AIAA 2002-1574

MODIFIED GALERKIN METHOD FOR VIBRATIONS OF BEAMS WITH VISCOUSDAMPINGC. L. Ko1

Oakland UniversityRochester, Michigan 48309-4478

ABSTRACT A semi-analytical method is used to solve freevibration problems of Euler-Bernoulli beams withviscous damping. Galerkin’s method is used toreduce the governing partial differential equationof a continuous system into series of ordinarydifferential equations in the same form as that of adamped single-degree-of-freedom system. Freevibrations of underdamped simply supported andcantilever beams due to the removal of staticuniformly distributed loads are used as examples.Serious of admissible orthogonal sinusoidalfunctions of the spatial variable are assumed to bethe partial solutions of the governing equation.Galerkins’ method is then applied to the partialdifferential equation to reduce it to series of time-dependent ordinary differential equations. Exactsolutions of these second-order ordinarydifferential equations are obtained and theapproximate vibration response is calculated byutilizing initial conditions and the orthogonalproperties of the trial functions. Reasonablyaccurate results are obtained for the case ofunderdamped simply supported beams. However,results for cantilever beams are not reasonabledue to the fact that using admissible orthogonalsinusoidal functions excludes certain dominantlower-order modes in the analysis. Nevertheless,the technique can still be useful for certain types ofbeams because the exact vibration analysis for adamped continuous system has not been shownfeasible in the literature to date.

INTRODUCTION The exact solution of the governing partialdifferential equation for vibrations of a dampedcontinuous system cannot be easily derived due

to the inclusion of an out-of-phase damping term inthe equation as well as due to the difficulty inidentifying orthogonal functions for imposing initialconditions. A numerical method, such as the finiteelement method, or the finite difference method,or the Runge-Kutta method, has to be used tosolve the governing partial differential equation1.Numerical methods can also have limitations due totheir complexity and stability in solving partialdifferential equations with out-of-phase dampingterms. As was shown by Dimarogonas2, theGalerkin method is an alternative analytical methodthat has been successfully employed to determinethe approximate natural frequencies and modalshapes of undamped beams. Wagner3 also utilizedthis method to solve large-amplitude freevibrations of an undamped beam. However, usingthe Galerkin method to successfully determine thetime response of a beam with viscous damping hasnever been reported in the literature. Therefore,the objective of the present analysis is to examinethe feasibility of modifying the conventionalGalerkin method to obtain the time response of adamped continuous system. The present modified Galerkin method is similarto the method employed by Pesterev et al.4 todetermine the response of an undampeddistributed system carrying moving linearconservative oscillators. However, instead ofapplying the Galerkin method to their analysis,Pesterev et al.4 utilized dynamic Green’s functionsof a distributed system as reported by Yang5 toreduce the partial differential equation to serious oftime-dependent ordinary differential equations.The Runge-Kutta method was then used to obtainnumerical solutions of their non-homogeneousordinary differential equations to satisfy modal

------------------------------------------------1Associate Professor, Department of mechanical Engineering, MemberCopyright C 2002 The American Institute of Aeronautics andAstronautics Inc. All rights reserved.

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43rd AIAA/ASME/ASCE/AHS/ASC Structures, Structural Dynamics, and Materials Con22-25 April 2002, Denver, Colorado

AIAA 2002-1574

Copyright © 2002 by the American Institute of Aeronautics and Astronautics, Inc. All rights reserved.

AIAA 2002-1574

initial conditions for each mode. The presentmodified Galerkin method reduces the governingpartial differential equation to series of time-dependent ordinary differential equations in thesame form as that of a damped single-degree-of-freedom system by utilizing the orthogonalproperties of the trial functions in this analysis.Transient vibrations due to the removal ofuniformly distributed loads on simply supportedand cantilever beams with underdamped viscousdamping forces are modeled to illustrate thismodified Galerkin method.

METHOD OF ANALYSIS In general, the problems considered can beformulated for both free and forced vibrations ofEuler-Bernoulli Beams with a uniform mass per unitlength, µ , a uniform damping coefficient per unit

length, c, and a uniform section modulus, EI. Thenon-dimensional governing partial differentialequation can be expressed as

∂4y∂x4 +s 1

∂y∂t +s 2

∂2y

∂t2=s 3 f1(x) f2(t) (1)

where s1 ≡ cL4

EI ,

s2 ≡ µL 4

EI , and

s3 ≡ ωo L3

EI . The

length of the beam and the maximum magnitude ofthe time-dependent impact distributed load perunit length are denoted as L and ω

o, respectively.

Here, y, x, and t are the transverse deflection ofthe beam non-dimensionalized by L, thelongitudinal coordinate non-dimensionalized by L,and time, respectively. Here, f1 and f2 are givenspatial and time functions to describe the spatialand time variations of the excitation. The maximumvalue of each function should be unity. For thecase of having a concentrated forced excitation,

ωo should be the maximum magnitude of the

concentrated force and f1 should be a spatialsingular function as described by Shigley6. Theboundary conditions of a cantilever beam can beexpressed as

(i) A tx=0 : y =0and∂y∂x

=0

(ii) A tx= 1:

∂2y∂x2 =0and

∂3y∂x3 = 0 (2)

For a simply supported beam, the boundaryconditions are

A tx=0andx =1: y= 0and

∂2y

∂x2=0 (3)

For the case of free vibration, the initial conditionscan be expressed as

At t=0: y=y og1(x) and∂y∂t

= Vog2(x) (4)

where, yo and Vo are the non-dimensionalizedmaximum initial displacement and velocity,respectively. Functions g1 and g2 are given spatialvariations of the initial displacement and velocity,respectively. The maximum value of each of thesetwo functions is unity. To obtain an approximatesolution of the governing differential equation,one can assume the solution to be

y = y iΣi =1

∞= A i ui(x) vi(t)Σ

i = 1

∞(5)

where Ai are undetermined coefficients and ui(x)are trial functions that satisfy the boundaryconditions of the beam. The infinitive number ofterms can be approximated by using only a fewdominating terms. Applying Galerkin’s method tothe problem by substituting Eq.(5) into thegoverning differential equation and thenmultiplying the equation by uj to integrate theresulting equation with respect to x from 0 to 1,one obtains

{A iΣ

i = 1

∞vi[uj

d3ui

dx30

1

–duj

dxd2ui

dx20

1+

d2uj

dx 2d 2ui

dx2 dx]}0

1

+ A iΣ

i = 1

∞[(s1

dvidt

+s2d 2vidt2 ) uj

0

1

u idx]=s 3f2 f1u jdx0

1

(6)

If admissible orthogonal functions are selected astrial functions, the following conditions should besatisfied:

u j

0

1u i dx=0 and

d2uj

dx2d2ui

dx2 dx= 0,0

1

ifi ≠j (7)

For a simply supported beam or a cantilever beam,the boundary conditions specify that

[u j

d3uidx3 ]

0

1= 0 and [

du jdx

d2u idx2 ]

0

1= 0 [8]

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AIAA 2002-1574

Therefore, Eq. (6) can be expressed as d2vi

d t2+2ζiωi

dvidt

+ ωi2 vi = qi f2(t) (9)

where, ωi ≡ β is2α i

, ζ i ≡s1

2s2ωi, q i ≡ s3γi

A is2α i,

α i ≡ (u i)2dx

0

1

,

β i ≡ (d2u i

dx2 )2dx0

1

, γ i ≡ f1u idx0

1

.

Eq. (9) is in the same form as the governingdifferential equation for the free and forcedvibrations of single-degree-of-freedom systemswith viscous damping. Three types of generalsolutions of the equation can possibly existdepending upon the value of the damping ratio,ζ i : the overdamped solution (ζ i > 1) , the criticallydamped solution (ζ i = 1) and the underdampedsolution ( ζ i < 1). For the free vibration problem, allthree types of solutions can be formulated in thesame way as those obtained for single-degree-of-freedom systems. The steady-state response ofthe beam with a periodic excitation and thetransient response of the beam with a non-periodicexcitation can also be formulated in the same wayas those of respective cases of a single-degree-of-freedom system without distinguishing the type ofdamping.

TRIAL FUNCTIONS In addition to satisfying boundary conditions, trialfunctions should also be mutually orthogonal sothat conditions expressed in Eq. (7) can be met.Since sinusoidal functions can easily be made intoorthogonal functions, the sinusoidal trial functionsfor a simply-supported beam can be derived to bethe following:

un =sin(nπx), n=1,2 , 3, ... (10)

The corresponding parameters αn and β n can be

evaluated as the following:

αn=12

and β n=n2π4

2, wherenis an integer. (11)

The angular frequency and the damping ratio ofthe n th trial function can be obtained as

ωn = nπ2

s2and ζn =

s1

2nπ2 s2 (12)

The first five trial functions for a cantilever beamcan also be obtained as follows:

u1 = 58

+ 1– 2

cos(πx)+ 1– 8

cos(2πx) (13)

u2=41cos(3πx)

32+

cos(4πx)– 2

+25cos(5πx)

– 32 (14)

u3=113cos(6πx)

56+

cos(7πx)– 2

+85cos(8πx)

– 56(15)

u4=221cos(9πx)

80+

cos(10πx)– 2

+181cos(11πx)

– 80 (16)

u5=365cos(12πx)

104+

cos(13πx)– 2

+313cos(14πx)

– 104(17)

Parameters αn and β n corresponding to these

trial functions can be evaluated as α1= 67

128, α2= 981

1024, α3=10389

3136, α4=41601

6400,

α5=11694910816

,β 1=π4

4,β 2=281161π4

1024,β 3=11653205π4

1568,

β 4=263849191π 4

6400, β 5=206354989π4

676(18)

The angular frequencies and damping ratios ofthese five trial modes can be determined to be

ω1=0.6911 π2

s2, ω2=16.93 π2

s2, ω3=47.36 π 2

s2,

ω4=79.64 π2

s2, ω5=168.0 π2

s2, ζ 1=

0.7235s1

π2 s2,

ζ 2=0.02953s1

π2 s2, ζ3=

0.01056s1

π2 s2, ζ4=

0.006277s1

π2 s2,

ζ 5=0.002977s1

π2 s2(19)

Trial functions specified in Eq. (10) consist of allpossible modes of a simply supported beam.However, The five trial functions for a cantileverbeams actually are close to modal shapes of thefirst, the fourth, the seventh, the tenth, and thethirteenth modes, because the number of nodalpoints between the two ends of the beamobtained from these functions are 0, 3, 6, 9, and12, respectively. The corresponding frequenciesincrease with the number of nodal point, whereascorresponding damping ratios decrease with the

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AIAA 2002-1574

increase of the number of nodal points. It isobvious that the infinitive number of terms in Eq.(5) can be reasonably approximated by thesummation of a finite number of terms, becausemagnitudes of higher-order term are ratherinsignificant.

TRANSIENT RESPONSE FOR FREEVIBRATIONS OF BEAMS DUE TO

REMOVAL OF STATIC LOADS A special type of problems considered are freevibrations of cantilever and simply-supportedbeams due to the removal of static uniformlydistributed loads from these beams. Onlyresponses of underdamped systems arepresented. However, responses for criticallydamped and overdamped systems can also beobtained by using the similar procedure. Thegeneral solution of the homogeneous case of Eq.(9) for an underdamped beam subjected to freevibrations can be obtained as

vi=e– ζiωit[sin(ωi 1 – ζ i2 t)+Bicos(ωi 1 – ζi

2 t)] (20)where, Bi are undetermined coefficients. Hence,the approximate solution of the homogeneouscase of Eq. (1) can be obtained from Eq. (5) as

y= A iΣi = 1

∞u ie

– ζiω it [sin(ωi 1 – ζi2 t)+Bicos(ωi 1 – ζ i

2 t)]

(21)Substituting Eq. (21) into the initial conditionsspecified in Eq. (4) and applying the orthogonalconditions of the trial functions to the integrationsof the resulting equations, one obtains

A iB i =

yoα i

g1(x)u i(x)dx0

1

(22)

A i =

Vo

ωiαi 1 – ζi2

g2 (x)u i(x)dx0

1

+

ζ iyo

αi 1 – ζ i2

g 1(x)u i(x)dx0

1

(23)

For the special case of undamped free vibrations,Eq. (21) reduces to

y = AiΣi = 1

∞u i[sin(ωit) + Bicos(ωit)] (24)

Eq. (22) remains applicable to the undamped case;however, Eq. (23) should be modified as

A i =

Voωiαi

g2(x)u i(x)dx0

1

+yoα i

g1(x)u i(x)dx0

1

(25)

Example problems considered are for freevibrations due to the removal of static loads from asimply supported beam and a cantilever beam. Afull-length uniformly distributed load with amagnitude ωo is considered to be the static load

on both beams. Utilizing the rigid body dynamics,the initial velocity distribution of the simplysupported beam can be derived as

g2(x) ={2x, for 0 ≤ x ≤ 1

22(1 – x), for 1

2 ≤ x ≤ 1 (26)

Similarly, the initial velocity distribution of thecantilever beam can be obtained as

g2(x) =x (27)

Exact solutions of the governing differentialequation, Eq. (1), for free vibrations with dampingare not easy to obtain because conditions fororthogonality are not explicitly identifiable.Nevertheless, the exact solutions for freevibrations of undamped simply-supported beamscan be found in many textbooks1,2. For a simply-supported beam, the general solution forundamped free vibrations can be obtained as

y= Cnsin(nπx)[sin(Σn = 1

∞n 2π2t

s2)+Dncos(n

2π2ts2

)] (28)

Undetermined coefficients Cn and Dn can bedetermined from initial conditions to be

CnDn = 2yo g1(x)sin(nπx)dx

0

1

(29)

Cn =

2Vo s2

n2π2 g2(x)sin(nπx)dx0

1

(30)

For a cantilever beam with the fixed end at x = 0,the general solution for undamped free vibrationscan be obtained as

y= Cn[sinhλnsin(λnx)+sinλnsinh(λnx)][sin(Σn = 1

∞ωnt)+

Dncos(ωnt)] (31)

where λn ≡ ωn s2 . Angular natural frequencies,

ωn , should satisfy

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AIAA 2002-1574

sinhλncosλn – sinλ ncoshλn =0 (32)

Undetermined coefficients Cn and Dn should bedetermined from initial conditions and from thecondition of orthogonality which is not easilyobtainable in this case. Therefore, the responsedescribed by Eq. (31) cannot be practicallydetermined.

Removal of a Uniformly Distributed StaticLoad ωo from a Simply Supported Beam The inial deflection can be determined as7

g1(x) =1 – 2x 2 +x 3 and

yo =ωoL

4

24EI (33)

The static deflection at the center of the beam is y1 = 5

8yo (34)

The coefficients for the trial functions can bedetermined as

A nBn =96yo

n5π5 ,if n=2k – 1 and AnBn =0,i f n=2k

An =96y os1

n 5π5 4n 2π4s2 – s12

+

16Vo

n2π2 4n2π4s2 – s12, if n =4k – 3

An =96y os1

n 5π5 4n 2π4s2 – s12

–

16Vo

n2π2 4n2π4s2 – s12, if n =4k – 1

A n =0, if n=2k (35)Here, k is a positive integer. For undampedvibrations, coefficients An reduce to

A n =96yo

n5π5 +8Vo s2

n3π4 , i fn =4k – 3

An =96y o

n5π5 –8Vo s2

n3π4 , if n =4k – 1

A n =0, if n=2k (36)

Removal of a Uniformly Distributed StaticLoad ωo from a Cantilever Beam

The initial deflection can be expressed as7

g1 (x)= (x4 – 4x3 + 6x 2) and yo =

ωoL3

24EI (37)

The deflection at the free end is y1 =3yo (38)

The undetermined coefficients for the fiveselected modes can be obtained from Eq. (22) andEq. (23) as

A1B1 =24yo

67(4+ 10

π2 + 1π4) = 1.799yo

A2B2 =– 16yo

981(328

π2 + 13079675π4 ) = –0.545yo

A3B3 =9605yo

187002( 1π2 + 649

4704π 4) =0.00513y o

A4B4 =– 884y o

407731401(6155

π2 – 3011550481675π4 )

=– 0.00136yo

A5B5=2y o

116949(114245

441π2 – 83501670587646104π4)

=0.000447y o (39)

A1 =128Vo

67ω1 1 – ζ 12

( 516

+ 1π2) +

ζ 1A1B1

1 – ζ 12

=0.791Vo

ω1 1 – ζ 12

+ζ 1A1B1

1 – ζ 12

A2 =– 2048Vo

8829π2ζ 2 1 – ζ 22

+ζ 2A2B2

1 – ζ 22

=– 0.0235Vo

ω2 1 – ζ22

+ζ 2A2B2

1 – ζ 22

A3 =10389Vo

153664π2ω3 1 – ζ32

+ζ3A3B3

1 – ζ32

=0.00685Vo

ω3 1 – ζ32

+ζ3A3B3

1 – ζ32

A4 =– 1932800Vo

37066491π2ω4 1 – ζ42

+ζ4A4B4

1 – ζ42

=0.00528Vo

ω4 1 – ζ42

+ζ4A4B4

1 – ζ42

A5 =116949Vo

1827904π2ω5 1 – ζ 52

+ζ 5A5B5

1 – ζ 52

=0.00528Vo

ω5 1 – ζ52

+ζ5A5B5

1 – ζ52

(40)

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AIAA 2002-1574

NUMERICAL EXAMPLES ANDCONCLUSIONS

Free vibrations of cantilever and simply-supportedbeams due to the removal of static loads areconsidered for numerical calculations. Aluminumbeams of 0.9144 meter (3 feet) in length, 0.1016meters (4 inch) in width, and 1.588 mm (0.0625inch) in thickness are modeled as examples. TheYoung’s modulus and the density of aluminum areapproximated to be 68.95 Gpa (10.0 x106 Psi) and2768 kg/m3 (0.100 lbm/in3), respectively. Thedistributed damping coefficient c is assumed to be10.0 N-sec/m2 (0.209 lbf-sec/ft2) and the mass ofthe beam per unit length is calculated to be 0.4464kg/m (0.0250 lbm/in). Parameters s1 and s2 arecalculated to be 2.993 sec and 0.1335 sec2,respectively. The transient response due to theremoval of a uniform distributed static load isconsidered for both the simply supported beamand the cantilever beam. Admissible sinusoidal orthogonal functions areused as trial functions in this analysis. For a simplysupported beam, these orthogonal functionsinclude all possible modal shapes. However, forcantilever beams, these orthogonal functionsexclude two middle modes for every series of fourmodes. Hence, only the fundamental and thefourth modes can be modeled among the first fourmodes as a result of satisfying the requirement ofall trial functions being mutually orthogonal. Fig. 1 shows time variations of the ratios of thedisplacements at the center and at a distance of14

beam length from the support to the static

deflection at the center of the simply supportedbeam due to the removal of the uniformlydistributed load. The damping ratio for thefundamental mode is calculated to be 0.208, whichis the largest value among all possible modes.Since the exact analysis of the transient responseof damped beams does not exist, comparison ofthe result cannot be made to that obtained fromsuch an analysis. Nevertheless, the natural periodof the fundamental mode of the undamped simplysupported beam with the same properties can becalculated to be 1.16 seconds, which is in goodagreement with the period shown in Fig. 1. Theinitial displacements are also identical to the static

deflection of the beam under the application of auniformly distributed load. Therefore, one canconclude that the present method can yieldaccurate results for simply supported beams withuniformly distributed viscous damping forces. Fig. 2 shows time variations of the ratios of thedisplacements at the center and at the free end ofthe cantilever beam to the static deflection at thefree end of the cantilever beam due to the removalof a uniformly distributed load. The damping ratioof the fundamental mode is calculated to be 0.300.Based on the exact analysis for an undampedcantilever beam, the fundamental natural period ofundamped vibration of the beam is calculated to be3.26 seconds, whereas the natural period shownin Fig. 2 is about 1.75 seconds. In addition, at theinstant of removing the uniformly distributed load,the displacement at the free end of the beam failsto coincide with the static deflection under the loadat the same location. Since trial functions for thesecond and the third modes cannot be included inthe analysis due to the requirement of their beingorthogonal to other modes, the effects of thesetwo dominant modes are absent in the resultsshown in Fig. 2. Therefore, It is possible that theunreasonable results are due to this limitation.Hence, including all dominant modal shapes in theadmissible and mutually orthogonal trial functionscan be essential for the successful application ofthis modified Galerkin method.

REFERENCES1. Thomson, W. T., Theory of Vibration with

Applications, Prentice Hall, Englewood Cliffs,New Jersey, 1993, fourth edition, chapters 9and 10.

2. Dimarogonas, A., Vibration for Engineers,Prentice Hall, Upper Saddle River, NewJersey, 1996, chapter 9, pp. 498-500.

3. Wagner, H., “Large-Amplitude FreeVibrations of a Beam,” ASME Journal ofApplied Mechanics, Vol. 32, December 1965,pp. 887-893.

4. Pesterev, A, V., Young, B., Bergman, L. A.,and Tan, C. -A., “Response of Elastic

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AIAA 2002-1574

Continuum Carrying Multiple MovingOscillators,” ASCE Journal of EngineeringMechanics, Vol. 127, no. 3, 2001, pp. 260-265.

5. Yang, B., “Integral formulas for Non-Self-Adjoint Distributed Dynamic Systems,”AIAA Journal, Vol. 34, 1996, pp. 2132-2139.

6. Shigley, J. E., Mechanical EngineeringDesign , McGraw-Hill, New York, 1977, chapter2, pp. 40-43.

7. Gere, J. M. And Timoshenko, S. P., Mechanicsof materials, PWS-KENT, Boston, 1990, thirdedition, appendix G.

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0 0.5 1 1.5 2 2.5 3-1

-0.5

0

0.5

1

1.5

time (second)

non-

dim

ensi

onal

dis

plac

emen

ts

x=0.5

x=0.25

Figure 1 time variations of displacements non-dimensionalized by the static deflection at the center of the simply supported beam

0 0.5 1 1.5 2 2.5 3-0.4

-0.2

0

0.2

0.4

0.6

0.8

time (second)

non-

dim

ensi

onal

dis

plac

emen

ts

x=1

x=0.5

Figure 2 time variations of displacements non-dimensionalized by the static deflection at free end of the cantilever beam

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