The analysis of torsional vibrations is important in rotary drives involving large inertia. A major application of torsional vibration is in diesel engine drives with fluctuating loads. It is important to note that the operating speed of the engineer should nowhere be near to the natural frequency of the system.

This chapter deals with the determination of the frequency of torsional vibration Of the various systems such as:

1] Natural frequency of free torsional vibrations.

2] Effect of inertia on torsional vibrations.

3] Free torsional vibrations of a single rotor system.

4] Free torsional vibration of a two rotor system.

5] Free torsional vibration of a three rotor system and

6] Free torsional vibration of a geared system.

Case 2: Effect of inertia on torsional vibrations:

Prove that for finding the natural frequency of a torsional system, the mass moment of inertia of the shaft can be taken into account by adding one third of its inertia to the disc inertia.

Solution:

Let ‘L’ be the length of the shaft and let $\rho$ be the mass of the shaft per unit length. Mass of shaft, M = $\rho$L

Mass moment of inertia of the shaft will be:

$I_s = M K^2 \ = \rho L K^2$

Where k is the radius of gyration.

Consider a small element ‘dy’ of the shaft at a distance ‘y’ from the support.

The mass of this element is ‘dM’ , where dM = $\rho$ dy.

The mass moment of inertia of this shaft section about the torsional axis would be:

$dI_s \ = \ dM \times k^2$

$= \rho^2 \ dy$

Considering that the free end of the shaft has moved by ‘0’, the displacement at the fixed point will be zero. From similar triangles, at any point, at a distance ‘y’, the displacement is given by $’\phi’$ such that:

$\frac{\phi}{y} \ = \ \frac{θ}{L}$

$\therefore$ $\phi \ = \ \frac{yθ}{L}$

Hence, angular velocity of shaft section having mass moment of inertia $dI_s$ will be $\phi$

$\phi = \frac{y\dot{θ}}{L}$

The kinetic energy of the system will be due to the inertia of the disc 1 and the shaft having a moment of inertia $I_s$

$KE \ = \ KE_{shaft} \ + \ KE_{disc}$

$= \int_0^L \ \frac{1}{2} \ dI_s \ \phi^2 \ + \ \frac{1}{2} \ I \ \dot{θ}^2$

$= \frac{1}{2} I \theta^2 + \frac{1}{2} \int^L_0 \ pk^2 dy \times (\frac{y \dot{θ}}{L})^2$ (substituting $\phi$)

$= \frac{1}{2} I \theta^2 + \frac{1}{2} \int^L_o pk^2 \times \frac{y^2 \dot{θ}^2}{L^2} dy$

$= \frac{1}{2} I \theta^2 + \frac{1}{2} \frac{pk^2 \theta^2}{L^2} \int^L_0 y^2 dy$

$= \frac{1}{2} I \theta^2 + \frac{1}{2} \frac{pk^2 \theta^2}{L^2} [\frac{y^3}{3}]_0^L$

$= \frac{1}{2} I \theta^2 + \frac{1}{2} \frac{pk^2 \theta^2}{L^2} \frac{L^3}{3}$

$= \frac{1}{2} I \theta^2 + \frac{1}{6} p L k^2 \theta^2 $

$= \frac{1}{2} I \theta^2 + \frac{1}{6} I_s \theta^2$ (as $I_s = p L k^2$)

$KE \ = \ \frac{\theta^2}{2} ( I + \frac{I_s}{3})$

$PE \ = \ \frac{1}{2} K_T \theta^2$

U = KE + PE

$U \ = \ \frac{\theta^2}{2} ( I + \frac{I_s}{3}) + \frac{1}{2} K_T \theta^2$

Differentiating and using energy method, we get,

$\frac{dU}{dt} \ = \ \frac{2}{2} \dot{θ} \ddot{θ} (I + \frac{I_s}{3}) + \frac{2}{2} K_T \dot{θ} \ddot{θ} = 0$

$\therefore \ddot{θ} (I + \frac{I_s}{3}) + K_T \theta \ = 0 $

$w_n = \sqrt{ \frac{K_T}{( I + \frac{I_s}{3})}}$

$\therefore \ddot{θ} + \frac{K_T}{(I + \frac{I_s}{3})} \theta \ = \ 0$

$f_n \ = \ \frac{1}{2 \pi } \sqrt{ \frac{K_T}{( I + \frac{I_s}{3})}}$

Which shows that moment of inertia of the shaft can be taken into account by adding one third of its inertia to the disc inertia.

Case 3: FREE TORSIONAL VIBRATION OF A SINGLE ROTOR SYSTEM.

LET $I_B$ be the mass MI of single rotor at B.

I = Polor M.I. of shaft.

L = Length of the shaft.

$K \ = \ \frac{T}{\theta} \ = \ \frac{CJ}{l} =$ Torsional stiffness of the shaft where $\theta = $ angular displacement at any given instant which is harmonic in nature.

Let $\theta \ = \ \beta \ sin \ wt$ (assume)

Where $\beta$ = Amplitude of vibration,

W = natural frequency of vibration of rotor.

Then considering force acting on rotor.

By Newton’s 2nd law of motion, we can write

$I \ddot{θ} = \ - K \ \theta$

i.e. $I \ddot{θ} + \ K \theta \ = \ 0$

From above equation, natural frequency of vibration.

$w_n \ = \ \sqrt{ \frac{K}{I}}$

i.e. $w_n \ = \ \sqrt{ \frac{C J }{I l}}$