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number of elements to use for one structural element

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mophie

Aerospace
Jan 4, 2003
9
I have to model a truss and perform a dynamic a analysis in nastran/patran. Since the real structure is built from metal tubes welded together, I will use bar-elements for (almost) all structural elements.

My question: Does it make any difference whether I use one bar-element for an entire tube, or should I splitt up each (long) tube in a number of bar-elements?

Rob

 
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mophie,

Personally I would use beam elements for this case (by saying "bar" elements, do you mean beam elements?). These are tried, tested and trusted, and can give predictable and easily checkable results. Moreover, in the case of static/dynamic analysis, it should not matter if you use only a single beam (for a single tube), as the model should pick up the features of the mode/displaced shape. However, if you feel that the mode shape looks slightly "coarse", you could always introduce a few more - this introduction will not increase run-time by any significant amount (beam eleemnts are very simple in formulation, and require very little CPU time), but in the case of static results the more beams you have for a single tube the more the results will be a pain in the u-know-what to check and post-pro! You would only ever need to use more beams if you were intending to perform some form of dynamic contact analysis (mesh size here = fine mesh, then quadruple it!) where propagation wave times become important.

Hope this helps
-- drej --
 
I have to disagree with Drej as the number of elements will affect the frequency calculation. The fewer the number of elements then the structure behaves more stiff, and hence the frequencies will be incorrect. If you only have a single element then you won't pick up all the mode shapes as these will depend on the number of nodes you have, and if you have no nodes in the centre of the beam, for example, then you'll miss the first mode shape. If in doubt run the analysis with a single element then double the number of elements to see the difference. You should see a convergence to the correct solution. There comes a point, of course, where the difference in solutions is negligible and you can be confident of your results.
 
Thanx for your replies!

I was somewhat unclear about the elements I use. I started out with beam element, but I got the tip to use rods, because the tubes in the structure are basically not designed to carry lateral loads. What will be the differences in a static and normal modes analysis?


 
I'm not sure what you mean by a beam element and a rod element as a beam element can have the properties of any section. I guess you mean a truss element which can only carry loads along its length and has no bending stiffness. Make sure you are using the correct element as a support could be pinned at its end but have mode shapes that a truss element wouldn't pick up. If it is a truss that represents the structure correctly then a single element would be correct or otherwise the trusses/rods would be pinned at the nodal positions, which would be wrong for both static and modal analysis. I'd use a beam element with the correct bending properties of the tube but pin it at its end where it connects to the rest of the structure.
 
The ROD element has only area and torsional constant and will therefore support only axial and torsional loads. The unsupported degrees-of-freedom will be condensed from the stiffness matrix by AUTOSPC. It shouldn't matter if you use one or multiple ROD elements in your model.

pj
 
mophie

Mode of vibration of truss and beam is almost different. The natural vibration modes of truss are only governed by its axial stiffness/mass while those of beam model (Timoshenko beam) are governed by its axial, bending and shearing stiffness. It's clear that axial stiffness of element of the same cross section is significantly higher than beam element. So its natural frequency is significantly higher for truss model than beam of the same properties. In my view, stiffness of truss element sujected to nodal load is exact since its shape functions come from complementary solution of differential equation (EAu')' = 0 ; u = A+Bx (linear). Therefore no matter how we increase the number of elements, the results should be the same.

 
sorry correct the 3rd line

It's clear that axial stiffness of element of the same cross section is significantly higher than its flexural stiffness.

 
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