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# Meshfree approximation schemes

Wed, 2006-11-08 00:01 - N. Sukumar

An overview on meshfree approximation schemes that I recently posted can be found here. Following Zhigang's note indicating the limited accessibility of blogspot.com, a local version of the article is also provided. I am also attaching a PDF version of the html file. The conversion was done using PDF Online. An article that provides more details is also available online. JAVA applets for plotting basis functions can also be accessed [1D] [2D ] [3D ].

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## Comments

## Blogspot.com is not accessible in some parts of the world

Suku: The Meshfree Methods Blog uses blogspot.com, which is not accessible in some parts of the world. This is one of several reasons that caused us to migrate Applied Mechanics News to iMechanica.

## Sukumar's notes on the meshfree method

Dear Suku: Thank you very much for creating the link to your notes on the meshfree method. I can access it readily, and hope more people will benefit from your work.

## How to interpolate?

Dear Sukumar: For the plane strain problems in the couple stress theory, any node has three degrees of freedom, i.e. the displacement

andu, the rotation vectorvThe shape functionωz.ψwill be adopted is the C1 natural neighbor interpolant,in which ψ3I-2 ,ψ3I-1, ψ3I for node I are directly associated with the three degree nodal degrees of freedom wI , θIx and θIy respectively.But how to interpolate the,u,vusing the shape functionωzψ? Thank you!## Re: Interpolation in couple-stress theory

The C1 shape functions are like the counterpart of Hermites in two dimensions. In one dimension (beam), each Hermite element has 4 degrees of freedom (2 at each node), and the C1 is just an extension of the same to 2D. So, for n nodes in 2D you would have 3n degrees of freedom. Assuming you require independent interpolation for your displacement vectors and rotation and the first two should be C1 and the rotation is C0 then one can use C1 nn for the displacements and C0 nn for the rotation. The interpolation order/smoothness that is needed will stem from what appears in the weak/variational form and the essential boundary conditions that must be met by the trial functions you adopt. Must ensure that the trial functions are admissible within the variational formulation; of course, one will end-up with more unknowns at a node given the demands on smoothness and nodal interpolation that have been placed on the trial functions. Not sure if this answers your question, but hope it points you in the right direction?

P.S.:There are quite a few applications of meshfree methods in gradient (different theories) elasticity. You might wan't to refer to Tang et al. (2003), Computer Modeling in Engineering Sciences, Vol. 4, No. 1., pp 177-196 where meshless local Petrov-Galerkin (MLPG ) is implemented for the Fleck-Hutchinson theory. Yes, if only first derivatives of the trial/test function appear in the variational statement then C0 basis suffices. If second derivatives are present in the weak form, then you w'd need C1 basis functions.## Re:Interpolation in couple-stress theory

Dear Sukumar:Thank for your useful advice.The rotation vector

discussedωzis microrotation which equals to the macrocopical rotation,so the interpolation for displacement vectors and microrotation isn't independent.In this case,the interpolation puzzled me for a long time.## How to interpolate?

Dear Sukumar:In the general couple stress theory,the micro-rotation is treated as an independent kinematic quantity with no direct dependence upon the displacement.Thus,there are only the first-order derivatives of the displacement and micro-rotation appearing in the variational equation,so C0 nn are required for displacement and micro-rotation.Is my thought right? thank you!

## matrix equation

Dear Sukumar:

(1)For the matrix equation

ku=b,stiffness matrixkshould be a 6n*6n matrix when considering the x-directional and y-directional derivatives of the shape functions of the c1 natural neighbor interpolant. Here n is the number of the nodes. Generalized displacementsuis a 6n*1 matrix. It is obvious thatbis also a 6n*1 matrix, but if we ignore the body forces, boundary integralbis a 4N *1 matrix, N (N<n) represents the number of the associated boundary nodes. How to explain the contradiction?(2)When assembling the external force vector,how to impose the essential BCs which is composed of the nodal function and the nodal gradients?

Thank you!

## Dear Sukumar i am using

Dear Sukumar

i am using EFG in my thesis. i got confuse in one stage. please clear my

doubt.in this method D_m= D_max*C_i is there. where D_m is the radious of

influence.It has been written in many papers D_max is generally taken as

2-4 forstatic analysis. but i have a doubt in that. for convergence study

when i am increasing the no of nodes (more than 1000),i am not getting

good

convergent result when i have taken D_max=3. now when i am increasing the

value of D_max say 5,6,7,8 then i am getting better convergent

value(Energy norm is reducing on increasing the no of nodes). Am i right

or not ? So how should i take the value of D_max. what is the maximum

value of D_max can i take ? please help me.

So what is the appropriate value of D_max which i should take?

thanks

## Hi

Hi Vivek,

Theoretically, the more nodes is used, the better result will be obtained. But it's not correct in your situation! It may relate to A matrix in MLS, when large number of nodes is used, A rapidly becomes ill-condition. This results in inaccuracy solution if invert form of A is used. As a solution to this problem, LU decomposition can be employed in determing shape function and its derivatives.

No maximum of D_max is ruled, but D_max should be chosen so that do not lose local character of shape function.

Hope this helps.

Cheers,

Canh Le

Canh V. Le, PhD Student

Department of Civil and Structural Engineering

The University of Sheffield

Mappin Street,

Sheffield, S1 3JD, UK.

Phone: GB +44 1142225724

http://www.cladu.shef.ac.uk/new

## Re: Dear Sukumar i am using

Dear Vivek,

Theoretically we need np=(n+p)!/n!/p! (n ane p are respectively the physical dimension and consistency) no of points within the support of the kernel function in order to make the moment matrix non-singular. However from numerical perspective we generally take ‘slightly’ bigger support. But ‘arbitrarily’ large support may lead to an exaggerated smoothness in the approximation. Consequently system matrix may lose diagonal dominancy and becomes ill-conditioned.

Now the question is how big the support should be? That depends on many things such as particle density, the function you are approximating (solution of PDE), how the particles are distributed within the support etc. Best way (in the context of EFGM) of choosing an ‘optimum’ support size is numerical experiments.

It is relevance to mention that there exist some mesh-free techniques which does not rely on any user defined support size. For instance NEM (you can find a brief introduction and references on sukumar’s webpage), NURBS-based mesh-free methods (doi:10.1016/j.cma.2007.11.024).

Hope it helps.

Cheers!!