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Determination of "static" bending line

Determination of "static" bending line

Determination of "static" bending line

(OP)
Hy,

how can I determine the "static" bending line of a vertical shaft inside 3 journal bearings, including a static force (For the determination of the bearing-loads and the rotordynamic coefficients). I can get the static deflection curves of the bearings x=x(Fstat) and y=y(Fstat) but I get confused about the displacement not in force-direction..
Thank you for comments
Michael

RE: Determination of "static" bending line

Hi,
a fluid-film bearing carries the load with an eccentricity at an angle (this comes from the Reynolds equation). Thus, given a load in one single direction, say X, the equilibrium position will not be in that same direction, it will have both X and Y components. The same holds for a load in the Y direction. So you will have four compliance coefficients for a fluid-film bearing:
Cxx is the displacement in X direction caused by a unit force in X direction; Cxy is the displacement in Y direction caused by a unit force in X direction; Cyx id displ in X caused by unit Fy; Cyy is displ in Y dir caused by unit Fy.
All this is valid at a fixed regime, because the equilibrium position of the shaft in a fluid-film bearing depends on the rotation speed.
However, if you have these compliance (or, equivalently, stiffness) values for a given regime, you will have to solve the equilibrium equation as usual, but in matricial form. This is not exactly trivial for an hyperstatic shaftline as you describe, but with some (!!!) patience it can be done with handcalcs.

Regards

RE: Determination of "static" bending line

(OP)
Hi cbrn,

I thought, the coefficients cxx cxy cyx and cyy do not determine the equilibrium position, but the dynamic behaviour at (!) equilibrium powition. Is that correct ?

My question concernded the determination of the equilibrium position...

Regars

RE: Determination of "static" bending line

Hi,
No.
Think of the simplest schematization of "something" supported by a spring: single mass, single spring. It's true that the C of the spring influences the natural frequency of the oscillating system and by consequence its response to an external force (dynamic behaviour), but it also influences the equilibrium position of the mass (the spring will compress until its reaction force F=displ/C equals the weight force W=Mg!
With fluid-film bearings, it's the exact same thing, but you have to write the equation in matricial form since the equilibrium is out-of-plane (except the trivial case of zero rotating speed, where the fluid film doesn't work at all and the shaft directly lies on the white metal part of the sleeve). Because you have a vertical shaft, the influence of weight is null but the influence of your force remains! With machine stopped, the shaft position in the sleeves will be shifted of a quantity equal to the radial clearance; in motion, the equilibrium position will be dictated by F, Cxx, Cxy, Cyx and Cyy through the reaction forces at the three bearings.
Just write the foundamental equation of equilibrium, in matricial form, including ALL the pertaining factors...
If you get confused about that, take one good book of rotor design and look for keywords "flexible rotors on flexible supports".

Regards

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