natural frequencies of rectangular plate … · natural frequencies of circular plate bending ......
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1
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.
Figure 1.
Let Z represent the out-of-plane displacement.
r
Y
X
0
2
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, Wiley-
Interscience, 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.
3
APPENDIX A
The total strain energy V of the plate is
ddrr
2Z
r
1
r12
2
0
R
0 2
Z2
2r
1
r
Z
r
1
r2
Z212
2
2
Z2
2r
1
r
Z
r
1
2r
Z2
2
eDV
(A-1)
Note that the plate stiffness factor De is given by
2112
3EheD (A-2)
where
E = elastic modulus
h = plate thickness
= Poisson's ratio
For a displacement which is symmetric about the center,
0),r(Z
(A-3)
0),r(Z2
2
(A-4)
Substitute equations (A-3) and (A-4) into (A-1).
4
2
0
R
0ddrr
r
Z
r
1
r2
Z212
2
r
Z
r
1
2r
Z2
2
eDV
(A-5)
2
0
R
0ddrr
r
Z
r
1
r2
Z222
2
r
Z
r
1
r
Z
r
1
2r
Z22
2
2r
Z2
2
eDV
(A-6)
The total strain energy equation for the symmetric case is thus
2
0
R
0ddrr
r
Z
r
1
r2
Z22
2
r
Z
r
12
2r
Z2
2
eDV
(A-7)
The total kinetic energy T of the plate bending is given by
2
0
R
0
22
ddrrZ2
hT (A-8)
where
= mass per volume
= angular natural frequency
5
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
0),a(Z (B-1)
0r
Z
ar2
2
(B-2)
The following function satisfies the geometric boundary conditions.
a2
rcosZ),r(Z o (B-3)
The partial derivatives are
0),r(Z
(B-4)
0),r(Z2
2
(B-5)
a2
rsin
a2Z),r(Z
ro (B-6)
a2
rcos
a2Z),r(Z
r
2
o2
2
(B-7)
6
The total kinetic energy T of the plate bending is given by
2
0
a
0
2
o
2
ddrra2
rcosZ
2
hT (B-8)
2
0
a
0
2o
2
ddrra
rcos1
4
ZhT (B-9)
2
0
a
0
2o
2
ddra
rcosrr
4
ZhT (B-10)
Evaluate equation (B-9) using the integral table in Appendix C
2
0
a
02
222o
2
da
rcos
a
a
rsin
ra
2
r
4
ZhT (B-11)
2
0 2
2
2
222o
2
d0cosa
cosa
2
a
4
ZhT (B-12)
2
0 2
222o
2
da2
2
a
4
ZhT (B-13)
2
0
2
2
22o
2
d48
aZhT (B-14)
2
0
2
2
22o
2
d48
aZhT (B-15)
24
8
aZhT 2
2
22o
2
(B-16)
7
44
aZhT 2
22o
2
(B-17)
22
o
2 aZh4671.0T (B-18)
Again, the total strain energy for the symmetric case is
2
0
R
0ddrr
r
Z
r
1
r2
Z22
2
r
Z
r
12
2r
Z2
2
eDV
(B-19)
2
0
R
0ddrr
r
Z
r
1
r2
Z22
2
eD
2
0
R
0ddrr
2
r
Z
r
1
2
eD
2
0
R
0ddrr
2
2r
Z2
2
eDV
(B-20)
2
0
a
0ddrr
a2
rsin
a2
rcos
r
13
a222
oZ2
eD
2
0
a
0ddrr
a2
r2sin
2
a22r
12oZ
2
eD
2
0
a
0ddrr
a2
r2cos4
a2
2oZ
2
eDV
(B-21)
8
2
0
a
0
32o
2
0
a
0
22
2
2o
2
0
a
0
242
o
ddrra2
rsin
a2
rcos
r
1
a22
2
DZ
ddrra2
rsin
a2r
1
2
DZ
ddrra2
rcos
a22
DZV
(B-22)
2
0
a
0
32o
2
0
a
0
2
2
22o
2
0
a
0
242
o
ddrra2
rsin
a2
rcos
r
1
a22
2
DZ
ddrra2
rsin
r
1
a22
DZ
ddrra2
rcos
a22
DZV
(B-23)
2
0
a
0
32o
2
0
a
0
222
o
2
0
a
0
42o
ddra
rsin
a22
DZ
ddra2
rsin
r
1
a22
DZ
ddrra
rcos1
2
1
a22
DZV
(B-24)
The first and third integrals are evaluating using the tables in Appendix C.
9
2
0
a
0
32o
2
0
22o
2
0
a
02
2242o
da
rcos
a
a22
DZ
d8242.0a22
DZ
da
rcos
a
a
rsin
ra
2
r
2
1
a22
DZV
(B-25)
2
0
32o
2
0
22o
2
0 2
2242o
da2
a22
DZ
d8242.0a22
DZ
da2
2
a
2
1
a22
DZV
(B-26)
2
0
22o
2
0
22o
2
0 2
2242o
da22
DZ
d8242.0a22
DZ
da2
2
a
2
1
a22
DZV
(B-27)
10
2a22
DZ
28242.0a22
DZ
2a2
2
a
2
1
a22
DZV
22o
22o
2
2242o
(B-28)
22
o
22
o
2
2242
o
a2DZ
8242.0a2
DZ
a2
2
a
2
1
a2DZV
(B-29)
2
a28242.0
2
a22
2a2
2
2a
2
14
a2D2
oZV
(B-30)
22
2
224232
oa2
18242.0
a2
1a2
2
a
2
1
a2
1DZV
(B-31)
11
222
22
4
232o
a4
18242.0
a4
1a2
2
a
2
1
a16
1DZV
(B-32)
2222
232o
a4
18242.0
a4
12
2
1
2
1
a16
1DZV
(B-33)
4
8242.04
1
2
2
2
1
2
1
16
122a
13D2oZV
(B-34)
4
8242.04
12
2
1
32
1
a
1DZV
2
2
2
32o
(B-35)
4
8242.04
12
232
1
a
1DZV
2
2
32o
(B-36)
4
8242.04
14
64
1
a
1DZV 2
2
32o
(B-37)
168242.0164a64
1DZV 2
2
32o
(B-38)
168242.0164a64
1DZV 2
2
32o
(B-39)
12
160568.19a64
1DZV
2
32o
(B-40)
25.02978.0a
1DZV
2
32o
(B-41)
Now equate the total kinetic energy with the total strain energy per Rayleigh's method.
25.02978.0
a
1DZ4
4
aZh
2
32o
222
o2
(B-42)
25.02978.0
a
1D4
4
ah
2
3222
(B-43)
25.02978.0
a
1D4
4
h
4
322
(B-44)
25.02978.0a
1D4
4
h
4
422
(B-45)
25.02978.0a
4D4h
4
422 (B-46)
25.02978.04h
1
a
4D
24
42 (B-47)
1911.14h
1
a
1D
24
42 (B-48)
Let = 0.3, which is the typical Poisson's ratio.
13
3.01911.1
4h
1
a
1D
24
42
(B-49)
4911.1
4h
1
a
1D
24
42
(B-50)
4911.1
4h
1
a
1D
24
42
(B-51)
4911.1
4h
1
a
1D
24
4
(B-52)
h
D
a
9744.4
2 (B-53)
The natural frequency fn is
2
1fn (B-54)
h
eD
2a2
9744.4nf
(B-55)
14
APPENDIX C
Integral Table
Equation (C-1) is taken from Reference 1.
2b
bxcos
b
bxsinxdxbxcosx (C-1)
Now consider
dra
rcos1
r
1
2
1dr
a2
rsin
r
1 a
0
a
0
2
(C-2)
Nondimensionalize,
a
rx
(C-3)
rxa
(C-4)
dra
dx
(C-5)
drdxa
(C-6)
dxxcos1x
1
a2
adr
a2
rsin
r
1
0
a
0
2
(C-7)
dxxcos1x
1
2
1dr
a2
rsin
r
1
0
a
0
2
(C-8)
15
Recall the series
!12
x
!10
x
!8
x
!6
x
!4
x
!2
x1xcos
12108642
(C-9)
dx!12
x
!10
x
!8
x
!6
x
!4
x
!2
x11
x
1
2
1dr
a2
rsin
r
1
0
12108642a
0
2
(C-10)
dx!12
x
!10
x
!8
x
!6
x
!4
x
!2
x
x
1
2
1dr
a2
rsin
r
1
0
12108642a
0
2
(C-11)
dx!12
x
!10
x
!8
x
!6
x
!4
x
!2
x
2
1dr
a2
rsin
r
1
0
1197531a
0
2
(C-12)
0
12108642a
0
2
!1212
x
!1010
x
!88
x
!66
x
!44
x
!22
x
2
1dr
a2
rsin
r
1
(C-13)
6483.12
1dr
a2
rsin
r
1a
0
2
(C-14)
8242.0dra2
rsin
r
1a
0
2
(C-15)
16
APPENDIX D
Solution of Differential Equation via Bessel Functions
The governing equation is taken from References 5 and 6.
0),r(Z4),r(Z4 (D-1)
eD
h24 (D-2)
4/1
eD
h2
(D-2)
2112
3EheD (D-3)
2
2
2r
1
rr
1
2r
2
2
2
2r
1
rr
1
2r
2224 (D-4)
The governing equation may be written as
0),r(Z2222 (D-5)
Thus the equation is satisfy by
0),r(Z22 (D-6)
Separate variables
)()r(R),r(Z (D-7)
17
By substitution
0)()r(R22
2
2r
1
rr
1
2r
2
(D-8)
0R2R2
2
2r
1R
rr
1R
2r
2
(D-9)
0R22d
2d
2r
R
dr
dR
r
1
2dr
R2d
(D-10)
Similarly,
0R22d
2d
2r
R
dr
dR
r
1
2dr
R2d
(D-11)
Thus,
0R22d
2d
2r
R
dr
dR
r
1
2dr
R2d
(D-12)
2d
2d
2r
RR2
dr
dR
r
1
2dr
R2d
(D-13)
2d
2d
2r
12
dr
dR
rR
1
2dr
R2d
R
1
(D-14)
2d
2d12r2
dr
dR
r
1
2dr
R2d
R
2r
(D-15)
18
The equation can be satisfied if each expression is equal to the same constant .2k
2k2d
2d12r2
dr
dR
r
1
2dr
R2d
R
2r
(D-16)
Thus
2k2r2
dr
dR
r
1
2dr
R2d
R
2r
(D-17)
0R2r
2k2
dr
dR
r
1
2dr
R2d
(D-18)
Define a new variable
rj
r (D-19)
2r2
2r22
(D-20)
2
2
2r
2 (D-21)
Chain rule
drd (D-22)
dr
d
d
d
dr
d
(D-23)
d
d
dr
d (D-24)
19
0R2r
2k
2r
2
d
dR
2r2d
R2d
2r
2
(D-25)
0R2k2
d
dR
2d
R2d2
(D-26)
0R2
2k1
d
dR1
2d
R2d
(D-27)
Equation (D-27) is Bessel’s equation of fractional order.
The solution for circular plates that are closed in the direction is
nKGnYFnIDnJC)(R (D-28)
Equation (D-28) represents Bessel of the first and second kind and modified Bessel of the
first and second kind.
Both nY and nK are singular at = 0.
Thus for a plate with no hole, F = G = 0.
nIDnJC)(R (D-29)
Furthermore, from equation (D-16),
2k2d
2d1
(D-30)
02k2d
2d
(D-31)
20
The solution for circular plates that are closed in the direction is
ksinBkcosA , k = n = 0, 1, 2, 3, ….. (D-32)
Or equivalently
)(kcosA .. (D-33)
The total solution is thus
)(kcosAnIDnJC),(Z (D-34)
Set the phase angle =0.
)kcos(AnIDnJC),(Z (D-35)
Set A =1. Note that the mass normalization will be performed using the C and D
coefficients.
)kcos(nIDnJC),(Z (D-36)
21
APPENDIX E
Simply Supported Plate, Bessel Function Solution
The boundary conditions are
0),a(Z (E-1)
0rM at r = a (E-2)
02
Z2
at r = a (E-3)
Note that
2
2
22
2
erZ
r
1
r
Z
r
1
r
ZDM (E-4)
Boundary condition (E-3) requires that
r
Z
rr
ZDM
2
2
er at r = a (E-5)
)()r(R),r(Z (E-6)
)ncos(rIDrJC),r(Z nn (E-7)
0)ncos(aIDaJC),a(Z nn (E-8)
0anIDanJC (E-9)
22
a r at )ncos(rIDrJCrr
1D
)ncos(rIDrJCr
DM
nne
nn2
2
er
(E-10)
a r at rIdr
d
r
1DrJ
dr
d
r
1C)kcos(D
rIdr
dDrJ
dr
dC)kcos(DM
nne
n2
2
n2
2
er
(E-11)
0arrM
(E-12)
0rIdr
d
a
1DrJ
dr
d
a
1CrI
dr
dDrJ
dr
dC nnn2
2
n2
2
, at r = a
(E-13)
0rIdr
d
arI
dr
dDrJ
dr
d
arJ
dr
dC nn2
2
nn2
2
, at r = a
(E-14)
Let
r (E-15)
23
0Id
d
aI
d
dDJ
d
d
aJ
d
dC nn2
22
nn2
22
,
at a
(E-36)
0Id
d
aI
d
dDJ
d
d
aJ
d
dC nn2
2
nn2
2
,
at a
(E-37)
Recall equation (E-9).
0nIDnJC , at a (E-38)
nJCnID , at a (E-39)
nI
nJCD , at a (E-40)
By substitution,
0Id
d
aI
d
d
I
JCJ
d
d
aJ
d
dC nn2
2
n
nnn2
2
,
at a
(E-41)
0)(Id
d
aI
d
d
I
JJ
d
d
aJ
d
dnn2
2
n
nnn2
2
, at a
(E-42)
24
0I
d
d
aI
d
d
I
1J
d
d
aJ
d
d
J
1nn2
2
nnn2
2
n
,
at a
(E-43)
Note the following identities:
n1nn1nn Jn
JJn
JJd
d (E-44)
nJd
dn1nJ
d
dnJ
2d
2d (E-45)
nJ
n1nJ
n1nJ
1nnJnJ
2d
2d (E-46)
nJ
n1nJ
n1nJ
1nnJnJ
2d
2d (E-47)
1nJ
1nJ
2
2n1nJ
2d
2d (E-48)
25
Analyze the first term of equation (E-42).
a
n
2
2n1
anJ
1nJ
a
n
2
2n1
nJ
1nJ
anJ
1nJ
nJn
1nJnJ
1
a1nJ
1nJ
2
2n1
nJ
1
nJd
d
anJ
2d
2d
nJ
1
(E-49)
Consider the following identities:
n1nn1nn In
IIn
IId
d (E-50)
nId
dn1nI
d
dnI
2d
2d (E-51)
nI
n1nI
n1nI
1nnInI
2d
2d (E-52)
1nI
1nI
2
2n1nI
2d
2d (E-53)
26
Analyze the second term of equation (E-42).
a
n
2
2n1
nI
1nI
a
a
n
2
2n1
nI
1nI
anI
1nI
nIn
1nInI
1
a1nI
1nI
2
2n1
nI
1
nId
d
anI
2d
2d
nI
1
(E-54)
By substitution,
,0I
d
d
aI
1I
d
d
I
1
Jd
d
aJ
1J
d
d
J
1
nn
n2
2
n
nn
n2
2
n
at a
(E-55)
0a
n
2
2n1
nI
1nI
aa
n
2
2n1
anJ
1nJ
,
at a
(E-56)
27
02nI
1nI
aanJ
1nJ
, at a (E-57)
02nI
1nI
aanJ
1nJ
, at a (E-58)
2nI
1nI
aanJ
1nJ , at a (E-59)
a
2
nI
1nI
nJ
1nJ , at a (E-60)
a
1
a
2
nI
1nI
nJ
1nJ , at a (E-61)
1
a2
nI
1nI
nJ
1nJ , at a (E-62)
1
2
nI
1nI
nJ
1nJ , at a (E-63)
1
2
nI
1nI
nJ
1nJ , at a (E-64)
The following form is better suited for numerical root-finding purposes.
nInJ
1
21nInJ1nJnI , at a (E-65)
28
The roots of equation (E-65) for 3.0 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 (E-66)
e
24
D
h (E-67)
4/1
eD
h2
(E-68)
4/1
e
2
D
ha
(E-69)
4e
42
ah
D
(E-70)
h
D
a
e2
2
(E-71)
h
D
a
4.9351 e2
(E-72)
29
The mode shapes are defined by
)ncos(rnIDrnJC),r(Z (E-72)
Recall
nI
nJCD , at a (E-73)
)ncos(rnInI
nJCrnJC),r(Z
(E-74)
)ncos(rnInI
nJrnJC),r(Z
(E-75)
Again,
a/ (E-76)
30
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
)ncos(rII
JrJC),r(Z n
n
nnknnk
(F-1)
Again,
a/ (F-2)
Normalize the eigenvectors as
1drdr),r(Zha
0
2
0
2nk
(F-3)
1drdr)n(cosrII
JrJCh
a
0
2
0
22
nn
nn
2kn
(F-4)
1drdr)ncos(2
1
2
1rI
I
JrJhC
a
0
2
0
2
nn
nn
2kn
(F-5)
1drdrrII
JrJhC
a
0
2
0
2
nn
nn
2kn
for n=0 (F-6)
31
1)nsin(n2
1
2rdrrI
I
JrJhC
2
0
a
0
2
nn
nn
2kn
for n>1
(F-7)
1rdrrII
JrJhC
a
0
2
nn
nn
2kn
for n>1 (F-8)
For k=0, n=0,
1drdrrII
JrJhC
a
0
2
0
2
nn
nn
2kn
(F-9)
1drdrrII
JrJhC2
a
0
2
0
2
00
00
200
(F-10)
1drrrII
JrJhC2
a
0
2
00
00
200
(F-11)
1drrrII
JrJhC2
a
0
2
00
00
200
(F-12)
The eigenvalue is
= 4.9351 2.2215 for k=0, n=0 (F-13)
0
0
I
J 0.03686 for = 2.2215 (F-14)
32
By substitution
1drrrI0.03686rJhC2a
0
200
200 (F-15)
Recall
a/ (F-16)
By substitution,
1drra/rI0.03686a/rJhC2a
0
200
200 (F-17)
1drra/r 2.2215I0.03686a/r2.2215JhC2a
0
200
200 (F-18)
Let
u = a/r2.2215 (F-19)
r = a u / 2.2215 (F-20)
dr = a du / 2.2215 (F-21)
r dr = a2
u du / 4.9351 (F-22)
By substitution,
1du u uI0.03686uJ 4.9351
ahC2
2.2215
0
200
22
00
(F-23)
The mass is
2ahm (F-24)
10.65254.9351
m2C 2
00 (F-25)
33
10.2644mC 200 (F-26)
m
1.945C00 (F-27)
The eigenvector for k=0, n=0 is
a/r 2.2215I0.03686a/r2.2215Jm
1.945Z 0000 (F-28)
Participation Factors
The participation factor kn is
a
0
2
0 nn
nnknkn drdr)ncos(rI
I
JrJCh (F-29)
Note that
0kn for n > 1 (F-30)
The participation factor for k=0, n=0, is
a
0
2
0 knkn drdr),r(Zh (F-31)
a
0
2
0 0000 drdra/r 2.2215I0.03686a/r2.2215Jm
1.945h
(F-32)
a
0
2
0 0000 drdra/r 2.2215I0.03686a/r2.2215Jm
1.945h (F-33)
34
a
0
2
0 00200 drdra/r 2.2215I0.03686a/r2.2215Jma
m1.945 (F-34)
a
0 00200 drra/r 2.2215I0.03686a/r2.2215Ja
m1.9452 (F-35)
a
0 00200 drra/r 2.2215I0.03686a/r2.2215Ja
m3.890 (F-36)
Recall
u = a/r2.2215 (F-37)
r dr = a2
u du / 4.9351 (F-38)
By substitution,
2.2215
0
2
00200 du4.9351
u a uI0.03686uJ
a
m3.890 (F-39)
2.2215
0 0000 du4.9351
u uI0.03686uJm3.890 (F-40)
2.2215
0 0000 duuuI0.03686uJm0.7881 (F-41)
1.0682m0.788100 (F-42)
m0.841900 (F-43)
35
The effective modal mass is
200 m0.8419M (F-44)
m0.7088M00 (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
36
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
)ncos(rIDrJC),r(Z nn (G-1)
The boundary conditions are
0Z
r
1
r
Z
r
1
r
ZDM
2
2
22
2
er
at r = a (G-2)
0Z
r
1
r1
r
1Z
r
1
rr
1
rrDV
2
2
22
2
er
at r = a
(G-3)
The moment is
)ncos(rIDrJCr
1
rr
1
rDM nn2
2
22
2
er
(G-4)
37
)ncos(rIDrJCr
1Dn
)ncos(rIDrJCdr
d
r
1D
)ncos(rIDrJCdr
dDM
nn2e2
nne
nn2
2
er
(G-5)
)ncos(DrIr
nn
dr
d
r
1
dr
dD
)ncos(CrJr
n
dr
d
r
1
dr
dDM
n2
22
2
2
e
n2
2
2
2
er
(G-6)
Let
r (G-7)
/r (G-8)
drd (G-9)
dr
d
d
d
dr
d
(G-10)
d
d
dr
d (G-11)
2
22
2
2
d
d
dr
d
(G-12)
38
)ncos(DIn
nd
d1
d
dD
)ncos(CJn
d
d1
d
dDM
n2
22
2
22
e
n2
2
2
22
er
(G-13)
Now consider the shear boundary condition.
0Z
r
1
r1
r
1Z
r
1
rr
1
rrDV
2
2
22
2
er
(G-14)
Zrr
1
r
11
r
1D
Zrr
1
r
2
rr
1
rr
1
rDV
2
3
2
2
2e
2
3
22
2
32
2
23
3
er
(G-15)
)ncos(rIDrJCrr
1
r
11
r
1D
)ncos(rIDrJCrr
1
r
2
rr
1
rr
1
rDV
nn2
3
2
2
2e
nn2
3
22
2
32
2
23
3
er
(G-16)
39
)ncos(rIDrr
1
r
11
r
1D
)ncos(rJCrr
1
r
11
r
1D
)ncos(rIDrr
1
r
2
rr
1
rr
1
rD
)ncos(rJCrr
1
r
2
rr
1
rr
1
rDV
n2
3
2
2
2e
n2
3
2
2
2e
n2
3
22
2
32
2
23
3
e
n2
3
22
2
32
2
23
3
er
(G-17)
)ncos(rIDrr
11
r
1D
)ncos(rJCrr
11
r
1D
)ncos(rIDrr
1
r
2
rr
1
rr
1
rD
)ncos(rJCrr
1
r
2
rr
1
rr
1
rDV
n2
3
2
2
2e
n2
3
2
2
2e
n2
3
22
2
32
2
23
3
e
n2
3
22
2
32
2
23
3
er
(G-18)
40
)ncos(rIDrr
11
r
1
rr
1
r
2
rr
1
rr
1
rD
)ncos(rJCrr
11
r
1
rr
1
r
2
rr
1
rr
1
rD
V
n2
3
2
2
22
3
22
2
32
2
23
3
e
n2
3
2
2
22
3
22
2
32
2
23
3
e
r
(G-19)
)ncos(rIDrr
11
r
1
rr
1
r
2
rr
1
rr
1
rD
)ncos(rJCrr
11
r
1
rr
1
r
2
rr
1
rr
1
rD
V
n2
3
2
2
22
3
22
2
32
2
23
3
e
n2
3
2
2
22
3
22
2
32
2
23
3
e
r
(G-20)
Let
r (G-21)
41
)ncos(ID1
111211
D
)ncos(JC1
111211
D
V
n2
3
2
2
22
3
22
2
32
2
23
33
e
n2
3
2
2
22
3
22
2
32
2
23
33
e
r
(G-22)
)ncos(ID1
1nnn211
D
)ncos(JC1
1nnn211
D
V
n2
2
2
2
3
2
2
2
23
33
e
n2
2
2
2
3
2
2
2
23
33
e
r
(G-23)
42
APPENDIX H
Fixed Plate, Bessel Function Solution
The boundary conditions are
0),a(Z (H-1)
0r
Z
at r = a (H-2)
The displacement is
)ncos(rIDrJC),r(Z nn (H-3)
The slope is
)ncos(rIDrJCrr
Znn
(H-4)
)ncos(rIrd
dDrJ
rd
dC
rd
dZnn
(H-5)
Let
r (H-6)
d
d
dr
d (H-7)
The displacement is
)ncos(IDJC),r(Z nn (H-8)
The slope is
)ncos(Id
dDJ
d
dC
rd
dZnn
(H-9)
43
Apply the boundary conditions
0)ncos(IDJC nn at a (H-10)
0IDJC nn at a (H-11)
0)ncos(Id
dDJ
d
dC nn
at a (H-12)
0Id
dDJ
d
dC nn
at a (H-13)
Assemble the equations in matrix form.
0
0
D
C
Id
dJ
d
d
IJ
nn
nn
at a (H-14)
The roots are found by setting the determinant of the coefficient matrix equal to zero.
0Jd
dII
d
dJ nnnn
at a (H-15)
0Jn
JIIn
IJ n1nnn1nn
at a (H-16)
0IJIJ n1n1nn at a (H-17)
44
The roots of equation (H-17) for 3.0 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 (H-18)
e
24
D
h (H-19)
4/1
eD
h2
(H-20)
4/1
e
2
D
ha
(H-21)
4e
42
ah
D
(H-22)
h
D
a
e2
2
(H-23)
h
D
a
10.2158 e2
(H-24)
45
The mode shapes are defined by
)ncos(rnIDrnJC),r(Z (H-25)
Recall
nI
nJCD , at a (H-26)
)ncos(rnInI
nJCrnJC),r(Z
(H-27)
)ncos(rnInI
nJrnJC),r(Z
(H-28)
Let
r (H-29)
)ncos(rII
JJC),(Z n
n
nn
(H-30)