NPTEL – Mechanical Engineering – Nonlinear Vibration

Module 2
Development of Equation of Motion for Nonlinear vibrating systems
In this module following points will be discussed for deriving the governing equation of
motion of a system

 Force and moment based approach

• Newton’s 2nd Law
• Generalized d’Alembert’s Principle
 Energy based Approach
• Lagrange Principle
• Extended Hamilton’s Principle
 Temporal equation using Galerkin’s method for continuous system
 Ordering techniques, scaling parameters, book-keeping parameter
 Examples of Commonly used nonlinear equations: Duffing equation, Van der Pol’s
oscillator, Mathieu’s and Hill’s equations

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Lecture M2 L01
Force and Momentum based Approach

In this approach one uses Newton’s second law of motion or d’Alembert’s principle to
derive the equation of motion. This is a vector based approach in which first one has to
draw the free body diagrams of the system and then write the force and moment
equilibrium equations by considering the inertia force and inertia moment of the system.

According to Newton’s second law when a particle is acted upon by a force it moves so
that the force vector is equal to the time rate of change of the linear momentum vector.

Consider a body of mass m positioned at a distance r from the origin of the coordinate
system XYZ as shown in Figure 2.1.1 is acted upon by a force F . According to
Newton’s 2nd Law, if the body has a linear velocity v , linear momentum vector p = mv ,
the external force is given by the following equation.

 dp d (mv )
=
F = (2.1.1)
dt dt

F
Z m

v
r

Figure 2.1.1: A body moving in XYZ plane under the action of a force F


Considering r to be the absolute position vector of the particle in an inertial frame, the
absolute velocity vector can be given by

 dr 
=v = r (2.1.2)
dt
the absolute acceleration vector is given by

 d 2 r 
a= 2 =r (2.1.3)
dt
Assuming mass to be time invariant,
 d  dr  d 2r


Hence= F  m=  m= ma (2.1.4)
dt  dt  dt 2

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     
Equation (2.1.4) can also be written as , F= − ma 0 or , = F + Fi 0 where Fi = −ma is
the inertia force. This is d’ Alembert’s principle which states that a moving body can be

brought to equilibrium by adding inertia force Fi to the system. In magnitude this inertia
force is equal to the product of mass and acceleration and takes place in a direction
opposite to that of acceleration. Now two examples are given below to show the
application of Newton’s 2nd law or d’ Alembert’s principle to derive the non linear
equation of motion of some systems.

Example 2.1.1: Use Newton’s 2nd law to derive equation of motion of a simple pendulum

Y
ĵ
T

X
θ
l θ

iˆ
m mg

Figure 2.1. 2: (a) simple pendulum (b) Free body diagram

Solution: Figure 2.1.2 (a) shows a simple pendulum of length l and mass m and Figure
2.1.2(b) shows the free body diagram of the system. The acceleration of the pendulum
can be given by lθ ˆj − lθ 2iˆ . From the free body diagram total external force acting on
the mass is given by
(2.1.5)
 
Now using Newton’s second law’s of motion i.e., F = ma

F =(−T + mg cos θ )iˆ − mg sin
= θ ˆj m(lθ ˆj − lθ 2iˆ) (2.1.6)
Now equating the real and imaginary parts one can get the equation of motion and the
expression for the tension. The equation of motion is given by
g
mlθ + mg sin θ =0 or θ + sin θ =0 (2.1.7)
l
and the expression for tension can be given by
(2.1.8)

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θ3 θ5 θ7
Taking sin θ =θ − +
− , the nonlinear equation of motion of the system up to
3! 5! 7!
7th order nonlinear term can be given by
 g θ3 θ5 θ7 
θ + θ − + − +  = 0 (2.1.9)
l 3! 5! 7! 
g θ3 θ5 θ7 
Or, θ +  θ − + − +  = 0 (2.1.10)
l 6 120 5040 
It may be noted that for higher power of θ , the coefficient become very small and hence
the higher order terms can be neglected.

Keeping up to 5th order, the equation can be written as

 g θ3 θ5 
θ + θ − + = 0 (2.1.11)
l 6 120 
which is a form of Duffing equation with cubic and quintic nonlinearities.

One may derive the same equation using the fact that the moment of a force about a fixed

point M 0 is equal to the time rate of change of the angular momentum about poin H 0 . In
 
mathematical form it can be written as M 0 = H 0 . Refereeing to Figure 2.1.2(b)

 
M 0= r × F (2.1.12)

Or, M 0 = (liˆ) × ( mg cos θ − T ) iˆ − mg sin θ ˆj  (2.1.13)

=
Now, H 0 =
d
dt
( )
ml 2θ kˆ ml 2θkˆ (2.1.14)

Or, − mgl sin θ kˆ = ml 2θ kˆ (2.1.15)

Or , ml 2θ + mgl sin θ = 0 (2.1.16)
g
Or, θ + sin θ = 0 (2.1.17)
l
Keeping up to cubic nonlinearity Eq. (2.1.17) can be written as

g g θ3
θ + θ − =
0 (2.1.18)
l l 6

Taking the length of the pendulum 1 m and acceleration due to gravity as 10 m/s2 the
equation of motion can be written as

θ + 10θ − 1.6667θ 3 =

0 (2.1.19)

It may be noted that the coefficient for the cubic order term is very less than that of the
linear term. A MatLab code is given below to obtain the variation of restoring force

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with θ which will give an idea regarding the approximation one has to take while
writing the equation of motion.
The equation is similar to a Duffing equations with soft type cubic nonlinearity.

Matlab code for restoring force plot

%Plot for restoring Force of a simple pendulum

% Written by S. K. Dwivedy on 30th May 2012
% th= theta
%L= length of pendulum

L=1;
g=9.8;

th=-pi:pi/100:pi;
f=(g/L)*(th-(1/factorial(3))*th.^3); %upto cubic order
f3=(g/L)*sin(th); % Actual
f1=10*th; % linear approximation
f5=(g/L)*(th-(1/factorial(3))*th.^3+(1/120)*th.^5); %uto quintic order
f7=f5-(g/L)*(1/factorial(7))*th.^7; %uto 7th order

plot(th,f,th,f1,'r',th,f3,'v',th,f5,'g',th,f7,'b')
grid on

xlabel('\theta')
ylabel('Restoring force')

linear

7th order

Fig. 2.1.3: Different approximation of the restoring force.

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Figure 2.1.3 shows the restoring force for actual, linear, cubic, 5th order and 7th order
approximation. It may clearly be noted that depending on the range of θ one may take
the approximation accordingly.

Example 2.1.2: Derive equation of motion for a nonlinear spring-mass-damper system

as shown below. Consider the spring force in the form of f = kx + α x3 and damping
s
force equal to f = cx + β x 2
d
x(t)

m X Fs
f(t)
Fi f (t )
m

Fd
(a) Free body diagram (b)

Figure 2.1.4 (a) Nonlinear spring-mass-damper system (b) free body diagram

Solution: Taking unit vector along positive X direction as iˆ , if a small displacement x(t )
is given to the mass m, as shown in Fig. 2.1.4(b), spring force, damping force and inertia
force will act in a direction opposite to that of the external force f (t ) . Now applying d’
Alembert’s principle one can write the following equation.
 
∑ F + F=i f (t )iˆ − ( Fs + Fd + Fi )=
iˆ 0 (2.1.20)

Or, (2.1.21)

Or, mx + cx + kx + α x3 + β x 2 =f (t ) (2.1.22)

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Example 2.1.3: Derive the equation of motion of a pendulum of length l mass m which is
attached to a mass less moving support as shown in Figure 2.1.5.

T − mlθ 2
Y (t )

ĵ mlθ

iˆ
m m ( g − Y)

Figure 2.1.5: (a) Simple pendulum attached to a periodically translating support, (b) free body diagram of the mass.

Solution: Considering the free body diagram as shown in Fig. 2.1.5 (b), the body is under
dynamic equilibrium under the action of tension, apparent weight and inertia force.
Fixing unit vector iˆ and ĵ as shown in Figure 2.1.5(b) and applying Newton’s 2nd Law
one can write

(
F =(−T + m ( g − Y) cos θ )iˆ − m ( g − Y) sin θ ˆj =m lθ ˆj − lθ 2iˆ ) (2.1.23)

Separating the ith and jth component of the forces and equating them to 0 one obtains the
expression for the tension and the governing equation of motion as given below.

m ( g − Y) cos θ + mlθ 2

T= (2.1.24)
−m ( g − Y) sin θ =
mlθ
(2.1.25)
 g Y 
Or, θ +  −  sin θ = 0 (2.1.26)
l l 
Taking the oscillation to be very small, sin θ ≈ θ and hence Eq.(2.1.26) reduces to

g Y ( t )
θ + θ − θ=
0 (2.1.27)
l l
It may be noted from the 3rd term in Eq. (2.1.27) that the coefficient of the response θ is
a time varying parameter. Hence this type of system is known as parametrically excited
system and this equation is known as Mathieu-Hill type of equation. Taking cubic order
nonlinear term, this equation will become the equation of a parametrically excited system
with cubic nonlinearities.

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g Y (t )   θ3 
θ +  −  θ −  (2.1.28)
l l  6 

Exercise Problems

Problem 2.1.1 Derive the equation motion of a compound pendulum.

Problem 2.1.2 Derive the equation of motion a tuned vibration absorber considering the
primary spring force equal to F=s
k1 x + k2 x 3
Problem 2.1.3 Derive the equation motion of the following system. Consider the spring to
be a nonlinear spring having the spring force= Fs K ( x − ε x 3 ) . Assume other elements to
be linear.

F(t)
K
m
L
• M
C
a b

Figure 2.1.6 System for exercise problem 2.1.3.

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Lecture M 2 L02

Derivation of Equation of motion for Multi-degree of freedom systems

In this lecture the nonlinear governing equation of motions of multi-degree of freedom

nonlinear systems will be derived by using Newton’s 2nd Law or d’Alembert’s principle.
The approach is similar to that of the single degree of freedom system. One can derive
the equation of motion by drawing the free body diagrams and then writing the force or
moment equilibrium equations by including the inertia force. Let us consider following
simple examples to derive the equation of motions.

Example 2.2.1: Derive the equation motion of system shown in Fig. 2.2.1. Consider the
last spring to be nonlinear where the spring force is given by =
Fs k3 x + k4 x . Consider 2

other spring and damper behaviour to be linear.

k3 , k 4
m1 m2 m3

Figure 2.2.1. A multi degree of freedom system

Solution
Considering the equilibrium of the mass m1 ,

k1x1 k2 ( x1 − x2 )
m1 m1
x1
c1x1 c2 ( x1 − x2 )
Figure 2.2.2: Free body diagram of part with mass m1

k2 ( x2 − x1 ) k3 ( x2 − x3 ) + k4 ( x2 − x3 )
2

m2 m2 
x2
c2 ( x2 − x1 ) c3 ( x2 − x3 )
Figure 2.2.3: Free body diagram of part with mass m2

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Equating the forces acting on mass m1 as shown in Fig. 2.2.2 one obtains

x1 + k1x1 + c1x1 + k2 ( x1 − x2 ) + c2 ( x1 − x2 ) =0.

m1 (2.2.1)

Similarly considering the free body diagram for the 2nd mass the equation of motion can
be written as

(2.2.2)

From the free body diagram shown in Fig. 2.2.4, the equation of motion for the 3rd mass
can be given by

=0 (2.2.3)

k3 ( x3 − x2 ) + k4 ( x3 − x2 )
2

m3 m3
x3
c3 ( x3 − x2 )
Figure 2.2.4: Free body diagram of part with mass m2

It may be noted that as the last spring is connected to both second and third masses, the
obtained second and third equations are nonlinear. So the equation of motions of the
system can be written as

x1 + k1x1 + c1x1 + k2 ( x1 − x2 ) + c2 ( x1 − x2 ) =

m1 0 (2.2.4)

(2.2.5)

(2.2.6)

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Exercise Problems:

Prob. 2.2.1: Derive the equation of motion of the nonlinear vibration absorber as shown
in Fig. 2.2.3. Consider spring k1 and k2 to be nonlinear with linear, quadratic and cubic
nonlinear coefficients. [Ref: Y.A. Amer, A.T. EL-Sayed, Vibration suppression of non-
linear system via non-linear absorber, Communications in Nonlinear Science and
Numerical Simulation, Volume 13, Issue 9, November 2008, Pages 1948-1963]

Figure 2.2.5: Nonlinear Vibration absorber.

Answer:

(2.2.7)

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Prob. 2.2.2: Find the equation of motion of the following system which models the
vibration control of ultrasonic cutting via dynamic absorber [Ref: Y.A. Amer, Vibration
control of ultrasonic cutting via dynamic absorber, Chaos, Solitons & Fractals, Volume
33, Issue 5, August 2007, Pages 1703-1710]

Figure 2.2.6: System for vibration control of ultrasonic cutting via dynamic absorber.

Prob. 2.2.3: Derive the equation of motion of a two stage gear system with mesh
stiffness fluctuation, bearing flexibility and backlash [Ref: Lassâad Walha, Tahar
Fakhfakh, Mohamed Haddar, Nonlinear dynamics of a two-stage gear system with mesh
stiffness fluctuation, bearing flexibility and backlash, Mechanism and Machine Theory,
Volume 44, Issue 5, May 2009, Pages 1058-1069 ]

Figure 2.2.7: Two stage gear considering nonlinear coupling

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Prob. 2.2.4: Derive the equation of motion of a pair of spur gear considering backlash.
The system is shown in Fig. 2.2.7) [Ref: Hamed Moradi, Hassan Salarieh, Analysis of
nonlinear oscillations in spur gear pairs with approximated modelling of backlash
nonlinearity, Mechanism and Machine Theory, Volume 51, May 2012, Pages 14-31]

Figure 2.2.8: Modeling of a pair of spur gear considering backlash

Answer:
( )
I1θ1 + cr1 r1θ1 − r2θ2 + kr1 ( r1θ1 − r2θ 2 ) =
T1
(2.2.8)
I θ + cr ( rθ − r θ ) − kr ( rθ − r θ ) =
2 2 2 1 1 2 2 2 1 1 2 2T 2

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Lecture M 2 L03
Derivation of the equation of motion of continuous system using d’Alembert’s
principle.

In this lecture, with help of example we will derive the governing equation of motion of a
continuous or distributed mass system using d’Alembert’s principle. It may be noted that
in previous two lectures we considered discrete system in which the governing equation
of motions are in the form of ordinary differential equations. But in continuous system
the governing equations are in the form of partial differential equation as the state vector
(e.g., displacement) depends not only on time but also on the space co-ordinates. For
example in case of axial vibration of a bar the axial displacement of the bar depends on
the time and location of the point on the bar at which the displacement has to be
measured. Also, it may be noted that, unlike discrete system where the natural
frequencies of the system has a definite value, in case of continuous system the system
has infinite number of natural frequencies. Depending on particular applications, one may
convert the analysis of continuous system to that of a multi-degree of freedom system by
considering finite participating modes in the analysis.

Example 2.3.1: Figure 2.3.1 shows a roller-supported base excited cantilever beam with
tip mass. In practical application it can be a single-link flexible Cartesian manipulator
with a payload of mass M. The left end of the manipulator is roller-supported which is
subjected to harmonically varying support motion Y = b (t ) Z cos Ω1t . The right end of the
manipulator is subjected to a sinusoidally varying axial force P(t) =+ P0 P1 cos Ω 2t . The
motion of the manipulator is considered to be in the vertical plane. Derive the governing
equation of motion using d’Alembert’s principle.
P(t)
Y
M

v s θ

u
X

=
Yb Z cos Ω1t
Figure 2.3.1 Schematic diagram of a single-link Cartesian manipulator with payload subjected to harmonically varying axial
force.

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Solution:
Here, the Cartesian manipulator with payload is modeled as a roller-supported Euler–
Bernoulli beam with a tip mass. The thickness (h) of the beam is considered to be very
small in comparison to the length of the beam (L). Hence, the effects of the shear
deformation and rotary inertia of the beam are neglected. The transverse vibration (v) of
the beam is assumed to be purely planar. The torsional mode of the beam is neglected in
this analysis. Payload mass is considered as a point mass which is symmetrically placed
with respect to the centerline of the beam. The harmonically varying tip load is acting
always in the tangential direction of the elastic line and the amplitude of the axial force is
taken less than the critical buckling load.
The governing equation of motion of the present system is derived using d’ Alembert’s
principle. Considering a small element at a distance s from the roller-supported end
(Fig.2 3.1) along the elastic line of the beam, the bending moment M ( s ) of the beam can

be expressed as:
 1 
M ( s ) ≈ E I  vss + vs2 vss  . (2.3.1)
 2 
Here, v is the transverse displacement of the beam. ( ) is the first derivative with respect
s
to s along the beam. Following Zavodney and Nayfeh (1989), and Cuvalci (1996, 2000),
one may write the inextensibility condition of the beam in terms of the longitudinal
displacement u (ξ , t ) and the transverse displacement v(ξ , t ) as:
ξ 1
v + (1 + us ) = or, u ( ξ, t ) = ξ − ∫ (1 − vη2 ) 2 d η .
2 2
s 1. (2.3.2)
0

Here, ξ, η are the integration variables. Considering the inertia forces per unit length of

the beam ρ Au, and ρ A ( v + Yb ) , in longitudinal and transverse directions, respectively

and inertia forces of the tip mass in longitudinal and transverse directions as
M u and M ( v + Yb ) , respectively, one may write equation (2.3.1) as follows:

M (s) − Mξ (s) − M L (s) =

0. (2.3.3)

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Here, M ξ ( s ) is the moment due to inertia force at a distance ξ from the roller-supported

end and M L ( s ) is the moment due to inertia force for the payload at the tip of the

manipulator and their expressions are given below:

L ξ L ξ

M ξ ( s ) =− ∫ ρ Au∫ sin θ d η d ξ − ∫ ρ A ( v + Yb ) ∫ cos θ d η d ξ , (2.3.4)

s s s s

L L L
and, M L ( s ) =− M u∫ sin θ d ξ − M ( v + Yb ) ∫ cos θ d ξ − P ( t ) ∫ sin θ d ξ . (2.3.5)
s s s

Considering equivalent viscous damping force cd v due to interaction of the system with
the environment and by differentiating Eq. (2.3.3) twice with respect to s, using the
Leibniz’s rule and applying the binomial expansion, one may obtain the following
governing differential equation of motion.

(
EI vssss +
1 2
2
3
)
s

0
(  )
vs vssss + 3vs vss vsss + vss + ρAvs ∫  vξ2 + vξ vξ d ξ + M ( v+Y
 b ) vs vss + vs vss

( ) L 
ξ s

( ξ ξ) ( ξ ξ ξ) +
L

∫
ρAYb ( L − s ) + ∫ ( ρ Av+ cd v ) d η − vss  ρA ∫ v
 + v v
 d ξdη + M ∫ v
 + v v
 d ξ
2 2
ξ
s
s 0 0 
 1 2
1 − vs  ( ρA ( v+Y
 b ) + cd v ) + ( P (t ) vs ) s =
0. (2.3.6)
 2 
Example- 2.3.2
Derive the equation motion of a string fixed at one end and attached by a nonlinear spring
at the other end.
Solution

 ∂T ( x )   ∂w ( x, t ) ∂ 2 w ( x, t )  ∂w ( x, t )
 ( )
T x + dx   + dx  + f ( x, t ) dx − T ( x )
 ∂x   ∂x ∂x 2
 ∂x
∂ 2 w ( x, t )
= ρ ( x ) dx , 0< x<L (2.3.7)
∂t 2
∂  ∂w ( x, t )  ∂ 2 w ( x, t )
T ( x )  + f ( x, t ) =
ρ ( x ) dx , 0< x<L (2.3.8)
∂x  ∂x  ∂t 2
Subjected boundary conditions
∂w ( x, t )
w ( x, t ) = 0, at x = 0 and T ( x ) + K1w ( x, t ) + K 2 w3 ( x, t ) =0, at x = L (2.3.9)
∂x

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Equation (2.3.8) is partial differential equation of motion and (2.3.9) are the boundary
conditions.

To develop different nonlinear equations of motion for string systems, one may refer the
following papers on nonlinear vibration of strings.

1. G.S.S. Murthy, B.S. Ramakrishna, Non-linear character of resonance in stretched

strings, J. Acoust. Soc. Am., 38 (1965), p. 461
2. J.W. Miles, Stability of forced oscillations of a vibrating string, J. Acoust. Soc.
Am., 38 (1965), p. 855
3. G.V. Anand, Non-linear resonance in stretched strings with viscous damping, J.
Acoust. Soc. Am., 40 (1966), p. 1517.
4. E.W. Lee, Non-linear forced vibration of a stretched string, Br. J. Appl. Phys., 8
(1957), p. 411
5. D.W. Oplinger, Frequency response of a non-linear stretched string, J. Acoust. Soc.
Am., 32 (1960), p. 1529
6. G.F. Carrier, On the non-linear vibration problem of the elastic string, Q. Appl.
Math., 3 (1945), p. 157

Exercise Problems

Prob. 2.3.1. Derive the equation of motion of a base excited cantilever with an attached
mass at arbitrary position as shown in Fig. 2.3.2. (Ref: Zavodney and Nayfeh(1989),
Prob.2.3.2: Derive the equation of motion of a dynamic vibration absorber as shown in
Figure 2.3.3.
Prob. 2.3.3: Derive the equation motion of the moving belt system shown in Fig. 2.3.4
[C.A. Jones, P. Reynolds, A. Pavic, Vibrational power flow in the moving belt passing
through a tensioner, Journal of Sound and Vibration, Volume 330, Issue 8, 2011, Pages
1531-156 ]

Answer:

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Lecture M2 L04
Derivation of equation of motion using Extended Hamilton’s Principle

The purpose of this lecture is to use extended Hamilton’s principle to derive the equation
of motion of different systems. According to this method, for a system with kinetic
energy T, potential energy U and virtual work done by the non-conservative force δ Wnc ,
the governing equation motion can be obtained by using the following equation.
t2

∫ (δ (T − U ) + δ W )dt = 0, δ r ( t ) = δ r ( t ) = 0, i = 1,2,....n
t1
nc i 1 i 2 (2.4.1)

Here, t1 and t 2 are the time at which it is assumed that the virtual displacements δ ri for a
system represented by n physical co-ordinates ( ri ) vanishes. Using Lagrangian
( L= T − U ) and m generalized co-ordinates qi of the system, the above equation can be
written as
t2

∫ (δ L + δ W )dt =
t1
nc 0, δ q ( t ) =
δ q (t ) =
i 1 0, i =
1,2,....m
i 2 (2.4.2)

Equation (2.4.1) and (2.4.2) are the equation for Extended Hamilton’s principle. For a
conservative system as δ Wnc = 0 , Eq. (2.4.2) reduces to
t2

qk ( t1 ) δ=
∫ δ Ldt 0, δ=
= qk ( t 2 ) 0 (2.4.3)
t1

which is known as the Hamilton’s Principle.

This method is particularly useful for continuous systems where one can obtain both
governing equation of motion and boundary conditions of the system.

Let us derive the equation of motion of few linear and nonlinear systems to get
familiarize with the application of this method.

Example 2.4.1: Derive the equation of motion of a simple pendulum using extended
Hamilton’s principle.

Solution: In this case the kinetic energy T and potential energy U of the system can be
given by

1
( )
m lθ =
2
=T , U mgl (1 − cos θ ) (2.4.4)
2
1
( )
So, L =T − U = m lθ − mgl (1 − cos θ )
2
(2.4.5)
2
Now, applying Hamilton’s principle one can write

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t2

∫=
δ Ldt 0, δθ
= ( t1 ) δθ=
( t2 ) 0 (2.4.6)
t1

t2
1 
∫ δ  2 m ( lθ )
2
− mgl (1 − cos θ ) dt =
0,
t1 
t2
1 
or, ∫  2 m2l ( lθ ) δ (θ ) − mgl sinθδθ dt =
t1
0,

t2
 d 
or, ∫  ml 2θ (δθ ) − mgl sinθδθ dt =
0
t1  dt 

(2.4.7)
The first term (marked in red colour) tends to zero as δθ
= ( t1 ) δθ=
( t2 ) 0 . As the virtual
displacement δθ is arbitrary, hence the coefficient of δθ dt term should vanish. Therefore
one obtains
ml 2θ + mgl sinθ =0 (2.4.8)
th
as the equation of motion of the simple pendulum. Taking up to 5 order terms this
equation can be written as
g θ3 θ5 
θ +  θ − + = 0 (2.4.9)
l  6 120 

Example 2.4.2: Derive the equation of motion for the transverse vibration of an Euler-
Bernoulli beam with fixed-free boundary condition subjected to axial periodic load as
shown in Fig. 2.4.1.
Solution:
Let us first derive the equation of motion of the system considering small displacement of
the system.

The kinetic energy T of the beam can be given as follows:

L
1
T = ∫ mw,2t dx (2.4.10)
20

where, m is the mass of the beam per unit length and ( ),t represents the differentiation
with respect to time.

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P (t )

Fig: 2.4.1: Schematic diagram of a cantilever beam under transverse vibration due to application of a periodic axial load

The potential energy of the system is due to the strain energy of the system and is given
by
L
1
U = ∫ EIw,2xx dx (2.4.11)
20
L L
1 1
Hence the Lagrangian of the system L = T − U = ∫ mw,2t dx − ∫ EIw,2xx dx (2.4.12)
20 20

Assuming inextensible beam condition, there will be no elongation in the axial direction
along the neutral axis of the beam. The longitudinal deformation u of the beam due to
transverse deformation w can be expressed as
2 2
 du   dw 
1 +  +   =
1 (2.4.13)
 dx   dx 
Rearranging Eq. (2.4.13) and using a first order Taylor series expansion, the following
relationship can be obtained.
2
1  dw 
L
u = − ∫  dx (2.4.14)
2 0  dx 

The work done due to the nonconservative axial force can be given by

1 L 2 
δ Wnc =
− Pδ u = Pδ  ∫ w′ dx  (2.4.15)
2 0 
Using Hamilton’s principle

∫ δ ( L + W )dt = 0, δ w ( t1 ) = δ w ( t2 ) = 0
t2
nc (2.4.16)
t1

Using Eq (2.4.12) and Eq. (2.4.15) in Eq. (2.4.16) one can write

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 1 L 1
L
 1  L 2 
∫ (δ L + δ=
W )dt ∫
t2 t2
δ  ∫ mw,t dx − ∫ EIw, xx dx  + Pδ  ∫ w, x dx  dt
2 2
nc (2.4.17)
  2 0 20  2 0  
t1 t1

 ∂w   ∂w  t2    ∂2w   ∂2w  ∂w  ∂w   
L L
t2
or, ∫ ∫0  ∂t   ∂t 
m δ dxdt − ∫t1  ∫0   ∂x 2   ∂x 2  ∂x δ  ∂x   dx dt =
  EI   δ   − P 0 (2.4.18)
   
t1

L  t2  ∂w  ∂  t2  
L
 ∂ 2 w  ∂   ∂w   ∂w ∂  
or, ∫0  ∫t  ∂t  ∂t ( )  ∫t1  ∫0   ∂x 2  ∂x  δ  ∂x   − P ∂x ∂x (δ w)  dx dt (2.4.19)
 m δ w dt dx − EI
1     
Using integration by parts Eq.(2.4.19) can be written as,
L
 ∂w   t2  ∂ 2 w 
L 
∫0 m  ∂t  ( δ w )t dx − ∫  ∫ m  2  δ wdt dx
t2

1 
0  t1  ∂t  
= 0 as per defination

  ∂ 2 w    ∂w   L
t2
 ∂3w   ∂w  
− ∫  EI  2   δ  ( δ ) ( δ )
L L
 + EI   w − P   w  dt
  ∂x    ∂x  0  ∂x   ∂x 
3

0 0
t1  
 
Boundary conditions

  ∂4w
t2 L
∂2w  
− ∫  ∫  EI 4 + P 2 δ wdx  dt =
0 (2.4.20)
t1  0 
∂x ∂x  

  ∂ 2 w    ∂w   L
t2
 ∂3w   ∂w  
− ∫  EI  2   δ  ( δ ) ( δ )
L L
 + EI  3 w − P   w  dt
t1 
  ∂x    ∂x    ∂x 
0
 ∂x 
0

  
0

Boundary conditions
Or,
 
  ∂2w 
t2 L
∂ w4
∂ w
2
− ∫ ∫ m  2  + EI 4 + P 2 δ wdxdt = 0

t1 0   ∂t 
  ∂x ∂x 
 =0, 
 Equation of motion  (2.4.21)

As the virtual displacement δ w is arbitrary, hence the right hand side of the equation will
be zero only if
 ∂2w  ∂4w ∂2w
m  2  + EI 4 + P 2 =0 (2.4.22)
 ∂t  ∂x ∂x

which is the equation of motion of the system. The boundary conditions can be obtained
from the term marked in blue colour in Eq. (2.4.21). Now taking the periodic axial load
as
P =+
P0 P1 cos Ωt , Eq. (2.4.22) can be written as

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∂2w ∂4w ∂2w

m + EI + ( P + P cos Ωt ) =
0 (2.4.23)
∂t 2 ∂x 4 ∂x 2
0 1

Due to the presence of a periodic term (marked in pink colour) as the coefficient of the
term containing the response (marked in blue colour), the system is a parametrically
excited system.

Exercise Problems:

Problem 2.4.1: Derive the equation of motion of a cantilever beam subjected to magnetic
field using extended Hamilton’s principle.

Fig. 2.4.2: Schematic diagram of a cantilever beam subjected to magnetic field.

Hints: The expression for kinetic and strain energy of the system can be taken similar to
that taken in example 2.4.2.
L
1
T = ∫ mw,2t dx (2.4.24)
20

L
1
(U t )bending = ∫ Et It w,2xx dx (2.4.25)
20
Considering conductive material, the magnetoelastic load applied to the beam is
equivalent to the horizontal force n and the distributed moment m which are expressed
in terms of the longitudinal displacement ( u ) and transverse displacement ( w ) as (Zhou
and Wang [1] ), Moon and Pao [2]
B0 2bh
n= u,xx and (2.4.26)
µe

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 
B 2
bh  π h x  B0 2bh3
=mm 0
 u − w ln + w,x  − w, xxx . (2.4.27)
µ0  2 ln x , x 2π , xx L − x  12 µe
 L−x 
Here, µ0 and µe are respectively the permeability of the free space and the beam
materials.
The non-conservative work done due to the applied axial periodic load and the above
mentioned magnetoelastic loads and moments can be given by
L L
1
W= ∫ + ∫  ntδ ut + nbδ ub + mtmδ w, x + mbmδ w, x  dx
2
nc Pw,x dx (2.4.28)
20 0

Problem. 2.4.2: Derive the equation motion of a base excited cantilever beam with
arbitrary mass position using extended Hamilton’s principle. (Refer: Kar and Dwivedy
1999 for the derivation using d’Alembert’s principle)

References for Further Reading

[1] G. Y. Zhou, Q. Wang, Use of Magnetorheological Elastomer in an Adaptive

Sandwich Beam with Conductive Skins. Part I: Magnetoelastic Loads in Conductive
Skins, International Journal of Solids and Structures 43, 5386-5402, 2006.
[2] F.C. Moon and Y. H. Pao, 1969, Vibration and dynamic instability of a beam-plate in
a transverse magnetic field, Journal of Applied Mechanics 36, 92–100, 1969
[3] R. C. Kar and S. K. Dwivedy, Non-linear dynamics of a slender beam carrying a
lumped mass with principal parametric and internal resonances. International Journal of
Nonlinear Mechanics, 34 (3)515-529, 1999.

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Lecture M2 L05
Derivation of Equation of motion using Lagrange Principle
Both Hamilton’s principle and Lagrange principle are based on energy principle for deriving
the equation of motion of a system. As energy is a scalar quantity, the derivation of equation
of motion is more straight forward unlike the derivation based on Newton’s 2nd Law or d’
Alembert’s principle which are vector based approach. In the Newton’s or d’Alembert’s
approach, with increase in degrees of freedom of the system it is very difficult and time
consuming to draw the free body diagrams to find the equation of motion using force or
moment equilibrium. Hence it is advantageous to go for energy based approach. While in
Hamilton’s principle one uses a integral based approach, in Lagrange principle a differential
approach is followed. Hence, use of Lagrange principle is easier than the Hamilton’s
principle. Though all these methods in principle can be applied to any system, however it is
better to use Newton’s 2nd Law or d’Alembert’s principle for single or two degree of freedom
systems, Lagrange principle for multi degree of freedom and extended Hamilton’s principle
for continuous systems.

In Lagrange principle, generally the equations of motion are derived using generalized
coordinates. Let us consider a system with N physical coordinates and n generalized
coordinates. The kinetic energy T for a system of particles can be given by
T = T ( r1,r2,...........rN ,r1,r2,........rN , ) (2.5.1)
Where ri , and ri are the position and velocity vector of a typical particles of mass mi (i=1,2,..,
N). Considering qk and q k as the displacement and velocity in terms of kth generalized
coordinates, one may write,
n
∂r
ri = ∑ i q k (2.5.2)
k =1 ∂qk

So using generalized coordinate one may write,

T = T ( q1, q2 ......qn,q1, q 2 ......q n, ) (2.5.3)
n
 ∂T ∂T  
Hence, δ T
= ∑  ∂q δ qk +
∂q k
δ qk  (2.5.4)
 k
k =1 

The virtual work ( δ W )performed by the applied force Fi can be written in terms of
generalized forces and virtual displacement or

n
δ W = ∑ Qk δ qk (2.5.5)
k =1
 ∂r
where, Qk ∑
= = Fi . i , k 1,2,....n . (2.5.6)
∂qk
The over bar in δ W shows that the work done is a path function. Substituting (2.5.4) and
(2.5.5) into the extended Hamilton’s Principle,

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t2

∫ (δT + δ W )dt =
t1
q (t )
0 , δ= k 1 qk ( t 2 ) 0 , k=1,2,…..n
δ= (2.5.7)

one obtains the following equation.

t2
 n  ∂T ∂T   n 
 ∑
∫t k 1= δ q +
∂q k
δ q  + ∑ Q δ q dt =
0 (2.5.8)
  ∂qk
k k k k
1  k1 

Now,
∂T  ∂T d ∂T d  ∂T 
t2 t2 t2

∫t ∂q= δ q dt ∫t ∂qk dt ( δ ) δ ∫t dt  ∂qk  δ qk dt

t2
= q dt q −
k k k
∂q k  t1
1 1
=0 1
(2.5.9)
d  ∂T 
t2

= −∫   δ qk dt
t1
dt  ∂q k 

Substituting (2.5.9) in (2.5.8) we have

t2 n
 d  ∂T  ∂T 
∫ ∑ − dt  ∂q
t1 k =1 
+
k  ∂qk
+ Qk δ qk dt =
0 (2.5.10)


Considering the arbitrariness of the virtual displacement δ qk , equation ( 2.5.10) will be

satisfied for all values of δ qk provided
d  ∂T  ∂T
 − =Qk , k =1,2,…..n (2.5.11)
dt  ∂q k
 ∂qk
Equation (2.5.11) is known as Lagrange’s equation .
Considering both conservative force Qkc and nonconservative force Qnkc , the total generalized
force Qk can be written as
Q=k Qkc + Qknc (2.5.12)
and recalling potential energy depends on coordinates alone, the work done by the
conservative force Wc is equal to the negative of the potential energy V . Hence, one may
write

∂v n n
δ Wc = −∑
−δ U = δ qk = ∑ Qkcδ qk . (2.5.13)
= ∂qk
k 1= k 1

So the conservative generalized forces have the form

∂U
Qkc = − , k=1, 2,…..n (2.5.14)
∂qk

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Substituting Eq. (2.5.12) and Eq. (2.5.13) in Eq. (2.5.10) we have

d  ∂T  ∂T ∂U
 − + = Qknc , = k 1,2,, n. (2.5.15)
dt  ∂q k  ∂qk ∂qk
As the potential energy does not depend on velocity, using Lagrangian L= T − U , Eq.
(2.5.15) can be rewritten as
d  ∂L   ∂L 
 − =Qknc (2.5.16)
dt  ∂qk   ∂qk 

Using dissipation energy D, this equation further can be written as

d  ∂L   ∂L   ∂D 
 − + =Qknc (2.5.17)
dt  ∂qk   ∂qk   ∂q 
Using both external forces and moments one may write the generalized force as

 ∂r ∂ω
Qk = ∑i Fi . ∂qi + ∑i Mi . ∂q i , i =
1,2 N, k =
1,2,...., n (2.5.18)
k k

Mi is the vector representation of the externally applied moments, ωi is the system angular
velocity about the axis along which the considered moment is applied.

− Lagrange equation can be used for any discrete system whose motion
lends itself to a description in terms of generalized coordinates, which
include rigid bodies.
− can be extended to distributed parameter system, but such system, they
are not as versatile as the extended Hamilton’s Principle

Let us take some examples to derive the equation of motion using Lagrange principle.

Example 2.5.1: Derive the equation of motion of a spring-mass-damper system with spring
force given by F=
s kx − α x 3 and damping force given by F=
d cx − β x 2 x . The external force
=
acting on the system is given by F f1 sin ω1t + f 2 sin ω2t . Consider mass of the system as m
and displacement from the static equilibrium point as x .

Solution: In this single degree of freedom system one can take x as the generalized co-
ordinate. From the given expressions for different forces acting on the system, the expressions
for kinetic energyT, potential energy V, dissipation energy D can be given by the following
expressions.

1 2
T= mx ,
2 (2.5.19)
V =∫ Fs dx = ( )
1 2 1 4
∫ kx − α x dx =2 kx − 4 α x
3

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1 2 1 2 1 4
L = T −V = mx −  kx − α x  (2.5.20)
2 2 4 
=
D dx ∫ ( cx − β x x )dx
∫ F=
d
2
(2.5.21)
Qknc= F= f1 sin ω1t + f 2 sin ω2t (2.5.22)

Using Lagrange equation (2.5.17)

d  ∂L   ∂L   ∂D 
 − + =Qknc (2.5.23)
dt  ∂qk   ∂qk   ∂q 

  1 2  1 2 1 4     1 2  1 2 1 4  
 ∂  mx −  kx − α x     ∂  mx −  kx − α x   
d  2 2 4     2 2 4  
− +F =
F (2.5.24)
dt  ∂x   ∂x  d
   
   

d
or, ( mx ) − ( − Fs ) + Fd =F (2.5.25)
dt

(
or, mx + kx − α x 3 + cx − β =
x 2 x)f1 sin ω1t + f 2 sin ω2t (2.5.26)

Example 2.5.2
Use Lagrange Principle to derive equation of motion of the following system.

a x

A
M O F1 sin ω1t , iˆ

θ m, L
P F2 sin ω2t
Figure 2.5.1: Vibration of a spring-mass system with a pivoted link
Solution:
Let A is the position of the mass M at time t =0 when the link is in vertical position. Now it
has come to position marked O after some time t. The motion can be completely described in
terms of a physical coordinate system fixed at the fixed end. Also, one may use translation x
of mass M and rotation θ of the link as the generalized coordinates. Here,
=q1 x= and q2 θ . iˆ and ĵ are the unit vector along the horizontal and vertical direction as
shown in the figure. To find the kinetic energy of the link, first we have to determine the
velocity of the mass center of the link. The position vector of the mass center of the link is
  L  L
rc =  a + x + sinθ  iˆ − cos θ jˆ (2.5.27)
 2  2

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So the velocity
  L  ˆ L 
vc =  x + 2 cos θ θ  i + 2 θ sinθ j .
ˆ (2.5.28)
 

dr1 d ( a + x ) iˆ
Similarly velocity of mass= M is = xi ˆ (2.5.29)
dt dt
Hence, kinetic energy of the system which is due to the kinetic energy of the mass M and the
kinetic energy of link with mass m is
1   1   1 2
=T Mr1 ⋅ r1 + mv c .v c + Ι cθ (2.5.30)
2 2
  2

Translational KE Rotational KE

1 1 
= ( M + m ) x 2 + mLxθ cos θ + mL2θ2  (2.5.31)
2 3 
The potential energy of the system is due to the spring element and also due to the change in
height of the link. Considering a hard spring with cubic nonlinearity, the potential energyV of
the system can be given by the following equation.
1 1 L
V = kx 2 + α x 4 + mg (1 − cos θ ) (2.5.32)
2 4 2
As two forces are acting on the system, to find the generalized force first we have to find the
position vector of the point where the forces are acting. For the force F1 sin ω 1t the position

vector from the fixed coordinate system is r=1 ( a + x ) iˆ . Similarly, for the second force which

is acting on the pivoted link is r2 = ( a + x + L sinθ ) iˆ + ( −L cos θ ) jˆ . So the generalized
forces can be obtained by using Eq. (2.5.18) as follows.
2 

∂ri
Qknc = ∑ Fi . (2.5.33)
l =1 ∂qk

=

Q1nc F1 sin ω 1tiˆ.
∂ ( a + x ) iˆ
+ F2 sin ω 2 tiˆ.
(
∂ ( a + x + L sinθ ) iˆ + ( −L cos θ ) jˆ ) 
 ∂x ∂x  (2.5.34)
 
= F1 sin ω 1t + F2 sin ω 2 t

=


Q2 nc F1 sin ω 1tiˆ.
∂ ( a + x ) iˆ
+ F2 sin ω 2 tiˆ.
(
∂ ( a + x + L sinθ ) iˆ + ( −L cos θ ) jˆ ) 
θ  ∂θ 
 ∂ 
(2.5.35)
=0

=0 + F2L sin ω 2 t cos θ = F2L sin ω 2 t cos θ

Now using Lagrange Principle

d  ∂L  ∂L
 − =
Qknc (2.5.36)
dt  ∂q k  ∂qk

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Where
1 1  1 1 L 
L = ( M + m ) x 2 + mLxθ cos θ + mL2θ2  −  kx 2 + α x 4 + mg (1 − cos θ ) (2.5.37)
2 3  2 4 2 

For k=1
d
 ( M + m ) x + mLθ cos θ  + kx +=
α x 3 F1 sin ω 1t + F2 sin ω 2 t (2.5.38)
dt 
For k=2
d  1  1 l
 mlx cos θ + mL2θ  + mlxθ sin θ + mg sin θ = F2L cos θ sin ω 2 t (2.5.39)
dt  3  2 2
d   1
or,
dt  
1
3  2
(
mlx cos θ + mL2θ  + ml xθ + g sin θ = )
F2L cos θ sin ω 2 t (2.5.40)

Example 2.5.2: Using Lagrange Principle to find the equation of motion of the system shown
in Figure 2.5.2 . Spring K1 is under pretension T for small amplitude of vertical Oscillation
i.e., . Spring K2 is a soft spring with cubic nonlinearity.

Figure 2.5.2: Vibration of a spring mass system with additional pre-tensioned horizontal spring.

Solution
As spring K1 is under pretension T0 which is produced by an initial extension of the spring
by an amount δ 0 , one may write
T0 = K1δ 0 (2.5.41)
1
The kinetic energy of the system is T = mx 2 (2.5.42)
2
The potential energy of the system is due to the potential energy of the nonlinear spring K 2
and due to the linear spring K1 . Considering oscillations about the static equilibrium position,
the potential energy can be obtained as follows.
1 1 1
V (x) K1 (δ 0 + ∆l ) + K 2 x 2 − ε K 2 x 4
2
= (2.5.43)
2 2 4

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where ∆L is the change in length of the spring with stiffness K1 due to the motion x of the
mass. The coefficient of the cubic nonlinear term is assumed to be ε K 2 . The negative sign is

due to the soft spring assumption. For , from Fig. 2.5.3 one may write

( L)
2
∆L= L2 + x 2 − L= L 1 + x −L (2.5.44)

 2
1 x  1 x 
4
1 x 
6
 1  x 2 1  x  4 1  x 6 
≈ L  1 +   −   +    − L= L    −   + 
 2  L  8  L  16  L  
(2.5.45)
 2  L  8  L  16  L  
   
Hence the expression for potential energy is L + ∆L
x

L
Figure 2.5.3

2
1   1  x 2 1  x  4 1  x 6   1 1
V (x) = K1  δ 0 + L    −   +     + K2 x − ε K2 x
2 4
(2.5.46)
2    2  L  8  L  16  L    2 4
 
Taking the generalized coordinate q = x , the Lagrangian of the system can be written as
2
1 1   1  x 2 1  x  4 1  x 6    1 1 

L = T − V = mx − K1  δ 0 + L    −   +     −  K 2 x 2 − ε K 2 x 4 
2

2 2   2  L  8  L  16  L     2 4 
 
(2.5.47)
As no external force is acting on the system, the Lagrange Equation can be given by
d  ∂L  ∂L
− =
dt  ∂x  ∂x
0 (2.5.48)

Neglecting the two higher order terms marked in blue in Eq. (2.5.47) and applying (2.5.48)
one can get the following equation.
d 1   1  x 3 L 2x  1 1 
3

 m. 2 x  − − K1  2 + 2δ 0 . . 2  − K 2 . 2 x + ε K 2 .4 x  = 0 (2.5.49)
dt  2   2  L 2 L  2 4 
 1  x 3 x 

Or, mx − − K1  2 + 2δ 0  − K 2 x + ε K 2 x 3  =0 (2.5.50)
 2  L L 
 K  K 
Or, mx +  δ 0 1 + K 2  x +  12 − ε K 2  x 3 =
0 (2.5.51)
 L   2L 
T K  K 
Or, mx +  0 1 + K 2  x +  12 − ε K 2  x 3 = 0 (2.5.52)
 1
K L   2L 
T  K 
Or, mx +  0 + K 2  x +  12 − ε K 2  x 3 =
0 (2.5.53)
L   2L 

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Exercise Problems

Problem 2.5.1 Use Lagrange equation to derive the equation of motion of the following
system. Here, mass m2 is subjected to a periodic force f sin ωt . Also, it is connected to a
nonlinear spring in the right side.

f sin ωt
x1 x2
m1 m2

Figure 2.5.4: Multi degree of freedom system with nonlinear spring.

Hints:
=
Kinetic energy: T
1
2
(
m1q12 + m2q 22 )
1 2 1 
k1q1 + k 2 ( q2 − q1 ) + k3q22 + k 4q24 
2
Potential energy: V= 
2 2 

Rayleigh’s dissipation function can be written

1  2
c1q1 + c2 ( q 2 − q1 ) + c3q 22 
2
=
D
2 

Problem 2.5.2: Derive the equation of motion of the following system using Lagrange
fs k ( x + 0.1x 3 ) and the damping force as
principle. Consider the spring force as =
(
c x + 0.1x 2 x + 0.1x 3 .
fd = )

Figure 2.5.5: Vibration isolator with cubic nonlinear spring and damper
(Ref: Zhenlong Xiao, Xingjian Jing, Li Cheng, The transmissibility of vibration isolators
with cubic nonlinear damping under both force and base excitations, Journal of Sound and
Vibration, 332(5),1335-1354, 2013. )

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Problem 2.5.3: Derive the equation of motion of a vibration isolator modeled by a linear
spring and nonlinear damping. The nonlinear damping force can be given by
fd =c ( x + 0.1x 2 x + 0.01x 3 )

Figure 2.5.6: vibration isolators with cubic nonlinear damping under both force and base excitations

Answer: Governing equation of motion

mx1 = k ( u − x1 ) + c ( u − x1 ) + 0.1c ( u − x1 ) ( u − x1 ) + 0.01c ( u − x1 )

2 3

(Ref: Zhenlong Xiao, Xingjian Jing, Li Cheng, The transmissibility of vibration isolators
with cubic nonlinear damping under both force and base excitations, Journal of Sound and
Vibration, 332(5),1335-1354, 2013. )

Problem 2.5.4: Using Lagrange principle, derive the equation of motion of the shown system.
The variation of spring force with displacement (x) of the mass M is given by
F=
s 5 x + 0.5 x3 kN, the damping force is given by Fd = 0.2 x kN and the external applied
=
force Ft 2sin 5t + 5sin 4t kN.
Take M= 10 kg, L= 1m, a= b= 0.25m . Write the equation of the system using book-
keeping parameter. Consider the beam to be of negligible mass

K Ft
m
L
• M
C
a b

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Lecture M2 L06
Development of temporal equation of motion using Galerkin’s method for continuous
system

In this lecture one will learn the development of temporal equation of motion using
generalized Galerkin’s method for continuous system. It may be noted that unlike discrete
system where the equations are ordinary differential equations, in case of continuous or
distributed mass system the governing equations are partial differential equation as they
depend on both time and space variables. Hence it is required to reduce the partial differential
equation to ordinary differential equation for finding the solution of the system easily. In case
of vibrating system these equations are generally reduced to their temporal form by using
Galerkin’s method. In this method following steps have to be followed.

• Assume an approximate function for the mode shape of the continuous system. Here
one may take single or multi-mode approximation.
• Substitute the mode shape(s) in the governing partial differential equation of motion to
obtain the residue.
• Minimize the residue by using a weight function and equate it to zero to obtain the
temporal equation of motion.

One may take orthogonal functions for mode shapes and weight function to simplify the
integration to obtain the coefficients of the temporal equation. In nonlinear systems with many
terms, one may use symbolic software like Mathematica and Mapple to derive the equation of
motion. One may write a Matlab program having inbuilt integration schemes to obtain the
coefficients. The method is illustrated with the help of the following example.

Example 2.6.1: Consider the transverse vibration of a beam with roller supported at one end
and attached mass and periodically varying load at the other end. The roller supported end is
subjected to periodic motion. The governing equation of motion using d’Alembert’s principle
is given in Eq. (2.3.6). We have to derive the temporal equation of motion of the system.

P(t)
Y
M

v s
θ
u
X

=
Yb Z cos Ω1t
Fig. 2.6.1: Schematic diagram of a roller supported beam with tip mass and transverse follower load.

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Solution
Figure 2.6.1 shows the system with a payload of mass m at the tip where a compressive force
P = P0 + P1 cos Ω2t is applied. Also this system is subjected to a harmonic base

excitation Yb (t ) = Z cos Ω1t at the roller supported left end. Here Z and Ω1 are the amplitude

and frequency of the base excitation, P0 , P1 are the static and dynamic force amplitude, and

Ω 2 is the frequency of the periodic force acting at the free end of the manipulator. The
motion is considered to be in the vertical plane.
Using d’Alembert principle the equation of this system can be given by

3  + ρAv  v 2 + v v 
 1 s
EI  v ssss + v s2 v ssss + 3v s v ss v sss + v ss  +v v
s ∫ ξ ξ ξ  s ss
 2  0 
 L ξ 
L
( )dη + ρAYb (L − s ) + M (v + Yb )− vss   ∫ ρA ∫ vξ2 + vξ vξ dξdη + M  ∫ vξ2 + vξ vξ dξ 
s
 ∫ ρ A v + c v
s d
   s 0  0  


 2 
( ( b d ) 0 1 )
1 − 1 v2  ρA v + Y + c v + (P + P cos Ω t )v = 0
s 2 ss
(2.6.1)

Here, E , I , ρ , A, L and cd are the Young modulus, moment of inertia, mass density, area of
cross-section, length of the cantilever beam and damping factor of the system and
ζ,η respectively, are used as integration variables. To determine the temporal equation of
motion, one may discretize the governing equation of motion (2.6.1) by using following
assumed mode expression.
n
v ( s, t ) = ∑ r ψ i ( s ) qi ( t ) . (2.6.2)
i =1

Here, r is the scaling factor; qi ( t ) is the time modulation of the ith mode and ψ i ( s ) is the

eigenfunction of the cantilever beam with tip mass which is given by

 sin βi L + sinh βi L 
ψ i (s) =
−  ( cos βi s − cosh βi s ) + ( sin βi s − sinh βi s ) (2.6.3)
 cos βi L + cosh βi L 
One may determine βL from the following equation.

1 + cos β L cosh β L + mβ L ( cos β L sinh β L − sin β L cosh β L ) =

0 (2.6.4)

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The following non-dimensional parameters are used in the further analysis.

s Ω Ω r M EI Z Z
x= , τ = ωt , ω1 = 1 , ω2 = 2 , λ = , m = , χ= , r = , and Z = .
L ω ω L ρAL ρAL 4
r L
Substituting the above mentioned nondimensional parameters and equation (2.6.2) into
equation (2.6.1) one may obtain the residue equation R. Now taking ψ i ( s ) as the weight

function and using the generalized Galerkin’s method, one may write the following equation.

∫ Rψ i ( s ) dx = 0 (2.6.5)

This equation can be written in the following form which is the non-dimensional temporal
equation of motion of the system.
 m m m 
qn + q + 2 ε ζ qn + ε  ∑∑∑ ( α1nijk qi q j qk + α 2 nijk qi q j qk + α 3nijk qi q j qk )  +
=i 1 =j 1 =k 1  (2.6.6)
 m m 
ε  ∑∑ α 4 nijk ω12cos ( ω1 τ ) qi q j + α5 nijk ω12 cos ( ω2 τ ) + α 6 nijk cos ( ω2 τ ) q  =0
=i 1 =j 1 
[Derivation of only one/two terms are shown below. Taking only the first two terms in Eq.
(2.6.1) and substituting (2.6.2) one may obtain the residue equation
 1 
EI  vssss + vs2vssss +   (2.6.7)
 2 
m
Substituting v ( s, t ) = ∑ r ψ i ( s ) qi ( t ) in the above equation
i =1

 n 
1 m
( ) ( ) ( )
m m
EI  ∑ r ψ i ( s ) qi ( t ) + ∑ r ψ i ∑ r ψ j ∑ r ψ kiv 
iv ' '
i 1 2 i 1=j 1 =k 1 
=  =  (2.6.8)
 n 
( )
3 m m m
r
EI  ∑ r ψ iiv ( s ) qi ( t ) + ∑∑∑ ψ i'ψ 'jψ kiv qi q j qk 
=i 1 2 =i 1 =j 1=k 1 
 
Taking weight function as ψ n , multiplying ψ n in the above equation and integrating over the
domain one obtains
 ml 
∫ =∑
EI  r ψ iv
i ( s ) qi ( t ) +
r3 m m m
∑∑∑
2 =i 1 =j 1=k 1
( )
ψ i'ψ 'jψ kiv qi q j qk ψ n ds

0 i 1 
l  m 
( )
3 m m m
r
∫ EI =∑ r ψ iiv ( s )ψ n qi ( t ) + ∑∑∑ ψ i'ψ 'jψ kiv ψ n qi q j qk ds (2.6.9)
2 =i 1 =j 1=k 1 
0 i 1 
r3 m m m *
= rEIh1 qn + ∑∑∑ α1ijk qi q j qk
*
2 =i 1 =j 1=k 1

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Where
 
l
 m iv 
l

=h1* ∫  ∑ ∫  ψ
ψ i=
ψ n ds1 ψ n +ψ 2 ψ n +ψ 3 ψ n +  +ψ n ψ n +  ds
iv iv iv iv

0 i =1 0  
 = 0 due to orthogonality principle  (2.6.10)
l l
1 1
= ∫ψ niv =
( s )ψ n ( s ) ds =4 ∫ n n
ψ ivψ ldx 3 h1
0
l 0 l

 m m m m
l 
= ∫ EI  ∑∑∑∑ ψ i'ψ 'jψ kiv ψ n ds
α1*ijk
=i 1 =i 1 =j 1 =k 1  ( )
0  
1  
( )
m m m m
1
= 5 ∫ EI  ∑∑∑∑ ψ i' ( x )ψ 'j ( x )ψ kiv ( x ) ψ n ( x ) dx (2.6.11)
l 0 =i 1 =i 1 =j 1 =k 1 

1
= 5 α1ijk
l

It may be noted that while h1* is in dimensional form h1 is in the nondimensional form.
Similar procedures have to be followed to find all other terms.
Considering single mode discretization i.e. by substituting m=1, the above equation reduces to

q + q + 2 ε ζ q + ε  α1 q 3 + α 2 q q + α3 q 2 q  +
2

 
( )
(2.6.12)
ε α 4 ω1 cos ( ω1 τ ) q + α5 ω1 cos ( ω2 τ ) + α 6cos ( ω2 τ ) q =0
2 2 2

Eq. (2.6.12) is the required temporal equation of motion. The coefficients used in this
equation are described below.

The natural frequency ( ωe ) of the lateral vibration of an elastic beam

ωe = EI  h14  + P0  h21  = χ h14 + P  h21  (2.6.13)

ρAL4  h2  ρA L2  h2 
0 
h2  h2 

Cd
Damping ratio ( ζ ) = , (2.6.14)
2 ε ρ A ωe

Coefficient of the nonlinear geometric term q 3

χλ2  h19 h18 h 

= α1 = 2
 + + 3 20  (2.6.15)
εωe  h2 2h2 h2 

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Coefficient of the nonlinear inertia term q 2 q

λ2  h3 h4 h h h h 
= α2 =  + + m 5 − 6 − m 7 − 8  , (2.6.16)
ε  h2 h2 h2 h2 h2 h2 

Coefficient of the nonlinear inertia term q 2 q

λ2  h11 h12 h 
= α3 =  − − m 13  (2.6.17)
ε  h2 h2 h2 

Coefficient of the term q 2 cos(ω1τ)

Z λ  h15 h h 
= α4 =  + m 16 − 17  (2.6.18)
ε  h2 h2 2h2 

r  h1 
Coefficient of the direct forced term (cos(ω2τ )) , α5 =  , (2.6.19)
ε  h2 
 h21  h 
Coefficient of the parametric excitation ( q cos(ω2τ ) ) , α 6 =
P1
  = P1  21  . (2.6.20)
M ω L  h2 
2 2
e  h2 
Where
1  dψ (x )  x  dψ(ξ )
 
1 2

h1 = ∫ ψ nivψ n ds ,h2 = ∫ [ψ ( x )] dx , h3 = ∫ 
1
2
∫ ξ ψ( x )dx ,
 d
0  dx  0  d ξ   
 
0
0  
1  dψ ( x ) d 2 ψ ( x )  1  1  dψ ( x ) d ψ y ( x ) 
2
h4 = ∫  2 ∫
ψ ξ d ξ()

 ψ ( x )d x , h 5 = ∫  [ψ ( x )]2
dx ,
0  dx dx  x  0 
 d x d x 2


1  d 2ψ
h6 = ∫ 
(x )  1 η  dψ (x )
ξ η


2
ψ ( )

 =
1  d 2 ψ ( x )  x  dψ ξ 
∫  
2

ξ

 ψ ( )
 ()
dx ,
∫ ∫
 x 0  d ξ  2 ∫ 
d d x d x , h d x
0  dx
2  
7
0  dx 0  dξ   
       
1  dψ (x ) 2 [ψ(x )]2 dx , 1  dψ ( x ) d 2 ψ ( x )  1
()
 
h8 = ∫   h9 = ∫   ∫ ψ ξ d ξ  ψ ( x )dx ,
0  dx  0  dx dx  x
2
 

1  dψ (x ) 2 [ψ(x )]2 dx , 1  dψ
( x )  x  dψ ξ ()
2 


h10 = ∫   h11 = ∫  ∫  d ξ ψ ( x )dx ,
0  dx  0  dx  0  d ξ  
   

1  d 2ψ
h12 = ∫ 
(x )  1 η  dψ (ξ ) 2 dξd η ψ(x )dx , 1  d 2ψ
h13 = ∫ 
(x )  x  dψ(ξ )
ξ

 ψ ( )
 2
dx ,
2 ∫ ∫  2 ∫  d x
0  dx x 0  dξ    0  dx 0  dξ   
       

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1  d 4ψ ( x )  1 dψ( x ) d 2ψ( x ) 
h14 = ∫  4
ψ ( x )d x  , h15 = ∫( 1 − x ) 2
ψ ( x )dx ,
0  dx  0 dx dx 

1  dψ ( x ) d 2 ψ ( x )  1  dψ ( x )  1  dψ ( x )  d 4 ψ ( x ) 
2 2
h16 = ∫  ψ ( x )d x , h17 = ∫  ψ ( x )d x , h18 = ∫   ψ( x )dx ,
0  dx  0  dx 
2 4
0  dx dx   dx 

1  d 2ψ ( x )  1  dψ ( x ) d 2 ψ ( x ) d 3ψ ( x ) 
h19 = ∫  2
ψ ( x )d x , h20 = ∫ 2 3
ψ ( x )dx ,
0  dx  0  dx dx dx 
1  d 2ψ (x )
and h21 = ∫  ( ) .
2 ψ x dx
0  dx 
One may find that the non-dimensional temporal equation (2.6.12) has a linear forced
term (α 5ω12 cos ω1τ ) , a linear parametric term (α 6 cos ( ω2 τ ) q ) and a nonlinear parametric

excitation term (( α ω cos ω τ ) q )

4
2
1 1
2
along with cubic geometric ( a1q3 ) and inertial

(α 2
q 2 q + α q 2 q
3
) nonlinear terms. Here the system is subjected to a two-frequency

excitation. One may note that the temporal equation of motion contains many nonlinear terms
and it is very difficult to find the exact solution. Hence one may go for approximate solution
by solving equation (2.6.12) using perturbation method.

Exercise Problems:

Problem 2.6.1:
Derive the temporal equation of motion of a micro-beam system whose spatio-temporal
equation is given below [1]

Fig 2.6.2: Schematic diagram of a clamped-clamped micro-beam.

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 Ebh l 2  bεV 2  1 2w ( x, t ) 3 w ( x, t )
2
4 w ( x, t )
3 
ρ bhwtt + EIwxxxx −  Ni +
 2l ∫0
wx dx  wxx −
 2
 2
 d gap
+ 3
d gap
+ 4
d gap
+ 5
d gap
+  

 
εβV 2  1 w ( x, t ) w ( x, t ) w ( x, t ) 
2 3
−  + 2 + + +  =
0
2  d gap d gap 3
d gap 4
d gap 
 

Answer:

d 2q
+ β1q (τ ) + 1 β 2 ( q (τ ) )2 + β3 ( q (τ ) )3 + β 4 ( q (τ ) )4 + β5  =
0
dτ 2

Problem 2.6.2:
The equation of motion of a base excited cantilever beam with arbitrary mass position can be
given by the following equation. Derive the temporal equation of motion using single mode
approximation.

Fig 2.6.3: Schematic diagram of a base excited cantilever beam with arbitrary mass position.

(
EI vssss +
1 2
2
3
) 2
( )
vs vssss + 3vs vss vsss + vss + 1 − 0.5vs {( ρ+ mδ( s − d ) )v+ cv} +

{( ρ + mδ ( s − d ) ) v + cv}d ζ − ∂∂s  J δ( s − d )( vs )tt  − ∂∂s ( Nvs ) =0

L
vs vss ∫
s

Where
 ζ ( v 2 ) d η d ζ − 1 m L δ ( ζ − d )  ζ ( v 2 ) d η d ζ
L
∫s  ∫0 s tt  2 ∫s  ∫0 s tt 
1
N ρ
2

+ m (   s
z − g ) ∫ δ ( ζ − d ) d ζ + ρL 1 −  ( z − g)
L

s
 L
Answer:
q + 2 ε ζ q + ω2 (1 − εf cos Ωt ) q + ε ( α1 q 3 + α 2 q 2 q + α3 q 2 q ) = 0

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Problem 2.6.3:
For the same system given in problem 2.6.2, carryout two mode approximation to derive the
temporal equation of motion. [3]

Answer:

qn + 2 ε ζ qn + ωn2 qn − ε∑ f nm qm cos Ωt + ε∑∑∑ ( α klm qk ql qm + γ nklm qk ql qm ) = 0
2 2 2 2
n
qk ql qm + βklm
n

=
m 1 =
k 1 =l 1 =
m 1

where n = 1, 2.

References

1. H. M. Sedighi, K. H. Shirazi, Vibration of micro-beam actuated by an electric field via

Parameter Expansion Method, Acta Astronautica, 85, 19-24, 2013.

2. L.D. Zavodney, A.H. Nayfeh, the nonlinear response of a slender beam carrying a lumped
mass to a principal parametric excitation : theory and experiment, International Journal of
Nonlinear Mechanics, 24,105-125, 1989.

3. R.C. Kar and S.K. Dwivedy, Nonlinear dynamics of a slender beam carrying a lumped
mass with principal parametric and internal resonances, International Journal of Nonlinear
Mechanics, 34, 515-529, 1999.

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Lecture M2 L07
Ordering and scaling technique in nonlinear equations

In the previous lectures we learned about the derivation of equation of motion of both discrete
and distributed mass system. In the later case the equation has been reduced to its temporal
form. In these equations the coefficients of different terms used in the differential equations
may not be of the same order and hence sometimes some terms get neglected in comparison to
other terms. But for accurate solution one should take as many term as possible and hence it
is required to know the ordering and scaling techniques. So in this lecture following points
will be discussed with the help of examples.

• Ordering techniques,
• scaling parameters,
• Book-keeping parameter.
• Commonly used nonlinear equations: Duffing equation, Van der Pol’s oscillator, Mathieu’s and
Hill’s equations

Let us consider the equation we have derived for the simple pendulum. It can be written as
g
θ + sin θ = 0 (2.7.1)
l
Keeping up to quintic nonlinearity Eq. (2.7.1) can be written as

g g θ3 g θ5
θ + θ − + =
0 (2.7.2)
l l 6 l 120

Taking the length of the pendulum 1 m and acceleration due to gravity as 10 m/s2 , the
equation of motion can be written as

θ + 10θ − 1.6667θ 3 + 0.0083θ 5 =

0 (2.7.3)

In Eq. (2.7.3), the coefficient of the linear term θ is 10, the coefficient of cubic nonlinear
term is -1.667 and the coefficient of quintic term is 0.008. As the coefficients of quintic and
cubic terms are very very less than the linear term, one can neglect these terms to obtain the
approximate solution. But to obtain the accurate solution one should consider these terms.
One can use scaling parameter and book-keeping parameters to make the coefficient of
nonlinear and linear terms of the same order so that the effect of these nonlinear terms can be
taken into account.

To use scaling factor, let us take θ = py and substitute this in Eq. (2.7.3). Now the resulting
equation can be written as

py + 10 py − 1.6667 p 3 y 3 + 0.0083 p 5 y 5 =

0 (2.7.4)

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Or, 
y + 10 y − 1.6667 p 2 y 3 + 0.0083 p 4 y 5 =
0 (2.7.5)

Now by taking different values of p , the coefficient of the nonlinear terms can be changed
significantly without changing the coefficient of the linear part. For example, taking p = 10,
the above equation becomes

y + 10 y − 166.67 y 3 + 83 y 5 =
0 (2.7.6)
Taking p = 5, Eq. (2.7.5) can be written as

y + 10 y − 41.667 y 3 + 5.1875 y 5 =
0 (2.7.7)

While in Eq. (2.7.6) the coefficient of linear and non-linear terms have large differences, in
Eq. (2.7.7), these coefficients are closer to each other. Hence by suitably choosing the value of
p , it is possible to bring the coefficient of the linear and nonlinear terms to the same order
and in that case, instead of neglecting the higher order terms, one can consider these terms and
solve the equation to obtain more accurate response.

Considering Eq. (2.7.3), as the coefficients of the cubic and quintic order terms are very very
small in comparison to the coefficient of the linear term, one can use a book-keeping
parameter ε ( ) to order the coefficients. In this case one may write Eq. (2.7.3) as

 1.6667  3 3  0.0083  5
θ + 10θ − ε  θ + ε  θ = 0 (2.7.8)
 ε   ε 
3

Taking ε = 0.1 , Eq. (2.7.8) can be rewritten as

θ + 10θ − ε 16.667θ 3 + ε 3 8.3θ 5 =

0 (2.7.9)

In Eq. (2.7.9) now the numerical part of the coefficients (16.667 and 8.3) are approximately
same orders as that of the linear terms (i.e. 10). So in this way one can use the book-keeping
parameter to order the nonlinear terms in a given nonlinear differential equation of motion.

Commonly used nonlinear equation of motion

Duffing equation (Free vibration with quadratic nonlinear term)

d 2u
2
+ ω02u + εα u 2 =
0 (2.7.10)
dt
Duffing equation (Free vibration with cubic nonlinear term)
d 2u
2
+ ω02u + εα u 3 =
0 (2.7.11)
dt

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Duffing equation (Free vibration with both quadratic and cubic nonlinear terms)
d 2u
2
+ ω02u + εα1u 2 + εα 2u 3 =0 (2.7.12)
dt
Duffing equation with damping and weak forcing terms
x + ωn2 x + 2εζωn x + εα x 3= ε f cos Ωt
 (2.7.13)
Duffing equation with damping and strong forcing terms
x + ωn2 x + 2εζωn x + εα x 3 = f cos Ωt
 (2.7.14)
Duffing equation with multi-frequency excitation
x + ωn2 x + 2εζωn x + εα=
 x 3 f1 cos Ω1t + f 2 cos Ω 2t + f3 cos Ω3t +  (2.7.15)
Rayleigh’s equation
d 2u
2
+ ω02u − ε (u − u 3 ) =
0 (2.7.16)
dt
Substituting v = 3u in Eq. (2.7.16) and differentiating the resulting equation with respect to
time one will obtain the van der Pol’s equation as follows
d 2v dv
2
+ ω02 v =ε (1 − v 2 ) (2.7.17)
dt dt
Hill’s equation

x + p (t ) x = 0 (2.7.18)
Mathieu’s equation
x + (ωn2 + 2ε f cos Ωt ) x =
 0 (2.7.19)
Mathieu’s equation with cubic nonlinearies and forcing terms
x + (ωn2 + 2ε f1 cos Ω1t ) x + εα x=
 3
ε f 2 cos Ω 2t (2.7.20)
Lorentz equation
x = σ ( y − x )
y = rx − y − xz (2.7.21)
=
z xy − bz
Here σ , r , b > 0 are parameters
Generic equation for one dimensional pitchfork bifurcation
x= µ − x 2
Generic equation for saddle-node bifurcation
x µ x + α x 3
=
Generic equation for transcritical bifurcation
x µ x − x 2
=
Equation for Hopf bifurcation
r µ r + α r 3
=
θ= ω + β r 2

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Exercise problems
Problem 2.7.1:
Use scaling parameter to order the following equation.

( i ) 3
x + 30 x − 0.1x 2 + 0.05 x3 =
0
( ii ) 
x + 20 x − 0.5 x 2 + 0.05 x3 =
10sin 5t
( iii ) 
x + 50 x − 0.5 x 2 + 0.3 x3 =0.1sin 2t
( iv ) 
x + 50 x − ( 0.1sin 2t ) x =0
( v ) 
x + 50 x − 0.25 x 2 − ( 0.1sin 2t ) x =
0
( vi ) 
x + 50 x + 0.25 x3 + ( 0.1sin 2t ) x =
0

Problem 2.7.2:

Figure 2.7.1 shows a two-stage nonlinear vibration isolation system whose equation of motion
is given below. Using book-keeping parameter, write the equation of motion by taking
different values of m, k1 , k3 , kv1 , kh1 , c1 , ω .

Fe cos (ωt )
mx + c1 x + k1 x + k3 x 3 =
 l 
Fe kv1 x + 2kh1 1 − 2 0 2
where = x
 x +l 

Fig. 2.7.1: Two-stage nonlinear vibration isolation system [1]

Problem 2.7.3:

Write the equation of motion for system with (a) fractional order, (b) time delay, (c) piece-
wise nonlinearity, (d) random excitation, (e) gyroscopic effect, (f) contact, (g) backlash (h)
friction and wear.

Problem 2.7.4:

Study the nonlinear systems given in the references [2-13]. Taking numerical values and
using ordering and scaling parameters write the equation of motion. Use Matlab to solve the
temporal equation of motion in each case.

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Reference for further reading

1. Z. Lu, M. J. Brennan, T. Yang, X. Li, and Z. Liu, An investigation of a two-stage

nonlinear vibration isolation system, Journal of Sound and Vibration, 332 (6), 1456-1464,
2013.
2. A. Carrella, M. J. Brennan, T. P. Waters, and K. Shin, Introducing stiffness nonlinearity
using magnets to improve vibration isolation, Journal of Sound and Vibration, 315 (3), 712–
720, 2008.
3. R.A. Ibrahim, Recent advances in nonlinear passive vibration isolators, Journal of Sound
and Vibration, 314(3–5), 371–452, 2008. (Review paper)
4. H. M. Sedighi, K. H. Shirazi, Vibrations of micro-beams actuated by an electric field via
Parameter Expansion Method, Acta Astronautica, 85, 19-24, 2013.
5. A. D. S. Barr and D. C. McWhannel, Parametric instability in structures under support
motion, Journal of Sound and Vibration 14, 491—509,1971.
6. V. V. Bolotin, The Dynamic Stability of Elastic Systems, Holden-Day, Inc,1964.
7. M.P. Cartmell, The equations of motion for a parametrically excited cantilever beam,
Journal of Sound and Vibration 143, 395-405,1990.
8. G. Chakraborty and A.K. Mallik, Parametrically excited non-linear traveling beams with
and without external forcing, Nonlinear Dynamics 17, 301–324, 1998.
9. G. Chakraborty, A.K. Mallik, and H. Hatwal, Non-linear vibration of a traveling beam.
International Journal of Non-Linear Mechanics 34, 655–670, 1999.
10. O. Cuvalci and A. Ertas, Pendulum as vibration absorber for flexible structures:
experiments and theory, Journal of Vibration and Acoustics: Trans. of ASME, 118, 558-566,
1996.
11. S.K. Dwivedy and R.C. Kar, Nonlinear response of a parametrically excited system using
higher order method of multiple scales, Nonlinear Dynamics 20, 115–130, 1999.
12. S.A. Emam and A.H. Nayfeh, Nonlinear response of buckled beams to subharmonic-
resonance excitations, Nonlinear Dynamics 35,105–122, 2004.
13. H.P. Lee, Stability of a cantilever beam with tip mass subjected to axial sinusoidal
excitations, Journal of Sound and Vibration 183, 91–98,1995.
14. S.N. Mahmoodi, S.E. Khadem, and M. Kokabi, Non-linear free vibrations of Kelvin–
Voigt visco-elastic beams, International Journal of Mechanical Science 49, 722–732, 2007.

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