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!)
Solved! Go to Solution.
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.
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() !
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
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