Hi,everyone
I'm trying to analyze a column under variable axial load and lateral load.
Is it possiable to control displacement and load at the same time in the model?that is to say,which kind of "Intergrator" should I take?
Thanke for your help.
Can I control displacement and load at the same time?
Moderators: silvia, selimgunay, Moderators
can you be a bit more specific. you could for example use loadControl
and in 1 pattern place the loads which use one path and in another specify the displacements which follow some other path .. this is IMPOSED
DISPLACEMENTS and NOT DISPLACEMENT CONTROL
Most people want displacement cotrol for pushover curves so be carefull.
and in 1 pattern place the loads which use one path and in another specify the displacements which follow some other path .. this is IMPOSED
DISPLACEMENTS and NOT DISPLACEMENT CONTROL
Most people want displacement cotrol for pushover curves so be carefull.try this with example1.tcl
I assume that you want the loading in a particular direction and at a fixed ratio. Menwhile, you also want to limit the magnitude of displacements. Try this with example1.tcl, where the force applied is 100 in x, -50 in y at node 4. With the full load, nodal displacements at node 4 would be
0.53 and -0.1779, respectively.
Now comment the following line, and add displacement control under it:
#integrator LoadControl 1.0
integrator DisplacementControl 4 1 0.1 1
Futher increase analysis to three steps with:
#analyze
analyze 3
We will get force on node 4 to be 56.6 and -28.3, respectively. (Notice that the 2 to 1 ratio is maintained.) The displacements are 0.3 and -.1, since we force displacement to be 0.1 in X each time, and run the analysis three times. This elastic problem can also be confirmed: i.e., 0.3/0.53*100=56.6; -0.1/-0.1779*50=28.3
Now further increase the steps way beyond that is possible with the load, i.e, change the following line to
analyze 10
Becuae the load is not large enough to cause 10*0.1 or 1.0 x-displacement, the analysis stops at force 100 and -50, and gives the exact same displacements as the loadcontrol.
Thus, if you want to force the displacement to a certain value, then put much larger loads on. If you want limit on both , then put the correct magnitude and correct steps on.
0.53 and -0.1779, respectively.
Now comment the following line, and add displacement control under it:
#integrator LoadControl 1.0
integrator DisplacementControl 4 1 0.1 1
Futher increase analysis to three steps with:
#analyze
analyze 3
We will get force on node 4 to be 56.6 and -28.3, respectively. (Notice that the 2 to 1 ratio is maintained.) The displacements are 0.3 and -.1, since we force displacement to be 0.1 in X each time, and run the analysis three times. This elastic problem can also be confirmed: i.e., 0.3/0.53*100=56.6; -0.1/-0.1779*50=28.3
Now further increase the steps way beyond that is possible with the load, i.e, change the following line to
analyze 10
Becuae the load is not large enough to cause 10*0.1 or 1.0 x-displacement, the analysis stops at force 100 and -50, and gives the exact same displacements as the loadcontrol.
Thus, if you want to force the displacement to a certain value, then put much larger loads on. If you want limit on both , then put the correct magnitude and correct steps on.