Question

A 4-cylinder engine coupled to a propeller are approximated to a three rotor system in which the engine is equivalent to a rotor of moment of inertia 800 kgm2 , the flywheel second rotor 320 kgm2 and the propeller to a 3-rd rotor 20 kgm2 . The first and second rotors being connected by 50 mm diameter 2 meter long shaft and the second and third rotors connected by 25 mm diameter and 2 meter long shaft. Neglecting the inertia of the shaft taking the modulus of rigidity 80 GN/m2 , determine

1. Natural frequency of torsional oscillations, and

2. The position of nodes.

Answer 1

Given IA = 800 kg-m2; IB = 320kg -m2; d1 = 50 mm = 0.05 m; l1 = 2; d2 = 25 mm = 0.025 m; I2 = 2 m; C = 80 x 109N/m2

1. Natural frequency of torsional oscillations,

First of all, replace the original system, as shown in fig (a) by n equivalent system as shown in fig (b). It is assumed that the diameter of equivalent shaft is d1 = 50 mm = 0.05 m

We know that length of equivalent shaft,

Now let IA = Distance of node N1 from rotor A, and

Ic = Distance of node N2 from rotor C.

we know that

We see that when IC = 34.42 m, then IA = 0.86 m. This gives the position of single node for IA = 0.86 m. The value of IC = 20.88 m and corresponding value of IA = 0.52 m gives the position of two nodes, as shown in Fig. (c).

We know that polar moment of inertia of the equivalent shaft,

Natural frequency of torsional vibrations for a single node system,

Similarly natural frequency of torsional vibrations for a two node system,

2. The position of nodes.

We have already calculated that for a two node system on an shaft, IC = 20.88 m from the propelle.

Corresponding value of IC in an original system from the propeller.

Therefore one node occurs as a distance of IA = 0.52 m from the engine and the other node at a distance of IC = 1.3 m from the propeller.