torsional natural frequency - beams
torsional natural frequency - beams
(OP)
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.
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.





RE: torsional natural frequency - beams
Tobalcane
"If you avoid failure, you also avoid success."
RE: torsional natural frequency - beams
Cheers
Greg Locock
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: torsional natural frequency - beams
If you have a beam clamped at one end and supporting J at the other end, of course the resonant frequency is w = sqrt(K/J).
If the beam is not massless but instead has inertia j, you can calculate as above, except subsitute
Jeffective = J + 0.3 *j (similar to the more familiar Meffective = M+0.24*m)
If you have a disk on each end of a massless shaft (not clamped at either end), the resonant frequency can be calculated using:
Jeffectve = J1*J2/(J1+J2)
Formula for torsional equivalent of 2-mass/2-spring system is shown in Harris' Shock and Vib Handbook page 1.11 (too long for me to reproduce... but analogous to the formula for linear vibration 2mass/2spring system.
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RE: torsional natural frequency - beams
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RE: torsional natural frequency - beams
Actually, it is not always the zero's. Could be either one. Need to look closely at the specific system and transfer function to interpret the poles and zero's.
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RE: torsional natural frequency - beams
Anyway. Holzer method, for shaft/flywheel systems, or Rayleigh Ritz if we are feeling funky.
Cheers
Greg Locock
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: torsional natural frequency - beams
You need only 6 simple rules in order to solve any sinsoidal steady state lumped mechanical vibration problem using superior
1 - mechancial force F plays the role of electrical current i
2 - mechanical velocity difference V plays the role of electrical voltage difference v
3 - Let mechanical impedance be defined as Z=F/V
(this is reciprocal of Harris, but provides a direct analogy to electrical impedance).
4 - A mechanical spring (k) acts like an electrical inductor of inductance L=1/k
The impedance is Zk = j*w/*k
5 - A mechanical damper (c) acts like an electrical resistor of resistance R = 1/c
The impedance is Zc = 1/c
6 - A mechanical mass (m) acts like an electrical capacitor of capacitance C =m **
The impedance is Zm = 1/(j*w*k)
** one terminal of this mass/capacitance is connected to ground.
This analogy also preserves the definition of instantaneous power.
Pmech(t) = V(t)*F(t) vs Pelec(t) = v(t) i(t)
Also if you look carefully at the max stored energy in the spring and mass elements, you will find they are exactly what is predicted by electical calc of Einductor = 0.5*L*i^2 and Ecapacitor = 0.5*C*v^2.
Derivation is available upon request.
As an example of the power of the method, the linear-motion version of the Jeq =J1*J2/(J1+J2) cited above for two masses on the end of a shaft with no connection to ground is effortlessly derived as shown in attached.
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RE: torsional natural frequency - beams
The impedance is Zm = 1/(j*w*m)
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RE: torsional natural frequency - beams
For torsional, there is no analogy to pinned. Harris' Shock and Vib Handbook section 7 has solutions for simple circular beam with boundary conditions: free/clamped. There might be enough info in there to figure out the solutions for free/free and clamped/clamped, but I'm not sure.
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RE: torsional natural frequency - beams
RE: torsional natural frequency - beams
clamped/free
wn = (2n-1)*pi *sqrt(G/rho) / (2*L)
free/free:
wn = n*pi *sqrt(G/rho) / L
I don't think these can be translated for non-circular cross sections simply by substituting the material properties G and rho.
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RE: torsional natural frequency - beams
3 - Let mechanical impedance be defined as Z=v/f
Using that definition, the derivation of the mechanical impedances and their electrical equivalencies:
Spring:
f = k x
f = k (v/jw)
Zk = v/f = j*w / k
By comparison to ZL = j*w*L, we find L = 1/k
Mass
f = m a
f = m* (jw * v)
Zm = v/f = 1/ (jw*m)
By comparison to ZC = 1/(j*w*C), we find C = m
Damping
f = c v
Zc = v/f = 1/C
By comparison to ZR = R, we find R = 1/c
(ok, I think I'm done).
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RE: torsional natural frequency - beams
Yes - page 431.
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RE: torsional natural frequency - beams
Rao's Mechanical Vibrations 3rd edition page 520 addresses all three cases of boundary conditions for cylindrical rods:
FIXED/FREE: wn=(2*n+1)*pi*c/(2*L) n=0,1,2...
FREE/FREE: wn = n*pi*c/L, n=0,1,2
FIXED/FIXED: wn=n*pi*c/L, n=1,2,3...*
(*note the FIXED/FIXED has different choice of n)
where c = sqrt(G/rho)
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RE: torsional natural frequency - beams
velocity-velocity
mass=mass
damping=damping
spring=spring
derivation available on request.
(grins)
Cheers
Greg Locock
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: torsional natural frequency - beams
But as Greg has explained before, the very first mode of that free/free is really 0 hz. It corresponds to constant speed rotation of the whole beam. (or we have the same phenomenon for longitudinal vib of free/free beam where 0 hz corresponds to translatio of the beam). Rao communicates the same thing by telling us that n starts at 0 which computes to 0 for the first free free mode.
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RE: torsional natural frequency - beams
Cheers
Greg Locock
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: torsional natural frequency - beams
I have yet to see anyone mechanical who said: "Wow - that is really neat.". But for me when I read it and started using it, my reaction was that it really was really neat. Somehow it clicks better. It is no longer equations on paper but graphical understandable representation and vector relationships.
To each his own, I guess.
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RE: torsional natural frequency - beams
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RE: torsional natural frequency - beams
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
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: torsional natural frequency - beams
thread384-155945: trick question on beams
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RE: torsional natural frequency - beams
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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RE: torsional natural frequency - beams
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
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: torsional natural frequency - beams
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RE: torsional natural frequency - 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.
CODE
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
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: torsional natural frequency - beams
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.