torsional natural frequency - beams 1

[40818]

Hi, with simple structures such as beams (fixed,simply supported, cantilevers etc) its pretty well documented how to calculate the natural frequencies of the beams for different end restraints.
Now, from FE analysis on many real structures, i am also getting an abundance of torsional modes, and i have no method of calculating these frequencies by hand, and havn't managed to find any simple equations to use.
Does anybody have any information on how to calculate torsional frequencies for the common beam solutions.

Thanks.

 

[GregLocock]

Sorry I meant the effect of restraining a free-free bar. I suppose what it is is that when we add constraints to a system we expect the frequencies to go up, and gut feeling to anyone who plays with this stuff in the lab is that RBMs aren't 'the same' as flexural modes.

The trick is to creep up slowly on the problem - starting from free free, add very soft springs, so the behaviour is still free-free like, and then gradually make them stiffer. At some point the behaviour will flip and the bending mode will occur before the resonance on the springs.

Haven't actually done that, I might do it tonight.

For extra points, what happens if I take a free free beam and restrain it via springs at the nodes of the first flexural mode?





Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[electricpete]

I was satisfied with the explanation you had given before.
thread384-155945

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[electricpete]

I was trying to figure out if the formula's presented above for torsional beams (for example Rao's formulas as posted 20 Aug 08 21:10) apply to both circular and non-circular cross section beams (as long as cross seciton is uniform along the length)

My gut says it applies to both. We have something like w ~ sqrt(J/K). In the numerator we have something like
J = Ip * rho where Ip is polar moment of inertia.
K = Ip * G where same Ip
So it seems plausible that the frequency is dependent only on material properties G and rho, but not on geometry factors since the Ip cancels out.

And yet, every single reference I looked at presented these formulas as formulas for "cylindrical beams" or "rods" or "shafts". They didn't exactly say it was limited to that cylindrical, but I'm not sure why they all chose that terminology instead of simply calling it a beam. Two possibilities:
1 - There is something more to it than my analysis above where Ip cancelled out
2 - They just use the word shaft, cylindrical etc because those are the typical applications where we are concerned about torsional.

What do you think? 1 or 2?



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[GregLocock]

It definitely depends on geometry, I think you are muddling first and second moments of inertia.

More accurately - the elastic torsional constant of an arbitrary cross section is NOT related to the polar moment of inertia.

For circular solid or hollow shafts it is, but not for any other shape. Working out the elastic torsional constant for complex shapes is rather difficult, there are several approaches, the easiest in my experience is just to set the thing up in FEA and measure it.

http://en.wikipedia.org/wiki/Torsion_(mechanics)

has some useful pointers to analytical approaches



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[electricpete]

Thanks Greg. I think I did mix up the moments. Star for you.

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[GregLocock]

Back to the free free beams

An experiment

Here's a beam with a spring at each end. The bar is 525 mm long, entire thing is constrained to xy plane.

Property 2 - Beam
  Type BEAM            Color 110   Layer 1        Material 1        #Elem 21   
End A    Area 1164.          ShearF, K1 0.             ShearF, K2 0.          
           I1 1827000.               I2 1827000.              I12 0.          
  NS Mass/Len 0. 
                                                 J 2738000.    
 Material 1 - Steel_Nmm
  Type ISOTROPIC       Color 104   Layer 1               #Prop 2       
          Density 0.00000786     Damping 0.           Ref Temp 294.        
STIFFNESS       E 210000.              G 80770.             Nu 0.3         
STRENGTH  Tension 0.            Compress 0.              Shear 0.          
THERMAL     Alpha 0.0000108            K 4.32684      SpecHeat 1.25798    


					spring	mode 1	mode 2	mode 3	mode 4
					N/mm	Hz	Hz	Hz	Hz			
Free Free				0.0	0	0	83	228
					500	2	3	83	228
					5000	7	12	84	228
some bending at 20 Hz			50000	20	39	95	232
pretty much pure bending at 33 Hz	500000	33	101	169	276
					5000000	36.5	142	301	488
					5e6	37	147	329	581

So, as we add constraints to the system the frequencies all increase, just like we'd have hoped.

Of course mounting the springs at the node of the free-free flexural mode might be a lot more interesting.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Dinosaur]

Getting back to the OP, the differential equation for torsional springs and mass behavior is the same first order equation as the ones for axial displacement and mass.

Axial: delta = PL/AE

Torsional: theta = TL/GJ

Look up a solution to an analogous axial vibration problem, and you can substitute the letters and get a torsional solution. The real problem is computing the warping energy. The solutiion only works well for St. Venant torsion problems.