Why does a catenary receive bending moments below a certain hight? Shouldn’t it only show normal normal forces? Is there something obvious I am missing here?
The example is using AnalyzeThI.

btw. this is also the case for the parabola from the Otto Wagner Karamba webinar (however applying “global projected” loads)

Furthermore, I analyzed the catenary as an extruded shell. There I used the ShellForces to get corresponding principal forces and moments for each shell element. After applying sigma_f = F/t and sigma_m = 6M/t² the plan was compare stresses resulting from normal forces vs. stresses resulting from bending moments and divide these values (sigma_m / sigma_f) to derive a single value and to be able to judge how “well” the shell is performing (“well” meaning largely based on normal forces and not moments).

Here is a plot of these values - the maximum s set to 1 in this case, so every value larger than 1 is set to 1:

According to this quotient the stresses in this shell are largely due to bending moments in the blue area. However this does not make much sense to me as this is a catenary shell…
There has to be an error in my thinking, but I am not sure where. Any ideas?

Hi @rudolf.neumerkel,
did you orientate the supports in such a way that the resultant reaction force is parallel to the shell plane and the beam middle axis respectively?
– Clemens

Hi @rudolf.neumerkel,
I tried out an example with beam-loads and gravity (see here: CatenaryBeam_cp.gh (34 KB)
). I have to check the reason for the difference between the results with beam-loads and gravity.
– Clemens

In the 2D version (see file attached above) the catenary is oriented along the x-axis and the supports are using the default xy-plane. Movement is prevented in x,y and z axes respectively. So I guess they are oriented in parallel?

In the 3D (shell) version the supports are also oriented parallel to the shell plane:

There is a difference since gravity loads equal beam loads equivalent to the self weight as far as I understand. So that would mean total mass (=113kg) divided by the catenary length (=10.4m), which would be equal to 0.108kN/m. In the example you attached, a beam load of 1kN/m is applied, so there should a be an obvious difference.

Thanks a lot for your suggestions… However the strange problem regarding the bending moments still exists.

Furthermore, in the file you attached the catenary is supported in a very delicate way (for example only in local x direction on the curve ends), only allowing normal forces to be transmitted to the supports. However the bending moments are still there. Can you explain why you chose this way of supporting the catenary?

Transverse supports at arches and shells lead to shear forces there, which entail bending moments. This is the reason why I chose the specific way of support.
When checking the above example I found the reason for the bending moments which I initially overlooked: the axial deformation of the arch. With increasing cross section area A the bending moments diminish ( CatenaryBeam2_cp.gh (33.2 KB)). Another source of bending moments is the approximation of the catenary with straight beam pieces. This can be remedied by using shorter and shorter segments.
A third source of the bending moments is the limited accuracy of double precision numbers.
– Clemens

Very interesting! My understanding was that deformations are only taken into account with second order theory… (AnalyzeThII) Thanks for your clarification!

However I still don’t understand the behaviour of the catenary extruded shell. Where stresses due to bending moments seem to be very large…

It is due to the missing boundary conditions along the free edges: CatenaryShell3_cp.gh (50.2 KB).
I think the small bending moments that still exist near the bottom supports are caused by a missing symmetry condition which I haven’t figured out yet or by an inaccuracy of the geometry, or…
– Clemens