1. Ahmed H. Zubydan - Inelastic Large Deflection Analysis of Space Steel Frames Including H-shaped...

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Inelastic large deection analysis of space steel frames includingH-shaped cross sectional members

Ahmed H. ZubydanFaculty of Engineering, Port-Said University, Port-Said, Egypt

a r t i c l e i n f o

Article history:Received 11 April 2012Revised 15 August 2012Accepted 21 September 2012Available online 24 November 2012

Keywords:InelasticLarge deectionSpace frameNonlinear analysisResidual stressesBeamcolumn

a b s t r a c t

This paper presents an efcient inelastic and large deection analysis of space frames using spread of plasticity method. New accurate formulae are proposed to describe the plastic strength surface for steelwide-ange cross sections under axial force and biaxial bending moments. Moreover, empirical formulaeare developed to predict the tangent modulus for cross sections under the combined forces. The tangentmodulus formulae are extended to evaluate the secant stiffness that is used for internal force recovering.The formulae are derived for steel sections considering the residual stresses as recommended by Euro-pean Convention for Construction Steelwork (ECCS). A nite element program based on stiffness matrixmethod is prepared to predict the inelastic large deection behavior of space frames using the derivedformulae. The nite element model exhibits good correlations when compared with the ber modelresults as well as previous accurate models. The analysis results indicate that the new model is accurateand computational efcient.

2012 Elsevier Ltd. All rights reserved.

1. Introduction

In recent years, there were numerous researches on the simula-tion of the nonlinear behavior of beamcolumns in space steelframes [16] . In general, the nonlinear behavior of steel framecan be predicted by using nite element method in which framemembers are modeled by using solid, plate or shell elements[7,8] . This method could successfully capture the nonlinear behav-ior of the structure but it is too time-consuming because of thegreat number of elements required for this type of analysis. More-over, the model processing of this analysis type is not easy at all. Inthe other direction of nonlinear analysis of steel frames, a line ele-ments approach is widely used. These studies may be categorizedinto two main types: plastic hinge analysis and spread of plasticityanalysis. The plastic hinge formulation is the most direct approachfor representing inelasticity in a beamcolumn element [912] . Inplastic hinge approach, the effect of material yielding is lumpedinto a dimensionless plastic hinge. Generally, this type of analysisis limited by its ability to provide the correct strength assessmentof beamcolumns that fail by inelastic buckling. This is because theplastic hinge analysis assumes that the cross-section behaves aseither elastic or fully plastic, and the element is fully elastic be-tween the member ends [1315] . In this model, the effect of resid-ual stresses between hinges is not accounted for either. Theadvantages of this method are its simplicity in formulation as well

as implementation and the least elements needed for membermodeling. The stability functions may be introduced to considergeometric nonlinearities using only one beamcolumn elementto dene the second-order effect of an individual member so it isan economical method for frame analysis [16,17] . This method ac-counts for inelasticity but not the spread of yielding through thesection or between the plastic hinges. For slender members inwhich failure mode is dominated by elastic instability, the plastichinge method compares well with spread of plasticity solutions.However, for stocky members that suffer signicant yielding, itoverestimates the capacity of members due to neglect of gradualreduction of stiffness as yielding progresses through and alongthe member. The so-called rened plastic-hinge analysis, basedon simple renements of the plastic hinge model, was proposedfor frames analysis in order to overcome disadvantages of the elas-ticplastic hinge method [1820] .

On the other hand, the spread of plasticity method uses thehighest renement for predicting the inelastic behavior of framedstructures. In the spread of plasticity method, the gradual spreadof yielding is allowed through the volume of the members. In thismethod, a frame member is divided into subelements, and thecross-section of each element is subdivided into many bers[2125] . The internal forces are calculated by integrating the crosssectional subelement forces. In such case, residual stress in each -ber can be explicitly considered, so, the gradual spread of yieldingcan be traced [2631] . Because of considering the spread of plastic-ity and residual stresses in a direct way, a spread of plasticitysolution is considered an exact method. Although the spread of

0141-0296/$ - see front matter 2012 Elsevier Ltd. All rights reserved.http://dx.doi.org/10.1016/j.engstruct.2012.09.024

E-mail address: [email protected]

Engineering Structures 48 (2013) 155165

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Engineering Structures

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http://dx.doi.org/10.1016/j.engstruct.2012.09.024mailto:[email protected]://dx.doi.org/10.1016/j.engstruct.2012.09.024http://www.sciencedirect.com/science/journal/01410296http://www.elsevier.com/locate/engstructhttp://www.elsevier.com/locate/engstructhttp://www.sciencedirect.com/science/journal/01410296http://dx.doi.org/10.1016/j.engstruct.2012.09.024mailto:[email protected]://dx.doi.org/10.1016/j.engstruct.2012.09.024http://crossmark.dyndns.org/dialog/?doi=10.1016/j.engstruct.2012.09.024&domain=pdf


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plasticity solution may be considered exact, it is still toocomputationally intensive and too costly. Among these two typesof analysis the co-called ber hinge method was developed in anattempt to take the advantages of the two methods [32] . In which,the element is divided into three segments, two end-ber hingesegments and an interior elastic segment, to simulate the inelasticbehavior of the material according to the concentrated inelastic

approximation. The mid-length of end hinge segment is dividedinto bers so that the uniaxial stressstrain relationships of the -bers can be monitored during the analysis process.

Recently, a new simplied model was proposed by the Authorbased on the spread of plasticity method in an attempt to elimi-nate the need of cross section discretization [33,34] . In this model,closed form formulae were derived to predict the tangent modulusof steel cross sections subjected to combined axial force anduniaxial bending moment about major or minor axis consideringthe residual stresses. Due to eliminating the integration of internalforces on the cross section level, a lot of consumed computationaltime could be saved. In the present paper tangent modulus of wide-ange steel cross sections subjected to axial compressionforce and biaxial bending is derived. New formulae are derivedby simulation of the results obtained from the ber model. Priorto the derivation of tangent modulus, new plastic strength surfacesfor H-shaped cross sections subjected to axial force and biaxialbending moment are proposed. The model achieves the accuracyof the spread of plasticity method but in an easy and a directway. The research aims to eliminate complex calculations and sominimizing the consumed running time and the cost. Theupdated-Lagrangian method is applied in the formulation of theincremental matrix equilibrium equations of the proposedbeamelement model [35,36] . The minimum residual displace-ment combined with NewtonRaphson method is used tosatisfy the convergence when solving the nonlinear equilibriumequations.

2. Numerical model

2.1. Basic assumptions

The following assumptions are made in the formulation of thebeamcolumn element:

(1) A plane cross section remains plane after deformation.(2) Local buckling and lateral torsional buckling are not

considered.(3) Small strains but large displacements and rotations are

considered.(4) Only H-shaped sections are considered.(5) Strain hardening is not considered.(6) The effects of shear forces and torsional moment are not

considered when deriving the cross sectional plastic surfaceas well as the tangent modulus.

2.2. Cross-section plastic strength

The determination of cross-section plastic strength surface isvery essential in order to predict the nonlinear behavior of struc-tural members. The most common formulae that describe the fullplastication surface for cross sections are those proposed byAISCLRFD and Orbison. Recently, analytical plastic interaction cri-teria for steel I-sections under biaxial moment and axial force weredeveloped by Baptista [37] . Although the method requires manycalculations, it is considered as an exact method.

For cross sections subjected to axial force and biaxial bending

moments about both axes, Orbisons formula is given as [38]

1 :15 p2r m2ry m

4rz 3 :67 p

2r m

2ry 3 p

6r m

2rz 4 :65 m

2rz m

4ry a 1

while the AISCLRFD plastic surface formula is given as [39]

1

2 pr mrz mry a for pr 6

2

9m rz

2

9m ry 2 :a

pr 8

9m rz

8

9m ry a for p r >

2

9m rz

2

9m ry 2 :b

where a is a factor that equal unity at full plastication surface, pr isthe ratio of the applied normal force P to the yield value P y at theplastic strength envelope ( pr = P /P y), and m rz and m ry are the ratiosof the applied bending moments M z (about minor axis) and M y(about major axis) to the corresponding plastic moments M pz andM py, respectively, at the plastic strength envelope.

In the present paper, a new formula is derived to describe thewide ange cross section plastic strength surface based on the re-sults obtained from the analysis of many cross sections. The crosssections are analyzed using the ber model in which the cross sec-tion is discretized into small bers as shown in Fig. 1. The analyzedcross sections are selected to cover all popular universal columncross sections. Twenty universal column sections (H-shapedsection) are analyzed in which the ratios B/T = 5.522.4, D/t = 10

34.2 and D/B = 0.971.13, where D, B are the cross section depthand the anges breadth, respectively, and t and T are the thick-nesses of cross section web and anges, respectively.

The cross sections are analyzed using linear strain distributionalong their axes. For each cross section, curvatures with differentratios are gradually increased until reaching the maximum possi-ble bending moments at a xed value of axial force. The internalforces ( P , M y and M z ) are evaluated by accumulation of uniaxialstresses for all cross section discrete as follows:

y (major axis)

z (minor axis)

Fig. 1. Cross sectional ber model.

m rz

m ry

(m ryL, m rzL )

m ry *

m rz *

limiting points

p r1

p r2

(0,0) 1.0

Fig. 2. Proposed cross sectional plastic strength.

156 A.H. Zubydan/ Engineering Structures 48 (2013) 155165


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P Xn

i1a i f i 3 :a

M y Xn

i1a i f i z i 3 :b

M z

X

n

i1a i f i yi 3 :c

where ai is the steel ber area, yi and z i are coordinates of each ber, f i is the uniaxial stress at the steel ber and n is the number of steelbers. It is observed that, the plastic strength surface is not affectedby the presence of residual stresses. The proposed formula is graph-ically illustrated in Fig. 2 for two different values of P r . As shown inthe gure the formula consists of two groups of curves that inter-sect at limiting points of coordinates ( m ryL, m rzL). The proposed plas-tic strength surface formula for a cross section can be given asfollow:

m rz mrzLm rz mrzL

mrym ryL

R1

1 for M rz y P m rzL=m ryL 4 :a

m ry mryL

m ry mryL !

mrz

m rzL

R2

1 for M rz y < m rzL=m ryL 4 :b

where M rz _ y = M rz /M ry , M rz = M z /M pz , M ry = M y/M py and R1 and R2 arefactors that depend on the axial force ratio ( P r ) and they are given as

R1 0 :58 P r 2 :33 for P r 6 0 :4 5 :a R1 2 for P r > 0 :4 5 :b

R2 3 :25 P 2r 3 :24 P r 1 5 :c

m ryL and m rzL are coordinates of a limiting point that can be given as

m ryL 0 :2 P 2

r 0 :8 P r 1 6 :a

m rzL 1 :06 P 3

r 0 :41 P 2

r 1 :47 P r 6 :b

The values mrz and mry are the bending moment capacity ratiosabout z - and y-axis, respectively, for cross sections subjected to ax-ial force and uniaxial bending moment. These values are used asproposed by Zubydan for cross sections subjected to a bending mo-ment about major or minor axis in addition to an axial force [33,34] .For a cross section subjected to bending moment about major axis,m ry is evaluated using the following formulae:

p1 :5r mry 1 for pr 6 0 :2 7 :a

pr 7

8m ry 1 for pr > 0 :2 7 :b

Fig. 3. Plastic strength for UC steel sections.

Fig. 4. Plastic strength for HD 400 1086 steel section.

Fig. 5. ECCS residual stress for hot-rolled H and I sections.

A.H. Zubydan / Engineering Structures 48 (2013) 155165 157


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On the other hand, when the cross section is subjected to bendingmoment about minor axis, mrz is evaluated from the followingrelationships:

1

8 :55 pr mrz 1 for pr 6 0 :05 C

2

H 0 :43 C H 0 :37 8 :a

pC H r 0 :95 m rz 1 for pr > 0 :05 C 2

H 0 :43 C H 0 :37 8 :b

In which C H = 4.9( Aw/ A) + 1.33, Aw = (D 2T )t and A is the cross sec-tional area.The correlation of the proposed plastic surface formulae in Eq.

(4) with the ber model results are represented in Fig. 3. SinceAISCLRFD plastic formula is very conservative especially for sec-tions subjected to bending moment about minor axis, only Orbi-sons formula is compared to the derived one. As shown in Fig. 3,it is observed that the proposed formula correlates very well withthe ber model results. On the other hand, the Orbisons formulamay deviate from the ber model results at various axial force ra-tios. The proposed formula for plastic strength is also compared tothe results obtained from Baptistas method [37] for HD400 1086 steel section as shown in Fig. 4. It is clearly observedthat the proposed formula gives an excellent correlation with the

results obtained from Baptisas method. It should be mentionedthat the proposed formula is simpler and more direct than Baptis-tas method.

2.3. Tangent modulus for cross sections

It is known that the cross sectional tangent modulus exhibits apremature degradation when it is subjected to forces due to thepresence of residual stresses. In order to account for the effect of residual stresses, researchers [4042] use an empirical tangentialmodulus ratio depending on the formula proposed in AISCLRFD[39] to describe the tangent modulus of columns under axial force.This formula is given as follows:

E tr 1 for a 6 0 :5 9 :a E tr 4 a1 a for a > 0 :5 9 :b where E tr is the tangent modulus ratio ( E tr = E tang /E ), E is the elasticmodulus and a is a force-state parameter that measures the magni-tude of axial force and bending moments which may be calculatedfrom Eq. (1) or Eq. (2) .

In the present paper, a new tangent modulus is determined forH-shaped sections considering the effect the residual stresses.The residual stresses adopted in the present paper are based on

the recommendation by European Convention for ConstructionSteelwork (ECCS) [43] . The magnitude of residual stress ( r r ) is as-sumed to be dependent on depth/breadth ratio as shown in Fig. 5.For cross sections subjected to axial compression force and biaxialbending moments, the exural tangent modulus is evaluated byapplying the concepts of ber model. The selected cross sectionsare analyzed using the ber model in which the bending moments

with different ratios are incrementally applied to the cross sectionat constant values of axial compression forces. Through each loadincrement, the exural tangent modulus ratio for each directionis calculated as follows:

E try dM y=du y

EI y10 :a

E trz dM z =du z

EI z 10 :b

where EI y and EI z are the cross sectional elastic rigidities about y-and z -axis, respectively, and u y and u z are the curvatures aboutthe same axes. By plotting the relationships of the tangent modulusratios E try and E trz versus the moment ratios M ry and M rz , respec-tively, as shown in Fig. 6, two possible multilinear paths may be ob-

tained for each moment direction. When the value of axialcompression force ratio P r is less than P r 0, (P r 0 = 1 r r /r y [33,34] ), the relationships follow the path abcde for E try and abcdef for E trz .For such case, the tangent modulus ratios remain constant andequal to the elastic values until reaching M ry 0,rz 0 (point b) after theydecrease linearly with the shown slopes until they vanish whenreaching the plastic surface. On the other hand, when the axial com-pression force exceeds P r 0 the values E try and E trz follow the pathsa 0b0c 0d0e0 and a0b0c 0d0e0 f 0, respectively. In such case, the tangent mod-ulus ratios start with E try 1 and E trz 1 that are less than unity due tothe effect of high axial force. The tangent modulus ratios decreaseagain with the increase of bending moments until reaching theplastic surface.

The values M ry0 and M rz 0 are found to be related to each other bythe following relationship:

M rz 0M rz 0

M ry0M ry0

1 :0 11

where M ry0 P r 0 P r Z y=S y; M rz 0 P r 0 P r Z z =S z [33,34] , Z y, z andS y, z are elastic and plastic modulus of cross section about y- and z -axis, respectively.

The slopes S y1, z 1, S y2, z 2 and S y3, z 3 and the values E ty 0, tz 0, E ty 1,tz 1,E tz 2, E tz 3 and M ry 2, rz 2 depend on the biaxial moment ratio M rz _ y

Fig. 6. Inelastic tangent modulus ratios for cross section under biaxial bending moment and axial compression force.



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(where M rz _ y = M rz /M ry) and the axial force ratio P r as illustrated inthe Appendix A .

The proposed tangent modulus ratios are compared to those ob-tained from the ber model as shown in Figs. 7 and 8 for biaxialmoment ratios ( M rz _ y = 0.25, 0.5, 1 and 2). Fig. 7 illustrates E try ver-sus M ry while E trz versus M rz are illustrated in Fig. 8. For each biaxial

moment ratio, the axial force ratio varies from 0.2 to 0.8. It isclearly observed that the proposed formulae give very good corre-lations with the ber model results for various values of P r andM rz _ y. It should be mentioned that the proposed model is validfor all values of P r .

2.4. Cross section incremental secant modulus

For nonlinear analysis, the accuracy of results essentially de-pends on the accuracy of the internal forces calculation. The useof tangent stiffness for calculating the internal forces may slightlyoverestimate the structure strength, so the incremental secantstiffness is required in order to evaluate accurate values of internalforces. The same sequence followed by the Author [33] is used in

the present research to evaluate the secant stiffness for a loadincrement. For a load increment from steps j to j + 1, the curvature

changes from u j to u j+1 and the bending moment changes from M j

to M 0 j 1 when the initial tangent modulus is followed as shown inFig. 9. The value of M 0 j 1 can be calculated as

Fig. 7. Inelastic exural modulus ratios for cross sections about major axis. Fiber model; present model.

Fig. 8. Inelastic exural modulus ratios for cross sections about minor axis. Fiber model; present model.

Fig. 9. Moment-curvature curve for cross section.



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M 0 j 1 M j du E tr j EI 12

where du is the change in the curvature of the cross section fromstep j to step j + 1 and EI is the elastic exural rigidity of the crosssection. E tr j is the tangent modulus ratio at step j which can be cal-culated from Fig. 6 using the forces values at point j. As E tr changeslinearly with M r from point j j + 1, i.e. the following relationship canbe deduced:E tr c 1 M r c 2 13

where c 1 E 0tr j 1 E tr jM 0r j 1 M r j

and c 2 = E tr j c 1M r j. M 0 j 1 and E 0trj 1 are the

bending moment and the corresponding tangent modulus ratio atpoint j0 + 1 which is very close to point j + 1.

Based on Eq. (12) , the value of bending moment at j + 1(M j+1 )and the incremental secant modulus ratio ( E sr ) can be derivedand given as follow [33] :

M j 1 1

c 1M pek c 2 14

E sr 1

EI M j 1 M ju j 1 u j

15

where M p i s the cross section plastic moment andk c 1

u j 1 u ju y lnc 1 M r j c 2 . Eq. (15) can be applied for the bend-

ing moments about y- and z -axis to evaluate the corresponding se-cant modulus E sry and E srz , respectively.

2.5. Finite element model

A stiffness method for the analysis of space frames is developedconsidering both geometric and material nonlinearities. The equa-tion of equilibrium in terms of geometry of the deformed system isgiven as follows [44] :K K g fD D g fD F g 16

where { D F} and {D D} are the incremental force and displacementvectors, respectively. [ K] is the stiffness matrix of the structure con-sidering material nonlinearity and [ K g ] is the geometric stiffnessmatrix which represents the change in the stiffness that resultsfrom deformation effects. Consider a prismatic element of a sym-metric cross section about y and z axes. This element is subjectedto axial force N , torsional moment M x and bending moments M yand M z while the corresponding displacements are u,h x,h y and h z ,respectively as shown in Fig. 10. The shape functions for the axialdisplacement ( u), the axial rotation ( h x), the displacement in x yplan ( w) and the displacement in x z plan ( v ) are introduced con-sidering the shear effect as follow:

u 1 nu1 nu2 17 :a h x 1 nh x1 nh x2 17 :b

w1 U y=2 1 U y U yn 3 n2

2 n3

w 1 1 U y=2 2 U y=2 n n

2 xh z 1 U yn 3 n2 2 n

3 w2

U y=2 1 U y=2 n n2 xh z 2 17 :c

v 1 U z =2 1 U z U z n 3 n2 2 n3 v 1 1 U z =2

2 U z =2 n n2 xh y1 U z n 3 n

2 2 n3

v 2

U z =2 1 U z =2 n n2 xh y2 17 :d

In which n = x/L and L is the member length, U y = 12 EI z /GAQyL2,-U z = 12EI y/GAQz L2 , G is the shear modulus and AQy and AQz are theshear areas corresponding to y- and z -axis, respectively. Since the

matrices [ K] and [ K g ] are displacement dependent, Eq. (16) cannotbe directly solved. Various procedures can be used to solve the equi-librium equations. Generally, members are subdivided into subele-ments to produce satisfactory results. The modulus ratios E try ,trz orE sry ,srz are evaluated at each member ends and then average valuesat each direction are applied for each modulus ratio in the stiffnessmatrix [ k].

2.6. Internal force recovery

The employing of equilibrium equation in conjunction with theincremental analysis requires that the structural geometry in-cludes all accumulated deformations. For the current analysis,the node coordinates are updated after each iteration. That is, the

coordinates of each node are modied or updated to include thetranslational displacement components that occur during itera-tions. In updating the coordinates of nodes or element ends, thedeformed geometry of the structure is achieved by changing theposition and hence the orientation of each element with respectto the global coordinates system. For all the elements of a struc-ture, the element stiffness equation, as given in Eq. (16) , can beassembled to yield the stiffness equation of the structure for anincremental step. For an incrementaliterative nonlinear analysis,the element incremental displacement vector { D d} is used to cal-culate the incremental axial strain de, torsional strain d v and ex-ural strains du y and du z based on the assumed shape functions andso the secant modulus at element ends can be evaluated as illus-trated in Section 2.4 . The material stiffness matrix [ k] for each ele-

ment is reformulated using the average secant exural modulusratios E sry and E srz instead of the tangent modulus. The secant mod-ulus for axial force and torsional moment can be modied by usingthe average of E sry and E srz . On the other hand, the geometric stiff-ness matrix [ k g ] is also formulated again using a new axial force P which is calculated by adding the incremental axial force to the to-tal previous value as follows:

P P prev ious deE sr EA 18

The increments of internal force vector { D f int } for an element can becalculated as

f D f int g k k g fD dg 19

By summing the element forces at the structural nodes and compar-

ing them with the applied loads, the unbalanced forces for thestructure can be obtained. Finally, by treating the unbalanced forcesas applied loads, other iterations can be repeated.

3. Numerical solution

The nonlinear analysis algorithm consists of four basic steps;the formulation of the initial stiffness matrix, the solution of theequilibrium equations for the displacement increments, the deter-mination of the new updated stiffness and member forces usingthe cross sectional model, and the check of conversion. Since theglobal stiffness matrix of the structure depends on the displace-ment increments, the solution of the equilibrium equations istypically accompanied by an iterative method through the conver-

gence check. In the present model, the Newton Raphson method isused by updating the tangent stiffness matrix at each iterationFig. 10. Space frame element.

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[45] . Also, the minimum residual displacement method is used inorder to trace the post-peak path [46] . A full nonlinear iterativesolution discussed by Zubydan [33] is followed in the presentpaper.

4. Numerical analysis and results

A computer program is developed to predict the nonlinearbehavior of space frames using the derived model. The ber modelas well as the proposed simplied model is rst compared to well-established benchmark results. After that the analysis results ob-tained from the proposed and the previous simplied models arecompared to those obtained from the ber model.

4.1. Two-storey space frame

The two storey space steel frame shown in Fig. 11 was previ-ously analyzed by Kim and Lee [7] using nite element packageABAQUS [47] . For this study the frame was only subjected to thegravity loads and lateral loads in X -direction (i.e. the value of r = 0). All frame members are comprised of H150 160 10 6.5mm section. The frame has 2.5 m wide in Y -direction and3 m long in X -direction. The height of the second oor is 1.76 mfrom the column base while the roof elevates 2.2 m from the sec-ond oor. The yield stress, Youngs modulus, and shear modulusof material are 320 MPa, 221 GPa, and 85 GPa, respectively. Theframe columns have out-of-plumbness imperfection as listed inTable 1 . The ABAQUS shell element S4R was employed to modelframe members. The total numbers of shell elements used to mod-el the structure by ABAQUS are 49,840. The shown applied loadswere proportionally increased until failure of the structure. Forsuch case of loading ( r = 0), the frame is reanalyzed using thepresent ber model and also using the proposed simplied model.For the analyzed frame, each frame member is discretized into 10equal elements. The comparisons of the obtained and the pub-

lished analysis results are illustrated in Fig. 12. It is observed thatthe results obtained from the proposed model correlates well withthe ber model results as well as the results obtained fromABAQUS.

The frame is also analyzed using the both the simplied and theber models with different values of r (r = 0.25, 0.5, 1.0). Compar-ison of the analysis results are shown in Fig. 13 which illustrates

the resultant displacement at point A versus the load P. It is clearlyobserved that the results obtained from simplied modelcorrelates very well the results obtained from the ber model. Itis observed that the frame capacity is greatly reduced with the in-crease of lateral loads in Y -direction due to the increase of bendingmoments produced about cross sectional minor axes.

4.2. Six storey space frame

The six-story space steel frame shown in Fig. 14 was rst ana-lyzed by Orbison et al. [38] and later by many researchersFig. 11. Two storey space frame.

Table 1

Out-of-plumbness imperfection of two-storey space frame.

Level Imperfect ion (mm)

Column 1 Column 2 Column 3 Column 4

x y x y x y x y

Roof 4.51 11.08 5.49 11.41 8.17 6.58 4.31 12.04Second

oor

1.39 6.88 0.68 6.77 5.11 2.11 3.96 6.19

Base 0 0 0 0 0 0 0 0

Fig. 12. Load-displacement relationships for the two storey frame ( r = 0).

Fig. 13. Load-displacement relationships for the two storey frame ( r = 0.25, 0.5 and1.0).



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[11,31,48,49] . The yield strength of frame members is 250 MPa,Youngs modulus E = 206,850 MPa and shear modulusG = 79,293 MPa. The frame is subjected to proportional gravityand lateral loads. Uniform oor pressure of 9.6 kN/m 2 is convertedinto equivalent concentrated loads on the top of columns. Windloads are simulated by point loads of 53.376 kN in the Y -directionat every beamcolumn joints. Jiang et al. analyzed the frame usingber plastic zone model in which the cross section ange is dis-credited into bers and the frame member into nine elements[31] . The frame is reanalyzed using the present model with/with-out the shear effect. Since most plastic deformation is concentratedat member edges, each frame member is discretized into three ele-ments with two short edge elements of length = 0.1 L. The tangentmodulus for I-shaped cross sections (as in case of frame girders) iscalculated as proposed by Zubydan [33] . Comparisons of the pro-posed simplied model with Jiangs analysis results are shown in

Fig. 15 . It is clearly observed that the present model results withshear effect correlate very well the Jiang results. On the other hand,when the shear effect is neglected, the model slightly overesti-mates the frame capacity.

4.3. Single storey space frame

The frame shown in Fig. 16 is assumed to be made of steel withyield strength r y = 250 Mpa and elastic modulus E = 200,000 MPa.The frame is subjected to incremental lateral loads H in X -directionand rH in Y -direction in addition to constant value of vertical loadsP . The frame is analyzed using the ber model and also by using thederived model neglecting the shear effect. For all analysis types,each frame member is divided into 10 elements to accumulatethe plastic deformation through the whole member. The tangentmodulus proposed by researchers (Eq. (9) ) accompanied with theOrbisons plastic surface formula is also used to analyze the frame.The cross section of the frame members is assumed to be

UC203 203 60 and the frame is analyzed using different valuesof r (r = 0.25, 0.5, 1.0 and 2.0). For each value of r , vertical load

Fig. 14. Six storey space frame.

Fig. 15. Load-displacement relationships for the six storey frame.

Fig. 16. Single storey space frame.



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ratios P r = 0.2, 0.4 and 0.6 are used. Fig. 17 shows comparisons be-tween the results obtained from different models. The gure illus-trates the relationships between the lateral load H and the lateralresultant displacement at A. It is clearly observed that the proposedmodel gives excellent correlations with the ber model for all val-ues of r and P r . On the other hand, the previously used formula in

Eq. (9) mostly overestimates the stiffness and the capacity of theframe compared to the ber model. Table 2 shows the normalized

capacities of the frame using the proposed model and that fromEqs. (9) and (1) with respect to the ber model results. As shownin Table 2 , the diversion of the analysis results obtained fromEqs. (9) and (1) from the ber model increases with the increaseof r and P r . The overall capacity of the frame obtained from the pre-vious model more (Eqs. (9) and (1) ) may be than twice the capacityobtained from the ber model.

5. Conclusions

A new simplied model was developed to predict the largedeection inelastic behavior of space steel frames. New plasticstrength surfaces for H-shaped cross sections under axial forcesand biaxial bending moment were derived. Moreover, the tangentmodulus of cross sections subjected to the combined forces waspredicted. The residual stress distributions were considered as rec-ommended by European Convention for Construction Steelwork(ECCS). The internal forces were recovered by using derived incre-mental secant stiffness. The derived cross sectional model wasimplemented into a nite element program based on stiffnessmethod to predict the full nonlinear behavior of steel space frames.The simplied model correlated very well the ber model withoutthe need of cross section discretization. The proposed model could

successfully simplify the plastic zone analysis and save a lot of computational time and data storage by the elimination of

Fig. 17. Load-displacement relationships for the single storey frame.

Table 2

Normalized ultimate loads for single storey space frame.

P r = 0.2 P r = 0.4 P r = 0.6

r = 0.25Present model 0.9977 0.9917 1.0523Eqs. (9) & (1) 1 .0823 1.0198 1.1377






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iterations on the cross sectional level. From the analyzed examples,it was found that the average run time of the frame using the bermodel ranged from 8 to 10 times the run time required using thesimplied model. These run times included the formulation andthe solving of nonlinear equilibrium equations and they also in-cluded the time consumed to equilibrate the bers on the crosssectional level which was saved by using the proposed simplied

model.

Appendix A

Values of S y1 , S y2, S y3 , E ty 0, E ty 1 and M ry 1:

S y1 0 :38 M 2

rz y 0 :05 M rz y 0 :74 f 1 for M rz y 6 1 :0 A:1a S y1 0 :88 M rz y 0 :26 f 1 for M rz y > 1 :0 A:1b

where f 1 = 0 for P r 6 0.5 and f 1 7:81 P 2r 6:84 P r 1:47 for P r > 0.5

For M rz _ y 6 1.5:

S y2 10 :82 M 2

rz y 17 :87 M rz y 4 :03 for P r 6 0 :2 A:2a

S y2 11 :45 M 2

rz y 15 :65 M rz y 7 :37 for 0 :3 6 P r 6 0 :5 A:2b

S y

2 17 :57 M 2rz y

21 :43 M rz y

13 :67 for P r

P 0 :6 A:2c

For M rz _ y > 1.5:

S y2 1 :4 M rz y 5 :12 A:2d

S y3 0 :05 M 2

rz y 0 :36 M rz y 0 :96 A:3

E ty0 0 :65 P 2

r 0 :22 P r 1 :05 A:4

E ty1 1 :3 P 2

r 0 :45 P r 1 :1 A:5

M ry1 f 2

1 M rz y6 m ry A:6

where M ry1 6 1.05 P r + 1.07; M ry1 6 0.98, f 2 = 0.4P r + 1.015 forP r 6 0.3 and f 2 3:89 P

3r 4:44 P

2r 2:27 P r 1:53 for P r > 0.3

Values of S z 1, S z 2 , S z 3 , E tz 0, E tz 1, E tz 2, E tz 3 and M rz 1

S z 1 1 :37 M 2

ry z 0 :94 M ry z 1 :52 f 3 for M ry z 6 1 :0 A:7a S z 1 2 :38 M ry z 0 :49 f 4 f 3 for M ry z > 1 :0 A:7b

where S y1 P 0:9; M ry z M ry =M rz ; f 3 4:65 P 3r 1:87 P

2r

1:03 P r 0:19 ; f 4 0 for P r 6 0.5 and f 4 = (10.74 P r 5.4) M ry_ z forP r > 0.5.

S z 2 13 :59 M 2

ry z 6 :11 M ry z 2 :54 for M ry z 6 1 :0 A:8a S z 2 12 :77 M ry z 2 :65 for M ry z > 1 :0 A:8b

S z 3 0 :94 M ry z 0 :55 6 1 :5 A:9 E tz 0 1 :9 P r 1 :94 A:10

E tz 1 2 :5 P 2

r 5 :45 P r 3 :1 A:11

For P r 6 0.5:

E tz 2 0 :4 M ry z 0 :65 for M ry z 6 5 :0 A:12a E tz 2 0 :007 M ry z 0 :44 for M ry z > 5 :0 A:12b

For P r > 0.6:

E tz 2 0 :007 M ry z 0 :44 P r 0 :6 A:12c

E tz 3 0 :11 0 :35 f 5 M ry z f 5 for M ry z 6 3 :0 A:13a E tz 3 0 :3 for M ry z > 3 :0 A:13b

where f 5 = 0.53 P r + 0.54

M rz 1 f 6

1 M ry z 6 m rz A:14

where M rz 1 6 0:95 P 3r 0:16 P 2r 0:07 P r 0:95 ; f 6 0:75 P

2r 0:4P r

1:07 for P r 6 0.3 and f 6 1:86 P 2r 0:79 P r 1:05 for P r > 0.3.

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