Natural Frequencies of Circular Plate Bending Modes
29
1 NATURAL FREQUENCIES OF CIRCULAR PLATE BENDING MODES Revision A By Tom Irvine Email: [email protected] August 6, 2009 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. θ r Y X 0
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Transcript of Natural Frequencies of Circular Plate Bending Modes
1
NATURAL FREQUENCIES OF CIRCULAR PLATE BENDING MODES
Revision A
By Tom Irvine
Email: [email protected]
August 6, 2009
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.
θ
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
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
( )
( ) θ
θ∂
∂
∂
∂µ−+
π
θ∂
∂+
∂
∂
∂
∂µ−−
θ∂
∂+
∂
∂+
∂
∂= ∫ ∫
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
µ−
=2112
3EheD (A-2)
where
E = elastic modulus
h = plate thickness
µ = Poisson's ratio
For a displacement which is symmetric about the center,
0),r(Z =θθ∂
∂ (A-3)
0),r(Z2
2
=θθ∂
∂ (A-4)
Substitute equations (A-3) and (A-4) into (A-1).
4
( )∫ ∫π
θ
∂
∂
∂
∂µ−−
∂
∂+
∂
∂=
2
0
R
0ddrr
r
Z
r
1
r2
Z212
2
r
Z
r
1
2r
Z2
2
eDV
(A-5)
( )∫ ∫π
θ
∂
∂
∂
∂µ+−+
∂
∂+
∂
∂
∂
∂+
∂
∂=
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
∫ ∫π
θ
∂
∂
∂
∂µ+
∂
∂+
∂
∂=
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
∫ ∫π
θΩρ
=2
0
R
0
22
ddrrZ2
hT (A-8)
where
ρ = mass per volume
Ω = 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 =θ (B-1)
0r
Z
ar2
2
=∂
∂
=
(B-2)
The following function satisfies the geometric boundary conditions.
π=θ
a2
rcosZ),r(Z o (B-3)
The partial derivatives are
0),r(Z =θθ∂
∂ (B-4)
0),r(Z2
2
=θθ∂
∂ (B-5)
π
π−=θ
∂
∂
a2
rsin
a2Z),r(Z
ro (B-6)
π
π−=θ
∂
∂
a2
rcos
a2Z),r(Z
r
2
o2
2
(B-7)
6
The total kinetic energy T of the plate bending is given by
∫ ∫π
θ
πΩρ=
2
0
a
0
2
o
2
ddrra2
rcosZ
2
hT (B-8)
∫ ∫π
θ
π+
Ωρ=
2
0
a
0
2o
2
ddrra
rcos1
4
ZhT (B-9)
∫ ∫π
θ
π+
Ωρ=
2
0
a
0
2o
2
ddra
rcosrr
4
ZhT (B-10)
Evaluate equation (B-9) using the integral table in Appendix C
∫π
θ
π
π+
π
π+
Ωρ=
2
0
a
02
222o
2
da
rcos
a
a
rsin
ra
2
r
4
ZhT (B-11)
( ) ( )∫π
θ
π−π
π+
Ωρ=
2
0 2
2
2
222o
2
d0cosa
cosa
2
a
4
ZhT (B-12)
∫π
θ
π−
Ωρ=
2
0 2
222o
2
da2
2
a
4
ZhT (B-13)
[ ]∫π
θ−ππ
Ωρ=
2
0
2
2
22o
2
d48
aZhT (B-14)
[ ]∫π
θ−ππ
Ωρ=
2
0
2
2
22o
2
d48
aZhT (B-15)
[ ][ ]π−ππ
Ωρ= 24
8
aZhT 2
2
22o
2
(B-16)
7
[ ]44
aZhT 2
22o
2
−ππ
Ωρ= (B-17)
( ) 22
o
2 aZh4671.0T Ωρ= (B-18)
Again, the total strain energy for the symmetric case is
∫ ∫π
θ
∂
∂
∂
∂µ+
∂
∂+
∂
∂=
2
0
R
0ddrr
r
Z
r
1
r2
Z22
2
r
Z
r
12
2r
Z2
2
eDV
(B-19)
∫ ∫
∫ ∫
∫ ∫
πθ
∂
∂
∂
∂µ+
πθ
∂
∂+
πθ
∂
∂=
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)
( )∫ ∫
∫ ∫
∫ ∫
π
θ
π
π
πµ+
π
θ
π
π+
π
θ
π
π+=
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
( )∫ ∫
∫ ∫
∫ ∫
π
π
π
θ
π
π
πµ+
θ
π
π+
θ
π
π+=
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)
( ) ∫ ∫
∫ ∫
∫ ∫
π
π
π
θ
π
π
πµ+
θ
π
π+
θ
π
π+=
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)
∫ ∫
∫ ∫
∫ ∫
π
π
π
θ
π
πµ+
θ
π
π+
θ
π+
π+=
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
∫
∫
∫
π
π
π
θ
π
π
πµ−
θ
π+
θ
π
π+
π
π+
π+=
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)
∫
∫
∫
π
π
π
θ
π
πµ+
θ
π+
θ
π−
π+=
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)
10
∫
∫
∫
π
π
π
θ
πµ+
θ
π+
θ
π−
π+=
2
0
22o
2
0
22o
2
0 2
2242o
da22
DZ
d8242.0a22
DZ
da2
2
a
2
1
a22
DZV
(B-27)
( )
( ) ( )
π
πµ+
π
π+
π
π−
π+=
2a22
DZ
28242.0a22
DZ
2a2
2
a
2
1
a22
DZV
22o
22o
2
2242o
(B-28)
( )
πµπ+
ππ+
π−
ππ+=
22
o
22
o
2
2242
o
a2DZ
8242.0a2
DZ
a2
2
a
2
1
a2DZV
(B-29)
11
( )
πµ+
π+
π−
ππ+=
2
a28242.0
2
a22
2a2
2
2a
2
14
a2D2
oZV
(B-30)
( )
µ+
+
π−
ππ+=
22
2
224232
oa2
18242.0
a2
1a2
2
a
2
1
a2
1DZV
(B-31)
( )
µ+
+
π−
ππ+=
222
22
4
232o
a4
18242.0
a4
1a2
2
a
2
1
a16
1DZV
(B-32)
( )
µ+
+
π−
ππ+=
2222
232o
a4
18242.0
a4
12
2
1
2
1
a16
1DZV
(B-33)
( )
µ+
+
π−
π
π+=4
8242.04
1
2
2
2
1
2
1
16
122a
13D2oZV
(B-34)
( )
µ+
+
π−
π
π+=4
8242.04
12
2
1
32
1
a
1DZV
2
2
2
32o
(B-35)
( )
µ+
+
−
π
π+=4
8242.04
12
232
1
a
1DZV
2
2
32o
(B-36)
12
[ ] ( )
µ+
+−π
π+=4
8242.04
14
64
1
a
1DZV 2
2
32o
(B-37)
[ ] ( ) µ++−π
π+= 168242.0164a64
1DZV 2
2
32o
(B-38)
[ ] ( ) µ++−π
π+= 168242.0164a64
1DZV 2
2
32o
(B-39)
µ+
π+= 160568.19a64
1DZV
2
32o
(B-40)
µ+
π+= 25.02978.0a
1DZV
2
32o
(B-41)
Now equate the total kinetic energy with the total strain energy per Rayleigh's method.
[ ] µ+
π+=−ππ
Ωρ25.02978.0
a
1DZ4
4
aZh
2
32o
222
o2
(B-42)
[ ] µ+
π=−ππ
Ωρ25.02978.0
a
1D4
4
ah
2
3222
(B-43)
[ ] µ+
π=−ππ
Ωρ25.02978.0
a
1D4
4
h
4
322
(B-44)
[ ] µ+
π=−πΩρ
25.02978.0a
1D4
4
h
4
422
(B-45)
13
[ ] µ+
π=−πΩρ 25.02978.0a
4D4h
4
422 (B-46)
[ ] µ+
−πρ
π=Ω 25.02978.04h
1
a
4D
24
42 (B-47)
[ ] µ+
−πρ
π=Ω 1911.14h
1
a
1D
24
42 (B-48)
Let µ = 0.3, which is the typical Poisson's ratio.
[ ] 3.01911.14h
1
a
1D
24
42 +
−πρ
π=Ω (B-49)
[ ] 4911.14h
1
a
1D
24
42
−πρ
π=Ω (B-50)
[ ] 4911.14h
1
a
1D
24
42
−πρ
π=Ω (B-51)
[ ] 4911.14h
1
a
1D
24
4
−πρ
π=Ω (B-52)
h
D
a
9744.4
2 ρ=Ω (B-53)
The natural frequency fn is
Ωπ
=2
1fn (B-54)
h
eD
2a2
9744.4nf
ρπ= (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∫∫
π−=
π (C-2)
Nondimensionalize,
a
rx
π= (C-3)
rxa
=π
(C-4)
dra
dxπ
= (C-5)
drdxa
=π
(C-6)
[ ] dxxcos1x
1
a2
adr
a2
rsin
r
1
0
a
0
2∫∫
π−
π
π=
π (C-7)
[ ] dxxcos1x
1
2
1dr
a2
rsin
r
1
0
a
0
2∫∫
π−=
π (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 ∫∫π
+−+−+−−≈
π
(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 ∫∫π
−+−+−≈
π
(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 ∫∫π
−+−+−≈
π
(C-12)
π
⋅−
⋅+
⋅−
⋅+
⋅−
⋅≈
π∫
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 ≈
π∫ (C-14)
8242.0dra2
rsin
r
1a
0
2 ≈
π∫ (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 =θβ−θ∇ (D-1)
eD
h24 ρω
=β (D-2)
4/1
eD
h2
ρω=β (D-2)
µ−
=2112
3EheD (D-3)
θ∂
∂+
∂
∂+
∂
∂
θ∂
∂+
∂
∂+
∂
∂=∇∇=∇
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 =θβ−∇β+∇ (D-5)
Thus the equation is satisfy by
0),r(Z22 =θβ±∇ (D-6)
Separate variables
)()r(R),r(Z θΘ=θ (D-7)
17
By substitution
0)()r(R22
2
2r
1
rr
1
2r
2=θΘ
β+
θ∂
∂+
∂
∂+
∂
∂ (D-8)
0R2R2
2
2r
1R
rr
1R
2r
2=Θβ+
Θ
θ∂
∂+Θ
∂
∂+Θ
∂
∂ (D-9)
0R22d
2d
2r
R
dr
dR
r
1
2dr
R2d=Θβ+
θ
Θ+Θ+Θ (D-10)
Similarly,
0R22d
2d
2r
R
dr
dR
r
1
2dr
R2d=Θβ−
θ
Θ+Θ+Θ (D-11)
Thus,
0R22d
2d
2r
R
dr
dR
r
1
2dr
R2d=Θβ±
θ
Θ+Θ+Θ (D-12)
2d
2d
2r
RR2
dr
dR
r
1
2dr
R2d
θ
Θ−=Θβ±
Θ+Θ (D-13)
2d
2d
2r
12
dr
dR
rR
1
2dr
R2d
R
1
θ
Θ
Θ−=β±
+ (D-14)
2d
2d12r2
dr
dR
r
1
2dr
R2d
R
2r
θ
Θ
Θ−=β±
+ (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=
θ
Θ
Θ−=β±
+ (D-16)
Thus
2k2r2
dr
dR
r
1
2dr
R2d
R
2r=β±
+ (D-17)
0R2r
2k2
dr
dR
r
1
2dr
R2d=
−β±++ (D-18)
Define a new variable
β
β=ξ
rj
r (D-19)
β−
β=ξ
2r2
2r22 (D-20)
β−
β=
ξ2
2
2r
2 (D-21)
Chain rule
drd β=ξ (D-22)
dr
d
d
d
dr
d ξ
ξ= (D-23)
ξβ=
d
d
dr
d (D-24)
19
0R2r
2k
2r
2
d
dR
2r2d
R2d
2r
2=
−
ξ+
ξ
ξ+
ξ
ξ (D-25)
0R2k2
d
dR
2d
R2d2 =
−ξ+
ξξ+
ξξ (D-26)
0R2
2k1
d
dR1
2d
R2d=
ξ−+
ξξ+
ξ (D-27)
Equation (D-27) is Bessel’s equation of fractional order.
The solution for circular plates that are closed in the θ direction is
( ) ( ) ( ) ( )ξ+ξ+ξ+ξ=ξ 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 ( )ξnY and ( )ξnK are singular at ξ = 0.
Thus for a plate with no hole, F = G = 0.
( ) ( )ξ+ξ=ξ nIDnJC)(R (D-29)
Furthermore, from equation (D-16),
2k2d
2d1=
θ
Θ
Θ− (D-30)
02k2d
2d=Θ+
θ
Θ (D-31)
20
The solution for circular plates that are closed in the θ direction is
θ+θ=Θ ksinBkcosA , k = n = 0, 1, 2, 3, ….. (D-32)
Or equivalently
[ ])(kcosA φ−θ=Θ .. (D-33)
The total solution is thus
( ) ( ) [ ] )(kcosAnIDnJC),(Z φ−θξ+ξ=θξ (D-34)
Set the phase angle φ =0.
( ) ( ) )kcos(AnIDnJC),(Z θξ+ξ=θξ (D-35)
Set A =1. Note that the mass normalization will be performed using the C and D
coefficients.
( ) ( ) )kcos(nIDnJC),(Z θξ+ξ=θξ (D-36)
21
APPENDIX E
Simply Supported Plate, Bessel Function Solution
The boundary conditions are
0),a(Z =θ (E-1)
0rM = at r = a (E-2)
02
Z2=
θ∂
∂ at r = a (E-3)
Note that
θ∂
∂+
∂
∂µ+
∂
∂−=
2
Z2
2r
1
r
Z
r
1
2r
Z2DrM (E-4)
Boundary condition (E-3) requires that
∂
∂µ+
∂
∂−=
r
Z
r2r
Z2DrM at r = a (E-5)
)()r(R),r(Z θΘ=θ (E-6)
( ) ( )[ ] )kcos(rnIDrnJC),r(Z θβ+β=θ (E-7)
( ) ( )[ ] 0)kcos(anIDanJC),a(Z =θβ+β=θ (E-8)
( ) ( ) 0anIDanJC =β+β (E-9)
22
( ) ( )[ ]
( ) ( )[ ] a r at )kcos(rnIDrnJCrr
1D
)kcos(rnIDrnJC2r
2DrM
=
θβ+β
∂
∂µ−
θβ+β
∂
∂−=
(E-10)
( ) ( )
( ) ( ) a r at rnIdr
d
r
1DrnJ
dr
d
r
1C)kcos(ED
rnI2dr
2dDrnJ
2dr
2dC)kcos(EDrM
=
β+βθµ−
β+βθ−=
(E-11)
0arrM =
= (E-12)
( ) ( ) ( ) ( ) 0rIdr
d
a
1DrJ
dr
d
a
1CrI
dr
dDrJ
dr
dC nnn2
2
n2
2
=
β+βµ+
β+β , at r = a
(E-13)
( ) ( ) ( ) ( ) 0rIdr
d
arI
dr
dDrJ
dr
d
arJ
dr
dC nn2
2
nn2
2
=
β
µ+β+
β
µ+β , at r = a
(E-14)
Let
rβ=λ (E-15)
23
( ) ( ) ( ) ( ) 0Id
d
aI
d
dDJ
d
d
aJ
d
dC nn2
22
nn2
22 =
λ
λ
µβ+λ
λβ+
λ
λ
µβ+λ
λβ ,
at aβ=λ
(E-36)
( ) ( ) ( ) ( ) 0Id
d
aI
d
dDJ
d
d
aJ
d
dC nn2
2
nn2
2
=
λ
λ
µ+λ
λβ+
λ
λ
µ+λ
λβ ,
at aβ=λ
(E-37)
Recall equation (E-9).
( ) ( ) 0nIDnJC =λ+λ , at aβ=λ (E-38)
( ) ( )λ−=λ nJCnID , at aβ=λ (E-39)
( )( )λ
λ−=
nI
nJCD , at aβ=λ (E-40)
By substitution,
( ) ( )( )( )
( ) ( ) 0Id
d
aI
d
d
I
JCJ
d
d
aJ
d
dC nn2
2
n
nnn2
2
=
λ
λ
µ+λ
λβ
λ
λ−
λ
λ
µ+λ
λβ ,
at aβ=λ
(E-41)
( ) ( )( )( )
( ) 0)(Id
d
aI
d
d
I
JJ
d
d
aJ
d
dnn2
2
n
nnn2
2
=
λ
λ
µ+λ
λβ
λ
λ−
λ
λ
µ+λ
λβ , at aβ=λ
(E-42)
24
( )( ) ( )
( )( ) ( ) 0I
d
d
aI
d
d
I
1J
d
d
aJ
d
d
J
1nn2
2
nnn2
2
n=
λ
λ
µ+λ
λβ
λ−
λ
λ
µ+λ
λβ
λ ,
at aβ=λ
(E-43)
Note the following identities:
( ) ( ) ( ) ( ) ( )λλ
+λ+−=λλ
−λ−=λλ
nJn
1nJnJn
1nJnJd
d (E-44)
( ) ( ) ( )λλλ
+λ+λ
−=λλ
nJd
dn1nJ
d
dnJ
2d
2d (E-45)
( ) ( ) ( ) ( ) ( )
λ
λ+λ+−
λ+
λ+
λ
++λ−=λ
λnJ
n1nJ
n1nJ
1nnJnJ
2d
2d (E-46)
( ) ( ) ( ) ( ) ( )
λ
λ+λ+−
λ+
λ+
λ
++λ−=λ
λnJ
n1nJ
n1nJ
1nnJnJ
2d
2d (E-47)
( ) ( ) ( )λ+
λ+λ
λ+−=λ
λ1nJ
1nJ
2
2n1nJ
2d
2d (E-48)
25
Analyze the first term of equation (E-42).
( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( )
( )( )
( )( ) λ
µ+
λ+−β+
λ
β−
µ
λ
λ+−=
λ
µ+
λ+−β+
λ
λ+µ−
λ
β
λ
λ+=
λ
λ+λ+−
λ
µ+
λ+
λβ+λ
λ+−β
λ=
λ
λ
µ+λ
λβ
λ+
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:
( ) ( ) ( ) ( ) ( )λλ
+λ+=λλ
−λ−=λλ
nIn
1nInIn
1nInId
d (E-50)
( ) ( ) ( )λλλ
+λ+λ
=λλ
nId
dn1nI
d
dnI
2d
2d (E-51)
( ) ( ) ( ) ( ) ( )
λ
λ+λ+
λ+
λ+
λ
+−λ=λ
λnI
n1nI
n1nI
1nnInI
2d
2d (E-52)
( ) ( ) ( )λ+
λ−λ
λ+=λ
λ1nI
1nI
2
2n1nI
2d
2d (E-53)
26
Analyze the second term of equation (E-42).
( )( ) ( )
( )( ) ( )
( )( ) ( )
( )( )
( )( )
( )( ) λ
µ−
λ+β−
λ
λ+
λ
β−
µ−=
λ
µ−
λ+β−
λ
λ+µ−
λ
λ+λ
β=
λ
λ+λ+
λ
µ−
λ+
λβ−λ
λ+β
λ−=
λ
λ
µ+λ
λβ
λ−
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
=
λ
λ
µ
λ−
λ
λλ−
λ
λ
µ
λ+
λ
λλ+
at aβ=λ
(E-55)
( )( )
( )( )
0a
n
2
2n1
nI
1nI
aa
n
2
2n1
anJ
1nJ=
λ
µ−
λ+β−
λ
λ+
λ
β−
µ−
λ
µ+
λ+−β+
λ
β−
µ
λ
λ+− ,
at aβ=λ
(E-56)
27
( )( )
( )( )
02nI
1nI
aanJ
1nJ=β−
λ
λ+
λ
β−
µ−
λ
β−
µ
λ
λ+− , at aβ=λ (E-57)
( )( )
( )( )
02nI
1nI
aanJ
1nJ=β+
λ
λ+
λ
β−
µ+
λ
β−
µ
λ
λ+ , at aβ=λ (E-58)
( )( )
( )( )
β−=λ
λ+
λ
β−
µ+
λ
β−
µ
λ
λ+ 2nI
1nI
aanJ
1nJ , at aβ=λ (E-59)
( )( )
( )( )
λ
β−
µ
β−=
λ
λ++λ
λ+
a
2
nI
1nI
nJ
1nJ , at aβ=λ (E-60)
( )( )
( )( )
−
µ
β−=
λ
λ++λ
λ+
a
1
a
2
nI
1nI
nJ
1nJ , at aβ=λ (E-61)
( )( )
( )( ) µ+−
β−=
λ
λ++λ
λ+1
a2
nI
1nI
nJ
1nJ , at aβ=λ (E-62)
( )( )
( )( ) µ+−
λ−=
λ
λ++λ
λ+1
2
nI
1nI
nJ
1nJ , at aβ=λ (E-63)
( )( )
( )( ) µ−
λ=
λ
λ++λ
λ+1
2
nI
1nI
nJ
1nJ , at aβ=λ (E-64)
The following form is better suited for numerical root-finding purposes.
( ) ( ) ( ) ( ) ( ) ( )[ ]λλµ−
λ=λλ+λλ ++ nn1nn1nn IJ
1
2IJJI , at aβ=λ (E-65)
28
The roots of equation (E-65) for 3.0=µ are
k n=0 n=1 n=2 n=3
0 4.9351 13.8982 25.6133 39.9573
1 29.9844 48.7391 70.1170 95.2930
2 74.9211 103.4057 134.4289 168.8374
3 138.7787 176.8456 218.2026 264.0849
The roots were determined using the secant method.
The fundamental natural frequency is thus
aβ=λ (E-66)
e
24
D
hρω=β (E-67)
4/1
e
2
D
ha
ρω=λ (E-68)
4e
42
ah
D
ρ
λ=ω (E-69)
h
D
a
e2
2
ρ
λ=ω (E-70)
h
D
a
4.9351 e2 ρ
=ω (E-71)
The mode shapes are defined by
( ) ( )[ ] )kcos(rIDrJC),r(Z nn θβ+β=θ (E-72)
29
Recall
( )( )λ
λ−=
n
n
I
JCD , at aβ=λ (E-73)
( )( )( )
( ) )kcos(rII
JCrJC),r(Z n
n
nn θ
λ
λ−+=θ (E-74)
( )( )( )
( ) )kcos(rII
JrJC),r(Z n
n
nn θ
λ
λ−=θ (E-75)
