Unit 3
Two Degree of Freedom Systems – Undamped Vibrations
Two degree of freedom system
The System which require two independent co-ordinates to specify its motion at configuration at instant is called two degree of freedom system. – Examples:
There are two equations of motion for a 2DOF system, one for each mass (more precisely, for each DOF). Such equations are called coupled differential equation.
They are generally in the form of couple differential equation that is, each equation involves all the coordinates.
Undamped free longitudinal vibration on 2 DOF spring and mass system
Consider a system with 2 masses and 2 springs having 2 DOF as shown in fig.
FBD of each mass is
From FBD
Rearranging the terms
These are the two equations of motion
Solution for 
and 
are
= Amplitude of vibration of mass 
= Amplitude of vibration of mass 
A system with two degree of freedom vibrates in two principal mode of vibrations corresponding to its two natural frequencies.
Mode shape is nothing but the ratio of amplitude, 
From two equations of motion, we get two mode shapes.
Differentiating two times, we get acceleration
Substituting values of displacement and acceleration in equation of motion for first mass
Substituting values of displacement and acceleration in equation of motion for first mass
Equating both mode shapes
The above equation is a quadratic equation in 
and gives two values of 
i.e. two positive and two negative values of 
Two positive values of 
gives two natural frequencies 
and 
of the system.
Therefore, the above equation is called as frequency equation.
Mode shapes and natural frequency
A system with two degree of freedom vibrates in two principal mode of vibrations corresponding to its two natural frequencies.
Mode shape is nothing but the ratio of amplitude, 
From two equations of motion, we get two mode shapes.
We have, frequency equation as,
If 
Now, if 
These are two natural frequencies of the system.
To find first mode shape, i.e. first ratio of amplitude
Substituting, 
; 
and
This is the first mode shape of system corresponding to first natural frequency 
Now,
To find second mode shape, i.e. second ratio of amplitude
Substituting, 
; 
and 
This is the second mode shape of system corresponding to second natural frequency 
Steps to find natural frequency and mode shapes of a 2 DOF spring coupled system
Step I: Draw FBD for each mass
Step II: Write equation of motion for each mass from FBD. There are 2 equations in 2 DOF system.
Step III: The solution for two equations is in the form
Find acceleration by differentiating the solution two times.
Step IV: Substitute displacement and acceleration in both equation of motions
Step V: Find amplitude ratio 
from each equation and equate them.
Step VI: After equating, we get a quadratic equation in 
which gives two values of 
i.e. two positive and two negative values of 
Two positive values of 
gives two natural frequencies 
and 
of the system.
Step VII: Solve the quadratic equation and find two natural frequencies 
and 
of the system
Step VIII: Substitute both the natural frequencies simultaneously in the equation of any one amplitude ratio 
. This gives mode shapes. One gets two mode shapes corresponding to two natural frequencies.
Consider a same example of two degree of freedom system.
FBD of each mass is
From FBD
Rearranging the terms
Again, rearranging the terms
Above equation can be written in matrix form as
Solution for 
and 
is
Therefore, acceleration is given by
Where,
= Eigen value
And,
Eigen Vector
Substituting displacement and acceleration in equation of motion
Where,
Eigen Vector
= Eigen value
The non-zero solution for eigen vector 
will occur when, 
is a singular matrix
i.e. 
From the above equation, eigen value 
can be obtained
One gets two values of 
Now, 
From two values of one get two natural frequencies.
Now, to find mode shapes,
Substitute two eigen values simultaneously in the following equation
Now, Find amplitude ratio i.e. mode shapes for each eigen value.
Free Torsional Vibration for 2 DOF system
Consider a torsional system as shown in Fig.
(a) Torsional system (b) FBD of disc
The differential equations of motion are derived from fig (b) as
The solution for this torsional system is similar to that of longitudinal system.
Free Torsional Vibration of two rotor system
A two-rotor system consists of a shaft with two rotors A and B at its end as shown in fig.
Node Point
There is a point or a section of the shaft which remains untwisted. This point or section where amplitude of vibration is zero is known as node point or nodal section.
Two rotor system with shafts AB carrying rotors A and B at ends can be considered as equivalent to two single rotor system with:
Let,
and 
= Mass moment of Inertia of rotors A and B
and 
= Torsional stiffness of shaft NA and NB
and 
= Amplitude of vibration of rotor A and rotor B
= Polar Moment of Inertia of shaft
Circular natural frequency for single rotor system is given by
Therefore, circular natural frequency for shaft NA and rotor A
And, circular natural frequency for shaft NB and rotor B
To find position of node i.e. 
The circular frequencies of shaft are same
Total length of shaft is
Now, once 
and 
are known, both the natural frequencies can be determined
Ratio of Amplitude:
From fig (b)
Or
Zero frequency vibration of two rotor system
If two rotors A and B rotate in same direction with same speed, then the shaft is said to vibrate with zero frequency. In this condition, amplitude of vibration at both ends will be in same direction. Such behavior is called as zero frequency behavior.
Torsionally Equivalent system
In two rotor and three rotor system, it is assumed that the diameter of shaft is uniform. But in actual practice, the shaft may have different diameters for different lengths. If the shafts of different diameters are replaced by a theoretically equivalent shaft of a uniform diameter, as shown in fig, such shafts are called as Torsionally equivalent shaft.
Let 
, 
and 
be the angles of twist for shafts of length 
, 
and 
with diameters 
, 
and 
respectively.
Let 
, be the angle of twist fortorsionally equivalent shaft of length 
with diameter 
.
Total Angle of twist for actual shaft
Angle of twist for equivalent shaft
Angle of twist of equivalent shaft =Total Angle of twist for actual shaft
Now,
Generally, diameter of equivalent shaft is taken as one of the diameters of actual shaft. For, e.g. 
In many applications, the body is subjected to combined rectilinear and angular motions. For example, when brakes are applied on a moving car, two motions of car occur simultaneously.
Consider a system as shown in fig
Let ‘I’ be the MI of a body with mass ‘m’ about CG
be the angular displacement at any instant and
be the linear displacement at that instant.
From fig (b) compression of spring can be given as
FBD of the system is shown below
From fig., the differential equation of motion for the system are
And
Or
The above equations have both terms 
and 
. Such equations are called as coupled equations.
Solution for these two equations under steady state conditions are given by
Where 
and 
are amplitudes of vibration for linear and angular motion respectively
Therefore,
Substituting values of 
, 
, 
and 
in below equation
Now, substituting values of 
, 
, 
and 
in below equation
Equating both equations
Solving above equation, we get
The above frequency equation is the quadratic equation in 
Solving the above equation for 
one get two positive and two negative values for 
.
These two positive values of 
are two natural frequencies for the system.
Many a times, torsional vibration is observed in a geared system. The geared system is shown in figure (a).
The shaft 1 carries a rotor A at one end and pinion to another end. The shaft 2 carries a gear meshing with pinion at one end and rotor B at another end.
The gear system may be replaced by an equivalent two rotor system, if the inertia is neglected. This system consists of continuous shaft with rotor A at one end and rotor B’ at another end. (fig. b)
If inertia is not neglected, then the equivalent system will be a three rotor system as shown in fig (c).
If Inertia of gear is neglected.
The geared system will b equivalent to two rotor system if:
Consider,
KE of Geared system = KE of Equivalent System
KE of section 
+ KE of section 
= KE of section 
+KE of section
KE of section 
= KE of section
Where,
gear ratio
Now consider,
Strain Energy of Geared system = Strain Energy of Equivalent system
Strain Energy of section 
+ Strain Energy of section 
= Strain Energy of section 
+Strain Energy of section
Strain Energy of section 
= Strain Energy of section
But,
Let,
polar moment of Inertia of shaft 2 = 
polar moment of Inertia of shaft 2 = 
Total length of equivalent shaft is
If moment of Inertia is considered,
An additional rotor C’ must be placed at distance 
from the rotor A on the shaft.
Such that,
Where,
and 
are the MI of pinion C and gear D
The system will act as three rotor system with rotor A, rotor B’ and rotor C’
References
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2. Bell, L. H. and Bell, D. H., Industrial Noise Control – Fundamentals and Applications, Marcel Dekker Inc.
3. Alok Sinha, Vibration of Mechanical System, Cambridge university Press , India
4. Debabrata Nag, Mechanical Vibrations, Wiley India Pvt. Ltd, New Delhi.
5. Kelly S. G., Mechanical Vibrations, Schaums outlines, Tata McGraw Hill Publishing Co.Ltd., New Delhi.
6. Meirovitch, L., Elements of Mechanical Vibrations‖, McGraw Hill.
7. Ver, Noise and Vibration Control Engineering, Wiley India Pvt. Ltd, New Delhi.
8. Bies, D. and Hansen, C., Engineering Noise Control - Theory and Practice, Taylor and Francis.
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