Rotational Springs

silvia · Post by **silvia** » Tue Apr 22, 2008 6:30 am

the results are consistent with the model.
make sure you check your sign conventions....

dobi3 · Post by **dobi3** » Mon May 12, 2008 9:08 am

Dr. Silvia,

Here are few reasons why the results are questionable:
1) OpenSees does not seem to recognize the properties of the zero length element at all. I drastically changed the values of My and the initial slope to check how these chances will reflect in the results; however, the displacement was still the same, namely -14.4 in (rather large value for a 2 ft long column). I even changed the uniaxial Material from Steel02 to Elastic, but the results did not chance at all.

2) The deflection varies linearly with a chance in the horizontal load. That is, if I use 10 time higher value for the horizontal load, the deflection is also increased 10 times.

3) The deflection is inversely proportional to the vertical load. If I increase the vertical load 10 times, the deflection is reduces 10 times. This observation does not make sense, since increasing the vertical load should result in larger P-Delta effect. Thus, the deflection at the top of the column should increase, but instead it decreases.
I do not understand what I am doing wrong. I followed the procedure other users described in their massages to you. Do you know what might be the reason for these questionable results?
The original script follows:
wipe;
model basic -ndm 2 -ndf 3;
file mkdir data;
# define GEOMETRY _____________________________________________
# nodal coordinates
node 1 0 0;
node 2 0 144;
node 3 0 0;
# single point coordinates - Boundary Conditions
#nodal masses:
mass 2 0.259 0.0 0.0;
fix 1 1 1 1;
# define ELEMENTS _____________________________________________
geomTransf PDelta 1; # geom tranf.
#connectivity
element elasticBeamColumn 1 3 2 28.2 29000 833 1;
# Semi-Rigid Conection
equalDOF 1 3 1 2;
uniaxialMaterial Steel02 1 1989 786732 0.1 20 0.925 0.15;
element zeroLength 2 1 3 -mat 1 -dir 3;
# define RECORDERS ____________________________________________
recorder Node -file Data/DNode2.out -time -node 2 -dof 1 disp;
#defne GRAVITY ________________________________________________
pattern Plain 1 Linear {
load 2 0.0 -100 0.0;
}
constraints Plain;
numberer Plain;
system BandGeneral;
test EnergyIncr 1.0e-8 10;
algorithm Newton;
integrator LoadControl 0.1;
analysis Static;
analyze 10;
loadConst -time 0.0;
# define Lateral Load ________________________________________
pattern Plain 2 Linear {
load 2 10.0 0.0 0.0;
}
# pushover: displacement analysis
integrator LoadControl 0.1;
analyze 10;
puts "Done!"

Your help is highly appreciated!

silvia · Post by **silvia** » Mon May 12, 2008 10:32 am

are things working when you have the simple cantilever without the base connection?

dobi3 · Post by **dobi3** » Mon May 12, 2008 10:45 am

Yes, and the displacement in this case is 0.424 in.

silvia · Post by **silvia** » Mon May 12, 2008 10:55 am

it looks fine to me, please check it again.

civily · Post by **civily** » Thu Feb 05, 2009 9:57 am

I'm currently having the same problem..I wanted to define translational and/or rotational springs at supports. I've been following the discussions and wondering if you guys have figured it out...
I'd appreciate if you let me know..

dobi3 · Post by **dobi3** » Thu Feb 12, 2009 9:10 am

Instead of using the following:

# Semi-Rigid Conection
equalDOF 1 3 1 2;
uniaxialMaterial Steel02 1 1989 786732 0.1 20 0.925 0.15;
element zeroLength 2 1 3 -mat 1 -dir 3;

Try defining a section as follows:

# Semi-Rigid Conection
equalDOF 1 3 1 2;
uniaxialMaterial Steel02 1 1989 786732 0.1 20 0.925 0.15;
section Uniaxial 1 1 Mz;
element zeroLengthSection 13 1 3 1;

I hope this helps.

silvia · Post by **silvia** » Thu Feb 12, 2009 9:11 am

also, as mentioned in a recent thread, use dof 6 for rotation of the zero-length spring.