IGCAR FEM , syllabus

7/27/2019 IGCAR FEM , syllabus

1/91

FINITE ELEMENTS ANALYSIS

lectures at IGCAR

Lecture-1

by

Dr.C.Jebaraj

Professor of Mechanical Engineering

AU-FRG Institute for CAD/CAM

CEG Campus, Anna University

Chennai-25

[email protected]


2/91

1. Why CAD/CAM/CAE/CAPP.CAx

2. SAFE DESIGN

3. OPTIMUM DESIGN1. In shape

2. In weight

3. In size4. In cost


3/91

Example of an Optimum Design

2

max

2

wLPLM

Z

MyI

MfyI

Mf

R

E

y

f

I

M

maxmaxmax;

Bending formula

BMDQuadratic

Pw/Length

L


4/91

Z-variation

L

d

b

261

2

3121

bdd

bd

y

IZ


5/91

STRAIGHT BAR SUBJECTED TO AN AXIAL LOAD

AE

PLU max

L

P

P= applied load

L= length of rod

E= youngs modulus of material

A= area of cross section

dx

AE

Pdx

du

dx

duE

A

P

E

E

(Elongation ofsmall segment dx)

AE

PLU

AE

PdxxU

L

L

max

AE

Px

)(

0

0

Total elongation

P

dx


6/91

Umax

)()( LINEARAE

PxxU

STRAIGHT BAR SUBJECTED TO AN AXIAL LOAD.


7/91

TAPER ROD SUBJECTED TO AXIAL LOAD

0

log

min

max

0

0

at xU

Lat xU

A(x)

E

P

A(x)EPdxU(x)

L

L

L

P

P


8/91

TAPERED ROD SUBJECTED TO AXIAL LOAD

WITH SELF WEIGHT

f(x)dxE

A(x)

(x)dxP

EA(x)E

(x)dxPU(x)

ht (x)Self weigPAxial Load(x)PTotal load

1

1

NOTE:

i. We may not have explicitly integration formulae

ii.The solution is highly complex than a

logarithmic solution.

P


9/91

Trapezoidal method (Linear)

h)......fff(fff

f(x)dx

n

1432

21

2

NUMERICAL INTEGRATION

ERROR IS SMALL

L0

h 0.05L

f(x)

f1 f2 f3 f4 fn+1 fn

h 0.1 L

f(x)


10/91

SIMPSONS METHOD (QUADRATIC)

Error is small

2...)(2

42

2

9753

86421

h

ffff

.....)fff(fff

f(x)dx

N

h 0.1 L

f(x)


11/91

Summary

1. In the absence of an explicit integration

formulae available, we accept numericalintegration formulation which is an

approximation only

2. Approximation is improved byi) Increasing the No of segments

ie., by decreasing h

ii) Increasing the order of polynomial ofrequirements.

3. Error is minimized and there is convergence

towards exact solution


12/91

Formulation of Problem for Approximation

solution

System L is under equilibriumSubsystem dx is also under

equilibrium

+ d P

AdxAd

AdxAdAFv

0

0

L

dx


13/91

0)(

0)(

Adx

duEA

dx

d

AdxAdx

duEd

AdxAEd

Domain : 0 < x < L

0)()(:..

xA

dx

duxEA

dx

dEDG

BCs : at x=0 , u=0

at x=l , R=P

Pdx

duEA

Boundary

value

problem

P

L


14/91

Approximate Solution

0)()(:..

xAdx

du

xEAdx

d

EDG

Domain : 0 < x < L

BCs : at x=0 , u=0

at x=l , R=P

1

nnxa...........xaxaxaxaa

solutionroximationbe the appuLet

4

43

32

210U*

U*


15/91

3

3

2

210 xaxaxaau*

Assume,

Apply BCs

23120* aau

Substituting 2 into 1 we get

0

):(

)(0)()(*

i

i

da

dR

axR

RESIDUExRxAdx

udxEA

dx

d

Should be minimum

2


16/91

ILLISTRATIVE PROBLEM

Consider the equation

2

2

Xdx

dU

xdx

d

in the domain 1< X < 2

With BCs as U(1) =2 and

21

2 Xdx

dUX

CONSTRUCTION OF TRIAL FUNCTION

)x(..........)x()x()x( 110 nnaaU

2)1(..........)1()1()1( 110 nnaaU

if .....n1,2,3.....i0)1(then2)1(0 i


17/91

2

1..................

22

11

2

0

2

X

nn

XXXdx

dxa

dx

dxa

dx

dx

dx

UdX

1

if

2

1

2

0

Xdx

d

x

then ....n1,2,3.....i02

X

i

dx

dx

Let

2aaaa(x)U

xaxaxaa(x)U

4321

3

4

2

321

or 4321 aaa2a

2

1)12a4a2(a

dx

UdX 432

2X

2

432 12a4a

4

1a


18/91

Substituting for a1& a2in the expression for , we have

xa)x(a

11xx1xa3x1xa1x4

12xU

22110

2

43

11xx1x23x1x1

1x4

1

20

2

It can be easily seen that the above trail function satisfiesthe conditions imposed on the boundary. Thus the construction of

trial function is over.

xU

when


19/91

WRM APPLICATIONConsider the equation

0

22

xdx

xdUxdx

d

Substituting the trial solution xU for xU , this equation is unlikely to be satisfied

022

xdxxUdx

dxd

This is called as a Residue and is a measure of the error involved.

222

1x2a43x3a1x4

41xR

42

31

aa

aa

i.e., the RHS is a non zero function R(X)

i.e R(x)


20/91

COLLOCATION METHODFor all

ia choose a point ix in the domain and at each such ix

force the residual to be exactly zero

i.e R(x1) = 0R(x2) = 0

R(xn) = 0

The chosen points are called collocation point.

They may be located any where on the boundary or in the domain .

For the present problem we have 2 parameter

Therefore select any two collocation point say1a &. 2a

X1=4/3 & X2=5/3Substituting in the expression for R(x) we have

100

9713

3

8

8

114

3

4

21

21

aa

aa


21/91

1a =2.0993 &

2a = -0.356

11xx1x0.3563x1-x2.09931x412xU 2

therefore

Solving the simultaneous equation


22/91

THE SUB- DOMAIN METHODFor each undetermined parameter

1a choose an intervalx , in the domain.

Then force average of the residual in the each interval to be zero.

x11

0R(x)dxx

1

.

.

.

0R(x)dxx

1

x22

xnn

0R(x)dxx

1

1

X1 i

R(x)dxx

1 iX

x1

x2

x3

x4


23/91

which yields n system of residual equations which can be solved for1

a

The intervalsix are called the sub-domain.

They may chosen in any fashion.

Taking

1

x = 1< X < 1.5,2

x = 1.5 < X


24/91

LEAST SQUARE METHOD

2

1

2 0)( dxxRai for i. 1,2,3n

2

10/ dxaddRXR i i=1,2,3,n

this yield the results

1= 2.3155

2= -0.3816


25/91

THE GALERKIN METHOD

For each parameter 1 we required that a weighted

average of R(x) over the entire domain be zero.

xi

associated with .a i

Weighting functions are the trial function

.

.

.

0dx(x)R(x)

2

1

1

0dx(x)R(x)2

1

i this yields 1= 2.13782 = - 0.3477


26/91


lectures at IGCAR

Lecture-2&3

by

Dr.C.Jebaraj

Professor of Mechanical Engineering

AU-FRG Institute for CAD/CAM


Chennai-25

[email protected]


27/91

Comparison of WRM

Collocation method = R(x) = 0

Ritz method

0.)( FdxWxR

0)(2

dxda

dR

xRi

Sub domain method 0)( dxxR

Least square method

0)(

0.)(

dxxR

FdxWxR

i

Galerkin method

RITZ VARIATIONAL METHOD (weak formulation)


28/91

RITZ VARIATIONAL METHOD (weak formulation)

Staring with equation

in0xfdx

dU

xdx

d

The WR became

0

dxxf

dx

dUx

dx

dxW

Xb

Xa

where W(x)= weighing function

0 dxxWxRie

0)()(

dxxWxfdxxWdxdU

xdx

d

RITZ VARIATIONAL METHOD (contd )


29/91

RITZ VARIATIONAL METHOD (contd ..)

Xb

Xb

Xb

Xa

Xb

Xa

dxdx

dW

dx

dUx

dx

dUxxW

dxdx

dUx

dx

dxW

Integration by parts for first term:

vduuvudv

xUxW

xU

In Ritz method we take

where

is specified, as at the boundary, W(x) = 0.


30/91


31/91

0

xAdx

du

XEAdx

d

L

dxxAdx

duxEA

dx

dxw

0

0

The governing equation is

The WR formulation is

where w(x) is the weighting function and

u(x) is the trial solution.

Integrating by parts and re arranging we get


32/91

LPLwPwdxxwxA

dxdx

dw

dx

duXEA

L

L

000

0

Since u(0) = 0 (specified), uw at x = 0 vanishes

LPwdxxwxAdxdx

dw

dx

duXEA

L L

0 0

Integrating by parts and re-arranging we get

)( dvdxxEA Strain energy stored

External

work done

LAW OF CONSERVATION OF ENERGY!!

LHS =

Solution Procedure


33/91

Solution Procedure

,,, ii

(x) wj j

j

a

xaxaxau(x)

a) u(

xaxaxaaLet u

321

3

1

33

221

000

3

3

2

210

Th V i ti l E ti


34/91

The Variational Equation

L(w)B(u,w)

Pw(L)A(x)wdxdxdx

dw

dx

duxEA

1,2,3iw

au

i

3

1j

jj

Now substituting

ii

iii

j

jj

raK

LPdxxAdxdxd

dxdxEAa

)()(..)(3

1


35/91

L

ij

ij dxdx

d

dx

dxEAk

0

)(0

LPdxxxAr i

L

ii

3

3

2

2

1

x

x

x

23

2

1

3

2

1

xdx

d

xdx

d

dx

d


36/91

300

0

1111 dx

dx

d

dx

dxEAk = E(80-0.2x) .1.1 dx

=1.5x104E

dxdxd

dx

dxEAk 2112

= E(80-0.2x) .1. 2x dx

= 3.6x106E

dxdx

d

dx

dxEAk 3113

= E ( 80 - 0.2 x) .1. 3x2dx

= 9.45 x 108 E

Similarly k21=..

k22=

.

.

k33=


37/91

511 103773.12.080)( dxxxdxxAr

7222 104.22.080)( dxxxdxxAr

9333 10598.42.080)( dxxxdxxAr

xPL(L)PP 710311

10x9L)( 9222 PLPP

1027 11333 xPL(L)PP

864 10x9 4510x3 6x1051 a


38/91

119

97

75

3

2

1

14118

1196

10x2710x4.598

10x910x2.4

10x310x.371

10x1.32210x3.8810x9.45

10x2.8810x1.210x3.6

10x9.4510x3.6x105.1

a

a

a

On solving

a1 = 6.6762 x 10-5

a2 = -4.946 x 10-8

a3 = 6.4736 x 10-10

U(x) = a1x + a2x2 + a3x

3

U(X=300) = uL= a1 (300) + a2 (300)2+ a3 (300)

3 = 0.033056 cm

But Exact Value is 0.0378 cm.

CONVERGENCE STUDY fi t


39/91

L

U(x)

0.0378

10Th Order

0

6Th

Order

0.03306

CONVERGENCE STUDY-p refinementincreasing the order of approximate polynomial

NODAL APPROXIMATION METHOD


40/91

NODAL APPROXIMATION METHOD

he

i je

Be(u ,w ) = L e(w )

e

he he

e

e

ee

PiwdxxAdxdx

dw

dx

du

XAE 0 0

Ue= a0+ a1 x

=

1

0

a

a

he

xAAAxA jii


41/91

Ui=

1

0

a

a

Uj=

1

0

a

a

1

0

1

01

a

a

heU

U

j

i

j

i

U

U

hea

a1

1

0

1

01

101

U


42/91

.

1

011

j

i

U

U

hexU

j

i

U

U

he

x

he

x1

2

1j

jjNu

1,2i, iNw

.he

x

N,he

x

N 21 1

;

1

,

1 21

hedx

dN

hedx

dN

hex

AAAxAjii


43/91

Characteristic of shape function

hexN 11

hexN 2

N1 N2

N1= 1 at x=0

N1=0 at x=he

N2= 0 at x=0

N2= 1 at x=he

N1+N2=1

h


44/91

2

11.

.)(

21

211

he

0

1111

AAheE

dxhehehe

AAAE

dxdx

dN

dx

dNxEAk

2

11.

21

21

12112

AA

he

E

dxhehehe

AAAEkk


45/91

xAAhe


46/91

1

2

0

2

36P

heA

heA

Pdxhe

x

he

AAAr

ji

ji

i

On substituting the value

7070

7070

100

1 Ek

0

6

2006

220

1001 Rr


47/91

For element 2

50505050

100

2

Ek

O

Or

6

1406

160

1002


48/91

For element 3

3030

3030

100

3 Ek

P

O

r680

6

100

100

3


49/91

On assembling

3

2

1

k

k

k

4

3

2

1

u

u

u

u

r1

r2

r3

=

17070 U


50/91

4

3

2

1

303030305050

50507070

7070

100UU

U

U

EX

6

806

1006

1406

160

6

2006

220

100

P

R

0

0

17070 U


51/91

4

3

2

1

3030308050

5012070

7070

100UU

U

U

EX

6

806

240

6

3606

220

100

P

R

0

0

A li Gl b l b d diti U 0


52/91

Appling Global boundary condition U1= 0

P

OO

U

UUE

80

240360

6

100

3030

30805050120

1004

3

2

U4= 0.0355,

U3= 0.0188,U2= 0.0088.

Umaxexact = 0.0378

CONVERGENCE STUDY h improvement


53/91

CONVERGENCE STUDY- h improvement

increasing No of element

L0.0378

0

0.0355

3 element

6 element

10 element

U(x)

C i f fi


54/91

L

U(x)

0.0378

10Th Order

0

6Th

Order

0.03306

Comparison of refinement


55/91

L0.0378

EXACT

0

0.0355

3 element

6 element

10 element

U(x)

Comparison of refinement (contd)



56/91


lectures at IGCAR

Lecture-4

by

Dr.C.Jebaraj

Professor of Mechanical EngineeringAU-FRG Institute for CAD/CAM


Chennai-25

[email protected]


57/91

HEAT TRANSFER BYCONDUCTION & CONVECTION

Heat transfer Problems using FEM


58/91

Heat transfer Problems using FEM

K=conductivity

coeff..

h=convection coeff

T= Ambient Temp

P= Perimeter

A=area of cross

section


59/91


60/91


61/91

Heat in = Heat out

)()( TThPdxAdqqqA

f luxheatdx

dTKq

0)()(

)(0

TTdxhPdx

dTKAd

TTdxhPdqA

dxby 0)()( TThPdx

dTKA

dx

d

q + dqq

dx

MATHEMATICAL FORUMULATION


62/91

MATHEMATICAL FORUMULATION

0)()(

TThPdx

dTKA

dx

d

0 < X < L

@ x = 0, T=TO

@ x=L,)(

TThAdx

dTKA

BCs:

If the end is open toatmosphere

0

dx

dTKA If the end is insulated

GDE:

DOMAIN:


63/91

Approximate Solution Procedure

Let the App solution be, T* = a0+ a1x + a2x2

+ a3x3

Apply B.Cs and get 23120* aaT

Substituting in to GDE*T

RESIDUExRTThP

dx

TdKA

dx

d )(0)()(

*

*

Minimisation of residue by RITZ METHOD

RITZ FORMULATION


64/91

RITZ FORMULATION

0.)( dxFWxR

Let (x)be the weighting function

0)()()( dxxTThP

dx

dTKA

dx

d

0)()()()( dxxhPTdxxhPTdxxdxdT

KAdx

d

Integration by parts vduuvudvdv

u


65/91

ddT


66/91

0

).().()(

)(

xLx dxdTKAx

dxdTKAxdxxhPT

dxxhPTdxdx

d

dx

dTKA

)(),( LTB

3

3

2

20 1 xaxaxaaT*

Q)(

)(

dxxhPT

dxxhPTdxdx

d

dx

dTKA

A

3


67/91

321 ,,j j On applying eqn B into eqn A we get

e

j

ee

he

i

ee

hejiee

i

QdxATh

dxxTPhdxdx

d

dx

dAKa

)(00

;aTi

ii

*

1 B

ii raK

Where i=1,2,3 & j=1,2,3 and we can calculate Kij

and similarly ri

RITZ E ti f Fi it l t


68/91

RITZ Equation for a Finite element

O heI Je

)(),( ee LTB

he

eee

he

eeeehe

ee

QdxxTAh

dxxTphdxdx

d

dx

dTAK

0

00


69/91

T should be expressed in term of e


70/91

J

I

T

T

T should be expressed in term of

nodal variables

eJI

0 he

heaahe, Tx

aa, T@ x

J

I

10

10

00

Putting it in matrix form


71/91

Putting it in matrix form

1

0

1

01

a

a

heT

T

J

I

J

I

T

T

hea

a1

1

0

1

01

matrix form cont*


72/91

J

I

T

T

he

T

1

*

1

01 x1

J

I

T

T

he

x

he

x1

J

I

JI

T

TNN

xxaaT* 110

1

0

a

a

2 T


73/91

321

2

1

,,iNxfunctionweighting

T

Tre q wheqNT

i

J

I

j

j

jj

*

II

he

eee

heeee

eheee

QdxxTAh

dxxTphdxdx

d

dx

dTAK

0

0

*

0

I

Repeating the RITZ Eqn. I have

On substituting II in I


74/91

On substituting II in I

he

i

ee

he

ij

eeijhe ee

j

j

dxNTPh

dxNNphdxdx

dN

dx

dNAKq

0

00

2

1

;;12121 21hex N

hex-; N,; j,i


75/91


76/91

heKA


77/91

322

hehP

he

KAK

21

12

611

11 hehP

he

KAKe

dxNhPTrly 11lll

21he

hPTdxhe

x

hPT

hehPTdxNhPTr


78/91

2 22 hPTdxNhPTr

11

2hehPTre

eJ

Ie rT

TK

1

1

221

12

611

11 hehPT

T

ThehP

he

KA

J

I

On Appling the element stiffness matrix and load


79/91

vectors and introducing the continuity condition

we get

3

2

1

kk

k

4

3

2

1

T

T

T

T

2

1

r

r

2

1

r

r

2

1

r

r=

On applying the BCs and solving for the

remaining system of equations we get the nodal

values

ILLUSTRATION BROBLEM


80/91

ILLUSTRATION BROBLEM80OC

8 cm

1 cm4cm

K = 3 w/cm 0c

h = 0.1 w/cm20cT= 20

0c

51 2 3 4

421 3

Let the element be of equal length he = 2 cm


81/91

The element matrices are


82/91

hA

hehP

he

KA

K

e

0

00

21

12

6

11

11

ThA

ThehPf e

0

1

1

2

The element matrices for ELEMENT 1, 2 & 3 are

20

20;

666.6667.5

667.5666.6ee fK

The element matrices for ELEMENT 4 is


83/91

28

20

;066.7667.5

667.5666.6ee

fK

The assemblies of the element matrices are

28

40

40

40

20

006.7667.5000

667.5337.13667.500

0667.533.13667.50

00667.5333.13667.5

000667.5667.6

5

4

3

2

1

T

T

T

T

T

Appling the boundary condition T1= 80oC


84/91

1

28

4040

80667.540

006.7667.500

667.5333.13667.500667.5333.13667.5

00667.5337.13

5

4

3

2

T

TT

T

3.308.329.399.5380T

Exact solution is

6.302.332.403.5480T


85/91


86/91


87/91

Description Solidmechanic

Equation


88/91

mechanic

= Variable U=axial

displacement

(x) = property

associated with

the DERIVATIVE

of the variable.

E=youngs

modulus

= Property

associated with

the variable0

f(x) = Load not

at all associated

with the variablerA(x)

0xrAdx

dyxEA

dx

d:GDE

Description Fluidmechanic

Equation


89/91

mechanic

0Qdx

dK

dx

d:GDE XX

f(x) = Load not

at all associated

with the variable

Q=volume of flow

rate

= Property

associated with

the variable

h = convec co-eff

(x) = property

associated with

the DERIVATIVE

of the variable.

Kxx= Permeability

Co-eff

= Variable=fluid head

Description Electromagnetic

Equation


90/91

magnetic

0:

dx

dv

dx

dGDE

= Variable V=Electric

potential

= Property

associated with

the variable

0

(x) = property

associated with

the DERIVATIVE

of the variable. permissivity

f(x) = Load not

at all associated

with the variable

=Volume

charge

Density


91/91

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IGCAR FEM , syllabus

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