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GENERAL PROCEDURE OF FE SOFTWARE FE SOFTWARE PROCEDURE PRE - - PowerPoint PPT Presentation

GENERAL PROCEDURE OF FE SOFTWARE FE SOFTWARE PROCEDURE PRE PROCESSING SOLVING POST PROCESSING SOLVING GENERAL PROCEDURE ENGINEERING MATHEMATICS (PDE) FEM DERIVATION COMPUTER PROGRAMMING CODE VALIDATION FREE VIBRATION


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SLIDE 1

GENERAL PROCEDURE OF FE SOFTWARE

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SLIDE 2

POST – PROCESSING PRE – PROCESSING SOLVING

FE SOFTWARE PROCEDURE

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SLIDE 3

SOLVING

ENGINEERING MATHEMATICS (PDE) COMPUTER PROGRAMMING CODE VALIDATION

GENERAL PROCEDURE

FEM DERIVATION

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SLIDE 4

COLOR GRAPHICS COMPUTER PROGRAMMING

FREE VIBRATION

FINITE ELEMENT METHOD ENGINEERING MATHEMATICS

C C ELEMENT CONDUCTION MATRIX: C DO 100 I=1,3 DO 100 J=1,3 AKC(I,J) = 0. DO 110 K=1,2 AKC(I,J) = AKC(I,J) + BT(I,K)*B(K,J) 110 CONTINUE AKC(I,J) = TK*AREA*THICK*AKC(I,J) 100 CONTINUE DO 120 I=1,3 DO 120 J=1,3 AKE(I,J) = AKE(I,J) + AKC(I,J) 120 CONTINUE C C ELEMENT CONVECTION MATRICES: C IF(LTYPE(IE,3).NE.1) GO TO 300 FAC = H*AREA/12. DO 230 I=1,3 DO 230 J=1,3 AKH(I,J) = FAC 230 CONTINUE DO 240 I=1,3 AKH(I,I) = 2.*FAC 240 CONTINUE FAC = H*AREA*TI/3. DO 250 I=1,3 QH(I) = FAC 250 CONTINUE DO 260 I=1,3 QE(I) = QE(I) + QH(I) DO 260 J=1,3 AKE(I,J) = AKE(I,J) + AKH(I,J) 260 CONTINUE 300 CONTINUE

 

2 2 2 2 2 2

d x d y d z m k x y z dt dt dt              

     

2

K M        

mx kx   

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SLIDE 5
  • Free vibration system

k

x(t)

m

Newton’s 2nd law of motion

 



x

F ma mx

   mx kx

ENGINEERING MATHEMATICS (PDE)

FREE VIBRATION ANALYSIS

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SLIDE 6

   mx kx

Mass matrix Stiffness matrix Displacement ( 3 dimensions )

FEM DERIVATION

FREE VIBRATION ANALYSIS      

K M     

  • Finite element equation
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SLIDE 7

     

2

M K        

FEM DERIVATION

FREE VIBRATION ANALYSIS

Circular frequency Mode shapes

  • Harmonics motion

time, t

Applitude,

Period, T = 1/ f

x

( ) sin( ) t t     2 f   

 Finite element equation for free vibration

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SLIDE 8

FREE VIBRATION ANALYSIS

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SLIDE 9

FREE VIBRATION ANALYSIS

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SLIDE 10

COMPUTER PROGRAMMING

FREE VIBRATION ANALYSIS

Example in FORTRAN

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SLIDE 11

      

T e

K V B C B 

 

1 1 1 1 2 2 (1 )(1 2 ) 1 2 2 1 2 2                                    E C              

 

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

1 6                       b b b b c c c c d d d d B c b c b c b c b V d c d c d c d c d b d b d b d b

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SLIDE 12

CODE VALIDATION

FREE VIBRATION ANALYSIS

 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 0.8 0.6 0.4 0.2 1.0

Present Exact solution

u

2 90 4 180 6 8 10 270 360 Yoon, et al. Present

Nu 

  • 0.4
  • 0.6

20 10

  • 0.2

30 0.0 0.2 0.4 40 50 60 0.6 70 Present Malan et al. Sampaio

v t

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SLIDE 13

HEAT TRANSFER ANALYSIS SOLVING

FE ANALYSIS MUST SATISFY ENERGY EQUATION FOURIER’S LAW

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SLIDE 14

COLOR GRAPHICS COMPUTER PROGRAMMING

HEAT TRANSFER

FINITE ELEMENT METHOD ENGINEERING MATHEMATICS

C C ELEMENT CONDUCTION MATRIX: C DO 100 I=1,3 DO 100 J=1,3 AKC(I,J) = 0. DO 110 K=1,2 AKC(I,J) = AKC(I,J) + BT(I,K)*B(K,J) 110 CONTINUE AKC(I,J) = TK*AREA*THICK*AKC(I,J) 100 CONTINUE DO 120 I=1,3 DO 120 J=1,3 AKE(I,J) = AKE(I,J) + AKC(I,J) 120 CONTINUE C C ELEMENT CONVECTION MATRICES: C IF(LTYPE(IE,3).NE.1) GO TO 300 FAC = H*AREA/12. DO 230 I=1,3 DO 230 J=1,3 AKH(I,J) = FAC 230 CONTINUE DO 240 I=1,3 AKH(I,I) = 2.*FAC 240 CONTINUE FAC = H*AREA*TI/3. DO 250 I=1,3 QH(I) = FAC 250 CONTINUE DO 260 I=1,3 QE(I) = QE(I) + QH(I) DO 260 J=1,3 AKE(I,J) = AKE(I,J) + AKH(I,J) 260 CONTINUE 300 CONTINUE

t T c Q z q y q x q

z y x

                     

           

T K K K T C

r h c

   

  

 

 

  

r h q Q c

Q Q Q Q Q     

           

T K K K T C

r h c

   

  

 

 

  

r h q Q c

Q Q Q Q Q     

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SLIDE 15

STRUCTURAL ANALYSIS SOLVING

FE ANALYSIS MUST SATISFY EQUILIBRIUM EQUATION STRESS – STRAIN REL. STRAIN – DISPLACEMENT REL.

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SLIDE 16

COLOR GRAPHICS COMPUTER PROGRAMMING

SOLID MECHANICS

FINITE ELEMENT METHOD ENGINEERING MATHEMATICS

DO 20 I=1,3 DO 30 J=1,6 B(I,J) = B(I,J)/(2.*AREA) BT(J,I) = B(I,J) 30 CONTINUE 20 CONTINUE C C ELASTICITY MATRIX: C FAC = ELAS/(1.-PR*PR) C(1,1) = FAC C(1,2) = FAC*PR C(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. C C ELEMENT STIFFNESS MATRIX: C DO 100 I=1,3 DO 100 J=1,6 DUMA(I,J) = 0. DO 200 K=1,3 DUMA(I,J) = DUMA(I,J) + C(I,K)*B(K,J) 200 CONTINUE 100 CONTINUE

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SLIDE 17

FLUID DYNAMICS ANALYSIS SOLVING

FE ANALYSIS MUST SATISFY CONSERVATION OF MASS CONSERVATION OF MOMENTUM CONSERVATION OF ENERGY

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SLIDE 18

LOW-SPEED INCOMPRESSIBLE FLOW

COLOR GRAPHICS COMPUTER PROGRAMMING ENGINEERING MATHEMATICS FINITE ELEMENT METHOD

DO 110 I=1,6 DO 110 J=1,3 DO 110 K=1,3 DO 110 L=1,6 CXX = 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 CONTINUE SXX(IA,IB) = 2.*ANEW*CXX + ANEW*CYY SXY(IA,IB) = ANEW*CXY SYX(IA,IB) = ANEW*CYX SYY(IA,IB) = ANEW*CXX + 2.*ANEW*CYY 100 CONTINUE C C COMPUTE [HX] AND [HY] MATRICES: C DO 150 IA=1,3 DO 150 IB=1,6 CX = 0. CY = 0. DO 160 I=1,6 DO 160 J=1,3 CX = CX + A(IB,I)*B(I,J)*G(J,IA) CY = CY + A(IB,I)*C(I,J)*G(J,IA) 160 CONTINUE HX(IA,IB) = CX/DEN HY(IA,IB) = CY/DEN 150 CONTINUE

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SLIDE 19

HIGH-SPEED COMPRESSIBLE FLOW

COLOR GRAPHICS COMPUTER PROGRAMMING ENGINEERING MATHEMATICS FINITE ELEMENT METHOD

DO 110 I=1,6 DO 110 J=1,3 DO 110 K=1,3 DO 110 L=1,6 CXX = 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 CONTINUE SXX(IA,IB) = 2.*ANEW*CXX + ANEW*CYY SXY(IA,IB) = ANEW*CXY SYX(IA,IB) = ANEW*CYX SYY(IA,IB) = ANEW*CXX + 2.*ANEW*CYY 100 CONTINUE C C COMPUTE [HX] AND [HY] MATRICES: C DO 150 IA=1,3 DO 150 IB=1,6 CX = 0. CY = 0. DO 160 I=1,6 DO 160 J=1,3 CX = CX + A(IB,I)*B(I,J)*G(J,IA) CY = CY + A(IB,I)*C(I,J)*G(J,IA) 160 CONTINUE HX(IA,IB) = CX/DEN HY(IA,IB) = CY/DEN 150 CONTINUE

         {U} {E} {F} t x y

                                                            

2 2

u v u u p uv {U} ;{E} ; {F} v uv v p u pu v pv

 

    

      

n * i i u

M U t C uU R

 

 

      

2 n k us j i us

t u K u U R 2

  

          

2 n k ps ps

t u K p R 2