Linear buckling with an unsymmetric T - extrusion
Altair EmployeeHi there,
i tried to model a linear buckling analysis with an unsymmetric T - extrusion. The solution is far away from the solution calculated with hand:
HM solution with load applied to the shear center: 5543 kN
HM solution with load applied to the extrusion center: 5543 kN
Calculation by hand: 8980 kN for euler case 2
Calculation by hand: 2328 kN for 'drill-buckling' (don't know how this is called in english-> german: Biegedrill-Knicken)
So why the solutions are so off and why the solutions for shear- and extrusion center are the same? If the load isn't applied to the shear center the unsymmetric extrusion should also drill-buckle which leads into much less buckle load.
I added the 2 FE-models (shear center and extrusion center)
Thank you for your help /emoticons/default_smile.png' srcset='/emoticons/smile@2x.png 2x' title=':)' width='20'>
Answers
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Thank's for investigating. Please look at the Last discharge point. In one model it's the shear center and in the other model it's the extrusion center.
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Altair Forum User said:
Hi,
Can you try these models?
i opened these models but don't know what to do with it? You only changed the extrusion?
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Hi,
I saw you use PBEAM property. You should not use it.
Could you try defining property with 'PBEAML' card image and rerun?
and, to change load position, you should use offsets on cbeam elems
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Altair Forum User said:
opened these models but don't know what to do with it? You only changed the extrusion?
I have given an offset for both the cases on beam elements so that load is applied on respective centers
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Thank you very much
/emoticons/default_smile.png' srcset='/emoticons/smile@2x.png 2x' title=':)' width='20' /> I will try this later today and report. Could you please tell whats the difference between PBEAM and PBEAML ?
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