general procedure of fe software...computer programming color graphics heat transfer engineering...

GENERAL PROCEDURE

OF FE SOFTWARE

POST – PROCESSING

PRE – PROCESSING

SOLVING

FE SOFTWARE PROCEDURE

SOLVING

ENGINEERING MATHEMATICS (PDE)

COMPUTER PROGRAMMING

CODE VALIDATION

GENERAL PROCEDURE

FEM DERIVATION

COLOR GRAPHICSCOMPUTER PROGRAMMING

FREE VIBRATION

FINITE ELEMENT METHODENGINEERING MATHEMATICS

CC ELEMENT CONDUCTION MATRIX:C

DO 100 I=1,3DO 100 J=1,3AKC(I,J) = 0.DO 110 K=1,2AKC(I,J) = AKC(I,J) + BT(I,K)*B(K,J)

110 CONTINUEAKC(I,J) = TK*AREA*THICK*AKC(I,J)

100 CONTINUEDO 120 I=1,3DO 120 J=1,3AKE(I,J) = AKE(I,J) + AKC(I,J)

120 CONTINUECC ELEMENT CONVECTION MATRICES:C

IF(LTYPE(IE,3).NE.1) GO TO 300FAC = H*AREA/12.DO 230 I=1,3DO 230 J=1,3AKH(I,J) = FAC

230 CONTINUEDO 240 I=1,3AKH(I,I) = 2.*FAC

240 CONTINUEFAC = H*AREA*TI/3.DO 250 I=1,3QH(I) = FAC

250 CONTINUEDO 260 I=1,3QE(I) = QE(I) + QH(I)DO 260 J=1,3AKE(I,J) = AKE(I,J) + AKH(I,J)

260 CONTINUE300 CONTINUE

2 2 2

2 2 2 0d x d y d zm k x y zdt dt dt

2 0K M 0mx kx

Free vibration system

k

x(t)

mNewton’s 2nd law of motion

xF ma mx

0 mx kx

ENGINEERING MATHEMATICS (PDE)

FREE VIBRATION ANALYSIS

0 mx kx

Mass matrix

Stiffness matrix

Displacement( 3 dimensions )

FEM DERIVATION


0KM

Finite element equation

2 0MK

FEM DERIVATION


Circular frequency Mode shapes

Harmonics motion

time, tApplitude,

Period, T = 1/ f

x ( ) sin( )t t

2 f

Finite element equation for free vibration

COMPUTER PROGRAMMING


Example in FORTRAN

TeK VB C B

1 0 0 01 0 0 0

1 0 0 01 20 0 0 0 0

2(1 )(1 2 )1 20 0 0 0 0

21 20 0 0 0 0

2

EC

1 2 3 4

1 2 3 4

1 2 3 4

1 1 2 2 3 3 4 4

1 1 2 2 3 3 4 4

1 1 2 2 3 3 4 4

0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 01

0 0 0 060 0 0 0

0 0 0 0

b b b bc c c c

d d d dB

c b c b c b c bVd c d c d c d c

d b d b d b d b

CODE VALIDATION


Exact Solution

Numerical Solution (FDM , FEM, FVM, etc.)

Commercial Software (ANSYS, NASTRAN, etc.)

Experimental Result

0.0

0.8

0.6

0.4

0.2

1.0

0.0 0.80.60.40.2 1.0

PresentExact solution

u

0

2

90

4

180

6

8

10

270 360

Yoon, et al. Present

0

Nu

0

-0.4

-0.62010

-0.2

30

0.0

0.2

0.4

40 50 60

0.6

70

PresentMalan et al.Sampaio

v

t

HEAT TRANSFER ANALYSIS

SOLVING

FE ANALYSIS MUST SATISFY

ENERGY EQUATION

FOURIER’S LAW


HEAT TRANSFER

FINITE ELEMENT METHODENGINEERING MATHEMATICS

CC ELEMENT CONDUCTION MATRIX:C

DO 100 I=1,3DO 100 J=1,3AKC(I,J) = 0.DO 110 K=1,2AKC(I,J) = AKC(I,J) + BT(I,K)*B(K,J)

110 CONTINUEAKC(I,J) = TK*AREA*THICK*AKC(I,J)

100 CONTINUEDO 120 I=1,3DO 120 J=1,3AKE(I,J) = AKE(I,J) + AKC(I,J)

120 CONTINUECC ELEMENT CONVECTION MATRICES:C

IF(LTYPE(IE,3).NE.1) GO TO 300FAC = H*AREA/12.DO 230 I=1,3DO 230 J=1,3AKH(I,J) = FAC

230 CONTINUEDO 240 I=1,3AKH(I,I) = 2.*FAC

240 CONTINUEFAC = H*AREA*TI/3.DO 250 I=1,3QH(I) = FAC

250 CONTINUEDO 260 I=1,3QE(I) = QE(I) + QH(I)DO 260 J=1,3AKE(I,J) = AKE(I,J) + AKH(I,J)


tTc Q

zq

yq

x

q zyx

T K K K T C rhc

rhqQc Q Q Q Q Q

T K K K T C rhc

rhqQc Q Q Q Q Q

STRUCTURAL ANALYSIS

SOLVING


EQUILIBRIUM EQUATION

STRESS – STRAIN REL.

STRAIN – DISPLACEMENT REL.


SOLID MECHANICS

FINITE ELEMENT METHODENGINEERING MATHEMATICSDO 20 I=1,3DO 30 J=1,6B(I,J) = B(I,J)/(2.*AREA)BT(J,I) = B(I,J)


CC ELASTICITY MATRIX:C

FAC = ELAS/(1.-PR*PR)C(1,1) = FACC(1,2) = FAC*PRC(1,3) = 0.C(2,1) = C(1,2)C(2,2) = C(1,1)C(2,3) = 0.C(3,1) = 0.C(3,2) = 0.C(3,3) = FAC*(1.-PR)/2.

CC ELEMENT STIFFNESS MATRIX:C

DO 100 I=1,3DO 100 J=1,6DUMA(I,J) = 0.DO 200 K=1,3DUMA(I,J) = DUMA(I,J) + C(I,K)*B(K,J)


FLUID DYNAMICS ANALYSIS

SOLVING


CONSERVATION OF MASS

CONSERVATION OF MOMENTUM

CONSERVATION OF ENERGY

LOW-SPEED INCOMPRESSIBLE FLOW


ENGINEERING MATHEMATICS FINITE ELEMENT METHODDO 110 I=1,6DO 110 J=1,3DO 110 K=1,3DO 110 L=1,6CXX = CXX + A(IA,I)*B(I,J)*A(IB,L)*B(L,K)*G(J,K)CYY = CYY + A(IA,I)*C(I,J)*A(IB,L)*C(L,K)*G(J,K)CXY = CXY + A(IA,I)*C(I,J)*A(IB,L)*B(L,K)*G(J,K)CYX = CYX + A(IA,I)*B(I,J)*A(IB,L)*C(L,K)*G(J,K)

110 CONTINUESXX(IA,IB) = 2.*ANEW*CXX + ANEW*CYYSXY(IA,IB) = ANEW*CXYSYX(IA,IB) = ANEW*CYXSYY(IA,IB) = ANEW*CXX + 2.*ANEW*CYY

100 CONTINUECC COMPUTE [HX] AND [HY] MATRICES:C

DO 150 IA=1,3DO 150 IB=1,6CX = 0.CY = 0.DO 160 I=1,6DO 160 J=1,3CX = CX + A(IB,I)*B(I,J)*G(J,IA)CY = CY + A(IB,I)*C(I,J)*G(J,IA)

160 CONTINUEHX(IA,IB) = CX/DENHY(IA,IB) = CY/DEN

150 CONTINUE

HIGH-SPEED COMPRESSIBLE FLOW


ENGINEERING MATHEMATICS FINITE ELEMENT METHODDO 110 I=1,6DO 110 J=1,3DO 110 K=1,3DO 110 L=1,6CXX = CXX + A(IA,I)*B(I,J)*A(IB,L)*B(L,K)*G(J,K)CYY = CYY + A(IA,I)*C(I,J)*A(IB,L)*C(L,K)*G(J,K)CXY = CXY + A(IA,I)*C(I,J)*A(IB,L)*B(L,K)*G(J,K)CYX = CYX + A(IA,I)*B(I,J)*A(IB,L)*C(L,K)*G(J,K)

110 CONTINUESXX(IA,IB) = 2.*ANEW*CXX + ANEW*CYYSXY(IA,IB) = ANEW*CXYSYX(IA,IB) = ANEW*CYXSYY(IA,IB) = ANEW*CXX + 2.*ANEW*CYY

100 CONTINUECC COMPUTE [HX] AND [HY] MATRICES:C

DO 150 IA=1,3DO 150 IB=1,6CX = 0.CY = 0.DO 160 I=1,6DO 160 J=1,3CX = CX + A(IB,I)*B(I,J)*G(J,IA)CY = CY + A(IB,I)*C(I,J)*G(J,IA)

160 CONTINUEHX(IA,IB) = CX/DENHY(IA,IB) = CY/DEN

150 CONTINUE

{U} {E} {F} 0

t x y

2

2

u vu u p uv

{U} ;{E} ; {F}v uv v p

u pu v pv

n*

i i uM U t C uU R

2

nk us j i us

t u K u U R2

2

nk ps ps

t u K p R2

general procedure of fe software...computer programming color graphics heat transfer engineering...

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