Undamped Single Degree of Freedom System
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Transcript Undamped Single Degree of Freedom System
Mechanical Vibrations
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Introduction
1
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Examples
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Examples
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Equation of motion
Equation of
motion
Newton's 2nd
law of
motion
Other
methods
D’Alembert’s
principle
Virtual
displacement
Energy
conservation
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Newton's 2nd law of motion
Draw the
free-body
diagram
Apply
Newton s
second law
of motion
Determine the static
equilibrium configuration
of the system
Select a
suitable
coordinate
The rate of change of
momentum of a
mass is equal to the
force acting on it.
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Newton's 2nd law of motion
..
d d xt
d 2 xt
F t
m
m
mx
2
dt
dt
dt
..
..
F t kx m x m x kx 0
Energy conservation
Kinetic energy
Potential energy
1 .2
T mx
2
U
T U constant
d
T U 0
dt
1 2
kx
2
..
m x kx 0
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Vertical system
..
m x k x st mg ; mg k st
..
m x kx 0
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Solution to the equation of motion:
mx kx 0
(1), it a homogenuos 2nd order
ordinary differenti al equation. Equation (1) can be written as,
x n2 x 0
where n
k / m is the known as the natural frequency
solution t o equation (1) may has the solution of the type :
x (t ) Be st where B is constant and S are the roots of solution.
substitute the solution into eqation (1) yields :
( s 2 n2 ) Be st 0
.
or
s 2 n2 0
e st 0
s j n
i.e
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The solution t o equation (1) can now be written as :
x(t ) B1e jnt B2 e jnt
Note that :
e jnt cos nt j sin nt ,
Therefore
x(t ) ( B1 B2 ) cos nt j ( B1 B2 ) sin nt
Put
.
A1 B1 B2 and A2 j ( B1 B2 )
Then
x(t) A 1cosω n t A 2 sinω n t
(2)
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The solution t o equation (1) can also be written in the form :
x(t) Acos( ω n t φ)
Where A
tan
1
(3)
A12 A22 is the amplitude of motion and
A2
is the phase angle.
A1
we can say that t he free vibration of the system consists of a harmonic motion
of an amplitude A and frequancy n .
Note : The constants A1 and A2 (or A and ) can be obtained from
applying initial co nditions of the motion.
Inatial conditions :
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Let at t 0 ; x(t ) X 0 and x(t ) X 0 . Sustitutin g these
conditions into the general solution (2) or (3) yields :
.
A1 X 0 and A2
X0
n
then the general solution becomes :
.
X0
x(t ) X 0 cos nt
sin n t ........(4)
ω
n
The amplitude A and the phase angle φ become :
2
.
.
X0
2
X0
1
A X0
and
tan
X 0ωn
ω n
Equation (4) can be used to obtain the value of displaceme nt of the
system under considerat ion at any time t.
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Graphical representation
x(t ) = A cos(ωnt – ϕ )
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Torsional vibration
A system has a shaft of torsional stiffness
of moment of ineria J D as shown. Note that the stiffness
of the shaft is calculated as;
d4
Kt
where I p
is the polar moment of ineria.
L
32
Applying Newtons 2 nd law :
M 0 J D
K J
J K 0 is the equation of motion
GI p
t
D
D
t
and from which the natural frequency can be obtained as
n
.
K t and a disk
Kt
rad/sec.
JD
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Natural Frequency (ωn) :
It is a system property. It depends, mainly, on the stiffness and the mass
of vibrating system.
It has the units rad./sec, or cycles/sec. (Hz)
It is related to the natural period of oscillation (τn) such that, τn = 2π/ωn
and
ωn = 2 π fn where fn is the natural frequency in Hz.
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Example
2.1
The column of the water tank shown in Fig.
is 90m high and is made of reinforced
concrete with a tubular cross section of
inner diameter 2.4m and outer diameter
3m. The tank mass equal 3 x 105 kg when
filled with water. By neglecting the mass of
the column and assuming the Young’s
modulus of reinforced concrete as 30 Gpa.
determine the following:
the natural frequency and the natural
period of transverse vibration of the water
tank
the vibration response of the water tank
due to an initial transverse displacement of
tank of 0.3 m and zero initial velocity.
the maximum values of the velocity and
acceleration experienced by the tank.
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Example 2.1 solution:
Initial assumptions:
1.
the water tank is a point mass
2.
the column has a uniform cross section
3.
the mass of the column is negligible
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Example 2.1 solution:
a. Calculation of natural frequency:
1. Stiffness: k 3EI
, But: I
3
l
d
64
4
o
d i4
3
64
4
2.4 4 2.3475 m 4
3x30 x109 x 2.3475
289,812 N / m
So: k
3
90
2. Natural frequency : n
k
289,812
0.9829 rad / s
5
m
3x10
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Example 2.1 solution:
b. Finding the response:
1. x(t ) = A cos (ωn t - ϕ )
2
.
Xo
A X o2
X o 0.3m
n
So,
.
.
0
1 X o
1
0
tan
tan
X on
X on
x(t ) = 0.3 cos (0.9829 t )
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Example 2.1 solution:
c. Finding the max. velocity:
.
xt 0.30.9829sin 0.9829t
x (t ) is maximum when 0.9829t /2
then t /(2 * 0.9829)
.
x max 0.30.9829 0.2949m / s
Finding the max. acceleration :
..
2
xt 0.30.9829 cos0.9829t
..
xt is maximum when 0.9829t
then t / 0.9829
..
2
x max 0.30.9829 0.2898m / s 2
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Example 2.2 :
Free vibration of single degree of freedom system
Example 2.2 : solution
Let the displaceme nt of the centres of pulleys 1 and 2
are x1 and x 2 respective ly. The free body diagrams of the pulleys
show that :
2w k1 x1
and 2w k 2 x 2 where w mg. Note that
2w 2w
x 2
k2
k1
The system is equivalent to a mass spring system of mass m
x 2x1 2x 2 or
and stiffness k eq such that
w k eq x, then by substituti ng
for x the equivalent stiffness of the system becomes :
k eq
1 k1k 2
, the equation of motion is then
4 k1 k 2
mx k eq x 0, and the natural frequency is n
k eq
m
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Simple
pendulum
Governing equation:
..
M O J O J O
..
J O mgl sin 0
Assume θ is very small
sin
..
J O mgl 0
Natural frequency (ωn)
JO
mgl
n
2
JO
mgl
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Solution
t A1 cosnt A2 sin nt
t 0 A1 o
.
t 0
.
o
A2n o , A2
n
.
.
o
t o cosnt sin nt
n
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Example
2.3
Any rigid body pivoted at a
point other than its center of
mass will oscillate about the
pivot point under its own
gravitational force. Such a
system is known as a compound
pendulum (as shown). Find the
natural frequency of such a
system.
Solution
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the governing equation is found as:
b
..
r
J O Wd sin 0
a Assume small angle of vibration:
t
..
i
J O Wd 0
o
n So:
s
Wd
mgd
n
.
JO
JO
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Example
2.5: Q2.45
Draw the free-body diagram and derive the equation of motion
using Newton s second law of motion for each of the systems
shown in Fig