Frequency Equation of Flexural Vibrating Cantilever Beam

Research ArticleFrequency Equation of Flexural Vibrating Cantilever BeamConsidering the Rotary Inertial Moment of an Attached Mass

Binghui Wang1 Zhihua Wang2 and Xi Zuo3

1School of Architecture and Civil Engineering Jiangsu University of Science and Technology 2Mengxi Road Zhenjiang Jiangsu China2Institute of Geotechnical Engineering Nanjing Tech University 200 North Zhongshan Road Nanjing China3Institute of Architectural Engineering Jinling Institute of Technology Nanjing China

Correspondence should be addressed to Zhihua Wang wzhnjut163com

Received 13 December 2016 Accepted 5 February 2017 Published 22 February 2017

Academic Editor Rahmat Ellahi

Copyright copy 2017 Binghui Wang et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Themajor goal of this paper is to address the derivation of the frequency equation of flexural vibrating cantilever beam consideringthe bendingmoment generated by an additional mass at the free end of beam not just the shear force It is a transcendental equationwith two unambiguous physical meaning parameters And the influence of the two parameters on the characteristics of frequencyand shapemode was madeThe results show that the inertial moment of the mass has the significant effect on the natural frequencyand the shape mode And it is more reasonable using this frequency equation to analyze vibration and measure modulus

1 Introduction

The cantilever beam is a simple structure and it is animportant simplified model for many engineering problemin the fields of mechanical engineering civil engineeringand so forth However there is a vast number of papersconcerned with the determination of the eigenfrequencies ofthe cantilever beam subject to various boundary conditionswhich can be found in the classic book [1] And more andmore issues of cantilever beam with complicated boundaryconditions [2] or external load conditions [3 4] have beenstudied by theoretical deduction [5] or numericalmethod [6]or it is application to determine Youngrsquos modulus [7]

The cantilever beam model is also used in geotechnicalearthquake engineering as a simplified model [8] for theground responses due to earthquake It is also a basic prin-ciple of measuring the soilrsquos dynamic shear modulus whichis an indispensable parameter for analyzing the earthquakeresponse of site caused by far-field ground The apparatusused to obtain the dynamic shear modulus is well knownas resonant column apparatus The soil column installedin this apparatus is driven by electromagnetic force at freeend [9] producing its torsional vibration If flexural vibration

of the soil column occurs the dynamic Young moduluscan be obtained which is also an important parameter fordynamic analysis of site suffered by near-field groundmotionEssentially the frequency equation of flexural vibrating can-tilever beam with an additional mass is needed Howeverthe vibrating frequency and shape mode of soil column areeffected by not only the shear force but also themoment forcegenerated by the motion of the additional mass attachedat the free end of the soil column So the influence ofthese two forces on the vibrating of soil column must bediscussed Cascante et al [10] derived a frequency equationof this vibrational problemusing Rayleighrsquosmethod assumingthat the mode shape is a third-order polynomial Laura [11]derived the frequency equation of a cantilever beam attachingan additional mass which is considered as shear force actedon the free end of beam but did not consider the momentforce generated by the mass And recently Gurgoze [12ndash14]studies the eigenfrequencies of a cantilever beam carrying atip mass or spring-mass

The paper tried to derive the frequency equation of thecantilever beam in order to obtain more accurate solution forthe soil column and to analyze the frequency characteristics

HindawiMathematical Problems in EngineeringVolume 2017 Article ID 1568019 5 pageshttpsdoiorg10115520171568019

2 Mathematical Problems in Engineering

120601(x)

x

L

m1 J1EI m = constants

Figure 1 Diagram of the cantilever beam with an additional mass

of this system effected by the bending moment generated bythe additional mass

2 Flexural Motion Equation of CantileverBeam with an Additional Mass

The flexural cantilever beam to be considered is the straightuniform beamwith an additional rigidmass attached by fixedconnection The significant physical properties of this beamare assumed to be the flexural stiffness 119864119868 and the mass perunit length 119898 both of which are constant along the span 119871The transverse displacement response 119910(119909 119905) is a functionof position and time The free vibration equation of motionfor the system as shown in Figure 1 is easily formulated bydirectly expressing the equilibrium of all forces acting on thedifferential segment of beam And it becomes [15]

119864119868 1205974119910 (119909 119905)1205971199094 + 119898 1205972119910 (119909 119905)1205971199052 = 0 (1)

Since 119864119868 and 119898 are constant one form of solution of thisequation can be obtained by separation of variables using

119910 (119909 119905) = 120601 (119909) 119884 (119905) (2)

which indicates that the free vibration motion is of a specificshape 120601(119909) having a time-dependent amplitude 119884(119905) Substi-tuting this equation into (1) can yield two ordinary differentialequations

(119905) + 1205962119884 (119905) = 01206011015840101584010158401015840 (119909) minus 1205734120601 (119909) = 0 (3)

in which 1205734 = 1205962119898119864119868 and 120596 is the circular frequencyTheseequations have the solution separately as follows

119884 (119905) = 1198611 sin (120596119905) + 1198612 cos (120596119905) (4)

120601 (119909) = 1198601 cos120573119909 + 1198602 sin120573119909 + 1198603ch120573119909 + 1198604sh120573119909 (5)

in which constants 1198611 and 1198612 depend upon the initialdisplacement and velocity conditions and real constants 119860 119894must be evaluated so as to satisfy the known boundaryconditions at the ends of the beam

The cantilever beam considered has a fixed end so its twoknown boundary conditions are

120601 (0) = 01206011015840 (0) = 0 (6)

Making use of (5) and its first partial derivative with respectto 119909 from (6) one obtains 1198603 = minus1198601 and 1198604 = minus1198602 So (5)becomes

120601 (119909) = 1198601 (cos120573119909 minus ch120573119909) + 1198602 (sin120573119909 minus sh120573119909) (7)

An additional rigid mass 1198981 having a rotary mass moment ofinertia 1198691 is attached by fixed connection to its free end as alsoshown in Figure 2These internal force components are alongwith the translational and rotary inertial force components11989811205962120601(119871) and 119869112059621206011015840(119871) respectively So the force andmoment equilibrium of the additional mass requires theboundary conditions

119872 (119871) = 11986411986812060110158401015840 (119871) = minus12059621206011015840 (119871) 1198691119881 (119871) = 119864119868120601101584010158401015840 (119871) = minus1205962120601 (119871) 1198981 (8)

Making use of (5) and its first second and third partialderivative with respect to 119909 and substituting them into (8)yield

[[[[

1205732 (cos120573119871 + ch120573119871) + 12059621198691119864119868 120573 (sin120573119871 + sh120573119871) 1205732 (sin120573119871 + sh120573119871) minus 12059621198691119864119868 120573 (cos120573119871 minus ch120573119871)1205733 (sin120573119871 minus sh120573119871) + 12059621198981119864119868 (cos120573119871 minus ch120573119871) minus1205733 (cos120573119871 + ch120573119871) + 12059621198981119864119868 (sin120573119871 minus sh120573119871)

]]]]

11986011198602 = 00 (9)

For coefficients 1198601 and 1198602 to be nonzero the determinantof the square matrix in this equation must equal zero thusgiving the frequency equation

[1205732 (cos120573119871 + ch120573119871) + 12059621198691119864119868 120573 (sin120573119871 + sh120573119871)]sdot [minus1205733 (cos120573119871 + ch120573119871) + 12059621198981119864119868 (sin120573119871 minus sh120573119871)]

minus [1205732 (sin120573119871 + sh120573119871) minus 12059621198691119864119868 120573 (cos120573119871 minus ch120573119871)]

sdot [1205733 (sin120573119871 minus sh120573119871) + 12059621198981119864119868 (cos120573119871 minus ch120573119871)]= 0

(10)

Mathematical Problems in Engineering 3

V(L) = EI120601998400998400998400 (L)

M(L) = EI120601998400998400(L)

minus1205962120601(L)m1

minus1205962120601998400(L)J1

Figure 2 Diagram of force equilibrium analysis at the free end

which can reduce to the following form making use of1205962119864119868 = 1205734119898119869111989811198982 1205734 (1 minus cos120573119871ch120573119871)

+ 1198981119898 120573 (sin120573119871ch120573119871 minus cos120573119871sh120573119871)= 1198691119898 1205733 (sin120573119871ch120573119871 + cos120573119871sh120573119871)

+ (1 + cos120573119871ch120573119871)

(11)

Setting

119911 = 120573119871 = 4radic 1205962119898119864119868 119871 (12)

is a power function of the frequency of the flexural vibratingcantilever beam And setting 120572119869 = 11986911198981198713 = 11986911198981198791198712 is aratio of the rotary mass moment of inertia for the additionalmass and for a rigid mass of 119898119879 having a rotary arm length of119871 and 120572119898 = 1198981119898119871 = 1198981119898119879 is a ratio of the mass betweenthe additionalmass and the cantilever beam So the frequencyequation becomes

1205721198691205721198981199114 (1 minus cos 119911ch119911) + 120572119898119911 (sin 119911ch119911 minus cos 119911sh119911)= 1205721198691199113 (sin 119911ch119911 + cos 119911sh119911) + (1 + cos 119911ch119911) (13)

This frequency equation is a transcendental equation thatcontained two parameters with unambiguous physical mean-ing The solution of this equation which is the frequency ofthe system of the flexural vibrating cantilever beam can beonly obtained by numerical method for now

3 Solution of the Frequency Equation

In order to solve the frequency equation the term of the rightside of (13) can bemoved to the left side and then suppose theleft side equals 119891(119911 120572119869 120572119898) And 119891 = 0 is equivalent to (13)Suppose the parameters 120572119869 and 120572119898 are constant and then therelationship between 119891(119911) and 119911 can be calculated and drawnas shown in Figure 3 in which the parameters 120572119869 = 10 and120572119898 = 1 However the value of 119891(119911) rises abruptly So the 119910-axis alters to log119891(119911) in order to show the curve is up anddown across the 119909-axis obviously The intersection points of

0

10

20

30

40

0 10 20 30 40 50

z

log(f(z

))

minus10

minus20

minus30

minus40

120572j = 10120572m = 1

Figure 3 Relationship between 119891(119911) and 119911

0

05

1

15

2

25

3

0001 001 01 1 10 100 1000 10000120572J

120572m = 1000

120572m = 100

120572m = 10

120572m = 1

120572m = 01

120572m = 001

z min

Figure 4 Varied 119911 with 120572119869

the curve and 119909-axis as marked in Figure 3 with circle pointsare the roots of the equation 119891(119911) = 0 where its approximateroots can be obtained using numerical methods for solvingsystem of nonlinear equations such as method of bisectionand Newtonrsquos method

For a specific system where the 120572119869 and 120572119898 are deter-mined the solution of the frequency equation (13) can beobtained And the roots of the parameter 119911 are variedwith theparameters120572119869 and120572119898 Figures 4 and 5 show their effectsWiththe increment of the parameter 120572119869 the root 119911 is increasedidentically with a sigmoid function as shown in Figure 4It means that the frequency of the system will increase ifthe rotary mass moment of inertia for the additional massis increased And the rate of the increment is larger around120572119869 = 120572119898 than others For example the increment of 119911 atthe range of 120572119869 = 01sim10 is fastest while 120572119898 = 1 With theincrement of the parameter 120572119898 the root 119911 is decreasing asshown in Figure 5 which means that the frequency of thesystem will decrease with the additional mass increasing


Table 1 The first five natural frequencies of the cantilever beam considering the bending moment

Frequencynumber (i)

Nondimensional natural frequency 1205961015840119894 = radic1198641198681198981198714119911119894 1205961015840119894lowast120572119898 = 100120572119869 = 1120572119898 = 10120572119869 = 1

120572119898 = 1120572119869 = 100120572119898 = 1120572119869 = 10

120572119898 = 1120572119869 = 1120572119898 = 1120572119869 = 01

120572119898 = 1120572119869 = 001120572119898 = 1120572119869 = 0001

120572119898 = 1120572119869 = 0120572119898 = 1(ref [11])

1 01749603 06054297 2950242 291158 255044 170226 1571098 155867 1557298 15572982 2230366 2247413 2393841 2393129 2386031 2317405 1913831 1657872 1625009 16250093 6166046 6183822 6343232 6342955 6340190 631286 6072695 5350208 5089584 50895844 12090684 1210850 1227265 1227251 12271062 1225668 1212014 1132826 1051983 10519835 19986943 2000480 2017208 20171991 20171107 2016229 2007639 1942422 1792320 1792320lowastNotes The values of this column were calculated by authors using the frequency equation from [11]And the results were carefully checked by comparing the values of 119911119894 (using the character 119910119894 in this reference)

001 01 1 10 100 10000

05

1

15

2

25

3

120572J = 100

120572J = 10

120572J = 1

120572J = 01

120572J = 001

120572m

z min


4 Natural Frequencies and ShapeModes Characteristics

Thenatural frequency120596119894 of the flexural vibration of cantileverbeam considering the bending moment generated by theattached mass can be calculated using the transformation of(12) Table 1 lists the results of the first five natural frequenciesof this system with the varied parameters values 120572119895 and120572119898 The natural frequency increases with the parameter 120572119898decreasing which means the quality of the additional massreduction compared to that of the cantilever beam This isconsistent with the result of others [7 16] And the frequencyalso increases with the increment of the rotary mass momentof inertia for the additional mass compared to the beamwhich is defined as the parameter 120572119895 In order to makecomparisons the results of frequencies when the parameters120572119895 = 0 and 120572119898 = 1 are also listed which means the free endof beam has no moment force generated by the additionalmass As expected these natural frequency values are thesame as that obtained from [11] in which only the shear forceis considered generated by the attached mass And the larger

value of 120572119895 causes the higher frequency of the cantilever beamas shown in Table 1

However the interval between the two frequencies for120572119895 equal to zero and for 120572119895 unequal to zero is considerableespecially that between the twohigher frequency numbers Sothe ignored rotary mass moment of inertia may be too idealfor engineering problem and cause obviously error

The vibration shape mode can also be obtained bysubstituting the ratio of 1198601 and 1198602 into the shape equation(7) And Figure 6(a) shows the first four mode shapes ofthe cantilever beam considering the moment force with theparameters 120572119898 = 1 and 120572119869 = 10 The first four mode shapesof the cantilever beam are shown without considering themoment force in [11] It is clearly shown that the vibrationshape is effected by the moment force from the additionalmass significantly However the existence of the additionalmass restrains the development of displacement at the freeend of beam

In order to analyze the influence of 120572119869 on themode shapethe 3rd to 5th mode shapes of considering the moment forcewith 120572119898 = 1 and 120572119869 = 0001 and without consideringsituation are given as shown in Figure 6(b) Even if theparameter 120572119869 is of very small value the 5th mode shape ofthese two situations has obviously divergence though thesmaller the serial number of mode shape is the less obviousthe divergence becomes

5 Conclusion

The frequency equation of cantilever beamwith an additionalmass exciting flexural vibration was derived consideringthe rotary inertial moment of inertia of an attached massincluding the shear force It is a transcendental equationand it contains two parameters with unambiguous physicalmeaning which can be defined as the ratio of rotary massmoment of inertia and the ratio of the mass respectively

These two parameters effect both the natural frequencyand the shape mode of the beam As the ratio of rotary massmoment of inertia increases the natural frequency climbsEven a little increment of the ratio may cause higher variancebetween considering and not considering the rotary massmoment of inertia especially for the high natural frequencyAnd the ratio of the rotary mass moment of inertia also


1st mode2nd mode

3rd mode4th mode

Without considering moment force at free end

1st mode2nd mode

3rd mode4th mode

00 02 04 06 08 1000051015202530

Disp

lace

men

t

Normalized lengthminus05

minus10

minus15

minus20

minus25

Considering moment force at free end (120572m = 1 120572j = 10)

(a)

00 02 04 06 08 1000051015202530

Disp

lace

men

t


minus10

minus15

minus20

minus25



3rd mode4th mode

5th mode

3rd mode4th mode

5th mode

(b)

Figure 6The first five mode shapes of considering the moment force with 120572119898 = 1 and 120572119869 = 10 and without considering situation (a) and the3rd to 5th mode shapes of considering the moment force with 120572119898 = 1 and 120572119869 = 0001 and without considering situation (b)

effects the mode shape of this system The higher the serialnumber of mode shape investigated is the more obvious thedivergence becomes

Competing Interests

There are no competing interests regarding this paper

Acknowledgments

This paper is supported by National Science Foundation ofChina (NSFC 51309121) and Natural Science Foundation ofJiangsu Province (BK20130463)

References

[1] S S Rao Mechanical Vibrations Pearson Education UpperSaddle River NJ USA 4th edition 2004

[2] Y K Rudavskii and I A Vikovich ldquoForced flexural-and-torsional vibrations of a cantilever beam of constant crosssectionrdquo International AppliedMechanics vol 43 no 8 pp 912ndash923 2007

[3] D Zhou and T Ji ldquoEstimation of dynamic characteristics of aspring-mass-beam systemrdquo Shock and Vibration vol 14 no 4pp 271ndash282 2007

[4] D Zhou and T Ji ldquoDynamic characteristics of a beam anddistributed spring-mass systemrdquo International Journal of Solidsand Structures vol 43 no 18-19 pp 5555ndash5569 2006

[5] C A Rossit and P A A Laura ldquoFree vibrations of a cantileverbeamwith a spring-mass systemattached to the free endrdquoOceanEngineering vol 28 no 7 pp 933ndash939 2001

[6] J R Banerjee ldquoFree vibration of beams carrying spring-mass systemsmdasha dynamic stiffness approachrdquo Computers andStructures vol 104-105 pp 21ndash26 2012

[7] R M Digilov and H Abramovich ldquoFlexural vibration test ofa beam elastically restrained at one end a new approach forYoungrsquos modulus determinationrdquo Advances in Materials Scienceand Engineering vol 2013 Article ID 329530 6 pages 2013

[8] M I Idriss and B H Seed ldquoSeismic response of horizontal soillayersrdquo Journal of the Soil Mechanics and Foundation DivisionAmerican Society of Civil Engineers vol 94 no 4 pp 1003ndash10311968

[9] V P Drnevich Resonant-Column TestingmdashProblems and Solu-tions ASTM Denver Colo USA 1978

[10] G Cascante C Santamarina and N Yassir ldquoFlexural excitationin a standard torsional-resonant column devicerdquo CanadianGeotechnical Journal vol 35 no 3 pp 478ndash490 1998

[11] P A A Laura J L Pombo and E A Susemihl ldquoA note on thevibrations of a clamped-free beam with a mass at the free endrdquoJournal of Sound amp Vibration vol 37 no 2 pp 161ndash168 1974

[12] M Gurgoze ldquoOn the representation of a cantilevered beamcarrying a tip mass by an equivalent spring-mass systemrdquoJournal of Sound and Vibration vol 282 no 1-2 pp 538ndash5422005

[13] M Gurgoze ldquoOn the eigenfrequencies of a cantilever beamcarrying a tip spring-mass systemwithmass of the helical springconsideredrdquo Journal of Sound andVibration vol 282 no 3-5 pp1221ndash1230 2005

[14] MGurgoze ldquoOn the eigenfrequencies of a cantilever beamwithattached tip mass and a spring-mass systemrdquo Journal of Soundand Vibration vol 190 no 2 pp 149ndash162 1996

[15] R Clough and J PenzienDynamics of StructuresMcGraw-Hill1993

[16] L E Monterrubio ldquoFree vibration of shallow shells using theRayleighmdashRitz method and penalty parametersrdquo Proceedingsof the Institution of Mechanical Engineers Part C Journal ofMechanical Engineering Science vol 223 no 10 pp 2263ndash22722009

Submit your manuscripts athttpswwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


120601(x)

x

L

m1 J1EI m = constants

Figure 1 Diagram of the cantilever beam with an additional mass

of this system effected by the bending moment generated bythe additional mass

2 Flexural Motion Equation of CantileverBeam with an Additional Mass

The flexural cantilever beam to be considered is the straightuniform beamwith an additional rigidmass attached by fixedconnection The significant physical properties of this beamare assumed to be the flexural stiffness 119864119868 and the mass perunit length 119898 both of which are constant along the span 119871The transverse displacement response 119910(119909 119905) is a functionof position and time The free vibration equation of motionfor the system as shown in Figure 1 is easily formulated bydirectly expressing the equilibrium of all forces acting on thedifferential segment of beam And it becomes [15]

119864119868 1205974119910 (119909 119905)1205971199094 + 119898 1205972119910 (119909 119905)1205971199052 = 0 (1)

Since 119864119868 and 119898 are constant one form of solution of thisequation can be obtained by separation of variables using

119910 (119909 119905) = 120601 (119909) 119884 (119905) (2)

which indicates that the free vibration motion is of a specificshape 120601(119909) having a time-dependent amplitude 119884(119905) Substi-tuting this equation into (1) can yield two ordinary differentialequations

(119905) + 1205962119884 (119905) = 01206011015840101584010158401015840 (119909) minus 1205734120601 (119909) = 0 (3)

in which 1205734 = 1205962119898119864119868 and 120596 is the circular frequencyTheseequations have the solution separately as follows

119884 (119905) = 1198611 sin (120596119905) + 1198612 cos (120596119905) (4)

120601 (119909) = 1198601 cos120573119909 + 1198602 sin120573119909 + 1198603ch120573119909 + 1198604sh120573119909 (5)

in which constants 1198611 and 1198612 depend upon the initialdisplacement and velocity conditions and real constants 119860 119894must be evaluated so as to satisfy the known boundaryconditions at the ends of the beam

The cantilever beam considered has a fixed end so its twoknown boundary conditions are

120601 (0) = 01206011015840 (0) = 0 (6)

Making use of (5) and its first partial derivative with respectto 119909 from (6) one obtains 1198603 = minus1198601 and 1198604 = minus1198602 So (5)becomes

120601 (119909) = 1198601 (cos120573119909 minus ch120573119909) + 1198602 (sin120573119909 minus sh120573119909) (7)

An additional rigid mass 1198981 having a rotary mass moment ofinertia 1198691 is attached by fixed connection to its free end as alsoshown in Figure 2These internal force components are alongwith the translational and rotary inertial force components11989811205962120601(119871) and 119869112059621206011015840(119871) respectively So the force andmoment equilibrium of the additional mass requires theboundary conditions

119872 (119871) = 11986411986812060110158401015840 (119871) = minus12059621206011015840 (119871) 1198691119881 (119871) = 119864119868120601101584010158401015840 (119871) = minus1205962120601 (119871) 1198981 (8)

Making use of (5) and its first second and third partialderivative with respect to 119909 and substituting them into (8)yield

[[[[

1205732 (cos120573119871 + ch120573119871) + 12059621198691119864119868 120573 (sin120573119871 + sh120573119871) 1205732 (sin120573119871 + sh120573119871) minus 12059621198691119864119868 120573 (cos120573119871 minus ch120573119871)1205733 (sin120573119871 minus sh120573119871) + 12059621198981119864119868 (cos120573119871 minus ch120573119871) minus1205733 (cos120573119871 + ch120573119871) + 12059621198981119864119868 (sin120573119871 minus sh120573119871)

]]]]

11986011198602 = 00 (9)

For coefficients 1198601 and 1198602 to be nonzero the determinantof the square matrix in this equation must equal zero thusgiving the frequency equation

[1205732 (cos120573119871 + ch120573119871) + 12059621198691119864119868 120573 (sin120573119871 + sh120573119871)]sdot [minus1205733 (cos120573119871 + ch120573119871) + 12059621198981119864119868 (sin120573119871 minus sh120573119871)]

minus [1205732 (sin120573119871 + sh120573119871) minus 12059621198691119864119868 120573 (cos120573119871 minus ch120573119871)]

sdot [1205733 (sin120573119871 minus sh120573119871) + 12059621198981119864119868 (cos120573119871 minus ch120573119871)]= 0

(10)


V(L) = EI120601998400998400998400 (L)

M(L) = EI120601998400998400(L)

minus1205962120601(L)m1

minus1205962120601998400(L)J1




+ (1 + cos120573119871ch120573119871)

(11)

Setting

119911 = 120573119871 = 4radic 1205962119898119864119868 119871 (12)






0

10

20

30

40

0 10 20 30 40 50

z

log(f(z

))

minus10

minus20

minus30

minus40

120572j = 10120572m = 1


0

05

1

15

2

25

3

0001 001 01 1 10 100 1000 10000120572J

120572m = 1000

120572m = 100

120572m = 10

120572m = 1

120572m = 01

120572m = 001

z min






Frequencynumber (i)


120572119898 = 1120572119869 = 100120572119898 = 1120572119869 = 10

120572119898 = 1120572119869 = 1120572119898 = 1120572119869 = 01

120572119898 = 1120572119869 = 001120572119898 = 1120572119869 = 0001

120572119898 = 1120572119869 = 0120572119898 = 1(ref [11])


001 01 1 10 100 10000

05

1

15

2

25

3

120572J = 100

120572J = 10

120572J = 1

120572J = 01

120572J = 001

120572m

z min








5 Conclusion




1st mode2nd mode

3rd mode4th mode


1st mode2nd mode

3rd mode4th mode

00 02 04 06 08 1000051015202530

Disp

lace

men

t


minus10

minus15

minus20

minus25


(a)

00 02 04 06 08 1000051015202530

Disp

lace

men

t


minus10

minus15

minus20

minus25



3rd mode4th mode

5th mode

3rd mode4th mode

5th mode

(b)



Competing Interests


Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










V(L) = EI120601998400998400998400 (L)

M(L) = EI120601998400998400(L)

minus1205962120601(L)m1

minus1205962120601998400(L)J1




+ (1 + cos120573119871ch120573119871)

(11)

Setting

119911 = 120573119871 = 4radic 1205962119898119864119868 119871 (12)






0

10

20

30

40

0 10 20 30 40 50

z

log(f(z

))

minus10

minus20

minus30

minus40

120572j = 10120572m = 1


0

05

1

15

2

25

3

0001 001 01 1 10 100 1000 10000120572J

120572m = 1000

120572m = 100

120572m = 10

120572m = 1

120572m = 01

120572m = 001

z min






Frequencynumber (i)


120572119898 = 1120572119869 = 100120572119898 = 1120572119869 = 10

120572119898 = 1120572119869 = 1120572119898 = 1120572119869 = 01

120572119898 = 1120572119869 = 001120572119898 = 1120572119869 = 0001

120572119898 = 1120572119869 = 0120572119898 = 1(ref [11])


001 01 1 10 100 10000

05

1

15

2

25

3

120572J = 100

120572J = 10

120572J = 1

120572J = 01

120572J = 001

120572m

z min








5 Conclusion




1st mode2nd mode

3rd mode4th mode


1st mode2nd mode

3rd mode4th mode

00 02 04 06 08 1000051015202530

Disp

lace

men

t


minus10

minus15

minus20

minus25


(a)

00 02 04 06 08 1000051015202530

Disp

lace

men

t


minus10

minus15

minus20

minus25



3rd mode4th mode

5th mode

3rd mode4th mode

5th mode

(b)



Competing Interests


Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Frequencynumber (i)


120572119898 = 1120572119869 = 100120572119898 = 1120572119869 = 10

120572119898 = 1120572119869 = 1120572119898 = 1120572119869 = 01

120572119898 = 1120572119869 = 001120572119898 = 1120572119869 = 0001

120572119898 = 1120572119869 = 0120572119898 = 1(ref [11])


001 01 1 10 100 10000

05

1

15

2

25

3

120572J = 100

120572J = 10

120572J = 1

120572J = 01

120572J = 001

120572m

z min








5 Conclusion




1st mode2nd mode

3rd mode4th mode


1st mode2nd mode

3rd mode4th mode

00 02 04 06 08 1000051015202530

Disp

lace

men

t


minus10

minus15

minus20

minus25


(a)

00 02 04 06 08 1000051015202530

Disp

lace

men

t


minus10

minus15

minus20

minus25



3rd mode4th mode

5th mode

3rd mode4th mode

5th mode

(b)



Competing Interests


Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1st mode2nd mode

3rd mode4th mode


1st mode2nd mode

3rd mode4th mode

00 02 04 06 08 1000051015202530

Disp

lace

men

t


minus10

minus15

minus20

minus25


(a)

00 02 04 06 08 1000051015202530

Disp

lace

men

t


minus10

minus15

minus20

minus25



3rd mode4th mode

5th mode

3rd mode4th mode

5th mode

(b)



Competing Interests


Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Documents

Frequency Equation of Flexural Vibrating Cantilever Beam