Any idea how to calculate the axial forces?
Using Hooke’s law. F=kx, where k is the strength parameter of the length goal, and x is the difference between rest length and final length
And what about the anchor forces? Remember you gave me a def which showed the force amplitude. I can’t get it to divide the loads between the two anchors.
I have a definition somewhere to output reactions at anchors, I’ll try and post it tomorrow - but if there’s only one cable attached to the anchor, the reaction must be exactly the opposite of the force in the last segment
Ok The cable is anchored at two points.
I got the results of your last suggestion. What are the units?
Btw I have managed to eliminate all differences in the inputs between the Kangaroo def and the structural program. The results are pretty much the same, and any differences I have managed to find were totaly negligable. Thank you very much for all that help these few days. You have made my life a lot easier.
I will wait for the reaction forces definition, because I just can´t get it to do it myself.
But this example you gave me here reised another interesting question. Since it is possible to materialize cables with Kangaroo, could I do the same for PVC membranes or ETFE cushions? I have a very specific kind of load which I need to iterate great many times to get a valid result. If I can do the same kind of materialization in real units with Kangaroo, I have the rest of the definition ready.
I must say Kangaroo proves to be invaluable, I have a bit of trouble understanding all but with your help and examples I gain ground quickly. Thank you again for all this.
Glad to hear the results are matching up now
There are often misconceptions about using real materials and units in Kangaroo, so it’s nice to show what is possible.
Here’s a basic example of getting reactions at anchors.reactions_example.gh (10.8 KB)
Again, the length of the vector should be the strength in Newtons.
Membrane elements still need work though to support real material values. With the soapfilm triangles it is possible to set a specific uniform tension value, but this is for formfinding and quite different from analysing a fabric, where the tensions would not generally be uniform, and there would also be shear resistance.
Making a proper membrane analysis element has been on the to-do list for a while. It’s definitely possible.
I can’t open it because I can’t make Kangaroo 2.5 work on my pc. Do you have it for Rhino 5? I am missing the Grab component only.
Because we have put enormous force in the calculation for the anchor points there is no difference between the input and the output. Hence the vector length is zero and I can’t define the direction this way, also the Zombie solver doesn’t work with trees in the same way as the normal one so I used closest point to find the corresponding vertices.
Ah, I wondered if that might happen. I’ll look tomorrow at adding something to handle this case.
Remember the def you gave me about the shape optimization? There I get a vector direction.
catenary_min.gh (16.3 KB)
As I mentioned earlier-
if there’s only one cable attached to the anchor, the reaction must be exactly the opposite of the force in the last segment
so if this is the case, it’s simply a question of checking the tension in the member connected to the anchor.
Only one cable to each anchor point. Later point load is applied. I need tho get the values in Newtons.
How can I check it?
Hi Daniel, I was looking at your post about the comparison between “rhino centenary” and kangaroo centenary. All the units part and strength part I understand. However, I did not understand that why I change young’s modulus or gravity largely the shape that kangaroo gave does not change. It still shows “converged”. If it does change, it is giving some invalid result…
The deformation in this example is extremely small. As you’d expect - a 40m long, 10mm diameter steel cable hanging under only its self weight stretches such a tiny amount it isn’t visible (less than 0.2mm over the whole cable), and for most practical purposes the result would probably be considered identical to the standard grasshopper catenary component (which also has it’s own tiny errors, due to the way it is discretized).
If you change the Young’s modulus to, say, 2e+7, to make it a much more stretchy material, you’ll see the difference more clearly.