Mechanical Vibrations

Transcription

Kingdom of Saudi Arabia
Ministry of Higher Education
Umm Al-Qura University
College of Engineering & Islamic Architecture
Mechanical Engineering Department
Mechanical Vibrations
804 420 – 3
Lecture No. 9
Dr. Mohammad S. Alsoufi
BSc, MSc PhD
Room No.: 1080
Tel.: 00966 (012) 5270000 Ext.: 1163
E-mail: [email protected]
4. Harmonic Motions
9. Two Degree of Freedom Systems
9.1 Introduction
 Systems that require two independent coordinates to describe their motion are called “twodegree-of-freedom systems”.
 Some examples of systems having two degrees of freedom were shown in Figure.
Two-degree-of-freedom systems (2DoF)





Consider the automobile shown in Figure (a).
For the vibration of the automobile in the
vertical plane, a two-degree-of-freedom model
shown in Figure (b) can be used.
The body is idealized as a bar of mass m and
mass moment of inertia
supported on the
rear and front wheels (suspensions) of stiffness
and .
The displacement of the automobile at any
time can be specified by the linear coordinate
x(t) denoting the vertical displacement of the
C.G. of the body and the angular coordinate
( ) indicating the rotation (pitching) of the
body about its C.G.
Alternately, the motion of the automobile can
be specified using the independent coordinates,
( )
( ) of points A and B.




Multistory building subjected to an earthquake
Automobile
Consider the motion of a multistory building
under an earthquake. For simplicity, a twodegree-of-freedom model can be used as shown in
Figure.
The building is modeled as a rigid bar having a
mass m and mass moment of inertia .
The resistance offered to the motion of the
building by the foundation and surrounding soil is
approximated by a linear spring on stiffness k and
a torsional spring of stiffness .
The displacement of the building at any time can
be specified by the horizontal motion of the base
x(t) and the angular motion ( ) about the point
O.



Consider the system shown in Figure (a), which illustrates the packaging of an instrument of
mass m.
Assuming that the motion of the instrument is confined to the xy-plane, the system can be
modeled as a mass m supported by springs in the x and y directions, as indicated in Figure
(b).
Thus the system has one point mass m and two degrees of freedom, because the mass has
two possible types of motion (translations along the x and y directions).
Packaging of an instrument.
 The general rule for the computation of the number of degrees of freedom can be stated as
follows:
(Number of masses in the system)
Number
×
=
of degrees of freedom of the system
(Number of possible types of motion of each
mass)
9.2 Equations of Motion for Forced Vibration
 Consider a viscously damped two-degree-of-freedom spring-mass system, shown in Figure
(a).



( ), which
The motion of the system is completely described by the coordinates ( )
define the positions of the masses m1 and m2 at any time t from the respective equilibrium
positions.
( ), act on the masses m1 and m2 respectively.
The external forces ( )
The free-body diagrams of the masses m1 and m2 are shown in Figure (b).

The application of Newton’s second law of motion to each of the masses gives the equations
of motion:
(
) ̇
(
)
̇
̈
̈





̇
(
,⏟-
⃗̇ ( )
,⏟-



)
[
]
⃗( )
⃗( )
Where
, -

(
It can be seen that 1st equation contains terms involving x2 (namely,
̇ and
).
nd
It can be seen that 2 equation contains terms involving x1 (namely,
̇ and
).
Hence, they represent a system of two coupled “second-order differential equations”.
The motion of the mass m1 will influence the motion of the mass m2 and vice versa.
Both equations can be written in “matrix form” as:
,⏟- ⃗̈ ( )

) ̇
, -
0
1
, -
[
]
⃗( ) and ⃗( ) are called the “displacement and force vectors”, and are given by
⃗( )
{
( )
( )
⃗( )
{
( )
( )
It can be seen that [m], [c], and [k] are all 2×2 matrices whose elements are the known
masses, damping coefficients, and stiffnesses of the system, respectively.
These matrices can be seen to be “symmetric”, so that
, -
, -
, -
, -
, -
, -
Where the superscript T denotes the transpose of the matrix.
9.3 Free-Vibration Analysis of an Undamped System
 The solution of 1st and 2nd equations involves four constants of integration (two for each
equation).
 We shall first consider the “free vibration solution”.
 For the free-vibration analysis of the system shown in Figure (a), we set
( )

If the damping is disregarded

The1st and 2nd equations of motion reduce to
̈ ( )
(
) ( )
̈ ( )




( )
( )
(
) ( )
We are interested in knowing whether m1 and m2 can oscillate harmonically with the same
frequency and phase angle but with different amplitudes.
Assuming that it is possible to have harmonic motion of m1 and m2 at the same frequency
and the same phase angle . The solutions becomes as
( )
(
)
( )
(
)
Where
and
are constants that denote the maximum amplitudes of ( ) and
is the phase angle.
( )
(
) and
( )
(
) into
Substituting
(
) ( )
( )
( ) (
) ( )
and
̈ ( )
,*
,

( )
(
*
)+
(
)+
(
-
(
( ) and
̈ ( )
)
)
Since the above equation must be satisfied for all values of the time t, the terms between
brackets must be “zero”. This yields…
*
(
)+
*
(
)+






Which represent two simultaneous homogenous algebraic equations in the unknowns
.
For “trivial solution”
, this implies that there is no vibration.
For “nontrivial solution” of
, the determinant of the coefficients of
be zero:
*
(
)+
[
]
*
(
)+
Or
(
)
*(
)
(
) +
*(
)(
)
+
(
)
(
)

8
(
)(
(
)
9]
)
The values of
and
remain to be determined.
These values depend on the natural frequencies
and .
We shall denote the values of
and
( )
 Corresponding to
as
and
( )
 Corresponding to
as
and
Since *
(
)+
homogenous, only the ratios

9
The roots are called “natural frequencies” of the system.
This shows that it is possible for the system to have a nontrivial harmonic solution of the
form of
( )
(
)
( )



must
This equation is called the “frequency” or “characteristic” equation because its solution
yields the frequencies or the characteristic values of the system.
The roots are given by
(
)
(
)
8
9
[8


and
For
(
( )
) and
( )
*
and
( )
( )
(
( )
( )
(
) can be found.
and
( )
(
(
)
)
( )
(
(
)
)
(
(
)
(
)
(
)
)
)+
is



The normal modes of vibration corresponding to
as
)
⃗(
)
{
( )
)
{
( )
( )
⃗ ( )( )

( )
{
( )
}
( )
}
{
( )
}
{
( )
( )
( )
( )
( )
( )
}
{
(
( )
(
( )
}
{
)
)
(
( )
)
(
)
}
}
⃗ ( )( )
⃗ ( )( )
Where and are constants.
Since ⃗ ( ) ( ) and ⃗ ( ) ( ) already involve the unknown constants
choose
with no loss of generality.
The components of the vector ⃗( ) can be expressed as
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
Where the unknown constants
,
conditions:
(
)
( )
(

{
( )
The resulting motion can be obtained by a linear superposition of the two normal modes
( )

}
( )
⃗( )


( )
can be expressed, respectively,
The vectors ⃗ ( ) and ⃗ ( ) which denote the normal modes of vibration are known as the
“modal vectors” of the system.
The free-vibration solution or the motion in time can be expressed as
⃗ ( )(

( )
⃗(
and
)
Then,




(
̇ (
(
̇ (
)
)
)
)
( )
(
)
( )
(
)
( )
,
and
( )
( )
( )
( )
(
̇ (
)
̇ ( )
( )
( )
 we can
)
can be determined from the initial
̇ ( )
( )
and
)
)
( )
( )
(
̇ (
( )
( )
( )
( )


The solution can be expressed as
( )
8
( )
8
( )
( )
̇ ( )
(
̇ ( )
9
)
( )
9
( )
8
( )
8
̇ ( )
(
From which we obtain the desired solution:
( )
(
)
)
( )
3
( )
6*
( )
(
[2
[2
6*
*
( )+
( )
3
( )
( )+
( )
2
2
̇ ( )
( )
*
( )
}
8
,
( )
{
( )
}
8
̇ ( )+
,
7
3 ]
̇ ( )
( )
{
3 ]
̇ ( )
( )
̇ ( )+
7
̇ ( )
9
( )-
̇ ( )
̇ ( )
9
( )
( )-
Example No.1: Free-Vibration Response of a Two-Degree-of-Freedom System
Find the free-vibration response of the system shown in Figure with:
 k1 = 30, k2 = 5 and k3 = 0,
 m1 = 10 and m2 = 1
 c1 = c2 = c3 = 0
For the initial conditions:
( )
̇ ( )
( )
̇ ( )
( )
9
̇ ( )
9
)
Solution No.1:
 For the given data
[
]{
[

By
setting
]{
the
[
determinant
]{
}
}
of
coefficient
matrix
in
2 3 to zero, we obtain the frequency equation
from which the natural frequencies can be found as

leads to
, while
thus, the normal modes (or eigenvectors) are given by
( )
( )
⃗(
yields
( )
⃗(
)
{
( )
}
2 3
( )
)
{
( )
}
2
3
( )
( )
,
( )
( )
The free-vibration responses of the masses m1 and m2 are given by
( )
( )

2 3
2 3
the


}
( )
(
)
( )
(
)
Where the unknown constants
conditions:
(
)

(
)

)
 ̇ (
)
 ̇ (
( )
,
( )
,
( )
( )
and
( )
(
)
(
)
can be determined from the initial
( )
( )
( )
( )
( )
( )
( )

The
solution
( )
of
(
( )
)
( )
The solution of ̇ (
( )
( )
( )
and ̇ (
yields
( )
( )
So,
( )

( )
Thus the free-vibration responses of masses m1 and m2 are given by
( )
( )

)
( )
)
( )

(
and
yields
( )

( )
The graphical representation of the free-vibration responses is shown below
)
9.4 Torsional System
 Consider a torsional system consisting of two discs
mounted on a shaft, as shown in Figure.
 The three segments of the shaft have rotational spring
constants kt1, kt2 and kt3 as indicated in the figure.
 Also shown are the discs of mass moments of inertia J1
and J2, the applied torques Mt1 and Mt2 and the
rotational degrees of freedom and
 The differential equations of rotational motion for the
discs J1 and J2 can be derived as
̈
(
̈

(
)
Which upon rearrangement become
̈
(
)
̈

)
(
)
For the free-vibration analysis of the system, the above
equation reduces to
̈
(
̈
Torsional system with discs
mounted on a shaft
)
(
)
Example No.2: Natural Frequencies of a Torsional System
Find the natural frequencies and mode shapes for the torsional
system shown in Figure for:




Torsional system
Solution No.2:
 From the given data
̈
̈

Rearranging and substituting the harmonic solution
( )
(

gives the frequency equation:

The solution of
gives the natural frequencies
√

)
(
√
)
√
(
√
)
The amplitude ratios are given by
( )
(
( )
)
√
( )
(
√
)
( )
9.5 Forced-Vibration Analysis
 The equations of motion of a general two-degree-of-freedom system under external forces
can be written as
̈
1{ }
̈
0

Equations 
̈
̈
(
̇
̇
1{ }
̇
0
) ̇
(
̇
) ̇
[
(
]2 3
(
)
{ }
)
 Can be seen to be special cases of the equations of motion of a
general two-degree-of-freedom system under external forces, with
and
and

We shall consider the external forces to be harmonic
( )

Where
is the forcing frequency. We can write the steady-state solutions as
( )

Where
and
are, in general, complex quantities that depend on and the system
parameters. Substitution of ( )
and ( )
into equations of motion of
a general two-degree-of-freedom system under external forces leads to
[

(
(
)
)
)
]{
)
(
we define the mechanical impedance
(

(
(
)-
[
(
(
, (
)- ⃗
)
)
(
(
⃗
Equation , (
)- ⃗
)
]
)
{
}
{
}
, (
)-
⃗
Where the inverse of the impedance matrix is given by
, (

⃗
⃗ can be solved to obtain
⃗

}
) as
⃗

{
)
Where
, (
}
)-
(
)
(
)
(
)
(
)
(
The solution
(
(
)
)
( )
(
(
)
)
)
(
)
(
)
[
( )
( )
(
)
( )
( )
(
)
Example No.3: Steady-State Response of Spring-Mass System
Find the steady-state response of the system shown in Figure,
when the mass m1 is excited by the force ( )
.
Also, plot its frequency-response curve.
A two-mass system subjected to harmonic force
( )
( )
( )
]
( )
Solution No.3:
 The equations of motion of the system can be expressed as
̈
1{ }
̈
0









( )
( )
2
3
( )
(
( )
Since
8
(
)
(
)
(
)
(
)(
)
(
)
(
)(
)
)
[8

12 3
( )
( )

0
(
(
)
)
9
(
)
9
8
(
)(
)
9]
Therefore,
{
( )
.
/ }
[.
/
.
/ ][
.
/ ]
[.
/
.
/ ][
.
/ ]
( )
Frequency-response curves
9.6 Vibration Absorbers
 The vibration absorber, also called dynamic vibration absorber, is a mechanical device used
to reduce or eliminate unwanted vibration.
 It consists of another mass and stiffness attached to the main (or original) mass that needs to
be protected from vibration.
 Thus the main mass and the attached absorber mass constitute a two-degree-of-freedom
system; hence the vibration absorber will have two natural frequencies.
 The vibration absorber is commonly used in machinery that operates at constant speed,
because the vibration absorber is tuned to one particular frequency and is effective only over
a narrow band of frequencies.
 Common applications of the vibration absorber include reciprocating tools, such as sanders,
saws, and compactors, and large reciprocating internal combustion engines which run at
constant speed (for minimum fuel consumption).
 In these systems, the vibration absorber helps balance the reciprocating forces.
 Without a vibration absorber, the unbalanced reciprocating forces might make the device
impossible to hold or control.
 Vibration absorbers are also used on high-voltage transmission lines.
 In this case, the dynamic vibration
absorbers, in the form of dumbbellshaped devices (see Figure), are hung
from transmission lines to mitigate the
fatigue effects of wind induced vibration.
 A machine or system may experience excessive vibration if it is acted upon by a force
whose excitation frequency nearly coincides with a natural frequency of the machine or
system.
 In such cases, the vibration of the machine or system can be reduced by using a vibration
neutralizer or dynamic vibration absorber, which is simply another spring-mass system.
 The dynamic vibration absorber is designed such that the natural frequencies of the resulting
system are away from the excitation frequency.
 We shall consider the analysis of a dynamic vibration absorber by idealizing the machine as
a single-degree-of-freedom system.
9.6.1 Undamped Dynamic Vibration Absorber
 When we attach an auxiliary mass m2 to a
machine of mass m1 through a spring of stiffness
k2, the resulting two-degree-of-freedom system
will look as shown in Figure.
 The equations of motion of the masses m1 and m2
are
̈
̈
(
)
(
)
Undamped dynamic vibration absorber

By assuming harmonic solution
( )

we can obtain the steady-state amplitudes of the masses
(


)
(
)(
)
(
)(
)
We are primarily interested in reducing the amplitude of the machine
In order to make the amplitude of
zero, the
(
(
)
)(
)
numerator
of
should be set equal to zero. This gives

If the machine, before the addition of the dynamic vibration absorber, operates near it
resonance,
. Thus if the absorber is designed such that

The amplitude of vibration of the machine, while operating at its original resonant
frequency, will be zero. By defining
(

as the natural frequency of the machine or main system, and
(

)
)
As the natural frequency of the absorber or auxiliary system, equations
rewritten as
.
[
.
/ ][
/
.
/ ]
and
can be
[

.
/ ][
.
/ ]
Figure shows the variation of the amplitude of
vibration of the machine
with the machine
speed








The two peaks correspond to the two natural
frequencies of the composite system.
As seen before,
at
At this
frequency, equation
gives
This shows that the force exerted by the
auxiliary spring is opposite to the impressed
) and neutralizes it, thus
force (
reducing
to zero.
The size of the dynamic vibration absorber can
be found
Effect of undamped vibration absorber
on the response of machine
The values of
and
depend on the
allowable value of
It can be seen from Figure that the dynamic vibration absorber, while eliminating vibration
at the known impressed frequency
introduces two resonant frequencies
and
at
which the amplitude of the machine is infinite.
The operating frequency must therefore be kept away from the frequencies
and
The values of
and
can be found by equating the denominator of ( )
to
zero. Noting that
(

.
and setting the denominator of
(

) (
)
[
(
) 6
The two roots of this equation are given by
.
(
/ ][
)
/
.
/ ]
)(
to zero leads to
) 7
(
)
(
)
{[
.
/.
/ ]
8[
.
.
}
/.
/ ]
.
/ 9 }
/
Example No.4: Vibration Absorber for Diesel Engine
A diesel engine, weighing 3,000 N, is supported on a pedestal mount. It has been observed that the
engine induces vibration into the surrounding area through its pedestal mount at an operating speed
of 6,000 rpm. Determine the parameters of the vibration absorber that will reduce the vibration
when mounted on the pedestal. The magnitude of the exciting force is 250 N, and the amplitude of
motion of the auxiliary mass is to be limited to 2 mm.
Solution No.4:
 The frequency of vibration of the machine is

Since the motion of the pedestal is to be made equal to zero, the amplitude of motion of the
auxiliary mass should be equal and opposite to that of the exciting force
| |
(
(
) (
) (
Example No.5: Absorber for Motor-Generator Set
A motor-generator set, shown in Figure, is designed to
operate in the speed range of 2000 to 4000 rpm.
However, the set is found to vibrate violently at a
speed of 3000 rpm due to a slight unbalance in the
rotor. It is proposed to attach a cantilever mounted
lumped-mass absorber system to eliminate the
problem. When a cantilever carrying a trial mass of 2
kg tuned to 3000 rpm is attached to the set, the
resulting natural frequencies of the system are found to
be 2500 rpm and 3500 rpm. Design the absorber to be
attached (by specifying its mass and stiffness) so that
the natural frequencies of the total system fall outside
the operating-speed range of the motor-generator set.
)
)
Motor-generator set
Solution No.5:
 The natural frequencies
of the motor-generator set and
√

of the absorber are given by
√
. /
The resonant frequencies
and
of the combined system are given by equation
}.
. /
Since the absorber (m = 2 kg) is tuned,
rpm). Using the notation
.
(corresponding to 3,000
√.
/
/

Since
and
are known to be 261.80 rad/s (or 2500 rpm) and 366.52 rad/s (or 3500
rpm), respectively, we find that

Hence,
.

/
√.
Or
4

Since
lower limit of

/
, then
5
and
. The specified
is 2000 rpm or 209.44 rad/s, and so
With this value of
, then
and
(
)
. With
these values, the second resonant frequency can be found from
.

/
√.
/
Which gives
larger than the specified upper limit of 4,000 rpm. The
spring stiffness of the absorber is given by
(
) (
)
Problem No.1:
Find the natural frequencies of the system shown in Figure, with
m1 = m, m2 = 2m, k1 = k and k2 = 2k. Determine the response of
the system when k = 1,000 N/m, m = 20 kg, and the initial values
of the displacements of the masses m1 and m2 are 1 and -1,
respectively.
Solution No.1:
Problem No.2:
 One of the wheels and leaf springs of an
automobile, traveling over a rough road, is shown
in Figure (a).
 For simplicity, all the wheels can be assumed to be
identical and the system can be idealized as shown
in Figure (b).
 The automobile has a mass of m1 = 1,000 kg and
the leaf springs have a total stiffness of k1 = 400
kN/m.
 The wheels and axles have a mass of m2 = 300 kg
and the tires have a stiffness of k2 = 500 kN/m.
 If the road surface varies sinusoidally with
amplitude of Y = 0.1 m and a period of l = 6 m,
find the critical velocities of the automobile.
(a)
Solution No.2:
(b)

Mechanical Vibrations

Transcription

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