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    ininvestigation are proposed for the calculation of the effective buckling length coefficient for different levels of frame sway ability. In addition, a

    complete set of rotational stiffness coefficients is derived, which is then used for the replacement of members converging at the bottom and topends of the column in question by equivalent springs. All possible rotational and translational boundary conditions at the far end of these members,featuring semi-rigid connection at their near end as well as the eventual presence of axial force, are considered. Examples of sway, non-sway andpartially-sway frames with semi-rigid connections are presented, where the proposed approach is found to be in excellent agreement with thefinite element results, while the application of codes such as Eurocode 3 and LRFD leads to significant inaccuracies.c 2005 Elsevier Ltd. All rights reserved.

    Keywords: Buckling; Effective length; Stiffness coefficients; Multi-story sway; Non-sway and partially-sway frames; Semi-rigid connections

    1. Introduction

    Nowadays, the buckling strength of a member can beevaluated using engineering software based on linear or alsonon-linear (in terms of large displacements and/or materialyielding) procedures with analytical or numerical methods [15].Nonetheless, the large majority of structural engineers stillprefer analytical techniques such as the effective length andnotional load methods [26]. These two methodologies areincluded in most modern structural design codes (for example,Eurocode 3 [9], LRFD [23]).

    The objective of this work is to propose a simplifiedapproach for the evaluation of critical buckling loads of multi-story frames with semi-rigid connections, for different levelsof frame sway ability. To that effect, a model of a column in amulti-story frame is considered as individual. The contribution

    column is taken into account by equivalent springs. Namely,the restriction provided by the other members of the frameto the rotations of the bottom and top nodes is modeledvia rotational springs with constants cb and ct , respectively,while the resistance provided by the bracing system to therelative transverse translation of the end nodes is modeledvia a translational spring with constant cbr . This is shownschematically in Fig. 1. The rotational stiffness of the springsmust be evaluated considering the influence of the connectionnon-linearity. This model has been used by several investigators(for example, Wood [27], Aristizabal-Ochoa [1], and Cheong-Siat-Moy [6]) for the evaluation of the critical buckling load ofthe member, and is adopted by most codes.

    The stiffness of the bottom and top rotational springsis estimated by summing up the contributions of membersJournal of Constructional Steel Re

    Buckling strength of multi-story swaywith semi-rigi

    Georgios E. MageiroLaboratory of Metal Structures, Department of Structural Engineering, Nation

    Athens

    Received 2 August 2005; a

    Abstract

    The objective of this paper is to propose a simplified approach to therigid connections. To that effect, analytical expressions and correspondof members converging at the bottom and top ends of the

    Corresponding author. Tel.: +30 210 9707444; fax: +30 210 9707444.E-mail addresses: [email protected] (G.E. Mageirou),

    [email protected] (C.J. Gantes).1 Tel.: +30 210 7723440; fax: +30 210 7723442.

    0143-974X/$ - see front matter c 2005 Elsevier Ltd. All rights reserved.doi:10.1016/j.jcsr.2005.11.019rch 62 (2006) 893905www.elsevier.com/locate/jcsr

    non-sway and partially-sway framesconnections

    , Charis J. Gantes1

    Technical University of Athens, 9 Heroon Polytechniou, GR-15780, Zografou,reece

    epted 30 November 2005

    valuation of the critical buckling load of multi-story frames with semi-g graphs accounting for the boundary conditions of the column underconverging at the bottom and top ends, respectively:

    cb =

    icb,i , ct =

    j

    ct, j . (1)

    A frame is characterized as non-sway if the stiffness cbr ofthe bracing system is very large, as sway if this stiffness isnegligible, and as partially-sway for intermediate values of this

  • on894 G.E. Mageirou, C.J. Gantes / Journal of C

    Notations

    A, B, C, D integration constantsE modulus of elasticityG distribution factor at the end nodes of the column,

    according to LRFDI moment of inertiaK effective buckling length coefficientL span length of adjoining membersM bending momentN axial force of adjoining membersP compressive load factor of slope deflection methoda factor of slope deflection method, factor for the

    effect of the boundary condition at the far endnodes of the member

    c stiffness coefficientc ratio of flexural stiffness to spanc# dimensionless rotational stiffnessd factor for the effect of the axial forceh column heightk non-dimensional compressive load effective buckling length coefficient, according to

    EC3n ratio of members compressive force to Eulers

    buckling loadx longitudinal coordinatez dimensionless distribution factor at the end nodes

    of the columnw transverse deflection relative transverse deflection between the end

    nodes of the member distribution factor at the end nodes of the column,

    according to EC3 rotation at the end nodes of the member

    Subscripts:

    A bottom end node of the columnB top end node of the columnE EulerEC3 Eurocode 3FEM Finite Element MethodLRFD Load Resistance Factor Designc columncr criticalb bottombm beambr bracing systemi member in noder rigid connectiont topstiffness. Eurocode 3 and LRFD provide the effective lengthK h of columns in sway and non-sway frames via graphs orstructional Steel Research 62 (2006) 893905

    Fig. 1. (a) Multi-story steel frame; (b) proposed model of column underinvestigation.

    analytical relations as functions of the rotational boundaryconditions without considering the connection non-linearityand the partially-sway behaviour of the frame. The criticalbuckling load is then defined as:

    Pcr = 2 E Ic

    (K h)2(2)

    where E Ic is the flexural resistance.The main source of inaccuracy in the above process lies

    in the estimation of the rotational boundary conditions. LRFDmakes no mention to the dependence of the rotational stiffnessof members converging at the ends of the column underconsideration on their boundary conditions at their far end ortheir axial load. Annex E of EC3 is more detailed in accountingfor the contribution of converging beams and lower/uppercolumns, but ignores several cases that are encountered inpractice, and are often decisive for the buckling strength. Bothcodes ignore the partially-sway behaviour of the frames as wellas the connection non-linearity.

    This problem has been investigated by several researchers.The work of Wood [27] constituted the theoretical basis ofEC3. Cheong-Siat-Moy [5] examined the k-factor paradoxfor leaning columns and drew attention to the dependenceof buckling strength not only on the rotational boundaryconditions of the member in question but also on the overallstructural system behavior. Bridge and Fraser [4] proposedan iterative procedure for the evaluation of the effectivelength, which accounts for the presence of axial forces inthe restraining members and thus also considers the negativevalues of rotational stiffness. Hellesland and Bjorhovde [11]proposed a new restraint demand factor considering the verticaland horizontal interaction in member stability terms. Kishiet al. [14] proposed an analytical relation for the evaluation ofthe effective length of columns with semi-rigid joints in swayframes. Essa [8] proposed a design method for the evaluationof the effective length for columns in unbraced multi-storyframes considering different story drift angles. Aristizabal-Ochoa examined the influence of uniformly distributed axialload on the evaluation of the effective length of columns in

    sway and partially-sway frames [2]. He then examined thebehavior of columns with semi-rigid connections under loads

  • truc

    e, a

    form of a mmaximum loproperties), oet al. [21] pdata, in termcomparativedifferent conand Haldaranalysis algoframes withhistories. Thorder to defiis the ratiocarry accord

    frame, shownby replacingnoting by wntiation withing the signolumn in itsn differential

    (3)ultiple linear regression relationship between thead and various parameters (frame and sectionbtained from numerous analyses of frames. Liew

    roposed a comprehensive set of moment-rotations of stiffness and moment capacity, so that aassessment of the frame performance due to

    nection types could be undertaken. Reyes-Salazar[24], using a nonlinear time domain seismicrithm developed by themselves, excited three steelsemi-rigid connections by thirteen earthquake timeey proposed a parameter called the T ratio inne the rigidity of the connections. This parameter

    Consider the model of a column in a non-swayin Fig. 2(a), resulting from the model of Fig. 1(b)the translational spring with a roller support. Dethe transverse displacement and by the differerespect to the longitudinal coordinate x , and usconvention of Fig. 2(d), the equilibrium of this cbuckled condition is described by the well-knowequation:

    w (x) + k2w (x) = 0where:G.E. Mageirou, C.J. Gantes / Journal of Cons

    Fig. 2. Model of column in (a) non-sway frame, (b) sway fram

    such as those produced by tension cables that always passthrough fixed points or loads applied by rigid links [3]. Whatis more, Kounadis [16] investigated the inelastic buckling ofrigid-jointed frames.

    Christopher and Bjorhovde [7] conducted analyses of aseries of semi-rigid frames, each with the same dimensions,applied loads and member sizes, but with different connectionproperties, explaining how connection properties affectmember forces, frame stability, and inter-story drift. Jaspartand Maquoi [12] described the mode of application of theelastic and plastic design philosophies to braced frameswith semi-rigid connections. The buckling collapse of steelreticulated domes with semi-rigid joints was investigated byKato et al. [13] on the basis of a nonlinear elasticplastic hingeanalysis formulated for three-dimensional beamcolumns withelastic, perfectly plastic hinges located at both ends andmid-span for each member. Lau et al. [17] performed ananalytical study to investigate the behavior of subassemblageswith a range of semi-rigid connections under different testconditions and loading arrangements. They showed thatsignificant variations in the M response had a negligibleeffect on the load carrying capacity of the column and thebehavior of the subassemblage. A method for column designin non-sway bare steel structures which takes into accountthe semi-rigid action of the beam to column connectionswas proposed by Lau et al. [18]. In [19], closed-formsolutions of the second order differential equation of non-uniform bars with rotational and translational springs werederived for eleven important cases. A simplified methodfor estimating the maximum load of semi-rigid frames wasproposed by Li and Mativo [20]. The method was in theof the moment the connection would have toing to the beam line theory and the fixed endtional Steel Research 62 (2006) 893905 895

    nd (c) partially-sway frame, and (d) the sign convention used.

    moment of the girder. In [25], the equilibrium path was tracedfor braced and unbraced steel plane frames with semi-rigidconnections with the aid of a hybrid algorithm that combinesthe convergence properties of the iterative-incremental tangentmethod, calculating the unbalancing forces by considering theelement rigid body motion. Yu et al. [28] described the details ofa test program of three test specimens loaded to collapse and thetest observations for sway frames under the combined actionsof gravity and lateral loads.

    However, all these studies mention nothing about thedependence of the rotational stiffness of the membersconverging on the column under consideration, from theboundary conditions at their far ends and from their axial loads.This dependence is investigated in the present work for multi-story frames with semi-rigid connections for different levelsof lateral stiffness cbr . Easy to use analytical relations andcorresponding graphs are proposed for the estimation of thecolumns effective length for sway, non-sway and partially-sway frame behaviour. Furthermore, analytical expressions arederived for the evaluation of the rotational springs stiffnesscoefficients for different member boundary conditions andaxial loads accounting for the connection non-linearity. Resultsobtained via the proposed approach for sway, non-sway andpartially-sway frames with semi-rigid connections are found tobe in excellent agreement with finite element results, while theapplication of design codes such as Eurocode 3 and LRFD leadsto significant inaccuracies.

    2. Buckling strength of columns in multi-story frames

    2.1. Non-sway framesk = PcrE Ic

    = K h

    . (4)

  • ns

    The general solution of

    w (x) = A sin (kx) + BThe integration constanapplying the boundary c

    Transverse displacem

    w(0) = 0.Moment equilibrium at

    E Icw (0) = cbw (Moment equilibrium at

    E Icw (h) = ctw (h)Transverse displacemen

    w (h) = 0.The four simultaneoussolution for the fourdeterminant of the coefyields the buckling equa

    32K 3 (zt 1) (zb 1)

    [16K 2)]sin

    ( K

    )= 0 (10)

    obtained by the non-l stiffnesses cb and ctfness cc:

    (11)

    (12)

    or the effective lengthEq. (2) to provide thee upper left graph ofed.

    sway frame, shown inFig. 3. Effective buckling length factor K for different levels of frame-sway ability.

    this differential equation is given by:

    cos (kx) + Cx + D. (5)ts A, B, C , and D can be obtained byonditions at the two column ends:ent at the bottom:

    (6)the bottom:

    0) . (7)the top:

    . (8)t at the top:

    (9)equations (6)(9) have a non-trivialunknowns A, B, C , and D if the

    ficients is equal to zero. This criteriontion for the effective length factor K :

    + (zt + zb 2zt zb) 2]

    cos(

    K

    )+

    + 20K 2 (zt + zb) + zt zb(2 24K 2

    where zb and zt are distribution factorsdimensionalization of the end rotationawith respect to the columns flexural stif

    zb = cccc + cb , zt =

    cc

    cc + ctwhere:

    cc = 4E Ich .Eq. (10) can be solved numerically f

    factor K , which is then substituted intocritical buckling load. Alternatively, thFig. 3, obtained from Eq. (10), can be us

    2.2. Sway framesThe simplified model of a column in a896 G.E. Mageirou, C.J. Gantes / Journal of Co 4K[8K 2 (zt 1) (zb 1)tructional Steel Research 62 (2006) 893905Fig. 2(b), is considered, resulting from the model of Fig. 1(b)by omitting the translational spring. Three boundary conditions

  • truc

    Furthermore, there is no rotation at node B:G.E. Mageirou, C.J. Gantes / Journal of Cons

    are described by Eqs. (6)(8), while a fourth conditionexpresses horizontal force equilibrium at the top:

    E Icw (h) Pw (h) = 0. (13)Thus the buckling equation for the effective length factor K

    is derived, following the same procedure as above:

    4 [zt (2zb 1) zb] K cos(

    K

    )+[

    zt zb(

    K

    )2 16 (zt 1) (zb 1)]

    sin(

    K

    )= 0. (14)

    Alternatively, the bottom right graph of Fig. 3, obtained fromEq. (14), can be used.

    2.3. Partially-sway frames

    Finally, consider the simplified model of a column ina partially-sway frame, shown in Fig. 2(c). Similarly, theboundary conditions are described by Eqs. (6)(8) and thefollowing equation, representing horizontal force equilibriumat the top:

    E ICw (h) Pw (h) = cbrw (h) . (15)The effective length factor K of a column in a partially-sway

    frame is then obtained from the following equation:

    32K 5cbr (zt 1) (zb 1) + 4K[8K 4cbr (zt 1) (zb 1)

    + K 2cbr (zt + zb 2zt zb) 2 + (zt zb + 2zt zb) 4]

    cos(

    K

    )+

    [4K 4cbr (4 5zt 5zb + 6zt zb)

    16K 22 (1 zt zb + zt zb) K 2cbr2zt zb + 4zt zb

    ]sin

    ( K

    )= 0 (16)

    where:

    cbr = cbr h3

    E I. (17)

    Easy to use graphs such as those presented in Fig. 3 areobtained from the above equation for several values of cbr .

    3. Stiffness distribution factors

    The rotational stiffness of each member converging at thetop or bottom node of the column in question is derived usingthe slope-deflection method [22]. The moments MAB and MB Aat the two ends of a member AB with span L and flexuralstiffness E I , without axial force or transverse load (Fig. 4), canbe obtained as a function of the end rotations A and B and therelative transverse deflection of the end nodes from:

    MAB = 2E IL(

    2A + B + 3L)

    ,( )

    MB A = 2E IL 2B + A +

    3L

    . (18)tional Steel Research 62 (2006) 893905 897

    Fig. 4. Undeformed (dotted line) and deformed (continuous line) state of amember AB , and the sign convention of the slope-deflection method.

    If, in addition, the member is subjected to a compressiveaxial force P , then Eq. (18) becomes:

    MAB = 2E IL[

    anA + a f B +(an + a f

    ) L

    ]

    MB A = 2E IL[

    anB + a f A +(an + a f

    ) L

    ](19)

    where:

    an = n2(2n 2f

    ) , a f = f2(2n 2f

    ) (20)n = 1 kL cot kLk2L2 , f =

    1k2L2

    (kL

    sin kL 1

    ). (21)

    Using the above equations, the rotational stiffness expres-sions have been derived for members with all possible boundaryconditions at the far end and a semi-rigid connection at the closeend, with or without axial force, and are shown in Table 1. Thederivation of the rotational stiffness factors is described next fortwo characteristic cases: one for a member without and one fora member with axial force.

    3.1. Member with a fixed support at the far end and a semi-rigid connection at the near end, without axial force

    Consider the member AB of Fig. 5(a), with span Li andflexural stiffness Ei Ii , where A refers to the bottom or topnode of the column under investigation, while B is the farnode, attached to a fixed support. The connection at node Ais considered as semi-rigid with a rotational stiffness cn .

    The slope-deflection equations are given by (18), withindices i referring to the specific member. Firstly, theconnection at node A is considered as rigid. The rotationalstiffness cr,i of the member AB with a rigid connection wasevaluated in previous work by the authors [10].

    The moment at node A of the member with rigid connectionsis given by the equation:

    MAB = 2Ei IiLi (2A + B) . (22)B = 0. (23)

  • 898 G.E. Mageirou, C.J. Gantes / Journal of Constructional Steel Research 62 (2006) 893905Table 1Converging members rotational stiffness expressions for different boundary conditions at the far end and for a semi-rigid connection at the near end, with or withoutaxial force

    Rotational boundary conditions of far end Without axial force With axial force

    Fixed support ci = 4ci cn4ci +cn ci =4ci cn (10.33ni )cn+4ci 1.32ci ni

    Roller fixed support ci = ci cnci +cn ci =ci cn (10.82ni )cn+ci 0.82ci ni

    Pinned support ci = 3ci cn3ci +cn ci =3ci cn (10.66ni )cn+3ci 1.98ci ni

    Simple curvature ci = 2ci cn2ci +cn ci =2ci cn (10.82ni )cn+2ci 1.64ci ni

    Double curvature ci = 6ci cn6ci +cn ci =6ci cn (10.16ni )cn+6ci 0.96ci ni

    Roller support ci = 0ci cn0ci +cn ci =ci cn (00.97ni )

    cn+0ci 0.97ci ni

    Rotational spring support ci = ci cnc#

    (ci +cn )c#+cn ci =ci cn [c#(1.047ni +1.773)ni ]

    cn(0.591ni c#+c#+1)+ci [c#(1.047ni +1.773)ni ]

    Pinned and rotational spring support ci = 4ci cn (c#+3)

    4ci (c#+3)+cn (c#+4) ci =2ci cn [(c#(c#+9)+24)ni 230(c#+3)(c#+4)]

    2ci (c#(c#+9)+24)ni 215(c#+4)[4ci (c#+3)+cn (c#+4)]

  • trucG.E. Mageirou, C.J. Gantes / Journal of Cons

    Fig. 5. Model of a member with a fixed support at the far end B and a semi-rigid connection at the near end A: (a) without axial force, (b) with axial force.

    Then, by virtue of (23), Eq. (22) becomes:

    MAB = 4Ei IiLi A. (24)

    Thus, the rotational stiffness representing the resistance ofmember AB to the rotation of node A is given by:

    cr,i = 4Ei IiLi (25)

    which can be written as

    cr,i = 4ci (26)where:

    ci = Ei IiLi . (27)

    Two rotational springs with rotational stiffness cr,i and cnin series are considered, in order to evaluate the rotationalstiffness ci of the member AB with a semi-rigid connection.The total rotation is the sum of the rotations of the two springs.Therefore:

    i = r,i + n . (28)Considering that the springs have the same moment, Eq. (29)

    is written:1ci

    = 1cr,i

    + 1cn

    . (29)

    Thus, the rotational stiffness ci of member AB with a semi-rigid connection is evaluated from Eq. (29):

    ci = cr,i cncr,i + cn . (30)

    By substituting (26) into (30), the rotational stiffness ci ofthe member AB with a semi-rigid connection is evaluated:ci = 4ci cn4ci + cn . (31)tional Steel Research 62 (2006) 893905 899

    3.2. Member with a fixed support at the far end and a semi-rigid connection at the close end, with axial force

    Now consider the member of Fig. 5(b), with span Li andflexural stiffness Ei Ii , where A again refers to the bottom ortop node of the column under investigation, while B is thefar node, rotationally fixed. The member AB is subjected toa compressive axial force Ni . The rotational stiffness ci of themember AB with a semi-rigid connection is evaluated in thesame manner as above.

    Firstly, the member AB is considered having rigidconnections. The slope-deflection equations are given by (19),with indices i referring to the specific member. The moment atnode A of the member with rigid connections is given by theequation:

    MAB = 2Ei IiLi(n,iA + f,iB

    ). (32)

    As there is no rotation at node B (B = 0), the previousequation is rewritten:

    MAB = 2Ei IiLi n,iA. (33)Therefore, the rotational stiffness representing the resistance

    of member AB to the rotation of node A is given by:

    cr,i = 2Ei IiLi n,i (34)

    which, by means of (20) and (21) and k2i = N/Ei Ii , NE,i =2 Ei Ii/L2i , becomes:

    cr,i = 4Ei IiLi

    ni

    ni cot (ni ) 1

    4ni 8 tan(

    12

    ni) (35)

    where ni is the ratio of the members compressive force toEulers buckling load NE,i :

    ni = NiNE,i . (36)

    A Taylor series expansion of Eq. (35) gives:

    cr,i = 4Ei IiLi(

    1 2

    30ni 11

    4

    25 200n2i

    6

    108 000n3i

    5098

    2328 480 000n4i . . .

    ). (37)

    Keeping the first two terms, Eq. (38) is obtained:

    cr,i = 4Ei IiLi(

    1 2

    30ni

    )(38)

    which can be written as

    cr,i = 4ci (1 0.33ni) . (39)By substituting (39) into (30), the rotational stiffness ci of themember AB with a semi-rigid connection is evaluated:ci = 4ci cn(1 0.33ni)cn + 4ci 1.32cini . (40)

  • ons900 G.E. Mageirou, C.J. Gantes / Journal of C

    Fig. 6. The frame of example 1.

    4. Examples

    This section gives four examples of a simple frame thatconsists of a column and a beam with a variety of supports atthe far end namely, a three-story, single-bay sway frame; athree-story, single-bay non-sway frame; and a one-story, single-bay sway, non-sway and partially-sway frame for which theproposed approach is demonstrated and its results are comparedto (i) buckling loads obtained via linearized buckling analysisof finite element models, considered as exact, for verificationpurposes, and (ii) buckling loads calculated by applying thepertinent procedures of Eurocode 3 and LRFD.

    4.1. Example 1

    Consider the frame of Fig. 6 with a single span L = 20 mand height h = 10 m, having a column with HEB360 cross-section (Ic = 43 190 cm4) and a beam with IPE400 cross-section (Ibm = 23 130 cm4). The column is considered to bepinned at the base. A concentrated load P is imposed on thebeamcolumn joint. The beamcolumn joint is considered to besemi-rigid, with a rotational stiffness of cn = 150 kN m/rad.The beam has restricted translation and a rotational springsupport with a rotational stiffness of c = 500 kN m at thefar end. The frame is made of steel with Youngs modulusE = 210 000 000 kN/m2.

    Firstly, a linearized buckling analysis of the frame isconducted using the commercial finite element software MSC-NASTRAN. The critical buckling load obtained from thisanalysis is Pcr,FEM = 8981.02 kN. Secondly, the bucklingstrength is evaluated by using the proposed methodology. Inorder to do so, the frame is substituted by the equivalent modelof Fig. 2(b). The rotational stiffness of the beam considering thesemi-rigid connection is evaluated from the last row of Table 1:

    cB B =4cB B cn(c#B B + 3)

    4cB B (c#B B + 3) + cn(c#B B + 4)= 147.02 kN m (41)

    where:

    cB B = E IbmL = 2428.65 kN m (42)

    c#B B =

    c

    cB B = 0.206. (43)tructional Steel Research 62 (2006) 893905

    Applying the proposed method, the distribution factor zb isequal to 1 due to the pinned support, while zt is obtained from:

    zt = cccc + cB B = 0.996 (44)

    where:

    cc = 4E Ich = 36 279.60 kN m. (45)Then, the evaluation of the buckling length coefficient is

    conducted by means of Eq. (10), giving K = 0.998. Thus, theEuler buckling load is equal to:

    Pcr,prop = 2 E Ic

    (K h)2= 8981.01 kN. (46)

    Therefore, the results of the proposed approach are inexcellent agreement with the results of the finite elementmethod. The previous procedure is followed for the same framewith different supports at the far end of the beam as well asfor a single span, one-story sway and non-sway frame for theverification of the rotational stiffness of Table 1. The results ofthe proposed approach and the FEM analysis are presented inTable 2 and are practically the same.

    4.2. Example 2

    Consider the three-story sway frame of Fig. 7 with a singlespan L = 20 m and uniform story height h = 10 m,having columns with HEB360 cross-section (Ic = 43 190 cm4)and beams with IPE400 cross-section (Ibm = 23 130 cm4).The columns are considered to be pinned at the base. Equalconcentrated loads P/3 are imposed on all beamcolumnjoints. The beamcolumn joints are considered to be semi-rigidwith a rotational stiffness of cn = 150 kN m. The frame is madeof steel with Youngs modulus E = 210 000 000 kN/m2.

    At first, a buckling analysis of the frame is conducted usingthe same finite element software. The first buckling modeis obtained from this analysis, and the corresponding criticalbuckling load is 22.02 kN. In order to verify the proposedrotational stiffness coefficients, the frame is then substituted bya series of equivalent models. The first among them, denoted asequivalent model 1a, is illustrated in Fig. 7(b). It is obtained bysubstituting the beams at the three levels by rotational springs.Assuming that, in the first buckling mode, the beams deformwith a double curvature, the stiffness of the springs is obtainedfrom the corresponding row of Table 1:

    cbm = 6cbmcn6cbm + cn = 148.47 kN (47)

    where:

    cbm = E IbmL = 2428.65 kN m. (48)The first buckling mode of the equivalent model 1a is

    obtained from FEM analysis, and the corresponding critical

    buckling load is also 22.02 kN, thus verifying the correctnessof this substitution.

  • G.E. Mageirou, C.J. Gantes / Journal of Constructional Steel Research 62 (2006) 893905 901Table 2Critical loads according to the proposed and finite element methods for a variety of structural systems with semi-rigid connections of example 1

    Frame Pcr,FEM (kN) Pcr,prop (kN) Pcr,propPcr,FEMPcr,FEM (%)

    8981.58 8981.16 0.005

    8979.83 8979.86 0.001

    9027.06 9027.30 0.003

    10.98 10.97 0.091

    8981.02 8981.01 0.0001

  • ons

    ale

    h The first buckling mode of the equivalent model 1c iscal buckling, as well as

    l model, arereement.umn AB isRFD.distribution

    rt, while thefrom beam

    and columnfor the swayoefficient is

    (58)

    s of LRFD,the pinnedis equal toc#C D =cbmcC D

    = 0.016 (51)

    NC D = 13 22.02 kN = 7.34 kN. (52)Alternatively, the axial design load can be used for the NC D ,

    without any significant influence on the results:

    NE,C D = 2 E Ich2

    = 8942.56 kN (53)

    nC D = NC DNE,C D = 0.001. (54)

    Then, the total rotational stiffness at node C of model 1b is:

    cC D = cC D + cbm = 218.65 kN m. (55)The critical buckling load of the first buckling mode of the

    equivalent model 1b, obtained from Nastran, is 22.00 kN.The third equivalent model, denoted as 1c, is illustrated in

    Fig. 7(d). It is obtained by substituting column BC of model1b by a rotational spring with stiffness cBC calculated similarlyfrom the second last row of Table 1:

    obtained from Nastran, and the corresponding critiload is 21.96 kN. The critical loads of all modelstheir deviations from the critical load of the fulsummarized in Table 3, demonstrating excellent ag

    In addition, the critical buckling load of colevaluated according to the provisions of EC3 and L

    Following the procedure of Annex E of EC3, thefactor 1 at node A is 1 due to the hinged suppodistribution factor 2 at node B has a contributionB B assumed to deform with a double curvatureBC , and is found to be equal to 2 = 0.833. Then,buckling condition, the effective buckling length cfound to be L = 3.996.

    Thus, the Euler buckling load is calculated as:

    Pcr,EC3 = 2 E Ic[(L)

    h]2 = 560.03 kN.

    In the same manner, following the provisionthe distribution factor G A at node A is 10 due tosupport, and the distribution factor G B at node B902 G.E. Mageirou, C.J. Gantes / Journal of C

    Fig. 7. (a) The three-story sway frame of example 2, (b) equiv

    The second equivalent model, denoted as 1b, is illustrated inFig. 7(c). It is obtained by substituting the top column CD by arotational spring with stiffness cC D calculated from the secondlast row of Table 1:cC D

    =cC D cn,C D

    [c#C D (1.047nC D + 1.773)nC D

    ]cn,C D

    (0.591nC D c#C D + c#C D + 1

    )+ cC D

    [c#C D (1.047nC D + 1.773)nC D

    ]= 70.18 kN m (49)

    where:

    cC D = E Ic = 9069.90 kN m (50)cBC = 100.98 kN m. (56)tructional Steel Research 62 (2006) 893905

    nt model 1a, (c) equivalent model 1b, (d) equivalent model 1c.

    Table 3Critical loads for model 1 and its equivalent models 1a, 1b, 1c of Example 2

    Pcr (kN) Pcr Pcr,model1Pcr,model1 (%)

    Model 1 22.02428 0Model 1a 22.01921 0.02Model 1b 22.00429 0.09Model 1c 21.96301 0.28

    Then, the total rotational stiffness at node B of model 1c is:

    cBC = cBC + cbm = 249.45 kN m. (57)7.469. Assuming uninhibited side-sway behavior, the effectivebuckling coefficient is calculated to be K = 1.820.

  • trucG.E. Mageirou, C.J. Gantes / Journal of Cons

    Table 4Critical loads according to different methodologies of column AB of example 2

    Pcr (kN) Pcr Pcr,FEMPcr,FEM (%)

    FEM 22.02428 0EC3 560.60 2 445.37LRFD 2702.46 12 170.40Proposed 21.9399 0.38

    Fig. 8. The three-story non-sway frame of example 3.

    The Euler buckling load is equal to:

    Pcr,LRFD = 2 E Ic

    (K h)2= 2702.46 kN. (59)

    Applying the proposed method, the distribution factor zb isequal to 1 due to the pinned support, while zt is obtained from:

    zt = cccc + cBC = 0.994. (60)

    Then, the evaluation of the buckling length coefficient isconducted by means of Eq. (14) and gives K = 20.940. Thus,the Euler buckling load is equal to:

    Pcr,prop = 2 E Ic

    (K h)2= 21.94 kN. (61)

    The above results are summarized in Table 4, and comparedto the results of the linearized buckling analysis, which areconsidered to be exact. The proposed method is in very goodagreement with the numerical results.

    4.3. Example 3

    Next, consider the same three-story frame of example 2, butwith inhibited side-sway at all stories, shown in Fig. 8.

    The same procedure is followed for the verification of

    the proposed approach. The critical buckling load of columnAB is evaluated according to the code provisions as above.tional Steel Research 62 (2006) 893905 903

    Table 5Critical loads according to different methodologies of column AB of example 3

    Pcr (kN) Pcr Pcr,FEMPcr,FEM (%)

    FEM 11 237.75 0EC3 9 358.89 16.72LRFD 11 745.60 4.52Proposed 11 274.80 0.33

    Fig. 9. The frames of example 4 with (a) partially-sway, (b) non-sway and (c)sway behaviour.

    The proposed method is in very good agreement with thenumerical results, while EC3 is overconservative and LRFD isunderconservative but with much smaller deviations than in thesway-case (Table 5).

    4.4. Example 4

    Lastly, consider the one-story, partially-sway, non-sway and

    sway frames of Fig. 9 (a), (b) and (c), respectively. The frameshave a single span L = 20 m and a story height h equal to 10 m,

  • ons

    -s

    lengths of columns. J Struct Eng 1987;113:134156.behaviour consideration give overconservative results while thenon-sway behaviour consideration leads to underconservativeresults (Table 6). The results of the proposed methodology andthe design codes are presented in Tables 68 for the frames ofFig. 9 (a), (b) and (c), respectively.

    5. Summary and conclusions

    A simplified method for the evaluation of the criticalbuckling load of multi-story sway, non-sway and partially-

    [5] Cheong-Siat-Moy F. K -factor paradox. J Struct Eng 1986;112:174760.[6] Cheong-Siat-Moy F. An improved K -factor formula. J Struct Eng 1999;

    125:16974.[7] Christopher JE, Bjorhovde R. Response characteristics of frames with

    semi-rigid connections. J Constr Steel Res 1998;46:2534.[8] Essa HS. Stability of columns in unbraced frames. J Struct Eng 1997;123:

    9527.[9] Eurocode 3. Design of steel structures Part 1.1: General rules and rules for

    buildings. CEN Brussels 2004, CEN Document EN 1993-1-1:2004.[10] Gantes C, Mageirou G. Improved stiffness distribution factors for

    evaluation of effective buckling lengths in multi-story sway frames. EngStruct 2005;27:111324.904 G.E. Mageirou, C.J. Gantes / Journal of C

    Table 6Critical loads according to different methodologies of column AB of the partially

    FEMEC3 (lower limit assuming sway behaviour)EC3 (upper limit assuming non-sway behaviour)LRFD (lower limit assuming sway behaviour)LRFD (upper limit assuming non-sway behaviour)Proposed

    Table 7Critical loads according to different methodologies of column AB of the non-sway frame of example 4

    Pcr (kN) Pcr Pcr,FEMPcr,FEM (%)

    FEM 8 980.67 0EC3 9 980.74 11.14LRFD 11 821.70 31.64Proposed 8 980.67 0

    Table 8Critical loads according to different methodologies of column AB of the swayframe of example 4

    Pcr (kN) Pcr Pcr,FEMPcr,FEM (%)

    FEM 14.77 0EC3 898.78 5 983.56LRFD 3441.23 23 192.60Proposed 14.77 0

    having columns with HEB360 cross-section (Ic = 43 190 cm4)and beams with IPE400 cross-section (Ibm = 23 130 cm4). Thecolumns are considered to be pinned at the base. A compressiveconcentrated load P is imposed on the beamcolumn joints.The beamcolumn joints are considered to be semi-rigid witha rotational stiffness of cn = 150 kN m. A translation springwith a stiffness of cbr = 1000 kN/m simulates the partially-sway behaviour of the first frame. The frame is made of steelwith Youngs modulus E = 210 000 000 kN/m2.

    The same procedure is followed for the verification of theproposed approach. Moreover, the critical buckling load ofcolumn AB is evaluated according to the code provisions,considering the first frame firstly to be sway and secondly tobe non-sway. The proposed method is in excellent agreementwith the numerical results, while EC3 and LRFD with the swaysway frames with semi-rigid connections has been presented.Firstly, three analytical expressions for the effective bucklingtructional Steel Research 62 (2006) 893905

    way frame of example 4

    Pcr (kN) Pcr Pcr,FEMPcr,FEM (%)

    5 000.636 0898.78 82.03

    9 980.74 99.593 441.23 31.18

    11 821.70 136.405 000.01 0.01

    length coefficient as a function of the end nodes distributionfactors, as well as accompanying graphs, have been proposedfor different levels of sway ability. The rotational stiffness ofthe members (columns and beams) converging at the bottomand top ends of the column with semi-rigid connectionsdepend heavily on the boundary conditions at their far endand on the existence of axial force in them. Thus, analyticalexpressions of the stiffness distribution factors accountingfor these issues have been derived. Examples of sway, non-sway and partially-sway structures with semi-rigid connectionsand comparisons to finite element results have been usedto establish the improved accuracy of the above mentionedprocedure compared to the pertinent code provisions. It isbelieved that the proposed approach maintains the inherentsimplicity of the effective length method by not significantlyincreasing the required workload, but at the same time improvesits accuracy a lot and could thus be considered to be acompetitive alternative for practical applications.

    Acknowledgments

    Financial support for this work is provided by thePythagoras: Support of Research Groups in Universities. Theproject is co-funded by the European Social Fund (75%) andNational Resources (25%) (EPEAEK II)PYTHAGORAS.

    References

    [1] Aristizabal-Ochoa JD. K -factor for columns in any type of construction:Non-paradoxical approach. J Struct Eng 1994;120:127390.

    [2] Aristizabal-Ochoa JD. Stability of columns under uniform axial load withsemi-rigid connections. J Struct Eng 1994;120:321222.

    [3] Aristizabal-Ochoa JD. Elastic stability of beamcolumns with flexuralconnections under various conservative end axial forces. J Struct Eng1997;123:1194200.

    [4] Bridge RQ, Fraser DJ. Improved G-factor method for evaluating effective[11] Hellesland J, Bjorhovde R. Improved frame stability analysis witheffective lengths. J Struct Eng 1996;122:127583.

  • G.E. Mageirou, C.J. Gantes / Journal of Constructional Steel Research 62 (2006) 893905 905

    [12] Jaspart J, Maquoi R. Guidelines for the design of braced frames with semi-rigid connections. J Constr Steel Res 1990;16:31928.

    [13] Kato S, Mutoh I, Shomura M. Collapse of semi-rigidly jointed reticulateddomes with initial geometric imperfections. J Constr Steel Res 1998;48:14568.

    [14] Kishi N, Chen WF, Goto Y. Effective length factor of columns in semi-rigid and unbraced frames. J Struct Eng 1997;123:31320.

    [15] Kounadis A, Simitses G, Giri J. Nonlinear analysis of portal frames. Int JNumer Methods Eng 1981;17:12332.

    [16] Kounadis A. Nonlinear inelastic buckling of rigid-jointed frames underfinite displacements. Acta Mech 1987;67:191207.

    [17] Lau S, Kirby P, Davison J. Appraisal of partially restrained steel columnsin non-sway frames. J Struct Eng 1997;123:8719.

    [18] Lau SM, Kirby PA, Davison JB. Semi-rigid design of partially restrainedcolumns in non-sway steel frames. J Constr Steel Res 1997;50:30528.

    [19] Li QS. Buckling analysis of non-uniform bars with rotational andtranslational springs. Eng Struct 2003;25:128999.

    [20] Li G-Q, Mativo J. Approximate estimation of the maximum load of semi-rigid steel frames. J Constr Steel Res 2000;54:21325.

    [21] Liew JYR, Yu CH, Ng YH, Shanmugan NE. Testing of semi-rigid

    unbraced frames for calibration of second-order inelastic analysis.J Constr Steel Res 1997;41:15995.

    [22] Livesley RK, Chandler DB. Stability functions for structural framework.Manchester University Press; 1956.

    [23] L.R.F.D. Load and resistance factor design specification for structuralsteel buildings. Chicago: American Institute of Steel Construction Inc;1999.

    [24] Reyes-Salazar A, Haldar A. Non-linear seismic response of steelstructures with semi-rigid and composite connections. J Constr Steel Res1999;51:3759.

    [25] Rodrigues F, Saldanha A, Pfeil M. Non-linear analysis of steel planeframes with semi-rigid connections. J Constr Steel Res 1998;46:13.947.

    [26] Task Committee on Effective Length, Effective length and notional loadapproaches for assessing frame stability: Implications for American SteelDesign. New York: ASCE; 1997.

    [27] Wood RH. Effective lengths of columns in multi-story buildings. StructEng 1974;52: 23544, 295302, 3416.

    [28] Yu CH, Liew JYR, Shanmugam NE, Ng YH. Collapse behaviour of swayframes with end-plate connections. J Constr Steel Res 1998;48:16988.

    Buckling strength of multi-story sway, non-sway and partially-sway frames with semi-rigid connectionsIntroductionBuckling strength of columns in multi-story framesNon-sway framesSway framesPartially-sway frames

    Stiffness distribution factorsMember with a fixed support at the far end and a semi-rigid connection at the near end, without axial forceMember with a fixed support at the far end and a semi-rigid connection at the close end, with axial force

    ExamplesExample 1Example 2Example 3Example 4

    Summary and conclusionsAcknowledgmentsReferences

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Journal of Constructional Steel Research 62 (2006) 893–905 www.elsevier.com/locate/jcsr Buckling strength of multi-story sway, non-sway and partially-sway frames with semi-rigid connections Georgios E. Mageirou , Charis J. Gantes 1 Laboratory of Metal Structures, Department of Structural Engineering, National Technical University of Athens, 9 Heroon Polytechniou, GR-15780, Zografou, Athens, Greece Received 2 August 2005; accepted 30 November 2005 Abstract The objective of this paper is to propose a simplified approach to the evaluation of the critical buckling load of multi-story frames with semi- rigid connections. To that effect, analytical expressions and corresponding graphs accounting for the boundary conditions of the column under investigation are proposed for the calculation of the effective buckling length coefficient for different levels of frame sway ability. In addition, a complete set of rotational stiffness coefficients is derived, which is then used for the replacement of members converging at the bottom and top ends of the column in question by equivalent springs. All possible rotational and translational boundary conditions at the far end of these members, featuring semi-rigid connection at their near end as well as the eventual presence of axial force, are considered. Examples of sway, non-sway and partially-sway frames with semi-rigid connections are presented, where the proposed approach is found to be in excellent agreement with the finite element results, while the application of codes such as Eurocode 3 and LRFD leads to significant inaccuracies. c 2005 Elsevier Ltd. All rights reserved. Keywords: Buckling; Effective length; Stiffness coefficients; Multi-story sway; Non-sway and partially-sway frames; Semi-rigid connections 1. Introduction Nowadays, the buckling strength of a member can be evaluated using engineering software based on linear or also non-linear (in terms of large displacements and/or material yielding) procedures with analytical or numerical methods [15]. Nonetheless, the large majority of structural engineers still prefer analytical techniques such as the effective length and notional load methods [26]. These two methodologies are included in most modern structural design codes (for example, Eurocode 3 [9], LRFD [23]). The objective of this work is to propose a simplified approach for the evaluation of critical buckling loads of multi- story frames with semi-rigid connections, for different levels of frame sway ability. To that effect, a model of a column in a multi-story frame is considered as individual. The contribution of members converging at the bottom and top ends of the Corresponding author. Tel.: +30 210 9707444; fax: +30 210 9707444. E-mail addresses: [email protected] (G.E. Mageirou), [email protected] (C.J. Gantes). 1 Tel.: +30 210 7723440; fax: +30 210 7723442. column is taken into account by equivalent springs. Namely, the restriction provided by the other members of the frame to the rotations of the bottom and top nodes is modeled via rotational springs with constants c b and c t , respectively, while the resistance provided by the bracing system to the relative transverse translation of the end nodes is modeled via a translational spring with constant c br . This is shown schematically in Fig. 1. The rotational stiffness of the springs must be evaluated considering the influence of the connection non-linearity. This model has been used by several investigators (for example, Wood [27], Aristizabal-Ochoa [1], and Cheong- Siat-Moy [6]) for the evaluation of the critical buckling load of the member, and is adopted by most codes. The stiffness of the bottom and top rotational springs is estimated by summing up the contributions of members converging at the bottom and top ends, respectively: c b = i c b,i , c t = j c t , j . (1) A frame is characterized as non-sway if the stiffness c br of the bracing system is very large, as sway if this stiffness is negligible, and as partially-sway for intermediate values of this 0143-974X/$ - see front matter c 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.jcsr.2005.11.019
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