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Posted:
2 decades ago
2009年11月5日 GMT-5 03:48
Hi Lucia,
I guess that if you have imposed a sinusoidal load at a fixed frequency (i.e. a load which varies in time) you have performed a time domain simulation.
If it is your case, maybe you have to "wait" enough time to make the system reach the steady state.
On the other hand, a frequency domain analysis gives you directly the steady state response, discarding any transient behavior.
Please pay attention to the fact that in the frequency domain analysis you have just to set the amplitude (and in case the phase too) of your load as boundary condition.
I hope this help, hi!
Alessandro
Hi Lucia,
I guess that if you have imposed a sinusoidal load at a fixed frequency (i.e. a load which varies in time) you have performed a time domain simulation.
If it is your case, maybe you have to "wait" enough time to make the system reach the steady state.
On the other hand, a frequency domain analysis gives you directly the steady state response, discarding any transient behavior.
Please pay attention to the fact that in the frequency domain analysis you have just to set the amplitude (and in case the phase too) of your load as boundary condition.
I hope this help, hi!
Alessandro
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Posted:
2 decades ago
2009年11月5日 GMT-5 06:08
Hi Alessandro,
thank you for your answer.
The problem is that I get the correct (= predicted by theory) answer when I use the sinusoidal load. In my simple case (which is a first approximation of more complex behaviours), the response in displacement is periodic, of the same frequency of the incoming load: as I am in the time domain simulation, I first tried a time corresponding to "one period" (of the incoming load), but I obviously got the same result (in amplitude and phase) when using a time corresponding to 10 periods.
For the frequency domain analysis I just set the amplitude and fixed the "behaviour" of the sweep. But as "y-displacement" and "x-displacement" I get half of the correct result. What does "-displacement" mean in the frequency domain analysis? Does it have a different meaning with respect to the same variable name in the time domain?
Thank you in advance.
Hi Alessandro,
thank you for your answer.
The problem is that I get the correct (= predicted by theory) answer when I use the sinusoidal load. In my simple case (which is a first approximation of more complex behaviours), the response in displacement is periodic, of the same frequency of the incoming load: as I am in the time domain simulation, I first tried a time corresponding to "one period" (of the incoming load), but I obviously got the same result (in amplitude and phase) when using a time corresponding to 10 periods.
For the frequency domain analysis I just set the amplitude and fixed the "behaviour" of the sweep. But as "y-displacement" and "x-displacement" I get half of the correct result. What does "-displacement" mean in the frequency domain analysis? Does it have a different meaning with respect to the same variable name in the time domain?
Thank you in advance.
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Posted:
2 decades ago
2009年11月5日 GMT-5 11:34
Dear Lucia,
I think that the right variables you have to choose in the frequency domain approach are "Disp. amplitude x-dir" and "Disp. amplitude y-dir".
When you run a frequency domain analysis, displacements (i.e. the usual "y-displacement" and "x-displacement" variables) become complex numbers because of the formulation of the problem in frequency domain.
"Disp. amplitude x-dir" and "Disp. amplitude y-dir" variables are post-processing variables which are obtained
from the usual displacement variable as (I think) as for instance:
"Disp. amplitude y-dir" = sqrt(real(v)^2+imag(v)^2)
Whereas, for what concern the phase you have:
"Phase y-dir" = atan(imag(v)/real(v))
I hope this help.
Hi!
Alessandro
Dear Lucia,
I think that the right variables you have to choose in the frequency domain approach are "Disp. amplitude x-dir" and "Disp. amplitude y-dir".
When you run a frequency domain analysis, displacements (i.e. the usual "y-displacement" and "x-displacement" variables) become complex numbers because of the formulation of the problem in frequency domain.
"Disp. amplitude x-dir" and "Disp. amplitude y-dir" variables are post-processing variables which are obtained
from the usual displacement variable as (I think) as for instance:
"Disp. amplitude y-dir" = sqrt(real(v)^2+imag(v)^2)
Whereas, for what concern the phase you have:
"Phase y-dir" = atan(imag(v)/real(v))
I hope this help.
Hi!
Alessandro
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Posted:
2 decades ago
2009年11月5日 GMT-5 12:05
Thank you again, Alessandro.
You are right, I also got to the same conclusion on my own. The problem is that both the values of "Disp. amplitude x-dir" and "Disp. amplitude y-dir" are not correct. They are smaller than the expected values (which were confirmed by the "single frequency" analysis).
"Disp. amplitude y-dir"_theoretical = 148e-7 m
"Disp. amplitude x-dir"_theoretical = 2.1e-7 m
"Disp. amplitude y-dir"_COMSOL= 107e-7 m
"Disp. amplitude x-dir"_COMSOL= 1.69e-7 m
Now I'm trying to use a bigger amplitude of the force, in order to see if the ratio between theoretical and simulated values remains the same. I alsoDo you have any other suggestion? The results seems to be "mesh independent".
Thak you very much for the help.
Thank you again, Alessandro.
You are right, I also got to the same conclusion on my own. The problem is that both the values of "Disp. amplitude x-dir" and "Disp. amplitude y-dir" are not correct. They are smaller than the expected values (which were confirmed by the "single frequency" analysis).
"Disp. amplitude y-dir"_theoretical = 148e-7 m
"Disp. amplitude x-dir"_theoretical = 2.1e-7 m
"Disp. amplitude y-dir"_COMSOL= 107e-7 m
"Disp. amplitude x-dir"_COMSOL= 1.69e-7 m
Now I'm trying to use a bigger amplitude of the force, in order to see if the ratio between theoretical and simulated values remains the same. I alsoDo you have any other suggestion? The results seems to be "mesh independent".
Thak you very much for the help.