Attached is numerical analysis of your rotor. The big assumption here is that any motion of support mass neglected (that’s an assumption, don’t really know anything about what is below this fan).
It uses numerical transfer matrix method (Myklestad method). Assumptions listed at the bottom of the instructions tab. I used to have a companion website for this spreadsheet until I changed ISP’s. It included test cases that have been used to validate the spreadsheet. I can put that website back up online if anyone knows of free website.
Tab Main is the basic options for the calculations. Note that I used Gryoscopic option 1 which includes gyroscopic effects (more later).
RotorSection and RotorGeometryGraph show the model of the rotor used (tried to recreate your drawing). Units have been converted to SI (much easier calculations that way imo).
Outsheet is a summary of the results. Table at the top has columns corresponding to calculated 1st, 2nd, 3rd etc natural frequencies. The rows of the table corresond to a “bearing stiffness multiplier” which is the factor by which the bearing stiffness in “main” (1E7 N/m) is multiplied for the calculation (to see effect of a range of bearing stifnesses).
Below the table in outsheet is a “critical speed map”. It shows how each natural frequency changes as a function of bearing stiffness.
I chose further examination of bearing stiffness multipliers 0.1 and 1000.
For bearing stiffness multiplier 0.1, we see from the critical speed map that the frequencies of the first and 2nd modes are increasing as sqrt(Kb) (slope of ½ on log log plot), indicating the behavior of the bearing spring dominates in this stiffness range for these modes (rigid rotor modes, flexible supports).
For bearing stiffness multiplier 1000, we see from the critical speed map that the frequencies of the first three modes are constant, indicating the behavior of the bearing springs are irrelevant in this stiffness range (flexible rotor modes, rigid supports).
The tabs labeled 0.1 and 1000 show the first three calculated modes which confirm earlier observations about rigid and flexible modes. You’ll also notice as the calculated natural frequency goes up, the left end of modeshape tends to approach displacement and slope of zero, indicating the gyroscopic effect is dominating to create that type of boundary condition.
You’re probably interested in rigid support (mult = 1000), first mode. It has low enough frequency that the gryoscopic effect is not dominating as discussed above, so we can’t replace it with simple slope=0, displacement = 0 boundary condition like you could for the higher modes. How important is the gyroscopic effect for this rigid support (mult = 1000), first mode? Change the calculation mode from 1 (critical speed including gryo) to 3 (no disk effects) on the main tab, re-run the calcuation and you’d see the calculated rigid-support first critical decreases from 49 (with gyroscopic effects) to 44 (without gyroscopic effects). So more than 10% change in natural frequency means gyroscopic effect is still pretty important even though you wouldn’t have guessed it from the modeshape.
The next step if you wanted to improve the model would be to model the support mass (not an easy thing with transfer matrix approach, but can be done with other numerical approaches).
I’m suspecting as natural you’d rather start with simpler models to build and understanding / feel for things. I am short on ideas of simple model to capture essential elements of model so far.
Do you feel comfortable to neglect the mass of the shaft?
Then you can use the mechanics of materials approach described in my earlier post to give you the 2x2 stiffness coefficients of a 2DOF system (mass on each end of the rotor.. one for the fan and one for the sheave).
K*X = -M*X’’ for 2*2 matrix.
K*X = w^2*X solve for w using eigenvalue problem
If you want to add gyropscopic effects, that’s trickier. Can still be done with 2*2 matrices. Something like
K*X + M*X’’ + w*G* X’ = 0
K*X – w^2* M*X + w^2*G* X = 0
Again 2x 2 matrices
where G is a skew symmetric matrix.... only off-diagonal terms are zero and they are opposite signs. Magnitude is the polar moment of inertia for thin disks. Figuring the signs always requires some thought or looking in a book.
I’m sure there is analytical treatment in textbooks somewhere if that’s what you’re interested in. In fact I do remember reading something like that but I’m not going to go digging for it. Again if you post in the vib forum you might get more input.
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(2B)+(2B)' ?
