natural frequencies of circular plate bending modes

NATURAL FREQUENCIES OF CIRCULAR PLATE BENDING MODES

Revision A

By Tom Irvine

Email: tomirvine@aol.com

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.

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,

APPENDIX A

The total strain energy V of the plate is

( ) θ

∂µ−+

∂µ−−

∂= ∫ ∫

Note that the plate stiffness factor De is given by

3EheD (A-2)

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

=θθ∂

∂ (A-4)

Substitute equations (A-3) and (A-4) into (A-1).

( )∫ ∫π

∂µ−−

( )∫ ∫π

∂µ+−+

The total strain energy equation for the symmetric case is thus

∫ ∫π

∂µ+

The total kinetic energy T of the plate bending is given by

∫ ∫π

θΩρ

ddrrZ2

hT (A-8)

ρ = mass per volume

Ω = angular natural frequency

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)

The following function satisfies the geometric boundary conditions.

rcosZ),r(Z o (B-3)

The partial derivatives are

0),r(Z =θθ∂

∂ (B-4)

0),r(Z2

=θθ∂

∂ (B-5)

π−=θ

a2Z),r(Z

ro (B-6)

π−=θ

a2Z),r(Z

The total kinetic energy T of the plate bending is given by

∫ ∫π

πΩρ=

ddrra2

hT (B-8)

∫ ∫π

ZhT (B-9)

∫ ∫π

rcosrr

ZhT (B-10)

Evaluate equation (B-9) using the integral table in Appendix C

ZhT (B-11)

( ) ( )∫π

π−π

d0cosa

ZhT (B-12)

ZhT (B-13)

[ ]∫π

θ−ππ

aZhT (B-14)

[ ]∫π

θ−ππ

aZhT (B-15)

[ ][ ]π−ππ

Ωρ= 24

aZhT 2

(B-16)

aZhT 2

−ππ

Ωρ= (B-17)

( ) 22

2 aZh4671.0T Ωρ= (B-18)

Again, the total strain energy for the symmetric case is

∫ ∫π

∂µ+

(B-19)

∫ ∫

∂µ+

(B-20)

( )∫ ∫

∫ ∫

r2cos4

(B-21)

( )∫ ∫

∫ ∫

ddrra2

(B-22)

( ) ∫ ∫

∫ ∫

ddrra2

(B-23)

∫ ∫

(B-24)

The first and third integrals are evaluating using the tables in Appendix C.

πµ−

d8242.0a22

(B-25)

d8242.0a22

(B-26)

d8242.0a22

(B-27)

( ) ( )

28242.0a22

(B-28)

πµπ+

ππ+=

8242.0a2

(B-29)

ππ+=

a28242.0

(B-30)

ππ+=

224232

18242.0

(B-31)

ππ+=

18242.0

(B-32)

ππ+=

18242.0

(B-33)

8242.04

13D2oZV

(B-34)

8242.04

(B-35)

8242.04

(B-36)

[ ] ( )

+−π

8242.04

1DZV 2

(B-37)

[ ] ( ) µ++−π

π+= 168242.0164a64

1DZV 2

(B-38)

[ ] ( ) µ++−π

π+= 168242.0164a64

1DZV 2

(B-39)

π+= 160568.19a64

(B-40)

π+= 25.02978.0a

(B-41)

Now equate the total kinetic energy with the total strain energy per Rayleigh's method.

[ ] µ+

π+=−ππ

Ωρ25.02978.0

(B-42)

[ ] µ+

π=−ππ

Ωρ25.02978.0

(B-43)

[ ] µ+

π=−ππ

Ωρ25.02978.0

(B-44)

[ ] µ+

π=−πΩρ

25.02978.0a

(B-45)

[ ] µ+

π=−πΩρ 25.02978.0a

422 (B-46)

[ ] µ+

−πρ

π=Ω 25.02978.04h

42 (B-47)

[ ] µ+

−πρ

π=Ω 1911.14h

42 (B-48)

Let µ = 0.3, which is the typical Poisson's ratio.

[ ] 3.01911.14h

−πρ

π=Ω (B-49)

[ ] 4911.14h

−πρ

π=Ω (B-50)

[ ] 4911.14h

−πρ

π=Ω (B-51)

[ ] 4911.14h

−πρ

π=Ω (B-52)

9744.4

2 ρ=Ω (B-53)

The natural frequency fn is

1fn (B-54)

9744.4nf

ρπ= (B-55)

APPENDIX C

Integral Table

Equation (C-1) is taken from Reference 1.

bxsinxdxbxcosx +=∫ (C-1)

Now consider

2∫∫

π−=

π (C-2)

Nondimensionalize,

π= (C-3)

= (C-5)

[ ] dxxcos1x

2∫∫

π (C-7)

[ ] dxxcos1x

2∫∫

π−=

π (C-8)

Recall the series

x1xcos

12108642

+−+−+−≈ (C-9)

12108642a

2 ∫∫π

+−+−+−−≈

(C-10)

12108642a

2 ∫∫π

−+−+−≈

(C-11)

1197531a

2 ∫∫π

−+−+−≈

(C-12)

⋅−

⋅≈

12108642a

(C-13)

( )6483.12

π∫ (C-14)

8242.0dra2

π∫ (C-15)

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)

h24 ρω

=β (D-2)

ρω=β (D-2)

3EheD (D-3)

∂=∇∇=∇

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)

By substitution

0)()r(R22

2=θΘ

∂ (D-8)

2=Θβ+

∂+Θ

∂ (D-9)

R2d=Θβ+

Θ+Θ+Θ (D-10)

Similarly,

R2d=Θβ−

Θ+Θ+Θ (D-11)

R2d=Θβ±

Θ+Θ+Θ (D-12)

Θ−=Θβ±

Θ+Θ (D-13)

Θ−=β±

+ (D-14)

2d12r2

Θ−=β±

+ (D-15)

The equation can be satisfied if each expression is equal to the same constant .2k

2d12r2

Θ−=β±

+ (D-16)

2r=β±

+ (D-17)

−β±++ (D-18)

Define a new variable

r (D-19)

2r22 (D-20)

2 (D-21)

Chain rule

drd β=ξ (D-22)

ξ= (D-23)

d (D-24)

ξ (D-25)

R2d2 =

−ξ+

ξξ (D-26)

ξ−+

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),

Θ− (D-30)

2d=Θ+

Θ (D-31)

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)

APPENDIX E

Simply Supported Plate, Bessel Function Solution

The boundary conditions are

0),a(Z =θ (E-1)

0rM = at r = a (E-2)

∂ at r = a (E-3)

Note that

∂µ+

∂−=

Z2DrM (E-4)

Boundary condition (E-3) requires that

∂µ+

∂−=

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)

( ) ( )[ ]

( ) ( )[ ] a r at )kcos(rnIDrnJCrr

)kcos(rnIDrnJC2r

θβ+β

∂µ−

θβ+β

∂−=

(E-10)

( ) ( )

( ) ( ) a r at rnIdr

1C)kcos(ED

rnI2dr

2dDrnJ

2dC)kcos(EDrM

β+βθµ−

β+βθ−=

(E-11)

0arrM =

= (E-12)

( ) ( ) ( ) ( ) 0rIdr

dC nnn2

β+βµ+

β+β , at r = a

(E-13)

( ) ( ) ( ) ( ) 0rIdr

dC nn2

µ+β+

µ+β , at r = a

(E-14)

rβ=λ (E-15)

( ) ( ) ( ) ( ) 0Id

dC nn2

µβ+λ

λβ ,

at aβ=λ

(E-36)

( ) ( ) ( ) ( ) 0Id

dC nn2

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−=

nJCD , at aβ=λ (E-40)

By substitution,

( ) ( )( )( )

( ) ( ) 0Id

dC nn2

λβ ,

at aβ=λ

(E-41)

( ) ( )( )( )

( ) 0)(Id

λβ , at aβ=λ

(E-42)

( )( ) ( )

( )( ) ( ) 0I

at aβ=λ

(E-43)

Note the following identities:

( ) ( ) ( ) ( ) ( )λλ

+λ+−=λλ

−λ−=λλ

1nJnJn

1nJnJd

d (E-44)

( ) ( ) ( )λλλ

+λ+λ

−=λλ

2d (E-45)

( ) ( ) ( ) ( ) ( )

λ+λ+−

++λ−=λ

1nnJnJ

2d (E-46)

( ) ( ) ( ) ( ) ( )

λ+λ+−

++λ−=λ

1nnJnJ

2d (E-47)

( ) ( ) ( )λ+

λ+−=λ

2d (E-48)

Analyze the first term of equation (E-42).

( )( ) ( )

( )( )

( )( ) λ

λ+−β+

λ+−=

λ+−β+

λ+µ−

λ+λ+−

λβ+λ

λ+−β

(E-49)

Consider the following identities:

( ) ( ) ( ) ( ) ( )λλ

+λ+=λλ

−λ−=λλ

1nInIn

1nInId

d (E-50)

( ) ( ) ( )λλλ

+λ+λ

2d (E-51)

( ) ( ) ( ) ( ) ( )

λ+λ+

+−λ=λ

1nnInI

2d (E-52)

( ) ( ) ( )λ+

λ−λ

λ+=λ

2d (E-53)

Analyze the second term of equation (E-42).

( )( ) ( )

( )( )

( )( ) λ

λ+β−

µ−=

λ+β−

λ+µ−

λ+λ+

λβ−λ

λ−=

(E-54)

By substitution,

( )( )

( )( ) ,0I

λλ−

at aβ=λ

(E-55)

( )( )

λ+β−

λ+−β+

λ+− ,

at aβ=λ

(E-56)

( )( )

1nJ=β−

λ+− , at aβ=λ (E-57)

( )( )

1nJ=β+

λ+ , at aβ=λ (E-58)

( )( )

β−=λ

λ+ 2nI

1nJ , at aβ=λ (E-59)

( )( )

β−=

λ++λ

( )( )

β−=

λ++λ

( )( )

( )( ) µ+−

β−=

λ++λ

( )( )

( )( ) µ+−

λ−=

λ++λ

( )( )

( )( ) µ−

λ++λ

The following form is better suited for numerical root-finding purposes.

( ) ( ) ( ) ( ) ( ) ( )[ ]λλµ−

λ=λλ+λλ ++ nn1nn1nn IJ

2IJJI , at aβ=λ (E-65)

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)

hρω=β (E-67)

ρω=λ (E-68)

λ=ω (E-69)

λ=ω (E-70)

4.9351 e2 ρ

=ω (E-71)

The mode shapes are defined by

( ) ( )[ ] )kcos(rIDrJC),r(Z nn θβ+β=θ (E-72)

Recall

( )( )λ

λ−=

JCD , at aβ=λ (E-73)

( )( )( )

( ) )kcos(rII

JCrJC),r(Z n

λ−+=θ (E-74)

( )( )( )

( ) )kcos(rII

JrJC),r(Z n

λ−=θ (E-75)

natural frequencies of circular plate bending modes

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