Dear Itasca team, I try to simulate the soil structure interaction in 3DEC under seismic loading, considering hysteretic damping of the soil mass. Can we apply hysteretic damping (associated with shear modulus reduction according to shear strain levels) in 3DEC? Thanks for your support.
Hysteretic damping is not available in 3DEC. Usually people are using 3DEC to model rock, so it is not something we have added.
Thank for your answer. By the way, I wrote a fish function to modify shear modulus according to the shear reduction curve, which is approximated by a sigmoid function with Massing rule for unloading and reloading phase according to information provided from FLAC document related to hysteretic damping section. This function will be called by the command ‘fish callback’ to the calculation cycle. It works quite well when I test with a block with some zone elements at some first reversal cycles. When the number of reversal cycle increases, its behavior becomes wrong. In addition, when I refined the mesh, resulting in more zone elements, the shear stress-strain loop looks very weird. I try to change the execution point in the callback command, but shear stress-strain is still wrong. Can your team help check and suggest the reason to me? Thank you so much for your kind support.
This is the hysteretic loop when mesh size is 0.1.
This is the hysteretic loop when mesh size is 0.03
This is the instruction how to detect the reversal point in hysteretic damping model in FLAC documentation, which is adopted to build the fish function. Actually, I do not understand why using these terms to detect reversal points. Can you help to explain? (γ is shear strain and e is deviatoric strain, right?, why need to divide the term by square root of 6?)
Thank you so much for your kind support.
@cheng any thoughts on this?
The reversal is based on the Massing rule.
𝑒 represents strain (not deviatoric strain).
The square root of 6 is included because the second invariant of deviatoric strain, J, is defined as follows: J = 4* {[(e11-e22)^2+((e22-e33)^2+(e33-e11)^2]/6 + (e12)^2 + (e23)^2 + (e31)^2},
Thus, J is simply the square norm of the defined 2𝛾 (note that the first three terms are divided by 6 in J).
We have no idea what caused the issue in the second figure since the scripts are not available, and it is generally due to some details.
Thank you so much for your clear explanation. I have just sent my script to your email for your checking. Because it is quite long, I do not upload on the forum.
In addition, if J is the square norm of 𝛾, it should be 𝛾4 = 𝛾4 + ( Δe11-Δ22)^2/6^0.5, why in the instruction it is 𝛾4 = 𝛾4 + ( Δe11-Δ22)*2/6^0.5, and it should be 𝛾1 = 𝛾1 + ( Δe12)^2, why in the instruction it is 𝛾1 = 𝛾1 + ( Δe12)*2. It is so confused. Thanks.
It is not wrong. However, I slightly corrected my reply. there is a factor of 4.
Thank for your clear explanation.