Chapter 12 vibration of two degree of freedom system

• Unlike single degree of freedom system, where only one co-ordinate and hence one equation of motion is required to express the vibration of the system, in two-dof systems minimum two c

CHAPTER 12 TWO-DEGREE- OF-FREEDOM-SYSTEMS

Introduction to two degree of freedom systems:

• The vibrating systems, which require two coordinates to describe its motion, are called two-degrees-of –freedom systems

• These coordinates are called generalized coordinates when they are independent

of each other and equal in number to the degrees of freedom of the system

• Unlike single degree of freedom system, where only one co-ordinate and hence one equation of motion is required to express the vibration of the system, in two-dof systems minimum two co-ordinates and hence two equations of motion are required to represent the motion of the system For a conservative natural system, these equations can be written by using mass and stiffness matrices

• One may find a number of generalized co-ordinate systems to represent the motion of the same system While using these co-ordinates the mass and stiffness matrices may be coupled or uncoupled When the mass matrix is coupled, the system is said to be dynamically coupled and when the stiffness matrix is coupled, the system is known to be statically coupled

• The set of co-ordinates for which both the mass and stiffness matrix are uncoupled, are known as principal co-ordinates In this case both the system equations are independent and individually they can be solved as that of a single-dof system

• A two-dof system differs from the single dof system in that it has two natural frequencies, and for each of the natural frequencies there corresponds a natural state of vibration with a displacement configuration known as the normal mode Mathematical terms associated with these quantities are eigenvalues and eigenvectors

• Normal mode vibrations are free vibrations that depend only on the mass and stiffness of the system and how they are distributed A normal mode oscillation is defined as one in which each mass of the system undergoes harmonic motion of same frequency and passes the equilibrium position simultaneously

• The study of two-dof- systems is important because one may extend the same concepts used in these cases to more than 2-dof- systems Also in these cases one can easily obtain an analytical or closed-form solutions But for more degrees of

freedom systems numerical analysis using computer is required to find natural frequencies (eigenvalues) and mode shapes (eigenvectors)

The above points will be elaborated with the help of examples in this lecture

Few examples of two-degree-of-freedom systems

Figure 1 shows two masses with three springs having spring stiffness

free to move on the horizontal surface Let

1 2and

1, 2 and

mass m1 and m2 respectively

Using d’Alembert principle for mass m1, from the

free body diagram shown in figure 1(b)

Important points to remember

• Inertia force acts opposite to the

direction of acceleration, so in both the

free body diagrams inertia forces are shown

Figure 1 (b), Free body diagram towards left

• For spring , assuming k2 x2 >x1,

The spring will pull mass m1 towards right by k x2( 2−x )1 and it is stretched by x2−x1

(towards right) it will exert a force of k x2( 2−x )1 towards left on mass Similarly assuming

2

m

x >x , the spring get compressed by an amount x2−x1 and exert tensile force

of k x2( 1− )x2 One may note that in both cases, free body diagram remain unchanged

Now if one uses Lagrange principle,

The Kinetic energy = 2

Now writing the equation of motion in matrix form

e x is the linear displacement of point C and c θc the rotation about point C, the equation of motion of this system can be obtained by using d’Alember’s principle Now summation of all the forces, viz the spring forces and the inertia forces must be equal to zero leads to the following equation

00

Now depending on the position of point C, few cases can are studied below

Case 1 : Considering , i.e., point C and G coincides, the equation of motion can be written as

Hence in this case the system is dynamically coupled but statically uncoupled

Case 3: If we choose , i.e point C coincide with the left end, the equation of

motion will become

Here the system is both statically and dynamically coupled

Normal Mode Vibration

Again considering the problem of the spring-mass system in figure 1 with ,

, , the equation of motion (9) can be written as

We define a normal mode oscillation as one in which each mass undergoes harmonic motion of the same frequency, passing simultaneously through the equilibrium position For such motion, we let

02

Now from equation (1)., it may be observed that for these frequencies, as both the

equations are not independent, one can not get unique value of So one should find a normalized value One may normalize the response by finding the ratio of

ω =λ = in equation (22) and (23) yields the same values,

as both these equations are linearly dependent Here,

Free vibration using normal modes

When the system is disturbed from its initial position, the resulting free-vibration of the system will be a combination of the different normal modes The participation of different modes will depend on the initial conditions of displacements and velocities So for a system the free vibration can be given by

Here A and B are part of participation of first and second modes respectively in the

resulting free vibration and ψ1 and ψ2 are the phase difference They depend on the initial conditions This is explained with the help of the following example

Example: Let us consider the same spring-mass problem (figure 4) for which the natural frequencies and normal modes are determined We have to determine the resulting free vibration when the system is given an initial displacement x1(0)=5,x2(0) 1= and initial velocity x1(0)=x2(0)=0

Any free vibration can be considered to be the superposition of its normal modes For each of these modes the time solution can be expressed as:

2.731

sin1.00

x

t x

x

t x

Normal modes from eigenvalues

The equation of motion for a two-degree-of freedom system can be written in matrix form as

M x+K x=0 (28) where M and K are the mass and stiffness matrix respectively; is the vector of

generalized co-ordinates Now pre-multiplying

where { } 2

From equation (31) it is apparent that the free vibration problem in this case is reduced to

that of finding the eigenvalues and eigenvectors of the matrix A

Example: Determine the normal modes of a double pendulum

θθ

2 2

θθ

2 2

2

2 2

The normal modes can be determined from the eigenvalues

The corresponding principal modes are obtained as

12(2 2 2)

g l g l g l g l

It may be noted that while in the first mode

Both the pendulum moves in the same direction, Figure 6

In the second mode they move in opposite direction

One may solve the same problem by taking x1 and x as the generalized coordinates Here x1 is the horizontal distance moves by mass m1 and x2 is the distance move by mass m2 Figure 7 show the free body diagram of both the masses

1 1

m x

Figure 7

From the free body diagram of mass m2,

Hence

2 2

3

2.41420.4142

l g

g X

The different modes are as shown in the above figure

Example Determine the equation of motion if the double pendulum is started with initial conditions x1(0)=x2(0)=0.5,x1(0)=x2(0)=0

Solution:

The resulting free vibration can be considered to be the superposition of the normal modes For each of these modes, the time solution can be written as

Damped-free vibration of two-dof systems

Consider a two degrees of freedom system with damping as shown in figure

As in the previous case, here also the solution of the above equations can be written as

which is a fourth order equation in s and is known as the characteristic equation of the

system This equation is to be solved to get four roots The general solution of the system can be given by

Example: Find the response of the system as shown in figure 9 considering

2 2 2

00

1

2 2

amplitude ratio is inversely proportional to the mass ratio the system

Similarly one may show for the two-rotor system,

the ratio of angle of rotation inversely proportional to the moment of inertia of the rotors

Forced harmonic vibration, Vibration Absorber

Consider a system excited by a harmonic forceF1sinωtexpressed by the matrix equation

sin0

So it may be observed that the system will have maximum vibration

whenω ω= 1or,ω ω= 2 Also it may be observed that X1=0, when ω2 =2 /k m

Tuned Vibration Absorber

Consider a vibrating system of mass m1, stiffness k1, subjected to a force Fsinωt As studied in case of forced vibration of single-degree of freedom system, the system will have a steady state response given by

which will be maximum when ω ω= n. Now to absorb this

vibration, one may add a secondary spring and mass

system as shown in figure 13

Figure 13

The equation of motion for this system can be given by

= − − (57) where λ1and λ2 are the roots of the characteristic equation Z( )ω = of the free-0vibration of this system., which can be given by

= = Hence, by suitably choosing

the stiffness and mass of the secondary spring and mass system, vibration can be

completely eliminated from the primary system For 2 2

2,

k m

Centrifugal Pendulum Vibration Absorber

The tuned vibration absorber is only effective when the frequency of external excitation equals to the natural frequency of the secondary spring and mass system But in many cases, for example in case of an automobile engine, the exciting torques are proportional

to the rotational speed ‘n’ which may vary over a wide range For the absorber to be effective, its natural frequency must also be proportional to the speed The characteristics

of the centrifugal pendulum are ideally suited for this purpose

Placing the coordinates through point O’, parallel and normal to r, the line r rotates with angular velocity (θ φ  ) +

ˆj

ˆi

O′

r R

O

The acceleration of mass m

( ) /

sin( / )

r So it may be noted that for the centrifugal pendulum the gravity

field is replaced by the centrifugal field Rn2

Torque exerted by the pendulum on the wheel

With the component of equal to zero, the pendulum force is a tension along , given by times the component of

sin( / )

Hence, as the length of the pendulum

(4 )n n R/

small it will be difficult to design it To avoid this one may go for Chilton bifilar

design

Exercise problems

1 In a certain refrigeration plant, a section of pipe carrying the refrigerant vibrated violently

at a compressor speed of 232 rpm To eliminate this difficulty, it was proposed to clamp a

cantilever spring mass system to the pipe to act as an absorber For a trial test, for a 905

gm Absorber tuned to 232 cpm resulted in two natural frequencies of 198 and 272 cpm

If the absorber system is to be designed so that the natural frequencies lie outside the

region 160 to 320 cpm, what must be the weight and spring stiffness?

2 Derive the normal modes of vibration of a double pendulum with same length and

mass of the pendulum

3 Develop a matlab code for determination of free-vibration of a general two-degree

of freedom system

the systems If the system is started with the following initial conditions: x1(0) =x2(0) = X,

5 A centrifugal pump rotating at 500 rpm is driven by an electric motor at 1200 rpm

through a single stage reduction gearing The moments of inertia of the pump impeller

and the motor are 1600 kg.m2 and 500 kg.m2 respectively The lengths of the pump shaft

and the motor shaft are 450 and 200 mm, and their diameters are 100 and 50 mm

respectively Neglecting the inertia of the gears, find the frequencies of torsional

oscillations of the system Also determine the position of the nodes