Energy methods for damped systems

Section 1.4 Modeling and Energy
Methods
• Provides 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

• Very useful for more complicated systems to
be discussed later (MDOF and distributed
mass systems)

College of Engineering @ProfAdhikari, #EG260 1/53
© Eng. Vib, 3rd Ed.

Potential and Kinetic Energy
The potential energy of mechanical
systems U is often stored in “springs” x=0 x0

(remember that for a spring F = kx) k
x0 x0 M
1 2
U spring = ∫ F dx = ∫ kx dx = kx0
0 0
2 Mass Spring

The kinetic energy of mechanical systems T is due to the
motion of the “mass” in the system

1 2 1 2

Ttrans = mx , Trot = Jθ
2 2
College of Engineering College of Engineering 2/53

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

T + U = constant
d
or (T + U ) = 0
dt
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 − U 2 = T2 − T1
and
U max = Tmax

Deriving the equation of motion
from the energy
x=0 x
k
M

Mass Spring

d d !1 2 1 2$

(T +U) = # mx + kx & = 0
dt dt " 2 2 %
⇒ x ( m + kx ) = 0
 x

Since x cannot be zero for all time, then
m + kx = 0
x

Determining the Natural frequency
directly from the energy
If the solution is given by x(t)= Asin(ωt+ϕ) then the maximum
potential and kinetic energies can be used to calculate the natural
frequency of the system

1 2 1
U max = kA Tmax = m(ωn A)2
2 2
Since these two values must be equal
1 2 1 2
kA = m(ωn A)
2 2
2 k
⇒ k = mωn ⇒ ωn =
m

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."

1 2 1 2

Trot = Jθ and Ttrans = mx
2 2


Solution continued

x = rθ ⇒ x = rθ⇒T = J 
1 x2
Rot
2 r2
The max value of T happens at vmax = ω n A
1 (ω n A)2 1 1" J% 2
⇒ Tmax = J + m(ω n A)2 = $ m + 2 'ω n A 2
2 r2 2 2# r &
The max value of U happens at xmax = A
1
⇒ U max = kA 2 Thus Tmax = U max ⇒
2
1" J% 2 2 1 2 k
$ m + 2 'ω n A = kA ⇒ ω n =
2# r & 2 " J%
$m + 2 '
# r &
Effective mass


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

l
θ

2 m!
J = ml
mg


Now write down the energy

1 2 1 2 2
T = J 0θ = m θ
2 2
U = mg(1− cosθ ), the change in elevation
is (1− cosθ )
d d " 1 2 2 %
(T +U) = $ m θ + mg(1− cosθ )' = 0
dt dt # 2 &


 2 
m θθ + mg(sin θ )θ = 0
⇒θ  m 2θ + mg(sin θ ) = 0
 ( )
 2
⇒ m θ + mg(sin θ ) = 0
 g
⇒ θ (t) + sin θ (t) = 0

 g g
⇒ θ (t) + θ (t) = 0 ⇒ ωn =
 

Example 1.4.4 The effect of including the mass of
the spring on the value of the frequency.

y
y +dy

m s, k
l
m
x(t)

m !
mass of element dy : s dy #
 #
" assumptions
y
velocity of element dy: vdy = x (t), #

#
 $
 2
1 ms & y )
Tspring = 
∫  (  x+ dy (adds up the KE of each element)
2 0
' *
1 , ms / 2
= . 1x 
2- 3 0
1 2 & 1 , ms / 1 ) 2 1, ms / 2 2
 
Tmass = mx ⇒ Ttot = ( . 1 + m+ x ⇒ Tmax = . m + 1ω n A
2 '2 - 3 0 2 * 2- 30
1 2
U max = kA
2
k
⇒ ωn = • This provides some
m
m+ s simple design and
3
modeling guides"

What about gravity?
kΔ" mg − kΔ = 0, from FBD,
and static equilibrium
m!
k
+x(t)"
0"
mg
1
U spring = k(Δ + x)2
m! Δ

2
+x(t) U grav = −mgx
1 2

T = mx
2

d
Now use (T +U) = 0
dt
d $1 2 1 2'

⇒ & mx − mgx + k(Δ + x) ) = 0
dt % 2 2 (
⇒ mx − mgx + k(Δ + x) x
x  
⇒ x (m + kx) + x (kΔ − mg) = 0
 x 

 
0 from static
equilibiurm

⇒ m + kx = 0
x • 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.
College of Engineering College of Engineering
© Eng. Vib, 3rd Ed. 14/53

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

d " ∂L % ∂L
$ '− =0 (1.63)

dt # ∂q & ∂q
Which becomes

d " ∂T % ∂T ∂U
$ '− + =0 (1.64)

dt # ∂q & ∂q ∂q
© Eng. Vib, 3rd Ed. Here q is a generalized coordinate

Example 1.4.7 Derive the equation of motion
of a spring mass system via the Lagrangian
1 2 1
 and U = kx 2
T = mx
2 2
Here q = x, and and the Lagrangian becomes
1 1

L = T −U = mx 2 − kx 2
2 2
Equation (1.64) becomes

d " ∂T % ∂T ∂U d
$ '− + 
= ( mx ) − 0 + kx = 0

dt # ∂x & ∂x ∂x dt
⇒ m + kx = 0
x


Example l
x = sin θ
2


k 2 k

θ 2

h = l (1 − cos θ )
m
1 2 1 2
U = kx + kx + mgl (1 − cos θ )
2 2
kl 2
= sin θ + mgl (1 − cos θ )
2
College of Engineering
4 College of Engineering 17/53

1 2 1 2 2
The Kinetic energy term is: T = J 0θ = m θ
2 2
Compute the terms in Lagrange’s equation:
d " ∂T % d  
$ '=
dt # ∂θ & dt
(
m 2θ = m 2θ )
∂T
=0
∂θ
∂U ∂ " k 2 2 % k 2
= $ sin θ + mg(1− cosθ )' = sin θ cosθ + mgsin θ
∂θ ∂θ # 4 & 2

Lagrange’s equation (1.64) yields:
d " ∂T % ∂T ∂U  k 2
$ '− + = m 2θ + sin θ cosθ + mgsin θ = 0

dt # ∂q & ∂q ∂q 2


Does it make sense:
2
2  k
m θ + sin θ cosθ + mgsin θ = 0
2 
  
0 if k=0
Linearize to get small angle case:
 k 2
m 2θ + θ + mgθ = 0
2
 + " k + 2mg %θ = 0
⇒θ $ '
# 2m &
k + 2mg
⇒ ωn =
2m
What happens College of Engineering
if you linearize first?
19/53

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@TheSandy36

Hashtag EG-260


1.5 More on springs and stiffness

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


Figure 1.21 Torsional Stiffness

•  Jp is the polar
moment of inertia of
the rod
GJ p •  J is the mass
Jp
k= moment of inertia of
l the disk
0
•  G is the shear
J! θ(t)

modulus, l is the
length


Example 1.5.1 compute the frequency of a shaft/mass
system {J = 0.5 kg m2}

From Equation (1.50)

 
∑ M = Jθ ⇒ Jθ (t) + kθ (t) = 0
(t) + k θ (t) = 0
⇒θ Figure 1.22
J
k GJ p πd4
⇒ ωn = = , Jp =
J J 32
For a 2 m steel shaft, diameter of 0.5 cm ⇒
GJ p (8 ×1010 N/m 2 )[π (0.5 ×10 −2 m)4 / 32]
ωn = =
J (2 m)(0.5kg ⋅ m 2 )
= 2.2 rad/s

Fig. 1.22 Helical Spring

d = diameter of wire
2R= diameter of turns
2R " n = number of turns
x(t)= end deflection
x(t) "
G= shear modulus of
spring material"
"
4
Gd
Allows the design of springs
to have specific stiffness k= 3
64nR
College of Engineering 24/53

Fig 1.23 Transverse beam stiffness

•  Strength of materials
f and experiments
yield:
m 3EI
k= 3
l
x With a mass at the tip:
3EI
ωn =
ml 3 25/53

Example for a Heavy Beam
Consider again the beam of Figure 1.23 and what
happens if the mass of the beam is considered.

P = applied static load
Much like example 1.4.4
M = mass of beam where the mass of a spring
m was considered, the
procedure is to calculate the
kinetic energy of the beam
y x(t) itself, by looking at a
differential element of the
From strength of materials the static beam and integrating over
deflection of a cantilever beam of the beam length
length l is: Py 2
x ( y) = ( 3l − y )
6 EI
Pl 3
Which has maximum value of (at x =l ): x max =
College of Engineering College of Engineering 3EI

Next integrate along the beam to compute the
beam’s kinetic energy contribution
1 
Tmax = ∫ (mass of differential element)i(velocity of differential)2
2 0
1 " 2 M 2
1 M xmax 
 %
= ∫ # x ( y)$ dy = 6 ∫0 (
3y 2 − y 3 ) dy
2  2  4

0

Mass of
element dy

1 ' 33 * 2
= ) 
M , xmax
2 ( 140 +
33
Thus the equivalent mass of the beam is: M eq = M
140
And the equivalent mass of the beam- mass system is:
33
msystem = M +m
140

With the equivalent mass known the
frequency adjustment for including the
mass of the beam becomes

3EI
k l3
ωn = =
meq 33
m+ M
140
3EI
= rad/s
⎛ 33 ⎞
l 3 ⎜ m + M ⎟
⎝ 140 ⎠


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 the book for more examples.


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
E, I m! frequency of the system
changes as the fuel is
l used up
x(t) "


Mass of pod 10 kg empty 1000 kg full
 = 5.2x10-5 m4, E =6.9x109 N/m,  = 2 m

•  Hence the
natural 3EI 3(6.9 × 10 9 )(5.2 × 10 −5 )
ωfull = =
frequency ml 3
1000 ⋅ 23
changes by an = 11.6 rad/s=1.8 Hz
order of
3EI 3(6.9 × 10 9 )(5.2 × 10 −5 )
magnitude ωempty = =
while it ml 3
10 ⋅ 23
empties out = 115 rad/s=18.5 Hz
fuel.

This ignores the mass of the wing

Example 1.5.3 Rolling motion of a ship


Jθ (t) = −W GZ = −Whsin θ (t)
For small angles this becomes

Jθ (t) + Whθ (t) = 0
hW
⇒ ωn =
J


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

A k1 B k2 C •  Equivalent Spring

1
series: k AC =
k1 1 1
a b +
k1 k2
k2 parallel: kab = k1 + k2
This is identical to the combination of capacitors in
electrical circuits


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 ωn
•  Check static deflection

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

k2
•  Let m = 10 kg
k1
•  Compare a series and
m parallel combination
•  a) k1 =1000 N/m, k2 = 3000
k3 N/m, k3 = k4 =0
•  b) k3 =1000 N/m, k4 = 3000
k4
N/m, k1 = k2 =0


Case a) parallel connection:
k3 = k4 = 0, keq = k1 + k2 = 1000 + 3000 = 4000 N/m
keg 4000
⇒ ω parallel = = = 20 rad/s
m 10
Case b) series connection:
1 3000
k1 = k2 = 0, keq = = = 750 N/m
(1 k3 ) + (1 k4 ) 3 + 1
keg 750
⇒ ωseries = = = 8.66 rad/s
m 10
Same physical components, very different frequency"
Allows some design ﬂexibility in using off the shelf components"

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

k 3 k4
k1+k2+k5 k1 + k2 + k5 +
k3 + k 4
k1 k2
m m
m
k3
k3 1 k3 k4
= =
k5 k4 1 1 k3 + k4
+
k3 k4
k4

k1k3 + k2 k3 + k5 k3 + k1k4 + k2 k4 + k5 k4 + k3 k4
ωn = 37/53
m ( k3 + k 4 )

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 k1 =1000 N/m and k2 =3000 N/m with m = 10 kg
yields:
1 750 N/m
keq = = 750 N/m ⇒ ω series = = 8.66 rad/s
1 / 1000 + 1 / 3000 10 kg

A parallel connect of k1 =1000 N/m and k2 =3000 N/m with m = 10 kg
yields:

4000 N/m
keg = 1000 N/m + 3000 N/m = 4000 N/m ⇒ ω par = = 20 rad/s
10 kg

Same components, very different frequency

Static Deflection
Another important consideration in designing with springs is the
static deflection

mg
Δk = mg ⇒ Δ =
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


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


Measuring moments of inertia
using a Trifilar suspension system
gT 2r02 ( m0 + m )
J= − J0
4π l 2

T is the measured period
g is the acceleration due to gravity


Stiffness Measurements
From Static Deflection:

Force or stress
Linear Nonlinear

F = k x or σ = E ε
F
⇒k=
x
Deflection or strain
From Dynamic Frequency:

k 2
ωn = ⇒ k = m ωn
m


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

Figure 1.24  = 1 m, m = 6 kg, I = 10-9 m4 , and measured T = 0.62 s
ml 3
T = 2π = 0.62 s
3EI
3
4π ml 2 3
4π 2 ( 6 kg )(1 m )
⇒E= = = 2.05 × 1011 N/m 2
3T 2 I (
3(0.62 s)2 10 −9 m 4 )

Damping Measurement (Dynamic only)
Define the Logarithmic Decrement:

x(t)
δ = ln (1.71)
x(t + T )
Ae−ζω n t sin(ω d t + φ )
δ = ln −ζω n (t +T )
Ae sin(ω d t + ω dT ) + φ ) (1.72)
δ = ζω nT

c δ δ
ζ= = = 2 2
ccr ωnT 4π + δ
(1.75)

Section 1.7: Design Considerations

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


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 < ωn < 10 rad/s
•  For initial conditions: x0 = 0, v0 < 300
mm/s
•  Choose a c so response is always < 25
mm


Solution:
•  Write down x(t) for 0
initial displacement
•  Look for max 1

amplitude
0.5
•  Occurs at time of first

Amplitude
peak (Tmax) 0
•  Compute the
amplitude at Tmax -0.5

•  Compute ζ for
A(Tmax)=0.025 -1
0 0.5 1 1.5 2
Time(sec)


v0 −ζωnt
x(t) = e sin(ω d t)
ωd
 
 
Amplitude

⇒ worst case happens at smallest ω d ⇒ ω n = 8.16 rad/s
⇒ worst case happens at max v0 = 300 mm/s
With ω n and v0 fixed at these values, investigate how varies with ζ
First peak is highest and occurs at
d
( x(t)) = 0 ⇒ ω d e−ζωnt cos(ω d t) − ζω n e−ζωnt sin(ω d t) = 0
dt
1 1 # 1− ζ 2 &
−1 ω d −1
Solve for t = Tmax ⇒ Tm = tan ( )= tan % % ζ (
(
ωd ζω n ω d $ '
ζ 1−ζ 2
v0 − tan −1 (
ζ
) # 1− ζ 2 &
1−ζ 2 −1
Sub Tmax into x(t) : Am (ζ ) = x(Tm ) = e sin(tan %
% ζ ()
(
2
ω n 1− ζ $ '
ζ 1−ζ 2
− tan −1 ( )
v 1−ζ 2 ζ
Am (ζ ) = 0 e
ωn

To keep the max value less then 0.025 m solve
Amax (ζ ) = 0.025 ⇒ ζ = 0.281
Using the upper limit on the mass (m = 3 kg)
yields
c = 2mω nζ = 2 ⋅ 3⋅ 8.16 ⋅ 0.281= 14.15 kg/s

v0
FYI, ζ = 0 yields Amax = = 37 mm
ωn

€

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

Given ζ =1, m=1361 kg, Δ=0.05 m, compute c, k .
k 2 mg
ωn = ⇒ k = 1361ω n , mg = kΔ ⇒ k =
m Δ
mg 9.81
⇒ ωn = = = 14 rad/s ⇒
mΔ 0.05
k = 1361(14)2 = 2.668 × 10 5 N/m
ζ =1 ⇒ c = 2mω n = 2(1361)(14) = 3.81 × 10 4 kg/s


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

m = 1361 + 290 = 1651 kg
mg 1651⋅ 9.8
⇒Δ= = 5
≈ 0.06 m
k 2.668 × 10
g 9.8
⇒ ωn = = = 12.7 rad/s
Δ 0.06
c 3.81 × 10 4 So some oscillation"
ζ= = = 0.9 results at a lower"
ccr 2mω n frequency."

Section 1.8 Stability
Stability is defined for the solution of free
response case:

Stable: x(t) < M, ∀ t > 0

Asymptotically Stable: lim x(t) = 0

t→∞
Unstable:
if it is not stable or asymptotically stable

52/53

Examples of the types of stability
Stable

Asymptotically Stable

x(t) x(t)

t

t

x(t) x(t)

t

t

Divergent instability Flutter instability


Example: 1.8.1: For what values of the
spring constant will the response be
stable?
Figure 1.37

! k 2 $ k 2

m 2θ + # 
sin θ & cosθ − mgsin θ = 0 ⇒ m 2θ + θ − mgθ = 0
" 2 % 2

⇒ 2mθ + ( k − 2mg)θ = 0 (for small θ )

⇒ k l > 2mg for a stable response

54/53

Energy methods for damped systems

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