Stress results in cylindrical coordinate system

Tim_Gowing

Hi
I see the formula function in the RMB drop down from Solutions in the tree; please give me a hint on how to translate orthogonal stresses into cylindrical stresses.
I've searched around for a means to create a new coordinate system (rect or cyl), is this possible? Can solver results be translated to a cylindrical system that is not at the global origin, i.e. offset, if there were several cylindrical features for which cylindrical results were of interest?
Tim

Sergio

Hi Tim, in CCX with the *TRANSFORM card you can create a new coordinate system (rectangular or cilindrical, related to an specific node set), don't know if t will serve to your matters. Also the *NODAL FILE card (used to store the nodal results in the .frd file) has a modifier to specify the coordinated system (GLOBAL or LOCAL). Maybe reading the documentation for that cards you can solve your problem.

I have experience with TRANSFORM, but not paired with the NODAL FILE / LOCAL.

Regards

Tim_Gowing

Hi Sergio,
Thank you for your suggestion; I was working directly in Mecway but yes see that I could work in CCX and certainly view the results in CGX. I should try to read the .frd with cylindrical results (using *TRANSFORM) back into Mecway.
I'm keen to explore the post-processing, between coordinate systems, within Mecway.
Regards, Tim

Sergio

You can insert those cards directly using the Mecway inteface (custom CCX data) and see the results on Mecway directly. I don´t have experience with results with the coordinated system changed, I use the TRANSFORM to apply cilindrical bondary conditions only.

What is the reason to choose that kind of results?

Tim_Gowing

Sergio,
I'm looking at a ring beam so the stresses of interest are on planes that run radially from the axis that passes through the ring centre and is normal to the plane of the ring. I have calculated the transverse shear force, bending moment and twisting moment within the ring (annular) beam, that is supported in 6 places, equi-spaced; loading is a constant line load around the ring centreline. The direct and shear stresses calculated are in the radial plane (that is also perpendicular to the tangent of the circular centreline of the ring beam). I would like to reconcile these hand calculations with the FEA testcase before moving on to a more complex geometry. The hand calculation is important in sizing of the ring beam cross-section (a more desirable situation than adjusting the FEA cross-section multiple times in the search for a near optimal design).
Tim

Victor

You could technically do it using formulas, but they'll be fairly big. You'd be expanding out two 3x3 matrix multiplications as 6 scalar equations.

The formula for rotating the stress tensor is well documented, such as:

http://www2.mae.ufl.edu/haftka/adv-elast/lectures/Section2-4.pdf

http://web.mit.edu/course/3/3.11/www/modules/trans.pdf

Tim_Gowing

Victor,
Thank you for the stress tensor links.
The radial component is defined as (x*u.x + y*u.y) / sqrt(x^2+y^2) in the manual. My poor transformation skill leads to ur = (u.x^2+u.y^2)^0.5 but this losses the sign (sense) of the displacement whereas your transformation succeeds.
What is the corresponding transformation for the theta component of displacement?
(it's not atan(u.y/u.x))
Can the option of a cylindrical coordinate system be considered as an enhancement?
An arbitrary coordinate system in space may not be necessary but a cylindrical coordinate system positioned at the origin would perhaps be in keeping with the Mecway workflow.
An alternative enhancement would be include the solution results derived in the cylindrical coordinate system. This avoids dealing with BCs and loading on a cylindrical basis, only post-processing.
Tim

Victor

I got the formula from the dot product of the displacement with the unit radial vector. From the meaning of dot product, that gives the component of displacement parallel to the radial vector. Both your formulas look like they're for a cylindrical coordinate system centered on the node so all radial displacements would be positive.

I would imagine you can do the same for the tangential component - dot product of displacement with the unit tangential vector (r^ cross z^ ?), or start from the rotation matrix formula and expand it into scalar terms.

Cylindrical coordinates come up form time to time, which was part of the intention behind the formula tool, but it's not very convenient still being limited to scalars. I may have to add a special cylindrical stress option.

Tim_Gowing

Thank you Victor, some textbook revision required here.

Victor

I've just worked out the matrix equations for stress transformation and it ends up quite simple. For cylindrical coordinates centered on the origin with Z as the axial direction, the radial and tangential components of stress are, expressed in Mecway formula notation:

stress_rr = ( x^2*s.xx + 2*x*y*s.xy + y^2*s.yy ) / ( x^2+y^2 )

stress_tt = ( y^2*s.xx - 2*x*y*s.xy + x^2*s.yy ) / ( x^2+y^2 )

Attached is an example of a uniform ring with hoop stress. You can see the formulas match the XX and YY components at locations where they are X and Y axes are aligned with the radial and tangential axes.

I still have to do more thorough validation, then I'll include the formulas in the manual.

65mustang

Victor,
Thanks much for including the two pdf files above. I have some studying to do....

What program do I need to open the .liml file you posted on March12, above?

Thanks for you efforts.

Andrea

I opened it with Mecway8...OK at warning message