02 Damping - SPC408 - Fall2016

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Vibration Damping
Maged Mostafa
Energy Method

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Vibration Damping
Maged Mostafa
Objective
• Estimate the Keinitic and Potential Energy
of Mass-Spring System
• Utilize the energy method to estimate the
natural frequency of SDOF

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Vibration Damping
Maged Mostafa
Modeling and Energy Methods
•An alternative way to determine the equation
of motion and an alternative way to calculate
the natural frequency of a system
•Useful if the forces or torques acting on the
object or mechanical part are difficult to
determine

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Vibration Damping
Maged Mostafa
Potential and Kinetic Energy
The potential energy of mechanical systems U is often stored in
“springs” (remember that for a spring F=kx)
Uspring  F
0
x0
 dx  kx
0
x0
 dx 
1
2
kx0
2
The kinetic energy of mechanical systems T is due to the motion
of the “mass” in the system
M
k
x0
Mass Spring
x=0
)(
2
1
)(
2
1 2
txmtmvTtrasn
 )(
2
1
tJTrot &

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Vibration Damping
Maged Mostafa
Conservation of Energy
T  U  constant
or
d
dt
(T  U)  0
For a simple, conservative (i.e. no damper), mass spring system
the energy must be conserved:
At two different times t1 and t2 the increase in potential energy
must be equal to a decrease in kinetic energy (or visa-versa).
U1 U2  T2  T1
and
Umax  Tmax

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Vibration Damping
Maged Mostafa
Deriving the equation of motion
from the energy
M
k
x
Mass Spring
x=0
0
thent,allforzerobecannot
0)(
0)
2
1
2
1
()( 22



kxxm
xSince
kxxmx
kxxm
dt
d
UT
dt
d





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Vibration Damping
Maged Mostafa
Determining the Natural frequency
directly from the energy
Umax 
1
2
kA2
Tmax 
1
2
m(n A)2
Since these two values must be equal
1
2
kA2

1
2
m(n A)2
 k  mn
2
 n 
k
m
Assuming the solution is given by
x(t)= ASin(t+f)
then the maximum potential and kinetic energies can be used to
calculate the natural frequency of the system

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Vibration Damping
Maged Mostafa
Vibration Damping of SDOF

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Vibration Damping
Maged Mostafa
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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Vibration Damping
Maged Mostafa
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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Vibration Damping
Maged Mostafa
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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Vibration Damping
Maged Mostafa
Shock Absorbers

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

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

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Vibration Damping
Maged Mostafa
Critical damping cont’d
• No oscillation occurs
t
n
n
etxvxtx 
 
 ])([)( 000
0 1 2 3 4
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
Time (sec)
Displacement(mm)
k=225N/m m=100kg and =1
x0 =0.4mm v0 =1mm/s
x0 =0.4mm v0 =0mm/s
x0 =0.4mm v0 =-1mm/s

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Vibration Damping
Maged Mostafa
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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Vibration Damping
Maged Mostafa
The over-damped response
0 1 2 3 4
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
Time (sec)
Displacement(mm)
k=225N/m m=100kg andz=2
x0
=0.4mm v0
=1mm/s
x0
=0.4mm v0
=0mm/s
x0
=0.4mm v0
=-1mm/s

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

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Vibration Damping
Maged Mostafa
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


f



f


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

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Vibration Damping
Maged Mostafa
Stability
Stability is defined for the solution of free response
case:
Stable:
Asymptotically Stable:
Unstable:
if it is not stable or asymptotically stable
x(t)  M,  t  0
0)(lim 

tx
t

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Vibration Damping
Maged Mostafa
x t sin .2 t y t .e
.0.1 t
x t z t e
.0.1 t
r t .z t x t
0 5 10
1
1
x t
t
0 5 10
1
1
y t
t
0 5 10
1
2
3
z t
t
0 5 10
4
2
2
4
r t
t
Examples of the types of stability

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Vibration Damping
Maged Mostafa
Example: 1.8.1: Find values of spring stiffness
that gives stable response?
 kl  2mg for a stable response
  022
1cos,sin
,
0sincossin
2
2
2
2













mglklml
for
mgl
l
kml



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Vibration Damping
Maged Mostafa
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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Vibration Damping
Maged Mostafa
Example 1.7.1: Design Problem
• Mass 2 kg < m < 3kg and k > 200 N/m
• For a possible frequency range of
8.16 rad/s < n < 10 rad/s
• For initial conditions: x0 = 0, v0 < 300 mm/s
• Choose a c so response is always < 25 mm

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Vibration Damping
Maged Mostafa
Solution: Design Problem
• Write down x(t) for 0 initial displacement
• Look for max amplitude
• Occurs at time of first peak (Tmax)
• Compute the amplitude at Tmax
• Compute  for A(Tmax)=0.025

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Vibration Damping
Maged Mostafa August 30, 1999
30Mechanical Engineering at Virginia Tech













00
01
2
0
2
00
2
tan
)()(
1
frequencynaturaldamped,1
)sin()(
xv
x
xxvA
tAetx
n
d
dn
d
nd
d
tn


f



f
Recall the Underdamped response

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Vibration Damping
 
)sin(
1
)(
:
)sin()(
00tan
,
0
2
0
0
1
0
0
t
e
v
tx
OR
te
v
tx
v
A
x
dt
n
d
t
d
d
n
n




f












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Vibration Damping
Solution:
sec,/300
sec/16.8,200,3
0 mmv
radkm n

 
Maximum amplitude, will occur at:
• Smallest Natural Frequency
• Smallest Spring Stiffness
• Largest Mass
• Largest initial velocity

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Vibration Damping
Solution:
281.0
025.0)(
1
tansin)(
2
1
2
1
1
tan
1
0
2
1
1
tan
1
0

















 








 









 
















d
n
d
n
e
v
tx
OR
e
v
tx
n
d
Maximum amplitude, will occur at:
• First Peak

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Vibration Damping
Solution:







 




















 
2
1
02
1
tan
1
0)sin(
1
)cos()(
max
d
x
t
dd
t
evtttx n

Substitute in x(t):

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Vibration Damping
Maged Mostafa
1. Use the given data to plot the response of the
SDOF system
2. Solve the equation
And plot the response
HW #2
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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Vibration Damping
Maged Mostafa
HW #2
• Homework is due next week:
02/10/2016

02 Damping - SPC408 - Fall2016

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