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TRANSVERSE VIBRATION OF A BEAM VIA THE FINITE ELEMENT METHOD

Revision E

By Tom Irvine

Email: [email protected]

November 18, 2008

_______________________________________________________________________

Introduction

Many structures are too complex for analysis via classical method. Closed-form solutionsare thus unavailable for these structures.

For example, a structure may be composed of several different materials. Some of thematerials may be anisotropic. Furthermore, the structure may be an assembly of plates,

beams, and other components.

Consider three examples:

1. A circuit board has numerous chips, crystal oscillators, diodes, connectors,capacitors, jump wires, and other piece parts.

2. A large aircraft consisting of a fuselage, wing sections, tail section, engines, etc.3. A building has plies, foundation, beams, floor sections, and load-bearing walls.

The finite element method is a numerical method that can be used to analyze complexstructures, such as the three examples.

The purpose of this tutorial is to derive for a method for analyzing beam vibration usingthe finite element method. The method is based on Reference 1.

Theory

Consider a beam, such as the cantilever beam in Figure 1.

Figure 1.

where

E is the modulus of elasticity.

I is the area moment of inertia.

EI,

L

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L is the length.

is mass per length.

The product EI is the bending stiffness.

The vibration modes of the cantilever beam can be found by classical methods.Specifically, the fundamental frequency is

1

187510 2=

.

L

EI (1)

This problem presents a good opportunity to compare the accuracy of the finite element

method to the classical solution.

Let y(x,t) represent the displacement of the beam as a function of space and time.

The free, transverse vibration of the beam is governed by the equation:

( ) ( )2

2

2

2

2

2

t

)t,x(yx)t,x(y

x

dxEI

x

=

(2)

Equation (2) neglects rotary inertia and shear deformation. Note that it is also independentof the boundary conditions, which are applied as constraint equations.

Assume that the solution of equation (1) is separable in time and space.

)t(f)x(Y)t,x(y = (3)

( ) ( )2

2

2

2

2

2

t

)t(f)x(Yx)t(f)x(Y

xxEI

x

=

(4a)

( ) ( )2

2

2

2

2

2

t

)t(fx)x(Y)x(Y

xxEI

x)t(f

=

(4b)

The partial derivatives change to ordinary derivatives.

( )2

2

2

2

2

2

td

)t(fdx)x(Y)x(Y

dx

dEI

dx

d)t(f =

(5)

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( ) 2

2

2

2

2

2

td

)t(fd

)t(f

1)x(Y

dx

dEI

dx

d

x)x(Y

1=

(6)

The left-hand side of equation (6) depends on x only. The right hand side depends on t

only. Both x and t are independent variables. Thus equation (6) only has a solution if both

sides are constant. Let 2 be the constant.

( )2

2

2

2

2

2

2

td

)t(fd

)t(f

1)x(Y

dx

dEI

dx

d

x)x(Y

1==

(7)

Equation (7) yields two independent equations.

0)x(Y)x()x(Ydxd)x(EI

dxd 22

2

2

2

=

(8)

0)t(f)t(ftd

d 22

2

=+ (9)

Equation (8) is a homogeneous, forth order, ordinary differential equation.

The weighted residual method is applied to equation (8). This method is suitable forboundary value problems. An alternative method would be the energy method. The

energy method is introduced in Appendix A.

There are numerous techniques for applying the weighted residual method. Specifically,

the Galerkin approach is used in this tutorial.

The differential equation (8) is multiplied by a test function )x( . Note that the testfunction )x( must satisfy the homogeneous essential boundary conditions. The essentialboundary conditions are the prescribed values of Y and its first derivative.

The test function is not required to satisfy the differential equation, however.

The product of the test function and the differential equation is integrated over the domain.

The integral is set equation to zero.

0dx)x(Y)x()x(Ydx

d)x(EI

dx

d)x(

22

2

2

2

=

(10)

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The test function )x( can be regarded as a virtual displacement. The differential equationin the brackets represents an internal force. This term is also regarded as the residual.

Thus, the integral represents virtual work, which should vanish at the equilibrium

condition.

Define the domain over the limits from a to b. These limits represent the boundary pointsof the entire beam.

0dx)x(Y)x()x(Ydx

d)x(EI

dx

d)x(

b

a

22

2

2

2

=

(11)

{ } 0dx)x(Y)x()x(dx)x(Ydx

d)x(EI

dx

d)x(

b

a

2b

a 2

2

2

2

=

(12)

Integrate the first integral by parts.

{ } 0dx)x(Y)x()x(

dx)x(Ydx

d)x(EI

dx

d)x(

dx

ddx)x(Y

dx

d)x(EI

dx

d)x(

dx

d

ba

2

b

a 2

2b

a 2

2

=

(13)

{ } 0dx)x(Y)x()x(

dx)x(Ydx

d)x(EI

dx

d)x(

dx

d)x(Y

dx

d)x(EI

dx

d)x(

b

a

2

b

a 2

2b

a

2

2

=

(14)

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{ } 0dx)x(Y)x()x(

dx)x(Ydx

d)x(EI

dx

d)x(

dx

d)x(Y

dx

d)x(EI

dx

d)x(

b

a

2

b

a 2

2b

a

2

2

=

(15)

Integrate by parts again.


d)x(EI)x(

dx

d

dx)x(Ydx

d)x(EI)x(

dx

d

dx

d)x(Y

dx

d)x(EI

dx

d)x(

b

a

2b

a 2

2

2

2

b

a 2

2b

a

2

2

=

+

(16)


d)x(EI)x(

dx

d

)x(Y

dx

d)x(EI)x(

dx

d)x(Y

dx

d)x(EI

dx

d)x(

b

a

2b

a 2

2

2

2

b

a

2

2b

a

2

2

=

+

(17)

The essential boundary conditions for a cantilever beam are

0)a(Y = (18)

0dx

dY

ax

==

(19)

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Thus, the test functions must satisfy

0)a( = (20)

0dx

d

ax

=

= (21)

The natural boundary conditions are

0)x(Ydx

d)x(EI

dx

d

bx2

2

=

=

(22)

0)x(Ydx

d)x(EI

bx2

2

=

= (23)

Equation (23) requires

0)x(dx

d)x(EI

bx2

2

=

=

(24)

Apply equations (20), (21), and (24) to equation (17). The result is


d)x(EI)x(

dx

d b

a

2b

a 2

2

2

2

=

(25)

Note that equation (25) would also be obtained for other simple boundary condition cases.

Now consider that the beam consists of number of segments, or elements. The elementsare arranged geometrically in series form.

Furthermore, the endpoints of each element are called nodes.

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Figure 2.

Represent each L coefficient in terms of a cubic polynomial.

3

4i

2

3i2i1ii ccccL +++= , i =1, 2, 3, 4

(32)

{ }{ }{ } 10,hcccc

ycccc

hcccc

ycccc)(Y

j3

442

342414

j3

432

332313

1j3

422

232212

1j3

412

312111

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{ }{ }

{ } 10,hc3c2cyc3c2c

hc3c2c

yc3c2c)('Y

j2

443424

j2

433323

1j2

422322

1j2

413121

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The displacement equations becomes

{ }

{ }{ } 10,hccc

yccc1

hccc

yccc)(Y

j3

442

3424

j34323323

1j3

422

2322

1j3

412

3121

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The displacement equations becomes

{ }{ }{ } 10,hcc

ycc1

hcc

ycc)(Y

j3

442

34

j3

432

33

1j

3

42

2

23

1j3

412

31

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0cc1 4333 =++ (60)

4333 c1c = (61)

0cc1 4434 =++ (62)

4434 c1c = (63)

The displacement equation becomes

[ ]

[ ]{ }[ ]{ }

[ ]{ } 10,hcc1ycc11

hcc

ycc1)(Y

j3

442

44

j3

432

43

1j3

422

42

1j3

412

41

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The slope equation becomes

{ }

[ ]{ }[ ]{ } j4444j4343

1j4242

1j4141

hc3c121yc3c12

hc3c2

yc3c12)1('Y

+++++

++

++=

(66)

Boundary condition (38) requires

{ }

[ ]{ }[ ]{ } 1jj4444j4343

1j4242

1j4141

hhc3c121

yc3c12

hc3c2

yc3c12

=+++++

++

++

(67)

{ }[ ]{ }

[ ]{ } 1jj4444j4343

1j4242

1j4141

hhc3c221

yc3c22

hc3c2

yc3c22

=+++

++

++

++

(68)

0c2 41 =+ (69)

2c 41 = (70)

1c 42 = (71)

0c2 43 =+ (72)

2c 43 = (73)

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0c1 44=+ (74)

1c44 = (75)

The displacement equation becomes

( )[ ]

( )[ ]{ }[ ]{ }

( )[ ] ( ){ } 10,h111y2211

h11

y221)(Y

j32

j32

1j32

1j32

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{ } { }

10,h/xj,jhxh)1j(

,h2y231

hy23)x(Y

j32

j32

1j32

1j32

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[ ]Tjj1j1j hyhya = (90)

The derivative terms are

h/xj,jhxh)1j(,a'Lh

1)x(Y

dx

d T =

= (91)

h/xj,jhxh)1j(,a"Lh

1)x(Y

dx

d T22

2

=

= (92)

Note that primes indicate derivatives with respect to .

In summary.

+

+

=

32

32

32

32

2

231

23

L

(90)

+

+

=

2

2

2

2

341

66

3266

'L

(91)

+

=

64

126

62

126

"L

(92)

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Recall

0dx)x(Y)x(dx)x(Ydx

d)x(

dx

dEI 2

2

2

2

2

=

(93)

The essence of the Galerkin method is that the test function is chosen as

)x(Y)x( = (94)

Thus

[ ] 0dx)x(Ydx)x(Ydx

d)x(Y

dx

dEI

22

2

2

2

2

=

(95)

Change the integration variable using equation (79b). Also, apply the integration limits.

[ ] 0d)x(Yhd)x(Ydx

d)x(Y

dx

dEIh

1

0

221

0 2

2

2

2

=

(96)

[ ][ ] 0daLaLh

da"Lh

1a"L

h

1EIh

1

0

TT2

1

0

T2

T2

=

(97)

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[ ] [ ]{ }

[ ][ ] 0daLaLh

da"La"LEIh

1

1

0TT2

1

0

TT3

=

(98)

[ ] [ ]{ }

[ ][ ] 0daLLah

da"L"LaEIh

1

1

0

TT2

1

0

TT3

=

(99)

{ } { } 0daLLahda"L"LaEIh

1 1

0

TT21

0

TT3

=

(100)

{ } { } 0adLLhd"L"LhEIa1

0T2

1

0T

3T =

(101)

{ } { } 0dLLhd"L"Lh

EI 1

0

T21

0

T3

=

(102)

For a system of n elements,

n,...,2,1j,0MK j2

j == (103)

where

{ }

= d"L"L

h

EIK

1

0

T3j

(104)

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( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )

( )( ) ( )( )( )( )

+++

+

+

=

6464

64126126126

6462126626262

6412612612662126126126

"L"LT

(109)

+

++

+++

++++

=

2

22

222

2222

T

364816

72842414414436

3636872601236244

7284241441443672601214414436

"L"L

(110)

+

++

+++

++++

=

d364816

72842414414436

3636872601236244

7284241441443672601214414436

h

EI

K

1

0

2

22

222

2222

3

j

(111)

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1

032

3232

323232

32323232

3

j

122416

244224487236

1218824301212124244224487236243012487236

h

EI

K

+

++

+++++++

=

(112)

+

++

+++

++++

=

122416

244224487236

1218824301212124

244224487236243012487236

h

EI

K

3

j

(113)

=

4

612

264

612612

h

EIK

3j

(114)

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32323232

32

32

32

32

T

223123

2

231

23

LL

++

+

+

=

(115)

( ) ( )( ) ( )( ) ( )( )

( ) ( )( ) ( )( )( ) ( )( )( )

+

+++++

++

=

232

3232232

32323232232

323232323232232

T

2

22312312231

223231232323

LL

(116)

=

44a

34a33a

24a23a22a

14a13a12a11a

LL T

(117)

654 412911a += (118)

65425312a += (119)

6543241292313a += (120)

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6543278314a ++= (121)

654 222a += (122)

65432 25323a += (123)

65433324a ++= (124)

65432412946133a +++= (125)

654322782234a +++= (126)

65432 46444a ++= (127)

Recall

{ } =1

0

Tj dLLhM

(128)

{ } += d4129h,M1

0

65411j

(129)

1

0

76511j

7

42

5

9h,M

+= (130)

+=7

42

5

9h,M 11j

(131)

=

3513h,M 11j (132)

=420

156h,M 11j

(133)

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{ } ++= d2783h,M1

0

654314j

(139)

1

0

765414j

72

67

58

43h,M

+

+

= (140)

+

+

=

7

2

6

7

5

8

4

3h,M 14j

(141)

=420

13h,M 14j

(142)

{ } += d2h,M1

0

65422j

(143)

1

0

76522j

7

1

3

1

5

1h,M

+

= (144)

+

=7

1

3

1

5

1h,M 22j

(145)

=105

1h,M 22j

(146)

=

420

4h,M 22j

(147)

{ } += d253h,M1

0

6543223j

(148)

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=420

156h,M 33j

(159)

{ } +++= d27822h,M1

0

6543234j

(160)

1

0

76543234j

7

2

6

7

5

8

2

1

3

2

2

1h,M

+

+

+

=

(161)

+

+

+

=

7

2

6

7

5

8

2

1

3

2

2

1h,M 34j

(162)

=420

22h,M 34j

(163)

{ } ++= d464h,M1

06543244j (164)

1

0

7654344j

7

1

3

2

5

6

3

1h,M

+

+

= (165)

+

+

=

7

1

3

2

5

61

3

1h,M 44j

(166)

=420

4h,M 44j

(167)

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Recall

{ }

= d"L"L

h

EIK

1

0

T3j

(168)

{ } =1

0

Tj dLLhM

(169)

=

4

612

264

612612

h

EIK

3j

(170)

=

4

22156

3134

135422156

420

hMj

(171)

Example 1

Model the cantilever beam in Figure 1 as a single element using the mass and stiffness

matrices in equations (170) and (171). The model consists of one element and two nodes

as shown in Figure 3.

Figure 3.

Note that h=L.

N1 N2

E1

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The eigen problem is

=

2

2

1

1

2

2

2

1

1

3

h

y

h

y

422313

221561354

313422

135422156

420

L

h

y

h

y

4626

612612

2646

612612

L

EI

(172)

=

2

2

1

1

2

2

1

1

h

y

h

y

422313

221561354

313422

135422156

h

y

h

y

4626

612612

2646

612612

(173)

where

24

EI420

L

= (174a)

= 4L

EI420

(174b)

The boundary conditions at node 1 are

0y1= (175)

01= (176)

The first two columns and the first two rows of each matrix in equation (173) can thus be

struck out.

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The resulting eigen equation is thus

=

2

2

2

2

h

y

422

22156

h

y

46

612

(177)The eigenvalues are found using the method in Reference 2.

=

8846.2

029715.0

2

1 (178a)

=

6984.1

1724.0

2

1 (178b)

The finite element results for the natural frequencies are thus

=

6984.1

1724.0

L

EI420

42

1 (179)

=

807.34

5331.3

L

EI

42

1 (180)

The finite element results are compared to the classical results in Table 1.

Table 1. Natural Frequency Comparison, 1 Element

Index

Finite ElementModel

EI

L4

ClassicalSolution

EI

L4

1 3.5331 3.5160

2 34.807 22.034

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The local stiffness matrix for element 2 is

( )

3

3

2

2

3

hy

h

y

4626612612

2646

612612

2/L

EI (182)

The local mass matrix for element 1 is

( )

2

2

1

1

h

y

h

y

422313

221561354

313422

135422156

420

2/L (183)

The local mass matrix for element 2 is

( )

3

3

2

2

h

y

h

y

422313

221561354

313422

135422156

420

2/L (184)

The global eigen problem assembled from the local matrices is

( )

( )

=

3

3

2

2

1

1

2

3

3

2

2

11

3

h

y

h

y

h

y

42231300

22156135400

31380313

135403121354

00313422

00135422156

420

2/L

h

y

h

y

h

y

462600

61261200

268026

612024612

002646

00612612

2/L

EI

(185)

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Again, the boundary conditions at node 1 are

0y1= (186)

01= (187)

The first two columns and the first two rows of each matrix in equation (185) can thus be

struck out.

( )

( )

=

3

3

2

2

2

3

3

2

2

3

h

y

h

y

422313

221561354

31380

13540312

420

2/L

h

y

h

y

4626

612612

2680

612024

2/L

EI

(188)

Let

( )

( )

=

3

2

2/L

EI

420

2/L

(189)

( )( ) 23

EI4202/L2/L

= (190)

24

EI6720

L

= (191a)

=

4L

EI6720 (191b)

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The eigen problem in equation (188) becomes

=

3

3

2

2

3

3

2

2

hy

h

y

422313221561354

31380

13540312

hy

h

y

4626612612

2680

612024

(192)

The eigenvalues are found using the method in Reference 2. Equation (192) yields four

eigenvalues.

=

0810.7

84056.0

073481.0

0018414.0

4

3

2

1

(193a)

=

6610.2

91682.0

27107.0

042912.0

4

3

2

1

(193b)

The finite element results for the first two natural frequencies are thus

=

27107.0

042912.0

L

EI6720

42

1 (194)

=

221.22

5177.3

L

EI

42

1 (195)

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Table 2. Natural Frequency Comparison, 2 Elements

Index

Finite Element

Model

EI

L4

Classical

Solution

EI

L4

1 3.5177 3.5160

2 22.221 22.034

Excellent agreement is obtained for the first two roots.

The next step would be to solve for the eigenvectors, which represent the mode shapes. Agreater number of elements would be required to obtain accurate mode shapes, however.

Example 3

Repeat example 1 with 16 elements. Let each element have equal length. The global

stiffness and mass matrix are omitted for brevity.

The eigenvalue scale factor is

= 4

4

LEI)16(420 (196)

The model yields 32 eigenvalues. The first four are

=

0.00053121

0.0001383

005-1.7639e

007-4.4913e

4

3

2

1

(197)

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36

=

0.023048

0.01176

0.0041999

0.00067017

4

3

2

1

(198)

=

0.023048

0.01176

0.0041999

0.00067017

L

EI)16(420

4

3 4

42

1

(199)

=

92.120

698.61035.22

5160.3

4L

EI

4

32

1

(200)


Table 3. Natural Frequency Comparison, 16 Elements

Index

Finite ElementModel

EI

L4

ClassicalSolution

EI

L4

1 3.5160 3.5160

2 22.035 22.034

3 61.698 61.697

4 120.92 120.90

Excellent agreement is obtained for the first four roots.

The corresponding mode shapes are shown in Figures 5 through 8. Note that each mode

shape is multiplied by an arbitrary amplitude scale factor. The absolute amplitude scale is

thus omitted from the plots.

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38

0

0 L

x

Displacement

CANTILEVER BEAM, 16 ELEMENT MODEL, MODE 3

Figure 7.

0

0 L

x

Displacement

CANTILEVER BEAM, 16 ELEMENT MODEL, MODE 4

Figure 8.

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40

APPENDIX A

Energy Method

The total strain energy P of a beam is

=

L

0dx

2

2dx

y2dEI

2

1P (A-1)

The total kinetric energy T of a beam is

[ ] =L

0

22n dxy2

1

T (A-2)

Again let

h/xj,jhxh)1j(,aL)x(YT == (A-3)

h/xj,jhxh)1j(,a'Lh

1)x(Y

dx

d T =

= (A-4)

h/xj,jhxh)1j(,a"Lh

1)x(Y

dx

d T22

2

=

= (A-5)

h/dxd = (A-6)

Assume constant mass density and stiffness.

The strain energy is converted to a localized stiffness matrix as

{ }

= d"L"L

h

EIK

1

0

T3j

(A-7)

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41

The kinetic energy is converted to a localized mass matrix as

{ } =1

0

Tj dLLhM

(A-8)

The total strain energy is set equal to the total kinetic energy per the Rayleigh method. Theresult is a generalized eigenvalue problem.

For a system of n elements,

n,...,2,1j,0MK j2

j == (A-9)

where

=

4

612

264

612612

h

EIK

3j

(A-10)

=

4

22156

3134135422156

420

hMj

(A-11)

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42

APPENDIX B

Beam Bending - Alternate Matrix Format

The displacement matrix for beam bending is

2

2y

1

1y

(B-1)

The stiffness matrix for beam bending is

=

2h4

h612

2h2h62h4

h612h612

3h

EIjK (B-2)

The mass matrix for beam bending is

=

2h4

h22156

2h3132h4

h1354h22156

420

hjM (B-3)

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43

APPENDIX C

Free-Free Beam

Repeat example 1 from the main text with a single element but with free-free boundary

conditions.

The eigen problem is

=

2

2

1

1

2

2

2

1

1

3

h

y

h

y

422313

221561354

313422

135422156

420

L

h

y

h

y

4626

612612

2646

612612

L

EI

(C-1)

=

2

2

1

1

2

2

1

1

h

y

h

y

422313

221561354

313422

135422156

h

y

h

y

4626

612612

2646

612612

(C-2)

where

24

EI420

L

= (C-3)

=

4L

EI420 (C-4)

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44

The eigenvalues are found using the method in Reference 2. Equation (C-1) yields four

eigenvalues.

=

20

1.7143

0

0

4

3

2

1

(C-5)

=

4.4721

1.3093

0

0

4

3

2

1

(C-6)

The finite element results for the first four natural frequencies are thus

=

4.4721

1.3093

0

0

4L

EI420

4

3

2

1

(C-7)

=

91.652

26.8330

0

4L

EI

4

32

1

(C-8)

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The finite element results are compared to the classical results in Table C-1.

Table C-1. Natural Frequency Comparison, 2 Elements

Index

Finite ElementModel

EI

L4

ClassicalSolution

EI

L4

1 0 0

2 0 0

3 26.833 22.373

4 91.652 61.673

beam_fem

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