Research Article Flexural Free Vibrations of …downloads.hindawi.com/journals/mpe/2016/7314280.pdfResearch Article Flexural Free Vibrations of Multistep Nonuniform Beams GuojinTan,WenshengWang,andYuboJiao

Research ArticleFlexural Free Vibrations of Multistep Nonuniform Beams

Guojin Tan Wensheng Wang and Yubo Jiao

Department of Road and Bridge College of Transportation Jilin University Changchun 130022 China

Correspondence should be addressed to Yubo Jiao jiaoybjlueducn

Received 28 June 2016 Revised 20 September 2016 Accepted 11 October 2016

Academic Editor Salvatore Caddemi

Copyright copy 2016 Guojin Tan 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

This paper presents an exact approach to investigate the flexural free vibrations of multistep nonuniform beams Firstly one-stepbeam with moment of inertia and mass per unit length varying as 119868(119909) = 1205721(1 + 120573119909)119903+4 and 119898(119909) = 1205722(1 + 120573119909)119903 was studied Byusing appropriate transformations the differential equation for flexural free vibration of one-step beam with variable cross sectionis reduced to a four-order differential equation with constant coefficients According to different types of roots for the characteristicequation of four-order differential equation with constant coefficients two kinds of modal shape functions are obtained andthe general solutions for flexural free vibration of one-step beam with variable cross section are presented An exact approachto solve the natural frequencies and modal shapes of multistep beam with variable cross section is presented by using transfermatrix method the exact general solutions of one-step beam and iterative method Numerical examples reveal that the calculatedfrequencies and modal shapes are in good agreement with the finite element method (FEM) which demonstrates the solutions ofpresent method are exact ones

1 Introduction

Beams with nonuniform cross section are widely used invarious engineering fields such as bridges tall buildings andhelicopter rotor blades A large number of studies can befound in literature about the free vibrations of nonuniformbeams Vibration problems of beams with nonuniform crosssection are often described by partial differential equationsand in most cases it is extremely difficult to find their closedform solutions Consequently a wide range of approximateand numerical solutions such as Rayleigh-Ritz Galerkinfinite difference finite element and spectral finite elementmethods have been used to obtain the natural vibrationcharacteristics of variable-section beams [1ndash4]

Huang and Li investigated the free vibration of axiallyfunctionally graded beams with variable flexural rigidity andmass density [5] A novel and simple approach was pre-sented to solve modal shapes and corresponding frequenciesthrough transforming traditional fourth-order governing dif-ferential equation into Fredholm integral equation Koplow etal proposed an analytical solution for the dynamic responseof Euler-Bernoulli beams with step changes in cross section

which was verified by experimental tests and receptance cou-pling methods [6] Firouz-Abadi et al studied the transversefree vibrations of a class of variable-cross section beams usingWentzel-Kramers-Brillouin (WKB) approximation [7] Thegoverning equation of motion for the Euler-Bernoulli beamincluding axial force distribution was utilized to obtain a sin-gular differential equation in terms of the natural frequencyof vibration and a WKB expansion series was applied to findthe solution Inaudi andMatusevich investigated longitudinalvibration problems of variable-cross section rods using animproved power series method [8] This method introduceddomain partition implementation in matrix formulation asan alternative to other power series techniques in vibrationanalysis Therefore the method solved linear differentialequations efficiently up to a desired degree of accuracy andremedies two limitations of the conventional power seriesmethod The Adomian decomposition method (ADM) isemployed to investigate the free vibrations of tapered Euler-Bernoulli beams with a continuously exponential variation ofwidth and a constant thickness along the length under variousboundary conditions [9] Duan and Wang demonstrated thefree vibration of beams with multiple step changes using the

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2016 Article ID 7314280 12 pageshttpdxdoiorg10115520167314280

2 Mathematical Problems in Engineering

modified discrete singular convolution (DSC) [10] The jumpconditions at the steps were used to overcome the difficultyin using ordinary DSC for dealing with ill-posed problemsA transfer matrix method and the Frobenius method wereadopted by J W Lee and J Y Lee to solve the free vibrationcharacteristics of a tapered Bernoulli-Euler beam and obtainthe power series solution for bending vibrations [11]

Nevertheless besides all advantages of such numericalmethods exact solutions can provide adequate insight intothe physics of the problems and convenience for parametricstudies The other advantage of exact solutions is their signif-icance in the field of inverse problems An exact solution canbemore useful than numerical solutions to design the charac-teristics and damage identification of a structure

Wang derived the closed form solutions for free vibrationof a flexural bar with variably distributed stiffness but uni-formmass [12] Using a systematic approach Abrate obtaineda closed form solution of longitudinal vibration for rodswhose cross section varies as 119860(119909) = 1198600(1 + 119886[119909119871])2 [13]Kumar and Sujith found the exact solutions for longitudinalvibration of nonuniform rods whose cross section varies as119860 = (119886 + 119887119909)119899 and 119860 = 1198600 sin2(119886 + 119887119909) [14] Li presentedan exact approach for free longitudinal vibrations of one-stepnonuniform rod with classical and nonclassical boundaryconditions [15] The approach assumed that the distributionof mass is arbitrary and distribution of longitudinal stiffnessis expressed as a functional relation with mass distributionand vice versa Li et al obtained exact solutions of flexuralvibration for beam-like structures whose moment of inertiaandmass per unit length vary as119864119868(119909) = 119886(1+120573119909)119899+4119898(119909) =120572(1 + 120573119909)119899 and 119864119868(119909) = 119886 sdot exp(minus119887119909) 119898(119909) = 120572 sdot exp(minus119887119909)respectively [16 17]

The components whose moment of inertia 119868(119909) and massper unit length119898(119909) satisfy 119868(119909) = 1205721(1 + 120573119909)119903+4 and119898(119909) =1205722(1 + 120573119909)119903 are widely used in civil engineering such as thebridgewith cross section height varying as ℎ(119909) = ℎ0(1+120573119909)2This kind of beam-like structure is usually solved by dividingit into several segments whose 119868(119909) and119898(119909) satisfy differentdistributions The paper derived the exact general solutionof one-step nonuniform beam firstly and then obtained theexact solution of multistep nonuniform beam by combiningthe transfer matrix method and exact solutions of one-stepbeam The exact solutions of present method not only canprovide convenience for parametric studies but also are veryuseful for inverse problems such as damage identification

2 Modal Shape Function of One-Step Beam

21 Beams with Uniform Cross Section The governing differ-ential equation for undamped free flexural vibration of beamwith uniform cross section can be written as [18]

1198641198681205974119910 (119909 119905)1205971199094 + 1198981205972119910 (119909 119905)1205971199052 = 0 (1)

where 119864 is Youngrsquos modulus 119868 is moment of inertia 119898 ismass per unit length and 119910(119909 119905) is transverse displacementat position 119909 and time 119905

Assuming the beam performs a harmonic free vibrationat equilibrium position that is

119910 (119909 119905) = 120601119906 (119909) 119890119895120596119905 (2)

where 120601119906(119909) is mode shape function of uniform beam 120596 iscorresponding natural frequency

Inserting (2) into (1) obtains

1198641198681198894120601119906 (119909)1198891199094 minus 1205962119898120601119906 (119909) = 0 (3)

Equation (3) leads to the modal shape function of beamwith uniform cross section

120601119906 (119909) = 4sum119894=1

119861119894119878119906119894 (119909) (4)

where 1198781199061(119909) = sinh 119896119909 1198781199062(119909) = cosh 119896119909 1198781199063(119909) = sin 1198961199091198781199064(119909) = cos 119896119909 are 119861119894 (119894 = 1 2 3 4) are integration constants1198964 = 120596211989811986411986822 Beams with Variable Cross Section The governing differ-ential equation for undamped free flexural vibration of beamwith variable cross section can be written as [19]

12059721205971199092 [119864119868 (119909) 1205972119910 (119909 119905)1205971199092 ] + 119898 (119909) 1205972119910 (119909 119905)1205971199052 = 0 (5)

where 119864 is Youngrsquos modulus 119868(119909) is bendingmoment of iner-tia at position 119909 119898(119909) is mass per unit length at position 119909and119910(119909 119905) is transverse displacement at position119909 and time 119905


119910 (119909 119905) = 120601 (119909) 119890119895120596119905 (6)

where 120601(119909) is mode shape function of beam 120596 is naturalfrequency

Substituting (6) into (5) arrives at

11988921198891199092 [119864119868 (119909) 1198892120601 (119909)1198891199092 ] minus 1205962119898(119909) 120601 (119909) = 0 (7)

Themoment of inertia 119868(119909) andmass per unit length119898(119909)of beam are assumed to vary as

119868 (119909) = 1205721 (1 + 120573119909)119903+4 119898 (119909) = 1205722 (1 + 120573119909)119903

(1205721 1205722 120573 = 0) (8)

where 1205721 1205722 and 120573 are arbitrary constants but not equal to 0and 119903 is a positive integer

Mathematical Problems in Engineering 3

Substituting (8) into (7) yields

0 = 1198641205721 (1 + 120573119909)4 1198891206014 (119909)1198891199094+ 21198641205721120573 (119903 + 4) (1 + 120573119909)3 1198891206013 (119909)1198891199093+ 11986412057211205732 (119903 + 4) (119903 + 3) (1 + 120573119909)2 1198891206012 (119909)1198891199092minus 12059621205722120601 (119909)

(9)

Let

119911 = ln (1 + 120573119909) 119863 = 119889119889119911

(10)

It can be derived that

119889120601119896 (119909)119889119909119896 (1 + 120573119909)119896= 120573119896119863 (119863 minus 1) (119863 minus 2) sdot sdot sdot (119863 minus 119896 + 1) 120601 (119911)

(119896 = 1 2 119899) (11)

Introducing (11) into (9) one arrives at

0 = 11986412057211205734119863 (119863 minus 1) (119863 minus 2) (119863 minus 3) 120601 (119911)+ 211986412057211205734 (119903 + 4)119863 (119863 minus 1) (119863 minus 2) 120601 (119911)+ 11986412057211205734 (119903 + 4) (119903 + 3)119863 (119863 minus 1) 120601 (119911)minus 12059621205722120601 (119911)

(12)

Equation (12) can be simplified as

0 = [1198634 + (2 (119903 + 4) minus 6)1198633+ ((119903 + 4) (119903 + 3) minus 6 (119903 + 4) + 11)1198632+ (4 (119903 + 4) minus (119903 + 4) (119903 + 3) minus 6)119863 minus 1205962120572211986412057211205734]sdot 120601 (119911)

(13)

Let

1198861 = 2 (119903 + 4) minus 61198862 = (119903 + 4) (119903 + 3) minus 6 (119903 + 4) + 111198863 = 4 (119903 + 4) minus (119903 + 4) (119903 + 3) minus 61198864 = minus 1205962120572211986412057211205734

(14)

Equation (13) can be written as

(1198634 + 11988611198633 + 11988621198632 + 1198863119863 + 1198864) 120601 (119911) = 0 (15)

The characteristic equation of (15) is

1198894 + 11988611198893 + 11988621198892 + 1198863119889 + 1198864 = 0 (16)

The four roots of (16) are

11988912 = 119892 minus 11988612 plusmn radic(11988612 minus 119892)2 minus 4 (1199052 minus 119891)2 11988934 = minus119892 minus 11988612 plusmn radic(11988612 + 119892)2 minus 4 (1199052 + 119891)2

(17)

where

119905 = 3radicminus1199022 + radic11990224 + 119901327 + 3radicminus1199022 minus radic11990224 + 119901327 + 11988623 119905 is a real root

119901 = 11988611198863 minus 1311988622 minus 41198864119902 = 13119886111988621198863 minus 22711988632 minus 119886211198864 + 8311988621198864 minus 11988623 119891 = 1003816100381610038161003816100381610038161003816100381610038161003816(

11990524 minus 1198864)121003816100381610038161003816100381610038161003816100381610038161003816 119892 = (11988611199054 minus 11988632)119891

(18)

Themodal shape function of one-step beamwith variablecross section can be written as

120601 (119909) = 11988811198901198891119911 + 11988821198901198892119911 + 11988831198901198893119911 + 11988841198901198894119911 (19)

where 1198881 1198882 1198883 and 1198884 are four undetermined coefficientsSince 1198864 lt 0 then119891 gt 1199052 and 1199052minus119891 lt 0 it can be found

that 1198891 and 1198892 are two real roots in the first equation of (17)From the second equation in (17) it can be found that when(11988612+119892)2 gt 4(1199052+119891) 1198893 and 1198894 are also two real roots andwhen (11988612 + 119892)2 lt 4(1199052 + 119891) 1198893 and 1198894 are two conjugatecomplex roots

Modal shapes of one-step beam are determined by theroots as follows

(1) When 1198891 1198892 1198893 and 1198894 are all real roots substituting119911 = ln(1 + 120573119909) into (19) one obtains120601 (119909) = 1198611 (1 + 120573119909)1198891 + 1198612 (1 + 120573119909)1198892

+ 1198613 (1 + 120573119909)1198893 + 1198614 (1 + 120573119909)1198894 (20)

(2)When 1198891 and 1198892 are real roots while 1198893 and 1198894 are twoconjugate complex roots let

1199041 = minus119892 minus 119886122 1199042 =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816radic4 (1199052 + 119891) minus (11988612 + 119892)22

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816 (21)


1 k minus 1 k k + 1 m minus 1 m

Figure 1 Multistep beam

the roots 1198893 and 1198894 can be expressed as

1198893 = 1199041 + 11990421198951198894 = 1199041 minus 1199042119895 (22)

where 119895 = radicminus1Substituting 119911 = ln(1 + 120573119909) and (22) into (19) yields

120601 (119909) = 1198611 (1 + 120573119909)1198891 + 1198612 (1 + 120573119909)1198892+ 1198613 (1 + 120573119909)1199041 cos 1199042 ln (1 + 120573119909)+ 1198614 (1 + 120573119909)1199041 sin 1199042 ln (1 + 120573119909)

(23)

In (20) and (23) 119861119894 (119894 = 1 2 3 4) have the same meanings asthose in (4) and they are integration constants The solutionsof (20) and (23) are the exact general solutions only at the con-dition of 120573 = 0This is because 119911will be equal to zero in (10) iffor 120573 = 0 And the derivation of constant term for the secondequation in (10) has no mathematical meaning Meanwhilefrom the numerical point of view small value of 120573 (120573 rarr 0) isadopted and the uniform beam is approached then it couldbe solved by (20) or (23) However compared to solutions ofuniform beam by (4) the solutions of approximate uniformbeam are not exact general solutions but numerical solutions

Equations (20) and (23) can be expressed with a uniformstyle as

120601 (119909) = 4sum119894=1

119861119894119878119894 (119909) (24)

when the modal shape function is (20) 119878119894(119909) = (1 + 120573119909)119889119894for 119894 = 1 2 3 4 and when it is (23) 119878119894(119909) = (1 + 120573119909)119889119894 for119894 = 1 2 and 1198783(119909) = (1 + 120573119909)1199041 cos 1199042 ln(1 + 120573119909) 1198784(119909) =(1 + 120573119909)1199041 sin 1199042 ln(1 + 120573119909)3 Transfer Relationship for

Undetermined Coefficients of MultistepBeam Modal Shapes

As shown in Figure 1 the multistep beam is divided into 119898segments For an arbitrary beam segment 119896 (119896 = 1 2 119898)with length 119897119896 a local Cartesian coordinated system isestablished with the origin locating at the left end of segment119868119896(119909) is bending moment of inertia for the 119896th segment and120601119896(119909) is mode shape function for the 119896th segment where 119909 isdefined as belonging to the 119896th segment in the local Cartesiancoordinated system

Taking the 119896th segment for example the continuity ofdeformations equilibrium of moments and shear forces are

satisfied at the right end of the 119896th segment and left end of the(119896 + 1)th segment that is

120601119896 (119909)1003816100381610038161003816119909=119897119896 = 120601119896+1 (119909)1003816100381610038161003816119909=0 1206011015840119896 (119909)10038161003816100381610038161003816119909=119897119896 = 1206011015840119896+1 (119909)10038161003816100381610038161003816119909=0 119864 [119868119896 (119909) sdot 12060110158401015840119896 (119909)]10038161003816100381610038161003816119909=119897119896 = 119864 [119868119896+1 (119909) sdot 12060110158401015840119896+1 (119909)]10038161003816100381610038161003816119909=0 119889 [119864119868119896 (119909) sdot 12060110158401015840119896 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=119897119896= 119889 [119864119868119896+1 (119909) sdot 12060110158401015840119896+1 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=0

(25)

where prime denotes the derivative with respect to localcoordinate 119909

Substituting modal shape functions of the 119896th and (119896 +1)th segments ((4) or (24)) into (25) yields

4sum119894=1

119861119896119894 119878119896119894 (119897119896) = 4sum119894=1

119861119896+1119894 119878119896+1119894 (0) 4sum119894=1

119861119896119894 1198781015840119896119894 (119897119896) = 4sum119894=1

119861119896+1119894 1198781015840119896+1119894 (0) 119868119896 (119897119896) 4sum

119894=1

119861119896119894 11987810158401015840119896119894 (119897119896) = 119868119896+1 (0) 4sum119894=1

119861119896+1119894 11987810158401015840119896+1119894 (0) 1198681015840119896 (119897119896) 4sum

119894=1

119861119896119894 11987810158401015840119896119894 (119897119896) + 119868119896 (119897119896) 4sum119894=1

119861119896119894 119878101584010158401015840119896119894 (119897119896)= 1198681015840119896+1 (0) 4sum

119894=1

119861119896+1119894 11987810158401015840119896+1119894 (0)+ 119868119896+1 (0) 4sum

119894=1

119861119896+1119894 119878101584010158401015840119896+1119894 (0)

(26)

where 119861119896119894 denotes the 119894th undetermined coefficient of modalshape function of the 119896th segment and 119878119896119894 is the 119894th term inmodal shape function of the 119896th segment

Equation (26) can be rewritten into a matrix form

H119896U119896 = H119896+1U119896+1 (27)

where

U119896 = 1198611198961 1198611198962 1198611198963 1198611198964119879 U119896+1 = 119861119896+11 119861119896+12 119861119896+13 119861119896+14 119879 H119896 = [H1198961 H1198962 H1198963 H1198964]

H119896+1 = [H(119896+1)1 H(119896+1)2 H(119896+1)3 H(119896+1)4] (28)


in which

H119896119894 =[[[[[[[

119878119896119894 (119897119896)1198781015840119896119894 (119897119896)119868119896 (119897119896) 11987810158401015840119896119894 (119897119896)1198681015840119896 (119897119896) 11987810158401015840119896119894 (119897119896) + 119868119896 (119897119896) 119878101584010158401015840119896119894 (119897119896)

]]]]]]]

H(119896+1)119894 =[[[[[[[

119878119896+1119894 (0)1198781015840119896+1119894 (0)

119868119896+1 (0) 11987810158401015840119896+1119894 (0)1198681015840119896+1 (0) 11987810158401015840119896+1119894 (0) + 119868119896+1 (0) 119878101584010158401015840119896+1119894 (0)

]]]]]]]

(29)

Let

T119896 = Hminus1119896+1H119896 (30)

then

U119896+1 = Hminus1119896+1H119896U119896 = T119896U119896 (31)

Equation (31) represents the transfer relationship of fourundetermined coefficients for modal shapes of the 119896th and(119896 + 1)th segments T119896 is the transfer matrix In accordancewith (31) transfer relationship between the first segment andthe last segment can be expressed as

U119898 = T119898minus1T119898minus2 sdot sdot sdot T2T1U1 = TU1 (32)

where T = T119898minus1T119898minus2 sdot sdot sdot T2T14 Solutions of Natural Frequencies

and Modal Shapes

Based on the modal shape function of one-step beam ((4) or(24)) and transfer relationships for undetermined coefficientsof modal shapes for each segment ((32)) natural frequenciesand modal shapes of multistep beam can be determinedby boundary conditions of the beam The cantilever andsimply supported boundaries are considered in this paperrespectively

For cantilever multistep beam the boundary conditionsare

1206011 (0) = 012060110158401 (119909)10038161003816100381610038161003816119909=0 = 0

119864119868119898 (119909) sdot 12060110158401015840119898 (119909)10038161003816100381610038161003816119909=119897119898 = 0119864119889 [119868119898 (119909) sdot 12060110158401015840119898 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=119897119898 = 0(33)

Inserting the modal shape functions into (33) and utiliz-ing transfer relationship in (32) one arrives at

[ C1C119898T

]U1 = 0 (34)

where

C1 = [11987811 (0) 11987812 (0) 11987813 (0) 11987814 (0)119878101584011 (0) 119878101584012 (0) 119878101584013 (0) 119878101584014 (0)] C119898 = [119878101584010158401198981 (119897119898) 119878101584010158401198982 (119897119898) 119878101584010158401198983 (119897119898) 119878101584010158401198984 (119897119898)1198781015840101584010158401198981 (119897119898) 1198781015840101584010158401198982 (119897119898) 1198781015840101584010158401198983 (119897119898) 1198781015840101584010158401198984 (119897119898)]

(35)

For simply supported multistep beam the boundaryconditions are

1206011 (0) = 01198641198681 (119909) sdot 120601101584010158401 (119909)10038161003816100381610038161003816119909=0 = 0

120601119898 (119897119898) = 0119864119868119898 (119909) sdot 12060110158401015840119898 (119909)10038161003816100381610038161003816119909=119897119898 = 0

(36)

Similar to (34) one obtains

[ C10158401C1015840119898T

]U1 = 0 (37)

where

C10158401 = [ 11987811 (0) 11987812 (0) 11987813 (0) 11987814 (0)1198781015840101584011 (0) 1198781015840101584012 (0) 1198781015840101584013 (0) 1198781015840101584014 (0)] C1015840119898 = [ 1198781198981 (119897119898) 1198781198982 (119897119898) 1198781198983 (119897119898) 1198781198984 (119897119898)119878101584010158401198981 (119897119898) 119878101584010158401198982 (119897119898) 119878101584010158401198983 (119897119898) 119878101584010158401198984 (119897119898)]

(38)

Equations (34) and (37) can be written into a uniformstyle

CU1 = [[[[[[

11988811 11988812 11988813 1198881411988821 11988822 11988823 1198882411988831 11988832 11988833 1198883411988841 11988842 11988843 11988844]]]]]]

11986111119861121198611311986114

= 0 (39)

where the elements 119888119894119895 (119894 119895 = 1 2 3 4) in C are functions ofnatural frequencies 120596 for multistep beam

Existence of nontrivial solutions of (39) leads to thefollowing frequency equation

|C| = 0 (40)

If the frequency region [120596119886119894 120596119887119894] for the 119894th naturalfrequency 120596119894 of multistep beam and the expressions of modalshape functions for each segment ((20) or (23)) can bedetermined then the expressions of elements in matrix Ccan be uniquely determined The 119894th natural frequency 120596119894 ofmultistep beam can be calculated by (40) directly and unde-termined coefficients inmodal shape function of the first seg-ment can be solved by (39) then the undetermined coeffici-ents for other segments can be obtained by using transferrelationship in (32)


x

y

A B C5 5

hA hB hC

Figure 2 Two-step beam

If the frequency region or expressions of modal shapefunction for each segment ((20) or (23)) cannot be deter-mined the natural frequencies and modal shapes can besolved by the half-interval method [20] The detailed processfor determining natural frequencies is as follows Firstly aninitial value of the 119894th natural frequency 120596119886119894 is assumed andthe expressions for modal shape function of each segmentcorresponding to 120596119886119894 can be determined ((20) or (23)) thenthe coefficient matrix C corresponding to 120596119886119894 is obtainedand the determinant of matrix C is calculated (denoted as119863119886119894 = |C(120596119886119894)|) Then a new value 120596119887119894 = 120596119886119894 + Δ120596 withΔ120596 (eg Δ120596 = 05) representing the increment of 120596 isassumed and the same calculations are repeated to determinethe new determinant corresponding to 120596119887119894 (denoted as119863119887119894 =|C(120596119887119894)|) If 119863119886119894 and 119863119887119894 have opposite signs there is at leastone natural frequency in the interval [120596119886119894 120596119887119894] otherwiselet 120596119886119894 = 120596119887119894 and continue the above procedures until 119863119886119894and 119863119887119894 have opposite signs For the next step let 120596119888119894 =(120596119886119894 + 120596119887119894)2 if119863119886119894 and119863119888119894 have the same sign let 120596119886119894 = 120596119888119894otherwise let 120596119887119894 = 120596119888119894 a new interval [120596119886119894 120596119887119894] is obtainedThe same calculations are repeated to determine the newinterval [120596119886119894 120596119887119894] until the rank of matrix C is equal to threeThe accurate values of 120596 are obtained respectively using thehalf-interval method

5 Numerical Examples

To verify the correctness of present analyticalmethod severalnumerical examples are studied in this paper The materialsused in these examples are all the same that is Yongrsquos modu-lus is 325 times 1010 Pa and density is 2500 kgm3 Poissonrsquos ratiois ] = 03 in a two-dimensional FEM (2D-FEM) analysis

51 Reliability of the Proposed Method

511 Validation by Finite Element Method In order toillustrate the proposed method a two-step beam with 10mspan as shown in Figure 2 is used as numerical model andsolved by one-dimensional FEM (1D-FEM) 2D-FEM andthe present analytical method respectively

Cross section of this beam is rectangular with constantwidth 03m The heights of sections at 119860 119861 and 119862 areexpressed by ℎ119860 ℎ119861 and ℎ119862 respectively ℎ119860 = ℎ119862 ℎ119861 =07ℎ119860 And the height of each step varies nonlinearly as

119860 to 119861ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 (0 le 119909 le 5)

120573119860119861 = radicℎ119861ℎ119860 minus 15 = radic710 minus 15 (41)

Figure 3 A two-dimensional ANSYS model for 2D-FEM analysis(119871ℎ119860 = 10)

119861 to 119862ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 (120585 = 119909 minus 5 5 le 119909 le 10)


For 1D-FEM The finite element method with two-nodesbeam element Each node of the beam element has twodegree-of-freedoms that is transverse displacement androtation angle The cubic Hermite polynomial is adoptedas shape function for beam element Therefore the shapefunctions119873119894 (119894 = 1 2 3 4) can be expressed as

1198731 = (1198973 minus 31198971199092 + 21199093)1198973 1198732 = (1198972119909 minus 21198971199092 + 1199093)1198972 1198733 = (31198971199092 minus 21199093)1198973 1198734 = (1199093 minus 1198971199092)1198972

0 le 119909 le 119897

(43)

In (43) 119897 is the length of beam element and 119897 = 05mLet N = 1198731 1198732 1198733 1198734 stiffness matrix and mass

matrix of the beam element can be evaluated as

K119890=int1198970B119879119864119868 (119909)B 119889119909

M119890 = int1198970119898(119909)N119879N 119889119909

(44)

where B = 1198892N1198891199092 is the strain matrix of beam elementUsing above stiffness matrix and mass matrix finite elementanalysis is carried on by the codes written in MATLAB

For 2D-FEM A finite element analysis is carried out usingANSYS software A 2D-FEMmodel of two-step beam is builtwith 8-node PLANE183 elements as shown in Figure 3

The ratios of span and height at 119860 (119871ℎ119860) vary from 10 to100 ℎ119860 ℎ119861 and ℎ119862 are variables with 119871ℎ119860 and 120573119860119861 and 120573119861119862are constants Relative errors of the first two-order naturalfrequencies obtained by 1D-FEM and 2D-FEM with respectto present analytical method are shown in Figure 4

As can be seen from Figure 4 the relative errors between1D-FEM and present method remain at a very small leverall the time no matter how 119871ℎ119860 changes and the relative


10 20 30 40 50 60 70 80 90 100

Relative error of 2nd frequency between 1D-FEM andpresent methodRelative error of 2nd frequency between 2D-FEM andpresent method

Relative error of 1st frequency between 2D-FEM andpresent method


minus50

minus45

minus40

minus35

minus30

minus25

minus20

minus15

minus10

minus05

00

05

Rela

tive

erro

r (

)

LhA

Figure 4 Relative error of 1D-FEM and 2D-FEM

errors between 2D-FEM and the present method decreaserapidly with 119871ℎ119860 increasingThis is because 1D-FEMand thepresent method are employed without considering the effectof shear deformation and the stiffness and mass matrixes areformed by integration method for the variable moment ofinertia and mass in 1D-FEM modeling However the sheardeformation is inevitable for a two-dimensional FEMmodelWhen 119871ℎ119860 reaches a certain lever the error between 2D-FEM and the present method becomes smaller For examplewhen 119871ℎ119860 = 60 the frequencies obtained by 2D-FEM andthe present method are very close and relative error of thefirst two order frequencies is about minus01 With regard tobeam-like structural free vibration it can be treated as planestress problem so a 2D-FEM analysis is more accurate thana 1D-FEM analysis in theory For easy of calculation thebeam-like structure can be treated as Euler-Bernoulli beamwhen 119871ℎ119860 reaches a certain lever and the effect of sheardeformation could be ignored And the object of this paperis not considering the effect of shear deformation so thesolutions of the proposed method are exact ones based onEuler-Bernoulli beam theory This leads to different relativeerrors of 1D-FEM and 2D-FEM shown in Figure 4 It alsoreveals that the present method possesses favorable accuracyfor the larger ratio of span and sectional height

512 Numerical Simulation for Small Value of 120573 From thenumerical point of view small value of 120573 (120573 rarr 0) isadopted and the uniform beam is approached In order toverify the explanations a one-step beam with small valueof 120573 (shown in Figure 5) and a one-step uniform beam(shown in Figure 6) are used for numerical simulation by the

x

y

A B15

hA hB

Figure 5 One-step beam with small value of 120573

x

15A B

y

h

Figure 6 One-step uniform beam

present method ((20) and (23) for nonuniform beams (4) foruniform beams)

One-Step Beamwith Small Value of 120573The span is 15m Crosssection of this beam is rectangular with constant width 03m

The heights of sections at 119860 and 119861 are ℎ119860 = 025m andℎ119861 = 0249m respectively and the height varies nonlinearlyfrom 119860 to 119861 as

ℎ (119909) = ℎ119860 (1 + 120573119909)2 (0 le 119909 le 15) 120573 = radicℎ119861ℎ119860 minus 115 = radic249250 minus 115 (120573 997888rarr 0) (45)

Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (46)

One-Step Uniform Beam The beam is also 15m span withconstant width 03m and the height is 0249m

The one-step beams with different boundary conditionshave been analyzed

Case 1 A cantilever beam model

Case 2 A simply supported beam model

The first three natural frequencies of one-step beam withsmall value of 120573 (120573 rarr 0) calculated by (20) or (23) arecompared with those of one-step uniform beam calculated by(4) the results are listed in Tables 1 and 2

As can be seen from the results the relative errorsbetween nonuniformbeamwith small value of120573 and uniform


Table 1 Calculation results of natural frequencies for Case 1

Beam type One-step beam with small value of 120573 One-step uniform beam with height 0249m Relative error ()1st (rads) 40680 40499 minus044512nd (rads) 254507 253805 minus027553rd (rads) 712273 710663 minus02261


Beam type One-step beam with small value of 120573 One-step uniform beam with height 0249m Relative error ()1st (rads) 113912 113684 minus020022nd (rads) 455647 454734 minus020033rd (rads) 1025205 1023152 minus02003beam are very small It reveals that the uniform beam isapproached when 120573 is small enough (120573 rarr 0) However forsmall value of 120573 representing a uniform beam the solutionsare not exact general solutions but numerical solutions

52 One-Step Beam A one-step beam with 15m span asshown in Figure 5 is used for numerical simulation Crosssection of this beam is rectangular with constant width 03mThe heights of sections at 119860 and 119861 are ℎ119860 = 025m andℎ119861 = 0175m respectively and the sectional height variesnonlinearly from 119860 to 119861 as

ℎ (119909) = ℎ119860 (1 + 120573119909)2 (0 le 119909 le 15) 120573 = radicℎ119861ℎ119860 minus 115 = radic710 minus 115 (47)

Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (48)

The one-step beam with different boundary conditionshas been analyzed



The first three natural frequencies of one-step beam withdifferent boundary conditions calculated by the proposedmethod and FEM are listed in Tables 3 and 4 modal shapesare shown in Figures 7 and 8

As can be seen from the results calculated by the proposedmethod and FEM relative errors for the first three naturalfrequencies of the one-step beam with different boundaryconditions are very small It reveals that the solutions forone-step beam with variable cross section by the proposedmethod are exact ones The reasons lie in that only thedifferential equations of motion for one-step beam withvariable cross section are used to make the solutions and noother assumptions are introduced

2nd mode-2D-FEM1st mode-present

3rd mode-present2nd mode-present

1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts0 3

Length (m)6 9 12 15

Figure 7 Modal shapes of Case 3

0 3 6 9 12 15Length (m)



1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts


53 Multistep Beam To verify the correctness of presentmethod for free vibration analysis of multistep beams a two-step simply supported beam with variable cross section (asshown in Figure 9) is used for numerical simulation firstlyThe cross section of this beam is rectangular The segments


x

y

A B C10 12

hA hB hC

(a)

A

A

B

B

C

C

bA bB bC

(b)

Figure 9 Two-step simply supported beam (a) elevation graph (b) plane graph


Methods Present (rads) 1D-FEM (rads) 2D-FEM (rads) Relative error ()1D-FEM 2D-FEM

1st 42234 42229 42232 minus00120 minus000472nd 228615 228680 228406 00283 minus009143rd 613271 613522 611800 00410 minus02399



1st 95349 95398 95316 00511 minus003502nd 383529 383716 383029 00487 minus013033rd 862536 862964 859980 00495 minus02964119860119861 and 119861119862 have variable heights and widths In the two-dimensional FEM analysis a fine mesh of elements is usedand the average thickness of element is adopted for elementthickness

For segment119860119861 the heights of sections119860 and 119861 are ℎ119860 =036m and ℎ119861 = 024m respectively the widths of sections119860 and 119861 are 119887119860 = 03m and 119887119861 = 02449m respectively Theheight and width from 119860 to 119861 vary as

ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 119887119860119861 (119909) = 119887119860 (1 + 120573119860119861119909)

(0 le 119909 le 10) 120573119860119861 = radicℎ119861ℎ119860 minus 110 = radic23 minus 110

(49)

Therefore

119868119860119861 (119909) = 119887119860119861 (119909) sdot ℎ3119860119861 (119909)12 = 119887119860ℎ3119860 (1 + 120573119860119861119909)712 119898119860119861 (119909) = 120588119887119860119861 (119909) sdot ℎ119860119861 (119909) = 120588119887119860ℎ119860 (1 + 120573119860119861119909)3

(0 le 119909 le 10) (50)

For segment 119861119862 the heights of sections 119861 and 119862 are ℎ119861 =024m and ℎ119862 = 038m respectively the widths of sections 119861

and 119862 are 119887119861 = 02449m and 119887119862 = 03082m respectively Theheight and width from 119861 to 119862 vary as

ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 119887119861119862 (120585) = 119887119861 (1 + 120573119861119862120585)

(120585 = 119909 minus 10 10 le 119909 le 22) 120573119861119862 = radicℎ119862ℎ119861 minus 112 = radic1912 minus 112

(51)

Therefore

119868119861119862 (120585) = 119887119861119862 (120585) sdot ℎ3119861119862 (120585)12 = 119887119861ℎ3119861 (1 + 120573119861119862120585)712 119898119861119862 (120585) = 120588119887119861119862 (120585) sdot ℎ119861119862 (120585) = 120588119887119861ℎ119861 (1 + 120573119861119862120585)3

(0 le 120585 = 119909 minus 10 le 12) (52)

The first three natural frequencies of two-step simplysupported beam are calculated by the proposed method andFEM which are listed in Table 5Themodal shapes are shownin Figure 10

Secondly a three-step simply supported beam with vari-able cross section as shown in Figure 11 is used for numericalsimulation Cross section of this beam is rectangular withconstant width 03mThe segments119860-119861 and119862-119863 are variablesections and 119861-119862 is uniform cross section


Table 5 Calculation results of natural frequencies of two-step simply supported beam


1st 57666 57712 57570 00805 minus016562nd 251551 251761 251157 00838 minus015663rd 571504 572114 569501 01068 minus03490

Table 6 Calculation results of natural frequencies of three-step simply supported beam


1st 52110 52132 52089 00414 minus004022nd 216022 215989 215683 minus00155 minus015713rd 502561 503032 501738 00938 minus01638

0 2 4 6 8 10 12 14 16 18 20 22Length (m)

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM

Figure 10 Modal shapes of two-step simply supported beam

For segment119860119861 the heights of sections119860 and 119861 are ℎ119860 =03m and ℎ119861 = 02m respectivelyThe height of section from119860 to 119861 varies as

ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 (0 le 119909 le 5) 120573119860119861 = radicℎ119861ℎ119860 minus 15 = radic23 minus 15 (53)

Therefore

119868119860119861 (119909) = 119887ℎ3119860119861 (119909)12 = 119887ℎ3119860 (1 + 120573119860119861119909)612 119898119860119861 (119909) = 120588119887ℎ119860119861 (119909) = 120588119887ℎ119860 (1 + 120573119860119861119909)2

(0 le 119909 le 5) (54)

x

y

A B C5 10 5

D

hA hB hC hD

Figure 11 Three-step simply supported beam

For segment119862119863 the heights of sections119862 and119863 are ℎ119862 =02m and ℎ119863 = 03mTheheight of section from119862 to119863 variesas

ℎ119862119863 (120585) = ℎ119862 (1 + 120573119862119863120585)2 (0 le 120585 = 119909 minus 15 le 5) 120573119862119863 = radicℎ119863ℎ119862 minus 15 = radic32 minus 15 (55)

Therefore

119868119862119863 (120585) = 119887ℎ3119862119863 (120585)12 = 119887ℎ3119862 (1 + 120573119862119863120585)612 119898119862119863 (120585) = 120588119887ℎ119862119863 (120585) = 120588119887ℎ119862 (1 + 120573119862119863120585)2

(0 le 120585 = 119909 minus 15 le 5) (56)

The first three natural frequencies of three-step simplysupported beamcalculated by the proposedmethod andFEMare listed in Table 6 and the modal shapes are shown inFigure 12

From Tables 5 and 6 and Figures 10 and 12 for two-step and three-step simply supported beams it can be foundthat the first three natural frequencies and modal shapescalculated by the proposed method are in good agreementwith those calculated by FEM It reveals that the solutions formultistep beam with variable cross section by the proposedmethod are exact ones This is because only transfer matrixmethod and exact solutions of one-step beam with variablecross section are used to solve the equations of vibrationfor multistep beam with variable cross section and no otherassumptions are introduced


0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM

Figure 12 Modal shapes of three-step simply supported beam

6 Conclusions

In this paper an exact approach to investigate the flexuralfree vibrations of multistep nonuniform beams is presentedThe differential equation for flexural free vibration of one-step beam with 119868(119909) = 1205721(1 + 120573119909)119903+4 and 119898(119909) = 1205722(1 +120573119909)119903 is reduced to a four-order differential equation withconstant coefficients by using appropriate transformationsAnd the general solutions of one-step beam are obtainedwhose modal shape functions are found to have two kindsof expressions Then combining transfer method iterativemethod and the general solutions of one-step beam thesolving method for natural frequencies and modal shapesof multistep beam is formulated Numerical simulations onmultistep nonuniform beam are used to verify the feasibilityThe following conclusions can be obtained

(1) The comparison of relative errors among the presentanalytical method and finite element methods revealsthat the solutions of proposed method are exact onesbased on Euler-Bernoulli beam theory The presentmethod is suitable for the larger ratio of span andsectional height and possesses favorable accuracy

(2) The relative errors between nonuniform beam withsmall value of 120573 and uniform beam are very smallIt reveals that for small value of 120573 representing auniform beam the solutions are not exact generalsolutions but numerical solutions

(3) Numerical studies of one-step cantilever beam andsimply supported beam indicate that natural frequen-cies and modal shapes calculated by the proposedmethod are very close to the FEM results whichdemonstrates the solutions of one-step beam are exactones And numerical examples of two-step and three-step simply supported beams show that the calcu-lated frequencies and modal shapes are also in goodagreementswith FEMresults which also demonstratethe solutions of presentedmethod formultistep beamwith variable cross section are exact onesThe reasons

lie in that only transfer matrix method and generalsolution of one-step beam are used to obtain thesolution equation of natural vibration characteristicsand no other assumptions are introduced

The exact solutions of proposedmethod can provide ade-quate insight into the physics of problems and conveniencefor parametric studies Furthermore they are very useful forinverse problems such as structural damage identificationTherefore it is always desirable to obtain exact solutions forsuch problemsHowever when the ratio of span and sectionalheight is small relative errors between FEM and presentmethod may be rather larger The deficiency of this paperis not considering shear deformation in vibration analysis ofnonuniform beams leading to larger error for the small ratioof span and sectional height This also has become the topicwhich the author further deliberated from now on

Competing Interests

The authors declare that they have no competing interestsregarding the publication of this paper

Acknowledgments

The authors express their appreciation for financial supportof the National Natural Science Foundation of China underGrants nos 51478203 and 51408258 China PostdoctoralScience Foundation funded project (nos 2014M560237 and2015T80305) Fundamental Research Funds for the CentralUniversities and Science (JCKYQKJC06) and TechnologyDevelopment Program of Jilin Province

References

[1] D Zhou and Y K Cheung ldquoThe free vibration of a type oftapered beamsrdquo Computer Methods in Applied Mechanics andEngineering vol 188 no 1 pp 203ndash216 2000

[2] N M Auciello ldquoOn the transverse vibrations of non-uniformbeams with axial loads and elastically restrained endsrdquo Interna-tional Journal of Mechanical Sciences vol 43 no 1 pp 193ndash2082001

[3] J Murın and V Kutis ldquo3D-beam element with continuousvariation of the cross-sectional areardquo Computers amp Structuresvol 80 no 3-4 pp 329ndash338 2002

[4] GWang and NMWereley ldquoFree vibration analysis of rotatingblades with uniform tapersrdquo AIAA Journal vol 42 no 12 pp2429ndash2437 2004

[5] Y Huang and X-F Li ldquoA new approach for free vibrationof axially functionally graded beams with non-uniform cross-sectionrdquo Journal of Sound and Vibration vol 329 no 11 pp2291ndash2303 2010

[6] M A Koplow A Bhattacharyya and B P Mann ldquoClosedform solutions for the dynamic response of Euler-Bernoullibeams with step changes in cross sectionrdquo Journal of Sound andVibration vol 295 no 1-2 pp 214ndash225 2006

[7] R D Firouz-Abadi H Haddadpour and A B Novinzadeh ldquoAnasymptotic solution to transverse free vibrations of variable-section beamsrdquo Journal of Sound and Vibration vol 304 no3ndash5 pp 530ndash540 2007


[8] J A Inaudi and A E Matusevich ldquoDomain-partition powerseries in vibration analysis of variable-cross-section rodsrdquoJournal of Sound and Vibration vol 329 no 21 pp 4534ndash45492010

[9] Q B Mao and S Pietrzko ldquoFree vibration analysis of a typeof tapered beams by using Adomian decomposition methodrdquoAppliedMathematics andComputation vol 219 no 6 pp 3264ndash3271 2012

[10] G Duan and X Wang ldquoFree vibration analysis of multiple-stepped beams by the discrete singular convolutionrdquo AppliedMathematics and Computation vol 219 no 24 pp 11096ndash111092013

[11] JW Lee and J Y Lee ldquoFree vibration analysis using the transfer-matrixmethod on a tapered beamrdquoComputersamp Structures vol164 pp 75ndash82 2016

[12] G Y Wang Vibration of Building and Structures Science andTechnology Press 1978

[13] S Abrate ldquoVibration of non-uniform rods and beamsrdquo Journalof Sound and Vibration vol 185 no 4 pp 703ndash716 1995

[14] B M Kumar and R I Sujith ldquoExact solutions for the longi-tudinal vibration of non-uniform rodsrdquo Journal of Sound andVibration vol 207 no 5 pp 721ndash729 1997

[15] Q S Li ldquoExact solutions for free longitudinal vibrations of non-uniform rodsrdquo Journal of Sound and Vibration vol 234 no 1pp 1ndash19 2000

[16] L Qiusheng C Hong and L Guiqing ldquoAnalysis of freevibrations of tall buildingsrdquo Journal of Engineering Mechanicsvol 120 no 9 pp 1861ndash1876 1994

[17] Q S Li ldquoFlexural free vibration of cantilevered structuresof variable stiffness and massrdquo Structural Engineering andMechanics vol 8 no 3 pp 243ndash256 1999

[18] M Attar ldquoA transfer matrix method for free vibration analysisand crack identification of stepped beams with multiple edgecracks anddifferent boundary conditionsrdquo International Journalof Mechanical Sciences vol 57 no 1 pp 19ndash33 2012

[19] Q S Li ldquoFree vibration analysis of non-uniform beams with anarbitrary number of cracks and concentrated massesrdquo Journalof Sound and Vibration vol 252 no 3 pp 509ndash525 2002

[20] H-Y Lin and Y-C Tsai ldquoFree vibration analysis of a uniformmulti-span beam carrying multiple spring-mass systemsrdquo Jour-nal of Sound and Vibration vol 302 no 3 pp 442ndash456 2007

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


modified discrete singular convolution (DSC) [10] The jumpconditions at the steps were used to overcome the difficultyin using ordinary DSC for dealing with ill-posed problemsA transfer matrix method and the Frobenius method wereadopted by J W Lee and J Y Lee to solve the free vibrationcharacteristics of a tapered Bernoulli-Euler beam and obtainthe power series solution for bending vibrations [11]

Nevertheless besides all advantages of such numericalmethods exact solutions can provide adequate insight intothe physics of the problems and convenience for parametricstudies The other advantage of exact solutions is their signif-icance in the field of inverse problems An exact solution canbemore useful than numerical solutions to design the charac-teristics and damage identification of a structure

Wang derived the closed form solutions for free vibrationof a flexural bar with variably distributed stiffness but uni-formmass [12] Using a systematic approach Abrate obtaineda closed form solution of longitudinal vibration for rodswhose cross section varies as 119860(119909) = 1198600(1 + 119886[119909119871])2 [13]Kumar and Sujith found the exact solutions for longitudinalvibration of nonuniform rods whose cross section varies as119860 = (119886 + 119887119909)119899 and 119860 = 1198600 sin2(119886 + 119887119909) [14] Li presentedan exact approach for free longitudinal vibrations of one-stepnonuniform rod with classical and nonclassical boundaryconditions [15] The approach assumed that the distributionof mass is arbitrary and distribution of longitudinal stiffnessis expressed as a functional relation with mass distributionand vice versa Li et al obtained exact solutions of flexuralvibration for beam-like structures whose moment of inertiaandmass per unit length vary as119864119868(119909) = 119886(1+120573119909)119899+4119898(119909) =120572(1 + 120573119909)119899 and 119864119868(119909) = 119886 sdot exp(minus119887119909) 119898(119909) = 120572 sdot exp(minus119887119909)respectively [16 17]

The components whose moment of inertia 119868(119909) and massper unit length119898(119909) satisfy 119868(119909) = 1205721(1 + 120573119909)119903+4 and119898(119909) =1205722(1 + 120573119909)119903 are widely used in civil engineering such as thebridgewith cross section height varying as ℎ(119909) = ℎ0(1+120573119909)2This kind of beam-like structure is usually solved by dividingit into several segments whose 119868(119909) and119898(119909) satisfy differentdistributions The paper derived the exact general solutionof one-step nonuniform beam firstly and then obtained theexact solution of multistep nonuniform beam by combiningthe transfer matrix method and exact solutions of one-stepbeam The exact solutions of present method not only canprovide convenience for parametric studies but also are veryuseful for inverse problems such as damage identification

2 Modal Shape Function of One-Step Beam

21 Beams with Uniform Cross Section The governing differ-ential equation for undamped free flexural vibration of beamwith uniform cross section can be written as [18]

1198641198681205974119910 (119909 119905)1205971199094 + 1198981205972119910 (119909 119905)1205971199052 = 0 (1)

where 119864 is Youngrsquos modulus 119868 is moment of inertia 119898 ismass per unit length and 119910(119909 119905) is transverse displacementat position 119909 and time 119905


119910 (119909 119905) = 120601119906 (119909) 119890119895120596119905 (2)

where 120601119906(119909) is mode shape function of uniform beam 120596 iscorresponding natural frequency

Inserting (2) into (1) obtains

1198641198681198894120601119906 (119909)1198891199094 minus 1205962119898120601119906 (119909) = 0 (3)

Equation (3) leads to the modal shape function of beamwith uniform cross section

120601119906 (119909) = 4sum119894=1

119861119894119878119906119894 (119909) (4)

where 1198781199061(119909) = sinh 119896119909 1198781199062(119909) = cosh 119896119909 1198781199063(119909) = sin 1198961199091198781199064(119909) = cos 119896119909 are 119861119894 (119894 = 1 2 3 4) are integration constants1198964 = 120596211989811986411986822 Beams with Variable Cross Section The governing differ-ential equation for undamped free flexural vibration of beamwith variable cross section can be written as [19]

12059721205971199092 [119864119868 (119909) 1205972119910 (119909 119905)1205971199092 ] + 119898 (119909) 1205972119910 (119909 119905)1205971199052 = 0 (5)

where 119864 is Youngrsquos modulus 119868(119909) is bendingmoment of iner-tia at position 119909 119898(119909) is mass per unit length at position 119909and119910(119909 119905) is transverse displacement at position119909 and time 119905


119910 (119909 119905) = 120601 (119909) 119890119895120596119905 (6)

where 120601(119909) is mode shape function of beam 120596 is naturalfrequency

Substituting (6) into (5) arrives at

11988921198891199092 [119864119868 (119909) 1198892120601 (119909)1198891199092 ] minus 1205962119898(119909) 120601 (119909) = 0 (7)

Themoment of inertia 119868(119909) andmass per unit length119898(119909)of beam are assumed to vary as

119868 (119909) = 1205721 (1 + 120573119909)119903+4 119898 (119909) = 1205722 (1 + 120573119909)119903

(1205721 1205722 120573 = 0) (8)

where 1205721 1205722 and 120573 are arbitrary constants but not equal to 0and 119903 is a positive integer



0 = 1198641205721 (1 + 120573119909)4 1198891206014 (119909)1198891199094+ 21198641205721120573 (119903 + 4) (1 + 120573119909)3 1198891206013 (119909)1198891199093+ 11986412057211205732 (119903 + 4) (119903 + 3) (1 + 120573119909)2 1198891206012 (119909)1198891199092minus 12059621205722120601 (119909)

(9)

Let

119911 = ln (1 + 120573119909) 119863 = 119889119889119911

(10)



(119896 = 1 2 119899) (11)



(12)



(13)

Let


(14)


(1198634 + 11988611198633 + 11988621198632 + 1198863119863 + 1198864) 120601 (119911) = 0 (15)


1198894 + 11988611198893 + 11988621198892 + 1198863119889 + 1198864 = 0 (16)



(17)

where



11990524 minus 1198864)121003816100381610038161003816100381610038161003816100381610038161003816 119892 = (11988611199054 minus 11988632)119891

(18)


120601 (119909) = 11988811198901198891119911 + 11988821198901198892119911 + 11988831198901198893119911 + 11988841198901198894119911 (19)





+ 1198613 (1 + 120573119909)1198893 + 1198614 (1 + 120573119909)1198894 (20)


1199041 = minus119892 minus 119886122 1199042 =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816radic4 (1199052 + 119891) minus (11988612 + 119892)22

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816 (21)





1198893 = 1199041 + 11990421198951198894 = 1199041 minus 1199042119895 (22)


120601 (119909) = 1198611 (1 + 120573119909)1198891 + 1198612 (1 + 120573119909)1198892+ 1198613 (1 + 120573119909)1199041 cos 1199042 ln (1 + 120573119909)+ 1198614 (1 + 120573119909)1199041 sin 1199042 ln (1 + 120573119909)

(23)



120601 (119909) = 4sum119894=1

119861119894119878119894 (119909) (24)






120601119896 (119909)1003816100381610038161003816119909=119897119896 = 120601119896+1 (119909)1003816100381610038161003816119909=0 1206011015840119896 (119909)10038161003816100381610038161003816119909=119897119896 = 1206011015840119896+1 (119909)10038161003816100381610038161003816119909=0 119864 [119868119896 (119909) sdot 12060110158401015840119896 (119909)]10038161003816100381610038161003816119909=119897119896 = 119864 [119868119896+1 (119909) sdot 12060110158401015840119896+1 (119909)]10038161003816100381610038161003816119909=0 119889 [119864119868119896 (119909) sdot 12060110158401015840119896 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=119897119896= 119889 [119864119868119896+1 (119909) sdot 12060110158401015840119896+1 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=0

(25)



4sum119894=1

119861119896119894 119878119896119894 (119897119896) = 4sum119894=1

119861119896+1119894 119878119896+1119894 (0) 4sum119894=1

119861119896119894 1198781015840119896119894 (119897119896) = 4sum119894=1

119861119896+1119894 1198781015840119896+1119894 (0) 119868119896 (119897119896) 4sum

119894=1

119861119896119894 11987810158401015840119896119894 (119897119896) = 119868119896+1 (0) 4sum119894=1

119861119896+1119894 11987810158401015840119896+1119894 (0) 1198681015840119896 (119897119896) 4sum

119894=1

119861119896119894 11987810158401015840119896119894 (119897119896) + 119868119896 (119897119896) 4sum119894=1

119861119896119894 119878101584010158401015840119896119894 (119897119896)= 1198681015840119896+1 (0) 4sum

119894=1

119861119896+1119894 11987810158401015840119896+1119894 (0)+ 119868119896+1 (0) 4sum

119894=1

119861119896+1119894 119878101584010158401015840119896+1119894 (0)

(26)



H119896U119896 = H119896+1U119896+1 (27)

where

U119896 = 1198611198961 1198611198962 1198611198963 1198611198964119879 U119896+1 = 119861119896+11 119861119896+12 119861119896+13 119861119896+14 119879 H119896 = [H1198961 H1198962 H1198963 H1198964]

H119896+1 = [H(119896+1)1 H(119896+1)2 H(119896+1)3 H(119896+1)4] (28)


in which

H119896119894 =[[[[[[[

119878119896119894 (119897119896)1198781015840119896119894 (119897119896)119868119896 (119897119896) 11987810158401015840119896119894 (119897119896)1198681015840119896 (119897119896) 11987810158401015840119896119894 (119897119896) + 119868119896 (119897119896) 119878101584010158401015840119896119894 (119897119896)

]]]]]]]

H(119896+1)119894 =[[[[[[[

119878119896+1119894 (0)1198781015840119896+1119894 (0)

119868119896+1 (0) 11987810158401015840119896+1119894 (0)1198681015840119896+1 (0) 11987810158401015840119896+1119894 (0) + 119868119896+1 (0) 119878101584010158401015840119896+1119894 (0)

]]]]]]]

(29)

Let

T119896 = Hminus1119896+1H119896 (30)

then

U119896+1 = Hminus1119896+1H119896U119896 = T119896U119896 (31)




and Modal Shapes



1206011 (0) = 012060110158401 (119909)10038161003816100381610038161003816119909=0 = 0

119864119868119898 (119909) sdot 12060110158401015840119898 (119909)10038161003816100381610038161003816119909=119897119898 = 0119864119889 [119868119898 (119909) sdot 12060110158401015840119898 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=119897119898 = 0(33)


[ C1C119898T

]U1 = 0 (34)

where

C1 = [11987811 (0) 11987812 (0) 11987813 (0) 11987814 (0)119878101584011 (0) 119878101584012 (0) 119878101584013 (0) 119878101584014 (0)] C119898 = [119878101584010158401198981 (119897119898) 119878101584010158401198982 (119897119898) 119878101584010158401198983 (119897119898) 119878101584010158401198984 (119897119898)1198781015840101584010158401198981 (119897119898) 1198781015840101584010158401198982 (119897119898) 1198781015840101584010158401198983 (119897119898) 1198781015840101584010158401198984 (119897119898)]

(35)


1206011 (0) = 01198641198681 (119909) sdot 120601101584010158401 (119909)10038161003816100381610038161003816119909=0 = 0

120601119898 (119897119898) = 0119864119868119898 (119909) sdot 12060110158401015840119898 (119909)10038161003816100381610038161003816119909=119897119898 = 0

(36)


[ C10158401C1015840119898T

]U1 = 0 (37)

where

C10158401 = [ 11987811 (0) 11987812 (0) 11987813 (0) 11987814 (0)1198781015840101584011 (0) 1198781015840101584012 (0) 1198781015840101584013 (0) 1198781015840101584014 (0)] C1015840119898 = [ 1198781198981 (119897119898) 1198781198982 (119897119898) 1198781198983 (119897119898) 1198781198984 (119897119898)119878101584010158401198981 (119897119898) 119878101584010158401198982 (119897119898) 119878101584010158401198983 (119897119898) 119878101584010158401198984 (119897119898)]

(38)


CU1 = [[[[[[

11988811 11988812 11988813 1198881411988821 11988822 11988823 1198882411988831 11988832 11988833 1198883411988841 11988842 11988843 11988844]]]]]]

11986111119861121198611311986114

= 0 (39)



|C| = 0 (40)



x

y

A B C5 5

hA hB hC








119860 to 119861ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 (0 le 119909 le 5)



119861 to 119862ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 (120585 = 119909 minus 5 5 le 119909 le 10)




0 le 119909 le 119897

(43)



K119890=int1198970B119879119864119868 (119909)B 119889119909

M119890 = int1198970119898(119909)N119879N 119889119909

(44)






10 20 30 40 50 60 70 80 90 100




minus50

minus45

minus40

minus35

minus30

minus25

minus20

minus15

minus10

minus05

00

05

Rela

tive

erro

r (

)

LhA




x

y

A B15

hA hB


x

15A B

y

h






Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (46)














Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (48)








1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts0 3

Length (m)6 9 12 15


0 3 6 9 12 15Length (m)



1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts




x

y

A B C10 12

hA hB hC

(a)

A

A

B

B

C

C

bA bB bC

(b)









ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 119887119860119861 (119909) = 119887119860 (1 + 120573119860119861119909)


(49)

Therefore

119868119860119861 (119909) = 119887119860119861 (119909) sdot ℎ3119860119861 (119909)12 = 119887119860ℎ3119860 (1 + 120573119860119861119909)712 119898119860119861 (119909) = 120588119887119860119861 (119909) sdot ℎ119860119861 (119909) = 120588119887119860ℎ119860 (1 + 120573119860119861119909)3

(0 le 119909 le 10) (50)



ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 119887119861119862 (120585) = 119887119861 (1 + 120573119861119862120585)


(51)

Therefore

119868119861119862 (120585) = 119887119861119862 (120585) sdot ℎ3119861119862 (120585)12 = 119887119861ℎ3119861 (1 + 120573119861119862120585)712 119898119861119862 (120585) = 120588119887119861119862 (120585) sdot ℎ119861119862 (120585) = 120588119887119861ℎ119861 (1 + 120573119861119862120585)3

(0 le 120585 = 119909 minus 10 le 12) (52)










0 2 4 6 8 10 12 14 16 18 20 22Length (m)

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM




Therefore

119868119860119861 (119909) = 119887ℎ3119860119861 (119909)12 = 119887ℎ3119860 (1 + 120573119860119861119909)612 119898119860119861 (119909) = 120588119887ℎ119860119861 (119909) = 120588119887ℎ119860 (1 + 120573119860119861119909)2

(0 le 119909 le 5) (54)

x

y

A B C5 10 5

D

hA hB hC hD




Therefore

119868119862119863 (120585) = 119887ℎ3119862119863 (120585)12 = 119887ℎ3119862 (1 + 120573119862119863120585)612 119898119862119863 (120585) = 120588119887ℎ119862119863 (120585) = 120588119887ℎ119862 (1 + 120573119862119863120585)2

(0 le 120585 = 119909 minus 15 le 5) (56)




0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM


6 Conclusions







Competing Interests


Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











0 = 1198641205721 (1 + 120573119909)4 1198891206014 (119909)1198891199094+ 21198641205721120573 (119903 + 4) (1 + 120573119909)3 1198891206013 (119909)1198891199093+ 11986412057211205732 (119903 + 4) (119903 + 3) (1 + 120573119909)2 1198891206012 (119909)1198891199092minus 12059621205722120601 (119909)

(9)

Let

119911 = ln (1 + 120573119909) 119863 = 119889119889119911

(10)



(119896 = 1 2 119899) (11)



(12)



(13)

Let


(14)


(1198634 + 11988611198633 + 11988621198632 + 1198863119863 + 1198864) 120601 (119911) = 0 (15)


1198894 + 11988611198893 + 11988621198892 + 1198863119889 + 1198864 = 0 (16)



(17)

where



11990524 minus 1198864)121003816100381610038161003816100381610038161003816100381610038161003816 119892 = (11988611199054 minus 11988632)119891

(18)


120601 (119909) = 11988811198901198891119911 + 11988821198901198892119911 + 11988831198901198893119911 + 11988841198901198894119911 (19)





+ 1198613 (1 + 120573119909)1198893 + 1198614 (1 + 120573119909)1198894 (20)


1199041 = minus119892 minus 119886122 1199042 =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816radic4 (1199052 + 119891) minus (11988612 + 119892)22

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816 (21)





1198893 = 1199041 + 11990421198951198894 = 1199041 minus 1199042119895 (22)


120601 (119909) = 1198611 (1 + 120573119909)1198891 + 1198612 (1 + 120573119909)1198892+ 1198613 (1 + 120573119909)1199041 cos 1199042 ln (1 + 120573119909)+ 1198614 (1 + 120573119909)1199041 sin 1199042 ln (1 + 120573119909)

(23)



120601 (119909) = 4sum119894=1

119861119894119878119894 (119909) (24)






120601119896 (119909)1003816100381610038161003816119909=119897119896 = 120601119896+1 (119909)1003816100381610038161003816119909=0 1206011015840119896 (119909)10038161003816100381610038161003816119909=119897119896 = 1206011015840119896+1 (119909)10038161003816100381610038161003816119909=0 119864 [119868119896 (119909) sdot 12060110158401015840119896 (119909)]10038161003816100381610038161003816119909=119897119896 = 119864 [119868119896+1 (119909) sdot 12060110158401015840119896+1 (119909)]10038161003816100381610038161003816119909=0 119889 [119864119868119896 (119909) sdot 12060110158401015840119896 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=119897119896= 119889 [119864119868119896+1 (119909) sdot 12060110158401015840119896+1 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=0

(25)



4sum119894=1

119861119896119894 119878119896119894 (119897119896) = 4sum119894=1

119861119896+1119894 119878119896+1119894 (0) 4sum119894=1

119861119896119894 1198781015840119896119894 (119897119896) = 4sum119894=1

119861119896+1119894 1198781015840119896+1119894 (0) 119868119896 (119897119896) 4sum

119894=1

119861119896119894 11987810158401015840119896119894 (119897119896) = 119868119896+1 (0) 4sum119894=1

119861119896+1119894 11987810158401015840119896+1119894 (0) 1198681015840119896 (119897119896) 4sum

119894=1

119861119896119894 11987810158401015840119896119894 (119897119896) + 119868119896 (119897119896) 4sum119894=1

119861119896119894 119878101584010158401015840119896119894 (119897119896)= 1198681015840119896+1 (0) 4sum

119894=1

119861119896+1119894 11987810158401015840119896+1119894 (0)+ 119868119896+1 (0) 4sum

119894=1

119861119896+1119894 119878101584010158401015840119896+1119894 (0)

(26)



H119896U119896 = H119896+1U119896+1 (27)

where

U119896 = 1198611198961 1198611198962 1198611198963 1198611198964119879 U119896+1 = 119861119896+11 119861119896+12 119861119896+13 119861119896+14 119879 H119896 = [H1198961 H1198962 H1198963 H1198964]

H119896+1 = [H(119896+1)1 H(119896+1)2 H(119896+1)3 H(119896+1)4] (28)


in which

H119896119894 =[[[[[[[

119878119896119894 (119897119896)1198781015840119896119894 (119897119896)119868119896 (119897119896) 11987810158401015840119896119894 (119897119896)1198681015840119896 (119897119896) 11987810158401015840119896119894 (119897119896) + 119868119896 (119897119896) 119878101584010158401015840119896119894 (119897119896)

]]]]]]]

H(119896+1)119894 =[[[[[[[

119878119896+1119894 (0)1198781015840119896+1119894 (0)

119868119896+1 (0) 11987810158401015840119896+1119894 (0)1198681015840119896+1 (0) 11987810158401015840119896+1119894 (0) + 119868119896+1 (0) 119878101584010158401015840119896+1119894 (0)

]]]]]]]

(29)

Let

T119896 = Hminus1119896+1H119896 (30)

then

U119896+1 = Hminus1119896+1H119896U119896 = T119896U119896 (31)




and Modal Shapes



1206011 (0) = 012060110158401 (119909)10038161003816100381610038161003816119909=0 = 0

119864119868119898 (119909) sdot 12060110158401015840119898 (119909)10038161003816100381610038161003816119909=119897119898 = 0119864119889 [119868119898 (119909) sdot 12060110158401015840119898 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=119897119898 = 0(33)


[ C1C119898T

]U1 = 0 (34)

where

C1 = [11987811 (0) 11987812 (0) 11987813 (0) 11987814 (0)119878101584011 (0) 119878101584012 (0) 119878101584013 (0) 119878101584014 (0)] C119898 = [119878101584010158401198981 (119897119898) 119878101584010158401198982 (119897119898) 119878101584010158401198983 (119897119898) 119878101584010158401198984 (119897119898)1198781015840101584010158401198981 (119897119898) 1198781015840101584010158401198982 (119897119898) 1198781015840101584010158401198983 (119897119898) 1198781015840101584010158401198984 (119897119898)]

(35)


1206011 (0) = 01198641198681 (119909) sdot 120601101584010158401 (119909)10038161003816100381610038161003816119909=0 = 0

120601119898 (119897119898) = 0119864119868119898 (119909) sdot 12060110158401015840119898 (119909)10038161003816100381610038161003816119909=119897119898 = 0

(36)


[ C10158401C1015840119898T

]U1 = 0 (37)

where

C10158401 = [ 11987811 (0) 11987812 (0) 11987813 (0) 11987814 (0)1198781015840101584011 (0) 1198781015840101584012 (0) 1198781015840101584013 (0) 1198781015840101584014 (0)] C1015840119898 = [ 1198781198981 (119897119898) 1198781198982 (119897119898) 1198781198983 (119897119898) 1198781198984 (119897119898)119878101584010158401198981 (119897119898) 119878101584010158401198982 (119897119898) 119878101584010158401198983 (119897119898) 119878101584010158401198984 (119897119898)]

(38)


CU1 = [[[[[[

11988811 11988812 11988813 1198881411988821 11988822 11988823 1198882411988831 11988832 11988833 1198883411988841 11988842 11988843 11988844]]]]]]

11986111119861121198611311986114

= 0 (39)



|C| = 0 (40)



x

y

A B C5 5

hA hB hC








119860 to 119861ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 (0 le 119909 le 5)



119861 to 119862ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 (120585 = 119909 minus 5 5 le 119909 le 10)




0 le 119909 le 119897

(43)



K119890=int1198970B119879119864119868 (119909)B 119889119909

M119890 = int1198970119898(119909)N119879N 119889119909

(44)






10 20 30 40 50 60 70 80 90 100




minus50

minus45

minus40

minus35

minus30

minus25

minus20

minus15

minus10

minus05

00

05

Rela

tive

erro

r (

)

LhA




x

y

A B15

hA hB


x

15A B

y

h






Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (46)














Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (48)








1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts0 3

Length (m)6 9 12 15


0 3 6 9 12 15Length (m)



1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts




x

y

A B C10 12

hA hB hC

(a)

A

A

B

B

C

C

bA bB bC

(b)









ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 119887119860119861 (119909) = 119887119860 (1 + 120573119860119861119909)


(49)

Therefore

119868119860119861 (119909) = 119887119860119861 (119909) sdot ℎ3119860119861 (119909)12 = 119887119860ℎ3119860 (1 + 120573119860119861119909)712 119898119860119861 (119909) = 120588119887119860119861 (119909) sdot ℎ119860119861 (119909) = 120588119887119860ℎ119860 (1 + 120573119860119861119909)3

(0 le 119909 le 10) (50)



ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 119887119861119862 (120585) = 119887119861 (1 + 120573119861119862120585)


(51)

Therefore

119868119861119862 (120585) = 119887119861119862 (120585) sdot ℎ3119861119862 (120585)12 = 119887119861ℎ3119861 (1 + 120573119861119862120585)712 119898119861119862 (120585) = 120588119887119861119862 (120585) sdot ℎ119861119862 (120585) = 120588119887119861ℎ119861 (1 + 120573119861119862120585)3

(0 le 120585 = 119909 minus 10 le 12) (52)










0 2 4 6 8 10 12 14 16 18 20 22Length (m)

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM




Therefore

119868119860119861 (119909) = 119887ℎ3119860119861 (119909)12 = 119887ℎ3119860 (1 + 120573119860119861119909)612 119898119860119861 (119909) = 120588119887ℎ119860119861 (119909) = 120588119887ℎ119860 (1 + 120573119860119861119909)2

(0 le 119909 le 5) (54)

x

y

A B C5 10 5

D

hA hB hC hD




Therefore

119868119862119863 (120585) = 119887ℎ3119862119863 (120585)12 = 119887ℎ3119862 (1 + 120573119862119863120585)612 119898119862119863 (120585) = 120588119887ℎ119862119863 (120585) = 120588119887ℎ119862 (1 + 120573119862119863120585)2

(0 le 120585 = 119909 minus 15 le 5) (56)




0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM


6 Conclusions







Competing Interests


Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













1198893 = 1199041 + 11990421198951198894 = 1199041 minus 1199042119895 (22)


120601 (119909) = 1198611 (1 + 120573119909)1198891 + 1198612 (1 + 120573119909)1198892+ 1198613 (1 + 120573119909)1199041 cos 1199042 ln (1 + 120573119909)+ 1198614 (1 + 120573119909)1199041 sin 1199042 ln (1 + 120573119909)

(23)



120601 (119909) = 4sum119894=1

119861119894119878119894 (119909) (24)






120601119896 (119909)1003816100381610038161003816119909=119897119896 = 120601119896+1 (119909)1003816100381610038161003816119909=0 1206011015840119896 (119909)10038161003816100381610038161003816119909=119897119896 = 1206011015840119896+1 (119909)10038161003816100381610038161003816119909=0 119864 [119868119896 (119909) sdot 12060110158401015840119896 (119909)]10038161003816100381610038161003816119909=119897119896 = 119864 [119868119896+1 (119909) sdot 12060110158401015840119896+1 (119909)]10038161003816100381610038161003816119909=0 119889 [119864119868119896 (119909) sdot 12060110158401015840119896 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=119897119896= 119889 [119864119868119896+1 (119909) sdot 12060110158401015840119896+1 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=0

(25)



4sum119894=1

119861119896119894 119878119896119894 (119897119896) = 4sum119894=1

119861119896+1119894 119878119896+1119894 (0) 4sum119894=1

119861119896119894 1198781015840119896119894 (119897119896) = 4sum119894=1

119861119896+1119894 1198781015840119896+1119894 (0) 119868119896 (119897119896) 4sum

119894=1

119861119896119894 11987810158401015840119896119894 (119897119896) = 119868119896+1 (0) 4sum119894=1

119861119896+1119894 11987810158401015840119896+1119894 (0) 1198681015840119896 (119897119896) 4sum

119894=1

119861119896119894 11987810158401015840119896119894 (119897119896) + 119868119896 (119897119896) 4sum119894=1

119861119896119894 119878101584010158401015840119896119894 (119897119896)= 1198681015840119896+1 (0) 4sum

119894=1

119861119896+1119894 11987810158401015840119896+1119894 (0)+ 119868119896+1 (0) 4sum

119894=1

119861119896+1119894 119878101584010158401015840119896+1119894 (0)

(26)



H119896U119896 = H119896+1U119896+1 (27)

where

U119896 = 1198611198961 1198611198962 1198611198963 1198611198964119879 U119896+1 = 119861119896+11 119861119896+12 119861119896+13 119861119896+14 119879 H119896 = [H1198961 H1198962 H1198963 H1198964]

H119896+1 = [H(119896+1)1 H(119896+1)2 H(119896+1)3 H(119896+1)4] (28)


in which

H119896119894 =[[[[[[[

119878119896119894 (119897119896)1198781015840119896119894 (119897119896)119868119896 (119897119896) 11987810158401015840119896119894 (119897119896)1198681015840119896 (119897119896) 11987810158401015840119896119894 (119897119896) + 119868119896 (119897119896) 119878101584010158401015840119896119894 (119897119896)

]]]]]]]

H(119896+1)119894 =[[[[[[[

119878119896+1119894 (0)1198781015840119896+1119894 (0)

119868119896+1 (0) 11987810158401015840119896+1119894 (0)1198681015840119896+1 (0) 11987810158401015840119896+1119894 (0) + 119868119896+1 (0) 119878101584010158401015840119896+1119894 (0)

]]]]]]]

(29)

Let

T119896 = Hminus1119896+1H119896 (30)

then

U119896+1 = Hminus1119896+1H119896U119896 = T119896U119896 (31)




and Modal Shapes



1206011 (0) = 012060110158401 (119909)10038161003816100381610038161003816119909=0 = 0

119864119868119898 (119909) sdot 12060110158401015840119898 (119909)10038161003816100381610038161003816119909=119897119898 = 0119864119889 [119868119898 (119909) sdot 12060110158401015840119898 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=119897119898 = 0(33)


[ C1C119898T

]U1 = 0 (34)

where

C1 = [11987811 (0) 11987812 (0) 11987813 (0) 11987814 (0)119878101584011 (0) 119878101584012 (0) 119878101584013 (0) 119878101584014 (0)] C119898 = [119878101584010158401198981 (119897119898) 119878101584010158401198982 (119897119898) 119878101584010158401198983 (119897119898) 119878101584010158401198984 (119897119898)1198781015840101584010158401198981 (119897119898) 1198781015840101584010158401198982 (119897119898) 1198781015840101584010158401198983 (119897119898) 1198781015840101584010158401198984 (119897119898)]

(35)


1206011 (0) = 01198641198681 (119909) sdot 120601101584010158401 (119909)10038161003816100381610038161003816119909=0 = 0

120601119898 (119897119898) = 0119864119868119898 (119909) sdot 12060110158401015840119898 (119909)10038161003816100381610038161003816119909=119897119898 = 0

(36)


[ C10158401C1015840119898T

]U1 = 0 (37)

where

C10158401 = [ 11987811 (0) 11987812 (0) 11987813 (0) 11987814 (0)1198781015840101584011 (0) 1198781015840101584012 (0) 1198781015840101584013 (0) 1198781015840101584014 (0)] C1015840119898 = [ 1198781198981 (119897119898) 1198781198982 (119897119898) 1198781198983 (119897119898) 1198781198984 (119897119898)119878101584010158401198981 (119897119898) 119878101584010158401198982 (119897119898) 119878101584010158401198983 (119897119898) 119878101584010158401198984 (119897119898)]

(38)


CU1 = [[[[[[

11988811 11988812 11988813 1198881411988821 11988822 11988823 1198882411988831 11988832 11988833 1198883411988841 11988842 11988843 11988844]]]]]]

11986111119861121198611311986114

= 0 (39)



|C| = 0 (40)



x

y

A B C5 5

hA hB hC








119860 to 119861ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 (0 le 119909 le 5)



119861 to 119862ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 (120585 = 119909 minus 5 5 le 119909 le 10)




0 le 119909 le 119897

(43)



K119890=int1198970B119879119864119868 (119909)B 119889119909

M119890 = int1198970119898(119909)N119879N 119889119909

(44)






10 20 30 40 50 60 70 80 90 100




minus50

minus45

minus40

minus35

minus30

minus25

minus20

minus15

minus10

minus05

00

05

Rela

tive

erro

r (

)

LhA




x

y

A B15

hA hB


x

15A B

y

h






Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (46)














Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (48)








1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts0 3

Length (m)6 9 12 15


0 3 6 9 12 15Length (m)



1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts




x

y

A B C10 12

hA hB hC

(a)

A

A

B

B

C

C

bA bB bC

(b)









ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 119887119860119861 (119909) = 119887119860 (1 + 120573119860119861119909)


(49)

Therefore

119868119860119861 (119909) = 119887119860119861 (119909) sdot ℎ3119860119861 (119909)12 = 119887119860ℎ3119860 (1 + 120573119860119861119909)712 119898119860119861 (119909) = 120588119887119860119861 (119909) sdot ℎ119860119861 (119909) = 120588119887119860ℎ119860 (1 + 120573119860119861119909)3

(0 le 119909 le 10) (50)



ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 119887119861119862 (120585) = 119887119861 (1 + 120573119861119862120585)


(51)

Therefore

119868119861119862 (120585) = 119887119861119862 (120585) sdot ℎ3119861119862 (120585)12 = 119887119861ℎ3119861 (1 + 120573119861119862120585)712 119898119861119862 (120585) = 120588119887119861119862 (120585) sdot ℎ119861119862 (120585) = 120588119887119861ℎ119861 (1 + 120573119861119862120585)3

(0 le 120585 = 119909 minus 10 le 12) (52)










0 2 4 6 8 10 12 14 16 18 20 22Length (m)

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM




Therefore

119868119860119861 (119909) = 119887ℎ3119860119861 (119909)12 = 119887ℎ3119860 (1 + 120573119860119861119909)612 119898119860119861 (119909) = 120588119887ℎ119860119861 (119909) = 120588119887ℎ119860 (1 + 120573119860119861119909)2

(0 le 119909 le 5) (54)

x

y

A B C5 10 5

D

hA hB hC hD




Therefore

119868119862119863 (120585) = 119887ℎ3119862119863 (120585)12 = 119887ℎ3119862 (1 + 120573119862119863120585)612 119898119862119863 (120585) = 120588119887ℎ119862119863 (120585) = 120588119887ℎ119862 (1 + 120573119862119863120585)2

(0 le 120585 = 119909 minus 15 le 5) (56)




0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM


6 Conclusions







Competing Interests


Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










in which

H119896119894 =[[[[[[[

119878119896119894 (119897119896)1198781015840119896119894 (119897119896)119868119896 (119897119896) 11987810158401015840119896119894 (119897119896)1198681015840119896 (119897119896) 11987810158401015840119896119894 (119897119896) + 119868119896 (119897119896) 119878101584010158401015840119896119894 (119897119896)

]]]]]]]

H(119896+1)119894 =[[[[[[[

119878119896+1119894 (0)1198781015840119896+1119894 (0)

119868119896+1 (0) 11987810158401015840119896+1119894 (0)1198681015840119896+1 (0) 11987810158401015840119896+1119894 (0) + 119868119896+1 (0) 119878101584010158401015840119896+1119894 (0)

]]]]]]]

(29)

Let

T119896 = Hminus1119896+1H119896 (30)

then

U119896+1 = Hminus1119896+1H119896U119896 = T119896U119896 (31)




and Modal Shapes



1206011 (0) = 012060110158401 (119909)10038161003816100381610038161003816119909=0 = 0

119864119868119898 (119909) sdot 12060110158401015840119898 (119909)10038161003816100381610038161003816119909=119897119898 = 0119864119889 [119868119898 (119909) sdot 12060110158401015840119898 (119909)]119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816119909=119897119898 = 0(33)


[ C1C119898T

]U1 = 0 (34)

where

C1 = [11987811 (0) 11987812 (0) 11987813 (0) 11987814 (0)119878101584011 (0) 119878101584012 (0) 119878101584013 (0) 119878101584014 (0)] C119898 = [119878101584010158401198981 (119897119898) 119878101584010158401198982 (119897119898) 119878101584010158401198983 (119897119898) 119878101584010158401198984 (119897119898)1198781015840101584010158401198981 (119897119898) 1198781015840101584010158401198982 (119897119898) 1198781015840101584010158401198983 (119897119898) 1198781015840101584010158401198984 (119897119898)]

(35)


1206011 (0) = 01198641198681 (119909) sdot 120601101584010158401 (119909)10038161003816100381610038161003816119909=0 = 0

120601119898 (119897119898) = 0119864119868119898 (119909) sdot 12060110158401015840119898 (119909)10038161003816100381610038161003816119909=119897119898 = 0

(36)


[ C10158401C1015840119898T

]U1 = 0 (37)

where

C10158401 = [ 11987811 (0) 11987812 (0) 11987813 (0) 11987814 (0)1198781015840101584011 (0) 1198781015840101584012 (0) 1198781015840101584013 (0) 1198781015840101584014 (0)] C1015840119898 = [ 1198781198981 (119897119898) 1198781198982 (119897119898) 1198781198983 (119897119898) 1198781198984 (119897119898)119878101584010158401198981 (119897119898) 119878101584010158401198982 (119897119898) 119878101584010158401198983 (119897119898) 119878101584010158401198984 (119897119898)]

(38)


CU1 = [[[[[[

11988811 11988812 11988813 1198881411988821 11988822 11988823 1198882411988831 11988832 11988833 1198883411988841 11988842 11988843 11988844]]]]]]

11986111119861121198611311986114

= 0 (39)



|C| = 0 (40)



x

y

A B C5 5

hA hB hC








119860 to 119861ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 (0 le 119909 le 5)



119861 to 119862ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 (120585 = 119909 minus 5 5 le 119909 le 10)




0 le 119909 le 119897

(43)



K119890=int1198970B119879119864119868 (119909)B 119889119909

M119890 = int1198970119898(119909)N119879N 119889119909

(44)






10 20 30 40 50 60 70 80 90 100




minus50

minus45

minus40

minus35

minus30

minus25

minus20

minus15

minus10

minus05

00

05

Rela

tive

erro

r (

)

LhA




x

y

A B15

hA hB


x

15A B

y

h






Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (46)














Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (48)








1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts0 3

Length (m)6 9 12 15


0 3 6 9 12 15Length (m)



1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts




x

y

A B C10 12

hA hB hC

(a)

A

A

B

B

C

C

bA bB bC

(b)









ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 119887119860119861 (119909) = 119887119860 (1 + 120573119860119861119909)


(49)

Therefore

119868119860119861 (119909) = 119887119860119861 (119909) sdot ℎ3119860119861 (119909)12 = 119887119860ℎ3119860 (1 + 120573119860119861119909)712 119898119860119861 (119909) = 120588119887119860119861 (119909) sdot ℎ119860119861 (119909) = 120588119887119860ℎ119860 (1 + 120573119860119861119909)3

(0 le 119909 le 10) (50)



ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 119887119861119862 (120585) = 119887119861 (1 + 120573119861119862120585)


(51)

Therefore

119868119861119862 (120585) = 119887119861119862 (120585) sdot ℎ3119861119862 (120585)12 = 119887119861ℎ3119861 (1 + 120573119861119862120585)712 119898119861119862 (120585) = 120588119887119861119862 (120585) sdot ℎ119861119862 (120585) = 120588119887119861ℎ119861 (1 + 120573119861119862120585)3

(0 le 120585 = 119909 minus 10 le 12) (52)










0 2 4 6 8 10 12 14 16 18 20 22Length (m)

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM




Therefore

119868119860119861 (119909) = 119887ℎ3119860119861 (119909)12 = 119887ℎ3119860 (1 + 120573119860119861119909)612 119898119860119861 (119909) = 120588119887ℎ119860119861 (119909) = 120588119887ℎ119860 (1 + 120573119860119861119909)2

(0 le 119909 le 5) (54)

x

y

A B C5 10 5

D

hA hB hC hD




Therefore

119868119862119863 (120585) = 119887ℎ3119862119863 (120585)12 = 119887ℎ3119862 (1 + 120573119862119863120585)612 119898119862119863 (120585) = 120588119887ℎ119862119863 (120585) = 120588119887ℎ119862 (1 + 120573119862119863120585)2

(0 le 120585 = 119909 minus 15 le 5) (56)




0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM


6 Conclusions







Competing Interests


Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










x

y

A B C5 5

hA hB hC








119860 to 119861ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 (0 le 119909 le 5)



119861 to 119862ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 (120585 = 119909 minus 5 5 le 119909 le 10)




0 le 119909 le 119897

(43)



K119890=int1198970B119879119864119868 (119909)B 119889119909

M119890 = int1198970119898(119909)N119879N 119889119909

(44)






10 20 30 40 50 60 70 80 90 100




minus50

minus45

minus40

minus35

minus30

minus25

minus20

minus15

minus10

minus05

00

05

Rela

tive

erro

r (

)

LhA




x

y

A B15

hA hB


x

15A B

y

h






Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (46)














Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (48)








1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts0 3

Length (m)6 9 12 15


0 3 6 9 12 15Length (m)



1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts




x

y

A B C10 12

hA hB hC

(a)

A

A

B

B

C

C

bA bB bC

(b)









ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 119887119860119861 (119909) = 119887119860 (1 + 120573119860119861119909)


(49)

Therefore

119868119860119861 (119909) = 119887119860119861 (119909) sdot ℎ3119860119861 (119909)12 = 119887119860ℎ3119860 (1 + 120573119860119861119909)712 119898119860119861 (119909) = 120588119887119860119861 (119909) sdot ℎ119860119861 (119909) = 120588119887119860ℎ119860 (1 + 120573119860119861119909)3

(0 le 119909 le 10) (50)



ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 119887119861119862 (120585) = 119887119861 (1 + 120573119861119862120585)


(51)

Therefore

119868119861119862 (120585) = 119887119861119862 (120585) sdot ℎ3119861119862 (120585)12 = 119887119861ℎ3119861 (1 + 120573119861119862120585)712 119898119861119862 (120585) = 120588119887119861119862 (120585) sdot ℎ119861119862 (120585) = 120588119887119861ℎ119861 (1 + 120573119861119862120585)3

(0 le 120585 = 119909 minus 10 le 12) (52)










0 2 4 6 8 10 12 14 16 18 20 22Length (m)

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM




Therefore

119868119860119861 (119909) = 119887ℎ3119860119861 (119909)12 = 119887ℎ3119860 (1 + 120573119860119861119909)612 119898119860119861 (119909) = 120588119887ℎ119860119861 (119909) = 120588119887ℎ119860 (1 + 120573119860119861119909)2

(0 le 119909 le 5) (54)

x

y

A B C5 10 5

D

hA hB hC hD




Therefore

119868119862119863 (120585) = 119887ℎ3119862119863 (120585)12 = 119887ℎ3119862 (1 + 120573119862119863120585)612 119898119862119863 (120585) = 120588119887ℎ119862119863 (120585) = 120588119887ℎ119862 (1 + 120573119862119863120585)2

(0 le 120585 = 119909 minus 15 le 5) (56)




0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM


6 Conclusions







Competing Interests


Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










10 20 30 40 50 60 70 80 90 100




minus50

minus45

minus40

minus35

minus30

minus25

minus20

minus15

minus10

minus05

00

05

Rela

tive

erro

r (

)

LhA




x

y

A B15

hA hB


x

15A B

y

h






Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (46)














Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (48)








1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts0 3

Length (m)6 9 12 15


0 3 6 9 12 15Length (m)



1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts




x

y

A B C10 12

hA hB hC

(a)

A

A

B

B

C

C

bA bB bC

(b)









ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 119887119860119861 (119909) = 119887119860 (1 + 120573119860119861119909)


(49)

Therefore

119868119860119861 (119909) = 119887119860119861 (119909) sdot ℎ3119860119861 (119909)12 = 119887119860ℎ3119860 (1 + 120573119860119861119909)712 119898119860119861 (119909) = 120588119887119860119861 (119909) sdot ℎ119860119861 (119909) = 120588119887119860ℎ119860 (1 + 120573119860119861119909)3

(0 le 119909 le 10) (50)



ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 119887119861119862 (120585) = 119887119861 (1 + 120573119861119862120585)


(51)

Therefore

119868119861119862 (120585) = 119887119861119862 (120585) sdot ℎ3119861119862 (120585)12 = 119887119861ℎ3119861 (1 + 120573119861119862120585)712 119898119861119862 (120585) = 120588119887119861119862 (120585) sdot ℎ119861119862 (120585) = 120588119887119861ℎ119861 (1 + 120573119861119862120585)3

(0 le 120585 = 119909 minus 10 le 12) (52)










0 2 4 6 8 10 12 14 16 18 20 22Length (m)

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM




Therefore

119868119860119861 (119909) = 119887ℎ3119860119861 (119909)12 = 119887ℎ3119860 (1 + 120573119860119861119909)612 119898119860119861 (119909) = 120588119887ℎ119860119861 (119909) = 120588119887ℎ119860 (1 + 120573119860119861119909)2

(0 le 119909 le 5) (54)

x

y

A B C5 10 5

D

hA hB hC hD




Therefore

119868119862119863 (120585) = 119887ℎ3119862119863 (120585)12 = 119887ℎ3119862 (1 + 120573119862119863120585)612 119898119862119863 (120585) = 120588119887ℎ119862119863 (120585) = 120588119887ℎ119862 (1 + 120573119862119863120585)2

(0 le 120585 = 119909 minus 15 le 5) (56)




0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM


6 Conclusions







Competing Interests


Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Therefore

119868 (119909) = 119887ℎ3 (119909)12 = 119887ℎ3119860 (1 + 120573119909)612 119898 (119909) = 120588119887ℎ (119909) = 120588119887ℎ119860 (1 + 120573119909)2

(0 le 119909 le 15) (48)








1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts0 3

Length (m)6 9 12 15


0 3 6 9 12 15Length (m)



1st mode-2D-FEM

3rd mode-2D-FEM

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts




x

y

A B C10 12

hA hB hC

(a)

A

A

B

B

C

C

bA bB bC

(b)









ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 119887119860119861 (119909) = 119887119860 (1 + 120573119860119861119909)


(49)

Therefore

119868119860119861 (119909) = 119887119860119861 (119909) sdot ℎ3119860119861 (119909)12 = 119887119860ℎ3119860 (1 + 120573119860119861119909)712 119898119860119861 (119909) = 120588119887119860119861 (119909) sdot ℎ119860119861 (119909) = 120588119887119860ℎ119860 (1 + 120573119860119861119909)3

(0 le 119909 le 10) (50)



ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 119887119861119862 (120585) = 119887119861 (1 + 120573119861119862120585)


(51)

Therefore

119868119861119862 (120585) = 119887119861119862 (120585) sdot ℎ3119861119862 (120585)12 = 119887119861ℎ3119861 (1 + 120573119861119862120585)712 119898119861119862 (120585) = 120588119887119861119862 (120585) sdot ℎ119861119862 (120585) = 120588119887119861ℎ119861 (1 + 120573119861119862120585)3

(0 le 120585 = 119909 minus 10 le 12) (52)










0 2 4 6 8 10 12 14 16 18 20 22Length (m)

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM




Therefore

119868119860119861 (119909) = 119887ℎ3119860119861 (119909)12 = 119887ℎ3119860 (1 + 120573119860119861119909)612 119898119860119861 (119909) = 120588119887ℎ119860119861 (119909) = 120588119887ℎ119860 (1 + 120573119860119861119909)2

(0 le 119909 le 5) (54)

x

y

A B C5 10 5

D

hA hB hC hD




Therefore

119868119862119863 (120585) = 119887ℎ3119862119863 (120585)12 = 119887ℎ3119862 (1 + 120573119862119863120585)612 119898119862119863 (120585) = 120588119887ℎ119862119863 (120585) = 120588119887ℎ119862 (1 + 120573119862119863120585)2

(0 le 120585 = 119909 minus 15 le 5) (56)




0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM


6 Conclusions







Competing Interests


Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










x

y

A B C10 12

hA hB hC

(a)

A

A

B

B

C

C

bA bB bC

(b)









ℎ119860119861 (119909) = ℎ119860 (1 + 120573119860119861119909)2 119887119860119861 (119909) = 119887119860 (1 + 120573119860119861119909)


(49)

Therefore

119868119860119861 (119909) = 119887119860119861 (119909) sdot ℎ3119860119861 (119909)12 = 119887119860ℎ3119860 (1 + 120573119860119861119909)712 119898119860119861 (119909) = 120588119887119860119861 (119909) sdot ℎ119860119861 (119909) = 120588119887119860ℎ119860 (1 + 120573119860119861119909)3

(0 le 119909 le 10) (50)



ℎ119861119862 (120585) = ℎ119861 (1 + 120573119861119862120585)2 119887119861119862 (120585) = 119887119861 (1 + 120573119861119862120585)


(51)

Therefore

119868119861119862 (120585) = 119887119861119862 (120585) sdot ℎ3119861119862 (120585)12 = 119887119861ℎ3119861 (1 + 120573119861119862120585)712 119898119861119862 (120585) = 120588119887119861119862 (120585) sdot ℎ119861119862 (120585) = 120588119887119861ℎ119861 (1 + 120573119861119862120585)3

(0 le 120585 = 119909 minus 10 le 12) (52)










0 2 4 6 8 10 12 14 16 18 20 22Length (m)

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM




Therefore

119868119860119861 (119909) = 119887ℎ3119860119861 (119909)12 = 119887ℎ3119860 (1 + 120573119860119861119909)612 119898119860119861 (119909) = 120588119887ℎ119860119861 (119909) = 120588119887ℎ119860 (1 + 120573119860119861119909)2

(0 le 119909 le 5) (54)

x

y

A B C5 10 5

D

hA hB hC hD




Therefore

119868119862119863 (120585) = 119887ℎ3119862119863 (120585)12 = 119887ℎ3119862 (1 + 120573119862119863120585)612 119898119862119863 (120585) = 120588119887ℎ119862119863 (120585) = 120588119887ℎ119862 (1 + 120573119862119863120585)2

(0 le 120585 = 119909 minus 15 le 5) (56)




0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM


6 Conclusions







Competing Interests


Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















0 2 4 6 8 10 12 14 16 18 20 22Length (m)

minus1

minus08

minus06

minus04

minus02

002040608

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM




Therefore

119868119860119861 (119909) = 119887ℎ3119860119861 (119909)12 = 119887ℎ3119860 (1 + 120573119860119861119909)612 119898119860119861 (119909) = 120588119887ℎ119860119861 (119909) = 120588119887ℎ119860 (1 + 120573119860119861119909)2

(0 le 119909 le 5) (54)

x

y

A B C5 10 5

D

hA hB hC hD




Therefore

119868119862119863 (120585) = 119887ℎ3119862119863 (120585)12 = 119887ℎ3119862 (1 + 120573119862119863120585)612 119898119862119863 (120585) = 120588119887ℎ119862119863 (120585) = 120588119887ℎ119862 (1 + 120573119862119863120585)2

(0 le 120585 = 119909 minus 15 le 5) (56)




0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM


6 Conclusions







Competing Interests


Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 2 4 6 8 10 12 14 16 18 20Length (m)

minus1

minus08

minus06

minus04

minus02

0

02

04

06

08

1

Vert

ical

mod

e disp

lace

men

ts



1st mode-2D-FEM

3rd mode-2D-FEM


6 Conclusions







Competing Interests


Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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