02 damping - spc408 - fall2016

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http://WikiCourses.WikiSpaces.comVibration Damping

Maged Mostafa

Energy Method

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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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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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Maged Mostafa

Potential and Kinetic EnergyThe 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

2kx0

2

The kinetic energy of mechanical systems T is due to the motion

of the “mass” in the system

Mk

x0

Mass Spring

x=0

)(2

1)(

2

1 2 txmtmvTtrasn )(

2

1tJTrot &

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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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Maged Mostafa

Deriving the equation of motion

from the energy

Mk

x

Mass Spring

x=0

0

then t,allfor zero becannot

0)(

0)2

1

2

1()( 22

kxxm

xSince

kxxmx

kxxmdt

dUT

dt

d

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Maged Mostafa

Determining the Natural frequency

directly from the energy

Umax 1

2kA2 Tmax

1

2m(nA)2

Since these two values must be equal

1

2kA2

1

2m(nA)2

k mn

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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Maged Mostafa

Vibration Damping of SDOF

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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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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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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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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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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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Maged Mostafa

roots theof nature the

determines ,1nt discrimina theHere

equation quadratic a of roots thefrom

1

:in equation algebraican now iswhich

02

motion of eq. into subsitute & )(Let

2

2

2,1

22

nn

t

n

t

n

t

t

aeeaea

aetx

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Maged Mostafa

Three possibilities:

00201

21

,

:conditions initial theUsing

)(

221=

damped critically called

repeated & equal are roots1 )1

xvaxa

teaeatx

mkmcc

n

tt

ncr

nn

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Maged Mostafa

Critical damping cont’d

• No oscillation occurst

nnetxvxtx

])([)( 000

0 1 2 3 4-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

Time (sec)

Dis

pla

ce

me

nt (m

m)

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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Maged Mostafa

12

)1(

12

)1( where

)()(

1

:roots realdistinct two-damping-over called ,1 )2

2

0

2

02

2

0

2

0

1

1

2

1

1

2

2,1

22

n

n

n

n

ttt

nn

xva

xva

eaeaetx nnn

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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)

Dis

pla

ce

me

nt (m

m)

k=225N/m m=100kg and z=2

x0

=0.4mm v0

=1mm/s

x0

=0.4mm v0

=0mm/s

x0

=0.4mm v0

=-1mm/s

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Maged Mostafa

Most interesting Case!

2

2,1 1

:as formcomplex in roots write

pairs conjugate as rootscomplex Two

commonmost -motion dunderdampe called ,1 )3

jnn

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Maged Mostafa

Under-damping

00

01

2

0

2

00

2

1

2

1

1

tan

)()(1

frequency natural damped ,1

)sin(

)()(22

xv

x

xxvA

tAe

eaeaetx

n

d

dn

d

nd

d

t

tjtjt

n

nnn

f

f

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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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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

txt

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Maged Mostafa

x t sin .2 t y t .e.0.1 t

x t z t e.0.1 tr t .z t x t

0 5 10

1

1

x t

t

0 5 10

1

1

y t

t

0 5 101

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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Example: 1.8.1: Find values of spring stiffness

that gives stable response?

kl 2mg for a stable response

022

1cos ,sin

,

0sincossin2

2

22

mglklml

for

mgll

kml

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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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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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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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Maged Mostafa August 30, 1999

30Mechanical Engineering at Virginia Tech


00

01

2

0

2

00

2

tan

)()(1

frequency natural damped ,1

)sin()(

xv

x

xxvA

tAetx

n

d

dn

d

nd

d

tn

f

f

Recall the Underdamped response

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)sin(1

)(

:

)sin()(

00tan

,

0

2

0

0

1

0

0

te

vtx

OR

tev

tx

vA

x

dt

n

d

t

d

d

n

n

f

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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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Solution:

281.0

025.0)(

1tansin)(

21

21

1tan

1

0

21

1tan

1

0

dn

dn

ev

tx

OR

ev

tx

n

d

Maximum amplitude, will occur at:

• First Peak

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Solution:

21

02

1tan

1

0)sin(1

)cos()(

max

dx

t

dd

t

evtttx n

Substitute in x(t):

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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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Maged Mostafa

HW #2

• Homework is due next week:

02/10/2016