Natural frequency of a double cantilever 1

[geoffthehammer]

I have pair of parallel, flat,leaf springs arranged horizontally (stiff in the vertical direction, weak in the horizontal). The beams are fixed at one end (i.e cantivered). At the outer (free) end, the pair are fixed together by a rigid spacer that has mass. There is another fixed spacer (with mass) coupling the beams together at another point some distance 'x' from the fixed end. The beams are free to vibrate (intentionally) in the horizontal (weak) direction. How on earth do I calculate the natural frequency in this direction?

 

[GregLocock]

FEA would be quickest.

If you want to do it analytically then you probably need to read up on the theory of receptances, which tells you how to calculate the frequency response of systems that are joined together. Are there any simplifying approximations that you can make? Note that you are expecting two natural frequencies. Are the springs identical? If so the answer is much easier.

A big clue I think is that the springs aren't quite what you think they are, you have a 2DOF ground-spring-mass-spring-mass system, because teh rigid links force each pair of springs to operate in unison.









Cheers

Greg Locock

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[geoffthehammer]

Thanks Greg
I have not used FEA before (it wasn't invented when I was a student, come to that -neither were pc computers) Can you or anyone suggest a freeware (or at least reasonably priced) FEA program. Does anyone have any comments on the Cadre Lite 3.0 evaluation or Z88 (from Uni-bayreuth)programs, would either be suitable, especially for a beginner?

 

[GBor]

geoffthehammer,

I'm not familiar with the two programs that you reference, but most general FEA packages have demo versions. For your situation, it sounds like you may be able to do a very small, 2-D or Beam element model that would run under a couple of hundred nodes. Noran Engineering (

http://www.nenastran.com)

puts out a fully functional demo that allows under 300 nodes. Nastran is a good FEA solver to know as it is widely accepted in many industries. NENastran does a good job marketing their product. I'm not sure it is necessarily the best for you application, but I have run leaf springs in Algor (

http://www.algor.com)

and NENastran with success for frequency and for stress.

Garland E. Borowski, PE

 

[GregLocock]

geoff if you post the details of your setup I'll put a model together. Using beam elements it only needs 6 nodes, I think. I need the dimensions of the beams and the masses and the materials, and the general layout.

Z88 is not suitable, so far as I'm aware it only does static analysis, you need dynamic analysis. It is also a bit of a handful to learn (on my to do list - as it is free, fast, and capable of solving fairly large models).

Cadre Lite should be capable of solving this.

I was kind of hoping someone would come up with an analytical solution - it would be interesting to compare with FEA. It is directly analagous to a portal frame problem - don't the seismic engineers solve those?



Cheers

Greg Locock

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[GBor]

I just spent a little time looking through two books: Fundamentals of Vehicle Dynamics by Thomas Gillespie and the Spring Design Manual compiled by SAE (the Society of Automotive Engineers). There are several calculations for leaf springs. Once we get the particulars, I'll take a crack at it, but, admittedly, I'm "flying blind".

Garland E. Borowski, PE

 

[GregLocock]

The hard bit isn't the springs, it's the 2dof system.

Cheers

Greg Locock

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[MikeyP]

You could use Rayleigh-Ritz/Lagrange approach. It's not exactly analytical - but it would probably give you a model with the various dimensions as variables, meaning that you wouldn't have to keep changing it like you would with FE. It would also give you continuous mode shapes which look prettier than the descretised FE shapes!

M

--
Dr Michael F Platten

 

[GregLocock]

Good catch, yes, it is an ideal application for RR. How did you know I was going to build an ugly FEA model? (well I guess 6 nodes is bit of a giveaway)



Cheers

Greg Locock

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[sreid]

I think this problem can be simply turned into a problem with a known solution by the following visualization.

This is a two unequal mass, two unequal spring system. One spring (K1) has one end fixed, the other end attached to M1. The second spring (K2) has one end attached to M1 and the other end attached to M2. The spring constants (K1, K2) of the cantilever beams are determined by force vs. defelction for a cantilever beam with one end fixed, one end guided (times two, of course). The spring rates vary with the "X" length.

Blevin's "Natural Frequency and Mode Shape) gives a solution for this case.

 

[MikeyP]

I don't think that will work. It doesn't account for rotation of the two masses for one thing. The rotatory inertia of the two masses is likely to be significant compared to their mass.

M

--
Dr Michael F Platten

 

[sreid]

MikeyP,

Since the masses are supported by TWO flexures, the motion is translational at least for small angles.

 

[Denial]

The problem with sreid's visualisation is not the rotation of the linking masses. It is the fact that the leaf springs run continuously past M2. Thus his two "cantilever beams" are not independent of each other.

I believe an analytical solution along the lines he suggests is still possible, but there would be some tedious algebra involved in deriving it.

 

[sreid]

My interpretation is based on the statement in the original post;

"There is another fixed spacer (with mass) coupling the beams together at another point some distance 'x' from the fixed end."

I interpreted this to meaan that the intermediate mass is solidly "Fixed" to the cantilever beams. If the intermediate mass is "Pinned" to the beams then the beams would not be independent.

 

[GregLocock]

fxxxxxxxxxxxxxxxxxxxxxxxxx
fxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
            p                       p
            l                       l
            l                       l
            m                       m
            l                       l
            l                       l 
            p                       p
fxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
fxxxxxxxxxxxxxxxxxxxxxxxxxx

Is how I'm envisaging this. p is a pin joint, m is a mass, f is a fixed joint. x is some sort of leaf spring.








Cheers

Greg Locock

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[sreid]

Maybe GeofftheHammer can clarify this for us. Certainly if one or both mases are pinned the problem is "More Interesting."

 

[geoffthehammer]

Hello Guys, Sorry for the delay but I've been out of the office for a few days on holiday in the south of France. You have an interesting discussion going on!
Here is a description of the set-up:
If we look at Greg's last post, this is a good illustration of the set-up and shows a plan view of the situation.
F is the fixed end.
xxxxx are the two plain leaf springs. The dstance between the fixed support and the front end is 87mm. The springs are each 1.5 mm thick and 48mm high. The material is spring steel. The springs are 78mm apart.
At the front end there is a rectangular aluminium spacer (plus a small mass)78mm long x 48mm high and approx. 10mm thick. The spacer is machined out to reduce weight but retain much of the strength. The spacer is bolted rigidly to each spring. The spacer is the same height as the springs. Total mass of the spacer is approximately 200 grammes.
The intermediate spacer is positioned 60.5mm from the fixed (LH)end. This spacer is steel, 6.35mm diameter x 78 long and has a central mass. The spacer is bolted to each leaf spring with a single M3 screw through the leaf spring. The spacer is situated at mid-height (24mm) in the leaf springs. The total mass of this spacer is 95 grammes.
As 'Denial' points out, the two springs not independent of each other, there are no pin joints in the arrangement.
I hope that this description is fairly clear but let me know if otherwise.
Thanks again to all.
Geoff

 

[GregLocock]

I get 139 and 631 Hz. However, I'm not wildly happy with the model. Judging by the mode shapes anyone doing this analytically is probably safe to treat it as a 2DOF spring-mass-spring-mass system.



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Greg Locock

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[GregLocock]

OK, model sorted, 161 and 702 Hz

The dominant mass is the one at the end so 'approximately 200 grammes' is a major source of error.



Cheers

Greg Locock

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