Adaptive Wavelet Collocation for Elasticity Lu s Manuel Castro - - PDF document

Adaptive Wavelet Collocation for Elasticity

Lu´ ıs Manuel Castro Silvia Bertoluzza Instituto Superior T´ ecnico, Lisbon Istituto di Analisi Numerica del CNR, Pavia luis@civil.ist.utl.pt http://www.civil.ist.utl.pt/˜luis wavelet@dragon.ian.pv.cnr.it http://dragon.ian.pv.cnr.it/˜aivlis Wavelets in Numerical Analysis and Simulation Funchal, 11-12 March 2002

DECivil N´ ucleo de An´ alise de Estruturas - ICIST

2

Outline

• 1. Motivation
• 2. Adaptive collocation techniques
• 3. Plane elasticity problems
• 4. Numerical applications
• 5. Reissner-Mindlin plate bending problems
• 6. Numerical applications
• 7. Conclusions and further developments

3

Motivation Numerical Simulation of structural engineering problems

4

Adaptive Collocation techniques

• Bertoluzza, S., “An Adaptive Collocation Method

based on Interpolating Wavelets”, in Multi- scale Wavelet Methods for Partial Differen- tial Equations, edited by Dahmen, Kurdila and Oswald, Academic Press, 1997.

• Bertoluzza, S. and Naldi, G., “A wavelet col-

location method for the numerical solution

• f partial differential equations”, ACHA, 3,

1996.

• Bertoluzza, S., “Adaptive wavelet collocation

method for the solution of Burgers equation”, Transport Theory and Stat. Phys., 25, 1996. Interpolating wavelets

• Deslaurier, G. and Dubuc, S., “Symmetric it-

erative interpolation processes, Constructive Approximation, 5, 1989.

5

Deslaurier-Dubuc interpolating functions

• 0.2

0.2 0.4 0.6 0.8 1 1.2

• 5
• 4
• 3
• 2
• 1

1 2 3 4 5

θN(x) =

• φL(y)φL(y − x) dy

N = 2L + 1 Properties

• suppθ = [−N, N]
• θ is refinable
• θ(n) =

φL(y)φL(y − n) dy = δn0

• Polynomials up to order N can be represented

as a linear combination of the integer trans- lates of θ.

6

Classical Theory of Elasticity f =

  

fx fy

   t =   

txγ tyγ

  

Displacements u =

  

ux(x, y) uy(x, y)

  

Strains ε =

        

εxx(x, y) εyy(x, y) εxy(x, y)

        

Stresses σ =

        

σxx(x, y) σyy(x, y) σxy(x, y)

        

7

Classical Theory of Elasticity Definition of the stress field σxx = e1 ∂ ux ∂ x + e2 ∂ uy ∂ y σyy = e2 ∂ ux ∂ x + e1 ∂ uy ∂ y σxy = e3

∂ux

∂y + ∂uy ∂x

• e1 =

E 1 − ν2 e2 = ν E 1 − ν2 e3 = E 2 (1 + ν)

8

Classical Theory of Elasticity Problem 1 Find u = [ux, uy]T such that A u = f (Ω) u = g (Γu) B u = t (Γσ) A =

        

e1 ∂2 ∂ x2 + e3 ∂2 ∂ y2 (e2 + e3) ∂2 ∂ x ∂ y (e2 + e3) ∂2 ∂ x ∂ y e2 ∂2 ∂ x2 + e1 ∂2 ∂ y2

        

B =

       

e1 nx ∂ ∂ x + e3 ny ∂ ∂ y e2 nx ∂ ∂ y + e3 ny ∂ ∂ x e3 nx ∂ ∂ y + e2 ny ∂ ∂ x e3 nx ∂ ∂ x + e1 ny ∂ ∂ y

       

9

Numerical Applications

2 4 6 8 1 2 3 4 − 9 . 9 9 1 −3.3314 −1.1118 1 . 1 7 8 3.3274 5 . 5 4 7 7.7665 2 4 6 8 1 2 3 4 − . 8 6 2 4 2 − . 5 6 9 8 − . 1 5 9 6 −0.36002 2 4 6 8 1 2 3 4 −1.2631 − 2 . 3 2 8 −1.783 −1.523 −0.48343 − . 4 8 3 4 3 2 4 6 8 1 2 3 4

10

Numerical Applications

−10 −8 −6 −4 −2 2 4 6 8 10 0.5 1 1.5 2 2.5 3 3.5 4 Solução exacta Malha uniforme com Jmax=4 −1 −0.9 −0.8 −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 0.5 1 1.5 2 2.5 3 3.5 4 Solução exacta Malha uniforme com Jmax=4 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0.5 1 1.5 2 2.5 3 3.5 4 Solução exacta Solução com uma malha uniforme com Jmax=4

11

Numerical Applications

Discretisation N j0 jmax ngrid ndof TCP U A 4 3 3 81 162 0.04 B 4 3 4 289 578 0.72 C 4 3 5 1089 2178 52.63 Table 1: Discretisations involved in the analysis of the square cantilever

10

2

10

3

10

4

10

−2

10

−1

STRAIN ENERGY ERROR Uniform grid refinement Number of degrees of freedom error

12

Square cantilever

σ xx σ yy σ xy q V 3.000E+00

• 3.000E+00

0.000E+00 8.000E-01

• 1.000E+00

0.000E+00 0.000E+00

• 1.300E+00

= 2.50 p a / E σ xx σ yy σ xy q V 3.000E+00

• 3.000E+00

0.000E+00 8.000E-01

• 1.000E+00

0.000E+00 0.000E+00

• 1.300E+00

= 2.50 p a / E

13

Square cantilever

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 "malhai.res" 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 "malha2.res" 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 "malha3.res" 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 "malhaf.res"

10

2

10

3

10

4

10

−3

10

−2

10

−1

Adaptive refinement Uniform grid refinement Number of degrees of freedom STRAIN ENERGY ERROR error

14

Square cantilever

Mesh jmax ngrid ndof initial 4 289 578 2 5 597 1194 3 6 868 1736 final 7 917 1834 Table 2: Adaptive non-uniform grids used in the solution of the square plate

σ xx σ yy σ xy q V 3.000E+00

• 3.000E+00

0.000E+00 8.000E-01

• 1.000E+00

0.000E+00 0.000E+00

• 1.300E+00

= 2.50 p a / E

15

Numerical Applications

0.5 1 0.2 0.4 0.6 0.8 1 1.2 0.5 1 0.2 0.4 0.6 0.8 1 1.2 −1.2601 0.5 1 0.2 0.4 0.6 0.8 1 1.2 0.49277 −0.025409 0.5 1 0.2 0.4 0.6 0.8 1 1.2

16

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 0.49277 1.0109 − . 2 5 4 9 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 0.020447 − . 5 5 6 5 9 −0.13177 −0.36008 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 −0.0454 −0.13834 −0.41715 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2

17

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 −0.038184 −0.12615 −0.21412 −0.47802 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 −0.048228 −0.13607 −0.22391 −0.3996 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2 − . 3 5 4 3 1 −0.12469 −0.21394 − . 4 8 1 7 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 1.2

18

Square plate with a central crack

Mesh jmax ngrid ndof 1 4 289 578 8.1355 2 5 625 1250 0.6666 3 6 670 1340 0.6591 4 7 812 1624 0.6530 5 8 823 1646 0.6529 6 9 1082 2164 0.6529

0.5 1 0.2 0.4 0.6 0.8 1 1.2 −0.05394 0.5 1 0.2 0.4 0.6 0.8 1 1.2 0.97356 64782 0.5 1 0.2 0.4 0.6 0.8 1 1.2 − . 3 5 4 3 1 − . 1 2 4 6 9 −0.21394 0.5 1 0.2 0.4 0.6 0.8 1 1.2

19

Non-rectangular elements

20

Gravity dam

σ xx σ yy σ xy q V 1.000E+00

• 8.000E+00

0.000E+00 4.000E+00

• 1.000E+00

0.000E+00 5.000E+00

• 5.000E-01

= 16.6667 p a / E

0.5 1 1.5 2 2.5 3 1 2 3 4 5 "mesh.res"

21

Gravity dam

σ xx σ yy σ xy q V 0.000E+00

• 2.000E+01

0.000E+00

• 7.000E+01

1.500E+01

• 1.500E+01

0.000E+00 = 83.3333 p a / E

0.5 1 1.5 2 2.5 3 1 2 3 4 5 "mesh.res"

22

Gravity dam

σ xx σ yy σ xy q V 0.000E+00

• 2.000E+01

0.000E+00

• 7.000E+01

1.500E+01

• 1.500E+01

0.000E+00 = 83.3333 p a / E

0.5 1 1.5 2 2.5 3 1 2 3 4 5 "mesh.res"

23

Reissner-Mindlin plate bending theory f =

        

q

        

t =

        

mxγ myγ qγ

        

Displacements u =

        

θx(x, y) θy(x, y) w(x, y)

        

Strains ε =

                    

χxx(x, y) χyy(x, y) χxy(x, y) γx(x, y) γy(x, y)

                    

Stress resultants σ =

                    

mxx(x, y) myy(x, y) mxy(x, y) vx(x, y) vy(x, y)

                    

24

Reissner-Mindlin plate bending theory Definition of the stress resultant fields mxx = Df

∂ θx

∂ x + ν ∂ θy ∂ y

• myy = Df
• ν ∂ θx

∂ x + ∂ θy ∂ y

• mxy = D1

∂θx

∂y + ∂θy ∂x

• vx = D2
• θx + ∂w

∂x

• vy = D2
• θy + ∂w

∂y

• Df

= E h3 12(1 − ν2) D1 = G h3 12 D2 = 5 6 G h

25

Reissner-Mindlin plate bending theory Problem 2 Find u = [θx, θy, w]T such that A u = f (Ω) u = g (Γu) B u = t (Γσ)

A =

            

D2 ∂ ∂ x D2 ∂ ∂ y D2

∂2

∂ x2 + ∂2 ∂ y2

• Df

∂2 ∂ x2 + D1 ∂2 ∂ y2 − D2 ν Df ∂2 ∂ x ∂y + D1 ∂2 ∂ x ∂y −D2 ∂ ∂ x ν Df ∂2 ∂ x ∂y + D1 ∂2 ∂ x ∂y Df ∂2 ∂ y2 + D1 ∂2 ∂ x2 − D2 −D2 ∂ ∂ y

            

B =

            

D2 nx D2 ny D2 ∂ ∂ x nx + D2 ∂ ∂ y ny Df ∂ ∂ x nx + D1 ∂ ∂ y ny ν Df ∂ ∂ y nx + D1 ∂ ∂ x ny ν Df ∂ ∂ x ny + D1 ∂ ∂ y nx Df ∂ ∂ y ny + D1 ∂ ∂ x nx

            

26

Simply supported plate

0.5 1 0.5 1 −0.0034878 − . 1 5 1 2 −0.02456 0.5 1 0.5 1 − . 3 1 5 8 4 −0.02456 − . 1 4 2 4 − . 3 4 8 7 8 0.5 1 0.5 1 .0021006 . 6 3 3 4 7 0.012686 0.019037 0.5 1 0.5 1 . 5 2 8 8 0.10438 0.15668 0.5 1 0.5 1 . 5 2 9 1 . 1 4 3 8 0.13925 0.5 1 0.5 1 −0.10602 −0.082448 − . 4 7 8 3 − . 1 1 7 1 9 0.5 1 0.5 1 0.48974 0.24277 . 5 7 5 4 3 0.5 1 0.5 1 0.057041 . 2 4 2 4 2 0.48959 0.5 1 0.5 1

27

Simply supported rectangular plate

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 Diagrama de momentos Mx Solução exacta Solução com malha uniforme com Jmax=4 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.005 0.01 0.015 0.02 0.025 Solução exacta Solução com uma malha uniforme com Jmax=4 Campo de deslocamentos transversais 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 −0.14 −0.12 −0.1 −0.08 −0.06 −0.04 −0.02 Solução exacta Solução com malha uniforme comJmax=4 Diagrama de momentos torsores Mxy

28

Simply supported rectangular plate

0.5 1 0.5 1 0.5 1 0.5 1 0.5 1 0.5 1 0.5 1 0.5 1 . 2 9 7 7 0.16767 0.10453 0.041393 0.5 1 0.5 1 10264 0.065397 . 1 2 5 3 0.17566 0.5 1 0.5 1 −0.049106 − . 2 7 8 7 9 − . 6 6 5 1 9 0.5 1 0.5 1 0.5 1 0.5 1 0.5 1 0.5 1

29

Simply supported rectangular plate

0.5 1 0.5 1 0.5 1 0.5 1 0.5 1 0.5 1 0.5 1 0.5 1 . 2 6 4 4 0.084837 0.14903 0.19183 0.5 1 0.5 1 1 1 2 1 0.067062 0.12291 0.17876 0.5 1 0.5 1 0.066006 −0.043925 −0.014483 0.5 1 0.5 1 0.5 1 0.5 1 0.5 1 0.5 1

30

Simply supported rectangular plate

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 −0.05 −0.04 −0.03 −0.02 −0.01 0.01 Distribuição de momentos torsores Mxy Malha uniforme Malha refinada 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 −250 −200 −150 −100 −50 50 Distribuição de esforços transversos Vy Malha uniforme Malha refinada

31

Further developments

• General domains (domain decompositon)
• 3D problems
• Material behaviour