Help with Values

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UGMENTALCASE
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Registriert: Di 27. Sep 2016, 14:34

Help with Values

Beitrag von UGMENTALCASE »

Good afternoon,

I'm very interested, although a total NOOOOOOOB at this sort of thing. I'm looking at FEM in freecad as well, but struggling to understand what the value sections relate too? In FreeCAD if I wanted to add a weight to a face, I would use the weight in KG multiplied by 9.82 (g) and that would give me my value. However I'm struggling to find anything that gives me that information here. I may be looking in the wrong place, so please accept my apologies if I am. Any help would be appreciated.

I'm looking at basically (for now) adding a weight in KG to a face and looking at the results. I basically want to check back and make sure I'm doing it correctly.

Thanks in advance
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SHautsch
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Registriert: Mo 15. Apr 2013, 11:03

Re: Help with Values

Beitrag von SHautsch »

Hello UGMENTALCASE,

finite element analysis does not use units. The user himself has to decide which unit system is used by putting in the correct values corresponding to that self-chosen unit system. This is called consistency of units.

There is a good article about that on fem-helden.de, unfortunately only in German - but the table is quite useful:
http://fem-helden.de/einheiten-und-fem/

For example, if you use the Young's Modulus (E) with 206,000.00 N/mm^2, the unit for force is Newton (N) and the unit for length is millimeter (mm). So forces need to be applied in Newton, which means you multiply your weight/mass with the acceleration of gravity (around 9.81 m/s^2 or just use 10 as a good approximation). For example a part which weighs 5 kg produces 50 Newtons of force.
Don't mess this up with the actual mass unit, which would be tons (t) when using N and mm. To calculate the force, the use of kg in the example above comes from the definiton of the unit Newton, which is kg*m/s^2.

To specify the mass of a part, use its density in t/mm^3, for steel that is around 7.8E-9 t/mm^3.

Beware! Z88Aurora can NOT use gravity loads at the moment, so you would have to specify the gravity load by adding a surface load.

Kind regards,
SHautsch
UGMENTALCASE
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Registriert: Di 27. Sep 2016, 14:34

Re: Help with Values

Beitrag von UGMENTALCASE »

ok I think I get it, I've attached my part it doesn't behave how I would have expected to see the stresses, I'm guessing I'm going something wrong would you be able to help me? My part is a U shape item, I've attached an image. where there are two holes in the top edge, that's where I've put the fixed constraints, and where the weight is marked, that's where I want to replicate a force on around 25-30 kgs. So I've marked where I though it would stress in the corners, but like I say I'm probably looking at it wrong, or not doing something right. Would you be able to help me please?

Cheers
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SHautsch
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Re: Help with Values

Beitrag von SHautsch »

Dear UGMENTALCASE,

if we should help you with your problem, please send us the project folder of your Z88 project (zipped) to z88aurora-at-uni-bayreuth.de or attach it here in the forum (if that works).

In that specific case which you displayed in your attached image, you are looking at displacements in Z direction, which looks quite correct from here :mrgreen:
Better choose "total displacements" to display the part's behaviour / "real" deformation.

To look at stresses, please choose "stresses at ****" from the list in the post processor.

Kind regards,
SHautsch
UGMENTALCASE
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Re: Help with Values

Beitrag von UGMENTALCASE »

Thank you so much for your help, maybe I'm thinking it should be more complicated than it is? I'm not sure. I've attached the zip folder so you can see what I'm doing.
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SHautsch
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Re: Help with Values

Beitrag von SHautsch »

It looks pretty good.

Material is specified correctly.
You could use a slightly finer mesh, but thats OK for now.
Picking and boundary conditions look good as well.
The force is 294 N, which is around 30 kg.
Solver runs fine - so no errors here. Little hint: always use PARDISO if your RAM is large enough - it's much faster and shows wrong preprocessing like underconstained models...the iterative solvers often run fine and you get bad results (only zeros for example).

For your project: Total displacement is 12.1 mm in the lowest right corner of your part. The maximum von Mises stress is 3140 N/mm^2, which is much too high - this is way above linear elastic behaviour!
S235JR can stand up to 235 MPa of stress in the linear elastic area (yield point) and up to 510 MPa until it breaks ( tensile strength). Between those values we have a plastic area where your part will be deformed irreversible.
As your stress exceeds the 510 MPa, your part will surely break under load.
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UGMENTALCASE
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Re: Help with Values

Beitrag von UGMENTALCASE »

Ok thank you very much for checking it for me, I'm starting to get my head around it now. I'll try the other solver, just upgraded to 32GB of RAM so should handle it ok.

Thanks again, so nice to have some one help a Nooob out :-)
UGMENTALCASE
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Re: Help with Values

Beitrag von UGMENTALCASE »

Sorry just to add to this then, I wanted to understand what I was looking at between a little 1mm Radius in the corners and a 25mm radius. I've attached an image of it ran with the 25mm radius, so the von mises is a lot less, am I right in saying?

I might have a play around with it and try and understand more. I'm basically trying to replicate a screw force on that lower right face. So this thing is a clamp and the screw is coming through one of the top holes, and 'jacks' the part up so it clamps. So 30 kgs is probably a little excessive to be honest but I'm just experimenting :-)
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SHautsch
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Re: Help with Values

Beitrag von SHautsch »

Enlarging the radius clearly elimates the notch stress problem, which always occurs when you bend a part with a sharp corners. Good example:
https://en.wikipedia.org/wiki/Stress_co ... ration.jpg

The rest is due to the high load or small cross section of your part. As 1430 MPa is still too high, you should calculate the real load or experiment with other values so you can sort out which load can be endured.
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