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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)
@ProfAdhikari, #EG260
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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
20
0 0
1 2
x x
springU F dx kx dx kx
The kinetic energy of mechanical systems T is due to the motion of the “mass” in the system
Ttrans =12mx2, Trot =
12J θ 2
M k
x0
Mass Spring
x=0
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Conservation of Energy
+ =
+ =
constant
or ( ) 0
T Ud T Udt
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).
− = −
=
1 2 2 1
max max
andU U T T
U TCollege of Engineering
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Deriving the equation of motion from the energy
ddt
(T +U) = ddt
12mx2 +
12kx2!
"#
$
%&= 0
⇒ x mx + kx( ) = 0Since x cannot be zero for all time, thenmx + kx = 0
M k
x
Mass Spring
x=0
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Determining the Natural frequency directly from the energy
ω
ω
ω ω
= =
=
⇒ = ⇒ =
2 2max max
2 2
2
1 1 ( )2 2
Since these two values must be equal1 1 ( )2 2
n
n
n n
U kA T m A
kA m A
kk mm
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
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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."
Trot =12J θ 2 and Ttrans =
12mx2
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Solution continued
x = rθ ⇒ x = r θ ⇒ TRot =12J x
2
r2
The max value of T happens at vmax =ωnA
⇒ Tmax =12J (ωnA)2
r2 +12m(ωnA)2 =
12m+ J
r2
"
#$
%
&'ωn
2A2
The max value of U happens at xmax = A
⇒Umax =12kA2 Thus Tmax =Umax ⇒
12m+ J
r2
"
#$
%
&'ωn
2A2 =12kA2 ⇒ωn =
k
m+ Jr2
"
#$
%
&'
Effective mass
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Example 1.4.2 Determine the equation of motion of the pendulum using energy
θ
m !
mg
l
2lmJ =
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Now write down the energy
T = 12J0θ 2 =
12m2 θ 2
U =mg(1− cosθ ), the change in elevation is (1− cosθ )ddt
(T +U) = ddt
12m2 θ 2 +mg(1− cosθ )
"
#$
%
&'= 0
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m2 θ θ +mg(sinθ ) θ = 0
⇒ θ m2 θ +mg(sinθ )( ) = 0
⇒ m2 θ +mg(sinθ ) = 0
⇒ θ (t)+ g
sinθ(t) = 0
⇒ θ (t)+ gθ(t) = 0 ⇒ωn =
g
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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)
ms, k
y y +dy
m
l
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mass of element dy : ms
dy
velocity of element dy: vdy =yx(t),
!
"##
$##
assumptions
Tspring =12
ms
0
∫ yx
&
'()
*+
2
dy (adds up the KE of each element)
= 12ms
3,
-.
/
01 x2
Tmass =12mx2 ⇒ Ttot =
12ms
3,
-.
/
01+
12m
&
'(
)
*+ x
2 ⇒ Tmax =12m+ ms
3,
-.
/
01ωn
2A2
Umax =12kA2
⇒ωn =k
m+ ms
3• This provides some simple design and modeling guides"
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What about gravity?
m !
m !
+x(t)
0"
k
mg
kΔ"
+x(t)"
Δ Uspring =
12k(Δ+ x)2
Ugrav = −mgx
T = 12mx2
mequilibriustatic and FBD, from ,0=Δ− kmg
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Now use ddt
(T +U) = 0
⇒ddt
12mx2 −mgx + 1
2k(Δ+ x)2$
%&'
()= 0
⇒ mxx −mgx + k(Δ+ x) x⇒ x(mx + kx)+ x(kΔ−mg
0 from static equilibiurm
) = 0
⇒ mx + kx = 0 • 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 −UThen the equations of motion can be calculated from
ddt
∂L∂ q"
#$
%
&'−
∂L∂q
= 0 (1.63)
Which becomes
ddt
∂T∂ q"
#$
%
&'−
∂T∂q
+∂U∂q
= 0 (1.64)
Here q is a generalized coordinate College of Engineering
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Example 1.4.7 Derive the equation of motion of a spring mass system via the Lagrangian
T = 12mx2 and U =
12kx2
Here q = x, and and the Lagrangian becomes
L = T −U =12mx2 − 1
2kx2
Equation (1.64) becomes
ddt
∂T∂x"
#$
%
&'−
∂T∂x
+∂U∂x
=ddt
mx( )− 0+ kx = 0
⇒ mx + kx = 0
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Example
θ m
k k
θ= −l (1 cos )h
θ=l sin2
x
θ
θ θ
= + + −
= + −
l
l l
2 2
22
1 1 (1 cos )2 2
sin (1 cos )4
U kx kx mg
k mg17/53
2
2
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The Kinetic energy term is: T = 12J0 θ
2 =12m2 θ 2
Compute the terms in Lagrange’s equation: ddt
∂T∂ θ"
#$
%
&'=
ddt
m2 θ( ) =m2 θ∂T∂θ
= 0
∂U∂θ
=∂∂θ
k2
4sin2θ +mg(1− cosθ )
"
#$
%
&'=
k2
2sinθ cosθ +mgsinθ
ddt
∂T∂ q"
#$
%
&'−
∂T∂q
+∂U∂q
=m2 θ + k2
2sinθ cosθ +mgsinθ = 0
Lagrange’s equation (1.64) yields:
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Does it make sense:
m2 θ + k2
2sinθ cosθ
0 if k=0
+mgsinθ = 0
Linearize to get small angle case: m2 θ + k
2
2θ +mgθ = 0
⇒ θ + k+ 2mg2m
"
#$
%
&'θ = 0
⇒ωn =k+ 2mg
2mWhat happens if you linearize first?
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@TheSandy36
Hashtag EG-260
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1.5 More on springs and stiffness
• Longitudinal motion • A is the cross sectional
area (m2) • E is the elastic
modulus (Pa=N/m2) • is the length (m) • k is the stiffness (N/m)
x(t)"
m "
=lEAkl
l
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Figure 1.21 Torsional Stiffness
• Jp 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
Jp
J ! θ(t)
0
=lpGJ
k
l
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Example 1.5.1 compute the frequency of a shaft/mass system {J = 0.5 kg m2}
M∑ = J θ ⇒ J θ (t)+ kθ(t) = 0
⇒ θ (t)+ kJθ(t) = 0
⇒ωn =kJ=
GJpJ
, Jp =πd 4
32For a 2 m steel shaft, diameter of 0.5 cm ⇒
ωn =GJpJ
=(8×1010 N/m2 )[π (0.5×10−2 m)4 / 32]
(2 m)(0.5kg ⋅m2 ) = 2.2 rad/s
From Equation (1.50)
Figure 1.22
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Fig. 1.22 Helical Spring
2R "
x(t) "
d = diameter of wire 2R= diameter of turns n = number of turns x(t)= end deflection G= shear modulus of spring material""
k = Gd4
64nR3Allows 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:
ω
=
=
l
l
3
3
3
With a mass at the tip:
3n
EIk
EIm
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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
m
x(t)
Much like example 1.4.4 where the mass of a spring was considered, the procedure is to calculate the kinetic energy of the beam itself, by looking at a differential element of the beam and integrating over the beam length
M = mass of beam
From strength of materials the static deflection of a cantilever beam of length l is:
y
( )= −2
( ) 36Pyx y yEI
l
Which has maximum value of (at x =l ): =3
max 3PxEIl
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Next integrate along the beam to compute the beam’s kinetic energy contribution
Tmax =12
(mass of differential element)0
∫ i(velocity of differential)2
= 12
x y( )"# $%2
0
∫ Mdy
Mass ofelement dy
=
12Mxmax
2
46 3y2 − y3( )dy0
∫
= 12
33140
M'
()
*
+, xmax
2
Thus the equivalent mass of the beam is: =eq33140
M M
And the equivalent mass of the beam- mass system is:
= +system33140
m M mCollege of Engineering 27/53
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With the equivalent mass known the frequency adjustment for including the
mass of the beam becomes
ω = =+
=⎛ ⎞+⎜ ⎟⎝ ⎠
3
eq
3
3
33140
3 rad/s33
140
n
EIkm m M
EI
m M
l
l
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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 the book 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 m4, E =6.9x109 N/m, = 2 m
• Hence the natural frequency changes by an order of magnitude while it empties out fuel.
ω
ω
−
−
× ×= =
⋅=
× ×= =
⋅=
l
l
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
EIm
EIm
This ignores the mass of the wing College of Engineering
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Example 1.5.3 Rolling motion of a ship
J θ (t) = −WGZ = −Whsinθ(t)For small angles this becomesJ θ (t)+Whθ(t) = 0
⇒ωn =hWJ
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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
=+
= +1 2
1 2
1series: 1 1
parallel:
AC
ab
k
k kk k k
A B C
a b
k1
k1
k2
k2
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 ωn
• 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) k1 =1000 N/m, k2 = 3000
N/m, k3 = k4 =0 • b) k3 =1000 N/m, k4 = 3000
N/m, k1 = k2 =0
k1 k2
k3
k4
m
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ω
= = = + = + =
⇒ = = =
= = = = =+ +
3 4 1 2
1 23 4
Case a) parallel connection:0, 1000 3000 4000 N/m
4000 20 rad/s10
Case b) series connection:1 30000, 750 N/m
(1 ) (1 ) 3 1
eq
egparallel
eq
k k k k k
km
k k kk k
ω⇒ = = =750 8.66 rad/s10
egseries
km
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)
k1 k2
k3
k4
m
k5
k1+k2+k5
k3
k4
m m
€
=1
1k3
+1k4
=k3k4k3 + k4
€
k1 + k2 + k5 +k3k4k3 + k4
ω+ + + + + +
=+
1 3 2 3 5 3 1 4 2 4 5 4 3 4
3 4( )nk k k k k k k k k k k k k k
m k k
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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 k1 =1000 N/m and k2 =3000 N/m with m = 10 kg yields:
keq =1
1 /1000 +1 / 3000= 750 N/m ⇒ω series =
750 N/m10 kg
= 8.66 rad/s
A parallel connect of k1 =1000 N/m and k2 =3000 N/m with m = 10 kg yields:
keg = 1000 N/m + 3000 N/m = 4000 N/m ⇒ω par =4000 N/m
10 kg= 20 rad/s
Same components, very different frequency College of Engineering
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Static Deflection Another important consideration in designing with springs is the static deflection
Δk = mg⇒Δ =mgk
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
( )π
+= −
l
2 20 0
024gT r m m
J J
T is the measured period g is the acceleration due to gravity
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Stiffness Measurements
From Static Deflection:
Forc
e or
stre
ss
Deflection or strain
Linear Nonlinear
From Dynamic Frequency:
F = k x or σ = E ε
€
ωn =km⇒ k =mωn
2
€
⇒ k =Fx
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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
( )( )( )
π
ππ−
= =
⇒ = = = ×
l
l
3
322 311 2
2 2 9 4
2 0.62 s3
4 6 kg 1 m4 2.05 10 N/m3 3(0.62 s) 10 m
mTEI
mET I
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Damping Measurement (Dynamic only) Define the Logarithmic Decrement:
δ = ln x(t)x(t + T )
(1.71)
δ = ln Ae−ζωnt sin(ωdt + φ)Ae−ζωn (t+T ) sin(ωdt +ωdT ) +φ)
δ = ζωnT
€
ζ =cccr
=δωnT
=δ
4π 2 +δ2(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 < ωn < 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: • 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 ζ for
A(Tmax)=0.025 0 0.5 1 1.5 2 -1
-0.5
0
0.5
1
Time(sec)
Ampl
itude
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x(t) = v0
ωd
e−ζωnt
Amplitude
sin(ωdt)
⇒ worst case happens at smallest ωd ⇒ωn = 8.16 rad/s⇒ worst case happens at max v0 = 300 mm/sWith ωn and v0 fixed at these values, investigate how varies with ζFirst peak is highest and occurs atddt
x(t)( ) = 0 ⇒ωde−ζωnt cos(ωdt)−ζωne
−ζωnt sin(ωdt) = 0
Solve for t = Tmax ⇒ Tm =1ωd
tan−1( ωd
ζωn
) = 1ωd
tan−1 1−ζ 2
ζ
#
$%%
&
'((
Sub Tmax into x(t) : Am (ζ ) = x(Tm ) = v0
ωn 1−ζ 2e−
ζ
1−ζ 2tan−1( 1−ζ 2
ζ)
sin(tan−1 1−ζ 2
ζ
#
$%%
&
'(()
Am (ζ ) = v0
ωn
e−
ζ
1−ζ 2tan−1( 1−ζ 2
ζ)
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€
To keep the max value less then 0.025 m solveAmax (ζ ) = 0.025⇒ζ = 0.281Using the upper limit on the mass (m = 3 kg)yields c = 2mωnζ = 2 ⋅ 3 ⋅ 8.16 ⋅ 0.281=14.15 kg/s
FYI, ζ = 0 yields Amax =v0
ωn
= 37 mm
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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 ζ change with mass?
Given ζ=1, m=1361 kg, Δ=0.05 m, compute c, k .
ωn =km⇒ k = 1361ωn
2 , mg = kΔ⇒ k = mgΔ
⇒ωn =mgmΔ
=9.810.05
= 14 rad/s ⇒
k = 1361(14)2 = 2.668 ×105 N/mζ=1⇒ c = 2mωn = 2(1361)(14) = 3.81×104 kg/s
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Now add 290 kg of passengers and luggage. What happens?
m = 1361+ 290 = 1651 kg
⇒ Δ =mgk=
1651 ⋅9.82.668 ×105 ≈ 0.06 m
⇒ωn =gΔ=
9.80.06
= 12.7 rad/s
ζ =cccr
=3.81×104
2mωn
= 0.9 So some oscillation" results at a lower" frequency."
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Section 1.8 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
limt→∞ x(t) = 0
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Examples of the types of stability
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Stable Asymptotically Stable
Divergent instability Flutter instability
x(t)
x(t)
x(t)
x(t)
t t
t t
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Example: 1.8.1: For what values of the spring constant will the response be
stable? Figure 1.37
m2 θ + k2
2sinθ
!
"#
$
%&cosθ −mgsinθ = 0 ⇒ m2 θ + k
2
2θ −mgθ = 0
⇒ 2m θ + k− 2mg( )θ = 0 (for small θ )
⇒ >l 2 for a stable responsek mg54/53