Spring Reactions for Polar Coordinates in RISA 3D?

[sema79]

I am working on a RISA model in which spring constants need to be used in my boundary conditions. The spring constants I am recieving from an outside source are given as radial and tangetntial stiffnesses due to the cylinderical shape of the structure. As far as I can tell RISA only works in the rectangular coordinate system. My first reaction was to use a spreadsheet to convert each of the values to rectangular. A possible problem emerged in the combining of the x direction values with the y direction values from each stiffness. Specifically, because each stiffness value can be applied in two directions they are not vectors. If treated as vectors then two different outcomes are possible. Which outcome would be right? I could assume (depending on the direction of the lateral forces) that the springs are compression in some areas and tension in others but this will lead to a more complicated model and possibly multiple models. I would like to avoid that if possible.

I wonder how other engineers might approach the problem or if anyone knows of a way to enter polar values in RISA?

 

[weab]

There is no way to enter polar reaction in RISA, springs or rigid support. While I don't know what your model looks like, I'd try this.

At each support, add 2 members of about 1 inch long or so - one member oriented up radially and one tangentially. Set the member end conditions so that they are only capable of handling axial load. Set up your spring conditions in rectangular conditions and I think the reactions must then equate to the proper radial and tangential values. Beware of global stability issues.

 

[sema79]

Thanks. I ran a test run of two identical frames, one positioned so that both the loads and spring reaction are inline with each other, and the other frame at 45 degrees using the x and y components of the spring constant (0.707 x Spring constant). I expected the resultant reaction for the angled frame to be indentical to the reaction for the non-angled frame but it wasn't.

I then approached the problem from another perspective. I built 1" long members that were round bar with 1 square inch cross sectional area. I then assigned material properties to match the spring constants. Elastic modulus was equal to the spring constant for each. I positioned these members to be axial and tangential as needed with fixed connections on the ends with only axial loads allowed. I tested a frame along side an identical frame with an actual spring boundary condition. The reactions and defelctions were identical. Now instead of reporting the joint reactions I will report the loads in each "spring" member. Because only axial loads are allowed my results will be in the correct direction (radial or tangential). This approach seems to be working well as long as the "spring" members are 1 inch long with 1 inch cross sectional area. Any longer and the deflections will be exagerated.

 

[JoshPlumSE]

This is probably an issue better addressed in the RISA-3D forum, forum1419 .

Regardless, there is a good description about how to do "inclined supports" in the Modeling tips section of the help file / manual. The only draw back is that it is assuming a rigid inclined roller.... perfectly free in the roller direction and rigidly restrained perpendicular to the roller. But, we can still use the same basic technique.

Weab has it essentially correct. But, to me the steps would be:
1) Use a "rigid link" member oriented perpendicular to you radial grid or tank.

2) Connect this rigid link to a Fully restrained boundary condition.

2a) If you need to use a spring constant (rather than a fully fixed boundary condition), then adjust the AE (and possibly even the I) of the member used in step 1. The stiffness of this member will provide whatever spring stiffness you need.

3) Use the member end releases in the rigid link to dictate exactly what forces from your structure can be supported by this type of support.

4) The member forces that you get in the rigid link will be equal to the reactions in the inclined direction.


Lastly, I should point out that the method sema79 first describes should work (adjusting spring stiffnesses in the Global direction to provide equivalent spring stiffnesses in the other directions). However, it is dependent on setting the stiffnesses correctly. Plus, the other method (IMHO) is easier and will provide results that are very easy to interpret.