# MECHANICS OF MACHINES II (Note) ```MECHANICS OF MACHINES (MCE 322)
Ajayi, O.O
Mechanical Engineering Department,
Covenant University, Ota, Ogun State, Nigeria.
1.0 INTRODUCTION TO VIBRATIONS OF MACHINES
1.1 Introduction
Vibrations can be defined as the fluctuation of a mechanical or structural system about an
equilibrium position [1-3]. It is initiated when a part of the system is displaced from its
equilibrium position, thereby, making the system run to-and-fro about that position. In every
vibrating system, there is a restoring force or moment which tends to pull the system back
toward its equilibrium position creating a reversible conversion of potential energy to kinetic
energy. When there are no conservative forces, this transfer of energy is continuous, causing the
system to oscillate about its equilibrium position. Examples of vibratory motion include the
motion of a disturbed spring system and the pendulum bob.
Vibration analysis is becoming very important due to the current trend toward the use of higherspeed machines and higher structures. The existence of vibrations in these system can cause
damaging effects if not eliminated or reduced as much as possible by appropriate design .
1.2 Components of Mechanical Vibrations
The components of mechanical vibration include :
1. Period of vibration: this is the time interval required by the system to complete a full
cycle of motion. The S.I unit is second (s).
2. Frequency: This is the number of cycles of motion per unit time. The S.I unit is the Hertz
(Hz) or cycle/s.
3. Amplitude: this is defined as the maximum displacement of a vibratory system about the
equilibrium position. The S.I unit is meters (m).
1.3 Types of Vibrations
There are mainly two types of vibration [2, 3]:
1
1. Free vibration: this is the vibration in which the motion is maintained only by elastic
restoring or gravitational forces, e.g. motion of a pendulum bob and the motion of an
elastic spring system. However, this type of vibration can either be damped or undampedfree vibration. It is undamped when there are no frictional or non-conservative force
effects acting on the systems motion, hence, an undamped-free vibration is assumed to be
an un-ending (continuous) vibration of a system. The simplest type of vibratory motion is
the undamped-free vibration.
2. Forced vibration: this is the vibration in which the motion is caused by a periodic or
intermittent force. It can also be damped or undamped-forced vibration
1.4 Effects of vibration on mechanical systems
Most of the time, mechanical vibration is undesired in machines parts and components due to its
harmful effects. At times, it introduces some undesirable (annoying) outputs and leads to a
reduction in the quality and/or quantity of the desired/useful output. These effects include:
1. reduction of efficiency of machines
2. Undesirable noise
3. Wear and Tear of machine components or parts.
4. Vibration can also lead to fatigue failure of structures due to large dynamic stresses
introduced into it by an uncontrolled vibrating force.
1.5 Concepts from Vibration
To analyze effectively the vibratory motion of a mechanical system in a bid to eliminate or
reduce it, involves developing a simple model which will approximate the whole system
concisely. The simplest representation of a complex system in such a way that all useful
information regarding the systems characteristics is approximated is called a model. Before
modeling can be carried out, it is necessary to identify and characterize the various types of
system components, which implies establishing the excitation-response relation, or input-output
relation, for individual components or for groups of components, either from experience or
through testing. For vibrating system, the behaviour and characteristics are governed by
equations of motion. The derivation of these equations can be carried out by means of methods
of Newtonian mechanics or by methods of Lagragian mechanics .
Newtonian mechanics involves the use of the laws of physics to generate a mathematical
relation for the kinematics of vibratory motion (i.e. displacement, velocity and acceleration).
2
This mathematical formulation is always in the form of differential equations, of which solution
gives the system response. Newtonian mechanics require the use of free-body diagrams for each
mass in the system. This is regarded as a draw back of the method. However, the Lagrangian
method considers the system as a whole. It uses the methods of potential and kinetic energies,
and also the principle of virtual work by non-conservative forces. The principle of virtual work
in another hand, leads to the principle of virtual displacement (which relates a virtual
displacement to a force for virtual work to be done), and this leads to the development of
variation calculus. In deference to Newtonian mechanics, Lagrangian methods (also known as
the analytical mechanics) use a broader and more abstract approach, because the equations are
developed from the point of view of a generalized coordinate and force system. This makes it
independent of any system of coordinates.
References
1. Kelly, S.G. 2000. Fundamentals of Mechanical Vibrations, 2nd edn., McGraw-Hill, New
York
2. Hibbeler, R.C. 2004. Engineering Mechanics dynamics, 3rd edn., Prentice Hall,
Singapore. 605-639 (TA 352.H5)
3. Beer, F.P., Johnston, E.R. Jr., and Clausen, W.E. 2004. Vector Mechanics for Engineers:
Dynamics 7th edn., McGraw-Hill, New York. 1214-1265.
4. Meirovitch, L. 2001. fundamentals of Vibrations, McGraw-Hill, New York.
3
2.0 FREE VIBRATIONS
Free vibration is defined as the response of a system to initial excitations, consisting of initial
displacements and/or initial velocities . They are oscillation about a system’s equilibrium
position that occur in the absence of an external excitation, but are a result of a kinetic energy
imparted on a system or a displacement from the equilibrium position that leads to a difference in
potential energy from the system’s equilibrium position .
2.1 Degree-of-freedom systems
There are basically two types of degree-of- freedom systems. These are (1) Single (or one)degree-of-freedom and (2) Multiple-degree-of-freedom systems respectively.
The single-degree-of-freedom systems are those whose response are characterized by a
single variable or coordinate. Their systems’ response are governed by a single ordinary
differential equation, relating the displacement x (t) to the force of excitation F (t).
The multiple (or Multi)-degree-of-freedom systems are those with response characterized
by multiple variable or coordinates. Their systems’ response are governed by a system of n
differential equations, with n equal to the number of unknown (or multiple) variables. The
analysis of Multi-degree-of-freedom system is more complicated than a single-degree-offreedom system. The focus here shall be on the analysis of single-degree-of-freedom of
undamped and damped free vibrations.
2.2 Undamped Free Vibrations of Single-Degree-of-Freedom System
Consider the undamped system below and analyze the response of the vibrating system’s
displacement, velocity and etc.
x (t)
Fig. 1: Mass-spring system
If the mass of the box is m and each of the spring has spring constant = k 1 and k 2.
4
Since there is no imposed force, the system is undergoing free vibrations. The free body diagram
is given as:
F(t)
m
K1x
K2x
The equation of motion is given as:
mx + keq x = f ( t )
1
Since this is a free vibration, there are no imposed forces, hence:
mx + keq x = 0
2
The equivalent stiffness constant is derived from the situation of Parallel spring system in pages
25-29 of reference text , given by:
keq = k1 + k2
3
The equation 2 can be translated into simple familiar differential equation as:
m
d 2x
+ keq x = 0
dt 2
4
The solution of this Eq.4 after substitution of the values of m and keq gives the expression for the
displacement. Differentiating this gives the velocity, etc.
From Eq.3, dividing through by m gives:
x + ωn2 x = 0
5
Where ωn = natural frequency in rad/s, ωn = k
m
, which is dependent on the internal factors of
the system . Thus, the frequency and period of the free vibration can be expressed as:
T=
f =
2π
ωn
1
T
6
7
Where f = frequency of vibration in cycle/s or Hertz (Hz).
5
2.3 Damped Free Vibrations of Single-Degree-of-Freedom System
Consider the Fig. 1 above. If a viscous damping system is introduced to it, such as fig. 2 below:
x (t)
Since there is no imposed force, but a viscous damper, the system is performing free vibration
under a damping. The free body diagram is given as:
F(t)
m
K1x
K2x
cx
The equation of motion in Eq.1 is modified to include the damping parameter. Thus, the general
equation of motion for a damped free vibration is:
mx + cx + keq x = f ( t )
1
Because there is no imposed force, the right hand side becomes zero. Hence, Eq.1 becomes:
mx + cx + keq x = 0
2
The equation 2 can be translated into simple familiar differential equation as:
d 2x
dx
m 2 + c + keq x = 0
dt
dt
3
The solution of this Eq.3 after substitution of the values of m, c and keq gives the expression for
the displacement. Differentiating this gives the velocity, etc.
Where m = mass, c = damping coefficient and keq = equivalent spring constant for the parallel
spring system.
Dividing through by m gives:
6
x + 2ξωn x + ωn2 x = 0
Where ξ =
4
c
= non-dimensional viscous damping factor.
2mωn
The general solution for all differential equation is given as:
x ( t ) = Ae st
5
Inserting Eq.5 into 4 and dividing through by Ae st , we obtain the characteristic equation:
s 2 + 2ξωn s + ωn2 = 0
6
s1, s2 = −ξωn &plusmn; ξ 2 − 1ωn
7
This means that the solution of the response, x will be governed by the nature of ξ .
To find the solution of Eq.4 precisely means that boundary condition must be employed. This is
always used at when the time = 0 (i.e. when the initial displacement and velocity is zero), the
boundary condition is given as:
x ( 0 ) = xo
x ( 0 ) = vo
Therefore, by Eqs.5 and 7, the general solution to Eq.4 will be:
x ( t ) = A1e s1t + A2 e s2t
8
Where A1 and A2 are constants of integration, which depends on the initial conditions. The
constants can be determined by using the boundary conditions with Eq.8, thus:
x ( 0 ) = A1 + A2 = xo
9i
x ( 0 ) = s1 A1 + s2 A2 = vo
9ii
Solving the Eq.9 gives:
A1 =
vo − s2 xo
s x −v
and A2 = 1 o o
s1 − s2
s1 − s2
Inputting these values of As into Eq.8 gives:
x (t ) =
vo − s2 xo s1t s1 xo − vo s2t
e +
e
s1 − s2
s1 − s2
10
Substituting the required parameters gives the solution of the system’s displacement. Moreover,
to obtain a specific solution depends on the value of the damping factor, ξ . To understand this
7
we shall look at the effect of the damping factor on the nature of motion, after which we shall
apply such to Eq.10.
Effects of the Damping Factor on the Nature of Vibratory Motion
The different values of the damping factor according to Eq.7 ranges from 0 ≤ ξ ≤ 1 and ξ &gt; 1 .
Thus, when:
(1). ξ = 0 , the roots s1 and s2 are on the imaginary axis of the s-pane. The motion as a result of
this corresponds to (a) undamped free vibrations and (b) harmonic oscillations.
(2). 0 &lt; ξ &lt; 1 , the roots s1 and s2 are complex conjugates and corresponds to pairs of points in the
s-plane symmetrically located with respect to the real axis . The motion as a result of this
corresponds to (a) under-damped free vibrations and (b) oscillatory decay.
(3). ξ = 1 , the roots s1 and s2 merge at the point - ωn on the real axis of the s-plane. The motion as
a result of this corresponds to (a) critical damping and (b) aperiodic decay.
(4). ξ &gt; 1 , both roots s1 and s2 are located on the negative real axis of the s-plane. The motion as
a result of this corresponds to (a) over-damping and (b) aperiodic decay.
Going back to the Eq.10 above to obtain the response based on the value of the damping
factor.
Eq.10 is:
x (t ) =
vo − s2 xo s1t s1 xo − vo s2t
e +
e
s1 − s2
s1 − s2
10
Response for under-damped vibration
When 0 &lt; ξ &lt; 1
Let the value of the roots given in Eq.7 be re-written as:
s1, s2 = −ξωn &plusmn; ωd
11
Where ωd = ξ 2 − 1ωn is the frequency of the damped vibration.
Putting Eq.11 into 10 gives :
x (t ) =
e −ξωnt
⎡⎣ − ( −ξωn − iωd ) xo + vo ⎤⎦ eiωd t + ⎡⎣( −ξωn + iωd ) xo − vo ⎤⎦ e − iωd t
2iωd
{
8
}
e −ξωnt ⎡
=
(ξωn xo + vo ) eiωd t − e−iωd t + iωd xo eiωd t − e−iωd t ⎤⎦
⎣
2iωd
(
)
(
)
⎛ ξω x + v
⎞
= e −ξωnt ⎜ n o o sin ωd t + xo cos ωd t ⎟
ωd
⎝
⎠
= Ce −ξωnt cos (ωd t − φ )
The values of the C and phase angle are:
⎛ ξω x + v ⎞
C = x +⎜ n o o ⎟
ωd
⎝
⎠
2
2
o
and
⎛ ξωn xo + vo ⎞
⎟
⎝ ωd xo ⎠
φ = tan −1 ⎜
Where C = amplitude of vibration and φ = phase angle of the response.
Note that in arriving at the above solution, the relation was used.
eiθ − e − iθ = 2i sin θ
eiθ + e −iθ = 2i cos θ
Other types of damped response can be solved in the same way.
References
1. Meirovitch, L. 2001. fundamentals of Vibrations, McGraw-Hill, New York.
2. Kelly, S.G. 2000. Fundamentals of Mechanical Vibrations, 2nd edn., McGraw-Hill, New
York
9
3.0 DAMPING
Damping a vibratory motion is an act of introducing friction to the vibratory system. There
basically different types of damping which includes: Viscous damping, Coulomb damping, and
Hysteretic damping.
3.1 Viscous Damping
This is the type of damping caused by fluid friction when a rigid body moves in a fluid at low
and moderate speeds. It acts against motion and directly proportional to the velocity.
Mathematically it is described as:
F = −cx
Where F = Viscous friction force
c = coefficient of viscous damping and
x = Velocity.
The coefficient of viscous damping depends solely upon the physical properties of the fluid
and the construction of the dashpot .
3.2 Coulomb Damping
This is the type damping caused by dry friction between rigid bodies. It arises when bodies
slides on dry surfaces [1-3]. Just like viscous damping, the friction force due to this type is
directly proportional to the velocity and takes in the opposite direction to that motion, but
constant in magnitude. It is mathematically represented for a mass-spring system as:
N
k
Fd
Coulomb friction = Fd sgn ( x )
W
Where Fd = μk mg
μk = Coefficient of kinetic friction
m = mass
g = acceleration due to gravity
10
sgn ( x ) = signum function of x, which represents a function having the value of +1 if the
argument on x is positive and -1 if it is negative. It is mathematically given as:
sgn ( x ) =
x
/ x/
Thus,
Fd = − μk mg
x&gt;0
Fd = μk mg
x&lt;0
Summarily, the equation of motion for the mass-spring system undergoing free vibration is given
as:
mx + kx = − Fd
x&gt;0
mx + kx = Fd
x&lt;0
3.3 Hysteretic Damping
This is the type of damping caused by internal friction between molecules of elastic body. It
gives rise to energy dissipation from the system during each cycle of motion, thus, creating a
phenomenon of natural damping. The damping ratio for hysteretic damping is defined as:
ξ=
δ
2π
1
But for small hysteretic damping coefficient, h, equation 1 is given as:
ξ=
h
2
2
Since the logarithmic decrement δ is defined for small h, as:
δ =πh
3
The energy dissipated per cycle of motion is described mathematically as:
ΔE = π hX 2 k
4
Where X = Amplitude of motion during the cycle,
k = Spring constant
11
Other forms of damping
These include aerodynamic drag, radiation damping, and anelastic damping. These do not give
rise to exact solution in the governing differential equations. Approximate solutions can only be
obtained by developing equivalent viscous damping coefficient obtained by equating the energy
dissipated over one cycle of motion, assuming harmonic motion at a specific amplitude and
frequency, for that form of damping, to the energy dissipated over one cycle of motion because
of the force in a dashpot of the equivalent viscous damping coefficient . i.e. (see page 115 of
Ref. ):
For a harmonic motion of the form:
X (t ) = X sin ωt
1
The energy dissipated over one cycle of motion due to damping force is:
2π / ω
ΔE =
∫
2π / ω
FD xdt =
0
∫
FD X ω cos ωtdt
2
0
For viscous damping, equation 2 becomes:
2π / ω
ΔE =
∫
2π / ω
cx 2 dt =
0
∫
cω 2 X 2ω cos 2 ωtdt = cωπ X 2
0
By analogy, the equivalent viscous damping coefficient for another form of damping is:
ceq =
ΔE
πω X 2
For further reading see page 115-117 of ref .
Reference:
1. Beer, F.P., Johnston, E.R. Jr., and Clausen, W.E. 2004. Vector Mechanics for Engineers:
Dynamics 7th edn., McGraw-Hill, New York. 1214-1265.
2. Kelly, S.G. 2000. Fundamentals of Mechanical Vibrations, 2nd edn., McGraw-Hill, New
York
3. Meirovitch, L. 2001. Fundamentals of Vibrations, McGraw-Hill, New York.
12
4.0 TRANSVERSE VIBRATION OF BEAMS
When a simply supported beam loaded at the middle is allowed to oscillate vertically, the kind of
motion it will undergo is called transverse vibration. In which case, the beam is straight and bent
alternatively. Bending stresses are then induced in it as a result of this. Consider the Fig.1 below.
Provided the beam is elastic with its mass small compared to the mass of the particle attached to
it, the transverse vibration of the beam will be similar to the vibration of a mass-spring system
allowed to vibrate in a vertical x-direction.
x
L/2
L/2
(a)
k=
48EI
l3
x
(b)
Fig. 1: (a) Transverse vibration of a simply supported beam carrying a particle of mass m,
(b) Mass-Spring system vibrating in a vertical x-direction.
When the beam in the figure is deflected in the x-direction as shown, and suddenly released, it
will make transverse vibrations. This deflection is proportional to the load. However, if the beam
is deflected beyond its static equilibrium position, then the load will vibrate with simple
harmonic motion just like a spring system of fig.1 (b) will behave. Thus, the natural frequency of
a free transverse vibration can be estimated mathematically from the equilibrium method of
analysis as:
At equilibrium position, the weight (W) is balanced by the static force ( f s = kδ ) given by:
W = kδ
1
13
Where δ is the static deflection.
When the beam is displaced away from this equilibrium position to a new position given by
( δ + x ), a restoring force tends to bring it back to its original equilibrium position. The
magnitude of this restoring force (F) is given as:
F = W − k (δ + x ) = W − kδ − kx
2
Putting Eq.1 into 2 gives:
F = −kx
3
But force is mass x acceleration and given differentially as (after substituting Eq.3):
F = m&times;
d 2x
= −kx
dt 2
4
Hence, Eq.4 is given summarily as:
m&times;
d 2x
+ kx = 0
dt 2
5
This becomes:
d2x k
+ x=0
dt 2 m
6
As before:
x + ωn2 x = 0
7
The period of a transverse vibration is therefore given from Eq.7 as:
T=
2π
ωn
= 2π
m
δ
= 2π
k
g
(∵W = kδ = mg )
8
Hence, natural frequency of a transverse vibration is:
f =
1
2π
g
9
δ
The value of static deflection for a simply supported beam with a central load is given as:
δ=
Wl 3
48 EI
10
Knowing that the weight is the vertical force (i.e. W = F) on the beam gives:
14
l3
l3
δ=
W=
F
48 EI
48 EI
11
Re-arranging and comparing to spring force (F = k δ), shows that:
k=
48EI
l3
12
This gives a conclusion that a linear relationship exist between transverse displacement and static
load. Hence, if the mass of the beam is small compared to the attached load, the vibration of the
load can be modeled as the vertical motion of a particle attached to a spring (Fig.1 (b)) of
stiffness k given by Eq.12.
For further reading see Refs. [1-2]
Reference:
1. Kelly, S.G. 2000. Fundamentals of Mechanical Vibrations, 2nd edn., McGraw-Hill, New
York
2. Khurmi, R.S and Gupta, J.K. 2003. Theory of Machines, S. Chand, New Delhi, p 879937.
Tutorial Questions
1. A simply supported beam of length 1.0m is loaded eccentrically with a mass of 50kg at
0.3m from one end. Find the natural frequency of the resulting transverse vibration.
Assume that E = 110 x 109N/m2 and the diameter of the beam is d = 25mm. (HINT:
moment of inertia of the beam is I =
πd4
64
and δ =
Wa 2b 2
3EIl
2. If the beam above carries a simply supported load of 10kg/m, find the period of the
transverse vibration. (HINT: δ =
5 Wl 4
&times;
(at the centre))
384 EI
3. A mass-spring system is loaded as shown below. If the mass of the block is 5kg. The
spring constants and the coefficient of viscous damping are 0.2N/m and 0.05N.s/m
respectively. Find (a). The period and natural frequency of vibrations, (b) the damping
factor, (c) the response of the system to the free excitation. If there are no damper in the
system what will be the system’s response?
15
x
k1
k2
c
4. A damped system’s harmonic motion due to a damping force Fk is given by:
x ( t ) = X cos ωt
Where Fk = cx
(a). Estimate the energy dissipation over one cycle of motion due to the damping, of
coefficient 0.035N.s/m, ω = 0.02rad/s, t = 0.1s and X = 5mm. (b). If the damper is to be
replaced by a viscous damper, what will be the equivalent viscous damping coefficient, if its
ω and X for the viscous damper are 0.35rad/s and 3mm respectively?
16
5.0 FORCED VIBRATIONS
Forced vibrations are the vibrations in which the motion is caused by a harmonic or periodic
(or intermittent force). It arises when a body vibrates under the influence of external force.
This external force applied to the body imparts a periodic disturbance to the body and gives
the body its frequency . For one-degree-of-freedom systems, the periodic forces of
excitation of the forced vibrations causes work to be done on the system . Examples of
forced vibrations include the movement of the ground during earthquakes, unbalanced
reciprocating component of engines, movement of vehicles on rough roads, vibratory motion
of a rotating component such as turbines etc. However, forced vibrations can be due to either
harmonic or periodic excitation, in deference to transient excitations (which are basically
short response to initial excitations).
Harmonic excitations – These are forces that are proportional to trigonometric functions of
sin ωt or cos ωt , at times to a combination of the two trigonometric functions. Thus, the
general shape of a harmonic excitation is sinusoidal, with its characterization basically
dependent on frequency of excitation and amplitude. Time plays only a secondary role in this
type of excitation. This is why harmonic excitations are called steady-state excitations .
Periodic excitations – These belongs to a class of excitations whose functions repeat
themselves every time interval T (period of motion). It is similar to harmonic excitations in
that time also plays a secondary role, but whereas, harmonic functions are periodic, periodic
functions are not necessarily harmonic functions. They are expressed either as linear
combination of harmonic functions (Fourier series) or as trigonometric functions. Great deal
of information can be sort from a periodic function by plotting a curve of amplitude of a
Fourier series against the frequency [2-3].
5.1 Response of a single-degree-of-freedom system to harmonic excitations
There can be two situations represented under this – undamped and damped systems.
Consider Fig. 1 below representing the undamped and damped systems.
17
k1
F(t)
k1
k2
F(t)
k2
c
(a) Undamped mass-spring
System
(b) Damped mass-spring system
Fig. 1: Forced vibrations of a mass-spring system
The response of the undamped system is given by:
mx + keq x = F ( t )
1
Diving through by m gives:
x + ωn2 x =
F (t )
m
2
Representing the harmonic excitation with:
F ( t ) = Fo sin ωt
3
Thus, Eq.2 becomes:
x + ωn2 x =
Fo
sin ωt
m
4
Where m = mass, x = desired system’s response, Fo = amplitude of harmonic excitation, ω =
excitation frequency and ωn = natural frequency.
The solution of Eq.4 gives the undamped system’s response to the harmonic excitation.
However, the response of the damped system can be derived from the general equation:
mx + cx + keq x = F ( t )
5
This gives after dividing through by m and substituting Eq. 3:
x + 2ξωn x + ωn2 x =
Fo
sin ωt
m
6
Where c = coefficient of viscous damping.
Let Fo = kA = spring stiffness (k) x amplitude (A), then Eq.6 becomes:
x + 2ξωn x + ωn2 x =
k
A sin ωt
m
7
Thus, becoming:
18
x + 2ξωn x + ωn2 x = ωn2 A sin ωt
8
Assuming a general solution in the form:
x ( t ) = C1 sin ωt + C2 cos ωt
9
Putting Eq.9 into 8, rearranging, and comparing coefficients gives:
(ω
2
n
− ω 2 ) C1 − 2ξωωn C2 = ωn2 A
i
2ξωωnC1 − (ωn2 − ω 2 ) C2 = 0
ii
Solving by Cramer’s rule or any other method gives:
C1 =
C2 =
ω A (ω − ω
2
n
(ω
2
n
(ω
2
n
2
)
− ω 2 ) + ( 2ξωωn )
2
2
2
2ξωωn3 A
2
n
− ω 2 ) + ( 2ξωωn )
2
2
1 − ⎛⎜ ω ⎞⎟ A
⎝ ωn ⎠
=
2
⎡ ⎛ ω ⎞2 ⎤ ⎛ ω ⎞2
⎢1 − ⎜ ω ⎟ ⎥ + ⎜ 2ξ ω ⎟
n⎠ ⎦
n⎠
⎝
⎣ ⎝
=
2ξ ω
ωn A
2
⎡ ⎛ ω ⎞2 ⎤ ⎛ ω ⎞2
⎢1 − ⎜ ω ⎟ ⎥ + ⎜ 2ξ ω ⎟
n⎠ ⎦
n⎠
⎝
⎣ ⎝
Substituting these into Eq.9 gives:
2
⎧⎪ ⎡ ⎛
⎤
⎞
⎡ 2ξ ω ⎤ cos ωt ⎫⎪
ω
−
+
x (t ) =
1
sin
ω
t
⎜
⎟
⎨
⎬
⎢
⎥
2
ωn ⎥⎦
⎢⎣
⎪⎭
⎡ ⎛ ω ⎞2 ⎤ ⎛ ω ⎞2 ⎪⎩ ⎣ ⎝ ωn ⎠ ⎦
⎢1 − ⎜ ω ⎟ ⎥ + ⎜ 2ξ ω ⎟
n⎠ ⎦
n⎠
⎝
⎣ ⎝
A
10
Expressing Eq.10 in compound angle form gives:
x ( t ) = X cos (ωt − φ )
∵ sin φ =
1 − ⎛⎜ ω ⎞⎟
⎝ ωn ⎠
2
2
2⎫
⎧⎪ ⎡
⎛
⎞ ⎤ ⎛
⎞ ⎪
⎨ ⎢1 − ⎜ ω ω ⎟ ⎥ + ⎜ 2ξ ω ω ⎟ ⎬
n⎠ ⎦
n⎠
⎝
⎪⎩ ⎣ ⎝
⎪⎭
2
cos φ =
2ξ ω
ωn
1
2
2 2
⎧⎪ ⎡
⎤ ⎛ ω ⎞ 2 ⎫⎪
⎛
⎞
ω
⎨ ⎢1 − ⎜ ω ⎟ ⎥ + ⎜ 2ξ ω ⎟ ⎬
n⎠ ⎦
n⎠
⎝
⎪⎩ ⎣ ⎝
⎪⎭
19
1
2
And
A
Thus, X =
ωn2 − ω 2
and φ = tan
2ξωωn
−1
1
2
2 2
⎧⎪ ⎡
⎤ ⎛ ω ⎞2 ⎫⎪
⎛
⎞
ω
⎨ ⎢1 − ⎜ ω ⎟ ⎥ + ⎜ 2ξ ω ⎟ ⎬
n⎠ ⎦
n⎠
⎝
⎪⎩ ⎣ ⎝
⎪⎭
Where X = amplitude of the steady-state response and φ = phase angle of the steady-state
response.
N.B: The solution can also be arrived at using the exponential method and recognizing
the real and complex parts.
Assignments
1. Solve the Eq.4 for the system’s response of an undamped harmonic excitation, putting
Fo = kA.
2. Assuming A = 0.5m, ω = 0.2 rad , ωn = 0.15 rad , m = 5kg , c = 0.05 N .s . Go through
s
s
m
the computing steps of MATLAB and using it and Eq.10 above plot a graph x(t) against
time, for t = 1, 2, 3, 4, 5s.
Reference:
1. Khurmi, R.S and Gupta, J.K. 2003. Theory of Machines, S. Chand, New Delhi, p 879937.
2. Kelly, S.G. 2000. Fundamentals of Mechanical Vibrations, 2nd edn., McGraw-Hill, New
York
3. Meirovitch, L. 2001. Fundamentals of Vibrations, McGraw-Hill, New York.
20
6.0 WHIRLING OF ROTATING SHAFT
Loaded mechanical systems with rotors attached to flexible shafts and mounted on bearings are
capable of vibrating when the mass centre of the load does not coincide with the system’s
geometric centre (a phenomenon called eccentricity). The vibratory motion so produced is due to
the unbalance introduced to the system by this mis-alignment. When this shaft is rotated, it exerts
some centrifugal force, whose effect is to bend the shaft . The rotation of the plane containing
the bent shaft about the bearing axis is known as whirling .
j
y
x
shaft
c
m
o
l/2
l/2
Fig. 1a: Disc of mass m on a rotating massless shaft
y
a
o
x
i
Fig. 1b: Displacement
Of mass centre from the
Geometric centre, a.
e = distance ac, rc = distance oc,
c = mass centre.
Consider the Fig.1a above. The uniformly rotating shaft carries a disc of mass, m at its centre.
The motion of the system can be determined by the displacement x and y of the geometric centre
a, of the disc, leading to a 2-degree-of-freedom. Fig.1b shows the shifting of the mass centre, c
away from the geometric centre, a. the difference is called eccentricity, e.
To analyze the motion completely, the each of the two degrees of freedom is handled
independently. i.e. the motion along x-direction is handled independent from that of y.
Thus, the general equations of motion are given as:
max + cx + kx = 0 And ma y + cy + ky = 0
1
To determine the acceleration, the equation of displacement of the mass centre, c from the
geometric centre is given as:
rc = ( x + e cos ωt ) i + ( y + e sin ωt ) j
2
Differentiating to get acceleration:
ac = ( x − eω 2 cos ωt ) i + ( y − eω 2 sin ωt ) j
3
21
Breaking Eq.3 down gives:
acx = ( x − eω 2 cos ωt ) And acy = ( y − eω 2 sin ωt )
4
Putting Eq.4 into 1 gives:
m ( x − eω 2 cos ωt ) + cx + kx = 0
5
⇒ mx + cx + kx = meω 2 cos ωt
6
For the other coordinate:
m ( y − eω 2 sin ωt ) + cy + ky = 0
7
⇒ my + cy + ky = meω 2 sin ωt
8
Thus, Eqs.6 and 8 can be solved separately using the method of the last class. i.e. solving Eq.6:
Divide the Eq.6 by m gives:
x + 2ξωnx x + ωnx2 x = eω 2 cos ωt
9
Assume a general solution given as:
x ( t ) = C1 cos ωt + C2 sin ωt
i
Differentiating to get the velocity and acceleration gives:
x ( t ) = −C1ω sin ωt + C2ω cos ωt
ii
x ( t ) = −C1ω 2 cos ωt − C2ω 2 sin ωt
iii
Putting (i)-(iii) into Eq.9 and comparing coefficients gives:
C1 (ωnx2 − ω 2 ) + 2ξωnxωC2 = eω 2
a
−C1 2ξωnxω + (ωnx2 − ω 2 ) C2 = 0
b
This can be solved for C1 and C2 using Cramer’s rule, and the particular solution can be written
after substituting these values into the general solution.
Similarly, doing the same thing for the second coordinate equation should give:
C1 (ωny2 − ω 2 ) + 2ξωnyωC2 = 0
c
−C1 2ξωnyω + (ωny2 − ω 2 ) C2 = 0
d
Solving these also for C1 and C2 using Cramer’s rule should also result in the particular solution
after for this coordinate.
22
Class work: Having gone through the steps, get the amplitudes and phase angle expressions for
both coordinates.
References
1. Khurmi, R.S and Gupta, J.K. 2003. Theory of Machines, S. Chand, New Delhi.
2. Meirovitch, L. 2001. Fundamentals of Vibrations, McGraw-Hill, New York.
23
7.0 TORSIONAL VIBRATIONS
Equilibrium position
θ
Position after time, t seconds
Fig.1: Shaft-Disc system in a vertical position
Consider the Fig. 1 above. When the particles of the shaft (of negligible mass) move in a circle
about its axis, the vibration so produced is called Torsional vibrations. This causes the twisting
and untwisting of the shaft thereby producing what is known as torsional shear stresses in the
shaft .
7.1 Natural frequency of free Torsional Vibrations
From Fig. 1 above, the circular disc is being carried by a shaft of negligible mass. If the shaft is
displaced from its mean position through angle θ, after time t seconds, a restoring force tends to
bring it back to its original equilibrium position. Thus meaning that, the force of displacement is
countered by the restoring force.
Hence,
Force of displacement = restoring force.
Thus,
I&times;
d 2θ
= −q &times; θ
dt 2
1
Giving:
I&times;
d 2θ
+ q &times;θ = 0
dt 2
Dividing through by I gives:
24
d 2θ q
+ &times;θ = 0
dt 2 I
2
Eq.2 is represented with the fundamental equation of simple harmonic:
d 2θ
+ ω 2 &times;θ = 0
dt 2
3
⇒ θ + ω 2θ = 0
Comparing Eqs.2 and 3 gives:
q
I
ω=
4
Therefore, the period of torsional vibrations is:
T=
2π
ω
= 2π
q
I
5
And the natural frequency is:
f =
1
1
=
T 2π
I
q
6
Where I = Mass moment of inertial of the disc = m x k2 (kgm-2), k = radius of gyration (m),
m = mass of disc (kg), θ = Angular displacement of the shaft from the equilibrium position after
it has moved for a time t seconds and q = Torsional stiffness constant of the shaft (Nm).
However, the shaft’s stiffness constant can be determined by:
Restoring force, F = q &times; θ
(from Eq.1)
Hence,
q=
F
θ
But, the torsion equation is given as:
F C &times;θ
=
J
l
⇒q=
C&times;J
l
Where C = modulus of rigidity for the shaft material, J = polar moment of inertia of the crosssection =
π
32
d 4 , d = diameter of the shaft and l = length of shaft.
25
Reference
1. Khurmi, R.S and Gupta, J.K. 2003. Theory of Machines, S. Chand, New Delhi, p 938965.
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