Chapter 6 Distributed Parameter Systems

Chapter 6 Distributed Parameter Systems

Extending the first 5 chapters (particularly

Chapter 4) to systems with distributed mass and

stiffness properties:

Strings, rods and beams

Tacoma Narrows Bridge

The string/cable equation

• Start by considering a uniform string

stretched between two fixed boundaries

• Assume constant, axial tension τ in string

• Let a distributed force f(x,t) act along the

string

f(x,t)

y x

τ

2 2

1 1 2 2

( , )

sin sin ( , )

y

w x t

F A x

t

f x t x

 

   

  

   



Examine a small element of the string

• Force balance on an infinitesimal element

• Now linearize the sine with the small angle approximate sinx = tanx =slope of the string

• Writing the slope as the derivative: ^{tanx  }_^w_x

_

_

τ_

τ_

x₁ x₂ = x₁ +Δx

w(x,t) f (x,t)

Substitute into F=ma and use a Taylor expansion:

2 1

2 1 1

1

2

2

1

2

2

( , ) ( , ) ( , )

( , )

Recall the Taylor series of / about :

( )

( , )

( , )

x x

x x x

x

w x t w x t w x t

f x t x A x

x x t

w x x

w w w

x O x

x x x x

w x t w

x f x t x A

x x

  

  

  

 

   

  

   

    

 

       

   

   

           

     

     

     

 

  ²

2

2

( , )

( , ) ( , )

( , )

x t x t

w x t w x t

f x t A

x x t



    

  

 

   

 

 

Since τ is constant, and for no external force, the equation of motion becomes:

2 2

2

2 2

0 0

( , ) ( , )

,

( , 0) ( ), ( , 0) ( ) at 0

(0, ) ( , ) 0, 0

t

w x t w x t

c c

x t A

w x w x w x w x t

w t w t t

  

 ^  ^ 

  

  

(6.8)

, wave speed Second order in time and second order in space, therefore 4 constants of integration. Two from initial conditions:

And two from boundary conditions (e.g. a fixed-fixed string):

Each of these terms has a physical interpretation:

• Deflection is w(x,t) in the y-direction

• The slop of the string is w_x(x,t)

• The restoring force is τw_xx(x,t)

• The velocity is w_t(x,t)

• The acceleration is w_tt(x,t)

at any point x along the string at time t

Note that the above applies to cables as well as strings

There are Two Solution Types for Two Situations

• This is called the wave equation and if there are no boundaries, or they are sufficiently far away it is solved as a wave phenomena

– Disturbance results in propagating waves*

• If the boundaries are finite, relatively close together, then we solve it as a vibration phenomena (focus of Vibrations)

– Disturbance results in vibration

*focus of courses in Acoustics and in Wave Propagation

Solution of the Wave Equation:

• Interpret w(x,t) as a stress, particle velocity, or displacement to examine the propagation of waves in elastic media

– Called wave propagation

• Interpret w(x,t) as a pressure to examine the propagation of sound in a fluid

– Called acoustics

Solution of the “string equation” as a Wav e ) (

) (

) ,

( x t w

x ct w

x ct

w _{   }

A solution is of the form:

This describes on wave traveling forward and one wave traveling backwards (called traveling waves as the form of the wave moves along the media)

trough

crest The wave speed is c

Think of waves in a pool of water

Example 6.1.1 what are the boundary conditions for this system?

0

A force balance at yields

sin ( , ) 0

( , )

( , )

At 0, the BC is ( , ) 0

x y

x x

x

F kw t

w x t

k w x t x

x w x t

 





 



  

   



 



Fig 6.2

6.2 Modes and Natural Frequencies

2 2

2

2 2

2

2 2

2

2

( , ) ( ) ( )

( ) ( ) ( ) ( ) where = and =

( ) ( ) ( )

, ( ) ( ) 0

( ) ( ) ( )

( ) ( )

and 0

( ) ( )

w x t X x T t

d d

c X x T t X x T t

dx dt

c X x T t T t

T t T t

X x T t T t

d X x c X x

dx X x X x

 



 

  

        

 

 

   

 

 

2

2

( ) 2 ( ) 0 ( ) ( ) 0

X x X x X x X x

c

 

 

     

c c

     

(1)

(2)

Solving the Time Equation (1)

( )

( ) 0 ( ) sin cos

T t ^  T t ^{ } T t ^ A  t ^ B  t

Solving the Spatial Equation (2)

2

1 2

( ) ( ) 0 ( ) sin cos

X x ^   X x   X x  a  x a   x

This implies that T is oscillating with frequency ω

At this point we have 4 unknowns: A, B, a

1, and a

2

Use the Boundary Conditions to determine a

and a

1 2 2

1

(0, ) 0 (0) ( ) 0 (0) 0

sin(0) cos(0) 0 0

( , ) 0 ( ) ( ) 0 ( ) 0

sin( ) 0

, _n , 1, 2, 3,

w t X T t X

a a a

w t X T t X

a

n n n



   

    

    

    

 

    

( )

sin(

)

( )

sin(

), 1, 2,3, X x  a  x  X x  a  x n 

The X x

( ) are called eigenfunctions

(But n cannot be zero, why?)

Constructing the total solution by recombining X(x) and

( ) sin( ) cos( )

n n n n n n

c n c













( , ) ( ) ( ) sin sin cos sin

sin( ) sin( ) cos( ) sin( )

So there are solutions, since the system is linear we add them up:

( , ) sin( ) sin(

n n n n n n n n n

n n

n

w x t X x T t c t x d t x

n n n n

c ct x d ct x

n

n n

w x t c ct x

   

   

 

   

 



1

) _n cos( ) sin( )

n

n n

d  ct  x







We still do not know the constants c

n and d

n

but we have yet to use the initial conditions

Orthogonality is used to evaluate the remaining constants from the initial conditions

2 0

0

1

0 0 1

sin( ) sin( ) , (6.28)

0, 2

(proving this looks a lot like homework!) From the initial poistion:

( , 0) ( ) sin( ) cos(0)

( ) sin( ) sin( ) sin

nm

n n

n n

n m

n m

x x dx

n m

w x w x d n x

m n

w x x dx d x

  



 









 

    

  







 

0

( m )

 x dx _



0 0

0 0

0

1

0 0

2 ( ) sin( ) , 1, 2, 3

2 ( ) sin( ) , 1, 2, 3

( ) c sin( ) cos(0)

2 ( ) sin( ) , 1, 2, 3

m

n

n n n

n

d w x m x dx m

m n

d w x n x dx n

w x c n x

c w x n x dx n

n c





 









 

 

 



 









(6.31)

(6.32)

(6.33)

The Eigenfunctions become the vibration mode shapes

0

0

0 1

( ) sin , which is the first eigenfunction ( =1) ( ) 0, 0,

2 sin( ) sin( ) 0, 2, 3

1

( , ) sin( ) cos

n

n

w x x n

w x c n

d x n x dx n

d

w x t x c t



 

 



   

  

 

 

  

 



Causes vibration in the first mode shape

A more systematic way to generate the characteristic equation is write the boundary conditions (6.20) in matrix form

1 2

1 2

sin 0 and 0

sin 0 0

0 1 0

det( ) 0 sin 0

sin 0

0 1

a a

a a





 

 

      

     

     

     

     

 

 

 

This seemingly longer approach works in general and will be used to compute the characteristic equation in more complicated situations

The the solution of the boundary value problem results in an eigenvalue problem

2 2

For =1,2,3 , ( ) sin

Here the index results because of the indexed value of

( ) , ( ) 0, (0) ( ) 0

n n

n n n n n n

n X x a n x

n

X X X x X X

x





 



 

  

    

The spatial solution becomes:

(6.22)

(6.23)

The spatial problem also can be written as:

Which is also an eigenvalue, eigenfunction problem where λ =β² is the eigenvalue and X

n is the eigenfunction.

This is Analogous to Matrix Eigenvalue Problem

2

2 , plus boundary conditions, matrix operator ( ), eigenvector eigenfunction

( ) also an eigenfunction

orthoganality also results and the condition of normalizing plays the same role

eigen

i n

n

A x

X x X x

 





  

 

u

vectors become mode shapes, eigenvalues frequencies modal expansion will also happen



 



n x sin 2

Plots of mode shapes (fig 6.3)

0 0.5 1 1.5 2

1 0.5 0.5

1

X ,1 x X ,2 x X ,3 x

x

nodes

Example 6.2.2: Piano wire: l =1.4 m, τ =11.1x10

N,

4 1

110 g per 1.4 m = 0.0786 kg/m

11.1 10 N

1.4 1.4 0.0786 kg/m

2666.69 rad/s or 424 Hz A

c

A



   

 



   



Example 6.2.3: Compute the mode shapes and natural frequencies for the following system:

A cable hanging from the top and attached to a spring of tension τ and density ρ

w(x,t)

0

x

k

In this case the characteristic equation must be solved numerically

1 2

2 1

0 sin ( , ) 0

( , )

( , )

( ) sin cos ,

(0) 0 0 and ( ) sin

cos sin

tan

y x

x

F kw t

w x t

kw t x

X x a x a x

X a X x a x

k

k

 

 



 



  

 





    

 

 

   

  

  



The characteristic equation must be solved numerically for β_n

Fig 6.4

Note text uses σ not β

1000, 10, 2 1.497, 2.996, 4.501

(2 1) 6.013

2 sin( )

n

n n n

k

n

X a x









  







The Mode Shapes and Natural Frequencies are

• The values of β

n must be found numerically

• The eigenfunctions are again sinusoids

• The value shown for βn is for large n

• See Window 6.3 for a summary of method

Example: Compute the response of the piano wire to :

0 0

1

0

1

1

initial conditions: ( ) sin 3 , ( ) 0

Solution: ( , ) ( sin sin sin cos )

( ) 0 ( sin cos(0)) 0,

( , ) sin cos , at 0, Eq.(6.31)

2

n n

i

n n

i

n i

n

w x x w x

n n c n n c

w x t c x t d x t

n c n

w x c x c n

n n c

w x t d x t t

d



   

 

 













 

 

    

  









0

3

sin 3 sin , 1, 2, 3

= 0, for all except =3, 1

3 3

( , ) sin sin x m xdx m

m m d

w x t x c t

 

 









0 0

1 0

( ) sin

sin sin

2

n n

m

w x m xdx

n m

d x xdx

d



 











 

( , ) sin 6 sin 3 2

= 0

w t   c t

1.4 m, 11.89

(1.4 , ) 0.707 sin 80.4 4

c

w t t

  



3 3

( , ) sin sin

4 4

0.707 sin 3

w t c t

c t

 







Some calculation details:

Summary of Separation of Variables

• Substitute w(x,t)=X(x)T(t) into equation of motion and boundary conditions

• Manipulate all x dependence onto one side and set equal to a constant (-β²)

• Solve this spatial equation which results in eigenvalues β_n and eigenfunctions X_n

• Next solve the temporal equation to get T_n(t) in terms of β_n and two constants of integration

• Re form the product w_n(x,t)=X_n(x)T_n(t) in terms of the 2n constants of integration

• Form the sum

1

( , )

( )(

sin

cos

)

n

w x t

X x A  t B  t



  

• Use the initial displacement, initial velocity and the orthogonality of X_n(x) to compute A_n and B_n.

6.3 Vibration of Rods and Bars

x x +dx

w(x,t) x

dx

F+dF F

Equilibrium position

Infinitesimal element

0 l Figure 6.5

• Consider a small element of the bar

• Deflection is now along x (called longitudinal

vibration)

• F= ma on small element yields the following:

2 2

2 2 2

( , )

( ) (6.53)

( , ) ( , )

( ) ( )

( , ) ( , )

( ) ( ) (6.55)

( ) constant

w x t F dF F A x dx

t

w x t w x t

F EA x dF EA x dx

x x x

w x t w x t

EA x A x

x x t

E w

A x

 



  

  

   

  





  

 

    

  

 

 

  ( , )₂ ² ( , )₂

(6.56) At the clamped end: (0, ) 0, (6.57)

( , )

At the free end: 0 (6.58)

x

x t w x t

x t

w t w x t

EA x



 



 _







Force balance:

Constitutive relation:

Example 6.3.1: compute the mode shapes and natural frequencies of a cantilevered bar with uniform cross section.

2

2 2

2 2 2

( , ) ( ) ( ) ( , ) ( , )

( ) ( )

(6.59)

( ) ( )

( ) ( ) 0, (0) 0, ( ) 0 (i) ( ) ( ) 0, initial conditio

xx tt

w x t X x T t c w x t w x t X x T t

X x c T t

X x X x X AEX

T t c T t







  

    

     

   ns (ii)

From equation (i) the form of the spatial solution is ( )X x asin x bcos x





 

Next apply the boundary conditions in (i) to get the characteristic equation and the form of the eigenfunction:

Apply the boundary conditions to the spatial solution to get

sin(0) cos(0) 0

cos( ) sin( ) 0

0 1

0 and det 0

cos( ) sin( )

2 1

cos 0 , 1, 2, 3,

2

(2 1)

( ) sin( ), 1, 2, 3, 2

n

n n

a b

a b

b

n n

n x

X x a n

 

 

  



 

 

 

       

     

      



Next consider the time response equation (ii):

2

2

2 1

( ) ( ) 0

2

(2 1) (2 1)

( ) sin cos

2 2

(2 1) (2 1)

, 1, 2, 3 (6.63)

2 2

n

n n n

n

T t c n T t

n c n c

T t A t B t

n c n E

n



 

 

 

  

     

 

  

 

  

Example 6.3.2 Given v

(l)=3 cm/s, ρ =8x10

kg/m

and

N/m

, compute the response.

1

0 0

1

(2 1)

( , ) ( sin cos ) sin

2

2 (2 1)

( ) sin 0

2

(2 1) ( , 0) 0.03 ( ) cos(0) sin

2

n n n n

n

n

t n n

n

w x t c ct d ct n x

d w x n xdx

w x x c c n x

  



  









  

   

   







Multiply by the mode shape indexed m and integrate:

0

1 0

1

3 1 1

-6 9

(2 1)

0.03 (sin ) ( )

2

(2 1) (2 1)

sin sin

2 2

(2 1) 1 0.06( 1)

0.03sin

2 2 (2 1)

8 10 0.12( 1) ( 1)

7.455 10

210 10 (2 1) (

n n n

m m

m m

n n

n

m x x dx

m n

c c x xdx

m c

c c

E m

c n

 

  

 

 









 

  

 



 

   



  

  

 





 

1 -6

1

2 1) m

( 1) 2 1

( , ) 7.455 10 sin sin 512.348(2 1) m

(2 1) 10

n

n

n

w x t n x n t

n ^ 







   

 



Various Examples Follow From Other Boundary Conditions

• Table 6.1 page 524 gives a variety of different boundary conditions. The resulting frequencies and mode shapes are given in Table 6.2 page 525.

• Various problems consist of computing these values

• Once modes/frequencies are determined, use mode summation to compute the response from IC

• See the book by Blevins:Mode Shapes and Natural Frequencies for more boundary conditions

, from calculus ( , )

, from solid mechanics

=shear modulus

=polar moment of area cross section

d dx

x GJ x t

x G

J

 

 

 





2 2

2 2

2 2

Combining these expressions yields;

( , ) ( , )

, constant

( , ) ( , )

x t x t

GJ J GJ

x x t

x t G x t

t x

    

  

   

  

   

 

 



6.4 Torsional Vibrations

2 2

Summing moments on the element ( , )

Where is the shaft's mass density dx

dx J x t dx

x t

  

  

 



  

x,t) τ x

x,t)+dθ

dτ xdx

The initial and boundary conditions for torsional vibration problems are:

• Two spatial conditions (boundary conditions)

• Two time conditions (initial conditions)

• See Table 6.4 for a list of conditions and Equation (6.67) and Table 6.3 for odd cross section

• Clamped-free rod:

0 0

(0, ) 0 Clamped boundary (0 deflection) ( , ) 0 Free boundary (0 torque)

( ,0)= ( ) and ( ,0)= ( )

x

t

t

G t

x x x x





   





Example 6.4.1: grinding shaft vibrations

• Top end of shaft is connected to pulley (x = 0)

• J₁ includes collective inertia of drive belt, pulley and motor

Drive pulley, collective inertia

Shaft of stiffness GJ, length l J₁

θ(x,t) x

J₂

Grinding head inertia

Use torque balance at top and bottom to get the Boundary Conditions:

2

1 2

0 0

2

2 2

( , ) ( , )

at top

( , ) ( , )

at bottom

x x

x x

x t x t

GJ J

x t

x t x t

GJ J

x t

  

 

  

 

 

 



 

The minus sign follows from right hand rule.

Again use separation of variables to attempt a solution

2

2

2 2

( , ) ( ) ( )

( ) ( )

( ) ( )

( ) ( ) 0, ( ) ( ) 0

x t x T t

x T t

x G T t c G

x x T t T t

c G



 



 

  



  

       

   

 

     

 

The next step is to use the boundary conditions:

1

2 2 1

2 1

2 2

Boundary Condition at 0 (0) ( ) (0) ( )

(0) ( )

(0) ( )

(0) (0)

( ) ( )

x

GJ T t J T t

GJ T t

J T t c

J J

J J











 

    

     



    

   

Similarly the boundary condition at l yields:

The Boundary Conditions reveal the Characteristic

1 2 2

1 2 1

2

1 1

1 2

2 2

1 1

1 2 1 2

1 2

2 2

1 2

( ) sin cos (0)

( ) cos sin (0)

0

(0) (0)

( ) ( ) cos sin sin cos

( )( )

tan( )

( ) ( )

x a x a x a

x a x a x a

x

J J

a a

J J

x

J J

a a a a

J J

J J J

J J J

 

    

 

 

       

 

 

  

     

 

     

 

      

 

      

  

 THE CHARACTERISTIC EQUATION

(6.82)

Solving the for the first mode shape

1 2

2 2

1 2

1 1

( )( )

tan( ) has 0 as its first solution:

( ) ( )

Numerically solve for , 1, 2, 3, , and

Note for 1, 0 0 ( ) 0

( ) the rigid body mode of the shaft turning

n n n

J J J

J J J

n G

n T t

T t a bt

 

  

  



 

 



 

      

 

1 1 1 1

1 1 1

( ) 0, ( )

0 0 ( ) the first mode shape

x x a b x

x b x a

        

     

Solutions of the Characteristic Equation involve solving a transcendental equation

2

1 2

1 2 1 2

2 3

1 2

2 9

1 2

3 4

( ) tan

, ,

( )

10 kg m / rad, =2700 kg/m ,

=5 kg m / rad, =0.25 m 25 10 Pa

0 Hz, 38, 013 Hz,

76, 026 Hz, 114, 039 Hz,

bx a x x

J J

x a J b

J J J J J

J J J

G

f f

f f

 





 

  

 

  



 



 

 

Fig 6.9 Plots of each side of eq (6.82)

to assist in find initial guess for numerical routines used to compute the roots.

Next sum forces in the -direction (up, down) Sum moments about the point

Use the moment given from

stenght of materials Assume sides do not bend

(no shear deform y

Q

ation)

2 2

bending stiffness ( ) Youngs modulus

( ) cross-sect. area moment of inertia about

( , ) ( , ) ( )

EI x E

I x

z w x t M x t EI x

x













6.5 Bending vibrations of a beam

M(x,t)+M

x(x,t)dx

V(x,t)

V(x,t)+V_x(x,t)dx f(x,t)

w(x,t)

x x +dx

·^Q

f (x,t) w (x,t)

x

dx A(x)= h₁h₂ h₁

h₂

Summing forces and moments yields:

0 )

2 ( ) , ( )

, ) (

, ) (

, (

2 0 )

, (

) , ) (

, ( )

, ) (

, ) (

, (

) , ) (

( )

, ( )

, ) (

, ) (

, (

2

2 2

 



 



 

 



 



 









 



 



 



 



 





 



 



 

t dx x f x

t x dx V

t x x V

t x M

dx dx t x f

dx x dx

t x t V

x V t

x M x dx

t x t M

x M

t t x dx w

x A dx

t x f t

x V x dx

t x t V

x V



















 





A c EI

x

t x c w

t

t x w

t x x f

t x x w

x EI t

t x x w

A

t

t x dx w

x A dx

t x f x dx

t x M

x t x t M

x V

 

















 



 















 



 



 













, ) 0

, ( )

, (

) , ) (

, ) (

) ( , ) (

(

) , ) (

( )

, ) (

, (

) , ) (

, (

4 4

2 2

2

2 2

2 2 2

2

2 2

2 2

Substitute into force balance equation yields:

Dividing by dx and substituting for M yields

Assume constant stiffness to get:

0 force

shear

0 slope

end Sliding

2 2

 



 



 





x EI w x x

w













2 2

Pinned (or simply supported) end deflection 0

bending moment= 0

w

EI w x





 



The possible boundary conditions are (choose 4):

0 force

shear

0 moment

bending end Free

2 2

2 2

 



 



 





x EI w x

x EI w













Clamped (or fixed) end deflection 0

slope 0

w w

x





 

 

Solution of the time equation yields the oscillatory nature:

2 2

2

0 0

( ) ( )

( ) ( )

( ) ( ) 0

( ) sin cos Two initial conditions:

( , 0) ( ),

( , 0) ( )

X x T t

c X x T t

T t T t

T t A t B t

w x w x w x w x





 

    

  

 

 