Static Analysis; Beam "not rigid"

Static Analysis; Beam "not rigid"

armeno_sam
Explorer Explorer
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Static Analysis; Beam "not rigid"

armeno_sam
Explorer
Explorer

Hi to all,

I did a simple beam analysis of a custom I-profile, the results of the displacement are not as expected: >20%, when I expect the same result, as beam analysis --> beam theory.

Boundery condition:
fixed constrain one side
Distributed load 113.03 N/mm
lenght: 1899 mm
I-propertiesI-properties

situation.png
displacement.png

Displacement: 9.571 mm
Hand calculation = AutoCAD calculation = 8.05 mm

By changing the beam section, keeping the value of the moment of inertia I, I did the analysis with a square section, side 190 mm. Result= 8.05 mm

So ... I did a solid, raw analysis ... to help me understand this result.
solid.png

From this I understand that:
- the neutral line shifts by 8,376 mm,

- the flange by 9.99 mm.

So I deduce that the beam analysis also considers the deformation of the flange, which I didn’t expect at all, I always believed that the element was considered rigid. Did I get that right?

If the spelling is correct, how do I see / ask for the result on the neutral axis?

I tried to "play" with the reference points in defining the cross section, but I didn’t get anything interesting.

I am curious to know if I read the results correctly and my conclusions are correct.

Thanks in advance for your time.

Greetings
Sam

P.S. Workinh with Inventor Nastran 2024 and 2025.

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John_Holtz
Autodesk Support
Autodesk Support
Accepted solution

Hi @armeno_sam . Welcome to the Inventor Nastran forum.

 

The flange of the beam element is not displacing like the solid model. The only location the displacement is calculated is at the line end points. (That could be the centroid or shear center or ___ as set on the Idealization.)

 

It is more likely that Nastran is including the deflection due to shear (because the shear factors are not zero) whereas your hand calculation is only considering the bending. A quick test would be write down the calculated A, Iyy, Izz and enter those values using an cross-section input of Properties (instead of a cross-section). When using properties, you can leave the two shear coefficients blank to avoid calculating the shear displacement. The Nastran results should match your hand calculation.

 

John



John Holtz, P.E.

Global Product Support
Autodesk, Inc.


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Message 3 of 3

armeno_sam
Explorer
Explorer

Hi John,

thanks for your explication.

I definitely hadn’t thought about the shear.
The order of magnitude of the displacement on the flange had led me to speculate wrongly.

armeno_sam_2-1742969455919.png

armeno_sam_0-1742969428781.png

 

Thank you for this very fast support and especially thank you for the global support you give!
Thank you!


Best regards
Sam


Thank you very much!

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