NATURAL FREQUENCIES OF CIRCULAR PLATE BENDING MODES Revision F By Tom Irvine

NATURAL FREQUENCIES OF CIRCULAR PLATE BENDING MODES
Revision F
By Tom Irvine
Email: [email protected]
February 17, 2012
Introduction
The Rayleigh method is used in this tutorial to determine the fundamental bending
frequency. The method is taken from References 1 through 3. In addition, a Bessel
function solution is given in Appendices D and E.
A displacement function is assumed for the Rayleigh method which satisfies the
geometric boundary conditions. The assumed displacement function is substituted into
the strain and kinetic energy equations.
The Rayleigh method gives a natural frequency that is an upper limit of the true natural
frequency. The method would give the exact natural frequency if the true displacement
function were used. The true displacement function is called an eigenfunction.
Consider the circular plate in Figure 1.
Y
r

X
0
Figure 1.
Let Z represent the out-of-plane displacement.
1
Table 1.
Appendix
Topic
A
Strain and kinetic energy
B
Simply Supported Plate, Rayleigh Method
C
Integral Table
D
Solution of Differential Equation via Bessel Functions
E
Simply Supported Plate, Bessel Function Solution
F
Mass Normalization of Eigenvectors
G
Completely Free Circular Plate
H
Fixed Plate, Bessel Function Solution
References
1. Dave Steinberg, Vibration Analysis for Electronic Equipment, Wiley-Interscience,
New York, 1988.
2. Weaver, Timoshenko, and Young; Vibration Problems in Engineering, WileyInterscience, New York, 1990.
3. Arthur W. Leissa, Vibration of Plates, NASA SP-160, National Aeronautics and
Space Administration, Washington D.C., 1969.
4. Jan Tuma, Engineering Mathematics Handbook, McGraw-Hill, New York, 1979.
5. L. Meirovitch, Analytical Methods in Vibrations, Macmillan, New York, 1967.
6. W. Soedel, Vibrations of Shells and Plates, Third Edition, Marcel Dekker, New York,
2004.
2
APPENDIX A
The total strain energy V of the plate is
2

D e 2 R   2 Z 1 Z 1  2 Z 
 2 Z  1 Z 1  2 Z 
V


 2 1   

2 r  r  r r 2  2 
2 0 0   r 2 r  r r 2   2 





   1 Z 
 2 1    

  r  r  
2
 r dr d


(A-1)
Note that the plate stiffness factor De is given by
De 
Eh 3
12 1   2 


(A-2)
where
E = elastic modulus
h = plate thickness
 = Poisson's ratio
For a displacement which is symmetric about the center,

Z(r, )  0

2
 2
(A-3)
Z(r, )  0
(A-4)
Substitute equations (A-3) and (A-4) into (A-1).
3
2


D e 2 R   2 Z 1 Z 
 2 Z  1 Z 

 r dr d
V

 2 1   
2 r  r  r 
2 0 0   r 2 r  r 






(A-5)
2

2 Z  1 Z 
  2 Z  1 Z   1 Z  2
D e 2 R   2 Z 



  
   2  2 

 r dr d
V
 2








2
2 r r   r r 
2 0 0
 2 r  r  r 
  r 
  r 


(A-6)
The total strain energy equation for the symmetric case is thus
2


2
 1 Z 
D e 2 R   2 Z 
 2 Z  1 Z 
  2

 r dr d
V
 
2
2 0 0   r 2 
r

r
r

r



 r




(A-7)
The total kinetic energy T of the plate bending is given by
T
h  2 2  R 2
Z r dr d
2
0 0
 
(A-8)
where
 = mass per volume
 = angular natural frequency
4
APPENDIX B
Simply Supported Plate
Consider a circular plate which is simply supported around its circumference. The plate
has a radius a. The displacement perpendicular to the plate is Z. A polar coordinate
system is used with the origin at the plate's center.
Seek a displacement function that satisfies the geometric boundary conditions.
The geometric boundary conditions are
Z(a, )  0
 2Z
 2r r a
(B-1)
0
(B-2)
The following function satisfies the geometric boundary conditions.
 r 
Z(r, )  Z o cos 
 2a 
(B-3)
The partial derivatives are

Z(r, )  0

2
 2
(B-4)
Z(r, )  0
(B-5)

    r 
Z(r, )   Z o   sin 
r
 2a   2a 
2
(B-6)
2
 
 r 
Z(r, )   Z o   cos 
 2a 
 2a 
 r2
5
(B-7)
The total kinetic energy T of the plate bending is given by
T
2
h  2 2  a 
  r 
Z
cos
  r dr d
 o
2
0 0 
 2a 
(B-8)
T
h  2 Z o 2 2  a 
  r 
1  cos  r dr d

4
0 0 
 a 
(B-9)
h  2 Z o 2 2 a 
  r 
r  r cos  dr d

4
0 0 
 a 
(B-10)
 
 
 
T
Evaluate equation (B-9) using the integral table in Appendix C
a
2
h  2 Z o 2 2  r 2 a r
 r  a
  r 
T

sin

cos  d

 
4

0 2
 a  2
 a 

0

T

h  2 Z o 2 2   a 2 a 2
a2
cos 
cos 0 d
 
4
0  2

2
2


T
h  2 Z o 2 2  a 2 2a 2 


 d
2 
4
0  2




T
T
(B-12)
(B-13)

h  2 Z o 2 a 2 2  2
  4 d
0
8 2
(B-14)
2  4 02 d
(B-15)

h  2 Z o 2 a 2
T

(B-11)
8 2
h  2 Z o 2 a 2
8 2
2  42
(B-16)
6


(B-17)
 0.4671  h  2 Zo 2 a 2
(B-18)
h  2 Z o 2 a 2 2
T
 4
4
T
Again, the total strain energy for the symmetric case is
2


2
 1 Z 
D e 2 R   2 Z 
 2 Z  1 Z 
  2

 r dr d
V
 
2
2 0 0   r 2 
r  r 
r

r



 r




(B-19)
2

D e 2 R   2 Z  
V
r dr d
2 0 0   r 2  

 


2

D e 2 R  1 Z  

 r dr d

2 0 0  r  r  


D 2 R 
 2 Z  1 Z 

 r dr d
 e    2
2
2 0 0 
r

r


 r

(B-20)
D e 2 a 2    4
  r 
2
V 
Z o   cos   r dr d
 2a 
2 0 0
 2a 



D e 2 a 2  1
Zo
r2
2 0 0

2

 
2 r 
  sin   r dr d
 2a 
 2a 

3

D e 2 a 2 
   1   r    r 


Z
2

cos
sin
    r dr d
 
o

2 0 0
 2a  r
 2a   2a 

(B-21)
7
V 
4

Z o 2 D 2 a   
2 r 

 r dr d
cos




2 0 0  2a 
2a 



Z 2 D 2 a  1

 o
2 0 0  r 2

2

 
2 r 
  sin   r dr d
 2a 
 2a 
3
Z o 2 D 2 a 
  1   r    r 

2    cos  sin   r dr d

2 0 0 
 2a  r
 2a   2a 

(B-22)
V 
4
Z o 2 D    2 a 
2   r 
     cos   r dr d
2  2a  0 0 
 2a 
2
Z o 2 D    2 a  1
2   r 

     2 sin   r dr d
2  2a  0 0  r
 2a 

3
2 a 
Zo 2 D
2     0 0  1 cos  r  sin  r  r dr d
2
 2a 
 2a   2a 
 r
(B-23)
Zo 2 D    4  1 
V 
   
2  2a   2 
2  a
  r 
0 0 1  cos a r dr d
Z o 2 D    2 2  a 1 2   r  

 
 sin   dr d
2  2a  0 0  r
 2a 
 
Z 2 D     3  2 a
  r 
 o
 
sin   dr d

2   2a   0 0 
 a 


 
(B-24)
The first and third integrals are evaluating using the tables in Appendix C.
8
Z 2 D   4  1 
V  o
   
2  2a   2 
a
2
ar
 r  a
  r 
sin   
cos  d
 

0 2
 a  2
 a 

0

2  r 2
Z o 2 D    2 2

0.8242 d
 
2  2a  0

a
Z 2 D     3   a  2   r 
 o
   
cos  d
2   2a      0
 a  0



(B-25)
Z 2 D   4
V  o
 
2  2a 
1
 
 2
2a 2 


 d
2
0  2
 


2  a 2
Z 2 D    2 2
 o
0.8242 d
 
2  2a  0

Z o 2 D     3   2a  2

   
d
2   2a      0



(B-26)
Z 2 D   4  1 
V  o
   
2  2a   2 
2a 2 


 d
2
0  2
 


2  a 2
Z 2 D    2 2
 o
0.8242 d
 
2  2a  0

Z o 2 D     2  2

 
d
2   2a   0



(B-27)
9
Z 2 D   4
V  o
 
2  2a 
2 2a 2 
 1  a


 2
 
 2   2
 2 
Z 2 D   2
 o
  0.8242  2
2  2a 
Z o 2 D     2 

   2
2   2a  


(B-28)
4
2 2a 2 
    1  a
V   Zo 2 D      


 2a   2   2
 2 
 
 Z o 2 D  
 2a 
2
0.8242
   2 
2
 Z o D     
  2a  


(B-29)

4
  
V   Z o 2 D   
 2a 


2
2 
2 2a 2 

 
 1  a
  
    0.8242      

 
  2a  
 2   2
 2   2a 



(B-30)
V
2
4

2 2a 2 
  1  2 
 2  1   1  a

1 
3
2
 Z o D       

    0.8242      


2
 2a   2   2
 2a 
  2a 



 



(B-31)
10

V   Z o 2 D 3  2

  1  
 1   1   a 2 2a 2   1 

   
0.8242    


 

4  2 2
2
2
2  
   4a 
 16a 
  4a  

(B-32)

  1  
  1   1  1 2   1 

       
0.8242    
 
V   Z o 2 D 3  2 
2
2
2
2

  4a  
  16a   2   2    4a 

(B-33)

1 

  1   1  1 2  1 

  
V   Z o 2 D 3    2      
   0.8242   
2
2

 4 
a 

  16   2   2    4 

(B-34)
 1 
 1  1 2  1 
  
V   Z o 2 D 3    2        0.8242   
 32   2  2   4 
 4 
a 2  
 1 
 1    1    2
  
V   Z o 2 D 3      
 2   0.8242   
 4 
 a 2    32   2
  4 


 1   1 
1 
  
V   Z o 2 D 3       2  4   0.8242   
2
 4 
 4
 a    64 
 1 
V   Z o 2 D 3 

 64a 2 
 2  4 160.8242  16
 1 
V   Z o 2 D 3 

 64a 2 
 2  4 160.8242  16
11
(B-35)
(B-36)
(B-37)
(B-38)
(B-39)
 1 
V   Z o 2 D 3 
  19.0568  16
 64a 2 
(B-40)
1 
V   Z o 2 D 3    0.2978  0.25
a 2 
(B-41)
Now equate the total kinetic energy with the total strain energy per Rayleigh's method.


h  2 Z o 2 a 2 2
1 
  4   Z o 2 D 3    0.2978  0.25
4
a 2 
(B-42)
h  2 a 2 2
 4 
4


1 
D 3    0.2978  0.25
a 2 
(B-43)
h  2 2
 4 
4


1
D 3    0.2978  0.25
a 4 
(B-44)


1 
D  4    0.2978  0.25
a 4 
(B-45)
h  2 2
 4 
4


h  2  2  4 
2 
2 
4
D  4    0.2978  0.25
a 4 
(B-46)

 4 
1


D  4  
  0.2978  0.25
4 
2
 a  h   4 

(B-47)

 1 
1


D  4  
  1.1911  
4 
2
 a  h   4 

(B-48)




Let  = 0.3, which is the typical Poisson's ratio.
12

 1 
1


D  4  
  1.1911
4 
2
 a  h   4 


2 

 0.3
(B-49)
2 

 1 
1


D  4  
  1.4911
4 
2

a

h


4
 


(B-50)
2 

 1 
1


D  4  
  1.4911
4 
2
 a  h   4 


(B-51)







 1 
1


D  4  
  1.4911
2


h


4
 a 4 



4.9744
a2

D
h

(B-52)
(B-53)
The natural frequency fn is
fn 
fn 
1

2
(B-54)
4.9744 D e
2 a 2 h
(B-55)
13
APPENDIX C
Integral Table
Equation (C-1) is taken from Reference 1.

x cos bx dx 
x sin bx cos bx

b
b2
(C-1)
Now consider
1 a
 r 
sin 2   dr 
2 0
0 r
 2a 

a1

1
  r 
1  cos  dr

r
 a 
(C-2)
Nondimensionalize,
r
a
(C-3)
a
xr

(C-4)
x

dr
a
(C-5)
a
dx  dr

(C-6)
dx 
a    1
 r 
1  cos x  dx
sin 2   dr 
 
2  a  0 x
0 r
 2a 
(C-7)
1  1
 r 
1  cos x  dx
sin 2   dr 
2 0 x
0 r
 2a 
(C-8)

a1


a1

14
Recall the series
cos x  1 
x10
x12
x 2 x 4 x 6 x8





2! 4!
6! 8! 10!
12!
(C-9)
2 x4
x10
x12  
1  1
x 6 x8
  x

 r 
sin 2   dr 





  dx
1  1 
2 0 x 
2! 4!
6! 8! 10!
12!  
0 r
 2a 

 

a1

(C-10)
x10
x12 
1  1  x 2 x 4 x 6 x8
 r 
sin 2   dr 






 dx
2 0 x  2! 4!
6! 8! 10!
12! 
0 r
 2a 



a1

(C-11)
x9
x11 
1   x1 x 3 x 5 x 7
 r 
sin 2   dr 




 
 dx
2 0  2! 4!
6! 8! 10!
12! 
0 r
 2a 



a1

(C-12)

x10
x12 
1  x2
x4
x6
x8
 r 
sin 2   dr 







2  2  2! 4  4! 6  6! 8  8! 10  10! 12  12!
0 r
 2a 

 0

a1
(C-13)
a1
0 r sin
a1
2   r  dr  1 1.6483
 
(C-14)
2   r  dr  0.8242
 
(C-15)
 2a 
0 r sin
2
 2a 
15
APPENDIX D
Solution of Differential Equation via Bessel Functions
The governing equation is taken from References 5 and 6.
 4 Z(r, )   4 Z(r, )  0
(D-1)
2h
4
 
De
(D-2)
1/ 4
 2h 


 D e 
De 
(D-2)
Eh 3
12 1   2 


(D-3)
 2 1 
1  2    2 1 
1  2 
 4   2 2  




 2 r r


r 2  2   r 2 r r r 2  2 
 r
(D-4)
The governing equation may be written as
2  2 2  2 Z(r, )  0
(D-5)
Thus the equation is satisfy by
2  2 Z(r, )  0
(D-6)
Separate variables
Z(r, )  R (r)()
(D-7)
16
By substitution
  2 1 

1  2 

2 R (r )()  0




 2

r r r 2  2 

r






(D-8)
 2

1
1 2

R 
R 
R    2 R  0
 2

r r
r 2  2
 r

(D-9)
 d 2R
1 dR R d 2  



  2 R  0


2
2
2
r dr r d
 dr

(D-10)
Similarly,
 d 2R
1 dR R d 2  



  2 R  0


2
r dr r 2 d 2
 dr

(D-11)
 d 2R
1 dR R d 2  



  2 R  0


2
2
2
r dr r d
 dr

(D-12)
 d 2R
1 dR 
R d 2
2


  R  

2
r dr 
dr
r 2 d 2


(D-13)
 1 d 2R
1 dR 
1 d 2


 2  
 R 2 R r dr 
 r 2 d 2
 dr

(D-14)
r 2  d 2 R 1 dR 
1 d 2

 2 r 2  
R  dr 2
r dr 
 d 2


(D-15)
Thus,
17
The equation can be satisfied if each expression is equal to the same constant k 2 .
r 2  d 2 R 1 dR 
1 d 2

 2 r 2  
R  dr 2
r dr 
 d 2


 k2
(D-16)
Thus
r 2  d 2 R 1 dR 

 2 r 2 
R  dr 2
r dr 


k2

d 2 R 1 dR
k2 
R  0

   2 
r dr

dr 2
r 2 
(D-17)
(D-18)
Define a new variable
 r

 j r
(D-19)
  2 r 2
2  
  2 r 2
(D-20)
 2   2

2
  2
r
(D-21)
Chain rule
d  dr
(D-22)
d
d d

dr d dr
(D-23)
d
d
 
dr
d
(D-24)
18

2 d 2R
 dR



r 2 d 2
r 2 d
2
2
k2 
R  0

r2
r 2 
d 2R
dR

  2  k 2  R  0

2

d

d

d 2R
1 dR
k2 
R  0

 1 
 d

d 2
 2 
(D-25)
(D-26)
(D-27)
Equation (D-27) is Bessel’s equation of fractional order.
The solution for circular plates that are closed in the  direction is
R()  C J n   D I n   F Yn   G K n 
(D-28)
Equation (D-28) represents Bessel of the first and second kind and modified Bessel of the
first and second kind.
Both Yn  and K n  are singular at  = 0.
Thus for a plate with no hole, F = G = 0.
R ()  C J n   D I n 
(D-29)
Furthermore, from equation (D-16),

1 d 2
 d 2
d 2
d 2
 k2
(D-30)
 k2  0
(D-31)
19
The solution for circular plates that are closed in the  direction is
  Â cos k  B̂ sin k ,
k = n = 0, 1, 2, 3, …..
(D-32)
Or equivalently
  A cos k(  )
..
(D-33)
The total solution is thus
Z(, )  C J n   D I n  A cos k(  ) 
(D-34)
Set the phase angle  =0.
Z(, )  C J n   D I n  A cos(k)
(D-35)
Set A =1. Note that the mass normalization will be performed using the C and D
coefficients.
Z(, )  C J n   D I n  cos(k)
20
(D-36)
APPENDIX E
Simply Supported Plate, Bessel Function Solution
The boundary conditions are
Z(a, )  0
(E-1)
M r  0 at r = a
(E-2)
2Z
 0 at r = a
 2
(E-3)
Note that
2Z
 1 Z 1  2 Z 

M r  D e 
 

2
2
2

r  r r  
  r


(E-4)
Boundary condition (E-3) requires that
 2Z
 Z 
M r  D e 


r r 
  r 2

at r = a
(E-5)
Z(r, )  R (r)()
(E-6)
Z(r, )  C J n  r   D I n  r cos(n)
(E-7)
Z(a, )  C J n  a   D I n  a cos(n)  0
(E-8)
C J n  a   D I n  a   0
(E-9)
21
 2


M r  D e 
C J n  r   D I n  r cos(n)
  r 2

1 
C J n  r   D I n  r cos(n)
 D e 
 r r

at r  a
(E-10)
 d2

d2
M r  D e cos(k) C
J n  r   D
I n  r 
 dr 2

dr 2
1d
 1d

 D e cos(k) C
J n  r   D
I n  r  at r  a
r dr
 r dr

(E-11)
Mr
r a
0
(E-12)
 d2

d2
1 d
 1 d

J n  r   D
I n  r    C
J n  r   D
I n  r   0 ,
C
a dr
 a dr

 dr 2

dr 2
at r = a
(E-13)
 d2

 d2

 d
 d
C
J n  r  
J n  r   D 
I n  r  
I n  r   0 ,
a dr
a dr
 dr 2

 dr 2

at r = a
(E-14)
Let
  r
(E-15)
22
 2 d2

 2 d2

 d
 d
C
J n   
J n    D  
I n   
I n    0 ,
a d
a d




d2
d2
at   a
(E-36)
 d2

 d2

 d
 d
C
J n   
J n    D  
I n   
I n    0 ,
2
2
a d
a d
 d

 d

at   a
(E-37)
Recall equation (E-9).
C J n   D I n   0 ,
at   a
(E-38)
D I n   C J n   ,
at   a
(E-39)
J  
D  C n
,
I n  
at   a
(E-40)
By substitution,
 d2


J n    d 2
 d
 d




C
Jn  
Jn    C
I n   
I n    0 ,

a d
I n    d2
a d
 d2



at   a
(E-41)
 d2
 J    d 2

 d
 d
J n   
J n    n
I n   
I n ()  0 , at   a


a d
a d
 d2
 I n    d2

(E-42)
23
 d2


 d
1  d2
 d
J n   
J n   
I n   
I n    0 ,


J n    d2
a d
I n    d2
a d




1
at   a
(E-43)
Note the following identities:
d
n
n
J n    J n 1    J n     J n 1    J n  
d


(E-44)
d2
d
n d
J n     J n 1   
J n  
d
 d
d2
(E-45)
d2
n 1
n

 n

J n     J n   
J n 1     J n 1    J n  



 

d2
(E-46)
d2
n 1
n

 n

J n     J n   
J n 1     J n 1    J n  



 

d2
(E-47)

d2
n2 
1
 J n      J n 1  
J n     1 
2
2
 

d
 
(E-48)
24
Analyze the first term of equation (E-42).

 d2

 d

J n   
J n  
J n    d2
a d


1

 
  1
n2 
n
1


  1 
 J n      J n 1   
 J n 1    J n  

J n    

 a J n   
 

2 
 

1
      J n 1   
J
n2  n

 n 1 




1

2
J n      a J n  

  a 

J n 1      
n2   n




  1
2
J n    a  

  a 
(E-49)
Consider the following identities:
d
n
n
I n    I n 1    I n    I n 1    I n  
d


d2
d
n d
I n   
I n 1   
I n  
d
 d
d2
(E-50)
(E-51)
d2
n 1
n

 n

I n    I n   
I n 1    I n 1    I n  
2



 

d
(E-52)
 n2 
d2
1
 I n      I n 1  
I n    1 
 
 2 
d2
(E-53)
25
Analyze the second term of equation (E-42).

 d2

I n   
I n    d2

1


 d
I n   
a d

  n2 
  1
n


1









1

I



I


I n 1    I n  


n
n

1



2
I n      

 

 a I n   

 
1
 I n 1    I n 1    n 2   n



 1 
 I n  
a I n  
a
 2 
 n2  n
    I n 1  

  
 1 
a
 a   I n  
 2 
(E-54)
By substitution,

 d2

1   d

J n    
J n  


2
J n    d
J n    a d





1  d2
1   d

I n    
I n    0 ,


2
I n    d

 I n    a d

1
at   a
(E-55)
 n2   n
      
J
n 2   n     I n 1  


 n 1       1 
  
 1 
 0,
J n    a  
a

 2 
2  a   a   I n  
at   a
(E-56)
26
          I  
J
 n 1        n 1
 2  0 , at   a
J n    a    a   I n  
(E-57)
J n 1           I n 1  



 2  0 , at   a
J n    a    a   I n  
(E-58)
J n 1           I n 1  



 2 , at   a
J n    a    a   I n  
J n 1   I n 1  
 2


J n  
I n  
  
a  


J n 1   I n 1  
 2


J n  
I n  
 1 
a  a


(E-59)
,
at   a
(E-60)
,
at   a
(E-61)
J n 1   I n 1    2a


, at   a
J n  
I n  
1 
(E-62)
J n 1   I n 1  
 2


, at   a
J n  
I n  
1 
(E-63)
J n 1   I n 1  
2


, at   a
J n  
I n  
1 
(E-64)
The following form is better suited for numerical root-finding purposes.
I n   J n 1   J n   I n 1   
2
J n   I n    ,
1 
27
at   a
(E-65)
The roots of equation (E-65) for   0.3 are expressed in terms of 2 as
k
n=0
n=1
n=2
n=3
0
4.9351
13.8982
25.6133
39.9573
1
29.7658
48.5299
70.1170
94.5490
2
74.2302
102.7965
134.2978
168.6749
3
138.3181
176.8012
218.2026
262.6244
The roots were determined using the secant method.
The fundamental natural frequency is thus
  a
4 
(E-66)
2h
De
(E-67)
1/ 4
 2h 


 D e 
1/ 4
 2h
a 
 D e
2 


(E-68)



(E-69)
4 D e
(E-70)
h a 4
2
a2
4.9351
a2
De
h
(E-71)
De
h
(E-72)
28
The mode shapes are defined by
Z(r, )  C J n  r   D I n  r cos(n)
(E-72)
Recall
J  
D  C n
,
I n  
at   a
(E-73)


J  
Z( r, )  C J n  r   C n
I n  r   cos(n)
I n  




 J  
Z( r, )  C J n  r    n  I n  r   cos(n)
 I n   


(E-74)
(E-75)
Again,
  /a
(E-76)
29
APPENDIX F
Note that the numerical calculations in this appendix are performed via Matlab script:
circular_SS.m.
Mass Normalization of Eigenvectors
The eigenvector equation is


 J  
Z k n ( r, )  C kn  J n  r    n  I n  r   cos(n)
 I n   


(F-1)
Again,
  /a
(F-2)
Normalize the eigenvectors as


a 2
Z k n ( r, ) 2 r d dr  1
0 0
h 

(F-3)
2


 J n  
a 2
2
2
h
C kn  J n  r   
 I n  r   cos (n) r d dr  1
0 0
 I n   
(F-4)
2
 1 1
 J n  
a 2 

2
C kn h
 J n  r   
 I n  r     cos(n)r d dr  1
0 0

 I n   

 2 2
(F-5)




2

 J n  
a 2 
2
C kn h
 J n  r   
 I n  r   r d dr  1
0 0
 I n   



30
for n=0
(F-6)
2 
2 


 J n  
a

  1
 
2
C kn h  J n  r   
sin(n)   1
 I n  r   rdr  
0
0 
 I n   



 2 2n



for n>1

(F-7)
2 


 J n  
a


2
 C kn h  J n  r   
I n  r   rdr   1

0
 I n   

for n>1
(F-8)
2

 J n  
a 2 
2
C kn h
 J n  r   
 I n  r   r d dr  1
0 0
 I n   
(F-9)




For k=0, n=0,


a 2
2C 002 h
0 0

2 C 00
2 h
2 C 00
2 h

2


 J 0  


 J 0  r   
 I 0  r   r d dr  1


 I 0   


(F-10)
2

 J 0  
a


 J  r   
 I 0  r   r dr  1
0 0


I


 0 
(F-11)
2

 J 0  
a


 J  r   
 I 0  r   r dr  1
0 0


I


 0 
(F-12)






The eigenvalue is
=
4.9351  2.2215 for k=0, n=0
 J 0  

  0.03686 for


I

 0 
 = 2.2215
31
(F-13)
(F-14)
By substitution
0  J 0  r   0.03686 I 0  r  
a
2 C 002 h
2 r dr  1
(F-15)
Recall
  /a
(F-16)
By substitution,
2 C 002 h
0  J 0  r / a   0.03686 I 0  r / a  
(F-17)
2 C 002 h
a
 J 0 2.2215 r / a   0.03686 I 0 2.2215 r / a  2 r dr  1
0
(F-18)
a
2 r dr  1

Let
u = 2.2215 r / a
(F-19)
r = a u / 2.2215
(F-20)
dr = a du / 2.2215
(F-21)
2
r dr = a u du / 4.9351
(F-22)
By substitution,
2 C 002
h a 2
4.9351
2.2215
 J 0 u   0.03686 I 0 u  2 u du
0

1
(F-23)
The mass is
m  h a 2
C 002
(F-24)
2m
0.6525  1
4.9351
(F-25)
32
C 002 m 0.2644  1
C 00 
(F-26)
1.945
(F-27)
m
The eigenvector for k=0, n=0 is
Z 00 
1.945
m
 J 0 2.2215 r / a   0.03686 I 0 2.2215 r / a  
(F-28)
Participation Factors
The participation factor kn is


 J  
a 2
C kn  J n  r    n  I n  r   cos(n) r d dr
0 0
 I n   
(F-29)
kn  0
(F-30)
kn  h 



Note that
for n > 1
The participation factor for k=0, n=0, is
a 2
Z kn (r, ) r d dr
0 0
kn  h 

(F-31)

a 2 1.945

J 0 2.2215 r / a   0.03686 I 0 2.2215 r / a   r d dr

0 0
m
00  h 



(F-32)
00  h
1.945 a 2

m 0 0
 J 0 2.2215 r / a   0.03686 I 0 2.2215 r / a   r d dr
33
(F-33)
00 
00 
00 
1.945 m a 2

a 2 m 0 0
21.945
 J 0 2.2215 r / a   0.03686 I 0 2.2215 r / a   r d dr
0  J 0 2.2215 r / a   0.03686 I 0 2.2215 r / a   r dr
m a
a 2
0  J 0 2.2215 r / a   0.03686 I 0 2.2215 r / a   r dr
3.890 m a
a2
(F-34)
(F-35)
(F-36)
Recall
u = 2.2215 r / a
(F-37)
2
(F-38)
r dr = a u du / 4.9351
By substitution,
00 
 J 0 u   0.03686 I 0 u  
a2 u
du
4.9351
(F-39)
 J 0 u   0.03686 I 0 u  
u
du
4.9351
(F-40)
3.890 m 2.2215
0
a2
00  3.890 m
2.2215
0
00  0.7881 m
 J 0 u   0.03686 I 0 u   u du
2.2215
0
(F-41)
00  0.7881 m 1.0682
(F-42)
00  0.8419 m
(F-43)
34
The effective modal mass is

M 00  0.8419 m
2
(F-44)
M 00  0.7088 m
(F-45)
Example
Assume a 64 inch diameter, 1 inch thick aluminum plate, with a simply-supported
perimeter. Calculate the fundamental frequency and mode shape.
Figure F-1. Fundamental Mode, fn = 45.6 Hz
35
APPENDIX G
This appendix is unfinished.
Completely Free Circular Plate
Consider a completely free circular plate. The plate has a radius a. The displacement
perpendicular to the plate is Z. A polar coordinate system is used with the origin at the
plate's center.
The mode shape is
Z(r, )  C J n  r   D I n  r cos(n)
(G-1)
The boundary conditions are
 2Z
 1 Z 1  2 Z 
  0

M r  D e
 

 r  r r 2  2 
  r 2


at r = a
    2 1 
1  2   1  
  1 Z  

  0


Vr  D e  


Z 
1



 r  r    
  r   r 2 r  r r 2  2   r  

 

(G-2)
at r = a
(G-3)
The moment is
 2
1 
1  2 
  C J n  r   D I n  r cos(n)
M r  D e 
 

 r  r r 2  2 
  r 2


36
(G-4)
M r  D e
d2
dr 2
 D e
C J n  r   D I n  r cos(n)
1d
C J n  r   D I n  r cos(n)
r dr
 n 2 D e
1
r2
C J n  r   D I n  r cos(n)
(G-5)
 d 2

1d
n2 
M r   D e 


 J n  r  C cos(n)
r dr
r 2 
 dr 2

 d 2

1d
n2 
 D e 

 n 2
 I n  r  D cos(n)
r dr
r 2 
 dr 2

(G-6)
Let
  r
(G-7)
r  /
(G-8)
d  dr
(G-9)
d
d d

dr d dr
(G-10)
d
d
 
dr
d
(G-11)
d2
dr 2
 2
d2
(G-12)
d2
37
 d 2

1 d
n2 
M r   D e  2 


 J n   C cos(n)
 d
2 
 d2

 d 2

1 d
n2 
 D e  2 

 n 2
 I n   D cos(n)
 d
2 
 d2

(G-13)
Now consider the shear boundary condition.
    2 1 
1  2   1  
  1 Z  





  0


Vr  D e


Z 
 1 
 r  r    
  r   r 2 r  r r 2  2   r  

 

(G-14)
 3
1  1 2
2 2
1 3 
Vr  D e 




Z
  r 3 r 2  r r  r 2 r 3  2 r 2  r 2 
 De
2
3 

1
1    1 2  2  1  2  Z
r
r  r  
  r  

(G-15)
 3
1  1 2
2 2
1 3 
Vr  D e 




C J n  r   D I n  r cos(n)
  r 3 r 2  r r  r 2 r 3  2 r 2  r 2 
 De
2
3 

1
1    1 2  2  1  2 C J n  r   D I n  r cos(n)
r
r  r  
  r  

(G-16)
38
 3
1  1 2
2 2
1 3 
Vr  D e 




C J n  r cos(n)
  r 3 r 2  r r  r 2 r 3  2 r 2  r 2 
 3
1  1 2
2 2
1 3 
 De 




D I n  r cos(n)
  r 3 r 2  r r  r 2 r 3  2 r 2  r 2 
 De
2
3 

1
1    1 2  2  1  2 C J n  r cos(n)
r
r  r  
  r  

 De
2
3 

1
1    1 2  2  1  2 D I n  r cos(n)
r
r  r  
  r  

(G-17)
 3
1  1 2
2 2
1 3 
Vr  D e 




C J n  r cos(n)
  r 3 r 2  r r  r 2 r 3  2 r 2  r 2 
 3
1  1 2
2 2
1 3 
 De 




D I n  r cos(n)
  r 3 r 2  r r  r 2 r 3  2 r 2  r 2 
 De
 De
r
 1 2
3 

D I n  r cos(n)
  r   2  r  2 
1   
2
1
r
 1 2
3 

C J n  r cos(n)
  r   2  r  2 
1   
2
1
(G-18)
39
Vr 
 3
 1 2
1  1 2
2 2
1 3
1
 3 

 De 





1   

  C J n  r  cos(n)
  r 3 r 2  r r  r 2 r 3  2 r 2  r 2 r 2
  r   2  r  2  
 3
 1 2
1  1 2
2 2
1 3
1
 3 

 De 





1   

  D I n  r  cos(n)
  r 3 r 2  r r  r 2 r 3  2 r 2  r 2 r 2
  r   2  r  2  
(G-19)
Vr 
 3
 1 2
1  1 2
2 2
1 3
1
 3 

  C J n  r  cos(n)
 De 





1   

 r   2  r  2  
  r 3 r 2  r r  r 2 r 3  2 r 2  r 2 r 2




 3
 1 2
1  1 2
2 2
1 3
1
 3 

  D I n  r  cos(n)
 De 





1   

 r   2  r  2  
  r 3 r 2  r r  r 2 r 3  2 r 2  r 2 r 2




(G-20)
Let
  r
(G-21)
40
Vr 
 3
 1 2
1  1 2
2 2
1 3
1
 3 

  C J n   cos(n)
 D e3 





1   

    2    2  
  3 2     2 3  2 2   2 2




 3
 1 2
1  1 2
2 2
1 3
1
 3 

  D I n   cos(n)
 D e3 





1   

    2    2  
  3 2     2 3  2 2   2 2




(G-22)
Vr 
 3
 1
1 
1 2
2n 2 n 2  n 2
 

 D e3 





1   

  C J n   cos(n)
  3 2     2
      
3
2   2
 3
 1
1 
1 2
2n 2 n 2  n 2
 

 D e3 





1   

  D I n   cos(n)
  3 2     2
      
3
2   2
(G-23)
41
APPENDIX H
Fixed Plate, Bessel Function Solution
The boundary conditions are
Z(a, )  0
(H-1)
Z
 0 at r = a
r
(H-2)
The displacement is
Z(r, )  C J n  r   D I n  r cos(n)
(H-3)
The slope is
Z 
 C J n  r   D I n  r cos(n)
r r
(H-4)

dZ  d
d
 C J n  r   D I n  r  cos(n)
dr  dr
dr

(H-5)
Let
  r
(H-6)
d
d
 
dr
d
(H-7)
The displacement is
Z(r, )  C J n   D I n  cos(n)
(H-8)
The slope is
 d

dZ
d
  C
J n    D
I n   cos(n)
dr
d
 d

42
(H-9)
Apply the boundary conditions
C J n   D I n cos(n)  0
at    a
C J n   D I n    0
at    a
 d

d
 C
J n    D
I n   cos(n)  0
d
 d

C
(H-10)
d
d
J n    D
I n    0
d
d
(H-11)
at    a
(H-12)
at    a
(H-13)
Assemble the equations in matrix form.

 J n  


 d J  
 d  n

I n     C  0
   
    
d
I n   D 0

d
at    a
(H-14)
The roots are found by setting the determinant of the coefficient matrix equal to zero.
J n  
d
d
I n    I n   J n    0
d
d
at    a
n
n




J n  I n 1    I n    I n   J n 1    J n    0






J n  I n 1   J n 1  I n    0
43
at    a
(H-15)
at    a
(H-16)
(H-17)
The roots of equation (H-17) for   0.3 are expressed in terms of 2 as
k
n=0
n=1
n=2
n=3
0
10.2158
21.2609
34.8770
51.0334
1
39.7711
60.8287
84.5837
111.0214
2
89.1041
120.0792
153.8151
190.3038
3
158.1842
199.0534
242.7285
289.1799
The roots were determined using the secant method.
The fundamental natural frequency is thus
  a
4 
(H-18)
2h
De
(H-19)
1/ 4
 2h 


 D e 
 2h
a 
 D e
(H-20)
1/ 4



(H-21)
4 D e
2
 
h a 4


2
a2
De
h
10.2158
a2
(H-22)
(H-23)
De
h
(H-24)
44
The mode shapes are defined by
Z(r, )  C J n  r   D I n  r cos(n)
(H-25)
Recall
J  
D  C n
,
I n  
at   a
(H-26)


J  
Z( r, )  C J n  r   C n
I n  r   cos(n)
I n  




 J  
Z( r, )  C J n  r    n  I n  r   cos(n)
 I n   


(H-27)
(H-28)
Let
  r
(H-29)


 J  
Z(, )  C J n     n  I n  r   cos(n)
 I n   


45
(H-30)