2103433
Introduction to Mechanical
Vibration
Nopdanai Ajavakom (NAV)
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Chapter 2: Free Vibration of SDOF Systems
• Introduction
• Undamped Free Vibration
• Damped Free Vibration
– Underdamped
– Overdamped
– Critically Damped
• Damping Measurement
• Design Consideration
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2.1 Introduction
• A system is said to undergo free vibration when it oscillates
only under an initial disturbance with no external forces
after the initial disturbance.
• Undamped vibrations result when amplitude of motion
remains constant with time (e.g. in a vacuum)
• Damped vibrations occur when the amplitude of free
vibration diminishes gradually overtime, due to resistance
offered by the surrounding medium (e.g. air)
Examples
• The oscillations of the pendulum of a clock (close enough)
• The vertical oscillatory motion felt by a bicyclist after hitting
a road bump
• The motion of a child on a swing under an initial push
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2.1 Introduction
Tranverse
Vibration
Torsional
Vibration
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2.1 Introduction
• EOM of a single degree of freedom system
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2.1 Introduction
• Review of Linear Second Order Differential Equation
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2.1 Introduction
• Three Solutions
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2.1 Introduction
• Three Solutions
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2.2 Free Undamped Vibration
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ด้
ตุ้
/crad/s)
ถู
ที่
natural frequency ท่ กกร ะ น วย in itial disturbance => fn =
2.2 Free Undamped Vibration
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2.2 Free Undamped Vibration
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2.2 Free Undamped Vibration
v0
x(t )  x0 cos nt 
sin nt
n
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2.2 Free Undamped Vibration
v0
x(t )  x0 cos nt 
sin nt
n
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2.2 Free Undamped Vibration
• Relationship between displacement, velocity, and
acceleration
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Assume
small
oscillation => 0
ม
ค่าน
Linearization
Fs
↓-FSLCOSE =
↳
mLig
ก
Fo
tar
ล้
มี
ด
บ
mgl cost =mgl =
แ
mant
woment vibration
มั
ทิ
* * Ms
7 ่ว
2.3 Free Damped Vibration
3 = /1
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2.3 Free Damped Vibration
=4m
m
C=
2
#
↓
Cr
3 : "f
23 =
23Wn = C
htt
25Wn =
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2.3 Free Damped Vibration
.
.
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=- BWn I W1 ;
-1 G
2.3.1 Underdamped Vibration
rects.
About sin (
wat·I units
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2.3.1 Underdamped Vibration
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2.3.2 Overdamped Vibration
42 -
, UI = - Bunt We
-SUnt
act) =
-wn
2 ( Ace
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1 ; 3
3.1 + + Ace
We
-1
t
(
23
2.3.3 Critically Damped Vibration
U, Be = - BWnI Un
แทน
B =1; VicU, = - Wa
e:: ect) = CA, + Act (
Unt
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2.3 Free Damped Vibration
Comparison of motions with different types of damping
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2.3 Free Damped Vibration
Comparison of motions with different types of damping
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2.3 Free Damped Vibration
Comparison of motions with different types of damping
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2.3 Free Damped Vibration
Example 1: spring-mass-damper system
A spring-mass-damper system has mass of 100 kg,
stiffness of 3000 N/m and damping coefficient of 300
kg/s. Calculate the (undamped) natural frequency, the
damping ratio and the damped natural frequency. Does
the solution oscillate? This system is given a zero initial
velocity and an initial displacement of 0.1 m. Calculate
the vibration response. [Inman1.40, 1.42]
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=e-
0.104
1. 29
1 =
0.104e "
3t
sin ( 5.268t + 1.29)
Bunt
ACOSWatt Basin Wat
2.3 Free Damped Vibration
Example 2: spring-mass-damper system
A Spring-mass-damper system has mass of 150 kg,
stiffness of 1500 N/m and damping coefficient of 200
kg/s. Calculate the undamped natural frequency, the
damping ratio and the damped natural frequency. Is the
system overdamped, underdamped or critically
damped? Does the solution oscillate? This system is
given an initial velocity of 10 mm/s and an initial
displacement of -5 mm. Calculate the vibration
response. [inman1.41, 1.43]
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2.3 Damping Measurement
How can we measure STIFFNESS of a spring?
What about DAMPING of a damper?
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2.3 Damping Measurement
Logarithmic Decrement
>หา ท
เว
ลา
2 ห ท
ที่
ั้
ที่
↓us 3 -> หา C
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เว
ลาน น
+ไ ปอีก
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2.3 Measurement
Logarithmic Decrement
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2.3 Measurement
Logarithmic Decrement
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2.3 Measurement
Example 3:
8.
On() = n(?) =
3=
8=
2.197
0.3
45 "+
C=
=
2
2 x 0.33
=36.15
N .s / m
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2.3 Measurement
Example 4:
a)
b)
we = = =
Wo =m# =
load Is ⑦
1
radls -
↓
Ig
้สมการ
แก
ร kom
ไ #=4;
K= 4
m -> # =
4m =
:: m =
และ
ด้
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1
M+
1 kg
K=N
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2.4 Design Consideration
• Design in vibration: adjusting the physical parameters of
a device to cause its vibration response to meet a
specified performance or other criteria e.g. max
displacement, max speed, max static deflection, natural
frequency within a range etc.
For example
– Design a m-c-k system to have the desired response.
• underdamped, overdamped, critically damped
– Design a system that has a given natural frequency.
• Select connection of springs (series or parallel)
• Use elastic elements as springs
• Consider acceptable static deflection
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2.4 Design Consideration
Example 5:
a)
b)
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2.4 Design Consideration
Example 6:
Consider modeling the vertical suspension system of a
small sports car, as a single-DOF system. The mass of the
automobile is 1361 kg. The static deflection of the spring
is 0.05 m. Calculate c and k of the suspension system of
the car to be critically damped. If there are passengers
and baggage of 290 kg in the car, how does this affect
the damping ratio?
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2.4 Design Consideration
Example 6:
Consider modeling the vertical suspension system of a small sports car, as a single-DOF
system. The mass of the automobile is 1361 kg. The static deflection of the spring is 0.05
m. Calculate c and k of the suspension system of the car to be critically damped. If there
are passengers and baggage of 290 kg in the car, how does this affect the damping ratio?
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2.5 Exercises
Exercise 1:
A cantilever beam carries a mass M at the free end as
shown in the figure. A mass m falls from a height h on
to the mass M and adheres to it without rebounding.
Determine the resulting transverse vibration of the
beam.
Ans:
x(t )  A sin(nt  )
1/2
2



v0 
2
A   x0    

 n  
 v 
  tan 1  0 
 x0 n 
n 
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k
3EI
 3
M m
l ( M  m)
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2.5 Exercises
Exercise 2:
Determine the natural frequency of the system shown in
the figure. Assume the pulleys to be frictionless and of
negligible mass.
1/2


k1k2
n  

4
m
(
k

k
)
1
2 

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rad/sec
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2.5 Exercises
Exercise 3: Shock Absorber for a Motorcycle
An underdamped shock absorber is to be designed for a
motorcycle of mass 200 kg. When the shock absorber is
subjected to an initial vertical velocity due to a road
bump, the resulting displacement-time curve is to be as
indicated…
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2.5 Exercises
Exercise 3: Shock Absorber for a Motorcycle
Find the necessary stiffness and damping constants of
the shock absorber if the damped period of vibration is
to be 2 s and the amplitude x1 is to be reduced to onefourth in one half cycle (i.e., x1.5 = x1/4). Also find the
minimum initial velocity that leads to a maximum
displacement of 250 mm.
Ans: k = 2358 N/m, c = 554.5 N.s/m, v0 = 1.43m/s
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2.5 Exercises
Exercise 3: Shock Absorber for a Motorcycle
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2.5 Exercises
Since
x1.5  x1 / 4,
x2  x1.5 / 4 , x1 / 16
Hence the logarithmic decrement becomes
 x1 
2
  ln   ln16  2.7726 
1  2
 x2 
(E.1)
From which ζ can be found as 0.4037. The damped
period of vibration given by 2 s. Hence,
2 d 
n 
2
d

2
n 1   2
2
2 1  (0.4037)
2
 3.4338 rad/s
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2.5 Exercises
The critical damping constant can be obtained:
cc  2mn  2(200)(3.4338)  1.373.54 N - s/m
Thus the damping constant is given by:
c  cc  (0.4037)(1373.54)  554.4981 N - s/m
and the stiffness by:
k  mn2  (200)(3.4338) 2  2358.2652 N/m
The displacement of the mass will attain its max value
at time t1, given by sin d t1  1   2
sin d t1  sin t1  1  (0.4037) 2  0.9149
This gives:
or
t1 
sin 1 (0.9149)

 0.3678 sec
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2103433 Intro to Mech Vibration, NAV
2.5 Exercises
The envelope passing through the max points is:
x  1   2 Xe  t
(E.2)
n
Since x = 250mm,
0.25  1  (0.4037) 2 Xe  ( 0.4037 )( 3.4338)( 0.3678)
X  0.4550 m
The velocity of mass can be obtained by differentiating
the displacement:
x(t )  Xe  t sin d t
n
as
x (t )  Xe nt (n sin d t  d cos d t )
(E.3)
When t = 0,
x (t  0)  x0  Xd  Xn 1   2  (0.4550)(3.4338) 1  (0.4037) 2
 1.4294 m/s
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Worksheet 2: Problem 1
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Worksheet 2: Problem 2
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Worksheet 2: Problem 3
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Worksheet 2: Problem 5
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Worksheet 2: Problem 6
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