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 I_A = 800 kg-m²; I_B = 320 k_{g -}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  I_A = Distance of node N₁ from rotor A, and

I_c = Distance of node N₂ from rotor C.

we know that

We see that when I_C = 34.42 m, then IA = 0.86 m. This gives the position of single node for IA = 0.86 m. The value of I_C = 20.88 m and corresponding value of I_A = 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, I_C = 20.88 m from the propelle.

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

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