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Direct cylic

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ssn3

ssn3

Structural
Apr 16, 2013
20
Hello,
I wanted to know if anyone here could explain the direct cyclic theory on abaqus. So I read the abaqus documentation and wanted to know more about the direct cyclic algorithm. I was wondering if anyone could explain to me what goes on and how abaqus obtaines a solution. I believe the documentation explains little on direct cylic in section 2.2.3 in the theory manual.
 
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I just needed to know the theory part a little as I am a little confused on how it works.
 
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I read the manual and it says for direct cylic method it solves for u by using an iterative method. I just wanted to know how exactly is it solving for u. Which are the knowns and unknowns. Is it solving for unknowns based on simultaneuous eequations or what? That is what I need to know. So far I know abaqus finds the coefficients of the residual vector R0, Rkc and Rks to find the corecction to the displacement coefficients c0,ckc and cks. These corrections are then added to the u0, ukc and uks for the next iteration.But I still am not sure if these are unknowns or knowns. The all seem like unknowns to me.
 
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They are all unknowns.

Two fields are at play: displacement and force. And, for convergence during an iterative algorithm, ABAQUS requires the displacement correction and force residual to be below a limit. And these increments are simply expressed as Fourier series expansions, whose coefficients must be determined for the iteration to proceed.

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So based on what you have said which ones are the knowns and unknowns? I'm guessing the displacements are the unknowns since that is what abaqus tries to obtain and the force are the knowns. Am I right?
 
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"They are all unknowns." Sorry, that's not correct. It depends on whether you are carrying out the loading under load or displacement control. So, if its load control in which the load is being applied, then the force residuals are the "known" variable.

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I think I am beginning to picture it now. So for the next iteration how does the force residuals R differ? My understanding is that the displacements u change depending on the correction to displacement c. So how is the R going to change for the next iteration? Technically abaqus is still tracking the response for the same cycle. This is the part that is not explained in the theory manual. That is why I got confused.
 
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I think your confusion has more to do with the iterative scheme than the direct cyclic analysis itself. Write a code for an iterative scheme and it will make sense. Nevertheless, continuing with the example, force residual is a "known" variable. The modified Newton algorithm will take a "small" load step resulting in a corresponding displacement increment (given its tangential stiffness), leaving a residual.

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You are right I am basically confused on how the iterative scheme works. My understanding is that you have a curve for one cycle given an iteration. It continues to iterate until it converges for that cycle. Converges meaning the R and c entries are small. I am mainly confused about the R entries. How would they vary from iterations? R is a function on time and this time is the step time. So the R repsonse is basically the load step at the different time steps. So how would it change for the next iteration? That is why I do not get how it works for the next iteration. I am sorry if I am not following you very well, I am a student and I am actually trying to understand the direct cycic process behind abaqus.
 
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Like I said, before you try to understand the direct cyclic algorithm, write a code for a simple iterative scheme (for a one-dimensional problem like y=sin(x)..or some non-linear function). Once you have it working and fully understood, then revisit the direct cyclic algorithm.

Now, direct cyclic analysis is NOT a transient analysis. It is, in fact, a quasi-static analysis. The displacement solution for the first increment is for ALL TIMES throughout the "cyclic" loading. So, you do not want to think in terms of cycles of loading.

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When you say increment what do you mean? Because from what I read increments are the time points used for the analysis. So if I use an increment size of 0.001, that would mean I have 1000 time points in a cycle.
 
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You are probably talking about iteration. In a direct cylic analysis iteration is where the displacement corrections are affected. Because I am pretty sure that increments are time point in abaqus. At least that is what Abaqus says in its manual. In static analysis then yes increment is pretty mucht he load step.
 
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Thank you for taking time to explain to me. I appreciate it. I get a picture of how it works but I need to dig more into it. I want to truly understand how it all adds up. Thanks again for helping.
 
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