Chapter 1 Introduction:

Chapter 1 Introduction:

Vibration and the Free Response

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 later (MDOF and distributed mass systems)

Potential and Kinetic Energy

The potential energy of mechanical systems U is often stored in

“springs” (remember that for a spring F = kx)

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

Conservation of Energy

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

constant

or ( ) 0 T U

d T U 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).

1 2 2 1

max max

and

U U T T

U T

  



Deriving the equation of motion from the energy approach

 

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

 

     

  

 

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

2 2

max max

2 2

2

1 1

( )

2 2

Since these two values must be equal

1 1

( )

2 2

n

n

n n

U kA T m A

kA m A

k m k

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 without slipping.

2 2

rot trans

1 1

and

2 2

T  J  T  mx

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 r x r T J x

r

T v A

A J

T J m A m A

r r

U x A

U kA T U

m J A kA

r

 



  

 

    



 

      



   

    

 

 

2 n

k m J

r

   

Effective mass

Example 1.4.2 Determine the equation of motion of the

pendulum using energy

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

 





 

 

 



 

        

Using the small angle approximation for sine:

 

2

2

2

(sin ) 0

(sin ) 0 (sin ) 0

( ) sin ( ) 0

( ) ( ) 0

m mg

m mg

m mg

t g t

g g

t t

  

  

 

 

  

 

  

  

  

    

Example 1.4.4 The effect of including the mass of the

spring on the value of the frequency.

2

0

2

2

mass of element :

assumptions velocity of element : ( ),

1 (adds up the KE of each element) 2

= 1

2 3

1 1 1

2 2 3 2

s

dy

s spring

s

s

mass tot

dy m dy

dy v y x t

m y

T x dy

m x

T mx T m m





 

 

  

 

 

 

 

    



2 2 2

max

2 max

1

2 3

1 2

3

s

n

n

s

x T m m A

U kA

k m m





      

   

 



 

 • This provides some simple design and modeling guides

What about gravity?

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

 

 

         

    

     

  

• 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.

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

Then the equations of motion can be calculated from

0 (1.62)

d L L

dt q q

     

   

 

Which becomes

0 (1.63)

d T T U

dt q q q

    

  

    

 

Here q is a generalized coordinate

L = T - U

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

d T T U d

mx kx

dt x x x dt

mx kx

  

       

    

 

  

Example

2 2

2

2

1 1

(1 cos )

2 2

sin (1 cos ) 4

U kx kx mg

k mg



 

   

  

The Kinetic energy term is :

2 2 2

0

1 1

2 2

T  J   m  Compute the terms in Lagrange’s equation:





2 2

2

0

sin (1 cos ) sin cos sin

4 2

d T d

m m

dt dt

T

U k k

mg mg

 





    

 

    

  

 

 



 

         

Lagrange’s equation (1.64) yields:

2

2 sin cos sin 0

2

d T T U k

m mg

dt ^ ^q ^ ^q  ^q        

Does it make sense:

2 2

0 if 0

sin cos sin 0

2

k

m  k   mg 



  

Linearize to get small angle case:

2

2 0

2

2 0

2 2

n 2

m k mg

k mg m

k mg m

  

 



  

  

   

  

What happens if you linearize first?

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)

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

Example 1.5.1

From Equation (1.50)

4

10 2 2 4

2

( ) ( ) 0

( ) ( ) 0

,

32 For a 2 m steel shaft, diameter of 0.5 cm

(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

t k t J

k GJ d

J J J

GJ J

  

 

 

  ^

   

  

   



 

 







2 rad/s

Figure 1.22

Fig. 1.22 Helical Spring

d = diameter of wire

2R= diameter of turns n = number of turns x(t)= end deflection G= shear modulus of spring material

Fig 1.23 Transverse beam stiffness

3

3

3

With a mass at the tip:

3

n

k EI

EI







• Strength of materials and experiments yield:

Example for a Heavy Beam

Consider the beam of Figure 1.28 and what happens if the mass of the beam is considered.

 

2

( ) 3

6

x y Py y

 EI 

Which has maximum value of (at y =• ):

3

max 3

x P

 EI

From strength of materials the static deflection of a cantilever beam of length l is:

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

Next integrate along the beam to compute the beam’s kinetic energy contribution

 

 

2

max 0

2 max2 2 3

0 6 0

Mass of element dy

2 max

1 (mass of differential element) (velocity of differential) 2

1 1

3

2 2 4

1 33

2 140 T

M M x

x y dy y y dy

M x



    

 

  



 

**

Thus the equivalent mass of the beam is: ^eq

33 M  140 M

And the equivalent mass of the beam- mass system is:

system

33

m  140 M  m

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

3

eq

3

3 33 140

3 rad/s

33 140

n

EI k

m m M

EI

m M

  



     

 

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.

Example 1.5.2 Effect of fuel on frequency of an airplane wing

x(t)

l

E, I m

• 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

Mass of pod 10 kg empty 1000 kg full l = 5.2x10

m

, E =6.9x10

N/m, l = 2 m

• 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

Example 1.5.3 Rolling motion of a ship

( ) sin ( )

For small angles this becomes

( ) ( ) 0

_n

J t W GZ Wh t

J t Wh t

hW J

 

 



   

 

 

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

This is identical to the combination of capacitors in electrical circuits

• Equivalent Spring

1 2

1 2

series: 1

1 1

parallel:

AC

ab

k

k k k k k





 

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 ω_n

• Check static deflection

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

3 4 1 2

1 2

3 4

Case a) parallel connection:

0, 1000 3000 4000 N/m

4000 20 rad/s 10

Case b) series connection:

1 3000

0, 750 N/m

(1 ) (1 ) 3 1

eq

eg parallel

eq

k k k k k

k m

k k k

k k



      

   

    

 

750 8.66 rad/s

10

eg series

k

 m

   

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)

1 3 2 3 5 3 1 4 2 4 5 4 3 4

3 4

( )

n

k k k k k k k k k k k k k k m k k

  ^{     }



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

750 N/m 8.66 rad/s

1/1000 1/ 3000 10 kg

eq series

k     



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

4000 N/m

1000 N/m + 3000 N/m = 4000 N/m 20 rad/s

10 kg

eg par

k    

Same components, very different frequency

Static Deflection

Another important consideration in designing with springs is the static deflection

k mg mg

     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

 

2 2

0 0

2 0

4

gT r m m

J J



  

T is the measured period

g is the acceleration due to gravity

Figure 1.33

Stiffness Measurements

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

Figure 1.24 l = 1 m, m = 6 kg, I = 10^-9 m⁴ , and measured T = 0.62 s

  

 

3

2 3

2 3

11 2

2 2 9 4

2 0.62 s

3

4 6 kg 1 m

4 2.05 10 N/m

3 3(0.62 s) 10 m

T m

EI E m

T I



 



 

    

Damping Measurement (Dynamic only)

Define the Logarithmic Decrement:

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 < ω_n < 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 amplitude

• Occurs at time of first peak (T_max)

• Compute the amplitude at T_max

• Compute ζ for A(T_max)=0.025

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, investigate how varies with First pea

nt

d d

d n

n

x t v 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

Sub into ( ) : ( ) ( ) sin(tan 1

1

nt nt

d d n d

d m

d n d

m m

n

d x t e t e t

dt

t T T

T x t A x T v e

 

 

 

   





   

 

 



 

 

 

 

   

  

 

   

 

 

  



2 1

2

tan ( 1 ) 0 1

)

_m( )

n

A v e





 



 

 

 

 

 

 

 



max

0 max

To keep the max value less then 0.025 m solve ( ) 0.025 0.281

Using the upper limit on the mass ( = 3 kg) yields

2 2 3 8.16 0.281 14.15 kg/s

FYI, =0 yields 37 mm

n

n

A

m

c m

A v

 

 

 

  

     

 

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, ie when the car more passengers and luggage?

2

2 5

4

Given = 0.3, =1361 kg, =0.05 m, compute , . 1361 ,

9.81 14 rad/s 0.05

1361(14) 2.668 10 N/m

= 0.3 2(0.3) 0.6(1361)(14) 1.143 10

n n

n

n

m c k

k mg

k mg k k

m

mg m

k

c m



 



 



      



    



  

     kg/s

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

5

4

1361 290 1651 kg 1651 9.8

0.06 m 2.668 10

9.8 12.7 rad/s 0.06

1.143 10 2 0.27

n

cr n

m

mg k

g

c

c m



 

  

     



   



   

A lower frequency and damping ratio results meaning the car takes longer to damp out

e

Section 1.8 Stability

Stability is defined for the solution of free response case:

Stable: x t ( )  M ,   t 0

Asymptotically Stable:

Unstable:

if it is not stable or asymptotically stable

Examples of the types of stability

( ) x t

Stable Asymptotically

Stable

t t

x(t)

Divergent instability Flutter instability

Figure 1.38 Figure 1.39

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

Figure 1.37

 

2 2

2 2

sin cos sin 0 0

2 2

2 2 0 (for small )

k k

m mg m mg

m k mg

      

  

 

       

 

   

1 0

1

( ) ( ) ( )

i lim i i

t

i i

dx t x t x t

dt t

t t t



 



 



  

1.9 Numerical Simulation

• Solving differential equations by numerical integration

• Euler, Runge-Kutta, etc.

• Available in Mathcad, Matlab, Mathematica and Maple (or in FORTRAN)

• Or use Engineering Vibration Toolbox

• Will use these to examine nonlinear vibration problems that do not have analytical expressions for solutions

First order differential equation

Example 1.9.1 solve dx/dt = -3x, x(0)=1

0

1 0 0

2 1 1

3, take 0.5 1

(0.5)( 3)( ) 0.5 (0.5)( 3)( ) 0.25 =

a t

x

x x x

x x x

   



    

   

0 0

3

( )

( ) 3 ( ) 3

3,

(0) 1 1 so that

( )

t

t t

t

x t Ae

x t x t Ae Ae

x x Ae A

x t e



 









     

 

    



Time step

• With time step at 0.5 sec the numerical solution oscillates about the exact solution

• Large errors can be caused by choosing the time step to be too small

• Small time steps require more computation

Numerical solution of the 2nd order equation of vibration:

It is necessary to convert the second order equation into two first order equations. To achieve this two new variables x

and x

are defined as follows.

1 2

0 Let , mx cx kx

x x x x

  

 

From this two first order differential equations can be written.

1 2

2 2 1

x x

c k

x x x

m m



   Called state space

The matrix form is

Combining these first order DEs in matrix form gives.

1 1

2 2

0 1

x x

x k c x

m m

      

     

     

       

 

The Euler numerical method can then be applied to the matrix form to give.

 

1

1

( ) ( ) ( )

i i i

i i

t t tA t

I tA





  

  

x x x

x x

x A x

Matlab Solutions ‘ode23’ and ‘ode45’

• Use Runge-Kutta. More sophisticated than the Euler method but more accurate

• Often picks Dt (i.e. if solution x(t) is rapidly changing Δt is chosen to be small and visa-versa

• Works for nonlinear equations too

function xdot=sdof(t,x) k=2;c=1;m=3;

A=[0 1;-k/m -c/m];

xdot=A*x;

Create Matlab function

» t0=0;tf=20;

» x0=[0 ; 0.25];

» [t,x]=ode45('sdof',[t0 tf],x0);

» plot(t,x)

In the command window

Resulting solution

Why use numerical simulation when we can compute the analytical solution and plot it?

• To have a tool that we are confident with that will allow us to solve for the response when an analytical solution cannot be found

• Nonlinear systems to not have analytical solutions, but can

be simulated numerically

Section 1.10 Coulomb Friction and the Pendulum

Nonlinear phenomenon in vibration analysis

Vibration of Nonlinear Systems

Sliding or Coulomb Friction ( ) 0

( ) 0 ( ) 0

( ) 0

c

N x t

f x x t

N x t





 



  

 



The force due to Coulomb friction opposes motion, hence the

‘sgn’ function is used. The force is proportional to the normal

force and independent of the velocity of the mass.

The free body diagram split depending on the direction of motion:

mass moving right ( ) 0

x t 

mass moving left ( ) 0

x t 

sgn( ) 0 (1.100) mx  mg x kx

   

Figure 1.44

0 5 10 15 20

-0.2 -0.1 0 0.1 0.2 0.3

Time (sec) Dis

pla ce me nt x

Does not settle at x=0 Linear

decay

The ‘sgn’ function is nonlinear

• Causes equation of motion to be nonlinear

• Can solve as piecewise linear (see text)

• Can solve numerically

• Has more than one equilibrium position

• Decay is linear rather then exponential

• Comes to rest when spring cannot overcome friction at the

Figure 1.45 shows the details of the free response of a

system with Coulomb damping

A General second order system can be written as a single first order equation

1 2

2 1 2

( , ) ( )

x x

x f x x

   

        

x F x

( )_e  0 F x

2 0 and ^smg 1 ^smg

x x

k k

 

   

The equilibrium position is defined:

For Coulomb friction this is defined as:

i.e. the positions where the force due to the spring can no longer overcome the sliding friction force

Example 1.10.2: Calculating the equilibrium position for nonlinear DEs

State space form:

1 2

2 2

2 1

(

1)

x x

x x  x



 

Equation of motion:

2 3

0 x x



Equilibrium positions:

2

2 2

1 1

0

( 1) 0

1 1

0 , ,

0 0 0

e

x

x x

x



 



  

   

     

      

        

Multiple equilibrium positions possible

The pendulum

Stable

Equilibrium θ = 0, 2π, 4π ...

Unstable Equilibrium θ = 0, π, 3π ...

????



m

mg

2 0 and 1 , 0,1, 2

x x n n

   

Example 1.10.2 Equilibrium of a Pendulum

Figure 1.44

1 2

1 2

2 1

( ) sin ( ) 0 ,

sin

t g t

x x

x x

x g x

 

 

 

 





 

2

1

2 1

( ) 0

sin

0 sin 0

x

g x

x x

   

 

 

 

 

  

 

 F x

Solution to the pendulum

• Can use numerical simulation to examine both linear and nonlinear response

• Let (g/L)=(0.1)² so that ω_n = 0.1

• a) use θ(0)=0.3 rad & initial vel: 0.3 rad/s

• b) change the initial position to: θ(0)= p rad which is near the unstable equilibrium

0 10 20 30 40 50

-10 0 10 20 30

Time (sec) Non-linear

Linear (b)

0 10 20 30 40 50

-1.5 -1 -0.5 0 0.5 1 1.5

Angle ( )

Non-linear Linear

(a)

Pendulum with friction added

sin 0

c g

     

0 20 40 60 80

0 5 10 15

Time (sec) Angle ()

Friction loss causes slow decay

After making a single =4π

loop the pendulum cannot make a second rotation and settles to the stable equilibrium position of θ=4π