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 seymours2571 (Mechanical) 22 Mar 06 14:12
 Hello All,I am trying to perform a modal analysis on a 3 piece bracket system with a cylindrical isolator.  I have test data for both the axial spring rate and the radial spring rate. The problem is: Even though I am using a cylindrical coordiant system, I cannot seem to simulate the radial spring movement correctly in Mechanica.  The spring properties dialog box calls for Kxx, Kyy, and Kzz for the spring rates.  There is no location to input a Krr (radial spring rate) even thought the spring is referencing a cylindrical coordinate system.I can attempt to resolve the radial spring rate into two spring rate vectors (Kxx and Kyy), but I don't believe this will yield very accurate results.Any help would be appreciated.Thanks,Steve
 JohnAndrews (Mechanical) 24 Mar 06 5:08
 Could you simulate the radial spring by using a point to point spring with the point from the bracket and isolator at the same location?
 prost (Structural) 26 Mar 06 13:36
 Can't you use sin(t), cos(t) to convert the Kxx, Kyy, Kzz to the radial-azimuthal-axial (r-t-z) coordinate system, much as you can do the same thing with Cartesian displacements, strains, and stresses? My only question is do these springs act like stresses (which require you convert for direction of stress and the plane where the stress acts) or like displacements (which require only correction for direction). For instance,  Kr=Kx*cos(t)+Ky*sin(t) Kt=-Kx*sin(t)+Ky*cos(t) or inverse Kx=Kr*cos(t)-Kt*sin(t) Ky=Kr*sin(t)+Kt*cos(t)  or does it behave like a stress? Kr=Kxx*(cos(t))^2+Kyy*(sin(t))^2+Kxy*sin(2t)...etc. or inverse Kxx=Krr*(cos(t))^2+Ktt*(sin(t))^2-0.5*(Krt+Ktr)*sin(2t) Kyy=Krr*(sin(t))^2+Ktt*(cos(t))^2+0.5*(Krt+Ktr)*sin(2t)Of course, it would take a little bit of experimenting.One other possibility; if the cylindrical coordinate system is chosen, AUTOMATICALLY the program might assume the springs are in cylindrical coordinate system (some FEA programs do this, and you have to read the manual to find this out).

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