I fixed the square tube. And i think the previous one was kinda wrong because it has two shells as you mentioned. I'll attach the updated one below and see if i fixed what you said was wrong.
Yes that looks correct. Have you tried to simulate this in implicit yet? I ask because I see that you have a contact defined between the blue impactor and the red tube. The difference in mesh size between the two is really different as well, which may give you issues in the future. But if it works then there is no reason to change anything.
not yet. i'm still trying to figure out which implicit keyword i should define. Btw, my case is considered as non-linear right. Below attached the force displacement graph that i got from my 17 hours (and still going) simulation of cylindrical tube. My displacement curve set for this was (0,0) and (280000,47.5) making the velocity equals to 0.00017m/s. The Peak force i got was 35.7kN while the past paper got 35,3kN. So i guess my simulation works well. However, for the square tube compression, i got 50kN while past paper was 30.69kN. Is this because i set the displacement curve as (0,0) and (47.5,47.5) which makes the velocity equals to 1m/s?
If that's the case i guess i have to run each simulation for 17 or more hours then?
not yet. i'm still trying to figure out which implicit keyword i should define. Btw, my case is considered as non-linear right. Below attached the force displacement graph that i got from my 17 hours (and still going) simulation of cylindrical tube. My displacement curve set for this was (0,0) and (280000,47.5) making the velocity equals to 0.00017m/s. The Peak force i got was 35.7kN while the past paper got 35,3kN. So i guess my simulation works well. However, for the square tube compression, i got 50kN while past paper was 30.69kN. Is this because i set the displacement curve as (0,0) and (47.5,47.5) which makes the velocity equals to 1m/s?
If that's the case i guess i have to run each simulation for 17 or more hours then?
Yes, that would mean that the square tube is rate dependent, meaning in this case that it has a stiffer response at higher rates. If you prescribe the correct velocity then I'm sure you will get similar results.
The only reason that they are taking 17 hours to run is because you are running a 100+ second simulation in explicit. It will not take long to run in implicit, you just have to figure out how to set it up though.
i'm watching this hour long video explaining about implicit analysis and she said something about IMPLICIT_GENERAL to activate the implicit solver. However, what does it mean by DTO and what do i set it as (see image attached)? take my square tube as example. Termination time 50ms, and displacement curve (0,0) (47.5,47.5). Btw, i use my square tube for trial and error since my cylindrical tube is still running..
As it says, DTO is a defined initial timestep. If you are not familiar with the concept of timesteps I suggest you read about it:
https://www.dynasupport.com/tutorial/ls-dyna-users-guide/time-step-size
The timestep in a simple sense, is determined by the minimum edge length of an element and the material of that element (Most influentially the mass). The smaller the timestep, the longer your model will take to run. If you manually define a timestep in LS-Dyna, then the mass of that element will be artificially increased during the simulation to achieve that timestep. This is what is called mass added during a simulation and can be found in the d3hsp or the message file of a simulation. Generally, you want to keep the mass added to less than 5% of the initial model mass (the compressed object, not the impactor). So I would look in your message file from the explicit simulation that you ran with the square tube to find what the calculated timestep was and define DTO as that times 1.05 and see how that works for you.
I still don't follow what you said above. Below is the normal termination of my simulation from d3hsp file. Which one is the calculated timestep?
In the D3hsp you should see a section labeled “100 smallest timestep” this lists the elemenents with the smallest timesteps in the model. So I’d find the smallest of those elements that is part of the hollow tube and look at the timestep value. Multiply that value by 1.05 and set that as the DTO