Itap
Mechanical
- Feb 1, 2013
- 10
Hi!
I have a problem with a simulation of the Hopkinson bar experiment. I am using Ls-Dyna with shel axisymmetric elements (tipe 15).
In the simulation there is a waves that propagates into the incident bar. It's a pressure wave along y-axis. This wave keeps the bar into the linear-elastic field. This bar is made of aluminium (E=71100Mpa), so, for the Hooke's law (I want to get 600 micro strains): s=e*E=600e-6*71100=42.66 MPa. So, the shape of the wave is trapezoidal with a maximum value of 42.66 MPa.
This is not the problem!
In the simulation I am using shell axisymmetric elements (type 15). The problem is this: when I use these elements with a 2 integration point (card SECTION_SHELL --> NIP) I obtain from the simulation, the rights results in terms of stress and strain (as you can see in the firsts 2 pictures, there are stress and strain takes from a node in the middle lenght of the bar) but these elements, have problem of hourglass.
Uploaded with ImageShack.us
Uploaded with ImageShack.us
So, in order to avoid this problem, I use the same elements, but with 4 integration points (fully integrated). In this way, I avoided the problem of hourglass, but I get wrong results in terms of strain (not for stresses).
Uploaded with ImageShack.us
In your opinion, where is the problem?
I think that from 2 to 4 integrations points, maybe the code could change the way wich it calculate the strains, but I am not sure about this.
If you want, I can upload the simulations files.
Sorry for my bad english.
Thank you.
I have a problem with a simulation of the Hopkinson bar experiment. I am using Ls-Dyna with shel axisymmetric elements (tipe 15).
In the simulation there is a waves that propagates into the incident bar. It's a pressure wave along y-axis. This wave keeps the bar into the linear-elastic field. This bar is made of aluminium (E=71100Mpa), so, for the Hooke's law (I want to get 600 micro strains): s=e*E=600e-6*71100=42.66 MPa. So, the shape of the wave is trapezoidal with a maximum value of 42.66 MPa.
This is not the problem!
In the simulation I am using shell axisymmetric elements (type 15). The problem is this: when I use these elements with a 2 integration point (card SECTION_SHELL --> NIP) I obtain from the simulation, the rights results in terms of stress and strain (as you can see in the firsts 2 pictures, there are stress and strain takes from a node in the middle lenght of the bar) but these elements, have problem of hourglass.
Uploaded with ImageShack.us
Uploaded with ImageShack.us
So, in order to avoid this problem, I use the same elements, but with 4 integration points (fully integrated). In this way, I avoided the problem of hourglass, but I get wrong results in terms of strain (not for stresses).
Uploaded with ImageShack.us
In your opinion, where is the problem?
I think that from 2 to 4 integrations points, maybe the code could change the way wich it calculate the strains, but I am not sure about this.
If you want, I can upload the simulations files.
Sorry for my bad english.
Thank you.
