Non-Singular Method of Fundamental Solutions and its Application to Two Dimensional Elasticity Problems

Transcription

1 on-singular Method of Fundamental Solutions and its Application to Two Dimensional Elasticit Problems Qingguo Liu Mentor: Prof. Dr. Božidar Šarler

2 Scope Motivation Introduction Previous Applications of MFS Governing Equations of Elasticit Solution Procedure MFS and MFS umerical Examples Conclusions 1

3 Motivation What we want to know: displacement of object subject to external forces Before deformation After deformation 2

4 Motivation Initial microstructure Final microstructure The problem will be solved b MFS-tpe solution procedure. 3

5 Trefftz-tpe technique Introduction Approximating the solution of the problem b a linear combination of fundamental solutions Singularities, so source points have to be located outside the domain Boundar conditions Determining the unknown coefficients and the coordinates of the source points 4

6 Introduction Method widel used in linear elasticit. Kupradze and Aleksidze, First used in solid mechanics Smrlis, Expand the result of Kupradze and Aleksidze Maharejin, Anisotropic media Berger and Karageorghis, Application to anisotropic problems Redekop and Thompson, Axismmetric problems Marin and Lesnic, Cauch problems 5

7 Introduction Traditional MFS Problems with the fictitious boundar with source points The source points can be placed on the real boundar directl Young et al. 2005, 2007 Chen K. H. et al Šarler, 2008, 2009 Chen and Wang, 2010 A modified method: Liu, 2010 Used for potential problems Integration of the fundamental solutions Area-distributed sources cover the source points Determining of the diagonal elements: Šarler, 2008,

8 Governing Equations of Elasticit The state of stress at a point can be described b stress components formed on an element with sides δx and δ Smmetr: x x The first suffix indicates the indication of the normal to the plane on which the stress acts. The second suffix indicates the indication in which the stress acts. x x xx 7

9 8 Governing Equations of Elasticit The equations of equilibrium for a two-dimensional sstem subjected to external forces and bod forces b x and b are given as xx x b 0 x x x b 0 x where b ma b ma x x m is qualit and a x and a are accelerations along directions of x-axis and - axis, rectivel. When there is no bod force (b x =0 and b =0): x xx x x x 0 0

10 Governing Equations of Elasticit We introduce liner elasticit b Hooke's law. ( ) 2 2 xx xx xx x x ( ) 2 2 xx x x where 2 /.(1 2 ), ν is Poisson's ratio, E/2(1 ) is the shear modulus of elasticit and E is a constant known as the modulus of elasticit or Young's modulus. The definition of strain is 1 u u i j ij i, j x, 2 xj x i 9

11 Governing Equations of Elasticit avier's equations without bod force for isotropic elasticit are u ( p) u ( p) ( ) x x u p px p 1 2 p p x u ( p) u ( p) ux( p) p p 1 2 p p x x p where p pxix pi and px, p are 2D Cartesian coordinates with base vectors i x,i 10

12 , Governing Equations of Elasticit, The boundar conditions are D u ( p) u ( p); p (Displacement B.C.) t i i i T ( p) t( p); p (Traction B.C.) i ix, 11

13 Governing Equations of Elasticit The definition of tractions is t n n i, j x, and j i i ii i ij j 12

14 Solution Procedure:MFS The Kelvin's fundamental solution of elastostatics is 1 1 ( pi si )( p j s j ) Uij ( ps, ) (3 4 )log( ) ij,,, 2 i j x 8 (1 ) r r where U ij (p, s) represents the displacement in the i direction at point p due to a unit point force acting in the j direction at s, distance between the point p and s. U U U xx x 1 1 ( ps, ) (3 4 )log( ) 8 (1 ) ( p s ) 2 x x 2 r r 1 ( p, s) Ux ( p, s) 8 (1 ) 1 1 ( ps, ) (3 4 )log( ) 8 (1 ) r ( p s )( p s ) x x 2 r 2 ( p s) 2 r ps, r ( p q ) ( p q ) 2 2 x x is the 13

15 Solution Procedure:MFS B using the definition of traction, we obtain 2 (1 ) U (, ) 2 (, ) (, ) (, ) xx p s Ux p s U U xx p s x p s T ( ps, ) n n 1 2 px 1 2 p p p x xx x 2 (1 ) U x ( p, s) 2 U ( p, s) U x ( p, s) U ( p, s) T ( ps, ) n n 1 2 px 1 2 p p p x x x Ux ( p, s) U (, ) 2 (1 ) (, ) xx p s Ux p s 2 Uxx ( p, s) T ( ps, ) n n px p 1 2 p 1 2 p x x x U ( p, s) U x ( p, s) 2 (1 ) U ( p, s) 2 U x ( p, s) T ( ps, ) n n px p 1 2 p 1 2 p x x 14

16 Solution Procedure:MFS In MFS, the displacement field is represented b fundamental solutions in source points s n, n=1,2,,. The displacements and the tractions are approximated b a linear combination of fundamental solutions as follows: u ( p) U ( p, s ) U ( p, s ) x xx n n x n n n1 n1 u ( p) U ( p, s ) U ( p, s ) x n n n n n1 n1 t ( p) T ( p, s ) T ( p, s ) x xx n n x n n n1 n1 t ( p) T ( p, s ) T ( p, s ) x n n n n n1 n1 15

17 Solution Procedure:MFS Let us place points p n, n=1,2,,. on the boundar u ( p) U ( p, s ) U ( p, s ) x xx n n x n n n1 n1 u ( p) U ( p, s ) U ( p, s ) x n n n n n1 n1 t ( p) T ( p, s ) T ( p, s ) x xx n n x n n n1 n1 t ( p) T ( p, s ) T ( p, s ) x n n n n n1 n1 We introduce boundar condition indicators D D 1, p T 1, p ( p) ( p) T 0, p 0, p T D 16

18 Solution Procedure:MFS The coefficients are calculated from a sstem of algebraic equations b k 1 ( ) b k 1, 2, kn n kn n k D T ( p ) U ( p, s ) ( p ) T ( p, s ) kn k ix k n k ix k n D T ( p ) U ( p, s ) ( p ) T ( p, s ) kn k i k n k i k n D T b ( p ) u ( p ) ( p ) t ( p ) k k i k k i k 1 b U xx U x u n x U U u x n 17

19 Solution Procedure: MFS In the non-singular method of fundamental solutions, we place distributed sources s n at points p n, n=1,2,,, on the boundar. 18

20 Solution Procedure: MFS The following two formulas satisf the governing equation u ( p) U ( p, s ) da( s ) U ( p, s ) da( s ) x A( p ) xx n n n n A( pn) x n n n n1 n1 u ( p) U ( p, s ) da( s ) U ( p, s ) da( s ) A( p ) x n n n n A( pn) n n n n1 n1 We are using integration over point sources for smoothing out the singularit. 19

21 Solution Procedure: MFS When p n is outside of A(p n ), we consider r a since R<<r. So 2 R 1 ( pi pni )( p j pnj ) U ( ) ij ( p, sn) da( sn) (3 4 )log( ) ij,, 2 i j x APn 8 (1 ) a a When p is inside of A(p n ), 1 2 R 1 R R 1 R log( ) da( s ) log( ) 2 log( ) log( ) A( ) n r rdrd r r rdr p n r 2 0 R ( p ) R x s nx r cos ( ) da( s ) A( ) 2 n p r r rdrd R sin( )cos( ) R ( px snx)( p sn ) 2 R r cos( )sin( ) R cos ( ) da s A( ) 2 n rdrd p ( ) 0 r r ( p sn ) 2 R r sin ( ) R sin( )cos( ) R da( s ) A( ) 2 n rdrd p r r

22 Solution Procedure: MFS xx p pn A( p ) xx p sn sn U (, ) U (, ) da( ) n 2 2 R 1 ( px pnx ) 2 (3 4 )log( ), a R 8 (1 ) a a (3 4 ) R (3 4 ) 1 R a R a ( px pnx ) log( ), a R 8 (1 ) R 16 (1 ) 16 (1 ) 8 (1 ) U ( p, p ) U ( p, p ) U ( p, s ) da( s ) x n x n A( p ) x n n 2 R ( px pnx )( p pn ) 2, 8 (1 ) a 0, a R U ( p, p ) U ( p, s ) da( s ) n A( p ) n n n n a R 2 2 R 1 ( p p ) n (3 4 )log( ), 2 a R 8 (1 ) a a R (3 4 ) R a R a ( p p ) log( ) 8 (1 ) R 16 (1 ) 16 (1 ) 8 (1 ) (3 4 ) 1 n, a R 21

23 Solution Procedure: MFS The displacements are approximated b a linear combination of Ũ ij (p, p n ) as follows: u ( p) U ( p, p ) U ( p, p ) x xx n n x n n n1 n1 u ( p) U ( p, p ) U ( p, p ) x n n n n n1 n1 22

24 Solution Procedure: MFS 2 (1 ) U ( p, p ) 2 U ( p, p ) U ( p, p ) U ( p, p ) t n n xx n x n xx n x n x( p) n x n1 1 2 px 1 2 p p px 2 (1 ) U x ( p, pn) 2 U ( p, pn) U x ( p, pn) U n nx n1 1 2 px 1 2 p p ( pp, n) n px U ( p, p ) U ( p, p ) 2 (1 ) U ( p, p ) 2 U ( p, p ) t n n x n xx n x n xx n ( p) n x n1 px p 1 2 p 1 2 px U ( p, pn) Ux ( p, pn) 2 (1 ) U ( p, pn) 2 Ux ( p, p ) n n nx n px p 1 2 p 1 2 p n1 x 23

25 Solution Procedure: MFS The coefficients are calculated from a sstem of algebraic equations b k 1 ( ) b k 1,2, kn n kn n k D T ( p ) U ( p, p ) ( p ) T ( p, p ) kn k ix k n k ix k n D T ( p ) U ( p, p ) ( p ) T ( p, p ) kn k i k n k i k n D T b ( p ) u ( p ) ( p ) t ( p ) k k i k k i k 24

26 Solution Procedure: MFS The method proposed b Šarler 2009 is applied to determine the diagonal coefficients of matrix [Φ, Ψ]. We first assume a constant solution, e.g ui ( p) c c everwhere. Then we can solve for the corresponding and, and use ui( p) ui( p) 0, i, j x, p p i then j u ( p) U ( p, p ) U ( pp, n) c n 0 x xx n c x n px n1 px n1 px u ( p) U ( p, p ) U ( pp, n) c n 0 x xx n c x n p n1 p n1 p u ( ) (, ) p U x p pn U ( p, pn) c c n n 0 px n1 px n1 px u ( p ) U (, ) U (, ) c p p p p 0 x n c n n p n1 p n1 p 25 n n n c

27 Solution Procedure: MFS We assume n1 n1 So when U U U ( pp, ) ij n c n pi Uij ( pp, n) c n p i p=p 0 0 i, j x, 1,2,, ( p, p ) 1 U ( p, p ), i, j x, jx m c jx n c n pi t n1 pi nm ( p, p ) 1 U ( p, p ), i, j x, j m c j n c n pi t n1 pi nm ow we have calculated all elements of the matrix and and. Uxx U u x n x U U u x n 26, m m n n

28 Example , E 1 / m Boundar conditions: 1 u p u p 4 x x Analtical solution: u u x p x 1 4 p 2 27

29 Example 1 R' 5d R d/3 28 MFS MFS

30 Example 1 The number of boundar nodes used is 100. A number M=19 of field points are selected inside the domain along the line =0.5 with 0.5 x The solutions for R d / ( 2,3,,11) at these field points are computed with MFS and compared with the analtical solutions. The relative errors of the numerical solutions are defined as. M 1 a a Err (( ui ui ) / ui ) M k 1 2 1/2 The solutions for the distance R' d( 2,3,,11) of the fictitious boundar from the true boundar are also computed with MFS and compared with the analtical solutions. 29

31 Example 1 With R' 5d, R d/3 30

32 Example 1 The number of boundar nodes used is from 100 to R=d/3 R =5d. We also use the 19 points we used just now to compute. The solutions are compared with the analtical solutions. We know the more nodes we get, the relative errors will be smaller with the MFS. In this case errors will be ver small with the MFS. 31

33 Example 1 The matrix becomes ill conditioned in MFS when the number of points is growing 32

34 Example E / m

35 Example 2 Boundar condition: 0 : u 0 u 0.05 x 1: u 0 u 0.05 x x 0 or 1: t 0 t 0 x R' 5d R d/3 34

36 Example 2 We put MFS and MFS solutions together and use red color to express the MFS results, blue color to denote the MFS results. And a number of M=19 field points are selected inside the domain along the line 0.05 x= we can see that the MFS results and the MFS results almost coincide and both on the boundar and in the square. So both methods can be used. 35

37 Example 3: shear deformation E / m 6 2 Material properties used are for steel. 36

38 Example 3: shear deformation Boundar condition : 0 : u 0 u 0.05 x 1: u 0 u 0.05 x x 0or1: t 0 t 0 x R' 5d R d/3 37

39 Example 3: shear deformation We put MFS and MFS solutions together and use red color to express the MFS results, blue color to denote the MFS results. And a number of M=19 field points are selected in side the domain along the line x=0.5 with We can see that the MFS resuilts and MFS resuilts are almost the same. So both methods can be used. 38

40 Conclusions Solving boundar-value problems with the nodes on the boundar onl First time used for elasticit problems Based on: modified MFS Integrated first Determining the diagonal elements: Šarler, 2008,

41 40 Future Work

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