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   2 2 2 d x d y d z                   m k x y z 0     2   2 2 2 K M 0   dt dt dt      mx kx 0 ENGINEERING MATHEMATICS FINITE ELEMENT METHOD 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 COLOR GRAPHICS COMPUTER PROGRAMMING

FREE VIBRATION ANALYSIS ENGINEERING MATHEMATICS (PDE)  Free vibration system Newton’s 2 nd law of motion m     F ma mx x k    mx kx 0 x(t)

FREE VIBRATION ANALYSIS FEM DERIVATION  Finite element equation    mx kx 0              M K 0 Stiffness Displacement Mass matrix ( 3 dimensions ) matrix

FREE VIBRATION ANALYSIS FEM DERIVATION  Harmonics motion Applitude,      ( ) t sin( t ) x time, t    2 f Period, T = 1/ f  Finite element equation for free vibration             2 K M 0   Circular frequency Mode shapes

FREE VIBRATION ANALYSIS

FREE VIBRATION ANALYSIS

FREE VIBRATION ANALYSIS COMPUTER PROGRAMMING Example in FORTRAN

      1 0 0 0          T      1 0 0 0   K B C B V       1 0 0 0 e     1 2   E    0 0 0 0 0 C   2     (1 )(1 2 )     1 2   0 0 0 0 0   2     1 2   0 0 0 0 0   2   b 0 0 b 0 0 b 0 0 b 0 0 1 2 3 4   0 c 0 0 c 0 0 c 0 0 c 0   1 2 3 4   0 0 d 0 0 d 0 0 d 0 0 d 1     1 2 3 4   B 6 V  c b 0 c b 0 c b 0 c b 0  1 1 2 2 3 3 4 4   0 d c 0 d c 0 d c 0 d c 1 1 2 2 3 3 4 4       d 0 b d 0 b d 0 b d 0 b 1 1 2 2 3 3 4 4

FREE VIBRATION ANALYSIS CODE VALIDATION  Exact Solution  Numerical Solution (FDM , FEM, FVM, etc.)  Commercial Software (ANSYS, NASTRAN, etc.)  Experimental Result 1.0 0.6  10 Exact solution Present Yoon, et al. 0.4 0.8 Malan et al. Present Present 8 Sampaio 0.2 0.6 6 v Nu u 0.0 0.4 4 -0.2 0.2 2 -0.4 0.0  0 t -0.6 0.0 0.2 0.4 0.6 0.8 1.0 0 40 50 60 70 10 20 30 0 90 270 360 180

SOLVING HEAT TRANSFER ANALYSIS FE ANALYSIS MUST SATISFY ENERGY EQUATION FOURIER’S LAW

HEAT TRANSFER                                     C C T T K K K K K K T T       q q q T c c h h r r         y x z Q c                                       x y z t Q Q Q Q Q Q Q Q Q Q c c Q Q q q h h r r ENGINEERING MATHEMATICS FINITE ELEMENT METHOD 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 COLOR GRAPHICS COMPUTER PROGRAMMING

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

SOLID MECHANICS ENGINEERING MATHEMATICS FINITE ELEMENT METHOD 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 COMPUTER PROGRAMMING COLOR GRAPHICS

SOLVING FLUID DYNAMICS ANALYSIS FE ANALYSIS MUST SATISFY CONSERVATION OF MASS CONSERVATION OF MOMENTUM CONSERVATION OF ENERGY

LOW-SPEED INCOMPRESSIBLE FLOW 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 COMPUTER PROGRAMMING COLOR GRAPHICS

HIGH-SPEED COMPRESSIBLE FLOW              n      *    M U t  C uU R  {U} {E} {F} 0 i i u    t x y  2     t u                u v n K u U R         k us j i us 2        2    u u p uv          {U} ;{E} ; {F}      2    2  t u v uv v p             n K p R                      k ps ps u pu v pv 2 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 COMPUTER PROGRAMMING COLOR GRAPHICS