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Linear buckling with an unsymmetric T - extrusion

User: "Altair Forum User"
Altair Employee
Updated by Altair Forum User

Hi 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'>

 

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