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Mechanics of machines; Vibration analysis. Formula and derivation; also contain diagrams and combined vibration analysis. This work was conducted by a professor from the University of Mauritius in 2012
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6
Vibration
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|2
IntroductionVibration plays an important role in nature and engineering.
Many engineering products (automobiles, jet engines, rockets,
bridges...) require a good insight in the concept of vibration for
efficient product design, development and performance evalua-
tion. Vibrations are fluctuations of a mechanical or structural
system about an equilibrium position.
Vibrations are initiated when an inertia element is displaced
from its equilibrium position due to energy being imparted to the
system through an external source (cf. Figure 1 and Figure 2).
A restoring force or moment then proceeds to pull the element
back towards equilibrium.
Examples:
Figure1
Figure2
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|3
When work is done on the block of Figure 1 to displace it from
its equilibrium position, potential energy is developed in the
spring. When the block is released, the spring force pulls the
block towards equilibrium with the potential energy being con-
verted to kinetic energy. In the absence of dissipative effects,
e.g. friction, this transfer of energy (potential-kinetic-potential) is
continual, causing the block to oscillate about its equilibrium
position indefinitely.
The study of vibration involves developing mathematical mod-
els for systems shown in Figure 1 and Figure 2 (among others),
and finding solutions to these mathematical models.
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|4
BasicConceptsVarious quantities, e.g. the position of a particle, undergo more or less regular changes over time. The processes in-
volved in these changes are called vibrations or oscillations.
Examples:
waves of the oceans
movement of a piston in an engine
vibrations in an electrical circuit
Frequently, a quantity will have the same value at regular time intervals (Figure 3), during an oscillatory process:
[1]
If the oscillatory process involves movement, the motion is
called a periodic vibration. The time is referred to as the PERIOD of the vibration.
Figure3
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|5
The quantity
1 [2]
is called the FREQUENCY of the vibration. It represents the num-
ber of cycles per unit time (a CYCLE is the motion completed
during 1 period). The dimension of frequency is 1/time. Its unit
is Hertz (abbreviated Hz):
1 1
Harmonic Vibrations
Harmonic vibrations are characterized by the fact that the quan-
tity is given by a cosine or sine function:
. or . [3]
Figure4
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|6
: Angular/circularfrequencyor : Amplitudeofvibration
Since 2 (cf. Figure 4) and 1 , it follows that:
2 2 [4]
Harmonic vibrations that are represented by pure cosine or
pure sine curves are subjected to special initial conditions.
For . , we have the initial conditions: 0 0 0
Similarly, for . the initial conditions are: 0 0 0
Harmonic vibrations with arbitrary initial conditions can always
be represented by
. [5]
: Amplitude : Phaseangle(cf.Figure 5)
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|7
Figure5
Harmonic vibrations [5] can also be obtained through a super-
position of two vibrations of the form of [3]. Using the trigono-
metric formula
. . . [6]
and setting
. and . [7]
we obtain
. . [8]
The representations of harmonic motions by equations [5] and
[8] are therefore equivalent and interchangeable.
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|8
A harmonic oscillation can be generated by a point (initial po-sition ) which moves on a circular path (radius C) with con-stant angular velocity (Figure 5). The projection on a verti-cal straight line (or on any diameter) performs a harmonic vibra-
tion.
Undamped vibration Vibration of constant amplitude
Damped vibration
Unstable vibration
Degrees of Freedom (DOF)
The degrees of freedom of a body are the set of variables nec-
essary to completely define the position and orientation of the
body in space.
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|9
FreeVibrationsFree vibrations occur naturally with no energy being added to
the vibrating system.
Undamped free vibrations
Consider a block (mass ) that moves on a smooth surface (Figure 6). It is connected to a wall with a linear spring (spring
constant ).
Figure6
The system has 1 DOF, given by the horizontal position of the
block.
Figure7:Freebodydiagram(FBD)showinghorizontalforces
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|10
To derive the equation of motion of the block, we introduce the
coordinate (cf. Figure 7). Let 0 be the equilibrium position of the block (unstressed spring).
1. Block is displaced ( units to the right from the equilibrium position
2. Draw a FBD showing the different forces acting on the block:
Neglecting friction from the surface with which the block is in
contact, the only horizontal force acting on the block is the
restoring force ( from the spring
3. Apply Newtons 2nd law ( ):
0 [9]
Setting
[10]
equation [9] is re-written as:
0 [11]
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|11
Equation [11] is a 2nd order, linear homogeneous differential
equation having constant coefficients. Its general solution is
given by
. . [12]
where and are integration constants.
4. Use initial conditions to determine integration constants:
0 0
Equation [12] becomes
. . [13]
Equation [12] is equivalent to
. [14]
With
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|12
According to equation [14], the block performs a harmonic vi-
bration with circular frequency . The natural fre-
quency () of free vibrations can be determined from the rela-tionship 2. The natural frequency of vibrations of a system is also commonly known as its eigenfrequency.
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|13
Problem 6.1
Find the equation of motion of a block of mass , suspended by a linear spring (spring constant ), which is displaced downwards from its equilibrium position and released with no
initial velocity.
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|14
Problem 6.2
A small mass "" is fastened to a vertical wire that is under tension . What will be the natural frequency of vibration of the mass if it is displaced laterally a slight distance and then re-
leased?
Figure8
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|15
Problem 6.3
A rod (length , with negligible mass) carries a mass at its upper end. It is supported by a linear spring (cf. drawing be-
low).
Describe the motion of the rod if it is displaced from its vertical
position (small displacement to the left) and then released with
no initial velocity.
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|16
Mass Moment of Inertia
The study of rotational dynamics requires the computation of
mass moments of inertia. Two situations normally arise:
Point mass rotating about an axis Rigid body rotating about an axis
Mass moment of inertia is an important property of a body. It is
involved in the analysis of any body which has rotational accel-
eration about a given axis.
Just as the mass of a body is a measure of the resistance to translational acceleration, the moment of inertia is a measure of resistance to rotational acceleration of the body.
The moment of inertia of an object is defined as the integral of
the "second moment" about an axis of all the elements of mass
which compose the body. For example, the body's moment of inertia about the in Figure 9 is
[15]
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|17
Figure9
The "" is the perpendicular distance from the to the arbitrary element . Since the formulation in-volves , the value of the moment of inertia is different for each axis about which it is computed.
For a point mass, the moment of inertia is calculated from
equation [16] (where is the perpendicular distance between the point mass and the axis of rotation).
. [16]
(for a point mass)
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|18
Parallel-axis theorem
If the moment of inertia of the body about an axis passing
through the body's mass center is known, then the moment of
inertia about any other parallel axis can be determined by using
the parallel-axis theorem.
Figure10
[17]
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|19
Perpendicular-axis theorem
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|20
Radius of gyration
The moment of inertia of a body about a specified axis is some-
times calculated using the radius of gyration, . This is a geo-metrical property which has units of length. For a given solid, if
its mass and radius of gyration are known, the bodys moment
of inertia is obtained using equation [18]
[18]
Composite bodies
If a body consists of a number of simple shapes such as disks,
spheres, and rods, the moment of inertia of the body about any
axis can be determined by adding algebraically the moments of
inertia of all the composite shapes computed about the axis.
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|21
Problem 6.4 Determine the moment of inertia for each of the following
situations:
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|22
Problem 6.5
Determine the moment of inertia for the slender rod. The rod's density and cross-sectional area are constant. Ex-press the result in terms of the rod's total mass .
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|23
Problem 6.6 The paraboloid shown is formed by revolving the shaded area
around the . Determine the radius of gyration . The density of the material is 5 .
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|24
Problem 6.7 The right circular cone shown is formed by revolving the
shaded area around the x axis. Determine the moment of iner-
tia and express the result in terms of the total mass of the cone. The cone has a constant density .
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|25
Problem 6.8
The body of arbitrary shape has a mass , mass center at , and a radius of gyration about of . If it is displaced a slight amount from its equilibrium position and released, determine the natural period of vibration.
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|26
Problem 6.9
The bent rod shown has a negligible mass and supports a
collar at its end. If the rod is in the equilibrium position shown,
determine the natural period of vibration for the system.
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|27
Spring constants of elastic systems
The force and the elongation of a linear spring are related by the relationship . .
The spring constant is therefore characterized by
[19]
Similar force-deformation relationships can be established for
many systems containing elastic components.
I. Massless bar (length l, axial rigidity EA)
A massless bar carries a mass
at its free end (Figure 11). Under the action of the mass,
the bar undergoes an elonga-
tion downwards. The bar provides a restoring force to oppose the downward dis-
placement of the mass and es-
tablish an equilibrium condi-
tion. A force equal in magni-
tude and opposite in direction
acts on the bar (action-
reaction).
Figure11
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|28
The force and the elongation are related by the equation
[20]
By analogy with the elastic spring, we can determine a
spring constant or stiffness for an elastic bar:
[21]
II. Massless cantilever beam (length l, flexural rigidity EI)
A massless cantilever beam
carries a mass at its free end (Figure 12). The mass causes the end of the initially
straight beam to move down-
wards. Under equilibrium con-
ditions, a restoring force pro-
vided by the cantilever beam
balances the weight of the
mass .
3 (cf. Deflection curves of beams)
Figure12
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|29
Thus, we obtain the spring constant for a cantilever beam as
3 [22]
III. Massless shaft (length l, torsional rigidity GJ)
The relationship between the
angle of twist and the torque
in a shaft is governed by the
following equation:
Figure13
The spring constant equivalent for this configuration is obtained
as
[23]
If a disk (moment of inertia ) is fixed to the end of the shaft and undergoes torsional vibrations, then the motion is de-
scribed by
0 [24]
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|30
Spring assembly Equivalent spring
Springs in parallel
The motion of a mass may cause elongations of several
springs in a system. Consider the case of two springs in parallel
(Figure 14).
Figure14
The two springs (spring constants and ) undergo the same elongation when the mass is displaced. They can be re-
placed by an equivalent single spring with the spring constant
.
[25]
Therefore, for a system of parallel springs
(spring constants ), the spring constant of the equivalent spring is given by the sum of the individual spring constants:
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|31
[26]
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|32
Springs in series
Consider two springs in series (Figure 15).
Figure15
Find the stiffness of the equivalent spring . In the case of arbitrarily many springs in series, the spring con-
stant of the equivalent spring is found from
The flexibility (compliance) of a spring is defined as the inverse
of the stiffness:
1 [27]
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|33
Problem 6.10
An elastic beam (flexural rigidity ) with negligible mass sup-ports a box (mass ) as shown below. Find the natural circu-lar frequency of the system.
K.A. | MECH4005Y | MechanicsofMaterials&MachinesIV Page|34
Problem 6.11 Determine the natural circular frequency of the system(s)
shown below.