Section1.4 Modeling and Energy Methods

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Nov 16, 2013 (3 years and 10 months ago)

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Section1.4 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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Potential and Kinetic Energy

The potential energy of mechanical
systems
U
is often stored in “springs”
(remember that for a spring
F=kx
)

0 0
2
0
0 0
1

2
x x
spring
U F dx kx dx kx
  
 
The kinetic energy of mechanical systems
T
is due to the
motion of the “mass” in the system

2 2
1 1
,
2 2
trans rot
T mx T J

 
M

k

x
0

Mass

Spring

x=0

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Conservation of Energy

T

U

c
o
n
s
t
a
n
t
o
r








d
d
t
(
T

U
)

0
For a simple, conservative (i.e. no damper), mass spring
system the energy must be conserved:

At two different times
t
1
and
t
2
the increase in potential
energy must be equal to a decrease in kinetic energy (or visa
-
versa).

U
1

U
2

T
2

T
1
a
n
d
U
m
a
x

T
m
a
x
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Deriving the equation of motion
from the energy



2 2
1 1
( ) 0
2 2
0
Since cannot be zero for all time, then
0
d d
T U mx kx
dt dt
x mx kx
x
mx kx
 
   
 
 
  
 
M

k

x

Mass

Spring

x=0

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Determining the Natural frequency
directly from the energy

U
m
a
x

1
2
k
A
2
T
m
a
x

1
2
m
(

n
A
)
2
S
i
n
c
e

t
h
e
s
e

t
w
o

v
a
l
u
e
s

m
u
s
t

b
e

e
q
u
a
l
1
2
k
A
2

1
2
m
(

n
A
)
2

k

m

n
2


n

k
m
If the solution is given by
x(t)
=
A
sin(

t+
f
) then the maximum
potential and kinetic energies can be used to calculate the natural
frequency of the system

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Example 1.4.1

Compute the natural frequency of
this roller fixed in place by a spring.
Assume it is a conservative system
(i.e. no losses) and rolls with out
slipping.

2
trans
2
rot
2
1

and

2
1
x
m
T
J
T





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Solution continued

2
Rot
2
max
2
2 2 2
max
2 2
max
2
max max max
2 2 2
2
1
2
The max value of happens at
( )
1 1 1
( )
2 2 2
The max value of happens at
1
Thus
2
1 1
2 2
n
n
n n
n
x
x r x r T J
r
T v A
A
J
T J m A m A
r r
U x A
U kA T U
J
m A kA
r
 


 
 
    

 
    
 
 

   
 
  
 
 
2
n
k
J
m
r

 

 
 
Effective mass

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Example 1.4.2 Determine the equation of
motion of the pendulum using energy




m


mg


2

m
J

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Now write down the energy

2 2 2
0
2 2
1 1
2 2
(1 cos ), the change in elevation
is (1 cos )
1
( ) (1 cos ) 0
2
T J m
U mg
d d
T U m mg
dt dt
 


 
 
 

 
    
 
 
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2
2
2
(sin ) 0
(sin ) 0
(sin ) 0
( ) sin ( ) 0
( ) ( ) 0
n
m mg
m mg
m mg
g
t t
g g
t t
 
  
 
 
  
 
  
  
  
    
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Example 1.4.4

The effect of including the mass of
the spring on the value of the frequency.


x(t)

m
s
, k


y


y +dy


m


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3






2
1
3
2
1
2
1
3
2
1
2
1
3
2
1
=

element)
each

of

KE

the
up

(adds

2
1
s
assumption

),
(

:
element

of
velocity
:
element

of

mass
2
max
2
2
max
2
2
2
2
0
s
n
n
s
s
tot
mass
s
s
spring
dy
s
m
m
k
kA
U
A
m
m
T
x
m
m
T
x
m
T
x
m
dy
x
y
m
T
t
x
y
v
dy
dy
m
dy
































































This provides some
simple design and
modeling guides

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What about gravity?


m

m




+
x
(
t
)

0


k

mg


k
D

+
x
(
t
)

D

2
2
1
( )
2
1
2
spring
grav
U k x
U mgx
T mx
 D
 

m
equilibriu

static

and

FBD,

from

,
0

D

k
mg
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2 2
0 from static
equilibiurm
Now use ( ) 0
1 1
( ) 0
2 2
( )
( ) ( ) 0
0
d
T U
dt
d
mx mgx k x
dt
mxx mgx k x x
x mx kx x k mg
mx kx
 
 
   D 
 
 
   D
   D 
  

Gravity does not effect the
equation of motion or the natural
frequency of the system for a linear
system as shown previously with a
force balance.

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Lagrange’s Method for deriving
equations of motion.

Again consider a conservative system and its energy.


It can be shown that if the Lagrangian
L

is defined as

L

T

U
Then the equations of motion can be calculated from

0 (1.63)
d L L
dt q q
 
 
 
 
 
 
Which becomes

0 (1.64)
d T T U
dt q q q
 
  
  
 
  
 
Here
q

is a generalized coordinate

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Example 1.4.7 Derive the equation of motion
of a spring mass system via the Lagrangian

2 2
1 1
and
2 2
T mx U kx
 
Here
q


=
x
,

and and the Lagrangian becomes

2 2
1 1
2 2
L T U mx kx
   
Equation (1.64) becomes



0 0
0
d T T U d
mx kx
dt x x x dt
mx kx
  
 
     
 
  
 
  
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1.5 More on springs and stiffness


Longitudinal motion


A

is the cross sectional
area (m
2
)


E

is the elastic
modulus (Pa=N/m
2
)



is the length (m)


k

is the stiffness (N/m)


x
(
t
)


m


k

E
A
l

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Figure 1.21 Torsional Stiffness


J
p

is the polar
moment of inertia of
the rod


J

is the mass
moment of inertia of
the disk


G

is the shear
modulus, is the
length

J
p


J



t
)

0

k

G
J
p
l
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Example 1.5.1
compute the frequency of a shaft/mass
system {
J

= 0.5 kg m
2
}

4
10 2 2 4
2
( ) ( ) 0
( ) ( ) 0
,
32
For a 2 m steel shaft, diameter of 0.5 c
m
(8 10 N/m )[ (0.5 10 m)/32]
(2 m)(0.5kg m )
2.
p
n p
p
n
M J J t k t
k
t t
J
GJ
k d
J
J J
GJ
J
  
 





   
  
   

 
 



2 rad/s
From Equation (1.50)

Figure 1.22

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Fig. 1.22 Helical Spring

2
R


x
(
t
)


d

= diameter of wire

2
R=

diameter of turns


n
= number of turns


x
(
t
)= end deflection

G
= shear modulus of


spring material


k

G
d
4
6
4
n
R
3
Allows the design of springs


to have specific stiffness

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Fig 1.23 Transverse beam stiffness

f



m

x




Strength of
materials and
experiments yield:



k

3
E
I
l
3
W
i
t
h

a

m
a
s
s

a
t

t
h
e

t
i
p
:

n

3
E
I
m
l
3
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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


See text for more examples.


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Example 1.5.2 Effect of fuel on
frequency of an airplane wing


Model wing as transverse
beam


Model fuel as tip mass


Ignore the mass of the
wing and see how the
frequency of the system
changes as the fuel is
used up


x
(
t
)


l


E, I m

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Mass of pod 10 kg empty 1000 kg full




= 5.2x10
-
5

m
4
,
E

=6.9x10
9

N/m,


㴠㈠2


Hence the
natural
frequency
changes by an
order of
magnitude
while it
empties out
fuel.

9 5
full
3 3
9 5
empty
3 3
3 3(6.9 10 )(5.2 10 )
1000 2
11.6 rad/s=1.8 Hz
3 3(6.9 10 )(5.2 10 )
10 2
115 rad/s=18.5 Hz
EI
m
EI
m




 
 


 
 


This ignores the mass of the wing

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Example 1.5.3 Rolling motion of a ship

( ) sin ( )
For small angles this becomes
( ) ( ) 0

n
J t WGZ Wh t
J t Wh t
hW
J
 
 

   
 
 
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Combining Springs:
Springs are usually only
available in limited stiffness values. Combing them
allows other values to be obtained


Equivalent Spring


s
e
r
i
e
s
:

k
A
C

1
1
k
1

1
k
2
p
a
r
a
l
l
e
l
:


k
a
b

k
1

k
2
A

B

C

a

b

k
1

k
1

k
2

k
2

This is identical to the combination of capacitors in
electrical circuits

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Use these to design from available
parts


Discrete springs available in standard
values


Dynamic requirements require specific
frequencies


Mass is often fixed or
+

small amount


Use spring combinations to adjust w
n


Check static deflection

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Example 1.5.5
Design of a spring mass system
using available springs: series vs parallel


Let
m

= 10 kg


Compare a series and
parallel combination


a)
k
1

=1000 N/m,
k
2

= 3000
N/m,
k
3

=
k
4

=0


b)
k
3

=1000 N/m,
k
4

= 3000
N/m,
k
1

=
k
2

=0



k
1



k
2



k
3



k
4



m

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rad/s

66
.
8
10
750



N/m

750
1
3
3000
)
1
(
)
1
(
1
,
0
:
connection

series

b)

Case
rad/s

20
10
4000



N/m

4000
3000
1000
,
0
:
connection

parallel

a)

Case
4
3
2
1
2
1
4
3






















m
k
k
k
k
k
k
m
k
k
k
k
k
k
eg
series
eq
eg
parallel
eq


Same physical components, very different frequency

Allows some design flexibility in using off the shelf components

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Example:
Find the equivalent stiffness
k
of the
following system (Fig 1.26, page 47)


k
1



k
2



k
3



k
4



m


k
5



k
1
+
k
2
+
k
5



k
3



k
4



m


m



1
1
k
3

1
k
4

k
3
k
4
k
3

k
4

k
1

k
2

k
5

k
3
k
4
k
3

k
4

n

k
1
k
3

k
2
k
3

k
5
k
3

k
1
k
4

k
2
k
4

k
5
k
4

k
3
k
4
m
(
k
3

k
4
)
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Example 1.5.5
Compare the natural
frequency of two springs connected to a mass in
parallel with two in series

A series connect of
k
1

=
1000 N/m and
k
2
=3000 N/m with
m

= 10 kg


yields:

k
e
q

1
1
/
1
0
0
0

1
/
3
0
0
0

7
5
0

N
/
m


s
e
r
i
e
s

7
5
0

N
/
m
1
0

k
g

8
.
6
6

r
a
d
/
s


A parallel connect of
k
1

=
1000 N/m and
k
2
=3000 N/m with
m

= 10 kg


yields:

k
e
g

1
0
0
0

N
/
m

+

3
0
0
0


N
/
m

=

4
0
0
0

N
/
m


p
a
r

4
0
0
0

N
/
m
1
0

k
g

2
0

r
a
d
/
s
Same components, very different frequency

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Static Deflection

Another important consideration in designing with springs is the


static deflection

D
k

m
g

D

m
g
k
This determines how much a spring compresses or sags due


to the static mass (you can see this when you jack your car up)

The other concern is “rattle space” which is the maximum


deflection
A

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Section 1.6 Measurement


Mass
: usually pretty easy to measure using
a balance
-

a static experiment


Stiffness
: again can be measured statically
by using a simple displacement
measurement and knowing the applied force


Damping
: can only be measured
dynamically

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Measuring moments of inertia
using a Trifilar suspension system

J

g
T
2
r
0
2
m
0

m


4

2
l

J
0
T

is the measured period


g

is the acceleration due to gravity

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Stiffness Measurements

From Static Deflection:

Deflection or strain

Linear


Nonlinear


From Dynamic Frequency:

F
=
k x

or
s

=
E

e



n

k
m

k

m

n
2


k

F
x
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Example 1.6.1
Use the beam stiffness
equation to compute the modulus of a material

Figure 1.24


㴠ㄠ1Ⱐ
m

=

6 kg,
I
= 10
-
9
m
4
, and measured
T

= 0.62 s

T

2

m
l
3
3
E
I

0
.
6
2

s

E

4

2
m
l
3
3
T
2
I

4

2
6

k
g


1

m


3
3
(
0
.
6
2

s
)
2
1
0

9

m
4



2
.
0
5

1
0
1
1

N
/
m
2
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Damping Measurement
(Dynamic only)

Define the Logarithmic Decrement:

( )
( )
ln (1.71)
( )
sin( )
ln
sin( ) )
n
n
t
d
t T
d d
n
x t
x t T
Ae t
Ae t T
T



 f

  f
 

 




 

2
2
4










T
c
c
n
cr
(1.75)

(1.72)

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Section 1.7: Design Considerations

Using the analysis so far to guide the
selection of components.

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Example 1.7.1


Mass 2 kg
<

m
<

3 kg and
k

>

200 N/m


For a possible frequency range of


8.16 rad/s
<


n

<

10 rad/s


For initial conditions:
x
0

= 0,
v
0

< 300
mm/s


Choose a
c

so response is always
<

25
mm

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Solution:


Write down
x
(
t
) for 0
initial displacement


Look for max
amplitude


Occurs at time of first
peak (
T
max
)


Compute the
amplitude at
T
max


Compute


for
A
(
T
max
)=0.025


0

0.5

1

1.5

2

-
1

-
0.5

0

0.5

1

Time(sec)

Amplitude

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0
Amplitude
0
0
( ) sin( )
worst case happens at smallest 8.16 rad
/s
worst case happens at max 300 mm/s
With and fixed at these values, invest
igate how varies with
First pea
n
t
d
d
d n
n
v
x t e t
v
v



 
 


  
 


2
1
2
2
1 1
max
1
2
tan ( )
1
1
0
max
2
k is highest and occurs at
( ) 0 cos( ) sin( ) 0
1
1 1
Solve for tan ( ) tan
1
Sub into ( ):( ) ( ) sin(tan
1
n n
t t
d d n d
d
m
d n d
m m
n
d
x t e t e t
dt
t T T
v
T x t A x T e
 




   


   


 

 
 




   
 

 
   
 
 

 

2
1
2
1
tan ( )
1
0
)
( )
m
n
v
A e











 
 
 
 

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mm

37

yields

0
=

FYI,
kg/s

15
.
14
281
.
0
16
.
8
3
2
2

yields
kg)

3

=
(

mass

on the
limit
upper

the
Using
281
.
0
025
.
0
)
(
solve

m

0.025
then
less

max value

the
keep

To
0
max
max











n
n
v
A
m
c
m
A






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© D. J. Inman

43
/44

Example 1.7.3
What happens to a good design when
some one changes the parameters? (Car suspension system). How
does


捨c湧攠ei瑨t浡獳?

kg/s

10
81
.
3
)
14
)(
1361
(
2
2
1
=
N/m

10
668
.
2
)
14
(
1361



rad/s

14
05
.
0
81
.
9

,
1361
.


,

compute

m,

0.05
=

kg,

1361
=
m

1,
=
Given
4
5
2
2











D


D


D




D
n
n
n
n
m
c
k
m
mg
mg
k
k
mg
k
m
k
k
c






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© D. J. Inman

44
/44

Now add 290 kg of passengers
and luggage. What happens?

9
.
0
2
10
81
.
3
rad/s

7
.
12
06
.
0
8
.
9
m

06
.
0
10
668
.
2
8
.
9
1651
kg

1651
290
1361
4
5






D







D




n
cr
n
m
c
c
g
k
mg
m



So some oscillation


results at a lower


frequency.