# Dynamics of Machine

**Topics:**Moment of inertia, Polar moment of inertia, Torque

**Pages:**56 (9363 words)

**Published:**December 24, 2012

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l

Theory of Machines

24

eatur tures Features

1. Introduction. 2. Natural Frequency of Free Torsional Vibrations. 3. Effect of Inertia of the Constraint on Torsional Vibrations. 4. Free Torsional Vibrations of a Single Rotor System. 5. Free Torsional Vibrations of a Two Rotor System. 6. Free Torsional Vibrations of a Three Rotor System. 7. Torsionally Equivalent Shaft. 8. Free Torsional Vibrations of a Geared System.

Torsional Vibrations

24.1. Introduction

We have already discussed in the previous chapter that when the particles of a shaft or disc move in a circle about the axis of a shaft, then the vibrations are known as torsional vibrations. In this case, the shaft is twisted and untwisted alternately and torsional shear stresses are induced in the shaft. In this chapter, we shall now discuss the frequency of torsional vibrations of various systems.

24.2. Natural Frequency of Free Torsional Vibrations

Consider a shaft of negligible mass whose one end is fixed and the other end carrying a disc as shown in Fig. 24.1. Let

θ

=

m I k q

= = = =

Angular displacement of the shaft from mean position after time t in radians, Mass of disc in kg, Mass moment of inertia of disc in kg-m2 = m.k2, Radius of gyration in metres, Torsional stiffness of the shaft in N-m.

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CONTENTS CONTENTS

Chapter 24 : Torsional Vibrations

∴

l

973

Restoring force

= q.θ

=I× d 2θ

... (i)

... (ii) dt 2 Equating equations (i) and (ii), the equation of motion is and accelerating force I× I× d 2θ dt 2 d 2θ dt 2 = − q .θ

or

+ q .θ = 0

Fig 24.1. Natural frequency of free torsional vibrations.

q + ×θ = 0 . . . (iii) ∴ 2 I dt The fundamental equation of the simple harmonic motion is + ω2 .x = 0 dt 2 Comparing equations (iii) and (iv), d 2θ

d 2θ

. . . (iv)

ω=

∴

q I 2π I = 2π ω q

1 1 q = t p 2π I

Time period,

tp =

fn =

and natural frequency ,

Note : This picture is given as additional information and is not a direct example of the current chapter.

A modern lathe can create an artificial hip joint from information fed into it by a computer. Accurate drawings of the joint are first made on a computer and the information about the dimensions fed is directly into the lathe.

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Note :

l

Theory of Machines

T C.θ = J l T C .J = θ l

T = q . . . ∵ θ

The value of the torsional stiffness q may be obtained from the torsion equation, or

∴

where

q=

C.J l

C = Modulus of rigidity for the shaft material, J = Polar moment of inertia of the shaft cross-section,

=

π 4 d ; d is the diameter of the shaft, and 32

l = Length of the shaft.

Example 24.1. A shaft of 100 mm diameter and 1 metre long has one of its end fixed and the other end carries a disc of mass 500 kg at a radius of gyration of 450 mm. The modulus of rigidity for the shaft material is 80 GN/m2. Determine the frequency of torsional vibrations. Solution. Given : d = 100 mm = 0.1 m ; l = 1 m ; m = 500 kg ; k = 450 mm = 0.45 m ; C = 80 GN/m2 = 80 × 109 N/m2 We know that polar moment of inertia of the shaft, J= π π × d 4 = (0.1) 4 = 9.82 × 10 −6 m 4 32 32

∴

Torsional stiffness of the shaft,

C.J 80 ×109 × 9.82 ×10−6 = = 785.6 ×103 N-m 1 l We know that mass moment of inertia of the shaft, q= 2 I = m.k 2 = 500(0.45)2 = 101.25 kg-m

∴

Frequency of torsional vibrations,

fn =

1 q 1 785.6 ×103 88.1 = = = 14 Hz Ans. 2π I 2 π 101.25 2π

Example 24.2. A flywheel is mounted on a vertical shaft as shown in Fig 24.2. The both ends of a shaft are fixed and its diameter is 50 mm. The flywheel has a mass of 500 kg and its radius of gyration is 0.5 m. Find the natural frequency of torsional vibrations, if the modulus of rigidity for the shaft material is 80 GN/m2. Solution. Given : d = 50 mm = 0.05 m ; m = 500 kg ; k = 0.5m; G = 80 GN/m2 = 84 × 109 N/m2 We know that polar moment of inertia of the shaft, J= π π × d 4 = (0.05) 4 m4 32 32

∴

= 0.6...

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