dimensionless analytical solution and new design formula for...

Research ArticleDimensionless Analytical Solution and New Design Formula forLateral-Torsional Buckling of I-Beams under Linear DistributedMoment via Linear Stability Theory

Wen-Fu Zhang12 Ying-Chun Liu2 Ke-Shan Chen2 and Yun Deng2

1School of Architecture amp Engineering Nanjing Institute of Technology Nanjing 211167 China2Northeast Petroleum University Daqing Heilongjiang 163318 China

Correspondence should be addressed to Wen-Fu Zhang zhang wenfu126com

Received 21 October 2016 Revised 27 December 2016 Accepted 18 January 2017 Published 31 December 2017

Academic Editor R Emre Erkmen

Copyright copy 2017 Wen-Fu Zhang et alThis 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

Even for the doubly symmetric I-beams under linear distributedmoment the design formulas given by codes of different countriesare quite different This paper will derive a dimensionless analytical solution via linear stability theory and propose a new designformula of the critical moment of the lateral-torsional buckling (LTB) of the simply supported I-beams under linear distributedmoment Firstly the assumptions of linear stability theory are reviewed the dispute concerning the LTB energy equation isintroduced and then the thinking of Plate-Beam Theory which can be used to fully resolve the challenge presented by Ojalvois presented briefly secondly by introducing the new dimensionless coefficient of lateral deflection the new dimensionless criticalmoment and Wagnerrsquos coefficient are derived naturally from the total potential energy With these independent parameters thenew dimensionless analytical buckling equation is obtained thirdly the convergence performance of the dimensionless analyticalsolution is discussed by numerical solutions and its correctness is verified by the numerical results given by ANSYS finally a newtrilinearmathematical model is proposed as the benchmark of formulating the design formula and with the help of 1stOpt softwarethe four coefficients used in the proposed dimensionless design formula are determined

1 Introduction

Structural stability has always been a key point in the designof steel structures [1ndash6] It is well known that an I-beam hasvarious buckling phenomena such as local buckling distor-tion buckling and lateral-torsional buckling in which thelateral-torsional buckling (LTB) is also known as the globalbuckling This may occur if the applied service loads exceedits LTB critical moment of the I-beam and hence in practicedesign it is important for the designers to obtain the accuratesolution of the critical moment as the upper limit of bucklingstrength of the steel I-beamsHowever until now even for thesimple-supported I-beams with doubly symmetric sectionssubjected to unequal endmoments the design formulas givenby the codesspecifications of different countries are quitedifferent (Figure 1) that makes the designers feel very con-fused In fact such problems also confuse those involved inthe formulation of the codesspecifications In order to avoid

greater controversy there is a trend in the specification that isthe initiative to delete the relevant provisions such as the newversion of the EC3 specification [7] However the deletion ofthe relevant provisions does not mean that such problems donot exist Therefore it is one of the objectives of this paper toprovide a more accurate design formula based on the dimen-sionless analytic solution which is also the main motivationof this paper

Due to the complexity of the LTB phenomenon of I-beams under nonuniform distributed moment so far onlyapproximate analytical solutions or numerical solutions havebeen published

Some approximate analytical solutions can be found inthe classical text books such as Chajes [2] Chen and Atsuta[3] Trahair [4] Timoshenko and Gere [5] and Bleich [6]

The numerical solutions can be obtained from finitedifference method [2 3 6] finite integral method [4] and

HindawiMathematical Problems in EngineeringVolume 2017 Article ID 4838613 23 pageshttpsdoiorg10115520174838613

2 Mathematical Problems in Engineering

12

10

20

18

16

14

08

06

04

02

00

eory in this paperGB50017-2003CAN-CSA-S161-1994BS5950-1-1990

AISCLRFD-2005BS5950-1-2000AS4100-1998

h H

bf1

bf2t f

t f

tw ℎw

Ly

z

M2 M1

minus05 00 05 10minus10

Figure 1 Discrepancies of EUMF for a simply supported beam under linear distributed moment

finite element method (FEM) [8] More recently Suryoat-mono and Ho [9] and Serna et al [10] have reported theresults of elastic LTB load analysis using finite differencetechnique to solve the governing differential equation Limet al [11] have conducted an extensive investigation intothe elastic LTB of I-beams using both the Bubnov-Galerkinapproach [1] and the finite element method [8] In theirpapers Vlasovrsquos differential equation governing [1] for lateral-torsional buckling of beams was used Park et al [12] usethe finite element program developed by Lim et al [11] toperformparametric studies to investigate the lateral-torsionalbuckling behavior of singly symmetric I-beams in which I-beams are modeled by the beam element with two nodes perelement and seven nodal degrees of freedom Greiner andLindner [13] Greiner et al [14] and Mohri et al [15] also usefinite element method to present a quite complete numericalstudy on lateral-torsional buckling of beams Kim et al [16]adopt the variation of the total potential energy to derive thestability equations for the lateral buckling analysis of an arbi-trarily laminated thin-walled composite beam and the exactanalytical stiffness matrix is presented based on power seriesexpansions

It is noteworthy that with the popularization andwidespread use of finite element software since the 2000ssome researchers have tried to use the general-purpose finiteelement software to simulate the buckling behavior or toassess the related design formula of the steel I-beams Forexample Mohri et al [17] adopt the beam elements (B31OS)and shell elements (S8R5) of ABAQUS [18] to model the lat-eral buckling phenomena The authors have found that shellelements could consider local buckling section distortionand the local effects of concentrated loads and boundaryconditions which were all ignored in the beam elementtheory Dolamune Kankanamge and Mahendran [19] use

ABAQUS [18] to investigate the LTB behavior of simplysupported cold-formed steel lipped channel beams subjectedto uniform bending The authors have found that Europeandesign rules are conservative while AustralianNorth Amer-ican design rules are unconservative Tong [20] uses ANSYS[21] to regress the design formula for the singly symmetricI-beams under linear distributed moment Sweedan [22]adopts ANSYS to numerically investigate the lateral stabilityof cellular steel beams subjected to equal end moments mid-span concentrated loads and uniformly distributed loadsSerna et al [10] use CosmosM27 [23] to simulate the lateral-torsional buckling of the I-beams

In theory FEM can be used not only to do research worksbut also to formulate design formulas However when youreally try to formulate the design formulas through a lot ofFEM analysis you will have to deal with some real troubleFirstly the finite element analysis must be a dimensionalanalysis [8 18 21 23] that is the section size and span(or slenderness) of the steel beams must be specified inadvance Hence there is an enormous amount of workloadin the early stage such as the design of hundreds of modelbeams and the relevant data preparation in the finite elementmodeling secondly one has to spend a lot of time and effortin later stages to select from a large number of data outputsand determine which results are useful LTB results finallyand most importantly our experience shows that becausethe meshing of FEM model will greatly affect the accuracyof finite element analysis hence the corresponding FEManalysis results lack of precision consistency and regularityThis may be the reason why the design formulas of differentcountries will be quite different

On the contrary not only does the dimensionless analyticsolution have the advantages of small input data amountand fast calculation speed but also the formula formulated

Mathematical Problems in Engineering 3

from these dimensionless results is more general than thoseobtained from FEM analysis To the authorrsquos knowledgeresearch on the dimensionless analytical solution is scareand began in 1970s with the Anderson and Trahairrsquos works[24] followed by Kitipornchai et al [25] and Kitipornchaiand Wang [26 27] to simulate the buckling behavior ofsingly symmetric I-beams and Tee beams under momentgradient However the dimensionless Wagnerrsquos coefficientand beam parameter defined in their work lack definitephysical and geometricmeaning andmore importantly noneof the dimensionless parameters is independent that is theyare not independent variables and cannot be used to get areasonable design formula Therefore it is a very importanttask to develop new dimensionless analytic solution and useit for design purpose

According to the theory and discussion in [28] it canbe proved that an exact dimensionless solution of the LTBproblem of the simply supported I-beam can be obtained bythe standard trigonometric series Recently based upon theauthorrsquos work [29 30] we have carried out a series of researchwork on the dimensionless analytical solutions and designformulas of the critical moment of the LTB of the I-beams[31ndash34] Recently Zhang et al [35] derived a dimensionlessanalytical solution and design formula of the critical momentof the LTB of the cantilever beam with tip lateral elastic braceunder uniform and concentrated load

As a continuation of the work of the author this studyshall present a dimensionless analytical solution via linearstability theory and design formula of the critical momentof the LTB of the simply supported I-beams under lineardistributed moment In Section 2 the basic assumptionsof the traditional linear stability theory and the disputeconcerning the LTB energy equation are introduced andthen the attempts to resolve the dispute and the Plate-Beam Theory are described briefly in Section 3 with themodal trial function expressed by the trigonometric seriesthe energy variation method is adopted in the derivation ofthe dimensionless analytical solution of the critical momentMoreover the definition of dimensionless critical momentand its applicability is discussed next the 1st-order and 6th-order approximate analytical solution are presented in Sec-tion 4 In Section 5 the numerical solution and convergenceperformance of the dimensionless analytical solution arediscussed and its correctness is verified by ANSYS software inSection 6 This part also presents a new technique to satisfythe rigid section hypothesis Finally a new mathematicalmodel of the design formula is presented and with the helpof 1stOpt software the four coefficients used in the proposeddimensionless design formula are determined

2 Theory Background of LTB Energy Equationfor I-Beams via Linear Stability Theory

As is known for the slender (119871 ge 10119889) thin-walled (119889 ge10119905) I-beams the LTB (ie overall buckling) will generallycontrol the final design In this case the simplified mechan-ical models that is differential equation model and energyequation model of the LTB problem of the thin-walled

beams can be established by introducing appropriate assump-tions

For the sake of completeness a brief review and com-ments of the basis of linear stability theory are presentedhereafter

21 Basic Assumptions of Traditional Linear Stability Theory

(1) Rigid section hypothesis that is the cross section isvery rigid that its original shape is retained duringbucklingThis is thewell-knownVlasovrsquos rigid sectionhypothesis which means that local and distortionalbuckling are excluded in the LTB analysis

(2) Euler-Bernoullirsquos hypothesis that is there is no sheardeformation in the middle surface of the cross sec-tionThis assumption is used to describe the in-planedeformation of a shellplate

(3) Kirchhoff rsquos hypothesis that is the shear deformationin the planes normal to themiddle surface of the crosssection is small and can be neglectedThis assumptionis used to describe the out-of-plane deformation of ashellplate

(4) Thematerial is an ideal isotropic material and followsHookersquos law

(5) Displacements and twist angle are assumed to besmall enough

The first two hypotheses are first explicitly proposedby Vlasov [1] and the last two are implicitly found in thetheoretical derivation of LTB problems [1 20 36ndash38]

Invoking assumption (1) the following transverse dis-placements were proposed by Vlasov119906 (119910 119911) = 119906 (119911) minus (119910 minus 1199100) 120579 (119911)

V (119909 119911) = V (119911) minus (119909 minus 1199090) 120579 (119911) (1)

where 119906(119911) V(119911) are the transverse displacements of the shearcenter 120579(119911) is the twist angle of the cross section and 1199090 1199100are the coordinates of the shear center

Invoking assumption (2) that is by setting the shearstrain of themiddle surface to zero the following longitudinaldisplacement was deduced119908 (119909 119911) = 119908 (119911) minus 1199091199061015840 (119911) minus 119910V1015840 (119911) minus 1205961205791015840 (119911) (2)

where 120596 is the ldquowarping functionrdquo [36 pg 11] but Vlasovcalled it the sectorial area [1 pg 16]

In Vlasovrsquos monograph (2) is called the law of sectorialarea Obviously since the condition about the shear strainof the middle surface is used in his derivation (2) is onlyapplicable to describing the longitudinal displacement of anarbitrary point in the middle surface rather than that outsidethe middle surface As a result the effect of St Venantrsquostorsion could not be deduced correctly by the law of sectorialarea Therefore in essence (2) cannot be directly used toderive the energy equation for LTB problem of thin-walledstructures


In short even though Vlasovrsquos theory is such invaluablethat almost all the literature on the theory of thin-walledbars will refer to his work thus making him an outstandingfigure among the pioneers in the field [36ndash38] his theoryis incomplete since in his work only the LTB differentialequations were established by the fictitious load method andno LTB energy equation was presented since his proposeddisplacement field could not be used to deduce correct energyequation This is precisely because of such basic defects andthe abuse of the law of sectorial area leading to a lot ofcontroversy in future generations

22 Dispute concerning the LTB Energy Equation The firstcorrect LTB energy equation is proposed by Bleich [6] in1952 (called traditional LTB energy equation) Howeverhis theoretical derivation was not rigorous That is it istheoretically incomplete The defects include the following(1) the longitudinal displacement of the cross section is notclearly defined (2) the strain energy corresponding to theSt Venant torsion cannot be derived naturally at the sametime and thus should be additionally added by invokingthe St Venant formula (3) the Wagner effect caused by thetransverse load cannot be obtainednaturally because only thelongitudinal potential energy is considered in his derivation

In addition bothVlasov andBleichrsquos theories do not applyto the combined cross sections (eg open-closedsolid-open)and composite cross sections (eg steel-concrete)

Because the LTB energy equation is the foundation of themodern mechanics analysis such as finite element methodand approximate analytic solution extensive research hasbeen carried out since the 1960s The related theoreticalderivation on the LTB energy equation for I-beams hasbeen detailed in [2ndash5 20 36ndash38] It is found that since thecorrectness of the LTB energy equation is often subject toa variety of academic challenges most of these efforts wereaimed at trying to explain the rationality of the following LTBenergy equation proposed by Bleich that is

Π = 12 int119871 [119864119868119910119906101584010158402 + 119864119868120596120579101584010158402 + (119866119869119896 + 2119872119909120573119909) 12057910158402+ 211987211990911990610158401015840120579 minus 1199021198861199021205792] 119889119911 (3)

where 119906 is the lateral deflection of the shear center 120579 is thetwist angle of the cross section 119864119868119910 is the minor axis flexuralrigidity119866119869119896 is the St Venant torsion rigidity119864119868120596 is theVlasovwarping torsion rigidity 119872119909 is the moment function aboutthe major axis 119871 is the length of the beam 119902 is the distributedload acting in the vertical plane of the I-beams 119886119902 is the loadposition parameter which is obtained by subtracting the 119910-coordinate of the load acting point from that of the shearcenter and 120573119909 is the Wagner coefficient and it is definedas 120573119909 = 12119868119909 int119860 119910 (1199092 + 1199102) 119889119860 minus 1199100 (4)

where119909119910 is the coordinates of an arbitrary pointwith respectto the origin (the centroid of the cross section) and 1199100 is thecoordinate of the shear center of the cross section

The most fatal challenge comes from the question of thevalidity of the Wagner hypothesis which is a part of the the-oretical basis for the traditional LTB theory Since 1981 thisproblem has been continuously questioned and challengedby Ojalvo [38ndash41] and considerable ongoing efforts [42ndash47]have been made by some famous scholars such as TrahairChen and Knag et al to confirm the validity of the Wagnerhypothesis in the framework of Vlasov theory Althoughthese in-depth series of discussions have led to a furtherunderstanding of theWagner hypothesis and the related LTBtheory the basis of the argument is still relatively inadequateand did not get everyonersquos recognition Concerning thisZiemian in his book [48] concludes that ldquoit should be notedthat a challenge to part of the theory of monosymmetric beamscalled the Wagner hypothesis was presented by Ojalvo (1981)At the present time (2009) this challenge still has not been fullyresolvedrdquo

23 Attempts to Resolve the Dispute and the Plate-BeamTheory From the appearance point of view the problem oftheWagner coefficient is only reflected in process of derivingthe load potential hence some attempts have been made toimprove the Vlasovrsquos displacement filed in order to obtain thecorrect Wagner coefficient For example based upon a morecomplex geometric analysis a longitudinal displacementwithtwo additional nonlinear terms was proposed by Trahair (see[4 pg 384])119908 (119909 119911) = 119908 minus 1199091199061015840 minus 119910V1015840 minus 1205961205791015840 + (minus119909V1015840120579 + 1199101199061015840120579) (5)

In addition even the concept of the rotation matrixwhich was suitable for the nonlinear analysis with largerotation [49] was introduced into the linear LTB analysis andthe transverse displacements with more additional nonlinearterms were proposed by Pi et al [50] as follows

119906 (119910 119911) = 119906 (119911) minus 119910120579 (119911) minus 119909 (11990610158402 + 1205792)2 minus 1199101199061015840V10158402minus 12059611990610158401205791015840V (119909 119911) = V minus 119909120579 (119911) minus 1199091199061015840V10158402 minus 119910 (V10158402 + 1205792)2 minus 120596V10158401205791015840119908 (119909 119911) = 119908 minus 1199091199061015840 minus 119910V1015840 minus 1205961205791015840 + (minus119909V1015840120579 + 1199101199061015840120579)

(6)

Obviously incorrect linear strain energy will be obtainedif the above displacements of Trahair and Pi et al were usedTherefore the above improvements are cumbersome andunnecessary

In fact The LTB problem is similar to the vibration prob-lem of the prestressed concrete beam and hence this problembelongs to the category of linear eigenvalue problem inessence In modern mechanics terms this problem is a smalldisplacement finite strain problem namely in such analysisassumptions (3) and (4) hold thus increasing the nonlinearterm contradiction with the above basic assumptions Inaddition the second variation criterion was used in theirderivation which has been discarded in modern variational


y

x

yL

z

M2 M1

M2 M1

S(0 y0)

C(0 0)

Figure 2 Calculation diagram of simply supported beam under linear distributed moment

principle [51ndash54] These problems were also criticized byOjalvo [41] as follows ldquo the underlying theorem on which thederivation must be based (the theorem of stationary potentialenergy) is misused Components of a finite strain tensor areused in the strain energy expression when infinitesimal strainexpressions are clearly called for by the theoremrdquo

In the authorrsquos opinion so far the problem of how toderive the linear strain energy correctly based only uponthe commonly used engineering mechanics theory has notbeen resolved in both the traditional theory and Ojalvorsquos newtheory [38] since in the whole process of the theoreticaldebates this problem has been deliberately avoided by bothsides of the debaters

In fact there is a common theoretical basis in bothBleichrsquostheory and Ojalvorsquos new theory that is they all tried to useengineering theory (beam or rod) to derive their LTB energyequation For example Navierrsquos plane section hypothesis isused in Bleichrsquos derivation of the linear strain energy Butthese thoughts were rarely mentioned in the published LTBliterature

This study considers a simply supported steel I-beamunder linear distributed moment as shown in Figure 2 Obvi-ously this I-beam is an open thin-walled beam composedof three flat plates (ie two flanges and one web plate)Therefore how to more rationally describe in-plane andthe out-of-plane deformation of a flat plate by the classicalengineering mechanics is the essence of solving the debateof the LTB problem reasonably Based on such a basicunderstanding the author created the Plate-Beam Theory[30 55ndash58] recently

In this new theory Vlasovrsquos hypotheses are preservedbut the warping function and the Vlasovrsquos longitudinaldisplacement are discarded The longitudinal displacementgeometrical equation strain energy and initial stress poten-tial energy of the in-plane bending and the out-plane bendingand torsion of each flat plate are described by the Euler beamtheory and Kirchhoff-Plate theory respectively (Figure 3)It is shown that this new theory combines the advantagesof Vlasov and Bleichrsquos theory and applies not only to open[56]closed cross sections [57] but also to open-closedsolid-open cross sections [58]

In short the Plate-Beam Theory provides a new way toresolve the dispute concerning the Wagner hypothesis com-pletely in which the rational elements of Bleichrsquos Vlasovrsquos

and Ojalvorsquos thoughts have been inherited and developedwhile the warping function of Vlasov was abandoned Fur-thermore the correctness of the traditional LTB energyequation such as (3) is reaffirmed via the commonly usedplate and beam theory

3 Dimensionless Analytical Solution

31 Moment Function and Modal Trial Functions For thesimply supported I-beam shown in Figure 2 it is easy toobtain the moment function as follows119872119909 (119911) = 1198961198721 + (1 minus 119896)1198721 119911119871 119896 isin [minus1 +1] (7)

where 119896 = 11987221198721 is the end moment ratio and11987221198721 arethe left and right end moment respectively and |1198722| le |1198721|

For the case of the simply supported I-beams the fol-lowing modal trial functions are proposed for the lateraldeflection 119906(119911) and twist angle 120579(119911) [29ndash35] which canbe expressed as the sum of trigonometric series as fol-lows

119906 (119911) = ℎinfinsum119899=1

119860119898 sin [119898120587119911119871 ] (119898 = 1 2 3 infin) (8)

120579 (119911) = infinsum119899=1

119861119899 sin [119899120587119911119871 ] (119899 = 1 2 3 infin) (9)

where 119860119898 and 119861119899 are the undetermined dimensionlesscoefficients for the lateral deflection 119906(119911) and twist angle 120579(119911)respectively ℎ is the distance between the centroid of top andbottom flange

From the perspective of structural dynamics 119860119898 and 119861119899can be regarded as the generalized coordinates or generalizeddegrees of freedom However their dimensions are differentso in order to remove the dimension of 119860119898 so that 119860119898and 119861119899 are both undetermined dimensionless coefficients weintroduce ℎ into the expression of 119906(119911) in this workThis ideais proposed by the author which is completely different fromother previous research [25ndash27] as discussed in Section 34and it has been used in LTB analysis of various steel beams[29ndash35]


Out-plane bending (torsion) at plate analogy

Kirchho-Plate theory (Zhang 2014)

In-plane bending beam analogy

Euler-beam theory (Zhang 2014)

ldquoPlate-Beam eoryrdquo (Zhang 2014)

Rigid section hypothesis (Vlasov 1961)

T

M Mu

u

T

Figure 3 Illustration of section deformation assumption and plate deformation analogy

Obviously the above trial functions are orthogonal andsatisfy the following geometric boundary conditions of thesimply supported beams that is119906 (0) = 11990610158401015840 (0) = 0119906 (119871) = 11990610158401015840 (119871) = 0120579 (0) = 12057910158401015840 (0) = 0120579 (119871) = 12057910158401015840 (119871) = 0

(10)

Moreover according to the theory and discussions givenby literature [28] it can be proved that trigonometric series isthe exact solution of the modal trial function for the problemshown in Figure 2 and hence the exact analytical expressionsof the total potential energy and the one for buckling equationcan be derived directly

32 Dimensionless Total Potential Energy Substituting themoment function (7) and the modal trial functions (8)-(9)into the potential energy functional (3) and integrating over

the beam length 119871 one can get the integration results of thetotal potential energy

Without loss of generality the following new dimension-less parameters [29ndash35] are introduced

0 = 1198721(12058721198641198681199101198712) ℎ 120573119909 = 120573119909ℎ 120578 = 11986811198682 119870 = radic12058721198641198681205961198661198691198961198712

(11)

where 0 is the dimensionless endmoment ℎ is the distancebetween the centroid of the top and bottom flange 119870 is thedimensionless torsional stiffness parameters 120578 is the ratio ofthe moment inertia about minor axes of the top flange to thatof the bottom flange 1198681 1198682 are the 2nd moment of inertia


about minor axes of the top and bottom flange respectively120573119909 is the dimensionless Wagnerrsquos coefficientIn the following derivation the energy equation of (3) is

first multiplied by a common factor 1198713(ℎ2119864119868119910) and then thesubstitution relationships of (11) are introduced For the sakeof clarity the final integration results are listed as follows inthree partsΠ1 = 12 int1198710 [119864119868119910119906101584010158402 + 11986411986812059612057910158402 + 11986611986911989612057910158402] 119889119911

= infinsum119898=1

1198602119898119898412058744 + infinsum119899=1

1198612119899 (1 + 11989921198702) 1198992120587412057841198702 (1 + 120578)2 Π2 = 12 int119871 211987211990912057311990912057910158402119889119911

= 0 infinsum119899=1

1198612119899 (1 + 119896)119898212058741205731199094+ 0 infinsum119904=1

infinsum119903=1119903 =119904

119861119904119861119903 4 (minus1 + 119896) 1205872119903119904 (1199032 + 1199042) 120573119909(1199032 minus 1199042)2 |119904 plusmn 119903| = even

Π3 = 12 int119871 [2119872119909119906101584010158401205791015840] 119889119911= minus0 infinsum119898=1

119860119898119861119898 (1 + 119896)119898212058744minus 0 infinsum119904=1

infinsum119903=1119903 =119904

119860 119904119861119903 4 (minus1 + 119896) 12058721199031199043(1199032 minus 1199042)2 |119904 plusmn 119903| = even

(12)

The total potential energy of this problem is the sum ofthe above equations that is

Π = 3sum119894=1

Π119894 (13)

33 Dimensionless Analytical Solution of Buckling EquationAccording to the principle of energy variation methodthe following stationary conditions that minimize the totalpotential energy are required120597Π120597119860119898 = 0 119898 = 1 2 3 infin120597Π120597119861119899 = 0 119899 = 1 2 3 infin (14)

Note that in the above equations 119898 and 119899 are arbitrary andindependent

From the condition 120597Π120597119860119898 = 0 we can get theequilibrium equation with respect to the lateral deflection asfollows

infinsum119898=1

119860119898119898412058742 minus 0 infinsum119898=1

119861119898 (1 + 119896)119898212058744

minus 0 infinsum119898=1

infinsum119903=1119903 =119898|119898plusmn119903|=even

119861119903 4 (minus1 + 119896) 12058721199031198983(1199032 minus 1198982)2 = 0(15)

For each fixed 119898 (corresponding to Line 119898) herein onehomogeneous equation corresponding to119860119898119861119898 and119861119903 willbe obtained

With the knowledge of the internal data structure of theabove infinite series equation and using the following thenotations listed as follows119860 = [1198601 1198602 1198603 sdot sdot sdot]119879 119861 = [1198611 1198612 1198613 sdot sdot sdot]119879 (16)

we can rewrite (15) in the following matrix form[0119877] 119860 + [0119878] 119861 = 0 ([1119877] 119860 + [1119878] 119861) (17)

0119877119898119898 = 119898412058742 119898 = 1 2 infin0119877119904119903 = 0 119903 = 119904 119904 = 1 2 infin 119903 = 1 2 infin (18)

0119878119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (19)

1119877119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (20)

1119878119898119898 = (1 + 119896)119898212058744 119898 = 1 2 infin1119878119904119903 = 4 (minus1 + 119896) 12058721199031199043(1199032 minus 1199042)2 119904 = 119903 |119904 plusmn 119903| = even 119904 = 1 2 infin 119903 = 1 2 infin

(21)

where 119860 and 119861 are the column vectors constituted by allundetermined coefficients119860119898 and 119861119898 respectively [0119877] and[1119877] are the coefficient matrixes for 119860 in the left and righthandof the above equilibriumequation respectively [0119878] and[1119878] are the coefficient matrixes for 119861 in the left and righthand of the above equilibrium equation respectively

Similarly from the condition 120597Π120597119861119899 = 0 we can getthe equilibrium equation with respect to the twist rotation asfollows


119860119899 (1 + 119896) 119899212058744minus 0 infinsum119899=1


119860119903 4 (minus1 + 119896) 12058721198991199033(1198992 minus 1199032)2+ 0 infinsum119899=1

119861119899 (1 + 119896) 119899212058741205731199092


+ infinsum119899=1

119861119899 (1 + 11989921198702) 1198992120587412057821198702 (1 + 120578)2+ 0 infinsum119899=1


119861119903 4 (minus1 + 119896) 1205872119903119899 (1199032 + 1198992) 120573119909(1199032 minus 1198992)2= 0

(22)

With the notations in (16) the above equation can also bearranged in the following matrix form[0119879] 119860 + [0119876] 119861 = 0 ([1119879] 119860 + [1119876] 119861) (23)

0119879119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (24)

1119879119904119903 = 1119878119903119904 119904 = 1 2 infin 119903 = 1 2 infin (25)

0119876119898119898 = (1 + 11989821198702)1198982120587412057821198702 (1 + 120578)2 119898 = 1 2 infin0119876119904119903 = 0 119903 = 119904 119904 = 1 2 infin 119903 = 1 2 infin (26)

1119876119898119898 = minus(1 + 119896)119898212058741205731199092 119898 = 1 2 infin1119876119904119903 = minus4 (minus1 + 119896) 1205872119903119904 (1199032 + 1199042) 120573119909(1199032 minus 1199042)2 119903 = 119904 |119904 plusmn 119903| = even 119904 = 1 2 infin 119903 = 1 2 infin

(27)

where [0119879] and [1119879] are the coefficient matrixes for 119860 inthe left and right hand of the above equilibrium equationrespectively [0119876] and [1119876] are the coefficientmatrixes for 119861in the left and right hand of the above equilibrium equationrespectively

If (17) and (23) are combined together the bucklingequation can be expressed as follows in the form of blockmatrix

[0119877 01198780119879 0119876]119860119861 = 0 [1119877 11198781119879 1119876]119860119861 (28)

where 0 is the dimensionless critical moment that is thedimensionless buckling strength 119877 119878 119879 and 119876 are thesubblock coefficient matrixes constructed from analyticalexpressions included in (18)ndash(21) and (24)ndash(27) respectively

Equation (28) is the dimensionless analytical bucklingequation of the LTBproblemof a simply supported steel beamunder linear distributed moment obtained in the presentwork

Obviously from the mechanical point of view the matrixat the left-hand side of the equation represents the linearstiffness matrix of the steel girder and is independent of

the load while the matrix on the left side of the equationdepends on the load pattern which in the present case can beconsidered as a geometric stiffness matrix due to the lineardistribution of moments

It is noted that the dimensionless analytical solution ofthe buckling equation presented in the paper not only has aclear data structure and but also is easy to program and solveIn addition unlike the FEM solutions published in literaturethis solution can easily be verified by anyone

34 Remarks on Definition of Dimensionless Critical MomentThere are different ways to define the dimensionless criticalmoment in which the well-known definition is the oneproposed by Kitipornchai et al [25] and Kitipornchai andWang [26 27] and can be expressed as follows

cr = 119872cr119871radic119864119868119910119866119869119896 (29)

Corresponding to this definition the following dimen-sionless Wagnerrsquos coefficient the load position parameterand the beam parameter are defined by Kitipornchai

120575 = 120573119909119871 radic(119864119868119910119866119869119896)120576 = 119886119871radic(119864119868119910119866119869119896) = radic1205872119864119868119910ℎ21198661198691198961198712 = 120587ℎ119871 radic119864119868119910119866119869119896

(30)

Obviously all these parameters lack clear physical andgeometric meanings More importantly since there is a com-mon factorradic119864119868119910119866119869119896 in (30) none of the parameters definedin (30) is independent that is they are not independentvariables thus it is difficult to use them to regress a rationaldesign formula

Instead the following dimensionless critical moment isused in this paper

cr = 119872cr(12058721198641198681199101198712) ℎ = 119872crℎ119875119864119910 (31)

This definition is first proposed by Professor Zhang [29]since 2008 and later used in the LTB analysis of various steelbeams [30ndash35]

It is the first finding in this paper that this definition canbe derived naturally from the total potential energy basedupon the new definition of displacement function that is (8)secondly it has a clear physical meaning as shown in the lastterm in (31) that is the product of Eulerrsquos bending bucklingload around the minor axis and the distance between thecentroid of the top and bottom flange is used to normalizethe actual critical moment


In addition in the case of the LTB of a doubly symmetricbeam under pure bending the definition of (31) also has adefinite geometric meaning that is

cr = 119872crℎ119875119864119910 = 119906120579 = 12radic1 + 119870minus2 (32)

That is the dimensionless critical moment is the dimen-sionless distance between the fixed axis of rotation and theshear center for a cross section It is clear that for the doublysymmetric beams the fixed axis of rotation always lies beyondthe tension flange due to 119870 gt 0

Finally and most importantly the other dimensionlessparameters defined in (11) also have definite physical mean-ings and are independent of each other and are thereforesuitable for the formulation of a new design formula

4 Approximate Analytical Solution

41 First-Order Approximation If 119898 = 119899 = 1 the followingsimple expression can be obtained from (28)

[[[[[12058742 00 (1 + 1198702) 120587412057821198702 (1 + 120578)2

]]]]]11986011198611

= 0 [[[[0 (1 + 119896) 12058744(1 + 119896) 12058744 minus(1 + 119896) 12058741205731199092

]]]]11986011198611

(33)

Under the conditions that the coefficient determinant iszero then the first-order approximation of the dimensionlesscritical moment of the I-beams under linear distributedmoment can be written in a compact form

cr = 1198621 [[(minus1198622119886 + 1198623120573119909)+ radic(minus1198622119886 + 1198623120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2)]]

(34)

where

1198621 = 21 + 119896 1198622 = 01198623 = 1(35)

or

cr = 21 + 119896 [[120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2)]] (36)

Obviously this 1st-order approximation only applies tocases where 119896 is greater than zero and at 119896 = minus1 the criticalmoment is infinite which is unreasonable

If 119896 = 1 which means that the two end moments areequal that is the beam is under pure bending then itsdimensionless critical moment can be expressed in terms of(36) as follows

cr = 120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2) (37)

It is easy to find that this expression is identical to theexact analytical solution in the textbook [1ndash6] This con-clusion indicates that the solution of the buckling equationpresented in the paper is indeed an accurate solution in thisspecial case

42 Sixth-Order Approximation If 119898 = 119899 = 6 the followinganalytical expression can be obtained from (28)

[0119877 01198780119879 0119876]

6times6

1198601198616times1

= 0 [1119877 11198781119879 1119876]6times6

1198601198616times1

(38)

where

0119878 = 0119879 = 1119877 = 0

0119877 =((((((((((((((

12058742 0 0 0 0 00 81205874 0 0 0 00 0 8112058742 0 0 00 0 0 1281205874 0 00 0 0 0 62512058742 00 0 0 0 0 6481205874

))))))))))))))


0119876 =(((((((((((((((((((((

(1 + 1198702) 120587412057821198702 (1 + 120578)2 0 0 0 0 00 (4 + 161198702) 120587412057821198702 (1 + 120578)2 0 0 0 00 0 (9 + 811198702) 120587412057821198702 (1 + 120578)2 0 0 00 0 0 (16 + 2561198702) 120587412057821198702 (1 + 120578)2 0 00 0 0 0 (25 + 6251198702) 120587412057821198702 (1 + 120578)2 00 0 0 0 0 (36 + 12961198702) 120587412057821198702 (1 + 120578)2

)))))))))))))))))))))

1119879119904119903 = 1119878119903119904 =(((((((((((((((((

14 (1 + 119896) 1205874 89 (minus1 + 119896) 1205872 0 16225 (minus1 + 119896) 1205872 0 24 (minus1 + 119896) 12058721225329 (minus1 + 119896) 1205872 (1 + 119896) 1205874 9625 (minus1 + 119896) 1205872 0 160441 (minus1 + 119896) 1205872 00 21625 (minus1 + 119896) 1205872 94 (1 + 119896) 1205874 43249 (minus1 + 119896) 1205872 0 89 (minus1 + 119896) 1205872256225 (minus1 + 119896) 1205872 0 76849 (minus1 + 119896) 1205872 4 (1 + 119896) 1205874 128081 (minus1 + 119896) 1205872 00 1000441 (minus1 + 119896) 1205872 0 200081 (minus1 + 119896) 1205872 254 (1 + 119896) 1205874 3000121 (minus1 + 119896) 1205872864 (minus1 + 119896) 12058721225 0 329 (minus1 + 119896) 1205872 0 4320121 (minus1 + 119896) 1205872 9 (1 + 119896) 1205874

)))))))))))))))))

1119876

=(((((((((((((((((

minus12 (1 + 119896) 1205874120573119909 minus409 (minus1 + 119896) 1205872120573119909 0 minus272225 (minus1 + 119896) 1205872120573119909 0 minus888 (minus1 + 119896) 12058721205731199091225minus409 (minus1 + 119896) 1205872120573119909 minus2 (1 + 119896) 1205874120573119909 minus31225 (minus1 + 119896) 1205872120573119909 0 minus1160441 (minus1 + 119896) 1205872120573119909 00 minus31225 (minus1 + 119896) 1205872120573119909 minus92 (1 + 119896) 1205874120573119909 minus120049 (minus1 + 119896) 1205872120573119909 0 minus409 (minus1 + 119896) 1205872120573119909minus272225 (minus1 + 119896) 1205872120573119909 0 minus120049 (minus1 + 119896) 1205872120573119909 minus8 (1 + 119896) 1205874120573119909 minus328081 (minus1 + 119896) 1205872120573119909 00 minus1160441 (minus1 + 119896) 1205872120573119909 0 minus328081 (minus1 + 119896) 1205872120573119909 minus252 (1 + 119896) 1205874120573119909 minus7320121 (minus1 + 119896) 1205872120573119909

minus888 (minus1 + 119896) 12058721205731199091225 0 minus409 (minus1 + 119896) 1205872120573119909 0 minus7320121 (minus1 + 119896) 1205872120573119909 minus18 (1 + 119896) 1205874120573119909

)))))))))))))))))

(39)

Equation (38) gives the sixth-order approximation of thedimensionless critical moment of a steel beam under thelinear distributed moment

It can be proved that the above result is exactly thesame as that obtained by using trigonometric series with sixterms as the modal function This shows that the infiniteseries solution that is the exact analytical solution derivedin previous section is correct

Our research results [34 35] have shown that in mostcases the sixth-order approximation can achieve satisfactoryresults However if one wants to obtain more accurate resultsof the dimensionless critical moment then more termsshould be included in the numerical computations Hencewe shall present the numerical algorithm to solve the dimen-sionless analytical buckling equation and the convergenceperformance is discussed in the next section

5 Numerical Solution andConvergence Performance

51 Matrix Expression of Eigenvalue Problem In order toobtain a numerical solution to the dimensionless analyticalsolution we shall take the number of trigonometric series in(8) and (9) as a finite number such as119898 = 119899 = 119873

If the common-used symbols used in traditional finiteelementmethod [8] are adopted then (28) can be abbreviatedas follows[1198700]2119873times2119873 1198802119873 = 0 [119870119866]2119873times2119873 1198802119873 (40)where 1198802119873 = [119860 119861]119879

[1198700]2119873times2119873 = [ 01198770119879 01198780119876 ]


295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1

Figure 4 Convergence performance

[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)

where 1198802119873 is the buckling mode which is consisted ofdimensionless undetermined coefficients (generalized coor-dinates) [1198700]2119873times2119873 is the linear stiffness matrix of the steelbeam [119870119866]2119873times2119873 is the geometric stiffness matrix of thesteel beam under the linear distributed moment 0 is thedimensionless critical moment

Mathematically the problem of solving 0 can ultimatelybe attributed to solving the generalized eigenvalue problemof (38) where 0 is the smallest eigenvalue and 1198802119873 is thecorresponding eigenvector (ie buckling mode)

Obviously the final form of the dimensionless bucklingequation (see (38)) has not only a clear data structure that iseasy to program but also a well-known form of the traditionalfinite element method and therefore has a definite physicalmeaning while that given by Kitipornchai et al [25ndash27]and McCann et al [59] is a Hessian matrix in which the

dimensionless critical moment is implicit in that matrixTherefore their expressions lack clear physical meaning andare not easily solved by the existing mathematical software

52 Convergence Performance and Numerical AlgorithmFirst the convergence performance of the trigonometricseries is presented Typical graphs of the relationship betweenthe dimensionless bucklingmoment and the number of termsof trigonometric series used in the modal trial functionsare shown in Figure 4 It can be seen that in generalthe trigonometric series is basically monotone convergencethe convergence rate for the negative end moment ratio isobviously lower than that for the positive end moment ratio

When 119896 changes from minus1 to 1 the convergence rate isgradually accelerated and the number of terms needed toobtain accurate numerical solutions is gradually reducedMoreover the results show that when 119896 = minus1 the conver-gence rate is the slowest and more than six terms are neededto obtain accurate numerical solutions However when


Figure 5 Simulation of rigid section hypothesis by ldquoCERIGrdquo command

119896 = 1 the convergence rate is the fastest and only one termis needed to get the exact solution

Next the numerical algorithm is discussed In fact anymethod used to solve the generalized eigenvalue problem isapplicable to LTB problem of this paper because it is differentfrom the large-scale finite element analysis program and thenumber of degrees of freedom involved in this paper is notlarge

It is noted that the eig(lowast lowast) function provided in theMATLAB [60] can be used to obtain the eigenvalue andeigenvector of a given matrix Even for matrixes with largedimensions very high accuracy results can be obtainedeffectively Therefore this paper uses the eig(lowast lowast) functionto develop the developed MATLAB program

In order to automatically obtain a numerical solution ofany desired accuracy it is necessary to perform an iterativeprocess in the developed MATLAB program in which thefollowing convergence criteria are used10038161003816100381610038161003816(119894+1)0 minus (119894)0 10038161003816100381610038161003816(119894)0 le tol (42)

where tol is the iteration tolerance When the eigenvalue 0is required to 2119904-digit accuracy [8] then tol = 10minus2119904 is used inthe subsequent analysis

In order to facilitate the dimensionless parameter analysisand to perform the comparative study between the numericalsolution from the exact dimensionless analytical solutionand the ANSYS finite element analysis a MATLAB programis developed according to the aforementioned numericalalgorithm

6 Verification by FEM Software

In this section the general-purpose finite element softwareANSYS is used to verify the correctness of the exact solutionand 6th-order approximations given previously

61 Description of FEM Model It is well known that theBEAM189 in ANSYS software cannot correctly simulate the

critical moment of LTB for the singly symmetric beamsTherefore the SHELL63 element is used in this paper tosimulate the buckling problem of the steel beams

The second reason is that since a typical steel I-beam iscomposed of three thin plates it is theoretically correct usingthe SHELL elements to simulate the buckling problem of theI-beamsThis is in accordance with Vlasovrsquos work [1] becausehis theory was established on the simplification of the shelltheory and the concept of ldquoreducedmodulusrdquo was used in hismonograph (see equation (55) in [1])

It is shown that only in this way all kinds of bucklingphenomena such as local buckling distortion bucklingand lateral-torsional buckling (LTB) can be simulated andcaptured However only the LTB phenomena is concerned inthis investigation Consequently when the SHELL63 elementis used to simulate the desired lateral-torsional buckling of asteel beam there is a problem that is for beams simulatedby SHELL63 element the Vlasovrsquos rigid section hypothesiscannot be automatically satisfied as the BEAM elementsTherefore special measures have to be taken in the FEMsimulation to ensure that the overall buckling that is LTBrather than the local buckling or the distortional bucklingappears first

Some approaches have been proposed such as themethod of adding stiffeners or the method of treating thestiffeners as membrane elements [20] to approximate therigid section hypothesis However it is found that for 119896 =minus1 no matter how many stiffeners are added no satisfactoryresults can be obtained In order to overcome this difficultythis paper presents an innovative simulation technique for theANSYS software to satisfy the rigid section hypothesis

After exploration and modeling practice we found thatin the ANSYS finite element model the ldquoCRIGrdquo commandis such a very effective command that you can be moreconfident to simulate the rigid section features of the buckledbeams without worrying about the failure of FEM numericalsimulation Figure 5 shows the ANSYS finite element modelafter the CERIG command is used Moreover each end of thesteel beam is simulated as the fork support as usual (Figure 6)and the end moment is simulated by applying a concentrated


Figure 6 Simulation of boundary conditions (fork supports)

Figure 7 Simulation of end moments

torque at the centroid of the cross section at each end of thesteel beam (Figure 7)

The material properties are taken as follows Youngrsquosmodulus 119864 = 206GPa Poisonrsquos ratio 120592 = 03 and the shearmodulus of elasticity is given by 119866 = 119864[2(1 + 120592)]62 Verification of FEM Model Although the analytical andnumerical solutions given in the previous section are dimen-sionless the finite element analysis must be a dimensionalanalysis in which the section size and span (or slenderness)of the steel I-beams must be specified in advance

Considering that the finite element analysis here is ofa verification nature only three typical cross sections areselected for the study purpose Section A is a doubly sym-metric section and both Section B and Section C are singlysymmetric sections in which Section B has a larger tensionflange whereas Section C has a larger compression flange Allpictures of the selected sections along with their geometricalproperties are shown in Table 1

The beam spans are chosen such that the span-to-depthratios lie between 15 and 30 Table 2 lists the dimensionlessparameters associated with the calculation of the LTB criticalmoments

The above-mentioned FEM model is used to calculatethe critical moment of the I-beams Since for a steel I-beamunder pure bending the critical moment has exact solutionthat is (37) then the FEM model of the beam with thesections listed in Table 1 is verified by this exact solution

Furthermore the sixth-order approximate analytical solutionand numerical solution of the critical moments are alsocalculated All the results are listed in Table 2

It is observed that (1) for all caseswhere doubly symmetricI-beams are studied the deviations from the FEM resultsare within 091 (2) for all cases where singly symmetric I-beams are studied the deviations from the FEM results arehigher than those of the doubly symmetric beams but thelargest deviation is not more than 185 (3) if three decimalplaces are retained the results of the 6th-order approximateanalytical solution and the numerical solution with 30 termsare exactly the same This proves the appropriateness of theANSYS FEMmodel developed in this paper

In addition since no exact solution exits for the beamsunder linear distributed moment the ANSYS FEM model isused to compare with the published data which is obtainedfrom newly developed FEM model Table 3 listed the resultsof ABSYS and those of B3DW [15]

It can be observed that (1) all the results of B3DWare higher than those of the ANSYS model developed inthis paper and the largest deviation is less than 13 (2)all the results given by the numerical solutions (referencedldquotheoryrdquo) are the lowest and closer to the those of ANSYSwith the maximum difference within 031 This providesan additional evidence of the appropriateness of the ANSYSFEM model and the analytical and numerical solutionsdeveloped in the current work are verified preliminarily

It should be pointed out that the BEAM189 in ANSYSsoftware can only be used to predict the critical moment for


Table 1 Section dimensions and geometric properties for the selected beams

Section A Section B Section C119867 = 400mm ℎ = 372mm 119867 = 400mm ℎ = 372mm 119867 = 400mm ℎ = 372mmℎ119908 = 344mm 119887119891 = 400mm ℎ119908 = 344mm 1198871198911 = 200mm ℎ119908 = 344mm 1198871198911 = 400mm119905119891 = 28mm 119905119908 = 18mm 1198871198912 = 400mm 119905119891 = 28mm 1198871198912 = 200mm 119905119891 = 28mm119868119910 = 2987 times 10minus4m4 119905119908 = 18mm 119868119910 = 1680 times 10minus4m4 119905119908 = 18mm 119868119910 = 1680 times 10minus4m4119869119896 = 6523 times 10minus6m4 119869119896 = 5059 times 10minus6m4 119869119896 = 5059 times 10minus6m4119868120596 = 10333 times 10minus6m6 119868120596 = 2296 times 10minus6m6 119868120596 = 2296 times 10minus6m6bf

bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2

Table 2 Critical moments for the selected beams under pure bending

Section 119870 119871m 120578 119896 120573119909 119873 = 6119872crTheory

N = 30119872crTheory119872crSHELL Diff1 119872crBEAM Diff2

A1063 6 1 1 0 0687 0687 0680 091 0689 minus0350797 8 1 1 0 0802 0802 0801 022 0808 minus0660531 12 1 1 0 1066 1066 1068 minus023 1076 minus096

B0569 6 0125 1 minus0322 0391 0391 0383 185 0641 61350427 8 0125 1 minus0322 0541 0541 0535 115 0809 46510284 12 0125 1 minus0322 0871 0871 0867 052 1162 3043

C0569 6 8 1 0322 1035 1035 1048 minus128 0641 minus39110427 8 8 1 0322 1185 1185 1204 minus155 0809 minus33110284 12 8 1 0322 1515 1515 1539 minus153 1162 minus2502

Note Diff1 = (119872crTheory minus119872crSHELL)119872crSHELL times 100 Diff2 = (119872crTheory minus119872crBEAM)119872crBEAM times 100

Table 3 Critical moments for the beams of HEA-200 under linear distributed moment

Number k SHELL103Nsdotm B3Dw103 Nsdotm Diff1 Theory103Nsdotm Diff2 BEAM103Nsdotm Diff3L-1 1 82064 8269 076 81872 minus023 82687 minus099L-2 075 93574 9479 130 93358 minus023 94285 minus098L-3 05 108090 10894 079 107853 minus022 108916 minus098L-4 025 126419 12745 082 126175 minus019 127393 minus096L-5 0 149142 15045 088 148935 minus014 150318 minus092L-6 minus025 175909 17761 097 175823 minus005 177353 minus086L-7 minus05 204151 20632 106 204317 008 205944 minus079L-8 minus075 225915 22838 109 226436 023 228096 minus073L-9 minus1 219686 22197 104 220378 031 221905 minus069

Section and its properties

bf

bf

t ft f

tw

ℎwh H

119867 = 190mm ℎ = 180mmℎ119908 = 170mm 119887119891 = 200mm119905119891 = 10mm 119905119908 = 65mm119871 = 8m 119860 = 5105 cm2119868119910 = 355923 cm4 119868119911 = 133333 cm4119869119896 = 148895 cm4 119868120596 = 108 times 103 cm6119864 = 210GPaNote Diff1 = (B3Dw minus SHELL)SHELL times 100 Diff2 = (Theory minus SHELL)SHELL times 100 Diff3 = (Theory minus BEAM)BEAM times 100


Table 4 Critical moments for the selected beams under linear distributed moment


119873 = 30119872crTheory119872crFEM Diff1

A

1063 6 1 05 0 0906 0906 0897 0991063 6 1 01 0 1179 1179 1164 1301063 6 1 0 0 1265 1265 1247 1451063 6 1 minus01 0 1357 1357 1336 1631063 6 1 minus05 0 1766 1766 1720 2691063 6 1 minus1 0 1872 1872 1809 348

B

0569 6 0125 05 minus0322 0512 0512 0502 1920569 6 0125 01 minus0322 0649 0649 0635 2180569 6 0125 0 minus0322 0689 0689 0673 2290569 6 0125 minus01 minus0322 0730 0730 0713 2410569 6 0125 minus05 minus0322 0908 0908 0883 2880569 6 0125 minus1 minus0322 1075 1075 1088 119

C

0569 6 8 05 0322 1365 1365 1385 minus1500569 6 8 01 0322 1767 1767 1784 minus0930569 6 8 0 0322 1889 1889 1900 minus0560569 6 8 minus01 0322 2017 2017 2029 minus0600569 6 8 minus05 0322 2263 2262 2228 1550569 6 8 minus1 0322 1126 1126 1090 329

Note Diff1 = (119872crTheory minus119872crFEM)119872crFEM times 100

a doubly symmetric beam but will yield erroneous results fora singly symmetric beam

In summary all the results show that the ldquoCRIGrdquo com-mand in ANSYS can be used to simulate the Vlasovrsquos rigidsection hypothesis and its simulation effect is more effectiveand hence better than the method of adding stiffeners

63 Verification of the Analytical and Numerical SolutionsThis paragraph will further to verify the accuracy of theanalytical and numerical solutions developed in this paper

For the steel I-beams with the section listed in Table 1under linear distributed moment their critical moments andbuckling modes are calculated by ANSY FEM model sixth-order approximation and numerical solution respectivelyThe results of the calculated critical moments are listed inTable 4 The predicted buckling modes for the beams withdoubly and singly symmetric sections are pictured in Figures8 and 9 respectively

It can be found that (1) if three decimal places are retainedthe results of the 6th-order approximate analytical solutionand the numerical solution with 30 terms are exactly thesame as shown in Table 2 (2) when 119896 changes from 1 tominus1 the deviations from the FEM results gradually increaseand the largest deviation occurs in the case of 119896 = minus1 andis less than 354 (3) there is close agreement between thebuckling modes of the present theory and those of ANSYSThis validates both the analytical and numerical solutionsdeveloped in the current work and it can be concludedthat the dimensionless analytical solution presented providesaccurate solutions for the LTB problem of the simply sup-ported steel I-beams

It is noted that for all I-beams studied in this paperthe results given by the 6th-order approximation are ofhigh accuracy Our other research results also support thisconclusion [34ndash36] Therefore if you are able to accept thecalculation results with three-decimal precision the sixth-order approximation proposed in this paper is a simple andquick calculationmethodHowever in order to obtain a high-precision design formula we propose to use the analyticalsolution with more terms that is use the numerical solutionproposed in previous sections

Finally it is noted that for the case of Section A and Sec-tion B SHELL solutions of ANSYS will provide lower criticalmoment predictions than the exact solution as expected butfor Section C the result is just the opposite Therefore forthe case of the beam under linear distributed moment FEMcannot be applied to formulate the design formula because itlacks precision consistency in the analysis

7 Proposed Dimensionless Design Formula

Nowadays the concept of equivalent uniform moment coef-ficient (EUMF) is widely accepted in design codes in manycountries However it is found that even for the case of thedoubly symmetric I-beams under linear distributedmomentthe design formulas given by the codesspecifications ofdifferent countries are quite different The correspondingcomparison between the design formulas [61ndash65] with theexact solutions developed in current work is depicted inFigure 1 Due to space constraints this paper does not intendto make too many evaluations of these results


00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60

FEM Analytical solution (N = 6)Numerical solution (N = 30)

Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1

Figure 8 Buckling mode graph of the beams with doubly symmetric section (Section A)

This paper attempts to use a new mathematical modelas the benchmark to solve this long-standing engineeringproblem with the above-mentioned dimensionless analyticalsolutions

71 Trilinear Mathematical Model Since the concept ofequivalent uniform moment factor (EUMF) proposed bySalvadori [66] is widely recognized in the engineering com-munity similar to others at first we have tried to use EUMFto describe the relationship between the criticalmoments andthe gradient moment factor k but we ultimately failed

In order to analyze the cause of failure through theparametric analysis we first study its change law of the data

Figures 10 11 and 12 show the trend of the curves of thecritical moments versus the gradient moment factor 119896 withvaried 120578 and 119870 respectively It can be seen that the trend ofall the curves are very different that no rules can be followedThis means that it is not possible to describe such a complexchange with a single EUMF In other words a single EUMFdoes not exist in the case of I-beams with unequal flangessubjected to unequal end moments hence we must breakthrough the shackles of traditional concepts so that we canget a breakthrough

With the aid of the idea of using a piecewise linear approx-imation to simulate an actual curve this paper proposes anew trilinear mathematical model used as the benchmark of


00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1

Figure 9 Buckling mode graph of the beams with singly symmetric section (Section C)

formulating the design formula This new idea is representedby the graphics shown in Figure 13 There are four controlpoints in this new model that is minus1119872cr

minus05119872cr0119872cr

+1119872crare the four control moments Since +1119872cr can be expressedby the existing exact solution that is (37) only the otherthree moments are undetermined parameters which shouldbe obtained by the nonlinear regression techniques

72 Parameter Regression Techniques and Comparison As weall know there are three key issues in regression technologynamely the choice of target formula how to generate dataand how to regress the unknown coefficient in the targetformula

Firstly according to the results of the previous parametricanalysis after a number of analyses observations and trieswe finally selected the following target formula

cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)

where 1198621 1198623 1198624 1198625 and 1198626 are undetermined regressioncoefficients and the later three are newly defined in this paper


0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K

Figure 10 Distribution pattern of critical moments with varied 119870and 119896

minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t

060300 09 12minus06minus09 minus03minus12

k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10

Figure 11 Critical moments versus gradient moment factor withvaried 120578 (119870 = 2)

In addition the study also found that these undeterminedregression coefficients for the cross sections with lager topflange are completely different from those of other crosssections in the range (minus10 le 119896 lt minus05) so we cannotexpect to describe all the cases with a set of unified regressioncoefficients by using the target formula (43) Therefore twosets of regression coefficients have to be used for 1198621 1198623 11986241198625 and1198626 in the proposed design formula for the case where119896 = minus10 and 119896 = minus05

Next theMATLAB program formulated in the paper canbe employed to generate the above-mentioned dimensionlessdata In the following data analysis the end moment ratio 119896

00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t

0300 06 09 12minus06minus09 minus03minus12k

K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ

Figure 13 Proposed trilinear model as the benchmark of designformula

is fixed for each control moment while 120573119909 120578 120573119909119870 are varied119870 is changed from 02 to 50 with the varied step size of 002sim005 120578 is changed from 01 to 100 with the varied step sizeof 005sim10 120573119909 is changed from minus04 to +04 with step size of001

Thirdly due to the highly nonlinear nature of the targetformula we chose the 1stOpt [67] software to do this regres-sion [30ndash35] 1stOpt is a software package for the analysis ofmathematical optimization problem The nonlinear curve fitof 1stOpt is powerful compared to any other similar softwarepackage available today The greatest feature is that it nolonger needs the end-user to provide or guess the initialstart-values for each parameter but randomly generated bymachine the probability for finding the correct solution ishigher than the others


0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ

Figure 14 Relationship of the target value and calculated value

Table 5

Parameter Best estimate1198881 22201658261198883 minus04586948911198884 72592702971198885 14823667651198886 2735838541

Taking a cross section having a larger top flange anda negative moment (119896 = minus1) as an example based onthe 475 sets of dimensionless data (120573119909 120578 119870 cr) obtainedfrom the dimensionless analytical solution using Levenberg-Marquardt algorithm in the 1stOpt software then after 39iterations the resulting regression coefficients are obtainedas shown in Table 5 and the corresponding correlation coef-ficient is 099545

The resulting relationship between the target value andthe calculated value is depicted in Figure 14 In this figurethe red line represents the calculated curve and the blue linerepresents the target curve It is found that quite satisfactoryregression results have been obtained

Using a similar approach we can get all the undeterminedcoefficients and the relevant results are listed in Table 6

It must be pointed out that the whole regression processtakes only a few minutes of which the calculation process ofnearly 500 sets of data only takes a few seconds If you intendto use the FEM to obtain such a large number of data sets itmay take at least a month This fully demonstrates that thedimensionless solution has an unparalleled advantage overthe finite element method in terms of efficiency and qualityin obtaining a large number of regression data

Finally the proposed dimensionless design formula isused to calculate the critical moments of the I-beams with the

Formula for Section AFormula for Section BFormula for Section C

eory for Section Aeory for Section Beory for Section C

00

05

10

15

20

25

minus05 05 10minus01 00k

M＝Ｌ

Figure 15 Comparison between theory and the proposed formula

sections listed in Table 1 and the results are comparedwith thetheoretical solution All results are plotted in Figure 15 andlisted in Table 7

The results show that the proposed formula is of highaccuracy and is applicable to both the doubly symmetricsection and the singly symmetric section under positive andnegative gradient moment

It must be noted that the trilinear mathematical modelcan also be converted to a cubic curve model or by addingmore points to obtain a quadratic or higher-order curvemodel Due to the limited space these will be discussedelsewhere


Table 6 Formula and parameters for I-beams under linear distributed moment119896 = minus1 119896 = minus05 119896 = minus0 119896 = +11198621 0234 (2220) 1387 (1095) 5983 11198623 4674 (minus0459) 0822 (0922) 0287 11198624 72762 (7259) 0615 (10860) 11468 11198625 0861 (1482) 0587 (1629) 11713 11198626 0566 (2736) 0273 (1154) 10446 1

minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ

M＝Ｌ0 = 1198721(12058721198641198681199101198712)ℎ 120573119909 = 120573119909ℎ 120578 = 11986811198682 119870 = radic12058721198641198681205961198661198691198961198712cr = 1198621[(1198623120573119909) + radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

Note (1) Parameters in parentheses for I-beams of singly symmetric sections with larger top flange only (2) applicable scope of the formula119870 = 02sim50 120578 =01sim100 120573119909 = minus04sim+04

Table 7 Comparison of the results of design formula with those of theory for I-beams under linear distributed moment


119873 = 30119872crTheory119872crRegression Diff

A

1063 6 1 1 0 0687 0687 0687 0001063 6 1 0 0 1265 1265 1255 minus0721063 6 1 minus05 0 1766 1766 1737 minus1681063 6 1 minus1 0 1872 1872 1919 254

B

0569 6 0125 1 minus0322 0391 0390 0391 0000569 6 0125 0 minus0322 0689 0689 0664 0510569 6 0125 minus05 minus0322 0908 0908 0915 5430569 6 0125 minus1 minus0322 1075 1075 1194 1108

C

0569 6 8 1 0322 1035 1035 1035 0000569 6 8 0 0322 1961 1961 1958 minus0190569 6 8 minus05 0322 2263 2262 2286 0880569 6 8 minus1 0322 1126 1126 1150 700

8 Conclusions

Because there is no reliable analytical solution available thisstudy is initially intended to develop an exact analyticalsolution to study the applicability of the design formulaspublished in literature However we found that results givenby the existing design formulas vary widely especially in therange minus10 le 119896 lt 0 so according to the idea that scientificresearch should serve for engineering a new dimensionlessdesign formula is proposed for practical engineering and thefollowing conclusions can be drawn

(1)There is a challenge to part of the theory of monosym-metric beams called the Wagner hypothesis was presentedby Ojalvo (1981) This challenge can be fully resolved by thePlate-BeamTheory put forward by the author in which onlythe commonly used plate and beam theory was used and theVlasovrsquos warping function was discarded

(2) For the case of Section A and Section B SHELLsolutions of ANSYS will provide lower critical momentpredictions than the exact solution as expected but forSection C the result is just the opposite Therefore for thecase of the beam under linear distributed moment FEMcannot be applied to formulate the design formula becauseit lacks precision consistency in the analysis

(3) By introducing a new dimensionless coefficient oflateral deflection new dimensionless critical moment anddimensionlessWagnerrsquos coefficient are derived naturally fromthe total potential energy These new dimensionless parame-ters are independent of each other and thus can be used asindependent variables in the process of regression formula

(4) The results show that in the case of the simple-supported I-beams under linearly distributed moments thedimensionless analytical solution presented in current workis exact in nature and has an unparalleled advantage over


the finite element method in terms of efficiency and qualityin obtaining a large number of regression data Thereforeit provides a more convenient and effective method for theformulation of a more accurate design formula than finiteelement method

(5) It is found that for all I-beams studied in thispaper the results given by the 6th-order approximation areof high accuracy Our other research results also supportthis conclusion [30ndash35] Therefore if you can accept thecalculated results of three-decimal precision then the sixth-order approximation presented in this paper is a simple andquick calculation method

(6) The dimensionless parameters defined by Kitiporn-chai lack clear physical and geometric meanings Moreimportantly none of these parameters is independent thusit is difficult to use them to regress a rational designformula

(7) The dimensionless analytical buckling equation isexpressed as the generalized eigenvalue problem as thatused in the traditional finite element method This not onlyfacilitates the solution in MATLAB programming but alsomakes the analytical solution have a clear physical meaningmore easy to understand and apply

(8) To satisfy Vlasovrsquos rigid section hypothesis theldquoCRIGrdquo command is used in the FEM model of ANSYS Theresults show that its simulation effect is more effective andhence better than the method of adding stiffeners

(9) BEAM189 in ANSYS software can only be used topredict the critical moment for a doubly symmetric beambut will yield erroneous results for a singly symmetricbeam

(10) It is found that the concept of equivalent uniformmoment factor (EUMF) is not applicable for the simple-supported I-beamswith unequal flanges subjected to unequalend moments that is a single EUMF does not exist in thiscase

(11) This paper proposes a new trilinear mathematicalmodel as the benchmark of formulating the design formulaThe results show that the proposed design formula has ahigh accuracy and is applicable to both the doubly symmetricsection and the singly symmetric section under positive andnegative gradient moment This model provides a new ideafor the formulation of more reasonable design formulas so itis suggested that the designers of the design specifications canformulate the corresponding design formulas with referenceto this model

Conflicts of Interest

The authors declare that they have no conflicts of interest

Acknowledgments

This work is supported by the National Natural ScienceFoundation of China (51578120 and 51178087) and the Scien-tific Research Foundation of Nanjing Institute of Technology(YKJ201617)

References

[1] V Z Vlasov Thin-Walled Elastic Beams Israel Program forScientific Translations Jerusalem Israel 1961

[2] A Chajes Principles of Structural Stability Prentice-Hall IncNew Jersey USA 1974

[3] W F Chen and T Atsuta Theory of Beam-Columns SpaceBehavior and Design vol 2 McGraw-Hill New York NY USA1977

[4] N S Trahair Flexural-Torsional Buckling of Structures Chap-man amp Hall London UK 1st edition 1993

[5] S P Timoshenko and J M Gere Theory of Elastic StabilityMcGraw-Hill New York NY USA 1961

[6] F Bleich Buckling Strength of Metal Structures McGraw-HillNew York NY USA 1952

[7] European Committee for Standardization EN 1993-1-1Eurocode 3 Design of steel structures Part 1-1 General rules andrules for buildings Brussels 2005

[8] K J Bathe Finite Element Procedures Prentice-Hall Inc NewJersey USA 1996

[9] B Suryoatmono and D Ho ldquoThe moment-gradient factor inlateral-torsional buckling on wide flange steel sectionsrdquo Journalof Constructional Steel Research vol 58 no 9 pp 1247ndash12642002

[10] M A Serna A Lopez I Puente and D J Yong ldquoEquivalentuniform moment factors for lateral-torsional buckling of steelmembersrdquo Journal of Constructional Steel Research vol 62 no6 pp 566ndash580 2006

[11] N-H Lim N-H Park Y-J Kang and I-H Sung ldquoElasticbuckling of I-beams under linear moment gradientrdquo Interna-tional Journal of Solids and Structures vol 40 no 21 pp 5635ndash5647 2003

[12] N-H Park Y-J Kang Y-M Jo and N-H Lim ldquoModificationof C-equation in the SSRCGuide for buckling ofmonosymmet-ric I-beams under transverse loadsrdquo Engineering Structures vol29 no 12 pp 3293ndash3300 2007

[13] R Greiner and J Lindner ldquoProposal for buckling resistanceof members Flexural and lateral torsional bucklingrdquo ECCS-Validation group report 6 1999

[14] R Greiner R Ofner and G Salzgeber ldquoLateral torsionalbuckling of beamndashcolumns Theoretical backgroundrdquo ECCS-Validation group report 5 1999

[15] FMohri NDamil andM Potier-Ferry ldquoLinear and non-linearstability analyses of thin-walled beams with monosymmetric Isectionsrdquo Thin-Walled Structures vol 48 no 4-5 pp 299ndash3152010

[16] N-I Kim D K Shin and M-Y Kim ldquoExact lateral bucklinganalysis for thin-walled composite beam under end momentrdquoEngineering Structures vol 29 no 8 pp 1739ndash1751 2007

[17] F Mohri A Brouki and J C Roth ldquoTheoretical and numericalstability analyses of unrestrainedmono-symmetric thin-walledbeamsrdquo Journal of Constructional Steel Research vol 59 no 1pp 63ndash90 2003

[18] ABAQUS Standard Userrsquos Manual vol 1 2 and 3 HibbittKarlsson and Sorensen Inc USA 68-1 edition 2008

[19] N Dolamune Kankanamge and M Mahendran ldquoBehaviourand design of cold-formed steel beams subject to lateraltor-sional bucklingrdquoThin-Walled Structures vol 51 pp 25ndash38 2012

[20] G S Tong Out-Plane Stability of Steel Structures PublishingHouse of Chinese Construction Industry Beijing China 2006(in Chinese)


[21] ANSYS Multiphysics Ansys Inc Canonsburg PA USA 12thedition 2006

[22] A M I Sweedan ldquoElastic lateral stability of I-shaped cellularsteel beamsrdquo Journal of Constructional Steel Research vol 67 no2 pp 151ndash163 2011

[23] CosmosM27 Structural Research and Analysis Corp (SARC)2002

[24] J M Anderson and N S Trahair ldquoStability of monosymmetricbeams and cantileversrdquo Journal of Structural Division ASCEvol 98 no ST1 pp 269ndash286 1972

[25] S Kitipornchai C Ming Wang and N S Trahair ldquoBuckling ofmonosymmetric I-beams under moment gradientrdquo Journal ofStructural Engineering (United States) vol 112 no 4 pp 781ndash799 1986

[26] S Kitipornchai and C MWang ldquoLateral buckling of tee beamsunder moment gradientrdquo Computers amp Structures vol 23 no 1pp 69ndash76 1986

[27] C M Wang and S Kitipornchai ldquoBuckling capacities ofmonosymmetric I-beamsrdquo Journal of Structural Engineering(United States) vol 112 no 11 pp 2373ndash2391 1986

[28] W Nowacki Research on Elastic Thin Plates and Shells SciencePress Beijing China 1956 in English Russian and Chinese

[29] W F Zhang EneRgy Variational Method for Flexural-TorsionalBuckling Analysis of Steel Beams Research Report DaqingPetroleum Institute Daqing China 2008 (in Chinese)

[30] W F Zhang Plate-Beam Theory And Analytical Solutions forFlexural-Torsional Buckling of Steel Beams Research ReportNortheast Petroleum University Daqing China 2015

[31] K Y Liu Elastic lateral buckling and applications to double spancontinuous beams [MS thesis] Daqing Petroleum InstituteDaqing China 2009

[32] W Zhang H Sui Z Wang and J Ji ldquoBuckling analysis of two-span continuous beams with lateral elastic brace under uniformloadrdquo Advanced Materials Research vol 163-167 pp 641ndash6452011

[33] W F Zhang Y Liu and J Ji ldquoLateral-torsional Buckling Analy-sis of tapered cantilever I-beamsrdquo LowTemperature ArchitectureTechnology vol 10 no 127 pp 19ndash21 2012

[34] G L Hou Stability Research of Steel Beams with Lateral ElasticBrace [MS thesis] Northeast Petroleum University DaqingChina 2012

[35] W-F Zhang Y-C Liu G-L Hou et al ldquoLateral-torsionalbuckling analysis of cantilever beamwith tip lateral elastic braceunder uniform and concentrated loadrdquo International Journal ofSteel Structures vol 16 no 4 pp 1161ndash1173 2016

[36] A Gjelsvik The Theory of Thin-Walled Bars Wiley New YorkNY USA 1981

[37] L W Lu S Z Shen et al Stability of Steel Structural MembersPublishing House of Chinese Construction Industry BeijingChina 1983 (in Chinese)

[38] MOjalvoThin-Walled Bars with Open ProfilesTheOlive PressColumbus Ohio 1990

[39] M Ojalvo ldquoWagner hypothesis in beam and column theoryrdquoJournal of the EngineeringMechanics Division vol 107 no 4 pp669ndash677 1981

[40] M Ojalvo ldquoBuckling of monosymmetric I-beams undermoment gradientrdquo Journal of Structural Engineering (UnitedStates) vol 113 no 6 pp 1387ndash1391 1987

[41] M Ojalvo ldquoBuckling of thin-walled open-profile barsrdquo Journalof Applied Mechanics vol 56 no 3 pp 633ndash638 1989

[42] N S Trahair ldquoDiscussion on Wagner hypothesis in beam andcolumn theoryrdquo Journal of the Engineering Mechanics Divisionvol 108 no 3 pp 575ndash578 1982

[43] S Kitipornchai C M Wang and N S Trahair ldquoBuckling ofmonosymmetric I-beams under moment gradientrdquo Journal ofStructural Engineering (United States) vol 113 no 6 pp 1391ndash1395 1987

[44] Y Goto andW F Chen ldquoOn the validity ofWagner hypothesisrdquoInternational Journal of Solids and Structures vol 25 no 6 pp621ndash634 1989

[45] Y J Knag S C Lee and C H Yoo ldquoOn the dispute concerningthe validity of theWagner hypothesisrdquo Compurers amp Slructuresvol 43 no 5 pp 853ndash861 1992

[46] W A M Alwis and C M Wang ldquoShould load remainconstant when a thin-walled open-profile column bucklesrdquoInternational Journal of Solids and Structures vol 31 no 21 pp2945ndash2950 1994

[47] W A M Alwis and C M Wang ldquoWagner term in flexural-torsional buckling of thin-walled open-profile columnsrdquo Engi-neering Structures vol 18 no 2 pp 125ndash132 1996

[48] R D Ziemian Guide to Stability Design Criteria for MetalStructures John Wiley amp Sons New Jersey USA 6th edition2010

[49] G J Hancock The behaviour of structures composed of thin-walled members [PhD thesis] University of Sydney SydneyAustralia 1975

[50] Y L Pi N S Trahair and S Rajasekaran ldquoEnergy equation forbeam lateral bucklingrdquo Journal of Structural Engineering (UnitedStates) vol 118 no 6 pp 1462ndash1479 1992

[51] H C Hu Variational Principle of Elasticity and Its ApplicationScience Press 1982 in Chinese

[52] J N Reddy Energy Principles and Variational Methods inApplied Mechanics Wiley 1984

[53] K Washizu Variational Methods in Elasticity and PlasticityPergamon Press Oxford UK 1968

[54] H Reismann and P S Pawlik Elasticity Theory and Applica-tions Pergamon 1974

[55] W F Zhang Plate-beam Theory for Thin-Walled StructuresNortheast Petroleum University Daqing China 2015 (in Chi-nese)

[56] W F Zhang ldquoNew Engineering Theory for Torsional Bucklingof Steel-concrete Composite I-Columnsrdquo in Proceedings of 11thInternational Conference on Advances in Steel and ConcreteComposite Structures vol 12 pp 225ndash232 Beijing China 2015

[57] W F Zhang ldquoNew theory of elastic torsional buckling for axialloaded box-shaped steel columnsrdquo in Proceedings of the 15thNational Symposium onModern Structure Engineering pp 793ndash804 2015

[58] W F Zhang ldquoVariational Model and its Analytical Solutionsfor the Elastic Flexural-torsional Buckling of I-Beams withConcrete-filled Steel Tubular Flangesrdquo in Proceedings of the 8thInternational Symposium on Steel Structures pp 1100ndash1108 JejuKorea 2015

[59] F McCannM AhmerWadee and L Gardner ldquoLateral stabilityof imperfect discretely braced steel beamsrdquo Journal of Engineer-ing Mechanics vol 139 no 10 pp 1341ndash1349 2013

[60] D Xue and Y Chen Solving Applied Mathematical Problemswith MATLAB CRC Press 2009

[61] GB50017-2003 Steel Structure Design Code Plans Press BeijingChina 2003


[62] British Standards Institution (BSI) BS5950 Part-1 BritishStandard for Steel Structures 1990

[63] Standards Australia (SA) AS 4100 Steel Structures SydneyAustralia 1998

[64] American Institute of Steel Construction (AISC) ANSIAISC360-05 Specification for Structural Buildings Chicago USA2005

[65] British Standards Institution (BSI) BS5950-1 Structural Use ofSteelwork in Buildings ndashPart1 Code of Practice for Design Rolledand Welded Sections England 2000

[66] M G Salvadori ldquoLateral Buckling of Eccentrically Loaded I-Columnrdquo Transactions of the American Society of Civil Engi-neers vol 121 pp 1163ndash1178 1956

[67] 1stOpt UserManual 7D SoftHigh Technology Inc China 2006

Submit your manuscripts athttpswwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Journal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


12

10

20

18

16

14

08

06

04

02

00

eory in this paperGB50017-2003CAN-CSA-S161-1994BS5950-1-1990

AISCLRFD-2005BS5950-1-2000AS4100-1998

h H

bf1

bf2t f

t f

tw ℎw

Ly

z

M2 M1

minus05 00 05 10minus10

Figure 1 Discrepancies of EUMF for a simply supported beam under linear distributed moment

finite element method (FEM) [8] More recently Suryoat-mono and Ho [9] and Serna et al [10] have reported theresults of elastic LTB load analysis using finite differencetechnique to solve the governing differential equation Limet al [11] have conducted an extensive investigation intothe elastic LTB of I-beams using both the Bubnov-Galerkinapproach [1] and the finite element method [8] In theirpapers Vlasovrsquos differential equation governing [1] for lateral-torsional buckling of beams was used Park et al [12] usethe finite element program developed by Lim et al [11] toperformparametric studies to investigate the lateral-torsionalbuckling behavior of singly symmetric I-beams in which I-beams are modeled by the beam element with two nodes perelement and seven nodal degrees of freedom Greiner andLindner [13] Greiner et al [14] and Mohri et al [15] also usefinite element method to present a quite complete numericalstudy on lateral-torsional buckling of beams Kim et al [16]adopt the variation of the total potential energy to derive thestability equations for the lateral buckling analysis of an arbi-trarily laminated thin-walled composite beam and the exactanalytical stiffness matrix is presented based on power seriesexpansions

It is noteworthy that with the popularization andwidespread use of finite element software since the 2000ssome researchers have tried to use the general-purpose finiteelement software to simulate the buckling behavior or toassess the related design formula of the steel I-beams Forexample Mohri et al [17] adopt the beam elements (B31OS)and shell elements (S8R5) of ABAQUS [18] to model the lat-eral buckling phenomena The authors have found that shellelements could consider local buckling section distortionand the local effects of concentrated loads and boundaryconditions which were all ignored in the beam elementtheory Dolamune Kankanamge and Mahendran [19] use

ABAQUS [18] to investigate the LTB behavior of simplysupported cold-formed steel lipped channel beams subjectedto uniform bending The authors have found that Europeandesign rules are conservative while AustralianNorth Amer-ican design rules are unconservative Tong [20] uses ANSYS[21] to regress the design formula for the singly symmetricI-beams under linear distributed moment Sweedan [22]adopts ANSYS to numerically investigate the lateral stabilityof cellular steel beams subjected to equal end moments mid-span concentrated loads and uniformly distributed loadsSerna et al [10] use CosmosM27 [23] to simulate the lateral-torsional buckling of the I-beams

In theory FEM can be used not only to do research worksbut also to formulate design formulas However when youreally try to formulate the design formulas through a lot ofFEM analysis you will have to deal with some real troubleFirstly the finite element analysis must be a dimensionalanalysis [8 18 21 23] that is the section size and span(or slenderness) of the steel beams must be specified inadvance Hence there is an enormous amount of workloadin the early stage such as the design of hundreds of modelbeams and the relevant data preparation in the finite elementmodeling secondly one has to spend a lot of time and effortin later stages to select from a large number of data outputsand determine which results are useful LTB results finallyand most importantly our experience shows that becausethe meshing of FEM model will greatly affect the accuracyof finite element analysis hence the corresponding FEManalysis results lack of precision consistency and regularityThis may be the reason why the design formulas of differentcountries will be quite different

On the contrary not only does the dimensionless analyticsolution have the advantages of small input data amountand fast calculation speed but also the formula formulated

















V (119909 119911) = V (119911) minus (119909 minus 1199090) 120579 (119911) (1)










Π = 12 int119871 [119864119868119910119906101584010158402 + 119864119868120596120579101584010158402 + (119866119869119896 + 2119872119909120573119909) 12057910158402+ 211987211990911990610158401015840120579 minus 1199021198861199021205792] 119889119911 (3)







(6)




y

x

yL

z

M2 M1

M2 M1

S(0 y0)

C(0 0)













119906 (119911) = ℎinfinsum119899=1

119860119898 sin [119898120587119911119871 ] (119898 = 1 2 3 infin) (8)

120579 (119911) = infinsum119899=1

119861119899 sin [119899120587119911119871 ] (119899 = 1 2 3 infin) (9)










T

M Mu

u

T



(10)





0 = 1198721(12058721198641198681199101198712) ℎ 120573119909 = 120573119909ℎ 120578 = 11986811198682 119870 = radic12058721198641198681205961198661198691198961198712

(11)





= infinsum119898=1

1198602119898119898412058744 + infinsum119899=1

1198612119899 (1 + 11989921198702) 1198992120587412057841198702 (1 + 120578)2 Π2 = 12 int119871 211987211990912057311990912057910158402119889119911


1198612119899 (1 + 119896)119898212058741205731199094+ 0 infinsum119904=1

infinsum119903=1119903 =119904



119860119898119861119898 (1 + 119896)119898212058744minus 0 infinsum119904=1

infinsum119903=1119903 =119904


(12)


Π = 3sum119894=1

Π119894 (13)




infinsum119898=1

119860119898119898412058742 minus 0 infinsum119898=1

119861119898 (1 + 119896)119898212058744



119861119903 4 (minus1 + 119896) 12058721199031198983(1199032 minus 1198982)2 = 0(15)





0119878119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (19)

1119877119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (20)


(21)




119860119899 (1 + 119896) 119899212058744minus 0 infinsum119899=1



119861119899 (1 + 119896) 119899212058741205731199092


+ infinsum119899=1

119861119899 (1 + 11989921198702) 1198992120587412057821198702 (1 + 120578)2+ 0 infinsum119899=1


119861119903 4 (minus1 + 119896) 1205872119903119899 (1199032 + 1198992) 120573119909(1199032 minus 1198992)2= 0

(22)


0119879119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (24)

1119879119904119903 = 1119878119903119904 119904 = 1 2 infin 119903 = 1 2 infin (25)



(27)



[0119877 01198780119879 0119876]119860119861 = 0 [1119877 11198781119879 1119876]119860119861 (28)







cr = 119872cr119871radic119864119868119910119866119869119896 (29)



(30)



cr = 119872cr(12058721198641198681199101198712) ℎ = 119872crℎ119875119864119910 (31)










[[[[[12058742 00 (1 + 1198702) 120587412057821198702 (1 + 120578)2

]]]]]11986011198611

= 0 [[[[0 (1 + 119896) 12058744(1 + 119896) 12058744 minus(1 + 119896) 12058741205731199092

]]]]11986011198611

(33)



(34)

where

1198621 = 21 + 119896 1198622 = 01198623 = 1(35)

or

cr = 21 + 119896 [[120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2)]] (36)



cr = 120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2) (37)



[0119877 01198780119879 0119876]

6times6

1198601198616times1

= 0 [1119877 11198781119879 1119876]6times6

1198601198616times1

(38)

where

0119878 = 0119879 = 1119877 = 0

0119877 =((((((((((((((

12058742 0 0 0 0 00 81205874 0 0 0 00 0 8112058742 0 0 00 0 0 1281205874 0 00 0 0 0 62512058742 00 0 0 0 0 6481205874

))))))))))))))


0119876 =(((((((((((((((((((((

(1 + 1198702) 120587412057821198702 (1 + 120578)2 0 0 0 0 00 (4 + 161198702) 120587412057821198702 (1 + 120578)2 0 0 0 00 0 (9 + 811198702) 120587412057821198702 (1 + 120578)2 0 0 00 0 0 (16 + 2561198702) 120587412057821198702 (1 + 120578)2 0 00 0 0 0 (25 + 6251198702) 120587412057821198702 (1 + 120578)2 00 0 0 0 0 (36 + 12961198702) 120587412057821198702 (1 + 120578)2

)))))))))))))))))))))

1119879119904119903 = 1119878119903119904 =(((((((((((((((((


)))))))))))))))))

1119876

=(((((((((((((((((



)))))))))))))))))

(39)







[1198700]2119873times2119873 = [ 01198770119879 01198780119876 ]


295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1


[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)







































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of




















V (119909 119911) = V (119911) minus (119909 minus 1199090) 120579 (119911) (1)










Π = 12 int119871 [119864119868119910119906101584010158402 + 119864119868120596120579101584010158402 + (119866119869119896 + 2119872119909120573119909) 12057910158402+ 211987211990911990610158401015840120579 minus 1199021198861199021205792] 119889119911 (3)







(6)




y

x

yL

z

M2 M1

M2 M1

S(0 y0)

C(0 0)













119906 (119911) = ℎinfinsum119899=1

119860119898 sin [119898120587119911119871 ] (119898 = 1 2 3 infin) (8)

120579 (119911) = infinsum119899=1

119861119899 sin [119899120587119911119871 ] (119899 = 1 2 3 infin) (9)










T

M Mu

u

T



(10)





0 = 1198721(12058721198641198681199101198712) ℎ 120573119909 = 120573119909ℎ 120578 = 11986811198682 119870 = radic12058721198641198681205961198661198691198961198712

(11)





= infinsum119898=1

1198602119898119898412058744 + infinsum119899=1

1198612119899 (1 + 11989921198702) 1198992120587412057841198702 (1 + 120578)2 Π2 = 12 int119871 211987211990912057311990912057910158402119889119911


1198612119899 (1 + 119896)119898212058741205731199094+ 0 infinsum119904=1

infinsum119903=1119903 =119904



119860119898119861119898 (1 + 119896)119898212058744minus 0 infinsum119904=1

infinsum119903=1119903 =119904


(12)


Π = 3sum119894=1

Π119894 (13)




infinsum119898=1

119860119898119898412058742 minus 0 infinsum119898=1

119861119898 (1 + 119896)119898212058744



119861119903 4 (minus1 + 119896) 12058721199031198983(1199032 minus 1198982)2 = 0(15)





0119878119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (19)

1119877119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (20)


(21)




119860119899 (1 + 119896) 119899212058744minus 0 infinsum119899=1



119861119899 (1 + 119896) 119899212058741205731199092


+ infinsum119899=1

119861119899 (1 + 11989921198702) 1198992120587412057821198702 (1 + 120578)2+ 0 infinsum119899=1


119861119903 4 (minus1 + 119896) 1205872119903119899 (1199032 + 1198992) 120573119909(1199032 minus 1198992)2= 0

(22)


0119879119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (24)

1119879119904119903 = 1119878119903119904 119904 = 1 2 infin 119903 = 1 2 infin (25)



(27)



[0119877 01198780119879 0119876]119860119861 = 0 [1119877 11198781119879 1119876]119860119861 (28)







cr = 119872cr119871radic119864119868119910119866119869119896 (29)



(30)



cr = 119872cr(12058721198641198681199101198712) ℎ = 119872crℎ119875119864119910 (31)










[[[[[12058742 00 (1 + 1198702) 120587412057821198702 (1 + 120578)2

]]]]]11986011198611

= 0 [[[[0 (1 + 119896) 12058744(1 + 119896) 12058744 minus(1 + 119896) 12058741205731199092

]]]]11986011198611

(33)



(34)

where

1198621 = 21 + 119896 1198622 = 01198623 = 1(35)

or

cr = 21 + 119896 [[120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2)]] (36)



cr = 120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2) (37)



[0119877 01198780119879 0119876]

6times6

1198601198616times1

= 0 [1119877 11198781119879 1119876]6times6

1198601198616times1

(38)

where

0119878 = 0119879 = 1119877 = 0

0119877 =((((((((((((((

12058742 0 0 0 0 00 81205874 0 0 0 00 0 8112058742 0 0 00 0 0 1281205874 0 00 0 0 0 62512058742 00 0 0 0 0 6481205874

))))))))))))))


0119876 =(((((((((((((((((((((

(1 + 1198702) 120587412057821198702 (1 + 120578)2 0 0 0 0 00 (4 + 161198702) 120587412057821198702 (1 + 120578)2 0 0 0 00 0 (9 + 811198702) 120587412057821198702 (1 + 120578)2 0 0 00 0 0 (16 + 2561198702) 120587412057821198702 (1 + 120578)2 0 00 0 0 0 (25 + 6251198702) 120587412057821198702 (1 + 120578)2 00 0 0 0 0 (36 + 12961198702) 120587412057821198702 (1 + 120578)2

)))))))))))))))))))))

1119879119904119903 = 1119878119903119904 =(((((((((((((((((


)))))))))))))))))

1119876

=(((((((((((((((((



)))))))))))))))))

(39)







[1198700]2119873times2119873 = [ 01198770119879 01198780119876 ]


295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1


[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)







































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of









Π = 12 int119871 [119864119868119910119906101584010158402 + 119864119868120596120579101584010158402 + (119866119869119896 + 2119872119909120573119909) 12057910158402+ 211987211990911990610158401015840120579 minus 1199021198861199021205792] 119889119911 (3)







(6)




y

x

yL

z

M2 M1

M2 M1

S(0 y0)

C(0 0)













119906 (119911) = ℎinfinsum119899=1

119860119898 sin [119898120587119911119871 ] (119898 = 1 2 3 infin) (8)

120579 (119911) = infinsum119899=1

119861119899 sin [119899120587119911119871 ] (119899 = 1 2 3 infin) (9)










T

M Mu

u

T



(10)





0 = 1198721(12058721198641198681199101198712) ℎ 120573119909 = 120573119909ℎ 120578 = 11986811198682 119870 = radic12058721198641198681205961198661198691198961198712

(11)





= infinsum119898=1

1198602119898119898412058744 + infinsum119899=1

1198612119899 (1 + 11989921198702) 1198992120587412057841198702 (1 + 120578)2 Π2 = 12 int119871 211987211990912057311990912057910158402119889119911


1198612119899 (1 + 119896)119898212058741205731199094+ 0 infinsum119904=1

infinsum119903=1119903 =119904



119860119898119861119898 (1 + 119896)119898212058744minus 0 infinsum119904=1

infinsum119903=1119903 =119904


(12)


Π = 3sum119894=1

Π119894 (13)




infinsum119898=1

119860119898119898412058742 minus 0 infinsum119898=1

119861119898 (1 + 119896)119898212058744



119861119903 4 (minus1 + 119896) 12058721199031198983(1199032 minus 1198982)2 = 0(15)





0119878119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (19)

1119877119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (20)


(21)




119860119899 (1 + 119896) 119899212058744minus 0 infinsum119899=1



119861119899 (1 + 119896) 119899212058741205731199092


+ infinsum119899=1

119861119899 (1 + 11989921198702) 1198992120587412057821198702 (1 + 120578)2+ 0 infinsum119899=1


119861119903 4 (minus1 + 119896) 1205872119903119899 (1199032 + 1198992) 120573119909(1199032 minus 1198992)2= 0

(22)


0119879119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (24)

1119879119904119903 = 1119878119903119904 119904 = 1 2 infin 119903 = 1 2 infin (25)



(27)



[0119877 01198780119879 0119876]119860119861 = 0 [1119877 11198781119879 1119876]119860119861 (28)







cr = 119872cr119871radic119864119868119910119866119869119896 (29)



(30)



cr = 119872cr(12058721198641198681199101198712) ℎ = 119872crℎ119875119864119910 (31)










[[[[[12058742 00 (1 + 1198702) 120587412057821198702 (1 + 120578)2

]]]]]11986011198611

= 0 [[[[0 (1 + 119896) 12058744(1 + 119896) 12058744 minus(1 + 119896) 12058741205731199092

]]]]11986011198611

(33)



(34)

where

1198621 = 21 + 119896 1198622 = 01198623 = 1(35)

or

cr = 21 + 119896 [[120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2)]] (36)



cr = 120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2) (37)



[0119877 01198780119879 0119876]

6times6

1198601198616times1

= 0 [1119877 11198781119879 1119876]6times6

1198601198616times1

(38)

where

0119878 = 0119879 = 1119877 = 0

0119877 =((((((((((((((

12058742 0 0 0 0 00 81205874 0 0 0 00 0 8112058742 0 0 00 0 0 1281205874 0 00 0 0 0 62512058742 00 0 0 0 0 6481205874

))))))))))))))


0119876 =(((((((((((((((((((((

(1 + 1198702) 120587412057821198702 (1 + 120578)2 0 0 0 0 00 (4 + 161198702) 120587412057821198702 (1 + 120578)2 0 0 0 00 0 (9 + 811198702) 120587412057821198702 (1 + 120578)2 0 0 00 0 0 (16 + 2561198702) 120587412057821198702 (1 + 120578)2 0 00 0 0 0 (25 + 6251198702) 120587412057821198702 (1 + 120578)2 00 0 0 0 0 (36 + 12961198702) 120587412057821198702 (1 + 120578)2

)))))))))))))))))))))

1119879119904119903 = 1119878119903119904 =(((((((((((((((((


)))))))))))))))))

1119876

=(((((((((((((((((



)))))))))))))))))

(39)







[1198700]2119873times2119873 = [ 01198770119879 01198780119876 ]


295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1


[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)







































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of





y

x

yL

z

M2 M1

M2 M1

S(0 y0)

C(0 0)













119906 (119911) = ℎinfinsum119899=1

119860119898 sin [119898120587119911119871 ] (119898 = 1 2 3 infin) (8)

120579 (119911) = infinsum119899=1

119861119899 sin [119899120587119911119871 ] (119899 = 1 2 3 infin) (9)










T

M Mu

u

T



(10)





0 = 1198721(12058721198641198681199101198712) ℎ 120573119909 = 120573119909ℎ 120578 = 11986811198682 119870 = radic12058721198641198681205961198661198691198961198712

(11)





= infinsum119898=1

1198602119898119898412058744 + infinsum119899=1

1198612119899 (1 + 11989921198702) 1198992120587412057841198702 (1 + 120578)2 Π2 = 12 int119871 211987211990912057311990912057910158402119889119911


1198612119899 (1 + 119896)119898212058741205731199094+ 0 infinsum119904=1

infinsum119903=1119903 =119904



119860119898119861119898 (1 + 119896)119898212058744minus 0 infinsum119904=1

infinsum119903=1119903 =119904


(12)


Π = 3sum119894=1

Π119894 (13)




infinsum119898=1

119860119898119898412058742 minus 0 infinsum119898=1

119861119898 (1 + 119896)119898212058744



119861119903 4 (minus1 + 119896) 12058721199031198983(1199032 minus 1198982)2 = 0(15)





0119878119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (19)

1119877119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (20)


(21)




119860119899 (1 + 119896) 119899212058744minus 0 infinsum119899=1



119861119899 (1 + 119896) 119899212058741205731199092


+ infinsum119899=1

119861119899 (1 + 11989921198702) 1198992120587412057821198702 (1 + 120578)2+ 0 infinsum119899=1


119861119903 4 (minus1 + 119896) 1205872119903119899 (1199032 + 1198992) 120573119909(1199032 minus 1198992)2= 0

(22)


0119879119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (24)

1119879119904119903 = 1119878119903119904 119904 = 1 2 infin 119903 = 1 2 infin (25)



(27)



[0119877 01198780119879 0119876]119860119861 = 0 [1119877 11198781119879 1119876]119860119861 (28)







cr = 119872cr119871radic119864119868119910119866119869119896 (29)



(30)



cr = 119872cr(12058721198641198681199101198712) ℎ = 119872crℎ119875119864119910 (31)










[[[[[12058742 00 (1 + 1198702) 120587412057821198702 (1 + 120578)2

]]]]]11986011198611

= 0 [[[[0 (1 + 119896) 12058744(1 + 119896) 12058744 minus(1 + 119896) 12058741205731199092

]]]]11986011198611

(33)



(34)

where

1198621 = 21 + 119896 1198622 = 01198623 = 1(35)

or

cr = 21 + 119896 [[120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2)]] (36)



cr = 120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2) (37)



[0119877 01198780119879 0119876]

6times6

1198601198616times1

= 0 [1119877 11198781119879 1119876]6times6

1198601198616times1

(38)

where

0119878 = 0119879 = 1119877 = 0

0119877 =((((((((((((((

12058742 0 0 0 0 00 81205874 0 0 0 00 0 8112058742 0 0 00 0 0 1281205874 0 00 0 0 0 62512058742 00 0 0 0 0 6481205874

))))))))))))))


0119876 =(((((((((((((((((((((

(1 + 1198702) 120587412057821198702 (1 + 120578)2 0 0 0 0 00 (4 + 161198702) 120587412057821198702 (1 + 120578)2 0 0 0 00 0 (9 + 811198702) 120587412057821198702 (1 + 120578)2 0 0 00 0 0 (16 + 2561198702) 120587412057821198702 (1 + 120578)2 0 00 0 0 0 (25 + 6251198702) 120587412057821198702 (1 + 120578)2 00 0 0 0 0 (36 + 12961198702) 120587412057821198702 (1 + 120578)2

)))))))))))))))))))))

1119879119904119903 = 1119878119903119904 =(((((((((((((((((


)))))))))))))))))

1119876

=(((((((((((((((((



)))))))))))))))))

(39)







[1198700]2119873times2119873 = [ 01198770119879 01198780119876 ]


295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1


[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)







































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of











T

M Mu

u

T



(10)





0 = 1198721(12058721198641198681199101198712) ℎ 120573119909 = 120573119909ℎ 120578 = 11986811198682 119870 = radic12058721198641198681205961198661198691198961198712

(11)





= infinsum119898=1

1198602119898119898412058744 + infinsum119899=1

1198612119899 (1 + 11989921198702) 1198992120587412057841198702 (1 + 120578)2 Π2 = 12 int119871 211987211990912057311990912057910158402119889119911


1198612119899 (1 + 119896)119898212058741205731199094+ 0 infinsum119904=1

infinsum119903=1119903 =119904



119860119898119861119898 (1 + 119896)119898212058744minus 0 infinsum119904=1

infinsum119903=1119903 =119904


(12)


Π = 3sum119894=1

Π119894 (13)




infinsum119898=1

119860119898119898412058742 minus 0 infinsum119898=1

119861119898 (1 + 119896)119898212058744



119861119903 4 (minus1 + 119896) 12058721199031198983(1199032 minus 1198982)2 = 0(15)





0119878119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (19)

1119877119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (20)


(21)




119860119899 (1 + 119896) 119899212058744minus 0 infinsum119899=1



119861119899 (1 + 119896) 119899212058741205731199092


+ infinsum119899=1

119861119899 (1 + 11989921198702) 1198992120587412057821198702 (1 + 120578)2+ 0 infinsum119899=1


119861119903 4 (minus1 + 119896) 1205872119903119899 (1199032 + 1198992) 120573119909(1199032 minus 1198992)2= 0

(22)


0119879119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (24)

1119879119904119903 = 1119878119903119904 119904 = 1 2 infin 119903 = 1 2 infin (25)



(27)



[0119877 01198780119879 0119876]119860119861 = 0 [1119877 11198781119879 1119876]119860119861 (28)







cr = 119872cr119871radic119864119868119910119866119869119896 (29)



(30)



cr = 119872cr(12058721198641198681199101198712) ℎ = 119872crℎ119875119864119910 (31)










[[[[[12058742 00 (1 + 1198702) 120587412057821198702 (1 + 120578)2

]]]]]11986011198611

= 0 [[[[0 (1 + 119896) 12058744(1 + 119896) 12058744 minus(1 + 119896) 12058741205731199092

]]]]11986011198611

(33)



(34)

where

1198621 = 21 + 119896 1198622 = 01198623 = 1(35)

or

cr = 21 + 119896 [[120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2)]] (36)



cr = 120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2) (37)



[0119877 01198780119879 0119876]

6times6

1198601198616times1

= 0 [1119877 11198781119879 1119876]6times6

1198601198616times1

(38)

where

0119878 = 0119879 = 1119877 = 0

0119877 =((((((((((((((

12058742 0 0 0 0 00 81205874 0 0 0 00 0 8112058742 0 0 00 0 0 1281205874 0 00 0 0 0 62512058742 00 0 0 0 0 6481205874

))))))))))))))


0119876 =(((((((((((((((((((((

(1 + 1198702) 120587412057821198702 (1 + 120578)2 0 0 0 0 00 (4 + 161198702) 120587412057821198702 (1 + 120578)2 0 0 0 00 0 (9 + 811198702) 120587412057821198702 (1 + 120578)2 0 0 00 0 0 (16 + 2561198702) 120587412057821198702 (1 + 120578)2 0 00 0 0 0 (25 + 6251198702) 120587412057821198702 (1 + 120578)2 00 0 0 0 0 (36 + 12961198702) 120587412057821198702 (1 + 120578)2

)))))))))))))))))))))

1119879119904119903 = 1119878119903119904 =(((((((((((((((((


)))))))))))))))))

1119876

=(((((((((((((((((



)))))))))))))))))

(39)







[1198700]2119873times2119873 = [ 01198770119879 01198780119876 ]


295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1


[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)







































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of







= infinsum119898=1

1198602119898119898412058744 + infinsum119899=1

1198612119899 (1 + 11989921198702) 1198992120587412057841198702 (1 + 120578)2 Π2 = 12 int119871 211987211990912057311990912057910158402119889119911


1198612119899 (1 + 119896)119898212058741205731199094+ 0 infinsum119904=1

infinsum119903=1119903 =119904



119860119898119861119898 (1 + 119896)119898212058744minus 0 infinsum119904=1

infinsum119903=1119903 =119904


(12)


Π = 3sum119894=1

Π119894 (13)




infinsum119898=1

119860119898119898412058742 minus 0 infinsum119898=1

119861119898 (1 + 119896)119898212058744



119861119903 4 (minus1 + 119896) 12058721199031198983(1199032 minus 1198982)2 = 0(15)





0119878119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (19)

1119877119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (20)


(21)




119860119899 (1 + 119896) 119899212058744minus 0 infinsum119899=1



119861119899 (1 + 119896) 119899212058741205731199092


+ infinsum119899=1

119861119899 (1 + 11989921198702) 1198992120587412057821198702 (1 + 120578)2+ 0 infinsum119899=1


119861119903 4 (minus1 + 119896) 1205872119903119899 (1199032 + 1198992) 120573119909(1199032 minus 1198992)2= 0

(22)


0119879119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (24)

1119879119904119903 = 1119878119903119904 119904 = 1 2 infin 119903 = 1 2 infin (25)



(27)



[0119877 01198780119879 0119876]119860119861 = 0 [1119877 11198781119879 1119876]119860119861 (28)







cr = 119872cr119871radic119864119868119910119866119869119896 (29)



(30)



cr = 119872cr(12058721198641198681199101198712) ℎ = 119872crℎ119875119864119910 (31)










[[[[[12058742 00 (1 + 1198702) 120587412057821198702 (1 + 120578)2

]]]]]11986011198611

= 0 [[[[0 (1 + 119896) 12058744(1 + 119896) 12058744 minus(1 + 119896) 12058741205731199092

]]]]11986011198611

(33)



(34)

where

1198621 = 21 + 119896 1198622 = 01198623 = 1(35)

or

cr = 21 + 119896 [[120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2)]] (36)



cr = 120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2) (37)



[0119877 01198780119879 0119876]

6times6

1198601198616times1

= 0 [1119877 11198781119879 1119876]6times6

1198601198616times1

(38)

where

0119878 = 0119879 = 1119877 = 0

0119877 =((((((((((((((

12058742 0 0 0 0 00 81205874 0 0 0 00 0 8112058742 0 0 00 0 0 1281205874 0 00 0 0 0 62512058742 00 0 0 0 0 6481205874

))))))))))))))


0119876 =(((((((((((((((((((((

(1 + 1198702) 120587412057821198702 (1 + 120578)2 0 0 0 0 00 (4 + 161198702) 120587412057821198702 (1 + 120578)2 0 0 0 00 0 (9 + 811198702) 120587412057821198702 (1 + 120578)2 0 0 00 0 0 (16 + 2561198702) 120587412057821198702 (1 + 120578)2 0 00 0 0 0 (25 + 6251198702) 120587412057821198702 (1 + 120578)2 00 0 0 0 0 (36 + 12961198702) 120587412057821198702 (1 + 120578)2

)))))))))))))))))))))

1119879119904119903 = 1119878119903119904 =(((((((((((((((((


)))))))))))))))))

1119876

=(((((((((((((((((



)))))))))))))))))

(39)







[1198700]2119873times2119873 = [ 01198770119879 01198780119876 ]


295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1


[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)







































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of





+ infinsum119899=1

119861119899 (1 + 11989921198702) 1198992120587412057821198702 (1 + 120578)2+ 0 infinsum119899=1


119861119903 4 (minus1 + 119896) 1205872119903119899 (1199032 + 1198992) 120573119909(1199032 minus 1198992)2= 0

(22)


0119879119904119903 = 0 119904 = 1 2 infin 119903 = 1 2 infin (24)

1119879119904119903 = 1119878119903119904 119904 = 1 2 infin 119903 = 1 2 infin (25)



(27)



[0119877 01198780119879 0119876]119860119861 = 0 [1119877 11198781119879 1119876]119860119861 (28)







cr = 119872cr119871radic119864119868119910119866119869119896 (29)



(30)



cr = 119872cr(12058721198641198681199101198712) ℎ = 119872crℎ119875119864119910 (31)










[[[[[12058742 00 (1 + 1198702) 120587412057821198702 (1 + 120578)2

]]]]]11986011198611

= 0 [[[[0 (1 + 119896) 12058744(1 + 119896) 12058744 minus(1 + 119896) 12058741205731199092

]]]]11986011198611

(33)



(34)

where

1198621 = 21 + 119896 1198622 = 01198623 = 1(35)

or

cr = 21 + 119896 [[120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2)]] (36)



cr = 120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2) (37)



[0119877 01198780119879 0119876]

6times6

1198601198616times1

= 0 [1119877 11198781119879 1119876]6times6

1198601198616times1

(38)

where

0119878 = 0119879 = 1119877 = 0

0119877 =((((((((((((((

12058742 0 0 0 0 00 81205874 0 0 0 00 0 8112058742 0 0 00 0 0 1281205874 0 00 0 0 0 62512058742 00 0 0 0 0 6481205874

))))))))))))))


0119876 =(((((((((((((((((((((

(1 + 1198702) 120587412057821198702 (1 + 120578)2 0 0 0 0 00 (4 + 161198702) 120587412057821198702 (1 + 120578)2 0 0 0 00 0 (9 + 811198702) 120587412057821198702 (1 + 120578)2 0 0 00 0 0 (16 + 2561198702) 120587412057821198702 (1 + 120578)2 0 00 0 0 0 (25 + 6251198702) 120587412057821198702 (1 + 120578)2 00 0 0 0 0 (36 + 12961198702) 120587412057821198702 (1 + 120578)2

)))))))))))))))))))))

1119879119904119903 = 1119878119903119904 =(((((((((((((((((


)))))))))))))))))

1119876

=(((((((((((((((((



)))))))))))))))))

(39)







[1198700]2119873times2119873 = [ 01198770119879 01198780119876 ]


295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1


[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)







































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of











[[[[[12058742 00 (1 + 1198702) 120587412057821198702 (1 + 120578)2

]]]]]11986011198611

= 0 [[[[0 (1 + 119896) 12058744(1 + 119896) 12058744 minus(1 + 119896) 12058741205731199092

]]]]11986011198611

(33)



(34)

where

1198621 = 21 + 119896 1198622 = 01198623 = 1(35)

or

cr = 21 + 119896 [[120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2)]] (36)



cr = 120573119909 + radic(120573119909)2 + 120578(1 + 120578)2 (1 + 119870minus2) (37)



[0119877 01198780119879 0119876]

6times6

1198601198616times1

= 0 [1119877 11198781119879 1119876]6times6

1198601198616times1

(38)

where

0119878 = 0119879 = 1119877 = 0

0119877 =((((((((((((((

12058742 0 0 0 0 00 81205874 0 0 0 00 0 8112058742 0 0 00 0 0 1281205874 0 00 0 0 0 62512058742 00 0 0 0 0 6481205874

))))))))))))))


0119876 =(((((((((((((((((((((

(1 + 1198702) 120587412057821198702 (1 + 120578)2 0 0 0 0 00 (4 + 161198702) 120587412057821198702 (1 + 120578)2 0 0 0 00 0 (9 + 811198702) 120587412057821198702 (1 + 120578)2 0 0 00 0 0 (16 + 2561198702) 120587412057821198702 (1 + 120578)2 0 00 0 0 0 (25 + 6251198702) 120587412057821198702 (1 + 120578)2 00 0 0 0 0 (36 + 12961198702) 120587412057821198702 (1 + 120578)2

)))))))))))))))))))))

1119879119904119903 = 1119878119903119904 =(((((((((((((((((


)))))))))))))))))

1119876

=(((((((((((((((((



)))))))))))))))))

(39)







[1198700]2119873times2119873 = [ 01198770119879 01198780119876 ]


295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1


[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)







































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of





0119876 =(((((((((((((((((((((

(1 + 1198702) 120587412057821198702 (1 + 120578)2 0 0 0 0 00 (4 + 161198702) 120587412057821198702 (1 + 120578)2 0 0 0 00 0 (9 + 811198702) 120587412057821198702 (1 + 120578)2 0 0 00 0 0 (16 + 2561198702) 120587412057821198702 (1 + 120578)2 0 00 0 0 0 (25 + 6251198702) 120587412057821198702 (1 + 120578)2 00 0 0 0 0 (36 + 12961198702) 120587412057821198702 (1 + 120578)2

)))))))))))))))))))))

1119879119904119903 = 1119878119903119904 =(((((((((((((((((


)))))))))))))))))

1119876

=(((((((((((((((((



)))))))))))))))))

(39)







[1198700]2119873times2119873 = [ 01198770119879 01198780119876 ]


295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1


[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)







































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of





295

300

305N

ondi

men

siona

l crit

ical

mom

ent

5 10 15 20 25 30 35 400Number of terms

k = minus10 K = 05 x = 002 = 16

(a) 119896 = minus1

5 10 15 20 25 30 35 400Number of terms

30

35

40

45

Non

dim

ensio

nal c

ritic

al m

omen

t

k = minus05 K = 05 x = 002 = 16

(b) 119896 = minus05

5 10 15 20 25 30 35 400Number of terms

1460

1465

1470

1475

1480

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 05 K = 05 x = 002 = 16

(c) 119896 = 05

5 10 15 20 25 30 35 400Number of terms

00

05

10

15

20

25

Non

dim

ensio

nal c

ritic

al m

omen

t

k = 10 K = 05 x = 002 = 16

(d) 119896 = 1


[119870119866]2119873times2119873 = [ 11198771119879 11198781119876 ]

(41)







































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of




































bf

t ft f

tw ℎwh H

t ft f

tw ℎwh H

bf1

bf2

t ft f

tw ℎw

h H

bf1

bf2











bf

bf

t ft f

tw

ℎwh H






A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of








A


B


C













00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of





00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

minus12

minus09

minus06

minus03

00

03

06

09

12

Disp

lace

men

t am

plitu

de

6543210


Span L (m)

(d) 119896 = minus1








00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of





00

02

04

06

08

10

12D

ispla

cem

ent a

mpl

itude

1 2 3 4 5 60


Span L (m)

(a) 119896 = 1

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(b) 119896 = 05

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de

1 2 3 4 5 60


Span L (m)

(c) 119896 = minus05

6543210minus02

00

02

04

06

08

10

12

Disp

lace

men

t am

plitu

de


Span L (m)

(d) 119896 = minus1



minus05119872cr0119872cr




cr = 1198621 [(1198623120573119909)+ radic(1198623120573119909)2 + 1198624120578(1198625 + 1198626120578)2 (1 + 119870minus2)]

(43)



0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of





0

2

4

6

8

10

12

14D

imen

sionl

ess c

ritic

al m

omen

t

05 10 15 20 25 30 35 40 4500K


minus02

02

00

04

06

08

10

12

14

16

18

Criti

cal m

omen

t


k

= 0001

= 01

= 02

= 03

= 04

= 05

= 06

= 07

= 08

= 09

= 10




00

05

10

15

20

25

30

35

40

45

Criti

cal m

omen

t


K = 03

K = 04

K = 05

K = 06

K = 07

K = 08

K = 09

K = 10

K = 11

K = 12

K = 13

K = 14

K = 15

K = 16

K = 17

K = 18

K = 19

K = 20


minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ

+1 M＝Ｌ

M＝Ｌ





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of





0 10 20 30

40

50

60

70 80 90

100

110

120

130

140

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

310

320

330

340

350

360

370

380

390

400

410

420

430

440

450

460

470

1

2

3

4

M＝Ｌ


Table 5









00

05

10

15

20

25


M＝Ｌ







minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of






minus05minus1 10 k

minus1 M＝Ｌ

minus05 M＝Ｌ

0 M＝Ｌ+1 M＝Ｌ






A


B


C


8 Conclusions

















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of















Acknowledgments


References













































































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of




























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





Journal of




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