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Mass matrix

by Hassan2
Tags: finite element, mass matrix
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Hassan2
#1
Mar27-12, 10:59 PM
P: 409
Hi all,

I need to do a dynamic structural analysis using finite element method and I have a question about the mass matrix.

Question: I have the force per nodes and I need to calculate the displacement of each node at a given time. For this purpose, it seems that I need to distribute the element mass between its nodes and then treat the while body as a finite number of interconnected point masses. If my approach is correct, how can I distribute the element mass to its nodes?

If that's not the right way, what do you suggest for such an analysis?

Thanks in advance.
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AlephZero
#2
Mar28-12, 06:38 PM
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I'm not sure what you are asking here. If you are using a finite element program for dynamics, it should calculate the element mass matrices automatically. You need to input the density for the material(s) in the model, of course. There will probably be a way to input additional "mass points" (including rotary inertias) at the grid points of the model.

If you need to formulate a mass matrix yourself, the basic requirement is that for any velocity vector ##v##, the mass matrix ##M## should give the kinetic energy correctly as ##v^TMv/2##.

In other words, you formulate an element mass matrix by integrating
[tex]\iiint N^TN \rho\,dx\,dy\,dz[/tex] over the volume of the element, where ##N## are the element shape functions.
Hassan2
#3
Mar28-12, 08:19 PM
P: 409
Quote Quote by AlephZero View Post
I'm not sure what you are asking here. If you are using a finite element program for dynamics, it should calculate the element mass matrices automatically. You need to input the density for the material(s) in the model, of course. There will probably be a way to input additional "mass points" (including rotary inertias) at the grid points of the model.

If you need to formulate a mass matrix yourself, the basic requirement is that for any velocity vector ##v##, the mass matrix ##M## should give the kinetic energy correctly as ##v^TMv/2##.

In other words, you formulate an element mass matrix by integrating
[tex]\iiint N^TN \rho\,dx\,dy\,dz[/tex] over the volume of the element, where ##N## are the element shape functions.
Many thanks. That was the answer I was looking for.

Yes, I'm writing the code by myself and I'm familiar with the finite element method because I did it 3-dimensional magnetic field analysis. When I moved to the dynamic analysis ( dynamic of an electric machine), I found that, since we are calculating the displacement of the nodes, we need the solve the dynamic equations for the nodes the, hence we need to assign some mass to each node. I think the integral does this for us. Thanks for the very helpful comment.

Hassan2
#4
Mar29-12, 08:18 PM
P: 409
Mass matrix

A question about the above integral:

In the above integral, is N a vector whose components are the shape functions of the element nodes?
AlephZero
#5
Mar30-12, 05:42 AM
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N is the matrix of the shape functions (which are functions of the variables that describe the position of a point inside the element). For a 3-dimensional "brick" element with n nodes, N would typically have 3n rows and 3 columns.

If x is the vector of nodal displacements for the element, then Nx is the displacement vector at a point inside the element.
Hassan2
#6
Mar30-12, 06:13 AM
P: 409
Aren't the shape functions of the element always the same for the three components of x? I mean the same interpolating function is used for each of the components.
AlephZero
#7
Mar30-12, 07:04 AM
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Often they are the same, and then you can partition the element mass matrix into
[tex]M = \begin{pmatrix} M_x & 0 & 0 \\ 0 & M_y & 0 \\ 0 & 0 & M_z \end{pmatrix}[/tex]
where [tex]M_x = M_y = M_z[/tex]

But the element variables might be arranged in a different order, so the partitioning is not so obvious. That depends on how the rest of the FE code is organized.
Hassan2
#8
Mar30-12, 07:27 AM
P: 409
So Mx=My=Mz=M, where M the element mass? At the beginning it was odd to me and I thought for a cubic element, Mx=My=Mz=M/8 . Now I understand it. It's clear that if force F is applied on one of the nodes only, and there rest see no force, the force F gives acceleration to the whole element.

Thanks.


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