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# Natural Frequency For Torsional Vibration

03/05/2017 3:35 PM

Hi All:

I am trying to calculate natural frequency for torsional vibration of fixed fixed beam using following formula in US system.

=1/2*3.142 sqrt(K*g/mr^2) hz

Where K=J*G/L

and G=Modulus of rigidity

g=acceleration due to gravity=386.4 in/se2

m=mass hanging on the beam at distance 'r' from the longitudenal axis

My question is how does the equation above (if its right) is affected if I change the beam from fixed fixed to fixed -free.

thanks

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Guru

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#1

### Re: natural frequency for torsional vibration

03/05/2017 4:15 PM
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#2

### Re: natural frequency for torsional vibration

03/05/2017 4:30 PM

Right off the top of my head, I would say that going from fixed-fixed to fixed-free would halve the restoring force so the frequency would be reduced to a factor 1/sqrt(2) or about 0.707.

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#3

### Re: natural frequency for torsional vibration

03/05/2017 4:35 PM

Hi thanks for the reply. But the website you pointed has equation for torsional vibration of shaft for cantiliver only. What about ifs its fixed fixed or simple -fixed for torsion (not bending which website has equations)

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#4

### Re: natural frequency for torsional vibration

03/05/2017 4:41 PM

Hi:

I am not getting how it will reduce by .707, because per my equation stiffness, K

K=J*G/L and this does not change based on support system (fixed-fixed pt fixed -free).

thanks

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#5

### Re: natural frequency for torsional vibration

03/05/2017 4:47 PM

This might help:

http://www.sciencedirect.com/science/article/pii/S0038080614001310

Table 1.

Effects of rotator (ψR) and torsional (ψT) inertias on frequency parameter Ωi of beams with different end constraints.

End constraintSwitch parameter
Frequency parameter

Ωi a and b

ψR

ψT

i=1

i=2

i=3

i=4

i=5

Free–free000.64230.65140.9098
100.64230.64550.8924
11

0.5774

0.64230.64550.8924

1.6317

Free–hinged000.64550.78391.5254
100.64470.77551.4885
110.64470.7755

0.9569

1.4885

2.3608

Free–clamped000.65320.90271.8175
100.65190.89151.7716
110.65190.8915

0.9569

1.7716

2.3608

Hinged–hinged000.70721.27382.4296
100.70441.25662.3695
110.70441.2566

1.6317

2.3695

3.1063

Hinged–clamped000.78391.52542.8086
100.78041.50402.7398
110.78041.5040

1.6317

2.7398

3.1063

Clamped–clamped000.90981.81663.2192
100.90571.79083.1427
110.9057

1.6317

1.7908

3.1063

3.1427

a

Results for a=0.2, t=0.1, s=0.43, Kw=500 and Ks=0.5.

b

Bold letters indicate the frequency parameters of beams vibrating in torsional mode.

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#6

### Re: natural frequency for torsional vibration

03/05/2017 5:02 PM
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#7

### Re: natural frequency for torsional vibration

03/05/2017 10:21 PM

My thought was if both ends were clamped, that the stiffness would be twice as much as if one end were clamped and the other free. The stiffness constant, K, is under the square root in the formula for frequency, which is where the sqrt(1/2) = .707 came from.

Maybe another way of looking at it is that a beam clamped at one end would be equivalent to a beam twice as long clamped at both ends.