Optimum Cantilever Shape
Optimum Cantilever Shape
(OP)
Here's the situation, and it is not a homework problem. I have a cantilever round hollow tapered beam. The length, small end diameter, and large end diameter are fixed.
What is the proper way to transition the diameters from one end to the next, assuming constant wall thickness, that will maximize the 1st natural frequency of vibration? Obviously a linear taper is the easiest to manufacture, but I have a feeling that something more like a exponential or logarthmic taper would give better results. I'm a little rusty with the variational calculus, so I'ms stuck with trial & error via FEA.
By the way, this is for a self supporting (un-guyed) steel stack. The diameter ideal variation would be appoximated by numerous cylindrical and conical elements. Meca Consulting, are you out there?
What is the proper way to transition the diameters from one end to the next, assuming constant wall thickness, that will maximize the 1st natural frequency of vibration? Obviously a linear taper is the easiest to manufacture, but I have a feeling that something more like a exponential or logarthmic taper would give better results. I'm a little rusty with the variational calculus, so I'ms stuck with trial & error via FEA.
By the way, this is for a self supporting (un-guyed) steel stack. The diameter ideal variation would be appoximated by numerous cylindrical and conical elements. Meca Consulting, are you out there?





RE: Optimum Cantilever Shape
That's the end of the good news! In that case I just built an Excel simulation of the tapered beam and let the Solver loose on it. Once the trend towards the final shape became obvious I just typed in the exact values and confirmed that it was an optimum.
My incination is to build the equivalent a Holzer type torsional vibration model of the beam, in discrete steps, and then optimise the resonant frequency of that. ie set KE=1/2*m*w^2*y^2 for each little ring, and equate it to 1/2*k*y^2. eyeballing the resultant distribution might lead to an optimimisd plot of r vs x.
Incidentally, is the big end or the little end free? grin. I'll assume the common sense answer.
The correct shape is likely to be a concave exponential(ish) curve, at a very rough guess, since the bending frequency of the base has an inordinately high effect on the tip stiffness. If you need the sort of shape that is likely look at the taper on the strength member of an orbital elevator
I'll have a play tonight and see if I can get neat model.
Incidentally, you might well find an analytical solution to this, it has that feel to it.
Cheers
Greg Locock
RE: Optimum Cantilever Shape
His other observation is that for complete convex exponentially tapered solid beams you will get an improvement in natural frequency of sqrt(6.3/4.8) (ie 15%)by going from an untapered beam to a fully tapered one.
From these two slightly related results I suspect that the frequency optimum is going to be fairly tolerant of deviations from the 'ideal' tapered form.
it seems to me that my proposed approach involves excessive iteration, since establishing the mode shape (and hence frequency) is not exactly a no-brainer.
Therefore I'm going to propose a simpler method: split the beam into two tapered tubes. Optimise the diameter of the common plane, and the length of the first tube. Then refine the technique, depending on the results.
I have a sneaking suspicion that an analytical technique could cope with this using the dreaded theory of receptances, but my uni books lie mouldering in the garage and brute force seems likely to work.
Cheers
Greg Locock
RE: Optimum Cantilever Shape
Cheers
Greg Locock
RE: Optimum Cantilever Shape
Developing a Holzer typebmodel for a lateral vibration is not quite as straight forward as it seems. The number of columns required is significantly higher than that required for the torsional model. If you have a reference solution for this class of problem I would appreciate any information.
Regards,
Gunnar
RE: Optimum Cantilever Shape
Thanks for taking an interest. The overall height is 60 ft. The small (free) end diameter is 18". The large end diameter is <or=60", and for obvious reasons, will be 60". Wall thicknes is 3/16" minimum (new & cold), or 1/16" minimum (corroded) at the small free end. Thickness can vary as needed in discreet steps of 1/16" for each individual shell course. Temperature is not a significant factor, as is will be <200 °F. Material is mild carbon steel with a nominal 30,000 PSI yield strength.
If the FEA software at my disposal was parametric, which it is NOT, :( my thought would be to vary the diameter using a function where the diameter is proportional to (x/L)^n. By varying the power of n from <<1 to >1, the taper ratio could be easily changed, and a good feel for a near optimum might be found by varying only the n parameter.
Due to having 1/8" corrosion allowance, buckling is a concern. My plan is to use the uncorroded weight and the corroded thickness (along with wind & seismic loads) as the governing buckling condition, as this would be the limiting case with localized corrosion. For uniform corrosion, the natural frequency shouldn't change much with thickness loss, so I don't think its a big factor in the original question, just in the buckling check.
RE: Optimum Cantilever Shape
First of all, I assumed the mode shape would be that of a uniform cantilever beam, given by
phi(x) = 2*(x/L)^2 - 4/3*(x/L)^3 + 1/3*(x/L)^4
This mode shape was determined based upon the boundary conditions of: 1.) No deflection at base, 2.) No slope at base, 3.) No moment at tip, 4.) No shear at tip, and 5.) Unit deflection at tip. When used for uniform beam, results agree to within <0.3% error of true solution.
The diameter as a function of length is given by:
D(x) = D1 + (D2-D1)*(x/L)^n
This was then used with Rayleigh's method to determine the natural frequency. It appears that an n value of 1.5 is the optimum.
Any comments on the validity of my methods or assumptions are welcome. I'm currently doing a more refined beam element FEA solution to check the solution.
RE: Optimum Cantilever Shape
Using 3/16 thickness throughout I get 4.6 Hz for a 60"tube, whereas the analytical result is 3.71 Hz. This is a much larger discrepancy than I was expecting, perhaps it is a local minimum, not a true one. A linear taper gives 4.9 Hz.
A linear taper to the half way point, followed by a cylinder is a disaster at 3.5 Hz, and a cylinder followed by a taper (which had always been the optimum in my trial runs) gives 5.8 Hz. This shape 'looks' very like the optimum for a buckling solution, and for an optimised solid cantilever for minimum weight, which I think are both parabolas in side view. It is a convex shape, whereas I guessed a concave shape might be better.
Stuffing around with the exact start point of the taper is unrewarding, for these particular end conditions.
Statement of the obvious - you will get much better results by using t=3/16 at the base and proportionately less as you go up the stack. Do I correctly understand that the tip must be 3/16 thick when new, and all the others are up for grabs at 1,2, or 3 /16? Are you trying to optimise the frequency when new, or when the tip is corroded to 1/16?
Fun problem, i'd always meant to write a rayleigh ritz solver, now I'll try and find out what the calibration problem is.
Cheers
Greg Locock
RE: Optimum Cantilever Shape
Yes, the minimum corroded thickness for all portions of the stack must be at least 1/16", giving a 3/16" original thickness. A preliminary buckling check (based upon wind loading) shows that a the base, it will have to be at least 1/4" original thickness. Shell bending stresses due to wind loading vastly outweigh either seismic or dead weight stresses.
Question: Using the form of equation you did for mode shape in Rayleigh-Ritz, how can you meet all 5 of the boundary conditions using 4 parameters? Am I missing something?
I definitely agree on the advantage of decreasing plate thickness as you go higher.
As to which case we are to optimize, if you assume uniform corrosion, and use thin shell theory, and uniform thickness, you will get the same result for new or corroded case, as both mass and stiffness are proportional to t to the 1st power.
RE: Optimum Cantilever Shape
Incidentally this type of problem would be good to study using Taguchi, or a genetic algorithm. It's a pity your constraints end up with such a boring solution - if the OD of the tube could be wider than the base you get some whacky shapes.
Cheers
Greg Locock
RE: Optimum Cantilever Shape
Your 'optimal' shape is not as good as my optimal shape, according to my model. Incidentally, according to my model once you take your parametric shape up to values of n above 4 then the frequency becomes invariant with n. Very odd! Fixing the mode shape as you have done will force you down one particular path.
Coding this stuff at 2 in the morning led to a bit of finger trouble, I think the calculated result for a cantilevered cylindrical tube 60 ft by 60" by 3/16" should be 5.43 Hz. Calibrating my model against this (adding an 8% fudge factor, of which I am not proud, but the result scales correctly with changes in thickness, diameter and length), a linear taper to 18" comes in at 6.1 Hz, and 60" to half way followed by a linear taper gives 7.2 Hz. For n=1.5 using your parametric formula I get 6.71 Hz, when allowed to find its own mode shape.
My quick and dirty FE model of the cylinder came in at 3.82 Hz .................! Second bending mode was at 22 Hz, which is in the right ratio, worryingly enough.
Cheers
Greg Locock
RE: Optimum Cantilever Shape
Thanks for all your help. I know that by forcing the mode shape, my answers will be a little (or maybe a lot) high. It started off as a very small taper ratio, at which point my assumed mode shape would be closer to right. As the taper ratio gets larger, my mode shape is more in error. For this case, I get 6.00 Hz, compared to your 5.43 Hz, which is a 10.5% error. I'm dusting off my Theory of Vibrations with Applications by Thomson to bone up on Rayleigh-Ritz a little better.
Your comment about the frequency being invariant with n>4 is not entirely surprising. If you plot the stack shape, you will see that as values of n get large in my formula, the stack shape approaches a uniform diameter until very near the tip, at which point it tapers rapidly. Sort of looks like a cigar for n>4.