different resuts with diffrnt dt in nonlinear dynmc analysis
Moderators: silvia, selimgunay, Moderators
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- Posts: 14
- Joined: Wed Dec 21, 2005 1:56 am
- Location: European Commission, Joint Research Centre, European Laboratory for Structural Assessment (ELSA)
different resuts with diffrnt dt in nonlinear dynmc analysis
Hi everyone,
My structure is a 20-story RC frame in 2D.
I've run nolinear dynamic analyses under three records with three analysis time steps (0.02, 0.005 and 0.001).
I've monitored the disp.s and acc.s of a node @ 20th story. The results are very different:
RECORD PROPERTIES ANALYSIS STRUCTURAL RESPONSES
Record/Component dt length (s) PGA (g) dt Duration (s) PeakJointAcc PeakJointDisp
ERZIKAN/ERZ-EW 0.005 21.32 0.5 0.02 30 8.27 1.9980
ERZIKAN/ERZ-EW 0.005 21.32 0.5 0.005 30 11.86 0.4585
ERZIKAN/ERZ-EW 0.005 21.32 0.5 0.001 30 4.85 0.0280
NORTHR/SCS142 0.005 39.995 0.9 0.02 30 9.58 3.0177
NORTHR/SCS142 0.005 39.995 0.9 0.005 30 15.76 0.6297
NORTHR/SCS142 0.005 39.995 0.9 0.001 30 10.98 0.0406
NORTHR/116360 0.02 39.98 0.13 0.02 30 3.24 0.0554
NORTHR/116360 0.02 39.98 0.13 0.005 30 1.26 0.0068
NORTHR/116360 0.02 39.98 0.13 0.001 30 1.31 0.0001
Someone has any idea..? (especially Frank and Silva)
Thanks,
Ufuk
My structure is a 20-story RC frame in 2D.
I've run nolinear dynamic analyses under three records with three analysis time steps (0.02, 0.005 and 0.001).
I've monitored the disp.s and acc.s of a node @ 20th story. The results are very different:
RECORD PROPERTIES ANALYSIS STRUCTURAL RESPONSES
Record/Component dt length (s) PGA (g) dt Duration (s) PeakJointAcc PeakJointDisp
ERZIKAN/ERZ-EW 0.005 21.32 0.5 0.02 30 8.27 1.9980
ERZIKAN/ERZ-EW 0.005 21.32 0.5 0.005 30 11.86 0.4585
ERZIKAN/ERZ-EW 0.005 21.32 0.5 0.001 30 4.85 0.0280
NORTHR/SCS142 0.005 39.995 0.9 0.02 30 9.58 3.0177
NORTHR/SCS142 0.005 39.995 0.9 0.005 30 15.76 0.6297
NORTHR/SCS142 0.005 39.995 0.9 0.001 30 10.98 0.0406
NORTHR/116360 0.02 39.98 0.13 0.02 30 3.24 0.0554
NORTHR/116360 0.02 39.98 0.13 0.005 30 1.26 0.0068
NORTHR/116360 0.02 39.98 0.13 0.001 30 1.31 0.0001
Someone has any idea..? (especially Frank and Silva)
Thanks,
Ufuk
Ufuk HANCILAR
ELSA - European Laboratory for Structural Assessment
European Commission - JRC
Ispra (VA) ITALY
ELSA - European Laboratory for Structural Assessment
European Commission - JRC
Ispra (VA) ITALY
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- Posts: 14
- Joined: Wed Dec 21, 2005 1:56 am
- Location: European Commission, Joint Research Centre, European Laboratory for Structural Assessment (ELSA)
[quote="silvia"]it makes sense that as you increase the time step you might miss critical peaks.
you should plot all three time-history responses and see.[/quote]
already done! no peaks missed.
last two time steps are too small, 0.001 and 0.005. but the results very different, esp. disp.s
you should plot all three time-history responses and see.[/quote]
already done! no peaks missed.
last two time steps are too small, 0.001 and 0.005. but the results very different, esp. disp.s
Ufuk HANCILAR
ELSA - European Laboratory for Structural Assessment
European Commission - JRC
Ispra (VA) ITALY
ELSA - European Laboratory for Structural Assessment
European Commission - JRC
Ispra (VA) ITALY
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- Posts: 14
- Joined: Wed Dec 21, 2005 1:56 am
- Location: European Commission, Joint Research Centre, European Laboratory for Structural Assessment (ELSA)
[quote="silvia"]what is the time step of the input motion?[/quote]
0.005 for two input records and 0.02 for the other one.
the record's dts and PGAs listed together with the results in my first post.
0.005 for two input records and 0.02 for the other one.
the record's dts and PGAs listed together with the results in my first post.
Ufuk HANCILAR
ELSA - European Laboratory for Structural Assessment
European Commission - JRC
Ispra (VA) ITALY
ELSA - European Laboratory for Structural Assessment
European Commission - JRC
Ispra (VA) ITALY
-
- Posts: 14
- Joined: Wed Dec 21, 2005 1:56 am
- Location: European Commission, Joint Research Centre, European Laboratory for Structural Assessment (ELSA)
[quote="silvia"]your analysis time step definitely has to be less than your ground-motion step.[/quote]
yes, my analyses time steps are less than or equal to input motion's dt but still produces very different results.
yes, my analyses time steps are less than or equal to input motion's dt but still produces very different results.
Ufuk HANCILAR
ELSA - European Laboratory for Structural Assessment
European Commission - JRC
Ispra (VA) ITALY
ELSA - European Laboratory for Structural Assessment
European Commission - JRC
Ispra (VA) ITALY
-
- Posts: 14
- Joined: Wed Dec 21, 2005 1:56 am
- Location: European Commission, Joint Research Centre, European Laboratory for Structural Assessment (ELSA)
[quote="horizon"][quote="silvia"]your analysis time step definitely has to be less than your ground-motion step.[/quote]
yes, my analyses time steps are less than or equal to input motion's dt but still produces very different results.[/quote]
hi all,
I think I've figured out the problem. dt of the input motion and dt of the analysis should be same.
OpenSees only reads the acceleration values from your input, it doesnt know the dt of your input motion.It just performs the dynamic analysis with the predefined time increments.
lets say I have an acceleration record wtih dt=0.005 and 40-sec length.this means my input motion consists of 40/0.005=8,000 points.
and I'm running the nonlinear analysis with dt=0.02 during 30 seconds. (30/0.02=1500 points)
in this case, the program gets the first 1500 acceleration values for the 30-sec. loading. it means that only the first 1500*0.005=7.5 sec-part of the input motion is applied to the structure during entire 30-sec loading.
and I was recording very big node displacements...
I've made the analysis dt smaller, 0.001 and again 30-sec analysis.
in this case, I got very small stuructural responses and the response was close to the linear.cause the program applied all the acc values from the input motion but only during 8,000*0.001=8 seconds.
the sturucture was loaded during the first 8 seconds of the 30-sec analysis and it responded almost linearly.
when I've performed the analysis with 0.005 time increments during 30 sec. under the input motion recorded at 0.005, I got the reasonable responses!
Ufuk
yes, my analyses time steps are less than or equal to input motion's dt but still produces very different results.[/quote]
hi all,
I think I've figured out the problem. dt of the input motion and dt of the analysis should be same.
OpenSees only reads the acceleration values from your input, it doesnt know the dt of your input motion.It just performs the dynamic analysis with the predefined time increments.
lets say I have an acceleration record wtih dt=0.005 and 40-sec length.this means my input motion consists of 40/0.005=8,000 points.
and I'm running the nonlinear analysis with dt=0.02 during 30 seconds. (30/0.02=1500 points)
in this case, the program gets the first 1500 acceleration values for the 30-sec. loading. it means that only the first 1500*0.005=7.5 sec-part of the input motion is applied to the structure during entire 30-sec loading.
and I was recording very big node displacements...
I've made the analysis dt smaller, 0.001 and again 30-sec analysis.
in this case, I got very small stuructural responses and the response was close to the linear.cause the program applied all the acc values from the input motion but only during 8,000*0.001=8 seconds.
the sturucture was loaded during the first 8 seconds of the 30-sec analysis and it responded almost linearly.
when I've performed the analysis with 0.005 time increments during 30 sec. under the input motion recorded at 0.005, I got the reasonable responses!
Ufuk
Ufuk HANCILAR
ELSA - European Laboratory for Structural Assessment
European Commission - JRC
Ispra (VA) ITALY
ELSA - European Laboratory for Structural Assessment
European Commission - JRC
Ispra (VA) ITALY
Dear silvia,
1) in the example manual (page 402) the ground-motion step is 0.01 and the analysis time step is used 0.02 (page 405).
now, why do you say that analysis time step definitely has to be less than ground-motion step?!
2) the manual says that the analysis time step does not have to be the same as the input ground motion; so i think the horizon's opinion is incorrect. any opinion?
1) in the example manual (page 402) the ground-motion step is 0.01 and the analysis time step is used 0.02 (page 405).
now, why do you say that analysis time step definitely has to be less than ground-motion step?!
2) the manual says that the analysis time step does not have to be the same as the input ground motion; so i think the horizon's opinion is incorrect. any opinion?
Iman Mansouri
PhD Candidate
University of Kerman
PhD Candidate
University of Kerman
the deltaT for the analysis and dT for the input motion do not have to be the same.
given the current time (T1) in the domain, which is equal to the last committed time plus deltaT, OpenSees will go to the load pattern and tell it to apply the loads for T1.
if the load pattern is a uniform excitation it will ask the TimeSeries object what is the current factor for T1. if T1 is some multiple of the dT's specified for the motion it will use the corresponding point. if not it will do a LINEAR INTERPOLATION of the two points on either side of T1.
this means that assuming the domain is at time 0.0 at start of transient analysis, if deltaT is equal to dT, then all points specifified in the input will be used, if deltaT is equal to 2*dT then only have the points specified will be used (EVERY SECOND POINT WILL BE USED and not just the first half),.
given the current time (T1) in the domain, which is equal to the last committed time plus deltaT, OpenSees will go to the load pattern and tell it to apply the loads for T1.
if the load pattern is a uniform excitation it will ask the TimeSeries object what is the current factor for T1. if T1 is some multiple of the dT's specified for the motion it will use the corresponding point. if not it will do a LINEAR INTERPOLATION of the two points on either side of T1.
this means that assuming the domain is at time 0.0 at start of transient analysis, if deltaT is equal to dT, then all points specifified in the input will be used, if deltaT is equal to 2*dT then only have the points specified will be used (EVERY SECOND POINT WILL BE USED and not just the first half),.