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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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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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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.
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