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# applying a axial force on axisymmetric model: F or F/2PI() ?

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 !)
5 REPLIES

## Re: applying a axial force on axisymmetric model: F or F/2PI() ?

Hello!,

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/

## Re: applying a axial force on axisymmetric model: F or F/2PI() ?

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 !)
Solution
Solution
Accepted by topic author selex_ct
‎12-03-2015 09:44 AM

## Re: applying a axial force on axisymmetric model: F or F/2PI() ?

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"

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

## Re: applying a axial force on axisymmetric model: F or F/2PI() ?

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 !)

## Re: applying a axial force on axisymmetric model: F or F/2PI() ?

Yes, a lot of changes in the subsequent two releases. SOL 401 wasn't even on the drawing board in NXN 8