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RAYLEIGH'S METHOD
Revision D
By Tom Irvine
Email: [email protected]
November 15, 2010
Introduction
Dynamic systems can be characterized in terms of one or more natural frequencies. The
natural frequency is the frequency at which the system would vibrate if it were given an
initial disturbance and then allowed to vibrate freely.
There are many available methods for determining the natural frequency.
examples are
1.
2.
3.
4.
Some
Newton’s Law of Motion
Rayleigh’s Method
Energy Method
Lagrange’s Equation
Note that the Rayleigh, Energy, and Lagrange methods are closely related.
Some of these methods directly yield the natural frequency. Others yield a governing
equation of motion, from which the natural frequency may be determined.
This tutorial focuses on Rayleigh’s method, which yields the natural frequency.
Rayleigh's method requires an assumed displacement function. The method thus reduces
the dynamic system to a single-degree-of-freedom system. Furthermore, the assumed
displacement function introduces additional constraints which increase the stiffness of the
system. Thus, Rayleigh's method yields an upper limit of the true fundamental
frequency.
Definition
Rayleigh's method can be summarized as
KEmax  PEmax
= total energy of the system
where
KE = kinetic energy
PE = potential energy
1
(1)
Note that potential energy is also referred to as strain energy for the case of certain
systems, such as beams.
Equation (1) can only be satisfied if the system is vibrating at its natural frequency.
References
1. T. Irvine, Natural Frequencies of Beam Bending Modes, Revision E, Vibrationdata,
1998.
2
APPENDIX A
Pendulum Example
Consider a conservative system. An example is the pendulum shown in Figure A-1.
g
L

m
Figure A-1.
The kinetic energy becomes zero when the pendulum reaches its maximum angular
displacement. The kinetic energy reaches its maximum value when the pendulum passes
through  = 0, which is also the "static equilibrium point."
On the other hand, the potential energy reaches its maximum level as the pendulum
reaches its maximum angular displacement. The potential energy reaches its minimum
value when the pendulum is at its static equilibrium point. For simplicity, the potential
energy can be considered as zero at the static equilibrium point.
Let
m = pendulum mass
L = length
 = angular displacement
Assume a small angular displacement.
3
The potential energy is
PE  mgL (1  cos )
(A-1)
The kinetic energy is
KE 
1
m(L ) 2
2
(A-2)
Assume a displacement equation of
(t )   sinn t 
(A-3)
 max  
(A-4)
The velocity equation is
 (t )   n cosn t 
(A-5)
 max   n
(A-6)
PEmax
 mgL (1  cos )
(A-7)
Consider the expansion
 2 4

cos   1 

 


2!
4!


(A-8)
Consider the maximum potential energy. Substitute equation (A-8) into (A-7).
PEmax
  2 4

 mgL 1  1 

 

2!
4!
 

 
PEmax
2 4 

 mgL 

4! 
 2!

(A-9)
(A-10)
4
PEmax
2 

 mgL 
 2 
(A-11)
Consider the maximum kinetic energy. Substitute equation (A-6) into (A-2).
KEmax

1
m L n 2
2
(A-12)
 2 
1

m L n 2  mgL 

2
2 


(A-13)
Simplifying,
n 2 
g
L
(A-14)
The pendulum natural frequency is thus
n 
g
L
(A-15)
5
APPENDIX B
Cantilever Beam with End Mass
Consider a mass mounted on the end of a cantilever beam, as shown in Figure B-1.
Assume that the end-mass is much greater than the mass of the beam.
EI
m
y
L
Figure B-1.
E
I
L
g
m
x
is the modulus of elasticity
is the area moment of inertia
is the length
is gravity
is the mass
is the displacement
The static stiffness at the end of the beam is
k
3EI
L3
(B-1)
Equation (B-1) is derived in Reference 1.
The potential energy is
PE 
1  3EI  2

y
2  L3 
(B-2)
1
m y 2
2
(B-3)
The kinetic energy is
KE 
Assume an end displacement of
y  A sin n t
(B-4)
6
The corresponding velocity is
y  n A cos n t
(B-5)
The maximum displacement is A. The maximum velocity is n A . Thus,
KEmax 
1
mn A 2
2
(B-6)
1  3EI  2

A
2  L3 
(B-7)
PEmax 
Applying Rayleigh’s method,
1
1  3EI  2
mn A 2  
A
2
2  L3 
(B-8)
The natural frequency of the end mass supported by the cantilever beam is thus
 3EI 

n 2  
 m L3 
n 
(B-9)
3 EI
m L3
(B-10)
7
APPENDIX C
Cantilever Beam with Internal Distributed Mass
Consider a cantilever beam with mass per length . Assume that the beam has a uniform
cross section. Determine the natural frequency.
EI, 
y
L
Figure C-1.
The governing differential equation is
 EI
4 y
2 y

x 4
t 2
(C-1)
The boundary conditions at the fixed end x = 0 are
y(0) = 0
(zero displacement)
dy
 0
dx x 0
(zero slope)
(C-2)
(C-3)
The boundary conditions at the free end x = L are
d2y
dx 2
 0
(zero bending moment)
(C-4)
x L
d 3y
dx 3
 0
(zero shear force)
x L
Propose a quarter cosine wave solution.
8
(C-5)

 x  
y( x)  y o 1  cos  
 2L  

(C-6)
dy
    x 
 y o   sin 
 2L   2L 
dx
(C-7)
d2y
  2
 x 
 y o   cos 
 2L 
 2L 
dx 2
(C-8)
d 3y
 x  3  x 
  y o   sin 
 2L 
 2L 
dx 3
(C-9)
The proposed solution meets all of the boundary conditions expect for the zero shear
force at the right end. The proposed solution is accepted as an approximate solution for
the deflection shape, despite one deficiency.
Again, Rayleigh’s method is used to find the natural frequency. The total potential
energy and the total kinetic energy must be determined.
The total potential energy P in the beam is
2
EI L  d 2 y 
P
dx
2 0  dx 2 
(C-10)
By substitution,
EI L
P
2 0
2
   2
 x  
y o   cos   dx
 2L 
  2 L 

EI     2 
y   
P
2  o  2 L  
EI     2 
y   
P
2  o  2 L  
2
2
(C-11)
2
L   x  
0 cos 2L   dx
(C-12)
L  1 
 x  
 2  1  cos L  dx


(C-13)
0
9
2
EI     2   1    L   x  
y   
P
x    sin 
2  o  2 L    2       L  
L
(C-14)
0
 4 
EI
2  
P
y
L
2 o  32 L4 


 
(C-15)
1 4  EI 
  y o 2
64  L3 
(C-16)
P
The total kinetic energy T is
T 
1
L 2
  2n   y dx
0
2
T 
1
L
  2n 
0
2
2
 
 x  
y
1

cos
dx


 o
 2 L  
 

(C-18)
T 
1
 x 
 x  
2 L
  2n y o  1  2 cos   cos2   dx
 2L 
 2L  
0 
2
(C-19)
T 
1
 x 
 x  
2 L
  2n y o  1  2 cos   cos2    dx
 2L 
 2L  
0 
2
(C-20)
T 
1
 x  1 1  x  
2 L
  2n y o  1  2 cos    cos  dx
 2L  2 2  L  
0 
2
(C-21)
(C-17)
 
 
 
1
 x 
 x  
2 L 3
  2n y o    2 cos   cos  dx
 2L 
 L 
0 2
2
(C-22)
L
1
 4 L   x   L   x  
2 3
2
T    n y o  x    sin     sin  
    2L      L  
2
2
0
(C-23)
T 
 
 
T 
1
 4L 
2 3
  2n y o  L    
  
2
2
 
(C-24)
10
T 
1
2   8
  2n y o L 3    
4
   
 
(C-25)
Now equate the potential and the kinetic energy terms.
1
2   8   1 4  EI 
2
  2n y o L 3     
   yo
4
     64
 L3 
 
 
(C-26)
 EI 
  8 1
  2n L 3       4  
     16  L3 
(C-27)

 EI  
 4   

 L3  
 2n  

16L 3   8   
    




(C-28)
1/2

 EI  
4
    

 L3  
n  

16L 3   8   
    




(C-29)
1/2

 EI  
4
    
 1 
 L4  
f n   

  8
 2  
16 3    

     

(C-30)
1/2

 EI  
4
    
 1 
 L4  
f n   

  8
 2  
16 3    

     

(C-31)
11

1/2


EI
 2 


 1 
f n   


2    8
 2  
 4L 
  3     

     

 1  3.664  EI
f n   

 2  L2  
(C-32)
(C-33)
12
APPENDIX D
Cantilever Beam with Internal Distributed Mass and End Mass
EI,
m
y
L
Figure D-1.
The following is based on a quarter cosine wave solution.
The kinetic energy is
T
  8  1
1 2
 n y o 2 L3     mn y o 2
4
    2
 1   8  1 
T  2n y o 2   L3     m
 4     2 
(D-1)
(D-2)
The potential energy is
P
1 4  EI 
  y o 2
64  L3 
(D-3)
Now equate the potential and the kinetic energy terms.
 1   8  1  1 4  EI 
2n y o 2   L3     m 
  y o 2
4

2
64
  
 L3 


13
(D-4)
   8 
 1
 EI 
2n  L3     2m   4  
    
 16  L3 
(D-5)
 EI 
4  
 L3 
2n 
   8 

16L3     2m
    

(D-6)
1/ 2


 EI 
4  


3


L

n  

16L3   8   2m 

     

 
(D-7)
1/ 2


 EI 
4  


3


L

n  

 32 1 L3   8   m 

  2    

 
(D-8)
1/ 2


 EI 
4  


3

 1 
L

f n   


 2    1   8  
32 L 3 
 m
  2    




fn 
1
2
3.0440 EI
(D-9)
(D-10)
0.2268 L  m L3
14
APPENDIX E
Rayleigh’s Quotient
Rayleigh’s method can also be applied to multi-degree-freedom-systems, as follows.
XT K X
2 
XT M X
(E-1)
where
K is the stiffness matrix
M is the mass matrix
X is an assumed mode shape with arbitrary scale
Equation (E-1) is essentially a numerical approximation. It overestimates the true
fundamental frequency. Thus, it should be used in a trial-and-error manner.
Note that the numerator in equation (E-1) is equal to twice the potential energy. The
denominator is equal to twice the kinetic energy if first multiplied by  2 .
As an example, consider the system defined in Figure E-1 and Table E-1.
15
x2
m2
k3
m1
k2
x1
k1
Figure E-1.
Table E-1. Parameters
Variable
Value
m1
2.0 kg
m2
1.0 kg
k1
1000 N/m
k2
2000 N/m
k3
3000 N/m
The homogeneous equation of motion is
 k3   x1  0
m1 0   x1  k1  k3

 0 m  x     k
k 2  k3  x 2  0
3
2  2 

16
(E-2)
The mass matrix is
 2 0
M 
 kg
0 1 
(E-3)
The stiffness matrix is
 4000  3000
K 
 N/m

3000
5000


(E-4)
The natural frequencies can be computed using the eigenvalue method. The eigenvalues
are the roots of the following equation.
det K  2M  0


(E-5)
Equation (E-5) can be solved exactly for systems with up to four degrees-of-freedom.
The first natural frequency is thus
1  30.03 rad / sec
(E-6)
The next task is to test the Rayleigh quotient method. Several candidate mode shape are
evaluated as shown in Table E-2.
Note that
x 
X   1
x 2 
(E-7)
17
Table E-2. Rayleigh Quotient Trials
X
 (rad/sec)
1
1

31.62
1
1.5
 
30.68
1 
2
 
31.62
1
 0 .5 
 
36.51
1
1
 
54.77
Again, the Rayleigh quotient overestimates the true fundamental frequency.
Thus, the best estimate for the natural frequency after five trials is
1  30.68 rad / sec
(estimate)
The estimated value is 2.2% higher than the exact value.
The Rayleigh quotient method thus gives very good results for this example.
18
(E-8)