energy methods for damped systems

College of Engineering © Eng. Vib, 3rd Ed.

1/53

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


2/53

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

College of Engineering


3/53

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


4/53

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



5/53

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



6/53

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



8/53

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

θ

m !

mg

l

2lmJ =



9/53

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



10/53

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"



13/53

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


College of Engineering © Eng. Vib, 3rd Ed. 14/53

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.



15/53

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 −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


16/53

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



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



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:

18/53 College of Engineering


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?

19/53 College of Engineering


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College of Engineering 20/44

@TheSandy36

Hashtag EG-260


21/53

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



23/53

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



25/53

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



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



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


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

College of Engineering 28/53


29/53

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.



30/53

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!



31/53

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


32/53

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



33/53

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



34/53

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



35/53

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



36/53

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"



37/53

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


38/53

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


39/53

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



40/53

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



41/53

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



42/53

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



43/53

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



44/53

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)



45/53

Section 1.7: Design Considerations

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



46/53

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



47/53

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



48/53

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)



49/53

€

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



50/53

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



51/53

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."




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


Stable Asymptotically Stable

Divergent instability Flutter instability

x(t)

x(t)

x(t)

x(t)

t t

t t

53/53



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

energy methods for damped systems

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