Forced vibration of two degrees of freedom system

Forced vibration of two degrees of freedom system Theory Learning objectives After completing this simulation experiment one should be able to  Model a given real system to an equiva

Forced vibration of two degrees of freedom system

Theory

Learning objectives

After completing this simulation experiment one should be able to

 Model a given real system to an equivalent simplified two degree system and studying its free vibration and forced vibration response with suitable assumptions / idealisations

 Calculate the natural frequency vibration and the amplitude of vibration for the give excitation force of given two degree freedom system

Forced vibration of 2 degree of freedom system

Introduction

Analysing real system as a single degree system is an approximate assumption Because real system has more than just one degree of freedom and are also very rarely linear It requires several degrees of freedom for a meaningful model Hence a simple 1DOF systems analysis

is not much helpful in studying the characteristics of the system A natural extension of single degree of freedom system is to consider the system as two degree of freedom models

A System that requires two independent coordinates to describe their motion is called two degree-of-freedom systems Some practical examples of two degree freedom systems shown in Fig 2, like as forging hammer anvil supported in its base, which in turn is supported on ground isolators, a reciprocating engine with a damper, a relatively light shaft with two disks as in a turbocharger rotor, a generator driving a motor through a coupling For

a two degree of freedom system there are two equations of motion, each one describing the motion of one of the degrees of freedom In general, the two equations are in the form of coupled differential equations

Fig 1: Two DOF systems modelling of a forging hammer

Fig 2: Two DOF system modelling of a motor generator set up

A simplified two degree of freedom system is shown in Fig 3 The motion of the

system is completely described by the coordinates and x 1 (t) and x 2 (t) which defines the

positions of the masses m 1 and m 2 at any time t from the respective equilibrium positions The external F 1 (t) and F 2 (t) respectively The free-body diagrams of the masses and are shown in

Fig 1(b)

Fig 3: A two degree of freedom spring mass system

Equations of motion of 2 degree of freedom system are given by:

1 1 1 1 2( 1 2) 1( )

2 2 2( 2 1) 2( )

Free vibration analysis:

For the free-vibration analysis of the system shown in Fig 1, we set F 1 (t) =F 2 (t) =0 The

equation 1 and 2 reduces to,

1 1 1 1 2( 1 2) 0

2 2 2( 2 1) 0

Let us assume the solutions for x 1 and x 2 under steady state conditions as harmonic excitations with the same frequency and phase angle but with different amplitudes and are given by,

1 1sin( )

x X  t and x2  X2sin( t )

By substituting equation 5 in 3 and 4 we have,

1 2( 1 2) 1 2 2 0

m k k k X k X

2

2 1 ( 2 2) 2 0

k X m k X

Natural frequency of vibration of the system found by solving equation 6 and 7 and is given

by,

1 2

1 2

1/ 2 2

2

1 ,

2

1

4 2

n n

k k m m k

m m

k k m m k k k k k

      

Resonance occurs when the exciting frequency coincides with any one of the natural frequency of the system

Forced vibration analysis:

For the forced-vibration analysis of the system shown in Fig 3, the system is excited by either

of harmonic forcing functions F 1 (t) and F 2 (t) or both But for the specific case, we assume

the system is excited by harmonic force F 1 (t) = F 0 sin ωt

The equation of motion of the two degree freedom system reduces to,

1 1 1 1 2( 1 2) 0sin

2 2 2( 2 1) 0

m x k x x 

When harmonic forcing function acts on the system, the solution consist of transient part and the steady state part In steady state part the vibration of any point in the system take place

at the frequency of excitation Let us assume , for the steady state , the solution as,

1 1sin( )

x X  t and x2  X2sin( t )

(10) Substituting equation 10 in 8 and 9, we will get,

1 2( 1 2) 1 2 2 0

m k k k X k X F

2

2 1 ( 2 2) 2 0

k X m k X

By solving equation 11 and 12 we get,

2

2 2

1 2 1 2 2( 1 2) 1 2

k m X

m m m k m k k k k







(5) and

 2 0 

1 2 1 2 2( 1 2) 1 2

k F X



The above two equations give the steady state amplitude of vibration of two masses

respectively as a function of forcing frequency ω

Let’s try to understand these equations by doing a few simple simulations, go to next tab procedure to find out how to run the simulation to EXPLORE (expR) and to EXPERIMENT (expT) A talking tutorial or a self-running demo with narration can be seen at EXPLAIN (expN)