Mechanical

Vibrations

Prof. Dr. Kenan Y. Şanlıtürk

sanliturk@itu.edu.tr

Content

1. Introduction to Vibration and Free response

2. Response to Harmonic Excitation
3. General Forced Response
4. Multi-Degree-of-Freedom systems
5. Design for Vibration Suppression

Page 1
 1. Introduction to Vibration and Free Response
• Viscous Damping
• Mass-Spring-Damper system
• Equation of motion, solution
• Critical Damping
• Overdamped Motion
• Underdamped Motion
• Sample problems

• Logarithmic decrement

Viscous damping
 Called: dashpot or damper

 c: Viscous damping coefficient.

 Unit: [N/m/s]
f c  cx(t )

oil

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Mass-Spring-Damper System

Equation of Motion:

mx(t )   f k  f c
  kx (t )  cx (t )
Frictionless surface

mx(t )  cx(t )  kx(t )  0

x(0)  x0 , x(0)  v0

Solution

x(t )  Cet

Frictionless surface (m 2  c  k )Cet  0
mx(t )  cx(t )  kx(t )  0 (m 2  c  k ) x(t )  0
x(0)  x0 , x(0)  v0 
m 2  c  k  0
Characteristic equation 
c k
2   0
m m

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 m 2  c  k  0
Solution
c k
2   0
m m

The roots of the characteristic equation:

Frictionless surface

c  c 2  4mk
mx(t )  cx(t )  kx(t )  0 1,2 
2m
x(0)  x0 , x(0)  v0
2
c  c  k
1,2      
2m  2m  m

Solution 2
c  c  k
1,2      
2m  2m  m

2
 c  k
Frictionless    0 or c 2  4mk  0
surface  2m  m
mx(t )  cx(t )  kx(t )  0 
x(0)  x0 , x(0)  v0 ccr  2 km  2m k / m  2mn

c c c
Viscous damping ratio:    
ccr 2 km 2mn

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Solution
mx(t )  cx(t )  kx(t )  0

c k
x(t )  x(t )  x(t )  0
m m
x(t )  2n x(t )  n2 x(t )  0
Frictionless surface

c  c  k
2  2  2n   n2  0
1,2      
2m  2m  m

 
c

c

c 1,2 in terms of  and n
ccr 2 km 2mn
c
 2n 1,2  n  n  2  1
m

1) Overdamped Motion(>1)
x(t )  C1e1t  C2e2t

1,2  n  n  2  1

2 distinct real roots 1 , 2

x(t )  ent (C1ent  2 1

 C2ent  2 1
)
mx(t )  cx(t )  kx(t )  0
x(0)  x0 , x(0)  v0 v0  (   2  1)n x0
C1 
2n  2  1
Imposing initial conditions
v0  (   2  1)n x0
C2 
2n  2  1

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The Response of Overdamped Systems (No Vibration!)

2) Critically damped system (1)

1,2  n  n  2  1

1  2  n

Frictionless surface
x(t )  C1e1t  C2te 2t

mx(t )  cx(t )  kx(t )  0

x(0)  x0 , x(0)  v0 x(t )  C1e nt  C2te nt

Imposing initial conditions: x(t )  [ x0  (v0  n x0 )t ]ent

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 2) Critically damped motion (1) (No Vibration!)
 No oscillatory motion
 Used for door closing systems, dials of analogue gauges etc.
nt
x(t )  [ x0  (v0  n x0 )t ]e

3) Underdamped Motion(<1)

1,2  n  n  2  1
1,2  n  jn 1   2
Two complex conjugate roots

1  n  jn 1   2
2  n  jn 1   2

1  n  jd
d  n 1   2
2  n  jd
d : Damped natural frequency

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 3) Underdamped Motion(<1)
x(t )  e nt (C1e jnt 1 2
 C2e  jnt 1 2
)
d  n 1   2
x(t )  e nt (C1e jd t  C2e  jd t )
Note: If C1 and C2 are complex conjugates,
one gets real A1 and A2. A2=C1+C2, A1=(C1-C2)j

x(t )  e nt ( A1 sin(d t )  A2 cos(d t ))

mx(t )  cx(t )  kx(t )  0
x(0)  x0 , x(0)  v0 Impose initial conditions:
Note: A2  xo
e jd t  cos(d t )  j sin(d t ) vo  n xo
A1 
d
 jd t
e  cos(d t )  j sin(d t )

3) Underdamped Motion(<1)

x(t )  e nt (C1e jnt 1 2

 C2e  jnt 1 2
)
d  n 1   2

x(t )  e nt (C1e jd t  C2e  jd t )

 Ae nt sin(d t   )

1
A (v0  n x0 ) 2  ( x0d ) 2
d
mx(t )  cx(t )  kx(t )  0
x(0)  x0 , x(0)  v0  x0d 
  tan 1  
 v0  n x0 

Page 8
x(t )  e nt ( A1 sin(d t )  A2 cos(d t )) x(t )  Ae nt sin(d t   )

A2  xo A  A12  A22
vo  n xo
A1    tan 1 (
A2
)
d A1

1
A (v0  n x0 ) 2  ( x0d ) 2
d

Underdamped Motion(<1)

 Many mechanical systems are in this category.

 Vibration amplitudes decay exponentially.
 The damped natural frequency is lower than
undamped natural frequency.

d  n 1   2

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 time

Higher damping results

in faster decay. m

k c1 + c2

d  n 1   2

The response of underdamped system

x(t )  Aent sin(d t   )

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Sample Problem
A 30 cm long spring with a spring constant of k = 857.8 N/m is fixed to a
rigid wall (ground) and a rigid block with 49.2 x 10-3 kg mass is attached
to the other end of the spring. Viscous damping coefficient is given as
c = 0.11 kg/s. Determine the viscous damping ratio for this system.

m  49.2 103 kg, k  857.8 N/m

ccr  2 km  2 49.2 103  857.8
 12.993 kg/s
c 0.11 kg/s
=   0.0085  This system is underdamped.
ccr 12.993 kg/s It vibrates if disturbed.

Sample Problem (Exercise)

The natural frequency of the system below without the damper

(dashpot) is known to be 20 Hz. It is also known that when the
damper is attached to the system, the viscous damping ratio of the
system becomes  = 0.224. Calculate the response of the mass to an
initial velocity of v0 = 0.6 m/s and zero initial displacement. What is
the maximum acceleration if there is no damping?

20 cycle 2 rad
n   125.66 rad/s
1 s cycle
d  125.66 1  ( 0.224 )  122.467 rad/s
2

Page 11
 vo  n xo vo
A  , d  122.66, v0  0.6, x0  0
d d
0.6
A =0.0049m,  =0
d
x(t )  0.0049e nt (sin d t )  0.0049e 28.148t sin(122.467t )
If no damping:

 0.6 
max ( x )  n2 A  n2    ( 0.6 ) (125.66 m/s )  75.4 m/s
2 2


 n

1g
Maximum Acceleration  75.40 m/s 2  7.68 g
9.81 m/s 2

Sample Problem
The general response of the underdamped system can also be
written as:
x(t )  ent ( A1 sin d t  A2 cos d t )
Determine the constants A1 and A2 using the initial conditions.

sin(a  b)  sin a cos b  cos a sin b 

A sin(d t   )  A sin(d t ) cos   A cos(d t ) sin 
A sin(d t   )  A1 sin(d t )  A2 cos(d t )

x(t )  Ae nt sin(d t   )  e nt ( A1 sin d t  A2 cos d t )

x(0)  x0  e0 ( A1 sin(0)  A2 cos(0))  A2  x0

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Sample Problem

x  n e nt ( A1 sin d t  A2 cos d t )

 d e nt ( A1 cos d t  A2 sin d t )
v0  n ( A1 sin 0  x0 cos 0)  d ( A1 cos 0  x0 sin 0)
v0  n x0
 A1  
d
 v  n x0 
x(t )  e nt  0 sin d t  x0 cos d t 
 d 

Damping Identification by Using Transient Measurement

x(t )  Aent sin(d t   )

Logarithmic decrement: d
Definition, derivation 

Page 13
 The response of underdamped system

Logarithmic decrement: d
x(t )  Aent sin(d t   )

 Xi 
d  ln  
 X i 1 

Logarithmic decrement: d  Xi 
d  ln  
Deplasman, [mm]
 X i 1 
Xi

X i 1 T X i (ti )  Ae nti
X i 1 (ti  T )  Ae n (ti T )
ti
Zaman, [s] Xi
 enT
X i 1
 Xi 
d  ln    nT
 X i 1 
 Xi 
If damping is small d  ln    nT  2
 X i 1 
1  Xi  d
d  ln    
n  X in  2

Page 14
Logarithmic decrement: d
Xi
 enT
X i 1
 Xi 
d  ln    nT
 X i 1 

2 2
If damping is NOT small T 
d n 1   2
 Xi  2
d  ln    nT  n
 X i 1  n 1   2

 Xi  2 d
d  ln    
 X i 1  1  2
(2 ) 2  d 2

Page 15