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 d1 and d2 = Diameters of the shafts C and D respectively,
l1 and l2 = Lengths of the shafts C and D,
IA and IB = Mass moment of inertia of the rotors A and B,
ωA and ωB = Angular speed of the rotors A and B,
d = Diameter of the equivalent shaft,
l = Length of the equivalent shaft, and
I’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 l1 or in the length l2 .
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 l1 or in the length l2 .
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 l1 + K.E. of section l3 = K.E. of section l1 + K. E. of section l2
K.E. of section l3 = K.E. of section l2
In order to satisfy the condition (2) for a given shaft diameter,
Strain energy of l1 and l3 = Strain energy of l1 and l2
Strain energy of l3 = Strain energy of l2
Where T2 and T3 = Torque on the sections l2 and l3, and
In order to satisfy the condition (2) for a given shaft diameter,
Strain energy of l1 and l3 = Strain energy of l1 and l2
Strain energy of l3 = Strain energy of l2
Where T2 and T3 = Torque on the sections l2 and l3, and
θ2 and θ3 = Angle of twist on sections l2 and l3.
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 l3 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.
No comments:
Post a Comment