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07-28-2016 08:20 AM

Hello again,

The problem kow is that in an liniar elastic analysis, with Chexa elements and glued surfaces I get huge differences between elemental and nodal-elemental stresses. That is, in nodal, I get 350 MPa, while in nodal-elemental, I get 950 MPa? The size of the elements has been reduced in the area of interest.

Many thanks

Ionut

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22 REPLIES

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07-29-2016 10:00 AM

Hi,

unlike the "usual" finite element modelling where displacements (not stresses!) are always continous over element borders, we do not have continuos displacements in areas of glue (unless we have coincidenting nodes and same element order). The underlying shape functions between the glued elements and therefore the displacements do not match exactly - this makes the stresses in the region of the glue "bumpy". It seems that this is what you experience. This is not an error and has to be considered when evaluating the stresses in glued regions.

Martin

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07-29-2016 10:19 AM

Hi Martin and thank you,

After posting the firts message, I meshed aplying a mesh mating between tethraedral elements. Now, teh differences between elemental and nodal elemental are no longer so big, but still important. That is, Nodal elemental stresses are almost double that the elemental ones. Both are averaged.

many thanks

Ionut

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08-04-2016 06:02 AM

Hi Ionut,

do you know the difference between elemental stress and nodal stresses in the first place?

Elemental stresses are already simplified may be by calculating only at the center of element and therefore in most cases e.g. with bending smaller than extrapolated nodal stresses. The quantity depends on size of elements.

Best wishes,

Michael

| *Production:* NX10; *Development:* VB, TCL/TK, FORTRAN; *Testing:* NX11 | *engelke engineering art GmbH, Germany *

| Kudos for good posts! And if my post answers your question, please mark it as an "Accepted Solution".

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08-04-2016 07:30 AM

Hi Michael,

Yes I know the difference and all the stuff. The problem is that the differences are huge!

I know that for a coorect solution the differecens between the two values must be less than 10%

%. The same model, in Solid Works, with tetra elements (of the same average size) are let's say normal.

Ionut

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08-04-2016 07:40 AM

Hi Ionut,

could you send pictures of displacement and stresses to get an imagination of your problem?

Best wishes

Michael

*Production:* NX10; *Development:* VB, TCL/TK, FORTRAN; *Testing:* NX11 | *engelke engineering art GmbH, Germany *

| Kudos for good posts! And if my post answers your question, please mark it as an "Accepted Solution".

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08-04-2016 09:11 AM

Hi Michael,

I have attached fourimages. The first, Model.jpg presents what I am analysing. It is a steel plate, simply supported on the bottom face, with an edge fixed, under a force applyed as can be seen. What is depicted is only half of the real structure. The symmetric condition is applied only on a part of the face because I want to model a crack. The analysis is a simple elastic one.

The other images presents the YY stresses, Nodal and element nodal, both averaged.

As you can see, the difference between the two stresses are important. The element size is 0.5 mm, Chexa20.

I have also done an analysis on an assembly, that includes also another part, a composite plate (picture Assembly.jpg). Here, the differences are even bigger.

Many thanks

Ionut

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08-04-2016 10:35 AM

Hi,

you are modelling an elastic crack - you do that with 20-noded elements which are the correct elements to use, so all looks fine. The whole analysis looks fine to me!

The nature of a crack with linear elastic material is that the stresses have a singularity: the closer you get to the crack tip, the higher the stresses. Next to singularities the stresss have a huge gradient: In your example the stresses at the crack tip are more than double then at the other side of the element. Extrapolating stresses from the inner of the elements to the nodes can give very different results.

Stresses at the crack tip are anyway of limited value. The finer the mesh, the higher the stresses - they are singular, means: unlimited at the crack tip. Run your model with a finer mesh and you get higher stresses!

So the question is: Why do you need the stresses are the crack tip? To invetigate possible crack propagation, the stresses at the crack tip are of no use because they depend on the element size anyway.

Best regadrs.

Martin

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08-04-2016 10:41 AM - edited 08-04-2016 10:44 AM

Hi Ionut,

after watching your figures I wonder if you try to model a crack and then I read it.

Can you please send unaveraged stresses, too? I think the effects will be more clear there and the gradients will increase.

If you can see the gradients in the last element before and after crack tip are gigantic. I think the strains in the frontier elements are gigantic, too. May be you will leave the range of use of linear elastic elements.

I think that is not the right way to model a crack tip because its a singularity in every mesh size.

Use smaller elements and a small radius at crack tip to avoid singularity.

Best wishes,

Michael

Edit: Oh, Mkuessner was faster :-)

*Production:* NX10; *Development:* VB, TCL/TK, FORTRAN; *Testing:* NX11 | *engelke engineering art GmbH, Germany *

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08-04-2016 10:50 AM

Hi Martin,

What I am trying to do is to evaluate the influence of the composite part on the stresses arround the crack (the more general issue is evaluating pipe repairs using composite wrappings). That is I will run the model with different crack lengths and asses the way the composite and the steel interact. If I correctly understant, the finner the mesh, the higher the stresses are, so I do not have a real possibility to evaluate the stresses arrond the crack.

Kind regards and thank you

Ionut

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