Single Degree of Freedom Systems

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Single Degree of Freedom Systems
Mohammad Tawfik
Introduction to Vibrations of
Structures

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Mohammad Tawfik
References
• M. Bismarck-Nasr, "Structural Dynamics in Aeronautical
Engineering," AIAA Educational Series, 1999
• D. Inman and E. Austin, “Engineering Vibration,” 2nd edition,
Prentice Hall, 2001
• A. A. Shabana, "Vibration of Discrete and Continuous Systems," 2nd
edition, Springer, 1997
• D. Thorby, “Structural Dynamics and Vibration in Practice” Elsevier,
2008
• A. G. Ambekar, “Mechanical Vibrations and Noise Engineering”
Prentice Hall – India, 2006
• Leonard Meirovitch, “Fundamentals of Vibrations,” 1st edition,
McGraw Hill, 2001

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Mohammad Tawfik
Single degree of freedom
systems

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Mohammad Tawfik
Objectives
• Recognize a SDOF system
• Be able to solve the free vibration equation
of a SDOF system with and without
damping
• Understand the effect of damping on the
system vibration
• Apply numerical tools to obtain the time
response of a SDOF system

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Mohammad Tawfik
Single degree of freedom systems
• When one variable can describe the
motion of a structure or a system of
bodies, then we may call the system a 1-D
system or a single degree of freedom
(SDOF) system. e.g. x(t), q(t) Z(t), y(x).

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Mohammad Tawfik
Stiffness
• From strength of materials recall:

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Newton’s Law
• Newton’s Law:
00 )0(,)0(
0)()(
)()(
vxxx
tkxtxm
tkxtxm







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Mohammad Tawfik
Solving the ODE
• The ODE is
• The proposed
solution:
• Into the ODE you get
the characteristic
equation:
• Giving:
0)()(  tkxtxm 
t
aetx 
)(
02
 tt
ae
m
k
ae 

m
k
2

m
k
j

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Mohammad Tawfik
Solving the ODE (cont’d)
• The proposed
solution becomes:
• For simplicity, let’s
define:
• Giving:
t
m
k
jt
m
k
j
eaeatx

 21)(
m
k

tjtj
eaeatx  
 21)(

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Mohammad Tawfik
Let’s manipulate the solution!

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Recall
   ajSinaCose ja

         bSinaCosbCosaSinbaSin 

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Mohammad Tawfik
Manipulating the solution
• The solution we have:
• Rewriting:
tjtj
eaeatx  
 21)(
    
    tjSintCosa
tjSintCosatx




2
1)(
       tSinaajtCosaatx  2121)( 
   tSinAtCosAtx  21)( 

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Mohammad Tawfik
Further manipulation
2
2
2
1 AAA 
   
A
A
Sin
A
A
Cos 12
&  
        tSinCostCosSinAtx  )(
   tASintx )(

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Mohammad Tawfik
Different forms of the solution
tjtj
eaeatx
tCosAtSinAtx
tASintx







21
21
)(
)(
)()(

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Mohammad Tawfik
NOTE!

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Natural Frequency of Oscillation
• In the previously obtained solution:
• The frequency of oscillation is 
• It depends only on the characteristics of the
oscillating system. That is why it is called the
natural frequency of oscillation
   tASintx )(
m
k
n 

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Mohammad Tawfik
Frequency
periodtheiss
2
Hz
2s2
cycles
rad/cycle2
rad/s
frequencynaturalthecalledisrad/sinis
n
nnn
n
n
T
f











We often speak of frequency in Hertz or
RPM, but we need rad/s in the arguments
of the trigonometric functions.

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Recall: Initial Conditions

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Amplitude & Phase from the ICs
    
Phase
0
01
Amplitude
2
02
0
0
0
tan,
yieldsSolving
cos)0cos(
sin)0sin(










v
xv
xA
AAv
AAx
n
n
nnn
n






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Mohammad Tawfik
Some useful quantities
peak valueA


T
T
dttx
T
x
0
valueaverage=)(
1
lim
valuesquaremeanroot=2
xxrms 
valuesquare-mean=)(
1
lim
0
22


T
T
dttx
T
x

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Mohammad Tawfik
Peak Values
Ax
Ax
Ax
2
max
max
max
:onaccelerati
:velocity
:ntdisplaceme







Maximum or peak (amplitude) values:

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Mohammad Tawfik

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Samples of Vibrating Systems
• Deflection of continuum (beams, plates,
bars, etc) such as airplane wings, truck
chassis, disc drives, circuit boards…
• Shaft rotation
• Rolling ships

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Wing Vibration

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Ship Vibration

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Effective Stiffness of
Structures

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Bars
• Longitudinal motion
• A is the cross sectional
area (m2)
• E is the elastic modulus
(Pa=N/m2)
• l is the length (m)
• k is the stiffness (N/m)x(t)
m

EA
k 
l

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Mohammad Tawfik
Rods
• Jp is the polar
moment of inertia of
the rod
• J is the mass
moment of inertia of
the disk
• G is the shear
modulus, l is the
length
Jp
J qt)
0

pGJ
k 

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Mohammad Tawfik
Helical Spring
2R
x(t)
d = diameter of wire
2R= diameter of turns
n = number of turns
x(t)= end deflection
G= shear modulus of
spring material
3
4
64nR
Gd
k 

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Mohammad Tawfik
Beams
f
m
x
• Strength of materials
and experiments
yield:
3
3
3
3


m
EI
EI
k
n 



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Mohammad Tawfik
Equivalent Stiffness

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Mohammad Tawfik
Summary
• Write down the equation of motion using
Newton’s law
• Solve the equation of motion for a SDOF
• Use initial conditions to determine the
amplitude and phase of vibration for a
SDOF
• Evaluate the effective stiffness of
structural members

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Mohammad Tawfik
1. The amplitude of vibration of an undamped
system is measured to be 1 mm. the phase
shift is measured to be 2 rad and the
frequency 5 rad/sec. Calculate the initial
conditions.
2. Using the equation:
evaluate the constant A1 and A2 in terms of
the initial conditions
HW #1

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Mohammad Tawfik
HW #1 (cont’d)
3. An automobile is modeled as 1000 kg
mass supported by a stiffness k=400000
N/m. When it oscillates, the maximum
deflection is 10 cm. when loaded with the
passengers, the mass becomes 1300 kg.
calculate the change in the frequency,
velocity amplitude, and acceleration if the
maximum deflection remain 10 cm.

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Mohammad Tawfik
Adding Damping

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Objectives
• Understand the damping as a force
resisting motion
• Adding viscous damping to the equation of
motion of a SDOF
• Understand the difference in the
responses of different systems depending
on the value of the damping

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Damping
• Damping is some form of friction!
• In solids, friction between molecules result
in damping
• In fluids, viscosity is the form of damping
that is most observed
• In this course, we will use the viscous
damping model; i.e. damping proportional
to velocity

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Viscous Damping
• A mathematical form
called a dashpot or
viscous damper
somewhat like a shock
absorber the constant c
has units: Ns/m or kg/s
)(txcfc


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Shock Absorbers

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Spring-mass-damper systems
• From Newton’s law:
00 )0(,)0(
0)()()(
)()()(
vxxx
tkxtxctxm
tkxtxctxm







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Mohammad Tawfik
Solution (dates to 1743 by Euler)
0)()(2)( 2
 txtxtx nn  
km
c
2
=
Where the damping Ratio
is given by: (dimensionless)
Divide the equation of motion by m

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Mohammad Tawfik




 

rootstheofnaturethe
determines,1ntdiscriminatheHere
equationquadraticaofrootsthefrom
1
:inequationalgebraicannowiswhich
02
motionofeq.intosubsitute&)(Let
2
2
2,1
22




nn
t
n
t
n
t
t
aeeaea
aetx

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Mohammad Tawfik
Three possibilities:
00201
21
,
:conditionsinitialtheUsing
)(
221=
dampedcriticallycalled
repeated&equalareroots1)1
xvaxa
teaeatx
mkmcc
n
tt
ncr
nn










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Mohammad Tawfik
Critical damping cont’d
• No oscillation occurs
t
n
n
etxvxtx 
 
 ])([)( 000

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12
)1(
12
)1(
where
)()(
1
:rootsrealdistincttwo-damping-overcalled,1)2
2
0
2
0
2
2
0
2
0
1
1
2
1
1
2
2,1
22

















n
n
n
n
ttt
nn
xv
a
xv
a
eaeaetx nnn

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The over-damped response

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Most interesting Case!
2
2,1 1
:asformcomplexinrootswrite
pairsconjugateasrootscomplexTwo
commonmost-motiondunderdampecalled,1)3




jnn

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Mohammad Tawfik
Under-damping















00
01
2
0
2
00
2
1
2
1
1
tan
)()(
1
frequencynaturaldamped,1
)sin(
)()(
22
xv
x
xxvA
tAe
eaeaetx
n
d
dn
d
nd
d
t
tjtjt
n
nnn









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Under-damped-oscillation
• Gives an oscillating response with exponential decay
• Most natural systems vibrate with an under-damped
response

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Summary
• Modeling viscous damping
• Solving the equation of motion involving
viscous damping
• Recognizing the different types of
response based on the level of damping

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1. Use the given data to plot the response of the
SDOF system
2. Solve the equation
And plot the response
HW #1 (cont’d)
8.0,6.0,4.0,2.0,1.0,01.0
/0,1sec,/2 00



 smvmmxradn
0,1
0
00 

vx
xxx 

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Mohammad Tawfik
HW #1 (cont’d)
• Homework is due next week:
26/9/2010

Single Degree of Freedom Systems

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