Consider a geared system as shown in Fig. a. It consists of a driving shaft C which carries a rotor A. It drives a driven shaft D which carries a rotor B, through a pinion E and a gear wheel F. This system may be replaced by an equivalent system of continuous shaft carrying a rotor A at one end and rotor B at the other end, as shown in Fig. b. It is assumed that

- The gear teeth are rigid and do not distort under the tooth loads ( are always in contact),
- There is no backlash in the gearing, and
- The inertia of the shafts and gears is negligible.

Let d

_{1}and d_{2}= Diameters of the shafts C and D respectively,
l

_{1}and l_{2}= Lengths of the shafts C and D,
I

_{A }and I_{B }= Mass moment of inertia of the rotors A and B,
ω

_{A }and ω_{B}= Angular speed of the rotors A and B,
d = Diameter of the equivalent shaft,

l = Length of the equivalent shaft, and

I

If the shafts are not strained beyond the limit of proportionality, each rotor in the geared system will oscillate with simple harmonic motion and there will be a node either in length l

^{’}_{B}_{ }= Mass moment of the inertia of the equivalent rotor B’.If the shafts are not strained beyond the limit of proportionality, each rotor in the geared system will oscillate with simple harmonic motion and there will be a node either in length l

_{1}or in the length l_{2}.
The following two conditions must be satisfied by an equivalent system:

- The kinetic energy of the equivalent system must be equal to the kinetic energy of the original system.
- The maximum strain energy of the equivalent system must be equal to the maximum strain energy of the original system.

In order to satisfy the condition (1) for a given load,

K.E. of section l

_{1}+ K.E. of section l_{3 }= K.E. of section l_{1}+ K. E. of section l_{2}
K.E. of section l

In order to satisfy the condition (2) for a given shaft diameter,

Strain energy of l

Strain energy of l

Where T

_{3}= K.E. of section l_{2}In order to satisfy the condition (2) for a given shaft diameter,

Strain energy of l

_{1 }and l_{3}= Strain energy of l_{1}and l_{2 }Strain energy of l

_{3}= Strain energy of l_{2}Where T

_{2}and T_{3}= Torque on the sections l_{2}and l_{3}, and
θ

_{2}and θ3 = Angle of twist on sections l_{2}and l_{3}.
Assuming that the power transmitted in the sections l3 and l2 is same, therefore,

Thus the single shaft is equivalent to the original geared system, if the mass moment of inertia of the rotor B

^{’}satisfies the equation (i) and the additional length of the equivalent shaft l_{3}satisfies the equation (viii).
Now the natural frequency of the torsional vibration of a geared system (two rotor system) may be determined as discussed below:

Let the node of the equivalent system lies at N as shown in Fig. c. then the natural frequency of torsional vibration of rotor A,

and natural frequency of the torsional vibration of rotor B

^{’},
From the equations (x) and (xi), the value of and may be obtained and hence the natural frequency of the torsional vibrations is evaluated.

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