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TWO DEGREE OF FREEDOM SYSTEM. INTRODUCTION. Systems that require two independent coordinates to describe their motion; Two masses in the system X two possible types of motion of each mass. Example: motor pump system.

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introduction
INTRODUCTION
  • Systems that require two independent coordinates to describe their motion;
    • Two masses in the system X two possible types of motion of each mass.
    • Example: motor pump system.
  • There are two equations of motion for a 2DOF system, one for each mass (more precisely, for each DOF).
  • They are generally in the form of couple differential equation that is, each equation involves all the coordinates.
equation of motion for forced vibration
Equation of motion for forced vibration
  • Consider a viscously damped two degree of freedom spring-mass system, shown in Fig.5.3.

Figure 5.3: A two degree of freedom spring-mass-damper system

equations of motion for forced vibration
Equations of Motion for Forced Vibration

The application of Newton’s second law of motion to each of the masses gives the equations of motion:

Both equations can be written in matrix form as

where [m], [c], and [k] are called the mass, damping, and stiffness matrices, respectively, and are given by

equations of motion for forced vibration1
Equations of Motion for Forced Vibration

And the displacement and force vectors are given respectively:

It can be seen that the matrices [m], [c], and [k] are all 2 x 2 matrices whose elements are known masses, damping coefficient and stiffnesses of the system, respectively.

equations of motion for forced vibration2
Equations of Motion for Forced Vibration
  • Further, these matrices can be seen to be symmetric, so that,

where the superscript T denotes the transpose of the matrix.

  • The solution of Eqs.(5.1) and (5.2) involves four constants of integration (two for each equation). Usually the initial displacements and velocities of the two masses are specified as

x1(t = 0) = x1(0) and 1( t = 0) = 1(0),

x2(t = 0) = x2(0) and 2 (t = 0) = 2(0).

free vibration analysis of an undamped system
Free Vibration Analysis of an UndampedSystem

By setting F1(t) = F2(t) = 0, and damping disregarded, i.e., c1 = c2 = c3 = 0, and the equation of motion is reduced to:

Assuming that it is possible to have harmonic motion of m1 and m2 at the same frequency ωand the same phase angle Φ, we take the solutions as

free vibration analysis of an undamped system1
Free Vibration Analysis of an UndampedSystem

Substituting into Eqs.(5.4) and (5.5),

Since Eq.(5.7)must be satisfied for all values of the time t, the terms between brackets must be zero. Thus,

free vibration analysis of an undamped system2
Free Vibration Analysis of an UndampedSystem

which represent two simultaneous homogenous algebraic equations in the unknown X1 and X2. For trivial solution, i.e., X1 = X2 = 0, there is no solution. For a nontrivial solution, the determinant of the coefficients of X1 and X2 must be zero:

or

slide10

Free Vibration Analysis of an Undamped System

which is called the frequency or characteristic equation. Hence the roots are:

The roots are called natural frequenciesof the system.

free vibration analysis of an undamped system3
Free Vibration Analysis of an UndampedSystem

To determine the values of X1 and X2, given ratio

The normal modes of vibration corresponding to ω12 and ω22 can be expressed, respectively, as

which are known as the modal vectorsof the system.

free vibration analysis of an undamped system4
Free Vibration Analysis of an UndampedSystem

The free vibration solution or the motion in time can be expressed itself as

Where the constants , , and are determined by the initial conditions. The initial conditions are

free vibration analysis of an undamped system5
Free Vibration Analysis of an UndampedSystem

The resulting motion can be obtained by a linear superposition of the two normal modes, Eq.(5.13)

Thus the components of the vector can be expressed as

where the unknown constants can be determined from the initial conditions:

free vibration analysis of an undamped system6
Free Vibration Analysis of an UndampedSystem

Substituting into Eq.(5.15) leads to

The solution can be expressed as

free vibration analysis of an undamped system7
Free Vibration Analysis of an UndampedSystem

from which we obtain the desired solution

example 5 3 free vibration response of a two degree of freedom system
Example 5.3:Free Vibration Response of a Two Degree of Freedom System

Find the free vibration response of the system shown in Fig.5.3(a) with k1 = 30, k2 = 5, k3 = 0, m1 = 10, m2 = 1 and c1 = c2 = c3 = 0 for the initial conditions

Solution: For the given data, the eigenvalue problem, Eq.(5.8), becomes

or

example 5 3 solution
Example 5.3 Solution

By setting the determinant of the coefficient matrix in Eq.(E.1) to zero, we obtain the frequency equation,

from which the natural frequencies can be found as

The normal modes (or eigenvectors) are given by

example 5 3 solution1
Example 5.3 Solution

The free vibration responses of the masses m1 and m2 are given by (see Eq.5.15):

By using the given initial conditions in Eqs.(E.6) and (E.7), we obtain

example 5 3 solution2
Example 5.3 Solution

The solution of Eqs.(E.8) and (E.9) yields

while the solution of Eqs.(E.10) and (E.11) leads to

Equations (E.12) and (E.13) give

example 5 3 solution3
Example 5.3 Solution

Thus the free vibration responses of m1 and m2 are given by

torsional system
TorsionalSystem

Figure 5.6: Torsional system with discs mounted on a shaft

Consider a torsional system as shown in Fig.5.6. The differential equations of rotational motion for the discs can be derived as

torsional system1
TorsionalSystem

which upon rearrangement become

For the free vibration analysis of the system, Eq.(5.19) reduces to

example 5 4 natural frequencies of torsional system
Example 5.4:Natural Frequencies of TorsionalSystem

Find the natural frequencies and mode shapes for the torsional system shown in Fig.5.7 for J1 = J0 , J2 = 2J0 and kt1 = kt2 = kt .

Solution: The differential equations of motion, Eq.(5.20), reduce to (with kt3 = 0, kt1 = kt2 = kt, J1 = J0 and J2 = 2J0):

Fig.5.7:Torsional system

example 5 4 solution
Example 5.4 Solution

Rearranging and substituting the harmonic solution:

gives the frequency equation:

The solution of Eq.(E.3) gives the natural frequencies

example 5 4 solution1
Example 5.4 Solution

The amplitude ratios are given by

Equations (E.4) and (E.5) can also be obtained by substituting the following in Eqs.(5.10) and (5.11).

coordinate coupling and principal coordinates
Coordinate Coupling and Principal Coordinates

Generalized coordinatesare sets of n coordinates used to describe the configuration of the system.

  • Equations of motion Using x(t) and θ(t).
coordinate coupling and principal coordinates1
Coordinate Coupling and Principal Coordinates

From the free-body diagram shown in Fig.5.10a, with the positive values of the motion variables as indicated, the force equilibrium equation in the vertical direction can be written as

and the moment equation about C.G. can be expressed as

Eqs.(5.21) and (5.22) can be rearranged and written in matrix form as

coordinate coupling and principal coordinates2
Coordinate Coupling and Principal Coordinates

The lathe rotates in the vertical plane and has vertical motion as well, unless k1l1 = k2l2. This is known as elastic or static coupling.

  • Equations of motion Using y(t) and θ(t).

From Fig.5.10b, the equations of motion for translation and rotation can be written as

coordinate coupling and principal coordinates3
Coordinate Coupling and Principal Coordinates

These equations can be rearranged and written in matrix form as

If , the system will have dynamic or inertiacoupling only.

Note the following characteristics of these systems:

coordinate coupling and principal coordinates4
Coordinate Coupling and Principal Coordinates
  • In the most general case, a viscously damped two degree of freedom system has the equations of motions in the form:
  • The system vibrates in its own natural way regardless of the coordinates used. The choice of the coordinates is a mere convenience.
  • Principal or natural coordinatesare defined as system of coordinates which give equations of motion that are uncoupled both statically and dynamically.
example 5 6 principal coordinates of spring mass system
Example 5.6:Principal Coordinates of Spring-Mass System

Determine the principal coordinates for the spring-mass system shown in Fig.5.4.

example 5 6 solution
Example 5.6 Solution

Approach: Define two independent solutions as principal coordinates and express them in terms of the solutions x1(t) and x2(t).

The general motion of the system shown is

We define a new set of coordinates such that

example 5 6 solution1
Example 5.6 Solution

Since the coordinates are harmonic functions, their corresponding equations of motion can be written as

example 5 6 solution2
Example 5.6 Solution

From Eqs.(E.1) and (E.2), we can write

The solution of Eqs.(E.4) gives the principal coordinates:

forced vibration analysis
Forced Vibration Analysis

The equations of motion of a general two degree of freedom system under external forces can be written as

Consider the external forces to be harmonic:

where ω is the forcing frequency. We can write the steady-state solutions as

forced vibration analysis1
Forced Vibration Analysis

Substitution of Eqs.(5.28) and (5.29) into Eq.(5.27) leads to

We defined as in section 3.5 the mechanical impedance Zre(iω) as

forced vibration analysis2
Forced Vibration Analysis

And write Eq.(5.30) as:

Where,

forced vibration analysis3
Forced Vibration Analysis

Eq.(5.32) can be solved to obtain:

where the inverse of the impedance matrix is given

Eqs.(5.33) and (5.34) lead to the solution

example 5 8 steady state response of spring mass system
Example 5.8:Steady-State Response of Spring-Mass System

Find the steady-state response of system shown in Fig.5.13 when the mass m1 is excited by the force F1(t) = F10cosωt. Also, plot its frequency response curve.

example 5 8 solution
Example 5.8 Solution

The equations of motion of the system can be expressed as

We assume the solution to be as follows.

Eq.(5.31) gives

example 5 8 solution1
Example 5.8 Solution

Hence,

Eqs.(E.4) and (E.5) can be expressed as

example 5 8 solution2
Example 5.8 Solution

Fig.5.14: Frequency response curves