Integration point 4

[ukd204]

What is an integration point for a finite element ? Please please give me a detail description.

 

[DanStro]

On the bottom of page 11 the pdf at the location below starts to explain the concept.

http://ocw.mit.edu/courses/material...hanics-of-materials-fall-1999/modules/fea.pdf

 

[EdR]

An integration point is the point within an element at which integrals are evaluated numerically. These points are chosen in such a way that the results for a particular numerical integration scheme are the most accurate. Depending on the integration scheme used the location of these points will vary. Check any numerical methods text for numerical integration methods and they will give you more information...

Since you asked for a detailed description consider that the stiffness of an element is:

K = integral (trans . [c] . dv) (over the volume of the element dv)

The question then becomes "How do I evaluate this volume integration in the most accurate manner". Selection of points to achieve this defines the "integration points".

Ed.R.

 

[flash3780]

To add to what others have said, it's important to note that finite element displacements are most accurate at the nodes. However, derived values (i.e. stresses and strains) tend to be most accurate at the integration points (and sometimes least accurate at the nodes). Of course, the accuracy of nodal strains is dependent on your mesh size, so with an adequate mesh it shouldn't matter.

On a (somewhat) related note, folks often compare elemental strains with averaged nodal strains to determine whether their mesh is adequate. If there is a significant difference, mesh refinement is required.

 

[3doorsdwn]

To add to what others have said, it's important to note that finite element displacements are most accurate at the nodes. However, derived values (i.e. stresses and strains) tend to be most accurate at the integration points (and sometimes least accurate at the nodes). Of course, the accuracy of nodal strains is dependent on your mesh size, so with an adequate mesh it shouldn't matter.

On a (somewhat) related note, folks often compare elemental strains with averaged nodal strains to determine whether their mesh is adequate. If there is a significant difference, mesh refinement is required.

You know, that's something I was thinking about today: software calculates stresses (typically) at the Gauss points of the element....but a variable in that is the displacements (which are calculated at the nodes [not the Gauss points]. So there is sort of a discontinuity there (if I am thinking about it correctly).

I guess it goes back to what you said about meshing properly.

 

[ukd204]

Thank you all of you guys.Now I understand the concept of integration point.On the basis of theory,results will be more accurate if integration points are more.Then what is the theory behind reduced integration.Those elements have been used to resist shear/volumetric locking for incompressible material.Deformation are large for the reduction in integration points.But anyhow,are those results accurate?

 

[flash3780]

You know, that's something I was thinking about today: software calculates stresses (typically) at the Gauss points of the element....but a variable in that is the displacements (which are calculated at the nodes [not the Gauss points]. So there is sort of a discontinuity there (if I am thinking about it correctly).

No discontinuity. Think about a simple link element (pictured below). The finite element solution provides displacements at the nodes (based on a weak form solution to the defining differential equations). Strains are a derivative of the displacements (du/dx), so you take the difference of the nodal values calculate the strains.

If you have some sort of non-uniform loading (say an acceleration load pulling things to the right), the real-life strains are not uniform within the element.

Even if the nodal displacements were exact, which they generally aren't, the linear shape function within the element could not produce anything but a constant strain [u(x)=C1x+C2 ==> du/dx = C1]. The constant strain results will generally be more accurate in the center of the elements (integration points), and least accurate at the nodes in this case.

Of course, you can see how having a lot of elements increases your accuracy, regardless. The idea being if you make the change in strain across the element small (d2u/dx2), the results become more accurate.

2 Undeformed Link Elements
o=====o=====o--->

2 Deformed Link Elements (du/dx used to compute strains)
o=======o=======o--->

Smarter elements (higher order) have shape functions that can better approximate the deformation of the element. If, for example, you had another node in the center of your link elements, you could have a quadratic shape function, which would handle a constant acceleration load (the example given) much more accurately. That additional accuracy comes at a price, though, since higher order elements are more computationally expensive.

So, there are two schools of thought in the finite element world:
1. Use lots of lower order elements in areas with high stress gradients to obtain an accurate solution (h-method), a-la ABAQUS or ANSYS
2. Use fewer, but higher-order (smarter) elements in areas of high stress gradients to obtain an accurate solution (p-method), a-la Pro/Mechanica

The best answer is probably some combination of the two, which is likely where (I think) FE software will go in the future.

 

[3doorsdwn]

No discontinuity. Think about a simple link element (pictured below). The finite element solution provides displacements at the nodes (based on a weak form solution to the defining differential equations). Strains are a derivative of the displacements (du/dx), so you take the difference of the nodal values calculate the strains.

I'm not sure I am following you here. Are you saying [some] software will use an interpolated displacement [between the nodes] to get a more accurate stress at the gauss points?

 

[flash3780]

All I'm saying is that you take the difference of the nodal positions and displacements to calculate the strain (du/dx); you get a constant value for strain. The constant strain value just so happens to be more accurate at the integration points (between the nodes) than at the nodes.