integrator load control value
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integrator load control value
what does mean the integrator load control value?
please explain by example:
suppose this is the applyed load pattern for 3-story frame:[1;2;3]
and we have following order:
integerator loadcontrol 10
if the factor in linear time serries would be one,
so the applied load in the next step is
1) [11;12;13]
or
2) [10;20;30]
which state is correct? 1 or 2?
please explain by example:
suppose this is the applyed load pattern for 3-story frame:[1;2;3]
and we have following order:
integerator loadcontrol 10
if the factor in linear time serries would be one,
so the applied load in the next step is
1) [11;12;13]
or
2) [10;20;30]
which state is correct? 1 or 2?
Mohaghegh University/Iran
Re: integrator load control value
Integrator LoadControl x
What is x? does it apply to base shear?
I want to increase base shear in amount of 100 N at each step, till base shear reach 600000 N in pushover analysis.
How can I do it?
Is the following correct?
Integrator LoadControl 0.000167 #100/600000=0.000167
…
Analyze 6000
I have a Linear times series with a factor of 1
What is x? does it apply to base shear?
I want to increase base shear in amount of 100 N at each step, till base shear reach 600000 N in pushover analysis.
How can I do it?
Is the following correct?
Integrator LoadControl 0.000167 #100/600000=0.000167
…
Analyze 6000
I have a Linear times series with a factor of 1
Last edited by Eli on Thu Jul 25, 2013 11:09 am, edited 2 times in total.
Re: integrator load control value
x is the amount the pseudio time is increased in the domain at each step. the load pattern passes the pseudo time to the time series to obtain the load actor to apply to the loads. if the only time series is a linear one with a facor of 1, then the load factor will equal the pseudo time.
in short: if the reference loads sum up to 1.0, then to increase the total base shear by 100 at each step you would specify an x of 100 to get an increment of 1.0. if the reference loads summed to 1.0, would would set x at 1.0.
in short: if the reference loads sum up to 1.0, then to increase the total base shear by 100 at each step you would specify an x of 100 to get an increment of 1.0. if the reference loads summed to 1.0, would would set x at 1.0.
Re: integrator load control value
Thanks dear fmk
i reached my goal by using:
set totalbaseshear 600000
set baseshearincrement 100
set Nsteps [expr int($totalbaseshear/$baseshearincrement)]
integrator LoadControl $baseshearincrement
set ok [analyze $Nsteps]
what should I do when Convergence Test failed in LoadControl pushover analysis?
Is there any convergence loop as DisplacementControl analysis?
i reached my goal by using:
set totalbaseshear 600000
set baseshearincrement 100
set Nsteps [expr int($totalbaseshear/$baseshearincrement)]
integrator LoadControl $baseshearincrement
set ok [analyze $Nsteps]
what should I do when Convergence Test failed in LoadControl pushover analysis?
Is there any convergence loop as DisplacementControl analysis?
Re: integrator load control value
you can switch to displacement control.
Re: integrator load control value
but it is necessary to use LoadControl. as I said i have to increase total base shear by 100 at each step.
Is there any way to fix convergence problem in LoadControl pushover Analysis?
Is there any way to fix convergence problem in LoadControl pushover Analysis?
Re: integrator load control value
Using the loadControl integrator is generally a poor choice for pushover analysis because it cannot pass the "peak" of your pushover curve. The reason is because you'll eventually prescribe a load that is above the peak (meaning the reference load does not cross the pushover curve, which is essential for the Newton-Raphson method to work). When this happens, the model will never converge because the residual will always be greater than the criterion specified.
Using displacement control will allow you to pass the peak of your pushover curve.
Using displacement control will allow you to pass the peak of your pushover curve.
Re: integrator load control value
Thanks dear fbeckwit
The problem is that: I could not increase base shear by a special amount at each step in displacement control analysis.
Any solution?
The problem is that: I could not increase base shear by a special amount at each step in displacement control analysis.
Any solution?
Re: integrator load control value
you can increase by a specified amount only in LoadControl .. as you have found after a certain point it will not arrive at a solution simply because you reach the peak, after that if you want to push further you have to use displacement control where you cannot control the increase and subsequent decrease in applied load. it is the only way however to go past the peak.