# Analysis of catenary

Any way to materialize a catenary into a cable with its material properties, such as cross-section area, E-modulus, etc… and then apply a point load at any given point and see deformation?

catenary_units.gh (17.4 KB)

Here’s an example of a 30m long 1cm diameter steel cable, hanging under self weight, showing how to use proper units for calculating deformation. You could add other point loads (in kg) as you like.

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I can’t seem to get it going. It says the Load component is from Kangaroo 2.5.0, but I can’t seem to make Kangaroo 2.5.0 work on my machine.

I think I have managed to do it. Any way to get the output only when the solver is converged?

You can use the ‘zombie’ solver.

Did it! It was a premature question , sorry.

Is there a way to do it with tha catenary component from Grasshopper and all the rest with Kangaroo?

I don’t follow, which part is ‘all the rest’?

I have checked the result from Kangaroo with the same geometry, same material and the same point load at the same location. I get way bigger deformation in Kangaroo than I get from my structural software.

On the other hand the catenary shape is perfect. So my guess is right now for some reason I allow for more deformation in Kangaroo.

The load here is in kN.

catenary_units_point_load.gh (20.7 KB)

Are you sure about your units? You show some dimensions in cm in your first image, while all dimensions in the Kangaroo definition need to be in m, and you say the load is in kN in the caption, but then in the definition you say it is in kg? (and sorry - I should have been clearer, the weight is in kg, but you still need to convert it to Newtons before you put it into the FV input of the Load component).
If you include the internalised geometry in your file I can take a look.

Oh sorry… forgot to internalize it!!! Ok and I put the load in N. I get the same deformation thoug… strange…

catenary_units_point_load.gh (15.9 KB)

In your other analysis software are you also starting from this curved initial state with uneven length divisions?

Yes. Exactly the same geometry.

I need to do the deformations properly, because I am trying to simulate a zipline.

I need to control the speed of arrival of a person who is 105 kg. I account for friction, air drag and all other loss.

However for accurate calculation I need to know the deformations at each point, once the person reaches it. This changes the angle of the slope all the time. I use a loop to iterate the results.

If I get the deformations right with Kangaroo… Baammm!

I notice the cable is only stretching by a tiny amount in your setup - 2cm over the whole 285m.
That’s probably why you see barely any change when you changed the load - the cable is pulling straight, but not stretching a very significant amount.

Also - I see in the definition you posted, you didn’t have the self weight of the cable itself connected. Once you add that, the effect of that point load is obviously much smaller.

The other thing that is different in this setup compared to the example I gave is that your segments vary in length. In the simpler one I was able to simply use the length of any 1 segment to calculate the self weight to apply for each point (since they were all the same). If they vary, to be more accurate you’d need to add half the weight of each of the adjacent segments to each point.

Yes! When I plug the self-weight this becomes orders of magnitude closer to reality.
So for the next step I need to add half the weight of each links connected to each point to its load for the self-weight? OK. I will try that

Man… you are my god I swear

I have made the segments the same length. Now the difference is that I have about 40cm less deformation in Kangaroo. But this is becoming a lot closer.

Any idea how I can reduce this difference?

catenary_units_point_load2.gh (20.1 KB)

For something simple like this I’d expect the difference to be much smaller than that. Are you doing a linear or non-linear analysis in your other software? Kangaroo always does non-linear. Still though I don’t think that would make a difference this big. I wonder if there are some other inputs different?

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Non-linear, both 2nd and 3rd order give the same result.

When I turn off the reduction of stiffness and the favorable effects in the other program, the difference drops to round 36cm. Kangaroo has smaller deformation.

I have found out that the way the mass is calculated in that other program is different in some way. If I plug in that mass per linear meter then the result is very close.