element Matrices

4

4

3

3

2

2

1

1

2

4

4

3

3

2

2

1

1

31122100

13122100

11300122

22031011

22013011

11100322

00211231

00211213

14

v

u

vu

v

u

v

u

Et

f

f

ff

f

f

f

f

y

x

y

x

y

x

y

x

Add the element matrices

What if triangles have midside nodes?

3

4

5

26

1

21211

210987

265

24321

),(

),(

ycxycxcycxccyxv

ycxycxcycxccyxu

What about a quadrilateral element?

3 4

21

xycycxccyxv

xycycxccyxu

8765

4321

),(

),(

What about arbitrary shapes?

For most problems, the element shapes are arbitrary, material properties are more general, etc.

Typical solution is to integrate stiffness solution numerically

Typically gaussian quadrature, 4 points

dVBDBkT

V