Mechanical Vibrations
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 Introduction
1
2
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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 xt  
d 2 xt 
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  cons tan t
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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Mathematical review
..
.
A1 t  xt   A2 t  xt   A3 t xt   F t 
F(t) = 0 → Homogenous
x(t) = C1x1 + C2x2
If A1,A2 and A3 are constants:
Characteristic equation:
A1s2+A2s+A3=0
.
s
 A2 
A22  4 A1 A3
2 A1
F(t) ≠ 0 → none homogenous
x(t) = C1x1 + C2x2 +xp
xp form is the same type
asF(t)
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s
Only one real root:
s1,2
x(t) = C1est + C2t est
 A2 
A22  4 A1 A3
2 A1
Two unequal real
root: s1,2
x(t) = C1es1t + C2 es2t
Complex root:
s1,2 = α±βi
x(t) = eαt {C1cos(βt)
+ C1sin(βt)}
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 Solution

..
m x kx  0 is 2nd order homogenous differential equation
where: A1 = m, A2=0 and A3=k
s


 4mk
k
 i
 in
2m
m
the solution : x(t) = eαt {C1cos(βt) + C1sin(βt)}
where α = 0 and β = ωn

x(t) = C1cos(ω nt) + C2sin(ω nt)
Natural
frequency
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
Solution

C1 and C2 can be determined from initial conditions (I.Cs). For
this case we need two I.Cs.


xt  0  C1  xo
.
.
xt  0  C2n  x o
The I.Cs for this case would be:
So: xt   xo cosnt  
.
xo
n
sin nt     Eq.1
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
Harmonic motion
Introduce Eq.2 into Eq.1: x(t ) = A cos(ωnt – ϕ ) = Ao sin(ωnt + ϕo)
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
Harmonic motion
xt   C1 cosnt   C2 sinnt 
xt   xo cos nt  
.
xo
n
sin nt 
Assume:
C1 = A cos(ϕ) --- Eq.2(a)
C2 = A sin(ϕ) --- Eq.2(b)
2
. 
 xo 
A  C12  C22  xo2     Amplitude
 n 
 
Ao  A
Harmonic functions in time.
Mass spring system is called
harmonic oscillator
 . 


C
xo 
  tan 1  2   tan 1 
 Phase angle

 xon 
 C1 


x 
o  tan  o. n 


 xo 
1
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
Harmonic motion
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
Natural Frequency (N.F)

A system property (i.e. depends of system parameters m
and k )

Unit: rad/sec

It is related to the periodic time (τ ) : τ = 2π/ωn

Periodic time is the time taken to complete one cycle
(i.e. 4 strokes)
.

The relation between the ωn and τ is inverse relation
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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
time period of transverse vibration of the
water tank
the vibration response of the water tank
due to an initial transverse displacement of
0.3m.
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
4.
the initial velocity of the water tank equal zero
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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

 di4 


3
64
4

 2.44  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 sin(ωnt + ϕ )
2
. 
 xo 
A  xo2     xo  0.3m
 n 
 
x 

1  xon 
o n 

  tan
 tan 

 . 
 0  2
 xo 
So, x(t ) = 0.3 sin (0.9829t + 0.5π )
.
1
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Example 2.1 solution:
c. Finding the max velocity:
.


xt   0.30.9829 cos 0.9829t  
2

.
x max  0.30.9829  0.2949m / s
Finding the max acceleration :
..


2
xt   0.30.9829 sin 0.9829t  
2

..
2
x max  0.30.9829  0.2898m / s 2
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Example2 :
Q2.13
Find the natural frequency of
the pulley system shown in Fig.
by neglecting the friction and
the masses of the pulleys.
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Example2 :
Q2.13
Solution:
1. Free body diagram
P
P
P
x1
x2
P
2. x = 2x1 +2x2 ---- Eq.1
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Example2 :
Q2.13
Solution:
3. Equilibrium for pulley_1 : 2P = k1 x1 = 2k x1 ---- Eq.2
4. Equilibrium for pulley_2 : 2P = k2 x2 = 2k x2 ---- Eq.3
 2P   2P 
1
1
4P
5. Substitute Eqs 2 and 3 in Eq.1: x  2   2   4 P   
 2k 2k  k
 k1   k2 
6.Let keq is the equivalent spring constant for the system: keq  P  k
x
..
7. Mathematical model: m x kx  0
.
8. Natural frequency: n 
keq
m

k
4m
4
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Rotational
system
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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Torsional system
Governing equation:
M
..
O
 J O  J O 
..
J O   kT  0
Natural frequency (ωn)
n 
JO 
JO
kT
   2
JO
kT
hD 4
32
WD 2

8g
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
Solution
 t   A1 cosnt   A2 sinnt 
 t  0  A1   o
.
.
 t  0  A2n   o
.
o
 t    o cosnt   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 (see the Fig). Find
the natural frequency of such a
system.
Solution
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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
2.4: Q2.12
Find the natural frequency of the system shown in Fig. with the
springs k1 and k2 in the end of the elastic beam.
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Example
2.4: Q2.12
Solution: F.B,D
Example
2.4: Q2.12
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Solution:
b
r
a keq is equivalent stiffness for the combination
t
3EI
i
kbeam  3
l
o
n k1 and k2 equivalent: apply energy concept
s
2
.
of k1, k2 and kbeam
1
1 2 1
 x1 
 x2 
2
2
keq ,1, 2 x  k1 x1  k2 x2  keq  k1    k2  
2
2
2
x
 x
2
Example
2.4: Q2.12
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b
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a Finding keq
t
i
1
1
o


n
keq keq ,1, 2
s
.
1
kbeam
 keq 
keq ,1, 2 kbeam
keq ,1, 2  kbeam
Example
2.4: Q2.12
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Solution:
b
r Finding natural frequency
a
keq
keq ,1, 2 kbeam
t
n 

i
m
m keq ,1, 2  kbeam
o
2
n
  x1  2
 x2  

k1    k 2   kbeam
s

.

 x
 x  

n 
2
  x1  2

x


2
m k1    k 2    kbeam 
 x

x




x1, x2 and x can
be found from
strength relation
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Example
2.5: Q2.7
Three springs and a mass are attached to a rigid, weightless bar PQ
as shown in Fig. Find the natural frequency of vibration of the
system.
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Example
2.5: Q2.7
Solution :
Assume small angular motion sin    
1
1
1
k1l12  k2l22
2
2
2
keq ,1, 2 l3   k1 l1   k2 l2   keq ,1, 2 
2
2
2
l32
Let keq is the equivalent stiffness for the whole system
keq ,1, 2 k3
1
1
1

  keq 
keq keq ,1, 2 k3
k3  keq ,1, 2
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Example
2.5: Q2.7
Solution :
Now find the natural frequency
n 
k k l k k l

m
m k l k l k l
keq

2
1 2 1
2
11
2
2 3 2
2
2
2 2
3 3

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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
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Example
2.5: Q2.45
Solution F.B.D
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Example
2.5: Q2.45
Equation of motion:
The distance: x  4r   o 
..
For mass m: mg  T  m x --- (1)
..
For pulley Jo: J o   Tr  4rk    o 4r  --- (2)
mg
According to static equilibrium: mgr  k 4r 4r  o   o 
--- (3)
16rk
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Example
2.5: Q2.45
Equation of motion [cont]:
Substitute equations 1 and 3 into equation 1:
.. 
.. 
mg 
2
J o    mg  m x r  16kr  

16rk 



..
..
..
..
2
J o   mgr  m x r  16kr   mgr  0  J o   m x r  16kr 2  0
..
..
Use the relation x  r  x  r  to relate the translational
motion with the rotational one:
..
2
J o  mr   16kr2   0
S
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P
End of chapter2 – part I