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12-03-2015 07:49 AM

To all,

I am trying to confirm if the axial load to apply to an axisymmetric model should be F (the total load) or F/2PI() ? The doc does not appear to give clear answer (none that I can find anyway!)

Thanks

Regards

Production: NX9.0.3.4, NX10.0.2.6

Development: VB.NET (amateur level !)

Development: VB.NET (amateur level !)

Solved! Go to Solution.

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12-03-2015 07:58 AM

Hello!,

**Before NX Nastran 10, the stiffness, mass, and loads for the CTRAX3, CQUADX4, CTRAX6, CQUADX8, CTRIAX, and CQUADX elements were based on a per radian section basis.**

Now with NX AdvSim V10 and FEMAP V11.2 both running NX NASTRN V10.X the stiffness, mass, and loads for these elements are based on a 2*PI section basis. For example, to apply a distributed load of 135.0 Newton/mm on a single grid where the radius is 0.5 mm:

The value entered on a FORCE entry = (Distributed force * 2 * π * Radius) = 135.0 N/mm * 2 * π * 0.5 mm = 424.115 Newtons

The **system cell 587** can optionally be set to 1 to revert to the per radian section basis. Note that the CTRAX3, CQUADX4, CTRAX6, CQUADX8, CTRIAX, and CQUADX elements are always on a per radian basis in solutions 601 and 701. In addition, the CTRAX3, CQUADX4, CTRAX6, and CQUADX8 elements are always on a 2*PI basis in a heat transfer solution.

In general, the section basis choice changes how force, strain energy, mass, volume, and work are computed. Displacement, pressure (surface tractions), stress, strain, and strain energy density will not change.

Best regards,

Blas.

Blas Molero Hidalgo, Ingeniero Industrial, Director

IBERISA • 48004 BILBAO (SPAIN)

WEB: http://www.iberisa.com

Blog Femap-NX Nastran: http://iberisa.wordpress.com/

IBERISA • 48004 BILBAO (SPAIN)

WEB: http://www.iberisa.com

Blog Femap-NX Nastran: http://iberisa.wordpress.com/

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12-03-2015 08:35 AM

Thanks. Read about the system cell. I am doing the analysis in SOL401.

I have created the force applied to an edge (Fx=constant), see attached but still don't know if I have to applied F or F/2PI() !

Production: NX9.0.3.4, NX10.0.2.6

Development: VB.NET (amateur level !)

Development: VB.NET (amateur level !)

Solution

Solution

Accepted by topic author selex_ct

12-03-2015
09:44 AM

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12-03-2015 08:46 AM

Remark 5 on the FORCE card states "For interpretation of a FORCE entry in an axisymmetric analysis, see the listing for the axisymmetric element type"

SOL 401 supports axisymmetric elements CTRAX3, CQUADX4, CTRAX6, and CQUADX8. If you are using CQUADX4, for example, Remark 10 on CQUADX4 states:

*A concentrated load (e.g., FORCE entry) at Gi is the total of the force around the circumference. Reaction force and applied load output are the same. For example, to apply a distributed load of 135.0 Newton/mm on a single grid where the radius is 0.5 mm:*

*The value entered on a FORCE entry = (Distributed force * 2 * π * Radius)*

*= 135.0 N/mm * 2 * π * 0.5 mm*

*= 424.115 Newtons*

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12-03-2015 09:00 AM

Thanks. Was looking at the NX 8 QRG __hard copy__ 'FORCE' card which does not have the remark 5 mentioned. Didn't think of looking at the CQUADX8 card. Looks it was added for the new release

total force it is then

Production: NX9.0.3.4, NX10.0.2.6

Development: VB.NET (amateur level !)

Development: VB.NET (amateur level !)

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12-03-2015 09:05 AM

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