inelastic buckling of pin-ended steel columns under longitudinal non-uniform temperature...

Journal of Constructional Steel Research 65 (2009) 132–141

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Journal of Constructional Steel Research

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Inelastic buckling of pin-ended steel columns under longitudinal non-uniformtemperature distributionK.H. Tan, W.F. Yuan ∗School of Civil and Environmental Engineering, Nanyang Technological University, 50 Nanyang Avenue, Singapore 639798, Singapore

a r t i c l e i n f o

Article history:Received 17 April 2007Accepted 2 July 2008

Keywords:Steel columnThermal restraintInelastic bucklingAnalytical analysis

a b s t r a c t

Columns under natural fire conditions are usually exposed to non-uniform temperature distribution inthe longitudinal direction. The motivation for this study stems from zone modeling of a compartmentfire where the gas layers are artificially divided into two zones, viz. the hotter upper zone and the coolerlower zone. However, for fieldmodeling of a compartment fire,more detailed information of temperaturedistribution can be obtained. The difference in temperature between the top and bottomends of a columncan be quite significant, particularly prior to flashover condition. Depending on the required accuracy,one example due to piece-wise step distribution in the longitudinal direction is analyzed in this paperand compared with experimental results. This represents more realistically the thermal response ofa column which experiences greater temperature variation with increasing height. In this paper, theinelastic stability of a pin-ended steel column under non-uniform temperature distribution is studiedanalytically. Across a column section, the temperature is assumed to beuniform. Two linear elastic springsconnected to the column ends simulate axial restraints from adjoining unheated structures.

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0. Introduction

The behavior of columns subjected to fire conditions is vastlydifferent from that under normal ambient temperature. Thermalrestraint from adjoining unheated structure plays a key role inthe stability of these columns. The structural response dependslargely on the temperature distribution in the cross-sectionaland longitudinal directions due to (i) thermal load which willchange the material properties of steel and (ii) under elevatedtemperature, the magnitude of thermal induced compressivestress arising from thermal restraint is of the same order asinitial applied stress at ambient temperature. The effect oftemperature variations in the cross-sectional direction has beenstudied by Ossenbruggen et al. [1]. However, this paper focuseson the analytical derivations of column stability subjected tolongitudinal temperature variations since there has not been anysignificant theoretical development on this aspect. The objectiveis to derive analytical solutions to enable engineers to quicklyascertain the column stability under non-uniform temperaturedistribution, without recourse to numerical methods. This methodis particularly useful when considering the effect of a local fire oncolumn stability. Temperature distribution may be obtained fromestablished fire plume models.

∗ Corresponding author. Tel.: +65 67904851.E-mail address:[email protected] (W.F. Yuan).

0143-974X/$ – see front matter© 2008 Elsevier Ltd. All rights reserved.doi:10.1016/j.jcsr.2008.07.010

At present, a numerical approach is very popular in the studyof structures under fire because a finite element program offers awide range of flexibility. A review of recent literature shows thatthe effects of axial restraint have been investigated numerically byNeves [2], and Shepherd et al. [3]. Besides, the effects of rotationalrestraint have also been studied numerically by Franssen et al. [4],Wang [5] and Valente et al. [6]. It was found that the criticaltemperatures of columns will be reduced by axial restraint butenhancedby rotational restraint. Huang andTan et al. [7] presenteda series of numerical studies conducted on thermally-restrainedsteel columns subjected to predominantly axial loads. A finiteelement program FEMFAN3Dwas developed for the fire resistanceanalysis and creep strain has been explicitly considered. Withincorporation of creep strain, the rate of increase in temperatureon mechanical response can be modeled.However, for design purposes, theoretical analysis that can be

performed manually is much needed as it enables engineers toquickly ascertain the column buckling loads, particularly underlocal fire scenarios. Culver et al. [8] proposed an approachto determine the buckling loads for pin-ended steel columnssubjected to a uniform temperature increase along the memberlength. In theirwork, the effects of residual stress and the influenceof temperature on the buckling strength in both the elastic andthe inelastic range were considered but thermal restraints werenot taken into account. On the other hand, it is acknowledgedthat the behavior of a steel column in fire is mostly affectedby the restraints of its adjoining structure [9]. Ali et al. [10]reported that axial restraint reduced the fire resistance of columns

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K.H. Tan, W.F. Yuan / Journal of Constructional Steel Research 65 (2009) 132–141 133

Nomenclature

A Area of cross section of the column;E Young’s modulus;E200 Elastic Young’s modulus at ambient temperature;µ Constant factor, Young’s modulus ratio;Giα Scalar, i = 1, 5;I Elastic bending inertia of the cross section;Iep Inelastic bending inertia of the cross section;ki Stiffness of springs, i = 1, 2;ke Equivalent stiffness of restraints;k200 Axial stiffness of a column under normal ambient

temperature;L Length of the column;PT Additional axial force due to thermal expansion;Pc Total internal compressive axial force acting on the

cross section of the column;P0 Service load;Pc−cr Critical total compressive axial load;PE critical compressive load of an elastic pin-ended

column under normal ambient temperature;T Temperature;T1 Temperature at segment 1;T2 Temperature at segment 2;ε Vector, total axial strain of the column;εep Vector, mechanical elasto-plastic strain;εT Vector, thermal induced strain;ε Scalar, total strain of the column;εep Scalar, mechanical elasto-plastic strain;εT Scalar, thermal induced strain;σ Axial stress in the column;σY Yield stress of material;σY−1 Yield stress of material at T1;σY−2 Yield stress of material at T2;β Thermal expansion ratio;η Ratio of critical compressive axial load;α Ratio of the length of segment 1 to the whole length

of a column;ϑ Ratio of the length of segment 2 to the whole length

of a column;ω Constant matrix;ωij Matrix element;γ Restraint stiffness ratio;

after a series of tests on 37 axially-restrained steel columnssubjected to quasi-standard fire. Tang et al. [11] proposed asimple approach based on the Rankine interaction formula toobtain a realistic estimate of column fire resistance. However,in that paper, columns were subjected to uniform temperaturedistribution. Huang and Tan [12,13] considered the axial restrainton an isolated heated column using a linear spring attached to thecolumn top end. They extended the traditional Rankine formulato predict the critical temperature of an axially-restrained steelcolumn. The proposed Rankine approach incorporating both theaxial restraint and creep strain yields very good agreement withthe finite element predictions.Although the stability of axially-loaded columns at elevated

temperature and subjected to elastic restraints has been studiedby some researchers, the assumption about uniform temperaturedistribution may give an overly conservative prediction, sincethe temperature distribution in the longitudinal direction underfire conditions is usually non-uniform. This is because throughthe convective process, the hottest layer of air will rise up tothe top with a relatively cooler layer at the bottom. Thus, based

Fig. 1. Column member under compressive load.

on uniform temperature assumption, for conservatism, engineersusually ascribe the hottest temperature at the column top as theuniform column design temperature. In 1972, Culver [14] analyzedthe stability of wide-flanged steel columns subject to elevatedtemperature using a finite difference approach. The bucklingloads were determined by solving the governing differentialequation based on the finite difference method. Various cases ofnon-uniform temperature distribution along the member lengthwere considered, but the influence of end restraints was notinvestigated. In this paper, as shown in Fig. 1, the temperature inthe longitudinal direction (x-axis) is assumed to be non-uniform.Two linear springs attached to the column ends simulate the linearrestraints from the adjoining unheated structure. To simplify theensuing derivations, the two elastic springs can be replaced by oneequivalent spring (ke) at the top end of the column. The critical loadis derived analytically using Galerkin’s method. Tan and Yuan [15]studied the stability of a pin-ended steel column subjected tovarying longitudinal temperature distribution but the analysis wasonly conducted for an elastic material model. In this paper, theinelastic buckling load is derived analytically.

1. Stability

1.1. Stress in the column

For a pin–pin ended column under non-uniform temperaturedistribution, Young’s modulus of steel is no longer constant,since temperature varies over the column height. Adopting thecoordinate system shown in Fig. 1, Young’s modulus can beexpressed by (1):

E = E(T ) = E(x). (1)

With the ends restrained by linear springs and without anyimposed external load, the total axial strain of a heated column canbe described by (2):

ε = −εep + εT (2)

where ε is the positive column strain in tension. The term εep isthemechanical elasto-plastic strain and εT is the thermally inducedstrain. In (2), it is assumed that the axial force is compressive andthe temperature is elevated.

134 K.H. Tan, W.F. Yuan / Journal of Constructional Steel Research 65 (2009) 132–141

Fig. 2. Bilinear strain–stress model.

Assuming the yield strength at temperature T is σY = σY (T ) =σY (x), εep in (2) can be calculated by (3) as illustrated in Fig. 2.

εep = ε1 + ε2

=µ− (µ− 1)H(σ − σY )

µE(x)· σ +

(µ− 1)H(σ − σY )µE(x)

· σY (3)

where σ is the compressive axial stress in the column, E is Young’smodulus in elastic phase, µ is a material constant and µE is thetangent value of the stress–strain curve when the material is inplastic phase. ε1 and ε2 are strains corresponding to stresses σYand σ , respectively, and H is the Heaviside function defined by:

H(x) ={01

(x ≤ 0)(x > 0). (4)

Denoting the shortening of the equivalent elastic spring by∆L,the external applied force P0 is resisted by the equivalent springand the column itself:

P0 = −ke∆L+ Aσ . (5)

From (2), one obtains:∫ L

0εdx = −

∫ L

0εepdx+

∫ L

0εTdx. (6)

The left-hand side of (6) represents the columnextensionwhichmust be equal to the shortening of the equivalent top spring ke, say∫ L0 εdx = ∆L. Hence, substituting (6) into (5), one obtains:∫ L

0εepdx−

∫ L

0εTdx =

P0 − Aσke

. (7)

The second term of the left-hand side of (7) can be calculatedfrom (8) according to the definition of thermal expansion.∫ L

0εTdx =

∫ L

0β(T )dx (8)

where β(T ) = β(x) = ∆(dL)dL is the thermal expansion ratio.

Thus, substituting (3) into (7), one obtains:∫ L

0

[µ− (µ− 1)H

µE(x)· σ +

(µ− 1)HµE(x)

· σY

]dx−

∫ L

0β(x)dx

=P0 − Aσke

. (9)

Fig. 3. The buckling mode of instability.

Hence, the total column internal compressive force is:

Pc = Aσ = PT + ϕ0P0 (10)

where ϕ0 = 1keAµ∫ L0

[µ+(ξ−1)(µ−1)H

E(x)

]dx+1and PT = keϕ0

∫ L0 β(x)dx. The

term ξ is defined as ξ = σYσ. Obviously, PT is the force induced

by the temperature and restraint effect. The term ϕ0 takes accountof the temperature effect, the strength reduction factor ξ , and thestiffness of the equivalent spring ke.

1.2. Buckling

In Fig. 3, the columndeflection is governed by the classical Eulerequation:

E(x)Iepd2ydx2+ Pcy = 0 (11)

where Iep is themoment of inertia of the cross section, and Pc is thetotal compression force acting on the cross section. In (11), Young’smodulus and Iep are dependant on x. It should be noted that Iep isalso dependent on Pc because plasticity is considered. In this paper,Iep is given by recourse of beam theory.

Iep = I[1− (1− µ)H] ={IµI[σ ≤ σY (x)][σ > σY (x)].

(12)

It must be noted that for inelastic buckling of columns, thetheory of reduced modulus of elasticity is widely used [16].According to this theory, the reduced Young’s modulus Er notonly depends on the material properties but also on the crosssection. However, Timoshenko and Gere [17] suggested that thestress–strain relation for the entire column can be defined by thetangent modulus Et . In this paper, Iep takes account of the inelasticbehavior of the column.It is difficult to directly solve (11) since Young’s modulus and

Iep are functions of x. To simplify the problem, the temperaturedistribution is assumed to be at the steady state. A trial functionfor deflection is selected as (13):

y =3∑i=1

aiNi = a1x(L− x)+ a2x2(L− x) (13)

where a1 and a2 are two constant coefficients. It should be notedthat the trial function automatically satisfies y(0) = y(L) = 0.Normally, y is not the true solution of (11). In this paper,

Gelerkin’smethod, which is one kind ofweighted residualmethod,


is used to minimize the difference between the trial function yand the exact solution. For convenience, a residual function R(x)is given as follows:

R(x) = E(x)Iepd2ydx2+ Pc y (14)

hence

R(x) = [F1(x)+ x(L− x)Pc]a1 + [F2(x)+ x2(L− x)Pc]a2 (15)

where F1(x) = −2E(x)Iep and F2(x) = (2L− 6x)E(x)Iep.According to the Gelerkin method, the following two equations

should be satisfied:∫ L

0N1R(x)dx =

∫ L

0x(L− x)R(x)dx = 0 (16)

and∫ L

0N2R(x)dx =

∫ L

0x2(L− x)R(x)dx = 0. (17)

Thus, substituting the expression of R(x) into (16) and (17), oneobtains:(C1 + C2Pc C3 + C4PcD1 + D2Pc D3 + D4Pc

)(a1a2

)=

(00

). (18)

Theoretically, the deflection becomes indefinite when the col-umn loses stability. At this situation, (19) is required mathemati-cally:∣∣∣∣C1 + C2Pc C3 + C4PcD1 + D2Pc D3 + D4Pc

∣∣∣∣ = 0. (19)

Hence, if Pc−cr denotes the critical internal axial load, it can bedetermined and expressed by (20):

Pc−cr =−KB ±

√K 2B − 4KAKC

2KA(20)

where KA = C2D4 − C4D2, KB = C1D4 − C4D1 + C2D3 − C3D2 andKC = C1D3 − C3D1. The subscript ‘−cr ’ denotes the critical state.From stability considerations, the lower Pc−cr (single-curvaturebuckling), calculated from Eq. (20), is taken as the critical internalload. Accordingly, the critical internal stress is defined as σcr =Pc−crA . It should be noticed that C1, C3, D1 and D3 are dependant onmaterial properties E(x) and Iep. They are defined by (A.1)–(A.8) inAppendix A.To simplify the derivation, (20) is rewritten as:

Pc−cr = η · PE = η · σeA (21)

where the term PE =π2E200 IL2

is the Euler buckling load of thecolumn under ambient temperature. η is a factorwhich varieswithtemperature. Its expression is given in Appendix C. σe =

PEA is

defined as the critical internal stress of the column at ambienttemperature.

1.3. Critical external load

It must be mentioned that the column also resists externalcompressive force as well as the thermal load, therefore the initialexternal load P0−cr at the critical state (denoted by subscript ‘−cr ’)can be determined from (22):

P0−cr =Pc−cr − PT

ϕ0=η · PE − PT

ϕ0. (22)

Fig. 4. Step temperature distribution.

2. Material model and temperature distribution

2.1. Material model

For steel under fire conditions, several material models havebeen published and used in design guides. This paper adoptsthe material stress–strain relationship by the modified ECCSmodel [18]. However, this model is further modified to simplifythe derivation. The new model is obtained by curve fitting, basedon the modified ECCS data. In the modified ECCS model, Young’smodulus is defined by a step function, while in the new modelYoung’s modulus is given as a continuous function of t:

E/E200 = c0 + c1t + c2t2+ c3t3 (23)

where t = T100 , c0 = 0.96483, c1 = 0.07922, c2 = −0.03622 and

c3 = 0.00192.The yield stress of steel under thermal conditions also adopts

the ECCS model:

σY/σ200 = d0 + d1t + d2t

2+ d3t3 (24)

where d0 = 0.957, d1 = 0.0877, d2 = −0.0488, d3 = 0.00295for 100 ◦C < T ≤ 500 ◦C and d0 = 1.919, d1 = −0.3806,d2 = 0.02344, d3 = −0.000421 for T > 500 ◦C.The thermal expansion is describedby theHarmathymodel [19]:

β(T ) =∆ll= a+ bT + cT 2 (25)

where a = −2.6601 × 10−4, b = 1.0923 × 10−5, c = 5.3006 ×10−9.

2.2. Step temperature distribution

As shown in Fig. 4, the temperature distribution throughoutthe column is assumed to follow a step function as given in(26). This assumption agrees well with experimental observationsbefore flashover occurs [20,21]. For a given fire size in a ventedcompartment with known thermal characteristics of walls andceilings, the evolvement of gas temperature with time can beobtained through either zone or field modeling. Using zonemodeling, the compartment gas temperature is only given at twolayers, viz. the upper hotter layer and the lower cooler layer.In each layer, the temperature is nearly uniform and the heightof each layer varies in different combustion stages. According toexperimental and numerical results, the temperature differencebetween the two layers is dependent on the fire size and maybe greater than 450 ◦C. For instance in [22], when a fire sizeis 500 kW, the experimental results show that at the steadystate, the lower layer temperature is 150 ◦C and the upper layertemperature is 600 ◦C. But when the fire size is 700 kW, the lowerlayer temperature is 175 ◦C and the upper layer temperature isup to 650 ◦C. Thus, it is beneficial to consider the non-uniform


Table 1Procedures for evaluation of critical buckling load

Values of variables Data source

Step 1. Data input

T1 = 300 ◦C, T2 = 350 ◦C, α =Ld1L = 0.5 Given conditions

σY−1 = 258.2 MPa, σY−2 = 237.8 MPa Eq. (24)PE = σeA = 314.8 kN, σe = 129.8 Pa Euler buckling load

Step 2. Evaluation of ϕ0 and PT

σI = 70.5 MPa, σII = 210.0 Pa, σIII = 231.2 Pa (B.1) in Appendix BH(σY−2 − σI ) = H(σY−1 − σII ) = 1,H(σIII − σY−1) = H(σII − σY−2) = 0ϕ0−I = 0.901, ϕ0−II = 2.686, ϕ0−III = −6.567 (B.2) in Appendix B:ϕ0 = 0.901 Eq. (32)PT = 170.9 kN Eq. (31)

Step 3. Evaluation of Pc−cr

G1α = 6.5,G2α = 6.5,G3α = 34.457,G4α = 34.457,G5α = −4.9141 Table C.1 in Appendix CηI = 0.895625, ηII = 0.127981, ηIII = 0.008956 (C.3) in Appendix CH(σY−1 − ηIIσe) = H(σY−1 − ηIIIσe) = 1H(ηIIIσe − σY−1) = H(ηIIσe − σY−1) = 0η1 = 0.895625, η2 = 0, η3 = 0 (C.2) in Appendix Cη = 0.895625, Pc−cr = η · PE = 282.0 kN Eq. (C.1) in Appendix C

Step 4. Evaluation of P0−cr

P0−cr =Pc−cr−PT

ϕ0=

282.0−170.90.901 = 123.2 kN Eq. (22)

temperature distribution in columns, particularly when they aresubjected to local fires.Assuming the temperature distribution is given by (26):

T (x) ={100a1T100a2T

(0 ≤ x ≤ L1)(L1 < x ≤ L)

(26)

where a1T =T1100 and a2T =

T2100 are two scalar constants, and

substituting (26) into (23)–(25), one obtains:

E(x) ={E1 = a1EE200E2 = a2EE200

(0 ≤ x ≤ L1)(L1 < x ≤ L)

(27)

with a1E = c0 + c1a1T + c2a21T + c3a31T and a2E = c0 + c1a2T +

c2a22T + c3a32T .

β(T ) = β(x) ={β1 = a+ 100ba1T + 10 000ca22Tβ2 = a+ 100ba1T + 10 000ca22T

(28)

and

σY (x) ={σY−1 = a1Yσ 20Yσ Y−2 = a2Yσ

20Y

(0 ≤ x ≤ L1)(L1 < x ≤ L)

(29)

with a1Y = d0 + d1a1T + d2a21T + d3a31T and a2Y = d0 + d1a2T +

d2a22T + d3a32T .

Without loss of generality, let T1 ≤ T2 where T1 and T2 arethe respective temperatures in segments 1 and 2, one obtains (30)according to (8).∫ L

0β(x)dx = SL (30)

with S = a+ b(T1α + T2ϑ)+ c(T 21α + T22ϑ).

Hence, substituting (30) into the definition of PT , one obtains:

PT = keϕ0SL. (31)

Denoting the terms σY−1 and σY−2 as the respective yieldstresses at temperature T1 and T2, the explicit expression for ϕ0 in(22) can be deduced as (32):

ϕ0 = ϕ0−IH(σY−2 − σI)+ ϕ0−IIH(σII − σY−2)H(σY−1 − σII)

+ϕ0−IIIH(σIII − σY−1) (32)

where the terms σI , σII , σIII , ϕ0−I , ϕ0−II and ϕ0−III can be found inAppendix B. It should be noted that σY−1 ≥ σY−2 since T1 ≤ T2.

The parameters C1, C3 and D1, D3 involved in Eq. (20) canbe evaluated after the material properties are given. During thecalculation, one should note that

∫ L0 (·)dx =

∫ L10 (·)dx +

∫ LL1(·)dx

since the temperatures in the two segments are different. Forstep temperature distribution, the factor η is given in Appendix C.Although only a two step temperature distribution is consideredin this section, the concept can be easily extended to the case of amultiple step temperature distribution.

3. Worked example

In order to demonstrate how to use the proposed approach tocalculate the inelastic buckling load of a column, aworked exampleis illustrated below:

3.1. Geometry and thermal load

Consider a typical steel column made of UB178 × 102 × 19, ithas the following dimensions and material properties:

Iz = 136.71 cm4, A = 24.26 cm2, L = 300 cm,E200 = 2.1× 10

5 MPa, σ 20Y = 3.0× 102 MPa, µ = 0.01.

The stiffnesses of the springs are assumed to be k1 = 0.1E200 AL

and k2 = 3.0E200 AL , which result in an equivalent stiffness ke =

0.097 E200 AL . The thermal load is: T1 = 300

◦C, T2 = 350 ◦C, α =Ld1L = 0.5.

3.2. Implementation

Once the geometry,material and thermal properties are known,the proposed approach can be implemented in the 4 steps given inTable 1:

4. Case study

A typical column made of UB178 × 102 × 19 is used for acase study. The cross section and the elastic material propertiesare the same as in Section 3.1. Assuming the equivalent spring

stiffness is ke = 0.097E200 AL , a series of studies are conducted for


Fig. 5a. Forces vs temperature, α = 0.0.

Fig. 5b. Forces vs temperature, α = 0.25.

different values of L (column length), α (segment length ratio) andµ (Young’s modulus softening ratio).When the bottom segment temperature T1 = 250 ◦C, the

thermally induced axial force PT and the critical internal load Pc−crare plotted in Figs. 5a–e for α = 0.0–1.0. It should be noted thatwhenα = 0 there is only oneuniform temperature T2 for the entirecolumn.While when α = 1 there is only one uniform temperatureT1. In the figures, the horizontal axis represents the column toptemperature T2 while the vertical axis represents the ratio of

imposed load to PE =π2E200 IL2, which is the critical compressive

force of a pin-ended column under uniform ambient temperature.In the figures, PY−1 = AσY−1 and PY−2 = AσY−2 are also definedto check whether the steel material enters the stage of plasticity inthe respective segment before the column buckles.Fig. 5a shows that the critical buckling load Pc−cr decreases as

T2 grows. Pc−cr and PT are independent of T1 because α = 0. It canbe seen that Pc−cr is always lower than PY−2, that means the steelis at elastic stage before the column buckles. Fig. 5a also showsthat the value of the critical external load P0−cr becomes zero attemperature T2 = 435 ◦C, which means the column will bucklewhen T2 ≥ 435 ◦C even without any external load. In this case,the buckling is due to the large thermal induced force. It shouldbe noted that the force PT will not always increase. When thetemperature T2 is 560 ◦C, the PT/PE ∼ T2 curve will coincide withthe Pc−cr/PE ∼ T2 curve and thereafter descendwith increasing T2.Fig. 5b shows that the column buckling is still in the elastic

stage when α = 0.25. However, the value of P0−cr becomes zero at

Fig. 5c. Forces vs temperature, α = 0.5.

Fig. 5d. Forces vs temperature, α = 0.75.

Fig. 5e. Forces vs temperature, α = 1.0.

temperature T2 = 475 ◦C, which is higher than the correspondingtemperature in Fig. 5a.Moreover, PT is lower comparedwith Fig. 5abecause the thermal expansion is smaller since the hotter segmentis shorter. Hence, the column capacity to withstand the externalload is slightly greater than the scenario in Fig. 5a. The PT/PE ∼ T2curve and the Pc−cr/PE ∼ T2 curve intersect at T2 = 585 ◦C.In Fig. 5c, the hotter segment is half of thewhole column height.

When T2 is 575 ◦C the value of P0−cr becomes zero. PT is lower than


Fig. 6. Comparison between results obtained by elastic and inelastic assumptions.

in scenario (b) since the hotter segment is even shorter. Similar toscenarios (a) and (b), whatever the service load is, plastic bucklingwill never occur because at T2 = 575 ◦C the column completelyloses the capacity to undertake external compressive load. On theother hand, it is interesting to find that the Pc−cr curve and PY−2curve overlap if T2 ≥ 725 ◦C, that means, plastic buckling can beobserved if the internal compressive stress achieves σY−2. It canalso be seen that the force PT increases as temperature T2 rises untilT2 = 625 ◦C where PT starts to decrease.In Fig. 5d, the hotter segment is only one quarter of the whole

column. The figure shows that P0−cr is always greater than zero.That means, the column will not buckle whatever the value of T2unless an external load is applied. It can also be seen that Pc−cr isgreater than PY−2 when T2 ≥ 675 ◦C. Theoretically, the columnmay not buckle even if the hotter segment is at the plastic stage.One finds that when T2 ≥ 675 ◦C, Pc−cr is nearly constant while PTdecreases as T2 grows because the hotter segment is in the plasticstage. In this case, P0−cr , which represents the column capacity toundertake external compressive load, has a positive growth whenT2 increases. The reason is: (i) The column cannot afford higherinternal compression since the steel is plastic, but (ii) the increaseof the external load is mostly resisted by the equivalent spring atthe top column end. However, it must be realized that a higherP0−cr does not mean a higher failure load because a large axialplastic deformationwill take place. This is also regarded as one kindof structural failure before the column buckles.If the hotter segment is very short, this part just behaves like

a plastic hinge when the steel within this area is plastic. Hence,the original problem turns out to be a pin-ended column undercompressive load at uniform temperature T1. This case can bereflected in Fig. 5e.For consistency, the change of T2 is still plotted in Fig. 5e but it

will not affect Pc−cr and PT because α = 1. It can be seen that Pc−cris lower than PY−1 since T1 is only 250 ◦C. That means that columnbuckling will occur at the elastic stage.The comparison between the results obtained by elastic (µ =

1.0) and inelastic (µ = 0.01) assumptions for the case α = 0.75is shown in Fig. 6. It is obvious that the difference is significant.Based on the inelastic assumption, PT will decrease even thoughT2 goes up, because the hotter segment is plastic (T2 > 675 ◦C).However, if plasticity is not considered, PT will continuouslyincrease until T2 > 850 ◦C. As for the critical axial load, theelastic assumption will result in a higher Pc−cr than the inelasticassumption when T2 > 600 ◦C. In other words, the elasticassumptionwill overestimate the bearing capacity of the particularcolumn. In Fig. 6 it can be observed that the elastic critical externalload P0−cr is lower than the plastic P0−cr if T2 > 740 ◦C. However,as discussed above, large plastic axial deformation, which is also

one kind of structural failurewill occur at this stage. The figure alsoshows that if T2 > 740 ◦C, the value of P0−cr for µ = 0.01 growsas T2 increases because the increase of the external load is mostlyresisted by the equivalent line elastic spring connected to the topcolumn end.In Fig. 7, the influence caused by strain hardening parameter

µ is depicted. The curves clearly show that at the plastic stage, alarger µ will result in a larger PT . This is understandable becausehardening at a higher stiffness will cause higher stress at the samestrain level. As for critical axial load, the curves show that a higherµwill lead to a higher Pc−cr .In order to investigate the effect of the column length, the

results based on L = 1.8m are plotted in Fig. 8 for T1 = 250 ◦C andT1 = 450 ◦C. It can be seen that both the two values of PY−1/PE aresmaller than 1.0, which means that the column will not buckle atthe elastic stage. For T1 = 250 ◦C, Pc−cr is always lower than PY−1but greater than PY−2 when T2 > 325 ◦C. The curves show thatwhen T2 ≤ 325 ◦C, the column buckles if the axial stress achievesthe yield stress of the hotter segment. However,when T2 > 325 ◦C,the columnmay not buckle even though the hotter segment yields.Certainly, large axial deformation will be observed at this stage.If the axial column shortening is large enough, the equivalentspring at the column top will resist almost all the increment ofexternal load because the column steel is plastic. For the case ofT1 = 450 ◦C, PY−1/PE is also lower than 1.0 and theoretically itshould be lower than PY−1/PE for T1 = 250 ◦C as shown in thefigure. It is obvious that elastic buckling will not occur because thecolumn is relatively very short, which results in an elastic bucklingload higher than the yielding force PY−1 (T1 = 450 ◦C). Therefore,the critical internal compressive load of the column is equal toPY−1 (T1 = 450 ◦C). It is very interesting to find whatever T2 is,Pc−cr(T1 = 450 ◦C) remains nearly constant because mu is verysmall. For both T1 = 250 ◦C and T1 = 450 ◦C, the increase of PTdue to T2 stops when the hotter segment yields.

5. Comparison with test results

Ali et al. [11] have conducted real tests for steel columns underelevated uniform temperature. For a steel column UB178× 102×19, the experimental results are compared with the proposedapproach as illustrated in Fig. 9. In the comparison, µ is assumedto vary from 0 (ideal elastic-plasticmaterial) to 1 (elasticmaterial).In the calculation, µ = 0 is replaced by µ = 1.0 × 10−6 to avoidsingularity. It should be noted that Ali et al. referenced the initialload level on the basis of design load. For purpose of consistency,the definitions of the initial load level in Ali et al.’s tests have beenconverted to the ratio P0/PE where PE is the critical compressiveforce of a pin-ended column under uniform ambient temperature.


Fig. 7. Comparison between results due to different µ values.

Fig. 8. Forces vs temperature, α = 0.75 and L = 1.8 m.

Fig. 9. Comparison between experimental and analytical results due to different µ.

Fig. 9 shows that an ideal elastic-plastic material model resultsin the lowest critical temperature for a given initial axial load. Onthe contrary, the elastic material model gives the greatest predic-tion. It can be seen that the temperature difference between thelower andupper bounds is up to 90 ◦C. The curves show that the an-alytical approach cannot predict all the experimental failure tem-peratures based on only one µ value. However, the experimental

results are in the range formed by the analytical lower and upperbounds. In fact, all the experimental results can be predicted accu-rately by the analytical approach if the value ofµ is allowed to varyfrom 0.8 to 0.9. On the other hand, one must realize that µ is notthe only factor that affects accuracy. Other factors such as initialimperfection will also significantly affect the column critical load.However, the related issues are not discussed in this paper.


6. Conclusions

The stability of a pin-ended steel column at elevated tempera-ture is studied analytically. In this model, two elastic springs con-nected to the ends of the member are considered to simulate thelinear restraints at the column ends. The temperature distributionalong the column is assumed to be non-uniform and a bilinear ma-terial model is employed. Comparing the results based on elasticand inelastic assumptions, it can be concluded that the effect oncritical load caused by plasticity is considerable especially at hightemperatures. The explicit formulation for step temperature distri-bution can be used conveniently to evaluate the critical load. Theresults can also be combined with fire modeling. For instance, inzone modeling, the compartment containing the fire is normallydivided into two layers. Hence, the fire modeling results can be di-rectly used as input in this study.It must be noted that the proposed results are not valid if the

temperature rises very slowly because Tan et al. [23,24] suggestedthat transient analysis should be carried out to consider the creepeffect. Furumura et al. [25] and Anderberg [26] also reachedsimilar conclusions. In future studies, the results of heat transfershould be adopted and the effect of rotation restraints will also beinvestigated.

Appendix A

C1 =∫ L

0[x(L− x)F1(x)]dx (A.1)

C2 =∫ L

0[x2(L− x)2]dx =

130L5 (A.2)

C3 =∫ L

0[x(L− x)F2(x)]dx (A.3)

C4 =∫ L

0[x3(L− x)2]dx =

160L6 (A.4)

D1 =∫ L

0[x2(L− x)F1(x)]dx (A.5)

D2 =∫ L

0[x3(L− x)2]dx =

160L6 (A.6)

D3 =∫ L

0[x2(L− x)F2(x)]dx (A.7)

D4 =∫ L

0[x4(L− x)2]dx =

1105L7 (A.8)

where F1(x) = −2E(x)Iep and F2(x) = (2L− 6x)E(x)Iep.

Appendix B

σI =S(

αE1+

ϑE2

)+

Ake

(B.1a)

σII =S + (1−µ)

µ

ϑσY−2E2(

αE1+

ϑµE2

)+

Ake

(B.1b)

σIII =S + (1−µ)

µ

(ασY−1E1+

ϑσY−2E2

)1µ

(αE1+

ϑE2

)+

Ake

(B.1c)

Table C.1The values of Giα for different values of α

α G1α G2α G3α G4α G5α

0.00 0.0000 13.0000 0.0000 64.0000 0.00000.05 0.3501 12.6499 0.1226 59.1008 4.77660.10 1.1805 11.8195 1.3926 48.7449 13.86250.15 2.2236 10.7764 4.9343 38.2175 20.84820.20 3.2874 9.7126 10.7549 30.4771 22.76800.25 4.2461 8.7539 17.8451 26.4076 19.74740.30 5.0306 7.9694 24.7968 25.5865 13.61670.35 5.6191 7.3809 30.3843 27.0181 6.59770.40 6.0275 6.9725 33.8881 29.6078 0.50400.45 6.3001 6.6999 35.1506 32.3693 −3.51990.50 6.5000 6.5000 34.4570 34.4570 −4.91410.55 6.6999 6.3001 32.3693 35.1506 −3.51990.60 6.9725 6.0275 29.6078 33.8881 0.50400.65 7.3809 5.6191 27.0181 30.3843 6.59770.70 7.9694 5.0306 25.5865 24.7968 13.61670.75 8.7539 4.2461 26.4076 17.8451 19.74740.80 9.7126 3.2874 30.4771 10.7549 22.76800.85 10.7764 2.2236 38.2175 4.9343 20.84820.90 11.8195 1.1805 48.7449 1.3926 13.86250.95 12.6499 0.3501 59.1008 0.1226 4.77661.00 13.0000 0.0000 64.0000 0.0000 0.0000

ϕ0−I =1

keLA

(αE1+

ϑE2

)+ 1

(B.2a)

ϕ0−II =1

keLA

(αE1+

ϑE2ξIIµ−ξII+1

µ

)+ 1

, ξII =σY−2

σII(B.2b)

ϕ0−III =1

keLA(ξIIIµ−ξIII+1)

µ

(αE1+

ϑE2

)+ 1

, ξIII =σY−1

σIII. (B.2c)

Appendix C

η = η1 + η2 + η3 (C.1)

η1 = ηIH(σY−2 − ηIσe) (C.2a)

η2 =

[ηIIH(ηIIσe − σY−2)H(σY−1 − ηIIσe)

+σY−2

σeH(σY−2 − ηIIσe)

]H(ηIσe − σY−2) (C.2b)

η3 =

[ηIIIH(ηIIIσe − σY−1)+

σY−1

σeH(σY−1 − ηIIIσe)

]×H(ηIσe − σY−2)H(ηIIσe − σY−1) (C.2c)

ηI =2π2

(G1αa1E + G2αa2E

−

√G3αa21E + G4αa

22E + G5αa1Ea2E

)(C.3a)

ηII =2π2

(G1αa1E + µG2αa2E

−

√G3αa21E + µ2G4αa

22E + µG5αa1Ea2E

)(C.3b)

ηIII = µηI (C.3c)with

σe =PEA, Ld1 = L1, Ld2 = L− L1, α =

Ld1L,

ϑ = 1− α.


ω =

0 0 165 −530 630 −252 0 0 0 0 013 0 −165 530 −630 252 0 0 0 0 00 0 0 0 27 225 −174 900 487 750 −749 700 663 705 −317 520 6350464 0 −2400 8320 17 145 −170 868 487 750 −749 700 663 705 −317 520 635040 0 2400 −8320 −44 370 345 768 −975 500 1499 400 −1327 410 635 040 −127 008

Box I.

Table C.2The values of aiE (i = 1, 2)

Ti (i = 1, 2) (◦C) 100 200 300 400 500 600 700 800 900 1000aiE (i = 1, 2) 1.00 0.994 0.928 0.825 0.695 0.551 0.403 0.264 0.144 0.055

In (C.2), ηI is independent of µ and it can be deduced byassuming the critical axial stress σcr is smaller than σY−2, while ηIIand ηIII are derived based on assumptions: (i) σY−2 < σcr ≤ σY−1and, (ii) σY−1 < σcr , respectively.In (C.3a) and (C.3b), Giα is given in (C.4) [15].

Giα =11∑j=1

ωijαj−1, (i = 1, 5) (C.4)

where ωij are given in Box I:According to (C.4) and Box I, the values ofGiα for different values

of α are calculated and given in Table C.1 in Appendix C. The valuesof aiE used in (C.3a) and (C.3b) are listed in Table C.2.

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inelastic buckling of pin-ended steel columns under longitudinal non-uniform temperature...

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