Joint challenge of strength and stability, octopus or galapagos


made a determination in karamba for the arch of optimal shape
using the stability node I find Buckling load factors, do I understand correctly that this coefficient shows how many times it is necessary to increase the load so that the system loses overall stability?


Further, through the Utilization node, I determine the utilization coefficient of each section by strength, should it be, as I understand it, less than 1?
Further, by transformations I get the value true or false if at least one element loses its strength, the coefficient is more than 1

I tried to combine the task of strength and stability through the galapagos, but the galapagos has one goal for optimization
I find the optimal value for the Buckling load factors, but at the same time do the strength elements fly out?

I tried to solve the octopus through the plugin, but I can’t correctly configure the octopus for this task

Can you please help?

arc (53.0 KB)

For octopus to run, you simply need to collect your list of objectives in one container and then feed this into Octopus, It works similar to Galapagos, but you can have more than one objective in the optimisation process.

Hi AlexWer,

It seems than 1 means 100%. For further reference you can take a look to Karamba3D online help:

Utilization numbers for beams rendered by this component (output-plug “Util” ) and the “ModelView” show differences – especially for compressive axial forces: The “ModelView” -component returns the ratio of stress to strength as the level of utilization, whereas the “Utilization of Elements”-component also includes buckling. See for example the two utilization entries on the in fig. The second load case (i.e. number “1”) is made up of an axial load acting in the middle of the beam. As both ends are axially fixed, one beam is in tension, one in compression. The absolute value of the normal force in both elements is the same. Yet the beam under compression has a utilization of 0.26, the one under tension only 0.05. “1” means 100 %.